Criterion A: Knowing and Understanding Unit Title: Stoichiometry Name: The questions in this test will be used to give you a criterion A level, it is divided into three sections. Section A: two questions that will assess how you explain your scientific knowledge. Section B: two questions that will assess your application of scientific knowledge and how you use it to solve problems. Section C: one questions that will assess the way you analyses, evaluate and judge scientific information. Each question is designed to be hierarchical, this means that as you answer each question you will find it will get progressively more difficult. Question Level 1-2 Section A Section B Section C 3-4 5-6 7-8 1 2 3 4 5 6 Section A: Strand One Question One This question is about Mole and Chemical Formula (a) State the missing words from this paragraph, using only the words from the list. You should only use a word once and you will need to use all of the words. [1-2] Mole Stoichiometry Same Particle Electron Proton 6,022 x 1023 Neutron Stoichiometry is a method for chemist do calculation. The heart of stoichiometry is mole. It is actually a unit of amount of particles . One mole contains of 6,0022 x1023 number of particles. The number of atoms or other particles in a mole is the same for all substances. In comparison, one mole of oