Chemistry - Oxford Fajar - SPM

OXFORD FAJAR ADVISORY BOARD The board consists of a team of experienced teachers who review our titles to ensure that the contents are in line with the current syllabus and examination requirements as set by the Examination Syndicate, Ministry of Education Malaysia. Success Chemistry SPM e-book was reviewed by • Ilani bte Ibrahim • Tan Sze Chuan • Tay Geok It ***************************************** Oxford Fajar Sdn. Bhd. (008974-T) (Formerly known as Penerbit Fajar Bakti Sdn. Bhd.) 4 Jalan Pemaju U1/15, Seksyen U1 Hicom-Glenmarie Industrial Park 40150 Shah Alam Selangor Darul Ehsan © Oxford Fajar Sdn. Bhd. (008974-T) 2013 First published 2013 ISBN 978 983 47 0756 9 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of Oxford Fajar Sdn. Bhd. (008974-T) Text set in 10 point ITC Giovanni by Leo & Libra Creative, Kuala Lumpur Word Power ii Contents Keywords v 4.7 Appreciating the Existence of Elements and Their Compounds 98 SPM Exam Practice 4 100 Introduction to Chemistry 1 CHAPTER 5 2 1.1 Chemistry and Its Importance 1.2 Scientific Method 1.3 Scientific Attitudes and Values in Conducting Scientific Investigations SPM Exam Practice 1 2 5 Chemical Bonds FORM 4 CHAPTER 1 5.1 5.2 5.3 5.4 Formation of Compounds Formation of Ionic Bonds Formation of Covalent Bonds The Properties of Ionic Compounds and Covalent Compounds SPM Exam Practice 5 7 9 CHAPTER 2 The Structure of the Atom 2.1 2.2 2.3 2.4 2.5 Matter The Atomic Structure Isotopes and Their Importance The Electronic Structure of an Atom Appreciating the Orderliness and Uniqueness of the Atomic Structure SPM Exam Practice 2 14 15 23 26 29 3.1 Relative Atomic Mass and Relative Molecular Mass 3.2 Relationship between the Number of Moles and the Number of Particles 3.3 Relationship between the Number of Moles of a Substance and Its Mass 3.4 Relationship between the Number of Moles of a Gas and Its Volume 3.5 Chemical Formulae 3.6 Chemical Equations 3.7 Scientific Attitudes and Values in Investigating Matter SPM Exam Practice 3 108 109 114 121 128 CHAPTER 6 2 Electrochemistry 134 6.1 6.2 6.3 6.4 6.5 6.6 6.7 Electrolytes and Non-electrolytes 135 Electrolysis of Molten Compounds 137 Electrolysis of Aqueous Solutions 142 Electrolysis in Industries 152 Voltaic Cells 158 The Electrochemical Series 165 Developing Awareness and Responsible Practices when Handling Chemicals used in the Electrochemical Industries 172 SPM Exam Practice 6 173 32 33 2 CHAPTER 3 Chemical Formulae and Equations 107 39 40 43 CHAPTER 7 2 45 Acids and Bases 48 51 57 181 60 62 7.1 Characteristics and Properties of Acids and Bases 182 7.2 The Strength of Acids and Alkalis 192 7.3 Concentration of Acids and Alkalis 196 7.4 Neutralisation 203 SPM Exam Practice 7 211 67 CHAPTER 8 2 68 74 75 82 89 93 Salts 218 8.1 Salts 8.2 Qualitative Analysis of Salts 8.3 Practising Systematic and Meticulous Methods when Carrying Out Activities SPM Exam Practice 8 219 240 2 CHAPTER 4 Periodic Table of Elements 4.1 4.2 4.3 4.4 4.5 4.6 Periodic Table of Elements Group 18 Elements Group 1 Elements Group 17 Elements Elements in a Period Transition Elements iii 255 256 2 CHAPTER 9 2 CHAPTER 3 Manufactured Substances in Industry 9.1 9.2 9.3 9.4 9.5 9.6 9.7 Sulphuric Acid Ammonia and Its Salts Alloys Synthetic Polymers Glass and Ceramics Composite Materials Appreciating Various Synthetic Industrial Materials SPM Exam Practice 9 262 Oxidation and Reduction 263 266 272 278 281 284 3.1 3.2 3.3 3.4 288 289 FORM 5 2 CHAPTER 4 Thermochemistry CHAPTER 1 Rate of Reaction 1.1 1.2 1.3 1.4 Rate of Reaction Factors that Affect the Rate of Reaction The Collision Theory Practising Scientific Knowledge to Enhance Quality of Life SPM Exam Practice 1 4.1 4.2 4.3 4.4 4.5 4.6 Energy Changes in Chemical Reactions Heat of Precipitation Heat of Displacement Heat of Neutralisation Heat of Combustion Appreciating the Existence of Various Energy Sources SPM Exam Practice 4 295 296 305 322 328 329 CHAPTER 2 Carbon Compounds 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 SPM 384 Redox Reactions 385 Rusting as a Redox Reaction 412 The Reactivity Series of Metals and Its Applications 418 Redox Reactions in Electrolytic Cell and Chemical Cell 430 3.5 Appreciating the Ability of the Elements to Change their Oxidation Numbers 441 SPM Exam Practice 3 444 452 453 463 469 474 482 489 492 2 CHAPTER 5 340 Carbon Compounds 341 Alkanes 343 Alkenes 346 Isomerism 352 Alcohols 357 Carboxylic Acids 362 Esters 366 Oils and Fats 370 Natural Rubber 371 Order in Homologous Series 376 The Variety of Organic Materials in Nature 376 Exam Practice 2 378 Chemicals for Consumers 499 5.1 5.2 5.3 5.4 SPM 500 510 515 519 521 Soaps and Detergents Uses of Food Additives Medicine Appreciating the Existence of Chemicals Exam Practice 5 SPM Model Test Answers Glossary iv 526 537 592 Key Words 1 Introduction to Chemistry conclusion – kesimpulan constant variable – pembolehubah yang dimalarkan manipulated variable – pembolehubah yang dimanipulasikan procedure – kaedah/prosedur responding variable – pembolehubah yang bergerakbalas scientific attitudes – sikap saintifik variable – pembolehubah denominator – penyebut element – unsur empirical formula – formula empirik ionic compound – sebatian ion mass – jisim molar volume – isipadu molar molecular formula – formula molekul numerator – pengangka product – hasil tindak balas reactant – bahan tindak balas reduced – diturunkan relative atomic mass – jisim atom relatif 4 Periodic Table of Elements 2 The Structure of the Atom charged particles – zarah bercas chemical reaction – tindakbalas kimia collision – perlanggaran compressibility – kemampatan condensation – kondensasi diffusion – peresapan duplet – duplet electronic configuration –susunan elektron electron shells – petala elektron forces of attraction – daya tarikan freezing point – takat beku half-life – setengah hayat isotope – isotop matter – jirim melting point – takat lebur non-renewable – tidak boleh diperbaharui nucleon number – nombor nukleon octet – oktet 3 Chemical Formulae and Equations anion (negatively-charged ion) – anion (ion bercas negatif) cation (positively-charged ion) – kation (ion bercas positif) compound – sebatian crucible – mangkuk pijar covalent compound –sebatian kovalen boiling point – takat didih chemical bonding – ikatan kimia covalent bond – ikatan kovalen double bond – ikatan ganda dua electrical conductivity – kekonduksian elektrik electrostatic force of attraction – daya tarikan elektrostatik inert gas – gas adi ionic bond – ikatan ion Lewis structure – struktur Lewis non-polar – tidak berkutub organic solvent – pelarut organik polar – berkutub shell – petala single bond – ikatan tunggal solubility – kelarutan triple bond – ikatan ganda tiga valence electron – electron valens volatility – kemeruapan 5 Chemical Bonds aqueous solution – larutan akueus giant molecules – molekul raksasa intermolecular force – daya tarikan antara molekul noble gas – gas adi 6 Electrochemistry alkaline cell – sel alkali anode – anod cathode – katod concentration – kepekatan v decomposition – penguraian discharge – nyahcas displacement reaction – tindak balas penyesaran dry cell – sel kering electrochemical series – siri elektrokimia electrochemistry – elektrokimia electrode – elektrod electrolysis – elektrolisis electrolyte – elektrolit electroplating – saduran elektrik half-reaction – tindak balas setengah non-electrolyte – bukan elektrolit non-rechargeable cell – sel yang tidak boleh dicas semula potential difference – beza keupayaan primary cell – sel primer rechargeable cell – sel yang boleh dicas semula secondary cell – sel sekunder 7 Acids and Bases basicity – kebesan degree of dissociation – darjah penceraian dilution – pencairan diprotic acid – asid dwibes end point – takat akhir molarity – kemolaran monoprotic acid – asid monobes hydroxide ion – ion hidroksida hydroxonium ion –ion hidroksonium neutralisation – peneutralan partial dissociation –penceraian separa standard solution – larutan piawai titration – pentitratan triprotic acid – asid tribes universal indicator –penunjuk semesta 8 Salts brown ring test – ujian cincin perang confirmatory test –ujian pengesahan continuous variation –perubahan berterusan crystal – hablur Key Words KEY WORDS FORM 4 decolourised – nyahwarna double decomposition –penguraian ganda dua evaporation – sejatan filtrate – hasil turasan impurities – benda asing insoluble salts –garam tak terlarutan precipitate – mendakan qualitative analysis – analisa kualitatif recrystallisation – penghabluran semula residue – baki soluble salts – garam terlarutan solution – larutan KEY WORDS 9 Manufactured Substances in Industry alloy – aloi biodegradable – terbiodegradasikan borosilicate glass – kaca borosilikat brass – loyang bronze – gangsa catalyst – mangkin ceramic – seramik coagulation – penggumpalan Contact process – proses sentuh corrosion – kakisan density – ketumpatan ductility – kemuluran explosive – bahan letupan fibre optic – gentian optik fused glass – kaca silika terlakur lead glass – kaca plumbum malleability – kebolehtempaan photochromic glass –kaca fotokromik polymerisation – pempolimeran refrigerant – bahan penyejuk rust – karat soda glass – kaca soda kapur solder – pateri stainless steel – keluli nirkarat superconductor – superkonduktor synthetic fibre – gentian sintetik FORM 5 energy profile diagram – rajah profil tenaga observable change – perubahan yang dapat diperhatikan rate of reaction – kadar tindak balas 2 Carbon Compounds addition – penambahan alkanes – alkana alkenes – alkena alkynes – alkuna combustion – pembakaran fractionating column –turus pemeringkat functional group – kumpulan berfungsi general formula – formula am homologous series – siri homolog hydration – penghidratan hydrogenation – penghidrogenan IUPAC nomenclature – sistem penamaan IUPAC saturated – tepu sootiness – kejelagaan straight chain – rantai lurus structural formula – formula struktur substitution – penukargantian unsaturated – tak tepu carboxylic acid – asid karboksilik coagulation – penggumpalan dehydration – pendehidratan distillation – penyulingan drying agent – agen pengontangan elasticity – kekenyalan esterification – pengesteran extraction – pengekstrakan fatty acid – asid lemak fermentation – penapaian hydroxonium ion – ion hidroksonium hydroxyl group – kumpulan hidroksil polyunsaturated fats –lemak poli tak tepu volatility – kemeruapan vulcanised – tervulkan 1 Rate of Reaction activation energy –tenaga pengaktifan average rate – kadar purata catalyst – mangkin collision frequency – frekuensi perlanggaran collision theory – teori perlanggaran effective collision –perlanggaran berkesan energy barrier – rintangan tenaga Key Words 3 Oxidation and Reduction blast furnace – relau bagas cast iron – besi tuangan chemical cell – sel kimia displacement reaction – tindak balas penyesaran electrolytic cell – sel elektrolisis extraction – pengekstrakan impurity – bendasing vi metal displacement –penyesaran logam oxidation state –keadaan pengoksidaan oxidising agent – agen pengoksidaan reactivity series – siri kereaktifan reducing agent – agen penurunan sacrificial metal – logam korban 4 Thermochemistry bond energy – tenaga ikatan endothermic reaction –tindak balas endotermik energy content –kandungan tenaga energy level diagram –gambar rajah aras tenaga exothermic reaction –tindak balas eksotermik fuel value – nilai haba bahan api heat of combustion – haba pembakaran heat of displacement –haba penyesaran heat of formation – haba pembentukan heat of neutralisation –haba peneutralan law of conservation of energy – hukum keabadian tenaga precipitation – pemendakan reversible reaction – tindak balas berbalik specific heat capacity –muatan haba tentu thermal dissociation –penceraian terma thermochemical equation – persamaan termokimia 5 Chemicals for Consumers additive – bahan tambahan analgesic – analgesik antioxidant – pengantioksida/ antipengoksida antipsychotic – antipsikotik biodegradable – terbiodegradasikan biological enzyme – enzim biologi codeine – kodeina detergent – detergen flavouring agent – agen perisa preservative – pengawet saponification – saponifikasi soap – sabun stabiliser – pengstabil thickening agent – agen pemekat FORM 4 THEME: Introducing Chemistry CHAPTER 1 Introduction to Chemistry SPM Topical Analysis 2008 Year Paper 1 Section Number of questions – 2009 2 A B C – – – 3 1 – – 2010 2 A B C – – – 3 1 – – 2011 2 A B C – – – 3 1 – – 3 2 A B C – – – ONCEPT MAP INTRODUCTION TO CHEMISTRY History of chemistry Importance of chemistry Meaning of chemistry The scientific method Science that studies the properties, composition and structure of substances and the changes they undergo In the fields of: • Food processing • Medicine • Agriculture • Transportation • Telecommunications • Daily usage of chemical products Methodology in chemistry Careers that need knowledge of chemistry: • Medicine and Dentistry • Pharmacy • Geology • Biochemistry • Engineering • Observe a situation • Identify all variables • Suggest a problem statement • Form a hypothesis • Select suitable apparatus • Carry out an experiment • Collect and tabulate data • Interpret the data • Write a report Scientific attitudes and values • • • • Attitudes Avoid wastage Maintain cleanliness Avoid accidents – 1 1.1 (a) Medicine: to fight diseases and prolong life. Chemistry and Its Importance 1 Human beings used chemical processes before 500 BC to extract metals such as copper and iron for making ornaments. They also found ways to make ceramics from clay. However, they could not explain the chemical processes that took place. 2 The next 1700 years of chemical history were dominated by a pseudo-science called alchemy (pseudo means not genuine or false). The word alchemy originated from the Arabic word ‘alkimiya’ (al === the; kimiya === art of changing). The alchemists in Egypt believed they could change cheap metals like lead into gold. Their efforts were unsuccessful but along the way they (a) discovered other substances like mercury, sulphur, antimony and phosphorus, (b) developed some reliable techniques of chemical manipulation, (c) learned to prepare some mineral acids such as sulphuric, hydrochloric and nitric acids. Medicine (b) Fertilisers and pesticides: increase crop yields. (c) Preservatives: prolong the storage of food. Food preservative (d) Materials used for making clothing such as cotton, silk and nylon. A chemist carrying out research 3 Modern chemistry originated from an Englishman named Robert Boyle. In 1661, he wrote a book called The Sceptical Chymist which introduced the modern concept of chemical elements. An element is a substance that cannot be broken down into simpler substances by chemical means. In the centuries that followed, many elements were discovered. Nylon (e) Building materials concrete and glass. such as cement, The Importance of Chemistry 1 Chemistry is the science concerned with the composition of substances, the basic forms of matter and the interactions between them. 2 Chemical substances or chemicals are very important in our lives. The following are a few examples of chemicals. Introduction to Chemistry Building 2 (b) how chemicals interact among each other, and (c) how to use the knowledge of the properties of these chemicals to produce new substances. (f) Components of automobiles and computers. 1 Many careers require knowledge of chemistry. For example, a dentist uses hydrogen peroxide gel (H2O2) to bleach teeth (making it whiter). For extraction of a tooth, the dentist will administer a local anaesthetic (procaine) before extracting the tooth of a patient. To fill a tooth, he/she will use amalgam which is an alloy of mercury and silver. Hydrogen peroxide, procaine and amalgam are all chemicals. (g) Consumer products such as soap and detergents. Consumer products 3 Table 1.1 shows the uses of some chemical substances in our daily life. 4 In chemistry, we study (a) the basic units that make up these materials, A dentist at work Table 1.1 Name of substance Uses Chemical formula Oxygen O2 Respiration and combustion Nitrogen N2 Manufacture of ammonia Carbon dioxide CO2 In photosynthesis and in making carbonated drinks Sodium chloride NaCl Food preservation, for example, salted fish Iron(II) sulphate FeSO4 Iron pills to treat anaemia Aspirin Calcium sulphate hemihydrate Copper-nickel alloy Urea Sulphuric acid Ethanol Sodium stearate Ethanoic acid (acetic acid) Calcium carbonate CH3COOC6H4COOH An analgesic drug to treat pain and fever 2CaSO4. H2O Used as a cast to support broken bones of accident victims 25% Nickel + 75% Copper CO(NH2)2 To make coins A nitrogenous fertiliser H2SO4 As an electrolyte in a lead-acid accumulator C2H5OH As a solvent and manufacture of industrial chemicals C17H35COONa Soap CH3COOH Preservation of fruits and manufacture of food flavourings CaCO3 Calcium supplement 3 Introduction to Chemistry 1 Chemistry Related Careers 2 A medical doctor needs a knowledge of chemistry to administer the correct amount of medicine to a patient. Categories of medicine include antibiotics, hormones, psychiatric medicine, analgesics, alkaloids and fungal creams. All these medicine are chemicals. Food processing 6 A farmer uses fertilisers to increase the yield of his crops. Pesticides, herbicides and fungicides are used to control pests. Therefore, even the farmer is required to have a knowledge of chemistry. 1 A doctor administering an injection 3 Pharmacy is a branch of science which deals with the interaction of medicine with the human body. It also finds ways to synthesise new drugs. Most medicine are organic compounds. Therefore a pharmacist must have an understanding of organic chemistry. A farmer spraying pesticides Chemical-based Industries in Malaysia and their Contributions Local chemical industries have contributed greatly to Malaysia’s economy. These industries not only provide job opportunities but also earn foreign exchange for the country when the chemicals produced are exported. Some notable chemical industries in Malaysia are: 1 Plants in Pasir Gudang, Johor and Gebeng, Pahang produce chemicals such as polyethylene. Polyethylene and polypropylene are used to make many household items such as chairs, raincoats, pails and basins. Table 1.2 shows the chemicals produced by the petrochemical plants. Pharmacist 4 The expertise of forensic chemists can help the police to solve crimes. The analyses and identification of samples of blood, drugs, semen, poison, weapons and a host of other items collected from the crime scene are used as evidence to convict criminals. Table 1.2 Chemicals produced by petrochemical plants in Malaysia Petrochemical plant Analysis of DNA 5 Many types of chemicals, namely, preservatives, colourings, antioxidants, flavour enhancers, food stabilisers and artificial flavourings are used in the food processing industry. Thus, food technologists require knowledge of chemistry to ensure the correct mixture of these chemicals. Introduction to Chemistry 4 Product BASF Petronas Chemicals Sdn Bhd Acrylic polymers Titan Petrochemicals (M) Sdn Bhd Polyethylene Petrochemicals (M) Sdn Bhd Expandable polystyrene of the water affect the solubility of sugar in water. (ii) A constant variable is the factor which is kept the same throughout the experiment. To study the effect of temperature on the solubility of sugar in water, the volume of the water used in the experiment must be kept constant. The volume of water is called the constant variable. (iii) A variable which is changed during the experiment is called the manipulated variable. An experiment can be carried out by heating the water to temperatures of 30 °C, 40 °C, 50 °C, 60 °C and 70 °C. The mass of sugar that dissolves at different temperatures of water is then measured. The temperature of water is called the manipulated variable. (iv) A responding variable is the variable that responds to the change made by the manipulated variable. The amount of sugar that dissolves in water at different temperatures is called the responding variable. Thus the variables are: Manipulated variable: Temperature of the water Responding variable: Amount of sugar that dissolves in water at different temperatures Constant variable: Volume of water (c) Suggesting a problem statement This is a question which identifies the problem related to the observation. For example, 1.1 1 Name two examples of chemicals used in each of the following fields. Field Chemicals used Agriculture Medicine Food processing 1.2 Scientific Method 1 Chemistry is an experimental science similar to Biology and Physics and requires scientific research. 2 There are some basic guidelines in approaching any scientific research. These guidelines are known as the scientific method. 3 The scientific method is a systematic approach to research. It consists of the following steps: (a) Making an observation about a situation A scientific research starts with an observation. For example, a student would have observed a situation as follows: Does the solubility of sugar increase proportionally with the increase in the temperature of water? When he adds 20 g of sugar to 100 cm3 of hot water and stirred, all the sugar dissolved. However, when 20 g of sugar is added to 100 cm3 of water at room temperature and stirred, some sugar remains undissolved in the water. This will lead to the forming of a hypothesis. (d) Forming a hypothesis A hypothesis is a proposition, idea, theory or any other statement used as a starting point for discussion, investigation or study. For example, to study the effect of the temperature of water on the amount of sugar that dissolves, a probable hypothesis would be: (b) Identifying variables (i) A variable is a factor which affects the results of the experiment. The factors that affect the solubility of sugar in water are called variables. It is found that the temperature and volume 5 Introduction to Chemistry 1 2 The Asean Bintulu Fertiliser (ABF) plant in Sarawak produces urea. This is a project undertaken by some Asean countries. Urea is a nitrogenous fertiliser. Lack of nitrogen in plants will cause chlorosis whereby the leaves of the plants turn yellowish. 3 Composite Technology Research of Malaysia (CTRM) in Malacca produces fibreglass used in the making of aircraft and boats. the scientist needs to draw a conclusion based on the experimental results. (j) Writing a report Lastly, the scientist has to write a report of his/her work. This will enable him/her to communicate with other scientists. The general format of a report: 1 The higher the temperature of the water, the greater the amount of sugar that can dissolve in it. (e) Apparatus and materials When planning an experiment, suitable apparatus and materials that are required to carry out the experiment are selected. (f) Listing a work procedure The procedure is the list of steps that needs to be taken to carry out an experiment. It is advisable to list the steps in point form. (g) Carrying out the experiment After planning the experiment, a scientist will carry out the experiment according to the procedure. (h) Data collection The scientist will then record the results of the experiment accurately. He or she should not change the results of the experiment and must be honest. (i) Data interpretation and conclusion After collecting the data, the scientist will analyse the results of his/her experiment. The results can be presented in various forms, such as a table, graph or calculation. Then, Title: Aim: Problem statement: Hypothesis: Variables: (a) Manipulated variable: (b) Responding variable: (c) Constant variable(s): Materials: Apparatus: Procedure: Data and observation: Interpreting data: Discussion: Conclusion: 4 The following is an example of an experimental report. Experiment 1.1 1.1 To investigate the effect of the temperature of water on the solubility of sugar Procedure Problem statement 1 100 cm3 of water is Does the amount of sugar that dissolves in water increase measured using a when the temperature of the water increases? measuring cylinder and Hypothesis is poured into a 250 The higher the temperature of the water, the greater cm3 beaker. the mass of sugar that dissolves in it. 2 The temperature of the Variables water is recorded using a thermometer. • Manipulated variable: Temperature of water 3 A 100 cm3 beaker is • Responding variable: Amount of sugar that filled with sugar. The dissolves at different temperatures Figure 1.1 beaker and its contents • Constant variable: Volume of water and size of are then weighed and recorded as a gram. sugar 4 The sugar is added a little at a time to the water Sugar and water. Materials in the beaker using a spatula. The mixture is then Apparatus stirred using a glass rod. 3 3 5 The process is continued until no sugar can 100 cm measuring cylinder, 250 cm beaker, 100 3 further dissolve in the water. cm beaker, electronic balance, Bunsen burner, 6 The beaker and its contents (sugar) are weighed tripod stand, wire gauze, spatula, thermometer and again and recorded as b gram. glass rod. Introduction to Chemistry 6 7 The amount of sugar that dissolved in the water at room temperature is (a – b) gram. 8 The experiment is repeated by heating the water to temperatures of 40 °C, 50 °C, 60 °C and 70 °C respectively. 9 The results are recorded in Table 1.3. Results Interpreting data Room temperature 40 50 60 70 Figure 1.2 Graph of mass of sugar dissolved against temperature Initial mass of beaker and its contents (g) a b c d e Final mass of beaker and its contents (g) b c d e f A graph of the mass of sugar dissolved against temperature is plotted as shown in Figure 1.2. (Note: Both axes must be labelled with their units and the title of the graph must be stated) Mass of sugar dissolved (g) (a – b) Temperature (°C) Conclusion (b – c) (c – d) (d – e) (e – f) The amount of sugar that dissolves in the water increases when the temperature of the water increases. The hypothesis is accepted. (Note: The unit of each reading must be stated:temperature in °C and mass in gram) 1.2 (iii) the constant variable of the experiment. (c) List the materials and apparatus needed to carry out the experiment. (d) Give a brief procedure of the experiment. (e) Tabulate your results. 1 You are required to investigate whether table salt dissolves in water and kerosene. (a) State a hypothesis for the experiment. (b) State (i) the manipulated variable, (ii) the responding variable, 1.3 1 A student must develop the following good laboratory practices. (a) Positive attitudes A student should (i) have an enquiring outlook, (ii) cooperate with other students while carrying out an experiment, (iii) be honest and not alter the results of an experiment. (b) Safety (i) Do not carry out an experiment without the supervision of the teacher. (ii) Do not taste any chemicals. (iii) Do not use burning paper to light a Bunsen burner. (iv) Always check the label of the chemical before using it. Scientific Attitudes and Values in Conducting Scientific Investigations Scientific Attitudes and Values Students carrying out an experiment 7 Introduction to Chemistry 1 Table 1.3 1 (e) Accidents (i) Any chemical spilled on the body, clothing or eyes must be washed immediately with plenty of water. (ii) Any chemical unintentionally ingested must be spat out immediately and the mouth must be washed with plenty of water. (v) Dispose of all toxic waste in a proper container. (vi) Do not play with electrical appliances. (c) Wastage (i) Do not waste chemicals. Take only whatever is necessary. (ii) Switch off the gas supply or electricity when it is not required. (d) Cleanliness After carrying out an experiment, (i) the apparatus must be cleaned and returned to the same place, (ii) the table must be wiped dry with a towel or rag, (iii) all solid waste must be thrown into the dustbin and not into the sink. 1.3 1 What are the safety precautions that must be taken when carrying out the following experiments? (a) Diluting concentrated acid. (b) Heating a solution in a test tube. (c) Carrying out an experiment that involves the release of a poisonous gas. (b) Responding variable: A variable that responds to the change of the manipulated variable. (c) Constant variable: The factor that is kept constant throughout the experiment. 5 After carrying out the experiment, you have to write a report which includes the following: (a) Write the aim or problem statement (b) State the hypothesis (c) List all the variables (d) List the chemicals and apparatus used in the experiment (e) Tabulation of your data (f) Interpret your result (g) Make a conclusion 1 The scientific method is a systematic approach to research. 2 The scientific approach begins with a hypothesis. A hypothesis is an intelligent guess relating a manipulated variable with a responding variable. 3 A variable is a factor that affects the result of a reaction. For example, the mass of salt that can dissolve in water depends on the volume and the temperature of the water. Volume and temperature are called variables. 4 There are three types of variables: (a) Manipulated variable: A variable that is changed during the experiment. Introduction to Chemistry 8 1 1.1 Chemistry and Its Importance 1 The chemical used to neutralise acidity in soil is A potassium nitrate B calcium hydroxide C copper(II) oxide D sodium carbonate 2 The chemical used in raising flour is A calcium carbonate B sodium nitrate C magnesium sulphate D sodium bicarbonate 3 Which of the following careers below do not need a knowledge of chemistry? A Geologist B Forensic scientist C Meteorologist D Pharmacist 4 The branch of chemistry that studies carbon compounds is A organic chemistry B polymer chemistry C inorganic chemistry D industrial chemistry 5 Chloroform has the formula of CHCl3. Which of the following statements are true about the chloroform molecule? I It is made up of three elements. II It is made up of five elements. III The molecule consists of five atoms. IV The molecule consists of four atoms. A I and III only B I and IV only C II and III only D II and IV only 6 The substance that cannot be broken down into simpler form is called A compound C molecule B element D particle 7 Which of the following chemicals is synthetic? A Neon B Protein C Sodium hydroxide D Citric acid 8 DDT is a chemical used as pesticide. It is made up of carbon, chlorine and hydrogen atoms. The molecular formula of DDT is CCl3CH(C6H4Cl)2. What is the total number of atoms in a DDT molecule? A 3 C 18 B 17 D 28 9 Chemical X is used as electrolyte in the accumulator. Chemical Y is used in soap making. What is chemical X and Y? X Y A Sulphuric acid Sodium hydroxide B Sodium hydroxide Sulphuric acid C Hydrochloric acid Sodium hydroxide D Sodium hydroxide Hydrochloric acid 1.2 Scientific Method 10 The factor that affects the result of an experiment is called a A solute C result B solution D variable For questions 11 – 13, use the information given below: Hydrogen peroxide decomposes as represented by the equation: 2H2O2 → 2H2O + O2 A student is required to study the effect of magnesium oxide and manganese(IV) oxide on the rate of decomposition of hydrogen peroxide. 9 11 What is the manipulated variable of the experiment? A Magnesium oxide and manganese(IV) oxide B Mass of magnesium oxide and manganese(IV) oxide C Temperature of hydrogen peroxide solution D Concentration of hydrogen peroxide solution 12 What is the constant variable of the experiment? I Volume of oxygen released II Mass of magnesium oxide and manganese(IV) oxide III Temperature of hydrogen peroxide solution IV Concentration of hydrogen peroxide solution A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV 13 The responding variable for the experiment is A the rate of release of oxygen gas. B the decreasing rate in volume of hydrogen peroxide. C the rate of increase in concentration of hydrogen peroxide. D the decreasing rate in mass of the metal oxide. 14 Magnesium ribbon reacts with hydrochloric acid as shown. Mg + 2HCl → MgCl2 + H2 If you are required to study the effect of concentration of hydrochloric acid on the rate of reaction above, what variables must be constant? I Time of reaction II Temperature of hydrochloric acid III Size of beaker IV Length of magnesium ribbon A IV only B II and IV only C I, II and IV only D II, III and IV only Introduction to Chemistry 1 Multiple-choice Questions 15 “The greater the quantity of sodium chloride added to ice, the lower its melting point”. If you are required to study the above hypothesis, what is the manipulated variable? A Mass of ice B Types of salt added C Mass of salt added D Temperature of ice 16 “Without water iron will not rust”. Which of the following is correct in carrying out the experiment to prove the statement above? 1 Manipulated variable Constant variable A Presence or absence of water Rusting of iron B Presence or absence of water Presence or absence of air C Presence or absence of water Presence of air D Presence or absence of air Presence of water 17 A student wants to find out the effect of temperature on the solubility of sugar in water. Which of the following is correct? Manipulated variable Responding variable Constant variable A Temperature of water Mass of sugar dissolved Volume of water B Volume of water Mass of sugar dissolved Temperature of water C Mass of sugar dissolved Temperature of water Volume of water D Temperature of water Mass of sugar dissolved Humidity of the air 18 An experiment is carried out to study the solubility of sodium chloride in water and in benzene. What is the (i) manipulated variable (ii) constant variable of the experiment? Manipulated variable Constant variable A Solvent Rate of stirring B Solvent Mass of sodium chloride C Solute Volume of solvent D Solute Rate of stirring 19 A hypothesis I is a law of science. II can be a true or false statement. III is a conclusion derived from the result of the experiment. IV is a statement that relates the manipulated variable and the responding variable. A I and III only B II and IV only C I, II and III only D II, III and IV only Introduction to Chemistry 10 20 Q Analyse data R Observe a situation S Make a hypothesis T Carry out the experiment U Collect data The steps above are the steps taken to carry out a scientific investigation. The correct order in carrying out the investigation is A S, R, T, U, Q B R, S, T, U, Q C R, T, U, Q, S D R, T, U, S, Q 21 During electrolysis, the mass of metal deposited at the cathode is dependent on the time and the amount of current passed through the electrolyte. If you are required to show that the mass of metal deposited is proportional to the current passed through the electrolyte, what are the variables in this experiment? Manipulated variable Constant variable A Time Types of electrodes used B Time Amount of current C Amount of current Types of electrodes used D Amount of current Time 1.3 Scientific Attitudes and Values 22 What precautions must you take when storing concentrated nitric acid? A Store it in a dark place B Store it in a fume cupboard C Store it in a locked cupboard D Store it away from any Bunsen burner 23 What precaution must you take when diluting concentrated sulphuric acid? A Add the concentrated sulphuric acid to water 25 A bottle of chemical has a label shown in the diagram. What does this label represents? A Flammable chemical B Corrosive chemical C Radioactive chemical D Oxidising chemical 24 A bottle of chemical has a label as shown in the diagram. What precaution must be taken when storing this chemical? A Store it in a dark place B Store it in a fume cupboard C Store it in a locked cupboard D Store it away from any Bunsen burner 26 When heating a solution in a boiling tube, what precaution must you take? A Never heat the solution too strongly B Never hold the boiling tube vertically C Never use a Pyrex boiling tube to heat the solution D Never direct the mouth of the boiling tube at your classmates 27 An experiment should be carried out in the fume cupboard if it involves A the release of poisonous gas. B the release of flammable gas. C the use of corrosive chemicals. D the use of oxidising chemicals. 28 Why is it important to understand the experimental procedures before carrying out the experiment? I To prevent wastage of chemicals II To prevent accidents from happening III To prevent repetition of the experiment IV To know what apparatus is needed to carry out the experiments A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV Structured Questions seconds. From the graph in (i), determine the concentration of the hydrochloric acid solution. [1 mark] 1 Table 1 shows the time taken for a 5 cm length of magnesium ribbon to dissolve in 50 cm3 of dilute hydrochloric acid of different concentrations. 2 Table 2 shows the mass of two salts P and Q that dissolved in 100 cm3 of water at different temperatures. Concentration of 0.1 0.2 0.3 0.4 0.5 hydrochloric acid (mol dm–3) Time taken for a 5 cm magnesium ribbon to dissolve (s) Temperature (°C) 30 26 22 18 14 Table 1 (a) State (i) the manipulated variable, (ii) the responding variable and (iii) the constant variable of the experiment above. [3 marks] (b) State a hypothesis for this experiment. (c) [1 mark] Solubility of salt (mass of salt soluble in 100 cm3 of water) Salt P (g) Salt Q (g) 30 5 7 40 10 14 50 15 21 60 20 28 70 25 35 Table 2 (i) Plot a graph of concentration of hydrochloric acid against time taken for the magnesium to dissolve. [5 marks] (ii) If a 5 cm length of magnesium ribbon is added to a hydrochloric acid solution of unknown concentration, the time taken for the magnesium ribbon to dissolve is 17 (a) State one other variable, besides temperature, that affects the solubility of salt. [1 mark] (b) In the experiment above, state (i) the manipulated variable (ii) the constant variable (iii) the responding variable 11 [3 marks] Introduction to Chemistry 1 B Add water to the concentrated sulphuric acid C Mix equal volumes of the concentrated sulphuric acid and water together D Mix one volume of concentrated sulphuric acid to three volumes of water together (c) Plot a graph of the solubility of salts P and Q against temperature on the same axis. [4 marks] (a) State a hypothesis for the experiment above. [1 mark] (d) What can you conclude from the graphs in (c)? (b) What is the (i) manipulated variable (ii) responding variable (iii) constant variable when carrying out the experiment? [3 marks] [2 marks] 1 3 The procedure below shows the sequence in carrying out an experiment to study the effect of the temperature of water on the mass of sugar that can dissolve. (c) Calculate the mass of sugar that dissolved in water at various temperatures and write your answer in the right column of the table. [2 marks] Procedure: • Initial mass of beaker P and sugar is taken (a gram). • 50 cm3 of water is poured into a separate 100 cm3 beaker. The water is heated to 30 °C. • Sugar is added to the 50 cm3 of water at 30 °C a little at a time while stirring the mixture until no more sugar can further dissolve. • The final mass of beaker P and sugar is taken (b gram). The mass of sugar that dissolved is (a – b) gram. • The experiment is repeated by dissolving the sugar in water heated to 40 °C, 50 °C, 60 °C and 70 °C. (d) Plot a graph of the mass of sugar that dissolved against the temperature of water. [3 marks] (e) From your graph, estimate the mass of sugar that can dissolve in water at 45 °C. [1 mark] 4 (a) Name four chemicals used in food processing. [4 marks] (b) Name six careers that need a knowledge of chemistry. [6 marks] The results are tabulated in Table 3. Initial mass Final mass Temperature of beaker P of beaker P of water (°C) and contents and contents (g) (g) 30 92.50 87.50 40 87.50 77.50 50 77.50 62.50 60 62.50 42.50 70 42.50 17.50 (c) Name three contributions of chemical industries to the country. [3 marks] Mass of sugar dissolved (g) (d) Name five scientific values that must be observed when carrying out scientific research. [5 marks] (e) Name two types of chemicals that can increase the yield of crops. [2 marks] Table 3 Essay Questions (b) The list below shows the steps involved in carrying out scientific research: 1 ‘Without air, an iron nail will not rust’. You are required to plan an experiment to verify the statement. (a) List the apparatus and materials needed to carry out the experiment. [3 marks] Making a hypothesis, making a conclusion, collecting data, making an inference, making an observation, carrying out an experiment, interpreting data, identifying variables and planning the procedure of the experiment. (b) State (i) the manipulated variable, (ii) the responding variable and (iii) the constant variable of the experiment above. [3 marks] (i) Arrange the steps in the correct order. (c) Briefly write the procedure for the experiment. [8 marks] [10 marks] (d) Tabulate your results. (ii) Explain the difference between inference and hypothesis. [4 marks] (iii) State three ways of presenting the experimental results. [6 marks] [4 marks] 2 (a) Explain the meaning of the scientific method. [2 marks] Introduction to Chemistry 12 Experiments 1 ‘The greater the volume of water, the higher the solubility of salt’. Plan an experiment to prove the statement. Your answer should include the following items: (a) Aim of experiment (b) Statement of hypothesis (c) All variables (d) List of materials and apparatus (e) Procedure (f) Tabulation of data [3 marks] [3 marks] [3 marks] [3 marks] [3 marks] [3 marks] Concentration of sodium hydroxide solution (mol dm–3) 0.1 0.2 0.3 0.4 0.5 pH value 13.0 13.3 13.5 13.6 13.7 1 2 The table shows the pH values of 25 cm3 sodium hydroxide solutions of different concentrations measured by a student using a pH meter. (a) State the variables of this experiment. [3 marks] (b) Suggest a hypothesis for the experiment. [3 marks] (c) Plot a graph of pH value against concentration of the NaOH solution. [3 marks] (d) Using the graph that you have plotted, determine (i) the pH value of a sodium hydroxide solution with a concentration of 0.35 mol dm–3. (ii) the concentration of NaOH solution with a pH value of 13.4. [3 marks] 3 Manganese(IV) oxide is a catalyst that speeds up the decomposition of hydrogen peroxide (H2O2) to form water and oxygen gas as represented by the equation: 2H2O2(l) → 2H2O(l) + O2(g) A student carried out an experiment by adding different amounts of manganese(IV) oxide to 50 cm3 of 0.2 mol dm–3 hydrogen peroxide solution. The table shows the results obtained by the student. Quantity of manganese(IV) oxide (g) 0.2 0.4 0.6 0.8 1.0 Time taken to collect 50 cm of oxygen (s) 30 25 20 15 10 3 (a) State the (i) manipulated variable, (ii) responding variable, (iii) constant variable of the experiment. [3 marks] (b) What can you conclude from the results of the experiment? 13 [3 marks] Introduction to Chemistry FORM 4 THEME: Matter Around Us CHAPTER 2 The Structure of the Atom SPM Topical Analysis 2008 Year 1 Paper 3 2 Section A Number of questions 1 — 2 5 2009 B – 1 2010 3 2 C A B C – 1 — 2 – – – 6 – 1 2 2011 2 3 A B C 1 – – 1 – 2 3 A B C 1 – – 3 – ONCEPT MAP MATTER Kinetic theory of matter Changes in states of matter Diffusion in a solid, liquid and gas Atomic structure Particles in matter: atom, molecule and ion Subatomic particles: proton, electron and neutron Determination of the melting and freezing points of naphthalene Electron arrangement in atoms and valence electrons Symbols of elements A Z X Isotopes 2.1 Matter (b) When the gas tap in the laboratory is turned on, the smell of the gas is immediately detected. This shows that the gas is also made up of particles in motion. 7 An element is a substance that cannot be made into anything simpler by means of a chemical reaction. 8 The particles in some elements are made up of atoms. For example, metals like gold, copper, iron, zinc are all made up of atoms. SPM ’08/P1 1 Substance 2 1 Chemistry is the study of matter, its composition and the changes it undergoes. 2 Matter is anything that occupies space and has mass. In other words, matter is anything that has volume and mass. 3 Examples of matter are books, pens, chairs, water, air and plants. Examples of non-matter are electricity and light. 4 The particle theory of matter states that matter is made of very tiny discrete particles. The particulate nature of matter is investigated in Activity 2.1. 5 Elementary particles that make up matter may be atoms, molecules or ions. ’04 Chemical formula Naphthalene C10H8 Iron Fe Sodium chloride NaCl Figure 2.1 Copper foil is made up of atoms 9 A compound is a substance that can be made into something smaller by means of a chemical reaction. 10 Compounds contain more than one element. The elements in a compound are not just mixed together. They are joined by strong forces called chemical bonds. Compounds do not have the same properties as the elements they contain. Compounds are (a) formed by chemical reactions, and (b) they have different properties from the elements they contain. 11 The particles in compounds may be molecules or ions. Molecules are made up of two or more atoms held together by chemical bonds. Molecules are particles that are not charged. 12 A molecule may consists of atoms of the same element, for example, oxygen molecules (O2), nitrogen molecules (N2), hydrogen molecules (H2) and sulphur molecules (S8) (Figure 2.2(a)). 13 A molecule may also consist of dissimilar atoms of two or more elements. For example, a water molecule (H2O) consists of one oxygen and two hydrogen atoms, and a carbon dioxide molecule (CO2) consists of one carbon and two oxygen atoms (Figure 2.2(b)). 14 Some molecules can be very large. For example, quinine which is a drug used to treat malaria patients has the formula C20H24N2O2. State the particles present in each of the above substances. Solution Naphthalene – molecules, iron – atoms, sodium chloride – ions. All metals and noble gases are made up of atoms. A compound formed between non-metallic elements (example: naphthalene, C10H8), is made up of molecules. A compound formed between a metal and a non-metal (example: sodium chloride, NaCl) is made up of ions. 6 The existence of these particles is supported by some observations. Some examples are: (a) When a drop of ink falls into a glass of water, the colour of the ink spreads throughout the water. This shows that ink is made up of particles in Dropping ink into a motion. glass of water 15 The Structure of the Atom (a) The particles (atoms, molecules or ions) possess kinetic energy. They are in constant motion and constantly collide with each other. (b) The velocities of the particles in the three physical states of matter—solid, liquid and gas—are different. (c) The higher the temperature, the higher the kinetic energy, as the velocity of the particles increases. (d) At a given temperature, the lighter particles move faster than the heavier ones. 2 In 1827, Robert Brown (a botanist) made an observation through a microscope. He found that pollen grains on the surface of water are in constant motion. He explained that the pollen grains are moving because the moving water molecules are constantly colliding with the pollen grains. The visible motion of these pollen grains is called the Brownian motion. 3 The Brownian motion gives the evidence that a liquid consists of particles in constant movement. (a) Model of nitrogen, oxygen and sulphur molecules (b) Model of carbon dioxide and water molecules 2 Figure 2.2 15 However, some compounds consist of atoms or a group of atoms that carry positive or negative charges. These charged particles are called ions. For example, table salt, NaCl, consists of sodium ions (Na+) and chloride ions (Cl–) (Figure 2.3). The rust on an iron nail consists of iron(III) ions (Fe3+) and oxide ions (O2–). Figure 2.3 Model of sodium chloride crystal 16 Ions which are positively-charged are called cations. For example, sodium ions (Na+) and iron(III) ions (Fe3+) are cations. 17 Ions which are negatively-charged are called anions. For example, chloride ions (Cl–) and oxide ions (O2–) are anions. 18 Generally, metals form positive ions and nonmetals form negative ions. Some examples of cations and anions are given in Table 2.1. Figure 2.4 Pollen grain being bombarded by water molecules 4 Another evidence of the movement of particles is diffusion. Diffusion is the random movement of particles from a region of high concentration to a region of low concentration. Table 2.1 Examples of positive ions (cations) You can smell perfume while you walk past cosmetic counters.The perfume particles have left the open perfume bottles and spread out through the air by diffusion. H+, K+, Cu2+, Al3+, NH4+, Mg2+, Ca2+, Zn2+, Pb2+ and Ag+ Examples of negative Br–, I–, OH–, NO3–, SO42–, CO32–, PO43–, O2–, S2– ions (anions) and S2O32– The Kinetic Theory of Matter 1 The kinetic theory is an extension of the particle theory of matter. According to the kinetic theory: The Structure of the Atom 5 There are three states of matter, namely, solid, liquid and gas. Table 2.2 shows the comparison between the three states of matter. 16 SPM Table 2.2 Comparison between the three states of matter solid liquid The particles are very closely packed. The particles are closely packed but there are more empty spaces between them compared to the solid state. The particles are very far apart from each other. Forces of The very strong forces attraction of attraction restrict the between particles movement of the particles. The particles in a solid are held in fixed positions. The forces of attraction are weaker than in the solid state. The particles are no longer held in fixed positions. The forces of attraction are very weak. The particles move randomly in all directions at great speed. Volume and shape Solids have fixed volumes and shapes. Liquids have fixed volumes. However, they do not have fixed shapes but take the shapes of the containers. Gases do not have fixed shapes or volumes. Types of movement Vibration and rotation Vibration, rotation and translation Vibration, rotation and translation Kinetic energy of particles The kinetic energy of the particles are low. The kinetic energy of the particles are high, on average. The kinetic energy of the particles are very high and they move at high speed. Compressibility Very difficult to be compressed because the particles are packed closely Not easily compressed because the particles are packed quite closely Easily compressed because the particles are very far apart Arrangement of particles Rate of diffusion Very low Average Diffusion gas Very high SPM ’08/P2 1 Diffusion refers to the process by which particles intermingle as a result of their kinetic energy of random motion. 2 Figure 2.5(a) shows a container that consists of gases A and B. The two gases are separated by a partition. The particles of both gases are in constant motion and make numerous collisions with the partition. 3 If the partition is removed as in Figure 2.5(b), the gases will mix because of the random motion of their particles. In time, a uniform mixture of gases A and B particles will be produced in the container. 4 The rate of diffusion depends on the temperature and the molecular mass of the particles. The higher the molecular mass, the lower the rate of diffusion. 17 The Structure of the Atom 2 States of matter ’11/P2 To investigate the diffusion of particles in a gas, liquid and solid SPM ’09/P1 Apparatus Two gas jars with plastic covers, beaker, teat pipette, boiling tube, spatula and rubber stopper. Materials Liquid bromine, potassium manganate(VII), KMnO4 crystals, water and hot jelly solution. Procedure (A) Diffusion in a gas 2 1 A few drops of liquid bromine are dropped into a gas jar using a teat pipette. 2 The gas jar is covered with a gas jar cover. 3 An empty gas jar is placed upside down on top of the first jar. 4 The cover is removed and any colour change is recorded. The time taken for the brown bromine vapour to spread into the second gas jar is recorded. Figure 2.6 (B) Diffusion in a liquid 2 filled with water. 1 A beaker is — 3 2 A few potassium manganate(VII) crystals are placed at the bottom of the water using a spatula. 3 Any colour change is recorded. The time taken for the purple manganate(VII) ions to spread throughout the water is recorded. Figure 2.7 (C) Diffusion in a solid 1 Some freshly cooked jelly solution is poured into a boiling tube until it is almost full. 2 The jelly is allowed to set. 3 A small potassium manganate(VII) crystal is placed on top of the jelly. 4 The boiling tube is then stoppered using a rubber stopper. 5 Any colour change is recorded. The time taken for the purple manganate(VII) ions to spread throughout the solid jelly is recorded. Figure 2.8 Results Activity 2.1 Experiment Observation A The brown bromine vapour spreads out into the upper gas jar. The time taken is very short. B After about 10 minutes, the purple colour of the manganate(VII) ions had spread throughout the water. C After a week, the purple colour of the manganate(VII) ions had spread throughout the solid jelly. The Structure of the Atom 18 Discussion 2 1 Diffusion has taken place in the gas (air in experiment A), liquid (water in experiment B) and solid (jelly in experiment C). 2 The rates of diffusion of the particles in the solid, liquid and gaseous states are different. It is highest in gases, lower in liquids and lowest in solids. 3 This shows that there are more and bigger spaces between particles in the gas. The spaces between liquid particles are smaller. The particles in the solid state are very close with little space between them. 4 The occurence of diffusion proves that matter (bromine and potassium manganate(VII)) consist of particles in constant motion. 5 The diffusion experiments show that because particles possess kinetic energy, they are in constant motion. SPM The Changes in the States of Matter ’10/P2 1 A substance can be changed from one state into another when it is heated or cooled. 2 The changes in the state of the substance can be explained using the kinetic theory model. Heating Heating Solid Liquid Gas 1 The particles in a solid are packed closely in a fixed pattern. 2 When the solid is heated, the particles receive heat energy. The kinetic energy of the particles increases and the particles vibrate faster. 3 At the melting point, the particles vibrate so much that they break away from their fixed positions. The solid becomes a liquid. 4 The temperature at which the solid changes into the liquid state is called the melting point. 1 When a liquid is continuously heated, the particles receive more energy and move even faster. They collide with each other more often. 2 At the boiling point, the particles receive enough energy to overcome the forces of attraction holding them together. The particles in the liquid state break loose to become the gaseous state. 3 When the liquid is cooled, the movement of the particles slows down. Stronger forces of attraction between the particles are formed. 4 The particles are arranged in an orderly manner in the solid state. The process whereby the liquid changes into a solid is called solidification. The temperature at which this process occurs is called the freezing point. 5 The melting point and the freezing point of a substance have the same value. 1 When a gas is cooled, the particles lose kinetic energy. The movement of the particles slows down. 2 The forces of attraction between the particles are formed which hold the particles together in the liquid state. 3 The process whereby the gas changes into a liquid is called condensation. 4 The temperature at which the gas condenses to the liquid state is the same as the boiling point. Cooling Cooling 19 The Structure of the Atom 3 Examples of substances that undergo sublima tion are iodine, ammonium chloride and solid carbon dioxide (dry ice). The process in which substances change directly from the gaseous to the solid state is also called sublimation. 2 Melting Point and Boiling Point 1 No two substances have the same melting and boiling points. We can thus identify a substance by its melting and boiling points. 2 The melting and boiling points of a substance will change when there is a small amount of impurity in it. For example, the melting point of pure water is 0 °C and its boiling point is 100 °C. A small amount of salt added to the water will decrease its melting point to –2 °C and increase its boiling point to 102 °C. 3 As the melting and boiling points of an impure substance will deviate slightly from its standard values, we can determine the purity of a substance by the melting and boiling points of the substance. When a state of matter gains or loses heat, it undergoes a change. A gain in heat is called an endothermic change. A loss in heat is called an exothermic change. Sublimation 1 Certain substances do not melt when heated. They change directly from the solid to the ’11/P1 gaseous state. 2 This process is called sublimation. SPM To determine the melting and freezing points of naphthalene (A) Heating of naphthalene 1 3 spatulas of naphthalene powder are placed in a boiling tube. 2 A 500 cm3 beaker is filled with water until it is 3 about — full. It is then placed on a tripod stand. 4 3 The boiling tube containing naphthalene is clamped in the beaker of water, making sure the naphthalene powder is below the water level of the water bath. 4 The water bath is heated until it reaches a temperature of about 65 °C as shown in Figure 2.9. The water is then heated with a low flame. 5 A stopwatch is started and the temperature of the naphthalene is recorded at 30-second intervals until the temperature reaches 90 °C. The naphthalene is stirred continuously during the experiment. 6 The results are recorded in a table. Apparatus Boiling tube, retort stand and clamp, tripod stand, Bunsen burner, wire gauze, thermometer (0 – 110 °C), 500 cm3 beaker, 250 cm3 conical flask, test tube holder and stopwatch. Materials Naphthalene and water. Procedure Figure 2.9 Heating of naphthalene Activity 2.2 Figure 2.10 Cooling of naphthalene The Structure of the Atom (B) Cooling of naphthalene 1 The boiling tube containing the molten naphthalene is removed from the hot water bath using a test tube holder. 2 It is immediately transferred into a conical flask to be cooled slowly as shown in Figure 2.10. 20 Discussion 1 In the heating of naphthalene, a water bath is used instead of direct heating. This is to ensure that an even heating process is carried out. 2 In the cooling of naphthalene, the boiling tube containing the liquid naphthalene is cooled inside a conical flask. This is to ensure that an even cooling process is carried out. 3 Stirring the naphthalene continuously also ensures even heating or cooling. 4 A water bath is suitable in this experiment because the melting point of naphthalene is below 100 °C, the maximum temperature that can be attained by the water bath. 5 If the melting point of the substance is above 100 °C, the water bath will have to be replaced by an oil bath or a sand bath. 6 Besides naphthalene, the other substance that is suitable for heating by water bath is acetamide. 7 The heating curve of naphthalene consists of three regions: AB, BC and CD as in Figure 2.11. Results (A) Heating of naphthalene Time (s) Temperature (°C) 0 30 60 90 120 150 180 210 (B) Cooling of naphthalene Time (s) Temperature (°C) 0 30 60 90 120 150 180 210 Region in the graph State of substance and the energy change Region AB Naphthalene is in the solid state. As napthalene is heated, heat energy is SPM ’04/07 converted to kinetic energy. Kinetic P2 energy increases and the molecules vibrate faster about their fixed positions. Temperature increases as the molecules receive more heat energy. Analysis of data Point B 1 A graph of temperature SPM against time is plotted ’10/P1 for the heating of naphthalene. The graph is shown in Figure 2.11. 2 A graph of temperature against time is plotted Figure 2.11 Heating curve Region BC for the cooling of of naphthalene SPM ’11/P1 naphthalene. The graph is shown in Figure 2.12. 3 When plotting a graph, make sure that: (a) The axes are labelled with their units. (b) The points are transferred correctly. Figure 2.12 Cooling curve (c) The curve is smooth. of naphthalene 21 As the kinetic energy of the molecules increases, the molecules vibrate faster. At point B, some molecules vibrate so much that they break away from their fixed positions. The solid naphthalene begins to melt. Naphthalene now consists of a mixture of solid and liquid. At this region the temperature remains constant because the heat energy supplied by the water bath is the same amount as the heat energy absorbed. Heat energy is absorbed to overcome the forces of attraction holding the naphthalene molecules together in the solid state. The heat absorbed to overcome the forces of attraction is called the latent heat of fusion. Latent heat of fusion of The Structure of the Atom 2 3 The stopwatch is started and the temperature of the naphthalene is recorded at 30-second intervals until it drops to about 70 °C. The naphthalene is stirred continuously during the experiment. 4 The results are recorded in a table. Region in the graph 2 Point C State of substance and the energy change Region in the graph State of substance and the energy change a substance is the heat required to convert a solid into a liquid without a change in temperature. moving except for small vibrations. At point Q, the liquid naphthalene begins to solidify or freeze. All the naphthalene has completely melted. Region QR Naphthalene now consists of a mixture of liquid and solid. At this region the temperature remains constant because the heat energy lost to the environment is the same amount as the heat energy released. Latent heat of fusion is released when forces of attraction are formed between the molecules as the liquid naphthalene solidify (or freezes). Region CD Naphthalene is in the liquid state. As the liquid naphthalene is heated, the molecules gain more heat energy. The temperature continues to increase. 8 The cooling curve of naphthalene consists of three regions: PQ, QR and RS as in Figure 2.12. Point R All the naphthalene has completely solidified. Region PQ Naphthalene is in the liquid state. The liquid naphthalene loses heat to the environment. The kinetic energy of the molecules decreases as the temperature decreases. Region RS Naphthalene is in the solid state. The solid naphthalene continues to lose heat to the environment and hence the temperature drops down to room temperature. Point Q Conclusion The melting point and the freezing point of naphthalene is 80 °C. Region in the graph State of substance and the energy change As the kinetic energy of the molecules decreases, the molecules move slower. At point Q, some molecules stop 2 ’09 A I and III only B II and IV only C I, II and III only D I, II and IV only Comment From time 0 to t1 the substance loses heat to the surroundings. Hence the temperature decreases. (Statement I is incorrect) From time t1 to t2 condensation takes place and heat energy is released. The kinetic energy of particles becomes lower and the forces of attraction become stronger. (Statement II is correct) The graph shows the cooling curve for gas X. Which of the following statements are true? I From time 0 to t1 heat energy is absorbed. II From time t1 to t2 forces of attraction between particles become stronger. III From time t2 to t3 the kinetic energy of particles increases. IV From time t3 to t4 heat energy released is equal to the heat lost to the surroundings. The Structure of the Atom From time t2 to t3 the particles continue to lose heat to the surroundings. Hence the kinetic energy of particles decreases. (Statement III is incorrect) From time t3 to t4 freezing takes place. The temperature of the substance remains constant because during freezing, heat energy released is equal to the heat lost to the surroundings. (Statement IV is correct) Answer B 22 2.1 1 State the type of particles (atoms, molecules or ions) that make up the substances below. (a) Ammonia gas (d) Potassium iodide (b) Sodium chloride (e) Copper wires (c) Iron nail (f) Cooking oil the heating curve for Figure 2.13 John Dalton and his model 3 However, Dalton’s atomic model had its weakness. It was found that: (a) The atom is not the smallest particle in an element. There are subatomic particles (proton, electron and neutron) in an atom. (b) A radioactive atom decomposes spon taneously, which means that an atom can be destroyed. A new atom can also be created by a process called transmutation. (c) Not all atoms of an element are alike. They may differ in atomic mass. For example, hydrogen has three isotopes 11H, 21H and 31H. 4 In 1897 J. J. Thomson discovered negativelycharged particles which he called electrons. Thomson then suggested that an atom is a positively-charged sphere with electrons embedded in it like a raisin pudding. (a) State the melting point of naphthalene. (b) What is the physical state of naphthalene at time t second? (c) Why does the temperature remain constant at region BC although heating is carried on? (d) Draw the cooling curve obtained when the molten naphthalene is cooled from T3 to room temperature. 2.2 The Atomic Structure The Historical Development of the Atomic Model 1 The concept of the atom originated from Democritus, a Greek philosopher. He proposed that if a piece of gold is divided repeatedly, it will reach a state whereby the smallest particle, which is indivisible, is obtained. He called the smallest indivisible particle atomos, which means ‘indivisible’ in Greek. 2 In 1808, John Dalton proposed the atomic theory. In this theory, Dalton proposed that: (a) All elements are made up of small indivisible particles called atoms. (b) Atoms are neither created nor destroyed in chemical reactions. (c) The atoms of an element are alike, but differ from the atoms of other elements. (d) When atoms combine, they do so in a simple ratio. (e) All chemical reactions result from the combination or separation of atoms. Figure 2.14 J. J. Thomson and his model 5 In 1911, Ernest Rutherford bombarded a thin gold foil with alpha particles (helium nuclei, He2+). (a) It was found that most of the alpha particles passed directly through the gold foil without deflection. Rutherford then suggested that most of the atom must be empty space. Figure 2.15(a) Rutherford’s experiment 23 The Structure of the Atom 2 2 The graph shows naphthalene. 7 In 1932, James Chadwick discovered rays of electrically neutral subatomic particles which he called neutrons. The neutron has a mass almost the same as that of a proton. Chadwick suggested that the nucleus of the atom contains protons and neutrons, and the nucleus is surrounded by a cloud of electrons. 2 Figure 2.15(b) Magnified view showing alpha particles deflected by the nuclei of gold atoms (b) However, some of the alpha particles were deflected at very acute angles. To explain the deflection of the alpha particles, Rutherford proposed that all the positive charge of an atom is concentrated in the nucleus, which repelled the positivelycharged alpha particles in the opposite direction. Further experimental studies led to the discovery of positive particles in the nucleus. Rutherford called the positively-charged particles protons. (c) Rutherford proposed that an atom consists of a positively-charged nucleus with a cloud of electrons surrounding the nucleus. Figure 2.18 James Chadwick and his model SPM 8 The atomic model in the present day is based on the contributions of the above scientists. In this atomic model: (a) The nucleus of an atom consists of protons and neutrons occupying a small space in the centre of the atom. (b) Electrons are moving around the nucleus in permissible orbits or electron shells (also known as quantum shells). ’08/P1 Figure 2.16 Ernest Rutherford and his model Subatomic Particles of an Atom 6 In 1913, Niels Bohr proposed that the electrons in the atom are arranged in permitted orbits called electron shells surrounding the nucleus. An atom is made up of three smaller particles which are called protons, neutrons and electrons. These particles are called subatomic particles. Table 2.3 shows the relative masses and charges of these particles. Table 2.3 The symbols, relative masses and the charges of subatomic particles Subatomic particle Figure 2.17 Niels Bohr and his model The Structure of the Atom 24 Symbol Relative mass Charge Proton p 1 +1 Electron e 1 — — — — 1840 –1 Neutron n 1 0 3 that the number of neutrons in phosphorus is 31 – 15 = 16. 6 The relative masses of the proton and neutron are almost similar. However, the relative mass of the electron is very small. So the mass of an atom is determined by the number of protons and neutrons in the atom. 7 The nucleon number and proton number of SPM an element is written in the following way: ’09/P2 ’05 The diagram shows a model of an atom. 2 Who introduced this model? A Niels Bohr B J. J. Thomson C John Dalton D Rutherford Answer A Niels Bohr. He proposed that electrons are arranged in shells surrounding the nucleus. Proton Number and Nucleon Number 1 Protons and neutrons are located in the nucleus and the electrons are arranged in electron shells surrounding the nucleus. 2 The nucleus is positively-charged because it contains protons, each of which carry a positive charge. 3 The proton number of an element is the number of protons in its atom. The proton number is also known as the atomic number. Each element has its own proton number. No two different elements can have the same proton number. For example, sodium, with a proton number of 11 means that it has 11 protons in its nucleus and an element with 11 protons in its nucleus must be sodium. 4 In a neutral atom, the proton number also tells us the number of electrons. For example, the proton number of magnesium is 12. Therefore, a magnesium atom has 12 protons and 12 electrons. The proton number of nitrogen is 7 and hence a nitrogen atom has 7 protons and 7 electrons. 5 The nucleon number (also known as the SPM mass number) of an element is the sum of the ’11/P2 number of protons and neutrons in its atom. A student need not memorise the proton number and nucleon number. It will be given in the examination. The proton number is smaller than the nucleon number. 4 ’03 State the number of protons, electrons and neutrons 37 in a chlorine atom, 17 Cl. Solution 17 protons, 17 electrons and 20 neutrons (37 – 17 = 20) Symbols of Elements 1 Each element is represented by a symbol, consisting of either one letter or two letters of the alphabet. 2 Some elements are represented by the first letter of its name. Examples are in the following table. Name of element Hydrogen Nitrogen Oxygen Fluorine Sulphur Nucleon Number of Number of = + number protons neutrons OR Nucleon Proton Number of = + number number neutrons Symbol H N O F S 3 The names of some elements start with the same letter. For example, the names of the elements Nitrogen, Neon, Nickel and Nobium start with the letter ‘N’. Therefore, a second letter is added to differentiate between these elements. The second letter used is always a small letter. Examples are in the following table. For example, a sodium atom has 11 protons and 12 neutrons; hence the nucleon number of sodium is 23. The proton number of phosphorus is 15 while its nucleon number is 31. This means 25 The Structure of the Atom and 53 stand for? How many protons, electrons and neutrons are there in an iodine atom? Symbol Si Ne Cl Ca Br Mg Name of element Silicon Neon Chlorine Calcium Bromine Magnesium 2 (a) A list of elements are represented by the letters given below: 11 5 Which two letters represent the same element? Explain your answer. (b) State four facts that you can derive from the nuclear symbol, 27 Al. 13 2 4 Some elements are represented by the letters of their Latin names. For example, Name of element Silver Copper Iron Gold Lead Tin Potassium Sodium Mercury Latin Name Symbol Argentum Cuprum Ferrum Aurum Plumbum Stannum Kalium Natrium Hydrargyrum Ag Cu Fe Au Pb Sn K Na Hg A, 126 B, 24 C, 23 D, 146 E and 147 F 12 11 2.3 Isotopes and Their Importance Isotopes 1 Isotopes are atoms of the same element with the same proton number but different nucleon ’10/P1 numbers. Alternatively, isotopes can be defined as atoms of an element with the same number of protons but different numbers of neutrons. 2 Many elements exhibit the phenomenon of isotropy, whereby an element can have more than one type of isotope. 3 The isotopes of an element have the same chemical properties because they have the same electron arrangement but their physical properties such as densities and melting points differ. 4 Table 2.4 shows examples of isotopes of some elements. SPM 2.2 1 (a) An atom of uranium (U) has 92 protons and 143 neutrons. What is the proton number and nucleon number? Write its atomic symbol. (b) Seaweed is rich in the element iodine, represented by 127 l. Lack of iodine in our diet 53 can cause goiter. What do the numbers 127 Table 2.4 Examples of isotopes of some elements SPM ’07/P2 Hydrogen, 11H Deuterium, 12H Tritium, 13H Proton number 1 1 1 Nucleon number 1 2 3 Number of protons 1 1 1 Number of neutrons 0 1 2 Percentage abundance 99.985% 0.015% Man-made isotope Carbon-12, 126C Carbon-13, 136C Carbon-14, 146C 6 6 6 12 13 14 6 6 6 6 7 8 98.1% 1.1% Trace amount 17 17 35 37 17 17 18 20 75.5% 24.5% 8 8 8 16 17 18 8 8 8 8 9 10 99.757% 0.038% 0.205% Element 35 Cl Chlorine-35, 17 37 Chlorine-37, 17 Cl Oxygen-16, 168O Oxygen-17, 178O Oxygen-18, 188O Same Different Same Different Nucleon number = Number of protons + Number of neutrons The Structure of the Atom 26 5 Some elements, such as fluorine, F, have only one isotope. However, most elements have more than one isotope. 6 The relative atomic mass of an element is based on the average mass of all the isotopes of the element. For example, the relative atomic mass of chlorine is 35.5 because chlorine has 75% of 35 Cl and 25% of 37 Cl. 17 17 7 In an element, some isotopes are stable while the rest are unstable isotopes. Unstable isotopes are radioactive isotopes. 8 Radioactive isotopes will undergo spontaneous decay to emit radioactive rays: alpha, beta and gamma. After radioactive decay, the proton Uses of Isotopes in Daily Life Isotopes are atoms of an element with the same number of protons but different numbers of neutrons. Alternatively, isotopes can be defined as atoms of an element with the same proton number but different nucleon numbers. SPM ’05/P2 Medicine 1 Cobalt-60 is a radioactive isotope of cobalt. It decays by giving out gamma radiation. In radiotherapy, malignant cancer cells are killed by directing a beam of gamma rays towards the cancer cells. to the thyroid gland. The radiation given out by the radioactive iodide ions will kill the malignant cancer cells without affecting the other parts of the body. SPM ’09/P1 Radiotherapy is used to kill cancer cells Patient suffering from thyroid cancer 2 Patients suffering from thyroid cancer are given a drink containing sodium iodide, (NaI) containing radioactive iodide ions. The radioactive iodide ions move preferen tially 3 Some medicine, surgical gloves, bandages, plastic hypodermic syringes are sterilised by using gamma radiation. These materials cannot be sterilised by boiling. Agriculture 1 Using the radioactive carbon-14 (14C) in carbon dioxide, the path of carbon during the photosynthesis process can be determined. The rate of absorption of phosphorus by the plant can be determined by adding radioactive phosphate ions (32PO3– ) to the ground. 4 2 Male pests can be attracted into traps using female hormones (pheromone). The male pests are then exposed to gamma radiation which can cause genetic mutation to the gametes (sperms). The male pests are then released to be allowed to mate with the females. The offsprings produced will have physical defects such as undeveloped digestive organs and wings. This will terminate the survival of the following generation. 27 The Structure of the Atom 2 number and nucleon number of the isotope may change. 9 There are many uses of radioisotopes, namely, in the field of medicine, agriculture, industry, archaeology, food preservation and electricity generation. Industry 1 Beta radiation is used to control the thickness of paper, plastic, metals and rubber made in industry. A radioactive source is located at the bottom of the material being produced. A detector is located on top of the material. Any change in the reading of the recorder signifies a change in thickness of the material. partially filled in which case a higher reading will be recorded. 2 Figure 2.20 Using radiation to detect if a container is fully filled 3 Radioisotopes are used to detect leaks in pipes carrying gas. A radioisotope (for example, sodium-24) is added to the gas so that it will be carried along by the gas flowing through the pipe. A detector is then moved along the external wall of the pipe. The detection of a high radioactive reading will signify the location of the leakage. Figure 2.19 Using radiation to control the thickness of materials 2 Gamma radiation is used to detect whether canned food or bottled drink is completely filled or only partially filled. A radioactive source emitting gamma radiation is directed to the bottled or canned food. More radiation will pass through if the container is only Figure 2.21 Using radiation to detect a leak Archaeology Carbon-14 is used to determine the age of archaeological artifacts. Plants take in carbon-14 in the form of carbon dioxide (14CO2) during photosynthesis. Carbon-14 is incorpo rated into animals or human beings when the plants are eaten. As long as the organism is alive, the amount of carbon-14 in it remains constant. This is because the intake of carbon-14 through food is offset by its spontaneous decay. However, when the organism dies, the intake of carbon-14 is stopped. The amount of C-14 ‘locked’ in the body will continue to decay. The amount of C-14 remaining (measured by its activity) is inversely proportional to the age of the artifacts. The age of bones dug out from a historical site can be estimated using carbon-14 dating. For very old bones, much of the C-14 would have decayed. The minute amount of C-14 left will show little radioactivity. A recent archaeological sample will have a high reading of C-14. Source: Jabatan Muzium Malaysia The Structure of the Atom SPM ’08/P1 28 Food preservation 1 Food such as vegetables, fruits and meat rot due to the activity of fungus and bacteria. These microorganisms can be killed by irradiating the food with gamma radiation of cobalt-60. The shelf-life of the food can be extended using this method. Irradiation is better than chemical preserva tives because it does not have adverse effects on health. 2 Irradiation can also slow down budding in potatoes and onions, thus extending their shelf-life. Gamma radiation can also slow the ripening of fruits to be exported. Radiation can be used to delay rotting of fruits and vegetables Nuclear energy is an alternative source of energy to replace fossil fuels such as petroleum, natural gas or coal. The nuclear fuel used is uranium-235. The uranium atoms become unstable when bombarded with fast neutrons. 5 This causes the uranium nuclei to split, producing heat energy. The heat energy released is used to produce steam from water. The steam drives the turbine of the generator, producing electricity. 2.4 ’05 Name an isotope and state its purpose for each of the following fields: (a) Medicine (c) Archaeology (b) Industry (d) Food preservation Solution (a) Cobalt-60 (gamma radiation from decay of Co-60 is used to kill cancer cells) (b) Sodium-24 (beta radiation from decay of Na-24 is used to detect leakages in pipes) (c) Carbon-14 (it is used to estimate the age of archaeological artifacts) (d) Cobalt-60 (gamma radiation from decay of Co-60 is used to kill fungus or bacteria that can cause food to rot) U, 234 92 U and 235 92 The Electronic Structure of an Atom Electron Arrangement in an Atom 1 Niels Bohr suggested that the electrons in an atom occupy orbits with definite energy levels. Each of these orbits or energy levels can hold a certain number of electrons. The electrons are not static but are moving around. 2 The electron orbits are also known as quantum shells. The shells are labelled first shell, second shell, third shell and so on, away from the nucleus. 3 The first shell is the one nearest to the nucleus and is filled first. It can hold a maximum of two electrons. 4 After the first shell is full, the remaining electrons are filled into the second shell. The second shell can hold a maximum of eight electrons. 2.3 1 Uranium has three isotopes: 2 Generation of electricity U. 238 92 What do you understand by the term isotopes? State the differences between these isotopes. 2 A list of elements are represented by letters of the alphabet as given below. Choose a pair of isotopes from these elements. Explain your answer. Figure 2.22 The electron shells of an atom is labelled away from the nucleus A, 127 B, 79 C, 131 D, 131 E, 55 F 53 34 55 53 25 23 11 5 After the first and second shells are full, the remaining electrons are filled into the third shell. The third shell can take a maximum of eight or 18 electrons. If the number of electrons of an 3 Give an example of a radioactive isotope of carbon. What is meant by radioisotope? Give a use of the isotope given in your example. 29 The Structure of the Atom atom is more than 20, the third shell will hold 18 electrons. If the number of electrons is 20 or less, the third shell will hold 8 electrons. 6 Table 2.5 shows the maximum number of electrons permitted in each shell. 7 The way in which the electrons are distributed in the shells of an atom is called the electron arrangement or electronic configuration of the atom. SPM ’11/P2 9 The electrons in the outermost occupied shell are called the valence electrons. Therefore, carbon atom has four valence electrons, chlorine atom has seven valence electrons and calcium atom has two valence electrons. 10 Elements with the same number of valence electrons have the same chemical properties. For example, lithium, sodium and potassium of Group 1 of the Periodic Table have the same chemical properties because each atom has one valence electron. Table 2.5 Maximum number of electrons permitted in each shell of an atom Maximum number of electrons Shell 2 Eight electrons are filled in the second shell. Eight electrons are filled in the third shell. Two electrons are filled in the fourth shell. The electron arrangement of calcium is 2.8.8.2. First 2 Second 8 Third 8 or 18 Fourth 32 8 The examples below show the electron arrangement of some elements: 1 The carbon atom, 126C has six protons. In a neutral atom, the number of electrons = the number of protons. Hence there are six electrons and are arranged as follows: Two electrons are filled in the first shell. Four electrons are filled in the second shell. The electron arrangement of carbon is 2.4. Electron arrangement Number of valence electrons Lithium Sodium Potassium 2.1 2.8.1 2.8.8.1 1 1 1 11 Group 17 elements have the same chemical properties because each element has seven valence electrons. 2 The chlorine atom, 3517Cl has 17 protons. In a neutral atom, the number of electrons = the number of protons. The 17 electrons are arranged as follows: Two electrons are filled in the first shell. Eight electrons are filled in the second shell. Seven electrons are filled in the third shell. The electron arrangement of chloride is 2.8.7. Group 17 element Electron arrangement Number of valence electrons Fluorine Chlorine Bromine Iodine 2.7 2.8.7 2.8.18.7 2.8.18.18.7 7 7 7 7 12 The inert or noble gases of Group 18 of the Periodic Table are very stable. They have filled outer shells of electrons. 3 Group 18 element Electron arrangement Number of valence electrons Helium Neon Argon 2 2.8 2.8.8 2 8 8 13 Helium has exactly two electrons in the first shell. It has attained the duplet electron arrangement which is stable. Neon and argon each has eight electrons in the outermost shell. It has attained the octet electron arrangement which is stable. The calcium atom, 40 Ca has 20 protons. 20 In a neutral atom, the number of electrons = the number of protons. The 20 electrons are arranged as follows: Two electrons are filled in the first shell. The Structure of the Atom Group 1 element 30 14 Table 2.6 shows the diagrammatic electronic structures and the electron arrangements of elements with proton numbers 1 to 20. SPM ’04,05 06/P1 ’08/P1 2 Table 2.6 The diagrammatic electronic structure of elements with proton numbers 1 to 20 6 X, 3517Y, 126Z. Write the electronic configuration of each of these elements. Solution 23 11 ’04 The atomic symbol of element X is 199X. Which of the following is true about the subatomic particles of element X? Proton number Number Number Electronic of of configuration protons electrons Proton number Nucleon number Electronic configuration X 11 11 11 2.8.1 A 9 19 2.7 Y 17 17 17 2.8.7 B 9 19 2.8.8.1 Z 6 6 6 2.4 C 19 9 2.7 D 19 9 2.8.8.1 Comment The proton number of X is 9. Hence it has 9 protons and 9 electrons. The 9 electrons are arranged as follows: Two electrons in the first shell and the remaining seven electrons are arranged in the second shell. Its electronic configuration is 2.7. Answer A 7 2.4 1 Write the electron arrangement and draw the atomic structures of carbon and magnesium atoms. [Proton number: C, 6; Mg, 12] 2 The diagram shows the atomic structure of an element X. (a) In an atom of X, how many of the following are there? (i) Valence electrons (ii) Protons (b) What is the nucleon number of X if it has 16 neutrons? (c) Write the atomic symbol of element X. ’04 The chemical symbols of three elements X, Y and Z are shown as follows: 31 The Structure of the Atom 2 2.5 Rutherford discovered the proton in 1911 and James Chadwick discovered the neutron in 1932. Niels Bohr explained the arrangement of the electrons in an atom. 3 We now know that the protons and neutrons are located at the center of the atom called the nucleus. The electrons are arranged in orbits around the nucleus. 4 The atomic structure of an atom can help us understand the chemical properties of the elements better and how they are bonded together to form compounds. Appreciating the Orderliness and Uniqueness of the Atomic Structure 1 John Dalton proposed the atomic theory about 200 years ago in 1807. Before that scientists thought that atoms were solid particles like marbles. 2 About 100 years later, other scientists discovered the subatomic particles. J.J. Thomson discovered the electron in 1897. Ernest 5 During freezing, the temperature remains constant because heat energy is released and the energy released is equals to the heat lost to the surrounding during cooling. 6 The proton number is the number of protons in the nucleus of an atom. 7 The nucleon number is the total number of protons and neutrons in the nucleus of an atom. 8 Isotopes are atoms of the same element which contain the same number of protons but different numbers of neutrons. 9 The protons and neutrons are enclosed in the nucleus whereas the electrons are arranged in shells surrounding the nucleus. (a) The first shell can hold a maximum of two electrons. (b) The second shell can hold a maximum of eight or 18 electrons. (c) The third shell can hold a maximum of 18 electrons. However for atoms with proton numbers 1 – 20, the atom attains stability when its third shell has eight electrons. (d) The valence electron is the electron in the outermost shell of the atom. For example, the electronic configuration of the calcium atom, 40 Ca is 2.8.8.2. 20 The calcium atom has two valence electrons. 1 There are three states of matter: solid, liquid and gas. 2 When a substance is heated or cooled it will change state. 3 The table shows the energy involved during the change in state: Change of state Process Change in energy Solid to liquid Melting Heat energy is absorbed Liquid to gas Boiling/ evaporation Heat energy is absorbed Solid to gas Sublimation Heat energy is absorbed Liquid to solid Freezing Heat energy is released Gas to liquid Condensation Heat energy is released Gas to solid Sublimation Heat energy is released 4 During melting the temperature remains constant because heat energy absorbed is used to overcome the forces of attraction between the molecules. The Structure of the Atom 32 2 Multiple-choice Questions Matter 1 What process and change in heat energy takes place when iodine ’11 crystals are heated at room temperature and pressure? Process A Melting B Melting C Sublimation D Sublimation Heat energy absorbed Heat energy released Heat energy absorbed Heat energy released 3 The diagrams show the spacing of the molecules of a substance at two different temperatures. at 85 °C What is the likely melting point and boiling point of the substance? Melting point Boiling point (°C) (°C) A –125 90 B –117 78 C –102 75 D –98 105 D The air particles diffuse out of the balloon at a faster rate at higher temperature. 5 Carbon dioxide(CO2), sulphur dioxide(SO2) and nitrogen dioxide(NO2) are three gases that cause acid rain. Which of the following lists the molecules in order of increasing average speed? [Relative atomic mass: C, 12; N, 14; O, 16; S, 32] Change in heat energy 2 Which statements below are true about a gas? I They move at low speed. II They are easily compressed. III They have a higher rate of diffusion compared to a liquid. IV They spread throughout the vessel in which they are contained. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV at –110 °C 4 An inflated balloon will shrink faster at higher temperature than at lower temperature. Which of the following is the best explanation for this observation? A The air particles liquefy at lower temperature. B The air particles react to form other compounds at higher temperature. C The air particles come closer together at lower temperature. Slowest Fastest A Sulphur Nitrogen Carbon dioxide dioxide dioxide B Sulphur Carbon dioxide dioxide Nitrogen dioxide C Nitrogen Sulphur Carbon dioxide dioxide dioxide D Carbon dioxide Sulphur Nitrogen dioxide dioxide 6 The table shows the changes in physical states and energies of four substances. Process Name of process Change of physical state Change of energy I Freezing Solid to liquid Heat is released II Melting Solid to liquid Heat is absorbed III Boiling Solid to gas Heat is absorbed IV Condensation Gas to liquid Heat is released Which of the following processes above are correct? A I and III only C I, II and IV only B II and IV only D II, III and IV only 7 Which of the following statements is true about pentane molecules when it is cooled to a temperature of –129 °C? [Melting point of pentane is –135 °C and its boiling point is 36 °C]. A The pentane molecules remain static. B The pentane molecules move randomly. C The pentane molecules are arranged closely together. D The distance between the pentane molecules increases. 33 The Structure of the Atom 2 2.1 8 The graph shows the temperature against time of a substance X when it is heated. Magnesium oxide ’07 Sodium Ammonia A Ions Atoms Molecules B Ions Molecules Molecules C Molecules Molecules Atoms D Ions Ions Ions 2.2 The Atomic Structure 11 Which of the following sets is correct? The scientists who discovered the electron, proton and neutron are 2 ’08 Which of the statements below are true about X ? I X starts to melt at Q. II The melting point of X is T1 °C. III X exists in gaseous state at region TU. IV At region RS, there is a mixture of solid and liquid X. A I and III only B II and IV only C I, II and III only D I, III and IV only 9 The diagram shows the graph of temperature against time for the heating of substance X. temperature (°C) 65 t1 t2 time (s) Which statements below are true about substance X? I It is a gas at room temperature. II It undergoes physical change at 65 °C. III It absorbs heat at time intervals t1 and t2. IV It exists as a mixture of liquid and solid at time intervals t1 and t2. A I and II only B I and III only C II and III only D II and IV only 10 State the particles in magnesium oxide, sodium and ammonia. The Structure of the Atom Electron Proton Neutron A Ernest Rutherford J.J. Thomson James Chadwick B J.J. Thomson Ernest Rutherford James Chadwick C J.J. Thomson Ernest Rutherford Niels Bohr D J.J. Thomson James Chadwick Ernest Rutherford 12 What can be deduced from the symbol 31 P? 15 I Phosphorus atom has five valence electrons. II Phosphorus atom has 15 protons and 31 neutrons. III Phosphorus atom has 16 neutrons. IV Phosphorus atom has proton number of 15 and nucleon number of 31. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV I It belongs to Group 15 in the Periodic Table. II It belongs to Period 3 of the Periodic Table. III It forms an ion with a charge of –3. IV It is a metal. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV 15 The diagrams show three models of the atom. 13 Two particles P and Q have the following compositions: Particle Electron Neutron Proton P 10 10 9 Q 10 12 11 It follows that A P and Q are both negativelycharged B P and Q have the same nucleon number. C P and Q are particles of the same element. D P is negatively-charged and Q is positively-charged. 14 An atom X has an electron arrangement of 2.8.5. Which of the following statements about X are correct? 34 Name the scientists who proposed these models? I II III A James J.J. Chadwick Thomson Ernest Rutherford B Niels Bohr John Dalton Ernest Rutherford C Niels Bohr J.J. Thomson Ernest Rutherford J.J. D Niels Bohr Ernest Rutherford Thomson 17 The atoms 126 C and 115 B have the same A number of protons B number of neutrons C physical properties D chemical properties 2.3 Isotopes and Their Importance 18 Two uranium isotopes are 235 U 92 and 238 U. Which of the following 92 statements below is true? A The 235 U atom has fewer 92 electrons than 238 U atom. 92 B The 235 U atom has 92 92 protons and 235 neutrons. C The 238 U atom has 92 92 protons and 146 neutrons. D The 235 U atom and 238 U atom 92 92 have the same number of neutrons. 19 Isotopes are different atoms with the same number of A protons but different number of neutrons. B electrons but different number of protons. C protons, electrons and neutrons. D protons but different number of electrons and neutrons. 20 Which of the following pairs are correct? Isotope Use I Uranium-235 To generate electricity II Iodine-131 To kill cancerous thyroid cells Isotope Use III Carbon-14 To estimate the age of archaeological artifacts IV Sodium-24 A B C D To detect leakages in pipes I, II and III only I, II and IV only II, III and IV only I, II, III and IV 21 Which of the following statements are true about isotopes? I They have the same chemical properties. II They have different physical properties. III The have a different number of neutrons. IV They have the same number of valence electrons. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV 22 Oxygen has the isotope 16 O, 17 O and 18 O. Which of the following oxygen gas has the lowest rate of diffusion? A 16 O = 16 O B 17 O = 17 O C 18 O = 18 O D 17 O = 18 O 23 X Y Which term describes the particles X and Y shown above? A Isotopes C Anions B Isomers D Cations 24 An element has two isotopes, which are represented by 127 X and 131 X. How does 127 X differ from 131 X ? A It has four less neutrons and three less electrons. B It has four less neutrons. C It has four less protons and three less electrons. D It has four less protons. 35 25 Which of the following comparisons between 79 Br and 35 81 Br are correct? 35 Bromine-79 Bromine-81 I Has 35 protons Has 35 protons II Has 35 electrons Has 35 electrons III Has 35 neutrons Has 35 neutrons IV Has 44 neutrons Has 46 neutrons A B C D I and IV only II and III only I, II and IV only I, III and IV only 26 The radioisotope that is used to kill cancerous cells is A uranium-235 B cobalt-60 C carbon-14 D phosphorus-32 2.4 The Electronic Structure of an Atom 27 Which of the following particles have eight valence electrons? I 168 W III 40 Y 18 23 + II 11 X IV 35 Z– 17 A I and III only B II and IV only C I, II and III only D II, III and IV only 28 Which of the following particles contains 18 electrons, 19 protons and 20 neutrons? A 39 X+ C 39 X– 19 18 B 40 20 D X 2+ 39 19 X 29 How many protons and neutrons are there in one tin atom with nucleon number 119? Protons Neutrons A 50 68 B 50 69 C 50 71 D 50 119 The Structure of the Atom 2 16 An atom has the symbol 11 X. Which of the following 5 statements about X are correct? I It has 5 valence electrons in its atom. II It has 6 neutrons in its atom. III It belongs to Group 13 of the Periodic Table. IV It belongs to Period 2 of the Periodic Table. A I, II and III only B I, II and IV only C II, III and IV only D I, II, III and IV 2 30 The symbol of an element X is 40 X. We can deduce that an atom 18 of element X I has eight valence electrons. II has 22 neutrons in its nucleus. III has three electron shells. IV has a total of 18 electrons in its atom. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV 31 Two particles X and Y have the following composition: Particle Electrons Neutrons Protons X 10 12 11 Y 11 12 11 Which of the following statements are true about X and Y? I Both X and Y are negativelycharged. II Both X and Y are positivelycharged. III Both X and Y have the same nucleon number. IV Both X and Y are particles of the same element. A I and III only B II and IV only C III and IV only D II, III and IV only 32 An element 39 X 19 A has one valence electron. B forms a positively-charged ion of charge +2. C is located in Group 17 of the Periodic Table. D has 19 protons and 39 neutrons. 33 Which of the following elements given below have the same number of valence electrons? 19 9 A B C D V ; 27 W; 35 X; 39 Y; 40 Z 13 17 19 20 Element W X Y Z Proton number 6 7 12 15 A B C D W and Y only X and Z only V and X only V and Y only 34 Which of the following list are the electron arrangements of all nonmetals? A 2.6 2.7 2.8.5 B 2 2.5 2.8.3 C 2.1 2.7 2.8.6 D 2.1 2.8.2 2.8.3 35 An atom 39 Y 19 I has 20 neutrons. II has 19 protons. III has one valence electron. IV has four electron shells. A I, II and III only B I, II and IV only C II, III and IV only D I, II, III and IV 36 What is the number of subatomic particles in 60 Co2+ ion? 27 Protons Neutrons Electrons A 27 33 27 B 27 33 25 C 33 27 27 D 33 27 25 37 The table shows the proton numbers of four elements. Which of the following pairs of elements has the same number of valence electrons? Y and Z only X and Z only X and Y only W and Z only 38 The electronic configuration of arsenic is 2.8.18.5. Which of the following statements is true? A Arsenic has three valence electrons. B Arsenic is in Group 14 of the Periodic Table. C The nucleon number of the arsenic atom is 33. D Arsenic is in the same group of the Periodic Table as an element with proton number 7. 39 The electronic configuration of the ion X– is 2.8.18.18.8. The ion X– has 74 neutrons. Determine the nucleon number of element X. A 127 B 128 C 129 D 130 40 The electronic configuration of the strontium ion, Sr2+ is 2.8.18.8. The Sr2+ ion has 49 neutrons. Determine the nucleon number of strontium. A 85 B 86 C 87 D 88 Structured Questions 1 Carbon has two isotopes as shown in Table 1 below. Atom Proton number Nucleon number (b) Draw the atomic structure of represent an electron. C 12 6 C 14 6 ’08 Table 1 (a) (i) Complete Table 1 with the proton numbers and nucleon numbers of the two different carbon isotopes. [2 marks] The Structure of the Atom (ii) What is the difference between the two isotopes 126C and 146C? [1 mark] 36 (c) Give one use of 146C. C using, x, to 14 6 [2 marks] [1 mark] (d) What is the number of valence electrons in both of the carbon atoms above? [1 mark] (g) (i) Explain the meaning of the term isotope. 2 Diagram 1 shows a graph of temperature against time of substance M when it is heated until it boils. [2 marks] (ii) State a pair of isotopes from the particles in Table 2. [1 mark] 4 Table 3 shows four substances and their respective formulae. ’04 Chemical formula Bromine Br2 Iron Fe Diagram 1 Naphthalene C10H8 (a) State the physical state of M at the region (i) PQ (iii) RS (ii) QR (iv) ST [3 marks] Sodium chloride NaCl (b) When does M begin to boil? [1 mark] (c) What is the melting point of M? [1 mark] 2 Substance Table 3 (a) State two substances that consist of molecules. [1 mark] (b) Which of the following substances has the highest melting point: bromine, iron or naphthalene? (d) Explain why the temperature of M remains constant from time t1 to t2. [1 mark] [1 mark] (e) Sketch the graph obtained when molten M is cooled from 450 °C to room temperature. [2 marks] (c) (i) State the substance that can conduct electricity in the solid state. [1 mark] (ii) Draw the arrangement of the particles of this substance. [1 mark] 3 Table 2 shows the proton numbers and nucleon numbers of five particles represented by the letters V, W, X, Y and Z. Particle Proton number Nucleon number V 6 12 W 8 16 X 8 18 Y 11 23 Z 16 32 Electron arrangement (d) Name the particles present in sodium chloride. [1 mark] (e) Diagram 2 shows the graph of temperature against time obtained when solid naphthalene is heated. Table 2 (a) Write the electron arrangements of all the particles in Table 2. [2 marks] (b) What is the number of valence electrons in particle V ? [1 mark] (c) Draw the atomic structure of particle Y. Diagram 2 [2 marks] (d) State the number of electron shells in particle Z. [1 mark] (e) Explain the meaning of nucleon number. [1 mark] (f) What is the number of neutrons in particle Y ? [1 mark] (i) State the melting point of naphthalene. [1 mark] (ii) Explain why there is no change in temperature from Q to R. [2 marks] (iii) State how the movement of naphthalene particles changes between R and S during heating. [1 mark] Essay Questions You are given two substances X and Y. They are either naphthol or naphthalene. You are required to carry out an experiment to identify X and Y. Design an experiment to determine X and Y. 1 (a) Compare the three physical states of matter in terms of particle arrangements, forces of attraction between the particles, kinetic energy of the particles and compressibility. [8 marks] (b) Table 1 shows the melting points of naphthol and naphthalene. 37 The Structure of the Atom Chemical 2 (a) Define the following terms: (i) Proton number (ii) Nucleon number (iii) Valence electron Melting point (°C) Naphthol 65 Naphthalene 80 Table 1 [3 marks] (b) (i) What are isotopes? [3 marks] (ii) Give an example of a pair of isotopes. [2 marks] (iii) Discuss six uses of isotopes. [12 marks] [12 marks] 2 Experiments 1 An experiment is carried out to determine the melting point of naphthalene. Solid naphthalene is heated and its temperature is recorded every 30 seconds. ’05 Diagram 1 shows the recorded temperature readings at 30-second intervals. Diagram 1 (a) Record the temperatures in the spaces provided in Diagram 1. [3 marks] (b) Draw a labelled diagram of the apparatus used to carry out the experiment. [3 marks] (c) Plot a graph of temperature against time for the heating of naphthalene. [3 marks] (d) State the melting point of naphthalene. [3 marks] (e) What is the physical state of naphthalene at time 90 seconds? [3 marks] (f) Explain why the temperature between time 60 s to 120 s remained constant. [3 marks] (g) Sketch a graph you expect to obtain if the molten naphthalene is cooled to room temperature. [3 marks] 2 ‘The melting point of a substance is lowered by the presence of impurities’. Using naphthalene and a mixture of naphthalene with some acetamide, describe an experiment to prove the statement above. Your answer should include the following items: (a) Aim of experiment [3 marks] (b) All variables involved [3 marks] (c) List of apparatus and materials used [3 marks] (d) Procedure of experiment [3 marks] (e) Tabulation of results [3 marks] The Structure of the Atom 38 FORM 4 THEME: Matter Around Us CHAPTER 3 Chemical Formulae and Equations SPM Topical Analysis 2008 Year 1 Paper 2009 3 2 Section A B C Number of questions 1 — 2 – – 4 – 1 9 2010 2 3 A B C 1 – – – 1 2011 2 4 3 A B C – – – 1 – 4 2 3 A B C – – – – ONCEPT MAP FORMULAE AND CHEMICAL EQUATIONS Atom Molecule Relative atomic mass Relative molecular mass Relative atomic mass, Ar in gram Relative molecular mass, Mr in gram Molar mass Mass of matter 3 Molar mass 4 Molar mass 4 Molar volume Volume of gas 3 Avogadro constant Number of moles 3 Molar volume Number of particles 4 Avogadro constant Reactants Chemical equation Products of reaction Empirical formula Chemical formula Molecular formula 3.1 Relative Atomic Mass and Relative Molecular Mass 1 It is impossible to weigh an atom in gram. So chemists compared how heavy one atom is to another atom which is taken as the standard. The comparison of the mass of an atom to another is called the relative atomic mass (r.a.m.). Relative Mass of one atom of the element atomic mass = — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — 1 of an element —  mass of one carbon-12 atom 12 3 For example, a sodium atom, Na is 23 times heavier than one-twelfth of the mass of one carbon-12 atom. Thus the relative atomic mass of Na is 23. Mass of one Na atom — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — = 23 1 —  mass of one carbon-12 atom 12 (Note: the mass of one carbon-12 atom is 12 units) 5 A molecule is a small group of atoms joined together. The simplest being diatomic molecules like O2, N2 and Cl2. Examples of triatomic molecules are CO2 and H2O. Some examples of larger molecules are ammonia (NH3), methane (CH4), sulphur (S8), phosphorus (P4) and ethanol (C2H5OH). 6 The relative molecular mass (Mr) of a compound is defined as the number of times one molecule of the compound is heavier than one-twelfth of the mass of a carbon-12 atom. The relative atomic mass of atoms or relative molecular mass of molecules can be determined using the mass spectrometer with carbon-12 as the standard 2 In 1961, scientists agreed to use carbon-12 as the standard. The mass of a carbon-12 atom is assigned a value of exactly 12 units. 3 Carbon is chosen as the standard because (a) the abundance of carbon-12 isotope is almost 99%. Carbon-13 and carbon-14 isotopes make up about only 1%. Thus the mass of a carbon atom using carbon-12 isotope or using the average mass of the three isotopes of carbon is still 12.00 units. (b) carbon is a solid at room temperature. Unlike hydrogen and oxygen which are gases, it does not require a container with a lid to contain it. (c) carbon is present in many organic substances, namely, wood, natural gas and petroleum. Thus carbon is easily available. Carbon can be obtained by burning these organic substances in a limited supply of oxygen. 4 The relative atomic mass (Ar) of an element SPM is defined as the number of times one atom ’11/P1 of the element is heavier than one-twelfth of the mass of a carbon-12 atom, that is: Chemical Formulae and Equations Relative Mass of one molecule molecular of the compound — — — — — — — — — — — — — — — — — — — — — — — — mass of a = — 1 —  mass of one compound 12 carbon-12 atom For example, a molecule of methane, CH4, is 16 times heavier than one-twelfth of the mass of a carbon-12 atom. Thus the relative molecular mass of CH4 is 16. A student need not memorise the relative atomic mass of elements. They will be given in the examination. However, one must know how to calculate the relative molecular mass of compounds from the Ar given. To determine the Mr of a molecule, we sum up the relative atomic mass of every atom present in the molecule. 40 Concept of relative atomic mass and relative molecular mass using analogy Apparatus Twin-pan balance 3 Ball bearings, iron nails, screws and nuts. Figure 3.1(a) Figure 3.1(b) 1 An iron nail is put into the pan on the right of the balance (Figure 3.1(a)). 2 Ball bearings are added to the pan on the left of the balance until it is balanced. 3 The number of ball bearings needed to balance the small nail is counted. 4 The procedure is repeated for a screw, a nut, a screw and a nut (Figure 3.1(b)) and a screw and two nuts. For each sample the number of ball bearings required to balance the object is counted and recorded in the table as follows: OR Mass of one iron nail ——————————————————— = 5 Mass of one ball bearing Hence, relative mass of one iron nail = 5 2 If we assume that a ball bearing represents onetwelfth of the mass of carbon-12 atom and the nail, screw and nut represent the atoms of other elements, then the relative atomic mass of these elements will be 5, 12 and 7 respectively. 3 If we assume that the screw and the nut form a molecule, relative molecular mass of the molecule = relative atomic mass of a screw + relative atomic mass of a nut = 12 + 7 = 19 4 A total of 26 ball bearings are required to balance the mass of one screw and two nuts. Hence, Mr of a screw and two nuts = r.a.m. of a screw + r.a.m. of two nuts = 12 + 7 + 7 = 26 Therefore to determine the relative molecular mass of a molecule, we sum up the relative atomic mass of each atom present in the molecule. 5 In this experiment, we do not need to know the actual mass of a ball bearing to determine the relative mass of other objects. Similarly, we do not need the actual mass of a carbon atom to determine the relative atomic mass of an element or the relative molecular mass of a molecule. Number of Relative mass ball bearings of object needed to (compared to balance object a ball bearing) Iron nail 5 5 Screw 12 12 Nut 7 7 Screw + nut 19 19 Screw + 2 nuts 26 26 (Assuming the relative mass of a ball bearing is 1 unit) 1 If we assume that the ball bearing has a mass of 1 unit, then the iron nail which is equivalent to five ball bearings has a relative mass of five units. 41 Chemical Formulae and Equations Activity 3.1 Object Mass of 1 iron nail = mass of five ball bearings Solution 1 24 (a) Z is heavier than Y by — — times = 1.5 times. 16 (b) Assume that n atoms of X has the same mass as the sum of 3 atoms of Y and 2 atoms of Z. 12n = 3(16) + 2(24) 12n = 96 96 n =— — 12 =8 3 Iridium is a very dense metal and was discovered in 1804 by Smithson Tennant. Determine how many carbon atoms will have the same mass as one iridium atom. [Relative atomic mass: C, 12; Ir, 192] Solution Assuming n carbon atoms has the same mass as one iridium atom. 12n = 192 192 n=— — — 12 = 16 2 5 Adrenaline is produced by the adrenal gland. Adrenaline has the formula C9H13NOx. If its r.m.m. is 183, determine the value of x. Then write the molecular formula of adrenaline. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16] SPM ’08/P1 Three cobalt atoms have the same mass as fifteen carbon atoms. Determine the relative atomic mass of cobalt. [Relative atomic mass: C, 12] Solution Relative molecular C9H13NOx= 183 Solution Assume that the relative atomic mass of Co = a mass of adrenaline is 9(12) + 13(1) + 14 + x(16) = 183 108 + 13 + 14 + 16x = 183 16x = 183 – 135 48 x=— — 16 x=3 Co + Co + Co = 15 C 3a = 15  12 15  12 a =— — — — — — — 3 = 60 The formula of adrenaline is C9H13NO3. 3 1 The mass of a rutherfordium (Rf) atom is equal to the sum of three sodium atoms and six sulphur atoms. What is the relative atomic mass of rutherfordium? [Relative atomic mass: Na, 23; S, 32] The relative formula mass of X3(PO4)2 is 310. Determine the relative atomic mass of element X. [Relative atomic mass: O, 16; P, 31] Solution Relative atomic mass of Rf = 3(23) + 6(32) = 69 + 192 = 261 Solution Assume that the relative atomic mass of the element X is p. Relative formula mass of X3(PO4)2 = 310 3(p) + 2[31 + 4(16)] = 310 3p + 2[31 + 64] = 310 3p + 190 = 310 3p = 310 – 190 120 p=— — 3 = 40 4 The relative atomic mass of elements X, Y and Z are 12, 16 and 24 respectively. (a) How much is an atom of Z heavier than an atom of Y? (b) How many atoms of X will have the same mass as the sum of 3 atoms of Y and 2 atoms of Z? Chemical Formulae and Equations ’09 42 1 (a) A platinum atom is five times heavier than a potassium atom. What is the relative atomic mass of platinum? [Relative atomic mass: K, 39] (b) Calculate the number of carbon atoms that has the same mass as one molybdenum atom. [Relative atomic mass: C, 12; Mo, 96] (c) Five aluminium atoms have the same mass as the sum of six lithium atoms and three phosphorus atoms. Determine the r.a.m. of phosphorus. [Relative atomic mass: Li, 7; Al, 27] (d) The relative atomic mass of elements W, X, Y and Z are 7, 39, 56 and 195 respectively. (i) One atom of thorium (Th) has the same mass as the sum of six W atoms, two X atoms and two Y atoms. What is the r.a.m. of thorium? (ii) How many W atoms will have the same mass as the sum of two X atoms, one Y atom and one Z atom? 2 Determine the relative molecular mass (or relative formula mass) of the following compounds: (a) Sodium stearate, C17H35COONa (Soap molecule) (b) Complex ion Cu(NH3)4SO4 [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Na, 23; S, 32; Cu, 64] 3 Borax is a compound used to kill cockroaches. Its molecular formula is X2B4O7. If the relative molecular mass of borax is 202, determine the relative atomic mass of the element X. Identify the element X from the list of elements given below. [Relative atomic mass: B, 11; C, 12; O, 16; F, 19; Na, 23; Mg, 24] 3.2 Amedeo Avogadro Amedeo Avogadro was a professor of physics at the University of Turin, Italy. In 1811, he proposed the hypothesis which states that under the same temperature and pressure, equal volumes of different gases contain equal numbers of molecules. He showed that 22.4 dm3 of any gas at a temperature of 0 °C and a pressure of 1 atmosphere contains 6.02  1023 molecules. Therefore the value of 6.02  1023 is called Avogadro’s number or the Avogadro constant in honour of him. Relationship between the Number of Moles and the Number of Particles Definition of the Mole Concept of the Mole SPM ’08/P1 1 One mole is the amount of substance which contains the same number of particles as there are in 12 grams of carbon-12. 2 The number of atoms in 12 grams of carbon-12 is 6.02  1023. 3 This number, 6.02  1023, is called Avogadro’s number or the Avogadro constant (NA). 4 The particles in matter can be atoms, ions or molecules. 5 For elements, the particles are atoms. For example, 1 mol of gold contains 6.02  1023 gold atoms. 1 The relative atomic mass of carbon atom is 12 and the relative atomic mass of helium atom is 4. This means that a carbon atom is three times heavier than a helium atom. 2 Thus, a sample containing 12 grams of carbon and four grams of helium will contain the same number of atoms, that is, number of atoms in 12 grams of carbon = number of atoms in 4 grams of helium 43 Chemical Formulae and Equations 3 3 Let us extend the reasoning to other elements. The number of atoms in a sample of any element with its relative atomic mass in grams is equal to the number of atoms in 12 g of carbon-12. For example, 1 g of hydrogen, 14 g of nitrogen, 23 g of sodium, 56 g of iron will all contain the same number of atoms as in 12 g of carbon-12. 4 Now the question that arises is: how many atoms are there in 4 g of helium, 1 g of hydrogen, 14 g of nitrogen, 23 g of sodium, 56 g of iron and 12 g of carbon-12? Through several experiments scientists have found that this number is 602 000 000 000 000 000 000 000 or 6.02  1023. 5 In Chemistry, the number 6.02  1023 is called one mole (or mol in short). 3.1 3 Conversion of the Number of Moles to the Number of Particles 6 For ionic compounds, the particles are ions. For example, 1 mol of magnesium ions contains 6.02  1023 Mg2+ ions. 1 mol of potassium iodide, KI contains 6.02  1023 K+ ions and 6.02  1023 I– ions. 1 mol of magnesium chloride, MgCl2 contains 6.02  1023 Mg2+ ions and 2  6.02  1023 Cl– ions. 7 For covalent compounds, the particles are molecules. For example, 1 mol of carbon dioxide contains 6.02  1023 CO2 molecules. 6 Calculate the number of particles in: (a) 0.75 mol of aluminium atoms, A1, (b) 1.2 mol of chloride ions, Cl–, (c) 0.07 mol of carbon dioxide molecules, CO2. [Assume NA = 6  1023 mol–1] Solution (a) 1 mol of aluminium contains 6  1023 Al atoms. 0.75 mol of aluminium contains Make sure that the 0.75 mol ———————  6  1023 Al atoms numerator and the 1 mol denominator have the same unit. = 4.5  1023 Al atoms (b) 1 mol of chloride ions contains 6  1023 Cl– ions. 1.2 mol of chloride ions contain 1.2 mol — — — — — —  6  1023 Cl– ions. 1 mol = 7.2  1023 Cl– ions (c) 1 mol of carbon dioxide contains 6  1023 CO2 molecules. 0.07 mol of carbon dioxide contains 0.07 mol — — — — — — —  6  1023 CO2 molecules 1 mol = 4.2  1022 CO2 molecules Conversion of the Number of Moles to the Number of Particles and Vice Versa 1 Since one mole of any substance contains 6.02  1023 particles, n moles of the substance will contain n  6.02  1023 particles. 2 Hence, number of particles = number of mole 3 NA (where NA = 6.02 3 1023) 3 If 6.02  1023 particles are found in 1 mol, then one particle is found in 1 particle — — — — — — — — — — — — — — — — — —  1 mol 6.021023 particles 1 =— — — — — — — — — — mol 6.021023 7 Therefore, x particles are found in x ————————— mol 6.02  1023 4 Thus, the Solution (a) 0.2 mol of SO2 contains 0.2  6  1023 molecules = 1.2  1023 molecules. 1 sulphur dioxide molecule (SO2) has 3 atoms (one sulphur and two oxygen atoms). Therefore the number of atoms = 3  1.2  1023 = 3.6  1023 atoms 5 Generally,  NA  NA mol (b) 0.125 mol of CH4 contains 0.125  6  1023 molecules = 7.5  1022 molecules. 1 methane molecule (CH4) has 5 atoms (one carbon and four hydrogen atoms). Therefore the number of atoms = 5  7.5  1022 = 3.75  1023 atoms A student need not memorise that the Avogadro constant is 6.02  1023. It will be given in the examination. However in most cases, the value of NA given is 6  1023 for easy calculation. Chemical Formulae and Equations SPM ’11/P1 Calculate the number of atoms in: (a) 0.2 mol of sulphur dioxide gas, SO2, (b) 0.125 mol of methane gas, CH4. [NA = 6  1023 mol–1] number of moles = number of particles  NA (where NA = 6.02  1023) number of particles SPM ’09/P1 44 Conversion of the Number of Particles to the Number of Moles 3 The mole-atom is the relative atomic mass of an atom expressed in gram. ’10/P1 [Relative atomic mass: C, 12; Al, 27; S, 32] SPM 1 = 1 = 1 = Calculate the number of moles of the following substances: (a) 6  1021 iron atoms, (b) 7.5  1023 carbon monoxide molecules. [NA = 6  1023 mol–1] mole-atom of carbon 12 g mole-atom of aluminium 27 g mole-atom of sulphur 32 g Each sample contains 6.02  1023 atoms 4 The mole-molecule is the relative molecular mass of a compound expressed in gram. [Relative molecular mass: H2O, 18; CO2, 44; NH3, 17; C2H5OH, 46; CH4, 16] Solution (a) 1 mol of iron contains 6  1023 atoms. Therefore 6  1021 iron atoms 6  1021 atom =— — — — — — — — — — —  1 mol 6  1023 atom = 0.01 mol (b) 1 mol contains 6  1023 molecules. Therefore 7.5  1023 CO molecules contain 7.5  1023 molecules — — — — — — — — — — — — — — — — —  1 mol 6  1023 molecules = 1.25 mol 1 mole-molecule of water, H2O = 18 g 1 mole-molecule of carbon dioxide, CO2 = 44 g 1 mole-molecule of ethanol, C2H5OH = 46 g Each sample contains 6.02  1023 molecules Conversion of the Number of Moles of a Substance to Its Mass 3.2 1 Since 1 mol of an element is the relative atomic mass in gram, x mol of the element has x  relative atomic mass in gram. 1 Calculate the number of particles in 1 (a) — mol of copper, 6 (b) 0.0625 mol of water molecule, H2O, (c) 1.3 mol of sodium ions, Na+. [NA = 6  1023 mol–1] Number of mole-atom mass in gram =— — — — — — — — — — — — — — — — — — — — — relative atomic mass 2 Calculate the number of atoms in (a) 0.012 mol of ethane gas, C2H6, (b) 1.1 mol of sulphur trioxide, SO3. [NA = 6  1023 mol–1] 2 Similarly, since 1 mol of a compound is the relative molecular mass in gram, x mol of the compound has x  relative molecular mass in gram. 3 Calculate the number of moles of the following substances: (a) 6  1022 sodium ions, (b) 1.8  1024 H2S molecules. [NA = 6  1023 mol–1] 3.3 SPM ’07/P2 Number of mole-molecule mass in gram =— — — — — — — — — — — — — — — — — — — — — — — — — relative molecular mass Relationship between the Number of Moles of a Substance and Its Mass 3 Generally, mole 1 The mass of a substance that contains one mole of the substance is called the molar mass. 2 One mole of substance contains 6.02  1023 particles. Therefore the molar mass of any substance contains 6.02  1023 particles.  Ar or Mr 4 Ar or Mr mass in gram number of moles 45 Chemical Formulae and Equations 3 8 Conversion of the Number of Moles of a Substance to Its Mass Conversion of the Number of Particles of a Substance to Its Mass and Vice Versa 1 Two steps are involved in the conversion of the mass of substance to the number of particles. Step 1: Mass in gram is converted to number of moles by dividing the mass by the relative atomic mass or relative molecular mass. Step 2: Number of moles is converted to number of particles by multiplying the number of moles by the Avogadro constant. 2 Two steps are involved in the conversion of the number of particles to mass. Step 1: Number of particles is converted to the number of moles by dividing the number of particles by the Avogadro constant. Step 2: Number of moles is converted to mass in gram by multiplying the number of moles by the relative atomic mass or relative molecular mass. 3 In general,  Ar  NA or Mr mass in number of mole particles gram  NA  Ar or Mr 3 9 Determine the mass for each of the following substances: 2 (a) — mol of aluminium atoms, 3 (b) 0.08 mol of ascorbic acid, C6H8O6, (c) 0.125 mol of magnesium hydroxide, Mg(OH)2. [Relative atomic mass: H, 1; C, 12; O, 16; Mg, 24; Al, 27; Cl, 35.5] Solution (a) 1 mol of Al = 27 g 2 2 — mol of Al = — 27 g = 18 g 3 3 (b) 1 mol of C6H8O6 = 6(12) + 8(1) + 6(16) g = 176 g 0.08 mol C6H8O6 = 0.08  176 g = 14.08 g (c) 1 mol of Mg(OH)2 = 24 + 2(16 + 1) g = 58 g 0.125 mol of Mg(OH)2 = 0.125  58 g = 7.25 g Conversion of the Mass of a Substance to the Number of Moles 10 Conversion of the Mass of a Substance to the Number of Particles SPM ’08/P1 11 Calculate the number of moles of the following substances: (a) 23.5 g of copper(II) nitrate, Cu(NO3)2, (b) 0.97 g of caffeine, C8H10N4O2 (a stimulant). [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Si, 28; S, 32; Cu, 64] Calculate the number of particles in: (a) 12.8 g of copper, (b) 8.5 g of ammonia, NH3. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Fe, 56; Cu, 64; NA = 6  1023 mol–1] Divide mass Solution (a) 1 mol of Cu(NO3)2 = 64 + 2[14 + 3(16)] g = 188 g 23.5 23.5 g of Cu(NO3)2 = — — — —  1 mol 188 = 0.125 mol (b) 1 mol of C8H10N4O2 = 8(12) + 10 + 4(14) + 2(16) g = 194 g in gram by the Solution relative atomic 12.8 to find the (a) 12.8 g of Cu = — — — mol = 0.2 mol mass number of moles 64 1 mol contains 6  1023 atoms. 0.2 mol contains 0.2  6  1023 atoms = 1.2  1023 atoms Number of moles is converted to (b) 1 mol of NH3 = 17 g number of particles by multiplying the 8.5 8.5 g of NH3 = — — mol = 0.5 mol number of moles 17 by the Avogadro 1 mol contains 6  1023 molecules. constant 0.5 mol contains 0.5  6  1023 molecules = 3  1023 molecules 0.97 g of C8H10N4 O2 0.97 = ———  1 mol 194 = 0.005 mol Chemical Formulae and Equations 46 Conversion of the Number of Particles of a Substance to Its Mass 3.3 12 Calculate the mass of the following substances: (a) 1.2  1022 zinc atoms, (b) 3  1023 ethanol (C2H5OH) molecules. [Relative atomic mass: H, 1; C, 12; O, 16; Zn, 65; NA = 6  1023 mol–1] Solution (a) 1 mol contains 6  1023 atoms. 1.2  1022 atoms Number of particles is converted to the number of 1.2  1022 moles by dividing the number =— — — — — — — — — mol 23 6  10 of particles by the Avogadro constant = 0.02 mol 1 mol of Zn = 65 g 0.02 mol of Zn = 0.02  65 g = 1.3 g Number of moles is (b) 1 mol contains converted to mass in 23 6  10 molecules. gram by multiplying 3  1023 C2H5OH molecules the number of moles by the relative atomic 3  1023 mass is contained in — — — — — — mol 6  1023 = 0.5 mol 1 mol of C2H5OH = 46 g 0.5 mol of C2H5OH = 0.5  46 g = 23 g 2 2 Calculate the number of moles of the following substances: (a) 2.8 g of iron, (b) 4.05 g of nicotine, C10H14N2 (an addictive substance in cigarette), (c) 1.49 g of ammonium phosphate, (NH4)3PO4 (a fertiliser), (d) 2.3 g of ethanol, C2H5OH. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; P, 31; Fe, 56] 3 Calculate the mass of the following substances: (a) 3  1023 titanium atoms, (b) 1.2  1024 argon atoms, (c) 7.5  1022 citric acid (C12H16O14) molecules. [Relative atomic mass: H, 1; C, 12; O, 16; Ar, 40; Ti, 48; NA = 6  1023 mol–1] 4 Calculate the number of particles in the following substances: (a) 4 g of sulphur, (b) 2.24 g of cadmium, (c) 36 g of glucose, C6H12O6. [Relative atomic mass: H, 1; C, 12; O, 16; S, 32; Cd, 112; NA = 6  1023 mol–1] ’04 The relative atomic mass of X and Y is 64 and 16 respectively. Which of the following is about the atoms of X and Y? The mass of one atom of Y is 16 g. The number of protons in X is 64. 4 mol of Y have the same mass as 1 mol of X. The density of one atom of X is 4 times that of an atom of Y. 5 Geranial is a compound found in lemon grass (daun serai). Its molecular structure is shown below. CH3 H H H CH3 H H ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ C == C –– C –– C –– C == C –– C == O ⎮ ⎮ CH3 H H Solution 16 The mass of one atom of Y is — — — — — — — g. 6  1023 (A is incorrect) The nucleon number of X is 64. (B is incorrect) Mass of 4 mol of Y = 4  16 g = 64 g (C is correct) Mass of 1 mol of X is 64 g. (a) Determine the mass of 1 mol of geranial. (b) Determine the mass of 0.02 mol of geranial. (c) Determine the number of molecules present in 7.6 g of geranial. (d) Determine the mass of 7.5  1022 geranial molecules. [Relative atomic mass: H, 1; C, 12; O, 16; NA = 6  1023 mol–1] Densities cannot be determined because the volume of the atom is not given. (D is incorrect) 47 Chemical Formulae and Equations 3 1 Calculate the mass of each of the following substances: (a) 1.25 mol of helium gas, 2 (b) — mol of cobalt, Co, 5 (c) 0.15 mol of hydrated copper(II) sulphate, CuSO4.5H2O, (d) 0.05 mol of potassium manganate(VII), KMnO4. [Relative atomic mass: H, 1; He, 4; O, 16; S, 32; K, 39; Mn, 55; Co, 59; Cu, 64] 3.4 Conversion of the Number of Moles of a Gas to Its Volume and Vice Versa Relationship between the Number of Moles of a Gas and Its Volume 1 Since 1 mol of any gas occupies 22.4 dm3 at s.t.p. (or 24 dm3 at r.t.p.), n mol of the gas will occupy n  22.4 dm3 at s.t.p. (or n  24 dm3 at r.t.p.) 2 Hence, volume of gas = number of moles of gas  molar volume where the molar volume is 22.4 dm3 at s.t.p. or 24 dm3 at room temperature. 3 If a volume of 22.4 dm3 (or 22 400 cm3) is occupied by 1 mol of gas, 1 1 cm3 of gas is occupied by — — — — — — ‑ mol. 22 400 4 Conversely, volume of gas number of moles of gas = — — — — — — — — — — — — — — — — — — — ‑ molar volume 3 Conversion of the Number of Particles of a Substance to Its Volume 1 The volume occupied by a gas depends on the temperature. As the temperature increases, the gas expands and occupies a larger volume. If the gas is cooled, the gas contracts and occupies a smaller volume. 2 The volume occupied by a gas also depends on the pressure. If a gas is compressed (with increased pressure), the volume of the gas decreases. If the pressure is decreased, the volume of the gas increases. 3 At the same temperature and pressure, equal volumes of all gases contain the same number of particles. Accordingly, one mole of any gas (which contains 6.02  1023 particles) will occupy the same volume at a particular temperature and pressure. 4 It is found that one mole of any gas at room temperature (25°C) and pressure of 1 atmosphere occupies a volume of 24 dm3 (or 24 000 cm3). 5 At standard temperature and pressure (s.t.p.), that is, at a temperature of 0 °C and pressure of 1 atmosphere, one mole of any gas occupies a volume of 22.4 dm3 (or 22 400 cm3). 6 The volume occupied by one mole of any gas SPM is called the molar volume. ’11/P1 Example • 1 mol of oxygen gas, O2(32 g) • 1 mol of carbon dioxide gas, CO2(44 g) A student need not memorise that molar volume is 22.4 dm3 at s.t.p. or 24 dm3 at r.t.p. It will be given in the examination. Conversion of the Number of Moles to Volume of Gas 13 Calculate the volume of 0.75 mol of nitrogen gas at s.t.p. [1 mol of gas occupies a volume of 22.4 dm3 at s.t.p.] Solution 1 mol of gas occupies a volume 22.4 dm3 at s.t.p. 0.75 mol of N2 gas occupies Make sure that the numerator and 0.75 mol 3 3 — — — — — — — —  22.4 dm = 16.8 dm denominator have the 1 mol same unit occupies 24 dm3 at room temperature or occupies 22.4 dm3 at s.t.p. [Relative molecular mass: O2, 32; CO2, 44] Conversion of Volume of Gas to Number of Moles 7 One mole of gas contains 6.02  10 molecules and therefore at s.t.p. 23 22.4 dm3 of oxygen gas, O2(32 g) 22.4 dm3 of carbon dioxide gas, CO2(44 g) Chemical Formulae and Equations SPM ’04/P2 14 Calculate the number of moles of the following gases at room temperature and pressure. (a) 4.8 dm3 of chlorine gas, (b) 1200 cm3 of methane gas. [1 mol of gas occupies a volume of 24 dm3 at room temperature] each contains 6.02  1023 molecules 48 Solution (a) 1 mol of gas occupies a volume of 24 dm3 at room temperature. 4.8 dm3 4.8 dm3 Cl2 gas contain — — — — — —  1 mol 24 dm3 = 0.2 mol (b) 1 mol of gas occupies a volume of 24 000 cm3 at room temperature. 1200 cm3 of CH4 gas contain 1200 cm3 — — — — — — — — —  1 mol 24 000 cm3 = 0.05 mol Solution 1 mol of NH3 gas = 17 g 3.4 g 3.4 g of NH3 = — — — — 1 mol = 0.2 mol 17 g 1 mol of gas occupies a volume of 22.4 dm3 at s.t.p. 0.2 mol of ammonia gas occupies a volume of 0.2  22.4 dm3 = 4.48 dm3. Conversion of Volume of Gas to Mass Conversion of the Volume of Gases to Mass and Vice Versa SPM Calculate the mass of the following gases at room temperature and pressure: (a) 7.2 dm3 of sulphur dioxide gas, SO2, (b) 600 cm3 of methane gas, CH4. [Relative atomic mass: H, 1; C, 12; O, 16; S, 32; 1 mol of gas occupies a volume of 24 dm3 at room temperature] ’06/P2 ’08/P2 1 Two steps are involved in the conversion of volume of gas to mass. Step 1: Volume of gas is converted to number of moles (by dividing the volume of gas by the molar volume). Step 2: Number of moles is converted to mass in gram (by multiplying the number of moles by the relative atomic mass or relative molecular mass of the gas). 2 Two steps are involved in the conversion of mass to volume of gas. Step 1: Mass in gram is converted to number of moles (by dividing the mass by the relative atomic mass or relative molecular mass). Step 2: Number of moles is converted to volume of gas (by multiplying the number of moles by the molar volume). 3 In general:  22.4 dm3 volume  22.4 dm3 (at s.t.p.) number of moles  Ar or Mr 3 16 Solution (a) 1 mol of gas occupies a volume of 24 dm3 at room temperature. 7.2 dm3 7.2 dm3 of SO2 = — — — — — — — — — — 1 mol = 0.3 mol 24 dm3 1 mol of SO2 = (32 + 32) g = 64 g 0.3 mol of SO2 = 0.3  64 g = 19.2 g (b) 1 mol of gas occupies a volume of 24 000 cm3 at room temperature. 600 cm3 600 cm3 of CH4 gas = — — — — — — — — —  1 mol 24 000 cm3 = 0.025 mol 1 mol of CH4 = 16 g 0.025 mol of CH4 = 0.025  16 g = 0.4 g Conversion of the Volume of Gases to the Number of Particles and Vice Versa mass in gram  Ar or Mr SPM ’04,06 /P2 1 Two steps are involved in the conversion of the volume of gas to the number of particles. Step 1: Volume of gas is converted to number of moles (by dividing the volume of gas by the molar volume). Step 2: Number of moles is converted to number of particles (by multiplying the number of moles by the Avogadro constant). 2 Two steps are involved in the conversion of the number of particles to volume of gas. Conversion of Mass to Volume of Gas 15 Calculate the volume occupied by 3.4 g of ammonia gas, NH3 at standard temperature and pressure. [Relative atomic mass: H, 1; N, 14; 1 mol of gas occupies a volume of 22.4 dm3 at s.t.p.] 49 Chemical Formulae and Equations Solution 1 mol contains 6  1023 molecules. 1.5  1023 molecules 1.5  1023 molecules =— — — — — — — — — — — — — — — — — —  1 mol = 0.25 mol 6  1023 molecules Step 1: Number of particles is converted to number of moles (by dividing the number of particles by the Avogadro constant). Step 2: Number of moles is converted to volume of gas (by multiplying the number of moles by the molar volume). 3 In general: 3 volume of gas  22.4 dm3 number of moles  22.4 dm3 (at s.t.p.)  NA  NA 1 mol of gas occupies a volume of 24 dm3 at room temperature. 0.25 mol of gas occupies a volume of 0.25  24 dm3 = 6 dm3 number of particles volume Conversion of Volume of Gas to Number of Particles  22.4 dm (or 24 dm3) 3 number of moles  (61023) 17  Mr  Mr mass in gram  (61023) number of particles Calculate the number of molecules present at s.t.p. in (a) 0.28 dm3 of N2 gas, (b) 448 cm3 of carbon monoxide, CO gas. [1 mol of gas occupies a volume of 22.4 dm3 at s.t.p., NA = 6  1023 mol–1] 3 ’03 Which of the following gases contain 6  1022 molecules? [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Avogadro constant = 6  1023 mol–1] I 1.0 g of hydrogen gas II 2.8 g of nitrogen gas III 4.4 g of carbon dioxide IV 1.8 g of water vapour I, II and III only II, III and IV only I, III and IV only I, II, III and IV Solution (a) 1 mol of gas occupies a volume of 22.4 dm3 at s.t.p. 0.28 dm3 0.28 dm3 of nitrogen gas = — — — — — — —  1 mol 22.4 dm3 = 0.0125 mol 1 mol of N2 contains 6  1023 molecules. 0.0125 mol of N2 contains 0.0125  6  1023 molecules = 7.5  1021 molecules (b) 1 mol of gas occupies a volume of 22 400 cm3 at s.t.p. 448 cm3 448 cm3 of CO gas = — — — — — — — — —  1 mol 22 400 cm3 = 0.02 mol 1 mol of CO contains 6  1023 molecules. 0.02 mol of CO contains 0.02  6  1023 molecules = 1.2  1022 molecules Solution 6  1022 6  1022 molecules = — — — — — — — ‑  1 mol = 0.1 mol 6  1023 I 1 mol of H2 = 2 g 1 1 g of H2 = — mol = 0.5 mol (I is incorrect) 2 II 1 mol of N2 = 28 g 2.8 2.8 g of N2 = — — mol = 0.1 mol (II is correct) 28 III 1 mol of CO2 = 44 g 4.4 4.4 g of CO2 = — — mol = 0.1 mol (III is correct) 44 IV 1 mol of H2O = 18 g 1.8 1.8 g of H2O = — — mol = 0.1 mol (IV is correct) 18 Answer Conversion of Number of Particles to Volume of Gas 18 Calculate the volume of 1.5  1023 molecules of ethane, C2H6 gas at room temperature and pressure. [1 mol of gas occupies a volume of 24 dm3 at r.t.p., NA = 6  1023 mol–1] Chemical Formulae and Equations  22.4 dm3 (or 24 dm3) 50 3.5 ’05 The activity of microorganisms on waste products at dump sites produces methane gas. If 180 dm3 of methane gas is collected, calculate the mass of methane obtained. [Relative atomic mass: H, 1; C, 12; 1 mol of gas occupies a volume of 24 dm3 at room temperature and pressure] Chemical Formulae SPM ’05/P2 ’08/P1 1 A chemical formula is used to represent a chemical compound. The chemical formula shows (a) the elements, denoted by their symbols, present in the compound. (b) the relative numbers, indicated by subscripts written after the symbols, of each element present in the compound. 2 For example, the chemical formula of sulphuric acid is H2SO4. The chemical formula indicates that (a) the elements present in sulphuric acid are hydrogen, sulphur and oxygen. (b) two hydrogen atoms, one sulphur atom and four oxygen atoms combine to form the compound. 3 Table 3.1 shows the chemical formulae of some covalent compounds. Solution 1 mol of gas occupies a volume of 24 dm3 at room temperature. 180 dm3 180 dm3 of CH4 gas = — — — — —  1 mol 24 dm3 = 7.5 mol 1 mol of CH4 = 16 g 7.5 mol of CH4 = 7.5  16 g = 120 g Table 3.1 The chemical formulae of some covalent compounds 3.4 Name of compound 1 Calculate the volume of 0.55 mol of oxygen gas at room temperature and pressure. [1 mol of gas occupies a volume of 24 dm3 at room temperature] Oxygen 2 Calculate the number of moles of 672 cm3 of carbon dioxide gas at s.t.p. [1 mol of gas occupies a volume of 22.4 dm3 at s.t.p.] 3 Calculate the volume occupied by 1.4 g of ethene gas, C2H4 at room temperature and pressure. [Relative atomic mass: H, 1; C, 12; 1 mol of gas occupies a volume of 24 dm3 at room temperature] Chemical formula O2 Number of each element in the compound 2 oxygen atoms Water H2O 2 hydrogen atoms and 1 oxygen atom Ammonia NH3 1 nitrogen atom and 3 hydrogen atoms Sulphuric acid H2SO4 2 hydrogen atoms, 1 sulphur atom and 4 oxygen atoms 4 The chemical formula of an ionic compound can be written if the charge of the cation (positively-charged ion) and the anion (negatively-charged ion) forming the ionic compound are known. 5 Table 3.2 shows the charges of some ions. 4 Calculate the mass of each of the following gases at standard temperature and pressure: (a) 16.8 dm3 of methane, CH4, (b) 6720 cm3 of carbon monoxide gas, CO. [Relative atomic mass: H, 1; C, 12; O, 16; 1 mol of gas occupies a volume of 22.4 dm3 at s.t.p.] Table 3.2 Charges of some cations and anions 5 Calculate the number of molecules present in: (a) 3.6 dm3 of N2 gas, (b) 1200 cm3 of ammonia gas at room temperature and pressure. [1 mol of gas occupies a volume of 24 dm3 at room temperature, NA = 6  1023 mol–1] Charge +1 6 Calculate the volume of 9  1021 molecules of CO at standard temperature and pressure. [1 mol of gas occupies a volume of 22.4 dm3 at s.t.p., NA = 6  1023 mol–1] 51 Cation Sodium ion Potassium ion Lithium ion Silver ion Copper(I) ion Hydrogen ion Ammonium ion Nickel(I) ion Symbol Na+ K+ Li+ Ag+ Cu+ H+ NH4+ Ni+ Chemical Formulae and Equations 3 4 Charge +2 3 +3 Cation Magnesium ion Calcium ion Zinc ion Iron(II) ion Copper(II) ion Manganese(II) ion Lead(II) ion Nickel(II) ion Mg2+ Ca2+ Zn2+ Fe2+ Cu2+ Mn2+ Pb2+ Ni2+ Iron(III) ion Aluminium ion Chromium(III) ion Fe3+ Al3+ Cr3+ Charge must be equal to the total negative charge of the anion. 7 To write the chemical formula of an ionic compound, the following steps can be used: (a) Write the formulae of the ions involved in forming the compound and their charges. (b) Then, balance the positive and negative charges. This can be done by writing the numerical charge of the cation next to the anion as a subscript and the numerical charge of the anion next to the cation as a subscript. (c) Finally write the chemical formula of the ionic compound without the charges. 8 Generally, if the ionic compound is formed by the ions Xm+ and Yn–, then the chemical formula of the compound is XnYm. If m = n, then the chemical formula is XY. Examples, (a) Chemical formula of sodium sulphate Charge of ion +1 –2 SO42– Formula of ion Na+ Ratio 2 1 Chemical formula is Na2SO4. (b) Chemical formula of iron(III) chloride Charge of ion +3 –1 Formula of ion Fe3+ Cl– Ratio 1 3 Chemical formula is FeCl3. (c) Chemical formula of magnesium oxide Charge of ion +2 –2 2+ Formula of ion Mg O2– Ratio 2 2 Chemical formula is MgO. Note that if the charges of the ions are the same, the ratio of the ions that combine is 1 : 1. 9 In general, the chemical formulae of compounds formed between the ions are summarised in Table 3.3. Symbol Anion Symbol –1 Fluoride ion Chloride ion Bromide ion Iodide ion Hydroxide ion Nitrate ion Nitrite ion Bicarbonate ion Permanganate ion Hydride ion F– Cl– Br– I– OH– NO3– NO2– HCO3– MnO4– H– –2 Oxide ion Sulphide ion Sulphate ion Sulphite ion Carbonate ion Thiosulphate ion Chromate(VI) ion Dichromate(VI) ion O2– S2– SO42– SO32– CO32– S2O32– CrO42– Cr2O72– –3 Phosphide ion Phosphate ion Nitride ion P3– PO43– N3– 6 Since a chemical compound is always electrically neutral, the total positive charge of the cation Table 3.3 Formulae of ionic compounds Cation Anion Chemical formula Examples X+ X+ X+ Y– Y 2– Y 3– XY X2Y X3Y Sodium chloride, NaCl Potassium dichromate(VI), K2Cr2O7 Ammonium phosphate, (NH4)3PO4 X 2+ X 2+ X 2+ Y– Y 2– Y 3– XY2 XY X3Y2 Calcium chloride, CaCl2 Copper(II) sulphate, CuSO4 Magnesium nitride, Mg3N2 X 3+ X 3+ X 3+ Y– Y 2– Y 3– XY3 X2Y3 XY Iron(III) chloride, FeCl3 Chromium(III) sulphate, Cr2(SO4)3 Aluminium nitride, AlN Chemical Formulae and Equations 52 Name of compound Cu+ Cu2+ O2– O2– Copper(I) oxide Copper(II) oxide Cu2O CuO Fe2+ Fe3+ Cl– Cl– Iron(II) chloride Iron(III) chloride FeCl2 FeCl3 Mn2+ Mn4+ O2– O2– Manganese(II) oxide Manganese(IV) oxide MnO MnO2 All transition elements are metals. Many of them like iron, manganese, copper, silver, gold, nickel and titanium are of major technological importance. A large number of the transition elements combine with each other to form useful alloys. 1 The empirical formula of a compound shows the simplest ratio of the atoms of the elements ’07/P2 that combine to form the compound. 2 The molecular formula of the compound shows the actual numbers of the atoms of the elements that combine to form the compound. 3 Table 3.5 shows the molecular and empirical formulae of some compounds. Compound Molecular formula C:H:O =1:2:1 CH2O C20H24N2O2 C : H : N : O C10H12NO = 10: 12 : 1 : 1 ’09 The table shows the mass of elements M and O in an oxide, and relative atomic mass of M and O. SPM Simplest ratio of the elements C6H12O6 Empirical formula 5 Empirical and Molecular Formulae Table 3.5 The molecular and empirical formulae of some compounds Glucose Simplest ratio of the elements 4 The following steps can be used to determine the empirical formula of a compound: Step 1: Write the mass or percentage of each element in the compound. Step 2: Calculate the number of moles of each element in the compound by dividing the mass or percentage of the element by the relative atomic mass of the element. Step 3: Next, divide each number by the smallest number to obtain the simplest ratio. Step 4: Finally write the empirical formula of the compound based on the ratio of the elements. Chemical formula Anion Molecular formula Quinine Table 3.4 Formulae of some compounds of transition elements Cation Compound Element M O Mass (g) 2.4 1.6 Relative atomic mass 48 16 What is the empirical formula of this compound? Solution Element SPM Step 1 Mass Step 2 Number of moles Step 3 Simplest ratio M O 2.4 g 1.6 g 2.4 —— mol 48 = 0.05 1.6 — — mol 16 = 0.1 0.05 — — — 0.05 =1 0.10 — — — 0.05 =2 ’09/P1 Empirical formula Water H2O H:O=2:1 H2O Ethene C2H4 C:H=1:2 CH2 Butane C4H10 C:H=2:5 C2H5 Ethane C2H6 C:H=1:3 CH3 Empirical formula of compound is MO2. 53 Chemical Formulae and Equations 3 10 Transition elements can form ions with different charges. In the IUPAC (International Union of Pure and Applied Chemistry) system, the standardised chemical nomenclature is denoted by Roman numerals in brackets to indicate the charge of the ion. For example, iron(II) represents Fe2+ ion and iron(III) represents Fe3+ ion. Table 3.4 shows examples of some compounds of transition elements. 6 Simplest ratio ’05 2.5 g of X combined with 4 g of Y to form a compound with the formula XY2. If the relative atomic mass of Y is 80, determine the relative atomic mass of X. 4.8 —— 1.6 =3 1.6 —— — 1.6 =1 4.84 ——— 1.6 =3 Empirical formula of boric acid is H3BO3. Solution Assume the relative atomic mass of X is a. Element X Y 2.5 4 2.5 —— mol a 4 — — mol 80 = 0.05 1 2 3 Mass Number of moles Simplest ratio 20 A gaseous hydrocarbon X contains 85.7% of carbon by weight. 4.2 g of the gas X occupies a volume of 3.36 dm3 at standard temperature and pressure. [Relative atomic mass: H, 1; C, 12; 1 mol of gas occupies a volume of 22.4 dm3 at s.t.p.] (a) Determine the empirical formula of X. (b) Determine the relative molecular mass of X. (c) What is the molecular formula of X? Since the empirical formula is XY2, the ratio of X : Y=1:2 2.5 — — a 1 — — —= — 0.05 2 2.5 0.05 — —= — — — a 2 2.5  2 a=— — — — — — 0.05 = 100 Solution 19 Boric acid is used to preserve prawns and fish. Chemical analysis of the compound shows that it contains 4.8% hydrogen, 17.7% boron and the rest is oxygen. Determine the empirical formula of boric acid. [Relative atomic mass: H, 1; B, 11; O, 16] Element H B O Percentage 4.8% 17.7% 100 – 4.8 – 17.7% = 77.5% Number of moles 4.8 17.7 —— — 100 g —— —100 g 100 100 = 4.8 g = 17.7 g 4.8 —— mol 1 = 4.8 17.7 ——— mol 11 = 1.6 Chemical Formulae and Equations Element C H Percentage 85.7% 100 – 85.7% = 14.3% Number of moles 85.7 — — — — mol 12 = 7.14 14.3 — — — mol 1 = 14.3 Simplest ratio 7.14 — — — 7.14 =1 14.3 — — — — 7.14 =2 (a) Empirical formula of X is CH2. (b) 4.2 g of the gas has a volume of 3.36 dm3 at s.t.p. 1 mol of the gas (22.4 dm3) has a mass of 22.4 dm3 — — — — — — —  4.2 g 3.36 dm3 = 28 g Solution Mass in 100 g of compound SPM ’04/P2 Therefore the relative molecular mass of X is 28. (c) Assume the molecular formula of X is (CH2)n. The relative molecular mass of X is (12 + 2)n = 28 28 n=— — 14 =2 77.5 ———  100 g 100 = 77.5 g 77.5 ——— mol 16 = 4.84 Thus the molecular formula of X is C2H4. 54 To determine the empirical formula of magnesium oxide by experiment combustion of the magnesium. Care is taken to prevent white smoke (magnesium oxide powder formed) from escaping by closing the lid immediately after it is lifted. 6 When the magnesium is completely burnt, the crucible is cooled and is weighed again with its lid. The weight is recorded. 7 The heating, cooling and weighing process is repeated until a constant mass is obtained. Procedure Results Mass of crucible + lid = a gram Mass of crucible + lid + magnesium = b gram Mass of crucible + lid + magnesium oxide = c gram Calculations Mass of magnesium used Mass of oxygen that combined with magnesium 1 The 20 cm length of magnesium ribbon is polished with sandpaper to remove the oxide layer on its surface. 2 The magnesium ribbon is rolled into a loose coil. 3 An empty crucible with its lid is first weighed and its weight is recorded. 4 The rolled magnesium ribbon is placed in the crucible. The crucible is weighed again and its weight recorded. 5 The crucible is placed on a clay pipe triangle and then heated strongly. The lid is lifted at intervals to allow the oxygen from the air to enter for the = (b – a) gram = (c – b) gram Element Mg O Mass (g) (b – a) (c – b) Number of moles Simplest ratio (b – a) — — — — mol 24 (c – b) — — — — mol 16 x y Conclusion The empirical formula is MgxOy. To determine the empirical formula of copper oxide Materials SPM ’09/P2 1 The combustion tube with a porcelain dish is weighed. The weight is recorded. 2 A spatula of copper oxide powder is placed in the porcelain dish. The combustion tube with its contents is weighed again. The weight is recorded. 3 Dry hydrogen gas is passed through the tube for a few minutes to expel all the air. 4 The copper oxide is then heated strongly and the hydrogen gas passing through the end of the combustion tube is lit. 5 When all the copper oxide is reduced to copper metal (the black copper oxide has all become brown), heating is stopped. 6 A continuous stream of hydrogen gas is allowed to pass through the tube until it is cooled. The empirical formula of copper oxide is determined by reducing the copper oxide using hydrogen gas. Apparatus 3 Materials Crucible with lid, tripod stand, sandpaper, Bunsen burner, clay pipe triangle, electronic balance and tongs. A magnesium ribbon of about 20 cm in length and oxygen. Porcelain dish, combustion tube and electronic balance. Copper oxide powder and dry hydrogen gas. Procedure 55 Chemical Formulae and Equations Activity 3.2 & 3.3 Apparatus 7 The combustion tube with its contents is weighed and the weight recorded. 8 The heating, cooling and weighing process is repeated until a constant weight is obtained. 3 Results Mass of combustion tube + empty porcelain dish = 54.31 g Mass of combustion tube + porcelain dish + copper oxide = 62.32 g Mass of combustion tube + porcelain dish + copper = 60.71 g Calculations Mass of copper obtained = (60.71 – 54.31)g = 6.40 g Mass of oxygen that combined with the copper = (62.32 – 60.71)g = 1.61 g Element Cu O Mass (g) 6.40 1.61 6.4 — — — mol = 0.1 64 1.61 — — — mol = 0.1 16 0.1 — — —= 1 0.1 0.1 — — —= 1 0.1 Number of moles Simplest ratio Precautions taken 1 Hydrogen gas must be allowed to pass through SPM the combustion tube for a few minutes before the ’11/P1 copper oxide is heated. This is to remove the air in the tube. A mixture of air and hydrogen can explode when lit. 2 The flow of hydrogen gas must be continued throughout heating. This is to ensure that air does not enter the combustion tube. Hence the hydrogen gas must be seen to be burning continuosly at the end of the combustion tube. 3 The hot copper metal is allowed to cool in a stream of hydrogen gas. This is to ensure that oxygen from the air does not oxidise the hot copper to copper oxide again. 4 The heating, cooling and weighing process is repeated until a constant weight is obtained. This is to ensure that all the copper oxide has been reduced. Conclusion The empirical formula of copper oxide is CuO. 3.5 1 Write down the chemical formulae of the following ionic compounds: (a) Silver nitrate (b) Sodium thiosulphate (c) Ammonium phosphate (d) Calcium hydroxide (e) Magnesium carbonate (f) Zinc phosphide (g) Iron(III) hydroxide (h) Aluminium oxide (i) Chromium(III) chloride (j) Copper(II) sulphate (k) Nickel(I) chloride (l) Magnesium nitride 4 10.2 g of vanadium metal combined with 8 g of oxygen to form a compound with empirical formula V2O5. Determine the relative atomic mass of vanadium. [Relative atomic mass: O, 16] 5 Hydrazine is used as a rocket fuel. It contains 87.5% nitrogen and 12.5% hydrogen. [Relative atomic mass: H, 1; N, 14] (a) Determine the empirical formula of the compound. (b) If the relative molecular mass of hydrazine is 32, determine its molecular formula. 6 x g of iron combined with 5.04 g of oxygen to form an oxide with empirical formula Fe2O3. Determine the value of x. [Relative atomic mass: O, 16; Fe, 56] 2 Silicon hydride contains 87.5% silicon by mass. Determine the empirical formula of silicon hydride. [Relative atomic mass: H, 1; Si, 28] 7 A gaseous hydrocarbon Q contains 20% hydrogen. 6 g of this hydrocarbon occupies a volume of 4.48 dm3 at s.t.p. [Relative atomic mass: H, 1; C, 12; 1 mol of gas occupies 22.4 dm3 at s.t.p.]. Determine: (a) The empirical formula of Q (b) Relative molecular mass of Q (c) Molecular formula of Q 3 Reduction of 7.55 g of tin oxide using hydrogen gas yields 5.95 g of tin metal. Determine the empirical formula of tin oxide. [Relative atomic mass: Sn, 119; O, 16] Chemical Formulae and Equations 56 3.6 Chemical Equations Remember that a chemical equation cannot be balanced by altering the formulae of the reactants or products. You can only balance the equation by putting a number in front of each of the formulae. Reactants and Products of Chemical Equations 1 A chemical reaction can be represented by a chemical equation. For example, when sodium hydroxide reacts with sulphuric acid, sodium sulphate and water are produced. The reaction can be represented symbolically by the equation: 1 Lets’s look at a chemical reaction to practice these rules. During combustion, magnesium reacts with the oxygen in air to produce a white powder, magnesium oxide. You might start by writing an equation for this reaction by writing the symbols for the reactants and products. 2NaOH + H2SO4 → Na2SO4 + 2H2O (a) The reactants are the chemicals that are reacting and they are written on the left-hand side of the equation. Hence, sodium hydroxide and sulphuric acid are the reactants. (b) The products are the chemicals formed in a reaction and they are written on the righthand side of the equation. Hence, sodium sulphate and water are the products. (c) 2 mol of NaOH react with 1 mol of H2SO4 to produce 1 mol of Na2SO4 and 2 mol of H2O. 2 Some symbols which appear in a chemical equation are: Symbol Meaning ∆ Heating of substance ↑ or (g) Gas evolved ↓ or (s) Precipitate formed reactants Mg + yield O2 ⎯→ product MgO 2 On the left-hand side of the equation there is 1 atom of magnesium and there is also 1 atom of Mg on the product side of the equation. However, there are 2 oxygen atoms on the left and only 1 on the right. To balance the number of atoms of each type you will need to add coefficients. If you multiply Mg on the left-hand side of the equation by 2 then you must do the same to the Mg atom on the right which will mean doubling the number of oxygen atoms on the right-hand side of the equation too since you may not change the ratio of atoms within a molecule because it will change the identity of the substance. The new equation will look like this: Reversible reaction 2Mg + O2 3 When writing a chemical equation, the following steps are followed: (a) Write the correct formulae of all reactants on the left-hand side of the equation. (b) Write the correct formulae of all products on the right-hand side of the equation. (c) The equation is then balanced. This involves making sure that the number of atoms of each element before and after the reaction are the same. (d) Finally the physical state of each of the reactants or products is written as: (s) – represents solid state (l) – represents liquid state (g) – represents gaseous state (aq) – represents aqueous state, that is, the substance is dissolved in water. ⎯→ 2MgO 3 Now the equation is balanced. There are the same number of Mg and O atoms on both sides of the equation. A balanced equation is the source of a great deal of information about both products and reactants during a chemical change. Examples of Chemical Equations 1 (a) When green copper(II) carbonate is heated, it decomposes to form black copper(II) oxide and carbon dioxide gas which turns limewater cloudy. CuCO3(s) ⎯→ CuO(s) + CO2(g) reactant 57 products Chemical Formulae and Equations 3 Writing Chemical Equations If 0.12 g of magnesium is added to excess hydrochloric acid, calculate (a) the mass of magnesium chloride salt formed. (b) the volume of hydrogen gas evolved at room temperature and pressure. [Relative atomic mass: Mg, 24; Cl, 35.5; 1 mol of gas occupies a volume of 24 dm3 at r.t.p.] (b) When calcium carbonate is reacted with hydrochloric acid, the products formed are calcium chloride, carbon dioxide and water. CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) (c)When an aqueous solution of potassium iodide is added to an aqueous solution of lead(II) nitrate, a yellow precipitate of lead(II) iodide is formed. Solution (a) Referring to the equation, 1 mol of Mg (24 g) produces 1 mol of MgCl2(95 g). Hence 0.12 g of Mg will produce 0.12 g Mg — — — — — — — — — — — — —  95 g MgCl2 24 g Mg 3 2KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2KNO3(aq) Interpreting Chemical Equations Qualitatively and Quantitatively = 0.475 g of MgCl2 SPM ’11/P1 (b) 1 mol of Mg (24 g) produces 1 mol of H2 gas (24 dm3) at r.t.p. Hence 0.12 g of Mg will produce 0.12 — — —  24 dm3 of H2 24 = 0.12 dm3 of H2 = 120 cm3 of H2 1 From a chemical equation, we can obtain information on (a) the reactants taking part in the reaction (on the left-hand side of the equation). (b) the products formed in the reaction (on the right-hand side of the equation). (c) the number of moles of each substance taking part in the reaction and the number of moles of the products formed (from the coefficients/numbers before the substances). (d) the physical states of all the reactants and products (from the symbols in bracket after the substance). For example: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) The chemical equation shows that: 2 mol of solid sodium react with 2 mol of liquid water to form 2 mol of aqueous sodium hydroxide and 1 mol of hydrogen gas. Solving Numerical Problems Using Chemical Equations 8 3.2 g of copper(II) oxide powder is reacted with excess dilute nitric acid. (a) Write a chemical equation for the reaction. (b) Calculate the mass of copper(II) nitrate salt formed in the reaction. [Relative atomic mass: N, 14; O, 16; Cu, 64] Solution (a) CuO(s) + 2HNO3(aq) → Cu(NO3)2(aq) + H2O(l) (b) 1 mol of CuO produces 1 mol of Cu(NO3)2. 80 g of CuO produces 188 g of Cu(NO3)2. Hence 3.2 g of CuO will produce 3.2 g CuO ——————  188 g of Cu(NO3)2 80 g CuO = 7.52 g of Cu(NO3)2 SPM ’08/P1 The quantity of reactants and products can be calculated using stoichiometric equations when numerical information is given (stoichiometry). 7 21 Ethanol burns in air as represented by the equation C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(l) Calculate the mass of ethanol burnt if 2.4 dm3 of carbon dioxide is produced at room temperature. [Relative atomic mass: H, 1; C, 12; O, 16; 1 mol of gas occupies a volume of 24 dm3 at room temperature] ’09 Magnesium reacts with hydrochloric acid as represented by the equation Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Chemical Formulae and Equations ’04 58 Solution Referring to the equation, 1 mol of C2H5OH (46 g) produces 2 mol (2  24 dm3) of CO2. Therefore 2.4 dm3 of CO2 is produced from 2.4 dm3 — — — — — — — — —  46 g of C2H5OH 2  24 dm3 = 2.3 g of ethanol, C2H5OH 3 Iron metal reacts with excess hydrochloric acid to produce iron(II) chloride and hydrogen gas. (a) Write a balanced chemical equation for the reaction. (b) If 2.8 g of iron metal is used in the reaction, calculate (i) the maximum mass of iron(II) chloride formed. (ii) the volume of hydrogen gas produced at room conditions. [Relative atomic mass: Cl, 35.5; Fe, 56; 1 mol of gas occupies a volume of 24 dm3 at r.t.p.] 22 The chemical name of baking powder is sodium bicarbonate (NaHCO3). When sodium bicarbonate is heated, it decomposes to sodium carbonate, carbon dioxide and water. (a) Write a balanced chemical equation for the decomposition of sodium bicarbonate when it is heated. (b) If 2.1 g of sodium bicarbonate is heated, calculate the volume in cm3 of CO2 produced at room temperature. [Relative atomic mass: H, 1; C, 12; O, 16; Na, 23; 1 mol of gas occupies a volume of 24 dm3 at room temperature] 4 Acetylene gas (C2H2) is used in metal welding. This gas can be prepared by reacting calcium carbide with excess water as represented by the equation CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g) If 4.8 g of calcium carbide is reacted with excess water, calculate (a) the volume of acetylene gas evolved at room temperature. (b) the mass of calcium hydroxide formed. [Relative atomic mass: H, 1; C, 12; O, 16; Ca, 40; 1 mol of gas occupies a volume of 24 dm3 at r.t.p.] Solution (a) 2NaHCO3(s) ⎯→ Na2CO3(s) + CO2(g) + H2O(l) ∆ (b) 2 mol of NaHCO3 (2  84 g) produces 1 mol of CO2 (24 dm3) at r.t.p. 2.1 g of NaHCO3 will produce 2.1 — — — — — —  24 dm3 of CO2 = 0.3 dm3 of CO2 2  84 = 300 cm3 of CO2 5 Hydrogen gas is prepared by reacting methane gas with steam using platinum as catalyst. The reaction is represented by the equation CH4(g) + H2O(g) → CO(g) + 3H2(g) If 60 dm3 of hydrogen gas is produced at room temperature and pressure, calculate (a) the mass of methane that is used in the reaction. (b) the number of carbon monoxide molecules released. [Relative atomic mass: H, 1; C, 12; 1 mol of gas occupies a volume of 24 dm3 at r.t.p; NA = 6 1023 mol–1] 3.6 1 Balance the following chemical equations: (a) Na(s) + H2O(l) → NaOH(aq) + H2(g) (b) Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g) (c) CuCO3(s) + HNO3(aq) → Cu(NO3)2(aq) + CO2(g) + H2O(l) (d) HCl(aq) + Na2S2O3(aq) → NaCl(aq) + SO2(g) + S(s) + H2O(l) (e) Pb(NO3)2(s) ⎯⎯→ PbO(s) + NO2(g) + O2(g) ∆ 6 Zinc phosphide (Zn3P2) is used as a rat poison. This substance can be prepared by reacting excess zinc powder with phosphorus. (a) Write a balance chemical equation for the reaction. (b) Calculate the mass of phosphorus needed to react with the excess zinc to produce 51.4 kg of zinc phosphide. [Relative atomic mass: P, 31; Zn, 65] 2 Write balanced equations for the following reactions: (a) Reaction between sulphur dioxide and oxygen to form sulphur trioxide. (b) Neutralisation reaction between potassium hydroxide and sulphuric acid to form potassium sulphate and water. (c) Burning of ethane (C2H6) in air to form carbon dioxide and water. 59 Chemical Formulae and Equations 3 (d) Displacement reaction between zinc metal and copper(II) sulphate solution to form copper metal and zinc sulphate salt. (e) Decomposition of zinc carbonate to zinc oxide and carbon dioxide gas when heated. (f) Reduction of solid lead(IV) oxide, PbO2 by hydrogen gas to form lead metal and water. 3.7 Scientific Attitudes and Values in Investigating Matter 3 1 In the 19th century, scientists introduced a simple way to represent an element. Each element is represented by a letter or two letters of the alphabet; the first letter is a capital letter and the second letter is a small letter. For example, the reaction between carbon and oxygen to form carbon dioxide can be represented by the following equation: to produce 2 mol of magnesium oxide, we need to burn 2 mol of magnesium in air. If we need to produce 1 mol of MgO (40 g), then we have to burn 1 mol of Mg (24 g) in air. Similarly, to produce 40 kg of MgO, the required amount of Mg needed is 24 kg. Any excess amount of magnesium used is considered a wastage. 5 The use of relative mass by scientists enabled the mass of atoms and molecules to be determined although they are too small to be measured by any weighing machines. The development of the standard to be used in the determination of relative atomic mass and relative molecular mass started with the use of the hydrogen atom, followed by the oxygen atom, and finally carbon-12 which is now used as the standard. 6 The Italian scientist, Amedeo Avogadro proposed the hypothesis that equal volumes of all gases at the same temperature and pressure contain the same number of molecules. With perseverance and ingenuity, he calculated that 22.4 dm3 of any gas contains 6.02  1023 particles at standard temperature and pressure. The number of 6.02  1023 is now known as Avogadro’s number or the Avogadro constant in honour of him. 7 Some products can be produced by different processes. For example, calcium oxide can be produced by (a) burning calcium in air: 2Ca(s) + O2(g) → 2CaO(s) or (b) heating calcium carbonate: CaCO3(s) → CaO(s) + CO2(g) C(s) + O2(g) → CO2(g) 2 The symbols of elements using letters of the alphabet are universal in that they are agreed upon by all scientists, regardless of the countries of origin. They become a tool of communication between chemists although the spoken languages may be different. 3 Many chemical reactions involve the formation of compounds. However, the names of some compounds are long. This makes the writing of equations in words cumbersome. For example, Sulphuric acid + sodium hydroxide → sodium sulphate + water To simplify the writing of the chemical equations, scientists used chemical formulae to represent compounds. The above reaction can then be represented by H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) Thus the writing of chemical equations is made easy. 4 In the formation of a desired product through a chemical reaction, it is important to calculate the correct amount of reactants required so as to prevent wastage. This is made possible by the mole concept used by chemists in the quantitative calculations of chemical reactions. For example, in the reaction between magnesium and oxygen, 8 The ideal process is one in which (a) a high percentage yield is obtained. For example, a process which produces 80% yield is superior to one which produces only 45% yield. The mole concept makes it possible to calculate the percentage yield. (b) the cost of reactants used is cheap. In general, the cheaper the reactants, the better the process is because the cost of producing the product is lower. Thus, the product can be priced competitively. 2Mg + O2 → 2MgO Chemical Formulae and Equations 60 3.7 2 Explain the meaning of the following chemical equations: (a) CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l) (b) 2Mg(NO3)2(s) ⎯→ 2MgO(s) + 4NO2(g) + ∆ O2(g) (c) H2(g) + PbO(s) → Pb(s) + H2O(l) (d) CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) 1 The relative atomic mass of an element is the number of times one atom of the element is heavier than one-twelfth the mass of one carbon-12 atom. 2 The relative molecular mass of a compound is the number of times one molecule of the compound is heavier than one-twelfth the mass of one carbon-12 atom. 3 One mole is the quantity of substance which contains the same number of particles as there are in 12.00 grams of carbon-12. 4 One mole of element is the relative atomic mass of the element expressed in gram. For example, 6 The molar volume of a gas is the volume occupied by one mole of the gas. At standard temperature and pressure, one mole of gas occupies a volume of 22.4 dm3. 7 The interconversion between mole, mass, volume and number of particles can be summarised below: 3 1 Write the chemical formulae for the following compounds: (a) Hydrobromic acid (b) Zinc hydroxide (c) Potassium nitrate (d) Sodium sulphate (e) Silver nitrate (f) Magnesium chloride  22.4 dm3 Volume 3 22.4 dm3 1 mol of C = 12 g 1 mol of Na = 23 g 3 (6 3 1023) All contain 6.02 3 1023 atoms 1 mol of S = 32 g 1 mol of CO2 = 44 g Mass in gram  RAM or RMM  (6 3 1023) Number of particles 5 One mole of compound is the relative molecular mass of the compound expressed in grams. For example, 1 mol of H2O = 18 g Number of moles 3 RAM or RMM All contain 6.02 3 1023 molecules 1 mol of C2H6 = 30 g 61 Chemical Formulae and Equations 3 Multiple-choice Questions 3 3.1 Relative Atomic Mass and Relative Molecular Mass 1 The relative formula mass of Mg(XO3)2 is 148. Determine the relative atomic mass of element X. [Relative atomic mass: O, 16; Mg, 24] A 10 C 14 B 12 D 18 2 A compound with formula M2S2O3.5H2O has a relative formula mass of 248. What is the relative atomic mass of M? [Relative atomic mass: H, 1; O, 16; S, 32] A 23 C 27 B 24 D 39 3 Veronal is a barbiturate used to induce sleep in psychiatric patients. The molecular formula of veronal is C4H2N2O3(C2H5)2. Determine the relative molecular mass of veronal. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16] A 160 C 186 B 184 D 196 4 The diagram shows the molecular structure of vanillin molecule which gives vanilla its taste. H C | C H H O H C C C H | C–O–C–H | H C | O–H Determine the relative molecular mass of vanillin. Chemical Formulae and Equations [Relative atomic mass: H, 1; C, 12; O, 16] A 136 B 140 C 151 D 152 3.2 Relationship between the Number of Moles and the Number of Particles 5 Three elements are represented by the letters X, Y and Z. One ’08 atom of Z is two times heavier than one atom of Y. One atom of Y is three times heavier than one atom of X. If the relative atomic mass of X is 39, what is the relative atomic mass of Z? A 78 C 195 B 117 D 234 8 Which of the following samples contain 3.0 3 1022 molecules? [Relative atomic mass: H,1; C, 12; N, 14; O,16; S, 32; 1 mol contains 6 3 1023 molecules] I 0.9 g of H2O II 0.85 g of NH3 III 1.4 g of C2H4 IV 1.6 g of SO2 A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV 6 Which compound in the table below is correctly matched with its relative formula mass? [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Na, 23; P, 31; S, 32; Cl, 35.5; Ca, 40; Co, 59] 9 Calculate the number of molecules in 0.88 g of vitamin C, C6H8O6. [Relative atomic mass: H, 1; C, 12; O, 16; NA = 6 3 1023 mol–1] A 3.0 3 1021 B 3.0 3 1022 C 7.5 3 1021 D 7.5 3 1022 Compound Relative formula mass I Ca3(PO4)2 310 II C14H18N2O5 294 III C17H35COONa 306 IV CoCl2.6H2O 170 A B C D I and III only II and IV only I, II and III only I, III and IV only 7 Calculate how many beryllium atoms that will have the same mass as one quinine molecule, C20H24N2O2 (an anti-malarial drug). [Relative atomic mass: H,1; Be,9; C,12; N,14; O,16] A 35 C 37 B 36 D 38 62 10 Calculate the number of atoms in 0.96 g of titanium. [Relative atomic mass: Ti, 48; NA = 6 3 1023 mol–1] A 1.2 3 1021 B 1.5 3 1021 C 1.2 3 1022 D 1.5 3 1022 11 Which of the following has the most number of molecules? [Relative atomic mass: H, 1; C, 12; O, 16; Br, 80] A 3.6 g of water, H2O B 4.0 g of methane, CH4 C 19.6 g of sulphuric acid, H2SO4 D 13.2 g of ethanoic acid, CH3COOH 12 The relative atomic mass of oxygen and sulphur are 8 and 32 respectively. Which of the following statements is/are true about oxygen and sulphur? 13 What is the total number of atoms present in 6.05 g dichlorodifluoromethane, CCl2F2. [Relative atomic mass: C, 12; F, 19; Cl, 35.5; NA = 6 3 1023 mol–1] C 1.2 3 1023 A 1.2 3 1022 B 3.0 3 1022 D 1.5 3 1023 14 Pyrethrin is an insecticide with molecular formula of C19H26O3. Calculate the number of pyrethrin molecules contained in a spray with 15.1 g of the compound. [Relative atomic mass: H, 1; C, 12; O, 16; NA = 6 3 1023 mol–1] C 1.5 3 1023 A 1.5 3 1022 B 3.0 3 1022 D 3.0 3 1023 15 If m is the number of atoms in 3 g of carbon, the number of atoms in 3 g of magnesium in terms of m is [Relative atomic mass: C, 12; Mg, 24] 1 A —m C m 2 1 B —m D 2m 3 3.3 Relationship between the Number of Moles of a Substance and Its Mass 16 Which of the following substances contain the same ’09 number of atoms as in 36 gram of carbon? [Relative atomic mass: H, 1; C, 12; O, 16; S, 32] I 3 g of hydrogen II 64 g of sulphur III 18 g of water IV 22 g of carbon dioxide A B C D I and III only I and IV only II and III only II and IV only 17 Calculate the mass of 7.5 3 1021 aspirin, C9H8O4 molecules. [Relative atomic mass: H, 1; C, 12; O, 16; NA = 6 3 1023 mol–1] A 1.25 g C 2.25 g B 1.44 g D 3.00 g 18 Caffeine is found in coffee beans. Its molecular formula is C4H5N2O. A pill contains 0.05 mol of caffeine. Determine the mass of the compound in the pill. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16] A 1.94 g C 4.85 g B 2.42 g D 9.70 g 19 A cockroach repellent has the formula CH3(CH2)5CHCHCHO. Determine the mass of 0.02 mol of this substance. [Relative atomic mass: H, 1; C, 12; O, 16] A 1.40 g C 3.50 g B 2.80 g D 5.60 g 20 Acetaminophen is a medicine used to relieve pain. 0.0002 mol of acetaminophen has a mass of 0.0302 g. Which of the following is the molecular formula of acetaminophen? [Relative atomic mass: H, 1; C, 12; N, 14; O, 16] C C8H9NO A C8H9NO2 D C8H9N2O B C8H8NO2 21 Ethyl ethanoate is a liquid used as nail varnish remover. If 0.025 mol of ethyl ethanoate has a mass of 2.20 g, determine the relative molecular mass of ethyl ethanoate. A 44 C 77 B 55 D 88 22 The diagram shows the molecular structure of allicin which is a compound obtained from garlic. Allicin have powerful antibiotic and antifungal properties. O i S C H C H2 C H2 S allicin 63 C H2 C H C H2 Calculate the number of moles in 4.86 g allicin. [Relative atomic mass: H, 1; C, 12; O, 16; S, 32] A 0.02 mol C 0.03 mol B 0.025 mol D 0.05 mol 23 Cocaine, C17H21O4N is a drug. Calculate the mass of 1.2 3 1021 cocaine molecules. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; NA = 6 3 1023 mol–1] A 0.202 g C 0.404 g B 0.303 g D 0.606 g 3.4 Relationship between the Number of Moles of a Gas and Its Volume 24 1.0 g of calcium carbonate is added into excess hydrochloric acid. CaCO3 + 2HCl → CaCl2 + H2O + CO2 Determine the volume of carbon dioxide gas evolved at room temperature. [Relative atomic mass: C, 12; O, 16; Ca, 40; 1 mol of gas occupies a volume of 24 dm3 at r.t.p.] C 180 cm3 A 120 cm3 B 150 cm3 D 240 cm3 25 Which of the following gases will occupy the same volume as 2.42 g dichlorodifluoromethane, CCl2F2? [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; F, 19; S, 32; Cl, 35.5] I 0.34 g of ammonia, NH3 II 0.88 g of carbon dioxide, CO2 III 1.28 g of sulphur dioxide, SO2 IV 0.40 g of methane, CH4 A I, II and III only B I, II and IV only C I, III and IV only D II, III and IV only 26 Which of the following gases occupy a volume of 600 cm3 at room temperature and pressure? [Relative atomic mass: H, 1; C, 12; O, 16; S, 32; 1 mol of gas occupies a volume of 24 dm3 r.t.p.] I 0.64 g of oxygen, O2 II 0.75 g ethane, C2H6 III 0.56 g of carbon monoxide, CO Chemical Formulae and Equations 3 I 4 g of oxygen and 8 g of sulphur contain the same number of atoms. II The sulphur atom has four times more neutrons than an oxygen atom. III The sulphur atom is four times denser than the oxygen atom. IV 8 g of oxygen contain two times more atoms than 8 g of sulphur. A I and III only B II and IV only C II and III only D IV only 3 IV A B C D 1.15 g nitrogen dioxide, NO2 I and III only II and IV only I, II and III only II, III and IV only 27 Which of the following gases has the heaviest mass at room temperature and pressure? [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; S, 32; 1 mol of gas occupies a volume of 24 dm3 r.t.p.] A 3 dm3 of sulphur dioxide, SO2 B 6 dm3 of nitrogen dioxide, NO2 C 15 dm3 of methane, CH4 D 96 dm3 of hydrogen, H2 28 Which of the following gases occupies the greatest volume at room temperature and pressure? [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; 1 mol of gas occupies a volume of 24 dm3 r.t.p.] A 0.64 g oxygen, O2 B 0.70 g nitrogen, N2 C 1.10 g carbon dioxide, CO2 D 0.51 g ammonia, NH3 29 What is the number of atoms in 0.05 mol of ammonia gas, NH3? [Avogadro number : 6 3 1023 mol–1] A 1.2 3 1022 B 3.0 3 1022 C 1.2 3 1023 D 3.0 3 1023 30 Which of the statements below are true? I 1.10 g of CO2 and 1.25 g of SO2 gases occupy the same volume at s.t.p. II 0.42 g of N2 and 0.66 g of CO2 gases occupy the same volume at s.t.p. III 2.24 dm3 of C2H6 and 1.12 dm3 of SO2 gases at s.t.p. have the same mass. IV 4.48 dm3 of O2 and 8.96 dm3 of CH4 gases at s.t.p. have the same mass. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; S, 32; 1 mol of gas occupies 22.4 dm3 at s.t.p.] A I and III only B II and IV only C I, II and IV only D II, III and IV only Chemical Formulae and Equations 3.5 Chemical Formulae 31 1.04 g of element X reacted with 0.48 g of oxygen to form an oxide with the empirical formula X2O3. Determine the relative atomic mass of element X. [Relative atomic mass: O, 16] A 24 C 52 B 48 D 56 32 When 3.64 g of a metal oxide of M is reduced, 2.04 g of the metal is obtained. Determine the empirical formula of the metal oxide. [Relative atomic mass: O, 16; M, 51] A MO2 C M3O2 B M2O3 D M2O5 33 2.75 g of metal M combines with 1.6 g of oxygen to form an ’10 oxide with the empirical formula of MO2. Determine the relative atomic mass of M. [Relative atomic mass: O, 16] A 48 C 55 B 52 D 56 34 x gram of antimony (Sb) combines with 0.48 g of oxygen to form an oxide with the empirical formula of Sb2O3. Determine the value of x. [Relative atomic mass: O, 16; Sb, 122] A 1.22 g B 2.44 g C 3.66 g D 4.88 g 35 An element E forms a fluoride compound with the formula EF3, which contains 16.2 % of E by mass. What is the relative atomic mass of E? [Relative atomic mass; F, 19] A 11 B 24 C 27 D 31 3.6 Chemical Equations 36 Calculate the mass of zinc required to react with excess ’09 nitric acid to produce 360 cm3 of hydrogen at room conditions. 64 [Molar volume: 24 dm3 mol–1 at room conditions; Relative atomic mass: Zn, 65] A 0.325 g B 0.650 g C 0.975 g D 4.333 g 37 0.31 gram of copper(II) carbonate is heated. Determine the volume of carbon dioxide gas released at room conditions. [Relative atomic mass: C, 12; O, 16; Cu, 64; Molar volume : 24 dm3 at room conditions] A 60 cm3 B 120 cm3 C 240 cm3 D 360 cm3 38 Calculate the mass of aluminium oxide, Al2O3 formed if 5.4 g of aluminium is heated in air. [Relative atomic mass: O, 16; K, 39] A 8.4 g B 8.6 g C 10.2 g D 11.8 g 39 Magnesium oxide reacts with nitric acid to form magnesium nitrate and water. If 8.0 g of magnesium oxide is reacted with excess nitric acid, calculate the mass of salt formed. [Relative atomic mass: N, 14; O, 16; Cu, 64] A 14.8 g B 17.2 g C 20.4 g D 29.6 g 40 Iron reacts with chlorine according to the equation ’09 below: 2Fe + 3Cl2 → 2FeCl3 If 1.68 gram of iron burns completely in chlorine, calculate the mass of product formed. [Relative atomic mass: Cl, 35.5, Fe, 56] A 2.438 g B 4.875 g C 6.555 g D 9.750 g Structured Questions (i) Determine the empirical formula of M oxide. 1 A hydrocarbon X contains 82.76% carbon by mass. 2.9 g of hydrocarbon X occupies a volume of 1.2 ’10 dm3 at room temperature and pressure. [Relative atomic mass: H, 1; C, 12; O, 16; 1 mol of gas occupies a volume of 24 dm3 at r.t.p.; NA = 6  1023 mol–1] (a) What is a hydrocarbon? [1 mark] [2 marks] (ii) Write a chemical equation for the reduction of M oxide to metal M using hydrogen gas. [2 marks] (e) Can the empirical formula of magnesium oxide be determined using the same arrangement of apparatus as above? Explain your answer. [2 marks] (b) Determine the empirical formula of hydrocarbon X. [2 marks] relative molecular mass 3 Table 1 shows the positive and negative ions in three salt solutions. of [2 marks] ’05 Name of salt (d) Determine the molecular formula of hydrocarbon X. [2 marks] Positive ion Copper(II) sulphate (e) Combustion of X in air produces carbon dioxide and water. Write a chemical equation for the reaction. [2 marks] (f) If 11.6 g of X is burnt, calculate (i) the mass of water formed. [1 mark] (ii) the number of carbon dioxide molecules produced at room temperature. [1 mark] Negative ion Cu SO42– 2+ Potassium iodide K+ I– Lead(II) nitrate Pb2+ NO3– Table 1 (a) What is another name for a positively-charged ion? [1 mark] (b) Write the formula of lead(II) nitrate. 2 The apparatus shown is used to determine the empirical formula of the oxide of metal M by reducing ’07 the metal oxide with dry hydrogen gas. [Relative atomic mass: O, 16; M, 55] [1 mark] (c) When aqueous lead(II) nitrate solution is added to aqueous potassium iodide solution, a yellow precipitate is formed. (i) Write a chemical equation for the reaction. [2 marks] (ii) Describe the chemical equation in (i). [1 mark] (iii) Name the yellow precipitate. [1 mark] (iv) If 0.04 mol of aqueous potassium iodide solution is added to excess lead(II) nitrate solution, calculate the maximum mass of the yellow precipitate formed. [Relative atomic mass: I, 127; Pb, 207] [2 marks] Diagram 1 (a) State one precaution that must be taken when carrying out the experiment. [1 mark] 4 Table 2 shows the descriptions and observations of two experiments, I and II. (b) How can you ensure that all the oxide of metal M has been reduced? [1 mark] (c) ’05 Experiment (i) Name two chemicals used to prepare hydrogen gas in the laboratory. [1 mark] (ii) Write an equation for the reaction in (i). Combustion of 1.2 g of magnesium powder in excess oxygen Magnesium burns brightly and a white powder is formed II Heating copper(II) carbonate strongly in a test tube Black solid X is formed and a gas P which turns limewater cloudy is evolved [1 mark] [1 mark] Mass of combustion tube + asbestos paper = 39.25 g Mass of combustion tube + asbestos paper + oxide of metal M before heating = 47.95 g Mass of combustion tube + asbestos paper + metal M after heating = 44.75 g Observation I (iii) Name a chemical used to dry hydrogen gas. (d) The information below shows the results of the experiment: Description Table 2 [Relative atomic mass: C, 12; O, 16; Mg, 24; Cu, 64; 1 mol of gas occupies a volume of 24 dm3 at room temperature and pressure] 65 Chemical Formulae and Equations 3 (c) Calculate the hydrocarbon X. 3 (a) Based on Experiment I: (i) The white powder formed is magnesium oxide. Write the chemical equation for the [2 marks] reaction that takes place. (ii) Calculate the mass of magnesium oxide formed if 3 g of magnesium is completely [2 marks] burnt in excess oxygen. (iii) State one use of magnesium oxide. [1 mark] (iv) The magnesium oxide is basic and reacts with nitric acid (HNO3) to form magnesium nitrate and water. Write a chemical equation [2 marks] for this reaction. (a) Write a balanced chemical equation for the decomposition of sodium bicarbonate on heating. [2 marks] (b) State a chemical test for carbon dioxide gas. [2 marks] (c) If 8.4 g of sodium bicarbonate decomposes, calculate (i) the volume of carbon dioxide gas envolved at room temperature and pressure. [3 marks] (ii) the mass of sodium carbonate formed. [3 marks] [Relative atomic mass: H, 1; C, 12; O, 16; Na, 23; 1 mol of gas occupies a volume of 24 dm3 at room temperature and pressure] (b) Based on Experiment II: (i) Name the black solid X and gas P formed when copper(II) carbonate is heated strongly. (d) Sodium bicarbonate reacts with nitric acid (HNO3) to form sodium nitrate, carbon dioxide and water. (i) Write a balanced chemical equation for this reaction. [2 marks] (ii) Calculate the mass of sodium bicarbonate that reacts with the excess acid to produce 17 g of sodium nitrate. [3 marks] (iii) State one use of sodium nitrate. [2 marks] [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Na, 23] [2 marks] (ii) Write the chemical equation for the reaction [1 mark] that takes place. (iii) If 6.2 g of copper(II) carbonate had reacted, calculate [1 mark] (a) the mass of solid X formed. (b) the volume of gas P formed at room [1 mark] temperature and pressure. 5 When sodium bicarbonate (NaHCO3) is heated, it decomposes to sodium carbonate, carbon dioxide and water. (e) (i) Name another chemical that reacts with nitric acid to form sodium nitrate. [1 mark] (ii) Write an equation for this reaction. [2 marks] Essay Questions (d) Describe a laboratory method of determining the empirical formula of lead oxide. Your answer should include (i) the procedure of the experiment [4 marks] (ii) tabulation of result [3 marks] (iii) calculation of the results obtained. [4 marks] [Relative atomic mass: O, 16; Pb, 207] 1 (a) Using a suitable example, explain the meaning of the following terms: (i) Empirical formula [2 marks] (ii) Molecular formula [2 marks] (b) State a suitable method that can be used to determine the empirical formula of lead(II) oxide. [2 marks] (c) Can the same method in (b) be used to determine the empirical formula of magnesium oxide? Explain your answer. [3 marks] Experiment 1 You are required to plan an experiment to determine the empirical formula of magnesium oxide. Your explanation should include the following: (a) Statement of the problem (b) All the variables (c) List of materials and apparatus (d) Procedure (e) Tabulation of data Chemical Formulae and Equations 66 [17 marks] FORM 4 THEME: Matter Around Us CHAPTER 4 Periodic Table of Elements SPM Topical Analysis 2008 Year 1 Paper 3 2 Section A Number of questions 1 — 2 4 2009 B – 1 2010 3 2 C A B C – 1 — 4 – – 1 3 – 1 3 2011 2 3 A B C 1 – – 1 1 6 2 3 A B C 1 – – – ONCEPT MAP PERIODIC TABLE Historical development of the Periodic Table Transition elements J. W. Dobereiner John Newlands Lothar Meyer Dmitri Mendeleev Henry G. J. Moseley Metallic properties of transition elements Group Physical and chemical properties of elements in: Group 1 Group 17 Inert property and uses of Group 18 elements Special characteristics of transition elements Period Identifying the group and period of an element based on the electron arrangement of the element • Changes in the properties of the elements and their oxides across Period 3 • Uses of semi-metals such as silicon and germanium in the microelectronics industry 4.1 3 This systematic method of classification of the elements will enable us to (a) study and generalise the chemical and physical properties of elements in the same group. (b) predict the position of an element in the Periodic Table from its properties. (c) identify and compare elements from different groups. (d) predict the chemical and physical proper ties of new elements in the same group. 4 Chemists such as Lavoisier, Dobereiner, Newlands, Meyer, Mendeleev and Moseley contributed to the development of the Periodic Table in use today. Periodic Table of Elements 4 Historical Development of the Periodic Table 1 Many of the elements known today were disco vered from the years 1800 to 1900. Chemists noted that certain elements have similar chemi cal properties. For example: chlorine, bromine and iodine; potassium and sodium; magnesium and calcium have similar chemical properties. 2 Chemists then tried to group the elements with the same chemical properties together. This led to the development of the Periodic Table. Contribution of Scientists to the Development of the Periodic Table Antoine Lavoisier 1 Antoine Lavoisier, a Frenchman, was the first chemist who attempted to classify the elements into four groups as in Table 4.1. 2 The four groups consisted of gases, metals, non-metals and metal oxides. 3 His classification was not accurate as light and heat, which are not elements, were included. Furthermore, some elements in each group did not have the same chemical properties. Table 4.1 Lavoisier’s Periodic Table Group I Group II Group III Group IV Oxygen Nitrogen Hydrogen Light Heat Sulphur Phosphorus Carbon Chlorine Fluorine Arsenic Bismuth Cobalt Lead Zinc, Nickel, Tin, Silver Calcium oxide Barium oxide Silicon(IV) oxide Magnesium oxide Aluminium oxide Antoine Lavoisier (1743–1794) Johann W. Dobereiner 1 Dobereiner classified the elements with the same chemical properties into groups of three called triads. Example: 2 Dobereiner discovered the relationship between the relative Element in the Relative atomic atomic mass (r.a.m.) of elements in triad mass (r.a.m.) each triad. He found that the r.a.m. Lithium (Li) 7 of the element in the middle of each Sodium (Na) 23 triad is approximately equal to the average of the total r.a.m. of the Potassium (K) 39 Johann W. Dobereiner other two elements. Average of the total r.a.m. of Li and K is (1780–1849) 3 However, this relationship did 7 + 39 not apply to most of the other ⎯⎯⎯ = 23 2 elements. Periodic Table of Elements 68 John Newlands 1 John Newlands arranged the elements in order of increasing nucleon number (mass number) in horizontal rows. Each row consisted of seven elements. 2 He found that the chemical properties of every eighth element are similar. This was known as the law of octaves. 3 Table 4.2 shows the arrangement of elements by John Newlands. Table 4.2 The classification of elements by Newlands (law of octaves) Be B C N O F Na Mg Al Si P S Cl K Ca John Newlands (1837–1898) 4 Li 4 The classification of elements by Newlands was not successful because: (a) The law of octaves was only accurate for the first 16 elements (from Li to Ca). (b) There were no positions allocated for elements yet to be discovered. 5 Nevertheless, Newlands was the first scientist who discovered the existence of periodicity in the elements. Lothar Meyer SPM ’09/P1 1 Meyer, a German chemist, Volume of an atom calculated the volume of mass of one mole-atom of the element an atom of an element = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ density of the element using the formula: 2 He plotted a graph of volume of atoms of elements against their relative atomic masses. The graph obtained is shown in Figure 4.1. Lothar Meyer (1830–1895) Figure 4.1 Lothar Meyer’s atomic volume curve 3 From the shape of the atomic volume curve, Meyer showed that elements occupying the corresponding positions of the curve exhibit similar chemical properties. For example: (a) Li, Na, K and Rb which are located at the peak of the curve show similar chemical properties. (b) Be, Mg, Ca and Sr which are located at positions after the maximum points also show similar chemical properties. 4 Just like Newlands, Meyer showed that the properties of the elements recur periodically. 69 Periodic Table of Elements Dmitri Mendeleev 1 Dmitri Mendeleev, a Russian chemist, arranged the elements in order of increasing atomic mass. 2 Table 4.3 below shows the Periodic Table suggested by Mendeleev. 4 Table 4.3 The partial Periodic Table by Mendeleev 1 2 3 4 5 6 I II H Li Na K Cu Rb Be Mg Ca Zn Sr III IV V VI B C N O Al Si P S () Ti V Cr () () As Se Y Zr Nb Mo ( ) represents unknown elements yet to be discovered VII VIII F Cl Mn Br () Fe, Co, Ni Dmitri Mendeleev (1834–1907) Ru, Rh, Pd 3 Dmitri Mendeleev was more successful for several reasons. (a) First, he left gaps for elements yet to be discovered. He even used the table to predict the existence and properties of undiscovered elements. Mendeleev correctly predicted the properties of the elements gallium, scandium and germanium which were only discovered much later. (b) Secondly, although the elements were arranged in order of increasing atomic mass, he changed the order of the elements if the chemical properties are not similar. Henry G. J. Moseley 1 Moseley was a British physicist. He bombarded different elements with high energy electrons and measured the frequency (f) of the X-ray emitted by the element. 2 He then plotted the square-root of the frequency of the X-ray ( ⎯ f ) against the proton number (atomic number) of the element and obtained a straight line graph (Figure 4.2). 3 Thus from the square-root of the frequency of the X-ray emitted by an unknown element, he could determine its proton number. 4 After obtaining the proton number of the elements, Moseley went on to arrange the elements in the Periodic Table in order of increasing proton number (atomic number). 5 Just like Mendeleev did, Moseley left gaps ( ) for elements yet to be discovered. He reasoned that there should be an element corresponding to each proton number. 6 Moseley successfully predicted the existence of four undiscovered elements from the atomic numbers. These elements were later determined to be technetium, promethium, hafnium and rhenium. With this prediction, Moseley was able to prove the separate existence of each element in the lanthanide series. Henry G. J. Moseley (1887–1915) Figure 4.2 Modern Periodic Table 2 In the modern Periodic Table, the elements are arranged in order of increasing proton number. This order is also related to the electron arrangement of the elements. 1 There are 117 discovered elements now. Most of these elements are naturally occurring elements. However, a few of these elements are made artificially in nuclear reactors. Periodic Table of Elements 70 (b) Periods 2 and 3 have eight elements each. The first three periods are called the short periods. (c) Periods 4 and 5 have 18 elements each. They are called the long periods. (d) Period 6 has 32 elements. Not all the elements can be listed on the same horizontal row. The elements with proton numbers 58 to 71 are separated and are grouped below the Periodic Table. These elements are called the Lanthanide Series. (e) Period 7 has 31 elements and not all can be listed on the same horizontal row. The elements with proton numbers 90 to 103 are grouped below the Periodic Table. They are called the Actinide Series. The Electron Arrangement of Elements in the Periodic Table 1 Table 4.4 shows the electron arrangement of the elements with proton numbers 1–20. 2 All members of the same group have the same number of valence electrons. Valence electrons are electrons in the outermost shell. For example: (a) Group 1 elements (Li, Na, K) each has one valence electron. (b) Group 2 elements (Be, Mg, Ca) each has two valence electrons. (c) Group 17 elements (F, Cl, Br, I, At) each has seven valence electrons. 3 (a) The number of valence electrons of Group 1 and Group 2 elements is the same as its group number. SPM ’11/P1, P2 Table 4.4 The electron arrangement of the first 20 elements in the Periodic Table Group Period 1 2 13 14 15 16 17 18 He 2 1 H 1 2 Li 2.1 Be 2.2 B 2.3 C 2.4 N 2.5 O 2.6 F 2.7 Ne 2.8 3 Na 2.8.1 Mg 2.8.2 Al 2.8.3 Si 2.8.4 P 2.8.5 S 2.8.6 Cl 2.8.7 Ar 2.8.8 4 K 2.8.8.1 Ca 2.8.8.2 71 Periodic Table of Elements 4 3 The Periodic Table is a classification of elements whereby elements with the same chemical properties are placed in the same group. This makes the study of the chemistry of these elements easier and more systematic. 4 The modern Periodic Table is shown on page 78. 5 The vertical columns of the Periodic Table are called groups. There are 18 groups in the Periodic Table. 6 Each member of a group shows similar chemi cal properties although their physical properties such as density, melting point and colour may show a gradual change when descending the group. • Group 1 elements (Li, Na, K, Rb, Cs, Fr) are called alkali metals. • Group 2 elements (Be, Mg, Ca, Sr, Ba, Ra) are called alkaline earth metals. • Group 17 elements (F, Cl, Br, I, At) are called halogens. • Group 18 elements (He, Ne, Ar, Kr, Xe, Rn) are called noble gases. 7 A block of elements called the transition elements separates Group 2 and Group 13. (a) The elements in Groups 1, 2, 13 and the transition elements are metals. (b) The elements in Groups 15, 16 and 17 are non-metals. (c) Group 14 consists of non-metals at the top of the group, followed by semi-metals (or metalloids) and metals lower down in the group. 8 The horizontal rows are called periods. There are seven periods. (a) Period 1 has two elements only: hydrogen and helium. 4 3 (b) Except for helium, the elements with more than 2 valence electrons (Groups 13 to 18), the group number = 10 + (number of valence electrons). 4 The period number is indicated by the number SPM of filled electron shells. For example: ’05/P2 (a) Elements in Period 1 (H and He) each has only one electron shell filled with electrons. (b) Elements in Period 2 (Li, Be, B, C, N, O, F, Ne) each has two electron shells filled with electrons. (c) Elements in Period 3 (Na, Mg, Al, Si, P, S, Cl, Ar) each has three electron shells filled with electrons. (d) Elements in Period 4 (K and Ca) each has four electron shells filled with electrons. 5 All elements in the same period have the same number of filled electron shells. 1 The proton number of element X is 20. Which of the following statements are true concerning the element X? I X can conduct electricity. II X belongs to Group 2 of the Periodic Table. III X belongs to Period 4 of the Periodic Table. IV X belongs to Period 3 of the Periodic Table. A II and III only B II and IV only C I, II and III only D I, II and IV only Comment X has 20 electrons. The electronic configuration of X is 2.8.8.2. X belongs to Group 2 of the Periodic Table because it has two valence electrons. (II is correct) Group 2 elements are metals and can conduct electricity. (I is correct) ’05 X belongs to Period 4 of the Periodic Table because it has four electron shells filled with electrons. (III is correct, IV is incorrect) Answer C Which of the following represents the electron arrangement of an element in Group 17? A B C D 4 Element Answer D (The element has seven valence electrons) 2 Electron arrangement ’03 R S 2.8.6 2.8.8 2 Determine the group in which Q, R and S belong to in the Periodic Table. In the Periodic Table, Y is below Z in the same group. If the proton number of atom Z is 11, what is the electron arrangement of atom Y ? A 2.2 C 2.8.3 B 2.7 D 2.8.8.1 Solution For elements with one or two valence electrons, Group number = Number of valence electrons Comment In an atom, the number of protons is equal to the number of electrons. Thus Z has 11 electrons. The electron arrangement of Z is 2.8.1. The elements in the same group have the same number of valence electrons. Therefore the electron arrangement of Y is 2.8.8.1 because each atom of Y and Z has one valence electron. Answer D Periodic Table of Elements Q For elements with more than two valence electrons, Group number = 10 + Number of valence electrons Therefore element Q belongs to Group 16 and element R belongs to Group 18 of the Periodic Table. S has two electrons in the first electron shell. It is helium and belongs to Group 18 of the Periodic Table. 72 1 Q R T X Z 3 15 18 19 35 Element Electron arrangement The proton numbers of elements Q, R, T, X and Z are given in the table above. Which of the following statements are true? I Elements Q and X belong to the same group in the Periodic Table. II Elements T and X belong to the same period. III Elements Q and X are metals. IV Elements R and Z are non-metals. A I and III only B II and IV only C I, III and IV only D II, III and IV only Q R 2.1 T X Z 2.8.5 2.8.8 2.8.8.1 2.8.18.7 Comment Elements Q and X belong to the same group in the Periodic Table because they have the same number of valence electrons. (Statement I is correct) Element T belongs to Period 3 (it has 3 electron shells), whereas element X belongs to Period 4 (it has four electron shells). (Statement II is wrong) Elements in Groups 1, 2 and 13 have one, two and three valence electrons and they are metals. (Statement III is correct) Elements with five, six, seven or eight valence electrons are non-metals. (Statement IV is correct) Answer C 4.1 3 Arsenic is represented by the symbol 75 As. In which 33 group and period does arsenic belong to in the Periodic Table? Name two elements that have the same chemical properties as arsenic. Explain your answer. 1 Explain the term valence electrons. State the number of valence electrons of the elements in the following groups in the Periodic Table. (a) Group 1 (b) Group 2 (c) Group 15 (d) Group 17 2 Element W X Y 4 The following table shows the proton numbers of 10 elements represented by the letters A to K: Element Z A Proton number 2 Electron 2.8.5 2.1 2.8.8.3 2.8.18.32.18.7 arrangement B C D E F G H J K 9 13 19 18 16 7 20 17 6 Electron arrangement The table shows the electron arrangement of four elements. (a) State the group of each element: (i) W (ii) X (iii) Y (iv) Z Group number Period number (a) Write the electron arrangement of each element. Then state the group number and period number of each element. (b) State the period of each element: (i) W (ii) X (iii) Y (iv) Z (b) Pick a pair of elements which have the same chemical properties. 73 Periodic Table of Elements 4 Element Proton number ’04 4.2 Group 18 Elements ’05 General He Ne Ar Kr Xe Rn 4 Figure 4.3 The elements of Group 18 in the Periodic Table • The noble gases are also known as the inert gases. They are elements of Group 18, at the far right of the Periodic Table (Figure 4.3). • The group consists of six elements: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn). • The atomic radius of the noble gases increases down the group. This is because as the number of filled electron shells increases down the group, the valence electron is further from the nucleus. Physical Properties This is because as the size of the atoms increases Physical properties of noble gases: down the group, the van der Waals forces of • All noble gases are insoluble in water and do attraction become stronger. not conduct electricity or heat. • All noble gases have low melting points and • The noble gases have very low densities. This is low boiling points. This is because the noble because the atoms are very far apart. The density of the noble gases increases when descending gases exist as monatoms which are attracted by the group. This is due to the increase in the very weak van der Waals forces of attraction. The melting points and boiling points of the relative atomic mass of the element. noble gases increase when descending the group. Chemical properties stable duplet electron arrangement. Chemical properties of noble gases: – The other noble gases have eight electrons in • Noble gases are chemically inert which means their outermost electron shells. They are said that they are unreactive in nature. They do not to have attained the stable octet electron react with any other elements. arrangement. • Noble gases are the only elements which exists • Therefore the noble gases do not need to as monatoms (as single atoms). accept, donate or share electrons with other • Noble gases are unreactive because they all elements. have filled outer shells of electrons which are a stable electron arrangement. (All chemical reactions involve either gaining, – Helium has only one filled electron shell and losing or sharing electrons.) it has exactly two electrons in this outermost electron shell. It is said to have attained the Table 4.5 The proton numbers, relative atomic mass, electron arrangement and physical properties of inert gases Element He Ne Ar Kr Xe Rn Proton Relative atomic Electron number arrangement mass 2 10 18 36 54 86 Periodic Table of Elements 4 20 40 84 131 222 2 2.8 2.8.8 2.8.18.8 2.8.18.18.8 2.8.18.32.18.8 74 Melting Atomic radius (nm) point (°C) 0.06 0.07 0.094 0.109 0.130 – –270 –248 –189 –156 –112 –71 Boiling point (°C) Density (g dm–3 ) –269 –246 –186 –152 –107 –62 0.17 0.84 1.66 3.54 5.45 – 4.2 ’04 1 2P, 11Q, 13R, 18S, 20T. The above is a set of elements with their proton numbers. Choose two elements that exist as monatomic gases. Explain your answer. A car distributor wants to use colourful electric lamps to attract customers. Which of the following substances A, B, C or D in the Periodic Table is suitable for use in the lamps? 2 Inert gases are the elements of Group 18 of the Periodic Table. Name two inert gases and state one use of each. 3 All the inert gases have low melting and boiling points. Explain why as we go down Group 18, the melting points and boiling points increase. Comment B is neon gas. Neon-filled electric bulbs produce an attractive bright red light which is used in advertising. Answer B 4.3 4 One of the characteristics of Group 18 gases is that they exist as monatoms. Explain why neon does not form compounds with other elements. Group 1 Elements General Figure 4.4 The elements of Group 1 in the Periodic Table • Group 1 elements are also known as the alkali metals. • The elements in Group 1 are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs) and francium (Fr). • All Group 1 elements are soft metals. The metals are grey in colour and are silvery when the surface is freshly cut, before being exposed to air. Physical Properties Physical properties of Group 1 elements: between the atoms becomes weaker down the • All Group 1 elements are metals; hence they group as the atomic radius increases. are good conductors of heat and electricity. • The electropositivity of the metals increases • The atomic radius increases down the group. down the group. Electropositivity is a measurement The reason is that as the number of filled of the ability of an atom to lose an electron electron shells increases when descending and form a positive ion. the group, the distance between the outermost M → M+ + e– (M = Li, Na, K, Rb, Cs, Fr) electron shell and the nucleus increases. • The density increases down the group. The densities of Li, Na and K is lower than that As the atomic radius becomes larger down of water, and hence these metals can float on the group, the force of attraction between water. the nucleus and the single valence electron • The melting point decreases when descending becomes weaker. Hence the elements lose the group. This is because the metallic bond the single valence electron more easily when descending the group. 75 Periodic Table of Elements 4 2 SPM 4 Reactivity ’07/P2, ’11/P2 Reactivity of Group 1 elements M(s) → M+ + e– (M = Li, Na, K, Rb, Cs, Fr) • All Group 1 elements are very reactive. The Li(s) → Li+ + e– reactivity increases down the group. 2.1 2 • The elements in Group 1 have one valence Na(s) → Na+ + e– electron each. During a chemical reaction, reactivity increases 2.8.1 2.8 an atom of a Group 1 element will donate a valence electron to form a univalent positive K(s) → K+ + e– ion to attain the stable duplet or octet in its 2.8.8.1 2.8.8 electron arrangement. • The reactivity of Group 1 elements depends on how easily it can donate its valence electron. The atomic radius of Group 1 elements increases down the group. This causes the force of attraction between the nucleus and the valence electron to become weaker, making it easier for a metal lower in the group to donate its valence electron. Therefore, the reactivity of Group 1 elements increases when descending the group. Chemical Properties Chemical properties of Group 1 elements • The elements in Group 1 have the same chemical properties because each has one valence electron. • Group 1 elements react with – cold water to produce hydrogen gas and alkalis, – oxygen to produce metal oxides, – halogen to produce metal halides. Table 4.6 Some physical properties of Group 1 elements SPM ’06/P2, ’08/P1 Element Colour Electron arrangement Atomic radius (nm) Density (g dm–3) Li Silvery 2.1 0.15 0.53 Conductor 181 Na Silvery 2.8.1 0.16 0.97 Conductor 98 K Silvery 2.8.8.1 0.23 0.86 Conductor 63 Ru Silvery 2.8.18.8.1 0.25 1.53 Conductor 39 Cs Silvery 2.8.18.18.8.1 0.26 1.87 Conductor 29 Fr Silvery 2.8.18.32.18.8.1 0.29 – Conductor 27 Electrical Melting Electropositivity conductivity point (°C) ⏐ ⏐ ⏐ Increases ⏐ ⏐ ⏐ ↓ 4.1 Experiment 4.1 To study the reaction of alkali metals with oxygen Problem statement How do lithium, sodium and potassium differ in reactivity with oxygen? Variables (a) Manipulated variable : The alkali metals used (b) Responding variable : Reactivity of the metals with oxygen (c) Constant variable : Excess supply of oxygen and the size of the metal piece used Materials Small pieces of Li, Na and K metals, oxygen gas, filter paper and phenolphthalein indicator. Hypothesis The alkali metals show similar chemical properties in their reactions with oxygen but the reactivity of the alkali metals with oxygen increases down the group (in the order from lithium, sodium to potassium). Periodic Table of Elements 76 3 The lithium metal is heated until its starts to burn. The spoon is then put into a gas jar containing oxygen gas. 4 The observation is recorded. 5 After the reaction has stopped, about 20 cm3 of water is added to the gas jar. The gas jar is shaken so that the product of the combustion is dissolved in the water. The solution is tested with a few drops of phenolphthalein indicator. The observation is recorded. 6 The experiment is repeated using sodium and potassium metals. Apparatus Pen knife, tongs, gas jar with cover, gas jar spoon and Bunsen burner. Safety precautions Alkali metals, especially sodium and potassium, are very reactive. Therefore we have to (a) use small pieces of each metal. (b) use goggles while carrying out the experiment. (c) ensure that we do not handle the metal with our bare hands. 4 Procedure 1 A piece of lithium metal is removed from the bottle with tongs. A small piece of the metal is cut using a pen knife. A piece of filter paper is used to absorb the paraffin oil from the piece of metal. 2 The lithium metal is then transferred onto a gas jar spoon using tongs. Figure 4.5 Reaction of Group 1 metals with oxygen Results Observation Metal Lithium The lithium metal burns with a red flame forming a white metal oxide. The metal oxide dissolves in water producing a solution which turns phenolphthalein indicator red. Sodium The sodium metal burns with a bright yellow flame forming a white metal oxide. The metal oxide dissolves in water producing a solution which turns phenolphthalein indicator red. Potassium The potassium metal burns with a very bright purplish flame forming a white metal oxide. The metal oxide dissolves in water producing a solution which turns phenolphthalein indicator red. Discussion 1 Alkali metals burn in oxygen to form white metal oxides 4Li(s) + O2(g) → 2Li2O(s) 4Na(s) + O2(g) → 2Na2O(s) 4K(s) + O2(g) → 2K2O(s) Reactivity increases down the group ⎯⎯→ SPM ’11/P1 The presence of hydroxide ions, OH– causes the solution to be alkaline. Conclusion All alkali metals burn in oxygen gas to produce white metal oxides. Group 1 elements show similar chemical properties. Their reactivity increases down the group from lithium, sodium to potassium. 2 These metal oxides dissolve in water to form alkaline solutions which turn phenolphthalein indicator red. 4M(s) + O2(g) → 2M2O(s), Li2O(s) + 2H2O(l) → 2LiOH(aq) Na2O(s) + 2H2O(l) → 2NaOH(aq) K2O(s) + 2H2O(l) → 2KOH(aq) where M = Group 1 element. 77 Periodic Table of Elements Periodic Table of Elements Periodic Table of Elements 4 Periodic Table of Elements Periodic Table of Elements 78 SPM 4.2 ’05/P2 ’08/P3 ’09/P1 ’10/P3 To study the reaction of alkali metals with water Problem statement How do lithium, sodium and potassium differ in reactivity with water? Figure 4.8 Reaction of potassium with water Hypothesis The alkali metals show similar chemical properties in their reactions with water but the reactivity of alkali metals with water increases down the group (in the order from lithium, sodium to potassium). Results Metal Observation The lithium metal moves slowly on the surface of the water (Figure 4.6). A colourless solution is produced. This solution turns red litmus paper blue. Sodium The sodium metal moves at a fast speed on the surface of the water with a ‘hissing’ sound. It is ignited during the reaction and burns with a yellow flame (Figure 4.7). A colourless solution is obtained. This solution turns red litmus paper blue. Potassium The potassium metal moves at a very fast speed on the surface of the water. It is ignited during the reaction and burns with a purple flame with a ‘pop’ sound (Figure 4.8). A colourless solution is obtained. This solution turns red litmus paper blue. Variables (a) Manipulated variable : The alkali metals used (b) Responding variable : Reactivity of the metal with water (c) Constant variable : Size of alkali metal and the temperature of water Materials Small pieces of lithium, sodium and potassium metals, basin filled with water, filter paper and red litmus paper. Apparatus Pen knife and tongs. Procedure 1 A piece of lithium metal is removed from the bottle. A small piece of the metal is cut using a pen knife. A piece of filter paper is used to absorb the paraffin oil from the piece of metal. 2 The lithium metal is then dropped into a basin of water carefully using a pair of tongs. 3 The observation is recorded. 4 The solution formed in the basin is tested with a piece of red litmus paper. 5 The experiment is repeated using small pieces of sodium and potassium metals. 4 Lithium 2 The metal hydroxides are alkaline and turn the colour of red litmus paper blue. Conclusion 1 Elements of Group 1 have similar chemical properties. All of them react with cold water to produce an alkaline solution and hydrogen gas. 2M(s) + 2H2O(l) → 2MOH(aq) + H2(g), where M = Group 1 element. 2 The reactivity increases down the group from lithium, sodium to potassium. Figure 4.6 Reaction of lithium with water Figure 4.7 Reaction of sodium with water 79 Periodic Table of Elements Experiment 4.2 Discussion 1 Alkali metals are reactive with water and can SPM displace hydrogen from water. They react with ’11/P2 water to produce hydrogen gas and aqueous solutions of metal hydroxides. 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) Reactivity 2Na(s) + 2H2O(l) → increases down 2NaOH(aq) + H2(g) the group 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g) react vigorously with water forming hydrogen gas and alkali solutions. If the size of the metal is quite large, it will ignite and start to burn giving out ‘pop’ sounds. When carrying out an experiment on the reaction of alkali metals with water, we must ensure that no flammable organic solvents are nearby because the fire from the burning alkali metals can spread to the organic solvents. Therefore alkali metals must be kept in paraffin oil in bottles to prevent them from reacting with water. When the Group 1 elements react with water they give off hydrogen gas. The heat generated by the chemical reaction sets the hydrogen gas alight and it burns with a coloured flame. Sodium burns with a yellow flame. Safety Precaution to be Taken When Handling Alkali Metals 4 Alkali metals are extremely reactive. With the exception of lithium, the rest of the alkali metals 4.3 SPM ’09/P1 To study the reaction of alkali metals with halogens Results Problem statement How do lithium, sodium and potassium differ in reactivity with chlorine gas? Metal Hypothesis The alkali metals show similar chemical properties in their reactions with chlorine but the reactivity of alkali metals with chlorine increases down the group (in the order from lithium, sodium and potassium). Variables (a) Manipulated variable : The alkali metals used (b) Responding variable : Reactivity of the metal with chlorine (c) Constant variable : Supply of chlorine gas and size of metal pieces used Lithium burns slowly with a reddish flame. A white solid is obtained. Sodium Sodium burns brightly with a yellowish flame. A white solid is obtained. Potassium Potassium burns very brightly with a purplish flame. A white solid is obtained. 2Li(s) + Cl2(g) → 2LiCl(s) 2Na(s) + Cl2(g) → 2NaCl(s) 2K(s) + Cl2(g) → 2KCl(s) Procedure 1 A small piece of lithium is cut and the paraffin oil on it is blotted using filter paper. 2 The lithium metal is then transferred onto a gas jar spoon using tongs. 3 The lithium metal is heated until it starts to burn. The spoon is then put into a gas jar containing chlorine gas. 4 The observation is recorded. 5 The experiment is repeated using sodium and potassium metals. ⎯⎯⎯→ Apparatus Pen knife, tongs, gas jar with cover, gas jar spoon and Bunsen burner. Reactivity increases down the group 2 If the experiments are repeated using bromine gas, the brown colour of bromine gas will disappear and white metal bromides will be formed. 2Li(s) + Br2(g) → 2LiBr(s) 2Na(s) + Br2(g) → 2NaBr(s) 2K(s) + Br2(g) → 2KBr(s) ⎯⎯→ Experiment 4.3 Lithium Discussion 1 All alkali metals react with chlorine gas to form white metal chlorides The reactivity of the metal increases down the group from Li, Na to K. Materials Small pieces of lithium, sodium and potassium metals chlorine gas and filter paper. Periodic Table of Elements Observation Reactivity increases down the group 3 Similarly, Group 1 elements will react with iodine vapour to produce white metal iodides. The purple colour of iodine vapour will disappear in the reactions. 80 4 Generally, all alkali metals react with halogens to produce metal halides. 2M(s) + X2(g) → 2MX(s) where M = Group 1 elements, X = halogen. 2M(s) + Cl2(g) → 2MCl(s) where M = Group 1 elements 2 The reactivity increases down the group from lithium, sodium to potassium. Conclusion 1 Elements of Group 1 have similar chemical properties. All react with chlorine to produce white metal chlorides. ’05 An atom of element X has three electron filled shells. When element X reacts with chlorine, it forms a compound with the formula XCl. Which of the following is element X? [Proton number of Li, 3; Na, 11; Mg, 12; K, 19] A Lithium C Magnesium B Sodium D Potassium Element Electron arrangement X 2.8.1 Y 2.8.8.1 The atomic radius of Y is larger than X. Hence Y can donate its valence electron more easily than X. Thus Y is more reactive. (I is correct) Both X and Y have one valence electron. Both are alkali metals which can conduct electricity. (II and III correct) Alkali metals react with water to form metal hydroxides and hydrogen gas. (IV is incorrect) Answer A Comment Alkali metals react with chlorine to form metal chloride with formula XCl. X is sodium because sodium has three electron shells. Answer B 4.3 4 ’03 1 Imagine that a new element is discovered. It is named Pentium (Pn) and is below sodium metal in Group 1 of the Periodic Table. (a) Predict three physical and three chemical properties of pentium. (b) Which is more reactive, pentium or sodium? Explain your answer. The table below shows the proton numbers of two elements: Element Proton number X 11 Y 19 2 List the alkali metals in order of decreasing reactivity. 3 How do each of the following properties of alkali metals change as we go down the group? (a) Melting point (b) Density (c) Hardness (d) Chemical reactivity Which of the following statements are true? I Y is more reactive than X. II Both X and Y can conduct electricity. III Both X and Y are in the same group of the Periodic Table. IV Both X and Y react with water to form metal oxides and hydrogen gas. A I, II and III only B I, II and IV only C II, III and IV only D I, II, III and IV 4 When a piece of burning sodium metal is placed in a gas jar containing bromine gas, the brown colour of bromine gas will disappear and white powder will be formed. Explain the above observation with a suitable equation. 81 Periodic Table of Elements 4 Comment 3 4 4.4 SPM Group 17 Elements ’08/P2 General Physical properties • The elements in Group 17 are also known as the halogens. • The elements in Group 17 are fluorine (F), chlorine (Cl), bromine (Br), iodine (I) and astatine (At). • Halogens are very reactive elements and most of them exist naturally as halide salts. • The halogen molecules exist as diatomic molecules: F2, Cl2, Br2, I2 and At2. Physical properties of the halogens • All Group 17 elements are non-metals. Hence, they are non-conductors of heat and electricity. • The atomic radius increases down the group. The reason is that as the number of filled electron shells increases down the group, the distance between the outermost electron shell and the nucleus increases. • The density increases down the group. This is due to the increase in relative molecular mass. • Halogens have low boiling points. The forces of attraction between the molecules are weak. • The melting points and boiling points of the halogens increase down the group. This is because the molecular size increases down the group. As the size increases, the van der Waals forces of attraction between the molecules become stronger. More heat is required to overcome the forces of attraction and therefore the melting points and boiling points increase. The first two elements (fluorine and chlorine) are gases at room temperature. Bromine is a liquid whereas iodine and astatine are solids at room temperature. • The colour of the halogen becomes darker down the group. Fluorine is a colourless gas; chlorine is a yellowish-green gas; bromine is a dark brown liquid and iodine is a black solid. • All halogens have high electronegativities. They are electronegative non-metals. Electronegativity is a measurement of the tendency of an element to attract electrons. The electronegativity decreases down the group from fluorine to iodine. As the atomic radius becomes larger down the group, the force of attraction between the nucleus and the electrons becomes weaker and thus electronegativity decreases. Figure 4.9 The elements of Group 17 in the Periodic Table Halothane (CHClBrCF3) is used as a general anaesthetic during a major operation. Can you name the halogens present in this compound? Answer Bromine, chlorine and fluorine Chemical properties Chemical properties of Group 17 elements • The elements in Group 17 have the same chemical properties because each has seven valence electrons. • Group 17 elements react with – water to produce acids, – metals such as iron to produce metal halides, – sodium hydroxide to produce salts and water. Periodic Table of Elements 82 SPM Reactivity Reactivity of Group 17 elements X2 + 2e– → 2X–, where X = F, Cl, Br or I • All Group 17 elements are very reactive. However, the Cl2 + 2e– → 2Cl– reactivity decreases down the group. 2.8.7 2.8.8 • The elements in Group 17 have seven valence electrons – each. During a chemical reaction, the atom of a Br2 + 2e → 2Br– Group 17 element will accept a valence electron to 2.8.18.7 2.8.18.8 form univalent negative ion to attain the stable octet – I2 + 2e → 2I– in its electron arrangement. 2.8.18.18.7 2.8.18.18.8 • The reactivity of Group 17 elements depends on its ability to gain an electron. The atomic radius of Group 17 elements increases down the group. Thus the force of attraction between the nucleus and the valence electrons become weaker. As a result, the halogen lower in the group has a lower tendency to attract an electron to form a negative ion. Therefore, the reactivity of halogens decreases down the group. Table 4.7 Some physical properties of three halogens 2.8.7 Atomic radius (nm) 0.099 Melting point (°C) –101 Boiling point (°C) –35 Physical state at room temperature Gas 2.8.18.7 2.8.18.18.7 0.114 0.133 –7 114 58 183 Liquid Solid Halogen Proton number Electron arrangement Chlorine 17 Bromine Iodine 35 53 4 ’06/P2, ’08/P2 SPM ’08/P2 Electro negativity 3.0 2.8 2.5 Colour Yellowishgreen gas Brown liquid Black solid 4.4 To study the reactions of chlorine, bromine and iodine with water Materials Chlorine gas, liquid bromine, iodine crystals and blue litmus paper. Problem statement How do chlorine, bromine and iodine react with water? Hypothesis The halogens show similar chemical properties when they react with water but the reactivity decreases down the group from chlorine to iodine. Apparatus Test tube, rubber stopper, test tube holder, delivery tube and teat pipette. Variables (a) Manipulated variable : Types of halogens used (b) Responding variable : The rate at which the halogen dissolves in water and products of reactions (c) Constant variable : Temperature of water Figure 4.10 Reaction of halogen with water (a) (b) (c) 83 Periodic Table of Elements Experiment 4.4 Safety precautions (a) Chlorine gas and bromine vapour are poisonous. The experiments should be carried out in a fume cupboard. (b) The chlorine gas and bromine vapour irritate the eyes. So goggles should be worn while carrying out the experiments. 4 Procedure (A) Reaction of chlorine with water 1 Chlorine gas is passed into a test tube containing water. 2 The solution produced is tested with blue litmus paper. (B) Reaction of bromine with water 1 A few drops of liquid bromine are added to some water in a test tube. Results 2 The test tube is tightly closed with a rubber stopper and then shaken. 3 The solution produced is tested with blue litmus paper. (C) Reaction of iodine with water 1 Some iodine crystals are added to some water in a test tube. 2 The test tube is tightly closed with a rubber stopper and then shaken. 3 The solution produced is tested with blue litmus paper. SPM ’11/P1 Observation Halogen Solubility Effect on litmus paper Chlorine Dissolves quickly in the water to form a light yellowish solution The solution first turns the blue litmus paper red, then it quickly decolourises it. Bromine Dissolves slowly in water to form a brown solution The solution first turns the blue litmus paper red. The red colour of the litmus takes a longer time to be decolourised. Iodine A little of the iodine crystals dissolves slightly in water to form a pale brown solution The solution turns the litmus paper from blue to red. The red litmus paper is not decolourised. Discussion 1 Chlorine, bromine and iodine dissolve in water to form acidic solutions which turn blue litmus paper red. The solubility of halogens decreases from chlorine to iodine. 2 Chlorine dissolves in water to form hydrochloric acid and hypochlorous(I) acid. Hypobromous(I) acid is a weak bleaching agent and takes a longer time to decolourise the red colour of litmus paper. 4 Iodine is only very slightly soluble in water. It forms hydroiodic acid and hypoiodous(I) acid. I2(s) + H2O(l) → HI(aq) + HOI(aq) hydroiodic hypoiodous(I) acid acid Cl2(g) + H2O(l) → HCl(aq) + HOCl(aq) hydrochloric hypochlorous(I) acid acid Hydroiodic acid, HI is an acid and it turns blue litmus red. Hypoiodous(I) acid has a very weak bleaching property. Hydrochloric acid, HCl is an acid and it turns blue litmus red. Hypochlorous(I) acid, HOCl is a strong bleaching agent. It decolourises the red colour of litmus paper quickly. 3 Bromine dissolves slowly in water to form hydrobromic acid and hypobromous(I) acid. Conclusion 1 Chlorine, bromine and iodine show similar chemical properties. They dissolve in water to form acidic solutions. 2 The solubility of halogens in water decreases down the group. 3 Aqueous chlorine and bromine solutions have bleaching properties. Aqueous iodine solution does not bleach the colour of litmus paper. Br2(l) + H2O(l) → HBr(aq) + HOBr(aq) hydrobromic hypobromous(I) acid acid Hydrobromic acid, HBr is an acid and it turns blue litmus red. Periodic Table of Elements 84 4.5 SPM To study the reactions of halogens with aqueous sodium hydroxide solution ’06/P2 Hypothesis The halogens show similar chemical properties when they react with sodium hydroxide solution but the reactivity decreases down the group from chlorine to iodine. Variables (a) Manipulated variable : Types of halogens used (b) Responding variable : The products of the reactions (c) Constant variable : Concentration of sodium hydroxide solution Halogen Observation Chlorine The greenish chlorine gas dissolves quickly in NaOH solution to form a colourless solution. Bromine The brownish liquid bromine dissolves steadily in NaOH solution to form a colourless solution. Iodine The dark iodine crystal dissolves slowly in NaOH solution to form a colourless solution. 4 Results Problem statement How do chlorine, bromine and iodine react with aqueous sodium hydroxide solution? Discussion 1 Chlorine gas reacts rapidly with sodium hydroxide solution to produce sodium chloride salt, sodium chlorate(I) salt and water. Cl2(g) + 2NaOH(aq) → NaOCl(aq) + NaCl(aq) + H2O(l) Materials Chlorine gas, liquid bromine, iodine crystals and sodium hydroxide solution. sodium chlorate(I) Apparatus Test tube, rubber stopper, test tube holder and teat pipette. sodium chloride 2 Bromine reacts moderately fast with sodium hydroxide solution to produce sodium bromide salt, sodium bromate(I) salt and water. Br2(l) + 2NaOH(aq) → NaOBr(aq) + NaBr(aq) + H2O(l) Procedure (A) Reaction of chlorine with aqueous sodium hydroxide solution 1 Chlorine gas is bubbled into aqueous sodium hydroxide solution. 2 The colour change of chlorine is recorded. sodium bromate(I) sodium bromide 3 Solid iodine reacts slowly with sodium hydroxide solution to produce the salts sodium iodide, sodium iodate(I) and water. I2(s) + 2NaOH(aq) → NaOI(aq) + NaI(aq) + H2O(l) (B) Reaction of bromine with aqueous sodium hydroxide solution 1 Two drops of liquid bromine are added to aqueous sodium hydroxide solution using a teat pipette. 2 The test tube is tightly closed with a rubber stopper and the mixture is shaken. 3 The colour change of bromine is recorded. sodium iodate(I) sodium iodide Conclusion 1 Chlorine, bromine and iodine react with sodium hydroxide solution to form two types of salts and water. (C) Reaction of iodine with aqueous sodium hydroxide solution 1 Some iodine crystals are added to aqueous sodium hydroxide solution. 2 The test tube is tightly closed with a rubber stopper and the mixture is shaken. 3 The colour change of iodine crystal is recorded. X2(g) + 2NaOH(aq) → NaX(aq) + NaOX(aq) + H2O(l), where X = Cl, Br, I 2 The reactivity of halogens with sodium hydroxide solution decreases down the group from chlorine to iodine. 85 Periodic Table of Elements Experiment 4.5 [Sodium chlorate(I), sodium bromate(I), sodium iodate(I) are also called sodium hypochlorite, sodium hypobromite and sodium hypoiodite respectively] 4.6 SPM To study the reactions of halogens with iron ’06/P2 Problem statement How do chlorine, bromine and iodine react with iron? (B) Reaction of bromine gas with iron wool 4 Hypothesis The halogens show similar chemical properties when they react with iron but the reactivity decreases down the group from chlorine to iodine. Figure 4.12 Reaction of bromine with iron wool Variables (a) Manipulated variable : Types of halogen used (b) Responding variable : Products of reactions and rate of the reactions (c) Constant variable : Iron wool 1 A small roll of iron wool is placed in the middle of a combustion tube and is heated strongly. 2 The liquid bromine is warmed up by using a Bunsen burner. 3 The bromine is vaporised and bromine gas passed through the heated iron wool. 4 The excess bromine gas is absorbed by the soda lime. Materials Chlorine gas, liquid bromine, iodine crystals, soda lime, potassium manganate(VII), concentrated hydrochloric acid and iron wool. (C) Reaction of iodine with iron wool Apparatus Combustion tubes, Bunsen burner, retort stand and clamp, conical flask and thistle funnel. Procedure (A) Reaction of chlorine gas with iron wool 1 A small roll of iron wool is placed in the middle of a combustion tube. The iron wool is then heated strongly. 2 Chlorine gas is prepared in the laboratory by adding concentrated hydrochloric acid to potassium manganate(VII). 3 The chlorine gas produced is allowed to pass through the heated iron wool. 4 The excess chlorine gas is absorbed by the soda lime. Figure 4.13 Reaction of iodine with iron wool 1 A few crystals of iodine are placed in a boiling tube. 2 A small roll of iron wool is then placed in the middle of a combustion tube. 3 The iron wool is heated strongly first, followed by the iodine crystals (sublimation will take place). 4 The iodine vapour produced is allowed to pass through the hot iron wool. Results Experiment 4.6 Halogen Chlorine Hot iron wool glows brightly when chlorine gas is passed over it. A brown solid is formed. Bromine Hot iron wool glows moderately bright when bromine gas is passed over it. A brown solid is formed. Iodine Hot iron wool glows dimly when iodine vapour is passed over it. A brown solid is formed. Figure 4.11 Reaction of chlorine with iron wool Periodic Table of Elements Observation 86 Discussion 1 The halogens react with hot iron wool to form iron(III) halides which are brown in colour but the reactivity of the halogen decreases down the group. reactivity decreases down the group Conclusion 1 Chlorine, bromine and iodine show the same chemical properties when they react with iron wool, producing brown iron(III) halides. 2 The reactivity of the halogen decreases down the group from chlorine to iodine. 2 The reaction between concentrated hydrochloric acid and potassium manganate(VII) produces chlorine gas: X and Y reacts with water to form two kinds of acids. (III is correct) Safety Precaution to be Taken When Handling Group 17 Elements X2 + H2O → HX + HOX 1 Fluorine, chlorine and bromine gases are very poisonous. In fact chlorine was used in the First World War to kill people. 2 Hence the experiments which involve the use of these gases should be carried out in a fume cupboard. 5 Answer C 4.4 1 Explain why a solution of chlorine is (a) acidic (b) able to bleach things. ’05 2 Aqueous bromine and iodine solutions are both brown. (a) How do you differentiate between the two solutions? (b) Carry out an experiment to show that bromine is more reactive than iodine. X Y Which of the following statements below are true concerning the elements X and Y in the above Periodic Table? I Y is more reactive than X. II X is more electronegative than Y. III They react with water to produce two kinds of acids. IV Both form univalent negative ions of charge –1 when reacted with sodium. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV 3 Iodine is an element below chlorine in Group 17 of the Periodic Table. (a) Does iodine show similar chemical properties as chlorine? Explain your answer. [Proton number of iodine is 53 and proton number of chlorine is 17] (b) How does the (i) density (ii) melting point of iodine compare to chlorine? (c) Write equations for the reaction of iodine with (i) aqueous sodium hydroxide solution (ii) iron wool Comment X has a smaller atomic radius. Thus it has a higher tendency to accept an electron to form a univalent negative ion of charge –1. Thus X is more reactive and more electronegative. (I is incorrect) 4 Explain why the reactivity of Group 17 elements decreases down the group. 5 Name five compounds containing halogens and state their uses. 87 Periodic Table of Elements 4 2Fe(s) + 3Br2(g) → 2FeBr3(s) iron(III) bromide 2Fe(s) + 3I2(g) → 2FeI3(s) iron(III) iodide 3 Soda lime is a mixture of calcium hydroxide and sodium hydroxide. It is used to absorb the excess halogen gas. The excess chlorine and bromine gas have to be absorbed because they are poisonous. ⎯⎯⎯⎯⎯⎯⎯→ 2Fe(s) + 3Cl2(g) → 2FeCl3(s) iron(III) chloride 2KMnO4(s) + 16HCl(aq) → 2KCl(aq) + 2MnCl2(aq) + 5Cl2(g) + 8H2O(l) 4 Modern Periodic Table (a) Elements are arranged in order of increasing proton numbers. (b) The vertical column is known as Group whereas the horizontal rows are called Periods. (c) The number of valence electrons in the element corresponds to the group the element is in. (d) The number of filled electron shells of an element corresponds to the period of the element. Group 18 elements (a) Group 18 elements are inert. (b) They have attained the duplet or octet electronic configuration. (c) Thus they do not need to share, donate or receive electrons from other elements. Group 1 elements (Alkali metals) (a) As we go down Group 1, (i) melting point decreases, (ii) density increases, (iii) reactivity and electropositivity increases. (b) Chemical properties of Group 1 elements: (i) Alkali metals react with chlorine to form metal halide salts. 2M(s) + Cl2(g) → 2MCl(s) (M = Li, Na, K, …) (ii) Alkali metals react with water to form metal hydroxides and hydrogen gas. 2M(s) + 2H2O(l) → 2MOH(aq) + H2(g) (M = Li, Na, K, …) (iii) Alkali metals burn in air to form metal oxides which are soluble in water. 4M(s) + O2(g) → 2M2O(s) (M = Li, Na, K, …) M2O(s) + H2O(l) → 2MOH(aq) Group 17 elements (Halogen) (a) As we go down group 17, (i) melting point and density increases, (ii) reactivity and electronegativity decreases. (b) Chemical properties of Group 17 elements: (i) Halogens dissolve in water to form two types of acids. X2(g) + H2O(l) → HX(aq) + HOX(aq) (X = F, Cl, Br, …) (ii) Halogens react with iron to form brown iron(III) halide salts. 2Fe(s) + 3X2(g) → 2FeX3(s) (X = F, Cl, Br, …) (iii) Halogens react with sodium hydroxide to form two types of salts and water. 2NaOH(aq) + X2(g) → NaX(aq) + NaOX(aq) + H2O(l) Periodic Table of Elements 88 (X = F, Cl, Br, …) period, other than Period 3, based on the changes in the properties of elements in Period 3. 4 Table 4.8 shows the trends across Period 3. Elements in a Period Elements in Period 3 1 The elements in Period 3 are sodium (Na), magnesium (Mg), aluminium (Al), silicon (Si), phosphorus (P), sulphur (S), chlorine (Cl), and argon (Ar). 2 The study of the elements in Period 3 will show a gradual change of physical and chemical properties across the period from left to right. 3 Period 3 is a typical period. Thus we can predict the trend of changes in properties across a Figure 4.14 The elements in Period 3 of the Periodic Table SPM Table 4.8 The trends across Period 3 Group ’11/P1 1 2 13 14 15 16 17 18 Element Na Mg Al Si P S Cl Ar Proton number 11 12 13 14 15 16 17 18 2.8.1 2.8.2 2.8.3 2.8.4 2.8.5 2.8.6 2.8.7 2.8.8 1 2 3 4 5 6 7 8 0.156 0.136 0.125 0.117 0.111 0.104 0.099 0.094 Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3.0 — Melting point (°C) 98 649 660 1410 590 119 –101 –189 Boiling point (°C) 883 1107 2467 2355 Ignites 445 –35 –186 Nature of elements Metal Metal Metal Formula of oxide Na2O MgO Al2O3 SiO2 P4O10 SO2 Cl2O7 None Character of oxide Basic Basic Amphoteric Acidic Acidic Acidic Acidic — Electron arrangement Number of valence electrons Atomic radius (nm) Metalloid Non-metal Non-metal Non-metal Non-metal Trends of Changes across Period 3 SPM ’05/P2 ’09/P2 Atomic Radius Valence Electrons The atomic radius decreases across the period. • All the elements in Period 3 have three filled electron shells but the proton number increases by one unit across the period. • As a result, the increase in the number of protons increases the electrostatic force between the nucleus and the valence electrons. • The valence electrons are pulled closer to the nucleus, causing the atomic radius to decrease. The number of valence electrons increases across the period. • As the proton number increases, the number of electrons increases. • The number of valence electrons increases by 1 from one element to the next across the period. 89 Periodic Table of Elements 4 4.5 4 Electronegativity Melting & Boiling Points The electronegativity increases across the period. Electronegativity is a measurement of the tendency of an atom to attract electrons. • The atomic radius decreases across the period. • The proton number increases across the period. • The increase in the number of protons (positive charge in the nucleus) and the decrease in the distance between the nucleus and the outermost electron shell across the period cause an increase in the force of attraction of the nucleus. The atoms will have a higher tendency to attract electrons. Therefore electronegativity increases. • Elements on the left side of the period tend to lose electrons to form positive ions. Elements on the right side of the period tend to gain electrons to form negative ions. • Thus the elements on the left side of the period (such as Na) are electropositive while the elements on the right side of the period (such as Cl) are electronegative. The melting points and boiling points of the elements increase from the left of the period to the middle of the period and then decrease again. • Sodium, magnesium and aluminium are metals with strong metallic bonds between the metal atoms. Hence they have high melting and boiling points. The strength of the metallic bonds increase with the increase in the number of valence electrons in the order: Na < Mg < Al. • Silicon has very high melting and boiling points. It has strong covalent bonds between atoms forming a 3-dimensional gigantic network. • Phosphorus, sulphur, chlorine and argon are non-metals with weak van der Waals forces of attraction between molecules. They are lowest at the right with chlorine and argon existing as gases at room temperature. Nature of Metals As the electronegativity of the elements increases, the elements change from metals to metalloid and finally to non-metals across the period. • The elements on the left of the period are metals (Na, Mg and Al). • Silicon has some metallic and some non-metallic properties. It is called a metalloid or semi-metal. • The elements on the right of the period are non-metals (P, S, Cl and Ar). Nature of Oxides SPM ’08/P1, ’10/P1 The oxides of the elements change from basic to • Acidic oxides react with alkali to form salts and amphoteric and then to acidic across the period. water. • Elements on the left of the period, which are For example, sulphur trioxide reacts with metals, form metal oxides. The metal oxides sodium hydroxide to form sodium sulphate are usually basic oxides. salt and water: • Basic oxides react with acids to form salts and water. SO3(g) + 2NaOH(aq) → Na2SO4(aq) + H2O(l) For example, magnesium oxide reacts with sulphuric acid to form a salt and water: • An amphoteric oxide can react with both acids and bases to form salt and water. Aluminium MgO(s) + H2SO4(aq) → MgSO4(aq) + H2O(l) oxide is an example of an amphoteric oxide. Aluminium oxide can react with both acids and • Elements on the right of the period are nonalkalis to form salts and water. metals. Non-metallic oxides are acidic oxides. • Argon as an inert gas, does not form oxide. Periodic Table of Elements 90 4.7 To determine the properties of the oxides of elements in Period 3 Problem statement How do the properties of the oxides of elements in Period 3 change across the period? Hypothesis The oxides change from basic to amphoteric and then to acidic across Period 3. 4 Variables (a) Manipulated variable : Oxides of Period 3 (b) Responding variable : Reaction with acid or alkali (c) Constant variable : Concentrations of sodium hydroxide and nitric acid solutions Figure 4.15 Reaction of oxides of Period 3 with (a) an acid, (b) an alkali 3 Two drops of universal indicator are added and the pH of the solution is noted. 4 The experiment is repeated with magnesium oxide, aluminium oxide, silicon(IV) oxide, phosphorus(V) oxide, sulphur dioxide and dichlorine heptoxide respectively in place of sodium oxide. (B) Reaction of the oxides of Period 3 elements with 2 mol dm–3 nitric acid and 2 mol dm–3 sodium hydroxide solutions 1 A little sodium oxide powder is put into two separate test tubes. 2 5 cm3 of nitric acid and 5 cm3 of sodium hydroxide are added separately to the contents in each test tube. 3 The contents in each test tube are heated slowly while being stirred with glass rods. 4 The solubility of sodium oxide in the two solutions is recorded. 5 The experiment is repeated with magnesium oxide, aluminium oxide, silicon(IV) oxide, phosphorus(V) oxide, sulphur dioxide and dichlorine heptoxide respectively in place of sodium oxide. Materials Sodium oxide, magnesium oxide, aluminium oxide, silicon(IV) oxide, phosphorus(V) oxide, sulphur dioxide, dichlorine heptoxide, nitric acid solution and sodium hydroxide solution of 2 mol dm–3, distilled water and universal indicator. Apparatus Test tube, test tube holder, Bunsen burner, rubber stopper and glass rod. Procedure (A) Reaction of the oxides of Period 3 elements with water 1 A little sodium oxide powder is added to some distilled water in a test tube. 2 The test tube is tightly closed with a rubber stopper and the contents are shaken. Results Experiment A Observation Solubility in water pH value of solution 13 – 14 Al2O3 Dissolves in water to form a colourless solution Slightly soluble in water to form a colourless solution Insoluble in water SiO2 Insoluble in water Na2O MgO Inference Solution obtained is a strong alkali. Sodium oxide is basic. Sodium shows metallic properties. Solution obtained is a weak alkali. Magnesium oxide is basic. Magnesium shows metallic properties. pH measured is pH of water as the oxide is insoluble in water. pH measured is pH of water as the oxide is insoluble in water. 8–9 7 7 91 Periodic Table of Elements Experiment 4.7 Oxide Oxide P4O10 SO2 Cl2O7 Observation Solubility in water pH value of solution Dissolves in water to form a colourless solution Dissolves in water to form a colourless solution Dissolves in water to form a colourless solution 2–3 Inference Solution obtained is acidic. P4O10 is acidic. Phosphorus shows non-metallic properties. Solution obtained is acidic. SO2 is acidic. Sulphur shows non-metallic properties. Solution obtained is a strong acid. Cl2O7 is acidic. Chlorine shows non-metallic properties. 2–3 1 4 Experiment B Oxide Reaction with 2 mol dm–3 NaOH Reaction with 2 mol dm–3 HNO3 Inference Na2O Does not react Reacts with nitric acid to form a colourless solution Sodium oxide is basic because it reacts with the acid. Sodium shows metallic properties. MgO Magnesium oxide does not dissolve Dissolves in nitric acid forming a colourless solution Magnesium oxide is basic because it reacts with the acid. Magnesium shows metallic properties. Al2O3 Dissolves in sodium Dissolves in nitric acid hydroxide forming a forming a colourless solution colourless solution SiO2 Reacts with sodium hydroxide solution Does not dissolve in nitric acid Silicon(IV) oxide is acidic because it reacts with an alkali. Silicon shows non-metallic properties. P4O10 Reacts with sodium hydroxide solution Does not react with nitric acid Phosphorus(V) oxide is acidic because it reacts with an alkali. Phosphorus shows non-metallic properties. SO2 Reacts with sodium hydroxide solution Does not react with nitric acid Sulphur dioxide is acidic because it reacts with an alkali. Sulphur shows non-metallic properties. Cl2O7 Reacts with sodium hydroxide solution Does not react with nitric acid Dichlorine heptoxide is acidic because it reacts with an alkali. Chlorine shows non-metallic properties. SiO2(s) + 2NaOH(aq) → Na2SiO3(aq) + H2O(l) Discussion 1 (a) Metallic oxides are basic. Metal oxides react with acids to form salts and water. P4O10(s) + 12NaOH(aq) → 4Na3PO4(aq) + 6H2O(l) SO2(g) + 2NaOH(aq) → Na2SO3(aq) + H2O(l) Na2O(s) + 2HNO3(aq) → 2NaNO3(aq) + H2O(l) Conclusion 1 Across Period 3 from left to right, (a) the element changes from being a metal, to a metalloid and a non-metal. (b) the oxides change from being basic to amphoteric and acidic. 2 Silicon is classified as a metalloid because it is a very weak conductor of electricity. However, its oxide is acidic. 3 Aluminium is classified as a metal because it is a very good conductor of electricity. However, its oxide is amphoteric. Aluminium oxide shows both metallic and non-metallic properties. MgO(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2O(l) (b) Aluminium oxide is amphoteric because it can react with both acids and alkalis. SPM ’05/P2 Al2O3(s) + 6HNO3(aq) → 2Al(NO3)3(aq) + 3H2O(l) Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4(aq) (sodium aluminate) Other examples of amphoteric oxides are lead(II) oxide and tin(II) oxide. (c) Non-metallic oxides are acidic. Non-metallic oxides react with alkalis to form salts and water. Periodic Table of Elements Aluminium oxide is amphoteric because it reacts with both acid and alkali. 92 Uses of Semi-metals (or Metalloids) 1 A semi-metal or metalloid is an element with properties intermediate between those of metals and non-metals. 2 For example, silicon is a non-metal and is a very poor conductor of electricity. However, the conductivity increases with temperature. It becomes a good conductor of electricity at high temperatures. 3 This type of substance is known as a semi-metal. Examples of semi-metals are silicon, germanium, boron, antimony, and arsenic. These elements are important industrial materials and are used to make semiconductors. 4 Adding of foreign elements (called doping) can increase the conductivity of semi-metals. (a) If silicon is doped with Group 13 elements such as boron, a p-type semiconductor is produced. (b) If it is doped with Group 15 elements such as arsenic or antimony, a n-type semiconductor is produced. 5 Semiconductors are very important in the microelectronic industry and are used to make transistors, diodes, rectifiers, thermistors and microprocessors. Hundreds of these electronic components can be built onto a crystal of silicon to make a microchip. Microchip wafer 4 Do you know that a silicon wafer can contain hundreds of microchips? Each microchip itself contains hundreds of electronic components. Our earth contains about 27.7% silicon and a large portion is found in sand. 4.6 Transition Elements Figure 4.16 The transition elements in the Periodic Table 1 The transition elements are elements in a block located in-between Group 2 and Group 13 in the Periodic Table. 2 There are 10 elements in each series. The first series is in Period 4 and consists of the elements: scandium(Sc), titanium(Ti), vanadium(V), chromium(Cr), manganese (Mn), iron(Fe), cobalt(Co), nickel(Ni), copper(Cu) and zinc(Zn). 3 Table 4.9 shows some physical properties of the transition elements in the first series in Period 4. 4 The transition elements are all metals with the following physical properties: (a) High density (b) High hardness (c) High electrical conductivity (d) High tensile strength (e) Silvery surface (f) Ductile and malleable (g) High melting point (h) High boiling point 5 The atomic radius and electronegativity of the transition elements are almost the same. 4.5 1 Predict the changes in the properties of the elements across Period 3 in the Periodic Table. (a) Atomic radius (b) Electronegativity (c) Acidic-base property of the oxides 2 The proton numbers of element X and Y are 3 and 9 respectively. (a) To which period of the Periodic Table do X and Y belong? (b) Which is more electronegative? Explain your answer. 3 Write the formula of the oxides of the elements in Period 3. State whether the oxide is basic, acidic or amphoteric. 4 Silicon is a semi-metal. State one difference between (a) silicon and sulphur, (b) silicon and iron. 93 Periodic Table of Elements Table 4.9 Some physical properties of the transition elements in the first series 4 Element Sc Ti V Cr Mn Fe Co Ni Cu Zn Atomic radius (nm) 0.162 0.147 0.134 0.130 0.135 0.126 0.125 0.124 0.128 0.138 Melting point (°C) 1539 1668 1900 1875 1245 1536 1495 1453 1083 419 Boiling point (°C) 2730 3260 3450 2665 2150 3000 2900 2730 2959 906 Density (g cm–3) 3.0 4.51 6.1 7.19 7.43 7.86 8.9 8.9 8.96 7.14 Electronegativity 1.3 1.5 1.6 1.6 1.5 1.8 1.8 1.8 1.9 1.6 Hardness (Mohs’ scale) soft 6.0 7.0 8.5 6.0 4.0 5.0 4.0 3.0 2.5 (Note: Hardness is measured using Mohs’ scale. Diamond, which is the hardest substance known, has a hardness of 10 on Mohs’ scale.) Titanium alloy is corrosion resistant and lightweight. It is used in orthopaedic and dental implants. Vanadium steel alloy is used in making gears and crankshafts of vehicles. Titanium alloy is light and have very high tensile strength. It is used to make aircraft engines. An engine of a A380 Airbus uses 11 tons of titanium. Uses of transition metals Cobalt is alloyed with iron, nickel and other metals to make Alnico, an alloy of unusual magnetic strength. Nickel is used in many industrial and consumer products, including stainless steel, magnets, coinage and special alloys. Chromium is used to make stainless steel and for the electroplating of iron. Periodic Table of Elements Iron is alloyed with carbon to make steel which is used in making cars, ships and in building industries. 94 Copper is used as electrical conductors and in piping. Special Properties of Transition Elements Transition elements have variable oxidation numbers 1 Unlike elements in the main group of the Periodic Table in which each has only one oxidation number, a transition element has more than one oxidation number in its compounds. 2 Oxidation number is the charge on the ion. In other words, a transition element can form ions with different charges. For example, magnesium (in Group 2), can form only Mg2+ ion, with an oxidation number of +2. Iron, however, as a transition element, can form Fe2+ ion (oxidation number of +2) and Fe3+ ion (oxidation number of +3). 3 Table 4.10 shows the different oxidation numbers of transition elements in their compounds. Table 4.10 The oxidation numbers of transition elements in their compounds Chromium(III) chloride Potassium dichromate(VI) CrCl3 K2Cr2O7 +3 +6 Manganese(II) chloride Manganese(IV) oxide Potassium manganate(VI) Potassium manganate(VII) MnCl2 MnO2 K2MnO4 KMnO4 +2 +4 +6 +7 Iron(II) sulphate Iron(III) chloride FeSO4 FeCl3 +2 +3 Nickel(II) sulphate Nickel(III) chloride NiSO4 NiCl3 +2 +3 Copper(I) oxide Copper(II) oxide Cu2O CuO +1 +2 Transition elements form coloured compounds 1 Unlike main group metal compounds which are usually white, transition elements can form compounds of different colours. 2 Unlike aqueous solutions of main group compounds or ions which are usually colourless, aqueous solutions of transition element compounds or their ions are coloured. 3 Table 4.11 shows the colours of some aqueous solutions of ions of transition elements. Formula of ion of transition element Colour of aqueous solution Manganese(II) ion Mn2+ Pink Chromium(III) ion 3+ Cr Green Nickel(II) ion Ni2+ Green 4 Precious stones are coloured due to the presence of compounds of transition elements. Table 4.12 shows the transition elements present which are responsible for the colours of some precious stones. Table 4.11 The colours of some aqueous solutions of ions of transition elements Name of ion of transition element Name of ion of transition element Formula of ion of transition element Colour of aqueous solution Chromate ion CrO Yellow Dichromate ion Cr2O Orange Permanganate ion MnO4– Purple Precious stone Colour Transition element present Iron(II) ion Fe2+ Green Ruby Red Chromium Iron(III) ion Fe Brown Sapphire Blue Iron and Titanium Copper(II) ion Cu2+ Blue Emerald Green Chromium Cobalt(II) ion Co2+ Pink Amethyst Purple Manganese 2– 4 2– 7 3+ Table 4.12 Examples of some precious stones and the transition elements which give them their distinctive colours 95 Periodic Table of Elements 4 Compounds of transition Chemical Oxidation elements formula number Transition metals or their compounds have catalytic properties 1 A catalyst is a substance that speeds up the rate of a reaction. A catalyst does not change chemically after a reaction. Catalysts are used in almost all chemical manufacturing plants. 2 Many catalysts are transition elements or their compounds. SPM ’08/P1/P2 3 Table 4.13 shows the uses of transition elements or their compounds as catalysts in industries. Table 4.13 The uses of transition elements or their compounds as catalysts in industries 4 Transition element or its compound Fine iron powder, Fe Industrial process which use the catalyst Haber process in the manufacture of ammonia. Iron catalyses the reaction between nitrogen and hydrogen gas to produce ammonia. Fe N2(g) + 3H2(g) ⎯⎯→ 2NH3(g) Vanadium(V) oxide, V2O5 SPM ’10/P1 Contact process in the manufacture of sulphuric acid. V2O5 catalyses the oxidation of sulphur dioxide to sulphur trioxide. V2O5 2SO2(g) + O2(g) ⎯⎯→ 2SO3(g) SO3 is used to manufacture sulphuric acid. Nickel Manufacture of margarine. Nickel catalyses the hydrogenation of unsaturated vegetable oil into saturated oil in the production of margarine. Platinum Ostwald process in the manufacture of nitric acid. Transition elements can form complex ions 1 A complex ion is a polyatomic cation or anion consisting of a central metal ion with other groups bonded to it. 2 Table 4.14 shows some examples of complex ions formed by transition elements. Table 4.14 Complex ions Reaction of Aqueous Solutions of Transition Element Compounds with Sodium Hydroxide and Ammonia Solutions Tetraamminecopper(II) Cu(NH3)2+ 4 Hexaamminechromium(III) Cr(NH3)3+ 6 Hexaaquocobalt(II) Co(H2O)2+ 6 Hexacyanoferrate(II) Fe(CN)64– Hexacyanoferrate(III) Fe(CN)63– 2 The ions of transition elements react with the hydroxide ions to form coloured metal hydroxide precipitates. 3 Table 4.15 gives some examples of the reactions of aqueous solutions of ions of transition elements with sodium hydroxide solution and ammonia solution. 1 The presence of ions of transition elements in a solution can be confirmed by using sodium hydroxide solution or ammonia solution. Periodic Table of Elements Formula 96 Table 4.15 Aqueous sodium hydroxide solution, NaOH(aq) Fe2+ Green precipitate of iron(II) hydroxide is formed. Precipitate is insoluble in excess NaOH solution. Green precipitate of iron(II) hydroxide is formed. Precipitate is insoluble in excess aqueous NH3 solution. Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s) (from NaOH) (green precipitate) Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s) (from NH3) (green precipitate) Brown precipitate of iron(III) hydroxide is formed. Precipitate is insoluble in excess NaOH solution. Brown precipitate of iron(III) hydroxide is formed. Precipitate is insoluble in excess aqueous NH3 solution. Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s) (from NaOH) (brown precipitate) Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s) (from NH3) (brown precipitate) Blue precipitate of copper(II) hydroxide is formed. Precipitate is insoluble in excess NaOH solution. Blue precipitate of copper(II) hydroxide is formed. Fe3+ Cu2+ Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s) (from NH3) (blue precipitate) Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s) (from NaOH) (blue precipitate) 6 Aqueous ammonia solution, NH3(aq) 4 Ions Precipitate is soluble in excess NH3 solution to form a complex ion which is dark blue in colour. Cu(OH)2(s) + 4NH3(aq) → Cu(NH3)2+ (aq) + 2OH–(aq) 4 (dark blue solution) Cu(NH3)2+ is a dark blue complex ion. Thus, Q is 4 copper(II) oxide. Answer A ’03 Q is a compound. Study the flowchart below and identify Q. Solid Q + HNO3(aq) +NaOH(aq) Blue solution is formed + NaOH(aq) Blue precipitate + excess ammonia solution A Copper(II) oxide B Iron(III) oxide Aqueous ammonia contains hydroxide ions: NH3 + H2O NH4+ + OH– Not soluble Dark blue solution C Copper(II) sulphate D Iron(II) sulphate Have high melting, boiling points and densities Comment Q is a base, since it dissolves in nitric acid. Metal oxides are bases. Since it forms a blue precipitate, Q must contain copper(II) ions. The above reactions can be represented by the equations CuO(s) + 2HNO3(aq) → Cu(NO3)2(aq) + H2O(l) Cu(NO3)2(aq) + 2NaOH(aq) → 2NaNO3(aq) + Cu(OH)2(s) (blue precipitate) Cu(OH)2(s) + 4NH3(aq) → Cu(NH3)2+ (aq) + 2OH–(aq) 4 Form coloured compounds Have more than one oxidation state Transition elements The presence of their ions can be confirmed using NaOH(aq) or NH3(aq) 97 Used as catalysts in chemical reactions Form complex ions Periodic Table of Elements in a table for easy studying. Scientists who played prominent roles in the development of the Periodic Table were J. W. Dobereiner, John Newlands, Lothar Meyer, Dmitri Mendeleev and Henry Moseley. 4.6 1 You are given two blue aqueous solutions; one containing Cu2+ ions and another containing a blue food colouring. Explain how you can differentiate between the two solutions. 2 The body of U-2 spy plane is made of titanium alloy. Give three reasons why titanium alloy is used? 4 4.7 Uses of the Elements and Compounds in Our Daily Life 1 An element is a substance that cannot be broken down into simpler substances. Table 4.17 shows the uses of some of these elements in daily life. 2 A compound is a substance made by chemically combining atoms of two or more elements. 3 Many compounds have been synthesised by chemists to improve our standard of living. Many items in our home are made up of elements. The shirt you are wearing may be made of Terylene, your sofa may be made of polyvinyl chloride, the cooking utensils in your kitchen are made of stainless steel, glass or ceramics, part of the motor of your ceiling fan is made of copper and the microprocessors in the computer are made of silicon and germanium. Elements are used to make vehicles, medicine and communication tools like the handphone. Imagine our life without these elements. Coloured compounds of transition elements are used as paint pigments and they make our homes more colourful. Table 4.18 shows the uses of some of these compounds. Appreciating the Existence of Elements and Their Compounds 1 In ancient times, gold, mercury, copper, iron, sulphur, tin, antimony, lead, diamond, and graphite were already discovered. Credit should be given to the scientists who, through their persistent efforts, discovered and isolated other elements. Table 4.16 below shows the scientists who isolated some of these elements. 2 Then about a century (100 years) later, scientists discovered the subatomic particles of atoms. Credit is given to J. J. Thomson, Ernest Rutherford, James Chadwick and Niels Bohr in helping us understand the atomic structure of atoms in elements. The atomic structure of atoms helps us understand how elements react to form compounds. 3 As more elements are isolated, there is a need to classify the elements. Elements with the same chemical properties are grouped together Table 4.16 Element Scientist Symbol Isolated in Hydrogen H Henry Cavendish (British) 1766 Nitrogen N Daniel Rutherford (British) 1772 Oxygen O Carl Wilhelm Scheele (Swedish) 1774 Sodium Na Sir William Ramsay (British) 1807 Potassium K Sir Humphry Davy (British) 1807 Magnesium Mg Sir Humphry Davy (British) 1808 Aluminium Al Hans Christian Orsted 1825 Uranium U Eugene Melchior Peligot (French) 1841 Helium He Sir William Ramsay (British) 1895 Radium Ra Pierre Curie and Marie Curie (French) 1898 Periodic Table of Elements 98 Table 4.17 Uses of some elements Formula Uses Hydrogen H2 Used in the hydrogenation of unsaturated oil to produce margarine. Used in the manufacture of ammonia. Aluminium Al Used in the production of duralumin alloy for use in the construction of aeroplane bodies. Silicon Si To make microchips Sulphur S To make matches, fireworks and for the manufacture of sulphuric acid Chlorine Cl2 To kill germs in drinking water Iron Fe Production of steel Copper Cu To make electrical wire, electrical motor, dynamo and coins (cupronickel alloy) Cobalt-60 Co The gamma rays emitted from this radioactive element is used to kill cancer cells. Table 4.18 Uses of some chemical compounds Chemical Formula Uses Magnesium oxide MgO In antacid drugs to treat gastric patients. Also used to treat acid poisonings. Tin fluoride SnF2 Is added to toothpaste. Fluoride ions can strengthen the teeth. Compound Sodium bicarbonate NaHCO3 Silver bromide AgBr Vinyl chloride CH2CHCl Ammonia Sodium hydroxide NH3 NaOH Used as a baking powder. Also used to treat acid burns. Used in the making of photographic films. To make PVC pipes, toys, raincoats and cushions To manufacture fertiliser, nitric acid and explosives To make soap Preventing Wastage pollution because many of these items are non-biodegradable. 3 We can minimise wastage in the school laboratory by practising the following: (a) Weigh or use the correct amount of chemicals that is required to carry out an experiment. (b) Read the label of the chemicals carefully, so that we do not take the wrong substance. (c) Read the procedure of the experiment and plan the experiment carefully before carrying them out, so as to avoid the need to repeat the experiment. 1 Most of the elements needed to keep life going are obtained from the Earth’s crust. A number of these elements are obtained from the oceans and the air. After extraction, these chemicals cannot be replaced and will be depleted one day. So it is important that we avoid wastage in using these elements. 2 Recycling of used materials is an alternative method. Used glass bottles, plastic bottles and aluminium cans should be separated and thrown into different bins provided by the government. This will also help to reduce 99 Periodic Table of Elements 4 Element 4 Cl2(g) + H2O(l) → HCl(aq) + HOCl(aq) (b) Halogens react with alkali metals to form halide salts, e.g. 2Na(s) + Cl2(g) → 2NaCl(s) (c) Halogens react with hot iron wool to form brown iron(III) halide salt, e.g. 2Fe(s) + 3Cl2(g) → 2FeCl3(s) 5 The reactivity of Group 17 element increases up the group because the atomic radius decreases. The smaller the atomic radius, the stronger the electrostatic force of attraction between the nucleus and the electron. Thus the elements higher in the group can accept electrons more easily. 6 Group 18 elements exist as monatoms because they have attained the duplet or octet electronic configuration. They do not need to donate, accept or share electrons with other elements. 7 Transition elements are a block of elements between Group 2 and Group 13 in the Periodic Table. The characteristics of transition elements are (a) they form coloured compounds (b) they have more than one oxidation number (c) they catalyse chemical reactions (d) they form complex ions 1 Chemical elements are classed into groups in the Periodic Table, with elements in the same group having the same chemical properties. 2 Group 1 elements are called alkali metals. (a) They react with cold water to produce an alkaline solution and hydrogen gas, e.g. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) (b) They react with halogen gas to form halide salts, e.g. 2Na(s) + Cl2(g) → 2NaCl(s) (c) They burn in air to form metal oxides, e.g. 4Na(s) + O2(l) → 2Na2O(s) All oxides of Group 1 metals can dissolve in water to form alkaline solutions. Na2O(s) + H2O(l) → 2NaOH(aq) 3 The reactivity of Group 1 elements increases down the group because the valence electron is further from the nucleus. The electrostatic force of attraction between the nucleus and the valence electron becomes weaker. Thus the elements lower in the group can release their valence electrons more easily. 4 Group 17 elements are called halogens. (a) Halogens dissolve in water to form acidic solutions, e.g. 4 Multiple-choice Questions 4.1 Periodic Table of Elements 1 The diagram below shows the electron arrangement of atom Y. 2 The structure of a compound containing element Q and hydrogen is shown as follows. H H x xx x x xx Y x x x x xx Which of the following is the position of atom Y in the Periodic Table? A B C D Group Period TC 52 2 3 3 3 2 13 3 13 Periodic Table of Elements xo H ox Q xo x x xo Q xo x oH H H To which group of the Periodic Table does element Q belong? A 14 C 16 B 15 D 17 3 Which of the following elements belong to the same group in the Periodic Table? 7 3 P 9 4 Q 27 13 A P and R B R and T R 35 17 S 40 20 C P and S D Q and T 100 T 4 Atom of element X has proton number of 19. Which of the following statements are true concerning X ? I X has one valence electron. II The oxide of element X is soluble in water. III X is in Period 4 of the Periodic Table. IV X has the same chemical properties as fluorine. A I and III only B II and IV only C I, II and III only D I, III and IV only 5 An atom of element Y has a nucleon number of 31 with 16 neutrons. 4.2 7 The symbols for four elements are shown below. 4 2 Period A 16 2 B 17 2 C 16 3 D 17 3 W 16 8 24 12 X Y 40 18 Z Which of the following statements is true? A Both elements W and Z are monatomic. B Element Z is more reactive than element X. C Elements W and Y react to form a compound with the formula YW. D Elements X and Y react to form a compound with the formula YX2. 6 An element X has a proton number of 9. In which group and period does X belong to in the Periodic Table? Group Group 18 Elements 8 Which of the inert gases below is used in a diver’s oxygen tank? A Neon C Helium B Argon D Krypton 9 Which element, as represented by the symbols below, exists as monatoms? A W B X C Y D Z 10 The element which does not form compound with other elements is likely to have a proton number of ’06 A 4 B 6 C 8 D 10 11 As I II A B we go down Group 18, the density increases. the reactivity decreases. I and III only II and IV only III IV C D the boiling point increases. the atomic size decreases. I, II and III only I, III and IV only 12 Argon does not form compound with chlorine because A it has three filled electron shells. B it has low melting and boiling points. C it has eight electrons in the outermost electron shell. D the atoms have the same number of protons and neutrons in the nucleus. 4.3 Group 1 Elements 13 Rubidium is below sodium in Group 1. Which statements below about rubidium are correct? ’08 I It is more reactive than sodium. II It burns in air to form rubidium oxide which is insoluble in water. III It is produced from the electrolysis of molten rubidium chloride. IV It reacts with cold water to form rubidium hydroxide and hydrogen. A I and III only C I, III and IV only B II and IV only D I, II, III and IV 101 14 Which of the following explains why sodium is more reactive ’06 than lithium? A Sodium has more protons than lithium. B Sodium has less valence electrons than lithium. C Sodium has a lower melting point than lithium. D Sodium can release its valence electron more easily. 15 An element X is burned in air. The product of combustion is then dissolved in water. The solution gives a pH value of 14. The element X is A potassium B phosphorus C aluminium D sulphur 16 Which statement below is true about sodium? A It burns in air to form sodium oxide of formula NaO which is soluble in water. B It burns in air to form sodium oxide of formula Na2O which is soluble in water. C It burns in air to form sodium oxide of formula NaO which is insoluble in water. D It burns in air to form sodium oxide of formula Na2O which is insoluble in water. 17 Which statement concerning the ions of Group 1 elements is correct? A Each contains more protons than electrons. B Each contains more electrons than protons. C Each has one electron in its outer electron shell. D Each contains the same number of protons and electrons. 18 As I II III IV A B C D we go down Group 1, the density increases the melting point increases the reactivity increases the electropositivity decreases I and II only I and III only II and III only II and IV only Periodic Table of Elements 4 Which of the following statements is true concerning element Y? A Atom of element Y has 16 protons. B Atom of element Y has two filled electron shells. C Atom of element Y has six valence electrons. D Element Y is in Group 15 of the Periodic Table. 4 19 The element X has a proton number of 19. Element X A forms a basic oxide that is insoluble in water. B reacts with cold water to form an acidic solution. C reacts with chlorine to form a compound with formula XCl2. D is more reactive than an element with a proton number of 11. 4.4 Group 17 Elements 20 Which of the following explains why chlorine is more reactive than bromine? A Chlorine is less dense than bromine. B Chlorine has a lower boiling point than bromine. C Chlorine atom accepts an electron more easily than bromine atom. D Chlorine atom contains less protons than bromine atom. 21 Which of the following halogens dissolve in water to produce an acidic solution which can decolourise the colour of litmus paper? I Chlorine II Bromine III Iodine IV Astatine A I and II only B I and III only C II and IV only D III and IV only 22 An element from Group X can dissolve in water to form an acidic solution. This resulting solution reacts with silver nitrate reagent to form a white precipitate. In which group of the Periodic Table does the element belong to? A 1 B 2 C 16 D 17 Periodic Table of Elements 23 Which of the statements below is not true concerning Group 17 elements? A Bromine is a gas at room temperature. B Chlorine is more reactive than bromine. C As we go down Group 17, the density increases. D Chlorine is more electronegative than bromine. 24 As I II III IV A B C D we go down Group 17, the reactivity decreases electronegativity increases the melting point increases the solubility of the element in water decreases I, II and III only I, III and IV only II, III and IV only I, II, III and IV 25 Which statement about ions of Group 17 elements is correct? A Each has eight valence electrons. B Each contains more protons than electrons. C Each contains more protons than neutrons. D Each has an odd number of electrons. 26 Which of the following statements are true about chlorine and bromine? I Both are gases at room temperature. II They react with sodium to form soluble salts. III They react with heated iron wool to form iron(II) halides IV They dissolve in water to form solutions with pH values of less than 7. A I and III only B II and IV only C I, II and IV only D II, III and IV only 27 An element X has a proton number of 35. Which statement about X is true? A X forms a positive ion during chemical reactions. B It belongs to Group 15 of the Periodic Table. 102 C Reacts with iron wool to form a compound with the formula FeX3. D Reacts with sodium to form a compound with formula NaX2. 28 Which of the following statements about the Periodic Table is true? A All inert gases have eight valence electrons. B All Group 1 elements are equally reactive. C The atomic radius increases in size across a period from left to right. D The electronegativity of Group 17 elements increases up the group. 4.5 Elements in a Period 29 The table shows the proton numbers of four elements. Element P Q R S T Proton 10 11 14 17 19 number Arrange the atomic radii of the elements in increasing order. A P, Q, R, S, T B P, S, R, Q, T C T, S, R, Q, P D T, Q, R, S, P 30 Which of the following metallic oxides can react with both acid ’06 and alkaline solutions? I Aluminium oxide II Tin(II) oxide III Lead(II) oxide IV Copper(II) oxide A I and III only B II and IV only C I, II and III only D I, III and IV only 31 X, Y and Z are elements in the same period of the Periodic Table. The oxide of X is acidic, the oxide of Y is basic and the oxide of Z is amphoteric. Arrange the elements X, Y and Z in increasing proton number. A Z, X, Y C Y, X, Z B X, Z, Y D Y, Z, X Metal oxides X oxide Y oxide Z oxide Observation Sodium hydroxide solution Nitric acid solution Dissolves to form colourless solution Dissolves to form colourless solution No change Dissolves to form colourless solution No change Dissolves to form colourless solution What is the correct arrangement in increasing proton number of the elements? A X, Y, Z B Y, X, Z C Z, Y, X D Z, X, Y 33 The graph shows the change in a property as we go across Period 3 of the Periodic Table. Element X Y Z Proton number 11 13 17 Which of the following statements is true? A X, Y and Z are all conductors of electricity. B All the elements above are made up of atoms. C The atomic radius decreases in the order Z, Y, X. D The electronegativity increases in the order X, Y, Z. 4.6 Transition Elements 35 Which of the following elements will form coloured compounds? I Cobalt III Aluminium II Nickel IV Manganese A I, II and III only B I, II and IV only C II, III and IV only D I, II, III and IV 36 The diagram shows part of a Periodic Table. Which of the following statements below are true about the elements in the Periodic Table? I Element W is inert. II Element Y has more than one oxidation state. III Element X reacts with cold water to form X oxide and hydrogen. IV Element Z reacts with iron to form a compound with formula FeZ3. A II and IV only B I, II and III only C I, II and IV only D I, II, III and IV 37 The properties of the transition metals include ’08 I they form white compounds. II they act as catalysts in chemical reactions. III they have more than one oxidation state. IV they form complex ions. A I, II and III only B I, II and IV only C II, III and IV only D I, II, III and IV 38 The table shows the properties of four elements. Which element is most likely to be a transition element? Element Electrical conductivity Melting point (°C) Density (g cm–3) 360 6.8 Good B 620 3.1 Good C 3500 3.5 Poor D 3400 8.1 Good A 39 An element X forms two oxides with the formula XO and X2O. Which of the following statements is true about the element X ? A It is a transition metal. B It is a Group 1 element. C It is a Group 17 element. D The oxides of element X are amphoteric. Which of the properties below corresponds to the change as shown in the graph above? A Atomic radius B Electropositivity C Electrical conductivity D Density 34 The table shows the proton numbers of three elements X, Y and Z. 40 Which of the following shows correctly the colour of the ions of the transition elements? Colour Fe2+ Cr2O72– MnO4– Co2+ Brown Orange Purple Pink B Green Orange Purple Pink C Brown Purple Orange Blue D Green Purple Orange Blue A 103 Periodic Table of Elements 4 32 X, Y and Z are elements in Period 3 of the Periodic Table. The table shows the properties of the oxides of X, Y and Z when reacted with sodium hydroxide and nitric acid solutions. Structured Questions 4 A list of elements represented by the letters with the nucleon numbers and proton numbers are given below: 1 Bromine is an element of Group 17 in the Periodic Table. (a) What is the physical state of bromine at room temperature? [1 mark] P, 126Q, 199R, 27 S, 35 T, 39 U 13 17 19 1 1 (b) Write an equation for the reaction between bromine and water. [1 mark] (a) Choose two elements from the list above that belong to the same group in the Periodic Table. 4 (c) Draw a labelled diagram for the apparatus that can be used to carry out a reaction between bromine and iron wool. [2 marks] [1 mark] (b) State the (i) group and (ii) period of element U. [2 marks] (d) What is the number of valence electrons in bromine? [1 mark] (c) Give one use of the element P. (d) Draw the atomic structure of element R. [2 marks] (e) Iodine is below bromine in Group 17. Which of the two elements, iodine or bromine, is more reactive? Explain your answer. [3 marks] (e) 2 Rubidium is placed below potassium in Group 1 in the Periodic Table. (i) Which of the elements in the list reacts with cold water to produce hydrogen gas? [1 mark] (ii) Write a balanced chemical equation for the reaction in (i). [1 mark] (a) Give two physical properties of rubidium. [2 marks] (f) Write the formula of the ion formed by the element U. [1 mark] (b) How is rubidium stored in the laboratory? [1 mark] (g) (c) Write the equations for the reactions of rubidium (Rb) with (i) water and (ii) chlorine gas. [2 marks] (i) What is electronegativity? (ii) Which is the more electronegative element between R and T? Explain your answer. [3 marks] (d) Explain why rubidium is more reactive than potassium. [2 marks] 5 A list of the symbols of the transition elements in Period 4 is given below: (e) Write the formula of (i) rubidium nitrate and (ii) rubidium sulphate. [2 marks] (f) What is the colour of rubidium nitrate? [1 mark] Ti, V, Cr, Mn, Fe, Co, Ni, Cu [1 mark] (a) Give two physical properties of the transition [2 marks] elements. 3 Diagram 1 shows a portion of the Periodic Table. The letters in the Periodic Table do not represent the ’05 actual symbols of the elements. (b) What is the colour of the aqueous Cu2+ ion solution? [1 mark] (c) Name a transition element in the list which is used as a catalyst in the (i) Haber process to manufacture ammonia. [1 mark] Diagram 1 (a) Choose two elements which are metals in the Periodic Table above. [2 marks] (ii) hydrogenation of unsaturated oil to make margarine. [1 mark] (d) Other than forming coloured compounds and having catalytic properties, name two other properties of transition elements. [2 marks] (b) Write the formula of the ions formed by the elements (i) E and (ii) Q. [2 marks] (c) Choose an element from the Periodic Table above that can form a coloured compound. [1 mark] (e) Name a reagent that can be used to differentiate between the ions of transition elements in the list above. [1 mark] (d) Choose an element that exists as monatoms. Give a use of this element. [2 marks] 6 The symbols of the elements in Period 3 of the Periodic Table are given below: (e) Which element from the Periodic Table above can form an acidic oxide? [1 mark] Na, Mg, Al, Si, P, S, Cl, Ar (f) Which is the more reactive element between E and R? Explain your answer. [2 marks] (a) Name an element that can conduct electricity in Period 3. [1 mark] Periodic Table of Elements 104 (b) How does the atomic radius change across the period? Explain your answer. [3 marks] (f) (c) Name an element that exists as monatoms in Period 3. Explain your answer. [2 marks] (d) Write the formulae of the oxides of the elements in Period 3. [3 marks] (g) The oxide of aluminium can react with both acid and alkali solutions. What is the term given to such an oxide? [1 mark] (i) Name a basic oxide of Period 3 that can dissolve in water. [1 mark] (ii) Write an equation for the reaction that takes place when its basic oxide in (i) is dissolved in water. [1 mark] (h) Give a use of the element silicon. [1 mark] 4 (e) (i) Name an acidic oxide of Period 3 that can dissolve in water. [1 mark] (ii) Write an equation for the reaction that takes place when the acidic oxide in (i) is dissolved in water. [1 mark] Essay Questions 1 (a) (b) Explain why Group 18 elements exist as monatoms. [5 marks] II (c) Iron(III) chloride Chlorine The reactivity of alkali metals increases down the group in the order: lithium < sodium < potassium Explain the statement given above. I [7 marks] Sodium chloride (b) Rubidium (Rb) is placed below potassium in Group 1 of the Periodic Table. Predict three physical properties and three chemical properties of rubidium. [10 marks] The flowchart shows the conversion of chlorine to sodium chloride and iron(III) chloride. Explain a method to carry out an experiment (i) to obtain sodium chloride from chlorine in conversion I. [5 marks] (ii) to obtain iron(III) chloride from chlorine in conversion II. [5 marks] (c) The nucleon number of sodium is 23 and its atom has 12 neutrons. The nucleon number of chlorine atom is 35 and it has 18 neutrons. Prove that sodium and chlorine belong to the same period in the Periodic Table. [3 marks] 4 (a) Using a suitable period in the Periodic Table as an example, explain the trend in the properties of the elements in terms of metals, non-metals and semi-metals. [5 marks] 2 (a) Explain the following statements: (i) The reactivity of Group 17 elements ’06 decreases down the group. [7 marks] (ii) The atomic radius of Period 3 elements decreases across Period 3 from left to right. (b) Using a suitable period in the Periodic Table as an example, explain why the electronegativity of the elements increases across a period from left to right. [5 marks] [5 marks] (b) With suitable examples, discuss the four properties of transition elements in the Periodic Table. (c) Write the formulae of all the oxides of the elements in Period 3. Describe a suitable experiment to show that the oxides change from basic to amphoteric and then to acidic across Period 3. [10 marks] [8 marks] 3 (a) Name three elements in Group 18 and state their uses. [5 marks] Experiments 1 The reactivity of a halogen in the reaction with iron wool depends on its position in Group 17. Diagram 1 shows the set-up of apparatus for an experiment to determine the reactivity of halogens in Group 17. Diagram 1 105 Periodic Table of Elements The experiment is carried out using bromine gas, chlorine gas and iodine vapour to react with heated iron wool respectively. Observation of the experiment is shown in Table 1 below. Halogen Observation Bromine Hot iron wool glows moderately bright when bromine gas is passed over it. Chlorine Hot iron wool glows brightly when chlorine gas is passed over it. Iodine Hot iron wool glows dimly when iodine vapour is passed over it. Table 1 (a) Complete Table 2 below based on the experiment. 4 Variables Action to be taken (i) Manipulated variable : (i) The way to manipulate variable : (ii) Responding variable : (ii) What to observe in the responding variable : (iii) Constant variable : (iii) The way to maintain the constant variable : Table 2 2 ’08 [6 marks] (b) State one hypothesis for the experiment. [1 mark] (c) Based on the observation, arrange bromine, chlorine and iodine in descending order of reactivity of halogens with iron wool. [1 mark] (d) The proton numbers of the halogens increase in the order: chlorine < bromine < iodine. Make a conclusion regarding the positions of halogens in Group 17 in relation to their reactivities. [1 mark] (e) Astatine is a halogen below iodine in Group 17. Predict the reactivity of astatine with iron wool. [1 mark] Lithium, sodium and potassium are in Group 1 of the Periodic Table. The reactivity of Group 1 elements increases down the group from lithium to potassium. You are required to design a laboratory experiment to prove the statements above. Your explanations should include the following: (a) Problem statement [3 marks] (b) Hypothesis [3 marks] (c) List of materials and apparatus [3 marks] (d) Procedure [3 marks] (e) Tabulation of data [3 marks] Periodic Table of Elements 106 FORM 4 THEME: Interaction between Chemicals CHAPTER 5 Chemical Bonds SPM Topical Analysis 2008 Year 1 Paper Section Number of questions 2009 3 2 5 A B C 1 — 3 – – – 1 3 2010 3 2 A B C 1 — 4 – – 1 – 6 2011 2 3 A B C 1 – – 1 – 5 2 3 A B C – – 1 — 2 – ONCEPT MAP CHEMICAL BONDS To attain the stable electron arrangement of the noble gases Ionic bonds Covalent bonds Formed by transfer of electrons from metal atoms to non-metal atoms Formed by sharing of electrons between non-metal atoms Metal atoms donate electrons Non-metal atoms share electrons Non-metal atoms accept electrons Example: CH4 Example: NaCl / / ¶ / 5H *S 5H * *S / / * / / Differences in physical properties Melting point and boiling point Solubility Electrical conductivity / 5.1 9 Helium has only one electron shell filled with 2 electrons (a duplet electron arrangement). All other noble gases have 8 electrons in the valence shell. This is known as an octet electron arrangement. 10 A duplet electron arrangement (as in helium) or an octet electron arrangment (as in the other noble gases) is very stable. As such, atoms of noble gases do not donate, accept or share electrons with other elements. Thus, atoms of noble gases do not combine with other elements or with itself. They exist as monatoms. Formation of Compounds 5 Stability of Noble Gases 1 A compound is a chemical substance that is formed by combining two or more elements chemically in fixed proportions. 2 Almost all chemical substances exist as compounds in nature. Examples of compounds are water (H2O), carbon dioxide (CO2), table salt sodium chloride (NaCl) and minerals such as metal silicates, metal oxides, metal carbonates and metal sulphides. 3 Only noble gases and a few minerals such as gold, diamond and platinum exist as pure elements. 4 The tendency of elements to combine with other elements to form compounds shows that compounds are more stable than elements. 5 The formation of compounds proves that chemical bonds hold atoms of elements together. 6 Noble gases are elements in Group 18 of the Periodic Table. They are also known as inert gases which consists of helium, neon, argon, krypton, xenon and radon. 7 Noble gases exist as elements. They are very stable and are inert, which means they are non–reactive chemically. 8 The stability of noble gases is due to their electron arrangements (or electronic configurations) as shown in Table 5.1. Conditions for the Formation of Chemical Bonds 1 Noble gases do not form chemical bonds because they have the stable duplet or octet electron arrangement. 2 Atoms of elements from Group 1 to Group 17: (a) Have less than 8 valence electrons (b) Each atom will tend to donate, accept or share electrons to achieve the stable duplet or octet electron arrangement as that of a noble gas 3 In the process of attaining the stable duplet or octet electron arrangement, chemical bonds will form between atoms of these elements. 4 The two types of chemical bonds are (a) ionic bonds (formed by transfers of electrons), and (b) covalent bonds (formed by sharing of electrons). 5 In the formation of chemical bonds, only valence electrons are involved in the donation, acceptance or sharing of electrons. Electrons in the inner shells are not involved. 6 The valence shell will then achieve an octet electron arrangement or a duplet electron arrangement (in the case where there is only one electron shell). Table 5.1 Electron arrangements of noble gases Noble gas Symbol Electron arrangement Helium He 2 Neon Ne 2.8 Argon Ar 2.8.8 Krypton Kr 2.8.18.8 Xenon Xe 2.8.18.18.8 Radon Rn 2.8.18.32.18.8 Chemical Bonds Neon is inert not just because it is a Group 18 element but because it has a stable octet electron arrangement with eight electrons in the outermost shell. 108 electrons respectively. In chemical reactions, these metal atoms tend to donate all their valence electrons to achieve the stable duplet or octet electron arrangement. 6 A negative ion (or anion) is formed when an atom accepts one or more electrons. The ion formed is negatively-charged because there are more electrons than protons. For instance, 5.1 1 The proton numbers of neon and argon are 10 and 18 respectively. Write the electron arrangements of neon and argon. Explain why these two elements exist as monatoms. 2 The electron arrangements of atoms P, Q and R are given in the table below. Electron arrangement P 2.8.2 Q 2.8.7 R 2.8.8 2– accepts 2 electrons 8p Oxygen atom (O) (8p, 8e) 3 State two types of chemical bonds. Formation of Ionic Bonds SPM ’08/P1 Formation of Ions 1 Atoms are neutral because the number of protons is the same as the number of electrons. 2 An ion is formed when an atom donates or receives one or more electrons. 3 A positive ion or cation is formed when an atom donates one or more electrons. The ion formed has less electrons than protons and is positively-charged. For example, 11p Sodium atom (Na) (11p,11e) donates an electron Oxide ion (O2–) (8p, 10e) 7 Generally, (a) non–metals usually form negative ions, (b) charge of negative ion = number of electrons received by an atom X + ne– → Xn– 8 Non–metal atoms from Groups 15, 16 and 17 in the Periodic Table have 5, 6 and 7 valence electrons respectively. In chemical reactions, non-metal atoms will accept electrons so that the ion formed achieves the stable octet electron arrangement. • The term electron arrangement is used interchangeably with electronic configuration. • Donate electrons can also be explained as lose electrons or release electrons. • Accept electrons can also be explained as gain electrons or receive electrons. • The name of a metal ion is the same as the metal atom. Examples: sodium ion (Na+), magnesium ion (Mg2+), aluminium ion (Al3+). • The name of the non-metal ion ends with -ide or -ate (when oxygen is attached to the non-metal). Examples: chloride ion (Cl–), sulphide ion (S2–), sulphate ion (SO42–), carbonate ion (CO32–). + 11e 8p 8e (a) Which atom is chemically inert? Explain your answer. (b) Which atom will take part in chemical bonding? Explain your answer. 5.2 10e 5 Atom 10e 11p Sodium ion (Na+) (11p,10e) 4 Generally, (a) metal atoms usually form positive ions, (b) charge of positive ion = number of electrons released by an atom M → Mn+ + ne– 5 Metal atoms from Groups 1, 2 and 13 in the Periodic Table have 1, 2 and 3 valence • The duplet or octet electron arrangement of noble gases are very stable. • Atoms form positive ions or negative ions so as to attain the electron arrangement as that of the noble gases. 109 Chemical Bonds 5 To prepare ionic compounds Materials Magnesium ribbon, sodium, chlorine gas, iron wool and sodium hydroxide solution. Apparatus Tripod stand, clay pipe triangle, Bunsen burner, crucible and lid, sandpaper, gas jar, gas jar spoon, combustion tube, filter funnel, retort stand, clamp and beaker. 2 The ignited sodium is placed in a gas jar filled with chlorine gas. Any changes that occur are recorded. (A) Preparation of magnesium oxide Procedure 1 A 5 cm length of magnesium ribbon is cleaned with a piece of sandpaper. 2 The magnesium ribbon is placed in the crucible. 3 The magnesium ribbon is heated strongly. Any changes that occur are recorded. Figure 5.2 Preparation of sodium chloride (C) Preparation of iron(III) chloride Procedure 1 A little iron wool is placed inside a combustion tube. 2 The end of the combustion tube is connected to a filter funnel inverted into a beaker with some sodium hydroxide solution. 3 The iron wool is heated strongly until it glows. 4 Chlorine gas is passed through the iron wool while being heated. Any changes that occur are recorded. Figure 5.1 Preparation of magnesium oxide (B) Preparation of sodium chloride Procedure 1 A small piece of sodium metal is placed in a gas jar spoon and is heated carefully until it begins to ignite. Figure 5.3 Preparation of iron(III) chloride Results Activity 5.1 Method Observation Inference Heating of magnesium in air • The magnesium ribbon burns with a bright flame • White powder is formed The white powder formed is magnesium oxide Burning of sodium in chlorine gas • Sodium burns with a bright yellow flame • The yellowish-green colour of chlorine gas is decolourised. • White fumes are produced and deposited as white powder The white powder formed is sodium chloride Heating of iron in chlorine gas • The iron wool continues to glow brightly in the chlorine gas • A brown powder is formed The brown powder formed is iron (III) chloride Chemical Bonds 110 2Mg(s) + O2(g) → 2MgO(s) 2 Magnesium oxide is an ionic compound that contains Mg2+ ions and O2– ions. 3 When sodium burns in chlorine gas, the white powder formed is sodium chloride. Conclusion 1 Generally, the reaction between metals and nonmetals produces ionic compounds. 2 Ionic compounds such as magnesium oxide, sodium chloride and iron(III) chloride can be prepared by direct combination of the metal and non-metal elements. 2Na(s) + Cl2(g) → 2NaCl(s) 4 Sodium chloride is an ionic compound that contains Na+ ions and Cl– ions. 5 When iron burns in chlorine gas, the brown powder formed is iron(III) chloride. Metal Magnesium + Sodium + Iron + 2Fe(s) + 3Cl2(g) → 2FeCl3(s) 6 Iron(III) chloride is an ionic compound that contains Fe3+ ions and Cl– ions. Formation of Ionic Bonds Non-metal oxygen → chlorine → chlorine → Ionic compound magnesium oxide sodium chloride iron(III) chloride SPM ’09/P1 (c) non–metal atoms receive electrons to form negative ions (anions). (d) positive ions and negative ions are then attracted to each other by the strong electrostatic force of attraction. The bond formed between ions of opposite charges is known as ionic bond or electrovalent bond. 1 Metals from Groups 1, 2 or 13 react with non–metals from Groups 15, 16 or 17 in the Periodic Table to form ionic compounds. 2 In the formation of an ionic bond, (a) electrons are transferred from a metal atom to a non–metal atom. (b) metal atoms donate valence electrons to form positive ions (cations). Formation of ionic bond in sodium chloride 1 A sodium atom with an electron arrangement of 2.8.1 achieves stability after it donates one valence electron to form a sodium ion, Na+. The electron arrangement of the sodium ion, Na+, is 2.8, with an octet of valence electrons. SPM ’06/P2 Cl + e– ⎯→ Cl– 2.8.7 2.8.8 3 Sodium ions, Na+ and chloride ions, Cl– with opposite charges are attracted to each other by the electrostatic force of attraction. This force of attraction is called the ionic bond. Na ⎯→ Na+ + e– 2.8.1 2.8 2 A chlorine atom with an electron arrangement of 2.8.7 achieves stability after it accepts one electron from a sodium atom – to form a chloride ion, Cl . The electron arrangement of the chloride ion, Cl–, is 2.8.8, with an octet of valence electrons. 111 Chemical Bonds 5 7 Sodium hydroxide solution is used to absorb the excess chlorine gas. Besides sodium hydroxide, soda lime can also be used. 8 A filter funnel is used to prevent the sodium hydroxide solution from being suctioned back into the combustion tube as chlorine gas is very soluble in water. Discussion 1 When magnesium is heated, magnesium atom combines with oxygen in the air to form magnesium oxide. Formation of ionic bond in magnesium oxide 1 A magnesium atom with the electron arrangement of 2.8.2 achieves the stable octet electron arrangement when it donates two valence electrons to form a magnesium ion, Mg2+. O + 2e– ⎯→ O2– 2.6 2.8 3 The electrostatic force of attraction that exists between the oppositely-charged magnesium ions, Mg2+ and oxide ions, O2– forms the ionic bond. Mg ⎯→ Mg2+ + 2e– 2.8.2 2.8 5 SPM ’09/P1 2 An oxygen atom with an electron arrangement of 2.6 achieves the stable octet electron arrangement when it accepts two electrons to form an oxide ion, O2–. Formation of ionic bond in potassium oxide electrons from two potassium atoms to form an oxide ion, O2–. 3 Hence each of the two potassium atoms donates one valence electron to be accepted by one oxygen atom. Ionic bonds are formed between the two potassium ions and the oxide ion. 1 A potassium atom with an electron arrangement of 2.8.8.1 achieves the stable electron arrangement of an octet when it donates one valence electron to form a potassium ion, K+. K ⎯→ K+ + e– 2.8.8.1 2.8.8 2 An oxygen atom with an electron arrangement of 2.6 achieves the stable octet electron arrangement when it accepts two valence How to Predict the Formula of an Ionic Compound When drawing the electron arrangements to show the formation of ionic bonds, • do not overlap the outermost electron shells of atoms. • the outermost shells of all ions must have eight electrons except Li+ and H+. • use ‘dots‘ or ‘crosses‘ to represent the electrons from different atoms. • show the charge of the ions clearly outside the brackets of the ions. The chemical formula of sodium chloride, NaCl, tells us that 1 mol of sodium ions combines with 1 mol of chloride ions and it is not 1 molecule of sodium chloride. Chemical Bonds 1 Metal atoms will donate their valence electrons to achieve the stable duplet or octet electron arrangement of the noble gases. 2 Non-metal atoms will accept electrons in order to achieve the stable octet electron arrangement of the inert gases. 3 For cations Mb+ and anion Xa–, the formula of an ionic compound formed between them is written as MaXb Number of electrons that will be received by X (or charge of X ion) Number of electrons that will be donated by M (or charge of M ion) 4 The overall positive charge of the cation must be equal to the overall negative charge of the anion in an ionic compound. Hence the formula of an ionic compound formed between them can also be derived as aMb++ bXa– → MaXb 112 1 2 Atoms in element J and element Q have proton numbers 12 and 17 respectively. Explain what type of bond will form between J and Q. Predict the formula of the compound formed. Element M is an element from Group 13 and element X is an element from Group 16 in the Periodic Table. What is the formula of the compound formed between element M and element X? Solution An atom of J, with a proton number of 12 and an electron arrangement of 2.8.2, will achieve the stable octet electron arrangement by donating two electrons to form a J 2+ ion. Solution M from Group 13 has three valence electrons and will form M 3+ ion. X from Group 16 will receive two electrons to form X 2– ion. The formula of the compound formed is Charge of X 2– ion Charge of M 3+ ion An atom of element Q with proton number 17 and an electron arrangement of 2.8.7 will achieve the stable octet electron arrangement by accepting one electron to form a Q– ion. 5.2 Q + e ⎯⎯→ Q 2.8.7 2.8.8 – 5 M2X3 J ⎯⎯→ J 2+ + 2e– 2.8.2 2.8 – 1 Give the formulae of the ions formed by the following elements: (a) Calcium (b) Phosphorus (c) Sulphur (d) Potassium (e) Nitrogen Two Q atoms will accept one electron each from the two electrons donated by every J atom. Ionic bonds are formed between J 2+ ions and Q – ions to produce a compound with formula JQ2. [Proton number: N, 7; P, 15; S, 16; K, 19; Ca, 20] 1 2 The proton numbers of element P and element Q are 9 and 20 respectively. (a) Write the equations that show the transfer of electrons and the formulae of the ions formed by atom P and atom Q respectively. (b) Draw the valence electron arrangement to show the formation of a compound formed between P and Q. ’02 R reacts with S to form an ionic compound with the formula R2S3. Which of the following electron arrangements of atoms R and atoms S is true? Electron arrangement of atom R Electron arrangement of atom S A 2.8.2 2.8.3 B 2.8.3 2.8.2 Group 1 2 16 17 C 2.8.2 2.5 Element E F G H D 2.8.3 2.6 3 The table shows the groups of four elements in the Periodic Table represented by the letters E, F, G and H. (a) Write the formulae of the ions formed by the elements E, F, G and H. (b) What is the formula of the compound formed from (i) E and G ? (ii) E and H ? (iii) F and G ? (iv) F and H ? Comments Since an ionic compound is formed, R must be a metal and S must be a non-metal. The formula of R2S3 shows that R will donate three electrons whereas S will accept two electrons to achieve the octet electron arrangement. Hence, R has three valence electrons and S has six valence electrons. Answer D 113 Chemical Bonds 3 A Lewis structure is a diagram which shows only the valence electrons of the atoms represented by dots. The Lewis structure of the hydrogen molecule is Formation of Covalent Bonds 1 A covalent bond is a bond that is formed from the sharing of valence electrons between ’08/P1 non-metal atoms to achieve the stable octet or a duplet electron arrangement. 2 Non-metals from Groups 15, 16 or 17 in the Periodic Table react with other non-metals of the same group or different groups to form covalent compounds. 3 Hydrogen is a non-metal element. It can form covalent bonds with other non-metal atoms from Groups 14, 15, 16 and 17. Examples are CH4, NH3, H2S and HCl. 4 Molecules with covalent bonds can be formed from: (a) Atoms of the same element Examples: Hydrogen (H2), chlorine (Cl2), oxygen (O2) and nitrogen (N2). (b) Atoms of different elements Examples: Water (H2O), ammonia (NH3), carbon dioxide (CO2), methane (CH4), tetrachloromethane (CCl4). 5 The types of covalent bond formed depend on the number of pairs of electrons shared between two atoms. There are three types of covalent bonds. (a) Single bond: One pair of electrons shared between two atoms. (b) Double bond: Two pairs of electrons shared between two atoms. (c) Triple bond: Three pairs of electrons shared between two atoms. H• + H• → H •• H (or H — H) Formation of a chlorine molecule, Cl2 1 A chlorine atom with an electron arrangement of 2.8.7, needs to share one electron to achieve the stable octet electron arrangement of 2.8.8. • • • • • • • • • • • • • • 2 Each chlorine atom contributes one valence electron to be shared. The sharing of one pair of electrons results in the formation of one single covalent bond between two chlorine atoms. 3 The Lewis structure showing the formation of chlorine molecule is • • 5 SPM • • 5.3 Cl • + • Cl •• → •• Cl •• Cl •• Formation of methane molecule, CH4 Formation of Single Covalent Bonds 1 A carbon atom with an electron arrangement of 2.4 needs to share four electrons to achieve the stable octet electron arrangement of 2.8. 2 A hydrogen atom needs to share one electron to achieve the stable duplet electron arrangement. 3 One carbon atom contributes four valence electrons to be shared with four hydrogen atoms respectively to form four single covalent bonds in the methane molecule, CH4. Formation of a hydrogen molecule, H2 1 A hydrogen atom with an electron arrangement of 1, needs to share one electron to achieve the stable duplet electron arrangement. 2 Each hydrogen atom will contribute one valence electron to be shared between two hydrogen atoms. One pair of electrons shared between two hydrogen atoms form a single covalent bond. Chemical Bonds 114 4 The Lewis structure for the formation of methane molecule is ⏐ 4H • + • C • → H •• C •• H or H ⎯ C ⎯H ⏐ H Formation of tetrachloromethane (carbon tetrachloride), CCl4 SPM ’10/P1, ’11/P1 1 A carbon atom with an electron arrangement of 2.4 needs to share four electrons to achieve the stable octet electron arrangement of 2.8. 2 A chlorine atom with an electron arrangement of 2.8.7 needs to share one electron to achieve the stable octet electron arrangement of 2.8.8. 3 Four chlorine atoms contribute one valence electron each to be shared with one carbon atom to form four single covalent bonds in the tetrachloromethane molecule, CCl4. 4 The Lewis structure showing the formation of ammonia molecule, NH3 is • • • • • ⎯ ⎯ 3H • + N → H •• N • H or H N H • • ⏐ H H • • • • H • • • Formation of water molecule, H2O 1 An oxygen atom with an electron arrangement of 2.6 needs to share two electrons to achieve the stable octet electron arrangement of 2.8. 2 A hydrogen atom needs to share one electron to achieve the stable duplet electron arrangement. 3 One oxygen atom contributes two valence electrons to be shared with two hydrogen atoms (which contributes one electron respectively) to form two single covalent bonds in the water molecule, H2O. 4 The Lewis structure showing the formation of water molecule, H2O is • • • • 4 The Lewis structure of the formation of tetrachloromethane molecule is • • • • • • • • H H ⏐ • • • • • • ⎯O • H O H or 2H • + •• O → • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Cl •• • Cl • ⏐ 4 •• Cl • + • C • → •• Cl •• C •• Cl •• or •• Cl ⎯ C⏐ ⎯ Cl •• • • • • • Cl • • Cl • • • 115 Chemical Bonds 5 1 A nitrogen atom with an electron arrangement of 2.5 needs to share three electrons to achieve the stable octet electron arrangement of 2.8. 2 A hydrogen atom needs to share one electron to achieve the stable duplet electron arrangement. 3 One nitrogen atom contributes three valence electrons to be shared with three hydrogen atoms (which contributes one electron respectively) to form three single covalent bonds in the ammonia molecule, NH3. H • • • H • Formation of ammonia molecule, NH3 Formation of Double Covalent Bonds Formation of carbon dioxide molecule, CO2 1 A carbon atom with an electron arrangement of 2.4 needs to share four electrons to achieve the stable octet electron arrangement of 2.8. 2 An oxygen atom with an electron arrangement of 2.6 needs to share two electrons to achieve the stable octet electron arrangement of 2.8. 3 One carbon atom contributes four valence electrons to be shared with two oxygen atoms (which contribute two electrons each) to form two double bonds in the carbon dioxide molecule, CO2 as shown in the diagram below. 1 An oxygen atom with an electron arrangement of 2.6 needs to share two electrons to achieve the stable octet electron arrangement of 2.8. 2 Each oxygen atom will contribute two valence electrons to be shared between two oxygen atoms. 3 The sharing of two pairs of electrons between two atoms results in the formation of a double covalent bond in an oxygen molecule, O2. 4 The Lewis structure below shows the formation of a carbon dioxide molecule, CO2. • • • • • • • • • • • • • •• •• 2O • + • C • → O • • C • • O or O == C == O • • • • • • • • • • • • Formation of Triple Covalent Bonds Formation of nitrogen molecule, N2 • • • • • • 1 Each nitrogen atom with an electron arrangement of 2.5 will contribute three valence electrons to be shared between two nitrogen atoms so as to achieve the stable octet electron arrangement of 2.8. 2 The sharing of three pairs of electrons results in the formation of a triple covalent bond in the nitrogen molecule, N2. In the formation of a covalent bond • an atom that requires one more electron to achieve the stable electron arrangement as in the noble gases will form one single covalent bond. For example: hydrogen and chlorine form single covalent bonds in H – H; Cl – Cl; H – Cl. • an atom (such as oxygen and sulphur) that requires two electrons to achieve the stable electron arrangement will form two single covalent bonds or one double covalent bond. For example: 1 double covalent bond per oxygen atom • Chemical Bonds 116 • • 3 The Lewis structure for nitrogen molecule, N2 is shown below. O=O •• N • + • N • → ••N •• •• N •• or •• N ≡ N •• • 2 single covalent bonds per oxygen atom ; • • H–O–H SPM ’10/P1 • • • • •• • == O •• O • + • O → O • • O or • O • • • • • • • • • 4 The Lewis structure showing the formation of oxygen molecule is • • 5 Formation of oxygen molecule, O2 SPM ’06/P2 ’08/P1 How to Predict the Formula of a Covalent Compound 2 Comments P, with 6 valence electrons, will need to share 2 electrons in order to achieve the stable octet electron arrangement. Q, with 4 valence electrons, will need to share 4 electrons to achieve the octet electron arrangement. Hence the formula of the compound is Table 5.2 Number of valence electrons (x) Valency (8–x) Group 14 4 4 Group 15 5 3 Group 16 6 2 Group 17 7 1 Q2P4 or QP2 2 electrons to be shared with Q by P Answer A 3 4 If a non-metal element M, with a valency of SPM x, combines with another non-metal element, ’09/P1 N, with a valency of y, the formula of the covalent compound formed will be MyNx. Element X is from Group 14, element Y is from Group 16 and element Z is from Group 17 of the Periodic Table. What is the formula of the compound formed between (a) element X and element Y? (b) element X and element Z? MyNx valency of N valency of M Solution (a) An atom of element X from Group 14 with 4 valence electrons needs to share 4 electrons in order to achieve the octet electron arrangement. An atom of element Y from Group 16 has 6 valence electrons and needs to share 2 electrons in order to achieve the octet electron arrangement. The formula of the compound formed is 5 Table 5.3 below shows the formulae of covalent compounds formed between elements from different groups. Table 5.3 Non-metal element, M Non-metal element, N Formula of covalent Examples compound MN4 CH4, CCl4, SiCl4 M2N4 or MN2 CO2, SiO2 Group 15 H or Group 17 (valency 3) (valency 1) MN3 NH3, PH3, PCl3 Group 16 H or Group 17 (valency 2) (valency 1) MN2 H2O, H2S, SCl2 Group 14 H or Group 17 (valency 4) (valency 1) Group 14 (valency 4) Group 16 (valency 2) 4 electrons to be shared with P by Q X2Y4 or XY2 2 electrons required by Y to be shared to achieve the octet electron arrangement 4 electrons required by X to be shared to achieve the octet electron arrangement (b) An atom of element Z from Group 17 requires 1 electron to be shared, hence the formula of the compound is X1Z4 or XZ4 Z shares 1 electron with X 117 X shares 4 electrons with Z Chemical Bonds 5 The electron arrangement of atom P is 2.8.6 and atom Q has four valence electrons. What is the formula of the compound formed between P and Q? A QP2 B QP4 C Q2P D Q4P 1 Covalent compounds are formed from the non-metal elements (from Group 14 to Group 17). 2 The number of electrons required by a nonmetal atom to achieve the stable octet electron arrangement as that of a noble gas is known as the valency. 3 The valency of a non-metal is (8–x) where x is the number of valence electrons. Table 5.2 shows the valency of non-metal elements. Element ’03 4 The proton numbers of element J and element Q are 16 and 17 respectively. Explain the type of bond and the formula of the compound formed between J and Q. The electron arrangement of an atom of Q is 2.8.7. Both J and Q are non-metals and they will form a covalent compound. An atom J needs to share 2 electrons and an atom Q needs to share 1 electron in order to achieve the stable octet electron arrangement. The formula of the compound formed is JQ2. Solution The electron arrangement of an atom of J is 2.8.6. 5 Guidelines in drawing the electron arrangement of a covalent compound formed by atom M and atom N. Step 1 Step 2 Determine the number of valence electrons and the valency of atom M and atom N, then determine the formula of the compound. Example Atom M has 5 valence electrons and valency is 3. Atom N has 7 electrons and valency is 1. Formula of compound is M1N3. [Proton number: M, 7; N, 9] Draw the positions of atom M and atoms N in the formula with atoms N (more number of atoms) surrounding atom M. Example: Step 4 Step 3 Determine the number of electrons to be shared (= 8 – x) where x = number of valence electrons. Draw the electrons to be shared as dot or crosses at the overlapped area. Example: Draw the valence electron shells overlapping between the two types of atoms. Step 5 Step 6 Draw the balance of the valence electrons not shared (x – number of electrons shared) on the shell outside the overlapped area. Draw the inner electron shells for atom M and atoms N and the electrons as dots and crosses. N has 7 valence electrons: 1 shared, remainder 6 unshared. Chemical Bonds 118 How to predict the type of chemical bonds formed and the formula of the compound formed from two elements, X and Y. Step 1 Step 2 Write the electron arrangement to determine the number of valence electrons of elements X and Y. 5 • If one of the elements (say X) has 1, 2 or 3 valence electrons and the other element (say Y) has 5, 6 or 7 valence electrons then X and Y form an ionic bond. • If both elements have 4, 5, 6 or 7 valence electrons, or one of the elements is hydrogen, X and Y form a covalent bond. Step 3 Determine the number of electrons X or Y needs to donate/ accept/share to achieve the stable octet electron arrangement. Ionic Bonding Covalent Bonding Formula is Xa Yb. Number of electrons that atom Y needs to accept to achieve an octet electron arrangement Formula is Xa Yb. Number of electrons that atom X needs to donate to achieve an octet electron arrangement Number of electrons that atom Y needs to share to achieve an octet electron arrangement Number of electrons that atom X needs to share to achieve an octet electron arrangement 1 Elements with 1, 2 or 3 valence electrons can only form ionic bonds with non-metals (elements with 5, 6 or 7 valence electrons). 2 Elements with 4 valence electrons can only form covalent bonds. 3 Elements with 5, 6 or 7 valence electrons can form both ionic and covalent bonds. 119 Chemical Bonds 3 5.3 ’04 1 Write the Lewis structures for the following compounds: (a) Hydrogen chloride, HCl (b) Water, H2O (c) Hydrogen cyanide, HCN (d) Phosphorus trichloride, PCl3 [Proton number: H, 1; C, 6; N, 7; P, 15; Cl, 17] 5 The diagram shows the electron arrangement of a compound formed between atoms X and Y. Which of the following statements is true about the compound? A The compound is formed by ionic bonds. B The compound is formed by electron transfer. C Atom X is a metal and atom Y is a non-metal. D It is a covalent compound. 2 The table shows four elements in different groups of the Periodic Table represented by the letters V, X, Y and Z. Comments The diagram shows the overlap of the outermost electron shells of atoms X and Y where the sharing of electrons is represented by dots and crosses. The diagram represents a covalent compound formed by the sharing of electrons between non-metal atoms. Answer D Similarity 15 16 17 Element V X Y Z 3 The proton numbers of elements P and Q are 6 and 9 respectively. Draw the electron diagrams for the formation of compounds between (a) Q and hydrogen (b) Q and P (c) Q and Q Covalent bonding Atoms achieve the stable (duplet or octet) electron arrangement after the formation of bonds. Difference Involves the transfer of electrons from metal atoms to non-metal atoms. Chemical Bonds 14 (a) Write the formula of the compound formed between (i) V and Z (ii) V and Y (iii) X and Z (b) Draw the Lewis structures for the compounds formed in (a). (c) What type of compounds are formed in (a)? Comparison between Ionic Bonding and Covalent Bonding Ionic bonding Group Involves the sharing of electrons between non-metal atoms. Positivelycharged ions and negativelycharged ions are formed. No charged ions are formed. Molecules are formed. Strong electrostatic force of attraction holds the oppositelycharged ions together. Van der Waals forces of attraction exist between the covalent molecules. CFCs (chlorofluorocarbon) are covalent compounds that consist of chlorine, fluorine and carbon atoms bonded by covalent bonds. Two examples of CFCs are CF2Cl2 and CFCl3. CFCs are colourless, odourless and non-toxic gases with low boiling points. CFCs are used in aerosol and as refrigerants in air-conditioners. However, since 1992, the usage of CFC was banned because it was believed to deplete the ozone layer which protects the earth from getting too much harmful ultraviolet rays from the sun. • Metals only form ionic bonding. • Non-metals form both ionic and covalent bondings. 120 8 The differences between the physical properties of ionic compounds and covalent compounds ’09/P1 are shown in Table 5.4. SPM The Properties of Ionic Compounds and Covalent Compounds Properties of Ionic Compounds and Covalent Compounds ’08/P1 SPM Table 5.4 Comparison of properties of ionic and covalent compounds Ionic compound Covalent compound (simple molecules) High Low (b) Volatility Non-volatile Volatile (can change to vapour when heated) (c) Solubility Usually soluble in water and polar solvents but insoluble in organic solvents Usually soluble in organic solvents such as benzene but insoluble in water (d) Electrical conductivity Conducts electricity in the molten state or aqueous solution Does not conduct electricity in any state SPM ’10/P2 Properties 1 The physical properties of ionic and covalent compounds are different because of the different types of bonds formed and the difference in the structures of the compounds. 2 An ionic compound is formed when ions of opposite charges are held by the strong electrostatic force of attraction. 3 The ions of ionic compounds are arranged in an orderly and compact manner and form large ionic structures. 4 Figure 5.4 shows the arrangement of ions in a three-dimensional network forming a large structure in an ionic compound. (a) Melting point and boiling point Figure 5.4 Three-dimensional network of ions 5 Most covalent compounds consist of simple molecules. 6 The covalent bond within a molecule is strong but the intermolecular forces (van der Waals forces of attraction) are weak. 7 Figure 5.5 shows the intermolecular forces (van der Waals forces of attraction) between methane molecules and the intramolecular bonds (covalent bonds) in the methane molecules. The dotted line represents the weak intermolecular forces (van der Waals forces). 5 5.4 Most ionic compounds are soluble in water. However there are some ionic compounds that are insoluble in water. For example, lead(II) bromide, lead(II) chloride and calcium carbonate are insoluble in water. There are also some covalent compounds that are polar and are soluble in water. For example, ethanol, ethanoic acid and sugar are soluble in water. • The low melting and boiling points of covalent compounds are due to the weak intermolecular forces between covalent molecules. • They are not due to the strength of covalent bonds inside the molecules. Figure 5.5 Types of bonds in methane molecules 121 Chemical Bonds To study the physical properties of ionic and covalent compounds (A) To investigate the melting point and volatility of compounds Apparatus 5 Materials SPM ’06/P2 3 Steps 1 and 2 are repeated with naphthalene and hexane respectively to replace magnesium chloride. Crucible, spatula, tripod stand, wire gauze and Bunsen burner. Magnesium chloride, naphthalene and hexane. Procedure 1 A spatula of magnesium chloride solid is placed in a crucible. 2 Magnesium chloride solid is heated slowly at first and then strongly. The change in the physical state of the compound is recorded. Figure 5.6 To investigate the melting point of a compound Results Compound Observation Physical state Magnesium chloride Solid No change High melting point Naphthalene Solid Melts rapidly Low melting point Hexane Liquid Vaporise rapidly Low melting point and very volatile Conclusion 1 Magnesium chloride is an ionic compound and has high melting and boiling points. Naphthalene and hexane are covalent compounds and they have low melting and boiling points. 2 Ionic compounds are non-volatile whereas covalent compounds are volatile. (B) To investigate the solubility of compounds Activity 4.2 Inference Action of heat Apparatus Boiling tubes and spatula Materials Magnesium chloride, naphthalene, hexane, water and cyclohexane. Procedure 1 A boiling tube is filled with 5 cm3 of distilled water. 2 Half a spatula of magnesium chloride solid is added to the distilled water and shaken. The solubility of magnesium chloride in water is noted. 3 Steps 1 to 2 are repeated with naphthalene and hexane to replace magnesium chloride. 4 Another boiling tube is filled with 5 cm3 of cyclohexane (an organic solvent). 5 Half a spatula of magnesium chloride is added to the cyclohexane and shaken. The solubility of magnesium chloride in cyclohexane is noted. 6 Steps 4 and 5 are repeated with naphthalene and hexane in turn to replace magnesium chloride. Chemical Bonds Results Compound Solubility In water In organic solvent Magnesium chloride Soluble Insoluble Naphthalene Insoluble Soluble Hexane Insoluble Soluble Conclusion 1 Magnesium chloride is an ionic compound and is soluble in water but insoluble in organic solvents. 2 Naphthalene and hexane are covalent compounds and are insoluble in water but soluble in organic solvents. 3 Hexane and water form two layers of liquid in a test tube. This shows hexane is insoluble in water. (C) To investigate the electrical conductivity of compounds Apparatus Crucible, spatula, graphite rods, batteries, light bulb, switch, connecting wires, tripod stand, clay triangle and Bunsen burner. Materials Lead(II) bromide, magnesium chloride, sugar and naphthalene. 122 Figure 5.7 To investigate the electrical conductivity of compounds Results Compound Lead(II) bromide Naphthalene Magnesium chloride Sugar State of compound Observation Inference Solid Light bulb does not light up Conducts electricity in the liquid but not in the solid state Molten Light bulb lights up Solid Light bulb does not light up Molten Light bulb does not light up Solid Light bulb does not light up Aqueous solution Light bulb lights up Solid Light bulb does not light up Aqueous solution Light bulb does not light up Does not conduct electricity in any state Conducts electricity in aqueous state but not in the solid state Does not conduct electricity in any state an electric current flows through the liquid or aqueous solution. 2 Lead(II) bromide has a lower melting point than magnesium chloride. Therefore, lead(II) bromide is more suitable for use in the investigation of the electrical conductivity of ionic compounds in the liquid state compared to magnesium chloride. 3 Lead(II) bromide is insoluble in water whereas magnesium chloride is soluble in water. Therefore, magnesium chloride is more suitable for the investigation of the electrical conductivity of ionic compound in an aqueous solution. 4 Naphthalene does not dissolve in water. Some covalent compounds such as sugar are soluble in water but they still do not conduct electricity. Conclusion 1 Ionic compounds such as lead(II) bromide and magnesium chloride do not conduct electricity in the solid state but conduct electricity in the molten state or in aqueous solution due to the presence of mobile ions. 2 Covalent compounds such as naphthalene and sugar do not conduct electricity in any state due to the absence of mobile ions. Discussion 1 Besides light bulbs, galvanometers and ammeters can be used to determine the electrical conductivity. The needle of a galvanometer will deflect if 123 Chemical Bonds 5 SPM Procedure ’04/P2 1 Three spatulas of lead(II) bromide solid is placed in a crucible. 2 Two graphite rods are dipped in the lead(II) bromide solid and the circuit is completed by connecting to the batteries and switch. 3 The switch is turned on and the bulb is checked if it lights up. 4 Lead(II) bromide is heated strongly until it melts. The switch is turned on again to check if the bulb lights up. 5 Steps 1 to 4 are repeated using naphthalene to replace lead(II) bromide. 6 Three spatulas of magnesium chloride solid is placed in a crucible. 7 The switch is turned on to check if the bulb lights up. 8 Water is added to the magnesium chloride. The mixture is stirred with a glass rod until all the magnesium solid dissolves in water. 9 The switch is turned on again to check if the bulb lights up. 10 Steps 6 to 9 are repeated using sugar to replace magnesium chloride. Explanation for the Differences between the Physical Properties of Ionic and Covalent Compounds consist of a covalent network of molecules forming a giant molecular structure. Some examples are carbohydrates, proteins and silicon dioxide. Strong covalent bonds hold the atoms together in these giant molecules. A lot of heat energy is required to overcome the strong covalent bonds of these giant molecules. 5 Melting points, boiling points and volatility 1 Ionic compounds have higher melting and boiling points than covalent compounds. This is because the electrostatic force of attraction between oppositely charged ions is very strong. A lot of heat energy is required to overcome this strong force of attraction. 2 There are two types of covalent compounds: (i) Simple molecules (ii) Giant molecules 3 Below are some examples of simple molecules and giant molecules. Most simple molecules exist as gases or liquids while giant molecules exist as solids. Simple covalent molecules Solubility 1 Ionic compounds are soluble in water because they can form bonds with water molecules but are insoluble in organic solvents. 2 Covalent compounds are soluble in organic solvents because they can form bonds with organic solvent molecules but are insoluble in water. 3 Some covalent molecules can dissolve in water because they form hydrogen bonds with water. Some examples of these covalent compounds are sugar, ethanol, acetone and carboxylic acids. 4 Most giant covalent molecules are insoluble in both water and organic solvents. Giant covalent molecules Carbon dioxide, CO2 Graphite Tetrachloromethane, CCl4 Diamond Ammonia, NH3 Silicon Hydrogen gas, H2 Silicon dioxide Oxygen gas, O2 Carbohydrates Water, H2O Protein Electrical conductivity 1 All ionic compounds can conduct electricity in the molten state and in aqueous solution because the charged ions can move freely. Ionic compounds in the solid state do not conduct electricity because the ions are held by the strong electrostatic force of attraction in the lattice structure and are not free to move. 2 Covalent compounds do not conduct electricity in any state because they consist of molecules. There are no freely moving charged particles. 3 A few covalent compounds can dissociate into ions when dissolved in water. So, they can conduct electricity in aqueous solution. Some examples are ammonia (NH3), hydrogen chloride (HCl) and sulphur dioxide (SO2). 4 Electrical conductivity in the molten or liquid state is the best physical property to differentiate between ionic compounds and covalent compounds. 4 Melting and boiling points of covalent compounds of simple molecules such as tetrachloromethane, naphthalene, chlorine, bromine and iodine are low because the intermolecular forces of attraction between molecules (van der Waals forces of attraction) are very weak. Little heat energy is required to overcome the weak intermolecular forces. 5 The low boiling points of the covalent compounds cause the covalent compounds to be volatile (easily changed to the vapour state when heated). Hence covalent compounds should be kept far away from a heat source. 6 Some covalent compounds have high melting and boiling points because they Chemical Bonds 124 1 The flowchart below helps us to determine the type of chemicals. Chemical Does it conduct electricity in the solid state? No Yes Metal Ionic compound or covalent compound Does it conduct electricity in the liquid state? No 5 Yes Ionic compound Covalent compound Does it have high melting and boiling points? No Yes Giant covalent compound Simple covalent compound Does it dissolve in water? No Yes Polar covalent compound Non-polar covalent compound 4 ’07 The table shows electrical condutivity and melting points of three substances X, Y, and Z. Substance Electrical conductivity Melting point (°C) Solid Molten Aqueous X No No No –117 Y No No No 80 Z No Yes Yes 801 (a) State the types of structure and bonding of substances X, Y and Z. (b) Explain why substance X has a low melting point. (c) Show how the bonds are formed in substance Y and substance Z. (d) Based on the information in the table, predict the solubility of substances X, Y and Z in water. Solution (a) Substances X and Y are covalent molecules with covalent bonds. Substance Z is an ionic compound with a giant network of ions held by ionic bonds. (b) Substance X consists of simple covalent molecules held by weak intermolecular forces of attraction that requires little energy to overcome them. (c) The bonds in substance Y are formed by the sharing of valence electrons to achieve the stable octet electron arrangement. The bonds in substance Z are formed by the transfer of electrons to achieve the stable octet electron arrangement. (d) Substances X and Y are insoluble in water. Substance Z is soluble in water. 125 Chemical Bonds 5 Uses of Covalent Compounds as Organic Solvents • Covalent bonds formed between atoms are very strong. When a covalent compound is heated, heat energy is used to overcome the weak intermolecular forces of attraction which is the van der Waals forces of attraction. The covalent bonds are not broken. • Some covalent compounds are polar (examples: ethanol, hydrogen chloride, ammonia) and can dissolve in water. • Some ionic compounds are insoluble in water (examples: Al2O3, PbCl2). 1 Organic solvents are liquids consisting of simple covalent compounds. 2 Examples of organic solvents are ethanol, ether, acetone, benzene, turpentine, petrol and tetrachloromethane. 3 Organic solvents are used (a) to clean or remove stains that cannot be cleaned by water. (b) to extract organic compounds. (c) as solvents for drugs in medicine or compounds used in cosmetics. 4 Some common useful organic solvents and their uses are given in Table 5.5. Property Table 5.5 Uses of some organic solvents Organic solvents Uses Petrol, kerosene and turpentine To remove grease and paint stains Ethanol, acetone To prepare medicated solution, perfume and ink Acetone, turpentine To dissolve varnish compounds such as shellacs, lacquer and paints Ether Ionic compound Simple molecule Giant molecule High Low High (b) Volatility Non-volatile Very volatile Non-volatile (c) Solubility Soluble in water, insoluble in organic solvents Soluble in organic solvents, insoluble in water Insoluble in both water and organic solvents (d) Electrical conductivity Conducts electricity in liquid and in aqueous solution Does not conduct electricity in any state (a) Melting and boiling points To extract organic compounds Covalent compound 5.4 2 The table shows the number of valence electrons of three elements represented by the letters Q, R and S. Aluminium oxide, silicon dioxide, bromine, ethanol, ammonium nitrate, barium sulphate, naphthalene, tetrachloromethane, hydrogen iodide, copper(II) chloride 1 Based on the chemicals in the list above, choose (a) two compounds with high melting points. (b) two compounds which can conduct electricity in the aqueous state. (c) one compound which dissolves in water but cannot conduct electricity. (d) one covalent compound with a high melting point. (e) two compounds which can be used as organic solvents. (f) one compound which cannot dissolve in water but can conduct electricity in the molten state. Chemical Bonds Element Q R S No. of valence electrons 2 4 7 Choose two elements which can combine to form a compound that (a) has low melting and boiling points, (b) conducts electricity in the liquid state, (c) is soluble in water. 3 Explain clearly why carbon dioxide is a gas whereas magnesium chloride is a solid at room temperature. Which compound can conduct electricity in the liquid state? Explain your answer. 126 (b) Double bond: Sharing of two pairs of electrons between two atoms. (c) Triple bond: Sharing of three pairs of electrons between two atoms. 14 For a covalent compound formed by a non-metal element P combined with another non-metal element, Q, the formula of the covalent compound formed will be PYQX. where 1 Inert gases are non-reactive because they have either a stable duplet or octet electron arrangement. 2 Inert gases do not release, accept or share electrons with other elements. 3 Atoms of elements from Group 1 to Group 17 have less than eight electrons. These elements will form chemical bonds. 4 Ionic bonds are formed between atoms of metals and atoms of non-metals. 5 Metal atoms from Groups 1, 2 and 13 will release their valence electrons to achieve the stable duplet or octet electron arrangement. 6 A positive ion (cation) is formed when an atom releases electrons. 7 Non-metal atoms from Groups 15, 16 and 17 will accept electrons to achieve the stable octet electron arrangement. 8 A negative ion (anion) is formed when an atom receives electrons. 9 The positive ions and negative ions are attracted to each other by strong electrostatic force of attraction in ionic bonds. 10 The total positive charge of the cation must be equal to the total negative charge of the anion in an ionic compound. 11 For an ionic compound consisting of cations Mb+ and anions Xa–, the formula of the ionic compound formed between them is written as Number of electrons required by Q to achieve duplet or octet electron arrangement Number of electrons required by P to achieve duplet or octet electron arrangement 15 The differences in physical properties between ionic compounds and covalent compounds: Property Melting point and boiling point MaXb Number of electrons that will be received by X (or charge of X ion) 5 PYQX Number of electrons that will be released by M (or charge of M ion) 12 A covalent bond is formed between atoms of nonmetals. 13 Non-metal atoms share valence electrons to achieve the stable duplet or octet electron arrangement in covalent compounds. There are three types of covalent bond. (a) Single bond: Sharing of one pair of electrons between two atoms. 127 Ionic compound Covalent compound (simple molecule) High Low Volatility Non-volatile Volatile Solubility Soluble in water, insoluble in organic solvents Soluble in organic solvents, insoluble in water Electrical conductivity Conducts electricity in molten form and in aqueous solution Does not conduct electricity in any state Chemical Bonds 5 Multiple-choice Questions 5 5.1 Formation of Compounds 1 Which of the statements below best explains why elements of Group 18 of the Periodic Table such as neon, argon and krypton are chemically inert? A Exists as monatoms B Have 8 valence electrons C Have 8 electrons in every electron shell D Have weak covalent bonds between atoms 2 Which of the following proton numbers belong to an element that does not form chemical bonds? I 8 II 10 III 18 IV 20 A I and II only B I and III only C II and III only D II and IV only 3 Which of the following statements best explains the formation of chemical bonds between elements? A The elements have high reactivities. B The atoms of the elements have less than eight valence electrons. C The atoms of the elements have too many electrons. D The elements are either electropositive or electronegative. 5.2 Formation of Ionic Bonds 4 Oxide and fluoride ions have the same number of ’03 [Proton number: O, 8; F, 9] A charge C neutrons B electrons D protons Chemical Bonds 5 Which of the following substances has particles bonded with very strong electrostatic forces? A Carbohydrate B Graphite C Napthalene D Magnesium sulphide 6 The proton numbers of three elements represented by the letters P, Q and R are given as follows: P: 4 Q: 13 R: 16 What will be the charges of the most stable ions of P, Q and R? P ion Q ion R ion A +4 +3 +6 B +2 +3 +6 C +2 +3 –2 D +4 –3 +2 7 The table shows the proton numbers for four elements R, S, T and U. Element R S Proton number 8 11 13 20 T U Which of the following ions have the same number of electrons as the sulphide ion? [Proton number of S = 16] C T 3+ A U 2+ + B S D R 2– 8 The proton number of sodium and oxygen are 11 and 8 respectively. ’07 Which of the following occurs when sodium metal is burned in air to produce sodium oxide? A One oxygen atom receives one electron from one sodium atom. B One oxygen atom receives two electrons from one sodium atom. C One oxygen atom receives two electrons, one from each sodium atom. D One sodium atom and one oxygen atom share two electrons. 128 9 Element P and element Q are located in Group 2 and ’10 Group 17 in the Periodic Table respectively. Element P reacts with element Q to form a compound. What is the chemical formula of the compound? A PQ B P2Q C PQ2 D P2Q3 10 Element P reacts with element Q to form an ionic compound with formula P2Q3. If the electronic configuration for Q is 2.8.6, what is the possible electron arrangement of P? A 2.8.2 C 2.8.5 B 2.8.3 D 2.8.6 11 An ionic nitride compound has formula X3N2. What is the possible proton number of atom X if nitrogen is in Group 15 of the Periodic Table? A 2 C 13 B 12 D 15 12 Which of the following represents the valence electron arrangement ’06 for the compound sodium oxide? [Proton number: O, 8; Na, 11] A B C D What is the nucleon number of element J ? A 18 C 38 B 20 D 40 14 G is an element in Group 2 and X is an element in Group 16. Which of the following is the formula of the compound formed by G and X? A GX C G3X D GX2 B G2 X6 5.3 Formation of Covalent Bonds 15 The diagram shows the electron arrangement of an atom of element X. 18 The following diagram shows the electron arrangements of atoms ’11 X and Y. X and Y are not the actual symbols of the elements. X Y Which pair of formula and the type of compounds is correct? Formula Type of compound A XY3 Ionic B XY3 Covalent C X3Y Ionic D X3Y Covalent 19The electronic configuration of the atoms of four elements represented by the letters P, Q, R and S are shown in the table below. Element P Q R S Electronic 1 2.7 2.8 2.8.8.2 configuration The atom of element X can form a covalent bond with another atom through the A acceptance of two electrons. B donation of two electrons. C sharing of two pairs of electrons. D sharing of six electrons. 16 An element forms a covalent compound with hydrogen. Atoms of this element also form covalent molecules themselves. This element may be I sodium III chlorine II oxygen IV nitrogen A I and II only B III and IV only C I, II and IV only D II, III and IV only 17 How many pairs of electrons are shared by the oxygen atoms in a molecule of oxygen gas? [Proton number: O, 8] A 1 C 3 B 2 D 4 Which of the following pairs of elements will form a covalent compound? A P and Q C Q and S B Q and R D P and R 20 Which of the following compounds have covalent double bonds? I O2 III CCl4 II N2 IV CO2 A I and II only B III and IV only C I and IV only D I, II and IV only 21 The diagram shows the electron arrangement in a compound ’05 formed by elements X and Y. X Y A 2 14 B 4 6 C 14 16 D 18 18 22 Element R reacts with element S to produce a covalent compound with the formula R2S3. Which of the following electron arrangements of atoms R and S are true? Electron Electron arrangement arrangement of atom R of atom S A 2.8.2 2.3 B 2.8.2 2.5 C 2.8.3 2.6 D 2.8.5 2.6 23 The electronic configuration of atom E is 2.8.6 and atom G has ’03 four valence electrons. What is the formula of the compound formed between E and G? A GE2 C G2E B G4E2 D G2E4 24 The diagram shows the symbols of atoms P and Q. 28 16 P 14 Q 8 Atom P reacts with atom Q to form a compound. Calculate the relative molecular mass for the compound formed. A 21 C 60 B 43 D 113 5.4 Properties of Ionic and Covalent Compounds 25 Element W and element Y have proton numbers 17 and 19 as shown below. Which of the following is the correct groups of X and Y in the Periodic Table of Elements? 129 Chemical Bonds 5 13 The following diagram shows the electron arrangement for the J 2+ ’07 ion. An atom of element J has 20 neutrons. 5 Which of the following are true of the compound formed between element W and element Y? I Has high melting point II Dissolves in organic solvents III Dissolves in water IV Conducts electricity in the solid state A I and III only B II and IV only C III and IV only D I, III and IV only 26 Which of the following elements will form an oxide that conducts electricity when dissolved in water? A Potassium B Silicon C Hydrogen D Aluminium 27 The electronic configuration of element X is 2.5. Which of the following statements are true of element X? I X forms a hydride with formula XH3. II Molecule X has a triple bond. III X forms an oxide with high melting point. IV Chlorine and X form a covalent compound. A I and III only B II and IV only C III and IV only D I, II and IV only 28 Which of the following statements best explains why bromine liquid is very volatile? A Bromine consists of diatomic covalent molecules. B Intermolecular forces of attraction between bromine molecules are weak. C Size of bromine molecules is small. D The covalent bond in bromine molecules is weak. 29 Covalent compounds do not conduct electricity because A they do not have free moving ions. B they have strong covalent bonds. Chemical Bonds C they are insoluble in water. D they can become vapour easily when heated. 30 Covalent compounds have low melting and boiling points because A they have weak covalent bonds. B they have weak intermolecular forces of attraction. C they are very volatile. D they cannot withstand strong heating. 31 Two elements represented by the letters X and Y have proton numbers as given in the table. Element X Y Proton number 16 20 From the information given in the table, it can be deduced that I X and Y will form a compound with formula XY2. II Y can form ionic compounds but X can form both ionic and covalent compounds. III element X is a non-conductor of electricity whereas element Y is a conductor of electricity. IV the compound formed between element X and element Y has a high melting point. A I and II only B III and IV only C I, II and III only D II, III and IV only 32 Both carbon and magnesium form compounds with oxygen. Which of the following is the difference between magnesium oxide and carbon dioxide? A Magnesium oxide is soluble in water whereas carbon dioxide is insoluble in water. B Magnesium oxide solid conducts electricity whereas carbon dioxide solid does not. C Magnesium oxide is basic whereas carbon dioxide is neutral. 130 D Magnesium oxide has a high melting point whereas carbon dioxide has a low melting point. 33 Element X and element Y form a compound that has a low melting point and does not conduct electricity. Element X and Y maybe I hydrogen and sulphur II chlorine and oxygen III sodium and sulphur IV potassium and oxygen A I and II only B I and III only C II and IV only D III and IV only 34 Which of the following compounds have a high melting point and are soluble in water? I Calcium chloride II Sulphur dichloride III Copper(II) chloride IV Tetrachloromethane A I and II only B III and IV only C I and III only D II and IV only 35 Element X has an electronic configuration of 2.8.2. Which of the following are true of element X? I It can form negative ions. II It is a metallic element. III It can form a covalent compound with chlorine. IV It can form a basic oxide. A I and II only B II and IV only C I and III only D III and IV only 36 Which of the following compounds can be used as solvents for covalent compounds? I Water II Ethanol III Benzene IV Tetrachloromethane A I and II only B II and IV only C III and IV only D II, III and IV only Structured Questions 1 Diagram 1 shows the Periodic Table of elements. The letters P, Q, R, S, T, U, V and W represent elements and are not the symbols of the actual elements. 1 2 13 14 15 16 17 18 1 2 P 3 R 4 V Q U S T W 5 5 6 Diagram 1 Answer the following questions with reference only to the letters given above. (a) (i) State the type of chemical bond found in the compound formed between R and T. (ii) Draw a diagram showing the electron arrangement of the compound formed in (i). (iii) State a physical property of the compound formed in (i). [1 mark] [2 marks] [1 mark] (b) (i) Write the electronic arrangement of an ion formed by element Q. (ii) Q exists as diatomic molecules. Draw a diagram that shows the electron arrangement of the diatomic molecule. [1 mark] [2 marks] (c) Can element U form a compound? Explain your answer. [2 marks] (d) Write the equation for the reaction between V and water. [1 mark] 2 Diagram 2 shows two elements represented by the letters P and Q. (a) Write the electron arrangement of atom P and atom Q. (b) State the groups of elements P and Q in the Periodic Table. 16 23 P [2 marks] Q 11 8 [2 marks] (c) Write an equation for the formation of (i) ion P. (ii) ion Q. Diagram 2 [2 marks] (d) P can combine with Q to form a compound. (i) Write the formula of the compound formed. (ii) What type of bond is formed in this compound? (iii) Draw the electron arrangement of this compound. [1 mark] [1 mark] [1 mark] (e) Q can combine with itself to form a compound. State a physical property of the compound formed. [1 mark] 3 Table 1 shows a few properties of substances P, Q, R and S. Conductivity Substance Melting point (°C) Solubility in water Solubility in organic solvent Solid state Liquid state P 114 Insoluble Soluble Non-conducting Non-conducting Q 782 Soluble Insoluble Non-conducting Conducting R 840 Insoluble Insoluble Conducting Conducting S 2750 Insoluble Insoluble Non-conducting (Does not exist in the liquid state) Table 1 Based on the information given in Table 1, answer the questions below. (a) What type of bond is found in the particles of (i) substance P? (ii) substance Q? (b) Explain why P has a low melting point. [1 mark] [1 mark] [2 marks] 131 Chemical Bonds (c) Name a substance that has the same physical property as Q. [1 mark] (d) (i) State two atoms in Table 2 that are elements in the same group in the Periodic Table. (d) Which of the above substances is a metal? (e) Z can react with heated iron to form a compound. (i) Write the formula of this compound. (e) Give reasons why Q conducts electricity in the liquid state but does not do so in the solid state. [2 marks] [1 mark] (ii) State one compound. 5 (f) Suggest one difference between the types of particles in substance R and in substance S. Give an example each for substance R and substance S. [2 marks] physical property of this [1 mark] 5 Diagram 3 shows the diagram of the valence electron arrangement of a compound formed by two types of atoms represented by the letters P and Q. 4 Table 2 shows the number of protons and neutrons for a few atoms represented by the letters V, W, X, Y and Z. Atoms [1 mark] [1 mark] (ii) Explain your answer in (i). [1 mark] Number of protons Number of neutrons V 3 4 W 6 6 X 6 7 Y 9 10 Z 17 18 Diagram 3 Table 2 (a) What is the formula of this compound? [1 mark] Use the information in Table 2 to answer the questions below. (b) Predict the groups to which the atoms P and Q belong to in the Periodic Table. [2 marks] (a) What is the nucleon number of element Y? [1 mark] (c) Name the type of bond formed in the compound shown in Diagram 3. [1 mark] (b) W can combine with Y to form a compound. (i) Name the type of chemical bond found in this compound. [1 mark] (ii) Write the formula of this compound. [1 mark] (iii) Predict the relative molecular mass of this compound. [1 mark] (d) State three physical properties of this compound. [3 marks] (e) Q is an element that exists as diatomic molecules. (i) Draw a diagram to show the valence electron arrangement of the molecule. [2 marks] (ii) Explain why element Q has a low boiling point. [1 mark] (c) Name two atoms that are isotopes. Explain your answer. [2 marks] Essay Questions 1 Diagram 1 shows the chemical symbols for three elements: carbon, sodium and chlorine: Diagram 1 (a) Construct a table to compare the three elements in terms of the number of protons, the number of neutrons, the electron arrangement and the number of valence electrons. [4 marks] (b) Carbon reacts with chlorine to form a covalent compound whereas sodium reacts with chlorine to form an ionic compound. Explain how these ionic and covalent compounds are formed. [8 marks] (c) State four differences in physical properties between the two types of compounds formed in (b). [4 marks] Chemical Bonds 132 2 Group Period W 1 3 Y 14 2 Z 17 3 Element Table 1 (a) With the help of an electron arrangement diagram, explain how two elements from Table 1 can combine to form (i) an ionic compound, and (ii) a covalent compound. [14 marks] (b) State the differences between an ionic compound and a covalent compound. [15 marks] (b) Explain why ionic compounds can conduct electricity in the liquid state and in aqueous solution whereas covalent compounds cannot. [5 marks] Experiments 1 A group of students carried out an experiment to determine the types of particles in three compounds X, Y and Z. The results obtained is recorded in Table 1. Chemical compound Physical state X Observation when shaken with Water Acetone Solid Forms a colourless solution X does not dissolve in acetone Y Liquid Forms two immiscible layers Y and acetone mix completely Z Liquid Z and water mix completely Z and acetone mix completely Potassium chloride Table 1 (a) If the experiment is repeated using potassium chloride, predict what will be observed and complete Table 1. [3 marks] (b) Deduce the solubility of X, Y and Z in water and in acetone. [3 marks] (c) Classify the four chemical compounds used in Table 1 into (i) ionic compound, (ii) covalent compound. [3 marks] (d) Potassium nitrate is used as a nitrogenous fertiliser in agriculture. Which of the compounds X, Y and Z in the experiment may be potassium nitrate? Explain your answer. [3 marks] (e) Compound Z is used as a solvent in medicine and perfume. Name a substance that may be compound Z. Explain your answer. [3 marks] 2 Covalent compounds and ionic compounds differ in their ability to conduct electricity in the liquid state. By referring to the statement above, design a laboratory experiment relating to the difference in electrical conductivity of ethanol and sodium nitrate. In designing your experiment, the following aspects must be included: (a) Problem statement (b) Hypothesis (c) All the variables (d) List of substances and apparatus (e) Experimental procedures (f) Tabulation of data [17 marks] 133 Chemical Bonds 5 [6 marks] 3 (a) Using suitable examples, explain the meaning of single covalent bond, double covalent bond and triple covalent bond. FORM 4 THEME: Interaction between Chemicals CHAPTER 6 Electrochemistry SPM Topical Analysis 2008 Year 1 Paper Section Number of questions 5 2009 2 A B C – 1 – 3 1 – 6 2010 2 A B C – 1 – 3 1 1 4 2011 2 A B C 1 – 1 3 1 – 5 3 2 A B C 1 – – – ONCEPT MAP ELECTROCHEMISTRY types of substances Electrolytes and non-electrolytes conversion of electrical energy into chemical energy conversion of chemical energy into electrical energy Electrolysis Electrolysis of molten compounds Electrolysis of aqueous solution products • Metals are formed at the cathode • Non-metals are formed at the anode Voltaic cells products Selective discharge of ions determined by: • Position of ions in the electrochemical series • Concentration of ions • Types of electrodes Uses of electrolysis: • Extraction of metals • Purification of metals • Electroplating of metals can be used to determine Various types of voltaic cells: • non-rechargeable cell and • rechargeable cell To construct the electrochemical series: • Potential difference between two different metals • Displacement reaction of metals Electrochemical series Uses of the electrochemical series: • Predict the products of electrolysis • Determine the terminals of cells • Prediction of displacement reactions 6 Metals which are conductors are not regarded as electrolytes because they are not decomposed by the passage of an electric current. 7 The electrical conductivity of any substance is based on the movement of charged particles. (a) In metals, electricity is conducted by freely moving electrons. Introduction 1 Electrochemistry is the study of the inter conversion of chemical energy and electrical energy. 2 The energy change in electrochemistry consists of the (a) conversion of electrical energy into chemical energy (in electrolysis), (b) conversion of chemical energy into electrical energy (in voltaic cells). 8 Most covalent substances are non-electrolytes because they consist of molecules, which do not form ions in aqueous solution. 9 Some covalent substances such as hydrogen chloride and ammonia are electrolytes because they react with water to produce freely moving ions. Examples: Electrolytes and NonSPM electrolytes ’08/P1 1 An electrolyte is a chemical compound which conducts electricity in the molten state or in an aqueous solution and undergoes chemical changes. 2 When an electrolyte conducts electricity, chemical changes occur and the electrolyte decomposes into its component elements at the electrodes. Electrolysis is said to have taken place. 3 Examples of electrolytes are acid solutions, alkali solutions, molten salts or aqueous salt solutions. 4 A non-electrolyte is a chemical compound which does not conduct electricity in any state. 5 Examples of non-electrolytes are metals and covalent substances such as naphthalene, latex and sugar solution. 6 6.1 (b) In electrolytes, electricity is conducted by freely moving ions: positive ions and negative ions. HCl(g) + H2O(l) → H3O+(aq) + Cl–(aq) NH3(g) + H2O(l) → NH4+(aq) + OH–(aq) 10 Ions are charged (positive or negative) particles. All metal ions and hydrogen ions are positive ions (also known as cations). All non-metal ions are negative ions (also known as anions). All electrolytes will dissociate into cations and anions in the molten states or aqueous solutions. 6.1 Problem statement How to identify electrolytes and non-electrolytes? Hypothesis Substances that, in the molten state or in aqueous solution, conduct electricity and then undergo chemical reactions are electrolytes. Substances that do not conduct electricity in any state are non-electrolytes. Apparatus Crucible, spatula, carbon (graphite) electrodes, batteries, light bulb, switch, rheostat, connecting wires, tripod stand, clay pipe triangle and Bunsen burner. Materials Glucose, naphthalene, lead(II) bromide and potassium iodide. Procedure (A) To investigate the electrical conductivity of substances in the solid state and in the molten state 1 A crucible is half-filled with lead(II) bromide solid. Variables (a) Manipulated variable : Types of compounds (b) Responding variable : Electrical conductivity (c) Constant variable : Numbers of batteries, type of light bulb and amount of substance used 135 Electrochemistry Experiment 6.1 To identify electrolytes and non-electrolytes 6 2 The crucible with its contents is placed on a clay triangle on a tripod stand. 3 Two carbon electrodes are dipped in the lead(II) bromide solid and are connected to the batteries, rheostat, switch and a light bulb with connecting wires (Figure 6.1). 4 The switch is turned on and the light bulb is checked if it lights up. 5 The lead(II) bromide solid in the crucible is heated up until it melts. The switch is turned on again to check if the light bulb lights up. 6 Steps 1 to 5 of the experiment are repeated using naphthalene in place of lead(II) bromide. (B) To investigate the electrical conductivity of substances in the solid state and in aqueous solutions 1 Three spatulas of potassium iodide solid are put in a beaker. 2 Two carbon electrodes are dipped in the potassium iodide solid and then connected to the batteries, rheostat, switch and a light bulb with connecting wires (Figure 6.2). 3 The switch is turned on and the light bulb is checked if it lights up. 4 Distilled water is added to the beaker and the mixture is stirred until all the potassium iodide has dissolved. 5 The switch is turned on again and the light bulb is checked if it lights up. 6 Steps 1 to 5 of the experiment is repeated using glucose in place of potassium iodide. Figure 6.1 To investigate the conductivity of substances in the solid state and in the molten state Figure 6.2 To investigate the conductivity of substances in the solid state and in aqueous solution Results Chemical substance Lead(II) bromide Naphthalene Potassium iodide Glucose Physical state Does the light bulb light up? Inference Solid No No noticeable change Non-electrolyte Liquid (molten) Yes Brown gas is evolved Electrolyte Solid No No noticeable change Non-electrolyte Liquid (molten) No No noticeable change Non-electrolyte Solid No No noticeable change Non-electrolyte Aqueous solution Yes Solution turns to a brown colour Electrolyte Solid No No noticeable change Non-electrolyte Aqueous solution No No noticeable change Non-electrolyte Disscussion 1 Electrolytes can conduct electricity because they have freely moving charged ions. 2 In the solid state, an ionic compound is not an electrolyte because the ions are held together by strong ionic bonds and are not free to move. 3 When an ionic compound is heated to its melting point, the heat energy supplied overcomes Electrochemistry Observation: Does reaction occur? the ionic bond. The ions are free to move in the molten state. Hence ionic compounds are electrolytes in the molten state. 4 When ionic compounds are dissolved in water, the water molecules separate the ions into freely moving ions. Hence aqueous solutions of ionic compounds are electrolytes. 136 5 Covalent compounds such as glucose and naphthalene do not conduct electricity because they consist of molecules which are uncharged particles. 6 When electricity passes through molten lead(II) bromide, the brown gas evolved is bromine gas. 7 When electricity passes through aqueous potassium iodide solution, the brown iodine solution is formed. 8 The presence of freely moving ions enable a molten compound or aqueous solution to conduct electricity. 6 Conclusion 1 Lead(II) bromide is an electrolyte in the liquid state but not in the solid state. 2 Potassium iodide is an electrolyte in aqueous solution but not in the solid state. 3 Lead(II) bromide and potassium iodide are ionic compounds. Ionic compounds are electrolytes in the molten state or aqueous solution but are non-electrolytes in the solid state. 4 Naphthalene and glucose are covalent compounds and are non-electrolytes in any state. 6.2 All liquid or aqueous solutions of ionic compounds are electrolytes. All covalent compounds (except those that can dissociate to form ions when dissolved in water) are non-electrolytes. An electrolyte can conduct electricity because it has freely moving ions. Electrolysis of Molten Compounds Meaning of Electrolysis and Electrolytic Cell • Electrolysis is a process of decomposition of an electrolyte by an electric current. • An electrolytic cell consists of batteries, a cathode, an anode and an electrolyte consisting of cations and anions. • During electrolysis, anions (negatively-charged ions) move towards the anode and cations (positively-charged ions) move towards the cathode. • Graphite or platinum is usually used as electrodes because they are inert; they do not react with the electrolyte or the products of electrolysis. 6.1 1 (a) (i) What is meant by the term electrolyte? Give two examples of electrolytes. (ii) What is meant by the term non-electrolyte? Give two examples of non-electrolytes. (b) State the difference between a conductor and an electrolyte. 2 Classify the ten substances below into electrolytes and non-electrolytes. Molten sulphur, molten zinc oxide, zinc oxide solid, aqueous zinc chloride solution, zinc metal, molten zinc, acetone, aqueous glucose solution, aqueous ethanoic acid solution, molten sodium chloride. 3 Explain why magnesium chloride solid cannot conduct electricity but becomes an electrolyte when it is in the molten state. Electrons flow from the anode to the cathode through the connecting wire in the external circuit. A rheostat can be used to control the quantity of electric current that flows through the electrolyte. An ammeter or a bulb can be used to indicate the flow of electric current. • Anode is the electrode connected to the positive terminal of the batteries. • At the anode, anions discharge by donating electrons. Example: 2Cl– → Cl2 + 2e– • Cathode is the electrode that is connected to the negative terminal of the batteries. • At the cathode, cations discharge by accepting electrons, Example: Na+ + e– → Na An electrolyte conducts electricity in aqueous solution or in the molten state as a result of the presence of freely moving cations (positive ions) and anions (negative ions). Figure 6.3 Electrolytic cell 137 Electrochemistry Electrolysis Process 1 In the solid state, the cations and anions of an electrolyte are unable to move freely because they are held together by strong ionic bonds in fixed positions in a lattice. 6 3 Generally, a molten compound produces Am+ cations and Bn– anions. SPM ’09/P1 2 When the solid is heated until it melts (in the molten form), the heat energy supplied is used to overcome the electrostatic force of attraction between the ions. Hence, the cations and the anions are free to move. AnBm AnBm(s) → nAm+(l) + mBn–(l) cation anion 4 Two steps occur during electrolysis. (a) Movement of ions to the electrodes Cations (positive ions) move towards the cathode (negative electrode) whereas anions (negative ions) move towards the anode (positive electrode). (b) Discharge of ions at the electrodes Cations are discharged by accepting (gaining) electrons. Generally: An+ + ne– → A Anions are discharged by donating (losing) electrons. Generally: Bn– → B + ne– Example: PbBr2(s) → Pb2+(l) + 2Br–(l) lead(II) ion bromide ion • Examples of cations are hydrogen ions, H+, ammonium ions, NH4+ and metal ions such as K+, Mg2+, Pb2+ and Al3+. • Examples of anions are Cl–, Br–, I–, OH–, SO42– and NO3–. 5 The electrons donated by anions at the anode are accepted by the cations at the cathode. The discharge of ions at the anode and cathode results in the (a) conduction of electricity by the electrolyte. (b) decomposition of the electrolyte into its component elements. Writing Half-equations for the Discharge of Ions at the Anode and the Cathode 4 Addition of the two half-equations at the anode and cathode gives the overall chemical equation for the reaction. 1 The reactions that take place at the anodes or the cathodes involve ions and electrons. The equations representing these reactions are known as half-equations. 2 The half-equation of the anode shows the discharge of the anions by the loss of electrons to produce neutral atoms. Half-equation at the anode: Half-equation at the anode: Bn– → B + ne– Half-equation at the cathode: Bn– → B + ne– An+ + ne– → A 3 The half-equation of the cathode shows the discharge of the cations by the gain of electrons to produce neutral atoms. Half-equation at the cathode: Electrochemistry ………… equation 1 ………… equation 2 Overall chemical reaction equation: An+ + Bn– → A + B …… equation 1 + equation 2 An+ + ne– → A 138 1 Step 2: Balance the number of atoms on both sides of the equation. Write the half-equation for the discharge of oxide ions, O2– at the anode. 2O2– → O2 Solution Step 1: Write the formulae of the reactant and the product of the reaction. Step 3: Balance the charge by adding in the right number of electrons. O2– → O2 6 2O2– → O2 + 4e– 2 When molten aluminium oxide, Al2O3 is electrolysed, aluminium metal and oxygen gas are produced. (a) Write half-equations for the reactions at the anode and cathode. (b) Write the overall chemical equation of electrolysis. At the cathode: Cations receive electrons Al3+ + 3e– → Al ……… (2) (Number of electrons donated at the anode = number of electrons accepted at the cathode) Solution At the anode: Anions lose electrons Equation (1)  3: 6O2– → 3O2 + 12e– Equation (2)  4: 4Al3+ + 12e– → 4Al Overall chemical equation: 4Al3+(l) + 6O2–(l) → 4Al(l) + 3O2(g) 2O2– → O2 + 4e– ……… (1) To investigate the electrolysis of molten lead(II) bromide SPM ’06/P2 Apparatus Crucible, spatula, graphite electrodes, batteries, light bulb, ammeter, switch, rheostat, connecting wires, tripod stand, clay pipe triangle and Bunsen burner. Procedure 1 A crucible is half-filled with lead(II) bromide solid. 2 The solid lead(II) bromide is heated strongly until it melts to a molten state. 3 Two carbon electrodes are dipped in the molten lead (II) bromide and are then connected to batteries, rheostat, switch and light bulb by the connecting wires (Figure 6.4). 4 Electric current is allowed to flow through for 15 minutes and the changes that occur at the light bulb, ammeter, cathode and anode are recorded. Figure 6.4 To investigate the electrolysis of lead(II) bromide 139 Electrochemistry Activity 6.1 Materials Lead(II) bromide. Results 6 Apparatus Observation Inference Light bulb Light bulb lights up Molten lead(II) bromide conducts electricity Ammeter Ammeter needle is deflected Anode Pungent brown gas that changes damp blue litmus paper to red is evolved Bromine gas is evolved Cathode Shiny grey metal is deposited Lead metal is formed 5 At the cathode, a lead(II) ion discharges by accepting 2 electrons to form a lead metal atom. Discussion 1 Solid lead(II) bromide consists of lead(II) ions, Pb2+ and bromide ions, Br– that are held by strong ionic bonds. 2 When heated until it melts to form the molten state, lead(II) ions and bromide ions are free to move. Pb2+ + 2e– → Pb 6 A larger quantity of the products would be formed at the anode and the cathode if the electroysis is carried out (a) for a longer time period, (b) with a larger current. PbBr2(s) → Pb2+(l) + 2Br –(l) 3 During electrolysis, bromide ions are attracted to the anode while lead(II) ions are attracted to the cathode. 4 At the anode, a bromide ion discharges by releasing one electron to form a bromine atom. Two bromine atoms combine to form a bromine molecule that exists as a brown gas. Conclusion 1 The lighting up of the bulb and the deflection of the ammeter needle shows that molten lead(II) bro mide is an electrolyte and can conduct electricity. 2 Electrolysis of molten lead(II) bromide produces bromine gas at the anode and lead metal at the cathode. 2Br – → Br2 + 2e– formed at the cathode while the non-metal component is formed at the anode. 2 Products of electrolysis of molten electrolytes can be predicted by the following steps. Prediction of the Products of Electrolysis of Molten Electrolytes 1 When molten compound is electrolysed, the metal component of the compound is Step 1 Step 2 Step 3 Identifying the cations and anions that are present in the molten electrolyte. Generally, a salt with formula AxBy will produce ions as below. Identify the cathode and anode and the movement of ions in the electrolyte to the electrodes. • Cations move to cathode (electrode connected to the negative terminal of the battery). • Anions move to anode (electrode connected to the positive terminal of the battery). Write the half-equations for the discharge of the ions at the electrode. • At the cathode: Cations discharge (losing the positive charge) by accepting electrons. AxBy → xAy+ + cation Electrochemistry yBx– anion 140 Ay+ + ye– → A • At the anode: Anions discharge (losing the negative charge) by donating electrons. Bx– → B + xe– Electrolysis of molten sodium chloride, NaCl Electrolysis of molten lead(II) oxide, PbO 1 Cations and anions produced from the molten lead(II) oxide are lead(II) ions, Pb2+ and oxide ions, O2–. 1 The cations and anions produced from molten sodium chloride are sodium ions, Na+ and chloride ions, Cl–. PbO → Pb2+ + O2– NaCl → Na+ + Cl– 2 Lead(II) ions are attracted to the cathode while the oxide ions are attracted to the anode. 3 At the cathode, a lead(II) ion accepts 2 electrons to form a lead metal atom (grey metal). 2 The sodium ions are attracted to the cathode while the chloride ions are attracted to the anode. 3 At the cathode, a sodium ion accepts an electron to form a sodium metal atom (grey metal). 6 Pb2+ + 2e– → Pb 4 At the anode, an oxide ion donates 2 electrons to form an oxygen atom. Two oxygen atoms combine to form an oxygen gas molecule. Na+ + e– → Na 2O2– → O2 + 4e– 4 At the anode, a chloride ion donates an electron to form a chlorine atom. Two chlorine atoms combine to form a chlorine molecule (greenish-yellow gas). The term electrolysis was introduced by Michael Faraday. ‘Lysis’ means loosening in Greek. Electrolysis means loosening by electric current. 2Cl– → Cl2 + 2e– 6.2 1 Write the formulae of the ions produced in the molten chemical compounds in the table below. Identify the ions that move to the anode and to the cathode respectively during electrolysis by completing the table below. Name of compound Ions that move to the Ions produced anode cathode (a) Zinc chloride (b) Magnesium oxide 2 For every compound below, write the half-equations that occur at the anode and the cathode, as well as the products formed during electrolysis. Name of compound Half-equation at the Products formed at the anode anode cathode cathode (a) Calcium oxide (b) Aluminium iodide 3 The figure shows the arrangement of apparatus used in an experiment. The switch is turned on for 20 minutes. (a) Predict the observations at (i) electrode P, (ii) electrode Q. (b) What are the products formed at (i) electrode P, (ii) electrode Q? (c) Write the half-equations for the reactions that take place at (i) electrode P, (ii) electrode Q. 141 Electrochemistry 6.3 Electrolysis of Aqueous Solutions (A) Positions of ions in the electrochemical series 1 The tendency of ions to be selectively discharged at electrodes depends on their positions in a series known as the electrochemical series. 2 The tendency of the ions to be discharged is shown below in ascending order. 6 Identifying Cations and Anions in Aqueous Solutions 1 A molten salt consists of one type of cation and one type of anion only. During electrolysis, the products formed are the component elements of the compound. 2 An aqueous solution is produced when a solute is dissolved in water. An aqueous solution of a salt consists of two types of cations (cations of the salt and hydrogen ions, H+) and two types of anions (anions of the salt and hydroxide ions, OH–). 3 H+ ions and OH– ions are always present together with the ions produced from the dissociation of salts in aqueous solutions. This is because water dissociates partially to form hydrogen ion and hydroxide ion. H2O Cation K+ Na+ Mg2+ Al3+ Zn2+ Fe2+ Sn2+ Pb2+ H+ Cu2+ Hg+ Ag+ 3 The lower the position of the ion in the electrochemical series, the easier the ion will be discharged. 4 For example, in the electrolysis of aqueous sodium sulphate (Na2SO4) solution, with carbon electrodes, H+ + OH– 4 For example, in an aqueous sodium chloride solution, 4 types of ions are present in the solution from the dissociation of NaCl and H2O. NaCl ⎯→ H2O cation Na+ + H+ + Na2SO4 → 2Na+ + SO42– H2O H+ + OH– anion Cl– OH– The cations present are Na+ ions and H+ ions. The anions present are SO42– ions and OH– ions. 5 Both the Na+ ions and the H+ ions are attracted to the cathode. However, only the H+ ions are discharged at the cathode because the H+ ion is lower in position than the Na+ ion in the electrochemical series. Half-equation at the cathode: The four types of ions present are Na+, Cl–, H+ and OH–. 5 During electrolysis of an aqueous salt solution, 2 types of cations will move to the cathode while 2 types of anions will move to the anode. For example, during electrolysis of aqueous sodium chloride solution, the cations that move to the cathode are Na+ ions and H+ ions. The anions that move to the anode are Cl– ions and OH– ions. However, only one type of cation and anion will be discharged at each electrode. 2H+ + 2e– → H2 Hence, hydrogen gas is produced at the cathode. 6 Both the SO42– ions and the OH– ions are attracted to the anode. However, only the OH– ions are discharged at the anode because the OH– ion is lower in position than the SO42– ion in the electrochemical series. Half-equation at the anode: Factors that Determine the Selective Discharge of Ions at the Electrodes The factors that determine the types of ions to be discharged at the electrodes are: (a) Positions of ions in the electrochemical series (b) Concentration of ions in the solution (c) Types of electrodes used Electrochemistry tendency to discharge increases Anion F– SO42– NO3– Cl– Br– I– OH– 4OH– → 2H2O + O2 + 4e– Hence, oxygen gas is produced at the anode. 142 (C) Effect of types of electrodes used (B) Effect of concentration of ions in the solution SPM ’04/05 P2 1 The common materials used as electrodes are carbon and platinum because they are inert. Both of these materials do not react with the electrolytes or the products of electrolysis. 2 The types of electrodes used can determine the types of ions discharged in electrolysis. 3 For example, in the electrolysis of aqueous copper(II) sulphate (CuSO4) solution, 1 When the concentration of a particular type of ion is high, that ion will more likely to be discharged in electrolysis irrespective of its position in the electrochemical series. 2 Usually, in the electrolysis of concentrated halide (Cl–/Br–/I– ions) solutions, the concentration of the halide ion is always higher than the hydroxide ion, OH–. Hence, halide ions will be selectively discharged at the anode. 3 For example, in the electrolysis of aqueous copper(II) chloride solution, CuCl2 with carbon electrodes, 6 CuSO4 → Cu2+ + SO42– H2O H+ + OH– both the anions, sulphate ions, SO42– and hydroxide ions, OH– are attracted to the anode. (a) If carbon is used as the electrodes, OH– ions are discharged at the anode because of the position of OH– ion in the electrochemical series. Half-equation at the anode: CuCl2 → Cu2+ + 2Cl– H2O H+ + OH– 4OH– → 2H2O + O2 + 4e– Hence, oxygen gas is produced at the anode. (b) If copper is used as the anode, both SO42– ions and OH– ions are not discharged. Instead the copper anode dissolves by releasing electrons to form copper(II) ions, Cu2+. Half-equation at the anode: Both types of anions, chloride ion, Cl– and hydroxide ions, OH– are attracted to the anode. (a) If a concentrated solution is electrolysed, chloride ions, Cl–, will be selectively discharged at the anode because the concentration of the chloride ions is higher than that of the hydroxide ions. Half-equation at the anode: Cu → Cu2+ + 2e– Hence, the mass of anode decreases. Copper acts as an active electrode here because it takes part in the chemical reaction during electrolysis. 2Cl– → Cl2 + 2e– Hence chlorine gas is evolved at the anode. (b) If a dilute solution is electrolysed, hydroxide ions, OH–, will be discharged at the anode because the concentration of the chloride ions is low and OH– ion is more easily discharged because of its position in the electrochemical series. Half-equation at the anode: Generally, in the electrolysis of a halide solution using carbon electrodes: • A concentration of more than 0.5 mol dm–3 halide solution is concentrated solution whereby the halide ions will be selectively discharged at the anode. • A concentration of less than 0.005 mol dm–3 halide solution is considered a dilute solution whereby the hydroxide ions will be selectively discharged at the anode. • Electrolysis of a solution with a concentration of between 0.005 mol dm–3 and 0.5 mol dm–3 may result in two types of products: halogen and oxygen to be produced at the anode. 4OH– → 2H2O + O2 + 4e– Hence, oxygen gas is evolved at the anode. 143 Electrochemistry 6.2 To investigate the effect of the positions of ions in the electrochemical series on the selective discharge of ions and the products of electrolysis of aqueous solutions 6 Problem statement How do the positions of ions in the electrochemical series determine the types of ions selectively discharged during electrolysis? Procedure 1 Aqueous copper(II) sulphate solution is put into an electrolytic cell with carbon electrodes. 2 Two test tubes, filled with copper(II) sulphate solution are inverted over the carbon anode and cathode respectively (Figure 6.5). Hypothesis If an aqueous solution consists of more than one type of ion, the lower the position of the ion in the electrochemical series, the higher the tendency it is for the ion to be discharged. Variables (a) Manipulated variable : Position of ions in the electrochemical series (b) Responding variable : Types of ions discharged at the anode and the cathode (c) Constant variable : Concentration of electro lytes, types of electrodes, duration of electrolysis Apparatus Batteries, electrolytic cell, carbon electrodes, ammeter, switch, connecting wires with crocodile clips and test tubes. Figure 6.5 Electrolysis of copper(II) sulphate solution 3 The switch is turned on and electric current is allowed to flow for 15 minutes. 4 Steps 1 to 3 of the experiment are repeated by replacing copper(II) sulphate solution with dilute sulphuric acid and sodium nitrate solution in turn. Materials Aqueous 0.5 mol dm–3 copper(II) sulphate, CuSO4 solution, 0.5 mol dm–3 dilute sulphuric acid, H2SO4 and 0.5 mol dm–3 sodium nitrate, NaNO3 solution. Results Electrolyte Observation Copper(II) sulphate At the cathode solution • Brown deposit is formed Inference Copper metal is deposited At the anode • Gas bubbles are formed Oxygen gas is produced • Gas produced lights up a glowing wooden splint Experiment 6.2 Colour of electrolyte • The blue colour of the solution becomes paler Dilute sulphuric acid and sodium nitrate solution At the cathode Hydrogen gas is produced • Gas bubbles are formed • When a lighted wooden splint is placed near the mouth of the test tube, a ‘pop’ sound is produced At the anode •Gas bubbles are formed •Gas lights up a glowing wooden splint Electrochemistry Concentration of Cu2+ ion decreases 144 Oxygen gas is produced 8 The ions present in the sodium nitrate solution are Na+ ions, NO3– ions, H+ ions and OH– ions. Discussion 1 The ions present in the aqueous copper(II) sulphate solution are Cu2+ ions, SO42– ions, H+ ions and OH– ions. CuSO4 → Cu2+ + SO42– H2O H+ + OH– 9 Both types of cations, H+ ions and Na+ ions are attracted to the cathode. H+ ions are selectively discharged at the cathode because the position of the H+ ion is lower than that of the Na+ ion in the electrochemical series. Hydrogen ions are discharged to form hydrogen gas. Half-equation at the cathode: 2 Both types of cations, Cu2+ ions and H+ ions are attracted to the cathode. Cu2+ ions are selectively discharged at the cathode because the position of the Cu2+ ion is lower than that of the H+ ion in the electrochemical series. A Cu2+ ion is discharged by accepting 2 electrons to form a copper atom at the cathode. Half-equation at the cathode: 2H+ + 2e– → H2 10 Both types of anions, NO3– and OH– ions are attracted to the anode. OH– ions are selectively discharged at the anode because of the position of the OH– ion in the electrochemical series. Hydroxide ions are discharged to form water and oxygen. Half-equation at the anode: Cu2+ + 2e– → Cu 3 Both types of anions, SO42– ions and OH– ions are attracted to the anode. OH– ions are discharged at the anode because the position of the OH– ion is lower than that of the SO42– ion in the electroche mical series. 4OH– ions are discharged by donating 4 electrons to form water and oxygen. Half-equation at the anode: 4OH– → 2H2O + O2 + 4e– 11 Electrolysis of dilute sulphuric acid and sodium nitrate solution is actually electrolysis of water because H+ ions and OH– ions are discharged. The decrease in the concentrations of H+ ions and OH– ions in the solution results in the increase of sulphuric acid and sodium nitrate concentration during electrolysis. 12 The ratio of H2 gas to O2 gas produced is 2 : 1. This is because the release of 4 electrons in the for ma tion of 1 molecule of O2 results in the formation of 2 molecules of H2 when these 4 electrons are accepted by 4 H+ ions. 4OH– → 2H2O + O2 + 4e– 4 The blue colour of the copper sulphate solution is due to the presence of copper(II) ion, Cu2+. The blue colour of the electrolyte becomes paler during electrolysis because the Cu2+ ion concen tration decreases when Cu2+ ions are discharged. 5 The ions present in the dilute sulphuric acid are SO42– ions, H+ ions and OH– ions. H2SO4 → 2H+ + SO42– H2O H+ + OH– Conclusion 1 Electrolysis of aqueous copper(II) sulphate solution using carbon electrodes produces copper at the cathode and oxygen gas at the anode. 2 Cu2+ ions and OH– ions are selectively discharged because of their lower positions in the electrochemical series. 3 Electrolysis of dilute sulphuric acid and sodium nitrate solution using carbon electrodes produces hydrogen gas at the cathode and oxygen gas at the anode. 4 H+ ions and OH– ions are selectively discharged because their positions are lower in the electrochemical series. The hypothesis is accepted. 6 Hydrogen ions are attracted to the cathode. H+ ion is discharged by receiving one electron to form a hydrogen atom. Two hydrogen atoms will combine to form a hydrogen gas molecule. Half-equation at the cathode: 2H+ + 2e– → H2 7 Both types of anions, SO42– ions and OH– ions are attracted to the anode. OH– ions are selectively discharged at the anode because of the position of the OH– ion in the electrochemical series. Four OH– ions are discharged to form water and oxygen. Half-equation at the anode: 4OH– → 2H2O + O2 + 4e– 145 Electrochemistry 6 NaNO3 → Na+ + NO3– H2O H+ + OH– 6.3 SPM ’09/P2 To investigate the effect of the concentration of ions on the selective discharge of ions and the products of electrolysis of aqueous solutions 6 Problem statement How does the concentration of ions determine the types of ions discharged during electrolysis? Hypothesis Ions of higher concentration will be selectively discharged during electrolysis. Variables (a) Manipulated variable: Concentration of ions in the solution (b) Responding variable: Types of ions to be discharged at the anode and cathode (c) Constant variables: Types of ions in the electrolyte, types of electrodes, duration of electrolysis. Figure 6.6 Electrolysis of copper(II) chloride solution Apparatus Batteries, electrolytic cell, carbon electrodes, ammeter, switch, connecting wires with crocodile clips and test tubes. Materials Aqueous 2.0 mol dm–3 copper(II) chloride, CuC12 solution and aqueous 0.001 mol dm–3 copper(II) chloride solution. Procedure 1 Concentrated aqueous copper(II) chloride solution of 2.0 mol dm–3 is put into an electrolytic cell with carbon electrodes. 2 Two test tubes, filled with copper(II) chloride solution are inverted over the carbon anode and cathode respectively (Figure 6.6). 3 The switch is turned on and electric current is allowed to flow for 15 minutes. 4 Any change in colour of the electrolyte and any other changes that occur around the carbon electrodes are recorded. 5 Steps 1 to 4 of the experiment are repeated using the dilute copper(II) chloride solution of 0.001 mol dm–3 to replace the concentrated copper(II) chloride solution. Results Electrolyte Experiment 6.3 Concentrated copper(II) chloride solution of 2.0 mol dm–3 Dilute copper(II) chloride solution of 0.001 mol dm–3 Electrochemistry Observation Inference At the cathode: Brown deposit is formed Copper metal is produced At the anode: Bubbles of pungent greenish-yellow gas are produced. The gas turns the damp blue litmus paper to red and then bleaches it Chlorine gas is produced Colour of electrolyte: The blue colour of the solution becomes paler Concentration of Cu2+ ions in copper(II) chloride solution decreases At the cathode: Brown deposit is formed Copper metal is produced At the anode: Bubbles of colourless gas are produced. The gas lights up a glowing wooden splint Oxygen gas is produced Colour of electrolyte: The blue colour of the solution becomes paler Concentration of Cu2+ ions in copper(II) chloride solution decreases 146 Cu2+ ions are discharged because the position of Cu2+ ions is lower than that of H+ ions in the electrochemical series. Hence copper metal is deposited. Discussion 1 Aqueous copper(II) chloride consists of Cu2+ ions, H+ ions, OH– ions and Cl– ions. CuCl2 → Cu2+ + 2Cl– H2O H+ + OH– Half-equation at the anode: (b) The blue colour of the solution becomes paler because the concentration of the Cu2+ ions decreases when Cu2+ ions are discharged at the cathode during electrolysis. Conclusion 1 In the electrolysis of concentrated aqueous copper(II) chloride solution, copper metal is produced at the cathode and chlorine gas is produced at the anode. At the anode, the Cl– ions are selectively discharged, producing chlorine gas because the con cen tration of Cl– ions is – higher than that of OH ions. 2 In the electrolysis of dilute aqueous copper(II) chloride solution, copper metal is produced at the cathode and oxygen gas is produced at the anode. At the anode, OH– ions are selectively discharged, producing oxygen gas because the concentration of Cl– ions is low. 3 The type of ions that is selectively discharged at the electrode is determined by the concentration of the ions. The hypothesis is accepted. Half-equation at the anode: 2Cl– → Cl2 + 2e– Hence chlorine gas is produced. 3 Electrolysis of dilute aqueous copper(II) chloride solution (0.001 mol dm–3): At the anode: 2 types of anions, Cl– ions and OH– ions are attracted to the anode. OH– ions are discharged because the concentration of Cl– ions is lower than that of OH– ions. Half-equation at the anode: 4OH– → 2H2O + O2 + 4e– Hence oxygen gas is produced. 4 (a) At the cathodes of both concentrated and dilute aqueous copper(II) chloride solution: 6.4 6 Cu2+ + 2e– → Cu 2 Electrolysis of concentrated aqueous copper(II) chloride solution (2 mol dm–3): At the anode: 2 types of anions, Cl– ions and OH– ions are attracted to the anode. Cl– ions are discharged because the concentration of Cl– ions is higher than that of OH– ions. SPM ’09/P3 Hypothesis The products of electrolysis of copper(II) sulphate solution with copper electrodes are different from that with carbon electrodes. Apparatus Batteries, electrolytic cell, carbon electrodes, copper electrodes, ammeter, switch, rheostat, connecting wires with crocodile clips and test tubes. Variables (a) Manipulated variable : Types of electrodes (b) Responding variable : Products of electrolysis Materials Aqueous 1.0 mol dm–3 copper(II) sulphate, CuSO4 solution 147 Electrochemistry Experiment 6.4 To investigate the effect of the types of electrodes on the selective discharge of ions and the products of electrolysis of aqueous solutions (c) Constant variables : Types of ions in the Problem statement electrolyte and the concen How do the types of electrodes determine the types tration of ions of ions discharged during electrolysis? 6 Procedure 1 Aqueous 1.0 mol dm–3 copper(II) sulphate solution is put into an electrolytic cell with carbon electrodes. 2 A test tube filled with copper(II) sulphate solution is inverted over the carbon anode (Figure 6.7). 3 The switch is turned on and the electric current is allowed to flow for 15 minutes. 4 Any change in colour of the electrolyte and any other changes that occur around the carbon electrodes are recorded. 5 Steps 1 to 4 of the experiment are repeated using copper electrodes to replace carbon electrodes. Figure 6.7 Electrolysis of copper(II) sulphate solution Results Type of electrodes Carbon Copper Observation At the cathode: Brown deposit is formed Copper metal is produced At the anode: Bubbles of colourless gas are produced The gas lights up a glowing wooden splint Oxygen gas is produced Colour of electrolyte: The blue colour of the solution becomes paler Concentration of Cu2+ ions decreases At the cathode: Formation of brown deposit makes the cathode thicker Copper metal is produced At the anode: Anode corrodes and becomes thinner Copper anode dissolves to form Cu2+ ions Colour of electrolyte: The blue colour of the solution remains unchanged Concentration of Cu2+ ions in copper(II) sulphate remains constant 4 The colour intensity of the blue solution decreases because the concentration of Cu2+ ions in the copper(II) sulphate decreases. 5 During electrolysis, the concentration of OH– ions decreases, leaving H+ ions behind. As a result, the solution becomes acidic. 6 If copper is used as the anode, both SO42– ions and OH– ions are not discharged. Instead, the copper anode dissolves by releasing electrons to form Cu2+ ions. Hence, the mass of the anode decreases and the anode becomes thinner. The types of electrodes (copper electrode is an active electrode) determine the product formed at the anode during electrolysis. Half-equation at the anode: Discussion 1 Aqueous copper(II) sulphate solution consists of Cu2+ ions, H+ ions, SO42– ions and OH– ions. 2 During electrolysis, OH– ions and SO42– ions move to the anode. If a carbon electrode is used as the anode, OH– ion is selectively discharged due to its position in the electrochemical series. Oxygen gas is produced. Half-equation at the anode: 4OH– → 2H2O + O2 + 4e– 3 Cu2+ ions and H+ ions move to the cathode. Cu2+ ion which is at a lower position than the H+ ion in the electrochemical series will be discharged. Copper metal is produced. Half-equation at the cathode: Cu → Cu2+ + 2e– 7 Cu2+ ions are still discharged at the cathode, producing copper metal. This causes the mass Cu2+ + 2e– → Cu Electrochemistry Inference 148 Conclusion 1 In the electrolysis of aqueous copper(II) sulphate solution: (a) If a carbon electrode is used as the anode, OH– ions are discharged and oxygen gas is produced. (b) If a copper electrode is used as the anode, both OH– ions and SO42– ions are not discharged. Instead the copper anode dissolves to produce Cu2+ ions. (c) Cu2+ ions are discharged at the cathode producing copper metal whether the cathode used is a carbon electrode or a copper electrode. 2 The types of electrodes used during electrolysis determine the types of ions discharged and the products of electrolysis. The hypothesis is accepted. Cu2+ + 2e– → Cu 8 The colour intensity of the blue solution does not change because the concentration of Cu2+ ions is constant. The number of moles of Cu2+ ions discharged at the cathode is the same as the number of moles of Cu2+ ions produced at the anode. 9 The mass of the electrodes can be weighed before and after electrolysis using an electronic balance. 10 The decrease in the mass of the copper anode is the same as the increase in the mass of the copper cathode. How to Predict the Products of Electrolysis of Aqueous Solutions 1 ’08 The diagram shows the arrangement of apparatus for the electrolysis of silver nitrate solution. Write the half-equation representing the reaction at electrode Q. Comments The ions present in the electrolyte are Ag+, NO3–, H+ and OH–. Electrode Q is the anode as it is connected to the positive terminal of the battery. OH– ions are discharged at the anode producing oxygen gas. (Factor: positions of ions in the electrochemical series) 4OH– → 2H2O + O2 + 4e– The cations at a higher position in the electrochemical series are very stable. These ions are unlikely to accept electrons to form neutral atoms. Hence K+ ions and Na+ ions are never discharged in an aqueous solution in electrolysis. The cations at the lower position of the electrochemical series are less stable. They are more likely to accept electrons to form neutral atoms. Similarly with anions. Anions at a higher position in the electrochemical series are never discharged in an aqueous solution in electrolysis. Hence F –, SO42– and NO3– ions are stable compared to the lower anions at a lower position in the electrochemical series. Cl–, Br –, I– or OH– ions will be discharged instead depending on their ionic concentration in the aqueous solution. 149 Electrochemistry 6 of the cathode to increase and the cathode becomes thicker. Half-equation at the cathode: How to predict the products of electrolysis of aqueous solutions. Step 1 Step 2 Identify the cations and anions that are present in the aqueous solution. Generally, an aqueous solution of MaXb will produce ions as follows: Identify the anode and cathode • Anode is the electrode connected to the positive terminal of the battery. Positive electrode attracts negative ions (anions). • Cathode is the electrode connected to the negative terminal of the battery. Negative electrode attracts positive ions (cations). Cation Anion M b+ , X a– H+ , OH– 6 From MaXb : From H2O : SPM ’10/P2, ’11/P2 Step 3 Step 4 Identify the movements of ions • M b+ ions and H+ ions move to the cathode. • X a– ions and OH– ions move to the anode. Identify the ions to be discharged at the cathode H+ ions are discharged at the cathode (producing H2 gas) except if Mb+ is Cu2+ or Ag+ (because these ions are lower than H+ ions in the electrochemical series). • H+ ions discharge by accepting electrons to form H2 gas. 2H+ + 2e– → H2 • Cu2+ ions or Ag+ ions discharge by accepting electrons to form metal atom. Cu2+ + 2e– → Cu Ag+ + e – → Ag Inert electrode Step 5 Identify the types of electrode as anode Inert electrode: carbon or platinum. Active electrode: Ag or Cu. SPM ’09/P1 OH – ions are discharged (producing O2 gas) except if X b– ions are Cl–/Br –/I – of high concentration. • OH– ions discharge by donating electrons to form water and O2 gas. 4OH– → O2 + 2H2O + 4e– • Cl–/Br –/I– of high concentration discharge by donating electrons to form halogen. Example: 2Cl– → Cl2 + 2e–. Electrochemistry 150 Active electrode Anions are not discharged. Instead anode dissolves to form ions. Examples: Ag → Ag+ + e– Cu → Cu2+ + 2e– Electrolysis of dilute acids and alkalis Electrolysis of aqueous copper(II) nitrate solution, Cu(NO3)2 with copper electrodes 1 Electrolysis of all dilute acids (HCl/H2SO4/ HNO3) and dilute alkali solutions (NaOH/ KOH) is actually the electrolysis of water. 2 At the cathode: H+ ions are discharged producing hydrogen gas. 1 At the cathode: Cu2+ ions are discharged. Copper metal is deposited. Mass of cathode will increase. (Factor: positions of ions in the electrochemical series). 2H+ + 2e– → H2 3 At the anode: OH– ions are discharged producing oxygen gas. 2 At the anode: Copper dissolves from the anode forming Cu2+ ions. Mass of the anode will decrease. (Factor: types of electrodes). 4OH– → 2H2O + O2 + 4e– Cu → Cu2+ + 2e– 4 The removal of H+ ions and OH– ions from the solution during electrolysis causes the concentration of the acid or alkali to increase. Electrolysis of aqueous silver nitrate solution, AgNO3 with carbon electrodes Electrolysis of concentrated aqueous sodium chloride solution, NaCl with carbon electrodes 1 At the cathode: Ag+ ions are discharged. Silver metal is deposited. (Factor: positions of ions in the electrochemical series). SPM 1 Cations present : Na+ ions, H+ ions ’08/P2 Anions present : Cl– ions, OH– ions 2 At the cathode: H+ ions are discharged producing hydrogen gas. (Factor: positions of ions in the electrochemical series). Ag+ + e– → Ag 2 At the anode: OH– ions are discharged producing oxygen gas. (Factor: positions of ions in the electrochemical series). 2H+ + 2e– → H2 3 At the anode: Cl– ions are discharged producing chlorine gas. (Factor: concentration of ions). 4OH– → 2H2O + O2 + 4e– 2Cl– → Cl2 + 2e– Electrolysis of aqueous silver nitrate solution, AgNO3 with silver electrodes Electrolysis of aqueous copper(II) nitrate solution, Cu(NO3)2 with carbon electrodes 1 Cations present : Cu2+ ions, H+ ions Anions present : NO3– ions, OH– ions 2 At the cathode: Cu2+ ions are discharged. Copper metal is deposited. (Factor: positions of ions in the electrochemical series). 1 At the cathode: Ag+ ions are discharged. Silver metal is deposited. Mass of silver cathode will increase. (Factor: positions of ions in the electrochemical series). Ag+ + e– → Ag Cu2+ + 2e– → Cu 3 At the anode: OH– ions are discharged producing oxygen gas. (Factor: positions of ions in the electrochemical series). 2 At the anode: silver anode dissolves forming Ag+ ions. Mass of silver anode electrode will decrease. (Factor: types of electrodes). 4OH– → 2H2O + O2 + 4e– Ag → Ag+ + e– 151 Electrochemistry 6 Cu2+ + 2e– → Cu such as sodium, calcium, magnesium and aluminium are extracted from their compounds using electrolysis. 2 Electrolysis is used because these reactive metals cannot be extracted from the minerals by reduction using carbon. 3 Examples are: (a) Extraction of aluminium from aluminium oxide, Al2O3. (b) Extraction of sodium from sodium chloride, NaCl. 6.3 1 Write the formulae of the ions present in the aqueous solutions below. Identify the ions that will be discharged at the anode and the cathode during electrolysis using carbon electrodes in every case. Ions present in the solution 6 Aqueous solution Ions discharged at the cathode Ions discharged at the anode (a) Aqueous nitric acid (b) Silver nitrate (c) Very dilute copper(II) chloride Extraction of aluminium metal from the mineral bauxite (Hall-Heroult process) 1 Bauxite is the major ore of aluminium consisting of aluminium oxide, Al2O3. 2 Cryolite (Na3AlF6) is added to aluminium oxide to lower its melting point from 2000°C to about 950°C. 3 Molten aluminium oxide is electrolysed using carbon as electrodes in the electrolytic cell as shown in Figure 6.8. 2 The figure above shows the arrangement of apparatus in an electrolysis experiment. (a) Name all the ions present in Cell I and Cell II. (b) Electrolysis is carried out for 20 minutes. (i) Predict the observations at electrodes P and Q. (ii) What will be the change in colour in Cell I? Explain your answer. (iii) Predict the observations at electrodes R and S. (iv) Write the half-equations at electrodes P and R. (c) What is the factor that determines the formation of the products in (i) electrode Q? (ii) electrode R? 6.4 Figure 6.8 Extraction of aluminium metal from aluminium oxide. 4 Molten aluminium oxide dissociates into aluminium and oxide ions as follows: Al2O3(l) → 2Al3+(l) + 3O2–(l) (a) At the cathode: Al3+ ions discharge to form aluminium metal Electrolysis in Industries Uses of Electrolysis in Industries Al3+ + 3e– → Al SPM (b) At the anode: O2– ions discharge to form oxygen gas ’09/P1 1 Electrolysis process is used widely in industries. 2 Some common industrial applications of electrolysis are (a) extraction of reactive metals (b) purification of metals (c) electroplating of metals 2O2– → O2 + 4e– Overall equation of electrolysis: 2Al2O3 → 4Al + 3O2 5 The carbon anode is required to be replaced from time to time because the oxygen gas generated oxidises the carbon anode to form carbon dioxide. (A) Extraction of Reactive Metals Using Electrolysis 1 Metals that are very reactive (placed at the top positions of the electrochemical series) Electrochemistry 152 Extraction of sodium metal from sodium chloride (Downs process) 4 Molten sodium chloride dissociates into ions as follows: 1 Sodium chloride is the most abundant and cheapest sodium compound. 2 Electrolysis of molten sodium chloride is carried out using iron as cathode and carbon as anode as in Figure 6.9. NaCl(l) → Na+(l) + Cl–(l) (a) At the cathode: Na+ ions discharge to form sodium metal. Sodium metal is less dense and floats on top of the electrolyte to be collected. chlorine gas sodium is produced here Na+ + e– → Na 6 molten sodium chloride cathode (iron) (b) At the anode: Cl– ions discharge to form chlorine gas. Chlorine gas is a useful by-product. anode (carbon) Figure 6.9 Extraction of sodium metal from molten sodium chloride 2Cl– → Cl2 + 2e– Overall chemical electrolysis: 3 Calcium chloride is added to lower the melting point of sodium chloride. equation of the 2NaCl → 2Na + Cl2 (B) Purification of Metals Using Electrolysis the anode to the cathode. After electrolysis, the mass of anode is reduced while that of the cathode is increased. 3 For example, in the purification of impure copper metal: (a) Anode: impure copper (b) Cathode: pure copper (c) Electrolyte: copper(II) ion solutions such as copper(II) sulphate To investigate the purification of copper metal using electrolysis SPM ’10/P1 2 A piece of impure copper plate is connected to the positive terminal of the batteries. This plate acts as the anode. 3 A piece of pure copper metal is connected to the negative terminal of the batteries. This plate acts as the cathode. 4 The circuit is completed using the connecting wires, rheostat and ammeter. The two copper plates are immersed in the copper(II) sulphate solution. The solution is electrolysed for 30 minutes (Figure 6.10). Apparatus Batteries, electrolytic cell, beaker, connecting wires with crocodile clips, ammeter and rheostat. Materials 1 mol dm–3 copper(II) sulphate solution, pure copper plate and impure copper plate. Procedure 1 About 200 cm3 of 1 mol dm–3 copper(II) sulphate solution is poured into a beaker. 153 Electrochemistry Activity 6.2 1 Impure metals containing impurities can be purified using electrolysis as outlined below. (a) The impure metal is used as the anode. (b) A piece of pure metal is used as the cathode. (c) The electrolyte is a solution containing the ions of the metal to be purified. 2 In the process of purification of a metal using electrolysis, metal is transferred from Discussion 1 At the anode: Copper dissolves from the anode by releasing electrons to form Cu2+ ions. The mass of the anode decreases. Half-equation: 2 At the cathode: Cu2+ ions are discharged by receiving electrons to form copper atoms. Copper metal is deposited on the surface of the cathode. As a result, the copper cathode becomes thicker. The mass of the cathode increases. 6 Figure 6.10 Purification of copper metal Results Observation At the anode: • The copper anode becomes thinner • Impurities are deposited below the anode Half-equation: Inference Copper metal is deposited at the cathode Colour of electrolyte: • Colour intensity of the blue solution does not change Concentration of Cu ions in the electrolyte does not change 2+ Cu2+ + 2e– → Cu 3 In this process, Cu2+ ions are transferred from the anode to the cathode and are deposited as pure copper metal. Impurities that are collected below the anode are known as anode mud. 4 The colour intensity of the blue solution does not change because the concentration of Cu2+ ions remains constant throughout electrolysis. The rate of formation of Cu2+ ions at the anode is the same as the rate of discharge of Cu2+ ions at the cathode. Copper anode dissolves to form Cu2+ ions At the cathode: • The copper cathode becomes thicker Cu → Cu2+ + 2e– Conclusion When copper(II) sulphate solution is electrolysed using pure copper as the cathode and impure copper as the anode, purification of copper takes place. Pure copper is deposited at the cathode. (C) Electroplating of Metals Using Electrolysis (a) The object to be electroplated is used as the cathode. (b) The anode is the electroplating metal. (c) The electrolyte is a solution that contains the electroplating metal ions. 4 An even and lasting layer of metal can be produced if (a) the surface of the object to be electroplated is first polished using sandpaper. (b) a low electric current is used so that electroplating is carried out slowly during the electroplating process. (c) the object to be plated is rotated steadily during electrolysis. 1 Electroplating is a process carried out to coat the surface of metal objects with a thin and even layer of another metal. 2 Two main aims of electroplating metals are (a) to prevent corrosion. For example, iron objects are plated with a thin layer of chromium or nickel metal to protect the iron from rusting. (b) to improve the appearance. For example, electroplating with gold, platinum and silver makes the surface of the objects appear shiny and more attractive. 3 In general, there are three conditions in the SPM electroplating of metal. ’08/P1 Electrochemistry 154 2 Comments At the anode, the silver foil dissolves by releasing electrons, thus becoming thinner: SPM ’10/P1 The diagram shows the set-up of the apparatus used to electroplate an iron key with silver. Ag → Ag+ + e– At the cathode, silver ions discharge by receiving electrons to form silver atoms, forming a shiny grey deposit on the iron key: Ag+ + e– → Ag Answer B A B C D Anode Cathode Shiny grey deposits are formed Silver foil becomes thinner Shiny grey deposits are formed Silver foil becomes thinner Silver foil becomes thicker Shiny grey deposits are formed Gas bubbles are released Gas bubbles are released 6 What is observed at the anode and cathode after 30 minutes? Presently, plastic electroplating is carried out to coat a thin layer of metal onto the surface of plastic objects. The object produced will have the advantanges of plastic: light, cheap, resistant to corrosion as well as having a shiny surface like a metal. As plastic is an electric insulator, a layer of graphite powder is coated onto the surface of the plastic, so that it can conduct electricity, before electroplating is carried out. 6.5 To investigate the electroplating of an iron spoon with copper using electrolysis Problem statement How is electrolysis used to electroplate an iron spoon with copper metal? Hypothesis Electroplating of an iron spoon with copper occurs if the iron spoon is used as the cathode, copper metal is used as the anode and aqueous copper(II) sulphate solution as the electrolyte. Variables (a) Manipulated variable : The position of the iron spoon as an electrode (b) Responding variable : The deposition of copper on the iron spoon (c) Constant variable : Type of electrolyte and arrangement of apparatus Materials 0.5 mol dm–3 copper(II) sulphate solution, copper plate and iron spoon. Procedure 1 About 200 cm3 of 0.5 mol dm–3 copper(II) sulphate solution is poured into a beaker. 2 An iron spoon is polished using sandpaper and is connected to the negative terminal of the batteries. The spoon acts as the cathode. 3 A piece of copper metal, as the anode, is connected to the positive terminal of the batteries. 155 Electrochemistry Experiment 6.5 Apparatus Batteries, electrolytic cell, beaker, connecting wires with crocodile clips, ammeter and rheostat. 4 The circuit is completed using the connecting wires, rheostat and ammeter. The iron spoon and the copper metal are immersed in the copper(II) sulphate solution. The solution is electrolysed for 30 minutes using a small current (0.5 A). 5 Steps 1 to 4 of the experiment are repeated by interchanging the positions of the iron spoon and copper metal, whereby the iron spoon is made the anode and the copper metal is made the cathode. 6 Set Set 1: Iron spoon as the cathode, copper metal as the anode Set 2: Copper metal as the cathode, iron spoon as the anode Figure 6.11 Electroplating of an iron spoon with copper Observation Inference At the cathode: A brown metal is deposited on the surface of the iron spoon The iron spoon is plated with copper metal At the anode: The copper anode becomes thinner The copper anode dissolves to form Cu2+ ions Colour of the electrolyte: Colour intensity of the blue solution does not change Concentration of Cu2+ ions in the electrolyte remains constant At the cathode: The copper plate becomes thicker Copper metal is deposited on the copper electrode At the anode: No noticeable change in the appearance of the iron spoon Electroplating of copper on the iron spoon does not take place Colour of the electrolyte: The blue colour of the solution becomes paler Concentration of Cu2+ ions in the electrolyte decreases 3 The colour intensity of the blue solution does not change because the concentration of Cu2+ ions remains constant throughout the electroplating process. 4 A slow electrolysis process using a small current will ensure that the layer of copper sticks firmly to the surface of the iron spoon. Discussion 1 The brown metal deposited on the iron spoon is copper metal. (a) At the anode: copper dissolves from the anode by releasing electrons to form Cu2+ ions. Half-equation: Cu → Cu2+ + 2e– Conclusion 1 In electroplating an iron spoon with copper using electrolysis, the iron spoon is made the cathode and a piece of copper metal is made the anode. 2 Copper metal is transferred from the copper anode to the iron spoon and is deposited as a thin layer of copper metal. 3 Electroplating does not take place if the iron spoon is made the anode. The hypothesis is accepted. (b) At the cathode: Cu2+ ions are discharged by accepting electrons to form copper atoms. Copper metal is deposited on the surface of the iron spoon. Half-equation: Cu2+ + 2e– → Cu 2 In this process, Cu2+ ions are transferred from the anode to the cathode (iron spoon) and are deposited as a thin and even layer of copper metal. Electrochemistry 156 Benefits and Harmful Effects of Electrolysis in Industries 1 Table 6.1 shows the advantages and disadvantages of using electrolysis. Advantages of using electrolysis Disadvantages of using electrolysis 1 Electrolysis is an effective method of extracting reactive metals from their compounds. Some chemical substances such as chlorine and sodium can be manufactured in large quantities using electrolysis. 2 In electroplating, the whole surface area of a metal such as iron is coated with a thin, even and valuable metal (such as gold, platinum and silver). This layer of metal protects iron from being exposed to air and water so as to prevent corrosion. Besides that, the layer of metal also gives an attractive appearance. Electroplating can also be performed using polymers as the coating material. This has been used to coat new cars with paint. The advantage of this process is that it can be done in water thereby eliminating the use of volatile organic solvents used in spray-paints. 3 Electrolysis process is used to purify metals such as zinc, silver, nickel, copper, lead and aluminium. This process is also known as electrorefining. 1 Electrolysis is a process that uses a large quantity of electricity. For example, the recycling of aluminium requires only 9% of the electrical energy used to produce the same quantity of aluminium in electrolysis. 2 The problem of environmental pollution especially in the electroplating process. (a) In the electroplating of iron by chromium and nickel, the waste chemicals contain chromium ions and nickel ions that can endanger human health as well as pollute water sources. (b) In silver electroplating, potassium silver cyanide, KAg(CN)2 solution is sometimes used as an electrolyte. The waste chemical of the electrolyte contains cyanide ions which are toxic. (c) Metal objects to be electroplated are cleaned by acids to remove the layer of metal oxide on the surface before electroplating. The used acid wastes will pollute water in the drains, rivers and lakes, thus destroying aquatic life. 2 Steps taken to overcome the problems of electrolysis in industries (a) Recycling such as the recycling of aluminium cans is encouraged to reduce the use of electrolysis in the extraction of aluminium. (b) Waste chemicals in electrolyte from electroplating are treated to remove the toxic substances before being drained out as effluent. 6.4 (a) What is a suitable metal that can be used as metal M? (b) State two observations that will be obtained in this experiment. (c) Write the half-equations for the reactions that take place at (i) the iron key (ii) the metal M (d) How will the concentration of silver nitrate solution change after electrolysis? Explain your answer. (e) Explain how the student can ensure that an even and lasting layer of silver metal stays on the surface of the iron key. 1 State three main uses of the electrolysis process in industries. 2 A student carried out an experiment to electroplate an iron key with silver using the apparatus as shown in the above figure. 157 Electrochemistry 6 Table 6.1 Advantages and disadvantages of using electrolysis. 6.5 Voltaic Cells 6 Simple Voltaic Cells (Chemical Cells) 5 SPM ’09/P2 1 A simple voltaic cell can be made by immersing two different types of metals in an electrolyte and connecting the two metals by wires in the external circuit. 2 In a simple voltaic cell, electrons flow from one metal to another metal through the connecting wires in the external circuit. 3 The more electropositive metal (metal that is at a higher position in the electrochemical series) will release electrons and thus acts as the negative terminal (anode) of the voltaic cell. 4 The less electropositive metal (metal that is at a lower position in the electrochemical 6 7 8 9 series) will accept electrons and acts as the positive terminal (cathode). A continuous flow of electrons from the negative terminal to the positive terminal of the cell through the external circuit produces an electric current. The flow of electric current can be detected by the lighting up of a light bulb or the deflection of a galvanometer needle. Voltaic cells are also known as galvanic cells or chemical cells. The potential difference (voltage) of the cell is the electromotive force (e.m.f.) that moves electrons and can be measured by a voltmeter. The further the distance between the positions of two metals in the electrochemical series, the bigger the voltage of the cell. For example, a magnesium/copper cell will produce a higher voltage than a zinc/copper cell. 6.6 To investigate the production of electricity from chemical reactions in a simple voltaic cell Problem statement How does a chemical reaction produce electrical energy in a simple voltaic cell? Procedure 1 A piece of magnesium plate and a piece of copper plate are polished with sandpaper. 2 Both pieces of the magnesium and copper plates are immersed in 200 cm3 of aqueous sodium chloride solution in a beaker as shown in Figure 6.12. 3 Both plates are connected by the connecting wire to a voltmeter. 4 The experiment is repeated using two pieces of copper plates as electrodes. Hypothesis Electric current is produced when two different metals connected by wires are immersed in an electrolyte. Variables (a) Manipulated variable : Pairs of different metals (b) Responding variable : Deflection of a voltmeter needle by the electric current produced (c) Constant variable : Types of electrolyte and arrangement of apparatus Experiment 6.6 Apparatus Voltmeter, beaker, connecting wires with crocodile clips and sandpaper. Materials 1 mol dm–3 sodium chloride solution, copper plates and magnesium plate. Electrochemistry SPM ’05/P2 Figure 6.12 158 Results Magnesium metal and copper metal Observation Inference • Voltmeter needle deflects but the deflection decreases after awhile • Magnesium metal corrodes • Bubbles of colourless gas are evolved around the copper metal Two pieces of copper metal • Voltmeter needle does not show a deflection • No noticeable change occurs at the copper electrode Discussion 1 The deflection of the voltmeter needle shows that an electric current is produced. The decreasing deflection indicates that the electric current decreases rapidly. 2 Magnesium metal is more electropositive than copper (at a higher position in the electrochemical series). Hence, it has a higher tendency to donate electrons than copper. 3 Magnesium atoms will donate electrons to form magnesium ions, Mg2+ in the solution, hence magnesium metal corrodes. Half-equation: • Electric current is not produced • No reaction occurs Figure 6.13 Movement of electrons and ions in a simple Mg/Cu cell using sodium chloride solution as the electrolyte Mg → Mg2+ + 2e– 8 If copper(II) solution is used as the electrolyte, Cu2+ ions will receive electrons and are discharged because its position is lower than H+ ions and Mg2+ ions in the electrochemical series. Copper metal is produced. The overall equation of the cell will be 4 Electrons accumulate at the surface of the magnesium metal. This makes magnesium act as the negative terminal (also known as the anode) of the cell. 5 The electrons flow through the external circuit from the magnesium metal (negative terminal) to the copper metal (positive terminal or cathode of the cell) producing electricity. 6 When sodium chloride solution is used as the electrolyte, H+ ions (from water), Na+ ions and Mg2+ ions move towards the copper metal. H+ ions will accept electrons from the copper metal and be discharged because its position is lower than Na+ ions and Mg2+ ions in the electro chemical series. Hydrogen gas is produced. Half-equation: • Electric current is produced. The voltage produced is not constant and decreases rapidly • Magnesium dissolves to form Mg2+ ions • Hydrogen gas is produced Mg + Cu2+ → Mg2+ + Cu Conclusion 1 An electric current is produced when a chemical reaction occurs in a simple voltaic cell consisting of two different metals, connected by wires externally and immersed in an electrolyte. 2 In a simple voltaic cell, chemical energy released from chemical reactions is converted into electrical energy. 3 No electric current will be produced if both electrodes are of the same material because there is no potential difference between them. The hypothesis is accepted. 2H+ + 2e– → H2 7 The overall chemical equation in the cell is: Mg + 2H+ → H2 + Mg2+ 159 Electrochemistry 6 Type of metal used as electrodes 3 ’04 The diagram shows the set-up of the apparatus of a simple chemical cell. Metal P Metal Q Iron Copper Zinc Lead Aluminium Magnesium Iron Magnesium A B C D What are metals P and Q? 6 Comments The diagram shows that electrons flow from metal P to metal Q. Metal P must be more electropositive (higher in position in the electrochemical series) than metal Q. This is because the more electropositive metal will release electrons to become the negative terminal of the chemical cell. Answer C SPM Different Types of Voltaic Cells ’10/P2 Voltaic cells can be divided into two categories as shown in Table 6.2. Table 6.2 Two categories voltaic cells Primary cells Secondary cells • Non-rechargeable cells (cells that cannot be charged again) • Rechargeable cells (cells that can be charged again) • Examples (a) Daniell cell (b) Dry cell (c) Mercury cell (d) Alkaline cell • Examples (a) Lead-acid accumulator (b) Nickel/cadmium cell Figure 6.14 Daniell cell using a salt bridge (b) A porous pot as shown in Figure 6.15. Figure 6.15 Daniell cell using a porous pot Daniell Cell 4 A salt bridge contains inert ions or salt that does not react with the electrolyte. Examples are sodium chloride, potassium chloride, potassium nitrate, ammonium chloride and dilute sulphuric acid. 5 A simple salt bridge can be made by immersing a piece of filter paper in sulphuric acid or in a salt solution. 6 A porous pot has fine pores that allow ions to flow through but can prevent the two different aqueous solutions from mixing. 1 A Daniell cell has copper metal as the positive terminal and zinc metal as the negative terminal. 2 The zinc metal is immersed in zinc sulphate solution and the copper metal is immersed in copper(II) sulphate solution. 3 The two solutions of the Daniell cell are connected using either of the following: (a) A salt bridge as shown in Figure 6.14. Electrochemistry 160 2 The electrolyte is ammonium chloride in the form of a paste. 3 The cross-section of a dry cell is shown in Figure 6.16. 7 The functions of salt bridges and porous pots are (a) to allow the flow of ions so that the ’11/P1 circuit is completed. (b) to prevent the two aqueous solutions from mixing. This will prevent displace ment reaction between a more electro positive metal and the salt solution of the less electropositive metal from taking place. 8 In a simple voltaic cell made by immersing both the zinc metal and copper metal in copper(II) sulphate solution, (a) zinc metal reacts directly with copper(II) sulphate solution in a displacement reaction. SPM 6 Figure 6.16 Dry cell 4 When the dry cell is in use, the zinc metal releases electrons and dissolves to form Zn2+ ions. Zn + CuSO4 → ZnSO4 + Cu As a result, the zinc metal will be coated by a layer of copper metal. (b) the electric current decreases rapidly. 9 The salt bridge or porous pot prevents the zinc from reacting directly with the copper(II) sulphate solution. 10 When the negative terminal (zinc) is connec ted to the positive terminal (copper), the highest voltage produced is 1.10 V if both zinc sulphate solution and copper(II) sulphate solution have a concentration of 1.0 mol dm–3. 11 The reactions that take place in a Daniell cell are as follows: At the negative terminal: Zn → Zn2+ + 2e– At the positive terminal: Cu2+ + 2e– → Cu At the negative terminal: Zn → Zn2+ + 2e– 5 Electrons flow from the zinc metal casing through the external circuit to the carbon rod, where NH4+ ions receive electrons to produce ammonia gas and hydrogen gas. At the positive terminal: 2NH4+ + 2e– → 2NH3 + H2 6 When the cell produces an electric current, zinc metal dissolves. When the zinc metal casing is perforated and the electrolyte starts to leak out, the dry cell can no longer be used. 7 Usually a dry cell produces quite a stable voltage of about 1.5 V. Overall chemical equation of cell: Zn + Cu2+ → Zn2+ + Cu 12 When the Daniell cell is in use, (a) the copper metal becomes thicker (mass increases), (b) the zinc metal becomes thinner (mass decreases), (c) the concentration of copper(II) sulphate solution decreases, hence the blue colour of the solution becomes paler, (d) the concentration of zinc sulphate solution increases. Dry cells of different sizes Dry Cell Dry cells that are out of electricity (old) need to be removed from the electrical appliance. This is because when the container is corroded, the electrolyte will leak out to damage the electrical appliance. 1 A dry cell consists of a carbon rod (positive terminal) and a metal casing made of zinc (negative terminal). 161 Electrochemistry Lead-acid Accumulator 6 In the recharging process, (a) reverse reactions occur at both electrodes. (b) lead(II) sulphate is converted back into lead(IV) oxide and hence lead(II) sulphate dissolves. (c) sulphuric acid is formed. 7 An accumulator normally consists of 6 pairs of plates and produces a voltage of 12 V. 1 A lead-acid accumulator is made of pieces of lead plates immersed in moderately concentrated sulphuric acid as shown in Figure 6.17. Other Types of Voltaic Cells 6 Mercury cell 1 A mercury cell consists of zinc (negative terminal), mercury(II) oxide, HgO (positive terminal) and a mixture of potassium hydroxide, KOH and zinc oxide, ZnO as electrolyte. Figure 6.17 Lead-acid accumulator is used as car batteries. 2 When the accumulator is used to produce current, the following changes occur. (a) At the negative terminal, a lead atom donates 2 electrons to form a Pb2+ ion. Pb → Pb2+ + 2e– (b) At the positive terminal, lead(IV) oxide accepts electrons and reacts with H+ ions in dilute sulphuric acid to form Pb2+ ions. Mercury cells are small PbO2 + 4H+ + 2e– → Pb2+ + 2H2O 2 Mercury cells are small and long-lasting, producing a constant voltage of 1.3 V. 3 Mercury cells are used in hearing aids, digital watches and heart pacemakers. (c) Lead(II) ions from both electrodes combine with SO42– ions in sulphuric acid to produce lead(II) sulphate, PbSO4. Pb2+ + SO42– → PbSO4 Alkaline cell The overall chemical reaction is represented by the equation below: Pb + PbO2 + 4H+ + 2SO42– → 2PbSO4 + 2H2O 3 In this reaction, sulphuric acid is used up and water is produced. Hence, sulphuric acid becomes more dilute and its density decreases. 4 Lead(II) sulphate is insoluble and exists as a white precipitate. When the precipitate covers the surface of both electrodes, further reaction is prevented and no electric current will be produced. 5 The accumulator can be recharged by passing an electric current in the opposite direction to renew the cell. Electrochemistry 1 Alkaline cells are non-rechargeable cells. 2 An alkaline cell consists of zinc (negative terminal), carbon rod (positive terminal) surrounded by manganese(IV) oxide, MnO2 and alkali (potassium hydroxide and sodium hydroxide) as the electrolyte. 162 Nickel-cadmium cell Nickel-cadmium cells are rechargeable cells A nickel-cadmium cell consists of cadmium (negative terminal), nickel(IV) oxide, NiO2 (positive terminal) and alkali, potassium hydroxide, KOH as the electrolyte. Other new types of cells include lithium ion, nickel hydride and polymeric cells. These cells are rechargeable cells. Unlike the lithium ion and nickel hydride cells which require battery casings, the polymeric cell is flexible and can be specifically shaped to fit the device it will power. 6 Advantages and Disadvantages of Various Types of Voltaic Cells Table 6.3 Advantages and disadvantages of various types of voltaic cells Type of cell Advantages Disadvantages 1 Daniell cell • Can be prepared easily in the laboratory • A type of wet cell, the electrolyte spills easily • Voltage is not constant 2 Dry cell • Cheap • No spillage as it is a dry cell • Produces a moderately constant current and voltage • Portable, can be carried around easily • Available in different sizes • Does not last long • Cannot be recharged • Zinc metal casing dissolves and the electrolyte that leaks out may corrode electrical instruments • Current and voltage produced is low 3 Alkaline cell • Lasts longer than a dry cell • Produces a higher and more stable voltage • Portable • Cannot be recharged • Cost more than a dry cell • If leakage occurs, electrolyte is corrosive 4 Mercury cell • Can be made into very small sizes • Produce a constant voltage for a long period • Can last for a long period of time • Expensive • Cannot be recharged • Mercury which is produced is toxic 5 Lead-acid accumulator • Can be recharged repeatedly • Produces a high current (175 A), suitable for heavy work such as starting a car engine • Produce a high voltage (12 V) for a long period • Acid may spill • Heavy and is difficult to be carried around • Loss of charge occurs if not used for a long time • Lead plates are easily corroded after a long period of usage 6 Nickel-cadmium cell • Can be charged repeatedly • No spillage occurs because it is a dry cell • The size is smaller that an accumulator • Expensive • Requires a transformer for the recharging process 163 Electrochemistry Comparison of Voltaic Cells and Electrolytic Cells SPM ’10/P2 1 Table 6.4 below shows several similarities and differences between an electrolytic cell and a voltaic cell. 6 Table 6.4 Electrolytic cell Voltaic cell Figure 6.18 Electrolytic cell Figure 6.19 Voltaic cell Similarities • Contains an electrolyte • Consists of an anode and a cathode • Electrons move from the anode to the cathode in the external circuit (connecting wires) Electrolytic cell • Positive ions and negative ions move in the electrolyte • Chemical reactions involve the donation (at the anode) or acceptance (at the cathode) of electrons Differences • A battery is required to supply electrical energy • Graphite (carbon) is usually used as electrodes • A battery is not required to supply electrical energy Basic structure • Electrodes are not made up of different metals • Electrical energy is converted into chemical energy • Anode (positive electrode): Anions (negative ions) lose electrons at the anode Energy conversion Transfer of electrons at the positive electrode Electrochemistry • Chemical energy is converted into electrical energy • Cathode (positive electrode): Oxidising agent accepts electrons from the cathode Cu+(aq) + 2e– → Cu(s) … reduction Transfer of electrons at the negative electrode Y 2+ + 2e– → Y ... reduction • Electrons flow from the anode (positive electrode) to the cathode (negative electrode) • Graphite (carbon) is not used as electrodes • Electrodes are made up of two different metals 2X – → X2 + 2e– ... oxidation • Cathode (negative electrode): Cations (positive ions) accept electrons from the cathode Voltaic cell Transfer of electrons in the external circuit 164 • Anode (negative electrode): Reducing agent releases electrons Zn(s) → Zn2+(aq) + 2e– … oxidation • Electrons flow from the anode (negative electrode) to the cathode (positive electrode) The table shows the differences between the terminals in voltaic and electrolytic cells as a result of the transfer of electrons. Transfer of electrons Type of cells Electrons are donated Electrons are accepted Voltaic cells At the negative terminal (anode) At the positive terminal (cathode) Electrolytic cells At the positive terminal (anode) At the negative terminal (cathode) (d) Predict the observations obtained after the voltaic cell is used for some time. 1 (a) Draw the circuit diagram for a simple voltaic cell consisting of iron metal, copper metal and copper(II) sulphate solution. Show the direction of the flow of electrons in the circuit diagram. (b) Which metal serves as the negative terminal? (c) Write the half-equations for the reactions that occur at both electrodes. 6.6 2 Compare the advantages and disadvantages of dry cells and alkaline cells. 3 Give an example of a voltaic cell that can be recharged. Explain how the reactions occur at the positive terminal and the negative terminal to produce an electric current. 4 The electrochemical series can be constructed by two methods. (a) The potential difference (voltage difference) between pairs of metals (b) The ability of a metal to displace another metal from its salt solution. The Electrochemical Series 1 The electrochemical series is an arrangement of elements according to their tendencies to form ions. 2 In the electrochemical series, a metal that has a higher tendency to ionise and form positive ions (by releasing electrons) is placed at a higher position in the series. Hence, metal ions at the upper positions of the electrochemical series are less likely to receive electrons to form metal atoms. 3 Part of the electrochemical series (for metal elements) is shown below. (A) To Construct the Electrochemical Series Based on the Potential SPM ’08/P1 Difference (Voltage Difference) ’09/P2 1 Metals are arranged in the electrochemical series according to their tendencies to donate electrons to form cations. 2 The electrochemical series can be constructed based on the measurement of the potential difference between two metals in voltaic cells. 3 When two different metals (immersed in their respective salt solutions) are connected in the external circuit through a voltmeter and a salt bridge: (a) The metal that serves as the negative terminal of the voltaic cell has a higher tendency to release electrons. Hence, that metal is placed at a higher position in the electrochemical series. Conversely, the metal that serves as the positive terminal is placed at a lower position in the electrochemical series. (b) The further apart the positions of two metals in the electrochemical series, the greater the potential difference (voltage). Metals Positive ions (cations) K ⎯⎯⎯→ K+ + e– Na ⎯⎯⎯→ Na+ + e– Ca ⎯⎯⎯→ Ca2+ + 2e– Tendency Tendency Mg ⎯⎯⎯→ Mg2+ + 2e– of cations of metal Al ⎯⎯⎯→ Al3+ + 3e– to accept atoms to Zn ⎯⎯⎯→ Zn2+ + 2e– electrons donate Fe ⎯⎯⎯→ Fe2+ + 2e– to form electrons Sn ⎯⎯⎯→ Sn2+ + 2e– metals to form Pb ⎯⎯⎯→ Pb2+ + 2e– increases ions increases H ⎯⎯⎯→ H+ + e– Cu ⎯⎯⎯→ Cu2+ + 2e– Ag ⎯⎯⎯→ Ag+ + e– 165 Electrochemistry 6 6.5 6.7 To construct the electrochemical series through the potential difference (voltage) of pairs of metals Procedure Problem statement How to construct the electrochemical series based on the measurement of the potential differences between pairs of metals in simple voltaic cells? 6 SPM ’06/07 P3 Hypothesis Two principles are used in the construction of the electrochemical series: (a) A metal that serves as the negative terminal of a cell is placed at a higher position in the electrochemical series. (b) The bigger the voltage differences of the voltaic cells, the further apart the positions of the two metals in the electrochemical series. Figure 6.20 1 Pieces of zinc, magnesium, iron, aluminium and silver metals are polished with sandpaper. 2 A piece of zinc metal and a piece of copper metal are connected to a voltmeter by the connecting wires with crocodile clips. 3 The two metals are then dipped in the sodium chloride solution in a beaker as shown in Figure 6.20. 4 The highest cell voltage obtained is recorded. 5 The direction of the flow of electrons is also noted to determine the terminals of the voltaic cell. Electrons flow from the negative terminal to the positive terminal. If the voltmeter reading shows a negative value, the metal pairs connected to the terminals of the voltmeter should be reversed. 6 Zinc metal is then replaced by other metals in turn: magnesium, iron, aluminium and silver. The highest cell voltage of every pair of metals is recorded. Variables (a) Manipulated variable : Pairs of metals as electrodes (b) Responding variable : Voltage values of voltaic cells (c) Constant variables : Type and concentra tion of electrolytes Apparatus Voltmeter, beaker, connecting wires with crocodile clips and sandpaper. Materials Sodium chloride solution of 1.0 mol dm–3, pieces of copper, zinc, magnesium, iron, aluminium and silver metals. Experiment 6.7 Results Pairs of metals Positive terminal Negative terminal Potential difference (V) Zn/Cu Copper Zinc 1.1 Mg/Cu Copper Magnesium 2.7 Fe/Cu Copper Iron 0.8 Al/Cu Copper Aluminium 2.0 Ag/Cu Silver Copper 1.1 2 Silver serves as the positive terminal when it is connected to copper. Hence, silver is placed at a lower position than copper in the electrochemical series. 3 The further apart the distance between the metals in the electrochemical series, the greater the potential difference (voltage). Conclusion 1 Copper metal serves as the positive terminal of the voltaic cells when paired with zinc, magne sium, iron and aluminium metal. Hence, copper is at a lower position than zinc, magnesium, iron and aluminium in the electrochemical series. Electrochemistry 166 4 The arrangement of the metals in the electrochemical series based on the voltage (potential difference) of the cell is as follows: Magnesium Aluminium Higher tendency to release electrons 2.0 V Zinc 1.1 V Iron 0.8 V Copper 2.7 V The bigger the voltage reading, the further the distance between the metals 1.1 V 6 Silver 4 SPM ’10/P1 The table shows information about three simple cells. Metal pairs Potential difference (V) Positive terminal P and Q 1.7 P Q and S 2.1 S R and S 0.6 R What is the potential difference of the metal pair P and R? Q Higher tendency to release electrons Comments • In the metal pair of P and Q, P is the positive terminal. Hence P is placed below Q in the electrochemical series. Similarly, S is placed below Q in the electrochemical series. • The potential difference between Q and S is bigger than that between Q and P. Thus S is placed below P. • In the metal pair R and S, R is the positive terminal. Hence R is placed below S. •The arrangement of the metals according to their increasing tendencies to form metal ions is as follows: 1.7 V 2.1 V P S 0.6 V The bigger the voltage reading, the further the distance between the metals. R Answer The potential difference between P and R = (2.1 – 1.7) + 0.6 = 1.0 V (a) metal M is more likely to release electrons than metal N. (b) metal M is more electropositive than metal N. (c) metal M is placed at a higher position than metal N in the electrochemical series. 3 Alternatively, if metal P is immersed in an aqueous Q2+ ion solution and no reaction takes place, then metal P is at a lower position than metal Q in the electrochemical series. (B) To Construct the Electrochemical Series from the Displacement Reactions of Metals 1 The electrochemical series can also be constructed based on the ability of a metal to displace another metal from its salt solution. 2 If metal M can displace metal N from an aqueous N salt solution, then 167 Electrochemistry 6.8 SPM To construct the electrochemical series from displacement reactions 6 Problem statement How to construct the electrochemical series based on the ability of a metal to displace another metal from its salt solution? ’08/P2, ’07/P1, ’04/P2 3 A piece of magnesium metal is placed in the solution of every test tube except that of its salt solution (Figure 6.21). Hypothesis A metal that can displace another metal from its salt solution is placed at a higher position in the electrochemical series. The greater the number of metals that can be displaced by a metal from their solutions, the higher its position in the electrochemical series. Variables (a) Manipulated variable : Different types of metal and their salt solutions (b) Responding variable : Deposition of metals or colour change in the salt solutions (c) Constant variable : Concentration of nitrate salt solutions Figure 6.21 Apparatus Test tubes, test-tube rack and sandpaper. 4 Observations are made after awhile to check if (a) there is any colour change in the solution, (b) there are any solid deposits on the magnesium metal, (c) magnesium metal dissolves. 5 If any of the above occurrences (a), (b) or (c) is observed, displacement reaction has taken place: a tick symbol, (✓) is marked in the table of results. 6 If there is no noticeable observation, a cross symbol, (✗) is marked at the table to indicate that displacement reaction did not take place. 7 The experiment is repeated using different metals and fresh solutions of ions. The results of the experiment are shown in the table below. Materials Pieces of magnesium, zinc, iron, tin, lead and copper metals, solutions of copper(II) nitrate, lead(II) nitrate, tin(II) nitrate, iron(II) nitrate, zinc nitrate and magnesium nitrate (concentration and volume of all salt solutions are 0.5 mol dm–3 and 10 cm3 respectively). Procedure 1 Pieces of magnesium, zinc, copper, tin, lead and iron metals are polished with sandpaper. 2 10 cm3 of 0.5 mol dm–3 solutions of copper(II) nitrate, lead(II) nitrate, tin(II) nitrate, iron(II) nitrate, zinc nitrate and magnesium nitrate are placed into separate test tubes. Results Solution Cu(NO3)2 Pb(NO3)2 Sn(NO3)2 Fe(NO3)2 Zn(NO3)2 Mg(NO3)2 Magnesium, Mg ✓ ✓ ✓ ✓ ✓ – Zinc, Zn ✓ ✓ ✓ ✓ – ✗ Iron, Fe ✓ ✓ ✓ – ✗ ✗ Tin, Sn ✓ ✓ – ✗ ✗ ✗ Lead, Pb ✓ – ✗ ✗ ✗ ✗ Copper, Cu – ✗ ✗ ✗ ✗ ✗ Experiment 6.8 Metal Electrochemistry 168 4 The result of the experiment shows that the order of the positions of the metals in the electrochemical series is: Conclusion 1 Metals can be arranged according to the number of tick symbols (3) recorded (or the number of metals displaced in reactions). The more (3) symbols, the more reactive the metal is and the position of the metal is placed higher in the electrochemical series. 2 Magnesium is placed at the highest position in the electrochemical series because it can displace all the other metals from their solutions. 3 Copper is placed at the lowest position in the electrochemical series because copper cannot displace any other metals in this experiment. Mg Zn Fe Sn Pb Cu Electropositivity of metal decreases 6 5 The electrochemical series can be constructed from displacement reactions. The hypothesis is accepted. The Uses of the Electrochemical Series To determine the terminals of voltaic cells To predict the ability of a metal to displace another metal from its salt solution 1 When two different metals are connected by wires and then immersed in an electrolyte, a simple voltaic cell is formed. The metal that is placed at a higher position in the electrochemical series will become the negative terminal of the cell. The metal that is placed lower in the electrochemical series will become the positive terminal of the cell. 2 The metal that is placed higher in the electrochemical series is more electropositive and has a higher tendency to release electrons. Electrons will flow from the negative terminal to the positive terminal. 3 For example, in the zinc/copper simple voltaic cell, zinc metal will become the negative terminal of the cell because zinc is above copper in the electrochemical series. Copper metal will become the positive terminal of the cell. To compare the standard voltage of the voltaic cell 1 A metal that is at a higher position in the electrochemical series can displace another metal that is lower than itself in the electro chemical series from its salt solution. 2 For example, aluminium is above iron in the electrochemical series, hence aluminium can displace iron from an iron(II) salt solution (such as iron(II) sulphate solution). To predict whether a metal can displace hydrogen from an acid 1 Hydrogen ion is placed between lead(II) ion and copper(II) ion in the electrochemical series. 2 All other metals that are placed at higher positions than hydrogen ion in the electrochemical series can displace hydrogen from acids. For example, zinc is above hydrogen in the electrochemical series; hence zinc can displace hydrogen from hydrochloric acid. SPM ’09/P1 1 The further the distance between two metals in the electrochemical series, the greater the cell voltage will be. 2 For example, the distance between magnesium and copper is further than that between zinc and copper in the electrochemical series. Hence the cell voltage produced by a magnesium/copper voltaic cell is greater than that from a zinc/copper voltaic cell. Zn + 2HCl → ZnCl2 + H2 3 Metals below hydrogen ion in the electro chemical series cannot displace hydrogen from acids. Examples are copper, mercury and silver. 169 Electrochemistry 6.9 To confirm the predictions of displacement reactions 6 Problem statement How can the prediction of the displacement reaction of a metal from its salt solution by another metal be confirmed? Materials Pieces of magnesium, iron and copper, solutions of copper(II) sulphate, iron(II) sulphate and magnesium sulphate (concentration and volume of all salt solutions are 0.5 mol dm–3 and 10 cm3 respectively). Hypothesis If metal X is at a higher position than metal Y in the electrochemical series, then metal X can displace metal Y from a salt solution of metal Y. Procedure 1 Pieces of magnesium, copper and iron metals are polished with sandpaper. 2 10 cm3 of 0.5 mol dm–3 solutions of copper(II) sulphate, iron(II) sulphate and magnesium sulphate are put into different test tubes. 3 Magnesium, copper and iron metals are placed in different salt solutions in the test tubes. 4 Observations are made after 20 minutes to check if there is (a) any change in colour of the solution, (b) any deposits of metal, (c) any corrosion of metal. 5 The result of the experiment is tabulated in the table below. Variables (a) Manipulated variable : Different types of metals and salt solutions (b) Responding variable : Deposits of metal or colour change in the salt solution (c) Constant variable : Concentration of salt solutions Apparatus Test tubes, test-tube rack and sandpaper. Experiment 6.9 Results Metal + salt solution Prediction Observation Magnesium + iron(II) sulphate solution Displacement occurs because magnesium is more electropositive than iron • Grey deposit is formed • The colour of the green solution becomes paler • Magnesium dissolves Magnesium + copper(II) sulphate solution Displacement occurs because magnesium is more electropositive than copper • Brown deposit is formed • The colour of the blue solution becomes paler • Magnesium dissolves Iron + magnesium sulphate solution Displacement does not occur because iron is less electropositive than magnesium No noticeable change Iron + copper(II) sulphate solution Displacement occurs because iron is more electropositive than copper • Brown deposit is formed • The blue coloured solution changes to green • Iron dissolves Copper + magnesium sulphate solution Displacement does not occur because copper is less electropositive than magnesium No noticeable change Copper + iron(II) sulphate solution Displacement does not occur because copper is less electropositive than iron No noticeable change Electrochemistry 170 Discussion 1 A more electropositive metal can displace a less electropositive metal from its salt solution. 2 Magnesium is at a higher position than iron and copper in the electrochemical series. Hence magnesium can displace iron from iron(II) sulphate solution and copper from copper(II) sulphate solution. Fe + Cu2+ → Cu + Fe2+ Conclusion The prediction that a metal at a higher position in the electrochemical series can displace a metal which is at a lower position from its salt solution is confirmed. The hypothesis is accepted. Mg + Fe2+ → Mg2+ + Fe Mg + Cu2+ → Mg2+ + Cu 3 Iron is at a higher position than copper in the electrochemical series. Hence iron can displace copper from copper(II) sulphate solution. Electrochemical series: A series of metals based on their tendencies to form metal ions Potassium, K Sodium, Na Calcium, Ca Magnesium, Mg Aluminium, Al Zinc, Zn Iron, Fe Tin, Sn Lead, Pb Hydrogen, H Copper, Cu Silver, Ag Gold, Au A metal placed higher in the series has a higher tendency to form positive ions (it is more electropositive). For example: Zn is more electropositive than Fe. A metal placed at a lower position in the series can be displaced by a metal above it. This series is used to: (a) predict the tendency of a metal to form ions. (b) predict the ability of a metal to displace another metal from its ionic solution. (c) determine the terminals and voltage of a voltaic cell. In a voltaic cell, the metal that becomes the negative terminal is the metal that is higher in position in the electrochemical series as it has a higher tendency to form ions. The further the distance between the metal pairs in a voltaic cell, the greater the cell voltage will be. 6.6 (i) Which metal will become the terminal of the cell? (ii) Predict the voltage of the cell. (c) Predict what will happen if (i) metal Y is immersed in a solution (ii) metal X is immersed in a solution (iii) metal W is immersed in a solution 1 The table below shows the voltages obtained from three voltaic cells using different pairs of metals. Voltaic cell Metal pairs Voltage (V) Positive electrode 1 X and Y 1.2 X 2 X and Z 0.9 X 3 Y and W 0.4 Y negative of Z salt. of Y salt. of X salt. 2 Silver is placed at a position lower than copper and magnesium in the electrochemical series. Predict the observation and reaction that will occur in the following experiment: (a) Silver in copper(II) sulphate solution, (b) Copper in silver nitrate solution, (c) Magnesium in silver nitrate solution. (a) Based on the observation above, arrange the metals W, X, Y and Z in an ascending order according to their electropositivity. (b) A voltaic cell is made from metal Z and metal W. 171 Electrochemistry 6 4 Copper cannot displace either magnesium or iron from their salt solutions because copper is below mag ne sium and iron in the electrochemical series. 6 6.7 2 However, a safe and systematic method of disposal of used batteries and industrial by-products in electrochemical industries is important to prevent environmental pollution. (a) Used batteries should be separated from other household disposal. They are required to be disposed off separately to prevent the chemicals of the batteries from leaking and polluting water sources. Parts of batteries that are useful should be recycled. (b) Chemical wastes from electrolytic industries should be treated to remove poisonous chemicals before being disposed as industrial waste. For example, (i) acids that are used to clean metals before electroplating should be diluted and neutralised before draining off as waste water. (ii) metal ions that are toxic and hazardous to human health such as cadmium ion, chromium ion and nickel ion need to be treated and removed from industrial effluent. Developing Awareness and Responsible Practices when Handling Chemicals used in the Electrochemical Industries 1 Electrochemical industries play an important role in our daily life by improving our quality of life. (a) For example, useful metals such as aluminium, sodium and magnesium are extracted from their minerals or compounds using electrolysis. (b) Useful chemical substances such as chlorine and sodium hydroxide are manufactured on a large scale using electrolysis. (c) Electroplating of iron with chromium protects the iron components of machinery from corrosion. Silver-plating is commonly used in the making of fine cutleries. (d) Various voltaic cells are used in different devices such as radio, torchlight, quartz watch, handphone and others. 1 An electrolyte is a chemical compound which conducts electricity in the molten state or in an aqueous solution and undergoes chemical changes. 2 A non-electrolyte is a chemical compound which does not conduct electricity in any state. 3 Electrolysis is the decomposition of an electrolyte (molten or in aqueous solution) by the passage of an electric current. 4 Graphite or platinum is usually used as electrodes because they are inert. 5 The anode is the electrode connected to the positive terminal of the batteries. 6 The cathode is the electrode connected to the negative terminal of the batteries. 7 Two steps occur during electrolysis. (a) Movement of ions to the electrodes: Cations (positive ions) move towards the cathode (negative electrode) whereas anions (negative ions) move towards the anode (positive electrode). (b) Discharge of ions at the electrodes: Cations discharge by receiving electrons. Generally: An+ + ne– → A Anions discharge by releasing electrons. Generally: Bn– → B + ne– Electrochemistry 8 The factors that determine the types of ions to be discharged at the electrodes are (a) positions of ions in the electrochemical series: The lower positioned ion will be discharged (b) concentration of ions in the solution: The more concentrated ion will be discharged (c) types of electrodes used 9 Uses of electrolysis in industries (a) Extraction of reactive metals. For example: Extraction of aluminium from molten bauxite (Al2O3) using carbon electrodes. (b) Refining of metals. For example: Purification of copper. (c) Electroplating of metals. For example: Copper plating or silver plating. 10 There are two types of voltaic cells: (a) Primary cells: Non-rechargeable cells (cells that cannot be charged again). (b) Secondary cells: Rechargeable cells (cells that can be charged again). 11 The electrochemical series is an arrangement of elements based on their tendencies to form ions. 172 6 6.1 Electrolytes and Nonelectrolytes 1 Which of the following can conduct electricity? A Ethanol B Solid lead(II) nitrate C Magnesium chloride solution D Liquid tetrachloromethane 2 Calcium carbonate powder does not conduct electricity because A it does not contain ions. B it contains covalent molecules. C it contains calcium ions and carbonate ions that are not free to move. D all the atoms in calcium carbonate are bonded by strong covalent bonds. 3 The diagram shows the set-up of apparatus to test the conductivity ’06 of the chemical in the beaker. It was found that there is no deflection on the ammeter needle. 4 Which of the following statements are true about an ’06 electrolyte? I It has ions that conduct electricity in the solid state. II It can conduct electricity in the molten state or in aqueous solution. III It is a compound with ionic bonds only. IV It can be decomposed by electric current. A I and II only B III and IV only C II and IV only D I, II, III and IV 5 Which of the following statements are true about electrolysis? I The cathode is the positive electrode. II Molten covalent compounds can be used as electrolytes. III Platinum can be used as inert electrodes. IV A compound is decomposed by electric current. A I and II only B III and IV only C II, III and IV only D I, III and IV only 6.2 Electrolysis of Molten Compounds 6 A Which of the following action will cause a deflection of the ammeter’s needle? A Add more dry cells in the circuit B Add water to glacial ethanoic acid C Add ethanol to glacial ethanoic acid D Substitute the platinum electrodes with carbon electrodes ? @ heating Which of the following occurs when molten zinc chloride is 6/9 electrolysed in the apparatus as shown in the diagram? 173 A Zinc metal is formed at electrode X. B Chlorine gas is formed at electrode Y. C Zinc ions are attracted to the anode. D Chloride ions are discharged at the positive electrode. 7 When molten lead(II) iodide solution is electrolysed using carbon electrodes, which of the following represents the half-equation that occurs at the cathode? A Pb2+ + 2e– → Pb B Pb → Pb2+ + 2e– C I2 → 2l– + 2e– D 2I– + 2e– → I2 8 Which of the following substances will produce aluminium metal when electrolysis is carried out using carbon electrodes? A Aqueous aluminium sulphate solution B Aqueous aluminium chloride solution C Solid aluminium oxide D Molten aluminium oxide 6.3 Electrolysis of Aqueous Solutions 9 Which of the following compounds produces oxygen gas and hydrogen gas during electrolysis? A Aqueous potassium hydroxide solution B Saturated sodium chloride solution C Aqueous copper(II) nitrate solution D Concentrated hydrochloric acid 10 The diagram below shows the apparatus set-up for the ’09 electrolysis of potassium nitrate solution, KNO3. Electrochemistry 6 Multiple-choice Questions carbon electrode Y potassium nitrate solution carbon electrode X What are the products formed at electrodes X and Y ? TC 54 6 X Nitrogen gas Hydrogen gas B Nitrogen dioxide gas Potassium C Oxygen gas Hydrogen gas D Hydrogen gas Oxygen gas 11 The products formed at the electrodes during the electrolysis of aqueous sodium sulphate solution using carbon electrodes are Anode A Sodium B Hydrogen Sulphur dioxide C Hydrogen Oxygen D Sodium Oxygen Sulphur 12 Electrolysis of dilute sodium chloride solution using carbon electrodes produces oxygen and hydrogen at the anode and cathode respectively. The products formed at the electrodes will change if A platinum is used as the cathode. B a bigger current flows through the circuit. C a concentrated sodium chloride solution is used. D the distance between the electrodes is reduced. 13 Which of the following is true about the electrolysis of aqueous copper(II) chloride solution using copper electrodes? A The mass of cathode decreases. Electrochemistry 14 The diagram shows the set-up of apparatus for the electrolysis ’03 of iron(II) nitrate solution. Y A Cathode B Chlorine gas is evolved at the cathode. C Copper metal deposits at the anode. D The intensity of the blue colour of the solution remains constant. What is formed at electrode X ? A Iron B Oxygen C Hydrogen gas D Nitrogen dioxide gas 15 When aqueous magnesium sulphate solution is electrolysed using graphite electrodes, A the mass of cathode increases. B the mass of anode decreases. C magnesium metal deposits at the cathode. D the concentration of magnesium sulphate solution increases. 16 Electrolysis of aqueous sodium iodide solution is carried out ’09 using carbon electrodes. Which half-equation shows the reaction at the cathode? A 2I– → I2 + 2e– B 4OH– → 2H2O + O2 + 4e– C Na+ + e– → Na D 2H+ + 2e– → H2 17 In an experiment, dilute aqueous potassium iodide solution is electrolysed using carbon electrodes. Which of the following statements are true about this experiment? I A gas that produces a small ‘pop’ sound when tested with a lighted wooden splint is produced at the cathode. II A gas that rekindles a glowing wooden splint is produced at the anode. III The solution around the anode changes to brown colour. IV The concentration of potassium iodide solution increases. A I and II only B II and IV only C I and III only D I, II and IV only 18 Metal Y is placed at a high position in the electrochemical series. When a dilute Y chloride solution is electrolysed using carbon electrodes, the product formed at the cathode is A hydrogen B oxygen C chlorine D metal Y 6.4 Electrolysis in Industries 19 Electrolysis is used to extract aluminium metal from molten aluminium oxide. Which chemical is used to lower the melting point of aluminium oxide to 900°C? A Cryolite B Bauxite C Silicon dioxide D Calcium carbonate 20 What are the suitable chemicals and apparatus used to electroplate a spoon with silver metal by electrolysis? Anode Cathode Electrolyte A Silver Spoon Silver chloride solution B Spoon Silver Silver nitrate solution C Carbon Spoon Silver nitrate solution D Silver Spoon Silver nitrate solution 174 22 The diagram shows the arrangement of apparatus to ’06 electroplate a metal key with chromium. It is found that electroplating does not occur. How would you change the arrangement of the apparatus in order to plate a layer of chromium on the surface of the key? A Replace chromium nitrate solution with chromium chloride solution B Change the supply of direct current to alternating current C Reverse the terminals of the batteries D Replace the chromium electrode with carbon 6.5 Voltaic Cells 24 The diagram below shows a simple chemical cell. Two ’11 different metals are used as electrodes. 0 1 2 3 4 5 26 Voltaic cells that are used in watches and calculators are A dry cells B alkaline cells C mercury cells D nickel-cadmium cells 27 The diagram shows the set-up of apparatus of a chemical cell. ’05 zinc plate copper plate sodium chloride solution 6 21 The presence of foreign metals in copper metal can reduce the conductivity of copper wire. Which of the following is suitable to be used as the cathode in the purification of copper by electrolysis? A Pure copper B Impure copper C Carbon D Platinum Which of the following metals can be used to replace the zinc plate to obtain the brightest light in the light bulb and the highest voltage reading? A Magnesium B Iron C Aluminum D Lead 25 When magnesium metal and copper metal are connected by wire and then immersed in copper(II) sulphate solution, which of the following does not happen? A Electron flows from copper metal to magnesium metal. B Mass of copper increases. C Mass of magnesium decreases. D The colour intensity of blue copper(II) sulphate solution decreases. Which of the following occurs in the chemical cell? A The magnesium rod becomes thicker. B The iron rod becomes thinner. C Electrons flow from iron to magnesium. D The green colour of iron(II) sulphate becomes paler. 28 The diagram shows the set-up of apparatus for an electrochemical cell. 23 The diagram shows the set-up of the apparatus used for the purification of a metal through electrolysis. Which of the following combinations is suitably used for the purification of copper metal? Electrode X Electrode Y Electrode Z A Pure copper Impure copper Copper(II) sulphate B Impure copper Pure copper Copper(II) nitrate C Pure copper Impure copper Sulphuric acid D Impure copper Pure copper Copper(II) carbonate 175 Which of the following observations are true for this experiment? I Zinc electrode becomes thinner. II Brown colour is formed around electrode X. III Gray deposit is formed at electrode Y. IV Intensity of blue colour in beaker M becomes paler. A I and III only B II and III only C II and IV only D I, II and IV only Electrochemistry 29 The diagram shows the set-up of apparatus for a simple cell. Which of the following pairs of metals gives the highest voltmeter reading? 6 Metal X Metal Y A Magnesium Iron B Zinc Copper C Aluminium Silver D Silver Copper 30 A voltaic cell is made using metal X and Y as the electrodes. If electrons flow from metal X to metal Y, metal X and metal Y may be Metal X Metal Y A Iron Silver B Silver Copper C Iron Magnesium D Copper Zinc 31 Which of the following statements is not true about lead-acid accumulator? A The electrolyte used is sulphuric acid. B Lead plate is the negative terminal. C Carbon is the positive terminal. D Lead(II) sulphate is formed when the cell is being used. 6.6 The Electrochemical Series 32 An experiment is carried out to measure the potential ’06 differences produced in voltaic cells made from metal electrode pairs Q-P, R-P, S-T or S-P metals. The results of the experiment is recorded in the table below. Electrochemistry Metal electrode pairs Negative terminal Potential difference (V) Q-P Q 2.7 R-P R 1.1 S-T S 1.3 S-P S 2.1 What is the potential difference of a voltaic cell made of metal electrode pair Q-T ? A 0.8 V B 1.4 V C 1.9 V D 3.5 V 33 If a piece of metal X is immersed in copper(II) sulphate solution, a brown deposit is formed. Metal X may be A copper B platinum C aluminium D silver 34 Two voltaic cells are constructed as shown in the diagram. The voltmeter reading of cell I is 1.1 V while that of cell II is 2.5 V. Which of the following is true of a voltaic cell constructed using metal Q and metal R? A Metal Q will be the negative terminal. B Electrons will flow from metal R to metal Q. C The cell will produce a reading of 3.6 V. D R ions and Q ions are formed. 35 A piece of zinc metal is immersed in a beaker containing a mixture of copper(II) sulphate and magnesium nitrate solution. Which of the following does not happen? A Zinc metal dissolves. 176 B Concentration of magnesium ion in the solution decreases. C A brown deposit is formed. D Zinc ions are formed. 36 The table below shows information about three voltaic ’10 cells. Metals P, Q, R and S are used as electrodes in the cells. Voltaic Negative Positive Voltage cell terminal terminal (V) I P Q 0.9 II R Q 1.3 III R S 2.1 What is the order of the metals from the most electropositive to the least electropositive? A P, Q, R, S B P, R, Q, S C R, P, Q, S D S, Q, P, R 37 Which of the following pairs can undergo a displacement reaction? A Magnesium and potassium chloride solution. B Calcium and zinc sulphate solution. C Iron and calcium nitrate solution. D Copper and magnesium nitrate solution. 38 When an iron nail is immersed in X solution, Fe2+ ions are produced. Solution X may be A magnesium sulphate B zinc nitrate C copper(II) nitrate D sodium chloride 39 Excess metal X powder is added to copper(II) sulphate solution and is stirred. After half an hour, the solution becomes colourless and brown deposit is formed. Metal X may be I calcium II aluminium III magnesium IV silver A I and II only B III and IV only C I, II and III only D II, III and IV only 40 The diagram shows four simple chemical cells. In each cell, copper is one of the electrodes. 6 ’05 In which cell does copper act as the negative terminal? A Cell I B Cell II C Cell III D Cell IV Structured Questions 1 In an experiment, different chemical substances are tested using the set-up of apparatus as shown in Diagram 1. 2 In an experiment, electrolysis of 0.001 mol dm–3 hydrochloric acid is carried out using a electrolytic cell as shown in Diagram 2. Gases are collected at both the electrodes. Diagram 1 Diagram 2 (a) When naphthalene is used as the chemical in the experiment, the light bulb does not light up. Explain this observation. [1 mark] (a) Write the formulae of all the ions present in hydrochloric acid. [1 mark] (b) When lead(II) bromide solid is used as the chemical in the experiment, the light bulb does not light up but lights up when lead(II) bromide is heated to the molten form. Explain this observation. [2 marks] (b) Name a suitable material that can be used as the electrodes in this experiment. [1 mark] (c) (i) Name gas X and gas Y. [2 marks] (ii) Write the half-equation for the reaction that occurs at the anode. [1 mark] (c) Predict the observation that will take place at the anode and the cathode when molten lead(II) bromide is used as the chemical in this experiment. [2 marks] (d) After the electrolysis is carried out for 50 minutes, the concentration of hydrochloric acid increases and a different gas is collected at the anode. (i) Explain why the concentration of hydrochloric acid increases. [2 marks] (ii) Name the new gas collected at the anode and explain why this gas is produced. [2 marks] (iii) Write the half-equation for the reaction that occurs at the anode in (ii). [1 mark] (d) Write the half-equations for the reactions that occur at the anode and the cathode in (c). [2 marks] (e) Predict the products that will be formed if molten zinc chloride is used instead of lead(II) bromide in this experiment. [2 marks] 177 Electrochemistry (a) Write the formula of all the cations present in the copper(II) sulphate solution. [1 mark] 3 Diagram 3 shows the arrangement of apparatus in an electrochemistry experiment. (b) State the direction of the flow of electrons in Cell Q. [1 mark] (c) (i) State the observation at the cathode of Cell P. [1 mark] (ii) Write a half-equation for the reaction that occurred at the cathode of Cell P. [1 mark] (iii) Predict the change of colour intensity of the copper(II) sulphate solution of cell P. [1 mark] 6 Diagram 3 (iv) Name the product formed at the anode if copper electrodes in Cell P are replaced by carbon electrodes. [1 mark] (a) What is the difference between the energy change in Cell A and Cell B? [2 marks] (b) Write half-equations for the reactions that occur at the (i) magnesium electrode in Cell A. [1 mark] (ii) copper electrode in Cell A. [1 mark] (c) (d) Based on cell Q: (i) State the observation on the zinc plate. [1 mark] (ii) Write the half-equations for the reaction that occurs at the zinc plate. [1 mark] (iii) Write an overall ionic equation for the reaction that has taken place. [1 mark] (iv) What happens to the cell voltage if the copper plate is replaced with a silver plate? (i) Name the electrode that serves as the negative terminal in Cell B. [1 mark] (ii) State the reason for your answer in (i). [1 mark] (iii) State the direction of the flow of electrons in Diagram 3. [1 mark] [1 mark] (d) In Cell B, (i) name the product formed at the carbon electrode Q and write an equation for the reaction that occurs. [2 marks] (ii) name the product formed at the carbon electrode P and write an equation for the reaction that occurs. [2 marks] 5 Diagram 5 shows a voltaic cell that is formed from copper metal and lead metal. (e) What would happen if the magnesium electrode in Cell A is replaced with a silver electrode? [1 mark] (f) What would happen if carbon electrodes P and Q are replaced with copper electrodes? [1 mark] Diagram 5 (a) State the positive terminal and the negative terminal of the voltaic cell. [2 marks] 4 Diagram 4 shows two types of cell. (b) Write ionic equations showing the reactions that occur at (i) the negative terminal of the cell. [1 mark] (ii) the positive terminal of the cell. [1 mark] (c) Write the overall ionic equation of the cell. [1 mark] (d) What is the function of the salt bridge? [1 mark] (e) The voltage of the above cell is 0.57 V. If magnesium is above lead in the electrochemical series, what would be the expected voltage produced from a magnesium/copper voltaic cell? [1 mark] 6 An electrolysis process is carried out using the arrangement of apparatus as shown in Diagram 6. Diagram 4 Electrochemistry 178 (c) Write the ionic equation that occurs at (i) electrode L (ii) electrode M [2 marks] (d) What is the product of electrolysis formed at (i) electrode R? (ii) electrode S? [2 marks] (e) Predict any colour change of the solution that may occur in beakers I and II after electrolysis has been carried out for an hour. [2 marks] Diagram 6 (a) Name the electrodes that serve as the anode. [2 marks] (f) (i) Name instrument Q in the diagram. (ii) What is the function of instrument Q? (b) Write the formulae of all the ions present in beaker I. [2 marks] 6 [2 marks] Essay Questions (b) Using a labelled diagram, describe an experiment to show how you can electroplate an iron spoon with another metal. In your description, give the observation and equations for the reactions that occur. [8 marks] 1 (a) What is meant by the term electrolysis? [2 marks] (b) Discuss in terms of ionic theory, the reasons why solid magnesium chloride (crystals) does not conduct electricity whereas molten magnesium chloride does. [4 marks] (c) You are supplied with magnesium chloride crystals and all the necessary apparatus. Describe an experiment to extract magnesium metal from magnesium chloride crystals using electrolysis. What would you observe in this experiment? Using ionic theory, explain how the products are formed at the cathode and the anode. 3 (a) What is the difference between an electrolytic cell and a voltaic cell? [4 marks] (b) You are supplied with metal P, metal Q, their nitrate salt solutions and all the necessary apparatus. Metal P is higher than metal Q in the electrochemical series and both metals have a valency of 2. Describe an experiment to show how you can produce an electric current from chemical reactions. Include a circuit diagram and show how you can detect the flow of electric current in your description. [12 marks] [14 marks] 2 (a) The products of electrolysis may be different even though the same type of electrolyte is used. Using a suitable electrolyte, explain how (i) the types of electrodes, (ii) the concentration of ions can determine the products of electrolysis of an aqueous solution. [12 marks] (c) Predict what will happen when a piece of metal P is placed in Q nitrate solution. Explain your answer. [4 marks] Experiments 1 A group of students carried out three experiments to determine the products of electrolysis of sodium hydroxide solution, potassium iodide solution and aqueous X solution using carbon electrodes. The results of the experiment obtained is tabulated in Table 1. Experiment Chemical substance I 0.1 mol dm–3 sodium hydroxide solution Observation at the cathode Colourless gas is evolved which produces a ‘pop’ sound when a lighted wooden splint is placed near the mouth of the test tube. 179 Observation at the anode Colourless gas is evolved which lights up a glowing wooden splint. Electrochemistry Experiment Chemical substance Observation at the cathode Observation at the anode II 0.5 mol dm–3 aqueous potassium iodide solution Colourless gas is evolved which produces a ‘pop’ sound when a lighted wooden splint is placed near the mouth of the test tube. A brown solution is formed. III 0.5 mol dm–3 aqueous X solution Brown deposit is formed. A brown gas is evolved which changes blue litmus paper to red and decolourises the litmus paper subsequently. 6 Table 1 (a) (i) In experiment I, name the products formed at the anode and cathode. (ii) What factor determines the type of ions discharged at the cathode? [3 marks] (b) (i) What is the product formed at the anode in experiment II? (ii) Suggest a test to identify the product of (i). (iii) What factor determines the type of ions discharged at the anode in experiment II? [3 marks] (c) Write ionic equations for the formation of the product(s) (i) at the anode and the cathode in experiment II. (ii) at the anode in experiment I. [3 marks] (d) In experiment III, aqueous X solution is blue in colour. (i) What is the brown deposit formed at the cathode? (ii) Name the brown gas produced at the anode. (iii) Suggest a chemical substance that may be X. [3 marks] 2 Diagram 1 and Diagram 2 show the set-ups of two electrolytic cells using copper(II) chloride solutions of different concentrations. Plan an experiment to investigate the factor that affects the products of electrolysis of aqueous solutions as shown in Diagrams 1 and 2. Diagram 1 Diagram 2 Your planning should include the following aspects: (a) Aim of experiment (b) Statement of hypothesis (c) All the variables (d) List of substances and apparatus (e) Procedure of the experiment (f) Tabulation of data Electrochemistry [17 marks] 180 FORM 4 THEME: Interaction between Chemicals CHAPTER 7 Acids and Bases SPM Topical Analysis 2008 Year Paper 1 Number of questions 2 A Section 5 2009 – 3 B C – 1 — 2 1 2 A – 5 2010 – 3 B C – 2 — 3 1 1 3 2011 2 3 A B C 1 – – 1 1 4 2 3 A B C 1 – 1 — 2 – ONCEPT MAP pH scale: measurement of the H+ ion concentration • Acids: pH < 7 • Alkalis: pH > 7 • pH value changes with the concentration and strength of acids/bases ACIDS AND BASES Acids: compounds that produce H+ ions in water • Strong acids: complete ionisation to form H+ ions in water • Weak acids: partial ionisation to form H+ ions in water Properties of Acids: • Colour change with indicators • React with bases • React with reactive metals • React with metal carbonates Concentration: units in • g dm–3 • mol dm–3 Relationship between pH values and concentration Bases: compounds that react with acids to form salts and water • Alkalis: soluble bases that produce OH– ions in water • Strong alkalis: complete ionisation to form OH– ions in water • Weak alkalis: partial ionisation to form OH– ions in water Neutralisation: • Reactions between acids and bases • Uses of acids/bases and neutralisation • Determination of end point in titration using acid/base indicators or a computer interface Properties of Bases: • Colour change with indicators • React with acids • React with ammonium salt on heating • React with metal ions to form metal hydroxide 7.1 5 Without the presence of hydrogen ions, a substance does not show any acidic property. Dry hydrogen chloride gas, HCl(g) dissolved in an organic solvent (such as methylbenzene), glacial ethanoic acid and solid ethanedioic acid do not show any acidic property. 6 Acids can be divided into two types: mineral acids and organic acids. Mineral acids are obtained from minerals and most do not contain the element carbon. Organic acids are extracted from living things and contain the element carbon. Characteristics and Properties of Acids and Bases 7 The Meaning of Acids 1 The definition of acids according to Arrhenius Theory: an acid is a chemical compound that produces hydrogen ions, H+ or hydroxonium ions, H3O+ when it dissolves in water. 2 A substance has acidic properties because of the formation of hydrogen ions or hydroxonium ions in water. 3 Dissociation of acids in water produces SPM hydrogen ions and anions. Examples: ’09/P1 Table 7.1 Examples of mineral acids and organic acids H2O HCl(g) ⎯⎯→ H+(aq) + Cl–(aq) (a) hydrogen chloride (b) hydrogen ion H2O HNO3(l) ⎯⎯→ H+(aq) + NO3–(aq) nitric acid (c) chloride ion hydrogen nitrate ion ion Type of acid Examples Mineral acid Hydrochloric acid, HCl, sulphuric acid, H2SO4 and nitric acid, HNO3 Organic acid Ethanoic acid (CH3COOH), methanoic acid (HCOOH), ethanedioic acid (H2C2O4), citric acid, tartaric acid, malic acid and ascorbic acid. H2O H2SO4(l) ⎯⎯→ 2H+(aq) + SO42–(aq) sulphuric acid (d)CH3COOH(l) ethanoic acid hydrogen sulphate ion ion H2O Malic acid is found in apples. Citric acid is found in citrus fruits such as oranges. Tartaric acid is found in grapes. Ascorbic acid is vitamin C. Ethanoic acid is found in vinegar. Lactic acid is found in sour milk. Tannic acid is found in tea leaves. H+(aq) + CH3COO–(aq) hydrogen ion ethanoate ion 4 In actual fact, the hydrogen ion, H+ does not exist individually but is combined with a water molecule (hydrated) to form a hydroxonium ion. H+ + H2O → H3O+ An acid is a substance that produces hydrogen ions in the presence of water. However, H3O+ is usually written as H+(aq) in the simplified way. Hydrochloric acid, nitric acid and sulphuric acid are acids that are usually used in the school laboratories. Our stomachs contain hydrochloric acid that is required for digestion of food. Aspirin, which is used as an analgesic (a type of medicine for reducing pain), is also a type of acid. Figure 7.1 The dissociation (ionisation) of a hydrogen chloride molecule to produce hydroxonium ion in water. Acids and Bases 182 6 A chemical substance has alkaline properties because of the formation of freely moving hydroxide ions, OH– in water. 7 In the presence of water, an alkali dissociates to hydroxide ions and cations. Examples: The Meaning of Bases and Alkalis 1 A base is defined as a chemical substance that can neutralise an acid to produce salt and water only. For example, HCl + NaOH → NaCl + H2O base salt (a) NH3(g) + H2O(l) → NH4+(aq) + OH–(aq) water ammonia 2 Examples of bases are metal oxides and metal hydroxides that contain oxide ions, O2– and hydroxide ions, OH– respectively. Examples: copper(II) oxide, magnesium hydroxide. 3 The reaction between an acid and a base is known as neutralisation. In neutralisation the O2– ions or the OH– ions of a base react with the H+ ions of an acid to form water. (b) KOH(s) potassium hydroxide potassium hydroxide ion ion (c) NaOH(s) ⎯→ Na+(aq) + OH–(aq) H2O sodium ion hydroxide ion H2O calcium hydroxide calcium ion hydroxide ion 8 A compound does not show any alkaline property in the absence of freely moving hydroxide ions. Examples: dry ammonia gas, ammonia gas dissolved in organic solvent (such as propanone), solid sodium hydroxide and solid potassium hydroxide do not show alkaline properties. Figure 7.2 Venn diagram for bases and alkalis Figure 7.4 The association (ionisation) of an ammonia molecule to produce a hydroxide ion Bases Zinc oxide, zinc hydroxide, copper(II) oxide, copper(II) hydroxide + OH–(aq) (d) Ca(OH)2(s) ⎯→ Ca2+(aq) + 2OH–(aq) 4 Most bases are not soluble in water. Bases that are soluble in water are known as alkalis. 5 An alkali is defined as a chemical compound that dissolves in water to produce freely moving hydroxide ions, OH–. examples H2O ⎯→ K+(aq) sodium hydroxide O2– + 2H+ → H2O OH– + H+ → H2O Bases that are insoluble in water ammonium hydroxide ion ion • An alkali is a compound that produces hydroxide ions in the presence of water. • A base is a compound that neutralises an acid and produces salt and water only. Bases that are soluble in water (alkalis) examples Sodium oxide, sodium hydroxide, potassium oxide, potassium hydroxide, calcium hydroxide, ammonia Theories on acids and alkalis: (a) Arrhenius theory: An acid is a compound that produces hydrogen ions when it dissolves in water. An alkali is a compound that produces hydroxide ions when it dissolves in water. (b) Brnsted–Lowry theory: An acid is a proton (hydrogen ion) donor. An alkali is a proton acceptor. Figure 7.3 Flowchart showing types and examples of bases 183 Acids and Bases 7 acid 1 ’07 7 Which of the following statements is true about all bases? A React with acids B Dissolve in water C Produces hydroxide ions D Change red litmus paper to blue Uses of Acids, Bases and Alkalis in Our Daily Life Comments All bases react with acids to form salts. Only soluble bases (alkalis) dissolve in water to produce hydroxide ions that change red litmus paper to blue. Answer A 3 Examples of bases and their uses are given in Table 7.3. SPM ’08/P2, ’09/P1 Table 7.3 Uses of bases 1 Acids and bases are widely used in our everyday life in agriculture, medicine, industry and in the preparation of food. 2 Examples of acids and their uses are given in Table 7.2. Base Sodium hydroxide To make soaps, detergents, bleaching agents and fertilisers Table 7.2 Uses of acids Acid Uses Ammonia Uses To make fertilisers, nitric acid, grease remover and to maintain latex in liquid form To make paints, detergents, polymers, fertilisers, as an electrolyte in lead-acid accumulators Calcium hydroxide To make cement, limewater and to neutralise the acidity of soil Hydrochloric acid To clean metals before electroplating Magnesium hydroxide To make toothpaste, gastric medicine (antacid) Nitric acid To make fertilisers, explosive substances (such as T.N.T.), dyes and plastics Aluminium hydroxide To make gastric medicine (antacid) Benzoic acid To preserve food Carbonic acid To make gassy (carbonated) drinks Ethanoic acid A component of vinegar Tartaric acid To make baking powder Sulphuric acid Cleaning agent contains ammonia Methanoic acid is used in the coagulation of rubber latex Fertilisers are made from acids and alkalis Acids and Bases Soaps and detergents are made from sodium hydroxide 184 7.1 SPM ’09/P2, ’10/P3, ’11/P2 Conclusion 1 Aqueous ethanoic acid turns blue litmus paper to red, indicating its acidic property. 2 Ethanoic acid in a dry condition or dissolved in organic solvents does not show any acidic property. 3 Ionisation of acids will only occur in the presence of water to produce hydrogen ions which are responsible for the acidic properties. 4 Water is essential for the formation of hydrogen ions which gives the acidic properties in an acid. The hypothesis is accepted. Problem statement Is water needed for an acid to show its acidic properties? Hypothesis An acid will only show its acidic properties when dissolved in water. Variables (a) Manipulated variable : Types of solvents-water and propanone (b) Responding variable : Change in the colour of blue litmus (c) Constant variable : Type of acid and blue litmus paper Apparatus Test tube and droppers. Materials Glacial (dry) ethanoic acid, aqueous ethanoic acid, ethanoic acid dissolved in dry propanone and blue litmus paper. Procedure 1 A piece of dry blue litmus paper is placed in a test tube. 2 A few drops of glacial ethanoic acid are placed onto the blue litmus paper using a dropper. 3 The effect of the glacial ethanoic acid on the blue litmus paper is recorded. 4 Steps 1 to 3 of the experiment are repeated using aqueous ethanoic acid and ethanoic acid dissolved in propanone to replace glacial ethanoic acid. 5 The observations are then tabulated. 7 To investigate the role of water in showing the properties of acids Discussion 1 In the presence of water, an acid dissociates into hydrogen ions that cause acidity in an acid. 2 Dry acids do not show any acidic properties in the absence of water because dry acids exist as covalent molecules. Hydrogen ions are not produced. 3 Solvents such as methylbenzene, propanone and trichloromethane cannot replace water for an acid to show its acidic properties. This is because an acid exists as covalent molecules in these organic solvents; H+ ions are not produced in these solutions. 4 Glacial ethanoic acid (CH3COOH) consists of acid molecules only. CH3COOH molecule is a covalent compound. 5 Figure 7.5 shows the types of particles that are present in ethanoic acid dissolved in propanone and in water. Condition of ethanoic acid Glacial (dry) Aqueous (dissolved in water) Dissolved in propanone Observation Inference No noticeable colour change in the litmus paper Blue litmus paper has changed to red No noticeable colour change in the litmus paper Does not show any acidic properties Shows acidic properties Does not show any acidic properties Figure 7.5 Particles in ethanoic acid dissolved in (a) propanone (b) water 185 Acids and Bases Experiment 7.1 Results 7.2 To investigate the role of water in showing the alkaline properties of alkali 7 Problem statement Is water essential for an alkali to show its alkaline properties? 2 The test tube must be stoppered immediately after the red litmus paper is put in. Hypothesis An alkali will only show its alkaline properties when dissolved in water. Results Variables (a) Manipulated variable : Types of solvents–water and propanone (b) Responding variable : Change in the colour of red litmus paper (c) Constant variable : Type of alkali and red litmus paper Apparatus Test tubes and droppers. Materials Dry ammonia gas stoppered in a test tube, ammonia gas dissolved in propanone, aqueous ammonia solution and red litmus paper. Condition of ammonia Experiment 7.2 Inference Dry No colour change in the red litmus paper Does not show alkaline property Aqueous (dissolved in water) Red litmus has changed to blue Shows alkaline properties Dissolved in propanone No colour change in the red litmus paper Does not show alkaline property Conclusion 1 Aqueous ammonia solution turns the red litmus paper to blue, indicating its alkaline property. 2 Dry ammonia gas or ammonia gas dissolved in organic solvents does not show any alkaline property. 3 An alkali shows its alkaline properties only in the presence of water. When water is present, ammonia ionises to produce OH– ions that are responsible for its alkaline properties. 4 Water is essential for the formation of hydroxide ions that cause alkalinity in an alkali. The hypothesis is accepted. Procedure 1 A piece of dry red litmus paper is put into a stoppered test tube of dry ammonia gas and the test tube is then stoppered back immediately (Figure 7.6). 2 The effect of the dry ammonia gas on the red litmus paper is recorded. 3 Another piece of dry red litmus paper is put in 5 cm3 of aqueous ammonia solution in a separate test tube. 4 Step 3 of the experiment is repeated using ammonia dissolved in propanone to replace aqueous ammonia solution. Discussion 1 In the presence of water, an alkali ionises to form hydroxide ions, OH– that change red litmus paper to blue. 2 Aqueous ammonia solution (ammonia dissolved in water) consists of NH4+ ions, OH– ions and NH3 molecules. An aqueous ammonia solution is alkaline due to the presence of hydroxide ions. NH3 + H2O Figure 7.6 Testing for the alkaline properties of ammonia gas NH4+ + OH– 3 Dry alkalis, solid alkalis (such as solid calcium hydroxide and barium hydroxide) and alkalis dissolved in organic solvents (such as propanone) do not show any alkaline properties. This is because the alkalis do not dissociate into hydroxide ions. Safety precautions 1 Ammonia gas is poisonous. This experiment involving dry ammonia gas should be carried out in a fume cupboard. Acids and Bases Observation 186 Glacial ethanoic acid is the pure and dry form of ethanoic acid. It is named ‘glacial’ because it appears as ice when it solidifies below its melting point. 1 Acids are sour in taste. 2 Acid solutions have pH values of less than 7. 3 Acids change colours of indicators as shown in Table 7.4. 4 Acids can react with (a) bases to produce salts and water, (b) metals to produce salts and hydrogen gas, (c) carbonates to produce salts, carbon dioxide gas and water. 1 If the electrical conductivity of ethanoic acid in pro panone and aqueous ethanoic acid is tested in turn, only the aqueous solution of acid conducts electri city (light bulb is lighted up or ammeter needle is deflected). 2 This shows the presence of freely moving ions in an aqueous solution of acid. CH3COOH(l) ethanoic acid CH3COO –(aq) + H+(aq) ethanoate ion hydrogen ion H+ ions and CH3COO– ions conduct electricity Table 7.4 Effects of acids on indicators H+ ions change blue litmus to red Colour of indicator in acidic solution Indicator 3 Dry acids do not conduct electricity. This is because there are no freely moving ions. Dry acid exists as covalent molecules. 4 Similarly, ammonia dissolved in propanone does not conduct electricity. It exists as covalent molecules. 5 An aqueous ammonia solution can conduct electricity, showing the presence of freely moving ions. Blue litmus paper Red Universal indicator Orange and red Methyl orange Red To investigate the chemical properties of acids Apparatus Test tube, test tube holder, spatula, Bunsen burner, delivery tubes with stopper and wooden splint. Materials 1.0 mol dm–3 sulphuric acid, copper(II) oxide, zinc powder, sodium carbonate powder and limewater. 7 Chemical Properties of Acids Procedure 1 A little copper(II) oxide is added to 5 cm3 of sulphuric acid in a test tube. The mixture is heated slowly (Figure 7.7) and any changes that occur are recorded. 2 A little zinc powder is added to 5 cm3 of dilute sulphuric acid in a test tube. The gas evolved is tested by placing a lighted wooden splint near the mouth of the test tube (Figure 7.8). 3 A little sodium carbonate powder is added to 5 cm3 of dilute sulphuric acid in a test tube. The gas evolved is tested with limewater (Figure 7.9). Figure 7.8 An acid with a metal Activity 7.1 Figure 7.7 An acid with a base SPM ’10/P2 Figure 7.9 An acid with a metal carbonate 187 Acids and Bases Results Test on acid Observation Inference Heating with copper(II) oxide • Black powder dissolved • Blue solution is formed Test with zinc powder • Effervescence occurred • Gas produced a ‘pop’ sound when it is tested with a lighted wooden splint • Zinc powder dissolved 7 Test with sodium carbonate • Copper(II) salt solution is formed • Effervescence occurred • Gas evolved turned limewater milky • White solid of sodium carbonate dissolved If the salt solution is evaporated until saturated, salt crystals will form upon cooling. Examples: (a) Black copper(II) oxide powder (a base) dissolves in dilute sulphuric acid to produce a salt, copper(II) sulphate (blue colour) and water. 3 A dilute acid will react with a metal carbonate to produce a salt, carbon dioxide gas and water. acid + metal carbonate → salt + carbon dioxide + water CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l) Examples: (a) Sodium carbonate reacts with dilute sulphuric acid to produce a salt, sodium sulphate, carbon dioxide gas and water. (b) Copper(II) oxide dissolves in ethanoic acid to form a salt, copper(II) ethanoate and water. CuO(s) + 2CH3COOH(aq) → Cu(CH3COO)2(aq) + H2O(l) Na2CO3(s) + H2SO4(aq) → Na2SO4(aq) + CO2(g) + H2O(l) (c) Nitric acid reacts with sodium hydroxide (an alkali) to produce a salt, sodium nitrate and water. (b) Calcium carbonate reacts with dilute hydrochloric acid to produce a salt, calcium chloride, carbon dioxide gas and water. HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l) Examples: (a) Zinc dissolves in sulphuric acid to form a salt, zinc sulphate and hydrogen gas. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) Acids and Bases • Carbon dioxide gas is produced • A salt solution is formed Mg(s) + 2CH3COOH(aq) → Mg(CH3COO)2(aq) + H2(g) Acid + base → salt + water acid + reactive metal → salt + hydrogen • A salt solution is formed (b) Magnesium dissolves in ethanoic acid to form a salt, magnesium ethanoate and hydrogen gas. Discussion 1 A dilute acid reacts with a base to produce salt and water only. 2 A dilute acid will react with a reactive metal to produce a salt and hydrogen gas. • Hydrogen gas is produced CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) Conclusion 1 Sulphuric acid reacts with a base (copper(II) oxide) to produce salt and water. 2 Sulphuric acid reacts with a reactive metal (zinc) to produce a salt and hydrogen gas. 3 Sulphuric acid reacts with a metal carbonate (sodium carbonate) to produce a salt, water and carbon dioxide gas. 188 Chemical Properties of Alkalis 1 Alkalis are bitter in taste and feel soapy. 2 Alkaline solutions have pH values of more than 7. 3 Alkalis change the colours of indicators as shown in Table 7.5 below. Non-reactive metals such as copper and silver do not react with dilute acid. Very reactive metals such as sodium and potassium will react with dilute acid vigorously and may produce an explosion. 2 Table 7.5 Effects of alkalis on indicators ’01 Colour of indicator in alkaline solution Indicator Which of the following compounds reacts with limestone powder to produce a gas that turns limewater milky? A Nitrogen dioxide gas B Hydrogen chloride gas dissolved in tetra chloromethane C Sulphur dioxide gas dissolved in propanone D Sulphur dioxide gas dissolved in water Blue 7 Red litmus paper Blue or purple Universal indicator Yellow Methyl orange 4 An alkali reacts with an acid to produce salt and water. For example: Comments An acidic gas must first dissolve in water before reacting with calcium carbonate (limestone) powder to produce carbon dioxide gas which turns limewater milky. KOH(aq) + HCl(aq) → KCl(aq) + H2O(l) 5 When an alkali is heated with an ammonium salt, ammonia gas is produced. For example: Answer D NH4+(aq) + OH–(aq) → NH3(g) + H2O(l) from ammonium salt Generally, (a) metal oxides and metal hydroxides are basic. For example: MgO + H2O → Mg(OH)2 magnesium oxide magnesium hydroxide from alkali 6 An aqueous alkali forms metal hydroxide as precipitate when added to an aqueous salt solution. For example: Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s) (b) non-metal oxides are acidic. For example, SO2, NO2 or CO2. SO2 + H2O → H2SO3 sulphur dioxide sulphurous acid from copper(II) salt solution from alkali copper(II) hydroxide as blue precipitate Apparatus Test tubes, test tube holder, spatula, Bunsen burner, delivery tubes with stopper and red litmus paper. Materials 2.0 mol dm–3 sodium hydroxide solution, benzoic acid powder, ammonium chloride powder, 1.0 mol dm–3 iron(III) sulphate solution. Procedure 1 A little benzoic acid powder is added to 5 cm3 of sodium hydroxide solution in a test tube. Any changes that occur are recorded. 2 A little ammonium chloride powder is added to 5 cm3 of sodium hydroxide solution in a test tube. The mixture is heated gently. The gas evolved is tested with a piece of damp red litmus paper. 3 5 cm3 of sodium hydroxide solution is added to 5 cm3 of iron(III) sulphate solution in a test tube. Any changes that occur are recorded. 189 Acids and Bases Activity 7.2 To investigate the chemical properties of alkalis Results 7 Test on sodium hydroxide Observation Inference 2 If the salt solution is evaporated in an evaporating dish until a saturated solution is formed, white crystals of sodium benzoate will be crystallised upon cooling. 3 When sodium hydroxide is heated with ammonium chloride (an ammonium salt), ammonia gas is produced. A salt solution is formed With benzoic acid powder added White powder dissolves and a colourless solution is formed Heating with ammonium chloride powder Ammonia gas A pungent gas that turns damp is produced red litmus paper blue is evolved With iron(III) A brown precipitate is sulphate solution added formed NaOH(aq) + C6H5COOH(s) → C6H5COONa(aq) + H2O(l) NH4Cl(s) + NaOH(aq) → NaCl(aq) + H2O(l) + NH3(g) Iron(III) hydroxide is formed In this reaction, ammonium ions react with hydroxide ions to produce ammonia gas. The ionic equation for this reaction is NH4+(aq) + OH–(aq) → NH3(g) + H2O(l) Conclusion 1 Sodium hydroxide reacts with benzoic acid to produce salt and water. 2 When sodium hydroxide is heated with ammonium chloride, ammonia gas which turns red litmus to blue is produced. 3 Sodium hydroxide solution reacts with an aqueous iron(III) solution to produce a brown precipitate, iron(III) hydroxide. 4 Sodium hydroxide solution hydroxide ions in water. to NaOH → Na+ + OH– Hydroxide ions combine with iron(III) ions from iron(III) sulphate solution to form insoluble iron(III) hydroxide as a brown precipitate. Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s) Discussion 1 Sodium hydroxide as an alkali reacts with benzoic acid, C6H5COOH to produce a salt, sodium benzoate and water in a neutralisation reaction. from iron(III) sulphate from sodium hydroxide CH3COOH(aq) Basicity of Acids 1 Basicity of an acid is the number of moles of OH– ions that are required to react with one mole of the acid. 2 Since one mole of OH– ions reacts with one mole of H+ ion, the basicity of an acid is also the number of moles of H+ ion that can be produced by one mole of the acid when it dissolves in water. 3 A monoprotic acid (or monobasic acid) is an acid that will produce one mole of H+ ion when one mole of the acid dissolves in water. For example, although ethanoic acid has four hydrogen atoms in the molecule, only one of the hydrogen dissociates to form H+ ion in water. Acids and Bases dissociates iron(III) hydroxide as brown precipitate CH3COO–(aq) + H+(aq) Three H atoms bonded to carbon do not dissociate Only one H atom dissociates to form H+ ion 4 A diprotic acid (or dibasic acid) is an acid that will produce two moles of H+ ions from one mole of the acid in water. For example: H2SO4(aq) → 2H+ (aq) + SO42–(aq) 1 mol sulphuric acid 2 mol hydrogen ions sulphate ion 5 A triprotic acid (or tribasic acid) is an acid that will produce three moles of H+ ions from one mole of the acid in water. For example: H3PO4(aq) 3H+(aq) + PO43–(aq) 1 mol phosphoric acid 190 3 mol hydrogen ions phosphate ion Table 7.6 Examples of monoprotic acid and diprotic acid Examples of monoprotic acid Hydrochloric acid (HCl), nitric acid (HNO3), ethanoic acid (CH3COOH) and methanoic acid (HCOOH) Basicity of an acid is not the same as the number of H atoms in the formula of the acid. Examples of diprotic acid Sulphuric acid (H2SO4), ethanedioic acid (H2C2O4), carbonic acid (H2CO3) and chromic acid (H2CrO4) Basicity is the number of moles of H+ ions produced by one mole of acid in water. 1 (a) Explain what you understand by the term (i) an acid (ii) a base (iii) an alkali (b) What is the effect of an acid and an alkali on moist litmus paper? 3 Identify the chemicals Q, R, X, Y and gas Z in the following reactions: (a) H2SO4 + Q → MgSO4 + H2O + CO2 (b) Ca(OH)2 + 2R → Ca(NO3)2 + 2H2O (c) 2Al + 6X → 2AlCl3 + 3H2 heat (d) Y + NH4NO3 ⎯⎯→ KNO3 + H2O + Z 2 Identify the correct uses of the following acids and bases. 4 Write equations to show the reactions between (a) sulphuric acid and magnesium oxide (b) nitric acid and aluminium metal (c) hydrochloric acid and calcium carbonate (d) ethanoic acid and sodium hydroxide (e) potassium hydroxide and ammonium chloride when heated Acids or bases: H2SO4, HNO3, CaO, Ca(OH)2, Mg(OH)2, NH3, NaOH Acids or bases Uses To make antacid 5 Effervescence occurs when magnesium powder is added to aqueous hydrochloric acid. However, no noticeable change takes place when magnesium powder is added to hydrogen chloride dissolved in methylbenzene. Explain why. To make fertiliser To make soap To neutralise acidity in soil Acid Alkali dissolves in water dissolves in water produces H+ ions produces OH– ions reacts with carbonate metal reacts with base acid ammonium salt metal ions heat salt + carbon dioxide + water salt + hydrogen salt + water 191 ammonia gas metal hydroxides Acids and Bases 7 7.1 1 The pH scale is a set of numbers used to indicate the degree of acidity or alkalinity of a solution. 2 The values of the pH scale range from 0 to 14. • pH < 7 ⇒ acidic solution • pH = 7 ⇒ neutral solution • pH > 7 ⇒ alkaline solution 3 pH is actually a measurement of the concen tration of hydrogen, H+ ions in a solution. 4 The higher the concentration of the H+ ions, the lower the pH value and the more acidic the solution. 5 The higher the concentration of the OH– ions, the higher the pH value and the more alkaline the solution. 6 The relationship between the pH scale, acidity or alkalinity and concentration of H+ ions is shown below. • All acids have pH < 7. • The lower the pH value, the higher the H+ ion concentration. • All alkalis have pH > 7. • The higher the pH value, the higher the OH– ion concentration. 7.2 The Strength of Acids and Alkalis 7 The pH Scale 4 A pH meter is an electric meter that is used to measure the pH value of a solution accurately. A pH meter will show the pH value when its probe is immersed in a solution pH meter to be tested. 5 With a computer interface, the exact pH value can be displayed on the computer screen when the pH meter is placed in the solution. Measurement of pH Value of a Solution 1 The pH value of a solution can be measured by using (a) universal indicator or pH paper (b) a pH meter (with or without a computer interface) 2 Universal indicator is a mixture of indicators that gives different colours corresponding to different pH values as shown in Table 7.7. 3 Universal indicator is used in the form of (a) solution, or (b) paper strips (also known as pH paper). Table 7.7 Colours of universal indicator pH value 0, 1, 2 Colour red 3 4 5 6 7 orange orange orange yellow green red yellow 8 9 10 11 12, 13, 14 greenishblue blue blue bluishpurple purple Activity 7.3 To measure the pH values of some solutions used in daily life Apparatus Beakers, universal indicator solution, dropper, standard colour chart of universal indicator. Acids and Bases Materials Soap solution, carbonated drink, tap water, orange fruit juice, distilled water, milk, tea, dilute sodium hydroxide and hydrochloric acid. 192 Procedure 1 About 10 cm3 of soap solution is placed in a small beaker. 2 Two drops of universal indicator solution are added to the soap solution. The solution is then stirred. 3 The colour of the solution produced is matched against the standard colour chart of universal indicator. The corresponding pH value of the colour is noted and recorded. 4 The experiment is repeated using carbonated drink, tap water, orange fruit juice, distilled water, milk, tea, dilute sodium hydroxide and hydrochloric acid in place of the soap solution. Results pH value Soap Carbonated Tap Orange solution drink water juice 10 5 6 Distilled water Milk Tea Dilute sodium hydroxide Dilute hydrochloric acid 7 6 5 13 1 4 7 Solution Conclusion 1 Different solutions have different pH values. 2 The pH value of a solution can be measured using the universal indicator solution. Degree of Dissociation HCl → H+ + Cl– HNO3 → H+ + NO3– H2SO4 → 2H+ + SO42– 1 The strength of an acid or an alkali depends on the degree of dissociation (also known as the degree of ionisation). 2 The degree of dissociation measures the percentage or fraction of molecules that dissociates into ions when dissolved in water. 3 For example, the degree of dissociation of hydrochloric acid is 100% or 1. This means that all the hydrogen chloride molecules in hydrochloric acid will ionise to form H+ ions and Cl– ions when dissolved in water. 4 In a 1.0 mol dm–3 aqueous ethanoic solution, only 4 out of 1000 molecules of ethanoic acid dissociate to form ions. Degree of dissociation of a 1.0 mol dm–3 aqueous ethanoic solution is 4 — — — — = 0.004 or 0.4%. 1000 (The one–way dissociation) → indicates complete 3 Complete dissociation (100%) in water by a strong acid produces a high concentration of H+ ions and hence a low pH. 4 Weak acids are chemicals that dissociate partially (incomplete dissociation) into hydrogen ions H+ in water. 5 Most of the organic acids such as ethanoic acid, ethanedioic acid, methanoic acid, citric acid and tartaric acid are weak acids. CH3COOH H2C2O4 5 Acids can be divided into 2 categories: strong acids and weak acids, depending on their degree of dissociation. 6 Alkalis can be divided into 2 categories: strong alkalis and weak alkalis, depending on their degree of dissociation. Strong and Weak Acids arrow (The two–way arrow reaction) CH3COO– + H+ 2H+ + C2O42– indicates reversible Examples of weak acids Ethanoic acid, CH3COOH Methanoic acid, HCOOH Ethanedioic acid, H2C2O4 Carbonic acid, H2CO3 Phosphoric acid, H3PO4 Chromic acid, H2CrO4 Nitrous acid, HNO2 Sulphurous acid, H2SO3 SPM ’08/P2, ’09/P1, ’10/P3 1 A strong acid is a chemical substance that dissociates completely (degree of dissociation is 100%) into hydrogen ions, H+ in water. 2 Mineral acids such as hydrochloric acid, nitric acid and sulphuric acid are strong acids. 193 Acids and Bases 6 In a weak acid solution, a big portion of the weak acid exists as molecules and only a small portion dissociates to ions. Strong and Weak Alkalis 1 A strong alkali is a chemical substance that dissociates completely (100%) to hydroxide ions, OH– in water. 2 Examples of strong alkalis are sodium hydroxide and potassium hydroxide. A concentrated acid does not mean that it is a strong acid. For example, concentrated ethanoic acid is still a weak acid. Consequently, a dilute acid does not mean that it is a weak acid. For example, dilute hydrochloric acid is still a strong acid. NaOH → Na+ + OH– KOH → K+ + OH– 7 (The one–way dissociation) Test 0.1 mol dm HCl Ca(OH)2 Magnesium ribbon Hydrogen gas evolves vigorously Solution Test CO2 gas evolves vigorously CO2 gas evolves slowly Electrical conductivity Light bulb lights up brightly Light bulb lights up dimly Conclusion • High • Low concentration concentration of H+ ions of H+ ions • HCl is a strong • CH3COOH is a acid weak acid Acids and Bases Ca2+ + 2OH– Table 7.9 Comparison of properties of a strong alkali (NaOH) with a weak alkali (NH3) Hydrogen gas evolves slowly Calcium carbonate NH4+ + OH– 5 Partial dissociation of a weak alkali results in a low concentration of OH– ions. Hence, the pH value of a weak alkali is lower than that of a strong alkali with the same concentration. 6 Table 7.9 shows the comparison of properties of a strong alkali (sodium hydroxide solution) with a weak alkali (aqueous ammonia solution). 0.1 mol dm CH3COOH Red colour, pH~1 Orange red colour, pH~4 complete (The two–way arrow indicates partial dissociation) –3 Universal Indicator → indicates NH3 + H2O Table 7.8 Comparison between a strong acid (HCl) and a weak acid (CH3COOH) –3 arrow 3 A weak alkali is a chemical substance that dissociates partially (incomplete dissociation) to hydroxide ions, OH– in water. 4 Examples of weak alkalis are aqueous ammonia, calcium hydroxide and magnesium hydroxide. 7 For two different acids of the same concentration, the acid with the lower pH value is the stronger acid. 8 Partial dissociation of a weak acid results in a low H+ ion concentration. 9 Strong and weak acids have the same chemical properties, but the rate of reaction and electrical conductivity of a weak acid is lower as shown in Table 7.8. Solution SPM ’08/P1, ’09/P1 194 0.1 mol dm–3 NaOH solution 0.1 mol dm–3 aqueous NH3 Universal Indicator Purple colour, pH~13 Blue colour, pH~10 Electrical conductivity Light bulb lights up brightly Light bulb lights up dimly Conclusion • High • Low concentration concentration of OH– ions of OH– ions • NaOH is a • Aqueous NH3 strong alkali is a weak alkali Decreasing pH 1 Increasing pH 7 Strong acids: • Complete ionisation • High H+ ion concentration • pH value: 1–2 Weak acids: • Partial ionisation • Low H+ ion concentration • pH value: 3–6 14 Weak alkalis: • Partial ionisation • Low OH– ion concentration • pH value: 8–12 Strong alkalis: • Complete ionisation • High OH– ion concentration • pH value: 13–14 SPM ’10/P1 7 To measure the pH values of solutions with the same concentration Materials 0.1 mol dm–3 hydrochloric acid, 0.1 mol dm–3 ethanoic acid, 0.1 mol dm–3 aqueous ammonia and 0.1 mol dm–3 sodium hydroxide solution. Procedure 1 About 15 cm3 of 0.1 mol dm–3 hydrochloric acid is placed in a small beaker. 2 The probe of a pH meter is rinsed with distilled water. 3 The pH meter probe is then immersed in the acid in the beaker. The reading registered on the pH meter after it has stabilised is recorded. 4 Steps 1 to 3 of the experiment are repeated using ethanoic acid, ammonia solution and sodium hydroxide solution to replace the hydrochloric acid. Results Solution of 0.1 mol dm–3 pH value Hydrochloric acid, HCl 1 • High concentration of H+ ions • A strong acid Ethanoic acid, CH3COOH 3 • Low concentration of H+ ions • A weak acid Aqueous ammonia, NH3 10 • Low concentration of OH– ions • A weak alkali Sodium hydroxide, NaOH 13 • High concentration of OH– ions • A strong alkali Inference Conclusion 1 Acids have pH values of less than 7. 2 Different acids with the same concentration have different pH values. The pH of hydrochloric acid is lower than ethanoic acid of the same concentration. 3 Alkalis have pH values of more than 7. 4 Different alkalis with the same concentration have different pH values. The pH of sodium hydroxide solution is higher than aqueous ammonia of the same concentration. Discussion 1 Hydrochloric acid, HCl, and ethanoic acid, CH3COOH, are both acidic as their pH values are less than 7. 2 However, the pH value of HCl is less than that of CH3COOH with the same concentration indicating that the concentration of hydrogen ions in HCl is higher than that in CH3COOH. 3 HCl is an example of a strong acid which undergoes complete ionisation. CH3COOH is an example of a weak acid which undergoes partial ionisation. 4 Ammonia, NH3, and sodium hydroxide solution are both alkaline as their pH values are more than 7. 5 However, the pH value of NaOH is more than that of NH3 with the same concentration indicating that the concentration of hydroxide ions in NaOH is higher than that in NH3. 6 NaOH is an example of a strong alkali which undergoes complete ionisation. NH3 is an example of a weak alkali which undergoes partial ionisation. 7 The strength of an acid (strong or weak) and the strength of an alkali (strong or weak) depends on the degree of dissociation. 195 Acids and Bases Activity 7.4 Apparatus 50 cm3 beaker and pH meter. 3 7.2 ’03 1 Six solutions A, B, C, D, E and F with concentration of 1.0 mol dm–3, have pH values as shown in the table below: Information about two solutions is given below: 7 Concentration of nitric acid = 1.0 mol dm–3 Concentration of carbonic acid = 1.0 mol dm–3 Which of the following statements are true about the two solutions given above? I Nitric acid is a stronger acid than carbonic acid. II The pH value of nitric acid is higher than carbonic acid. III The degree of dissociation of nitric acid in water is higher than that of carbonic acid. IV The concentration of H+ ions in nitric acid is higher than that in carbonic acid. A I and II only B III and IV only C I, III and IV only D I, II, III and IV B C D E F pH values 13 7.0 10 4.0 1.0 3.5 2 Using suitable examples, explain the terms strong acid and weak acid. Predict the difference in pH values of the two acids with the same concentration. 3 The degree of dissociation of ethanoic acid is higher than that of propanoic acid but is lower than that of methanoic acid. (a) Arrange the above three acids in ascending order of the strength of acidity. (b) If the pH value of 1.0 mol dm–3 ethanoic acid is 3, predict the pH value of 1.0 mol dm–3 methanoic acid and propanoic acid. ’07 7.3 Which of the following statements describe a strong alkali? I Has a high pH value II Ionises completely in water III Has a high concentration of hydroxide ions IV Exists as molecules in water A I and II only B III and IV only C I, II and III only D I, II, III and IV Concentration of Acids and Alkalis The Meaning of Concentration and Molarity, and Their Relationship 1 A solution is formed when a solute dissolves in a solvent. solute + solvent → solution For example, when sodium hydroxide (solute) dissolves in water (solvent), sodium hydroxide solution is formed. 2 Concentration and molarity are measurements of the amount of solutes dissolved in a given volume of solvent when a solution is formed. 3 The amount of a solute can be measured in the unit of ‘gram’ or ‘mole’. The quantity of a solution produced is usually measured in the unit of volume, dm3. Comments A strong alkali ionises completely in water to produce a high concentration of hydroxide ions in water and hence a high pH value. (I, II and III are correct) A strong alkali exists as ions in water. (IV is incorrect) Answer C Acids and Bases A (a) Which of the above solutions has (i) the highest concentration of H+ ions? (ii) the highest concentration of OH– ions? (b) Which of the above solutions is (i) a strong acid? (iii) a weak acid? (ii) a strong alkali? (iv) a weak alkali? (c) Which of the above solutions may be (i) sodium chloride solution? (ii) hydrochloric acid? (iii) aqueous ammonia? (iv) sodium hydroxide solution? Comments Nitric acid is a stronger acid (I correct), has a higher degree of dissociation (III correct) and hence a higher degree of H+ ions concentration (IV correct) but a lower pH value than carbonic acid, which is a weak acid (II incorrect). Answer C 4 Solution 196 Solution Mass of copper(II) sulphate = 5.00 g Convert volume from Volume of solution = 500 cm3 cm3 to dm3 500 =— — — — — dm3 = 0.5 dm3 1000 Hence, concentration of copper(II) sulphate solution 5.00 g =— — — — — — 0.5 dm3 Concentration (g dm–3) Mass of solute dissolved (g) = 10.0 g dm–3 =— — — — — — — — — — — — — — — — — — — — — — — — 4 The concentration of a solution is the mass (in grams) or the number of moles of solute dissolved in a solvent to form 1.0 dm3 (1000 cm3) of solution. Hence the concentration of a solution can be defined in two ways: Mass of solute dissolved (g) Concentration = — — — — — — — — — — — — — — — — — — — — — — — — — — Volume of solution (dm3) –3 (g dm ) Volume of solution (dm3) Number of moles of solute (mol) Concentration = — — — — — — — — — — — — — — — — — — — — — — — — — Volume of solution (dm3) –3 (mol dm ) 5 For example, • a 23.0 g dm–3 NaOH solution has 23.0 g of NaOH in 1.0 dm3 solution. • a 0.5 mol dm–3 NaOH solution has 0.5 mol of NaOH in 1.0 dm3 solution. 6 Concentration in terms of mol dm–3 is more commonly known as molarity. In chemistry, the measurement of concentration in mol dm–3 (molarity) is more useful because all changes in chemical reactions occur in terms of moles. 7 Concentration (in g dm–3) can be converted SPM to molarity by dividing concentration (in ’08/P1 g dm–3) by the molar mass. The molar mass is the mass of 1 mol of substance. 7 2 What is the mass of sodium carbonate required to dissolve in water to prepare a 200 cm3 solution that contains 50 g dm–3? Solution Volume of solution = 200 cm3 200 Convert volume =— — — — dm3 = 0.2 dm3 from cm3 to dm3 1000 Concentration mass of Na2CO3 dissolved (g) =— — — — — — — — — — — — — — — — — — — — — — — — (g dm–3) volume of solution (dm3) Mass of Na2CO3 required = 50 g dm–3  0.2 dm3 Mass = Concentration (g dm–3)  Volume of solution (dm3) = 10 g Concentration (g dm–3) Molarity (mol dm–3) = — — — — — — — — — — — — — — — — — — — — — Molar mass (g mol–1) 3 Calculate the number of moles of ammonia in 150 cm3 of 2 mol dm–3 aqueous ammonia. Solution M = molarity, V = volume in cm3 MV Number of moles = — — — — 1000 The relationship between the number of mols with molarity, M and volume, V can be represented by the formula below: MV Number of moles = — — — — — — — — — — 1000 150 Number of moles of ammonia = 2  — — — — = 0.3 1000 where M = molarity of solution (mol dm–3) V = volume of solution (cm3). 4 A 250 cm3 solution contains 0.4 mol of nitric acid. Calculate the molarity of the nitric acid. Calculations Involving Concentrations and Molarity Solution MV 250 Number of moles = — — — —= M  — — — — 1000 1000 1 0.4  1000 Molarity of nitric acid, M = — — — — — — — — — 250 = 1.6 mol dm–3 5.00 g of copper(II) sulphate is dissolved in water to form 500 cm3 solution. Calculate the concentration of copper(II) sulphate solution in g dm–3. 197 Acids and Bases 5 Concentration (g dm–3) Molarity = — — — — — — — — — — — — — — — — — — Molar mass (g mol–1) 16 Convert g dm–3 =— — — to mol dm–3 106 –3 = 0.15 mol dm Calculate the volume in dm of 0.8 mol dm sulphuric acid that contains 0.2 mol of H2SO4. 3 –3 Solution MV Number of moles = — — — — 1000 Number of moles 8 The concentration of a potassium hydroxide solution is 84.0 g dm–3. Calculate the number of moles of potassium hydroxide present in 300 cm3 of the solution. [Relative atomic mass: H, l; O, 16; K, 39] Molarity, M 7 0.2 mol Volume of sulphuric acid = — — — — — — — — — — 0.8 mol dm–3 = 0.25 dm3 6 Solution Molar mass of KOH = 39 + 16 + 1 = 56 Dilute hydrochloric acid used in school laboratories usually has a concentration of 2.0 mol dm–3. Calculate the mass of hydrogen chloride in 250 cm3 of the hydrochloric acid? [Relative atomic mass: H, l; Cl, 35.5] Molarity of KOH Molarity (mol dm–3) 84.0 Concentration (g dm–3) =— — — =— — — — — — — — — — — — — — — — — — — — 56 Molar mass (g mol–1) = 1.5 mol dm–3 Number of moles of KOH 1.5  300 =— — — — — — — — MV 1000 Number of moles = — — — — 1000 = 0.45 Solution Number of moles of HCl 2.0  250 MV =— — — — — — — — Number of moles = — — — — 1000 1000 = 0.5 Molar mass of HCl = 1 + 35.5 = 36.5 g mol–1 Mass of HCl = 0.5  36.5 g = 18.25 g 9 Calculate the number of moles of hydrogen ions present in 200 cm3 of 0.5 mol dm–3 sulphuric acid. Mass = Number of moles  Molar mass Solution Number of moles of H2SO4 0.5  200 =— — — — — — — — Number of moles 1000 = 0.1 7 4.0 g of sodium carbonate powder, Na2CO3, is dissolved in water and made up to 250 cm3. What is the molarity of the sodium carbonate solution? [Relative atomic mass: C, 12; O, 16; Na, 23] MV =— — — –— 1000 H2SO4 → 2H+ + SO42– Sulphuric acid is a diprotic acid, which means 1 mol of sulphuric acid will produce 2 mol of H+ ions. Hence, 0.1 mol of sulphuric acid will produce 0.1  2 = 0.2 mol of H+ ions. Solution Volume of sodium carbonate solution = 250 cm3 = 0.25 dm3 Convert volume from cm3 to dm3 Mass (g) Concentration = — — — — — — — — — — — Volume (dm3) 4.0 Convert mass to =— — — concentration 0.25 –3 =16 g dm Preparation of Standard Solutions 1 A standard solution is a solution with a known concentration. 2 A volumetric flask (also known as standard flask) is an apparatus with a known volume. Examples are: 100 cm3, 200 cm3, 250 cm3, 500 cm3 and 1000 cm3. Molar mass of Na2CO3 = (2  23) + 12 + (3  16) = 106 g mol–1 Acids and Bases 198 3 Volumetric flasks are used to prepare standard solutions. Beakers are not suitable for this purpose because volumes measured by beakers and measuring cylinders are not very accurate. 4 A volumetric flask can measure the volume of a liquid accurately, up to one decimal point. 1 Calculate the mass (m g) of the chemical required to prepare v cm3 of solution where v is the volume of the volumetric flask. 2 Weigh out the exact mass (m g) of the chemical accurately in a weighing bottle using an electronic balance. 3 Dissolve m g of the chemical in a small amount of distilled water. 4 Transfer the dissolved chemical into the volumetric flask. 5 Add enough water until the graduation mark. Figure 7.10 A 100 cm3 volumetric flask To prepare 100 cm3 of 2.0 mol dm–3 aqueous sodium hydroxide solution SPM ’06/P2 Q4 Materials Sodium hydroxide solid and distilled water. Procedure 1 The mass of sodium hydroxide (NaOH) required to prepare 100 cm3 of 2.0 mol dm–3 aqueous sodium hydroxide is calculated as follows: Mass of NaOH required = Number of moles  molar mass of NaOH MV = (— — — —)  (23 + 16 + 1) 1000 2.0  100 =— — — — — — — — 40 1000 = 8.0 g 2 8.0 g of sodium hydro xide, NaOH solid is weighed accurately in a weighing bottle using an electronic balance. 3 Sodium hydroxide solid is transferred to a small beaker. Sufficient dis tilled water is added to dissolve all the solid sodium hydroxide. 199 Activity 7.5 4 Using a filter funnel and a glass rod, the dissolved sodium hydroxide is transferred to a 100 cm3 volumetric flask. 5 The small beaker, the weighing bottle and the filter funnel are all rinsed with distilled water and the contents are transferred into the volumetric flask. 6 Distilled water is then added slowly until the water level is near the level mark of the volumetric flask. A dropper is then used to add water drop by drop to finally bring the volume of solution to the 100 cm3 graduation mark. 7 The volumetric flask is closed with a stopper. The volumetric flask is then shaken several times to mix the solution completely. The solution prepared is 100 cm3 of 2.0 mol dm–3 aqueous sodium hydroxide. Apparatus Electronic balance, 100 cm3 volumetric flask, filter funnel, dropper and washing bottle. 7 The steps involved in the preparation of a standard solution Acids and Bases 7 The Correct Techniques Used in the Preparation of Standard Solution 2 When a solution is diluted, the volume of solvent increases but the number of moles of solute remains constant. Hence the concentration of the solution decreases. 3 If a solution with volume of V1 cm3 and molarity of M1 mol dm–3 is diluted to become V2 cm3, the new concentration of the diluted solution, M2 mol dm–3 can be determined as follows: 1 The chemical is weighed in a weighing bottle and not on a piece of filter paper. Some chemicals such as sodium hydroxide can absorb moisture from the air and become wet and may stick to paper. 2 After transferring the dissolved solute to the volumetric flask, the weighing bottle, the small beaker that contained the solution as well as the filter funnel are rinsed with distilled water. The content is then transferred to the volumetric flask to ensure that all the mass of the chemical that has been weighed is transferred to the volumetric flask. 3 The addition of distilled water to the volumetric flask must be carried out carefully so that the level of the solution does not exceed the graduation mark of the volumetric flask. The last few cm3 of water should be added drop by drop using a dropper. 4 A volumetric flask and not a beaker must be used to prepare a standard solution because a volumetric flask is calibrated to a high degree of accuracy. 5 The volumetric flask is stoppered after the standard solution is prepared to prevent the evaporation of water which can change the concentration of the solution prepared. Number of moles of M1V1 =— — — — — solute before dilution 1000 Number of moles of M2V2 = ——— —— solute after dilution 1000 However, the number of moles of solute before dilution is the same as the number of moles of solute after dilution, M2V2 M1V1 ——— —— = ——— —— 1000 1000 or M1V1 = M2V2 The steps involved in the preparation of a standard solution by dilution 1 The volume of the stock solution, V1 required is calculated. 2 The required volume of stock solution is pipetted into a volumetric flask. 3 Enough distilled water is added to the volumetric flask to the required volume, V2. Preparation of a Solution with a Specified Concentration Using the Dilution Method The formula used in dilution is M1V1 = M2V2 where M1 M2 V1 V2 1 Dilution is a process of diluting a concentrated solution by adding a solvent such as water to obtain a more diluted solution. = = = = Initial molarity of solution Final molarity of solution Initial volume of solution Final volume of solution To prepare 100 cm3 0.2 mol dm–3 sodium hydroxide from a 2.0 mol dm–3 sodium hydroxide solution by the dilution method M1V1 = M2V2 Activity 7.6 Apparatus 100 cm3 volumetric flask, 10 cm3 pipette, pipette filler, filter funnel, dropper and washing bottle. 2.0 3 V1 = 0.2 3 100 0.2 3 100 V1 = — — — — — — — — = 10 cm3 1.0 Materials 2.0 mol dm–3 sodium hydroxide solution and distilled water. where M1 = Initial molarity of alkali M2 = Final molarity of alkali V1 = Initial volume of alkali V2 = Final volume of alkali Procedure (A) To calculate the volume of sodium hydroxide solution that is required for dilution Acids and Bases 200 3 The flask is stoppered and is inverted several times to mix the solution. The solution prepared is 0.2 mol dm–3 sodium hydroxide solution. Conclusion A 0.2 mol dm–3 sodium hydroxide solution can be prepared by diluting 10 cm3 of 2.0 mol dm–3 of sodium hydroxide solution to 100 cm3. (a) The higher the degree of dissociation of an acid, the lower the pH value of the acid. (b) The higher the degree of dissociation of an alkali, the higher the pH value of the alkali. 3 For an acid or alkali, its pH value depends on the molarity of the solution. (a) The higher the molarity of an acid, the lower the pH value. (b) The higher the molarity of an alkali, the higher the pH value. Relationship between pH Values and Molarities of Acids or Alkalis 1 The pH value of an acid or an alkali depends on two factors, that is (a) degree of dissociation and (b) molarity or concentration. 2 At the same concentration, the pH value of an acid or an alkali depends on the degree of dissociation. 7.3 7 (B) To prepare 100 cm3 0.2 mol dm–3 sodium hydroxide by the dilution method 1 Using a pipette and a pipette filler, 10.0 cm3 of 2.0 mol dm–3 sodium hydroxide solution is transferred to a 100 cm3 volumetric flask. 2 Using a washing bottle, distilled water is added to the alkali in the volumetric flask until near the graduation mark. A dropper is then used to add water slowly in the volumetric flask up to the graduation mark. SPM ’09/P3 To investigate the relationship between pH values and the molarity of an acid or an alkali Problem statement What is the relationship between pH values and the molarity of an acid or an alkali? 0.001 mol dm–3 hydrochloric acid as shown in Figure 7.11. 3 The pH value shown on the pH meter is recorded. 4 The pH values of hydrochloric acid solutions with Figure 7.11 Using a pH different molarities meter to measure the pH are measured one value by one in dry beakers as in Steps 1 to 3. 5 The experiment is repeated using sodium hydroxide solutions with different molarities to replace the hydrochloric acid. Hypothesis (A) When the molarity of an acid increases, its pH value decreases. (B) When the molarity of an alkali increases, its pH value increases. Variables (a) Manipulated variable : Molarity of acid or alkali (b) Responding variable : pH values (c) Constant variable : Type of acid or alkali used Apparatus pH meter, 100 cm3 beakers and 100 cm3 measuring cylinders. Materials Hydrochloric acids and sodium hydroxide solutions with molarities of 0.001 mol dm–3, 0.01 mol dm–3, 0.05 mol dm–3, 0.08 mol dm–3 and 0.10 mol dm–3. Molarities of 0.001 0.01 HCl (mol dm–3) pH values Procedure 1 30 cm3 of 0.001 mol dm–3 hydrochloric acid is put in a dry beaker. 2 The probe of a pH meter that has been washed with distilled water is immersed in 30 cm3 of the 3.0 2.0 0.05 0.08 0.10 1.3 1.1 1.0 Molarities of 0.001 0.01 0.05 0.08 0.10 NaOH (mol dm–3) pH values 201 11.0 12.0 12.7 12.9 13.0 Acids and Bases Experiment 7.3 Results 7 4 The graph of pH values versus molarity of an alkali is an increasing curve as shown in Figure 7.13. 5 When the molarity of an alkali increases, the concentration of OH– ions in the alkali increases and the solution becomes more alkaline. Hence the pH value increases. Discussion 1 The graph of pH values versus molarity of an acid is a decreasing curve as shown in Figure 7.12. 2 When the molarity of an acid increases, the concentration of H+ ions in the acid increases and the solution becomes more acidic. Hence the pH value decreases. 3 From the graph, we can predict (a) the pH value, if the concentration of H+ ions of the solution is known. (b) the concentration of H+ ions in the solution, if the pH value is known. Figure 7.13 Graph of pH versus molarity of NaOH Conclusion 1 The higher the molarity of hydrochloric acid, the lower the pH value. The pH value of an acid decreases with the increase in molarity. 2 The higher the molarity of an alkali, the higher the pH value. The pH value of an alkali increases with the increase in molarity. The hypothesis is accepted. Figure 7.12 Graph of pH versus molarity of HCl 0.8  250 M2 = — — — — — — — = 0.2 mol dm–3 1000 Molarity of potassium hydroxide solution produced is 0.2 mol dm–3. Numerical Problems Involving Molarity of Acids and Alkalis 1 The molarity of an acid will change when (a) water is added to it, (b) an acid of a different concentration is added to it, (c) an alkali is added to it. 11 What is the volume of distilled water required to be added to 60 cm3 of 2.0 mol dm–3 sulphuric acid to produce a 0.3 mol dm–3 sulphuric acid? 10 What is the molarity of the potassium hydroxide solution produced when 750 cm3 of distilled water is added to 250 cm3 of potassium hydroxide solution of 0.8 mol dm–3? Solution M1V1 = M2V2 2.0  60 = 0.3  V2 2.0  60 V2 = — — — — — — — — 0.3 3 = 400 cm Solution Final volume of alkali, V2 = 250 cm3 + 750 cm3 = 1000 cm3 M1V1 = M2V2 0.8  250 = M2  1000 Acids and Bases Calculate the total volume of acid produced Volume of distilled water needed to be added to 60 cm3 H2SO4 = (400 – 60) cm3 = 340 cm3. 202 12 500 cm3 of a solution that contains 2.0 mol sodium hydroxide is added to 1500 cm3 of a solution that contains 4.0 mol sodium hydroxide. Calculate the molarity of the sodium hydroxide solution produced. Two units for concentration mol dm–3  molar mass determined by determined by Mass of solute (g) — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — Volume of solution (dm3) Total volume of NaOH = (500 + 1500) cm3 = 2000 cm3 Calculate the total volume of alkali from the two solutions = 2 dm3 No. of moles of solute (mol) — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — Volume of solution (dm3) 7.3 Molarity of Number of moles =— — — — — — — — — — — — — — NaOH produced Volume (dm3) 6.0 mol =— — — — — — 2 dm3 = 3.0 mol dm–3 1 Calculate the mass of potassium hydroxide required to produce (a) 2.0 dm3 of 46.4 g dm–3 solution (b) 100 cm3 of 2.0 mol dm–3 solution [Relative atomic mass: H, 1; O, 16; K, 39] 2 2.12 g of sodium carbonate is dissolved in 500 cm3 of distilled water. What is the concentration of the solution in (a) g dm–3? (b) mol dm–3? [Relative atomic mass: C, 12; O, 16; Na, 23] 13 200 cm3 of 2.0 mol dm–3 hydrochloric acid is added to 300 cm3 of 0.5 mol dm–3 hydrochloric acid. Calculate the molarity of the hydrochloric acid produced. 3 The concentration of sodium hydroxide solution is 8.0 g dm–3. (a) What is the molarity of the solution? (b) What is the molarity of the solution produced when 100 cm3 of distilled water is added to 100 cm3 of this solution? (c) What is the molarity of the solution produced when 20 cm3 of 2.0 mol dm–3 sodium hydroxide solution is added to 100 cm3 of this solution? [Relative atomic mass: H, 1; O, 16; Na, 23] Solution Number of moles in 200 cm3 of 2.0 mol dm–3 HCl MV 2.0  200 =— — — —= — — — — — — — — 1000 1000 = 0.4 Calculate the total Number of moles in 300 cm3 of 0.5 mol dm–3 HCl MV 0.5  300 =— — — —= — — — — — — — — 1000 1000 = 0.15  molar mass g dm–3 number of moles of acid from the two solutions. 7.4 Total number of moles of HCl = 0.4 + 0.15 = 0.55 Neutralisation The Meaning of Neutralisation and the Equation for Neutralisation Total volume of solution = (200 + 300) cm3 = 500 cm3 Calculate the total volume of = 0.5 dm3 acid from the two solutions. 1 Neutralisation is the reaction between an acid and a base to produce salt and water only. 2 In a neutralisation reaction, the acidity of an acid is neutralised by an alkali. At the same time, the alkalinity of the alkali is neutralised by the acid. Salt and water are the only products of neutralisation. Molarity of Number of moles of HCl =— — — — — — — — — — — — — — — — — — — — HCl produced Volume of HCl 0.55 mol =— — — — — — — — = 1.1 mol dm–3 0.5 dm3 203 Acids and Bases 7 Solution Total number of moles of NaOH = 2.0 + 4.0 Calculate the total number of moles = 6.0 of alkali from the two solutions. 3 Some examples of neutralisation reactions are as follows: acid base salt The ionic equation for neutralisation between strong acids and strong alkalis is water H+(aq) + OH–(aq) → H2O(l) HCl + NaOH → NaCl + H2O H2SO4 + CuO → CuSO4 + H2O 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O 7 In a neutralisation reaction, H+ ions from the acid react with OH– ions from the base to produce water. The pH value for water is 7, and a neutral condition is achieved. 4 In the reaction between ammonia and acid such as hydrochloric acid, ammonium salt is formed. 5 7 NH3 + HCl → NH4Cl Which of the following pairs of compounds will react in a neutralisation reaction? I Hydrochloric acid and potassium hydroxide II Sulphuric acid and solid copper(II) oxide III Nitric acid and solid calcium carbonate IV Hydrochloric acid and zinc metal A I and II only B III and IV only C I, III and IV only D I, II, III and IV Although there is no water formed in the above equation, in actual fact there is a little water formed together with NH4Cl. This is because a portion of ammonia, NH3 exists as NH4+ ions and OH– ions in aqueous solution. OH– ions and H+ ions (from acid) react to produce water, H2O. 5 Acids, bases and salts dissociate to form free ions. Only water exists as molecules. For example, Comments Neutralisation is a reaction between an acid and a base to form salts and water only. Reactions in I and II are neutralisation because the two reactants react to form salts and water only. Reaction III is not neutrali sa tion because carbon dioxide gas is formed in addition to salt and water. Reaction IV is not neutralisation because hydrogen gas is formed in addition to salt. Answer A H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → from HCl from NaOH Na+(aq) + Cl–(aq) + H2O(l) from NaCl ’03 water molecule 6 Since Na+ ions and Cl– ions do not undergo any changes in the reaction, these ions can be omitted from the equation. Thus, the equation can be simplified. Applications of Neutralisation in Daily Life Neutralisation is used in various fields such as agriculture, health and industries. In agriculture 1 Controlling the acidity of water is important in the rearing of freshwater fish and prawns. Lime (consisting of calcium oxide, CaO) which produces calcium hydroxide in water is used to control the acidity in aqua farming. 2 Plants do not grow well in acidic soil or basic soil. Lime or calcium carbonate is used to neutralise acidic soil. Basic soil is treated with compost which can release acidic gas to neutralise the alkali in basic soil. Acids and Bases SPM ’09/P1, ’10/P1 Calcium oxide and calcium hydroxide are used to neutralise acidic soil 204 SPM In health produced. This will prevent the corrosion of teeth enamel. 2 Antacids are medicine which contain bases such as magnesium hydroxide, aluminium hydroxide, calcium carbonate and calcium bicarbonate. Magnesia milk contains magnesium hydroxide. Antacids and milk of magnesia are used to neutralise the excess hydrochloric acid in the stomachs of gastric patients. 3 Alkaline creams or baking powder are applied to cure bee stings and ant bites which are acidic. Vinegar which contains ethanoic acid is used to cure alkaline wasp stings. Toothpastes contain magnesium hydroxide to neutralise acids in teeth 1 Food trapped in gaps between teeth decompose into organic acids by bacteria. An alkaline compound such as magnesium hydroxide in toothpastes neutralises the organic acids In industries 1 Bacteria in latex produces organic acids which coagulate latex. Ammonia is used to neutralise the organic acids produced by bacteria to prevent coagulation, so that latex can remain in the liquid state. 2 Calcium carbonate is used as a base to remove acidic gas such as sulphur dioxide emitted by power stations and industries. 3 Effluent from factories which is acidic is treated with lime, which will neutralise the acids in it before being discharged. 4 Neutralisation reaction is also used in the industry to produce manufactured products such as fertilisers, soaps and detergents. Acidic gas emitted by factories must be removed before being discharged (a) The use of acid–base indicators such as methyl orange, phenolphthalein and litmus paper which changes colour at the end point. (b) Measurement of the pH values during titration using a computer interface. 5 Titration technique can be used to determine the concentration of an acid (or an alkali). Acid-base Titration 1 Titration is a quantitative analysis that involves the gradual addition of a chemical solution from a burette to another chemical solution of known quantity in a conical flask. 2 In acid-base titration, the volume of the alkali is measured using a pipette and transferred into a conical flask. The acid solution from a burette is then added slowly to the alkali in the conical flask until neutralisation occurs. 3 The end point of a titration is when neutralisation occurs, that is, when the acid has completely neutralised the alkali. 4 Since both the reactants (acid and alkali) and the products formed (salt and water) are all colourless, the end point of neutralisation is determined as follows: The Use of Acid-base Indicators in Titrations 1 Acid–base indicators are chemicals that show different colours when the pH value of the solution changes. 2 Table 7.10 shows the colour of three common types of indicators at different pH values. 205 Acids and Bases 7 ’08/P1 3 The pH meter records changes in pH values during titration. The information is linked to a computer by the interface. The change of pH values along the progress of titration is displayed on the computer screen. Table 7.10 Colours of three common types of indicators in alkaline, neutral and acidic conditions Indicator Colour in alkali Methyl orange Yellow Colour in Colour in neutral acid solution Orange Red Phenolphthalein Light pink Colourless Colourless 7 Litmus Blue Purple Red 3 Methyl orange shows a yellow colour in an alkaline solution. At the point of neutralisation, the colour changes from yellow to orange. 4 Phenolphthalein shows a light pink in an alkaline solution. The first drop of acid that decolourises the light pink colour of phenolphthalein indicates the end point of titration. In computer interface, the pH is displayed on a screen 4 A graph of pH value versus volume of alkali added is shown as in Figure 7.14. It is found that the pH value of the solution changes sharply at the end point of neutralisation. The neutralisation point can be determined from the midpoint of the sharp pH change. The Use of Computer Interface in Titration 1 When an alkaline solution is added slowly from a burette to an acid in a conical flask, the pH value of the mixture solution increases slowly. 2 In titrations using a computer interface, the probe of a pH meter, immersed in the solution to be titrated, is connected to the pH module of the computer interface. Figure 7.14 Graph of pH value versus volume of alkali added To find the end point of an acid-base titration during neutralisation using an acid-base indicator 3 A 50 cm3 burette is rinsed with distilled water and then rinsed with a little of the sulphuric acid. 4 The burette is then filled with sulphuric acid and is clamped to a retort stand. The initial burette reading is recorded. 5 The conical flask containing 25 cm3 of potassium hydroxide is placed below the burette. A piece of white tile is placed below the conical flask for clearer observation of the colour change (Figure 7.15). 6 Sulphuric acid is added slowly from the burette to the potassium hydroxide solution in the conical flask while swirling the flask gently. 7 Titration is stopped when the methyl orange changes colour from yellow to orange. The final burette reading is recorded. Apparatus 25 cm3 pipette, pipette filler, 50 cm3 burette, retort stand, retort clamp, conical flask, filter funnel and white tile. Activity 7.7 Materials Sulphuric acid of unknown concentration, 1.0 mol dm–3 potassium hydroxide and methyl orange. Procedure 1 A clean 25 cm3 pipette is rinsed with distilled water and then rinsed with a little of the potassium hydroxide solution. 2 25 cm3 of 1.0 mol dm–3 potassium hydroxide is transferred using the pipette to a clean conical flask. Three drops of methyl orange indicator are added to the alkali and the colour of the solution is noted. Acids and Bases SPM ’10/P2 206 8 Steps 1 to 7 are repeated until accurate titration values are obtained, that is, until the difference in the volumes of sulphuric acid used in two consecutive experiments is less than 0.10 cm3. Figure 7.15 Titration of sulphuric acid with potassium hydroxide Results Volume of sulphuric acid Rough Accurate Final burette reading (cm3) 21.00 40.95 20.15 Initial burette reading (cm ) 0.00 21.00 0.10 Volume of sulphuric acid used (cm3) 21.00 19.95 20.05 3 Conclusion 1 The volume of sulphuric acid used is calculated as follows: Volume of sulphuric acid used = Final burette reading – Initial burette reading 2 Average volume of sulphuric acid used Discussion 1 In this experiment, the pipette has to be rinsed with potassium hydroxide solution so that water droplets on the inner wall of the pipette do not dilute the concentration of the potassium hydroxide used. 2 The burette is rinsed with sulphuric acid so that water droplets at the inner wall of the burette do not dilute the concentration of the sulphuric acid used. 3 The conical flask does not need to be rinsed with potassium hydroxide so that the volume of the potassium hydroxide in the conical flasks will accurately be 25.0 cm3. Otherwise, droplets of potassium hydroxide in the conical flask may cause the volume of potassium hydroxide to exceed 25.0 cm3. 4 The end point of titration is when the colour of the indicator changes sharply. The colour of methyl orange is yellow in potassium hydroxide solution (because pH > 7). At the end point, the colour of methyl orange changes to orange (pH = 7). If methyl orange changes to a red colour, excess sulphuric acid has been added. 5 In acid-base titrations, only 2 or 3 drops of indicator should be used. This is because most of the indicators are weak acid or base that will affect the pH of the solution if used in excess. 7 19.95 + 20.05 =— — — — — — — — — — — = 20.00 cm3 2 Hence, 20.00 cm3 of H2SO4 is required to completely neutralise 25.0 cm3 of 1.0 mol dm–3 KOH. 3 A 50 cm3 burette is rinsed with distilled water and then with a little of the sodium hydroxide solution. 4 The burette is then filled with sodium hydroxide solution and is clamped to a retort stand. 5 A magnetic stirrer bar is placed in the beaker containing 25 cm3 of hydrochloric acid. The beaker is then placed on a magnetic stirrer below the burette. 6 A pH meter is connected to a computer using a computer interface. The pH meter probe is then dipped into the acid. The magnetic stirrer is switched on and the computer is set to record and display the pH (Figure 7.16). 7 Sodium hydroxide solution is added drop by drop from the burette at a constant rate, into the acid in the beaker. Apparatus 25 cm3 pipette, pipette filler, 50 cm3 burette, retort stand, retort clamp, 250 cm3 beaker, filter funnel, magnetic stirrer, magnetic stirrer bar, pH meter, computer interface and computer. Materials 1.0 mol dm–3 hydrochloric acid and 1.0 mol dm–3 sodium hydroxide. Procedure 1 A 25 cm3 pipette is rinsed with distilled water and then with a little of the hydrochloric acid. 2 25 cm3 of 1.0 mol dm–3 hydrochloric acid is transferred using the pipette to a clean beaker. 207 Acids and Bases Activity 7.8 To find the end point of acid–base titration during neutralisation using a computer interface 7 8 A graph of pH change against the volume of sodium hydroxide in cm3 is printed using the computer printer when 50 cm3 of sodium hydroxide is added to the beaker. Figure 7.16 Using a pH meter and a computer interface to measure pH changes during neutralisation Results 1 A graph of pH against the volume of sodium hydroxide in cm3 as displayed by the computer is shown in Figure 7.17. Conclusion 1 The pH value of hydrochloric acid is 1.0 at the beginning of titration. As sodium hydroxide is added to the acid, the pH value of the solution increases. 2 The pH value increases sharply at the end point of neutralisation. The midpoint of the sharp pH change is 7. At pH 7, the volume of sodium hydroxide used from the graph is 25.0 cm3. 3 When 1.0 mol dm–3 sodium hydroxide is titrated against 25.0 cm3 of 1.0 mol dm–3 hydrochloric acid, the end point of titration during neutralisation occurs at pH 7 when 25 cm3 of 1.0 mol dm–3 of sodium hydroxide solution has been added. 4 By using a computer interface to measure pH changes, the end point of titration during nuetralisation can be determined accurately. Figure 7.17 Graph of pH against the volume of sodium hydroxide in cm3 6 ’95 Describe an experiment to determine the concentration of a solution of sulphuric acid by titrating it with a 1.0 mol dm–3 potassium hydroxide solution. Solution A titration experiment similar to Activity 7.7 is carried out in which sulphuric acid of unknown concentration is titrated against 25.0 cm3 of 1.0 mol dm–3 potassium hydroxide using methyl orange (or phenolphthalein) as an indicator. Let's say the volume of sulphuric acid required to neutralise completely 25.0 cm3 of 1.0 mol dm–3 potassium hydroxide is V cm3. Number of moles of KOH in 25.0 cm3 of MV 1.0 3 25.0 1.0 mol dm–3 solution = ——— = ——————— = 0.025 1000 1000 Acids and Bases The equation for the neutralisation reaction between potassium hydroxide and sulphuric acid is 2KOH + H2SO4 → K2SO4 + 2H2O According to the equation, 2 mol of KOH requires 1 mol of H2SO4 for complete neutralisation. 0.025 mol KOH will require 1 0.025 3 — = 0.0125 mol H2SO4 2 MV Number of moles of H2SO4 = ——— 1000 Molarity of sulphuric acid, M = 0.0125 3 1000/V 12.5 = ——— mol dm–3 V 208 Calculation Involving Neutralisation Using Balanced Equations 14 SPM ’09/P2 SPM ’11/P1 In an experiment, 25.0 cm3 of a sodium hydroxide solution of unknown concentration required 26.50 cm3 of 1.0 mol dm–3 sulphuric acid for complete reaction in titration. Calculate the molarity of sodium hydroxide. 1 Say, in a balanced equation, a mol of acid reacts with b mol of base as represented by the equation below: aA + bB → products Solution b=2 MBVB 2 — — — —= — MAVA 1 MB  25.0 2 — — — — — — — — —= — 1.0  26.50 1 MAVA Number of moles of acid = — — — — 1000 where MB = molarity of NaOH MA = molarity of H2SO4 VB = volume of NaOH VA = volume of H2SO4 26.50 — — — — MB = 2  — 25.0 = 2.12 mol dm–3 MBVB Number of moles of base = — — — — 1000 Hence, the molarity of sodium hydroxide solution is 2.12 mol dm–3. In the stoichiometry equation, a mol of acid HA reacts completely with b mol of base, M(OH)X. Hence the mole ratio of acid to base is MAVA — — — — 1000 a — — — —= — or MBVB b — — — — 1000 a=1 SPM 15 ’10/P1, ’11/P2 What is the volume of 1.5 mol dm–3 aqueous ammonia required to completely neutralise 30.0 cm3 of 0.5 mol dm–3 sulphuric acid? MAVA a — — — —— = — MBVB b Solution 3 From the above relationship, the ratio of a and b can be obtained from the balanced equation. Any one of the four variables: MA, VA, MB, VB, can be determined if three of the other variables are known. b=2 2NH3 + H2SO4 → (NH4)2SO4 MBVB 2 — — — —= — MAVA 1 a=1 2 1.5  VB — — — — — — — —= — 0.5  30.0 1 1 The volume of a solution in a burette is read from the top to the bottom. 2 The accuracy of a burette reading is until 2 decimal places. 3 The accuracy of a pipette reading is until 1 decimal place. where MB = molarity of NH3 MA = molarity of H2SO4 VB = volume of NH3 VA = volume of H2SO4 2  30.0  0.5 VB = — — — — — — — — — — — — = 20 cm3 1.5 Hence, the volume of aqueous ammonia required is 20 cm3. 209 Acids and Bases 7 2NaOH + H2SO4 → Na2SO4 + 2H2O 2 Say, the molarity of an acid is MA mol dm and the molarity of a base is MB mol dm–3. If in a titration, VA cm3 of acid neutralises VB cm3 of base –3 16 Calculate the volume (cm3) of 2.0 mol dm–3 hydrochloric acid that is required to react completely with 2.65 g of sodium carbonate. [Relative atomic mass: C, 12; O, 16; Na, 23] • In an aqueous solution, the concentration of H+ ions in CH3COOH is lower than HCl of the same concentration. This is because CH3COOH dissociates partially in water. • However, both CH3COOH and HCl of the same concentration require the same amount of NaOH for complete neutralisation. This is because both are monoprotic (monobasic) acids. In the presence of NaOH, CH3COOH dissociates completely to react with NaOH. • In an acid-base titration, the preferred chemical to be put in the burette is the acid. This is because alkali can react with silica in glass to form silicate, thus dissolving the thin glass wall of the burette slowly. Solution Molar mass of Na2CO3 = (23  2) + 12 + (16  3) = 106 g mass Number of moles of Na2CO3 = — — — — — — — — — — molar mass 7 2.65 =— — — — = 0.025 106 Na2CO3 + 2HCl → 2NaCl + CO2 + H2O From the equation, 1 mol of Na2CO3 reacts with 2 mol of HCl. Hence, 0.025 mol of Na2CO3 will react with 0.025  2 = 0.05 mol of HCl. where MV Number of moles =— — — — 1000 M = molarity of HCl V = volume of HCl 2.0 V 0.05 = — — — — — — — 1000 0.05  1000 V=— — — — — — — — — — = 25 cm3 2.0 Hence, the volume of hydrochloric acid required is 25 cm3. 7.4 1 Complete the blanks in the equations below: (a) 2HCl + Mg → (b) H2SO4 + Zn(OH)2 → + (c) + NaOH → CH3COONa + H2O (d) + → CaSO4 + 2H2O 2 Nitric acid reacts with magnesium hydroxide solution to produce magnesium nitrate, Mg(NO3)2 and water. (a) Write a balanced equation for the reaction between nitric acid and magnesium hydroxide. (b) 10.0 cm3 of nitric acid is required to neutralise 0.001 mol of magnesium hydroxide. Calculate the concentration of the nitric acid in mol dm–3. 17 15 cm3 of an acid with the formula HaX of 0.1 mol dm–3 required 30 cm3 of 0.15 mol dm–3 sodium hydroxide solution for complete neutralisation. Calculate the value of a and hence determine the basicity of the acid. 3 Sodium carbonate reacts with hydrochloric acid as represented by the equation below: Solution HaX + aNaOH → aH2O + NaaX Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g) MAVA 1 — — — — —= — MBVB a Calculate the volume (cm3) of 1.25 mol dm–3 hydrochloric acid that is required to react completely with 25.0 cm3 of 1.0 mol dm–3 sodium carbonate. 0.1 3 15 1 — — — — — — — —= — 0.15 3 30 a 4 15.0 cm3 of sulphuric acid neutralises 25.0 cm3 of 2.0 mol dm–3 aqueous ammonia. Calculate the molarity of the sulphuric acid used. a=3 Hence HaX is a tribasic acid. Acids and Bases + 210 Acids Alkalis Acids are sour in taste Alkalis are bitter in taste and feel soapy pH values of less than 7 pH values of more than 7 Changes blue litmus paper to red Changes red litmus paper to blue 7 Chemical properties of an acid: (a) Reacts with a base to produce a salt and water. (b) Reacts with a reactive metal to produce a salt and hydrogen gas is evolved. MB = molarity of alkali VB = volume of alkali 7 Multiple-choice Questions 7.1 Characteristics and Properties of Acids and Bases 1 Chemical X is an acid. Which of the following may not be true about the property of X? A A pink colour is produced when phenolphthalein is added to a solution of X. B Hydrogen gas is evolved when zinc powder is added to a solution of X. C A solution of X reacts with an alkali to produce salt and water. D X dissolves in water to produce H+ ions. 2 Which of the following statements is true of all bases? ’07 A Dissolve in water B Contain hydroxide ions C Have a pH value of between 12 and 13 D Produce ammonia gas when heated with ammonium salts 3 The diagram shows the set-up of apparatus for the reaction ’06 between solution Z and magnesium ribbon. 211 Which of the following is solution Z? A Glacial ethanoic acid B Ethanoic acid in methylbenzene C Concentrated ethanoic acid D Hydrogen chloride dissolved in propanone Acids and Bases 7 (c) Reacts with a metal carbonate to produce a salt, carbon dioxide gas and water. 8 Chemical properties of an alkali: (a) Reacts with an acid to produce a salt and water. (b) When heated with an ammonium salt, ammonia gas is produced. 9 The pH scale is a set of numbers range from (0 to 14) used to measure acidity or alkalinity of a substance. 10 The pH value of an acid or alkali depends on (a) the degree of dissociation (strength of acid or alkali), (b) the concentration of the acid or alkali. 11 A standard solution is a solution of known concentration. 12 The concentration of a solution is measured in g dm–3 or mol dm–3. 13 Neutralisation is the reaction between an acid and a base to produce a salt and water only. 14 When a mol of acid reacts completely with b mol of alkali in a reaction: MAVA a M = molarity of acid ———– = —— , where A b VA = volume of acid MBVB 1 An acid is a chemical compound that produces hydrogen ions, H+ or hydroxonium ions, H3O+ when it dissolves in water. 2 Dry acids without water do not show any acidic property because they do not dissociate to H+ ions. 3 A base is defined as a chemical substance that can neutralise an acid to produce a salt and water. Most bases are not soluble in water. Bases that are soluble in water are known as alkalis. 4 An alkali is defined as a chemical compound that dissolves in water to produce free moving hydroxide ions, OH–. 5 Dry alkalis do not show alkaline property. 6 Physical properties of acids and alkalis: 4 The colours of indicator X in solutions of different pH values are shown below. pH value 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ←⎯→ ←⎯→ ←⎯⎯→ ←⎯→ ←⎯⎯⎯⎯⎯→ ←⎯→ ←⎯→ Colour red orange yellow green bluish-green blue purple 7 Indicator X will show a green colour in a solution of 0.1 mol dm–3 of A sodium hydroxide solution B sodium chloride solution C aqueous ammonia D ethanoic acid 5 Which of the following substances can change red litmus paper to blue when dissolved in water? A Carbon dioxide gas B Glacial ethanoic acid C Solid sodium oxide D Solid sodium sulphate 6 The acidity of hydrogen chloride gas cannot be shown when it dissolves in the following solvents: I Water II Ethanol III Methylbenzene IV Propanone A I and II only B III and IV only C I and III only D II, III and IV only 7 Zinc carbonate powder is added to liquid X. A gas which turns limewater milky is evolved. X could be A glacial ethanoic acid. B aqueous citric acid. C hydrogen chloride gas dissolved in methylbenzene. D hydrogen chloride gas dissolved in propanone. 8 The table shows the pH for four aqueous solutions, W, X, Y and Z. Aqueous solutions W X Y Z pH 2 7 12 13 When two of the solutions with the same volume are added together, which mixture will react Acids and Bases with magnesium powder to liberate hydrogen gas? A W and X B X and Y C X and Z D Y and Z 9 Sulphuric acid is known as a diprotic acid because A there are two hydrogen atoms in one molecule of sulphuric acid. B one mole of sulphuric acid contains two moles of hydrogen atoms. C one mole of sulphuric acid dissociates into two moles of hydrogen ions in water. D sulphuric acid can be neutralised by two types of bases. 10 Which of the following particles in a solution of ammonia is responsible for its alkaline properties? A NH3 B OH– C H+ D NH4+ 7.2 The Strength of Acids and Alkalis 11 A chemical that dissociates completely in water to produce hydroxide ions is a A strong acid B weak acid C strong alkali D weak alkali 12 The table shows the degree of dissociation of four solutions of acids which have the same concentration Solution Degree of dissociation W High X Medium Y Very high Z Low Which solution has the highest pH value? A W C Y B X D Z 212 13 Ethanoic acid is a weak acid because A it is an organic acid. B it dissolves slightly in water. C it is a weak conductor of electricity. D it ionises partially to form hydrogen ions in water. 14 Which of the following solutions has the highest pH value? A Ethanoic acid, 0.01 mol dm–3 B Hydrochloric acid, 0.01 mol dm–3 C Aqueous ammonia, 0.01 mol dm–3 D Sodium hydroxide solution, 0.01 mol dm–3 15 Which of the following statements is true of the two aqueous solutions shown above? A Both solutions are strong acids. B The pH of both solutions are equal. C Both solutions are strong electrolytes. D 25.0 cm3 of each solution requires 25.0 cm3 of 1.0 mol dm–3 sodium hydroxide to be neutralised. 16 Which of the following solutions contains the highest number of ’11 hydrogen ions? A 50 cm3 of 2 mol dm–3 ethanoic acid B 40 cm3 of 1 mol dm–3 sulphuric acid C 30 cm3 of 2 mol dm–3 nitric acid D 50 cm3 of 1 mol dm–3 hydrochloric acid 17 Which of the following is true when water is added to an aqueous sodium hydroxide solution? 18 An aqueous solution of Q has a pH value of 1. Q may be I 0.1 mol dm–3 hydrochloric acid II 0.001 mol dm–3 sulphuric acid III 0.1 mol dm–3 nitric acid IV 0.1 mol dm–3 ethanoic acid A I and III only B II and IV only C I, II and III only D I, III and IV only 19 Which of the following statements is true of both nitric acid and sulphuric acid? A Both are organic acids. B Both are diprotic acids. C Both undergo complete dissociation in water. D Both react with copper to produce hydrogen gas. 7.3 Concentration of Acids and Alkalis 20 Steps I to V below show the five steps that are involved in the preparation of a standard solution of sodium hydroxide, NaOH which may not be arranged in correct order. I Add distilled water until the graduation mark II Weigh the mass of sodium hydroxide III Transfer the solid sodium hydroxide into the volumetric flask IV Rinse the weighing bottle and pour the solution into the volumetric flask V Shake the volumetric flask Which of the following is the correct order of steps in the preparation? A I, II, III, IV, V B II, III, I, IV, V C II, III, IV, I, V D II, I, III, IV, V 21 Calculate the mass of potassium hydroxide that is required to prepare 250 cm3 of 2.0 mol dm–3 solution. [Relative atomic mass: H, 1; O, 16; K, 39] A 22.4 g C 56 g B 28 g D 112 g 22 0.2 mol dm–3 sulphuric acid is added slowly to a conical flask containing 20.0 cm3 of 0.2 mol dm–3 potassium hydroxide and 10 cm3 of water. What is the total volume (in cm3) of the solution in the flask when the solution is completely neutralised? A 30 C 50 B 40 D 60 23 The table shows four different test tubes P, Q, R and S containing different acids. Test tube Content P 15 cm of 0.5 mol dm–3 hydrochloric acid Q 15 cm3 of 1.0 mol dm–3 ethanoic acid R 10 cm3 of 1.0 mol dm–3 nitric acid S 10 cm3 of 0.5 mol dm–3 sulphuric acid 3 Which test tube will produce the biggest volume of hydrogen gas with excess magnesium? A P C R B Q D S 24 Which of the following sodium hydroxide solutions have a concentration of 0.5 mol dm–3? [Relative atomic mass: H, 1; O, 16; Na, 23] I 5 g NaOH in 250 cm3 of water II 20 g NaOH in 1 dm3 of water III 250 cm3 of 2 mol dm–3 NaOH to which distilled water is added until it becomes 1 dm3 IV 1 mol dm–3 NaOH diluted to twice its volume A I and III only B II and III only C III and IV only D I, II, III and IV 213 25 When 2.8 g of potassium hydroxide is dissolved in 250 cm3 of distilled water, which of the following are true about the solution produced? [Relative atomic mass: H, 1; O, 16; K, 39] I It has a molarity of 0.05 mol dm–3. II It contains 11.2 g in 1 dm3. III The solution produces 0.2 mol of hydroxide ions. IV It contains 2 mol in 10 dm3 of solution. A I and II only B I and III only C III and IV only D II and IV only 26 What is the mass of sodium hydroxide contained in 50 cm3 of 0.4 mol dm–3 sodium hydroxide solution? [Relative atomic mass: H, 1; O, 16; Na, 23] A 0.4 g B 0.8 g C 1.6 g D 3.2 g 27 Calculate the number of moles of hydroxide ions in 2 dm3 of calcium hydroxide solution with a concentration of 14.8 g dm–3. [Relative atomic mass: H, 1; O, 16; Ca, 40] A 0.20 B 0.26 C 0.40 D 0.44 28 Which of the following substances will react with glacial ethanoic acid? A Zinc metal B Ammonia gas C Potassium hydroxide solid D Aqueous sodium carbonate solution 7.4 Neutralisation 29 The ionic equation for the reaction between nitric acid and sodium hydroxide is represented by A 2H2 + O2 → 2H2O B H+ + OH– → H2O C 2H+ + O2– → H2O D Na+ + NO3–→ NaNO3 Acids and Bases 7 A The pH value decreases. B The degree of ionisation decreases. C The hydroxide ion concentration increases. D The alkalinity increases. 7 30 Plants do not grow well in acidic soil. Which of the following are used to neutralise acidic soil? I Sodium hydroxide II Calcium hydroxide III Potassium hydroxide IV Calcium oxide A I and III only B II and IV only C I, II and III only D I, II, III and IV 31 Antacid is used to neutralise excess acid in the stomach. Which of the following chemicals is found in antacid? A Sodium hydroxide B Potassium hydroxide C Magnesium hydroxide D Ammonia 32 20 cm3 of 0.5 mol dm–3 sulphuric acid is added to 20 cm3 of 0.5 mol dm–3 sodium hydroxide. Which of the following statements are true? I Neutralisation reaction takes place. II The solution produced is acidic. III 0.02 mol of water is produced. IV The solution contains sodium sulphate and water only. A I and II only B III and IV only C II and III only D I, III and IV only 33 The equation shows the reaction between calcium carbonate and hydrochloric acid. CaCO3 + 2HCl → CaCl2 + CO2 + H2O What is the mass of calcium carbonate required to react completely with 10 cm3 of 2.0 mol dm–3 hydrochloric acid? [Relative atomic mass: C, 12; O, 16; Ca, 40] A 0.5 g B 1.0 g C 2.0 g D 4.0 g Acids and Bases 34 Fe + 2HCl → FeCl2 + H2 Based on the equation above, calculate the mass of iron that will react with excess hydrochloric acid to produce 60 cm3 of hydrogen gas at room temperature. [Relative atomic mass: Fe, 56; 1 mol of gas occupies 24 dm3 at room temperature] A 0.14 g B 0.28 g C 3.36 g D 22.4 g 35 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) The equation above shows the neutralisation reaction between sodium hydroxide and sulphuric acid. Calculate the number of moles of sodium hydroxide that is required to neutralise 25 cm3 of 2.0 mol dm–3 sulphuric acid. A 0.05 mol B 0.10 mol C 0.50 mol D 1.00 mol 36 10.0 cm3 of a certain 0.5 mol dm–3 acid requires 50.0 cm3 of 0.3 mol dm–3 sodium hydroxide solution for complete neutralisation. Which of the following is the possible molecular formula for this acid? A HNO3 B H2SO4 C H3PO4 D CH3COOH 37 Which of the following reactions represent neutralisation? I CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) II H+(aq) + OH–(aq) → H2O(l) III Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l) 214 IV CuO(s) + H2(g) → Cu(s) + H2O(l) A I and III only B II and III only C I, II and III only D II, III and IV only 38 The reaction between dilute hydrochloric acid and calcium carbonate is represented by the equation as follows. CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) What is the minimum volume of 2 mol dm–3 hydrochloric acid that is required to react completely with 5 g of calcium carbonate? [Relative atomic mass: C, 12; O, 16; Ca, 40] A 5 cm3 B 25 cm3 C 50 cm3 D 100 cm3 39 In a titration process, 0.1 mol dm–3 sulphuric acid from a ’03 burette is added slowly to 20 cm3 of 0.1 mol dm–3 aqueous ammonia solution in a conical flask with methyl orange indicator until neutralisation occurs. What is the total volume in the conical flask at the end point of titration? A 10 cm3 B 20 cm3 C 30 cm3 D 40 cm3 40 The equation below represents the neutralisation reaction of ’10 aqueous W hydroxide and hydrochloric acid. W(OH)2 + 2HCl → WCl2 + 2H2O What is the volume of 0.5 mol dm–3 hydrochloric acid needed to neutralise 25 cm3 of 0.2 mol dm–3 aqueous W hydroxide? A 10 cm3 B 20 cm3 C 30 cm3 D 40 cm3 Structured Questions [Relative molecular mass of NaOH = 40] 1 Diagram 1 shows the arrangement of apparatus used to prepare hydrogen chloride in methylbenzene and in water respectively. [2 marks] (i) Explain if a measuring cylinder is suitable to be used to measure the volume of water in the preparation of the standard solution. [1 mark] (ii) Name a suitable apparatus that is required to be used in the preparation of the standard solution. [1 mark] (d) What are the two parameters that should be measured accurately to prepare the standard solution of sodium hydroxide? Parameter I: Parameter II: [2 marks] (e) State two steps that should be taken to ensure that the standard sodium hydroxide solution is exactly 100 cm3 with a molarity of 0.5 mol dm–3. [2 marks] 3 An experiment is carried out in the laboratory to determine the concentration of a strong diprotic acid, H2 A, by titration. A few drops of phenolphthalein indicator is added to 25.0 cm3 of 0.5 mol dm–3 potassium hydroxide solution and then titrated with the acid, H2 A, of unknown concentration. The results obtained are shown in Table 1. Diagram 1 (a) What is the purpose of using the filter funnels in Diagram 1? [1 mark] (b) (i) What is observed when a piece of magnesium ribbon is placed in beakers A and B respectively? [2 marks] (ii) State the reason for your answer in (i). (c) Name the particles present in (i) beaker A (ii) beaker B Experiment [2 marks] [2 marks] (d) The magnesium ribbon is removed. Water is added to the solution in beaker A and the mixture is then shaken. When sodium carbonate powder is added, effervescence occurs. (i) Name the gas and suggest a suitable test to identify the gas evolved. [1 mark] (ii) State the role of water in the reaction that caused the evolution of the gas. [1 mark] (iii) Write an ionic equation for the reaction involving the evolution of the gas. [1 mark] (iv) What is the conclusion that can be made from the observation? [1 mark] II III Final burette reading (cm3) 26.55 36.15 27.20 Initial burette reading (cm3) 0.50 10.00 1.10 Volume of H2 A used (cm ) … … … 3 Table 1 (a) What is meant by a diprotic acid? [1 mark] (b) Write an equation for the neutralisation reaction between the acid, H2 A and potassium hydroxide. [1 mark] (c) State the colour change of the phenolphthalein indicator at the end point of titration. [1 mark] 2 A student is required to prepare a standard solution of 100 cm3 sodium hydroxide solution with a molarity ’06 of 0.5 mol dm–3 in the laboratory. He is given all the necessary apparatus required. (a) State the meaning of (i) a standard solution. (ii) molarity of the solution. I Volume of H2 A (d) (i) Calculate the volume of the acid, H2 A used in the titration and complete Table 1. (ii) Calculate the average volume of the acid, H2 A used in the experiment. [1 mark] [1 mark] [1 mark] [1 mark] (e) Calculate the concentration of the acid H2 A used in the experiment. [2 marks] (b) Calculate the mass of sodium hydroxide that the student needs to prepare a 100 cm3 solution with a molarity of 0.5 mol dm–3. (f) Draw a diagram to show the arrangement of the apparatus used in the above experiment. [2 marks] 215 Acids and Bases 7 (c) 4 Experiment I Excess magnesium ribbon is put in 10.0 cm3 of 1.0 mol dm–3 sulphuric acid Experiment II Excess magnesium ribbon is put in 10.0 cm3 of 1.0 mol dm–3 ethanoic acid Table 2 7 Table 2 shows two experiments carried out. (a) The reactions in experiment I and II can be represented by the same ionic equation, involving a certain particle in both acids. (i) Write the formula of this particle. [1 mark] (ii) Write the ionic equation for the reactions that occur in both experiments. [1 mark] Diagram 2 (b) The rates of reactions are different in experiments I and II. (i) Which reaction is more vigorous? [1 mark] (ii) Give a reason for your answer in (i). (a) Mark the pH value on the graph in Diagram 2 when complete neutralisation occurs. What is the pH value? [2 marks] (b) Mark the volume of the nitric acid required for complete neutralisation on the graph in Diagram 2. What is this volume? [2 marks] [3 marks] (c) Compare and explain the difference between the pH values of sulphuric acid and ethanoic acid. [2 marks] (c) Write a balanced equation for the reaction between nitric acid and sodium hydroxide. (d) What will be the change in pH value if 10.0 cm of water is added to the sulphuric acid before the magnesium ribbon is added in experiment I? Explain your answer. [2 marks] 3 [1 mark] (d) If methyl orange is added to the sodium hydroxide solution at the initial stage of the experiment, what is the change in colour that will take place at the end point of titration? [1 mark] 5 An experiment was carried out to determine the end point of titration during neutralisation using a computer interface. Nitric acid is added 0.5 cm3 by 0.5 cm3 from a burette to 25.0 cm3 of 0.5 mol dm–3 sodium hydroxide in a beaker. The graph in Diagram 2 shows the change in pH value of the solution in the beaker against the volume of nitric acid used in cm3 as measured by the computer interface. (e) Calculate the concentration of nitric acid used in this experiment. [2 marks] (f) If the experiment is to be repeated by titrating sodium hydroxide against nitric acid, sketch a graph of the change in pH value against the volume of sodium hydroxide that will be obtained. [2 marks] Essay Questions the chemical formula of magnesium hydroxide and explain its function in antacid. Name another chemical found in antacid. [4 marks] 1 (a) Using suitable examples, explain what is meant by neutralisation. [4 marks] (b) Explain why sodium hydroxide solution and aqueous ammonia of the same concentration have different pH values. [6 marks] (b) Diagram 1 shows two beakers containing 0.1 mol dm–3 solution X and solution Y and their pH readings. (c) Explain how you would prepare 250 cm3 of 1.0 mol dm–3 potassium hydroxide, starting from solid potassium hydroxide. Subsequently, explain how you would prepare 250 cm3 of 0.1 mol dm–3 potassium hydroxide from the above solution. [Relative atomic mass: H, 1; O, 16; K, 39] [10 marks] 2 (a) Magnesium hydroxide is one of the chemical compounds found in antacid medicine. Write Acids and Bases Diagram 1 216 (i) Compare and contrast the two solutions X and Y in terms of their physical and chemical properties. Give a suitable example for each of the solutions X and Y. [10 marks] (ii) Predict and explain, with suitable ionic equations, what will happen when equal volumes of solution X and solution Y are mixed together. [4 marks] (iii) Calculate the concentration of the solution that will be produced when 80 cm3 of water is added to 20 cm3 of solution X. [2 marks] 3 (a) Using a suitable example, explain the role of water in causing the acidic properties of an aqueous solution of an acid. [8 marks] (b) Briefly describe three tests (other than the use of an indicator) that can be used to confirm an acidic solution. Explain your tests with suitable equations. [12 marks] 7 Experiments 1 An experiment is carried out to determine the relationship between the concentrations of H+ ions and the pH values of nitric acid solutions. The pH values of six nitric acid solutions with concentrations of 0.100 mol dm–3, 0.060 mol dm–3, 0.040 mol dm–3, 0.025 mol dm–3, 0.015 mol dm–3 and 0.010 mol dm–3 are each measured using a pH meter. The corresponding pH values and the concentrations of the nitric acid solutions are shown in Diagram 1. Diagram 1 (a) State the variables involved in this experiment. • Manipulated variable: • Responding variable: • Constant variable: [3 marks] (b) State the hypothesis for this experiment. [3 marks] (c) Construct a table to record the results of this experiment. [3 marks] (d) Based on the results of this experiment, draw a graph of pH value versus concentration of H ions on a graph paper. [3 marks] + (e) Using the graph you have drawn in (d), predict the pH value of a 0.020 mol dm–3 nitric acid solution. [3 marks] 2 The volume of a sample of sulphuric acid required to neutralise a potassium hydroxide solution can be determined by titration. Design a laboratory experiment to determine the volume of the sulphuric acid required to neutralise 25.0 cm3 of 0.5 mol dm–3 potassium hydroxide solution. In designing your experiment, the following items must be included. (a) Problem statement (b) All the variables involved (c) Statement of the hypothesis (d) List of materials and apparatus (e) Procedure (f) Tabulation of data 217 [17 marks] Acids and Bases FORM 4 THEME: Interaction between Chemicals CHAPTER 8 Salts SPM Topical Analysis 2008 Year 1 Paper 2 A Section Number of questions 2 2009 – 1 3 B C – 1 — 2 2 A – – 2010 – 3 B C – 1 — 3 – 1 3 2011 2 3 A B C 1 – – – 1 3 2 3 A B C – – – ONCEPT MAP SALTS preparation Soluble salts by Reaction of acids with (a) alkalis (b) metal oxides (c) metal carbonates (d) metals qualitative analysis Insoluble salts by Precipitation in double decomposition reactions Anions test by (a) Heating and identifying (i) the gases evolved (ii) the colour change of products formed (b) Reagents such as barium chloride, silver nitrate and brown ring test Cations test by (a) Sodium hydroxide (b) Aqueous ammonia (c) Specific reagents 1 Salts Generally, the formula of salt is The Meaning of Salts Salt = 1 Salt is an ionic compound that is formed when the hydrogen ion in an acid is replaced by a metal ion or ammonium ion (NH4+). 2 The chemical formula of a salt is comprised of a cation (other than hydrogen ion) and an anion (other than oxide ion and hydroxide ion). 3 The cations and the anions of a salt are bonded by strong ionic bonds. 3 Diprotic acids and triprotic acids contain more than one H+ ions that can be replaced. Hence, it is possible for these acids to form more than one type of salt as shown in Table 8.2. Table 8.2 Examples of diprotic and triprotic salts Examples of Salts 1 Examples of salts formed from their corresponding acids are shown in Table 8.1. Table 8.1 Examples of salts formed from their corresponding acids Acid General name of salt Cations Anions + (other than H+) (other than O2– or OH–) Type of acid Example of acid Types of salts that can be formed Diprotic acid H2SO4 2 NaHSO4, Na2SO4 Triprotic acid H3PO4 3 NaH2PO4, Na2HPO4, Na3PO4 Example of salt Example of salt NaCl, KCl, CuCl2, ZnCl2, NH4Cl Hydrochloric acid, HCl Chloride salts Nitric acid, HNO3 Nitrate salts NaNO3, KNO3, Mg(NO3)2, Pb(NO3)2, NH4NO3 Sulphuric acid, Sulphate H2SO4 salts Na2SO4, K2SO4, FeSO4, CaSO4, (NH4)2SO4 Carbonic acid, Carbonate H2CO3 salts Na2CO3, CaCO3, MgCO3, ZnCO3, PbCO3 A salt consists of cations and anions, but the anions must not be oxide ion or hydroxide ion. This is because if the anions are O2– or OH– ions, the compound is a base, not a salt. Generally, the type of salts can be classified according to the cation or anion. For example, sodium chloride is known as a type of sodium salt or alternatively it can also be classified as a chloride salt. 2 Chloride salts are formed when H+ ions in hydrochloric acid, HCl is replaced by a metal ion or ammonium ion (NH4+). KCl (potassium chloride) H+ replaced by K+ NaCl (sodium chloride) H+ replaced by Na + HCl (hydrochloric acid) H+ replaced by Mg2+ H+ replaced by NH + 4 NH4Cl (ammonium chloride) H+ replaced by Zn2+ MgCl2 (magnesium chloride) ZnCl2 (zinc chloride) 219 Salts 8 8.1 Uses of Salts in Daily Life In food preparation 1 Sodium chloride (NaCl), table salt, is used for seasoning food. 2 Monosodium glutamate (M.S.G.) is used to enhance the taste of food. 3 Self-raising flour contains sodium bicarbonate (NaHCO3) which helps breads and cakes to rise. 8 Sodium chloride is used to flavour food In food preservation so that food can be kept longer without spoiling. 1 Sodium chloride is used as a food preservative in food such as salted fish and salted eggs. 2 Sodium benzoate (C6H5COONa) is used as a food preservative in food such as tomato sauce, oyster sauce and jam. 3 Sodium nitrite (NaNO2) is used to preserve processed meat such as burgers, sausages and ham. Salts play an important role in our daily life. Here are few examples of salts and their uses. Sodium benzoate is used as a food preservative in sauce In agriculture: to increase the production of food. 1 Nitrate salts such as potassium nitrate (KNO3), sodium nitrate (NaNO3) and ammonium salts such as ammonium sulphate (NH4)2SO4, ammonium nitrate (NH4NO3), ammonium phosphate (NH4)3PO4 are nitrogenous fertilisers. 2 Salts such as copper(II) sulphate (CuSO4), iron(II) sulphate (FeSO4) and mercury(I) chloride (HgCl) are used as pesticides. Salts Copper(II) sulphate is a pesticide used to kill fungi Fertilisers used in agriculture are ammonium salts 220 In medicine patients to be seen clearly in X-ray films. 7 Iron pills containing iron(II) sulphate are taken to increase the supply of iron for anaemic patients. 8 1 Antacid medicine contain calcium carbonate (CaCO3), and calcium hydrogen carbonate Ca(HCO3)2 that are used to reduce acidity in the stomachs of gastric patients. 2 Smelling salts contain ammonium chloride (NH4Cl). 3 Plaster of Paris, used to support fractured bones, contains calcium sulphate. 4 Epsom salts (magnesium sulphate heptahydrate) and Glauber salt (sodium sulphate decahydrate) are used as laxatives to clear the intestines. 5 Potassium permanganate (KMnO4) is used as an antiseptic to kill germs. 6 Barium sulphate, BaSO4, enables the intestines of suspected stomach cancer Plaster of Paris consists of a calcium sulphate Other uses Salts play an important role in our daily life. Here are few examples of salts and their uses. 1 Fluoride toothpaste contains tin(II) fluoride, SnF2, that is used to prevent tooth decay. Silver bromide, AgBr, is used to produce black 2 and white photographic films. Sodium hypochlorite, NaOCl, is used as 3 a bleaching agent in soap powders and detergents. Fluoride salt in toothpaste prevents tooth decay Naturally occuring salts • Lead(II) sulphide (galena, PbS), calcium fluoride (fluorite, CaF2) and magnesium sulphate (Epsime, MgSO4) exist as minerals in the earth’s crust. • Corals, stalactites, stalagmites and limestone consist of calcium carbonate, (CaCO3). Stalactites and stalagmites 221 Salts Soluble Salts and Insoluble Salts 1 Solubility is the ability of a compound to dissolve in a solvent. Some salts are soluble in water while others are not. 2 The solubility of a salt in water depends on the types of cations and anions present as shown in Table 8.3. Table 8.3 Types of salt and their solubility in water 8 Type of salt Figure 8.1 Separation of a soluble salt and an insoluble salt by filtration Solubility in water Sodium, potassium and ammonium salts All are soluble Nitrate salts All are soluble Chloride salts All are soluble except PbCl2, AgCl and HgCl Sulphate salts All are soluble except PbSO4, BaSO4 and CaSO4 Carbonate salts All are insoluble except Na2CO3, K2CO3 and (NH4)2CO3 5 The methods of preparing salts depend on the solubility of salts. 6 Soluble salts can be prepared in the laboratory by four methods as follows: (a) Reaction between an acid and an alkali (b) Reaction between an acid and a metal (c) Reaction between an acid and a metal carbonate (d) Reaction between an acid and a metal oxide or hydroxide 7 Insoluble salts can be prepared by precipitation in double decomposition reactions. 3 Information on the solubility of salts is useful in (a) the separation of salts in a salt mixture. (b) choosing the methods to prepare a salt. (c) identifying the types of ions in a salt in the qualitative analysis of salts. 4 Filtration can be used to separate an insoluble salt (as the residue) from a soluble salt (as the filtrate) as shown in Figure 8.1. • If a salt is soluble in water, a solution is formed. The solution may be coloured if the cation is copper(II), iron(II) or iron(III). Otherwise, a colourless solution is formed. • If a salt is insoluble in water, a cloudy mixture is formed when stirred. The undissolved salt will settle down as a precipitate. 8.1 To study the solubility of nitrate, sulphate, carbonate and chloride salts Problem statement Are nitrate, sulphate, carbonate and chloride salts soluble in water? Materials Various types of salts and distilled water. Procedure 1 0.2 g of copper(II) nitrate is put in a test tube using a spatula. 2 5 cm3 of distilled water is added to the above test tube. The mixture is stirred and the solubility of the salt is noted. 3 Steps 1 and 2 are repeated using magnesium nitrate, zinc nitrate, lead(II) nitrate, calcium nitrate, copper(II) sulphate, magnesium sulphate, zinc sulphate, lead(II) sulphate, barium sulphate, calcium sulphate, copper(II) chloride, magnesium Experiment 8.1 Hypothesis Some salts are soluble in water while some are not. Variables (a) Manipulated variable : Types of salts (b) Responding variable : Solubility in water (c) Constant variable : Quantity of salts, volume and temperature of water Apparatus Test tubes, glass rods, spatulas and test tube holder. Salts 222 Type of salt Formula of salt Carbonate Na2CO3, K2CO3, (NH4)2CO3 Results Soluble CuCO3, MgCO3, ZnCO3 Type of salt Formula of salt Solubility in water Nitrate Cu(NO3)2, Mg(NO3)2, Zn(NO3)2, Pb(NO3)2, Ca(NO3)2 Soluble Sulphate CuSO4, MgSO4 and ZnSO4 Soluble PbSO4, BaSO4, CaSO4 Insoluble CuCl2, MgCl2, ZnCl2 Soluble PbCl2, AgCl, HgCl Insoluble Chloride Insoluble Conclusion 1 All nitrate salts are soluble in water. 2 All sulphate salts are soluble in water except lead(II) sulphate, PbSO4, barium sulphate, BaSO4 and calcium sulphate, CaSO4. 3 All chlorides salts are soluble in water except lead(II) chloride, PbCl2, silver chloride, AgCl, and mercury(I) chloride, HgCl. 4 All carbonates are insoluble in water except sodium carbonate, Na2CO3, potassium carbonate, K2CO3 and ammonium carbonate, (NH4)2CO3. 5 Some salts are soluble while some are not. The hypothesis is accepted. Table 8.4 Some examples of acids and alkalis used in the preparation of soluble salts Preparation of Soluble Salts The preparation of soluble salts can be divided into two categories. (a) Soluble salts of sodium, potassium and ammonium. (b) Soluble salts which are not salts of sodium, potassium and ammonium. Type of soluble salt Soluble Salts of K+, Na+ and NH4+ Type of alkali used Example of salt Type of acid used Salts of sodium NaCl HCl Sodium hydroxide, NaOH CH COONa CH COOH 3 3 Salts of potassium Potassium hydroxide, KOH Aqueous Salts of ammonium ammonia, NH3 1 The cation of a salt comes from the alkali while the anion comes from the acid. Hence the type of salt produced depends on the acid and alkali used. For example: K2SO4 H2SO4 KNO3 HNO3 NH4NO3 HNO3 (NH4)2SO4 H2SO4 5 Impure soluble salts can be purified by recrystallisation. When an impure soluble salt is dissolved in enough distilled water, the insoluble impurities can be removed by filtration. The filtrate is evaporated to remove excess water to form a saturated solution. When the saturated solution is cooled to room temperature, the salt will recrystallise to form pure crystals. 6 Soluble salts in a mixture of a few salts can also be purified by recrystallisation. This is because salts have different solubility in water. A salt with a lower solubility will recrystallise earlier than a salt with a higher solubility. The process of recrystallisation may be repeated a few times to obtain a pure salt. Na2SO4 from NaOH Solubility in water from H2SO4 2 Soluble salts of sodium, potassium and ammonium can be prepared from the reaction between an acid and an alkali, NaOH, KOH or NH3(aq) as in Table 8.4. 3 Titration method is used to ensure that all the acid is completely reacted with the alkali. 4 The flowchart in Figure 8.2 shows the steps involved in the preparation of soluble salts of sodium, potassium and ammonium. 223 Salts 8 chloride, zinc chloride, lead(II) chloride, silver chloride, mercury(I) chloride, sodium carbonate, potassium carbonate, ammonium carbonate, copper(II) carbonate, magnesium carbonate and zinc carbonate to replace the copper(II) nitrate. 1 Acid + alkali The alkali in the conical flask is titrated with acid in the burette 1 titration method Dilute salt solution 8 After titration, the dilute salt solution is heated to hasten evaporation 2 evaporation 2 Saturated salt solution The saturated salt solution is then cooled to precipitate out the salt 3 cooling 3 Salt crystals in saturated salt solution The salt crystals are then filtered out from the solution using filter paper 4 filtration 4 Salt crystals The salt crystals are then dried with more filter paper 5 drying Dry salt crystals 5 The resulting salt crystals of sodium, potassium or ammonium are produced Figure 8.2 Preparation of soluble salts of sodium, potassium and ammonium Salts 224 Preparation of Soluble Salts of Sodium, Potassium and Ammonium Materials 2 mol dm–3 hydrochloric acid and 2 mol dm–3 potassium hydroxide and phenolphthalein indicator. Procedure 1 25 cm3 of potassium hydroxide is pipetted into a clean conical flask. 2 Three drops of phenolphthalein indicator are added to the alkali and the colour of the solution is noted. 3 A 50 cm3 burette is then filled with hydrochloric acid and is then clamped to a retort stand. The initial burette reading is recorded. 4 Hydrochloric acid is added gradually from the burette to the potassium hydroxide solution in the conical flask while swirling the flask gently. 5 Titration is stopped when phenolphthalein changes from a light pink colour to colourless. The final burette reading is recorded. 6 The volume of hydrochloric acid used is calculated as follows: by dropping a drop of the solution on a piece of glass plate. If crystals are formed, then the solution is saturated. 9 The saturated solution is then cooled to allow crystallisation to occur. 10 The white crystals formed are then filtered, rinsed with a little distilled water and dried by pressing between filter paper. Figure 8.3 Titration of potassium hydroxide with hydrochloric acid Discussion 1 In the preparation of potassium chloride, the acid used is hydrochloric acid and the alkali used is potassium hydroxide. KOH + HCl → KCl + H2O V cm3 = Final burette – Initial burette reading reading 7 The experiment is repeated by adding V cm3 of hydrochloric acid to 25 cm3 of potassium hydroxide in a beaker without using phenolphthalein as an indicator. 8 The colourless solution in the beaker is evaporated to form a saturated solution (to about 1 — of the original volume). This can be tested 3 2 Phenolphthalein is used as an indicator at the beginning of the experiment to determine the volume of hydrochloric acid that is required to react with 25 cm3 of potassium hydroxide. However, the experiment is repeated without using phenolphthalein so that the salt prepared will not be contaminated by the indicator. 3 The salt solution is not heated until dry because the salt may decompose when heated strongly. In the preparation of soluble salts of Na+/K+/NH4+, titration is carried out to determine the exact amount of acid required to neutralise all the alkali to form a neutral salt using an indicator. Once the exact volumes of acid and alkali required are known, the indicator is not required to prepare the pure salt. 225 Salts Activity 8.1 Apparatus 25 cm3 pipette, pipette filler, 50 cm3 burette, retort stand, retort clamp, conical flask, filter funnel, filter paper, beaker, tripod stand, wire gauze and Bunsen burner. SPM ’04/P2 8 To prepare potassium chloride by the reaction between an acid and an alkali To purify potassium chloride by recrystallisation 3 The hot solution is filtered into a clean conical flask to remove the impurities. 4 The filtrate is evaporated until a saturated solution is formed. 5 The saturated solution is then allowed to cool to room temperature for crystallisation. 6 The filtered crystals are then rinsed with distilled water and dried between two pieces of filter paper. Apparatus Beaker, glass rod, Bunsen burner, conical flask, spatula, filter funnel and filter paper. 8 Materials Impure potassium chloride and distilled water. Procedure 1 Impure potassium chloride is placed in a beaker. 2 A little distilled water, enough to cover the crystals is added. The mixture is heated while stirring, and more distilled water is added slowly until all the crystals are dissolved. Conclusion Impure potassium chloride can be purified by recrystallisation. Soluble Salts which are not salts of Na+, K+, NH4+ ensure that all the acid is reacted completely, excess solids are used. The non-reacted excess solids can be removed by filtration. 1 Three methods are used to prepare soluble salts which are not salts of sodium, potassium and ammonium. This involves the reaction between (a) an acid and a metal (b) an acid and a metal carbonate (c) an acid and a metal oxide or hydroxide 2 Table 8.5 shows the chemicals suitable for the preparation of soluble salts which are not salts of sodium, potassium and ammonium. 3 In the reaction of an acid with a metal, metals that are less electropositive than hydrogen such as copper and silver do not react with dilute acids. 4 Metals, metal oxides and metal carbonates are solids that do not dissolve in water. Hence, to Table 8.5 Examples of some salts and chemicals used in their preparation Example of salt Type of acid used ZnCl2 Chemicals that react with the acid Metal oxide Metal Metal carbonate HCl ZnO Zn ZnCO3 Mg(NO3)2 HNO3 MgO Mg MgCO3 CuSO4 H2SO4 CuO _ CuCO3 Preparation of Soluble Salts which are Not Salts of Sodium, Potassium and Ammonium Activity 8.2 & 8.3 To prepare copper(II) nitrate by the reaction between an acid and a metal oxide SPM ’04/05 P2 Apparatus Beaker, glass rod, 100 cm3 measuring cylinder, wire gauze, tripod stand, Bunsen burner, conical flask, spatula, filter funnel and filter paper. Materials 1 mol dm–3 nitric acid and copper(II) oxide powder. Salts 226 Procedure 1 About 30 cm3 of 1 mol dm–3 nitric acid is put in a beaker and is heated. 2 Using a spatula, copper(II) oxide powder is added a little at a time, to the hot nitric acid while stirring continuously with a glass rod. The addition of copper(II) oxide is stopped when some black solids remain undissolved. 3 The mixture is filtered to remove the excess copper(II) oxide. 4 The filtrate is evaporated until a saturated solution is formed. 5 The saturated solution is then allowed to cool to room temperature. 6 The blue crystals formed are removed by filtration, rinsed with a little distilled water and dried between filter paper. 1 8 2 Discussion 1 Copper(II) oxide is a black powder. It dissolves in nitric acid to form a blue solution. The equation for the reaction is CuO(s) + 2HNO3(aq) → Cu(NO3)2 (aq) + H2O(l) 3 2 Neutralisation reaction takes place between copper(II) oxide and nitric acid. The ionic equation for the reaction is CuO(s) + 2H+(aq) → Cu2+(aq) + H2O(l) 3 The salt solution is not heated until dry because the salt may decompose when heated strongly. 4 The copper(II) nitrate crystals prepared may be purified by recrystallisation. 5 Copper(II) nitrate can also be prepared by the reaction between nitric acid and copper(II) carbonate. However, copper metal does not react with dilute nitric acid because copper is below hydrogen in the electrochemical series. 4 5 Conclusion Copper(II) nitrate can be prepared by the reaction between copper(II) oxide and nitric acid. 227 Salts To prepare iron(II) sulphate by the reaction of an acid and a metal Apparatus Beaker, glass rod, 100 cm3 measuring cylinder, Bunsen burner, conical flask, spatula, filter funnel and filter paper. 8 Materials Iron powder and 2 mol dm–3 sulphuric acid. Procedure 1 30 cm3 of 2 mol dm–3 sulphuric acid is put in a beaker. 2 Iron powder is gradually added to the sulphuric acid while stirring continuously with a glass rod, until a slight excess of iron is present. 3 The mixture is filtered to remove the excess iron powder. 4 The filtrate is evaporated until a saturated solution is formed. The saturated solution is then allowed to cool to room temperature. 5 The green crystals formed are removed by filtration, rinsed with a little distilled water and dried between filter papers. Discussion 1 Iron metal is grey in colour. It dissolves in sulphuric acid to form a green solution with effervescence. The gas evolved is hydrogen gas. The equation for the reaction is Fe(s)+ H2SO4(aq) → FeSO4(aq) + H2(g) 2 The ionic equation for the reaction between iron and H+ ion in acid is Fe(s)+ 2H+(aq) → Fe2+(aq) + H2(g) 3 Iron(II) sulphate can also be prepared by the reaction of sulphuric acid with iron(II) oxide or iron(II) carbonate. Conclusion Iron(II) sulphate can be prepared by the reaction between iron metal and sulphuric acid. Activity 8.4 & 8.5 To prepare magnesium chloride by the reaction of an acid and a metal carbonate Apparatus Beaker, glass rod, 100 cm3 measuring cylinder, Bunsen burner, conical flask, spatula, filter funnel and filter paper. Materials Magnesium carbonate powder and 2 mol dm–3 hydrochloric acid. Procedure 1 Magnesium carbonate powder is added a little at a time, to 30 cm3 of 2 mol dm–3 hydrochloric acid in a beaker while stirring continuously. The addition is stopped when there is no more effervescence, and a little magnesium carbonate powder remains undissolved. 2 The mixture is filtered to remove the excess magnesium carbonate powder. 3 The filtrate is evaporated until a saturated solution is formed. 4 The saturated solution is then allowed to cool to room temperature. 5 The white crystals formed is removed by filtration, rinsed with a little distilled water and dried between filter papers. Salts Discussion 1 Magnesium carbonate is white in colour. Effervescence occurs when it dissolves in hydrochloric acid to form a colourless solution. The gas evolved is carbon dioxide gas. The equation for the reaction is MgCO3(s) + 2HCl(aq) → MgCl2(aq) + CO2(g) + H2O(l) 2 The ionic equation for the reaction between magnesium carbonate and H+ ion in acid is MgCO3(s) + 2H+(aq) → Mg2+(aq) + CO2(g) + H2O(l) 3 Magnesium chloride can also be prepared by the reaction of hydrochloric acid with magnesium oxide or magnesium metal. Conclusion Magnesium chloride can be prepared from the reaction between magnesium carbonate and hydrochloric acid. 228 1 Insoluble salts can be prepared by precipitation in double decomposition reactions. 2 In the precipitation method, an insoluble salt is precipitated when two aqueous solutions containing the cations and the anions are mixed together. The precipitate is then obtained by filtration. 3 In double decomposition, one of the aqueous solutions contains the cations of the insoluble salt, while the other aqueous solution contains the anions of the salt. Can sodium nitrate be prepared by adding sodium chloride solution to nitric acid? Comments There is no reaction between sodium chloride solution and nitric acid. NaCl(aq) + HNO3(aq) NaNO3(aq) + HCl(aq) Solution consist of Na+, Cl–, H+ and NO3– ions Solution consist of Na+, Cl–, H+ and NO3– ions SPM ’10/P1 Sodium nitrate is usually prepared by the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH). H2O formed does not dissociate. Cation M+ (from a + soluble salt solution) HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l) Insoluble salt, Anion X– MX (from a soluble salt → (formed as precipitate) solution) 4 In double decomposition, the ions of the two aqueous solutions interchange to produce a new compound which is insoluble. 5 The general equation can be represented as follows Physical Characteristic of Salt Crystals 1 Salt crystals are formed when a saturated salt solution is cooled. 2 A salt is made up of positive ions and negative ions. When these ions are packed closely with a regular and repeated arrangement in fixed positions, a solid with definite geometry known as crystal lattice is formed. 3 The repeating basic unit in this orderly structure is called a unit cell. 4 All crystals have these physical characteristics: (a) Fixed geometrical shapes (e.g. cubic, hexagonal or rhombic). (b) Flat surface, straight edges and sharp angles. (c) Fixed angle between two adjacent surfaces. 5 All crystals of the same salt have the same shape although the sizes may be different. 6 The size of a crystal formed depends on the rate of crystallisation. Fast crystallisation (from fast cooling) will yield smaller crystals than slow crystallisation. 7 Crystals are hard and brittle, and can be cut into different shapes. This is because the particles of salt crystals are arranged in regular layers. MY(aq) + NX(aq) → MX(s) + NY(aq) solution solution precipitate solution 6 The precipitate produced is obtained by filtration. The residue is the insoluble salt, which is then rinsed with distilled water to remove any other ions as impurities. Example Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq) precipitate 7 Precipitation of lead(II) chloride can be simplified as follows: Pb2+(aq) + 2Cl–(aq) → PbCl2(s) Sodium ions, Na+ and nitrate ions, NO3– do not undergo any change in the reaction. They are known as spectator ions and can be ignored in the ionic equation. 229 Salts 8 Preparation of Insoluble Salts 1 8 The following guidelines show the steps used in writing an ionic equation for the formation of an insoluble salt. Step 1 Step 2 Step 3 Identify the insoluble salt and write it as the product on the right-hand side of the equation. Separate the cations and anions of the salt and write them as the reactants on the left-hand side of the equation. Balance the charges of the cations and anions by adding the correct coefficient as the number of moles reacting. Pb2+(aq) + Cl–(aq) → PbCl2(s) Pb2+(aq) + 2Cl–(aq) → PbCl2(s) 8 → PbCl2(s) 9 The following guidelines show the steps used in the selection of aqueous solutions in the preparation of an insoluble salt. Step 1 Step 2 Step 3 Identify the cation and anion of the insoluble salt. Example PbCl2: Cation = Pb2+ Anion = Cl– Identify a soluble salt that can supply the cation, example, a nitrate salt (all nitrate salts are soluble). Example Pb(NO3)2 solution or Pb(CH3COO)2 solution Identify a soluble salt that can supply the anion, example, a sodium or potassium salt (all sodium or potassium salts are soluble in water). Example NaCl or KCl solution PbCl2 from Pb(NO3)2 or Pb(CH3COO)2 from NaCl or KCl or HCl Table 8.6 Some examples of insoluble salts Barium salt Lead(II) salt Silver salt Name Formula Name Formula Name Formula Lead(II) chloride Lead(II) bromide Lead(II) iodide Lead(II) sulphate Lead(II) chromate(VI) Lead(II) carbonate PbCl2 PbBr2 PbI2 PbSO4 PbCrO4 PbCO3 Barium sulphate Barium chromate(VI) Barium carbonate BaSO4 BaCrO4 BaCO3 Silver chloride Silver bromide Silver iodide Silver carbonate AgCl AgBr AgI Ag2CO3 2 ’04 Can lead(II) sulphate be prepared by adding sulphuric acid to lead(II) oxide? Comments Both lead(II) sulphate and lead(II) oxide are insoluble in water. Hence, when lead(II) sulphate is formed, it cannot be separated from the mixture of lead(II) sulphate and lead(II) oxide. Lead(II) sulphate as an insoluble salt, is usually prepared from double decomposition reaction between Salts two aqueous solutions, one containing the lead(II) ions (example, lead(II) nitrate) and the other containing the sulphate ions (example, sodium sulphate). Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq) Similarly, lead(II) sulphate cannot be prepared by adding sodium sulphate solution to lead(II) chloride. Both lead(II) sulphate and lead(II) chloride are insoluble in water. 230 To prepare insoluble salts: lead(II) iodide, lead(II) chromate(VI) and barium sulphate by precipitation reaction Procedure (A) Preparation of lead(II) iodide 1 20 cm3 of 0.5 mol dm–3 potassium iodide solution is added to 20 cm3 of 0.5 mol dm–3 lead(II) nitrate solution in a beaker. 2 The mixture is stirred thoroughly with a glass rod. A yellow precipitate is formed immediately. 3 The mixture is filtered to obtain the yellow solids of lead(II) iodide as the residue. 4 The residue is rinsed with distilled water to remove any trace of other ions in it. 5 The yellow solid is dried by pressing between two pieces of filter papers. (B) Preparation of lead(II) chromate(VI) 1 20 cm3 of 0.5 mol dm–3 lead(II) nitrate solution is added to 20 cm3 of 0.5 mol dm–3 potassium chromate(VI) solution in a beaker. 2 The mixture is stirred thoroughly with a glass rod. A yellow precipitate is formed immediately. 3 The mixture is filtered to obtain the yellow solids of lead(II) chromate(VI) as the residue. 4 The residue is rinsed with distilled water and dried using filter papers. (C) Preparation of barium sulphate 1 20 cm3 of 0.5 mol dm–3 barium chloride solution is added to 20 cm3 of 0.5 mol dm–3 sodium sulphate solution in a beaker. 2 The mixture is stirred thoroughly with a glass rod. A white precipitate is formed immediately. 3 The mixture is filtered to obtain the white barium sulphate as the residue. Discussion 1 The chemical equation for the reaction that occur in the preparation of lead(II) iodide is Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) 8 Materials 0.5 mol dm–3 solutions of lead(II) nitrate, potassium iodide, potassium chromate(VI), sodium sulphate and barium chloride. 4 The residue is rinsed with distilled water and dried using filter papers. The ionic equation for the reaction is Pb2+(aq) + 2I–(aq) → PbI2(s) Lead(II) iodide is a yellow precipitate. 2 The chemical equation for the reaction that occurs in the preparation of lead(II) chromate(VI) is Pb(NO3)2(aq) + K2CrO4(aq) → PbCrO4(s) + 2KNO3(aq) The ionic equation for the reaction is Pb2+(aq) + CrO42–(aq) → PbCrO4(s) Lead(II) chromate(VI) is a yellow precipitate. 3 The chemical equation for the reaction that occurs in the preparation of barium sulphate is BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq) The ionic equation for the reaction is Ba2+(aq) + SO42–(aq) → BaSO4(s) Barium sulphate is a white precipitate. Conclusion Insoluble salts of lead(II) iodide, lead(II) chromate(VI) and barium sulphate can be prepared by precipitation in double decomposition reactions. Preparation of a Specified Salt 2 The following flowchart shows the procedure for the selection of the methods of preparing a specified salt (Figure 8.4). 1 The method of preparing a salt depends on (a) whether or not the salt is soluble in water, (b) whether the salt is a salt of sodium, potassium or ammonium for soluble salts. 231 Salts Activity 8.6 Apparatus Beakers, glass rods, conical flasks, filter funnels and filter paper. SPM ’08/P2 Method of preparing a salt Is the salt soluble in water? Yes No Is it the salt of sodium, potassium or ammonium? No Precipitation by double decomposition reaction Neutralisation using titration Reaction of an acid with • a metal • a metal oxide/metal hydroxide • a metal carbonate 8 Yes Select two aqueous solutions that can supply the cations and the anions of the salt Filtration to remove excess solids (keep the filtrate) Salt solution Filtration (keep the residue) 1 2 3 4 evaporation cooling (crystallisation) filtration recrystallisation (if necessary) Salt crystals 1 rinse with distilled water 2 dry with filter paper Pure salt crystals Figure 8.4 3 ’03 You are supplied with sodium carbonate solution, sulphuric acid and magnesium nitrate solution. Plan a scheme to prepare a sample of magnesium sulphate using the above chemicals. Write equations for the reactions involved. Comments Magnesium sulphate is a soluble salt which is not a salt of sodium, potassium and ammonium. Magnesium sulphate can be prepared by the reaction of sulphuric acid with magnesium metal/magnesium oxide/magnesium carbonate. Hence, a scheme of preparing magnesium sulphate is as follows: Salts Step 1 Magnesium nitrate is converted to magnesium carbonate by double decomposition between sodium carbonate solution and magnesium nitrate. Mg(NO3)2(aq) + Na2CO3(aq) → MgCO3(s) + 2NaNO3(aq) Step 2 Magnesium carbonate that is produced is then added to sulphuric acid until in excess. 232 MgCO3(s) + H2SO4(aq) → MgSO4(aq) + CO2(g) + H2O(l) Ionic Equations of Insoluble Salts Table 8.7 Formation of ionic equations from the mole ratio of ions No. of moles No. of moles of cation of anion 1 mol Pb2+ 2 mol Cl– Ionic equation Pb2+(aq) + 2Cl–(aq) → PbCl2(s) 1 mol Pb2+ 1 mol CrO42– Pb2+(aq) + CrO42–(aq) → PbCrO4(s) 2 mol Ag+ 1 mol CrO42– 2Ag+(aq) + CrO42–(aq) → Ag2CrO4(s) 4 The number of moles of the cation and anion can be calculated if the volume and molarities MV are known using the formula — — — —. 1000 Step 1 If the charge of cation M is b and the charge of anion X is a, the formula of the salt is MaXb, MaXb 1 charge of X charge of M 6.0 cm3 of 0.2 mol dm–3 Xn+ solution reacts completely with 4.0 cm3 of 0.1 mol dm–3 Ym– solution to form a salt XmYn. Write the ionic equation and hence determine the empirical formula of the salt in this reaction. For example, the formula of iron(III) carbonate is Fe2(CO3)3. Solution 0.2  6 Number of moles of X ions = ––––––– 1000 = 0.0012 n+ Step 2 0.1  4 Number of moles of Ym– ions = ––––––– 1000 = 0.0004 This show that a mol of M ions has combined with b mol of Xa– ions. b+ Fe2(CO3)3 shows that 2 mol of Fe3+ ions combines with 3 mol of CO32– ions. Mole ratio of Xn+ ions : = 0.0012 : 0.0012 : = –––––––– 0.0004 = 3 : MV — — — — 1000 MV — — — — 1000 Ym– ions 0.0004 0.0004 ––––––– 0.0004 1 Hence 3 mol of Xn+ react with 1 mol of Ym–. Step 3 Ionic equation is : 3Xn+ + 1Ym– → X3Y Thus the ionic equation for the formation of MaXb is Empirical formula of the salt is X3Y. aMb+(aq) + bXa–(aq) → MaXb(s) Constructing Ionic Equations Using the Continuous Variation Method The ionic equation for the formation of Fe2(CO3)3 is 1 The mole ratio of ions that react to form a salt can be determined from an experiment through the continuous variation method. 2 In this method, fixed volumes of a reactant X are added to varying volumes of a second reactant Y in different test tubes. If the salt formed is an insoluble salt, the amount of 2Fe3+(aq) + 3CO32–(aq) → Fe2(CO3)3(s) 3 The examples in Table 8.7 show the method of writing ionic equations based on the simplest mole ratio of cations to anions combined to form the salts. 233 Salts 8 1 An ionic equation for the formation of a salt can be written if (a) the formula of the salt is known (from the charges of cation and anion), (b) the number of moles of ions required to form the salt is known. 2 The following guidelines show the construction of the ionic equation for the formation of an insoluble salt from the charges of cation and anion. precipitate produced will increase until all of the ions in solution X have reacted completely. The height of the precipitate will remain constant despite the increasing volumes of solution Y. 3 The flowchart of Figure 8.5 shows the steps involved in the continuous variation method. To determine the ionic equation of the reaction between X ions and Y ions Carry out an experiment to investigate the reaction between • fixed volumes of solution X and • different and varying volumes of solution Y 8 Determine the volume of Y ions that reacts with all of the X ions Calculate the number of moles of X ions and the number of moles of Y ions that have reacted using the formula: MV Number of moles = — — — — 1000 Determine the simplest mole ratio of X ions to the Y ions in the reaction to construct the ionic equation Figure 8.5 Flowchart for the steps in the continuous variation method 8.2 SPM ’11/P3 Experiment 8.2 To construct a balanced ionic equation for the precipitation of lead(II) chromate(VI) using the continuous variation method Problem statement How to determine the ionic equation for the precipitation of lead(II) chromate(VI)? Hypothesis The height of precipitate will increase with the increase in volume of lead(II) nitrate solution until all the potassium chromate(VI) has reacted. Variables (a) Manipulated variable : Volumes of lead(II) nitrate solution (b) Responding variable : Height of yellow precipitate (c) Constant variable : Volume of potassium chromate(VI) solution and the size of test tubes Apparatus Test tubes of the same size, test tube rack, 50 cm3 burette, retort stand with clamp and ruler. Materials 0.5 mol dm–3 lead(II) nitrate solution and 0.5 mol dm–3 potassium chromate(VI) solution. Salts Procedure 1 A burette is filled with 0.5 mol dm–3 lead(II) nitrate solution and another burette is filled with 0.5 mol dm–3 potassium chromate(VI) solution. 2 Eight test tubes are labelled 1 to 8 and placed in a test tube rack. 3 5.00 cm3 of potassium chromate(VI) solution from the burette is placed in every test tube. Potassium chromate(VI) solution is yellow in colour. 4 Using another burette, 1 cm3 of 0.5 mol dm–3 of lead(II) nitrate solution is added to the first test tube. Progressively increase the volume of the lead(II) nitrate solution by 1 cm3 to the rest of the test tubes until 8 cm3 of lead(II) nitrate solution is added to the eighth test tube (Figure 8.6(a)). 5 Every test tube is well shaken in order to mix the solutions completely. The test tubes are then allowed to stand for 20 minutes for the yellow precipitate, lead(II) chromate(VI) to settle (Figure 8.6(b)). 234 6 The height of the precipitate formed in every test tube is measured accurately using a ruler. The colour of the solution above the precipitate is noted. 7 The result obtained is recorded in Table 8.8. 8 1.1 Figure 8.6 Continuous variation method Results Table 8.8 Test tube number Volume of potassium chromate(VI) solution (cm3) Volume of lead(II) nitrate solution (cm3) Height of precipitate (cm) Colour of solution 1 5.0 2 5.0 3 5.0 4 5.0 5 5.0 6 5.0 7 5.0 8 5.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 0.6 yellow 0.9 yellow 1.8 yellow 2.2 yellow 2.8 2.8 2.8 colourless 2.8 MV 0.5  5.0 = — — — — =— — — — — — — — = 2.5 3 10–3 1000 1000 Calculation 1 A graph showing the height of precipitate versus the volume of lead(II) nitrate solution is drawn (Figure 8.7). Number of moles of CrO42– ions in 5.0 cm3 of 0.5 mol dm–3 potassium chromate(VI) solution MV 0.5  5.0 = — — — — =— — — — — — — — = 2.5 3 10–3 1000 1000 4 Hence, 2.5  10–3 mol of Pb2+ ions react completely with 2.5  10–3 mol of CrO42– ions. ∴ 1.00 mol of Pb2+ ions will react completely with 1.00 mol of CrO42– ions. The ionic equation for the reaction is Figure 8.7 Graph of height of precipitate versus the volume of lead(II) nitrate solution Pb2+(aq) + CrO42–(aq) → PbCrO4(s) 2 From the graph, it is found that the height of precipitate increases as the volume of lead(II) nitrate increases. However, a constant height is reached when 5.0 cm3 of lead(II) nitrate solution is added. Thereafter, the height remains constant despite further increases in the volume of lead(II) nitrate solution. 3 This means that when 5.0 cm3 of 0.5 mol dm–3 lead(II) nitrate solution is used, all the chromate(VI) ions in 5 cm3 of 0.5 mol dm–3 potassium chromate(VI) solution has been precipitated. Number of moles of Pb2+ ions in 5.0 cm3 of 0.5 mol dm–3 lead(II) nitrate solution 5 Consequently, the balanced chemical equation for the reaction is 1 mol 1 mol 1 mol Pb(NO3)2(aq) + K2CrO4(aq) → PbCrO4(s) + 2KNO3(aq) Conclusion 1 Since the diameter of the test tubes are the same, the height of the precipitate is directly proportional to the mass of precipitate formed. 2 The ionic equation for the precipitate of lead(II) chromate(VI) is Pb2+ + CrO42– → PbCrO4. The hypothesis is accepted. 235 Salts Discussion 3 From test tubes 6 to 8, the heights of precipitate formed remains constant because all the chromate(VI) ions in the test tubes have been precipitated. There is an excess of Pb2+ ions in the test tubes. The clear solution above the precipitate which is colourless contains Pb2+ ions, K+ ions and NO3– ions. 8 1 From test tubes 1 to 4, the 2+ increase of Pb ions from the increase in volumes of lead(II) nitrate solution added, increases the mass of precipitate formed. There are excess (unreacted) CrO42– ions in the test tubes which produces the yellow colour of the solutions above the precipitate. The yellow solution contains CrO42– ions, K+ ions and NO3– ions. The yellow colour became paler as more CrO42– ions have reacted. 2 In test tube 5, the reaction is completed when the precipitate formed reaches a maximum height. All the chromate(VI) ions have reacted with all the lead(II) ions. The clear solution contains K+ ions and NO3– ions. 4 ’97 You are supplied with a lead(II) ions solution and a 0.1 mol dm–3 potassium chromate(VI) solution. Explain how you can determine the concentration of the lead(II) ions solution using the precipitation method. Comments An experiment using the continuous variation method of precipitating lead(II) chromate(VI), using a constant volume of potassium chromate(VI) solution and different volumes of lead(II) ions as in Experiment 8.2 is carried out. A graph of the height of precipitate against the volume of lead(II) ions solution will be obtained as follows: Salts From the graph, V cm3 of lead(II) ions solution is required to react completely with 5 cm3 of potassium chromate(VI) solution, when the height of the precipitate becomes constant. Calculation: 0.1 3 5 • Number of moles of CrO42– ions = ————— 1000 = 0.0005 M3V • Number of moles of Pb2+ ions = —————, 1000 where M is the concentration. • Ionic equation of the reaction Pb2+ + CrO42– → PbCrO4 • From the equation, 1 mol of Pb2+ ions react with 1 mol of CrO42– ions. Mole ratio of Pb2+ : CrO42– = 1:1, 0.0005 1 that is —————— = — 0.001MV 1 2+ • Concentration of Pb ions, M = 0.5/V mol dm–3 236 1 Hence, 0.2 mol of H2SO4 produce 0.2  — 3 = 0.067 mol of Al2(SO4)3. SPM Numerical Problems Involving ’09/P1 Calculation of Quantities of Reactants or Products in Stoichiometric Reactions 1 A balanced equation gives information regarding the number of moles of reactants in a reaction and the number of moles of products formed. 2 For example, the equation for the reaction between magnesium and hydrochloric acid is Type 2: Calculation involving quantities in mass If the quantities of reactants/products are given in terms of mass in gram, the quantity of solid can be converted to moles by the following relationship Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) 1 mol 2 mol 1 mol 1 mol The coefficients before the reactants and products indicate the number of moles of chemicals involved in the reaction. The equation above shows that 1 mol of magnesium reacts with 2 mol of hydrochloric acid to produce 1 mol of magnesium chloride and 1 mol of hydrogen gas. 3 A balanced chemical equation (stoichiometric reaction) can be used to calculate the stoichiometric quantities of the reactants or products in terms of: • Mass • Volume and concentration (of aqueous solution) • Volume (of gas) 4 Since the quantities of chemicals involved in a reaction are in terms of moles, the quantities of reactants or products (the quantity of chemicals in terms of volume of gas, volume of solution, mass, numbers of molecules or atoms), must first be converted to moles in the initial step of the calculation regarding quantities of reactants and products. SPM ’08/P1 3 2.0 g of sodium hydroxide reacts with excess sulphuric acid. What is the mass of sodium sulphate produced? [Relative atomic mass: H, l; O, 16; Na, 23; S, 32] Solution 2NaOH + H2SO4 → Na2SO4 + 2H2O Molar mass of NaOH = 23 + 16 + 1 = 40 g mol–1 2g 2 g of NaOH = — — — — — — — — ‑ 40 g mol–1 = 0.05 mol Step 2: Convert mass to mol. Step 3: Get the mole ratio of NaOH and Na2SO4 Type 1: Calculation involving quantities in moles 1 Hence, 0.05 mol of NaOH produce — 0.05 2 = 0.025 mol of Na2SO4 2 Calculate the number of moles of aluminium sulphate produced by the reaction of 0.2 mol of sulphuric acid with excess aluminium oxide. Molar mass of Na2SO4 = 2(23) + 32 + 4(16) = 142 g mol–1 Step 1: Write a balanced equation. Solution 3H2SO4 + Al2O3 → Al2(SO4)3 + 3H2O 1 mol Step 1: Write a balanced equation. From the equation, 2 mol of NaOH produces 1 mol of Na2SO4. Examples of Calculation 3 mol Mass (g) Number of moles = — — — — — — — — — — — — — — — — — — — Molar mass (g mol–1) Step 2: Get the mole ratio of H2SO4 and Al2(SO4 )3 . 0.025 mol of Na2SO4 = 0.025 mol  142 g mol–1 = 3.55 g From the equation, 3 mol of H2SO4 produce 1 mol of Al2(SO4)3 237 Step 4: Relate the number of moles of chemicals in the equation to that in the question. Step 5: Convert mol to mass. Salts 8 Step 3: Relate the number of moles of chemicals in the equation to that in the question. Type 3: Calculation involving volumes of gas If the quantities of reactants or products are given in terms of volumes of gas, the volume of gas can be converted to moles by the following relationship: Or: Number of moles = molarity of solution (mol dm–3)  volume of solution (dm3) 5 At s.t.p. (0°C and 1 atm): What is the mass of magnesium required to react with 20 cm3 of 2 mol dm–3 hydrochloric acid to produce 120 cm3 of hydrogen at room temperature? [Relative atomic mass: Mg, 24; 1 mol of gas occupies 24 dm3 at room temperature] Volume of gas (dm3) Number of moles = — — — — — — — — — — — — — — — — — 22.4 dm3 mol–1 8 At room conditions (25°C and 1 atm) : Solution Mg + 2HCl → MgCl2 + H2 Volume of gas (dm3) Number of moles = — — — — — — — — — — — — — — — — — 24.0 dm3 mol–1 1 mol of gas occupies 24 dm3. What is the volume of carbon dioxide gas evolved at s.t.p. when 2.1 g of magnesium carbonate reacts with excess nitric acid? [Relative atomic mass: C, 12; O, 16; Mg, 24; 1 mol of gas occupies 22.4 dm3 at s.t.p.] Step 3: Get the mole ratio of Mg and H2. Hence, 0.005 mol of H2 is produced by 0.005 mol of Mg Step 1: Write a balanced equation. Step 4: Relate the number of moles of chemicals in the equation to that in the question. MgCO3 + 2HNO3 → Mg(NO3)2 + CO2 + H2O Molar mass of MgCO3 = 24 + 12 + 3(16) = 84 g mol–1 2.1 2.1 g MgCO3 = — — = 0.025 mol 84 0.005 mol of Mg = 0.005  24 g = 0.12 g Step 2: Convert mass to mol. What is the volume of 2 mol dm–3 hydrochloric acid required to dissolve 10 g of marble (calcium carbonate)? [Relative atomic mass: H, 1; O, 16; C, 12; Ca, 40] Step 3: Get the mole ratio of MgCO3 and CO2. Hence, 0.025 mol of MgCO3 produce 0.025 mol of CO2. Step 4: Relate the number of moles of chemicals in the equation to that in the question. Solution Step 1: CaCO3 + 2HCl → CaCl2 + CO2 + H2O Step 5: Convert mol to volume. 10 10 Step 2: 10 g of CaCO3 = — — — — — — — — — — — —= — — — 40 + 12 + 3(16) 100 = 0.1 mol Hence, 0.025 mol of gas CO2 gas occupies 0.025  22.4 = 0.56 dm3 or 560 cm3. Step 3: From the equation, 1 mol of CaCO3 requires 2 mol of HCl for a complete reaction. Type 4: Calculation involving volumes and molarities of solutions If the quantities of reactants or products involves solutions, the quantity of chemicals in a solution can be converted to moles using the following relationship MV Number of moles = ———— 1000 Step 4: Hence, 0.1 mol of CaCO3 requires 0.1  2 = 0.2 mol of HCl for a complete reaction Step 5: Number of moles = molarity  volume (dm3) Volume of HCl Number of moles of HCl 0.2 mol =— — — — — — — — — — — — — — — — — — — —= — — — — — — — — — Molarity of HCl 2 mol dm–3 3 3 3 = 0.1 dm = 0.1  1000 cm = 100 cm Where M = molarity of solution (mol dm–3) V = volume of solution (cm3) Salts Step 5: Convert mol to mass. 6 From the equation, 1 mol of MgCO3 produces 1 mol of CO2. 1 mol of gas occupies 22.4 dm3. Step 2: Convert volume to mol. 120 — — — — — — — — mol = 0.005 mol 120 cm3 gas = — 24  1000 From the equation, 1 mol of H2 is produced by 1 mol of Mg. 4 Solution Step 1: Write a balanced equation. 238 7 SPM 5 ’08/P1 4.0 g of magnesium oxide is added to 30.0 cm of 2.0 mol dm–3 hydrochloric acid. What is the mass of magnesium oxide that does not dissolve in this reaction? [Relative atomic mass: O, 16; Mg, 24] ’02 3 Pb2+(aq) + 2I–(aq) → PbI2(s) What is the molarity of the potassium iodide solution? MV MV Number of moles = — — — — 1000 MgO + 2HCl → MgCl2 + H2O Number of moles = — — — — 1000 Comments 20.0 Number of moles of Pb2+ ions = 0.25  — — — — 1000 = 0.005 2.0  30 Number of moles of HCl = — — — — — — — — = 0.06 1000 From the equation, 2 mol of HCl will dissolve 1 mol of MgO. 8 Solution 5.0 cm3 of a potassium iodide solution requires 20.0 cm3 of 0.25 mol dm–3 lead(II) nitrate solution to react completely according to the equation below. Hence, 0.06 mol of HCl will dissolve From the equation, 1 mol of Pb2+ ions reacts with 2 mol of I– ions. 1 0.06  — = 0.03 mol of MgO. 2 Hence, 0.005 mol of Pb2+ ions react with (0.005  2) = 0.01 mol of I– ions. 1000 M = Number of moles 3 — — — — V 0.03 mol of MgO = 0.03  (24 + 16) g = 1.2 g 0.01  1000 Molarity of KI solution = — — — — — — — — — — = 2 mol dm–3 5 ∴ Mass of MgO that does not dissolve = Initial mass of MgO – mass of MgO dissolved = 4.0 g – 1.2 g = 2.8 g 8 Mass, m What is the mass of copper(II) carbonate that is produced when 60 cm3 of 1 mol dm–3 sodium carbonate is added to 50 cm3 of 2 mol dm–3 copper(II) sulphate? [Relative atomic mass H, 1; O, 16; Cu, 64] n  molar mass Mole, n Solution CuSO4 + Na2CO3 → CuCO3 + Na2SO4 Number of moles of CuSO4 2  50 =— — — — — = 0.1 1000 V  molar volume Calculate the number of moles of both reactants to check which of the reactants is used up in the reaction. The number of moles of product formed depends on the number of moles of reactant that is used up (the limiting reactant). V1  molarity n  molarity n  molar volume Volume of gas, V Number of moles of Na2CO3 1  60 =— — — — — = 0.06 1000 From the equation, 1 mol of CuSO4 reacts with 1 mol of Na2CO3 to produce 1 mol of CuCO3. Volume of solution, V1 8.1 1 Suggest suitable methods and reactants for the preparation of the following salts. (a) Na2SO4 (b) (NH4)2SO4 (c) Al2(SO4)3 (d) Pb(NO3)2 (e) ZnCl2 (f) PbSO4 (g) AgCl Hence, 0.06 mol of Na2CO3 will react completely, while 0.1 mol of CuSO4 is in excess. Thus, 0.06 mol of Na2CO3 will produce 0.06 mol of CuCO3. m  molar mass Na2CO3 is the limiting factor. 2 Suggest chemicals that can react with nitric acid to produce magnesium nitrate. Write equations for the reactions that take place. 0.06 mol of CuCO3 = 0.06  (64 + 12 + 3(16)) g = 0.06  124 g = 7.44 g 239 Salts 3 State the mole ratio of the ions in the following salts: (a) CaSO4 (b) Al(OH)3 Inference from the Colours of Salts or Salt Solutions 8 4 Write the ionic equations for the following reactions: (a) 2 mol of silver ions react with 1 mol of chromate(VI) ions (b) 0.3 mol of lead(II) ions react with 0.6 mol of bromide ions 1 Initial observation of the physical properties of a salt such as colour and solubility in water enables us to make inferences regarding the possible cations or anions present. However, the presence of the cations or anions needs to be confirmed by other tests. 2 Most salts are white in colour and when dissolved in water, will form colourless aqueous solutions. 3 Cations of transition elements have specific colours. 4 Table 8.9 below gives the colours of different cations in the solid form or in aqueous solutions. 5 5 cm3 of 0.2 mol dm–3 barium chloride solution reacts completely with 10 cm3 of 0.1 mol dm–3 sodium chromate(VI) solution. Calculate the mole ratio of the ions involved in the formation of barium chromate precipitate. Subsequently, write the balanced equation for the reaction that occurs. chemical 6 In an experiment, 10 cm3 of 0.5 mol dm–3 silver nitrate reacts completely with 5 cm3 of 0.5 mol dm–3 potassium carbonate. Determine the ionic equation for the precipitation above. Table 8.9 Colours of cations 7 Magnesium oxide reacts with excess phosphoric acid to produce 1.2 mol of magnesium phosphate. (a) Write a balanced equation for the reaction that occurs. (b) Calculate the number of moles of magnesium oxide that is used in the reaction above. 8.2 Qualitative Analysis of Salts The Meaning of Qualitative Analysis 1 Qualitative analysis is a chemical technique used to determine the identities of chemical substances present in a mixture but not their quantity. 2 Qualitative analysis of salt is a scheme of tests carried out to identify the cation and anion present in the salt. 3 The technique of qualitative analysis includes: (a) Observing the colour of the salt or colour of the aqueous salt solution. (b) Observing the solubility of the salt in water. (c) Observing the effect of heat on the salt. (d) Identifying the gas evolved when a test is performed on the salt. (e) Identifying the precipitate formed when a specific chemical reagent is added to the aqueous salt solution. (f) Carrying out confirmatory tests, which are specific chemical tests to confirm the identity of a cation or an anion present in a salt. Salts Solution Colour Solid White or colourless Salts of Na , K , NH4+, Mg2+, Ca2+, Ba2+, Al3+, Pb2+, Zn2+ (if the anions are colourless) Na , K+, NH4+, Mg2+, Ca2+, Ba2+, Al3+, Pb2+, Zn2+ Yellow PbO, PbI2, PbCrO4, BaCrO4 Fe3+, CrO42– Blue Hydrated Cu2+ salt Cu2+ Green Hydrated Fe2+ salt, CuCO3 and CuCl2 Fe2+ Black Cu2+, Fe2+ oxide or sulphide – Brown/ orange Hydrated Fe3+ salt Fe3+, CrO72– + + + 5 Table 8.10 shows the solubility of different types of salts in water. Table 8.10 Solubility of salts in water Type of Salt Solubility in water Salts of Na+, All are soluble K+, NH4+ 240 Nitrate All are soluble Sulphate All common sulphates are soluble except BaSO4, PbSO4 and CaSO4 Chloride All common chlorides are soluble except AgCl, HgCl and PbCl2(soluble in hot water) Tests of Gases Solubility in water Carbonate All common carbonates are insoluble except Na2CO3, K2CO3 and (NH4)2CO3 Oxide All oxides are insoluble except Na2O, K2O and CaO (slightly soluble) Hydroxide All hydroxides are insoluble except KOH, NaOH, Ca(OH)2 and Ba(OH)2 1 Certain gases may be evolved when a chemical substance is (a) heated, (b) reacted with a dilute or concentrated acid, (c) heated with an alkali. 2 Based on the gas evolved, information about the types of ions present can be deduced. For instance, if carbon dioxide gas is evolved in a reaction, carbonate ions are present in the salt. 3 The physical properties and chemical tests for a few gases are summarised in Table 8.11. Lead halides PbCl2, PbBr2 and PbI2 are insoluble in cold water but soluble in hot water Table 8.11 Physical properties and tests on gases Name of gas Colour of gas Smell of gas Oxygen, O2 Colourless No smell No effect When a glowing wooden splint is lowered into the test tube of oxygen, the glowing splint is lighted Hydrogen, H2 Colourless No smell No effect When a lighted wooden splint is placed near the mouth of the test tube of hydrogen, a ‘pop’ sound is produced Carbon dioxide, CO2 Colourless No smell Moist blue litmus turns to red When carbon dioxide gas is bubbled into limewater using a delivery tube, the limewater becomes milky Ammonia, NH3 Colourless Pungent Moist red litmus turns to blue When a glass rod dipped into concentrated hydrochloric acid is placed near the mouth of the test tube with ammonia, white fumes are formed Chlorine, Cl2 Greenishyellow Choking Decolourises moist red or blue litmus Hydrogen chloride, HCl Colourless Pungent Moist blue litmus turns to red When a glass rod dipped into concentrated ammonia is placed near the mouth of the test tube with hydrogen chloride, white fumes are formed Sulphur dioxide, SO2 Colourless Pungent Moist blue litmus turns to red When sulphur dioxide gas is bubbled into acidified potassium manganate(VII) solution, the purple colour is decolourised (or when it is bubbled into acidified potassium dichromate(VI) solution, the colour changes from orange to green) Nitrogen dioxide, NO2 Brown Pungent Moist blue litmus turns to red – Effect on damp litmus Confirmatory test on gas – (NH4)2CO3(s) → 2NH3(g) + H2O(l) + CO2(g) Heating Test on Salts (NH4)2SO4(s) → 2NH3(g) + H2SO4(l) 1 All ammonium, carbonate, nitrate and some sulphate salts will decompose when heated. 2 All ammonium salts liberate ammonia gas when heated. Examples NH4NO3(s) → NH3(g) + HNO3(g) 3 All carbonates except potassium carbonate and sodium carbonate produce carbon dioxide gas when heated. Table 8.12 shows the effect of heating on metal carbonates. 241 Salts 8 Type of Salt Table 8.12 Effect of heat on carbonate salts Carbonate salt Potassium carbonate Sodium carbonate Effect of heat Will not decompose on heating 8 Decompose to metal oxide and carbon dioxide gas Calcium carbonate CaCO3(s) → CaO(s) + CO2(g) Magnesium carbonate MgCO3(s) → MgO(s) + CO2(g) Aluminium carbonate Al2(CO3)3(s) → Al2O3(s) + 3CO2(g) Zinc carbonate ZnCO3(s) → ZnO(s) + CO2(g) Iron(III) carbonate Fe2(CO3)3(s) → Fe2O3(s) + 3CO2(g) Lead(II) carbonate PbCO3(s) → PbO(s) + CO2(g) Copper(II) carbonate CuCO3(s) → CuO(s) + CO2(g) Decompose to metal, carbon dioxide gas and oxygen gas Mercury(II) carbonate 2HgCO3(s) → 2Hg(l) + 2CO2(g) + O2(g) Silver carbonate 2Ag2CO3(s) → 4Ag(s) + 2CO2(g) + O2(g) Gold(I) carbonate 2Au2CO3(s) → 4Au(s) + 2CO2(g) + O2(g) Decompose to carbon dioxide gas, ammonia and water vapour without any residue Ammonium carbonate (NH4)2CO3(s) → 2NH3(g) + H2O(g) + CO2(g) 4 All nitrates decompose when heated. Table 8.13 shows the effect of heating on metal nitrates. (a) Sodium nitrate and potassium nitrate produce oxygen gas and nitrites when heated. (b) Other metal nitrates produce oxygen gas, nitrogen dioxide gas and metal oxides when heated. Table 8.13 Effect of heat on nitrate salts Nitrate Salt Effect of heat Decompose to metal nitrite and oxygen gas Potassium nitrate 2KNO3(s) → 2KNO2(s) + O2(g) Sodium nitrate 2NaNO3(s) → 2NaNO2(s) + O2(g) Decompose to metal oxide, oxygen gas and nitrogen dioxide gas Calcium nitrate 2Ca(NO3)2(s) → 2CaO(s) + 4NO2(g) + O2(g) Magnesium nitrate 2Mg(NO3)2(s) → 2MgO(s) + 4NO2(g) + O2(g) Aluminium nitrate 4Al(NO3)3(s) → 2Al2O3(s) + 12NO2(g) + 3O2(g) Zinc nitrate 2Zn(NO3)2(s) → 2ZnO(s) + 4NO2(g) + O2(g) Iron(III) nitrate 4Fe(NO3)3(s) → 2Fe2O3(s) + 12NO2(g) + 3O2(g) Lead(II) nitrate 2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g) Copper(II) nitrate 2Cu(NO3)2(s) → 2CuO(s) + 4NO2(g) + O2(g) Decompose to metal, nitrogen dioxide gas and oxygen gas Mercury(II) nitrate Hg(NO3)2(s) → Hg(l) + 2NO2(g) + O2(g) Silver nitrate 2AgNO3(s) → 2Ag(s) + 2NO2(g) + O2(g) Gold(I) nitrate 2AuNO3(s) → 2Au(s) + 2NO2(g) + O2(g) Decompose to nitrous oxide gas, water vapour without any residue Ammonium nitrate Salts NH4NO3(s) → N2O(g) + 2H2O(g) 242 5 Most sulphate salts do not decompose when heated. Only a few sulphates such as iron(II) sulphate, zinc sulphate and copper(II) sulphate decompose to sulphur dioxide or sulphur trioxide gas when heated. Examples 2FeSO4(s) → Fe2O3(s) + SO2(g) + SO3(g) Figure 8.8 Table 8.14 Deduction of types of ion present from gas produced Type of gas produced CuSO4(s) → CuO(s) + SO3(g) 6 All chloride salts are stable on heating except ammonium chloride. Ammonium chloride sublimes and decomposes to produce ammonia gas and hydrogen chloride gas. NH4Cl(s) → NH3(g) + HCl(g) 7 The deduction of the types of ions present based on the gas produced is shown in Table 8.14. 8 When a salt is heated, (a) the type of gas evolved has to be identified. This will give information to the type of anion (or cation, NH4+) present. (b) the colour change of the solid in the test tube must be recorded. This will give information regarding the type of cation present. Type of ion CO2 Carbonate ion, CO32– (except Na2CO3 and K2CO3) O2 Nitrate ion, NO3– NO2 and O2 Nitrate ion, NO3– (except NaNO3 and KNO3) SO2 Sulphate ion, SO42– NH3 Ammonium ion, NH4+ 8 ZnSO4(s) → ZnO(s) + SO3(g) 9 Most salts that decompose produced metal oxides as residue. The change of colour during heating gives a good indication towards the type of metal oxide formed as shown in Table 8.15. Table 8.15 Colour change of salts on heating Colour of residue after heating Metal oxide produced Cation present in salt White Yellow when hot, white when cold ZnO Zn2+ White Brown when hot, yellow when cold PbO Pb2+ Blue/green Black CuO Cu2+ Green/yellow Brown Fe2O3 Fe2+/Fe3+ To study the effect of heat on carbonate salts Apparatus Boiling tubes, test tubes, test tube holder, delivery tube with rubber stopper, spatula and Bunsen burner. Materials Potassium carbonate, sodium carbonate, calcium carbonate, magnesium carbonate, zinc carbonate, lead(II) carbonate, copper(II) carbonate and limewater. SPM ’10/P2 Procedure 1 One spatula of potassium carbonate powder is placed in a dry boiling tube and the colour of the solid is recorded. 2 The boiling tube is fitted with a stopper with a delivery tube. 3 The carbonate salt is heated slowly and then strongly. 243 Salts Activity 8.7 Original colour of salt 8 4 Any gas evolved is passed through the delivery tube into the limewater. The effect on limewater is recorded (Figure 8.9). 5 When there is no further change, the colour of the residue when it is hot is recorded. The colour of the residue when it is cooled to room temperature is also recorded. 6 Steps 1 to 5 of the experiment is repeated using other carbonate salts as shown in Table 8.16. stopped. Otherwise the limewater will be sucked back into the hot boiling tube. Precaution Make sure that the end of the delivery tube is removed from the limewater before heating is Figure 8.9 Heating test on carbonate salts Results Table 8.16 Heating test on carbonate salts Colour of residue Colour of salt before heating When hot When cold Potassium carbonate, K2CO3 White White White No visible change Sodium carbonate, Na2CO3 White White White No visible change Calcium carbonate, CaCO3 White White White Limewater turns milky Magnesium carbonate, MgCO3 White White White Limewater turns milky Zinc carbonate, ZnCO3 White Yellow White Limewater turns milky Lead(II) carbonate, PbCO3 White Brown Yellow Limewater turns milky Copper(II) carbonate, CuCO3 Green Black Black Limewater turns milky Carbonate salt 4 Note that if excess carbon dioxide gas is passed into the limewater, the white precipitate, CaCO3 formed will dissolve to form calcium hydrogen carbonate. The limewater will turn clear again. Discussion 1 In this experiment, limewater is used to test for the presence of carbon dioxide gas. Carbon dioxide gas turns limewater milky because calcium carbonate is formed as a white precipitate. CaCO3(s) + H2O(l) + CO2(g) → Ca(HCO3)2(aq) CO2(g) + Ca(OH)2(aq) → CaCO3(s) + H2O(l) 2 When zinc carbonate is heated, zinc oxide and carbon dioxide gas are produced. ZnCO3(s) → ZnO(s) + CO2(g) Zinc oxide is yellow when hot and white when cooled. 3 When lead(II) carbonate is heated, lead(II) oxide and carbon dioxide gas are produced. white precipitate colourless solution Conclusion 1 Potassium carbonate and sodium carbonate will not decompose on heating. 2 Other metal carbonates decompose on heating to produce metal oxides and carbon dioxide gas. A general equation representing the decomposition of carbonate salt by heat is MCO3(s) → MO(s) + CO2(g) PbCO3(s) → PbO(s) + CO2(g) Lead(II) oxide is brown when hot and yellow when cooled. Salts Effect on limewater metal carbonate 244 metal oxide To study the effect of heat on nitrate salts 4 When there is no further change, the colour of the residue is recorded when it is hot. The colour of the residue is recorded again when the residue is cooled to room temperature. 5 Steps 1 to 4 of the experiment are repeated using other nitrate salts as shown in Table 8.17. Apparatus Boiling tubes, litmus paper, test tube holder, wooden splint, spatula and Bunsen burner. Materials Potassium nitrate, sodium nitrate, magnesium nitrate, zinc nitrate, aluminium nitrate, lead(II) nitrate and copper(II) nitrate. 8 Procedure 1 One spatula of potassium nitrate is placed in a dry boiling tube and the colour of the solid is noted. 2 The nitrate salt is heated slowly and then strongly. 3 Any gas evolved is tested by a glowing wooden splint (Figure 8.10(a)) and moist blue litmus paper (Figure 8.10(b)). The results are recorded. Figure 8.10 Heating test on nitrate salt Table 8.17 Heating test on nitrate salts Test on gas Colour of residue When hot When cold Effect on Effect on Colour of glowing wooden moist blue gas splint litmus paper Potassium nitrate, KNO3 White White White Colourless Rekindles No change Sodium nitrate, NaNO3 White White White Colourless Rekindles No change Magnesium nitrate, Mg(NO3)2 White White White Brown Rekindles Turns red Aluminium nitrate, Al(NO3)3 White White White Brown Rekindles Turns red Zinc nitrate, Zn(NO3)2 White Yellow White Brown Rekindles Turns red Lead(II) nitrate, Pb(NO3)2 White Brown Yellow Brown Rekindles Turns red Copper(II) nitrate, Cu(NO3)2 Blue Black Black Brown Rekindles Turns red Discussion 1 The brown gas that changed moist blue litmus paper to red is nitrogen dioxide gas. 2 The gas that rekindles a glowing wooden splint is oxygen gas. Generally, 2MNO3(s) → 2MNO2(s) + O2(g), metal nitrate metal nitrite where M = K or Na. 2 Other metal nitrates decompose to metal oxides, nitrogen dioxide gas and oxygen gas when heated. A general equation representing the decomposition of other nitrate salts by heat is Conclusion 1 All nitrate salts decompose on heating. Potassium nitrate and sodium nitrate decompose to oxygen gas when heated. 2M(NO3)2(s) → 2MO(s) + 4NO2(g) + O2(g) metal nitrate 245 metal oxide Salts Activity 8.8 Nitrate salt Colour of salt before heating Test for carbonate ions, CO32– Test for nitrate ions, NO3–(brown ring test) 1 When dilute acid (hydrochloric acid, nitric acid or sulphuric acid) is added to an aqueous carbonate solution (or solid carbonate), effervescence occurs. 2 The gas evolved turns limewater milky. Carbon dioxide gas is produced, indicating the presence of carbonate ions. 1 When dilute sulphuric acid and iron(II) sulphate, FeSO4 solution are added to an aqueous nitrate solution, followed by concentrated sulphuric acid added slowly along the side of the test tube, a brown ring is formed in the middle section of the solution mixture. 2 The formation of the brown ring (a complex) indicates the presence of nitrate ions. 8 CO32–(aq) + 2H+(aq) → CO2(g) + H2O(l) Figure 8.11 Acid test for carbonate ions Figure 8.12 Brown ring test for nitrate ions Tests for the Presence of Anions in Aqueous Solutions SPM ’05/P2 Q8 The presence of anions, CO32–, NO3–, SO42–, and Cl– can be identified by conducting specific tests on the aqueous salt solution. Test for sulphate ions, SO42– SPM Test for chloride ions, Cl– ’10/P2 1 When dilute hydrochloric acid (or nitric acid) is added to an aqueous sulphate solution followed by barium chloride solution, BaCl2 (or barium nitrate solution, Ba(NO3)2), a white precipitate is formed. 2 The white precipitate is barium sulphate, BaSO4. 1 When dilute nitric acid is added to an aqueous chloride solution followed by silver nitrate solution, AgNO3, a white precipitate is formed. 2 The white precipitate is silver chloride, AgCl. Ag+(aq) + Cl–(aq) → AgCl(s) Ba2+(aq) + SO42–(aq) → BaSO4(s) 3 BaNO3 or BaCl2 solution provide the Ba2+ ions to react with the SO42– ions to produce the insoluble BaSO4 salt. 3 AgNO3 solution provides the Ag+ ions to react with the Cl– ions to produce the insoluble AgCl salt. H2SO4 followed by Ba(NO3)2 or BaCl2 solution is not a suitable test for the presence of SO42– ions. This is because the SO42– ions in H2SO4 will give a positive test with Ba(NO3)2 or BaCl2 solution. HCl followed by AgNO3 solution is not a suitable test for the presence of Cl– ions. This is because the Cl– ions in HCl will give a positive test with AgNO3 solution. Salts 246 To test for the presence of anions in aqueous salt solution Apparatus Test tubes, delivery tube with rubber stopper, spatula, test tube holder and Bunsen burner. sulphate solution, 2 mol dm–3 hydrochloric acid and nitric acid, concentrated sulphuric acid and 0.1 mol dm–3 silver nitrate solution. Materials 1 mol dm–3 sodium carbonate solution, sodium chloride solution, sodium sulphate solution, sodium nitrate solution, barium chloride solution and iron(II) Procedure The steps to test the aqueous salt solutions as planned in Table 8.18 are carried out. The observations and inferences are tabulated. 8 Results Table 8.18 Tests for the presence of anions in aqueous salt solutions Test 2 (a) About 2 cm3 of sodium chloride solution is placed in a test tube. (b) About 2 cm3 of dilute nitric acid is added to the chloride solution followed by about 2 cm3 of silver nitrate solution. 3 (a) About 2 cm3 of sodium sulphate solution is placed in a test tube. (b) About 2 cm3 of dilute nitric acid is added to the sulphate solution followed by about 2 cm3 of barium chloride solution. 4 (a) About 2 cm3 of sodium nitrate solution is placed in a test tube. (b) About 2 cm3 of dilute sulphuric acid is added to the nitrate solution followed by about 2 cm3 of iron(II) sulphate solution. The mixture is shaken to mix well. (c) Concentrated sulphuric acid is added care fully to the mixture along the wall of the tilted test tube without shaking the mixture. Inference Effervescence occurs. A gas is evolved. Limewater becomes milky. Carbon dioxide gas is evolved. Carbonate ion, CO32–, is present. A white precipitate is produced. Chloride ion, Cl–, is present. A white precipitate is produced. Sulphate ion, SO42–, is present. A brown ring is formed Nitrate ion, NO3–, is in the middle of the present. solution. Conclusion 1 The presence of carbonate ions can be identified by the addition of a dilute acid where carbon dioxide gas evolved will turn the limewater milky. 2 The presence of chloride ions can be identified by the addition of silver nitrate solution to an acidified solution when a white precipitate is produced. 3 The presence of sulphate ions can be identified by the addition of barium chloride solution to an acidified solution when a white precipitate is produced. 4 The presence of nitrate ions can be identified by the brown ring test. 247 Salts Activity 8.9 1 (a) About 2 cm3 of sodium carbonate solution is placed in a test tube. (b) About 2 cm3 of dilute hydrochloric acid is added to the carbonate solution. (c) Any gas produced is bubbled into limewater. Observation 8 Tests for Cations 1 The cations usually tested are: Al3+, Pb2+, Zn2+, Mg2+, Ca2+, Fe3+, Fe2+, Cu2+ and NH4+ ions. 2 An aqueous solution of the cation is prepared by (a) dissolving the salt in water (if the salt is soluble in water). (b) dissolving the salt in dilute acid and then filtering (if the salt is insoluble in water). The filtrate contains the cation. 3 The aqueous cation solution is then tested with (a) sodium hydroxide solution, NaOH, (b) aqueous ammonia solution, NH3(aq), (c) a specific reagent as a confirmatory test. 4 Sodium hydroxide and aqueous ammonia supply hydroxide ions, OH– to produce metal hydroxide as precipitate with cation solutions except Na+, K+ and NH4+ ions. 5 Transition element cations produce specific coloured metal hydroxide, whereas the other cations produce white precipitate as shown in Table 8.19. Cu(OH)2 Cu2+ Dirty green precipitate Fe(OH)2 Fe2+ Fe3+ – Na+, K+, NH4+ 9 All three Cu(OH)2, Fe(OH)2 and Fe(OH)3 do not dissolve in excess sodium hydroxide solution. However, Cu(OH)2 (a blue precipitate) dissolves in excess aqueous ammonia to form a dark blue solution. 10 A summary of the sodium hydroxide tests and aqueous ammonia tests for cations are shown in Table 8.20. SPM Blue precipitate Fe(OH)3 NH4+(aq) + OH–(aq) → NH3(g) + H2O(l) ’11/P1 Cation present Brown precipitate Some metal hydroxides are soluble in excess sodium hydroxide or aqueous ammonia to form complexes. 6 Al3+, Pb2+ and Zn2+ ions form metal hydroxides which are white precipitates that are soluble in excess sodium hydroxide. Furthermore, zinc hydroxide is also soluble in excess aqueous ammonia. 7 Mg2+ and Ca2+ ions form metal hydroxides which are white precipitates with sodium hydroxide, but only Mg2+ ions will produce white precipitate with aqueous ammonia. 8 Ammonium ion, NH4+ does not produce any precipitate with sodium hydroxide or aqueous ammonia. However, if a mixture of ammonium ion and sodium hydroxide is heated, ammonia gas is produced. Example: Pb2+(aq) + 2OH–(aq) → Pb(OH)2(s) Al3+(aq) + 3OH–(aq) → Al(OH)3(s) Formula of metal hydroxide Cation present No precipitate precipitate Observation Formula of metal hydroxide White precipitate Al(OH)3, Pb(OH)2, Al3+, Pb2+, Zn(OH)2, Mg(OH)2, Zn2+, Mg2+, Ca(OH)2 Ca2+ Generally, Mn+(aq) + nOH–(aq) → M(OH)n(s) Table 8.19 Colours of metal hydroxides Observation SPM Table 8.20 Hydroxide tests for cations Cation A little sodium hydroxide, NaOH(aq) Excess sodium hydroxide, A little aqueous NaOH(aq) ammonia, NH3(aq) No precipitate forms ’10/P1 Excess aqueous ammonia, NH3(aq) NH4+ No precipitate forms. No precipitate forms. NH3 NH3 gas evolves when gas evolves when heated heated Pb2+ White precipitate White precipitate soluble in White precipitate excess NaOH White precipitate insoluble in excess NH3(aq) Zn2+ White precipitate White precipitate soluble in White precipitate excess NaOH White precipitate soluble in excess NH3(aq) Al3+ White precipitate White precipitate soluble in White precipitate excess NaOH White precipitate insoluble in excess NH3(aq) Salts 248 No precipitate forms A little sodium hydroxide, NaOH(aq) Excess sodium hydroxide, A little aqueous NaOH(aq) ammonia, NH3(aq) Mg2+ White precipitate White precipitate insoluble in excess NaOH White precipitate White precipitate insoluble in excess NH3(aq) Ca2+ White precipitate White precipitate insoluble in excess NaOH No precipitate forms No precipitate forms Cu2+ Blue precipitate Blue precipitate insoluble in Blue precipitate excess NaOH Fe2+ Dirty green precipitate Dirty green precipitate insoluble in excess NaOH Fe3+ Brown precipitate Dirty green precipitate Excess aqueous ammonia, NH3(aq) Blue precipitate soluble in excess NH3(aq) to form a dark blue solution Dirty green precipitate insoluble in excess NH3(aq) Brown precipitate insoluble Brown precipitate Brown precipitate insoluble in excess NH3(aq) in excess NaOH Confirmatory tests for Fe2+, Fe3+,NH4+ and Pb2+ ions 1 Potassium hexacyanoferrate(II), K4Fe(CN)6 solution, potassium hexacyanoferrate(III), K3Fe(CN)6 solution and potassium thiocyanate, KSCN, solution can be used to confirm the presence of Fe2+ and Fe3+ ions. The observation is shown in Table 8.21. 2 Fe2+ ions can also be confirmed by acidified potassium manganate(VII), KMnO4 solution. If a few drops of KMnO4 solution acidified by dilute H2SO4 are added to a solution, and the purple colour of the manganate(VII) ion is decolourised, then the solution contains Fe2+ ions. 3 Pb2+ ions can be confirmed by adding (a) an iodide solution (e.g. KI), produces a yellow precipitate of PbI2. (b) a chloride solution (e.g. NaCl), produces a white precipitate of PbCl2. (c) a sulphate solution (e.g. H2SO4), produces a white precipitate of PbSO4. 4 Both lead(II) chloride, PbCl2 and lead(II) iodide, PbI2 are soluble in hot water and recrystallise when cooled. 5 Nessler reagent is a special reagent to test the presence of ammonium ion. This reagent forms a brown precipitate with ammonium ion. 6 A summary of the confirmatory tests for Pb2+, NH4+, Fe2+ and Fe3+ ions is shown in Table 8.21. SPM ’11/P1 Table 8.21 Confirmatory tests for Pb2+, NH4+, Fe2+ and Fe3+ ions Cation Specific reagent Lead(II) ions, Pb2+ KI, NaI Yellow precipitate, soluble in hot water and recrystallises when cooled KCl, NaCl, HCl White precipitate, soluble in hot water and recrystallises when cooled K2SO4, Na2SO4, H2SO4 White precipitate, insoluble in hot water Ammonium ions, NH4+ Nessler reagent Iron(II) ions, Fe 2+ Iron(III) ions, Fe3+ Observation Brown precipitate Potassium hexacyanoferrate(II), K4Fe(CN)6 Light blue precipitate Potassium hexacyanoferrate(III), K3Fe(CN)6 Prussian blue (dark blue) precipitate Acidified KMnO4 Purple colour decolourises Potassium thiocyanate, KSCN Blood red colour Potassium hexacyanoferrate(II), K4Fe(CN)6 Turnbull’s blue (dark blue) precipitate Potassium hexacyanoferrate(III), K3Fe(CN)6 Greenish-brown solution 249 Salts 8 Cation Flowchart for the analysis of cations in salts Unknown cation solution Test 1: Add NaOH(aq) solution White precipitate Coloured precipitate Zn , Al3+, Pb2+, Ca2+, Mg2+ 2+ Blue Cu2+ 8 Add excess NaOH(aq) solution Green Fe2+ No precipitate Brown Fe3+ NH4+ On heating Cation solution Precipitate dissolves Precipitate does not dissolve Zn2+, Al3+, Pb2+ Ca2+, Mg2+ Test 2: Add aqueous NH3 NH4+ Coloured precipitate Cation solution Test 2: White precipitate dissolves in excess NH3 Add excess aqueous NH3 White precipitate insoluble in excess NH3 Test 2: White precipitate is formed Mg2+ present Zn2+ present Blue Cu2+ Cation solution Add excess aqueous NH3 Add excess aqueous NH3 No precipitate Ca2+ present Al3+ or Pb2+ Green Fe2+ Brown Fe3+ Cation solution Cation solution Test 3: Blood red colour Cu present Fe present 3+ Cation solution Cation solution Test 3: Test 3: Add KI Add K3Fe(CN)6 Dark blue precipitate Salts No precipitate Yellow precipitate Al3+ present Pb2+ present Add KSCN Dark blue solution 2+ Fe2+ present 250 Gas turns red litmus to blue. NH3 gas evolves. Test 2 Add Nessler reagent Brown precipitate NH4+ present Flowchart for the analysis of anions in salts Unknown solid anion Test 1: Heating the solid Gas rekindles glowing wooden splint and a brown gas is evolved. O2 and NO2 gas evolve. Gas evolves and decolourises acidified KMnO4 solution. SO2 or SO3 evolves. No gas evolves CO32– ion NO3– ion SO42– ion SO42– or Cl– Solid salt or salt solution Anion solution Anion solution Anion solution 8 Gas evolves and turns limewater milky. CO2 gas is evolved. Test 2 Test 2 Add dilute acid Add FeSO4, dilute H2SO4 and then concentrated H2SO4 slowly Test 2 Add HNO3 and Ba(NO3)2 Brown ring White precipitate NO3– ion present SO42– ion present Gas evolves and turns limewater milky. CO2 evolves. Test 2 Add HNO3 and Ba(NO3)2 No precipitate. SO42– ion is not present. Anion solution CO32– ion present Test 3 Add HNO3 and AgNO3 White precipitate Cl– ion present Qualitative Analysis to Identify the Ions Present in a Salt 1 Two types of salt analysis: (A) To confirm the cation and anion present ’09/P2 in a named salt Example Compound X is lead(II) carbonate. How do you carry out tests to confirm the cation and anion in compound X? compounds. Carry out chemical tests to identify the anion and cation in solid Y and solution Z. 2 An example of problem A, to confirm the cation and anion in compound X, is shown in Activity 8.10 To confirm the cation, Pb2+ ions and the anion, CO32– ions in compound X. 3 An example of problem B, a planned analysis, is shown in Activity 8.11 To identify the cations and anions in unknown salt Y and salt Z. SPM (B) To identify the cation and anion present in an unknown salt Example You are supplied with solid Y and solution Z. Both Y and Z are ionic 251 Salts To confirm the cation, Pb2+ ions and the anion, CO32– ions in compound X Apparatus Test tubes, boiling tube, test tube holder, delivery tube with stopper, spatula and Bunsen burner. 8 Materials Solid lead(II) carbonate, 2 mol dm–3 nitric acid, 2 mol dm–3 sodium hydroxide solution, limewater and 0.5 mol dm–3 potassium iodide solution. Procedure 1 The steps in the experiment which are supplied in Table 8.22 are carried out. 2 The observations are recorded and inferences are made. 3 Care is taken to ensure that all test tubes and spatula are clean to prevent contamination. Results Table 8.22 Experiment 1 (a) A spatula of compound X is heated in a boiling tube, gently at first and then strongly. (b) The gas produced is passed into limewater in a test tube using a delivery tube. 2 5 cm3 of dilute nitric acid is added to a quarter spatula of compound X in a test tube. The gas evolved is tested with limewater. 3 The mixture in step 2 is filtered. The filtrate which contains the cation is divided into two portions in two test tubes. (a) For the first portion, sodium hydroxide solution is added gradually until in excess. (b) For the second portion, a little potassium iodide solution is added. (i) A little distilled water is added to the mixture and then heated. (ii) The mixture is allowed to cool under running water. Observation The residue is brown when hot and yellow when cold. The gas evolved turns limewater milky. The gas evolved turns limewater milky. Inference Lead(II) oxide is formed. Pb2+ ions may be present. Carbon dioxide gas is evolved. CO32– ions may be present. Carbon dioxide gas is evolved. CO32– ions are confirmed to be present. A white precipitate is produ ced which is soluble in excess sodium hydroxide solution. A yellow precipitate is produced. The yellow precipitate dissolves in hot water to form a colourless solution. Upon cooling, golden yellow crystals are reformed. Pb2+, Al3+ or Zn2+ ions may be present. Lead(II) iodide may be formed. The yellow precipitate is lead(II) iodide. Lead(II) ions, Pb2+ is confirmed to be present. Activity 8.10 & 8.11 Conclusion It is confirmed that compound X contains lead(II) ions and carbonate ions. To identify the cations and anions in unknown salt Y and salt Z Apparatus Test tubes, boiling tube, test tube holder, delivery tube with stopper, wooden splint and Bunsen burner. Salts Materials Unknown salt Y and salt Z, 2 mol dm–3 sodium hydroxide solution, aqueous ammonia, dilute sulphuric acid, dilute nitric acid, dilute hydrochloric acid, concentrated sulphuric acid, silver nitrate solution, barium chloride solution and iron(II) sulphate solution. 252 Procedure (A) Tests on salt Y Experiment Observation Inference A brown gas is produced. A gas that rekindles a glowing wooden splint is also produced. A white residue is formed. Nitrogen dioxide gas and oxygen gas are produced. Nitrate ion, NO3– is present. 2 The residue formed is cooled and then dissolved in dilute nitric acid. The resulting solution is divided into 3 portions. The white residue dissolves in nitric acid. The solution contains cations. (a) Sodium hydroxide solution is added to the first portion of solution Y until in excess. White precipitate that dissolves in excess sodium hydroxide is formed. Pb2+, Al3+ or Zn2+ ions may be present. (b) Aqueous ammonia is added to the second portion of solution Y until in excess. White precipitate that does not dissolve in excess aqueous ammonia is formed. Zn2+ ions are not present. Pb2+ ions or Al3+ ions may be present. (c) Potassium iodide solution is added to the third portion of solution Y. No noticeable change. Pb2+ ions are not present. Al3+ ions are present. A brown ring is formed. NO3– ions are present. 3 Salt Y is dissolved in distilled water. A little iron(II) sulphate and dilute sulphuric acid is added to solution Y. Concentrated sulphuric acid is then added slowly along the side of the test tube to the mixture. 8 1 Salt Y is heated strongly. (B) Tests on salt Z Experiment Observation Inference 1 Salt Z is dissolved in distilled water. The solution is divided into 4 portions. Z dissolves in water. Z is a soluble salt. 2 Sodium hydroxide solution is added to the first portion of solution Z until in excess. A white precipitate that does not dissolve in excess sodium hydroxide is formed. Mg2+ ions or Ca2+ ions may be present. 3 Aqueous ammonia is added to the second portion of solution Z until in excess. A white precipitate that does not dissolve in excess aqueous ammonia is formed. Mg2+ ions are present. 4 Dilute hydrochloric acid is added to the third portion of solution Z followed by barium chloride solution. No noticeable change. SO42– ions are not present. 5 Dilute nitric acid is added to the fourth portion of solution Z followed by silver nitrate solution. A white precipitate is formed. Cl– ions are present. Conclusion Compound Y contains Al3+ ion and NO3– ion. Compound Z contains Mg2+ ion and Cl– ion. 253 Salts 8 In qualitative analysis of unknown ions, you should not test for the ions present in the reagents used in analysis. For example, if the salt is dissolved in dilute hydrochloric acid, do not test for the presence of chloride ion. In the same way, if aqueous ammonia is added to the salt solution, do not test for ammonium ion. solution is added to Cl– ions solution followed by an acid, a white precipitate that is insoluble in acids will be formed. 3 Ba(NO3)2 or BaCl2 solutions produce insoluble salts with both CO32– ions and SO42– ions. However, BaCO3 is soluble in acids. Hence, the formation of white precipitate with Ba2+ ions solution in the presence of acids confirms the presence of SO42– ions. If Ba2+ ions solution is added to SO42– solution followed by an acid, a white precipitate that is insoluble in acids will be formed. 1 Pb(NO3)2 solution produces insoluble salts as precipitate with CO32– ions, SO42– ions and Cl– ions. Hence, Pb(NO3)2 solution is not a good test for the presence of the above three types of ions. However, a negative test may indicate the presence of NO3– ions. 2 AgNO3 solution produces insoluble salts with both CO32– ions and Cl– ions. However, Ag2CO3 is soluble in acids. Hence, the formation of white precipitate with AgNO3 solution in the presence of acids confirms the presence of Cl– ions. If AgNO3 6 ’05 Describe chemical tests that can be used to verify the cations and anions in beaker 1 and beaker 2. Comments • Add a little sodium carbonate powder to 5 cm3 of solution from beaker 1. The evolution of a gas that turns limewater milky will verify the presence of hydrogen ions in the acid. • Add 2 cm3 of barium nitrate solution to 5 cm3 of solution from beaker 1. The formation of a white precipitate will verify the presence of sulphate ions. • Add 1 cm3 of Nessler reagent to 5 cm3 of solution from beaker 2. The formation of a brown precipitate will verify the presence of ammonium ions. • Add a little of iron(II) sulphate and dilute sulphuric acid to 5 cm3 of solution from beaker 2, followed by concentrated sulphuric acid added slowly. The formation of a brown ring will verify the presence of nitrate ions. 8.2 2 Solid Y is not soluble in water but dissolves in dilute nitric acid and gives out a gas that turns limewater milky. The solution produced is yellow in colour and forms a brown precipitate when sodium hydroxide solution is added. (a) Give the name of salt Y. (b) Write an equation for the reaction between salt Y and dilute nitric acid. (c) Predict the reaction that would occur when salt Y is heated strongly. (d) Predict the observation that will occur when potassium thiocyanate solution is added to the yellow solution produced from the addition of nitric acid to salt Y. 1 The formulae of a few salts are given below: PbCl2, ZnSO4, Fe(NO3)3, ZnCO3, Na2CO3, Al2(SO4)3, CuCl2, CuCO3, Mg(NO3)2, Cu(NO3)2 Which of the above salts (a) is a solid that is insoluble in water? (b) is a white solid? (c) is soluble in water to produce a blue solution? (d) forms a white precipitate with barium chloride that is insoluble in acid? (e) forms a white precipitate that dissolves in sodium hydroxide solution? (f) forms a white precipitate with aqueous ammonia but is not soluble in excess aqueous ammonia? Salts 254 (e) Write an equation for the formation of the brown precipitate when sodium hydroxide solution is added to an acidic solution of salt Y. 4 You are given three types of acids: sulphuric acid, nitric acid and hydrochloric acid. Using suitable chemical tests, describe briefly how you can identify the three types of acids. 3 Dilute nitric acid is added to aqueous solutions of unknown salts P, Q and R respectively. It is found that solutions P and Q do not show any noticeable change, whereas effervescence occurs in solution R. The gas that is evolved from solution R turns limewater milky. When silver nitrate solution is added to solution P, a white precipitate is formed. When barium nitrate is added to solution Q, a white precipitate is produced. Identify the anions that are present in solutions P, Q and R. 5 You are given three types of salts: zinc nitrate, lead(II) nitrate and calcium nitrate. Using suitable chemical tests, describe briefly how you can differentiate between the three types of salts. 8 6 An experiment was carried out to identify the cation and anion present in an unknown salt Z. The tests and observations are tabulated below. Fill in the correct inferences in the table and deduce the cation and anion present in salt Z. Test Observation 1 NaOH solution is added gradually to a little Z solution until in excess. A white precipitate which is soluble in excess NaOH is produced. 2 Aqueous NH3 is added gradually to a little Z solution until in excess. A white precipitate which is soluble in excess aqueous NH3 is produced. 3 Solid Z is heated slowly and then strongly. A brown gas and a gas that rekindles a glowing wooden splint are produced. The residue formed is yellow when hot and white when cooled. 4 Dilute nitric acid followed by BaCl2 solution is added to a little Z solution. No noticeable change occurs. 8.3 Inference Practising Systematic and Meticulous Methods when Carrying Out Activities 1 Correct methods in preparing salt crystals (a) A salt solution is evaporated until saturated so that solid salt may be crystallised. The salt is not heated until dry to prevent decomposition of the salt. (b) The salt crystals formed are removed by filtration and then rinsed with distilled water to remove any foreign ions present in the salt. 2 Correct methods in salt analysis (a) All observations must be recorded carefully and immediately after every test. The inferences are then recorded as soon as possible. (b) During heating, do not direct the mouth of the test tube towards yourself or towards your fellow students. (c) Gases such as sulphur dioxide, nitrogen dioxide, chlorine, hydrogen chloride and ammonia are poisonous. Use suitable quantities as instructed. Do not use excess chem icals. This will cause wastage of chemicals and require a longer time to carry out the experiment. Making accurate observations will be difficult. Handle all apparatus and chemicals carefully in the laboratory. (d) Addition of a reagent to a salt solution should be carried out drop by drop while shaking the test tube until no further change occurs. 255 Salts 1 A salt is an ionic compound that is formed when the hydrogen ion in an acid is replaced by a metal ion or ammonium ion (NH4 +). 2 The solubility of a salt in water depends on the types of cations and anions present. 8 Type of salt 3 Filtration can be used to separate an insoluble salt (as the residue) from a soluble salt (as the filtrate). 4 The methods of preparing salts depend on the solubility of salts. 5 Insoluble salts can be prepared by double decomposition reaction. Two aqueous solutions containing the cations and the anions are mixed together. The precipitate is then obtained by filtration. 6 The mole ratio of ions that react to form a salt can be determined from an experiment through the continuous variation method. 7 Qualitative analysis of salt is a scheme of tests carried out to identify the cation and anion present in the salt. Solubility in water Sodium, potassium and ammonium and nitrate salts All are soluble Chloride salts All are soluble except PbCl2, AgCl and HgCl Sulphate salts All are soluble except PbSO4, BaSO4 and CaSO4 Carbonate salts All are insoluble except NaCO3, K2CO3 and (NH4)2CO3 8 Multiple-choice Questions 8.1 Salts 1 Which of the following salts is insoluble in water? A Lead(II) nitrate B Calcium chloride C Magnesium carbonate D Iron(III) sulphate 2 Which of the following chemicals is most suitable to ’06 react with HCl to prepare AgCl? A Silver carbonate B Silver nitrate C Silver metal D Silver oxide 3 Which following chemicals can be added to nitric acid to prepare copper(II) nitrate? A Copper metal B Copper(II) carbonate C Copper(II) chloride D Copper(II) sulphate Salts 4 Which is the best method to prepare ammonium nitrate? A Double decomposition B Neutralisation between an acid and an alkali C Reaction between an acid and a metal D Reaction between an acid and a metal carbonate 5 A sample of zinc(II) sulphate bought contained some impurities. The impure salt can be purified by a process known as A distillation B evaporation C crystallisation D recrystallisation 6 Which of the following salts can be prepared by double decomposition reaction? A Lead(II) nitrate B Aluminium sulphate C Silver chloride D Sodium sulphate 256 7 Which of the following pairs of solutions when added together will produce a precipitate? I Potassium carbonate and silver nitrate II Lead(II) nitrate and sodium chromate(VI) III Sodium carbonate and copper(II) sulphate IV Sodium sulphate and potassium carbonate A I and II only B II and III only C I, II and III only D I, III and IV only 8 Which of the following equations represent a double decomposition reaction? IMg(NO3)2(aq) + Na2CO3(aq) → MgCO3(s) + 2NaNO3(aq) II CaCO3(s) + H2SO4(aq) → CaSO4(aq) + CO2(g) + H2O(l) III BaCl2(aq) + H2SO4(aq) → BaSO4(s) + 2HCl(aq) 9 If 20 cm3 of 0.5 mol dm–3 aqueous sodium chloride solution is added to 20 cm3 of 1.0 mol dm–3 silver nitrate solution, which of the following ions are present in the solution produced? INa+ III NO3– II Ag+ IV Cl– A I and III only B II and III only C I, II and III only D I, II and IV only 10 4.2 g of magnesium carbonate reacts with excess hydrochloric acid to produce a salt. Which of the following are true about the reaction? [1 mol of gas occupies 24 dm3 at room temperature and pressure. Relative atomic mass: C, 12; O, 16; Mg, 24; Cl, 35.5] I Neutralisation reaction takes place. II 1.2 dm3 of gas is released. III Mass of salt formed is 4.75 g. IV 4.2 mol of water is formed. A I and II only B III and IV only C II and III only D I and IV only B Nitric acid and barium chloride solution C Nitric acid and silver nitrate solution D Sodium hydroxide solution 14 Aluminium sulphate solution and zinc sulphate solution can be differentiated by the A addition of silver nitrate solution. B addition of aqueous ammonia. C addition of sodium hydroxide solution. D addition of barium chloride solution. 15 When lead(II) nitrate solution is added to solution X, a white precipitate is produced. Solution X may be I H2SO4 II HNO3 Qualitative Analysis of Salts 11 When solid X is heated strongly, a gas that turns limewater milky is produced, leaving a white residue. Which of the following may be solid X? A Lead(II) carbonate B Sodium carbonate C Zinc carbonate D Magnesium carbonate 12 Which of the following salts will produce a brown gas on heating? A Lead(II) bromide B Ammonium nitrate C Potassium nitrate D Zinc nitrate 13 Which of the following is used to test for sulphate ions? A Iron(II) sulphate solution and concentrated sulphuric acid HCl BaCl2 I and III only II and IV only I, II and III only I, III and IV only 16 Solution M reacts with sodium hydroxide solution to form a white precipitate that is insoluble in excess sodium hydroxide solution. Solution M most probably contains I lead ion II calcium ion III aluminium ion IV magnesium ion A II and III only B I and III only C II and IV only D I, III and IV only 17 A student wants to identify cation X that is present in a salt solution. When ammonia solution is added into the salt solution, a green precipitate is ’11 formed. What is the next test that is needed and the expected observation to confirm cation X? Test Observation A Add potassium iodide solution Yellow precipitate is formed B Add potassium thiocyanate solution Blood red colour is formed C Add potassium hexacyanoferrate(II) solution Add Nessler reagent Light blue precipitate is formed D 8.2 III IV A B C D Brown precipitate is formed 18 When a gas Z is passed into copper(II) sulphate solution, a blue precipitate is produced. Gas Z may be A ammonia C hydrogen chloride B chlorine D sulphur dioxide 19 When aqueous iron(III) chloride solution is added to reagent X, a blood red colour is produced. Reagent X may be A ammonium sulphite C potassium thiocyanate B potassium iodide D potassium hexacyanoferrate(II) 20 Hydrochloric acid can be differentiated from sulphuric acid by adding I barium nitrate III silver nitrate II barium hydroxide IV sodium carbonate A III only C II and IV only B I and II only D I, II and III only 21 When lead(II) nitrate is heated strongly in a test tube, the following can be observed. I A brown gas is evolved. II A gas that rekindles a glowing wooden splint is evolved. III A gas that changes moist blue litmus to red is evolved. IV A white residue is formed. 257 Salts 8 IV 2Zn(NO3)2(s) → 2ZnO(s) + 4NO2(g) + O2(g) A I and II only C II and III only B I and III only D II and IV only 8 A B C D I and II only III and IV only I, II and III only I, II, III and IV 22 Which of the following ions will form a precipitate that dissolves in excess aqueous ammonia? I Copper(II) ions II Aluminium ions III Lead(II) ions IV Zinc ions A I and IV only B II and III only C II and IV only D II, III and IV only 23 Excess powdered carbonate of metal Z is added to sulphuric acid and stirred. After a few minutes, a light green solution is formed. Z could be a carbonate of A iron(II) C copper(II) B iron(III) D lead(II) 24 Which of the following reagents can be used to differentiate between sodium nitrate and potassium sulphate? I Lead(II) nitrate solution II Barium chloride solution III Silver nitrate solution IV Sodium hydroxide solution A I and II only B I and III only C II and IV only D I and IV only Solution X may be I sodium hydroxide II aqueous ammonia III Nessler reagent IV iron(III) nitrate A I and II only B III and IV only C I and IV only D I, II and III only B A white precipitate, which dissolves in excess ammonia, is formed when aqueous ammonia is added. C A white precipitate is formed when lead(II) nitrate solution is added. D A white precipitate is formed when silver nitrate solution is added. 26 When solid X is heated strongly, a brown gas that turned moist blue litmus paper to red is evolved and a black residue is formed. Which of the following may be solid X? A Copper(II) oxide B Sodium nitrate C Copper(II) nitrate D Magnesium nitrate 29 When solution X is added to sodium chloride solution, a white precipitate is formed. The precipitate dissolves when it is heated with a little distilled water. Which of the following will be observed when solution X is added to a solution of sodium iodide solution? A A white precipitate is formed. B A yellow precipitate is formed. C A brown solution is formed. D A purple solution is formed. 27 Lead(II) nitrate solution and aluminium sulphate solution can be distinguished respectively by adding I sodium hydroxide solution II potassium sulphate solution III barium nitrate solution IV sodium iodide solution A I and III only B II and III only C II and IV only D III and IV only 25 When solution X is added to iron(III) sulphate solution, a brown precipitate is produced. 30 Which of the following reagents can be used to differentiate Fe2+ ions from Fe3+ ions? I Potassium iodide solution II Potassium thiocyanate solution III Potassium hexacyanoferrate(III) solution IV Potassium manganate(VII) solution A I and II only B III and IV only C I, II and III only D II, III and IV only 28 Which of the following observations is true for both sodium chloride solution and zinc sulphate solution? A A white precipitate is formed when barium nitrate solution is added. Structured Questions 1 Diagram 1 is a flowchart showing a series of reactions starting from lead(II) oxide. Lead(II) oxide Precipitate F sodium hydroxide substance A Lead(II) nitrate heat Gas B + Gas C + Solid D potassium iodide Precipitate E Diagram 1 (a) (i) Name substance A that is used to react with lead(II) oxide to produce lead(II) nitrate. (ii) Write a chemical equation for the reaction that takes place in (i). Salts 258 [1 mark] [1 mark] (b) (c) (d) (e) Gas B is a brown gas while gas C is colourless. (i) Identify gas B and solid D. (ii) Suggest a test that you can use to test the presence of gas C. (i) Name precipitate E. What is the colour of precipitate E? (ii) Write the ionic equation for the formation of precipitate E. (i) Name precipitate F. (ii) Predict the observation that will occur if excess sodium hydroxide is added to precipitate F. Name another chemical that can replace lead(II) oxide to react with substance A to produce lead(II) nitrate. [2 marks] [1 mark] [1 mark] [2 marks] [1 mark] [1 mark] [1 mark] 2 Diagram 2 is a flowchart showing a series of reactions starting from magnesium oxide. Excess magnesium oxide 8 + Nitric acid stir and filter Residue A Filtrate Process 1: evaporate, cool and filter Process 2: heated strongly Crystal B Solid C + Gas D + Gas E Diagram 2 (a) Write a chemical equation showing the reaction between magnesium oxide and nitric acid. [1 mark] (b) If 50 cm3 of 2.0 mol dm–3 nitric acid is added to excess magnesium oxide, calculate the maximum mass of salt that can be produced. [Relative atomic mass: H, 1; N, 14; O, 16; Mg, 24] [3 marks] (c) The filtrate formed from the reaction above is colourless. (i) Predict what will be observed if aqueous sodium hydroxide is added to the filtrate gradually until in excess. [2 marks] (ii) Name crystal B. [1 mark] (d) In Process 2, the gas D produced is colourless while gas E is brown. (i) Identify gas E and name solid C. [2 marks] (ii) Suggest a test that you can use to test the presence of gas D. [1 mark] 3 The flowchart of Diagram 3 shows a series of reactions I to IV carried out to identify the ions present in compound A. Diagram 3 (a) (i) Identify gas C. (ii) What is the anion present in compound A? (b) Based on the observation obtained in reaction III and reaction IV, predict the cation present in compound A. (c) Based on (a) and (b), write a balanced equation for the action of heat on compound A. (d) Predict the colour of residue B when it is hot and when it is cooled. (e) (i) Write a balanced equation for the formation of solution D in reaction II. (ii) State an observation that can be noted in reaction II. 259 [1 mark] [1 mark] [1 [1 [1 [1 [1 mark] mark] mark] mark] mark] Salts (f) (i) Name the white precipitate F formed in reaction IV. (ii) Write an ionic equation for the formation of the white precipitate F. (g) Predict the observation that can be obtained when aqueous sodium hydroxide solution is added until in excess to solution E. [1 mark] [1 mark] [1 mark] 4 Diagram 4 shows the steps involved in the preparation of zinc carbonate. Zinc oxide Step 1 add hydrochloric acid Step 2 Salt solution P Zinc carbonate add solution Q 8 Diagram 4 (a) Write a balanced equation for the formation of salt solution P. (b) Explain briefly how you can obtain a solution of salt P. (c) (i) Name solution Q that is required to be added to salt solution P in Step 2 to produce zinc carbonate. (ii) Name the type of reaction involved in Step 2. (iii) Write an ionic equation for the formation of zinc carbonate. (d) 30 cm3 of 2.0 mol dm–3 hydrochloric acid is reacted with excess zinc oxide. [Relative atomic mass: C, 12; O, 16; Zn, 65] (i) Calculate the number of moles of salt P that is formed. (ii) Calculate the maximum mass of zinc carbonate that is produced. (e) Suggest how you would convert zinc carbonate back to zinc oxide. [1 mark] [2 marks] [1 mark] [1 mark] [1 mark] [2 marks] [2 marks] [1 mark] 5 The flowchart in Diagram 5 shows the result of a qualitative analysis that is carried out on a mixture of metal Q and a water soluble salt P. Salt P + metal Q Gas R add NaOH and heat Salt P add HNO3 + BaCl2 Process A Metal Q add HCl White precipitate S Gas T + Solution U Diagram 5 (a) Name the process A that is used to separate salt P and metal Q. (b) Gas R is a gas that can change red litmus paper to blue. Name gas R. Consequently what is the cation present in solution P? (c) Name the white precipitate S. What is the anion present in solution P? (d) When gas R is passed into solution U, a white precipitate is first formed but dissolves when excess gas R is passed through. Identify the cation present in solution U. (e) From your answer in (d), determine the identity of metal Q and gas T. [1 mark] [2 marks] [2 marks] [1 mark] [2 marks] Essay Questions 1 (a) The following are three examples of salts that can be prepared in the laboratory. ’07 • Sodium sulphate, Na2SO4 • Lead(II) chloride, PbCl2 • Magnesium nitrate, Mg(NO3)2 (i) From these examples, identify the soluble and insoluble salts. [3 marks] (ii) State the reactants for the preparation of the insoluble salt in (i). [3 marks] (b) With the aid of a labelled diagram, explain the crystallisation method for preparing a soluble salt from its unsaturated solution. [6 marks] (c) Table 1 shows the observations from tests carried out on salt X. Salts Test Procedure Observation I Heating of salt X solid A brown gas is given off and a residue that is brown when hot, yellow when cooled is formed. II Add excess aqueous sodium hydroxide solution to salt X solution until in excess A white precipitate which is soluble in excess sodium hydroxide solution is formed. Table 1 260 Describe a laboratory experiment to prepare the salt. In your description, include the chemical equations involved. [8 marks] (c) Three beakers with solutions labelled X, Y and Z may contain the following salt solutions: Based on the information in Table 1: (i) Identify an anion that is present in Test I and describe a chemical test to verify the anion. [4 marks] (ii) Identify three cations that are present in Test II. Describe a chemical test to verify the cation present in salt X based on Test I and Test II. [5 marks] • Zinc sulphate • Zinc nitrate • Magnesium sulphate 2 (a) What is meant by a salt? [2 marks] (b) You are required to prepare a sample of dry lead(II) carbonate salt. The following chemicals are supplied: • Sodium carbonate solution • Dilute nitric acid • Lead metal powder You are provided only with ammonia and barium nitrate solutions. Describe how you could differentiate between the 3 salt solutions by using the two reagents provided. Include your observations and conclusions. [10 marks] Experiment 1 Barium nitrate solution, Ba(NO3)2 and aqueous potassium chromate(VI), K2CrO4 solution react to produce a yellow precipitate with the formula BaCrO4. A student has carried out an experiment to measure the height of ’03 the precipitate produced by the reaction between 5.0 cm3 of 0.50 mol dm–3 aqueous barium nitrate solution and aqueous potassium chromate(VI) solution of unknown concentration at different volumes. The data of the experiment obtained is shown in Table 1. Test tubes Volume of barium nitrate solution (cm3) Volume of potassium chromate(VI) solution (cm3) Height of precipitate (cm) 1 5.0 1.0 0.5 2 5.0 2.0 1.0 3 5.0 3.0 1.4 4 5.0 4.0 1.9 5 5.0 5.0 2.4 6 5.0 6.0 2.4 7 5.0 7.0 2.4 Table 1 (ii) Calculate the number of moles of barium ions that is used in this experiment. (a) Suggest a suitable apparatus that can be used to measure the volume of barium nitrate solution and potassium chromate(VI) solution in this experiment. [3 marks] [3 marks] (e) Based on the formula of the yellow precipitate given, calculate the number of moles of chromate(VI) ions used in the volume in (d) (i). Hence calculate the molarity of the aqueous potassium chromate(VI), K2CrO4, solution used in this experiment. [3 marks] (b) State the manipulated variable, responding variable and constant variable in this experiment. [3 marks] (c) Draw a graph of the height of the precipitate against the volume of potassium chromate(VI) solution. [3 marks] (f) Explain why the height of the precipitate does not change when 5.0 cm3, 6.0 cm3 and 7.0 cm3 of potassium chromate(VI) solution is added to 5.0 cm3 of 0.5 mol dm–3 barium nitrate solution in Table 1. [3 marks] (d) (i) What is the minimum volume of potassium chromate(VI) solution required to react completely with 5.0 cm3 of 0.5 mol dm–3 barium nitrate solution? 261 Salts 8 FORM 4 THEME: Production and Management of Manufactured Chemicals CHAPTER 9 Manufactured Substances in Industry SPM Topical Analysis 2008 Year 1 Paper Section Number of questions 3 2009 2 A B C 1 – – 3 1 – 2 2010 2 A B C 1 – – 3 1 – 5 2011 2 A B C – – – 3 1 – 3 3 2 A B C – 1 – ONCEPT MAP MANUFACTURED SUBSTANCES IN INDUSTRY Sulphuric acid Uses: To make fertilisers, polymers, detergents, pigments Manufactured by the Contact process: • Temperature: 450–550 °C • Pressure: 1 atmosphere • Catalyst: V2O5 Ammonia Uses: To make fertilisers, nitric acid and as a cooling agent (refrigerant) Manufactured by the Haber process: • Temperature: 450–550 °C • Pressure: 200–500 atmospheres • Catalyst: Iron Synthetic polymers Examples and uses: • Polythene: To make plastic bags, plastic containers and toys. • Polypropene: To make plastic bottles, tables and chairs • PVC: To make water pipes, raincoats and wire casing Glass and ceramics Uses: Construction materials, household items, laboratory apparatus Types of glass: • Fused glass • Soda glass • Borosilicate glass • Lead glass Alloys Purposes: • To increase hardness and strength • To prevent corrosion • To improve the appearance Examples: Steel, bronze, brass, magnalium, pewter Composite materials Examples: • Reinforced concrete • Superconductor • Fibre optic • Fibreglass • Photochromic glass – 9.1 Sulphuric Acid 1 The manufacture of sulphuric acid is one of the most important chemical industries at the present time. 2 Sulphuric acid, H2SO4 is a non-volatile diprotic acid. 3 Concentrated sulphuric acid is a viscous colourless liquid. Manufacture of fertilisers 1 Calcium dihydrogen phos phate (super phosphate) is prepared from the reaction between sulphuric acid and tricalcium phosphate: ammonium sulphate 3 Potassium sulphate is prepared from the reaction between sulphuric acid and potassium hydroxide. 2H2SO4 + Ca3(PO4)2 → Ca(H2PO4)2 + 2CaSO4 calcium dihydrogen phosphate H2SO4 + 2KOH → K2SO4 + 2H2O 2 Ammonium sulphate is prepared from the reaction between sulphuric acid and aqueous ammonia. Manufacture of detergents (synthetic cleaning agents) Sulphuric acid reacts with hydrocarbon to produce sulphonic acid. Sulphonic acid is then neutralised with sodium hydroxide to produce the detergent. potassium sulphate Manufacture of white pigment in paint barium sulphate, BaSO4 The Uses of Sulphuric Acid in Daily Life Manufacture of synthetic fibres (polymers) Rayon is an example of a synthetic fibre that is produced from the action of sulphuric acid on cellulose. The Industrial Process in the Manufacture of Sulphuric Acid In school laboratories 1 As a strong acid 2 As a drying or dehydrating agent 3 As an oxidising agent 4 As a sulphonating agent 5 As a catalyst The neutralisation between sulphuric acid and barium hydroxide produces barium sulphate. sulphur or metal sulphide SPM ’07/P2 burned in air sulphur dioxide, SO2 1 Sulphuric acid is manufactured by the Contact process in industry. 2 The raw materials used in the Contact process are sulphur (or sulphide minerals), air and water. 3 The flowchart of the Contact process is shown in Figure 9.1. It describes how sulphur or metal sulphide is converted to concentrated sulphuric acid, H2SO4 through regulated steps in the process. (i) V2O5 as the catalyst (ii) temperature of 450 °C – 550 °C (iii) pressure of 1 atmosphere sulphur trioxide, SO3 dissolved in concentrated H2SO4 oleum, H2S2O7 diluted with equal volume of H2O concentrated sulphuric acid, H2SO4 Figure 9.1 Flowchart of the Contact process 263 Manufactured Substances in Industry 9 H2SO4 + 2NH3 → (NH4)2SO4 The Contact process involves three stages I SPM ’04/P2 II III sulphur ⎯→ sulphur dioxide ⎯→ sulphur trioxide ⎯→ sulphuric acid 9 Figure 9.2 shows the three stages in the manufacture of sulphuric acid by the Contact process in industry. Figure 9.2 The production of concentrated sulphuric acid in industry. SPM Stage I Stage II Production of sulphur dioxide gas, SO2 This can be done by two methods: 1 Burning of sulphur in dry air in the furnace. Conversion of sulphur dioxide to sulphur trioxide, SO3 1 The sulphur dioxide gas is dried and purified before being added to dry air to produce sulphur trioxide gas. This is (a) to remove water vapour in the air (the reaction of water with SO3 will produce heat that will vaporise the acid), (b) to remove contaminants such as arsenic compounds (found in the sulphur or sulphide minerals) that will poison the catalyst and make it ineffective. 2 Pure and dry sulphur dioxide with excess dry oxygen (from air) are passed through a converter. 3 A high percentage (98%) of sulphur dioxide is converted into suphur trioxide under the following conditions: (i) The presence of vanadium(V) oxide, V2O5, as a catalyst (ii) A temperature of between 450°C – 550°C (iii) A pressure of one atmosphere S + O2 → SO2 2 Burning of metal sulphide such as zinc sulphide or iron(III) sulphide in dry air. 2ZnS + 3O2 → 2SO2 + 2ZnO The production of sulphuric acid is known as the Contact process because the molecules of sulphur dioxide, SO2 and oxygen, O2 are in contact with the catalyst, V2O5. The use of V2O5 has replaced the use of platinum as a catalyst because (a) platinum is much more expensive, (b) platinum is easily poisoned (lose its catalyst effect) by arsenic compounds. Manufactured Substances in Industry 2SO2 + O2 264 ’09/P1 2SO3 Stage III Production of sulphuric acid 1 In the absorber, sulphur trioxide is dissolved in concentrated sulphuric acid to produce oleum, H2S2O7, a viscous liquid. SO3 + H2SO4 → H2S2O7 H2S2O7 + H2O → 2H2SO4 3 The two reactions in stage III are equivalent to adding sulphur trioxide to water. SO3 + H2O → H2SO4 However, sulphur trioxide is not dissolved directly in water to produce sulphuric acid. This is because (a) SO3 has a low solubility in water. (b) SO3 reacts violently in water, producing a large amount of heat which will vapourise sulphuric acid to form acid mist. The mist is corrosive, pollutes the air and is difficult to condense. The burning of fossil fuels causes environmental pollution, producing air pollutants such as carbon monoxide, nitrogen monoxide, nitrogen dioxide and sulphur dioxide. Scientists are searching for alternative sources of energy to replace fossil fuels. Other than wind energy, solar energy, geothermal energy (energy from earth’s internal heat) and nuclear energy, scientists are also researching into the production of fuels from natural products such as palm oil. Environmental Pollution by Sulphur Dioxide 1 Sulphur dioxide is the intermediate product of the Contact process. 2 Sulphur dioxide is also produced during volcanic eruptions. 3 However, the main source of sulphur dioxide is from the burning of fossil fuels such as petroleum. Most of the fossil fuels contain some sulphur. Hence sulphur dioxide is produced when fossil fuels are burned. 4 Waste gases from factories and extraction of metal from their sulphide ores also release sulphur dioxide into the atmosphere. 5 The burning of products manufactured from sulphuric acid such as rayon will also produce sulphur dioxide gas. 6 Sulphur is a poisonous and acidic gas that can cause environmental pollution. Inhaling sulphur dioxide affects the respiratory system. It can cause lung problems such as coughing, chest pains, shortness of breath and bronchitis. 7 Sulphur dioxide gas dissolves in atmospheric water to produce sulphurous acid, H2SO3 and sulphuric acid, H2SO4. The presence of these acids in rain water causes acid rain. 265 SO2 + H2O ⎯⎯→ H2SO3 2SO2 + O2 + 2H2O ⎯⎯→ 2H2SO4 8 The effects of acid rain are as follows: (a) Corrodes concrete buildings and metal structures (b) Destroys trees and plants in forest (c) Decreases the pH of the soil which becomes acidic, unsuitable for growth of plants and destroys the roots of plants (d) Reacts with minerals in the soil to produce salts which are leached out of the top soil; essential nutrients for plants growth are depleted (plants die of malnutrition and diseases) (e) Acid rain flows into lakes and rivers. This increases the acidity of water and may kill fish and other aquatic living things. 9 Two methods to reduce sulphur dioxide from the atmosphere: (a) Use low sulphur fuels to reduce the emission of sulphur dioxide in exhaust gases. (b) Remove sulphur dioxide from waste air by treating it with calcium carbonate before it is released. Manufactured Substances in Industry 9 2 Oleum is then diluted with an equal volume of water to produce concentrated sulphuric acid (98%). The industrial synthesis of sulphuric acid is represented by the flowchart. (a) Name this process. (b) Name gas X, gas Y and liquid Z. (c) Write balanced equations for Step 1 and Step 2. (d) State the optimum conditions involved in Step 2. (e) Why is gas Y not dissolved in water to produce sulphuric acid? 9 Manufacture by Contact process: • 2SO2 + O2 2SO3 • Temperature: 450 °C – 550 °C • Pressure : 1 atm • Catalyst : V2O5 Uses of Sulphuric Pollution by SO2: sulphuric acid: • Forms acid acid • Making paint rain pigments, detergents • Causes breathing and fertilisers problems and • As an electrolyte lung diseases in accumulators 9.2 1 Ammonia, NH3 is a very important compound in industry. 2 The main uses of ammonia: (a) To manufacture nitrogenous fertilisers such as ammonium sulphate, ammonium nitrate and urea (b) The liquid form is used as a cooling agent (refrigerant) in refrigerators (c) as a raw material for the manufacture of nitric acid in the Ostwald process (d) to be converted into nitric acid used for making explosives (e) as an alkali to prevent the coagulation of latex so that latex can remain in the liquid form (f) to produce ammonium chloride used as an electrolyte in dry cells (g) as a cleaning agent to remove grease (h) used in the manufacture of synthetic fibres such as nylon 3 The manufacture of nitrogenous fertilisers: SPM (a) Ammonium sulphate ’07/P2 Ammonia reacts with sulphuric acid by neutralisation to produce ammonium sulphate. 9.1 1 An important use of sulphuric acid is in the production of fertilisers. (a) Name the fertilisers produced and write the equations involved when sulphuric acid reacts with (i) aqueous ammonia (ii) potassium hydroxide (b) Barium sulphate, BaSO4, is a white pigment in paint. Write an equation for the formation of barium sulphate from the reaction between sulphuric acid and barium hydroxide. 2 Give three uses of sulphuric acid in daily life. 3 Concentrated sulphuric acid can be used as a dehydrating agent. (a) What is the function of a dehydrating agent? (b) When concentrated sulphuric acid is added to glucose, C6H12O6, a black residue is formed after a few minutes. (i) What is the black residue? (ii) Write a balanced equation for the reaction that has taken place. 4 2NH3 + H2SO4 ⎯⎯→ (NH4)2SO4 ammonium sulphate Sulphur + step 1 heating Gas X step 2 (b) Ammonium nitrate Ammonia reacts with nitric acid by neutralisation to produce ammonium nitrate. Gas Y Oxygen Sulphuric acid Manufactured Substances in Industry dilute with water Ammonia and Its Salts NH3 + HNO3 ⎯⎯→ NH4NO3 Liquid Z ammonium nitrate 266 (c) Urea Ammonia reacts with carbon dioxide at a temperature of 200 °C and a pressure of 200 atmospheres to produce urea. NH3 + H2O NH4+ + OH– 4 As ammonia is very soluble in water, an inverted filter funnel is used to prevent the suction of water (Figure 9.3). 2NH3 + CO2 ⎯→ (NH2) 2CO + H2O urea 9 4 Liquid ammonia is suitable for use as a cooling agent (refrigerant) in refrigerators because it has a low boiling point and is very volatile. 5 In the Ostwald process, ammonia is converted into nitric acid by the following steps: (a) Ammonia is oxidised to nitrogen monoxide gas in the presence of platinum as the catalyst. Figure 9.3 To dissolve ammonia gas in water platinum 5 Ammonia gas reacts with hydrogen chloride gas to form white fumes of ammonium chloride. (This is used as a test for ammonia gas). 4NH3 + 5O2 ⎯⎯→ 4NO + 6H2O (b) Nitrogen monoxide is further oxidised to nitrogen dioxide. NH3 + HCl ⎯⎯→ NH4Cl 2NO + O2 ⎯⎯→ 2NO2 6 Ammonia is alkaline in property and reacts with dilute acids in neutralisation to produce salts. For example: (c) Nitrogen dioxide is dissolved in water to produce nitric acid. 4NO2 + O2 + 2H2O ⎯⎯→ 4HNO3 2NH3 + H2SO4 → (NH4)2SO4 NH3 + HNO3 → NH4NO3 6 Nitric acid is used to make explosives such as TNT when nitric acid reacts with organic substances such as methylbenzene (common name: toluene). 7 Ammonia can neutralise the organic acids that are produced by microorganisms in latex. Thus it is used to maintain latex in the liquid form. 8 Ammonia reacts with hydrogen chloride to produce ammonium chloride which is used as the electrolyte in dry cells. 7 Aqueous solutions of ammonia produces OH– ions to react with metal ions (except Na+ ion, K+ ion and Ca2+ ion) forming precipitates of metal hydroxides. Fe3+ + 3OH– → Fe(OH)3 brown precipitate Mg2+ + 2OH– → Mg(OH)2 white precipitate NH3 + HCl → NH4Cl 8 Some metal hydroxides such as zinc hydroxide and copper(II) hydroxide dissolve in excess aqueous ammonia to form complexes. For example: The Properties of Ammonia 1 Ammonia is a colourless and pungent gas. It is less dense than air. 2 Ammonia changes moist red litmus paper to blue. Thus ammonia is an alkaline gas. 3 Ammonia dissolves in water to produce a weak alkali. A 0.1 mol dm–3 ammonia solution has a pH value of about 10. Zn(OH)2 + 4NH3 → [Zn(NH3)4]2+ + 2OH– Cu(OH)2 + 4NH3 → [Cu(NH3)4]2+ + 2OH– 267 Manufactured Substances in Industry To investigate the properties of ammonia gas 2 The mixture in the test tube is heated. The ammonia gas produced is tested in turn by using (a) a piece of moist red litmus paper, (b) a glass rod dipped in concentrated hydro chloric acid. 3 Ammonia gas is collected in inverted test tubes using downward displacement of air. The test tubes are then stoppered immediately. 4 A little distilled water is added to the test tube of ammonia gas which is then shaken. The solution formed is tested with a piece of pH paper. 5 A test tube of ammonia is inverted with its mouth below a beaker of water. The stopper of the test tube is then removed as shown in Figure 9.5. Observation made is recorded. Apparatus pH paper, red litmus paper, beaker, glass rod, test tubes, U-tube with soda lime, beaker and delivery tube. 9 Materials Ammonium chloride, calcium hydroxide, concen trated hydrochloric acid, distilled water. Procedure 1 A spatula of ammonium chloride and a spatula of calcium hydroxide are put in a test tube. The test tube is then connected to a U-tube with soda lime as shown in Figure 9.4. Figure 9.5 Testing the solubility of ammonia in water Figure 9.4 Preparation of ammonia gas Results Activity 9.1 Test Observation Inference 1 With moist red litmus paper The red litmus paper changed to a blue colour Ammonia gas is alkaline 2 With hydrogen chloride vapour White fumes are formed Ammonia gas forms white fumes of ammonium chloride with hydrogen chloride gas 3 With pH paper pH paper changed to a blue colour, corresponding to a pH value of about 10 Aqueous ammonia is a weak alkali 4 A test tube of ammonia is inverted in a beaker of water Water rushes in and fills up the whole test tube Ammonia gas is very soluble in water (b) The ionic equation for the reaction between ammonium salt and alkali is Discussion 1 Ammonia gas is produced when an ammonium salt such as ammonium chloride is heated with an alkali such as calcium hydroxide. (a) The equation for the reaction between ammonium chloride and calcium hydroxide is NH4+ + OH– → NH3 + H2O 2 Soda lime in the U-tube is used as a drying agent to dry the ammonia gas produced. In place of soda lime, anhydrous calcium oxide can also be used as a drying agent. However, concentrated sulphuric acid cannot be used to dry ammonia gas as it will react with ammonia in a neutralisation reaction. 2NH4Cl + Ca(OH)2 → CaCl2 + 2NH3 + 2H2O Manufactured Substances in Industry 268 3 Ammonia gas is collected in inverted test tubes using downward displacement of air because ammonia is less dense than air. 4 Ammonia gas is very soluble in water. As such it cannot be collected by a downward displacement of water. 5 Ammonia gas reacts with hydrogen chloride to form white fumes of ammonium chloride. Conclusion 1 Ammonia is an alkaline gas which turns red litmus paper to blue and dissolves in water to form a weak alkali. 2 Ammonia gas reacts with hydrogen chloride to form white fumes of ammonium chloride. 3 Ammonia gas is very soluble in water. The Industrial Process in the Manufacture of Ammonia 4 In the Haber process: (a) A mixture consisting of one volume of ’06/P2 nitrogen gas and three volumes of pure and dry hydrogen gas is compresssed to a pressure between 200 – 500 atmospheres. (b) The gas mixture is passed through a catalyst of powdered iron at a temperature of 450 – 550 °C. (c) At this optimum temperature and pressure, ammonia gas is produced. SPM 1 The Haber process is the industrial method used to prepare ammonia gas on a large scale using nitrogen gas and hydrogen gas. N2 + 3H2 2NH3 5 The gas mixture produced consists of about 17% ammonia gas. The ammonia gas is liquefied when the gas mixture is cooled. The unreacted nitrogen gas and hydrogen gas are pumped back to the catalytic column to be reacted again. In 1918, Fritz Haber was awarded the Nobel prize for his discovery of the industrial manufacture of ammonia gas from hydrogen gas and nitrogen gas 2 Nitrogen gas used in the Haber process is obtained from the fractional distillation of liquid air. 3 Hydrogen gas used in the Haber process can be obtained by two methods: (a) The reaction between steam and heated coke (carbon). H2O + C ⎯⎯→ CO + H2 This mixture is known as water gas (b) The reaction between steam and natural gas (consists mainly of methane, CH4). Figure 9.6 The manufacture of ammonia gas by the Haber process 2H2O + CH4⎯⎯→ CO2 + 4H2 269 Manufactured Substances in Industry 9 NH3 + HCl → NH4Cl (a) Based on the graph, what is the effect of temperature and pressure on the percentage of ammonia produced? (b) State one problem each that is accompanied with the effects mentioned in (a) when the factory wants to increase the percentage of ammonia produced. (c) Discuss how the problems in (b) may be overcome by using an optimum temperature and pressure. SPM ’06/P2 Q5 1 The reaction between nitrogen gas and hydrogen gas to produce ammonia gas is a reversible reaction. This means that not all the nitrogen gas will react with hydrogen gas completely. The optimum temperature and pressure are used to ensure a satisfactory yield of ammonia. 2 The percentage yield of ammonia gas depends on the temperature and pressure used, as shown in Figure 9.7 below. 9 Solution (a) Lower temperature will increase the percentage of ammonia produced. Higher pressure will increase the percentage of ammonia produced. (b) Lower temperature will lower the rate of reaction. Higher pressure will increase the cost of production. (c) Using an optimum temperature of 450-550 °C, the rate of reaction will not be too low and yet produce a high yield of ammonia. Using an optimum pressure of 200-500 atm, the cost of increasing the pressure and using stronger pipes will not be too high to produce a high yield of ammonia. Figure 9.7 Effects of temperature and pressure on the yield of ammonia (a) The lower the temperature, the higher the yield of ammonia. (b) The higher the pressure, the higher the yield of ammonia. (c) However, low temperature will lower the rate of reaction. High pressure will increase the cost of production. (d) Hence, the most suitable temperature and pressure to produce a high yield of ammonia is a temperature of 450 °C–550 °C and a pressure of 200–500 atmospheres. 1 Ammonium Fertilisers 1 Plants require nitrogen to produce protein. Nitrogen is absorbed by plants in the form of nitrates, NO3– which are soluble in water. 2 Ammonium fertilisers are chemical fertilisers added to the soil to replace the elements in soil used up by plants. 3 Ammonium fertilisers contain ammonium ions, NH4+, that can be converted into nitrate ions by bacteria living in the soil. 4 The effectiveness of ammonium fertilisers is determined by the percentage of nitrogen by mass in them. The fertiliser with a higher percentage of nitrogen is more effective. 5 The percentage of nitrogen by mass can be calculated using the following formula: ’06 A factory that is manufacturing ammonia carried out a test to determine the percentage of ammonia that can be produced at two different conditions: I and II. The results are shown in the graph below. Percentage of nitrogen by mass Mass of nitrogen = —————————————————————  100% Molar mass of fertiliser Manufactured Substances in Industry 270 1 SPM ’10/P1 Calculate the percentage by mass of nitrogen in ammonium sulphate, (NH4)2SO4. [Relative atomic mass: N,14; H.1; S,32; O,16] = 2[14 + 4 (1)] + 32 + 4(16) = 132 1 mol (NH4)2SO4 consists of 2 mol atoms of nitrogen. Percentage of nitrogen in 1 mol of (NH4)2SO4 2(14) =— — — — 100% = 21.2% 132 Solution Relative molecular mass of (NH4)2SO4 (B) To prepare ammonium sulphate crystals 1 25 cm3 of 2.0 mol dm–3 aqueous ammonia solution is pipetted into a clean conical flask. 2 (V2 – V1) cm3 of sulphuric acid is added from the burette to the aqueous ammonia solution. 3 The mixture in the conical flask is transferred to a beaker and is slowly evaporated until a saturated solution is formed. 4 The saturated solution is left to cool. White crystals of ammonium sulphate are produced. 5 The ammonium sulphate crystals are then removed by filtration, washed with distilled water and dried between filter papers. Figure 9.8 Titration of sulphuric acid with ammonia solution Discussion 1 The equation for the neutralisation of aqueous ammonia and sulphuric acid is 2NH3 + H2SO4 → (NH4)2SO4 2 The first titration (Experiment A) is carried out to determine the volume of sulphuric acid required to completely neutralise 25 cm3 of aqueous ammonia. 3 In the second titration (Experiment B), the methyl orange indicator is not used so that the salt produced is not contaminated with the indicator. 4 The ammonium sulphate solution produced is not evaporated until dry because ammonium sulphate solid will decompose when heated. 5 Usually, the mass of ammonium sulphate crystals obtained from the experiment is less than the theoretical value because not all ammonium sulphate crystals can be crystallised from the solution. Some ammonium sulphate remains dissolved in the solution. Conclusion Ammonium sulphate, an example of ammonium fertiliser, can be prepared by the neutralisation reaction between aqueous ammonia and dilute sulphuric acid. 271 Manufactured Substances in Industry Activity 9.2 Apparatus 25 cm3 pipette, 50 cm3 burette, dropper, retort stand with clamp and white tile. Materials 1 mol dm–3 sulphuric acid, 2 mol dm–3 aqueous ammonia solution and methyl orange indicator. Procedure (A) To determine the volume of sulphuric acid re quired to neutralise 25 cm3 of ammonia solution 1 25 cm3 of 2.0 mol dm–3 aqueous ammonia solution is transferred using a 25 cm3 pipette to a clean conical flask. Three drops of methyl orange indicator are added to the alkali and the colour of the solution is noted. 2 A 50 cm3 burette is filled with sulphuric acid and clamped to a retort stand. The initial burette reading (V1) is recorded. 3 The conical flask containing 25 cm3 of the aqueous ammonia is placed below the burette. A piece of white tile is placed below the conical flask for clearer observation of the change in colour (Figure 9.8). 4 Sulphuric acid is added slowly from the burette to the aqueous ammonia solution in the conical flask while the flask is gently swirled. 5 Titration is stopped when methyl orange changes colour from yellow to orange. The final burette reading (V2) is recorded. 6 The volume of sulphuric acid required to neutralise 25.0 cm3 of ammonia solution is (V2 – V1) cm3. 9 To prepare ammonium sulphate, (NH4)2SO4, an ammonium fertiliser Ammonium sulphate is an acidic salt. Hence, long term use of ammonium sulphate as a fertiliser will increase the acidity in the soil. This can be overcome by adding quicklime (calcium oxide) to neutralise the acidic soil. 9 Manufacture of ammonia: Haber Process N2(g) + 3H2(g) 2NH3(g) • Temperature: 450 °C–550 °C • Catalyst: iron powder • Pressure: 200–500 atm. Uses • Manufacture of fertilisers, nitric acid Ammonia, NH3 Tests • Turns moist red litmus paper blue • Forms white fumes with HCI gas Properties • Colourless • Pungent smell Reactions • Produce ammonium salts with acids • Produce metal hydroxide as precipitate • A weak alkali • Very soluble in water 9.2 The given diagram shows the production of a fertiliser, ammonium nitrate. Step 2 is known as the Ostwald Process. (a) Name gas A, gas B and acid C in the diagram. (b) Name the industrial process in the production of gas B in step 1. (c) State the source from which gas A is obtained. (d) What will be observed if gas B comes in contact with hydrogen chloride gas? (e) Write a balanced equation for step 3. (f) Calculate the percentage by mass of nitrogen in ammonium nitrate. [Relative atomic mass: H,1; N,14; O,16] 1 Ammonia is commercially produced by the Haber process. (a) Name the raw materials used in the production of ammonia gas. (b) Name the catalyst used in the Haber process. (c) State the optimum temperature and pressure used for this process. (d) Write a balanced equation for this process. (e) State two uses of ammonia in daily life. 2 Gas A + step 1 gas B H2 step 2 acid C Ostwald process step 3 ammonium nitrate 9.3 2 Pure metals are weak and soft because the arrangement of atoms in pure metals makes them ductile and malleable. 3 Arrangement of pure metal atoms A pure metal contains atoms of the same size arranged in a regular and organised closedpacked structure (Figure 9.9). Alloys Meaning and Purpose of Making Alloys 1 An alloy is a mixture of two or more elements with a certain composition in which the major component is a metal. Manufactured Substances in Industry 272 (b) In an alloy, these atoms of different sizes disrupt the orderly arrangement of the metal atoms and also fill up any empty spaces in the metal crystal structure. (c) Hence, the layers of metal atoms are prevented from sliding over each other easily. This makes the alloy harder and stronger, less ductile and less malleable than its pure metals. Figure 9.9 Arrangement of atoms in a pure metal 9 4 Weakness of pure metal (a) Ductility Pure metals are soft because the orderly arrangement of atoms of the same size enables the layers of atoms to slide over each other easily when an external force is applied on them. This makes the metals ductile and metals can be drawn to form long wires (Figure 9.10). Figure 9.12 Arrangement of atoms in alloys 6 The properties of a pure metal are thus improved by making them into alloys. There are three aims of alloying a pure metal: (a) To increase the hardness and strength of a metal (b) To prevent corrosion or rusting (c) To improve the appearance of the metal surfaces, with a better finish and lustre Figure 9.10 Ductility of pure metal (b) Malleability There are imperfections in the natural arrangement of metal atoms. Empty space exists in the structures of pure metals. When ham mered or pressed, groups of metal atoms may slide into new positions in these empty spaces. This makes metals malleable, able to be made into different shapes or pressed into thin sheets (Figure 9.11). Aluminium and copper wires can be made because of the ductility of metals Figure 9.11 Malleability of pure metal Source: Wikimedia; Rosmaniakos Golden drinking vessel from Iran was made in about 5 B.C., proving the use of metal alloys in early civilization Pure gold is too soft to make jewellery. 916 gold consiststs of 91.6% gold SPM ’08/P1, ’09/P2, 5 The making of alloys ’11/P2 (a) In the process of alloying, one or more foreign elements are added to molten metal. When the alloy hardens, the positions of some of the metal atoms are replaced by atoms of foreign elements, with sizes bigger or smaller than the original metal atoms. Pure metals have the following physical properties: (a) Ductile (can be drawn into a wire) (b) Malleable (can be shaped by hammering) (c) High melting and boiling points (d) High density (e) Good conductors of electricity 273 Manufactured Substances in Industry Gold is one of the most ductile and malleable metal. Pure gold is too soft and is not suitable for making any type of jewellery. Gold is thus usually alloyed with copper, silver or palladium. The carat unit is used to measure the purity of gold. The carat value is the number of parts of gold in 24 parts of the alloy. Hence, 24-carat gold is pure gold. 18-carat gold consists of 75% of gold by weight. 9 Aims of alloying To increase the hardness and strength Alloying improves the hardness and strength of a metal. 1 The addition of a little carbon to iron metal produces steel which is a very hard alloy of iron. 2 The addition of magnesium to aluminium metal produces an alloy called magnalium. Magnalium is harder than aluminium but still retains the low density of aluminium metal. 3 The addition of tin to copper metal produces bronze. Bronze is an alloy harder than both tin and copper. The body of airplanes are made of magnalium which is harder than pure aluminium Manufactured Substances in Industry To prevent corrosion Pure metals such as tin and iron are easily corroded in damp, polluted or acidic air. 1 The addition of carbon, nickel and chromium to iron metal produces stainless steel. Stainless steel is an alloy which can resist rusting. The chromium and nickel form chromium(III) oxide and nickel(IV) oxide which prevents the iron from rusting. 2 The addition of tin to copper produces bronze which is able to resist corrosion and tarnish. Stainless steel kitchenware resists rusting 274 To improve the appearance SPM ’05/P1 Metals are easily tarnished because of the formation of metal oxides on the metal surfaces. The process of alloying can maintain the lustre on the surface of the metal. 1 Stainless steel is more shiny than pure iron. 2 Adding a little copper and antimony to tin produces the alloy pewter which is harder and shinier, and not so easily tarnished. 3 Alloy wheels made from aluminium and other elements improve the look of vehicles. Malaysian pewter is suitable for making shiny and attractive ornamental objects 9.1 SPM ’05/P3 ’04/P2 To compare the hardness of a pure metal and its alloy 9 Problem statement Are alloys harder than pure metals? Hypothesis Bronze is harder than copper. When a weight is dropped onto a ball bearing placed on a metal block made of copper or bronze, a larger dent will be produced on the softer copper metal block than on the bronze block. Variables (a) Manipulated variable : Types of materials (copper or bronze) to make the metal block (b) Responding variable : Diameter of the dent made by a steel ball bearing (c) Constant variable : Size of steel ball bearing, mass of weight used, height from which it is dropped Materials Copper block, bronze block, ball bearing, 1 kg weight, metre ruler, retort stand with clamp, cellophane tape and thread. Procedure 1 A metre ruler is clamped to a retort stand, and a piece of copper block is placed on the base of the retort stand. 2 A steel ball bearing is placed on the copper block and a piece of cellophane tape is used to hold the ball bearing in place. 3 A 1 kg weight is hung at a height of 50 cm above the copper block (Figure 9.13). 4 The weight is dropped onto the ball bearing placed on the copper block. 5 The diameter of the dent made by the ball bearing is measured. 6 The experiment is repeated three times using different areas on the surface of the copper block. 7 The average diameter of the dent is calculated. 8 Steps 1 to 7 are repeated using a piece of bronze block. Figure 9.13 To compare the hardness of an alloy with its pure metal Results I II III Average Copper 3.2 3.3 3.2 3.23 Bronze 2.4 2.5 2.5 2.47 Discussion 1 The bigger the average diameter of the dents produced by the steel ball bearing on the metal means that it has been pressed deeper into the metal surface. 2 Thus copper is softer than bronze because the steel ball bearing has been pressed deeper into the surface of copper metal than that of bronze. 3 Bronze is a type of alloy formed from copper and tin. The tin atoms are larger than the copper atoms. They distort the orderly structure of the copper atoms so that the layers of copper atoms can no longer slide easily over one another. This makes bronze harder than copper. Conclusion 1 The average diameter of the dents made by the steel ball bearing on the copper block is bigger than that on the bronze block. 2 Hence, bronze, a type of alloy, is harder than pure copper metal. The hypothesis is accepted. 275 Manufactured Substances in Industry Experiment 9.1 Diameter of the dent (mm) Metal block 9.2 To compare the rates of rusting of iron, steel and stainless steel Procedure 1 Three test tubes are half-filled with jelly solution and are labelled as A, B and C. 2 1 cm3 of potassium hexacyanoferrate(III) solution is added to every test tube. 3 An iron nail, a steel nail and a stainless steel nail are polished with sandpaper to remove any rust formed. The nails are then placed in the three test tubes labelled A, B and C respectively. 4 All three test tubes are allowed to stand for 5 days before they are examined. Problem statement How do the rates of rusting of iron, steel and stainless steel differ? Hypothesis Pure iron rusts faster than steel while stainless steel does not rust easily. 9 SPM ’05/P3 Variables (a) Manipulated variable : Types of nails (iron, steel and stainless steel) (b) Responding variable : Rate of rusting (c) Constant variable : Size of nails, duration of rusting and conditions of experiment (temperature, supply of water and air) Materials Iron nail, steel nail, stainless steel nail, 5% jelly solution and potassium hexacyanoferrate(III) solution and sandpaper. Figure 9.14 To compare the rates of rusting of iron, steel and stainless steel Results Test tube Type of nail Observation A Iron nail Blue colour is formed around the nail Rusting occurs B Steel nail A slight blue colour is formed A little rusting occurs C Stainless steel nail No blue colour is observed No rusting occurs Discussion 1 When iron rusts, iron(II) ion, Fe2+ is produced. Conclusion 1 The formation of a blue colour shows that rusting of iron (corrosion) has occurred. 2 The presence of a blue colour shows that iron nail rusts easily (corroded), steel nail rusts slightly and stainless steel does not rust at all. The hypothesis is accepted. Fe → Fe2+ + 2e– Experiment 9.2 Inference 2 Potassium hexacyanoferrate(III) solution is used to test the presence of iron(II) ion. A dark blue colour will be formed. The intensity of the blue colour indicates the rate of rusting. 3 A stainless steel alloy which resists rusting, is produced by adding nickel and chromium to iron metal. Manufactured Substances in Industry 276 The Composition, Properties and Uses of Some Common Alloys Uses Properties Composition Carbon steel 99% iron 1% carbon Hard and strong Stainless steel 74% iron 18% chromium 8% nickel 90% copper 10% tin Shiny, strong and resists rusting Brass 70% copper 30% zinc Hard and shiny Magnalium 70% aluminium 30% magnesium 95% aluminium 3% copper 1% magnesium 1% manganese 97% tin 3% copper and antimony 50% tin 50% lead copper, nickel (percentage according to colour) Light, hard and strong Bronze Duralumin Pewter Solder Cupro-nickel Hard, strong and shiny Light, hard and strong Lustrous and strong Hard, shiny and with low melting point Hard, shiny and resists corrosion • Frameworks of buildings and bridges • In the making of tools, framework of heavy machinery and body of vehicles • In the making of cutlery and kitchenware • In the making of machine parts and surgical instruments • In the making of kitchenware and ships’ propellers • In the making of decorative ornaments, statues and art crafts • In the making of electrical connectors and musical instruments • In the making of kitchenware and decorative ornaments • In the making of aircraft body frames • In the making of rims of racing car tyres • In the making of the bodies of aircrafts and bullet trains • In the making of racing bicycles, fan blades, light electrical cable • In the making of mugs, candlesticks, decorative ornaments and souvenirs • In the making of solder for electrical wires and metal pipes • To make coins of 10 sen, 20 sen, 50 sen An alloy is a mixture of two or more elements in which the major component is a metal. Malaysian coins are made from cupro-nickel alloy (75% Cu, 25% Ni) 9.3 (a) Name the elements used to make the alloy pewter. (b) What are the advantages of pewter compared to its pure metal? (c) State a use of pewter. 1 Steel and stainless steel are two examples of iron alloys. (a) What is an alloy? (b) What is the element that is added to iron to form (i) steel? (ii) stainless steel? (c) Draw the arrangement of particles in (i) pure iron (ii) steel (d) Explain why stainless steel and not iron is used to make cutlery. 3 Copper is one of the metals used since ancient times. (a) Explain why copper alloys are more commonly used than its pure form. (b) Name two examples of copper alloys. (c) Pure copper is ductile and malleable. Explain this property in terms of the arrangements of atoms. 2 Pewter is an important alloy made in Malaysia. 277 Manufactured Substances in Industry 9 Alloy SPM ’10/P1, ’11/P1 9.4 The Meaning of Polymers 9 3 Protein is formed by the polymerisation of monomers known as amino acids. Synthetic Polymers polymerisation SPM amino acids ⎯⎯⎯⎯⎯⎯→ protein ’11/P1 (monomers) 1 The word polymer originated from the Greek polumeros which means ‘having many parts’. 2 Polymers are large molecules made up of many smaller and identical repeating units joined together by covalent bonds. These small molecules that are joined into chains are called monomers. 3 Polymerisation is the chemical process by which the monomers are joined together to form the big molecule known as the polymer. 4 Carbohydrates such as starch and cellulose consist of monomers known as glucose joined together chemically. polymerisation glucose ⎯⎯⎯⎯⎯⎯→ carbohydrate (monomers) A A A A A A ⎯→ –A–A–A–A–A–A– H CH3 H H H CH3H H ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ nC=C–C=C → —C–C=C–C — ⎮ ⎮ ⎮ ⎮ H H H H n polymer Or n A → (–A–)n where A = monomer n = a big number isoprene (monomer) natural rubber (polymer) 4 A polymer is a macromolecule (a very big molecule). Hence, the relative molecular mass of a polymer is large. 5 The properties of a polymer are different from its monomers. 6 Polymers can be divided into 2 types: (a) Naturally occurring polymers Polymers that exist in living things in nature (plants and animals) (b) Synthetic polymer Polymers that are man-made by chemical processes in the laboratories. Synthetic Polymers 1 Synthetic polymers are polymers made in the industry from chemical substances. 2 Through scientific research, scientists are able to copy the structures of natural polymers to produce synthetic polymers. 3 Many of the raw materials for synthetic polymers are obtained from petroleum, after the refining and cracking processes. 4 The types of synthetic polymers include (a) plastics (b) fibres (c) elastomers 5 Plastics (a) Thermoplastic is a polymer which, when subjected to heat, becomes soft so they can be moulded into various shapes. (b) The properties of plastics are: light, strong, inert to chemicals such as acids and alkali and are insulators of electricity and heat. (c) Examples of plastics are polythene (polyethylene), polyvinylchloride (PVC), polypropene (polypropylene), polystyrene, Perspex and Bakelite. 6 Synthetic fibres (a) Synthetic fibres are long chained polymers that withstand stretching. Naturally Occurring Polymers 1 Naturally occurring polymers exist in plants or animals. 2 Examples of naturally occurring polymers are (a) protein : in muscles, skin, silk, hair, wool and fur. (b) carbohydrates : in starch and cellulose. (c) natural rubber : in latex. Carbohydrates such as starch and cellulose are polymers Manufactured Substances in Industry (polymer) 5 Natural rubber found in latex consists of monomers known as isoprene (2-methylbuta1,3-diene) joined together chemically. polymerisation monomers (polymer) 278 (b) Condensation polymerisation occurs when the monomers with two functional groups combine to form the polymer through a condensation reaction. Examples of condensation polymers are nylon and Terylene. In general: nA monomer monomer A B H ⎮ n C = ⎮ CH3 H H ⎮ ⎮ C ⎯→ — C — ⎮ ⎮ H CH3 propene, C3H6 ’05 The diagram below shows a polymerisation process. H H H H ⎮ ⎮ n C = C ⎯→~C — C~ ⎮ ⎮ HCl H Cl n Substance X Substance Y Which of the properties is identical for substance X and Y? A Density C Relative molecular mass B Boiling point D Empirical formula Comments Substance X is a monomer while substance Y is a polymer. A polymer has a higher density, melting point and boiling point than its monomer. The relative molecular mass of a polymer is more than that of a monomer but the percentage composition and empirical formula are the same. Answer D H H H H ⎮ ⎮ ⎮ ⎮ n C = C ⎯→ — C — C — ⎮ ⎮ ⎮ ⎮ H H H H n polythene condensation polymer 2 There are two types of polymerisation. (a) Addition polymerisation occurs when the monomers with double bonds combine to form the polymer through an addition reaction. Examples of addition polymers are polythene, polypropene, PVC, polystyrene and Perspex. Examples: ethene → —( A – B — ) n + nH2O + nB 1 Nylon is a type of polyamide polymer, a polymer with the amide (–CONH–) group. 2 Terylene is a type of polyester polymer, a polymer with the ester (–COO–) group. 3 Neoprene is a type of synthetic rubber made from the monomer, chloroprene. 4 Styrene-butadiene rubber (SBR) is a type of synthetic rubber made from 2 types of monomers, styrene and butadiene. n = a big number H ⎮ C — ⎮ H n polypropene H H H H ⎮ ⎮ ⎮ ⎮ n C = C ⎯→ — C — C — ⎮ ⎮ ⎮ ⎮ Cl H Cl H n chloroethene, C2H3Cl The making of Terylene fibre polyvinylchloride (PVC) 279 SBR is used in making tyres Manufactured Substances in Industry 9 (b) Examples of synthetic fibres are nylon and Terylene. (c) Nylon is used to make ropes, fishing lines, stocking, clothing and parachutes. (d) Terylene is used to make clothing, sleeping bags and fishing nets. Clothes made from Terylene do not crease easily. 7 Elastomer (a) An elastomer is a polymer that can regain its original shape after being stretched or pressed. (b) Both natural rubber and synthetic rubber are examples of elastomers. (c) Examples of synthetic rubbers are neoprene and styrene-butadiene rubber (SBR). (d) SBR is used to make car tyres. 8 There are two types of polymerisation processes: (a) Addition polymerisation (b) Condensation polymerisation 9 Plastics such as polythene and PVC are produced by addition polymerisation, whereas synthetic fibres such as nylon and Terylene are made by condensation polymerisation. Table 9.1 Some examples of synthetic polymers, their monomers and uses 9 Synthetic Polymer Uses Monomer 1 Polyethylene (PE) IUPAC name: polythene Ethene, C2H4 Plastic bags, shopping bags, plastic containers, plastic toys, plastic cups and plates 2 Polypropylene (PP) IUPAC name: polypropene Propene, C3H6 Plastic bottles, bottle crates, plastic tables and chairs, car battery cases and ropes 3 Polyvinylchloride (PVC) IUPAC name: polychloroethene Chloroethene, C2H3Cl Water pipe, shoes, bags, rain clothes, artificial leather and wire casing 4 Polystyrene (PS) IUPAC name: polyphenylethene Phenylethene, C6H5CH = CH2 Packaging materials, heat insulators, toys, disposable cups and plates 5 Perspex (PP) IUPAC name: poly(methyl-2methylpropenoate) Methyl-2-methyl propenoate (methylmetacrylate) CH2 = C(CH3)CO2CH3 Safety glass, airplane windows, car lamps, traffic signs, lens, reflectors and toys Tetrafluoroethene, C2F4 6 Teflon (PTFE) IUPAC name: polytetrafluoroethene Coatings for non-stick frying pans and electrical insulators 7 Terylene Hexane-1, 6-diol and benzene-1, 4-dicarboxylic acid Clothing, sleeping bags, sails, ropes and fishing net. Clothes made from Terylene do not crease easily 8 Nylon Hexane-1, 6-diamine and hexane-1, 6-dioic acid Ropes, fishing lines, stocking, clothing, carpets and parachutes Issues of the Use of Synthetic Polymers in Every Day Life 2 1 Synthetic polymers have been used widely to replace natural materials such as metals, wood, cotton, animal skin and natural rubber because of the following advantages: (a) Strong and light (b) Cheap (c) Able to resist corrosion (d) Inert to chemical reactions (e) Easily moulded or shaped and be coloured (f) Can be made to have special properties according to specific needs 2 The use of synthetic polymers, however, results in environmental pollution problems. 3 4 5 Pollution Problem Caused by Synthetic Polymers 6 1 Most polymers are non-biodegradable, that is, they cannot be decomposed by bacteria or Manufactured Substances in Industry 280 other micro-organisms. This will cause disposal problems as the polymers will not decay like other organic garbage. Discarded plastic items may cause blockage of drainage systems and rivers thus causing flash floods. Plastic bottles and containers that are not buried in the ground will become breeding grounds for mosquitoes which will cause diseases such as dengue. Small plastic items that are thrown into the rivers, lakes and seas are sometimes swallowed by aquatic animals. These animals may die from choking. The open burning of polymers may release harmful and poisonous gases that will cause air pollution. For example, the burning of PVC will release hydrogen chloride gas which contributes to the acid rain problem. The burning of some polymers will release toxic gas such as hydrogen cyanide. The main source of raw materials for the making of synthetic polymers is petroleum. Petroleum is a non-renewable resource. Glass and Ceramics 1 The raw materials for making glass and ceramics are obtained from the Earth’s crust. 2 The main component of both glass and ceramics is silica or silicon dioxide, SiO2. 3 Both glass and ceramics are used widely in our daily life to replace metals because of the advantages above as well as their low cost of production. 4 Both glass and ceramics have the same properties as follows: (a) Hard but brittle (b) Inert to chemical reactions (c) Insulators of electricity (d) Poor conductors of heat and electricity (e) Withstand compression but not tension (stretching) (f) Can be easily cleaned 5 The use of glass and ceramics also depends on their differences. Table 9.2 shows the differences between glass and ceramic. 1 Reduce, reuse and recycle synthetic polymers (a) Reduce the use of non-biodegradable polymers. (b) Polymers are collected and reused or reprocessed to make new items. The biggest problem is the collection and separation. Not only must the plastics be separated from other types of solid waste but the different types of polymers must be separated from each other. 2 Develop biodegradable polymers These polymers can be decomposed by bacteria, other microorganisms or simply by sunlight (photodegradable). One type of biodegra dable polymer was developed by incorpora ting starch molecules into the plastic materials so that they can be decomposed by bacteria. However, biodegradable polymers are usually more expensive. Table 9.2 The differences in properties between glass and ceramic Glass 9.4 1 (a) What is a polymer? (b) Name two natural polymers that are used to make clothing. (c) Name two synthetic polymers that are used to make clothing. Ceramic Transparent Opaque Softens when heated High melting point, hence retains shape on heating Impermeable Usually porous except when glazed 2 Fill in the blanks below. Monomer Polymer In silicon dioxide, every silicon atom is bonded covalently to 4 oxygen atoms in a tetrahedral shape. Every oxygen atom is also bonded to two silicon atoms to form a giant covalent molecule (Figure 9.15). Ethene Chloroethene Polypropene Polystyrene 3 State two properties of synthetic polymers that will cause environmental pollution in the disposal of syn the tic polymers. State two methods to overcome these problems. Figure 9.15 Structure of silicon(IV) oxide 281 Manufactured Substances in Industry 9 9.5 Methods to overcome environmental problems of polymers Table 9.3 The uses of glass Property of glass Uses Examples Inert Household materials Lamp, bottles, glasses, plates, bowls and kitchen wares Transparent Building materials Mirrors and window glass Industrial materials Bulbs, glass tubes for radios, radars and televisions Scientific apparatus Lens, burettes, beakers, test tubes, conical flasks, glass tubes and prisms Inert and easily cleaned 9 Table 9.4 Uses of ceramics Property of ceramics Glass is transparent SPM ’05/P1 Uses Examples Hard and strong Building materials Long lasting and noncorrosive Materials for decorative Plates, bowls, cooking utensils, items porcelain and vases Electrical insulators To make electrical insulating parts Insulators in toasters and irons, spark plugs in car engines Inert and hard In surgical and dental apparatus Artificial hands, legs and teeth Semiconductor type of ceramics As microchips To make microchips in computers, radios and televisions Bricks, tiles and cement Ceramic is opaque and has higher melting point than glass the addition of chromium ions will give the glass a green hue, cobalt ions will give a blue hue while manganate ions will give a purple hue to the glass. Types, Properties, Composition and Uses of Glass 1 Fused glass is the simplest type of glass, which consists mainly of silica or silicon dioxide. Occasionally a little boron oxide is added. 2 Other types of glass are mainly metal silicates. 3 Various types of glass can be produced by changing the composition of glass. Different types of glass have different properties and they are used for various specific purposes. The chemical composition, specific properties and uses of four types of glass are summarised in Table 9.5. 4 Coloured glass is produced by adding traces of transition metal oxides to it. For example, Composition of Ceramics 1 Ceramic is a manufactured substance made from clay that is dried and then baked in a kiln at high temperature. 2 The main constituent of clay is aluminate, silica and feldspar. 3 Kaolinite is an example of high quality white clay that consists of hydrated aluminosilicate crystals. 4 Red clay contains iron(III) oxide which gives its red colour. Table 9.5 Properties, composition and uses of different types of glass Name of glass Properties Chemical composition Fused glass • Very high softening point (1700 °C), hence highly heat-resistant • Transparent to ultraviolet and infrared light • Difficult to be made into different shapes • Does not crack when temperature changes (very low thermal expansion coefficient) • Very resistant to chemical reactions SiO2 (99%) B2O3 (1%) Manufactured Substances in Industry 282 Examples of uses Telescope mirrors, lenses, optical fibres and laboratory glass wares Properties Chemical composition • Low softening point (700 °C), hence does not withstand heating • Breaks easily • Cracks easily with sudden temperature changes (high thermal coefficient of expansion) • Less resistant to chemical reactions • Easy to make into different shapes • Quite high softening point (800 °C), hence it is heat-resistant • Does not crack easily with sudden change in temperature • Transparent to ultraviolet light • More resistant to chemical reactions • Does not break easily SiO2 (70%) Na2O (15%) CaO (10%) Others (5%) Bottles, windowpanes, light bulbs, mirrors, flat glass, glass-plates and bowls. (The most widely used type of glass) SiO2 (80%) B2O3 (15%) Na2O (3%) Al2O3 (1%) Laboratory apparatus, cooking utensils, electrical-tubes and glass pipelines • • • • SiO2 (55%) PbO (30%) K2O (10%) Na2O (3%) Al2O3 (2%) Decorative items, crystal glasswares, lens, prisms and chandeliers Soda lime glass Borosilicate glass SPM ’09/P1, ’10/P1 Lead glass Low softening point (600 °C) High density High refractive index Reflects light rays and appears sparkling Examples of uses The Uses of Improved Glass and Ceramics for Specific Purposes Improved glass Examples of Improved Glass and Ceramics Photochromic glass • Photochromic glass is a type of glass that is sensitive to light intensity. The glass darkens when exposed to sunlight but becomes clear when light intensity decreases. • Photochromic glass is produced when a dispersion of silver chloride, AgCl or silver bromide, AgBr is added to normal glass. Conducting glass • Conducting glass is a type of glass that can conduct electricity. • Conducting glass is produced by embedding a thin layer of conducting material in glass. • A type of conducting glass is produced by adding a layer of indium tin(IV) oxide (ITO) that acts as an electrical conductor. This type of glass is used in the making of LCD (liquid crystal display) panel. • Another type of conducting glass is made by embedding thin gold threads in glass. Water condenses as ice on the window panes of aircraft at high altitudes and this obstructs the vision of the pilot. Hence, windows of aircraft are heated by passing electric current through the gold threads embedded in the glass. 283 Improved Ceramics Superconductor • Superconductors are a class of ceramics that conducts electricity without resistance and without loss of electrical energy. • Superconductor ceramics are used to make light magnets, electrical generators and electric motors. Ceramic car engine block • Clay heated with magnesium oxide produces a type of ceramic that has a high thermal resistance. • This type of ceramic is used for making car engine blocks because it can resist high temperatures. • At a higher temperature, the combustion of fuel becomes more efficient and produces more energy with less pollution. • Ceramic engines offer great advantages in terms of fuel economy, efficiency, weight savings and performance. Manufactured Substances in Industry 9 Name of glass 9 Main component is silica, SiO2 Glass Ceramic Types of glass • Fused glass (high heat resistance) • Soda lime glass (cannot withstand high temperatures) • Borosilicate glass (can withstand high temperatures) • Lead glass (high refractive index) Examples of ceramics • Tiles • Cement • Bricks • Porcelain Common properties • Hard but brittle • Inert to chemicals • Heat and electrical resistance • Resist compression • Can be easily cleaned 3 In the making of composites, substances (known as components) are combined to form new types of materials that can overcome the limitations of the original materials. 4 Most of the composite materials are comprised of two phases: a continuous phase (also known as the base) and the dispersed phase (also known as the matrix). 5 Composite materials are harder, stronger, lighter (lower density), more resistant to heat and corrosion and also made for specific purposes. 6 A few types of composite materials and their components are shown in Table 9.6. 9.5 1 Glass is a manufactured substance in industry. (a) What is the major component of glass? (b) State a cheap source of this component. (c) State four types of glass. 2 Glass and ceramics are both manufactured from materials in the Earth's crust. (a) What is the common component found in both glass and ceramics? (b) State the three similarities and three differences between glass and ceramics. (c) State five uses of glass. (d) State two examples of the use of ceramics in the building industry. Table 9.6 3 State one main difference between soda lime glass and borosilicate glass in terms of (a) composition and (b) property. 9.6 Differences • Glass is transparent, ceramic is opaque • Ceramic can withstand a higher temperature than normal glass Composite materials Reinforced concrete Concrete (cement, sand and small pebbles) and steel Superconductor Yttrium oxide (Y2O3), barium carbonate (BaCO3) and copper(II) oxide (CuO) Fibre optic Glass {silica (SiO2), sodium carbonate (Na2CO3) and calcium oxide (CaO)} with different refractive indices Fibreglass Glass fibre and polyester (a type of plastic) Photochromic glass Glass and silver chloride or silver bromide Composite Materials The Meaning of Composite Materials 1 A composite material is a structural material formed by combining two or more materials with different physical properties, producing a complex mixture. 2 The composite material produced will have different properties far more superior to the original materials. Manufactured Substances in Industry Components 284 (c) Glass and ceramics are brittle. (d) Plastic and glass cannot temperatures. 1 Most types of material used in our daily life have certain limitations. For example: (a) Metals can be easily corroded and are malleable and ductile. (b) Metals are good electrical conductors but the existence of resistance results in the loss of a big amount of electrical energy as heat. resist high 2 With knowledge of the compositions, structures and properties of these materials, chemists are able to develop new materials to suit specific purposes. Reinforced concrete 1 Concrete is a composite material made from a mixture of sand and small stones bound by cement. Concrete is strong in compression but brittle and weak in tension. Concrete cannot withstand vibrations and will crack under the action of bending forces. 2 Steel is strong in tension (high tensile strength). But using thick steel columns to support a heavy load is expensive. Furthermore, steel corrodes easily. 3 Reinforced concrete is made by adding the concrete mixture of cement, water, sand, chips and small stones into a frame of steel bars or steel wire netting (Figure 9.16). When set, a composite material is formed. 4 Reinforced concrete is a stronger building material as it combines the compressive strength of concrete and tensile strength of steel. In addition, it does not corrode easily. Reinforced concrete is also relatively cheap and can be moulded into any shape. 5 Steel and concrete have about the same coefficient of expansion. Hence they are good composite components and do not crack when mixed. 6 Reinforced concrete can withstand very high applied forces (high pressure) and SPM ’07/P1, ’08/P1 can support very heavy loads. It is used to construct framework for highways, brid ges, oil platforms and high-rise buildings. Figure 9.16 The formation of reinforced concrete Dams are constructed by reinforced concrete which is very strong Superconductors 1 In normal electrical conductors such as copper metal, the existence of resistance causes the loss of electrical energy as heat. Furthermore, resistance increases as temperature increases. 2 Superconductors can conduct electricity with zero resistance when they are cooled to extremely low temperatures. Thus, superconductors conduct electricity without any loss of energy. (known as the transition temperature). This low temperature can only be achieved using liquid helium which is expensive. 4 When a mixture of copper(II) oxide (CuO), barium oxide (BaO) and yttrium oxide (Y2O3) is heated up, a type of ceramic with the formula YBa2Cu3O7 is produced. This type of ceramic, known as perovskite or YBCO, can attain superconductivity at 90 K (–183°C). This temperature can easily be attained by using the cheaper liquid nitrogen. 3 Metals such as copper, can only achieve superconductivity at a very low temperature 285 Manufactured Substances in Industry 9 Comparison of the Properties of Composite Materials and Their Original Components 5 The metal oxides (CuO, Y2O3 and BaO) are all electrical insulators. However when they are combined to form a composite, the composite is a superconductor that can conduct very high current over long distance without any loss of energy. 6 Superconductors are used to make more efficient generators, magnetic energy-storage systems, transformers, electric cables, amplifiers and computer parts. They are also used in magnetic resonance imaging (MRI) – a type of medical imaging device. Superconductors are also used to make stronger, lighter and more powerful electromagnets. High – speed levitated trains (trains that float on the railway track) involve the use of electromagnets and superconductors. 9 Fibre optics (also known as optical fibres) 1 Optical fibres are bundles of glass tubes with very small diameters. They are finer than human hair and are very flexible. 2 Fibre optics is a composite material that can transmit electronic data or signals, voice and images in a digital format, in the form of light along the fine glass tubes at great speeds. 3 Fibre optics consists of a core of glass of higher refractive index enclosed by a glass cladding of lower refractive index. A light wave entering the fibre will travel along the glass tube due to total internal reflection (Figure 9.17). 4 In the field of telecommunications, fibre optics is used to replace copper wire in long distance telephone lines, mobile phones, video cameras and to link computers within local area networks (LAN). Fibre optics uses light instead of electrons to carry data. Fibre optics carry more data (higher transmission capacity) with less interference, has a higher chemical stability and a lower material cost compared to metal communication cables such as copper. Fibre optics can also send SPM ’11/P1 signals faster than metal cables and occupies less space. 5 In the field of medicine, a laser beam can be channelled through fibre optics in operations to remove unwanted Fibre optics tissues. Fibre optics is also used in endoscopes: instruments that are inserted into the body through the nose, mouth or ear, for doctors to examine the internal organs. 6 Fibre optics is also used in instruments to inspect the interior of manufactured products. Figure 9.17 Cross section of a fibre optic Fibreglass 1 Plastic is light (with a low density), elastic, flexible, but is brittle, not very strong and is inflammable (can catch fire). 2 Glass is hard and strong but is brittle, heavy (with a relatively high density) and has a low compressive strength. 3 When glass fibre filaments are embedded in polyester resin (a type of plastic), fibreglass which is light, strong, tough, resilient, inflammable, flexible with a high tensile strength is produced. It can also be easily coloured, moulded and shaped. A resilient material is one that returns to its original shape after bending, twisting, stretching and compression. Manufactured Substances in Industry Boats built from fibreglass is light and strong. 4 Fibreglass is an ideal material for making water storage tanks, boat hulls, swimming pool linings, food containers, fishing rods, car bodies, roofing, furniture and pipes. 286 Photochromic glass 7 Silver atoms and bromine gas recombine according to the following reaction 1 Glass is transparent and is not sensitive to light intensity. 2 Silver chloride or silver bromide is sensitive to light. When exposed to light, these compounds decompose to form dark silver particles. 3 In photochromic glass, silver chloride (AgCl) or silver bromide (AgBr) and a little copper(I) chloride is embedded into the structure of glass. 4 When exposed to ultraviolet light, the AgCl or AgBr decomposes to form silver and halogen atoms. The fine silver which is deposited in the glass is black and the glass is darkened. For example: Br2 + 2Cu+ → 2Br– + 2Cu2+ … (1) Cu2+ + Ag → Cu+ + Ag+ … (2) 8 The overall reaction is 9 Photochromic glass is used to make camera lens, car windshields, information display panels, light intensity meters and optical switches. uv light 2AgBr ⎯⎯⎯⎯→ 2Ag + Br2 5 Photochromic glass has the ability to change colour and become darker when exposed to ultraviolet light. 6 The photochromic glass will automatically become clear again when the light intensity is lowered, whereby silver is converted back to silver halides. Photochromic glass is used to make lenses that change from light to dark, eliminating the neccessity for a separate pair of sunglasses. Properties of composite materials compared to their components and the uses of composites Composite material Reinforced concrete Super conductor Component Properties of component Properties of composite Concrete Hard but brittle, with low tensile strength Steel Hard with high tensile strength but expensive and can corrode Copper(II) oxide, yttrium oxide and barium oxide Insulators of electricity 287 Uses of composites Stronger, higher tensile strength, not so brittle, does not corrode easily, can withstand higher applied forces and loads, relatively cheaper Construction of framework for highways, bridges and high-rise buildings Conducts electricity without resistance when cooled by liquid nitrogen To make more efficient generators, transformers, electric cable, amplifiers, computer parts, stronger and lighter electromagnets Manufactured Substances in Industry 9 2Ag + Br2 → 2AgBr Composite material Fibre optics Properties of component Properties of composite Transparent, does not reflect light rays Reflect light rays and allow light rays to travel along the fibre Transmit data in the form of light in telecommunications Glass Heavy, strong but brittle and nonflexible Polyester plastic Light, flexible, elastic but weak and inflammable Light, strong, tough, resilient and flexible, with high tensile strength, not inflammable Water and food storage containers, boats, swimming pool linings, fishing rods, car bodies and roofing Glass Transparent and not sensitive to light Silver chloride or silver bromide Sensitive to light Sensitive to light: darkens when light intensity is high, becomes clear when light intensity is low Photochromic optical lens, camera lens, car windshields, optical switches, information display panels and light intensity meters Component Glass of low refractive index Glass of higher refractive index 9 Fibreglass Photo chromic glass 2 New materials are required to overcome new challenges and problems we face in the changing world. 3 Synthetic materials are developed constantly due to the limitation and shortage of natural materials. 4 New needs and new problems will stimulate the development of new synthetic materials. For example, the use of new plastic compo site material will replace metal in the making of a stronger and lighter car body. This will save fuel and improve speed. Plastic compo site materials may one day be used to make organs for organ transplants in human bodies. New superconductors made from composite materials are developed. 5 The understanding of the interaction between different chemicals is important for both the development of new synthetic materials and the disposal of such synthetic materials as waste. 6 A responsible and systemic method of handling the waste of synthetic materials and their by– products is important to prevent environmental pollution. The recycling and development of environ mentally friendly synthetic material should be enforced. 9.6 1 (a) State what is meant by the term composite materials. (b) Give five examples of composite materials and name one use for each example. 2 (a) What is fibreglass? Explain how the properties of fibreglass are superior to that of its original components. (b) What is reinforced concrete? Give two reasons why reinforced concrete is a strong construction material. 3 What is the advantage of using photochromic glass in the making of spectacles? Briefly explain how this glass works. 9.7 Appreciating Various Synthetic Industrial Materials 1 Continuous research and development (R & D) is required to produce better materials used to improve our standard of living. Manufactured Substances in Industry Uses of composites 288 9 Multiple-choice Questions 9.1 Sulphuric Acid 1 The uses of substance X is given below. • To make fertilisers • To manufacture detergents • To make paints What of the following substances could be X? A Nitric acid B Sulphuric acid C Ammonia D Ammonium sulphate 2 Which of the following is a use of sulphuric acid? A As an electrolyte in dry cells B As a raw material in making explosives C As an electrolyte in the electroplating of metals D To clean the metal oxide layer of metals before electroplating 289 3 Which of the following is true about the manufacture of sulphuric acid by the Contact process? A Sulphur and vanadium(V) oxide are the raw materials. B Sulphur is converted into sulphur dioxide and then into sulphur trioxide. C Sulphur trioxide dissolves in water to form sulphuric acid. D A pressure of 200 atmosphere is used in the process. Manufactured Substances in Industry 9 9 The three aims of alloying are: (a) To increase the hardness and strength of a metal (b) To prevent corrosion or rusting (c) To improve the appearance of the metal surfaces 10 Polymers are large molecules made up of many smaller and identical repeating units (monomers) joined together by covalent bonds. 11 Some examples of synthetic polymers are polythene, polypropene, P.V.C., polystyrene, perspex, Terylene and nylon. 12 Synthetic polymers are strong and light, cheap, resist corrosion and inert to chemical attacks. However, they are nonbiodegradable and cause environmental pollution problems. 13 The main component of glass is silica or silicon dioxide, SiO2. The main constituents of ceramics are clay (aluminosilicate), sand (silica) and feldspar. 14 Both glass and ceramics have the following properties: (a) Hard but brittle (b) Inert toward chemicals (c) Insulators or bad conductors of heat and electricity 15 Some examples of glass are fused glass, soda lime glass, borosilicate glass and lead glass. 16 A composite material is a structural material formed by combining two or more materials with different physical properties to produce a complex mixture. 17 Some examples of composites are reinforced concrete, superconductors, fibre optic, fibreglass and photochromic glass. 1 Sulphuric acid is used to make other manufactured substances such as fertilisers, detergents, pesticides, polymers and paint pigments. 2 Sulphuric acid is manufactured by the Contact process using vanadium(V) oxide as a catalyst. The process involves three stages. I II Sulphur ⎯→ Sulphur dioxide ⎯→ Sulphur trioxide III ⎯→ Sulphuric acid 3 Sulphur dioxide gas can cause environmental pollution such as acid rain. 4 Ammonia is used to make nitrogenous fertilisers and nitric acid, and is used as a cooling agent in refrigerators. 5 Ammonia is produced in the industry by the Haber process with hydrogen gas and nitrogen gas and using iron powder as a catalyst. 6 An alloy is a mixture of two or more elements with a certain fixed composition in which the major component is a metal. 7 A pure metal is weak and soft because it contains atoms of the same size in an orderly arrangement. This enables the layers of atoms to slide over each other easily. 8 In an alloy, foreign atoms of different sizes disrupt the orderly arrangement of the metal atoms. This prevents the layers of metal atoms from sliding over each other easily. 9 4 The manufacturing of sulphuric acid in the Contact process ’06 involves several reactions. Which of the following is the reaction that requires a temperature of 450 – 550°C. A S + O2 → SO2 B 2SO2 + O2 2SO3 C SO3 + H2SO4 → H2S2O7 D H2S2O7 + H2O → 2H2SO4 5 In the Contact process, oleum is produced when A sulphur dioxide reacts with oxygen. B sulphur dioxide dissolves in water. C sulphur trioxide dissolves in concentrated sulphuric acid. D sulphur trioxide dissolves in water. 6 Which of the following gas dissolves in rainwater and consequently kills trees and corrodes concrete buildings? A Ammonia B Sulphur dioxide C Carbon monoxide D Carbon dioxide 9 Which of the following are true of ammonia gas? I It is a colourless and odourless gas. II It is lighter than air. III It is sparingly soluble in water. IV It produces white fumes with hydrogen chloride gas. A I and IV only B II and III only C II and IV only D III and IV only 10 Which of the following is a source of hydrogen for the Haber process? A The decomposition of water B Fractional distillation of liquid air C Reaction of coke or natural gas with steam D Reaction of zinc metal with sulphuric acid ’08 7 Which of the following are caused by the presence of sulphur dioxide in the atmosphere? I Certain lung diseases and bronchitis II An increase of acidity in blood III An increase of acidity in rain water IV Corrodes concrete buildings A I and II only B III and IV only C I, II and III only D I, III and IV only 9.2 Ammonia and its Salts 8 Which of the following chemicals is manufactured using ammonia in the industry? A Sodium hydroxide B Hydrochloric acid C Sulphuric acid D Nitric acid Manufactured Substances in Industry 11 Which of the following is the effect of using iron powder as a catalyst in the production of ammonia in the Haber process? A The quantity of ammonia produced is increased. B The temperature required for Haber process is reduced. C The pressure required for Haber process is reduced. D The rate of reaction between hydrogen gas and nitrogen gas is increased. 12 Which of the following compounds reacts with ammonia to produce urea, a type of fertiliser? A Carbon dioxide B Sulphuric acid C Phosphoric acid D Ethanoic acid 13 Which of the following is the function of iron powder in the Haber process? A To speed up the rate of production of ammonia B To increase the percentage yield of ammonia C To lower the cost of production of ammonia 290 D To lower the pressure required in the production of ammonia 14 Which of the following are the conditions in the production of ammonia by the Haber process? I A temperature of about 450 °C II A pressure of one atmosphere III Equal volumes of nitrogen gas and hydrogen gas IV The use of iron powder as a catalyst A I and II only B I and IV only C II and III only D I, II and IV only 15 Ammonium nitrate is used as a fertiliser. What is the ’10 percentage by mass of nitrogen in ammonium nitrate? [Relative atomic mass: H, 1; N, 14; O, 16] A 17.5% B 17.7% C 28.6% D 35.0% 9.3 Alloys 16 The alloying process increases the hardness of a metal. This is because the foreign atoms added to a metal in the alloying process A increases the orderliness of the arrangement of the metal atoms. B strengthens the bond between the metal atoms. C forms strong bonds between the metal atoms and the foreign atoms. D makes it difficult for the layers of metal atoms to slide over each other. 17 Which of the following is not the aim of alloying iron to form steel? A To make it harder B To make it stronger C To make it more resistant to rusting D To increase the melting point ’11 Alloy Uses A Duralumin Building of monuments B Brass Bodies of aeroplanes C Bronze Frameworks of buildings D Stainless steel Making of surgical instruments 19 The alloy produced by the addition of tin to copper metal is known as A bronze B brass C pewter D duralumin 20 Which of the following alloys is suitable for the making of an aircraft body? A Duralumin B Bronze C Brass D Cupro-nickel 21 Iron is alloyed to produce steel. Which of the following property is not true of steel compared to iron? A Harder B More resistant to rusting C More presentable D A better electrical conductor 9.4 Synthetic Polymers 22 The polymer formed from the polymerisation of phenylethene is known as A polythene B polypropene C polystyrene D polyvinyl chloride 23 Which of the following are the correct pairs of polymer and monomer? Polymer Monomer I Natural rubber Isoprene II Carbohydrate Sucrose III Polypropene Methylmethacrylate IV PVC Chloroethene A B C D II and III only I and IV only I, II and III only I, II, and IV only 24 The diagram shows the repeating unit of a synthetic polymer. H ⎮ — C – ⎮ CH3 H ⎮ C — ⎮ CH3 Which of the following are true of this synthetic polymer? I It is a type of addition polymer. II It dissolves easily in hot water. III It burns in air to produce carbon dioxide and water. IV It has a high relative molecular mass. A I and II only B III and IV only C II and III only D I, III and IV only 25 Which of the following are true about Terylene, a type of synthetic polymer? I It is easily biodegradable. II It burns easily. III It is a type of fibre. IV Its monomer is ethene. A I and IV only B II and III only C II, III and IV only D I, II and III only 26 Uncontrolled disposal of synthetic polymers will cause environmental pollution. Which of the following are the characteristics of synthetic polymers that causes this environmental pollution? I Polymers are nonbiodegradable. II Polymers increase the pH of the water when dissolved in water. 291 III Polymers promote excessive growth of algae in water. IV Polymers release toxic gases when burned. A I and IV only B II and III only C I, III and IV only D II, III and IV only 9.5 Glass and Ceramics 27 A transparent solid is formed when molten sand at high temperature is cooled quickly. What is this solid? A Ceramic B Fused glass C Soda lime glass D Borosilicate glass 28 Which of the following are true for both glass and ceramic? I They contain a common component, silica. II They are electrical insulators. III They are resistant to chemicals. IV They can resist compression. A I and II only B III and IV only C I, II and III only D I, II, III and IV 29 Material Y has the following properties: ’09 • Resistance towards chemical substances • Low coefficient of thermal expansion What is material Y ? A Bronze B Polystyrene C Borosilicate glass D Conducting glass 30 Which of the following glass has a low softening point and can be easily moulded into different shapes? A Fused quartz glass B Soda lime glass C Borosilicate glass D Photochromic glass 31 Lead glass is very suitable for making fine glassware and art objects because this type of glass Manufactured Substances in Industry 9 18 Which alloy is correctly matched to its uses? 9 A has a high refractive index. B has a high softening point. C is most transparent to ultraviolet and infrared rays. D is a good heat insulator. B Photochromic glass C Lead glass D Borosilicate glass 9.6 Composite Materials 32 What is the purpose of adding feldspar to kaolin in the making ’05 of porcelain? A To make kaolin softer B To make kaolin smoother C To make kaolin harder D To make porcelain that is inert towards chemicals 35 Which composite material is made of glass of different ’08 refractive index? A Fibreglass B Fibre optics C Photochromic glass D Borosilicate glass 33 The raw materials for making soda lime glass are I sodium carbonate II calcium carbonate III silicon dioxide IV boron oxide A I and II only B II and III only C I, II and III only D I, II, III and IV 36 The supporting pillars of flyovers of highways are made of ’06 substance X. Substance X has the following properties : strong, not brittle, can withstand erosion and can withstand weathering. Which of the following is substance X? A Concrete B Steel C Marble D Reinforced concrete 34 A decorative glassware manufacturer wants to make a display panel that is sensitive to the intensity of light. What is the most suitable material? A Conducting glass 37 The diagram shows the formation of a composite ’07 substance from its original components. Based on the diagram, why is reinforced concrete often used more to build buildings compared to concrete? A The steel bars cannot stretch and make it tough. B The concrete and the steel bars can slide over each other and make it flexible. C The steel bars fix the position of the concrete particles and make it hard. D The concrete particles are evenly dispersed among the steel bars and make it able to withstand vibrations. Structured Questions (c) (i) Name liquid Z. [1 mark] (ii) How is liquid Z formed from gas Y? 1 Diagram 1 shows a series of steps involved in the production of sulphuric acid in industry starting from sulphur. Sulphur Sulphuric acid I Oxygen IV Gas X Liquid Z [1 mark] (d) Sulphuric acid is also formed when gas Y dissolves in water. Why is this not done in the manufacturing of sulphuric acid in industry? [1 mark] II Oxygen III (e) State two uses of sulphuric acid. Gas Y [2 marks] 2 Diagram 2 shows how a type of fertiliser is produced. Diagram 1 ’07 (a) (i) Name gas X. [1 mark] (ii) Write a balanced equation for the reaction in step I. [1 mark] Process P Process Q Sulphuric acid Ammonia Fertilliser X Diagram 2 (b) (i) Name gas Y. [1 mark] (ii) State the conditions used in step II in order to produce a high percentage yield of gas Y. (a) Process P and process Q are industrial processes. What are the names of each of these processes? [2 marks] [2 marks] Manufactured Substances in Industry 292 [1 mark] (ii) State a natural source of silica. (iii) W is a an important component of [1 mark] borosilicate glass. What is W? (b) (i) Name compound X. [1 mark] (ii) Compound X can be used to make plastic chairs and tables. State one advantage of this type of material as compared to metals. [3 marks] (c) (i) Name fertiliser X. (ii) Write a balanced equation for the formation of fertiliser X. [2 marks] (d) Calculate the mass of ammonia that is required to react with 0.2 mol of sulphuric acid. [Relative atom mass: H, 1; N, 14] [2 marks] [1 mark] (e) Name another type of fertiliser that is produced when ammonia reacts with carbon dioxide. (c) (i) Identify component Y. [1 mark] (ii) Explain why magnalium is harder than pure [2 marks] aluminium. [1 mark] 3 The flowchart in Diagram 3 shows the conversion of nitrogen into various substances in steps I, II, III and IV. Nitrogen I II Ammonia + Substance X III (d) Z can withstand high pressures and can very heavy loads. What is Z? (e) (i) Identify the type of compound T. (ii) State a property of compound T. Nitric acid IV 5 Bronze, brass and duralumin are examples of alloys. + Substance Y (a) What is meant by alloy? Diagram 3 (c) What type of particles are present in pure [1 mark] copper? (a) Step I is an industrial process. State the conditions used for a good yield of ammonia in this process. (d) Draw a diagram to show the arrangement of particles in (i) pure copper (ii) bronze [2 marks] [3 marks] (b) Step II is also an industrial process. (i) Name this process. [1 mark] (ii) State the catalyst used in this process. (e) (i) Name the elements that are used to make [1 mark] the alloy duralumin. (ii) What is the advantage of duralumin compared [1 mark] to its main component? (iii) State one use of the alloy duralumin. [1 mark] (c) State one physical property and one chemical property of ammonia. Physical property: Chemical property: [2 marks] [1 mark] (d) Name a possible compound for substance X in step III and hence write a balanced equation that occurs in step III. [2 marks] (e) (i) Besides ammonia, name a compound that may be substance Y in step IV. [1 mark] (ii) Write a balanced equation for the reaction in step IV. [1 mark] 6 Polyethene and polyvinyl chloride are examples of synthetic polymers that are widely used in daily life. (a) Name the process in which monomers are [1 mark] joined together to form polymers. (b) (i) Name the monomer that is used for the [1 mark] making of polyethene. (ii) Give a use of polyethene. [1 mark] 4 Table 1 shows the examples and components of five different types of manufactured substances in industry. Glass Example Components Borosilicate glass Silica, sodium oxide and W X Polymer Alloy (c) The following diagram shows a part of the molecular structure of polyvinyl chloride. H Cl H Cl H Cl ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⎯ C ⎯ C ⎯ C ⎯ C ⎯ C ⎯ C ⎯ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ H H H H H H Propene Magnalium Composite material Z T Photochromic glass Aluminium and Y Concrete (cement, sand and small pebbles) and steel Glass and silver chloride (i) Draw the structure of its monomer. [1 mark] (ii) Polyvinyl chloride is widely used to make water pipes. State two advantages of PVC pipes compared with iron pipes. [2 marks] (iii) State two ways in which polyvinyl chloride can cause environmental pollution. [2 marks] (d) What is the source of the raw material used in producing polyethene and polyvinyl chloride? Table 1 (a) [1 mark] (b) Name the main element added to copper to form [1 mark] (i) brass: (ii) bronze: [1 mark] Nitrate salt Type support [1 mark] [1 mark] [1 mark] (i) Give the chemical name for silica. [1 mark] [1 mark] 293 Manufactured Substances in Industry 9 (b) State the conditions for process Q. Essay Questions 1 (a) Using polyethene as an example, explain the terms polymer and monomer. [4 marks] (b) Name three examples of (i) natural polymers and (ii) synthetic polymers. State a use for each type of polymer. What are composite materials? Use two suitable examples to explain the above statement. [10 marks] (b) Mr Vellu found that an art sculpture made of pure metal is easily dented in his workshop but if he were to use an alloy, the sculpture will not be dented. Using one suitable example, describe an experiment to show how you can compare the hardness of an alloy with that of a pure metal. [6 marks] (c) Explain briefly how sulphuric acid is manufactured in the industry. [10 marks] 9 2 (a) [10 marks] Composite materials are produced for the purpose of improving the original materials and to fulfill specific needs. Experiments (a) Measure the diameters of the two depressions accurately and record in the spaces provided in Diagram 2. [3 marks] 1 Brass is a copper alloy that is used to make souvenirs and decorative items. Diagram 1 shows the experimental set-up used to compare the hardness of pure copper and brass. (b) Construct a table to show all the data in the experiment. [3 marks] (c) State the operational definition for alloy. [3 marks] (d) What is the relationship between the diameter of the depression and the hardness of the materials? [3 marks] (e) Referring to results obtained from the experiment, state the conclusion that can be drawn from the experiment. [3 marks] (f) State three variables that must be kept constant in this experiment. [3 marks] Diagram 1 The 1 kg weight is dropped onto the steel ball bearing and the diameter of the depression formed on the block is measured. The experiment is repeated using a copper block to replace the brass block. 2 Iron nails that are used in the construction of buildings rust more than stainless steel nails when exposed to rain. The cross section of the diameter of depression of the two materials is shown in Diagram 2. Referring to the situation above, plan an experiment to compare the rate of rusting of an iron nail and a stainless steel nail. Your explanation should have the following items: (a) Statement of problem (b) All the variables (c) Statements of hypothesis (d) List of materials and apparatus (e) Procedure (f) Tabulation of data [17 marks] Diagram 2 Manufactured Substances in Industry 294 FORM 5 THEME: Interaction between Chemicals CHAPTER 1 Rate of Reaction SPM Topical Analysis 2008 Year Paper 1 Section Number of questions 2 2009 2 A B C 1 – – 3 1 – 6 2010 2 A B C 1 – – 3 1 1 5 2011 2 A B C – 1 – 3 1 1 4 3 2 A B C 1 – – 1 ONCEPT MAP Rate of reaction • Average speed is the amount of reactant used up or the product formed per unit time. 1 • Rate is proportional to ————————— . time taken Measuring the speed of reaction • from changes in the mass of reactant or product against time. • from changes in the volume of gas produced against time. Applications in daily activities • Combustion of charcoal • Keeping food in a refrigerator • Cooking food in a pressure cooker Concentration An increase in concentration increases the speed of reaction. Particle size A decrease in the particle size (larger total surface area) increases the speed of reaction. Concentration-time graph • The gradient of the graph indicates the rate of reaction. • The rate of reaction decreases as the reaction proceeds. RATE OF REACTION Temperature An increase in temperature increases the speed of reaction. Collision theory • Explains rate of reaction in terms of effective collisions between reactant particles. • For effective collisions, the particles must have energy equal to or greater than the activation energy. • Any factor that increases the rate of effective (successful) collisions will increase the speed of reaction. Pressure An increase in pressure increases the speed of reaction (applies only to gases). Catalyst Catalyst increases the rate of reaction. Uses of catalysts in industry • Iron in the Haber process N2 + 3H2 2NH3 • V2O5 in the Contact process 2SO2 + O2 2SO3 • Pt in the Ostwald process 4NH3 + 5O2 4NO + 6H2O 1.1 Reactants: CaCO3(s) + 2HCl(aq) → Products: CaCl2(aq) + CO2(g) + H2O(l) Rate of Reaction (b) During the reaction, the following observable changes take place. (i) The mass of calcium carbonate (the reactant) decreases. (ii) The concentration of hydrochloric acid (the reactant) decreases. (iii) The volume of carbon dioxide (the product) produced increases. (c) Thus, the rate of reaction between calcium carbonate and hydrochloric acid can be determined by measuring (i) the decrease in mass of calcium carbonate per unit time, or (ii) the increase in volume of carbon dioxide per unit time. That is, The Meaning of Rate of Reaction 1 During a chemical reaction, the reactants are used up as the products are formed. Example: CaCO3 + 2HCl → CaCl2 + H2O + CO2 Thus, the amounts of reactants decrease (Figure 1.1(a)) while the amounts of products increase as the reaction proceeds (Figure 1.1(b)). 1 Mass of CaCO3 reacted Reaction rate = — — — — — — — — — –— — — — — — — — — — — , or Time taken Figure 1.1 The graph of amount of substance (mol) against time (minutes) Volume of CO2 produced Reaction rate = — — — — — — — — — –— — — — — — — — — — — — — Time taken 2 Definition The rate of reaction is defined as the amount of a reactant used up or the amount of a product obtained per unit time. The gas produced during a reaction can be collected by using a burette or a gas syringe. Amount of reactant used up Rate of reaction = — — — — — — — — — — — — — — — — — — — — — — — — Time taken 5 The rate of reaction is inversely proportional to the time taken for the reaction to be completed. or 1 Reaction rate ∝ — — — — — — — — — Time taken Amount of product obtained Rate of reaction = — — — — — — — — — — — — — — — — — — — — — — — — — Time taken The reaction is fast if it takes a short time to complete. Conversely, the reaction is slow if it takes a long time for the reaction to complete. 6 Example of a reaction involving a change in colour (a) The reaction between potassium manganate(VII), KMnO4, and ethanedioic acid, H2C2O4, can be represented by the ionic equation below. 3 Methods of measuring reaction rates SPM The amount of a reactant used up or a product ’07/P1, ’09/P1, obtained can be measured in terms of ’11/P1 (a) changes in the mass or concentration of the reactant or product (b) volume of gas produced (c) changes in colour (d) formation of precipitate (e) changes in mass of the reaction mixture 4 Reaction between calcium carbonate and dilute hydrochloric acid (a) The reaction between calcium carbonate (marble chips) and dilute hydrochloric acid can be represented by the equation: Rate of Reaction 5C2O42–(aq) + 16H+(aq) + 2MnO4–(aq) ethanedioate ion manganate(VII) ion (purple) → 10CO2(g) + 8H2O(l) + 2Mn2+(aq) colourless 296 Table 1.1 Example of some fast reactions (b) Observable changes: When excess ethanedioic acid solution is added to an aqueous solution of potassium manganate(VII), KMnO4, the purple colour of KMnO4 decolourises slowly at room temperature. Type of reaction SPM ’09/P1 Example Neutralisation Reaction between an acid and an alkali. HCl + NaOH → NaCl + H2O 1 Reaction rate ∝ — — — — — — — — — — — — — — — — — — — — — — Time taken for the purple colour to disappear Reaction between silver nitrate Double decomposition solution and sodium chloride solution to form silver chloride precipitate. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) Combustion A concentrated solution of manganese(II) ions, Mn , is pink in colour. However, a very dilute solution of Mn2+ ions appears colourless. 2+ 7 Example of a reaction involving the formation of a precipitate (a) The reaction between sodium thiosulphate and dilute hydrochloric acid is a slow reaction. Burning fuel to form carbon dioxide and water. CH4 + 2O2 → CO2 + 2H2O Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + SO2(g) + S(s) 1 Other fast reactions include • burning of magnesium 2Mg + O2 → 2MgO • reaction of sodium or potassium with water 2Na + 2H2O → 2NaOH + H2 Table 1.2 Example of slow reactions yellow precipitate Example Type of reaction Iron rusting (b) Observable changes: When dilute hydrochloric acid is added to sodium thiosul phate solution, the solution becomes cloudy because sulphur is precipitated. Sulphur is a yellow solid, but in small quantities, it appears yellowish-white. SPM ’09/P1 Rusting takes place slowly in the presence of oxygen and water. 4Fe + 3O2 + 2H2O → 2Fe2O3•H2O rust Fermentation of In the presence of yeast, glucose solution fermentation of glucose solution produces alcohol and carbon dioxide. C6H12O6 → 2C2H5OH + 2CO2 1 Rate of reaction ∝ — — — — — — — — — — — — — — — — — — — Time taken for a given amount of sulphur precipitate to form glucose Photosynthesis 8 The units used for the rate of reaction will depend on the changes measured. For example, (a) cm3 per unit time (second or minute) for a gas evolved (b) g per unit time or mol per unit time for a solid reactant (c) mol dm–3 per unit time for a reactant in aqueous solution 9 Different chemical reactions take place at different rates. Some reactions occur rapidly and some slowly. Table 1.1 shows some examples of fast reactions. Table 1.2 shows some examples of slow reactions. alcohol (ethanol) During photosynthesis, carbon dioxide reacts with water to form glucose and oxygen gas. 6CO2 + 6H2O → C6H12O6 + 6O2 glucose • The reactions of Groups 1 and 2 metals with oxygen is a fast reaction. However, the reactions of other metals (such as copper) with oxygen are slow reactions. • The rate of decay of the radioactive carbon-14 is very low. For example, 1.0 g of carbon-14 takes 5730 years to disintegrate (decay) to 0.50 g. The rate of decay of carbon-14 is used in archaeology to estimate the age of ancient artifacts. This method is called carbon dating. 297 Rate of Reaction 2 A piece of magnesium ribbon weighing 0.1 g is added to dilute hydrochloric acid. After 5 seconds, all the magnesium had dissolved. What is the average rate of reaction? The rate of a reaction depends on various factors (Section 1.2). For example, the rate of rusting of iron is increased if the iron is exposed to acid (such as polluted air in industrial areas) or to the salt, sodium chloride in sea air. Solution Average rate of Mass of magnesium reacted = reaction Time taken 1 Measuring Reaction Rates 1 The rate of reaction can be expressed in two SPM ways: ’04/P1, ’10/P1, (a) the average rate of reaction over a period ’11/P1 of time, or (b) the rate of reaction at any given time. 2 The average rate of reaction is the average of the reaction rates over a given period of time. We can measure the average rate of reaction by measuring the change in amount (or concentration) of a reactant or a product over a period of time. 3 For example, the average rate of reaction between magnesium and hydrochloric acid can be determined by measuring the time taken for a piece of magnesium to dissolve completely in the acid. 1 = 0.1 g 5s = 0.02 g s–1 The value obtained is the average rate of reaction over a period of 5 seconds. 4 The average rate of reaction can also be determined from the graph. Based on Figure 1.2, the average rate of the first t1 second V1 =— — cm3 s–1 t1 Figure 1.2 The average rate of reaction from t1 to t2 (V2 – V1) =— — — — — — — cm3 s–1 (t2 – t1) SPM ’04/P1 Calcium carbonate reacts with dilute hydrochloric acid to form carbon dioxide. After 1.2 minutes, the volume of gas produced is 100 cm3. Calculate the average rate of reaction in units of (a) cm3 min–1, (b) cm3 s–1. 5 The rate of reaction at any given time is the actual rate of reaction at a given time. The reaction rate at any given time is also known as the instantaneous rate of reaction. 6 The rate of reaction at a given time can be determined by the following methods. (a) By measuring the gradient of the graph of mass of reactant against time (Figure 1.3). Solution Volume of CO2 produced (a) Average reaction rate = — — — — — — — — — — — — — — Time taken 100 cm3 = — — — — — — — 1.2 min = 83.3 cm3 min–1 Volume of CO2 produced (b) Average reaction rate = — — — — — — — — — — — — — Time taken 83.3 cm3 min–1 = — — — — — — — — — — — — 60 s min–1 Figure 1.3 Measuring the rate of reaction involving a change in mass at a given time = 1.39 cm3 s–1 Rate of Reaction 298 Determining the gradient of the tangent at time, t: The following steps are used to determine the gradient of the tangent at time, t (Figure 1.3). Analysing a reaction rate curve (i) The steeper the gradient, the higher the rate of reaction. Step 1 Draw the tangent XY at point P. Figure 1.5 Comparing the rates of reaction for a given reaction at different times Step 2 Complete the right-angled triangle XYZ. (ii) Figure 1.6 shows the graph of volume of hydrogen against time for the reaction between excess zinc powder and dilute hydrochloric acid. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) Step 3 1 Measure the lengths of XZ and ZY. Step 4 Find the gradient of the line XY. Figure 1.6 a Gradient of the line XY = — b = rate of reaction at time, t (g s–1) (b) By measuring the gradient of the graph of volume of gas produced (product) versus time (Figure 1.4). (a) Initially (t1), • the graph is steep, • the rate of reaction is high. (b) As the reaction proceeds (t2), • the graph is less steep, • the rate of reaction decreases because the concentration of hydrochloric acid decreases. (c) Finally (t3), • the graph becomes horizontal, • the gradient of the graph becomes zero, • the reaction stops because all the hydrochloric acid has reacted. Figure 1.4 Measuring the rate of reaction at a given time involving a change in the volume of a gas a Gradient = — b = rate of reaction at time, t (cm3 min–1) 299 Rate of Reaction 3 Consider the reaction between excess magnesium and dilute sulphuric acid: Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g) The reaction stops at 20 seconds. Plot the graph of (a) mass of magnesium against time, (b) concentration of sulphuric acid against time, (c) concentration of magnesium sulphate against time, (d) volume of hydrogen gas against time. Solution (a) (c) (d) 1 (b) There are some parameters which cannot be measured accurately to determine the instantaneous rate of reaction, for example the change in colour or the formation of precipitate. To find the reaction rates at (a) 90 s, (b) 180 s and (c) the average rate of the reaction between zinc and dilute sulphuric acid Procedure 1 The burette is filled with water and inverted over a basin of water. 2 Using a measuring cylinder, 20.0 cm3 of 0.3 mol dm–3 sulphuric acid is measured out and poured into a conical flask. 3 5.0 g of granulated zinc is then added to the sulphuric acid in the conical flask. 4 The conical flask is then closed and the hydrogen gas produced is collected in the burette by the displacement of water as shown in Figure 1.7. 5 The stopwatch is started immediately. 6 The volume of hydrogen gas collected in the burette is recorded at 30-second intervals. Figure 1.7 Apparatus Conical flask, measuring cylinder, delivery tube, burette, basin, retort stand, retort clamp and stopwatch. Materials Granulated zinc and 0.3 mol dm–3 sulphuric acid. Results Activity 1.1 Time (s) 0 30 60 90 120 150 180 210 240 270 300 330 360 Burette reading (cm3) 50.00 33.00 24.50 18.00 13.00 10.00 6.50 5.00 4.00 3.50 3.00 3.00 3.00 Volume of H2 released (cm3) 0.00 17.00 25.50 32.00 37.00 40.00 43.50 45.00 46.00 46.50 47.00 47.00 47.00 Based on the experimental results, a graph of the volume of hydrogen released against time is plotted. Rate of Reaction 300 Calculation (a) The rate of reaction at 90 s = gradient of the curve at 90 s YZ (52 – 20) cm3 = — — — =— — —— — –— — — — — — XY (180 – 30) s 32 cm3 = — — — — — = 0.213 cm3 s–1 150 s (b) The rate of reaction at 180 s = gradient of the curve at 180 s QR (48 – 30) cm3 = — — — =— — — — — — — — — — — PQ (240 – 18) s 18 cm3 = — — — — — = 0.081 cm3 s–1 222 s (c) The average rate of reaction Total volume of H2 produced = — — — — — — — — — — — — — — — — — — — — — — — — Total time taken 47 cm3 = — — — — — = 0.157 cm3 s–1 300 s 1 Conclusion The rate of reaction decreases as the reaction proceeds. To measure the rate of reaction between calcium carbonate (CaCO3) and excess hydrochloric acid Apparatus Conical flask, electronic balance, measuring cylinder and stopwatch. Materials Calcium carbonate (CaCO3) pieces, 2.0 mol dm–3 hydrochloric acid and cotton wool. Figure 1.8 Procedure 1 Using a measuring cylinder, 50 cm3 of 2 mol dm–3 hydrochloric acid is measured out and poured into a dry conical flask. The mouth of the conical flask is covered with some cotton wool. The cotton wool is inserted into the mouth of the conical flask to prevent liquid from splashing out during the reaction. 2 The conical flask is placed on the electronic balance as shown in Figure 1.8. 3 The mass of conical flask, calcium carbonate, hydrochloric acid and cotton wool is recorded. 4 The calcium carbonate is then transferred to the hydrochloric acid in the conical flask and the stopwatch is started immediately. 5 The mass of the conical flask (and its contents) is recorded at one-minute intervals. Results Mass of conical flask + contents (g) 0 1 2 3 4 5 6 7 8 60.0 59.1 58.3 57.9 57.4 57.0 56.8 56.5 56.3 Based on the experimental results, a graph of the mass of conical flask and its contents against time is plotted (Figure 1.9). 301 Rate of Reaction Activity 1.2 Time (min) 0.9 g =— — — — — –— 1.0 min = 0.9 g min–1 (b) The average rate of reaction between 1.4 minutes and 2.2 minutes. Rate of decrease in mass (58.8 – 58.3) g =— — — — — — — — — — — — From the graph (Figure 1.9) (2.2 – 1.4) min = 0.625 g min–1 (c) The reaction rate at the 5th minute a = gradient of the graph at the 5th minute = — b a = 57.5 – 56.4 = 1.1 g b = 7.0 – 3.4 = 3.6 minutes 1.1 g Gradient = ————— 3.6 min = 0.306 g min–1 1 Figure 1.9 Calculation (a) The average rate of reaction for the first minute. Decrease in mass = mass of carbon dioxide produced = (60.0 – 59.1) g See table of results. = 0.9 g Average rate of reaction for the first minute Mass of CO2 produced = — — — — — — — — — — — — — — — — — — – Time taken Conclusion The rate of reaction decreases as the reaction proceeds. Finally, the reaction will stop when all the calcium carbonate added have reacted. Solving Numerical Problems Involving Rate of Reaction SPM ’08/P2 The graph below shows the total volume of oxygen gas produced against time for the decomposition of hydrogen peroxide. 1 The rate of reaction can be stated in terms of (a) the average rate of reaction for a given period of time, or (b) the rate of reaction at any given time (instantaneous rate). 2 The average rate of reaction can be calculated (a) directly from the data given (Example 2) or (b) from the graph drawn (Example 4). 3 The reaction rate at a given time can only be obtained from the gradient of the graph at the given time (Example 5). At time, t, the maximum volume of oxygen is collected. The gradient of the curve at time, t is zero. Hence, the rate of reaction is zero, that is, the reaction has stopped at time, t. Rate of Reaction The rate of reaction is useful to a chemist because he is not satisfied with merely converting one substance to another. In most cases, he wants to obtain the products in the fastest and most economical way. 302 4 5 3.0 g of excess marble (CaCO3) are added to 100 cm3 of dilute hydrochloric acid. Figure 1.10 shows the graph of volume of carbon dioxide produced against time. Hydrogen peroxide decomposes as represented by the equation: 2H2O2(aq) → 2H2O(l) + O2(g) The results of an experiment on the decomposition of hydrogen peroxide are given below. Time (s) 0 15 30 45 60 90 Volume of O2 (cm3) 0 16 30 40 48 56 Calculate the rate of reaction at 40 seconds in units of (a) cm3 s–1, (b) cm3 min–1. 1 Solution (a) Figure 1.10 Calculate (a) the average rate of reaction, (b) the concentration of hydrochloric acid in mol dm–3. [1 mol of any gas occupies 24 dm3 at room conditions] Solution (a) Total volume of carbon dioxide evolved = 360 cm3 Time taken = 8.0 minutes 360 cm3 Average rate of reaction = — — — — — — — 8 min = 45 cm3 min–1 (b) Number of moles of CO2 evolved 360 cm3 =— — — — — — — — — — — — — — = 0.015 (24  1000) cm3 The rate of reaction at 40 s CaCO3 + 2HCl → CaCl2 + H2O + CO2 Mole ratio of HCl : CO2 =2:1 ? : 0.015 = gradient at 40 s a (49 – 21) cm3 =—=— — — — — — — — — — — b (58 – 18) s From equation According to the equation, number of moles of hydrochloric acid used = 2  0.015 = 0.03 mol. Concentration of hydrochloric acid Number of moles 0.03 mol =— — — — — — — — — — — — — — =— — — — — — — 3 Volume (in dm ) 0.1 dm3 = 0.3 mol dm–3 obtained from the graph = 0.70 cm3 s–1 (b) 1 minute = 60 seconds ∴Rate of reaction in cm3 min–1 = 0.70 cm3 s–1  60 s min–1 = 42 cm3 min –1 From the question: 100 cm3 = 0.1 dm3 303 Rate of Reaction 1 ’09 Which of the following is correctly matched with its rate of reaction? High reaction rate Low reaction rate A Combustion of fuels Respiration B Combustion of fuels Double decomposition between silver nitrate and sodium chloride solution C Rusting of iron Fermentation of glucose solution D Respiration Neutralisation reaction between an acid and an alkali 1 Comments Combustion, double decomposition and neutralisation are fast reactions. Rusting and respiration are slow reactions. Answer A 1.1 1 Which of the following reactions occur at (a) a high rate, (b) a low rate? I Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s) II 2Cu(s) + O2(g) → 2CuO(s) III S2O32–(aq) + 2H+(aq) → S(s) + H2O(l) + SO2(g) IV 4K(s) + O2(g) → 2K2O(s) 2 You are given the chemicals and apparatus as listed below. • A piece of zinc of mass 2.0 g • A beaker containing sulphuric acid • A stopwatch (a) Using the chemicals and apparatus given, describe an experiment to measure the rate of reaction between zinc and sulphuric acid. (b) State the units for the rate of reaction. (c) State two assumptions for this experiment. Figure 1.11 3 A student intends to study the rate of reaction between iron and dilute sulphuric acid. The equation for the reaction is as follows. (a) What is the total time required for the magnesium ribbon to react completely with hydrochloric acid? (b) Based on the graph, is the reaction rate at the first minute higher or lower than the reaction at the second minute? Explain your answer. (c) Is this a normal behaviour? Suggest one reason for this behaviour. (d) The reaction between hydrochloric acid and another metal produces 12 cm3 of hydrogen after 1.0 minute. Is this reaction rate higher or lower than the reaction between magnesium and hydrochloric acid? Explain your answer. Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g) Suggest two methods that he can use to measure the rate of reaction. 4 The graph in Figure 1.11 shows the results of an experiment to measure the rate of reaction of magnesium ribbon with an excess of dilute hydrochloric acid. Rate of Reaction 304 What is the average rate of reaction? (b) 4.0 g of magnesium is added to excess dilute sulphuric acid. If the average rate of reaction is 0.0030 mol s–1, what is the mass of magnesium unreacted after 0.5 minute? [Relative atomic mass of Mg = 24] 5 The table below shows the results for two experiments carried out under room conditions. Reaction Result I 1 g of nickel powder + 50 cm3 of 1 mol dm–3 hydrochloric acid Time taken to collect 60 cm3 of hydrogen gas = 120 s 1 g of zinc powder + 50 cm3 of 1 mol dm–3 hydrochloric acid Time taken to collect 45 cm3 of hydrogen gas = 56 s II 7 In the presence of manganese(IV) oxide, hydrogen peroxide decomposes according to the equation: MnO2 2H2O2(aq) ⎯⎯→ 2H2O(l) + O2(g) A sample of hydrogen peroxide decomposed in the presence of a catalyst and the volume of oxygen gas produced was collected at regular time intervals. The results of the experiment were recorded in the following table. Based on the information given in the table above, predict which metal is more reactive, nickel or zinc? 6 (a) The volume of hydrogen gas collected at regular intervals for the reaction between granulated zinc and dilute hydrochloric acid is shown below. 1.2 Time (s) Volume of H2 (cm3) 0 0 20 16 40 26 60 32 70 36 80 36 1 1— 2 1 2— 2 Time (min) 0 1 Volume of O2 (cm3) 0 32 46 56 64 69 74 74 2 3 4 5 1 Experiment Calculate (a) the average rate of reaction for the first 144 seconds. (b) the average rate of reaction for the overall reaction in cm3 s–1. (c) the average rate of reaction between the first minute and the 3rd minute. (d) the rate of reaction at the 150th second. Factors that Affect the SPM Rate of Reaction ’08/P1 Uranium is the radioactive isotope used for making nuclear bombs. Uranium decays slowly to form lead. The decay of uranium and other radioactive elements is unique. These nuclear reactions are not influenced by factors such as surface area, temperature and catalyst. 1 The rate of reaction is affected by the following factors: (a) Total surface area (or particle size) of the solid reactant (b) Concentration of reactant (c) Temperature of reaction (d) Use of catalyst (e) Pressure (for reactions involving gases) 2 When the condition of reaction changes, the rate of reaction also changes. 3 Table 1.3 explains briefly how these conditions of reaction affect the rate of reaction between zinc metal and dilute sulphuric acid. 4 Reaction involving gases (a) Changes in pressure will not affect reactions in aqueous solutions. (b) Changes in pressure will only affect reactions involving gases. (c) Increasing the pressure will compress more gas molecules into a given space. Hence the gaseous particles will collide more frequently and the rate of reaction increases. 305 Rate of Reaction Table 1.3 Surface area (particle size) • When zinc foil is broken into smaller pieces, the total surface area increases. • The smaller the size of zinc foil, the greater the total surface area exposed to the hydrogen ions. Hence the rate of reaction increases. Concentration of reactant 1 • In dilute acid, there are not so many hydrogen ions present. • In more concentrated acid, there are more hydrogen ions in the solution. Hence the rate of reaction increases. Temperature of reaction • When the temperature of a reaction increases, the particles move faster because they have higher kinetic energy. Hence the rate of reaction increases. Catalyst Some catalysts used in industry. The catalysts are in pellet form for larger surface area. SPM ’10/P2 • Manipulated variable: Size (total surface area) of magnesium • Responding variable: Time taken to collect 60 cm3 of hydrogen gas • Constant variables: Temperature, concentration and volume of sulphuric acid as well as mass of magnesium 2 The results of the experiments are shown below. Factors that Influence the Rate of Reaction Effect of Surface Area on the Rate of Reaction 1 Two experiments are carried out to study the rate of reaction between magnesium and dilute sulphuric acid under different conditions. Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g) Experiment Conditions of experiment I 1.0 g of magnesium ribbon and 50 cm3 of 1.0 mol dm–3 sulphuric acid 1.0 g of magnesium powder and 50 cm3 of 1.0 mol dm–3 sulphuric acid II The size (surface area) of magnesium is manipulated Rate of Reaction A catalyst will increase the rate of reaction. This will be explained in Section 1.3. 306 Time taken to collect 60 cm3 of H2 (s) 30 Average rate of reaction (cm3 s–1) 2 10 6 3 The results show that the time taken to collect 60 cm3 of gas using magnesium powder is shorter than using magnesium ribbon. This is due to the smaller size of particles (total surface area is greater) in magnesium powder than in magnesium ribbon. 1 mol dm–3 hydrochloric acid. This means that the higher the concentration of hydrochloric acid, the higher the rate of reaction. 4 (a) It is important to know • how to plot graphs (on the same axis), or • how to interpret graphs on rate of reaction if information on the conditions of reaction are given. (b) The two features on the graph to be considered are • the gradient of the graph which shows the rate of reaction, • the height of the graph which shows the total amount of product formed. Effect of Concentration of Reactant on the Rate of Reaction 1 Two experiments are carried out to study the rate of reaction between magnesium and ’04/P1 ’05/P3 hydrochloric acid. SPM ’06/P1 Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) • Manipulated variable: Concentration of hydrochloric acid • Responding variable: Time taken for magnesium to dissolve completely • Constant variables: Size of magnesium ribbon, volume of hydrochloric acid and temperature of experiment 2 The results of the experiments are shown below. Conditions of experiment Experiment 1 Two experiments are carried out to study the rate of reaction between zinc and sulphuric acid at different temperatures. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) • Manipulated variable: Temperature of sulphuric acid • Responding variable: Volume of hydrogen gas evolved • Constant variables: Mass of zinc, concentration and volume of sulphuric acid 2 The results are shown in Figure 1.12. Time taken for magnesium to dissolve completely (s) I 5 cm magnesium ribbon and 50 cm3 of 1 mol dm–3 hydrochloric acid 78 II 5 cm magnesium ribbon and 50 cm3 of 2 mol dm–3 hydrochloric acid 39 Concentration of acid is manipulated Figure 1.12 Comparing the rates of reaction at different temperatures 3 The shorter the time taken for a reaction to complete, the higher the rate of reaction. The results show that the time taken for magnesium to react completely in 2 mol dm–3 hydrochloric acid is shorter than that in 3 The results show that the higher the temperature of sulphuric acid, the steeper the graph and hence, the higher the rate of reaction. Experiment Mass of zinc Volume of sulphuric acid Concentration of sulphuric acid Temperature of sulphuric acid I 1.0 g of zinc powder 20 cm3 0.1 mol dm–3 28 °C II 1.0 g of zinc powder 20 cm 0.1 mol dm 35 °C 3 Conditions remain unchanged 307 –3 Temperature of acid is manipulated Rate of Reaction 1 Effect of Temperature on the Rate of Reaction 4 In general, the rate of reaction increases if the temperature of the reactants is increased. occurs. In this reaction, manganese(IV) oxide acts as a catalyst and catalyses the decomposition of hydrogen peroxide to give oxygen gas and water. Effect of Catalysts on the Rate of Reaction MnO2 1 Definition A catalyst is a substance that increases the rate of a reaction but is itself chemically unchanged at the end of the reaction. 2 In contrast, a substance that decreases the rate of a chemical reaction is called an inhibitor. 3 At room temperature, hydrogen peroxide decomposes very slowly. But when a very small amount of manganese(IV) oxide is added to hydrogen peroxide, a vigorous effervescence 1 The Characteristics of Catalysts Only a small amount of catalyst is needed to increase the rate of reaction. The physical appearance of a catalyst may change at the end of the reaction. For example, small pieces of catalyst may become fine powder after the reaction. 2H2O2(aq) ⎯⎯⎯→ 2H2O(l) + O2(g) 4 The reaction between zinc and dilute acid is a slow reaction. Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) When it is catalysed by copper(II) sulphate solution, the reaction speeds up. SPM ’06/P1, ’08/P1, ’10/P1, ’11/P1 • The catalyst remains chemically unchanged after the reaction. • Thus the chemical properties, mass and chemical composition of the catalyst remain unchanged at the end of the reaction. Characteristics of a catalyst A catalyst increases the rate of a chemical reaction but it does not change (increase or decrease) the yield of a chemical reaction. CuSO4 Zn(s) + H2SO4(aq) ⎯⎯⎯→ ZnSO4(aq) + H2(g) catalyst A catalyst lowers the activation energy of a reaction (see Section 1.3 – The Collision Theory) (a) In general, catalysts are highly specific. For example, iron catalyses the reaction: N2 + 3H2 2NH3 but not the reaction: 2SO2 + O2 2SO3 (b) However, some catalysts can catalyse several different reactions. For example, MnO2 can catalyse the following reactions: MnO2 2H2O2(aq) ⎯⎯⎯→ 2H2O(l) + O2(g) MnO2 2KClO3(s) ⎯⎯⎯→ 2KCl(s) + 3O2(g) Catalysts can be poisoned by impurities. When a catalyst is poisoned, its effectiveness as a catalyst is decreased. Rate of Reaction 308 Examples of Catalysts 1 Transition metals and compounds of transition metals are often used as catalysts for industrial processes. 2 Table 1.4 shows some examples of catalysts and the reactions catalysed by them. Table 1.4 Some common catalysts Type of reaction Catalyst used (a) Haber process for the manufacture of ammonia. N2(g) + 3H2(g) Iron, Fe 2NH3(g) (b) Contact process for producing sulphur trioxide. 2SO2(g) + O2(g) Vanadium(V) oxide, V2O5 2SO3(g) Sulphur trioxide is used for the manufacture of sulphuric acid. (c) Ostwald process for producing nitrogen monoxide. 4NH3(g) + 5O2(g) Platinum, Pt 4NO(g) + 6H2O(g) Nitrogen monoxide is used for the manufacture of nitric acid. Nickel, Ni (e) Cracking process When big alkane molecules are passed over a catalyst at 600 °C, a mixture of small alkane and alkene molecules is produced. This process is called catalytic cracking. (Refer Sections 2.2 and 2.3 on alkanes and alkenes) Aluminium oxide, Al2O3 or Silicon(IV) oxide, SiO2 2 1 (d) Manufacture of margarine In the presence of a catalyst at 200 °C, vegetable oils react with hydrogen to produce margarine. This process is called hydrogenation. Effect of Pressure on the Rate of Reaction ’06 1 The changes in pressure will only affect reactions involving gases. An increase in pressure increases the rate of reaction. In contrast, a decrease in pressure decreases the rate of reaction. In the following reactions involving gases, the rate of reaction increases if the pressure is increased. Which of the following statements about catalysts are true? I A catalyst is specific in its reaction. II A catalyst changes the quantity of product formed. III Only a small amount of a catalyst is needed to change the rate of reaction. IV The chemical properties of the catalyst remain unchanged at the end of a reaction. A I and II only C I, II and III only B II and IV only D I, III and IV only Answer D N2(g) + 3H2(g) 2NH3(g) (Haber process) 2SO2(g) + O2(g) 2SO3(g) (Contact process) 2 Pressure has no effect on reactions involving only solids or liquids. For example, the following reactions are not affected by changes in pressure. CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) A catalyst takes part in a chemical reaction. In actual fact, a catalyst combines with the reactants to form an unstable intermediate species. This species then decomposes to re-form the catalyst and to produce the products. 2H2O2(aq) → 2H2O(l) + O2(g) 309 Rate of Reaction 1.1 To investigate the effect of the surface area of a reactant on the rate of reaction SPM ’06/P2 Procedure Problem statement How does the surface area of a solid reactant affect the rate of reaction? 1 Hypothesis The smaller the size of the marble chips, that is, the larger the total surface area of the marble chips, the higher the rate of reaction. Variables (a) Manipulated variable : Size of the marble chips used (b) Responding variable : Volume of gas given off at 30-second intervals (c) Constant variables : Temperature of the experiment, mass of marble chips, concentration and volume of hydrochloric acid Apparatus Conical flask, delivery tube, retort stand and clamp, burette, measuring cylinder and stopwatch. Materials Marble chips, powdered marble and 0.08 mol dm–3 hydrochloric acid. Figure 1.13 1 A burette is filled with water and inverted over a basin containing water. The burette is clamped to the retort stand. The water level in the burette is adjusted and the initial burette reading is recorded. 2 5.0 g of marble chips are placed in a small conical flask. 3 50 cm3 of 0.08 mol dm–3 hydrochloric acid is added to the marble chips. 4 The conical flask is then stoppered and the stopwatch is started immediately (Figure 1.13). 5 The burette readings are recorded at 30-second intervals. Experiment I The rate of reaction using large marble chips Results Time (s) 0 30 60 90 120 150 180 210 240 Burette reading (cm3) 50.00 45.50 41.50 38.00 35.00 33.00 31.00 29.00 28.00 Volume of gas (cm3) 0.00 Experiment II 4.50 8.50 12.00 15.00 17.00 19.00 21.00 22.00 The rate of reaction using powdered marble Constant variable is also known as fixed variable or controlled variable. Experiment 1.1 Procedure 1 Steps 1 to 4 in Experiment I are repeated using 5.0 g of powdered marble. All other conditions such as temperature, volume and concentration of hydrochloric acid are kept constant. 2 The results of the experiment are recorded in the following table. Results Time (s) 0 30 60 90 120 150 180 210 240 Burette reading (cm3) 50.00 42.00 35.00 29.50 25.50 22.00 19.50 17.50 16.00 Volume of gas (cm3) 0.00 Rate of Reaction 8.00 15.00 20.50 24.50 28.00 30.50 32.50 34.00 310 3 4 5 Figure 1.14 Discussion 1 Figure 1.15 shows the graphs that will be obtained if the reactions in Experiments I and II are completed. 6 volume of the hydrochloric acid used in both the experiments are the same. The gradients of the graphs for Experiments I and II become less steep as the reactions proceed. This shows that the rates of reaction (a) are very high at the beginning of the reactions, (b) decrease as the reactions proceed, (c) become zero when the reactions are completed. At this time, the graphs become horizontal. The rate of reaction between the marble and hydrochloric acid decreases because (a) the mass of the remaining unreacted marble decreases, (b) the concentration of hydrochloric acid decreases. The reaction in Experiment I stops after t2 minutes while the reaction in Experiment II stops after t1 minutes, where t1 < t2. This shows that the rate of reaction for Experiment II (powdered marble) is higher than the rate of reaction for Experiment I (marble chips). The total volume of carbon dioxide collected in the burette is usually slightly less than the theoretical value (48 cm3 for the experiment above). This is because carbon dioxide is slightly soluble in water. To overcome this problem, a gas syringe is used to collect carbon dioxide released during the experiment (Figure 1.16). Figure 1.16 Measuring the volume of gas using a gas syringe Same maximum volume of CO2 collected because mass of CaCO3, concentration and volume of HCl are kept constant. Conclusion Graph II is steeper than graph I. This shows that the rate of reaction in Experiment II is higher than the rate of reaction in Experiment I as powdered marble is used in Experiment II. Thus, the rate is higher with powdered marble than with marble chips. Hence, we can conclude that the smaller the particle size, the larger the total surface area exposed for reaction and the higher the rate of reaction. The hypothesis is accepted. Figure 1.15 2 Figure 1.15 shows that both graphs level off at the same value. This indicates that the maximum volume of carbon dioxide collected at the end of reaction for both Experiments I and II are the same (that is, 44 cm3). This happens because the mass of the marble, concentration and 311 Rate of Reaction 1 Based on the results obtained, a graph of the total volume of carbon dioxide produced against time for each experiment is plotted on the same axes (Figure 1.14). 1.2 To study the effect of concentration on the rate of reaction between sodium thiosulphate solution and dilute sulphuric acid SPM ’07/P1, ’11/P2 Apparatus 10 cm3 and 100 cm3 measuring cylinders, 100 cm3 conical flask, white paper marked with a cross ‘X’ and stopwatch. Problem statement How does the concentration of a reactant affect the rate of reaction between sodium thiosulphate and dilute sulphuric acid? Materials 0.2 mol dm–3 sodium thiosulphate solution, 1.0 mol dm–3 sulphuric acid and distilled water. Procedure 1 50 cm3 of 0.2 mol dm–3 sodium thiosulphate solution is measured out using a 100 cm3 measuring cylinder. The solution is then poured into a clean, dry conical flask. 2 The conical flask is placed on a piece of paper with a cross ‘X’ marked on it (Figure 1.17). 3 5 cm3 of dilute sulphuric acid is measured out by using a 10 cm3 measuring cylinder. The acid is then quickly poured into sodium thiosulphate solution. The stopwatch is started immediately. 4 The reaction mixture is swirled once and the cross ‘X’ is viewed from above. A yellow precipitate will appear slowly in the conical flask. 5 The stopwatch is stopped as soon as the cross disappears from view and the time taken is recorded. 6 Steps 1 to 5 are repeated with different mixtures of sodium thiosulphate solution and distilled water as shown in the following table. 1 Figure 1.17 Hypothesis The more concentrated the sodium thiosulphate solution, the higher the rate of reaction. Variables (a) Manipulated variable: Concentration of sodium thiosulphate solution (b) Responding variable: Time taken for the cross ‘X’ to disappear (c) Constant variables: Concentration and volume of dilute sulphuric acid as well as the temperatures of the solutions Results Experiment 1 2 3 4 5 Volume of Na2S2O3 (cm ) 50 40 30 20 10 Volume of water (cm ) 0 10 20 30 40 Volume of H2SO4 (cm ) 5 5 5 5 5 0.20 0.16 0.12 0.08 0.04 24 30 42 62 111 3 3 3 Concentration of Na2S2O3 (mol dm–3) Time taken (s) Experiment 1.2 1 — — — — (s–1) Time 0.042 0.033 0.024 0.016 0.009 The ionic equation is as follows: Discussion 1 The following equation shows the reaction between sodium thiosulphate, Na2S2O3 and dilute sulphuric acid: Na2S2O3(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) + SO2(g) + S(s) Rate of Reaction Different volumes (V1) of Na2S2O3 solution are diluted with water to make up to 50 cm3 solution (V2). S2O32–(aq) + 2H+(aq) → S(s) + SO2(g) + H2O(l) The sulphur is precipitated as fine yellow particles that cause the solution to turn cloudy. 312 2 As the amount of sulphur increases, the cross ‘X’ becomes more and more difficult to see. Finally, the cross ‘X’ disappears from view when a certain mass of sulphur is precipitated. Hence, the time recorded for the disappearance of the cross ‘X’ is the time taken for the formation of a fixed mass of sulphur. Mass of sulphur produced 3 Rate of reaction = — — — — — — — — — — — — — — — — — — — — — Time taken 1 Hence, rate of reaction ∝ — — — — — — — — — — — — — — — — — — — Time taken for the cross ‘X’ to disappear 4 The concentration of sodium thiosulphate solution after mixing with water can be obtained by using the following formula: M1V1 Concentration of Na2S2O3 = — — — — — V2 7 The conical flask used for each experiment must have the same size (for example, 100 cm3 volume). If the conical flask of a larger size is used, the time, t, taken for the cross ‘X’ to disappear will increase. Conversely, if a smaller conical flask is used, the time taken for the cross to disappear will be shorter. 8 If the experiment is repeated with dilute sulphuric acid of different concentrations, but the concentration of sodium thiosulphate is kept constant, the rate of reaction will also be directly proportional to the concentration of the acid used. 0.2  Volume of Na2S2O3 used =— — — — — — — — — — — — — — — — — — — — — — — — — mol dm–3 50 5 Based on the results obtained, two graphs can be plotted. (a) The graph of concentration of sodium thiosulphate against time (Graph I, Figure 1.18). Conclusion 1 (a) From graph I, we can conclude that the higher the concentration of sodium thiosulphate, the shorter the time taken for a certain mass of sulphur to be precipitated, that is, for the cross ‘X’ to disappear from view. (b) This means that the higher the concentration of sodium thiosulphate, the higher the rate of reaction. 2 From graph II, it can be concluded that the concentration of sodium thiosulphate is directly 1 proportional to — — —. time 1 Concentration of Na2S2O3 ∝ — — — …(1) time 1 3 But the rate of reaction is ∝ — — — … (2) time Hence, combining equations (1) and (2), we have, 1 — — ∝ reaction rate concentration of Na2S2O3 ∝ — time That is, rate of reaction ∝ concentration of Na2S2O3 solution. The hypothesis is accepted. Figure 1.18 (b) The graph of concentration of sodium 1 thiosulphate against — — — (Graph II, Figure 1.19). time 6 Different volumes of distilled water are added to sodium thiosulphate solution so that the final volume of the diluted sodium thiosulphate solution is 50 cm3 in each experiment. Hence, the concentration of sodium thiosulphate solution is directly proportional to its volume before dilution. 313 Rate of Reaction 1 Figure 1.19 3 ’06 Which of the following reactants produces the highest rate of reaction with magnesium powder? A 50 cm3 of 0.5 mol dm–3 nitric acid B 50 cm3 of 0.5 mol dm–3 ethanoic acid C 50 cm3 of 0.5 mol dm–3 sulphuric acid D 50 cm3 of 0.5 mol dm–3 hydrochloric acid Comments Magnesium reacts with the hydrogen ions of the acids. Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) Ethanoic acid is a weak acid and produces very few H+ ions. HNO3, H2SO4 and HCl are strong acids. But each mole of H2SO4 produces two moles of H+ ions. Hence, 50 cm3 of 0.5 mol dm–3 sulphuric acid contains the highest concentration of H+ ions. Answer C Experiment 1.3 1 1.3 SPM To study the effect of temperature on the rate of reaction between sodium ’05/P1 thiosulphate solution and dilute sulphuric acid 5 The stopwatch is started immediately and the Problem statement conical flask is swirled gently. How does temperature affect the rate of reaction between 6 The cross ‘X’ is viewed from above. The stopwatch sodium thiosulphate solution and sulphuric acid? is stopped as soon as the cross disappears from view Hypothesis The higher the temperature of the and the time taken is recorded. reactant, the higher the rate of 7 The solution in the conical flask is poured out. reaction. The conical flask is washed thoroughly and dried. 50 cm3 of 0.1 mol dm–3 sodium thiosulphate Variables solution is poured into the conical flask. (a) Manipulated variable: The temperature of 8 The solution is heated over a wire gauze until the sodium thiosulphate solution temperature reaches about 45 °C (Figure 1.21). (b) Responding variable: The time taken for the cross ‘X’ to disappear (c) Constant variables: The concentrations and volumes of both sodium thiosulphate solution and dilute sulphuric acid Apparatus Conical flask, 10 cm3 measuring cylinder, thermometer, stopwatch, white paper marked with a cross ‘X’, wire gauze, tripod stand and Bunsen burner. Materials 0.1 mol dm–3 sodium thiosulphate solution and 1.0 mol dm–3 sulphuric acid. Figure 1.20 Procedure 1 50 cm3 of 0.1 mol dm–3 sodium thiosulphate solution is poured into a clean, dry conical flask. 2 The temperature of the sodium thiosulphate solution is measured with a thermometer. 3 The conical flask is placed on a white paper marked with a cross ‘X’ (Figure 1.20). 4 5 cm3 of 1 mol dm–3 sulphuric acid is quickly poured into the sodium thiosulphate solution. Rate of Reaction Figure 1.21 314 9 The hot conical flask is placed over a white paper marked with a cross ‘X’. 10 5 cm3 of 1 mol dm–3 sulphuric acid is measured out using a 10 cm3 measuring cylinder. 11 When the temperature of sodium thiosulphate solution falls to 40°C, the sulphuric acid is quickly poured into the thiosulphate solution. 12 The stopwatch is started immediately and the conical flask is swirled gently. 13 The cross ‘X’ is viewed from the top and the time taken for the cross to disappear from view is recorded. 14 Steps 7 to13 are repeated at higher temperatures as shown in the following table. Experiment 1 2 3 4 5 Temperature (°C) 30 40 50 55 60 Time (s) 52 27 16 13 10 1 — — — — (s–1) Time Discussion 1 The graph shows that the temperature of sodium thiosulphate solution is proportional (but not 1 linearly) to — — —. time 1 2 Temperature ∝ — — — … (1) time 1 But rate of reaction ∝ — — — … (2) time Combining equations (1) and (2), we have, Rate of reaction ∝ temperature 0.019 0.037 0.063 0.077 0.100 Based on the results of the experiment, a graph of temperature of sodium thiosulphate solution against 1 — — — is plotted (Figure 1.22). time Conclusion The higher the temperature of the experiment, the higher the rate of reaction. 4 ’05 The rate of reaction between sodium thiosulphate solution and dilute sulphuric acid can be determined by using the arrangement of apparatus as shown below. Which of the following conditions will cause the mark ‘X’ to take the shortest time to disappear from sight? Sulphuric acid A B C D Sodium thiosulphate solution Volume (cm3) Concentration (mol dm–3) Volume (cm3) Concentration (mol dm–3) Temperature (°C) 5 5 5 10 1.0 1.0 0.5 0.5 45 45 45 40 0.5 0.5 0.5 0.5 30 35 30 35 315 Rate of Reaction 1 1 Figure 1.22 Graph of temperature against — — — time Results Comments The shorter the time taken for the mark ‘X’ to disappear, the faster the reaction. The rate of reaction is affected by temperature and concentration. The higher the temperature, the faster the reaction. (Answer B or D is correct). The higher the concentration of sulphuric acid or sodium thiosulphate in the reaction mixture, the faster the reaction (Answer D is incorrect). Answer B 1.4 To study the effect of a catalyst on the rate of decomposition of hydrogen peroxide 1 Problem statement How do catalysts affect the rate of decomposition of hydrogen peroxide? Hypothesis Manganese(IV) oxide increases the decomposition of hydrogen peroxide. rate 4 0.5 g of manganese(IV) oxide, MnO2, is added to hydrogen peroxide and shaken. The changes that take place in the test tube and on the glowing splint are recorded. of Results Variables (a) Manipulated variable: The presence of manganese(IV) oxide (b) Responding variable: The release of oxygen gas (c) Constant variables: Volume and concentration of hydrogen peroxide Observation Experiment Apparatus Test tube and wooden splint Materials Hydrogen peroxide and manganese(IV) oxide Procedure 1 A test tube is half-filled with hydrogen peroxide. 2 A glowing splint is placed at the mouth of the test tube to test for the gas evolved (Figure 1.23). Inside the test tube On the glowing splint H2O2 without No effervescence MnO2 The glowing splint does not light up. H2O2 with MnO2 The glowing splint is rekindled and burns brightly. Bubbles of oxygen gas are produced Discussion 1 The following equation shows the decomposition of hydrogen peroxide: 2H2O2(aq) → 2H2O(l) + O2(g) Experiment 1.4 2 The glowing splint is rekindled in the presence of oxygen gas. Conclusion The rate of evolution of oxygen gas increases when manganese(IV) oxide is added to hydrogen peroxide. This proves that manganese(IV) oxide acts as a catalyst and speeds up the decomposition of hydrogen peroxide to water and oxygen. The hypothesis is accepted. Figure 1.23 The effect of a catalyst on the decomposition of hydrogen peroxide 3 The changes that take place inside the test tube and on the glowing splint are recorded. Rate of Reaction 316 reaction in Experiment II. We can therefore conclude that the higher the concentration of hydrogen peroxide, the higher the rate of reaction. (c) The maximum volume of oxygen gas produced in Experiment I is twice that produced in Experiment II. This is because the number of moles of hydrogen peroxide used in Experiment I is twice that used in Experiment II. The reaction mixture remaining after Experiment 1.4 can be filtered to obtain the manganese(IV) oxide. It is found that (a) the mass of manganese(IV) oxide before and after the experiment is the same (0.5 g), (b) the chemical properties of manganese(IV) oxide remain unchanged. SPM ’05/P1 Explaining the Effectiveness of Different Catalysts on the Rate of Decomposition of Hydrogen Peroxide 1 The graphs in Figure 1.24 show the effect of concentration of hydrogen peroxide on the rate of decomposition of hydrogen peroxide. 1 Figure 1.25 shows the results of an experiment carried out to study the effect of different catalysts (of the same mass) on the rate of decomposition of hydrogen peroxide. Graph I: more O2 produced and higher rate of reaction because larger volume and higher concentration of H2O2 is used. Maximum volumes of O2 collected are the same for Experiments I and II because the concentration and volume of H2O2 used are the same. Figure 1.24 The effect of concentration of hydrogen peroxide on the rate of decomposition of hydrogen peroxide Figure 1.25 The effect of different catalysts on the rate of reaction In Experiment I, 50 cm of 0.14 mol dm of hydrogen peroxide and 0.2 g of manganese(IV) oxide are used. In Experiment II, a solution containing 25 cm3 of the same hydrogen peroxide mixed with 25 cm3 of water and 0.2 g of manganese(IV) oxide are used. For both the experiments, the temperature is kept constant. 2 (a) For Experiment I Concentration of H2O2 = 0.14 mol dm–3 higher concentration 3 –3 In Experiment I, 50 cm3 of hydrogen peroxide and 0.5 g of manganese(IV) oxide are used. In Experiment II, 50 cm3 of hydrogen peroxide and 0.5 g of iron(III) oxide are used. For both the experiments, the concentration and volume of hydrogen peroxide as well as the temperature are kept constant. 2 Analysis of the reaction rate curve in Figure 1.25. (a) At any particular instant, the gradient of graph I is greater than the gradient of graph II. This means that the rate of reaction in Experiment I is higher than the rate of reaction in Experiment II. Thus, the experiment proves that manganese(IV) oxide is a more effective catalyst than iron(III) oxide in the decomposition of hydrogen peroxide. (b) The maximum volumes of oxygen gas collected in both the experiments are the same because the volume and concentration of hydrogen peroxide used are the same. This experiment shows that a catalyst does not change the yield of the products. For experiment 2, hydrogen peroxide is diluted. (M1V1)before dilution = (M2V2)after dilution Concentration of H2O2 after dilution 0.14 mol dm–3  25 cm3 =— — — — — — — — — — — — — — — — — — — — 50 cm3 lower concentration = 0.07 mol dm–3 (b) At any particular instant, the gradient of graph I is greater than the gradient of graph II. This means that the rate of reaction in Experiment I is higher than the rate of 317 Rate of Reaction 1 The Effect of Concentration of Hydrogen Peroxide on the Rate of Reaction Effect of surface area Effect of temperature Reaction of HCl(aq) with marble chips Reaction of HCl(aq) with zinc powder Graph I : Large marble chips Graph II : Small marble chips Graph III : Powdered marble Graph I : at 30°C Graph II : at 40°C mass of marble is the same Comment: The smaller the size of marble chips, the higher the reaction rate. mass of zinc (in excess), volume and concentration of HCl are kept constant Comment: When the temperature is raised, the rate of reaction also increases. Factors affecting the rate of reaction Effect of catalyst 1 Effect of concentration of reactant Reaction of HCl(aq) with magnesium Decomposition of hydrogen peroxide Graph I : 0.5 mol dm–3 HCl mass of Mg and Graph II : 1.0 mol dm–3 HCl volume of HCl (in excess) Graph III : 2.0 mol dm–3 HCl are kept constant Graph I : Fe2O3 used as catalyst Graph II : MnO2 used as catalyst Graph III : No catalyst Comment: When the concentration of hydrochloric acid increases, the rate of reaction also increases. Comment: A catalyst increases the reaction rate. MnO2 is a more effective catalyst than Fe2O3. Effect of concentration and volume of acid used Reaction of HCl(aq) with magnesium Reaction of HCI(aq) with magnesium Graph I : Mg in excess 20 cm3 of 0.2 mol dm–3 HCI Graph II : Mg in excess 20 cm3 of 0.1 mol dm–3 HCl Graph I : Mg in excess 10 cm3 of 0.2 mol dm–3 HCI Graph II : Mg in excess 30 cm3 of 0.1 mol dm–3 HCl Comment: Graph Reaction rate Amount of H2 released Rate of Reaction I II Higher Lower 0.002 mol (48 cm3) 0.001 mol (24 cm3) Comment: Graph Reaction rate Amount of H2 released 318 I II Higher Lower 0.001 mol (24 cm3) 0.0015 mol (36 cm3) is kept in a refrigerator will last longer because the decaying reaction that destroys the food can be slowed down. 4 In the supermarkets, fish, meat and other types of fresh foods are kept in deep-freeze compartments where the temperature is about –20 °C. This keeps the food fresh for a few months because the very low temperature slows down the chemical reactions that cause the food to decay. Applications of Factors that Affect Rates of Reaction in Daily Life and in Industrial Processes Combustion of Charcoal 1 Combustion of charcoal in excess oxygen produces carbon dioxide and water. Heat energy is released during combustion. 2 Large pieces of charcoal will not burn easily because the total surface area exposed to oxygen is small. 3 If small pieces of charcoal are used, they can burn easily. This is because the total surface area exposed to the air increases. Thus, the rate of reaction with oxygen (combustion) increases. 1 Pressure cookers are used to speed-up cooking. 2 In the pressure cooker, the higher pressure enables SPM water or oil to boil at a temperature higher than ’06/P1 their normal boiling points. Furthermore, an increase in pressure causes an increase in the number of water molecules or cooking oil molecules coming into contact and colliding with the food particles. 3 At a higher temperature and pressure, the rate of reaction becomes higher. Thus, food cooks faster in pressure cookers. Coal is mainly carbon. Coal mining is dangerous because coal dust present in the coal mine catches fire very easily. Because of this, serious accidents in coal mines can happen due to the explosion of coal dust. Human lives are often lost in such explosions. Uses of Catalysts in Industry Storing Food in Refrigerators 1 Catalysts do not increase the yields of reactions. However, catalysts are used widely in industrial processes to increase the rates of reactions so that the same amount of products can be obtained in a shorter time. As a result, the use of catalysts brings down the cost of production. 2 In the chemical industry, small pellets of solid catalysts are used instead of big lumps. This is to give a larger surface for catalytic reaction to occur and hence a faster reaction will result. 3 The table below summarises the raw materials and the conditions needed for the Haber, Contact and Ostwald processes. 1 The decomposition and decay of food is a SPM chemical reaction caused by the action of ’05/P2 /SB microorganisms such as bacteria and fungi. These microorganisms multiply very rapidly at the temperature range of 10–60 °C. 2 Room temperature is the optimum temperature for the breeding of microorganisms in food. As a result, food turns bad quickly at room temperature. 3 At low temperatures, for example, 5 °C (the normal temperature of a refrigerator), the activities of bacteria are slowed down. Hence, food that Industrial process Substances Manufacture of ammonia (Haber process) Nitrogen and hydrogen Optimum conditions/equation reaction Temperature: 450–500 °C Pressure: 250 atmospheres Catalyst: Finely divided iron (Fe) N2(g) + 3H2(g) Manufacture of sulphuric acid (Contact process) Sulphur (to make SO2), air and water 450 °C, 250 atm Fe (catalyst) 2NH3(g) Temperature: 400–450 °C Pressure: 1–2 atmospheres Catalyst: Vanadium(V) oxide, V2O5 319 Rate of Reaction 1 Cooking Food in Pressure Cookers Industrial process Substances Optimum conditions/equation reaction • The following reaction scheme shows the steps involved in the manufacture of sulphuric acid: oxidation oxidation step 1 step 2 S ⎯⎯→ SO2 ⎯⎯→ SO3 ⎯→ H2S2O7 ⎯→ H2SO4 step 3 step 4 • In step 2, sulphur dioxide is oxidised to sulphur trioxide. 450 °C, 1 atm 2SO2(g) + O2(g) Manufacture of nitric acid (Ostwald process) Ammonia, air and water 2SO3(g) V2O5 (catalyst) Temperature: 900 °C Pressure: 5 atmospheres Catalyst: platinum • The following reaction scheme shows the steps involved in the manufacture of nitric acid. oxidation oxidation oxidation step 1 step 2 step 3 1 NH3 ⎯⎯⎯→ NO ⎯⎯⎯→ NO2 ⎯⎯⎯→ HNO3 In step 1, ammonia is oxidised to nitric oxide. 4NH3(g) + 5O2(g) 900 °C, 5 atm Pt (catalyst) 4NO(g) + 6H2O(g) Solving Problems on Rate of Reaction 6 Curve I in Figure 1.26 is obtained by treating 5.0 g of granulated zinc with 2.0 mol dm–3 sulphuric acid (in excess) at 30 °C. Mass and nature of Zn C D Figure 1.26 A B Rate of Reaction Temperature 2 mol dm–3 40 °C 1 mol dm–3 30 °C Comments Curves I and II show that: (a) The total volume of hydrogen produced in Experiment II is the same as that produced in Experiment I. This means that the amount of zinc used is 5 g and not 2.5 g. Answer A is incorrect. (b) Reaction II is slower than reaction I. This means that zinc powder or a higher temperature of 40 °C is not used in Experiment II. Answers B and C are incorrect. The low rate is achieved by using sulphuric acid more dilute than 2 mol dm–3 (1 mol dm–3). Answer D Which of the following conditions will produce graph II? Concentration Mass and of H2SO4 nature of Zn 2.5 g of 2 mol dm–3 granulated Zn 5.0 g of Zn 2 mol dm–3 powder 5.0 g of granulated Zn 5.0 g of granulated Zn Concentration of H2SO4 Temperature 30 °C 30 °C 320 7 Two experiments were carried out to determine the rate of oxygen gas production during the decomposition of hydrogen peroxide. In Experiment I, 20 cm3 of 2 mol dm–3 hydrogen peroxide were used and the results of the experiment are shown on graph I in Figure 1.27. (b) Differences in terms of rate of reaction Graph II is steeper than graph I because the rate of reaction in Experiment II is expected to be higher than Experiment I. When the concentration of hydrogen peroxide is increased from 2 mol dm–3 to 4 mol dm–3, the rate of reaction also increases. Difference in terms of volume of oxygen released Step 1 To calculate the volume of oxygen produced in Experiment I 2H2O2(aq) → 2H2O(l) + O2(g) Figure 1.27 (a) Sketch a graph on the same axes to show the results of the experiment that will be obtained if 5 cm3 of 4 mol dm–3 hydrogen peroxide were used for the reaction. (b) Explain your answer in (a). (c) State the constant variables for both the experiments. Step 2 To calculate the volume of oxygen produced in Experiment II Number of moles of H2O2 used in Experiment II 45 =— — — — — = 0.02 1000 ∴ Volume of oxygen collected at room temperature in Experiment II 1 1 = —  0.02  24 000 = 240 cm3 (— V cm3) 2 2 Solution (a) (c) Constant variables: In both the experiments, the same mass of the catalyst and the same temperature of reaction are used. 1.2 1 State three ways that can be used to increase the rate of reaction of zinc powder with dilute sulphuric acid. 2 Excess calcium carbonate is added to hydrochloric acid at room temperature. The volume of carbon dioxide collected is recorded at regular time intervals. The results of the experiment are shown in Figure 1.28. (a) At what time does the reaction stop? (b) Why does the reaction stop at this particular time? (c) The experiment is repeated by using the same hydrochloric acid but at a lower temperature than room temperature. On the same axes, plot a graph to show the results of the second experiment. (d) State the constant variables for both the experiments. Figure 1.28 321 Rate of Reaction 1 Number of moles of H2O2 used in Experiment I 2  20 =— — — — — — = 0.04 1000 ∴ Volume of oxygen collected at room temperature in Experiment I 1 = —  0.04  24 000 = 480 cm3 (V cm3) 2 3 Hydrogen peroxide decomposes as represented by the equation: (c) Give one inference that can be made from the results in Experiment I. (d) Explain why the initial readings on the electronic balance are different for the three experiments. 2H2O2(aq) → 2H2O(l) + O2(g) 5 Four experiments are carried out to study the rate of reaction between zinc (in excess) and sulphuric acid at different conditions. In each experiment 80 cm3 of 0.1 mol dm–3 sulphuric acid is used. The time taken to collect 192 cm3 of hydrogen gas produced are shown below. (a) On the same axes, sketch two graphs of total amount (in mol) of oxygen gas given off against time to show the results of Experiments I and II under the conditions stated below. Experiment I: 100 cm3 of 1.0 mol dm–3 H2O2 Experiment II: 300 cm3 of 0.2 mol dm–3 H2O2 Experiment 1 (b) Explain your answer based on the graphs that you have sketched. 4 In an experiment carried out at room temperature (28 °C), 8.0 g of marble chips are added to 100 cm3 of dilute hydrochloric acid in a conical flask. The mass of the conical flask and its contents is determined using an electronic balance at the beginning of the experiment (that is, as soon as the marble chips are added) and then after 1 minute. Electronic Electronic balance balance Experiment Temperature reading at reading (°C) the after 1 beginning (g) minute (g) I 28 270.35 270.04 35 271.42 270.01 III 40 268.20 266.00 (a) State a hypothesis for the experiment. (b) State the constant variables for the experiment. 1.3 I Zinc + H2SO4 35 40 II Zinc + H2SO4 38 18 III Zinc + H2SO4 + 1 cm3 of CuSO4 35 12 IV Zinc + H2SO4 + 1 cm3 of Na2SO4 35 40 particles (atoms, molecules or ions) collide with each other. However, not all collisions will result in a chemical reaction to form the products of the reaction. It is likely that the particles collide and bounce back without producing any changes. The collisions that are successful in producing a chemical reaction are called effective collisions. 4 Collisions of particles that are unsuccessful in producing a chemical reaction are called ineffective collisions. 5 The collision theory states that for a chemical reaction to occur, the reacting particles must (a) collide with each other so that the breaking and formation of chemical bonds can occur. SPM The Collision Theory ’05/P1 1 According to the kinetic theory of matter, all matter is made up of tiny, discrete particles. These particles are continually moving and so have kinetic energy. 2 Based on the assumption that the particles in matter are moving all the time and collide with each other, the collision theory was introduced to explain (a) how chemical reactions occur, and (b) the factors (such as particle size, concentration, temperature, catalyst and pressure) affecting the rates of reactions. 3 When the reactants are mixed, the reactant Rate of Reaction Temperature Time (°C) (s) (a) Sketch the graphs of total volume of hydrogen released against time for Experiments I, II and III on the same axes. (b) Explain why the reaction rate for (i) Experiment I is different from that of Experiment II, (ii) Experiment II is different from that of Experiment III. (c) What conclusion can you make by comparing Experiments III and IV? (d) The reaction mixture in Experiment III is filtered. Excess sodium hydroxide is added to the filtrate. (i) Predict what you would observe. (ii) Write an ionic equation for the reaction. (e) Based on your answer in (d), what inference can you make with regards to the property of a catalyst? The experiment is repeated at 35 °C and 40 °C. The experimental results are shown below. II Substances 322 5 (a) If two molecules with sufficient energy (that is, energy equal to or more than the activation energy) collide in the correct orientation, the chemical bonds in the reactant molecules will break and reaction will occur to form new bonds in the product molecules (Figure 1.31). For example, Activation Energy 1 Activation energy is the minimum energy that the reactant particles must possess at the time of collision in order for a chemical reaction to take place. 2 The activation energy can also be considered as an energy barrier that must be overcome by the colliding particles in order that collision will result in the formation of product molecules. 3 Figure 1.29 shows the energy profile diagram for the exothermic reaction: A + B → C + D. An exothermic reaction is the reaction that releases heat energy (Refer Section 4.1). In the energy profile diagram, the y-axis represents the energy content of the reactants and products, while the x-axis represents the progress of the reaction (reaction coordinate). H2(g) + I2(g) 2HI(g) Figure 1.31 Effective collision (sufficient energy and correct orientation) (b) However, if two molecules, with energy equal to or more than the activation energy, but collide with each other in an incorrect orientation, then reaction will not occur. (c) If two reactant molecules, with energy less than the activation energy, collide in the correct orientation, then reaction will also not occur. The colliding molecules will simply rebound and move away from each other. SPM ’07/P2 The energy of the products is lower than the energy of the reactants. Therefore heat is released during the reaction. Relating the Frequency of Effective Collisions with Factors Influencing the Rate of Reaction Figure 1.29 Energy profile diagram for an exothermic reaction 4 Figure 1.30 shows the energy profile diagram for the endothermic reaction: E + F → G + H. An endothermic reaction is a reaction that absorbs heat energy. 1 Based on the collision theory, two important factors that determine the rate of a chemical reaction are (a) the frequency of effective collisions and (b) the magnitude of the activation energy. 2 Frequency of effective collisions For a given reaction, if the frequency of collisions between the reactant molecules is high, it follows that the frequency of effective collisions that causes a reaction to occur will also be high. As a result, the rate of reaction increases. 3 Magnitude of activation energy Reactions that have high activation energy will occur at a slow rate. This is because only a small fraction of the molecules possess sufficient energy to overcome the activation energy for the reaction to occur. In contrast, reactions that The energy of the products is higher than the energy of the reactants. Therefore heat is absorbed during the reaction. Figure 1.30 The energy profile diagram for an endothermic reaction 323 Rate of Reaction 1 (b) possess energy that is equal to, or more than the minimum energy called the activation energy. (c) collide in the correct orientation. Effect of Concentration on the Rate of Reaction possess low activation energy will occur at a high rate. This is because most of the molecules have sufficient energy to overcome the activation energy and this enables the reaction to occur. 4 In general, any factor that increases the rate of effective collisions will also increase the rate of reaction. 1 Magnesium reacts with dilute hydrochloric acid as represented by the equation Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) When the concentration of hydrochloric acid increases, the rate of reaction also increases. 2 Figure 1.33 shows the arrangement of particles in 1 mol dm–3 hydrochloric acid and 2 mol dm–3 (more concentrated) hydrochloric acid. Effect of Size of Reactant (Surface Area) on the Reaction Rate 1 1 The sodium chloride crystal as shown in Figure 1.32(a) has a total surface area of 16 cm2. When this crystal is divided into smaller crystals as shown in Figure 1.32(b), the total surface area is increased to 24 cm2. Total surface area of the NaCl crystal in Figure 1.32(a) = (1  2)  4 + (2  2)  2 = 16 cm2 Figure 1.33 Arrangement of particles in dilute and concentrated solutions When the concentration of hydrochloric acid increases, the number of particles per unit volume also increases and the particles are closer together. 3 When the number of particles increases, the frequency of collisions also increases. As a result, the frequency of effective collisions increases. This causes the rate of reaction to increase. Total surface area in Figure 1.32(b) = (1  1)  6  4 = 24 cm2 5 ’05 The decomposition of hydrogen peroxide produces oxygen gas. Curve P is obtained when 25 cm3 of 0.1 mol dm–3 hydrogen peroxide undergoes decomposition. Figure 1.32 The smaller the particle size, the greater the total surface area exposed for reaction to occur. 2 Dilute sulphuric acid reacts with magnesium as represented by the equation Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g) If small pieces of magnesium or magnesium powder are used, the rate of reaction between magnesium and sulphuric acid will increase. 3 The smaller the size of the solid, the larger the total surface area exposed for collisions. This means that the frequency of effective collisions (that is, collisions with the correct orientation and with energy equal to or greater than the activation energy) between reacting particles will increase. As a result, the rate of reaction also increases. Rate of Reaction If the experiment is repeated using another hydrogen peroxide solution, which solution will produce curve Q? A 10 cm3 of 0.25 mol dm–3 hydrogen peroxide B 15 cm3 of 0.15 mol dm–3 hydrogen peroxide C 20 cm3 of 0.20 mol dm–3 hydrogen peroxide D 25 cm3 of 0.15 mol dm–3 hydrogen peroxide Comments • Steepness of curves P and Q Curve Q is more steep than curve P. This means 324 that the rate of reaction is higher. That is, the hydrogen peroxide used for curve Q is more concentrated. • Maximum volume of oxygen gas produced in curve Q is less than that in curve P. This means that the number of moles of hydrogen peroxide used in curve Q is less than that in curve P. Number of moles of hydrogen peroxide concentration (mol dm–3)  volume (cm3) =— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — 1000 Relative concentration Number of moles of H2O2 For curve P A B C D 0.1 mol dm–3 More concentrated More concentrated More concentrated More concentrated 2.5  10–3 2.5  10–3 2.25  10–3 4.0  10–3 3.75  10–3 An increase in pressure will not increase the speed of the reacting particles. In actual fact, the increase in rate at high pressures is caused by the particles being squeezed closer together. This increases the frequency of effective collisions and hence the rate increases. Effect of Temperature on the Rate of Reaction 1 Calcium carbonate reacts with hydrochloric acid to form carbon dioxide as represented by the following equation CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) Answer B Figure 1.35 shows the graphs of total volume of carbon dioxide given off against the time taken for the reaction between calcium carbonate and dilute hydrochloric acid at 25 °C and 30 °C. Effect of Pressure on the Rate of Reaction 1 In chemical reactions involving gases, increasing the pressure increases the rate of reaction. Conversely, decreasing the pressure decreases the rate of reaction. For example, the rate of reaction between nitrogen and oxygen to produce nitrogen monoxide can be increased by increasing the pressure. SPM ’05/P2 ’06/P1 N2(g) + O2(g) → 2NO(g) Figure 1.35 Effect of temperature on the rate of reaction 2 At low pressures, the gaseous molecules are spread far apart (Figure 1.34(a)). When the pressure is increased, the volume of the gas is reduced (Figure 1.34(b)). 2 At low temperatures, particles of reactants move at a slower speed. However, when the temperature is increased, the particles absorb the heat energy. As a result, the kinetic energy of the particles increases. Hence, (a) the reacting particles move faster, and (b) the number of reacting particles with the activation energy required for the reaction increases. 3 Consequently, the frequency of effective collisions increases and hence, the rate of reaction also increases. 4 Temperature has a great effect on the rate of reaction. For most reactions, the rate of reaction approximately doubles when the temperature of reaction increases by 10 °C. Figure 1.34 Effect of pressure on gaseous molecules This means that at high pressures, (a) the number of gaseous molecules per unit volume is increased, and (b) the gaseous molecules are packed closer together. 325 Rate of Reaction 1 Solution 3 As a result, increasing the pressure causes the gaseous molecules to collide more frequently. Consequently, the frequency of effective collisions increases and the rate of reaction also increases. Effect of Catalysts on Reaction Rates 4 A catalyst provides an alternative reaction route (or pathway) for the reaction to occur. In the presence of a positive catalyst, this alternative route has a lower activation energy. In other words, a positive catalyst lowers the activation energy required for the reaction (Figure 1.37). As a result, more reacting particles possess sufficient energy to overcome the lower activation energy required for effective collisions. Hence, the frequency of effective collisions increases and the rate of reaction increases. 1 The decomposition of hydrogen peroxide to water and oxygen occurs very slowly at room temperature. 2H2O2(aq) → 2H2O(l) + O2(g) In the presence of a catalyst, the decomposition of hydrogen peroxide occurs rapidly. 2 Figure 1.36 shows the rate of evolution of oxygen for the decomposition of hydrogen peroxide without a catalyst and in the presence of a catalyst such as manganese(IV) oxide or iron(III) oxide. Ea = Activation energy without catalyst Ea’ = Activation energy with catalyst Ea’ < Ea 1 Figure 1.37 Effect of catalyst on the activation energy of a reaction Figure 1.36 The effect of a catalyst on the decomposition of hydrogen peroxide 3 A chemical reaction occurs when reactant particles collide with one another. In the presence of a catalyst, the reactant particles can collide with the catalyst and also with each other. This causes the reactants to react in a different way. Thus, the activation energy of the reaction can be increased or decreased depending on the type of catalyst used. Effective collisions: Collisions that produce a reaction. Reaction rate increases when the effective collisions increase. Enzymes are biological catalysts. Enzymes are protein molecules produced in living cells. The enzyme, catalase, is found in the liver. This enzyme can catalyse the decomposition of hydrogen peroxide (a toxic substance produced by metabolism in human bodies) to harmless substances, that is, water and oxygen. Enzymes are also used in detergents to remove protein stains (for example, food or bloodstains) on clothings. Collision theory: a reaction only occurs if the particles (a) have sufficient energy to overcome the activation energy and (b) collide in the correct orientation. Activation energy: the minimum energy the reacting substances must possess before reaction can occur. is used to explain the following factors particle size When the particle size is decreased, the total surface area exposed for reaction increases. concentration/pressure When the concentration/ pressure is increased, the number of particles per unit volume increases. temperature When the temperature is increased, the number of particles with the activation energy required increases. Frequency of effective collisions increases Rate of Reaction 326 catalyst A catalyst will lower the activation energy required for the reaction by providing an alternative route with a lower activation energy. Rate of reaction increases 6 ’03 Experiment Experimental set-up Time taken for all the zinc to dissolve (s) Temperature I II 30 12 35 °C 35 °C By using collision theory, explain why the time taken for Experiment II is different from that of Experiment I. Comments In these two experiments, the constant variables are • volume, concentration and temperature of sulphuric acid used, • mass and surface area of zinc used. Solution The time taken for Experiment II is shorter. This implies that the reaction for Experiment II is faster. Thus, copper(II) sulphate acts as the catalyst. • If a catalyst is added, the rate of reaction increases because the catalyst provides an alternative route with a lower activation energy for the reaction to occur. • Hence, the minimum energy required for the reaction is less. As a result, more reacting particles possess sufficient energy to overcome the lower activation energy required for effective collisions. Hence, the frequency of effective collisions increases and the rate of reaction increases. 1.3 3 Four experiments were carried out to study the rate of reaction between nitric acid and calcium carbonate of different sizes. In Experiment I, V cm3 of 1.0 mol dm–3 nitric acid is added to big lumps of excess calcium carbonate in a conical flask. The total volume of carbon dioxide produced is plotted against time taken (Figure 1.38). 1 Which of the following changes will increase the rate of reaction between sodium thiosulphate solution and sulphuric acid? I By using sulphuric acid of a higher concentration II By increasing the temperature of sodium thiosulphate solution used III By increasing the volume of sodium thiosulphate used IV By adding a small amount of sodium hydroxide solution to the reaction mixture Explain your answer in terms of the collision theory. 2 An experiment is carried out to study the decomposition of hydrogen peroxide. In this experiment, 2.0 g of manganese(IV) oxide is added to 30 cm3 of 0.2 mol dm–3 hydrogen peroxide. The volume of oxygen produced is recorded at 30-second intervals. (a) Calculate the maximum volume of oxygen gas produced in the experiment at room conditions. [1 mol of gas occupies 24 dm3 at room conditions] (b) (i) Sketch a graph of volume of oxygen produced against time. (ii) Explain the shape of the graph. (c) (i) What is the function of manganese(IV) oxide in this experiment? (ii) Based on collision theory, explain the effect of manganese(IV) oxide on the decomposition of hydrogen peroxide. Figure 1.38 (a)(i) What is the difference between the rate of reaction at the first minute and the rate of reaction at the second minute? (ii) Explain this difference in terms of collision theory. (b) The experiment was repeated three times by changing the reaction conditions for each experiment as shown below. 327 Rate of Reaction 1 Two experiments were carried out to study the rate of reaction between zinc and excess sulphuric acid at room temperature. The table below shows the results of the experiments. Experiment II Nitric acid at a lower temperature III Small lumps of calcium carbonate but of the same mass IV 2.0 mol dm–3 nitric acid but of the same volume (V cm3) 1 1.4 (i) Sketch the graphs for Experiments II, III and IV on Figure 1.38 and label each of these graphs. (ii) Explain the difference between the reaction rates for Experiments I and II in terms of collision theory. Change in conditions of reaction 4 With the aid of an energy profile diagram, explain how a negative catalyst (inhibitor) affects the rate of reaction. 4 Nowadays, enzymes are used extensively in industry to enable reactions to proceed rapidly at room temperature and pressure. 5 In our daily living, we face many social and environmental issues that threaten the quality of living. For example, air and water pollution, food shortages, diseases and so on. 6 We must use science and technology to overcome these problems in a rational and systematic way so that the quality of life can be improved. 7 We should be thankful for the contribution of scientists in enhancing the quality of life in modern living. Practising Scientific Knowledge to Enhance Quality of Life 1 In our homes, we require machines to increase the rates of reactions and machines to reduce the rates of reactions. Examples of such machines are microwave ovens and refrigerators respectively. 2 In the hospitals, oxygen tents are used to save lives. The high concentration of oxygen helps patients with difficulty in breathing to breathe normally. 3 In human bodies, enzymes (biological catalysts) are needed to catalyse complex biochemical reactions. 1 The rate of reaction is defined as the amount of a reactant used up or the amount of a product obtained per unit time. 2 The rate of a reaction is inversely proportional to the time taken for the reaction. 3 Different chemical reactions take place at different rates. A fast reaction takes a shorter time to complete than a slow reaction. 4 The rate of reaction can be determined in the school laboratory by measuring the • changes in the mass of the reactant, • changes in volume of gas produced, • time taken for formation of precipitate. 5 The rate of reaction can be expressed in terms of (a) the average rate of reaction over a period of time, or (b) the instantaneous rate of reaction at a given time. 6 The rate of a reaction is affected by the following factors: Rate of Reaction 7 8 9 10 328 • Particle size (surface area) of solid reactant • Concentration of reactants • Temperature of reactants • Presence of catalyst • Pressure (for reactions involving gases) Transition metals (Fe, Ni and Pt) and compounds of transition metals (MnO2 and V2O5) are often used as catalysts. According to the collision theory, a reaction will occur if the reacting particles • collide with each other • possess activation energy • collide in the correct orientation The collisions that are successful in producing a chemical reaction are called effective collisions. Any factor that increases the rate of effective collisions will also increase the rate of reaction. 1 Multiple-choice Questions 1.1 Rate of Reaction 1 Calcium carbonate reacts with dilute hydrochloric acid to produce carbon dioxide. The total volume of carbon dioxide collected during the reaction is ’11 shown below. Time (s) 0 5 10 15 20 25 30 35 4 The diagram below shows the apparatus set-up used to determine the rate of reaction. 40 The apparatus set-up is not suitable to be used for determining the rate of reaction for A Na2SO3(s) + H2SO4(aq) → Na2SO4(aq) + H2O(l) + SO2(g) What is the overall average rate of reaction? A 0.825 cm3 s–1 C 1.100 cm3 s–1 3 –1 B 0.943 cm s D 1.300 cm3 s–1 2 The average rate of reaction of calcium carbonate with hydrochloric acid is 0.0080 mol s–1. What is the time taken for 9.60 g of calcium carbonate to completely react with excess hydrochloric acid? [Relative atomic mass: C, 12; O, 16; Ca, 40] A 12 s B 24 s C 120 s D 240 s 3 Calcium carbonate reacts with dilute hydrochloric acid to produce carbon dioxide gas. The plot of volume of CO2 produced against time is shown as follows. 1 Volume of 0.00 12.00 20.00 26.50 31.00 32.50 33.00 33.00 33.00 CO2 (cm3) B Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) C NaHCO3(s) + HNO3(aq) → NaNO3(aq) + H2O(l) + CO2(g) MnO2(s) D 2H2O2(aq) ⎯⎯⎯→ 2H2O(l) + O2(g) 5 The graph shows the total volume of carbon dioxide evolved when 10.0 g of calcium carbonate (in excess) reacts with 20.0 cm3 of 1.0 mol dm–3 dilute hydrochloric acid. volume of gas(cm3) Z Y X 0 Which of the following can be deduced from the graph? I The average rate of reaction is 1.5 cm3 s–1. II The rate of reaction decreases with time. III The rate of reaction at 35 seconds is zero. IV The gradient of the curve decreases because the concentration of acid decreases. A I and II only C I, II and III only B III and IV only D II, III and IV only 329 t1 TC 55 t2 t3 time(s) Which of the following statements is correct? A The reaction is faster at point Y than at point X. B The reaction is fastest at point Z. C The reaction reaches completion at time t3. D The total volume of carbon dioxide evolved is the same if 12.0 g of calcium carbonate is used. Rate of Reaction 6 Two experiments on the decomposition of hydrogen peroxide were carried out. The graphs in the following diagram show the total volume of oxygen collected against time for each of the experiments. The rate of reaction is best determined by A measuring the volume of SO2 produced at regular time intervals. B measuring the concentration of hydrochloric acid at regular time intervals. C recording the time as soon as the ‘cross’ mark disappears. D recording the time as soon as precipitate appears. 1.2 Factors that Affect the Rate of Reaction 8 Ammonia is produced using the Haber process. The table shows the mass of ammonia produced at four conditions of temperature and pressure. 1 ’04 Which of the following graphs shows how the rates of reaction vary with time for the experiments? A Condition Mass of ammonia produced (kg) Time taken I II III IV 300 250 150 200 4 hours 2 ½ hours 8 minutes 12 minutes At which condition is the rate of production of ammonia the highest? A Condition I C Condition III B Condition II D Condition IV 9 Zinc powder reacts with hydrochloric acid according to the equation: B Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) In order to have the highest initial rate, which of the following solutions should be used for the reaction with zinc powder? A 30 g of HCl in 1000 cm3 of water C 15 g of HCl in 100 cm3 of water B 20 g of HCl in 1000 cm3 of water D 4.0 g of HCl in 50 cm3 of water 10 Two experiments were carried out at 25 °C to study the rate of reaction between magnesium carbonate powder (in excess) and an acid. The volume of carbon dioxide liberated was measured at regular intervals. C D Experiment Acid used I 100 cm3 of 0.5 mol dm–3 hydrochloric acid II 100 cm3 of 0.5 mol dm–3 sulphuric acid Which of the following graphs represents the results obtained in Experiments I and II? A C 7 The diagram shows the apparatus set-up for an experiment to ’07 determine the rate of reaction between sodium thiosulphate and hydrochloric acid D B S2O32–(aq) + 2H+(aq) → H2O(l) + SO2(g) + S(s) Rate of Reaction 330 Temperature (°C) Concentration (mol dm–3) Form of zinc A 30 0.5 Small pieces B 25 1.0 Powder C 35 1.0 Small pieces D 35 1.0 Powder 12 For the following reaction: Zn + H2SO4 → ZnSO4 + H2 , which factor does not affect the rate of reaction? ’08 A Surface area of zinc C Volume of sulphuric acid B Concentration of sulphuric acid D Temperature of sulphuric acid 13 Graph X in the diagram below shows the result of the decomposition of 10 cm3 of 0.4 mol dm–3 hydrogen peroxide. The experiment was carried out at 30 °C. Which of the following conditions produces graph Y if 0.1 g of manganese(IV) oxide was used as the catalyst for both experiments? Volume of H2O2 (cm3) Concentration of H2O2 (mol dm–3) Temperature (°C) A 10 0.25 30 B 12.5 0.40 30 C 20 0.25 30 D 20 0.40 40 14 The graph in the diagram below shows the changes in the concentration of hydrogen peroxide, H2O2, when powdered manganese(IV) oxide is ’04 added to it. The gradients of the graph at times t1 and t2 are different because the I concentration of hydrogen peroxide decreases. II volume of hydrogen peroxide decreases. 331 III temperature of hydrogen peroxide decreases with time. IV mass of manganese(IV) oxide decreases. A I only B I and II only C III and IV only D I, III and IV only 15 The following equation shows the reaction between powdered ’03 calcium carbonate and dilute hydrochloric acid CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) The production of carbon dioxide can be slowed down by I reducing the temperature of hydrochloric acid used. II adding distilled water to hydrochloric acid before the reaction. III using larger pieces of calcium carbonate. IV reducing the pressure on the reaction mixture. A I, II and III only B I, III and IV only C I, II and IV only D I, II, III and IV 16 A piece of magnesium ribbon is allowed to react with 100 cm3 of 1.0 mol dm–3 hydrochloric acid. Which of the following changes will increase the rate of reaction? I Increasing the temperature of hydrochloric acid. II Replacing the magnesium ribbon with magnesium powder. III Replacing the acid with 50 cm3 of 2.0 mol dm–3 hydrochloric acid. IV Adding 50 cm3 of 1.0 mol dm–3 hydrochloric acid. A I and II only B III and IV only C II and IV only D I, II and III only 17 What is the most suitable method for cooking 100 g of ’06 potatoes within a short time? A Steam the potatoes in a steamer B Fry the potatoes in a copper pot Rate of Reaction 1 11 2.5 g of zinc were allowed to react with 100 cm3 of hydrochloric acid under different conditions as shown below. Under which conditions will hydrogen gas be given off at the highest rate? C Boil the potatoes in a saucepan D Boil the potatoes in a pressure cooker 1 18 Iron(III) oxide is a brown solid and iron(III) salts are brown in colour. When iron(III) oxide is added to hydrogen peroxide solution in a test tube, a fast reaction occurs and oxygen gas is liberated. What is left in the test tube at the end of the reaction? A A brown solution only. B A brown solid and a brown solution. C A brown solid and a colourless solution. D A white solid and a colourless solution. 19 The diagram below shows the total volume of carbon dioxide given off when dilute hydrochloric acid reacts with calcium carbonate powder. As the reaction proceeds, the gradient of the graph becomes less steep because I the mass of calcium carbonate decreases. II the total surface area of calcium carbonate decreases. III the volume of hydrochloric acid decreases. IV the temperature of the mixture increases. A I and II only B I, II and III only C II, III and IV only D I, III and IV only 20 Which of the following reactions takes place in the Ostwald process? A N2(g) + 3H2(g) 2NH3(g) B 2SO2(g) + O2(g) 2SO3(g) C 2NH3(g) + O2(g) 2NO(g) + 3H2(g) D 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(l) Rate of Reaction 21 Which of the following pairs of catalyst and processes are correctly matched? Catalyst Process I Iron Manufacture of ammonia in the Haber process II Nickel Manufacture of nitric acid in the Ostwald process III Vanadium(V) oxide Manufacture of sulphuric acid in the Contact process IV Lead(IV) oxide Production of oxygen by the decomposition of hydrogen peroxide A I and II only B II and III only 22 Which of the following reactions require a catalyst to speed up the reaction? I 2H2O2 → 2H2O + O2 II 2SO2 + O2 → 2SO3 III SO3 + H2SO4 → H2S2O7 IV Na2S2O3 + H2SO4 → Na2SO4 + H2O + SO2 + S A I and II only B I and III only C I, III, and IV only D II, III and IV only 23 Which of the following reactions require a catalyst to speed up the reaction? I N2 + 3H2 → 2NH3 II SO3 + H2SO4 → H2S2O7 III 2H2O2 → 2H2O + O2 IV Na2S2O3 + 2HCl → 2NaCl + H2O + SO2 + S A I and II only B I and III only C I, III and IV only D II, III and IV only 24 The reaction between sulphuric acid and magnesium carbonate is carried out at different conditions. Which reaction is fastest? A B 332 C I, II and IV only D I, III and IV only C D 25 Magnesium ribbons of the same length are added separately to each of the following solutions of hydrochloric acid. In which solution will the magnesium ribbon disappear first? [The hydrochloric acid used is in excess] Volume Concentration Temperature of HCI of HCI of HCI (cm3) (mol dm–3) (°C) A 300 1.0 30 B 200 1.0 25 C 100 2.0 30 D 200 2.0 25 26 The energy profile diagram for an uncatalysed reaction is shown below. ∆H Ea A No change Decrease B Decrease No change C Decrease Decrease D Increase Increase 27 Carbon dioxide is produced when magnesium carbonate reacts with dilute hydrochloric acid. MgCO3(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) + CO2(g) Which of the following changes will increase the initial rate of carbon dioxide production? A Heat the reaction mixture B Increase the size of solid magnesium carbonate C Increase the volume of hydrochloric acid D Increase the pressure on the reaction mixture 28 Calcium carbonate is added to excess hydrochloric acid at 30 °C. The experiment is repeated at 40 °C. The volume of carbon dioxide released for each experiment is measured at room temperature and pressure. Which of the following graphs represents the results of these two experiments? A B C 1.3 The Collision Theory D 29 Three experiments are carried out to study the rate of decomposition of hydrogen peroxide by the catalyst, manganese(IV) oxide. In all these three experiments, the mass of the catalyst used is the same. The experimental results are shown in the following diagram. Solutions of hydrogen peroxide used Solution P: 50 cm3 of 2.0 mol dm–3 H2O2 Solution Q: 100 cm3 of 1.0 mol dm–3 H2O2 Solution R: 100 cm3 of 3.0 mol dm–3 H2O2 30 Based on the collision theory, what are the effects of a rise ’06 in temperature on the reactant particles? I The kinetic energy of the reactant particles increases. II The number of reactant particles per unit volume increases. III The frequency of collisions between reactant particles increases. IV The activation energy of the reactant particles increases. A I and II only B I and III only C II and IV only D III and IV only 31 When zinc powder is added to dilute sulphuric acid, gas bubbles ’11 are produced slowly. When a few drops of copper(II) sulphate are added to the reaction mixture, gas bubbles are produced vigorously. Which statement best explains the effect of copper(II) sulphate on the reaction? A It lowers the activation energy. B It increases the collision frequency between the reacting particles. C It increases the concentration of sulphate ions and hence increases the rate of reaction. D It causes the reacting particles to collide in the correct orientation. 32 The diagram below shows the energy profile for the following reaction Which of the following statements are correct? I The curve X is obtained using solution R. II The curve Y is obtained using solution P. III The curve Z is obtained using solution Q. IV The curve Z is obtained using solution R. A I and II only B II and IV only C I, II and III only D I, II and IV only 333 X(g) + Y(g) → Z(g) by by by by The curves, P and Q, represent two different paths for this Rate of Reaction 1 The reaction was repeated using a catalyst. What is the effect of the catalyst on the heat of reaction (∆H) and activation energy (Ea) for the reaction? reaction. What conclusion can be drawn based on the diagram? A The reaction by path P occurs at a higher temperature. B The reaction by path P occurs at a higher rate than by path Q. C The activation energy for path P is (x + y) kJ. D The activation energy for path Q is (x – y) kJ. 33 Consider the reaction between magnesium and dilute hydrochloric acid. 1 Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Which of the following will increase the frequency of collisions between the reactants? I Increase the concentration of hydrochloric acid. II Increase the temperature of reaction. III Use magnesium ribbon instead of magnesium powder. IV Remove the hydrogen gas produced. A B C D I and II only III and IV only I, II and III only II, III and IV only 34 The reaction between Fe3+ and SO32– is represented by the ’06 equation: 2Fe3++ SO32–+ H2O → 2Fe2+ + H2SO4 brown green It is found that the change of colour from brown to green occurs at a higher rate when the reaction mixture is heated. This is due to the I decrease in activation energy. II increase in the frequency of collisions between Fe3+ and SO32– ions. III increase in the kinetic energy of Fe3+ and SO32– ions. IV increase in the frequency of effective collisions. A I and II only B II and III only C I, III and IV only D II, III and IV only 35 Iron is used in the Haber process to manufacture ammonia from nitrogen and hydrogen. Why is iron used in this process? A To increase the rate of reaction between nitrogen and hydrogen B To absorb the smell of ammonia C To oxidise nitrogen to form ammonia D To increase the yield of ammonia 36 The rate of reaction between 1 mol dm–3 hydrochloric acid and 3 g of magnesium powder is higher than the rate of reaction between 1 mol dm–3 ethanoic acid and 3 g of magnesium powder. What is the explanation for this observation? A Hydrochloric acid is more soluble in water than ethanoic acid. B The kinetic energy of hydrochloric acid is higher than ethanoic acid. C Hydrochloric acid forms a soluble salt whereas ethanoic acid forms an insoluble salt. D The concentration of H+ ions in hydrochloric acid is higher than ethanoic acid. Structured Questions 1 Diagram 1 shows two experiments to investigate one factor that affects the rate of reaction between zinc ’06 and dilute sulphuric acid. Experiment II Experiment I Diagram 1 (a) What is the factor that affects the rate of reaction in Experiments I and II? [1 mark] (b) State two constant variables in both experiments. [2 marks] (c) The following equation represents the reaction that occurs in both the experiments. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) Rate of Reaction 334 recorded. The experiment is repeated using different acids as shown below. All experiments are carried out at the same temperature. Experiment Reactants I 50 cm of 1.0 mol dm HCl + 5.0 cm of magnesium ribbon t1 II 50 cm3 of 1.0 mol dm–3 CH3COOH + 5.0 cm of magnesium ribbon t2 (d) Graph 1 shows the results for both experiments. Graph 1 Based on Graph 1: (i) Which experiment has a higher rate of reaction? [1 mark] (ii) How do you come to this conclusion? [1 mark] (iii) Explain what happens after time t. [1 mark] (iv) Why are both curves at the same level after time t? [1 mark] 3 Time taken (s) –3 (i) Write the ionic equation for the reaction between magnesium and an acid. [1 mark] (ii) Which is shorter t1 or t2? Why? [3 marks] (c) Three experiments are carried out to investigate the factors that affect the rate of the following reaction: ’08 Mg + 2HCl → MgCl2 + H2 The conditions of this experiment are shown below. (e) State the conclusion for the experiments. [1 mark] (f) Another experiment was carried out using excess zinc powder and dilute sulphuric acid with different concentrations. Sketch the curve of concentration of dilute sulphuric acid against the time taken to collect a fixed quantiy of the product. [2 marks] 2 (a) Propane, C3H8, burns in excess oxygen to form carbon dioxide and water as represented by the equation Temperature (°C) Experiment Reactants I Excess magnesium powder + 25 cm3 1.0 mol dm–3 HCl 35 II Excess magnesium ribbon + 25 cm3 0.5 mol dm–3 HCl 25 III Excess magnesium ribbon + 25 cm3 1.0 mol dm–3 HCl 35 The results of this experiment are shown in Diagram 2. C3H8 + 5O2 → 3CO2 + 4H2O At time t, the rate of reaction of propane is 0.20 mol s–1. Calculate (i) the rate of consumption (using up) of oxygen, and (ii) the rate of production of carbon dioxide at time t. [2 marks] (b) 50 cm3 of 1.0 mol dm–3 of hydrochloric acid is poured into a conical flask. A piece of 5.0 cm magnesium ribbon is added to the acid. The time taken to dissolve the magnesium completely is Diagram 2 335 (i) Which curves (X, Y or Z) represent the results [3 marks] of Experiments I, II and III? Rate of Reaction 1 (i) Choose one of the products shown in the equation that is most suitably used to measure the rate of reaction. [1 mark] (ii) Give one reason for your answer in (i). [1 mark] (a) What is the average rate of reaction for Experiment II? [2 marks] (ii) Give one reason why the final volume of gas obtained in curve X is half the final volume of gas in curve Z. [1 mark] (b) Use the collision theory to explain why the time taken for Experiment II is shorter than for Experiment I. [3 marks] 3 Three experiments were carried out to investigate the factors influencing the rate of reaction. Three pieces of 0.12 g of magnesium ribbon are added separately to excess hydrochloric acid. The time taken for all the magnesium to dissolve is taken. Table 1 shows the results of the experiments. Experiment I II III Reactants 0.12 g Mg + excess HCl(aq) 0.12 g Mg + excess HCl(aq) 0.12 g Mg + excess HCl(aq) + CuSO4(aq) Temperature (oC) 30 35 35 Time (s) 60 32 10 (c) Explain why the time taken for Experiment III is longer than for Experiment II. [3 marks] (d) Suggest another method that can be used to increase the rate of decomposition of hydrogen in Experiment II. [1 mark] (e) Write a chemical equation for the catalytic decomposition of hydrogen peroxide. [1 mark] (f) Experiments II and III were allowed to continue until the decomposition of hydrogen peroxide was completed. Sketch a graph of total volume of gas against time for Experiments II and III on the same axes. [2 marks] 5 Dilute sulphuric acid reacts with sodium thiosulphate solution to produce sulphur. The presence of sulphur causes the solution to become cloudy. Five experiments were carried out to study the rate of reaction between dilute sulphuric acid and sodium thiosulphate solution. The reaction takes place in a conical flask placed over a white paper marked with a cross, ‘X’. The time taken for the cross to disappear from view was recorded. In each experiment, the volume and concentration of sodium thiosulphate solution were kept constant. The experimental results are shown in Table 3. 1 Table 1 (a) Write the chemical equation for the reaction between magnesium and hydrochloric acid. [1 mark] (b) Calculate the maximum volume of hydrogen gas produced in Experiment I. [2 marks] (c) Predict the maximum volume of hydrogen gas produced in Experiment III. Explain your answer. [2 marks] (d) Calculate the average rate of reaction in Experiment II. [1 mark] (e) (i) Which experiment has the highest rate? Justify your answer. (ii) Sketch the graphs for the volume of hydrogen gas against time for Experiments I, II and III on the same axes. [3 marks] [Relative atomic mass: Mg = 24; Molar gas volume, 24 dm3 at r.t.p.] Experiment 4 Three experiments were carried out to study the effect of iron(III) oxide, Fe2O3, on the rate of decomposition of 0.5 mol dm–3 hydrogen peroxide. Table 2 shows the mixtures of substances used and the time taken to collect 30 cm3 of the colourless gas given off in each experiment. Experiment Mixture of substances Time taken I 20.0 cm3 of 0.5 mol dm–3 H2O2 A few weeks II 20.0 cm3 of 0.5 mol dm–3 H2O2 + 0.2 g of Fe2O3 35 s III 20.0 cm3 of 0.5 mol dm–3 H2O2 + 25.0 cm3 of water + 0.2 g of Fe2O3 45 s Time taken (s) A 0.15 20 65 B 0.10 30 45 C 0.10 20 85 D 0.05 30 55 E 0.05 20 105 Table 3 (a) The reaction between dilute sulphuric acid and sodium thiosulphate (Na2S2O3) produces sulphur, sodium sulphate, sulphur dioxide and water. Complete the following equations: (i) Na2S2O3 + H2SO4 → ____ + ____ + ____ (ii) S2O32– + 2H+ → ____ + ____ + ____ [2 marks] (b) Table 2 Rate of Reaction Concentration Temperature of acid (°C) –3 (mol dm ) 336 (i) Which of the experiments shown above should be chosen to compare the effect of concentration of the acid on the rate of reaction? [1 mark] (ii) Give one reason why you chose these experiments. [1 mark] (iii) What conclusion can be made from the experiments in (i)? [1 mark] (c) (i) Which of the experiments should be chosen to compare the effect of temperature on the rate of reaction? [1 mark] (ii) Give one reason for your choice. [1 mark] (iii) What conclusion can be made from the experiments you have chosen in (i)? (a) Table 5 shows the conditions used for carrying out Experiment 1. Complete Table 5 to predict the conditions used for obtaining the results of Experiments 2, 3 and 4. In each case, state the constant variables and manipulated variables used and briefly explain your answer. Experi ment 1 Experi ment 2 Experi ment 3 Experi ment 4 6 Four experiments were carried out to investigate the decomposition of hydrogen peroxide to form water and oxygen gas. Volume of hydrogen peroxide (cm3) 40 40 … … MnO2 2H2O2(aq) ⎯⎯⎯→ 2H2O(l) + O2(g) Volume of water (cm3) 40 40 40 0 The total volume of oxygen evolved at one-minute intervals were recorded in Table 4. Temperature (°C) 30 30 32 30 Mass of MnO2 used (g) 1.0 … 1.0 1.0 Volume of oxygen gas released (cm3) Time (min) Experiment Experiment Experiment Experiment 1 2 3 4 0 0 0 0 0 1 18 0 25 34 2 33 0 35 60 3 36 0 37 69 4 37 0 38 75 5 38 0 38 76 6 38 0 38 76 Table 5 [5 marks] (b) Copper(II) oxide is a less effective catalyst than manganese(IV) oxide for the decomposition of hydrogen peroxide. If copper(II) oxide is used to replace manganese(IV) oxide in Experiment 1, what is the effect of this change on (i) the volume of oxygen collected at 1.0 minute? (ii) the volume of oxygen collected after the reaction has completed? Explain your answers. [2 marks] (c) ‘A catalyst remains chemically unchanged at the end of the reaction’. (i) What is meant by chemically unchanged? (ii) How would you prove that your answer in (i) is correct? [3 marks] Table 4 Essay Questions 1 (a) Two experiments are carried out to study the rate of reaction between iron and dilute acids. Experiment Reactants I 1.12 g of iron and 50 cm3 of 2.0 mol dm–3 sulphuric acid II 1.12 g of iron and 50 cm3 of 2.0 mol dm–3 hydrochloric acid The following graphs show the results of the experiments. 337 Rate of Reaction 1 [1 mark] (d) Explain your answer in (c)(iii) based on collision theory. [2 marks] Based on the graph: (i) Calculate the average rate of reaction for Experiment I. [2 marks] (ii) Explain the difference in the rate of reaction between Experiment I and Experiment II before 100 s. [6 marks] (c) Ethanedioic acid, H2C2O4, decolourises potassium manganate(VII) slowly at room temperature. The reaction is catalysed by manganese(II) sulphate, MnSO4. Describe how you would prove that the sulphate ions, SO42–, do not act as a catalyst in this reaction. [3 marks] (b) Describe an experiment to show that lead(IV) oxide is a more effective catalyst than copper(II) oxide for the decomposition of hydrogen peroxide. Your answer should include a labelled diagram on the apparatus set-up for the experiment. [12 marks] 4 (a) Two experiments are carried out to study the rate of reaction between zinc and two acids, R and T. ’07 The data for the experiments are shown below. ’07 Experiment Reactants Observation Products I Excess zinc and 25 cm3 of 1.0 mol dm–3 acid R The temperature of the reaction mixture increases Zinc sulphate and hydrogen II Excess zinc and 25 cm3 of 1.0 mol dm–3 acid T The temperature of the reaction mixture increases Zinc chloride and hydrogen 2 (a) Iron powder will dissolve in cold dilute hydrochloric acid while coarse iron filings do not dissolve until the acid is heated. Explain these observations. [7 marks] 1 (b) There is a high risk of explosions occurring in coal mines. Explain why this is so. [6 marks] (c) ‘Temperature is important in preserving food’. Give one example from your daily life to justify this statement. [4 marks] (d) When a drop of blood is added to hydrogen peroxide solution, a vigorous effervescence occurs. Explain this observation. [3 marks] 3 (a) (i) Define the term rate of reaction? [2 marks] (ii) At high temperatures and pressures, nitrogen reacts with hydrogen to form ammonia. N2(g) + 3H2(g) 2NH3(g) Use the collosion theory to explain how high pressure increases the rate of reaction between nitrogen and hydrogen. [5 marks] (b) Describe an experiment to demonstrate the effect of the concentration of sodium thiosulphate on the rate of reaction between hydrochloric acid and sodium thiosulphate. Draw the apparatus used for this experiment and state the hypothesis and variables in this experiment. [10 marks] (i) State the names of the acids used in experiments I and II. [2 marks] (ii) Write the chemical equation for the reaction that occurs in Experiment I. [2 marks] (iii) Draw the energy profile diagram for the reaction in Experiment II. On the energy profile diagram, show the • activation energy without catalyst, Ea, • activation energy with a catalyst, Ea’, • heat of reaction, ΔH. Explain the energy profile diagram. [10 marks] (b) Explain the factors that affect the rate of reaction in the following daily activities: (i) Combustion of charcoal (ii) Cooking food in a pressure cooker [6 marks] Experiments The experiment was repeated using sodium thiosulphate solutions at 30 °C, 35 °C, 40 °C and 45 °C. Diagram 1 shows the stopwatch readings for each of the experiments. 1 Sodium thiosulphate solution reacts with dilute sulphuric acid to produce a yellow precipitate of sulphur. 50 cm3 of 0.10 mol dm–3 sodium thiosulphate solution at 25 °C was measured into a 250 cm3 conical flask. The conical flask was placed on a white paper marked with the ‘X’ sign. 5 cm3 of 0.50 mol dm–3 sulphuric acid was added to the sodium thiosulphate solution and the mixture shaken. At the same time, the stopwatch was started. The time was taken as soon as the ‘X’ sign was no longer visible. Rate of Reaction 338 at 25 °C Time, t1 ______ s at 30 °C at 35 °C Time, t2 ______ s at 45 °C at 40 °C Time, t3 ______ s Time, t5 ______ s Time, t4 ______ s [3 marks] (b) State the variables in this experiment. Manipulated variable: Responding variable: Constant variable: [3 marks] 1 (c) Construct a table containing the information on temperature, time and ————— for the experiments. time [2 marks] 1 (d) (i) Draw a graph of temperature against ————— on a graph paper. time [4 marks] [3 marks] (ii) Based on the graph in (i), what conclusion can be drawn from this experiment? (e) Predict the time taken for the ‘X’ sign to disappear if the experiment is repeated at 50 °C. [3 marks] (f) State the hypothesis for this experiment. [3 marks] (g) Based on your hypothesis, explain why meat and fish are always kept in refrigerators. [3 marks] 2 An experiment was carried out to investigate the rate of reaction between granulated zinc and dilute hydrochloric acid. The results of the experiment are shown below. ’09 Time (s) 0 30 40 Burette reading 50.00 40.00 33.50 (cm3) …… …… 21.50 20.00 Volume of gas evolved (cm3) …… …… 28.50 30.00 32.50 32.50 0.00 10 20 10.00 16.50 50 60 70 80 17.50 17.50 Diagram 2 shows the burette readings at 30 seconds and 40 seconds respectively. 29 25 TC 56 28 24 At 30 s At 40 s Diagram 2 (a) Based on this experiment, what is meant by the rate of reaction? (b) Based on Diagram 2, what are the volumes of gas evolved at 30 seconds and 40 seconds? (c) State one conclusion, based on the experimental results. 339 [3 marks] [3 marks] [3 marks] Rate of Reaction 1 Diagram 1 (a) Record the readings of the stopwatch in the spaces provided in Diagram 1. FORM 5 THEME: Interaction between Chemicals CHAPTER 2 Carbon Compounds SPM Topical Analysis 2008 Year 1 Paper 2 A Section Number of questions 4 2009 1 B – 3 1 C – 2 A 1 2 2010 1 B 1 3 1 C – – 7 2011 2 3 A A C – 1 — 2 – 1 – 4 2 3 A B C – – 1 – ONCEPT MAP CARBON COMPOUNDS Organic carbon compounds: Produce CO2 and H2O on complete combustion Hydrocarbons (elements C and H only) Physical properties: Insoluble in water, low melting and boiling points, non-conductors of electricity Alkanes (saturated) Reactions: • Combustion • Substitution hydrogenation Alkenes (non-saturated) Reactions: • Combustion • Addition • Polymerisation Isomerism (same molecular formula, different structural formulae) Natural rubber (natural polymers) • Coagulation • Vulcanisation Non-hydrocarbons (Elements: C, H, O) hydration dehydration Alcohols Reactions: • Combustion • Oxidation • Dehydration • Esterification oxidation Carboxylic acids Reactions: • With metals/metal carbonates/alkalis to form salts • Esterification esterification Esters Physical properties: • Sweet fruity smell • Insoluble in water Fats and Oils • Fats: saturated, higher melting point • Oils: unsaturated, lower melting point 8 Almost all organic compounds contain the elements carbon and hydrogen. Hence the complete combustion of carbon compounds produces carbon dioxide and water. Carbon Compounds 1 Carbon compounds are compounds that contain the element carbon. 2 Carbon compounds can be classified into two groups: inorganic compounds and organic compounds. 3 Organic compounds are carbon compounds in which carbon is bonded to other elements by covalent bonds. Examples of organic compounds are hydrocarbons, alcohols, carboxylic acids and esters. 4 Most carbon compounds are derived from living organisms. Nowadays many organic compounds can be synthesised in laboratories. 5 Most inorganic compounds do not contain carbon. Examples of inorganic compounds containing the element carbon are carbonates, hydrogen carbonates, oxides of carbon and cyanides. 6 Carbon atom (proton number 6) has the electron arrangement: 2.4. Hence, each carbon atom can form four covalent bonds in organic compounds. 7 The covalent bonds in carbon compounds may be (a) single bond, (b) double bond or (c) triple bond. Carbon compounds Organic compounds Example: • Hydrocarbons • Alcohols • Carboxylic acids • Esters • Carbohydrates Inorganic compounds Example: • Hydrogen carbonates • Carbonates • Carbides • Oxides of carbon • Cyanides 2 2.1 Organic compounds are the largest group of chemicals we know today, numbering in thousands. All the food and medicines we consume are organic compounds as well as most of the synthetic products such as clothing and household materials. To investigate the products formed by complete combustion of organic compounds Apparatus Ethanol and palm oil, limewater, ice and water, anhydrous cobalt(II) chloride paper. Procedure Filter funnel, test tubes, delivery tubes, spirit lamp, suction pump and beaker. 1 A filter funnel is connected to the suction pump via test tubes A and B where test tube A is dipped in ice water and test tube B contains limewater. 2 A spirit lamp filled with ethanol is lit and placed under the filter funnel and the suction pump turned on. 3 The changes in test tubes A and B are noted. 4 The liquid collected in test tube A is tested with anhydrous cobalt(II) chloride paper. 5 Steps 1 to 4 are repeated using a spirit lamp with palm oil. Figure 2.1 Combustion of organic compounds 341 Carbon Compounds Activity 2.1 Materials Results Test tube Observation Inference A A colourless liquid is formed and it changes anhydrous cobalt(II) chloride paper from blue to pink The colourless liquid formed in test tube A is water B Limewater turns milky Carbon dioxide gas is produced Conclusion 1 The combustion of organic compounds such as ethanol and palm oil produces water and carbon dioxide. 2 During the combustion of an organic compound, (a) the carbon combines with oxygen to form carbon dioxide, (a) the hydrogen combines with oxygen to form water. excess oxygen (complete combustion), carbon dioxide and water are produced. 8 Incomplete combustion of hydrocarbons will produce water, carbon dioxide, carbon monoxide and carbon (as soot). 2 Hydrocarbons 1 Hydrocarbons are organic compounds that contain the elements carbon and hydrogen only. 2 Hydrocarbons that have only single covalent bonds between all the carbon atoms in the molecule are called saturated hydrocarbons. Example Propane H | H—C— | H H H | | C—C—H | | H H Hydrocarbon carbon-carbon single bond Saturated hydrocarbons have only single bonds between the carbon atoms. Examples: Ethane, propane 3 Hydrocarbons that have at least one carboncarbon double bond (C = C) or triple bond (C ≡ C) in the molecule are called unsaturated hydrocarbons. Example Propene H H H | | | H—C=C—C—H | H 1 carbon-carbon double bond SPM ’10/P1 The complete combustion of 0.1 mol of a hydrocarbon Z in excess oxygen produces 0.3 mol of carbon dioxide and 0.4 mol of water. Determine the molecular formula of hydrocarbon Z. 4 The main sources of hydrocarbons are: (a) Petroleum (crude oil) (b) Natural gas (c) Coal 5 Petroleum is a complex mixture of hydrocarbons. 6 Fractions of hydrocarbons are separated by a process called fractional distillation. The fractions are separated based on the difference in boiling points. The fractions with lower boiling points will be distilled off earlier. 7 Hydrocarbons contain carbon and hydrogen only. Thus when hydrocarbons are burnt in Carbon Compounds Unsaturated hydrocarbons have double or triple bonds between the carbon atoms. Examples: Ethene, propene Solution Since 0.1 mol of Z produces 0.3 mol of carbon dioxide and 0.4 mol of water, 1 mol of Z will produce 3 mol of carbon dioxide and 4 mol of water. C + O2 → CO2 The number of moles of carbon in 1 mol of Z = the number of moles of carbon dioxide = 3. 1 2H + —O → H2O 2 2 The number of moles of hydrogen in 1 mol of Z = 2 3 the number of moles of water = 2 3 4 = 8. Hence the molecular formula of Z is C3H8. 342 7 Table 2.1 shows the prefixes used to indicate the number of carbon atoms per molecule of an organic compound, the names and molecular formulae of the first ten alkanes. 2.1 1 Classify the following substances into organic compounds and inorganic compounds. Table 2.1 Prefixes used to indicate the number of carbon atoms, names and molecular formulae of alkanes 2 (a) What is meant by hydrocarbon? (b) State three sources of hydrocarbon. Number of carbon atoms per molecule Prefix Name of alkane 1 Meth Methane CH4 2 Eth Ethane C2H6 3 Prop Propane C3H8 4 But Butane C4H10 5 Pent Pentane C5H12 6 Hex Hexane C6H14 7 Hept Heptane C7H16 8 Oct Octane C8H18 9 Non Nonane C9H20 10 Dec Decane C10H22 3 State two main products that are formed when rubber is burnt in excess air. Explain your answer. 2.2 Alkanes 1 Alkanes are saturated hydrocarbons with the general formula Cn H2n+2, where n = 1, 2, 3… 2 Alkanes are called saturated hydrocarbons SPM because the molecules contain only single ’11/P1 covalent bonds between carbon atoms. 3 The molecular formula is a chemical formula that shows the actual number of atoms of each element present in one molecule of the substance. 4 The molecular formula of an alkane can be obtained by substituting n in the general formula Cn H2n+2 with the number of carbon atoms. 5 In the naming of alkanes according to the IUPAC system, all members of the alkane series have their names ending with -ane (IUPAC is the abbreviation for International Union of Pure and Applied Chemistry). 6 The first part (prefix) of the name of an alkane depends on the number of carbon atoms in the molecule. Molecular formula 8 The structural formula of an organic compound is the chemical formula that shows the arrangement of atoms and covalent bonds between atoms in a molecule of the compound. 9 Note that when writing the structural formula of alkanes, (a) each carbon atom should have four single covalent bonds. (b) each hydrogen atom should have one single covalent bond. (c) the carbon atoms are connected by single bonds. 10 The structural formulae of the first ten straight chain alkanes are shown as follows: Table 2.2 The structural formulae of the first ten straight chain alkanes H | H–C–H | H methane (CH4) H H | | H–C – C–H | | H H ethane (C2H6) H H H | | | H–C–C–C–H | | | H H H propane (C3H8) H H H H | | | | H–C–C–C–C–H | | | | H H H H butane (C4H10) H H H H H | | | | | H–C–C–C–C–C–H | | | | | H H H H H pentane (C5H12) H H H H H H | | | | | | H–C–C–C–C–C–C–H | | | | | | H H H H H H hexane (C6H14) 343 Carbon Compounds 2 Rubber, sugar, limestone, carbon dioxide, calcium carbonate, sand, sodium chloride, vinegar, polyvinylchloride, urea, ammonium sulphate. H H H H H H H | | | | | | | H–C–C–C–C–C–C–C–H | | | | | | | H H H H H H H heptane (C7H16) H H H H H H H H | | | | | | | | H–C–C–C–C–C–C–C–C–H | | | | | | | | H H H H H H H H octane (C8H18) H H H H H H H H H | | | | | | | | | H–C–C–C–C–C–C–C–C–C–H | | | | | | | | | H H H H H H H H H nonane (C9H20) H H H H H H H H H H | | | | | | | | | | H–C–C–C–C–C–C–C–C–C–C–H | | | | | | | | | | H H H H H H H H H H decane (C10H22) Physical Properties of Alkanes 1 On going down the alkane series, the physical properties change gradually as shown in Table 2.3. 2 Table 2.3 Physical properties of alkanes Name of alkane Molecular formula Relative molecular Melting point Boiling point mass (°C) (°C) Physical state Density (g cm–3) Methane CH4 16 –182 –162 Gas — Ethane C2H6 30 –183 –89 Gas — Propane C3H8 44 –188 –42 Gas — Butane C4H10 58 –138 –0.5 Gas — Pentane C5H12 72 –130 36 Liquid 0.626 Hexane C6H14 86 –95 69 Liquid 0.659 Heptane C7H16 100 –90 98 Liquid 0.684 Octane C8H18 114 –57 126 Liquid 0.703 Nonane C9H20 128 –54 151 Liquid 0.718 Decane C10H22 142 –30 174 Liquid 0.730 2 Melting and boiling points of alkanes (a) Alkanes exist as simple covalent molecules. Alkanes have low melting and boiling points because of the weak van der Waals forces between molecules. Little energy is required to overcome the weak forces of attraction. (b) When the number of carbon atoms per molecule of alkane increases, the relative molecular mass increases and the melting point and boiling point increase. This is because the larger the molecular size, the stronger the van der Waals forces of attraction between the molecules. 3 The physical states of alkanes The first four members are gases as their boiling points are below room temperature (25 °C). The alkanes from C5 Carbon Compounds Figure 2.2 The boiling points of alkanes increase with relative molecular mass 344 to C18 are liquids with the rest of the alkanes being solids. 4 Densitites of alkanes Alkanes are less dense than water. The density of alkanes increases gradually down the series as the molecular mass increases. 5 Solubility of alkanes (a) All alkanes are insoluble in water. When liquid alkane is shaken with water, two separate layers of liquids are formed. (b) Alkanes are soluble in organic solvents such as propanone. 6 Electrical conductivity of alkanes All alkanes do not conduct electricity because they are covalent compounds, consisting of molecules. SPM ’05/P1 CH4 + Cl2 → CH3Cl + HCl chloromethane CH3Cl + Cl2 → CH2Cl2 + HCl dichloromethane CH2Cl2 + Cl2 → CHCl3 + HCl trichloromethane CHCl3 + Cl2 → CCl4 + HCl tetrachloromethane Chemical Properties of Alkanes H Cl Cl | Cl2 | Cl2 | H — C — H ⎯→ H — C — H ⎯→ H — C — Cl | | | H H H CH3Cl CH2Cl2 1 Reactivity of alkanes Alkanes are saturated hydrocarbons and are less reactive compared to unsaturated hydrocarbons. 2 Combustion of alkanes (a) Alkanes undergo complete combustion in the presence of excess air or oxygen to produce carbon dioxide and water. SPM For example ’05/P1, CH4 + 2O2 → CO2 + 2H2O ’11/P2 1 H substituted 2 H substituted Cl Cl | Cl2 | ⎯→ H — C — Cl ⎯→ Cl — C — Cl | | Cl Cl CHCl3 CCl4 Cl2 methane 2C2H6 + 7O2 → 4CO2 + 6H2O ethane 3 H substituted (b) The combustion of alkanes is highly exothermic, that is, produces a lot of heat energy. Hence alkanes are used as fuels. (c) Incomplete combustion of alkanes will produce carbon (black smoke), carbon monoxide and water. (d) The larger the molecular size of the alkane molecule, (i) the smokier or sootier the flame, (ii) the more heat produced on complete combustion. 3 Substitution reactions SPM (a) When a mixture of alkane and chlorine ’10/P1 is exposed to sunlight or ultraviolet light, substitution reaction occurs slowly and a mixture of organic compounds and hydrogen chloride is produced. (b) In a substitution reaction, the hydrogen atoms in an alkane are replaced gradually by chlorine atoms. For example The reaction between methane and chlorine 4 H substituted The Effects of Methane on Everyday Life 1 Methane, commonly known as natural gas, is used as a fuel. 2 It is produced by the anaerobic decay of plants and organic matter by bacteria. Hence methane is found in landfills and peat swamps. 3 Methane is a greenhouse gas. It can trap radiation energy from the sun and also contribute to global warming. Methane is also known as marsh gas because it is found in marshes and stagnant ponds. It can cause fires in garbage landfills and peat swamps. 345 Carbon Compounds 2 In the substitution reaction of CH4 with Cl2, the carbon atom still has four covalent bonds, either bonded to H atom or Cl atom. 9 The names and molecular formulae of the first nine members of the alkene series are shown in Table 2.4. 2.2 1 Name and give the molecular formula of an alkane with (a) three carbon atoms (b) five carbon atoms (c) six carbon atoms Table 2.4 The names and molecular formulae of the first nine members of the alkene series Number of carbon atoms per molecule Prefix Name of alkene 2 Eth Ethene C2H4 3 Ethane reacts with chlorine under certain conditions to form chloroethane. (a) State the condition for reaction to take place. (b) Name the reaction that takes place. (c) Write an equation for the reaction that occurs. 3 Prop Propene C3H6 4 But Butene C4H8 5 Pent Pentene C5H10 4 A saturated hydrocarbon X has a relative molecular mass of 58. Identify X. [Relative atomic mass: H, 1; C, 12] 6 Hex Hexene C6H12 7 Hept Heptene C7H14 8 Oct Octene C8H16 9 Non Nonene C9H18 10 Dec Decene C10H20 2 2 Petrol used in cars consists mainly of octane. (a) Give the molecular formula of octane. (b) Write an equation for the complete combustion of octane. 2.3 Alkenes 1 Alkenes are hydrocarbons with the general formula CnH2n, where n = 2, 3, 4…. 2 Alkenes are unsaturated hydrocarbons and contain at least one double bond between carbon atoms. 3 The functional group in alkenes is the carboncarbon double bond (C=C). 4 The molecular formula of an alkene can be obtained by substituting n in the general formula CnH2n with the number of carbon atoms. 5 In the naming of alkenes according to the IUPAC system, all members of the alkene series have their names ending with -ene. 6 The first part (prefix) of the name of an alkene depends on the number of carbon atoms in the molecule. 7 The first member of the alkene series is ethene, C2H4 because there must be a minimum of two carbon atoms to form a carbon-carbon double bond (C=C). Hence methene does not exist. 8 Note that when writing the structural formula of alkenes, (a) there is a carbon-carbon double bond (C=C) in the chain. (b) each carbon atom forms four bonds (four single bonds or one double bond + two single bonds). (c) each hydrogen atom should have one single covalent bond. Carbon Compounds Molecular formula 10 The structural formulae of the first nine straight chain alkenes with one terminal double bond (double bonds at the end of a chain) are shown as follows: H H | | H–C=C–H ethene H H H | | | H–C–C=C–H | H propene H H H H | | | | H–C–C–C=C–H | | H H butene H H H H H | | | | | H–C–C–C–C=C–H | | | H H H pentene 346 2 Similarly, the carbon chain of an alkane drawn as a straight chain is the same as that drawn in a bent chain. H H H H H H | | | | | | H–C–C–C–C–C=C–H | | | | H H H H hexene H | H—C— | H H H H H H H H | | | | | | | H–C–C–C–C–C–C=C–H | | | | | H H H H H heptene H H H H H H H | | | | | | | C–C–C–C–C–C=C–H | | | | | H H H H H octene H | C— | H H | C — H is the same as | H H | H—C—H H | H—C—C— | | H H H | C—H | H Physical Properties of Alkenes 1 The physical properties of alkenes are the same as that of alkanes. Low melting and boiling points due to weak van der Waals forces between molecules H H H H H H H H H | | | | | | | | | H–C–C–C–C–C–C–C–C=C–H | | | | | | | H H H H H H H nonene Insoluble in water but soluble in organic solvents such as propanone Physical properties of alkenes H H H H H H H H H H | | | | | | | | | | H–C–C–C–C–C–C–C–C–C=C–H | | | | | | | | H H H H H H H H decene Do not conduct electricity because they consist of covalent molecules 1 In the drawing of the structural formula of an alkene, the hydrogen atom bonded to the carbon atom can be written as (a) bonded to the side or (b) to the top or (c) bottom of the carbon atom. This is because the single covalent bond can be rotated freely. 2 The physical properties change gradually on going down the alkene series as shown in Table 2.5. 3 Similar to that of alkanes, the melting points and boiling points of alkenes increase down the homologous series. When the number of carbon atoms per molecule increases, the molecular size increases. Thus the van der Waals forces of attraction between molecules increases, increasing the amount of heat needed to overcome the forces of attraction during melting or boiling. 4 Similar to that of alkanes, the density of alkenes increases gradually down the series as the molecular mass increases. H H H | | | H — C = C — H or C = | H H | C or H — C = C — H | | | H H H 347 Less dense than water hence will float on the surface of water Carbon Compounds 2 H | H–C– | H H | C— | H Table 2.5 Physical properties of alkanes Name of alkene Molecular formula Relative molecular mass Melting point (°C) Boiling point (°C) Density (g cm–3) Ethene C2H4 28 –169 –104 Gas – Propene C3H6 42 –185 –48 Gas – Butene C4H8 56 –130 –6 Gas – Pentene C5H10 70 –138 30 Liquid 0.64 Hexene C6H12 84 –140 64 Liquid 0.67 Heptene C7H14 98 –119 93 Liquid 0.70 Octene C8H16 112 –104 122 Liquid 0.72 Nonene C9H18 126 –94 146 Liquid 0.73 Decene C10H20 140 –66 171 Liquid 0.74 (c) Alkenes undergo addition reactions. During addition reactions, the carbon-carbon double bond breaks open to form two new single bonds. In this process, the unsaturated hydrocarbon is converted to a saturated compound. Chemical Properties of Alkenes 2 Physical state at room temperature 1 Combustion of alkenes (a) Alkenes burns in excess air or oxygen to ’07/P1 form carbon dioxide and water. Heat energy is released during combustion. SPM C2H4 + 3O2 → 2CO2 + 2H2O 2C3H6 + 9O2 → 6CO2 + 6H2O | | | —C=C—+X—Y→—C— alkene | (unsaturated) X (b) The combustion of an alkene is more luminous and smokier than an alkane with the same number of carbon atoms. This is because the percentage by mass of carbon in an alkene is higher than that of an alkane. Example Percentage by mass of carbon in hexene (Mr of C6H12 = 84) 6 3 12 =— — — — — — 3 100 = 85.7% 84 Percentage by mass of carbon in hexane (Mr of C6H14 = 86) 6 3 12 =— — — — — — 3 100 = 83.7% 86 (c) Incomplete combustion of alkenes produces carbon (black smoke) and carbon monoxide. 2 Addition reactions of alkenes (a) Addition reactions are reactions in which an unsaturated organic compound combines with another compound to form a single new saturated compound. (b) Alkenes contain a double bond between carbon atoms (C = C bond) which is very reactive. Hence alkenes are more reactive than alkanes. Carbon Compounds | C— | Y (saturated) one double bond two new single bonds 3 The following are examples of addition reactions between alkenes and other elements/ compounds: (a) Hydrogen (hydrogenation) (b) Halogens (halogenation) (c) Water (hydration) (d) Acidified potassium manganate(VII) solution (e) Hydrogen halides 4 Hydrogenation (a) Hydrogenation is the addition of a hydrogen molecule across a carbon-carbon double bond in the presence of nickel or platinum as a catalyst. An alkene is converted to an alkane. For example SPM ’07/P1 Ni or Pt C3H6 + H2 ⎯⎯⎯→ C3H8 propene propane (b) Hydrogenation is used to make margarine (in solid form) from vegetable oils (in liquid form). 348 | —C ⎯ | OH 5 Halogenation (a) The addition reactions between alkenes ’10/P1, ’11/P2 and halogens (chlorine and bromine) are called halogenation. (b) Bromination is used as a chemical test to distinguish alkanes from alkenes. Alkenes decolourise the brown colour of liquid bromine whereas alkanes do not decolourise the brown colour of liquid bromine. For example: When ethene is passed into liquid bromine, the brown colour of bromine is decolourised immediately and a colourless organic liquid is formed. SPM Br2 bromine room ⎯⎯⎯⎯→ temperature ethene (d) Like liquid bromine, acidified potassium manganate(VII) solution is used to distinguish between alkane and alkene compounds. An alkene decolourises the purple colour of acidified potassium manganate(VII) solution whereas an alkane does not. 8 Polymerisation Alkenes undergo polymerisation to form polymers. SPM (a) Ethene undergoes additional polymerisation ’07/P1, ’06/P1 to form polyethene. C2H4Br2 1,2-dibromoethane (saturated) 6 Hydration (a) Hydration occurs when a water molecule is added across the double bond between the atoms in the presence of phosphoric(V) acid with H3PO4 as a catalyst at 300 °C. SPM (b) An alkene is converted to an alcohol in ’07/P2 hydration. For example H H H H | | | | polymerisation nH — C = C — H ⎯⎯⎯⎯⎯→ — C — C — | | ethene H H n H3PO4 C2H4 + H2O ⎯⎯⎯⎯⎯→ C2H5OH 300 °C, 60 atm ethene ethane-1, 2-diol 2 ethene + (c) In the formation of a diol, two hydroxyl (–OH) groups are added across the double bond in the alkene molecule. For example: C2H4 + H2O + [O] → C2H4(OH)2 ethanol 7 Reaction with acidified potassium manganate(VII) solution (a) When an alkene reacts with acidified potassium manganate(VII) solution, the purple colour of potassium manganate(VII) is decolourised immediately and an organic compound called diol is formed. (b) A diol is a saturated alcohol with two hydroxyl (–OH) groups on adjacent carbon atoms. polyethene (b) Propene undergoes polymerisation to form polypropene. H H H | | | polymerisation nH — C = C — C — H ⎯⎯⎯⎯→ | H propene H | C — | H CH3 | C | H n polypropene To compare the chemical properties of alkanes and alkenes having the same number of carbon atoms Apparatus Porcelain dish, wooden splint, dropper and Bunsen burner. SPM ’11/P2 Procedure A Combustion of alkanes and alkenes in air EXUQLQJVSOLQW Materials Hexane, hexene, liquid bromine and acidified potassium manganate(VII) solution. ILOWHUSDSHU SRUFHODLQ GLVK KH[DQH RUKH[HQH KH[DQH RUKH[HQH Figure 2.3 Combustion of alkane and alkene 7& 349 Carbon Compounds Activity 2.2 C2H4 | C— | OH 3 The mixture is shaken gently. 4 The colour change that takes place in the test tube is recorded. 5 Steps 1 to 4 are repeated using hexene. 1 About 1 cm3 of hexane and hexene are placed separately into two separate porcelain dishes. 2 The organic liquids are ignited with a burning splint. 3 A piece of filter paper is held above the flames in each of the dishes to detect the amount of soot formed. 4 The sootiness of the flame and the amount of soot collected on the two pieces of filter papers are recorded. C Reactions with acidified potassium manganate(VII) solution 1 A few drops of potassium manganate(VII) solution are placed in a test tube, followed by about 1 cm3 of dilute sulphuric acid. 2 About 2 cm3 of hexane is then added to the acidified potassium manganate(VII) solution. 3 The mixture is shaken gently. 4 The colour change that occurs in the test tube is recorded. 5 Steps 1 to 4 are repeated using hexene. B Reactions with bromine 1 About 1 cm3 of liquid bromine is placed in a test tube. 2 About 2 cm3 of hexane is then added to the liquid bromine. Results Observations Test 2 Hexane Hexene Combustion Burns in air with a sooty flame Burns in air with a more sooty yellow flame Reaction with liquid bromine The brown colour of liquid bromine remains unchanged The brown colour of liquid bromine is decolourised Reaction with acidified potassium manganate(VII) solution The purple colour of potassium manganate(VII) solution remains unchanged The purple colour of potassium manganate(VII) solution is decolourised Conclusion The chemical properties of hexene are different from those of hexane: (a) Both hexane and hexene undergo combustion but the flame of hexene is sootier than that of hexane. (b) Hexene decolourises the brown colour of liquid bromine whereas hexane does not. (c) Hexene decolourises the purple colour of acidified potassium manganate(VII) solution whereas hexane does not. Discussion 1 The combustion of hexene produces more soot than the combustion of hexane. This is because the percentage by mass of carbon in hexene is higher than that of hexane. 2 Hexene undergoes addition reaction with bromine: C6H12 + Br2 → (brown) C6H12Br2 1, 2-dibromohexane (colourless) 3 Hexene undergoes addition reaction with acidified potassium manganate(VII) C6H12 + H2O + [O] → from KMnO4 (purple) Carbon Compounds C6H12(OH)2 hexane-1,2-diol (colourless) 350 Br H Br | | | C H—C—C—C—H | | | H H H H H H | | | D H — C — C — C — Br | | | H H H In the addition reaction of an alkene for example with chlorine, the product formed has the two Cl atoms bonded to the two C atoms adjacent to each other. Products with two Cl atoms bonded to the same C atom or across another C atom are not formed. Example: Addition of chlorine to propene cannot form the following products: H | C— | H Cl | C—H | Cl H | H—C— | Cl and 1 H | C— | H H | C—H | Cl Solution Propene undergoes addition reaction with Br2 at the carbon-carbon double bond, hence the product formed has two Br atoms attached to two adjacent C atoms. Answer: B Homologous Series ’05 1 A homologous series is a family of organic compounds with the same functional group and with similar chemical properties. 2 A functional group is an atom or a group of atoms that determines the chemical properties of an organic compound. 3 All members in the same homologous series have the same functional group and the same chemical properties. 4 All members in the same homologous series (a) have the same general formula (b) can be prepared using similar methods (c) show a gradual change in their physical properties (d) have similar chemical properties (e) differ from each other by a –CH2 group 5 General formulae of some homologous series are shown in Table 2.6. The following is the equation that represents the reaction between propene and bromine. Propene + Br2 → P Which of the following is the structural formula of P? Br H H | | | A H—C—C—C—H | | | Br H H H H Br | | | B H—C—C—C—H | | | H Br H Table 2.6 General formulae of some homologous series Homologous series General formula SPM ’07/P2 Functional group Alkanes CnH2n+2 Alkenes CnH2n – C = C – (double bond) Alcohols CnH2n+1OH – O – H (hydroxyl group) Carboxylic acids CnH2n+1COOH O i – C – O – H (carboxyl group) Esters CnH2n+1COOCmH2m+1 – 351 O i – C – O – (carboxylate group) Carbon Compounds 2 H | H—C— | H CnH2n+2 alkane Hydrogenation H2/Ni, 180 °C – (CnH2n)n – polymer Addition polymerisation CnH2n+1OH alcohol Addition of halogen, X2 CnH2n alkene Addition of hydrogen halide, HX Addition of acidified KMnO4 Addition of water, H3PO4, 180 °C, 60 atm CnH2nX2 CnH2n+1X 2 CnH2n(OH)2 diol 4 Isomers have different physical properties because they have different structural formulae. 5 For methane, ethane and propane, there is only one structure. Therefore methane, ethane and propane do not have isomers. All the other alkanes, have isomers. 6 There are two different ways of arranging the four carbon atoms and ten hydrogen atoms for butane, C4H10. Thus butane has two isomers as follows: 2.3 1 (a) What is the general formula of alkene? (b) Give the molecular formula of an alkene with (i) three carbon atoms (ii) five carbon atoms (iii) seven carbon atoms 2 Propane and propene are both hydrocarbons. (a) State two common physical properties between propane and propene. (b) State one common chemical property between propane and propene. 3 Write chemical equations for the reactions between propene with (a) chlorine (b) water (c) hydrogen (d) excess oxygen (e) acidified potassium manganate(VII) 2.4 Isomerism (a) Straight chain (all 4 C atoms form a straight chain) H H H H | | | | H—C—C—C—C—H | | | | H H H H (b) Branched chain (3 C atoms form a straight chain with 1 C atom forming a branch) H | H—C—H H H | | H—C—C—C—H | | | H H H SPM ’10/P1, ’10/P2, ’11/P2 1 Isomers are compounds which have the same molecular formula but with different structural formulae. 2 Isomerism is the existence of two or more compounds that have the same molecular formula but with different structural formulae. 3 Isomers will have the same chemical properties when they have the same functional group. Carbon Compounds 352 7 There are three different ways of arranging the five carbon atoms and twelve hydrogen atoms in pentane, C5H12. Thus C5H12 has three isomers. H H H H H | | | | | H—C—C—C—C—C—H | | | | | H H H H H (b) One branched chain (4 C atoms form a straight chain with 1 C atom forming a branch) H | H—C—H H H H | | | H—C—C—C—C—H | | | | H H H H (c) Two branched chains (3 C atoms form a straight chain with 2 C atoms forming two branches) H | H—C—H H H | | H—C—C—C—H | | H H H—C—H | H 8 All the alkenes above propene have isomers. Butene, C4H8 has three isomers as follows: (a) Straight chain with a double bond at the end of the chain (4 C atoms form a straight chain with a double bond at the first C atom) H H H H | | | | H—C=C—C—C—H | | H H (b) Straight chain with a double bond in the middle of the chain (4 C atoms form a straight chain with a double bond at the second C atom) H H H H | | | | H—C—C=C—C—H | | H H (c) Branched chain (3 C atoms form a straight chain with a double bond and 1 C atom forming a branch) H | H—C—H H H | | H—C=C—C—H | H 9 Naming of branched isomers of alkanes according to the IUPAC system Step 1 Find the longest continuous chain of carbon atoms in the molecule and name the longest chain as the parent chain. For example: H | H—C—H H H H | | | H—C—C—C—C—H | | | | H H H H The longest chain has four carbon atoms. The name of the parent chain is butane. 353 Carbon Compounds 2 (a) Straight chain (5 C atoms form a straight chain) Step 2 1 Name the branched chain attached to the parent chain as alkyl group. 2 The alkyl groups are named according to the number of carbon atoms present. Formula and name of alkyl groups Number of carbon atoms Formula Name 1 –CH3 Methyl 2 –C2H5 Ethyl 3 –C3H7 Propyl The alkyl group has the general formula CnH2n+1 where n = 1, 2, 3… H –CH3 (methyl group) | is attached to the H—C—H parent chain of four H H H carbon atoms | | | H—C—C—C—C—H | | | | H H H H 2 Step 3 Identify the position of the alkyl group that is attached to the parent chain by number. (a) This is done by numbering the carbon atom in the parent chain using the lowest number. (b) Use hyphens to separate words from numbers in the name, for example: 2 – methyl. For example: H | H—C—H –CH3 (methyl group) is attached to carbon H H H number 2 | | | H — C1 — C2 — C3 — C4 — H | | | | H H H H The name of the alkane is 2-methyl butane carbon no. 2 side chain is –CH3 parent chain has four carbon atoms Step 4 1 If there are more than one similar branch, use the following prefixes (a) di for two similar branched chains (b) tri for three similar branched chains (c) tetra for four similar branched chains 2 Name the positions of carbon atoms in the parent chain containing the branches. For example, the positions of two branches may be 2, 2 or 2, 3. For example: H | H—C—H H H H | | | H—C—C—C—C—H | | | H H H H—C—H | H 2,2–dimethyl butane both –CH3 branches attached to carbon no. 2 Carbon Compounds 354 two –CH3 branches parent chain has four carbon atoms Step 5 If there are more than one alkyl group, list the names of the alkyl groups in alphabetical order. For example: H H C2H5 CH3 H | | | | | H — C — C — C — C — C — | | | | | H H H H H 3-ethyl,4-methylhexane H | C — H | H SPM 10 Naming of alkenes according to the IUPAC system ’11/P1 For example Step 1 H H H H | | | | H — C1 — C2 == C3 — C4 — H || HH Correct name but-2-ene Step 2 1 Select the position of the double bond by choosing the smallest number for the carbon atom with the C=C bond. 2 Name the position of the double bond with a number followed by a hyphen, for example: 2-ene. parents chain has 4 carbon atoms 2 H | H—C—H H H | | H—C=C—C—H | longest chain with C=C H has four carbon atoms 1 Select the longest carbon chain with the double bond (C=C) as the parent alkene. 2 Name the parent alkene according to the number of carbon atoms. double bond present double bond at carbon no. 2 Wrong name but-3-ene 3, the larger number is not used Example Step 3 methyl group at carbon H no. 2 | H—C—H H H H | | | H—C—C=C—C—H | | H H Identify the alkyl group and its position in the parent chain as fixed in Step 2. 2-methylbut-2-ene –CH3 group at carbon no. 2 355 C=C at carbon no. 2 Carbon Compounds 1 To choose the longest carbon chain as the parent chain, count the number of carbon atoms which could be in a straight chain or a bent chain. For example: –C– This is the longest | – C – chain with 6 C atoms | –C–C=C–C–C– 2 The alkyl group bonded by a single covalent bond to a carbon atom can rotate in a molecule. It may be drawn as up or down or at the side in a 2-D structural formula. For example: These two structures below are not isomers, they are the same compound. H | C—H | H H H | | H—C—C=C— | H H—C—H | H 2 H | H—C—H H H | | H—C—C=C— | H H | C—H | H 3 The following alkene does not exist because the central carbon atom has five covalent bonds. C | C=C—C | C 4 The following alkene structures are the same, they are not isomers. (a) C = C — C — C — C is the same as C — C — C — C = C (b) C = C — C — C; C — C — C = C; C — C — C — C are the same isomer. | | i C C C 2.4 (c) H ⎯ 1 Give the IUPAC name of the following compounds: (a) H | H—C—H H H H | | | H—C—C=C—C—H | | HH (b) H | H H—C—H H H H | | | | | H ⎯ C ⎯ C ⎯ C ⎯ C = C ⎯ H | | | H H H—C—H | H Carbon Compounds H H H H | | | | C ⎯ C ⎯ C = C ⎯ C ⎯ H | | | | H H—C—H H—C—H H | | H H 2 Draw and name all the isomers for C5H10. 3 Ethane reacts with chlorine in the presence of sunlight to form a substituted product with the molecular formula of C2H4Cl2. Draw and name all the possible isomers of this product. Which of the isomer is also formed when ethene reacts with chlorine in the addition reaction? 356 Alcohols 7 All alcohols above ethanol have isomers. The position of the hydroxyl (–OH) group and the alkyl group are shown by numbering the carbon atoms from the end of the carbon chain which gives the smallest number to the –OH group. 8 Propanol, C3H7OH has two isomers: 1 Alcohols have the general formula CnH2n+1OH where n = 1, 2, 3… The CnH2n+1– group represents the alkyl group. 2 The functional group of alcohol is the hydroxyl (–OH) group. The hydroxyl (–OH) group is joined to the carbon atom in the alcohol molecule by a single covalent bond. 3 The molecular formula of an alcohol can be obtained by substituting n in the general formula CnH2n+1OH with the number of carbon atoms. 4 Based on the IUPAC system of naming straight chain alcohols, the letter e at the end name of alkane is replaced by the suffix ol. For example CH4 (methane) → CH3OH (methanol) C2H6 (ethane) → C2H5OH (ethanol) Alkane Alkane formula (CnH2n+2) Structural formula (a) Alcohol formula Methane CH4 Methanol CH3OH Ethane C2H6 Ethanol C2H5OH Propane C3H8 Propanol C3H7OH Butane C4H10 Butanol C4H9OH methanol H ⎮ H—C— ⎮ O ⎮ H H H H | | | H–C–C–C–H | | | H O H | H 3, the bigger number is not used The parent chain has three carbon atoms propan-2-ol position of the –OH group is at the second carbon atom The –OH group is at the second carbon atom 9 Butanol, C4H9OH has four isomers as follows: 5 Note that when writing the structural formula of alcohols, (a) each carbon atom should have four single covalent bonds. (b) each hydrogen atom should have one single covalent bond. (c) each oxygen atom has two single covalent bonds. (d) the carbon atoms are connected by single bonds. 6 Methanol and ethanol has one structural formula each. Hence they have no isomers. H ⎮ H—C—H ⎮ O ⎮ H H H H parent chain has three carbon atoms | | | H–C–C–C–H Correct name propan-1-ol | | | O H H position of the –OH group is at the first carbon atom | H Wrong name propan-3-ol The –OH group is at the first carbon atom (b) Alcohol IUPAC name Structural formula (a) H H H H | | | | H–C–C–C–C—H | | | | O H H H | H (b) H H H H ⎮ ⎮ ⎮ ⎮ H–C–C–C–C—H ⎮ ⎮ ⎮ ⎮ H O H H ⎮ H H ⎮ C—H ⎮ H ethanol 357 IUPAC name parent group has four carbon atoms butan-1-ol position of the –OH group is at the first carbon atom parent group has four carbon atoms butan-2-ol position of the –OH group is at the second carbon atom Carbon Compounds 2 2.5 Structural formula (c) Example IUPAC name H | H | H | H—C—C—C—H methyl group at second carbon atom H | H—C—H H H | | H–C–C–C–H | | | O H H | H methyl group at second carbon atom | | | O O O H H H | 2-methylpropan-1-ol parent group has three carbon atoms | | three-OH groups propan-1,2,3-triol position of the –OH group is at the first carbon atom parent group has three carbon atoms three-OH groups at 1st, 2nd and 3rd carbon atoms Industrial Production of Alcohols 2 (d) H | H — C — H H H | | H–C–C–C–H | | | H O H | H methyl group at second carbon atom 1 Ethanol (C2H5OH) can be produced industrially by two processes: (a) The hydration of ethene (b) The fermentation of sugar or starch 2 Hydration of ethene When a mixture of ethene and steam is passed over the catalyst, phosphoric(V) acid (H3PO4) at 300 °C and high pressure (65 atm), ethanol is produced. 3 Fermentation When yeast is added to sugar or starch, ethanol and carbon dioxide are produced. The enzyme called zymase, breaks down the glucose molecules to form ethanol and carbon dioxide. 2-methylpropan-2-ol parent group has three carbon atoms position of the –OH group at second carbon atom 10 In the naming of an alcohol with more than one –OH group: (a) di is used for two –OH groups (b) tri is used for three –OH groups SPM ’06/P1 yeast C6H12O6 ⎯⎯→ 2C2H5OH + 2CO2 glucose ethanol carbon dioxide To prepare ethanol in the laboratory by fermentation and distillation 3 The yeast paste is added to the glucose solution and the mixture is stirred well. 4 The conical flask is closed with a rubber stopper fitted with a delivery tube. The other end of the delivery tube is dipped in limewater. 5 The apparatus is left in a warm place (about 25 – 35 °C) for about one week. The changes that occur in the conical flask and the test tube are recorded from time to time. 6 After about one week, the products in the conical flask are filtered. The filtrate obtained is transferred into a distillation flask. 7 The filtrate is distilled in a flask fitted with a fractionating column and Liebig condenser. Apparatus Conical flask, boiling tube, thermometer, fractionating column, wire gauze, retort stand, distillation flask, rubber stopper, delivery tube, tripod stand and Bunsen burner. Activity 2.3 Materials Glucose, yeast, distilled water and limewater. Procedure 1 About 20 g of glucose is dissolved in 200 cm3 of distilled water in a conical flask. 2 A little warm water (35 °C) is added to about 5 g of yeast in a small beaker. The mixture is stirred well to form a paste. Carbon Compounds 358 Figure 2.4 Fermentation process Figure 2.5 Fractional distillation Conclusion Ethanol can be prepared in the laboratory by the fermentation of glucose or any other carbohydrates. 2 Result 1 During fermentation, ethanol and carbon dioxide are produced. 2 The carbon dioxide produced causes limewater to turn cloudy. 3 The concentration of ethanol produced in fermentation can be increased by fractional distillation. 2 Ethanol is a volatile liquid because it has a low boiling point at 78 °C. 3 Ethanol is very soluble in water because of the presence of the hydroxyl group. (a) The hydrocarbon part of alcohol is insoluble in water. (b) Hence alcohol with a large hydrocarbon chain is insoluble in water. (c) The solubility of alcohols in water decreases as the molecular size increases. 4 Alcohols are neutral and have a pH of 7. 5 Alcohols are covalent compounds. They do not conduct electricity. Yeast is a living organism. Ethanol is actually a by-product formed from the living process of yeast. The highest concentration of ethanol prepared by fermentation is only 14%. This is because yeast is killed when the ethanol formed exceeds 14%. Hence higher concentration of ethanol has to be obtained from fractional distillation. Physical Properties of Ethanol and Other Alcohols 1 Ethanol is a colourless liquid and has a characteristic odour. To investigate the chemical properties of ethanol Apparatus Evaporating dish, wooden splint, test tube, boiling tube, delivery tube with stopper, glass wool, porcelain chips, beaker, retort stand with clamp and test tube holder. Ethanol, concentrated sulphuric acid, potassium dichromate(VI), blue litmus paper and liquid bromine. (A) Combustion of ethanol in air Procedure 1 About 2 cm3 of ethanol is poured into an evaporating dish. 2 The ethanol is ignited using a lighted wooden splint. 3 The flammability of ethanol and the sootiness of the flame are recorded. 359 Carbon Compounds Activity 2.4 Materials 3 Acidified potassium dichromate(VI) solution is an oxidising agent. Result 2 Nature of combustion Observation Flammability Easily burned Colour of flame Blue Sootiness of flame Non-sooty Conclusion Ethanol burns readily in air. The combustion of ethanol produces a non-sooty, pale blue flame. The products of combustion are carbon dioxide and water. (B) Oxidisation of ethanol Procedure 1 About 1 cm3 of concentrated sulphuric acid and 5 cm3 of potassium dichromate(VI) solution are poured into a boiling tube. 2 About 5 cm3 of ethanol is added to the acidified potassium dichromate(VI) solution. 3 A rubber stopper fitted with a delivery tube is inserted into the boiling tube. The delivery tube is inserted into a test tube placed in a beaker half-filled with ice-cold water. 4 The mixture of ethanol and acidified potassium dichromate(VI) is heated gently. Any colour change in the mixture is noted. 5 The distillate is collected in the test tube and is tested with litmus paper. (C) Dehydration of ethanol Procedure 1 Some glass wool is placed in a boiling tube. 2 About 2 cm3 of ethanol is poured into the boiling tube to soak the glass wool. 3 Some porcelain chips are placed in the midsection of the boiling tube. 4 The boiling tube is closed with a rubber stopper fitted with a delivery tube. The other end of the delivery tube is placed under an inverted test tube filled with water in a beaker. 5 The porcelain chips are heated strongly until red hot. The Bunsen burner flame is then shifted to the glass wool to vaporise the ethanol absorbed in it. 6 The gas is collected in two test tubes. The gas produced is collected by displacement of water and tested with (a) a few drops of bromine water and shaken (b) a few drops of acidified potassium manganate(VII) solution and shaken SPM ’05/P1 Figure 2.7 Dehydration of ethanol Result Figure 2.6 Oxidation of ethanol Test on gas collected Result Compound Reactant mixture Distillate With bromine water Observation The acidified potassium dichromate(VI) solution changes from orange to green Brown colour of bromine is decolourised With acidified potassium Purple colour of manganate(VII) solution potassium manganate(VII) is decolourised A colourless liquid (a vinegary smell) which changes blue litmus paper to red Conclusion 1 When ethanol vapour is passed over heated porcelain chips (aluminium oxide), dehydration occurs and ethene is produced. 2 Ethene is confirmed to be present by the decolourisation of the brown colour of bromine water and the purple colour of acidified potassium manganate(VII) solution. Conclusion 1 When ethanol is boiled with acidified potassium dichromate(VI) solution, it is oxidised to ethanoic acid. 2 Ethanoic acid is a colourless liquid with a vinegary smell and turns blue litmus paper red. Carbon Compounds Observation 360 (b) Dehydration is carried out by (i) passing alcohol vapour over heated porcelain chips or (ii) refluxing alcohol with concentrated sulphuric acid (acts as a dehydrating agent). (c) Methanol does not undergo dehydration since there is no alkene with one carbon atom. Chemical Properties of Ethanol and Other Alcohols 1 (a) Alcohol undergoes complete combustion when it burns in excess air to produce carbon ’08/P2 dioxide and water. For example C2H5OH + 3O2 → 2CO2 + 3H2O SPM ethanol 9 C3H7OH + — O2 → 3CO2 + 4H2O 2 propanol The alcometer (breath tester) used for testing the breath of suspected drink-driver contains potassium dichromate(VI). The chemical will oxidise any ethanol present in the breath and produces an electric current. (b) Combustion of alcohol such as ethanol gives out a lot of heat energy. Hence ethanol is a good fuel. (c) If the supply of oxygen is insufficient, incomplete combustion of ethanol occurs. Carbon monoxide gas, carbon (black soot) and water are produced. 2 (a) Oxidation of alcohols produces the corresponding carboxylic acids. For example 1 Alcohols as fuels When alcohol is burn in air, carbon dioxide and water are produced, and a large quantity of heat energy is released. Ethanol is a clean fuel because it does not release toxic gases in combustion. 2 Alcohols as solvents Alcohols are good solvents for organic compounds such as shellac, varnishes, paints, perfumes and dyes. 3 Uses of alcohols in medicines (a) Ethanol is used as a mild antiseptic for skin infection and disinfection. (b) Propan-2-ol is used as rubbing alcohol to reduce fever. 4 Uses of alcohols in cosmetics (a) Ethanol is used to make aftershave lotion and nail polish. (b) Propan-1,2,3-triol (common name is glycerol) is used in moisturiser. (c) Alcohols are used as solvents for perfumes. 5 Alcohols as a source of chemicals (a) Ethanol is oxidised to make vinegar. (b) Methanol is used to make formalin and polymers. 6 The misuse and abuse of alcohols (a) Ethanol is a component of alcoholic drink. Excess drinking of alcohol increases the risk of heart disease, kidney disease and liver disease. (b) Alcoholism is an addiction caused by excessive drinking of alcohol for a prolonged period of time. K2Cr2O7/H2SO4 CH3CH2OH + 2[O] ⎯⎯⎯⎯⎯→ ethanol CH3COOH + H2O from K2Cr2O7 ethanoic acid CH3CH2CH2OH + 2[O] ⎯→ propanol CH3CH2COOH + H2O propanoic acid (b) Oxidation is carried out by heating the alcohols with oxidising agents such as acidified potassium manganate(VII) solution or acidified potassium dichromate(VI) solution. (c) When oxidation reaction occurs: (i) The colour of potassium dichromate(VI) changes from orange to green. (ii) The colour of potassium manganate(VII) changes from purple to colourless (decolourisation). 3 (a) Dehydration of alcohols (except methanol) produces the corresponding alkenes. For example SPM C2H5OH ⎯→ C2H4 + H2O ’07/P2, ’06/P1 ethanol ethene C3H7OH ⎯→ C3H6 + H2O propanol propene 361 Carbon Compounds 2 Uses of Alcohol in Everyday Life (c) Driving after drinking too much alcohol can cause road accidents. Number of carbon atoms Alkane Carboxylic acid 1 Methane Methanoic acid 2 Ethane Ethanoic acid 3 Propane Propanoic acid 2.5 1 Compound X has a molecular formula of C4H8O. When compound X is refluxed with acidified potassium dichromate(VI), the mixture changes colour from orange to green. (a) Name the funtional group of compound X. (b) What is the general formula of the homologous series in which compound X is a member. (c) Draw and name all the isomers of compound X. (d) Write the chemical equation for the reaction between compound X and acidified potassium dichromate(VI). 4 The name of the carboxylic acid will depend on the number of carbon atoms in its molecule. The carbon atom of the functional group is counted as part of the carbon chain. 5 The molecular formula of a carboxylic acid can be obtained by substituting n in the general formula CnH2n+1COOH with the number of carbon atoms. 6 In the writing of the structural formula of a carboxylic acid, (a) the –COOH group is always at the terminal carbon atom (at the end of the chain). (b) the carboxyl group consists of a carbon atom which forms a double bond with oxygen atom and a single covalent bond with the hydroxyl (–OH) group. 2 2 Identify the compounds P, Q, R, S and T from the reaction scheme given below: Glucose KMnO4/ yeast Compound H2SO4 P oxygen, heat Compound Q porcelain chips, heat Compound R + compound S O double bond i –C–O–H Compound T single bond 3 State the type of reactions that occur and give the molecular formulae of compounds W, X, Y and Z in the following equations. 7 The molecular and structural formulae of the first four members of carboxylic acids are shown in Table 2.7. concentrated H SO 2 4 (a) W ⎯⎯⎯⎯⎯⎯⎯⎯→ C3H6 + H2O Table 2.7 The names and formulae of the first four members of carboxylic acids phosphoric acid (b) X + H2O ⎯⎯⎯⎯⎯⎯→ C3H7OH K Cr O /H SO 2 2 7 2 4 (c) C4H9OH + 2[O] ⎯⎯⎯⎯⎯→ Y + H2O Name (d) Z + 6O2 ⎯→ 4CO2 + 5H2O 2.6 Methanoic HCOOH acid Carboxylic Acids Ethanoic acid 1 Carboxylic acids are organic acids that have the general formula CnH2n+1COOH, where n is 0, 1, 2, 3 …. 2 The functional group of carboxylic acids is the carboxyl group, –COOH. 3 Based on the IUPAC system of naming, a carboxylic acid is named by replacing the final letter e in the name of the corresponding alkane with oic acid. Carbon Compounds Molecular formula 362 CH3COOH Structural formula O i H — C — OH O i CH3 — C — OH Propanoic C2H5COOH acid O i CH3 — CH2 — C — OH Butanoic acid O i CH3 — CH2 — CH2 — C — OH C3H7COOH 2 Ethanoic acid has a vinegary smell. 3 Ethanoic acid is soluble in water. 4 On going down the homologous series of the carboxylic acids, (a) the solubility in water decreases, (b) the boiling point increases. Preparation of Carboxylic Acid 1 Carboxylic acid is prepared in the laboratory by the oxidation of the corresponding alcohol. Example oxidation Ethanol ⎯⎯⎯→ ethanoic acid oxidation Propanol ⎯⎯⎯→ propanoic acid Chemical Properties of Ethanoic Acid oxidation Butanol ⎯⎯⎯→ butanoic acid 1 Ethanoic acid is a weak acid because it undergoes partial ionisation in water. Only a small percentage of the ethanoic acid molecules ionise to form hydrogen ions, H+. Most of the ethanoic acid remains as molecules. CH3COOH CH3COO– + H+ 2 Aqueous ethanoic acid turns blue litmus paper red and is an electrolyte. 3 Ethanoic acid reacts with bases to form salts and water in neutralisation. The salts produced in the reaction are known as ethanoate. For example CH3COOH + NaOH → CH3COONa + H2O ethanoic acid sodium ethanoate 2CH3COOH + CuO → (CH3COO)2Cu + H2O ethanoic acid copper oxide (black) copper(II) ethanoate (blue) 4 Ethanoic acid reacts with metal carbonates to form salts, carbon dioxide and water. For ’05/P1 example 2CH3COOH + Na2CO3 ⎯→ 2CH3COONa + CO2 + H2O SPM K2Cr2O7/H2SO4 CH3CH2OH + 2[O] ⎯⎯⎯⎯⎯⎯→ CH3COOH + H2O sodium ethanoate 2CH3COOH + CaCO3 ⎯→ (CH3COO)2Ca + CO2 + H2O calcium ethanoate 5 Ethanoic acid reacts with reactive metals such as magnesium and zinc to form salts and hydrogen gas. For example 2CH3COOH + Mg ⎯→ (CH3COO)2Mg + H2 Figure 2.8 Preparation of ethanoic acid by reflux magnesium ethanoate 2CH3COOH + Zn ⎯→ (CH3COO)2Zn + H2 Physical Properties of Ethanoic Acid and Other Carboxylic Acids zinc ethanoate 6 Ethanoic acid reacts with an alcohol to form ester and water. For example Ethanoic acid + ethanol → ethyl ethanoate + water 1 Ethanoic acid is a colourless liquid at room temperature. Pure ethanoic acid is known as glacial ethanoic acid because it freezes to form colourless crystals which look like ice. (ester) 363 Carbon Compounds 2 2 In the laboratory, ethanoic acid is prepared by the oxidation of ethanol using an oxidising agent such as (a) acidified potassium dichromate(VI) or (b) acidified potassium manganate(VII) 3 The method of heating a mixture of ethanol and the oxidising agent in a flask fitted with an upright Liebig condenser is known as reflux. 4 Heating under reflux is used (a) to prevent volatile substances (ethanol and ethanoic acid) from escaping into the atmosphere, (b) to ensure that the reactants go to complete reaction. 5 In the oxidation of ethanol by acidified potassium dichromate(VI), (a) the colour of potassium dichromate(VI) solution changes from orange to green. (b) the ethanoic acid produced has a vinegary smell. To study the chemical properties of ethanoic acid Results Apparatus Test tubes, beakers, evaporating dish, delivery tube and wooden splints. Test Materials Ethanoic acid, sodium hydroxide solution, sodium carbonate, limewater, magnesium ribbon, glacial ethanoic acid and concentrated sulphuric acid. 2 Procedure 1 About 3 cm3 of dilute ethanoic acid is placed in a test tube. 2 About 3 cm3 of sodium hydroxide is added to the ethanoic acid and the mixture is shaken. 3 The reaction mixture is poured into an evaporating dish and is heated until it becomes dry. 4 The residue left in the evaporating dish is observed. Effervescence occurs and sodium carbonate dissolves (b) Gas released + limewater Gas produced turns limewater cloudy (C) Reaction between ethanoic acid and a metal SPM Procedure ’11/P2 1 About 5 cm3 of ethanoic acid is placed in a test tube. 2 A piece of magnesium ribbon is added to the ethanoic acid. 3 A lighted splint is placed near the mouth of the test tube to test the gas liberated. Results Observation A white powder was left in the evaporating dish. Test Conclusion Ethanoic acid reacts with a base, sodium hydroxide to produce a salt and water. (B) Reactions between ethanoic acid and a metal carbonate Observation (a) Ethanoic acid + magnesium Effervescence occurs and the magnesium ribbon dissolves (b) Gas released + lighted splint Gas burns with a ‘pop’ sound when the lighted splint is placed at the mouth of the test tube Conclusion Ethanoic acid reacts with magnesium, a metal to form a salt and hydrogen gas. Procedure 1 About 5 cm3 of ethanoic acid is placed in a test tube. 2 A spatula of sodium carbonate is added to the ethanoic acid. 3 The gas released is passed into limewater. Activity 2.5 (a) Ethanoic acid + sodium carbonate Conclusion Ethanoic acid reacts with sodium carbonate, a metal carbonate to form a salt, carbon dioxide and water. (A) Reaction of ethanoic acid with a base (D) Reactions between ethanoic acid and alcohol Procedure 1 2 cm3 of glacial ethanoic acid is placed into a test tube. 2 4 cm3 of pure ethanol is added to the ethanoic acid. 3 About 1 cm3 of concentrated sulphuric acid is added slowly and carefully to the mixture using a dropper. 4 The reaction mixture is shaken and heated slowly for about 3 minutes. 5 The content of the test tube is then poured into a beaker half-filled with water. Figure 2.9 Reaction with ethanoic acid and a metal carbonate Carbon Compounds Observation 364 2 An ester is an organic compound formed when a carboxylic acid reacts with an alcohol. 3 Esterification is the reaction between a carboxylic acid and an alcohol to produce ester and water. 4 Concentrated sulphuric acid acts as a catalyst to speed up esterification. 5 When ethanoic acid is warmed with ethanol in the presence of a few drops of concentrated sulphuric acid, esterification occurs. The ester known as ethyl ethanoate and water are produced. Observation 1 An oily product forms a layer on top of the water’s surface. 2 The product is colourless and has a sweet fruity smell. Conclusion 1 Ethanoic acid reacts with ethanol, an alcohol to form an ester and water. 2 Ester has a sweet fruity smell and is insoluble in water. conc. H2SO4 CH3COOH + C2H5OH ⎯⎯⎯⎯→ CH3COOC2H5 + H2O Discussion 1 The sweet smelling oily product is an ester. ethyl ethanoate Chemical properties of other carboxylic acids General 2RCOOH + CaCO3 → (RCOO)2Ca + H2O + CO2 Example 2HCOOH + CaCO3 → (HCOO)2Ca + H2O + CO2 2C2H5COOH + CaCO3 → (C2H5COO)2Ca + H2O + CO2 (d) react with alcohols to form esters and water. SPM General ’06/P1 RCOOH + R′OH → RCOOR′ + H2O Example HCOOH + C2H5OH → HCOOC2H5 + H2O C2H5COOH + C2H5OH → C2H5COOC2H5 + H2O 1 All members of the carboxylic acids have similar chemical properties because they have the same functional group, –COOH. 2 The general formula of a carboxylic acid can also be written as RCOOH where R is H or an alkyl group. 3 All members of this carboxylic group will (a) react with alkalis to form salts and water General RCOOH + NaOH → RCOONa + H2O Example HCOOH + NaOH → HCOONa + H2O C2H5COOH + NaOH → C2H5COONa + H2O (b) react with active metals to form salts and hydrogen gas General 2RCOOH + Mg → (RCOO)2Mg + H2 Example 2HCOOH + Mg → (HCOO)2Mg + H2 2C2H5COOH + Mg → (C2H5COO)2Mg + H2 (c) react with metallic carbonates to form salts, carbon dioxide and water Uses of Carboxylic Acids in Daily Life Carboxylic acid 365 Use Methanoic acid To coagulate latex Ethanoic acid To make vinegar Benzoic acid Used as a food preservative Carbon Compounds 2 Figure 2.10 Reaction of ethanoic acid and ethanol cracking Alkane CnH2n+2 hydrogenation Alkene CnH2n hydration dehydration combustion combustion Alcohol CnH2n+2OH combustion 2 2.6 SPM ’11/P1 1 Compound X with an empirical formula of CH2O has a relative molecular mass of 60. (a) Calculate the molecular formula of X and draw its structural formula. (b) Identify the functional group of X and give its general formula. (c) Write a balanced equation between X and calcium carbonate. Predict the observation that will take place. [Relative atomic mass: of H,1; C,12; O,16] 2 (a) Give the name and molecular formula for a carboxylic acid that has five carbon atoms. (b) Give the molecular formula of the organic compound formed when the carboxylic acid named in (a) reacts with (i) sodium hydroxide (ii) magnesium metal 3 The functional group of esters is the carboxylate group, –COO–. 4 The name of an ester is derived from the alcohol and the carboxylic acid used to prepare it. (a) The first part of the name of the ester is taken from the alkyl group of the alcohol. Example Q Step II R (a) Identify compound Q and compound R. (b) State the condition of reaction in Step II. (c) Name the types of reactions that have occurred in Steps I and II. 2.7 Alcohol Alkyl group Methanol Methyl Ethanol Ethyl Propanol Propyl Butanol Butyl Carboxylic acid Carboxylate group Methanoic acid Methanoate Ethanoic acid Ethanoate Propanoic acid Propanoate (c) In general, the names of esters are of the form ‘alkyl carboxylate’ where alkyl comes from the alcohol used to prepare the ester. Example Esters 1 The general formula for an ester is CnH2n+1COOCmH2m+1, where n = 0, 1, 2, 3, 4… and m = 1, 2, 3, 4…. 2 The general formula for an ester can also be written as R–COO–R′ where R and R′ are alkyl groups. Carbon Compounds esterification (b) The second part of the name comes from the carboxylic acid. The ending –oic of carboxylic acid is replaced by –oate. Example 3 Compound P, with a molecular formula of C3H8O is converted to compounds Q and R in the following reaction scheme. Step I: K2Cr2O7/H2SO4 Carboxylic Acid CnH2n+1COOH Ester RCOOR′ Carbon dioxide and water P oxidation 366 Alcohol Carboxylic acid Names of ester Methanol Methanoic acid Methyl methanoate Methanol Ethanoic acid Methyl ethanoate Ethanol Ethanoic acid Ethyl ethanoate Ethanol Propanoic acid Ethyl propanoate 5 In the writing of the structural formula of an ester using the general formula R–COO–R′, the part R– CO comes from the carboxylic acid and the part of O– R′ comes from the alcohol. 6 Generally, O i R – C – OR′ from carboxylic acid from alcohol ethanol O i CH3CH2C O – CH2CH3 O i CH3CH2C O–CH2CH3 + H2O from propanoic acid from ethanol Steps in naming an ester propanoic acid ethyl propanoate SPM ’07/P1, ’06/P1 Example: Step 1 the alcohol part (bonded to –O) is –CH3, hence the prefix is methyl O i CH3 — CH2 — C — O — CH3 Identify the alcohol part of the ester. The alcohol part is the alkyl part bonded to oxygen atom by single bond. Step 2 O i CH3 — CH2 — C — O — CH3 Identify the carboxylic acid part of the ester. The acid part is the alkyl part bonded to the carbon atom with a double bond with oxygen. the carboxylic part (with – C=O) has altogether 3 C atoms, hence it is from propanoic acid Step 3 Name of ester is methyl propanoate Combine the two parts to name the ester. The alcohol part is named first. 2 ’07 The molecular formula below represents an organic compound. Solution O i CH3 — C — O — CH2 — CH2 — CH3 this is the acid part as it is bonded to C=O this is the alcohol part as it is bonded to O atom O i CH3 — C — O — CH2 — CH2 — CH3 from ethanoic acid (ester is ethanoate) What is the name of the compound? from propanol (prefix is propyl) The name of the compound is propyl ethanoate. 367 Carbon Compounds 2 Example Hence, the name of the ester is ethyl propanoate. 7 The reaction between an alcohol and a carboxylic acid to produce an ester and water is known as esterification. General Alcohol + carboxylic acid → ester + water Example C2H5OH + CH3CH2COOH → Writing the structural formula of an ester SPM ’07/P1 To predict the formula of an ester prepared from a named alcohol and a named carboxylic acid. Example Write the structural formula of the ester produced from butanol and propanoic acid. General formula is Step 1 Write the general formula of ester in the form R – COO – R′ O i R – C – O – R′ Alcohol is butanol, with four C atoms. Hence R′ is –CH2CH2CH2CH3. Step 2 Write down the structural formula of the alcohol part to replace –R′(R bonded to O by single bond) 2 O i R – C – O – CH2CH2CH2CH3 Carboxylic acid is propanoic acid, with three C atoms. Hence RCO– is CH3CH2CO–. Step 3 Write down the structural formula of the acid part to replace R– (R bonded to C=O) O i CH3CH2C– Structural formula of ester is: O i CH3 — CH2 — C — O — CH2 — CH2 — CH2 — CH3 Step 4 Replace –OR′ with the alcohol part and the R–CO– with the carboxylic part in R – COO – R′ for the full structure of the ester. 3 Preparation of Ethyl Ethanoate ’07 1 Small quantities of ethyl ethanoate can be prepared by heating a mixture of glacial ethanoic acid with pure ethanol in the presence of a small quantity of concentrated sulphuric acid in a boiling tube. 2 To prepare large quantity of esters, the alcohol and carboxylic acid need to be heated under reflux. 3 Heating under reflux is necessary as ethanol, C2H5OH is very volatile. If the mixture is not heated under reflux, the ethanol C2H5OH will vaporise and escape into the atmosphere before it can react with ethanoic acid, CH3COOH. Write the chemical equation for the reaction between methanol and ethanoic acid. Solution Methanol, an alcohol reacts with ethanoic acid, a carboxylic acid to produce an ester and water. The ester will be methyl ethanoate. CH3OH + CH3COOH → CH3COOCH3 + H2O Carbon Compounds 368 SPM To prepare ethyl ethanoate ’04/P3 Apparatus Round bottomed flask, Liebig condenser, oil bath, porcelain chips Materials Pure ethanol, glacial ethanoic acid and concentrated sulphuric acid Procedure 1 About 30 cm3 of pure ethanol and 25 cm3 of glacial ethanoic acid are placed in a round bottom flask. Small pieces of porcelain chips are added to prevent bumping and to ensure smooth boiling. 2 About 5 cm3 of concentrated sulphuric acid is added to the reaction mixture. The mixture is shaken gently to ensure complete mixing. 3 The Liebig condenser is fitted vertically to the round bottom flask. The mixture is boiled under reflux for about 30 minutes. 4 After boiling, pure ethanol is obtained by distillation. Figure 2.11 Preparation of ethyl ethanoate by reflux Conclusion Ethyl ethanoate is produced when ethanoic acid and ethanol are heated in the presence of concentrated sulphuric acid as a catalyst. CH3COOH + C2H5OH → CH3COOC2H5 + H2O Natural Sources of Esters 2.7 1 Name the following esters and identify the alcohols and carboxylic acids required to prepare these esters. (a) HCOOCH3 (b) CH3COOC3H7 (c) C3H7COOCH3 1 Most of the simple esters exist naturally in flowers and fruits. For example, pentyl ethanoate is found in bananas, octyl ethanoate in lime and methyl butanoate in apples. 2 These volatile esters are responsible for the fragrant smell of flowers and fruits. 3 Vegetable oils and animal fats are esters with large molecules. For example, coconut oil and palm oil. 4 Waxes such as beeswax, wax found on leaves and candle wax are solid esters. 3 Methanol reacts with butanoic acid under certain conditions to produce an ester. (a) Write a balanced equation for the reaction that occurs. (b) State the conditions for the reaction that took place. (c) State two physical properties of the ester produced. Used as solvents for many organic compounds Uses of esters To make synthetic polymers Esters in oils are used to make soap Used as medicine, for example, aspirin 369 Carbon Compounds Activity 2.6 2 Name and draw the structural formulae of the organic compounds produced from the reactions between the following pairs of organic compounds. (a) Methanol + propanoic acid (b) Propan-1-ol + ethanoic acid (c) Propan-1-ol + methanoic acid Uses of Esters in Daily Life To make perfumes, cosmetics and artificial food flavourings 2 Observation A colourless liquid with a fragrant odour is obtained. 2.8 stearic acid is C17H35COOH or CH3(CH2)16COOH Oils and Fats 2 A fatty acid with a carbon-carbon double bond is an unsaturated acid. ’08/P1 For example: Oleic acid is C17H33COOH or CH3(CH2)7CH = CH(CH2)7COOH SPM 1 Oils and fats are naturally occurring esters and are found in animals and plants. 2 Fats (for example, butter) are found in animals. Oils are usually found in plants and fish. 3 Fats are solids at room temperature. In contrast oils are liquids at room temperature. Hence fats have higher melting points than oils. 4 When fats and oils are hydrolysed, glycerol SPM and long chain carboxylic acids are formed. ’08/P1 Hydrolysis means the decomposition (breaking up) of a chemical substance by water. The importance of oils and fats for body processes Sources of energy: Fats are high energy food, they provide energy for our bodies. 2 Ester + water → carboxylic acid + glycerol 5 The carboxylic acids produced from fats are also known as fatty acids. Fatty acids usually contain 16 or 18 carbon atoms per molecule. 6 Glycerol is an alcohol that contains three hydroxyl (–OH) groups per molecule. H | H—C— | OH H | C— | OH Source of nutrients: Fats are required to enable the human body to absorb vitamins A, D, E and K Importance of oils and fats in our bodies H | C—H | OH Thermal insulation: The layer of fat beneath the skin regulates body temperature propane-1,2,3-triol (glycerol) 7 There are two types of carboxylic acids, namely, saturated carboxylic acids and unsaturated carboxylic acids. (a) Saturated carboxylic acids do not contain double bonds. (b) Unsaturated carboxylic acids contain double bonds. 8 Animal fats usually contain a high percentage of saturated carboxylic acids whereas vegetable oils and fish oil contain a high percentage of unsaturated carboxylic acids upon hydrolysis. 9 Saturated fats are fats that contain saturated carboxylic acids. Unsaturated fats are fats that contain unsaturated carboxylic acids. 10 The presence of double bonds in unsaturated fats causes them to have lower melting points than saturated fats. Hence unsaturated fats (commonly known as oils) exist as liquids at room temperature. Protection: The layer of fats around the vital organs acts as a protective cushion Conversion of Unsaturated Fats to Saturated Fats 1 Vegetable oils can be converted to saturated fats by hydrogenation. 2 Hydrogenation is the chemical process in which hydrogen is added to the double bond between carbon atoms (C = C bond). 3 The hydrogenation process will change liquid vegetable oils to solid fats. 4 In the manufacture of margarine from vegetable oil, hydrogenation is carried out by passing hydrogen gas into palm oil at 200 °C and 4 atmospheres in the presence of nickel powder as catalyst. Ni Palm oil + hydrogen ⎯⎯⎯⎯⎯→ 200 °C, 4 atm margarine (fat) 1 A fatty acid with the general formula of CnH2n+1COOH is a saturated acid. For example: Palmitic acid is C15H31COOH or CH3(CH2)14COOH and Carbon Compounds 5 The hardness of margarine formed depends on the degree of hydrogenation. Partial hydrogenation will produce soft margarine. 370 Effects of Eating Food High in Fats on Health 2.8 1 State one similarity and one difference between fats and oils (a) in terms of their molecular structures. (b) in terms of their physical properties. 1 Food high in fats is high in calories. Hence high comsumption can lead to obesity. 2 Saturated fats contain a high percentage of cholesterol. Excess intake of saturated fats increase the risk of (a) hypertension (high blood pressure) (b) cardiovascular disease (heart disease) (c) stroke 2 Stearic acid is a saturated fatty acid whereas oleic acid is an unsaturated fatty acid. (a) Explain what is meant by a saturated fatty acid and an unsaturated fatty acid. (b) Give a test to differentiate stearic acid from oleic acid. 1 There are two types of oil extracted from fresh oil palm fruits. (a) Palm oil from the flesh of the fruit (b) Palm kernel oil from the kernel or seed 2 The flowchart shows the steps involved for the extraction of palm oil in the palm oil mill. Sterilisation Fruit branches are sterilised to kill fungus and bacteria Stripping Fruits are separated from the branches Digestion Fruits are heated to break down the oil-bearing cells Pressing Purification 2.9 Natural Rubber Natural Polymers 1 Polymers can be classified into two broad categories: natural polymers and synthetic polymers. 2 Examples of naturally-occurring polymers are natural rubber, carbohydrates and proteins. 3 The monomer of natural rubber is isoprene, C5H8. Hence natural rubber is polyisoprene. Natural rubber is produced by the addition polymerisation of isoprene. Oil is pressed out from fruits H CH3 H H | | | | addition polymerisation n(H — C = C — C = C — H) ⎯⎯⎯⎯⎯⎯⎯⎯→ Mixture is filtered to separate the oil. Oil is then dried. isoprene (monomer) H CH3 H H | | | | — ( C ⎯ C = C ⎯ C— )n | | H H Advantage of palm oil as a vegetable oil Rich in Vitamins A and E Lower the LDL or bad cholesterol and raise the HDL or good cholesterol in the body poly(isopropene) – natural rubber 4 The monomers for carbohydrates such as sugar, starch and cellulose is glucose. When starch is heated with dilute acid, glucose is produced. This reaction is called hydrolysis. Advantage of palm oil as a vegetable oil Withstand heat and resistant to oxidation, hence suitable to be used for deep frying – (C6H10O5)n– + nH2O → nC6H12O6 starch Highly competive in price. Cheaper than other types of vegetable oils glucose 5 The monomers of proteins are amino acids. Amino acids are joined together by peptide linkages. Hence proteins are polypeptides. 371 Carbon Compounds 2 3 (a) Name the process in which palm oil is converted to margarine. (b) State the conditions required for this conversion. (c) State one difference in physical property between palm oil and margarine. Industrial Extraction of Palm Oil (a) the hydrogen ions from the acid neutralise the negative charges on the membranes’ surfaces of the colloidal particles. (b) When the neutral rubber particles collide, the membranes will break, releasing the rubber polymers to form lumps. Hence the latex solidifies. 8 The coagulation of latex can occur without the addition of acid if the latex is exposed to air for a few days. This is because (a) the bacteria present in the latex produces organic acids. (b) the hydrogen ions from the acids produced neutralises the negative charges on the rubber particles. 9 Coagulation of latex can be prevented by the addition of aqueous ammonia because (a) the hydroxide ions from aqueous ammonia neutralise the acids produced by bacteria. (b) the negative charges at the membranes of rubber particles are maintained. 2 Coagulation of Latex 1 The milky fluid from tapped rubber trees is called latex. 2 The conversion of latex to the solid form is known as coagulation. 3 Latex is a colloidal solution containing an aqueous suspension of rubber particles. 4 Each rubber particle contains rubber polymers enclosed with a protein membrane with negative charge. 5 The negative charge on the membrane’s surface repel colloidal particles from one another, preventing the rubber polymers from combining together. Hence the latex remains in liquid form. 6 Coagulation of rubber can be speeded up by the addition of acids. 7 The addition of a weak acid on latex causes coagulation because add a weak coagulation acid of latex • Latex is coagulated by adding a weak acid such as methanoic acid or ethanoic acid. When an acid is added to latex, the hydrogen ions from the acid will neutralise the negative charges on the surfaces of the colloidal particles. As a result, the particles become neutral and can come closer and collide with one another. • The collisions between latex particles break open the protein membrane and releases the rubber polymers. • The rubber polymers can coalesce (combine) and form lumps of rubber. This process of forming lumps of rubber is called coagulation of latex. • The lumps of rubber are white solids and are quite elastic. Figure 2.12 The coagulation of latex Activity 2.7 To investigate the coagulation of latex and methods to prevent coagulation Apparatus Beaker, glass rod and dropper Procedure 1 About 50 cm3 of latex are placed in three beakers labelled A, B and C respectively. 2 Using a dropper, dilute ethanoic acid is added to the latex in beaker B. The mixture is stirred with a glass rod until the latex becomes acidic (blue litmus paper turns red). Materials Fresh latex, dilute ethanoic acid, dilute aqueous ammonia and blue litmus paper. Carbon Compounds SPM ’08/P3 372 3 Aqueous ammonia is slowly added to beaker C until the latex becomes alkaline (red litmus paper turns blue). 4 The three beakers are left overnight. The changes that occurred are recorded. Beaker Results Latex only B Latex + acid Coagulation of latex occurs rapidly C Latex + aqueous ammonia Latex does not coagulate (no visible changes) Observation Coagulation occurs slowly Properties of Natural Unvulcanised Rubber and Vulcanised Rubber 2 A Chemicals added to latex 6 Vulcanisation is the process of hardening rubber by heating it with sulphur or sulphur compounds. 7 Vulcanisation of rubber is carried out by (a) heating natural rubber with sulphur at about 140 °C, using zinc oxide as catalyst, (b) immersing rubber in a solution of disulphur dichloride (S2Cl2) in methylbenzene. 8 In vulcanised rubber, the sulphur atoms form cross-links between long chains of rubber polymers. SPM ’10/P1 1 Natural rubber has the following properties: (a) Quite elastic (b) Water repellent (c) Does not conduct electricity 2 Elasticity is the ability of an object to be stretched and then returned to its original shape when the stress is removed. 3 Natural rubber before treatment with sulphur is unvulcanised rubber. 4 Unvulcanised natural rubber has few practical uses because (a) it is not elastic enough (b) it becomes soft and sticky when heated (c) it becomes brittle and crack easily when oxidised by oxygen. 5 When unvulcanised rubber is stretched, the coiled rubber molecule is lengthened and straightened. c s s c rubber polymer c s s c c s s c sulphur cross-links TC 2/15 c c Figure 2.15 Vulcanised rubber has sulphur cross-links 9 Vulcanised rubber has many uses because of improved properties. Vulcanisation makes the rubber more elastic, stronger and more resistant to heat and oxidation. Improved properties of vulcanised rubber Stronger and harder polymer chain of rubber under ordinary conditions Observation Conclusion 1 Acids such as ethanoic acid speed up the coagulation of latex. 2 Alkalis such as aqueous ammonia slow down the coagulation of latex. 3 Coagulation of latex can occur by itself slowly. Figure 2.13 To investigate the coagulation of latex Beaker Chemicals added to latex polymer chain of rubber when straightened Figure 2.14 Unvulcanised rubber is not very elastic 373 Explanation The sulphur cross-links prevent the polymer chains from slipping past one another when stretched Carbon Compounds Improved properties of vulcanised rubber Improved properties of vulcanised rubber Explanation More elastic The sulphur cross-links pull the chains back to their original arrangement when released More resistant to heat The presence of sulphur increases the melting point of rubber More resistant to oxygen Explanation Sulphur cross-links reduce the number of double bonds in the molecules of vulcanised rubber 2 To produce vulcanised rubber and to compare the properties of vulcanised rubber and unvulcanised natural rubber (A) Preparation of natural rubber and vulcanised rubber Materials Vulcanised rubber and unvulcanised rubber. Apparatus Glass plate, beaker, a pair of tongs and razor blade. Procedure 1 A strip of vulcanised rubber is hung using a clip. 2 The original length of the vulcanised rubber strip is measured. 3 A 20 g weight is hung on the strip of vulcanised rubber. 4 The increase in length of the vulcanised rubber strip is measured. 5 The weight is removed and the final length of the vulcanised rubber strip is measured again. 6 Steps 1 to 5 are repeated using the unvulcanised rubber strip of the same length to replace the vulcanised rubber strip. Materials Disulphur dichloride in methylbenzene and rubber latex. Procedure 1 A small quantity of latex is poured on a glass plate. 2 A glass rod is used to level the latex to produce a flat, thin layer of latex about 1 mm thick. 3 The glass plate is put aside for one day for the latex to coagulate. 4 The rubber produced is cut into two strips of rubber of equal size using a razor blade. 5 One of the strips of rubber is dipped into a solution of disulphur dichloride in methylbenzene for 2-3 minutes to produce a strip of vulcanised rubber. 6 The strip of vulcanised rubber is then removed from the solution and dried with filter paper. (B) To compare the elasticity of vulcanised rubber and unvulcanised rubber Apparatus Clip, retort stand with clamp, metre rule and weight. FOLS UHWRUWVWDQG UXOHU WKUHDG JZHLJKW SPM ’06/P3 Figure 2.16 To compare the elasticity of rubber Results Activity 2.8 UXEEHUVWULS Original length (cm) Stretched length with weight (cm) Increase in length (cm) Final length after weight is removed (cm) Unvulcanised rubber X X1 X1 – X = Y1 Y3 Vulcanised rubber X X2 X2 – X = Y2 Y4 Type of rubber Carbon Compounds 374 Discussion 1 The increase in length of vulcanised rubber (stretched length) is less than the increase in length of unvulcanised rubber (that is, Y2 < Y1). This shows that vulcanised rubber is harder and stronger than unvulcanised rubber. 2 The vulcanised rubber has a higher ability to return to its original length after the weight is taken off (that is Y4 < Y3). This shows that the vulcanised rubber is more elastic than unvulcanised rubber. Conclusion Vulcanised rubber is harder, stronger and more elastic than unvulcanised rubber. Uses of Natural Rubber 2 Unvulcanised natural rubber is used for making adhesive (glue) and as crepe rubber in insulating blankets. 3 Vulcanised rubber has many uses in industries and home. Vehicle tyres Surgical gloves and protective gloves Shock absorbers Insulating layer for electric cables and equipment Shoe soles 2 1 Unvulcanised natural rubber has limited uses as it is soft, has poor heat resistance and does not wear well. It becomes soft and sticky when heated. It also becomes hard and brittle due to oxidation. Rubber hoses Things made from vulcanised natural rubber Conveyor belts Balloons Rubber mattresses Rubber bands Research on Natural Rubber in Malaysia 1 In Malaysia, the research and development on rubber is conducted by the (a) Rubber Research Institute of Malaysia (RRIM) (b) Malaysian Rubber Producer Research Association (MRPRA) (c) Rubber Board of Malaysia 2 The scope of research and development activities include (a) finding new uses of rubber and rubber products (b) improving the quality of natural rubber (c) automating the tapping system so as to overcome labour shortage. 375 Carbon Compounds 2.9 2.11 1 Complete the following table. Natural polymer Monomer Uses of Various Organic Materials in Everyday Life Protein Isoprene 1 Organic compounds were originally extracted from living things, products or remains of living things. The term organic actually means ‘derived from living organisms’. Living things are made of complex organic compounds that have structural, chemical or genetic functions. 2 However since the nineteenth century, organic compounds have been synthesised in the laboratory from inorganic materials. In 1828, the German chemist Friedrich Wohler was able to synthesise urea (an organic compound) from ammonium cyanate (an inorganic compound). 3 In present times, thousands of organic compounds are synthesised every year to fulfill our needs in the modern society. These synthetic organic compounds include synthetic polymers, vitamins, medicines, cosmetics, pesticides, paints, varnishes, glues, adhesives, synthetic fibers for clothing materials and others. 4 Organic synthesis is the preparation of specific and desired organic compounds from readily available resources. 5 Research and development towards natural organic compounds enable us to (a) simulate the structure of natural organic compounds and make useful synthetic organic compounds which are imitations of natural compounds. Examples: Dyes, food flavours, fragrances and medicines. (b) extract the active ingredients from traditional medicines. Cheaper and more effective medicines without side effects can then be made commercially. Generic medicines are made to lower the cost of medicines for consumers. (c) produce seeds of higher quality and more resistant towards pests, so as to increase the yield of food production. (d) find new uses for agriculture products. For example, oil palm waste is used to produce biomass fuel and make composite construction materials. 6 The economical development of our country depends heavily on products of organic Carbohydrate Starch 2 State the chemicals used in the following processes in rubber industries and their effects on the properties of rubber. (a) Coagulation (b) Vulcanisation 2 3 What causes latex to coagulate under natural conditions? Suggest a method to prevent this phenomenon. 2.10 Order in Homologous Series 1 Organic compounds are grouped in families called the homologous series to make the study of organic chemistry more systematic and orderly. 2 Alkanes, alkenes, alcohols, carboxylic acids and esters are examples of homologous series. 3 The chemical properties of members of a homologous series are the same as they have the same functional group. 4 The physical properties of members of a homologous series show a regular pattern and change gradually as the number of carbon atoms increases. Descending the homologous series as the number of carbon atoms increases • Relative molecular mass increases • Melting point increases • Boiling point increases • Volatility decreases • Density increases • Solubility decreases 5 The order in the physical properties of members in a homologous series enables us to predict the properties of an unknown member in the series. Carbon Compounds The Variety of Organic Materials in Nature 376 1 Organic carbon compounds include hydrocarbons such as alkanes and alkenes, alcohols, carboxylic acids, esters, fats and oils and natural rubber. 2 Physical properties of alkanes and alkenes: (a) Low melting and boiling points (b) Insoluble in water, soluble in organic solvents (c) Non-conductors of electricity 3 Chemical properties of alkanes (general formula: CnH2n+2): (a) Combustion in excess oxygen to form CO2 and H2O (b) Substitution reactions with halogen under ultraviolet light 4 Chemical properties of alkenes (general formula: CnH2n): (a) Combustion in excess oxygen to form CO2 and H2O (b) Addition reaction (i) with H2 (hydrogenation) to form alkanes (ii) with H2O (hydrolysis) to form alcohols (iii) with KMnO4 to form diols (iv) with Cl2 or Br2 to form dichloroalkanes or dibromoalkanes (c) Polymerisation to form polymers 5 Chemical properties of alcohols (general formula: CnH2n+1OH): (a) Combustion in excess oxygen to form CO2 and H2O (b) Oxidation by acidified KMnO4 or K2Cr2O7 to form carboxylic acids (c) Dehydration by heated porcelain to form alkenes (d) Esterification with carboxylic acids (with conc. H2SO4 as a catalyst) to form esters 6 Preparation of ethanol, C2H5OH: (a) Fermentation of glucose by yeast 7 8 9 10 11 12 377 (b) Hydrolysis of ethene (phosphoric acid as a catalyst) Chemical properties of carboxylic acids (general formula: CnH2n+1COOH): (a) Reacts with reactive metals to form salts and H2 gas (b) Reacts with metal carbonates to form salts and CO2 gas (c) Reacts with alkalis to form salts and H2O (d) Esterification with alcohols to form esters and H2O Physical properties of esters (general formula: RCOOR′): (a) Have sweet/ fragrant/ fruity smells (b) Insoluble in water Fats and oils are esters. (a) Fats are solids (with lower melting points) and are saturated (without C = C bonds). (b) Oils are liquids (with higher melting points) and are unsaturated (with C = C bonds). (c) Oils can be converted to margarine by hydrogenation. Hydrolysis of fats (or oils) will produced fatty acids (long chained carboxylic acids) and glycerol. Coagulation is the conversion of liquid latex to rubber solid. It (a) can be speeded by the addition of acids. (b) can occur slowly when the bacteria present produce acids. (c) can be prevented by the addition of aqueous ammonia. Vulcanisation of rubber is the conversion of natural rubber to vulcanised rubber by forming sulphur cross-links between rubber polymers. Vulcanised rubber is stronger, harder, more elastic and more resistant to heat and oxidation. Carbon Compounds 2 8 In 2012, universities and research institutes in collaboration with Agensi Innovasi Malaysia have successfully developed products such as disease-resistant chilli, lumber from oil palm, coconut body armour, biopesticide to control mosquito larvae and mosquito repellant gel. compounds such as petroleum, palm oil and natural rubber. 7 Presently, research and development of natural resources in our country is encouraged to produce new and useful organic compounds. New synthetic medicines are developed to combat disease and new polymers are synthesised to replace the use of metals and even replace organs in our bodies. 2 Multiple-choice Questions 2.1 Carbon Compounds 2 1 Which of the following is an organic compound? A Calcium carbonate B Glucose C Carbon monoxide D Copper(II) oxide 2 Which of the following products are formed from the complete combustion of all organic compounds? A Water and carbon dioxide B Carbon dioxide and carbon monoxide C Water, carbon dioxide and nitrogen dioxide D Carbon and water 2.2 Alkanes 3 Which of the following is a saturated hydrocarbon? ’06 A Alkane B Alkene C Alcohol D Carboxylic acid 4 Which of the following substances can undergo substitution reaction ’06 with chlorine in sunlight? A Ethane B Ethene C Ethanol D Ethanoic acid 5 What are the products formed when ethane is burnt with ’07 excess oxygen? A Carbon and hydrogen B Carbon dioxide and water C Carbon monoxide and water D Carbon monoxide and hydrogen Carbon Compounds 6 Which of the following produces the most soot when burned in air? A Methane B Ethane C Butane D Hexane 2.3 Alkenes 7 Which of the following substances can be used to ’04 differentiate propene from propane? A Limewater B Bromine water C Sodium hydroxide solution D Potassium dichromate(VI) solution 8 Which of the following is the structural formula of an ’05 unsaturated hydrocarbon? A H— H H H H | | | | C—C=C—C—H | | H H H H H H | | | | B H—C—C=C—C—H | | H OH C H— D H— H O | i C—C—O—H | H H | C— | H H | C— | H 378 H | C— | H H | C—H | H 9 Propene can be transformed to propane by the process of ’07 A hydration B oxidation C dehydration D hydrogenation 10 A hydrocarbon compound is burnt completely in air to form ’05 7.2 cm3 of carbon dioxide gas and 7.2 cm3 of water vapour. What is the molecular formula of the hydrocarbon compound? [Given that 1 mol of gas occupies 24 dm3 at room temperature] A C2H6 B C3H6 C C3H8 D C6H6 11 What is the product formed when propene is shaken with ’10 chlorine water? A 1, 1- dichloropropane B 2, 2 -dichloropropane C 1, 2- dichloropropane D 1, 3- dichloropropane 12 The diagram below shows the structural formula of a ’11 compound. H | H—C—H H H H | | | H—C=C—C—C—H | | H H What is the name of this compound? A Pentene B Methylbutene C 2-methylbut-1-ene D 2-methylbut-2-ene H | A —C— | H H | C— | H H | C— | H H | C— | H H | B —C— | H CH3 | C— | H H | C— | H CH3 | C— | H H | C —C— | H CH3 | C— | CH3 H | C— | H H | C— | H CH3 | D —C— | H H | C— | H H | C— | H H | C— | H 2.4 16 Which of the following pairs of compounds are isomers? A Butane and butene B Butane and 2-methylbutane C But-1-ene and 2-methylpropene D 2,2-dimethyl propane and 2-methylpropane 2.5 15 The structural formula of a compound is given below. H | H—C—H H H H | | | H—C—C—C—C—H | | | | H H H H Which of the following is the IUPAC name for the compound? H | H—C—H H H | | H—C—C=C—H | H Pentane Methylbutane 2-methylbutane 3-methylbutane Alcohols 17 The diagram shows the set-up of apparatus for the reaction of ’06 yeast with glucose solution. Isomerism 14 Which of the following is true of all the isomers of a hydrocarbon? A They have the same structural formula. B They have the same functional group. C They have the same chemical properties. D They produce the same number of moles of carbon dioxide and water on complete combustion. ’05 A B C D What is the name of the reaction that has taken place in the conical flask after a few days? A Oxidation B Hydration C Hydrolysis D Fermentation 18 The following chemical equation shows the conversion of ethanol ’04 to ethanoic acid. C2H5OH + 2[O] → CH3COOH + H2O What is the name of the process shown by the above equation? A Dehydration B Reduction C Oxidation D Fermentation 19 The diagram shows the structural formula of compound X. 379 What is the name of the compound formed when compound X reacts with potassium permanganate? A Butanol B 2-methylpropanol C Butan-1,2-diol D 2-methylpropan-1,2-diol 20 The diagram below shows the structural formula of ’04 pent-1-ene. H H H | | | H—C=C—C— | H H | C— | H H | C—H | H Which of the following are the possible alcohols that can produce pent-1-ene on dehydration? I Pentan-1-ol II Pentan-2-ol III Pentan-3-ol IV 2-methylbutan-1-ol A I and II only B II and IV only C I and IV only D I, II and III only 21 The diagram below shows the set-up of the apparatus for a ’05 reaction. glass wool soaked with ethanol porcelain chips compound X heat heat water Which of the following is compound X? A Ethane TC 2/18 B Ethene C Ethanoic acid D Carbon dioxide Carbon Compounds 2 13 Polypropene is a polymer that is formed from the combination ’07 of propene molecules. Which of the following represents part of the structure of polypropene? 22 CH3 | H3C – C – OH | H 2 Which of the following may be true of the compound represented in the figure above? A Decolourise the brown colour of bromine water. B Decolourise the purple colour of cold aqueous potassium permanganate. C Changes the orange colour of potassium dichromate to green when heated. D Produces a sweet-smelling liquid when heated with ethanol and concentrated sulphuric acid. 2.6 fermentation X oxidation Y Which of the following pairs of compounds may be compound X and compound Y? X 23 Compound X has the following properties: • Reacts with sodium carbonate to produce a gas that turns limewater milky. • Reacts with ethanol to produce a sweet-smelling compound. Which of the following compounds may be compound X? A Carbonic acid B Ethanoic acid C Propanol D Ethyl ethanoate 24 A liquid produced effervescence when reacted with zinc metal. ’05 Which of the following may be the molecular formula of the liquid? A HCOOH B HCOOCH3 C CH3OH D CH3COONa 25 The diagram shows the conversion of glucose to compound X and subsequently to compound Y. Carbon Compounds Y A Ethanol Ethanoic acid B Ethanol Ethene C Propanol Propanoic acid D Yeast Carbon dioxide 2.7 Carboxylic Acids Which of the following is the structural formula for compound Z? O i A C2H5 — C — O — CH3 Glucose Esters 26 Scented flowers contain naturally occurring esters. Which of the ’04 following is a property of an ester? A Soluble in water B Low boiling point C Higher density than water D Change blue litmus to red 27 The diagram below represents the structural formula of a ’04 carbon compound. O i CH3 — CH2 — C — O — CH2 — CH2 — CH2 — CH3 The compound is produced by the reaction between A ethanol and butanoic acid B butanol and ethanoic acid C butanol and propanoic acid D propanol and butanoic acid 28 The diagram below shows the process of producing compound ’05 Z. C3H6 O i B CH3 — C — O — C3H7 O i C C2H5 — C — O — C2H5 O i D C2H5 — C — O — C3H7 29 Which of the following equations can produce a product with a ’06 sweet fruity scent? I CH3COOH + NaOH → CH3COONa + H2O II CH3OH + C2H5COOH → C2H5COOCH3 + H2O III C2H5OH + 2[O] → CH3COOH + H2O IV C5H11OH + CH3COOH → CH3COOC5H11 + H2O A I and III only B II and IV only C III and IV only D I, II and IV only 30 The diagram shows the conversion of ethanol to compound X and subsequently to compound Y. acidified KMnO4 C2H5OH ⎯⎯⎯⎯⎯⎯→ X CH3OH, concentrated ⎯⎯⎯⎯⎯⎯⎯⎯⎯→ Y H2SO4, reflux + steam Ethanol + C2H5OH Y oxidation Compound Z 380 X Which of the following may be compound Y? A Ethanoic acid B Ethyl ethanoate C Ethyl methanoate D Methyl ethanoate A I only B III only Oils and Fats 31 Which of the following belong to the homologous series of esters? I Palm oil II Margarine IIISodium butanoate IV Glycerol A I and II only B I and III only C III and IV only D I, II and III only C I, II and III only D I, II, III and IV 34 The diagram shows the change in structure of natural rubber after process P. c c c c c c c c c c c c c s s c c s s c c c rubber X 32 Which of the following molecular formulae of fatty acids is formed from the hydrolysis of a nonsaturated fat? A C15H31COOH B C17H35COOH C C17H33COOH D C23H47COOH c s s c c c rubber Y Which of the following statements is true about the change? A Rubber X is more elastic than rubber Y. TCY.2/19 B Rubber X is stronger than rubber C Rubber X is more resistant to heat than rubber Y. D Rubber X is more easily oxidised than rubber Y. 35 A rubber tapper faces a problem of transporting latex to a glove-making factory in liquid form. To solve the problem he has to add a substance ’06 into the latex to prevent the coagulation of the latex. Choose the correct substance and the explanation to solve the problem. Substance 2.9 process P Natural Rubber 33 Which of the following chemicals can cause latex to coagulate? I Methanoic acid II Sulphuric acid III Ethanoic acid IV Nitric acid Explanation A Water To make the latex more dilute B Ethanoic acid Contains H+ ions that neutralise the negative charge on the membrane of the rubber particle C Ammonia solution Contains OH– ions that neutralise the H+ ions from the acids produced by bacteria D Sodium chloride solution As a preservative to maintain the original state of the latex Structured Questions (b) Complete Table 1 by filling in the molecular formula of propane and the name of C4H10. 1 Table 1 shows the names, molecular formulae, melting points and boiling points of a few straight chain members of a homologous series. Name Ethane Molecular formula Melting point (°C) Boiling point (°C) C2H6 –183 –89 –188 –42 C4H10 –138 –0.5 C5H12 –130 36 Propane Pentane [2 marks] (c) Draw the structural formula of the first member of this homologous series. [1 mark] (d) Predict the physical states of (i) propane (ii) pentane [1 mark] [1 mark] (e) Give the name and the molecular formula of the member of the same homologous series after pentane. [2 marks] (f) Write a balanced equation for the complete combustion of ethane. [1 mark] Table 1 (a) Give the name and the general formula of this homologous series. [2 marks] 381 Carbon Compounds 2 2.8 (b) State the function of concentrated sulphuric acid [1 mark] in this reaction. (c) (i) Name the reaction for the preparation of [1 mark] ethyl ethanoate. (ii) Write the chemical reaction for the reaction [1 mark] in (i). 2 Propene is an important hydrocarbon in the petrochemical industries. Diagram 1 shows the conversion of propene into other organic compounds. Polypropene process II (d) The experiment is repeated by replacing ethanol with methanol. [1 mark] (i) Name the ester formed. (ii) Draw the structural formula of the ester [1 mark] formed. (iii) State one physical property of the ester. process process Propene Propan-1-ol I III Propane process IV Compound X [1 mark] Diagram 1 (e) The flowchart shows the conversion of ethene to ethanol and then to ethanoic acid. Based on Diagram 1, answer the following questions. 2 (a) State the homologous series of propene. [1 mark] (b) Name process I. [1 mark] Ethene (c) Under certain conditions, propene reacts to form polypropene. Write an equation for the formation of polypropene in process II. [1 mark] (d) (i) Explain briefly how process III is carried out in industries. [2 marks] (ii) Draw the structural formula of propan-1-ol. [1 mark] (e) Acidified potassium manganate(VII) is added to propene in process IV. (i) Predict the observation that will take place. process I Ethanol process II Ethanoic acid Based on the flowchart, write the chemical equation for [1 mark] (i) process I (ii) process II [1 mark] (iii) State a suitable chemical that can be used [1 mark] to carry out process II. 4 Diagram 3 shows conversions I, II and III starting with glucose. [1 mark] (ii) Write a balanced equation for the reaction that has occurred. [1 mark] Glucose (f) Both propene and propane are combustible in air. Compare and explain the difference in the quantity of soot produced by the two compounds during combustion. [2 marks] yeast Liquid A II Liquid B reagent III concentrated H2SO4 asid Liquid C 3 Diagram 2 shows the set-up of apparatus for the preparation of ethyl ethanoate by heating ethanol and ethanoic acid under reflux. Diagram 3 (a) (i) Name conversion I. [1 mark] (ii) Draw the structural formula of liquid A. water out [1 mark] (b) Liquid B has a vinegary smell. (i) Name the type of reaction that takes place [1 mark] in conversion II. (ii) Write a chemical equation for conversion II. TC 2/20 water in xxxxxxxxxxxxxx [1 mark] (c) Give one chemical test that can be used to [2 marks] distinguish liquid B from liquid A. ethanol, ethanoic acid and concentrated sulphuric acid (d) heat Diagram 2 (i) Name the homologous series of which [1 mark] liquid C is a member. (ii) Name liquid C. [1 mark] (iii) State one use of liquid C. [1 mark] (e) Suggest another method to produce liquid A other than from glucose in conversion I. (a) Why is ethanol and ethanoic acid heated under reflux? [1 mark] Carbon Compounds I 382 [1 mark] (c) Name compound X. 5 Diagram 4 shows the conversion of latex to compound X. Latex process I Natural process II rubber c s s c c s s c c s s c (d) State two differences in physical properties between natural rubber and compound X. TC 2/21 [2 marks] (e) How is process II carried out in the industry? [1 mark] (f) (i) Name a chemical that can be used to retain [1 mark] latex in the liquid form. (ii) Explain the function of the chemical named [1 mark] in (i). c c compound X Diagram 4 (a) Name process I and process II. [1 mark] [2 marks] (b) Name a chemical that can be used to carry out process I. [1 mark] (g) Give one use of compound X. [1 mark] Essay Questions 2 (a) Diagram 1 shows the formation of carboxylic acids from alcohols. (b) The information below is referring to carbon compound X. • • • • • Alcohols Carboxylic acids 2 1 (a) Draw the structural formulae of two isomers of but-1-ene and give their IUPAC names. [4 marks] Diagram 1 Carbon 40.0% Hydrogen 6.7% Oxygen 53.3% Relative molecular mass = 60 Relative atomic mass of H = 1, C = 12 and O = 16 Using suitable reagents and with the help of a labelled diagram, describe how you can prepare a named carboxylic acid in the laboratory. Include the observation and a test to show that the product formed is an acid. Write a chemical equation for the reaction involved. [10 marks] Based on the information of the carbon compound X, (i) determine the molecular formula of X. (ii) draw the structural formula of X. (iii) name the carbon compound X. (iv) write the general formula for its homologous series. [8 marks] (b) Many artificial flavours used in the food industry are esters. Various types of esters can be formed from the esterification between an alcohol and a carboxylic acid. Name one possible ester that can be formed and describe how you can prepare the named ester in the laboratory. Name the alcohol and carboxylic acid that is used and the chemical equations involved. (c) Margarine can be made from palm oil. Compare and contrast margarine and palm oil in terms of their structures and physical properties. Briefly describe how palm oil can be converted to margarine. [8 marks] [10 marks] Experiment 1 Your (i) (ii) (iii) (iv) (v) (vi) Hexane is a saturated hydrocarbon whereas hexene is an unsaturated hydrocarbon. Both are colourless liquids. However they undergo different reactions toward addition reaction. You are required to plan an experiment to differentiate the two compounds. 383 planning should include the following: Statement of the problem All the variables Hypothesis List of materials and apparatus Procedure Tabulation of data [17 marks] Carbon Compounds FORM 5 THEME: Interaction between Chemicals CHAPTER 3 Oxidation and Reduction SPM Topical Analysis 2008 Year 1 Paper 2009 3 2 Section A B C Number of questions 1 — 6 – 1 5 – 1 2010 2 4 3 A B C – – 1 1 – 4 2011 2 3 A B C – – – – 1 3 2 3 A B C 1 – – – ONCEPT MAP Oxidising agent A substance that accepts (gains) electrons and is itself reduced in the process Reducing agent A substance that donates (loses) electrons and is itself oxidised in the process Oxidation • Gain of oxygen/increase in oxidation number • Loss of hydrogen/loss of electrons Reduction • Loss of oxygen/decrease in oxidation number • Gain of hydrogen/gain of electrons Rusting of iron Fe → Fe2+ + 2e– → Fe2O3.xH2O OXIDATIONREDUCTION (REDOX) REACTIONS Oxidation of Fe2+ ions: Fe2+ → Fe3+ + e– Reduction of Fe3+ ions: Fe3+ + e– → Fe2+ Combustion of metals Reactivity series of metals K > Na > Ca > Mg > Al > Zn > Fe > Sn > Pb > Cu > Ag > Au Prevention of rusting • Galvanising • Sacrificial metal • Alloying Displacement of halogens from the halides e.g. Cl2 + 2KBr → 2KCl + Br2 Displacement of metals from their salts e.g. Zn + CuSO4 → ZnSO4 + Cu Heating metal oxides with carbon and hydrogen Electrochemical series K > Na > Ca > Mg > Al > Zn > Fe > Sn > Pb > H > Cu > Ag Position of C and H in the reactivity series of metals K > Na > Ca > Mg > Al > C > Zn > H > Fe > Sn > Pb > Cu > Ag Electrolytic cell • Electrons flow from anode (+ve) to cathode (–ve) in the external circuit • Extraction of iron 2Fe2O3 + 3C → 4Fe + 3CO2 • Extraction of tin SnO2 + C → Sn + CO2 Chemical cell • Electrons flow from anode (–ve) to cathode (+ve) in the external circuit Redox Reactions Oxidation and Reduction Reactions Oxidation in Terms of Loss of Hydrogen 1 Oxidation can also be defined as the loss of hydrogen from a substance. If a substance loses hydrogen during a reaction, it is said to be oxidised. 2 When hydrogen sulphide gas is mixed with chlorine gas at room temperature, a yellow precipitate of sulphur is formed and hydrogen chloride gas is released. SPM ’08/P1, ’09/P1 1 Oxidation can be defined as (a) acceptance (gain) of oxygen, (b) donation (loss) of hydrogen, (c) loss of electrons and, (d) increase in the oxidation number of the element. 2 In contrast, reduction can be defined as (a) loss of oxygen, (b) gain of hydrogen, (c) gain of electrons and, (d) decrease in the oxidation number of the element. H2S(g) + Cl2(g) → 2HCl(g) + S(s) loss of hydrogen (oxidation) In this reaction, hydrogen sulphide loses hydrogen and is oxidised to sulphur. 3 When ammonia gas is passed over hot copper(II) oxide, the following reaction occurs. A newly cut apple turns yellow on exposure to air. This is due to the oxidation of apples by oxygen. The reaction is catalysed by the enzymes present in apples. 2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(l) oxidation Oxidation in Terms of Gain of Oxygen Ammonia undergoes oxidation because it loses hydrogen. In other words, ammonia is oxidised to nitrogen. 1 Oxidation is a chemical reaction in which oxygen is added to a substance. If a substance (element or compound) gains oxygen during a reaction, it is said to be oxidised. 2 When calcium burns in oxygen, the following reaction occurs: 2Ca(s) + O2(g) → 2CaO(s) Reduction in Terms of Loss of Oxygen 1 A reduction reaction is the reverse process of an oxidation reaction. Reduction is defined as the loss of oxygen from a substance. If a substance loses oxygen during a reaction, it is said to be reduced. 2 When a mixture of zinc powder and copper(II) oxide is heated, the following reaction occurs. addition of oxygen (oxidation) (a) This process is known as oxidation. (b) Calcium is oxidised to calcium oxide because it gains oxygen in this reaction. 3 Methane burns in air as represented by the equation: oxidation CH4(g) + 2O2(g) ⎯⎯⎯→ CO2(g) + 2H2O(g) oxidation In this reaction, (a) the carbon atom in methane is oxidised to carbon dioxide, (b) the hydrogen atoms in methane is oxidised to water. Therefore, combustion is an oxidation reaction. loss of oxygen (reduction) Zn(s) + CuO(s) → ZnO(s) + Cu(s) 3 In this reaction, copper(II) oxide has lost its oxygen. It is said to be reduced to metallic copper. 385 Oxidation and Reduction 3 3.1 oxidation Reduction in Terms of Gain of Hydrogen 1 Reduction can also be defined as the addition of hydrogen to a substance. If a substance gains hydrogen during a reaction, it is said to be reduced. 2 When a mixture of hydrogen and chlorine is exposed to sunlight, a vigorous reaction occurs and white fumes of hydrogen chloride are produced. Mg(s) + H2O(g) → MgO(s) + H2(g) reduction 4 In this reaction, magnesium has gained oxygen and is oxidised. In contrast, water has lost its oxygen and is reduced. addition of hydrogen (reduction) Respiration is a redox process. When respiration occurs, the food is oxidised and oxygen molecules accept electrons and are reduced to water. Photosynthesis is also a redox reaction. H2(g) + Cl2(g) → 2HCl(g) 3 In this reaction, chlorine has gained hydrogen. This means that chlorine has been reduced. 3 6CO2(g) + 6H2O(l) → C6H12O6(aq) + 6O2(g) sugar Electrons are removed from water molecules and are used to reduce carbon dioxide to sugar. Oxidising and Reducing Agents 1 An oxidising agent is a substance that brings about (causes) oxidation in another substance. In bringing about oxidation, the oxidising agent is itself reduced. 2 A reducing agent is a substance that brings about reduction in another substance and is itself oxidised. 3 The following are examples of oxidising and reducing agents. Antoine Lavoisier (1743–1794) Antoine Lavoisier is known as the Father of Modern Chemistry. He was the first chemist who explained the oxidation and reduction reactions that occur during combustion. Redox Reactions 1 Oxidation and reduction always take place together. A redox reaction is defined as a reaction in which both oxidation and reduction take place simultaneously. 2 In a redox reaction, when one substance in a reaction is oxidised, the other substance is reduced. 3 When steam is passed over heated magnesium, magnesium oxide and hydrogen are produced. Oxidation and Reduction SPM ’08/P1, ’09/P1 Oxidising agents Reducing agents • Chlorine and bromine • Acidified potassium manganate(VII) • Acidified potassium dichromate(VI) • Concentrated nitric acid • Metals such as sodium, magnesium, zinc and aluminium • Sulphur dioxide gas and hydrogen sulphide gas • Sodium sulphite and sodium thiosulphate 4 In a redox reaction involving A and B: if A is an oxidising agent, then B is the reducing agent and vice versa. 386 Reaction The reaction between copper(II) oxide and carbon: oxidised (gain oxygen) 2CuO(s) + C(s) → 2Cu(s) + CO2(g) Oxidising agent Reducing agent Copper(II) oxide • Copper(II) oxide oxidises carbon to carbon dioxide. • It is reduced to copper. Carbon • Carbon reduces copper(II) oxide to copper. • It is oxidised to carbon dioxide. Chlorine • Chlorine oxidises hydrogen sulphide to sulphur. • It is reduced to hydrogen chloride. Hydrogen sulphide • Hydrogen sulphide reduces chlorine to hydrogen chloride. • It is oxidised to sulphur. reduced (loss of oxygen) The reaction between chlorine and hydrogen sulphide: oxidised (loss of hydrogen) 3 Cl2(g) + H2S(g) → 2HCl(g) + S(s) reduced (gain hydrogen) 1 The equation for the reaction between iron(III) oxide and carbon monoxide is shown below. In many cases, oxidation or reduction reactions are accompanied by colour changes. Br2(brown) → Br–(colourless) … reduction MnO4–(purple) → Mn2+ (colourless) … reduction Cr2O72–(orange) → Cr3+ (green) … reduction Fe2+ (green) → Fe3+ (brown/yellow) … oxidation 2I– (colourless) → I2(brown) … oxidation Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) Identify the oxidising and reducing agents in this reaction. Solution Fe2O3 is an oxidising agent because it oxidises CO to CO2 and is itself reduced to Fe. oxidation Methylhydrazine, CH3NHNH2 is a powerful reducing agent. Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) 4CH3NHNH2 + 5N2O4 → 4CO2 + 9N2 + 12H2O It is used as rocket fuel for the Apollo 11 project which landed the first man on the moon on 21 July 1969. The compound is toxic and carcinogenic (cancer causing). It is also found in trace amounts in raw common mushrooms. reduction CO is a reducing agent because it reduces Fe2O3 to Fe and is itself oxidised to CO2. 387 Oxidation and Reduction 3.1 To identify oxidation and reduction processes in the reaction between metal oxides and carbon Problem statement In the reaction between metal oxide and carbon, which reagent undergoes oxidation and which reagent undergoes reduction? Materials Powdered carbon, powdered copper(II) oxide, iron(III) oxide and lead(II) oxide. 3 Hypothesis (a) Carbon undergoes oxidation to form carbon dioxide gas. (b) Copper(II) oxide, iron(III) oxide and lead(II) oxide undergo reduction to form copper, iron and lead respectively. Figure 3.1 Heating copper(II) oxide with carbon Procedure 1 A spatula of copper(II) oxide is placed in a crucible. 2 Another spatula of powdered carbon is added to the copper(II) oxide. 3 The two substances are mixed thoroughly and the mixture is then heated strongly. 4 The observations are recorded in the table given below. 5 Steps 1 to 4 are repeated using iron(III) oxide and lead(II) oxide in place of copper(II) oxide. Variables (a) Manipulated variable : Type of metal oxide (b) Responding variable : Reaction products (c) Constant variables : Carbon and the conditions of reaction Apparatus Crucible, clay-pipe triangle, tripod stand, spatula and Bunsen burner. Results Metal oxide Colour of metal oxide before heating Observation (a) Copper(II) oxide Black Brown spots (copper) are formed in the black mixture. (b) Iron(III) oxide Brown Grey solid (iron) is formed. (c) Lead(II) oxide Yellow Greyish-black solid (lead) is formed. Experiment 3.1 Discussion In all the reactions above, metal oxides have lost oxygen to form the metals. This shows that carbon has reduced metal oxides to the corresponding metals. Conclusion Carbon undergoes oxidation and the metal oxides undergo reduction. The hypothesis is accepted. The chemical equations for the redox reactions are (a) 2CuO(s) + C(s) → 2Cu(s) + CO2(g) (b) 2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g) (c) 2PbO(s) + C(s) → 2Pb(s) + CO2(g) Oxidation and Reduction 388 Table 3.1 Oxidation numbers of elements in the free state SPM ’08/P1 The oxidation number of an element is an arbitrary charge assigned to the element according to a set of rules. Oxidation number is also known as the oxidation state. Oxidation numbers of elements in ionic compounds 1 An ionic compound can contain monatomic ions (for example, Na+ and Cl– ions) or polyatomic ions (for example, NH4+ or SO42– ions). 2 For a monatomic ion in an ionic compound, the oxidation number is the charge on the ion. Magnesium oxide, MgO is an ionic compound. In magnesium oxide, magnesium exists as magnesium ions, Mg2+ and oxygen exists as oxide ions, O2–. Thus, magnesium is said to have the oxidation number of +2 and oxygen has the oxidation number of –2. Element Formula Oxidation number Hydrogen H2 0 Oxygen O2 0 Chlorine Cl2 0 Sulphur S 0 Iron Fe 0 Copper Cu 0 2 For monatomic ions, the oxidation number equals to the charge on the ion (Table 3.2). Table 3.2 Oxidation numbers of monatomic ions Simple ion Formula of ion Oxidation number Hydrogen ion H+ Sodium ion Na+ +1 Magnesium ion Mg +2 Aluminium ion Al3+ +3 Oxidation numbers of elements in covalent compounds Chloride ion Cl– –1 Oxide ion O2– –2 Carbon dioxide, CO2 is a covalent compound. However, when determining the oxidation numbers of carbon and oxygen, you will have to consider that this molecule exists as ions. • Each oxygen atom is considered as an oxide ion (O2–) and carries a charge of –2. So two oxide ions carry a total charge of –4. • As a result, each carbon ion carries a charge of +4 so that CO2 exists as a neutral molecule. Nitride ion N3– –3 +1 2+ 3 The sum of the oxidation states of all the atoms present in the formula of a compound is zero. The compound can be an ionic compound or a covalent compound. For example, (a) in calcium carbonate, CaCO3 CaCO3 CO2 (+2) + (+4) + 3(–2) = 0 (b) in aluminium nitrate, Al(NO3)3 (+4) + 2  (–2) = 0 Al(NO3)3 Therefore, the oxidation number of carbon is +4 and the oxidation number of oxygen is –2. (+3) + 3(+5) + 9(–2) = 0 4 For a polyatomic ion (that is, an ion that contains a few atoms), the sum of the oxidation numbers of all the atoms equals the charge on the ion. For example, (a) in a sulphate ion, SO42– Rules for Assigning Oxidation Numbers To work out the oxidation number, the following rules must be applied. 1 An atom or a molecule of an element in the SPM free state (that is, not combined with other ’06/P1 ’07/P1 elements) has an oxidation number of zero (Table 3.1). SO42– (+6) + 4(–2) = –2 389 sum of the oxidation numbers is the same as the charge on the polyatomic ion Oxidation and Reduction 3 Oxidation Number 3 8 The oxidation number of hydrogen in all its compounds is +1 except in metal hydrides. A metal hydride is a compound consisting of hydrogen and a metal only. The oxidation number of hydrogen in metal hydrides is –1. (a) Non-metallic hydride (b) in an ammonium ion, NH4+ NH4+ sum of the oxidation numbers is the same as the charge on the polyatomic ion (–3) + 4(+1) = +1 (c) in a nitrate ion, NO3– NO3– sum of the oxidation numbers is the same as the charge on the polyatomic ion (+5) + 3(–2) = –1 (d) in a dichromate(VI) ion, Cr2O72– Cr2O72– sum of the oxidation numbers is the same as the charge on 2(+6) + 7(–2) = –2 the polyatomic ion 5 In a given compound, the more electronegative atom is given a negative oxidation number and the less electronegative (or more electropositive) atom has a positive oxidation number. HCl (+1) NaH (+1) (+1) (+5) 3(–2) Oxidation and Reduction (+2) 2(–1) (+3) 3(–1) BaO (–2) (+2) H2O2 6 The oxidation number of fluorine remains unchanged in all its compounds and is always –1. This is because fluorine is the most electronegative element. F2O BrF3 Na3AlF6 2(–1) (+2) (+3) 3(–1) 3(+1) (+3) 6(–1) Notice that (a) the oxidation number of oxygen in F2O is +2 and not –2 (b) the oxidation number of bromine in BrF3 is +3 and not –1. 7 Chlorine, bromine and iodine usually have the oxidation number of –1 except when combined with a more electronegative element. For example, HClO Cl2O (+1) (+1) (–2) 2(+1) (–2) (+1) (–1) (–1) AlH3 (–2) (b) Peroxides electronegativity increases (–2) CaH2 H2O 2(+1) KIO3 2(+1) 9 The oxidation number of oxygen in all its compounds is –2 except in fluorine compounds (see point 6) and in peroxides. The oxidation number of oxygen in peroxides is –1. (a) Oxides KI (–1) (b) Metal hydrides I, Br, Cl, N, O, F H2S BaO2 2(+1) 2(–1) (+2) 2(–1) 10 Metals usually have positive oxidation numbers. For example, the oxidation number of a Group 1 element in a compound is always +1. The oxidation number of a Group 2 element in a compound is always +2. 11 Some metals show different oxidation numbers in their compounds. For example, manganese shows oxidation numbers of +2, +4, +6 and +7 (Table 3.3). Table 3.3 Oxidation numbers of manganese Compound MnSO4 Oxidation number of manganese +2 MnO2 K2MnO4 KMnO4 +4 +6 +7 12 Non-metals usually have negative oxidation numbers. However, chlorine, bromine, iodine and nitrogen can have positive or negative oxidation number (Table 3.4) depending on the elements which combine with them. 390 Table 3.4 Oxidation numbers of chlorine and nitrogen Chlorine compound Oxidation number of chlorine Nitrogen compound Oxidation number of nitrogen HCl HClO HClO2 ClO2 HClO3 HClO4 –1 +1 +3 +4 +5 +7 NH3 N2O NO NO2– NO2 NO3– –3 +1 +2 +3 +4 +5 Calculating the Oxidation Numbers of Elements in Compounds and Ions 2 4 What is the oxidation number of sulphur in the thiosulphate ion, S2O32–? Solution Let the oxidation number of manganese = x Oxidation number of K = +1 Oxidation number of O = –2 KMnO4 Solution Let the oxidation number of sulphur = x Oxidation number of O = –2 S2O32– 3 What is the oxidation number of manganese in the compound, KMnO4? 2(x) 3(–2) Sum of the oxidation numbers of all atoms in the polyatomic ion is equal to the charge on the ion. 2x + 3(–2) = –2 x = +2 The oxidation number of sulphur in S2O32– ion is +2. +1 x 4(–2) Sum of the oxidation numbers of all the elements in the neutral compound, KMnO4, is zero. (+1) + x + 4(–2) = 0 x = +7 The oxidation number of manganese in KMnO4 is +7. 3 1,1,1-trichloroethane (C2H3Cl3) is used as a solvent for halogens. The oxidation numbers of carbon, hydrogen and chlorine are shown below. Calculate the oxidation number of nitrogen in nitric acid, HNO3. C2H3Cl3 Solution Let the oxidation number of nitrogen = x Oxidation number of H = +1 Oxidation number of O = –2 HNO3 2(0) + 3(+1) + 3(–1) = 0 The oxidation number of an element in the free state is always zero. However, in some cases (for example, carbon in the C2H3Cl3), the oxidation number of the element in the compound can also be zero. +1 x 3(–2) Sum of the oxidation numbers of all the elements in the compound is zero. (+1) + x + 3(–2) = 0 x = +5 The oxidation number of nitrogen in HNO3 is +5. IUPAC Nomenclature of Inorganic Compounds 1 The IUPAC system is used to name inorganic compounds in order to avoid confusion that may arise due to elements having different oxidation numbers. 2 For example, there are two oxides of copper: copper(I) oxide, Cu2O, and copper(II) oxide, CuO. Copper(I) oxide is a brown powder whereas copper(II) oxide is a black powder. The Roman numerical figures (I) and (II) refer to the oxidation numbers of copper in the compound. The oxidation number of nitrogen in HNO3 is +5 and not 5. The oxidation number of the element must be accompanied by a positive or negative sign on the left of the number. 391 Oxidation and Reduction Oxidation Number and IUPAC Nomenclature Formula Oxidation of number of compound metal 3 1 (a) For an ionic compound or a covalent compound that contains a metal with more than one oxidation number, the Roman numerical figure is stated in brackets after the name of the metal to show the oxidation number of the metal. (b) For example, tin forms two types of chlorides, SnCl2 (ionic) and SnCl4 (covalent). SnCl2 is called tin(II) chloride and SnCl4 is called tin(IV) chloride. (c) Similarly, lead(II) oxide refers to the compound with the formula PbO whereas lead(IV) oxide refers to the compound PbO2. 2 Table 3.5 shows the formulae and the IUPAC names of some compounds containing metals. IUPAC name FeCl2 +2 Iron(II) chloride FeCl3 +3 Iron(III) chloride CuCl +1 Copper(I) chloride CuSO4 +2 Copper(II) sulphate Mn(NO3)2 +2 Manganese(II) nitrate MnO2 +4 Manganese(IV) oxide Oxidation and Reduction +1 Magnesium nitrate (NOT magnesium(II) nitrate) AlCl3 +3 Aluminium chloride (NOT aluminium(III) chloride) Formula Oxidation of number of metal compound in negative ion Table 3.6 Naming inorganic compounds containing elements of Groups 1, 2 and 3 K2SO4 +2 Table 3.7 Name of compounds containing metals in negative ions 3 The metallic elements in Groups 1, 2 and 3 of the Periodic Table always have the oxidation numbers +1, +2 and +3 respectively. According to the IUPAC nomenclature, the Roman numerical figure is not used in naming a compound if the metal shows only one oxidation state in its compounds. 4 Table 3.6 shows some examples of naming compounds containing elements of Groups 1, 2 and 3. Formula Oxidation of number of compound metal Mg(NO3)2 5 For a negative ion that contains a metal with more than one oxidation state, the Roman number is stated in brackets after the name of the metal, and the name of the metal ends with -ate. For example, (a) manganate(VII) refers to the negative ion containing the manganese metal with oxidation number +7, that is, MnO4–. (b) chromate(VI) refers to the negative ion containing chromium metal with oxidation number +6, that is, CrO42–. (c) dichromate(VI) refers to the negative ion containing two chromium atoms with oxidation number +6, that is, Cr2O72–. (d) hexacyanoferrate(III) refers to the negative ion containing six cyano (CN–) groups and iron metal with oxidation number +3, that is, [Fe(CN)6]3–. Table 3.7 shows the names of some compounds containing metals in negative ions. Table 3.5 Naming inorganic compounds containing metals Formula Oxidation of number of compound metal Name of compound K2MnO4 +6 Potassium manganate(VI) KMnO4 +7 Potassium manganate(VII) K2CrO4 +6 Potassium chromate(VI) K2Cr2O7 +6 Potassium dichromate(VI) K4Fe(CN)6 +2 Potassium hexacyanoferrate(II) K3Fe(CN)6 +3 Potassium hexacyanoferrate(III) Name of compound Potassium sulphate (NOT potassium(I) sulphate) 392 IUPAC name Notice that in K4Fe(CN)6, the negative ion is [Fe(CN)6]4– while in K3Fe(CN)6, the negative ion is [Fe(CN)6]3–. 6 Oxoanions are anions (negative ions) that consist of an oxygen atom and another nonmetallic atom. Examples of oxoanions are nitrate ion, NO3– and sulphate ion, SO42–. For a non-metal that shows more than one oxidation number in its oxoanion, the Roman number stated in brackets refers to the oxidation number of the non-metal. 7 Table 3.8 shows the common names and the IUPAC names for some compounds containing oxoanions. Table 3.8 Common names and IUPAC names for some compounds Oxidation number IUPAC name Na2SO3 +4 (for S) Sodium sulphate(IV) Sodium sulphite Na2SO4 +6 (for S) Sodium sulphate(VI) Sodium sulphate NaNO2 +3 (for N) Sodium nitrate(III) Sodium nitrite NaNO3 +5 (for N) Sodium nitrate(V) Sodium nitrate NaClO +1 (for Cl) Sodium chlorate(I) Sodium hypochlorite NaClO3 +5 (for Cl) Sodium chlorate(V) Sodium chlorate HNO2 +3 (for N) Nitric(III) acid Nitrous acid HNO3 +5 (for N) Nitric(V) acid Nitric acid H2SO4 +6 (for S) Sulphuric(VI) acid Sulphuric acid 1 Common name of compound 3 Molecular formula of compound (a) the changes in oxidation numbers or (b) the transfer of electrons; that is, acceptance (gain) or donation (loss) of electrons. 3 Using oxidation numbers to identify redox reactions (a) Oxidation is a process in which the oxidation number of the element is increased. (b) Reduction is a process in which the oxidation number of the element is decreased. 4 The following equation shows the reaction between iron and chlorine. ’06 What is the oxidation number for oxygen in the thiosulphate ion, S2O32–? A –2 B –3 C +2 D +3 Solution The oxidation number for oxygen in a polyatomic ion, such as NO3–, S2O32– or Cr2O72– is always –2. Answer A Comment The oxidation number of oxygen in covalent compounds, such as CO2 or SO3 is also –2. Oxidation and Reduction in Terms of Changes in Oxidation Numbers 1 Most redox reactions occur without involving hydrogen or oxygen. For example, the reaction between chlorine and iron(II) chloride is a redox reaction: (oxidation number increases) oxidation 0 0 +3 –1 2Fe(s) + 3Cl2 → 2FeCl3 reduction (oxidation number decreases) In this redox reaction, (a) the iron metal is oxidised to chloride because its oxidation increases from 0 to +3. (b) chlorine (Cl2) is reduced to ion (Cl–) because its oxidation decreases from 0 to –1. 2FeCl2(aq) + Cl2(g) → 2FeCl3(aq) 2 For reactions that do not involve hydrogen or oxygen, an oxidation or reduction reaction is discussed in terms of 393 iron(III) number chloride number Oxidation and Reduction (c) chlorine acts as the oxidising agent because it oxidises iron and is itself reduced. (d) iron acts as the reducing agent because it reduces chlorine and is itself oxidised. 5 When chlorine gas is passed into potassium bromide solution, the following reaction occurs: Sulphur dioxide usually acts as a reducing agent. However, in the following reactions, sulphur dioxide acts as an oxidising agent because hydrogen sulphide and carbon are stronger reducing agents than sulphur dioxide. (a) (oxidation number increases) oxidation 0 –1 –1 –2 0 0 SO2(g) + 2H2S(g) ⎯⎯⎯→ 2H2O(l) + 3S(s) Cl2(g) + 2KBr(aq) → 2KCl(aq) + Br2(aq) oxidation number of S increases (oxidation) +4 0 oxidation number of S decreases (reduction) reduction (oxidation number decreases) (b) oxidation 3 0 The oxidation number of potassium in the above reaction does not change because potassium does not take part in the reaction. +4 7 A reaction is not a redox reaction if the substances involved in the reaction do not undergo any changes in oxidation numbers. For example, +1 –2 +1 0 +1 –2 0 Oxidation and reduction always take place together. A redox reaction must have • a substance that undergoes oxidation and acts as the reducing agent, and • another substance that undergoes reduction and acts as the oxidising agent. reduction (oxidation number decreases) In this reaction, (a) ammonia is oxidised to nitrogen because the oxidation number of nitrogen increases from –3 to 0. (b) copper(II) oxide is reduced to copper because the oxidation number of copper decreases from +2 to 0. (c) copper(II) oxide acts as an oxidising agent and ammonia acts as a reducing agent. (d) the oxidation numbers of hydrogen and oxygen remain unchanged. Oxidation and Reduction +1 +6 –2 The reaction between sodium hydroxide (NaOH) and sulphuric acid (H2SO4) is a neutralisation reaction and not a redox reaction. As a result, the oxidation numbers of all the elements (sodium, oxygen, hydrogen and sulphur) are the same before and after the reaction. 2NH3(g) + 3CuO(s) → N2(g) + 3H2O(l) + 3Cu(s) +1 +6 –2 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) (oxidation number increases) oxidation +2 0 reduction In this reaction, (a) bromide ion (Br–) is oxidised to bromine because the oxidation number of bromine increases from –1 to 0. (b) chlorine (Cl2) is reduced to chloride ion (Cl–) because the oxidation number of chlorine decreases from 0 to –1. (c) chlorine acts as an oxidising agent and potassium bromide acts as a reducing agent. 6 Ammonia reacts with copper(II) oxide as represented by the equation: –3 +4 SO2(g) + C(s) ⎯⎯⎯→ CO2(g) + S(s) Oxidation and Reduction in Terms of Electron Transfer 1 In terms of electron transfer, (a) oxidation is defined as the loss of electrons SPM from a substance. If a substance loses ’05/P1 electrons during a reaction, it has been oxidised. 394 5 (b) reduction is defined as the gain of electrons by a substance. If a substance gains electrons, it has been reduced. 2 During a redox reaction, transfer of electrons occurs between the reactants. Write the half-equation for the reduction of acidified manganate(VII) ion (MnO4–) to manganese(II) ion (Mn2+) in the presence of acid. Solution Step 1: Write the reactants and products involved in the reaction. An oil rig is used for getting oil and gas out of the ground in the petroleum industry. Use ‘OIL RIG’ to help you remember oxidation and reduction in terms of electron transfer. OIL : OXIDATION IS LOSS OF ELECTRONS RIG : REDUCTION IS GAIN OF ELECTRONS MnO4– + H+ → Mn2+ + H2O Step 2: Balance the number of atoms on both sides of the equation: MnO4– + 8H+ → Mn2+ + 4H2O Step 3: Balance the number of charges on both sides of the equation: 3 The reactant that loses electrons undergoes oxidation and acts as a reducing agent. For example, Na(s) ⎯⎯⎯→ Na+(aq) + e– … (1) oxidation total charge total charge = (–1) + (+8) = +7 = +2 3 In this reaction, (a) sodium atoms undergo oxidation by losing electrons to form sodium ions (Na+). (b) sodium acts as the reducing agent. 4 A substance that accepts electrons undergoes reduction and acts as an oxidising agent. For example, MnO4– + 8H+ → Mn2+ + 4H2O to balance the charges, add 5e– to the left of the equation MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l) Cl2(g) + 2e– ⎯⎯⎯→ 2Cl–(aq) … (2) reduction 6 (a) Zinc reacts with hydrochloric acid as represented by the equation In this reaction, (a) each chlorine molecule (Cl2) accepts two electrons to form two chloride ions (Cl–). (b) chlorine acts as the oxidising agent and is itself reduced. 5 Balancing half-equations for oxidation and SPM reduction ’06/P1 Equations (1) and (2) as shown above are known as half-equations. Half-equations must be balanced in terms of (a) the number of atoms, and (b) the number of charges. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) The ionic equation for the reaction is oxidation (loss of electrons) Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) reduction (gain of electrons) (b) The transfer of electrons can be represented by the following half-equations: Photographic films are coated with silver bromide, AgBr. When the film is exposed to light, the following redox reaction occurs: Zn(s) → Zn2+(aq) + 2e– ... oxidation 2Ag+ + 2Br– → 2Ag + Br2 2H+(aq) + 2e– → H2(g) ... reduction reducing agent oxidising agent The amount of silver produced depends on how much light gets through the camera lens. In this reaction, silver ions are reduced to silver by the gain of electrons and bromide ions are oxidised to bromine by the loss of electrons. (c) In the reaction between hydrochloric acid and zinc, zinc is oxidised to zinc chloride whereas hydrochloric acid is reduced to hydrogen. 395 Oxidation and Reduction 9 Combustion of metals in chlorine Figure 3.3 shows the combustion of copper in chlorine. When the hot copper foil is placed in a gas jar of chlorine, a vigorous reaction occurs and a green precipitate of copper(II) chloride, CuCl2 is formed. (d) Hydrochloric acid acts as an oxidising agent by accepting electrons and is itself reduced. Conversely, zinc acts as the reducing agent by donating electrons and is itself oxidised. 7 In terms of transfer of electrons, oxidising agents are electron acceptors while reducing agents are electron donors. 8 If a coil of copper is placed in a solution of silver nitrate, the copper slowly dissolves and the solution turns blue. At the same time, the copper coil becomes coated with a layer of silver metal (Figure 3.2). Figure 3.3 Combustion of copper in chlorine 3 (a) In the reaction between copper and chlorine to form copper(II) chloride, a transfer of electron occurs between copper metal and chlorine gas. Figure 3.2 Oxidation of copper (a) The overall equation for the reaction is Cu(s) + Cl2(g) → CuCl2(s) Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s) reduction (gain of electrons) (b) The transfer of electrons can be represented by the half-equations as shown below: The reaction can be represented by the ionic equation: Cu(s) → Cu2+(s) + 2e– … oxidation oxidation (loss of electrons) Cl2(g) + 2e– → 2Cl– (s) ... reduction (c) In the reaction between copper and chlorine, the copper atom (Cu) (i) loses electrons (ii) undergoes oxidation (iii) is oxidised to copper(II) ion, Cu2+ (iv) acts as a reducing agent (d) Conversely, the chlorine molecule (Cl2) (i) gains electrons (ii) undergoes reduction (iii) is reduced to chloride ions, Cl– (iv) acts as an oxidising agent 10 Combustion of metals in oxygen When metals burn in oxygen, (a) the metals undergo oxidation by losing electrons to form metal ions, (b) the oxygen undergoes reduction by gaining electrons to form oxide ions (O2–). The combustion of lead in oxygen is studied in Activity 3.1. Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) reduction (gain of electrons) (b) In this reaction, each silver ion (Ag+) accepts one electron to form a silver atom (Ag). Ag+(aq) + e– → Ag(s) ... reduction An oxidising agent is an electron acceptor. Hence, silver ion acts as an oxidising agent in this reaction. (c) Conversely, each copper atom donates two electrons and are converted to copper(II) ion (Cu2+) in the aqueous solution. Cu(s) → Cu2+(aq) + 2e– ....oxidation Reducing agents are electron donors. Hence, copper acts as a reducing agent. Oxidation and Reduction oxidation (loss of electrons) 396 To investigate the combustion of metals in oxygen and chlorine Apparatus Gas jar, tongs, combustion spoon, gas jar and Bunsen burner. Materials Magnesium ribbon, sodium and chlorine. Procedure 2 The magnesium ribbon is held with a pair of tongs and lit in the Bunsen burner. 3 It is quickly placed into a gas jar filled with oxygen. 4 Any changes that occur are recorded. (B) Reaction of sodium with chlorine 1 A small piece of sodium metal is placed in a combustion spoon and heated. 2 When the sodium metal starts to burn, it is quickly placed in a gas jar filled with chlorine gas. 3 Any changes that occur are recorded. Figure 3.4 The combustion of magnesium in oxygen (A) Combustion of magnesium in oxygen 1 A piece of 5 cm magnesium ribbon is cleaned with sandpaper. 3 Figure 3.5 The combustion of sodium in chlorine Observation Experiment Observation Combustion of magnesium in oxygen • The magnesium ribbon burns with a bright white flame. • White fumes are produced. • A white powder is formed. Reaction of sodium with chlorine • The sodium metal burns with a yellow flame. • White fumes are produced. • A white powder is formed. 5 In this reaction, • magnesium acts as the reducing agent because it reduces oxygen to oxide ion. • oxygen acts as the oxidising agent because it oxidises magnesium to magnesium ion. (B) Combustion of sodium in chlorine 1 The combustion of sodium in chlorine produces sodium chloride (white powder). 2 Sodium is oxidised by losing electrons to form sodium ions, Na+. Half-equation: Na(s) → Na+(s) + e– oxidation (loss of electrons) 3 Chlorine is reduced by gaining electrons to form chloride ion, Cl–. 2Mg(s) + O2(g) → 2MgO(s) Half-equation: Cl2(g) + 2e– → 2Cl–(s) reduction (gain of electrons) 397 Oxidation and Reduction Activity 3.1 Discussion (A) Combustion of magnesium in oxygen 1 The combustion of magnesium in oxygen produces magnesium oxide (white powder). 2 Magnesium is oxidised by losing electrons to form magnesium ions, Mg2+. Half-equation: Mg(s) → Mg2+(s) + 2e– 3 Oxygen is reduced by gaining electrons to form oxide ion, O2–. Half-equation: O2(g) + 4e– → 2O2–(s) 4 The overall equation for the reaction is 4 The overall equation for the reaction is oxidation (loss of electrons) 2Na(s) + Cl2(g) → 2NaCl(s) reduction (gain of electrons) 5 In this reaction, • sodium acts as the reducing agent because it reduces chlorine to chlorine ion. • chlorine acts as the oxidising agent because it oxidises sodium to sodium ion. Conclusion 1 In the combustion of magnesium in oxygen, (a) magnesium undergoes oxidation to form Mg2+ ions, (b) oxygen undergoes reduction to form O2– ions. 2 In the combustion of sodium in chlorine, • sodium act as reducing agent by losing electrons, • chlorine act as oxidising agent by gaining electrons. 3 Conversion of Fe2+ Ions to Fe3+ Ions and Vice Versa There are some chemicals such as hydrogen peroxide (H2O2) and nitric(III) acid (nitrous acid, HNO2) which can act as an oxidising agent or a reducing agent depending on the conditions of the reaction. For example, in reaction (1), hydrogen peroxide acts as an oxidising agent, but in reaction (2), it acts as a reducing agent. 1 Iron metal (Fe) exhibits two oxidation states, +2 and +3. 2 Fe2+ ions can be converted to Fe3+ ions. Similarly, Fe3+ ions can be converted to Fe2+ ions. H2O2 + 2I– + 2H+ → I2 + 2H2O … (1) oxidising agent Oxidation of Fe2+ to Fe3+ reducing agent 1 Iron(II) ion, Fe2+, can be converted to iron(III) ions, Fe3+, by oxidation reaction. oxidation (loss of electrons) 5H2O2 + 2MnO4– + 6H+ → 2Mn2+ + 8H2O + 5O2 … (2) reducing agent SPM ’08/P2, ’09/P1 oxidising agent Fe2+(aq) → Fe3+(aq) + e– 2 Potassium manganate(VII) is an oxidising agent that can oxidise Fe2+ ions to Fe3+ ions. 3 (a) When acidified potassium manganate(VII) solution is added to a solution of iron(II) salt, decolourisation occurs. MnO4– ions are reduced to Mn2+ ions while Fe2+ ions (pale green) are oxidised to Fe3+ ions (brown). oxidation (loss of electrons) MnO (aq) + 8H (aq) + 5Fe (aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) – 4 + 2+ green reduction (gain of electrons) brown A catalytic converter Catalytic converters are fitted to the exhaust pipes of cars to reduce air pollution. In the catalytic converter, the following redox reactions take place to convert poisonous gases (NO, CO and petrol vapour) to nonpoisonous gases. For example, Dilute sulphuric acid is always used to acidify KMnO4 solution. (b) The half-equations for the reactions are: 2NO(g) + 2CO(g) → N2(g) + 2CO2(g) oxidising agent Fe2+(aq) → Fe3+(aq) + e– reducing agent Oxidation and Reduction (oxidation – loss of electrons) 398 MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l) reduction (gain electron) SO32– + H2O + 2Fe3+ → 2Fe2+ + H2SO4 (reduction – gain of electrons) brown 4 The formation of Fe3+ ions can be confirmed by using sodium hydroxide solution. When sodium hydroxide solution is added to the reaction product, a brown precipitate of iron(III) hydroxide, Fe(OH)3, insoluble in excess NaOH(aq) is obtained. Cl2(aq) + 2Fe (aq) → 2Fe (aq) + 2Cl (aq) SO32–(aq) + H2O(l) → SO42–(aq) + 2H+(aq) + 2e– (oxidation – loss of electrons) 3 The formation of Fe2+ ions can be confirmed by using sodium hydroxide solution. When sodium hydroxide solution is added to the reaction product, a dirty green precipitate of iron(II) hydroxide, Fe(OH)2, insoluble in excess NaOH(aq), is obtained. (c) Acidified potassium dichromate(VI) solution (acidified with dilute sulphuric acid) Cr2O72–(aq) + 14H+(aq) + 6Fe2+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) Fe2+(aq) + 2NaOH(aq) → Fe(OH)2(s) + 2Na+(aq) (d) Concentrated nitric acid 4 Other reducing agents that can be used to reduce Fe3+ ions to Fe2+ ions include the following. (a) Metals more reactive (electropositive) than iron. For example, zinc. HNO3(aq) + 3Fe (aq) + 3H (aq) → 3Fe3+(aq) + 2H2O(l) + NO(g) 2+ + (e) Acidified hydrogen peroxide Zn(s) + 2Fe3+(aq) → 2Fe2+(aq) + Zn2+(aq) H2O2(aq) + 2H (aq) + 2Fe (aq) → 2Fe3+(aq) + 2H2O(l) 2+ (b) Sulphur dioxide SO2(g) + 2H2O(l) + 2Fe3+(aq) → 2Fe2+(aq) + 2H+(aq) + H2SO4(aq) Reduction of Fe to Fe 2+ (c) Potassium iodide 1 Iron(III) ions, Fe3+, can be converted to iron(II) ions, Fe2+, by reduction. Br2(l) + 2Fe2+(aq) → 2Fe3+(aq) + 2Br–(aq) 3+ Fe3+(aq) + e– → Fe2+(aq) (reduction – gain of electrons) – (b) Liquid bromine + 2KI(aq) + 2Fe3+(aq) → 2Fe2+(aq) + 2K+(aq) + I2(s) reduction (gain electron) (d) Hydrogen sulphide Fe3+(aq) + e– → Fe2+(aq) H2S(aq) + 2Fe3+(aq) → 2Fe2+(aq) + 2H+(aq) + S(s) 2 (a) When sodium sulphite (Na2SO3) solution is added to iron(III) chloride, and the mixture is acidified with dilute sulphuric acid, the colour of the solution changes from brown to light green. (e) Tin(II) chloride solution Sn2+(aq) + 2Fe3+(aq) → 2Fe2+(aq) + Sn4+(aq) 399 Oxidation and Reduction 3 5 Other oxidising agents that can be used to oxidise Fe2+ to Fe3+ are as follows. (a) Chlorine gas or chlorine water 3+ oxidation (lose electron) Sodium sulphite acts as the reducing agent and reduces iron(III) ions to iron(II) ions and is itself oxidised to sulphate ions (SO42–). (b) The half-equations for the reaction are: Fe3+(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3Na+(aq) 2+ green 3 To study the oxidation of Fe2+ ions to Fe3+ ions and the reduction of Fe3+ ions to Fe2+ ions SPM ’09/P2 Apparatus Test tubes and droppers. Materials FeSO4, KMnO4, FeCl3, Na2SO3, dilute H2SO4 and dilute NaOH solution. 4 Sodium hydroxide solution is then added to the reaction mixture slowly until in excess. 5 The observations are recorded in the table below. Procedure (A) Conversion of Fe2+ ions to Fe3+ ions 1 About 2 cm3 of iron(II) sulphate solution is poured into a test tube. 2 About 2 cm3 of potassium manganate(VII) solution is poured into another test tube, followed by about 2 cm3 of dilute sulphuric acid. 3 Using a dropper, about 2 cm3 of the acidified potassium manganate(VII) solution is added slowly to the iron(II) sulphate solution. The mixture is shaken gently. (B) Conversion of Fe3+ ions to Fe2+ ions 1 About 2 cm3 of iron(III) chloride solution is added to a test tube. 2 Sodium sulphite (Na2SO3) solution is added to iron(III) chloride solution, followed by dilute sulphuric acid. The mixture is shaken gently. 3 Sodium hydroxide solution is then added slowly to the reaction mixture until in excess. 4 The observations are recorded in the table below. Observations Solution Test Observation FeSO4(aq) (a) Fe2+ ion + acidified KMnO4 • The light green iron(II) sulphate solution changes to yellow. • The purple colour of acidified KMnO4 solution turns colourless (decolourised). (b) Add excess NaOH(aq) to (a) • A brown precipitate, insoluble in excess NaOH(aq) is formed. (c) Fe3+ ion + Na2SO3(aq) • The colour of the solution changes from yellow to light green. (d) Add excess NaOH(aq) to (c) • A dirty green precipitate, insoluble in excess NaOH(aq) is obtained. FeCl3(aq) Conclusion 1 Fe2+ ions are oxidised to Fe3+ ions by the acidified KMnO4 solution. 2 Fe3+ ions are reduced to Fe2+ ions by the sodium sulphite (Na2SO3) solution. 3 Electropositive metals are strong reducing agents. In contrast, the metallic ions of electropositive metals are weak oxidising agents. Figure 3.6 shows that in the electrochemical series, (a) the strength of a metal as a reducing agent increases on going up the electrochemical series, (b) the strength of the metallic ion as an oxidising agent increases on going down the series. Displacement of Metals from Their Salt Solutions 1 The arrangement of metals according to their tendency to lose electrons to form positive ’04/P1 ’07/P1 ions is called the electrochemical series. 2 The higher the position of the metal in the electrochemical series, (a) the more electropositive the metal, (b) the more readily the metal donates electrons to form positive ions, (c) the more easily the metal will undergo oxidation. Activity 3.2 SPM Oxidation and Reduction 400 • Tendency of a metal to ionise (by donating electrons) increases. •Tendency of an ion to accept electrons increases. • Strength of a metal as a reducing agent increases. • Strength of an ion as an oxidising agent increases. 4 Consider the formation of sodium ions (Na+) from sodium metal (Na). position in the electrochemical series) will displace a less electropositive metal (lower position in the electrochemical series) from the salt solutions of the less electropositive metal. 6 Transfer of electrons occurs during displacement reactions. (a) The more electropositive metal donates electrons and acts as a reducing agent. The metal undergoes oxidation and is oxidised to its metal ions. (b) The metal ion (from the less electropositive metal) in aqueous solution acts as an oxidising agent. The metal ions undergo reduction and is reduced to its metal. Na metal has a strong tendency to lose an electron to form sodium ion ⎯⎯⎯⎯⎯⎯⎯⎯→ Na(s) → Na+(aq) + e– ←⎯⎯⎯⎯⎯⎯⎯⎯ + Na ion has a weak tendency to accept an electron to form Na metal (a) Sodium metal is placed at a high position in the electrochemical series. (b) This means that sodium metal donates electrons very easily. As a result, sodium is a strong reducing agent. (c) Conversely, sodium ions (Na+) have a weak tendency to accept electrons. Since oxidising agents are electron acceptors, sodium ions are weak oxidising agents. 5 A displacement reaction is a reaction in which one element (metal or non-metal) displaces another element (metal or non-metal) from its salt solution. In the displacement reactions of metals, the more electropositive metal (higher 3 Figure 3.6 Electrochemical series A more electropositive metal is also a more reactive metal. We can therefore state that a more reactive metal will displace a less reactive metal from the solution of its salts. 3.2 To study the redox reaction in terms of displacement reaction of a metal from its salt solution Problem statement How does redox reaction occur in a displacement reaction in which a metal is displaced from its salt solution? Variables (a) Manipulated variable : A pair of metals and salt solutions (b) Responding variable : Precipitation of metal and colour changes in the solutions (c) Constant variables : Volumes and concentrations of solutions containing the metal ions Apparatus Beakers 401 Oxidation and Reduction Experiment 3.2 Hypothesis (a) The metal that acts as a reducing agent will form metal ion. (b) The metal ion that acts as an oxidising agent will be precipitated as metal. Materials Copper(II) sulphate solution and silver nitrate solution, zinc plate and copper plate. Procedure 1 A strip of zinc plate and a strip of copper plate are cleaned with sandpaper. 2 The zinc plate is then immersed in copper(II) sulphate solution (beaker P) and the copper plate is immersed in silver nitrate solution (beaker Q). 3 The mixture is left aside for half an hour. 4 The changes that take place on the zinc plate, the copper plate, and in the copper(II) sulphate solution and the silver nitrate solution are recorded. Figure 3.7 The displacement of a metal from its salt solution Observation Reactants Observation Explanation • A section of the zinc plate dissolves. • Brown precipitate is deposited on the zinc plate. • The blue solution fades until it becomes colourless. • Zinc displaces copper metal (brown precipitate) from copper(II) sulphate solution. • Copper metal is deposited on the zinc plate. • The blue colour fades as the concentration of Cu2+ ions decreases. (b) Copper plate in silver nitrate solution • The copper plate dissolves. • A greyish-black precipitate is deposited on the copper plate. • The colourless solution turns blue. • Copper displaces silver metal (greyishblack) from the silver nitrate solution. • Silver metal is precipitated on the copper plate. • The colourless solution turns blue because of the formation of Cu2+ ions. 3 (a) Zinc plate in copper(II) sulphate solution Conclusion During the displacement reaction, the more electropositive metal will act as a reducing agent. It reduces the metal ion (oxidising agent) which is less electropositive to form metal. The hypothesis is accepted. Displacement of Copper by Zinc from Copper(II) Sulphate Solution The reaction can be represented by the following half-equations: (a) Zn(s) → Zn2+(aq) + 2e– SPM ’08/P2 1 The following equation shows the reaction between copper(II) sulphate solution and zinc. (oxidation – loss of electrons) Zinc acts as a reducing agent (electron donor) (b) Cu2+(aq) + 2e– → Cu(s) Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) (reduction – gain of electrons) Copper ion acts as an oxidising agent (electron acceptor) Zinc is more electropositive than copper. It displaces copper from its salt (that is, copper(II) sulphate). 2 A displacement reaction is a redox reaction. 3 When copper(II) ion is displaced, the concentration of Cu2+ ions in the solution decreases. This causes the blue colour to fade. oxidation (loss of electrons) Displacement of Silver by Copper from Silver Nitrate Solution Zn(s) + CuSO4(aq) ⎯⎯→ ZnSO4(aq) + Cu(s) 1 The displacement reaction between copper and silver nitrate solution is shown as follows. reduction (gain of electrons) Oxidation and Reduction 402 (a) Cu(s) → Cu2+(aq) + 2e– oxidation (loss of electrons) (oxidation – loss of electrons) Copper acts as a reducing agent (electron donor). Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s) (b) Ag+(aq) + e– → Ag(s) (reduction – gain of electrons) Silver ion acts as an oxidising agent (electron acceptor). reduction (gain of electrons) Copper is more electropositive than silver. It displaces silver from its salt (that is, silver nitrate). 2 The reaction can be represented by the following half-equations: 3 When copper dissolves in silver nitrate solution, the formation of copper(II) ion causes the solution to turn blue. The intensity of blue colour increases as more copper is dissolved. 2 ’07 Solution Metal AgNO3 Q Pb(NO3)2 FeSO4 MgSO4 No change No change No change No change No change X Silver is displaced Y Silver is displaced Lead is displaced Z Silver is displaced Lead is displaced What is the correct position of the metals, in ascending order, of the tendency of the metals to be oxidised? A Q, X, Y, Z C X, Y, Z, Q B Q, X, Z, Y D Y, Z, X, Q Comments The most reactive metal is the strongest reducing agent, that is, it has the highest tendency to form metal ions by losing electrons, that is to be oxidised. Iron is displaced No change Q is the least reactive. It has no reactions with Pb2+, Fe2+ and Mg2+. Y is the most reactive. It can displace three metals from their solutions. Z is more reactive than X. It can displace two metals from their salt solutions. Answer B Displacement of Halogens from Halide Solutions halogen will displace a less reactive halogen from the solution of its halide ions. 3 Hence, chlorine displaces bromine from an aqueous solution of bromide ions. It also displaces iodine from an aqueous solution of iodide ions. Similarly, bromine displaces iodine from an aqueous solution of iodide ions. 1 In general, the stronger the oxidising strength of the halogen, the weaker the reducing ’04/P1 ’05/P1 strength of the corresponding halide ion is. Thus, chlorine is a stronger oxidising agent than iodine, but the iodide ion is a stronger reducing agent than the chloride ion. Figure 3.8 shows the trend in the reactivity and oxidising strength of the halogens and the reducing strength of the halide ions. 2 The reactivity of the halogens can be used to predict whether the displacement reactions of halogens can occur or not. A more reactive SPM Cl2(aq) + 2KBr(aq) → 2KCl(aq) + Br2(aq) Cl2(aq) + 2KI(aq) → 2KCl(aq) + I2(aq) Br2(aq) + 2KI(aq) → 2KBr(aq) + I2(aq) 4 Conversely, bromine cannot displace chlorine from an aqueous solution of chloride ions and iodine cannot displace bromine or chlorine 403 Oxidation and Reduction 3 An experiment is carried out to determine the positions of the metals, Q, X, Y, Z in the electrochemical series. The results of the experiment on displacement reactions are shown in the table below. from an aqueous solution of bromide ions or chloride ions respectively. 1,1,1-trichloroethane. The colours of halogens in 1,1,1-trichloroethane are shown in Table 3.10. Table 3.10 The colours of halogens in 1,1,1trichloroethane Br2(aq) + KCl(aq) → No reaction I2(aq) + KCl(aq) → No reaction I2(aq) + KBr(aq) → No reaction 5 The colours of halogens in water are shown in Table 3.9. Halogen Colour Chlorine Colourless Bromine Brown Iodine Purple 3 Table 3.9 The colour of halogens in water Halogen Concentrated aqueous solution Dilute aqueous solution Chlorine Greenish-yellow Colourless Bromine Brown Yellow Iodine Brown Yellow The structural formula of 1,1,1-trichloroethane is: HCl ⎮ ⎮ H ⎯ C ⎯ C ⎯ Cl ⎮⎮ HCl 6 Halogens can be identified by adding 1,1,1trichloroethane (CH3CCl3) to its aqueous solution. Water and 1,1,1-trichloroethane are immiscible and two layers are formed. The upper layer is water and the lower layer is It is very volatile and is used as a solvent in paper correction fluid. It is produced as one of the organic products when ethane reacts in chlorine in the presence of sunlight. A The tendency of electrons being removed from halide ions to form halogens increases A Reactivity of halogens increases A Oxidising strength of halogens increases A The strength of halide ion as a reducing agent increases Figure 3.8 3.3 To study the displacement reactions between halogens and halide ions Problem statement How do redox reactions occur in displacement reactions between halogens and aqueous solutions of halide ions? Experiment 3.3 Hypothesis A more reactive halogen will displace a less reactive halogen from an aqueous solution of its halide ions. Variables (a) Manipulated variable : A pair of halogens and their halide ions (b) Responding variable : Changes in colour in 1,1,1-trichloroethane, CH3CCl3 (c) Constant variable : Volume of reaction mixture Apparatus Test tubes Oxidation and Reduction 404 2 2 cm3 of chlorine water is added to the potassium bromide solution. The mixture is shaken gently. 3 2 cm3 of 1,1,1-trichloroethane (CH3CCl3) is then added to the mixture obtained in step 2. The mixture is then shaken vigorously. 4 The test tube is allowed to stand for a few minutes and the colour of the 1,1,1-trichloroethane layer is recorded. 5 Steps 1 to 4 are repeated using the following mixtures. (a) Chlorine water and potassium iodide, KI solution (b) Liquid bromine and potassium chloride, KCl solution (c) Liquid bromine and potassium iodide solution (d) Iodine solution and potassium bromide solution (e) Iodine solution and potassium chloride solution Materials 1,1,1-trichloroethane, potassium bromide, KBr(aq), potassium chloride, KCl(aq), potassium iodide, KI(aq), chlorine water, liquid bromine and iodine solution. Figure 3.9 Displacement of bromine from potassium bromide solution Procedure 1 A test tube is filled with 2 cm3 of potassium bromide, KBr solution. 3 Observation Mixture Colour of CH3CCl3 layer Halogen in CH3CCl3 layer Has displacement reaction occurred? Cl2(aq) + KBr(aq) Brown Bromine Yes Cl2(aq) + KI(aq) Purple Iodine Yes Br2(l) + KCl(aq) Brown Bromine No Br2(l) + KI(aq) Purple Iodine Yes I2(aq) + KBr(aq) Purple *Iodine No I2(aq) + KCl(aq) Purple *Iodine No ψ The bromine present in the CH3CCl3 layer is due to the bromine added. *The iodine present in the CH3CCl3 layer is due to the iodine added. ψ Discussion 1 When chlorine water is added to potassium bromide solution, the colour of the solution changes from colourless to brown because chlorine displaces bromine from an aqueous solution of bromide ions. oxidation Cl2(aq) + 2KBr(aq) → 2KCl(aq) + Br2(l) reduction 4 The half-equations for the reactions are as follows: (a) 2Br–(aq) → Br2(l) + 2e– Cl2(aq) + 2KBr(aq) → 2KCl(aq) + Br2(l) 2 If 1,1,1-trichloroethane is added to the reaction mixture and shaken, two liquid layers are formed. The lower organic layer (1,1,1-trichloroethane) has a brown colour and shows the presence of bromine. This means that the bromide ions have been oxidised to bromine. 3 Displacement reactions can also be considered in terms of oxidation and reduction. Chlorine oxidises bromide ion (Br–) to bromine and chlorine is itself reduced to chloride ion (Cl–). (oxidation – loss of electrons) (b) Cl2 (aq) + 2e– → 2Cl–(aq) (reduction – gain of electrons) Chlorine accepts electrons and acts as an oxidising agent. Bromide ion (Br–) loses electrons and acts as a reducing agent. 5 Other displacement reactions that occur in this experiment are 405 Oxidation and Reduction (b) I2(aq) + KBr(aq) → No reaction I2(aq) + KCl(aq) → No reaction This is because iodine is less reactive than bromine and chlorine. oxidation Cl2(aq) + 2KI(aq) → 2KCl(aq) + I2(aq) Conclusion 1 Chlorine displaces bromine from potassium bromide solution and iodine from potassium iodide solution. Bromine displaces iodine from iodide solution but does not displace chlorine from chloride solution. Iodine does not displace chlorine from chloride solution or bromine from bromide solution. 2 The experimental results prove that a more reactive halogen can displace a less reactive halogen from its halide solution. The hypothesis is accepted. reduction oxidation Br2(l) + 2KI(aq) → 2KBr(aq) + I2(aq) reduction 6 The following displacement reactions do not occur. (a) Br2(l) + KCl(aq) → No reaction This is because bromine is less reactive than chlorine. 3 Redox Reactions by the Transfer of Electrons at a Distance In Figure 3.9 (Experiment 3.3), the aqueous layer contains KCl but the organic layer (CH3CCl3) does not contain KCl. This is because KCl is an ionic compound which is soluble in water but not in organic solvent. 3 1 If a solution containing an oxidising agent is separated from a solution containing a reducing ’06/P2, ’11/P2 agent by an electrolyte, the redox reaction can still occur by transfer of electrons at a distance. The apparatus set-up is shown in Figure 3.10. SPM ’04 Which of the following equations represent redox reactions? I Cu(s) + Cl2(g) → CuCl2(s) II Cu(s) + 2A

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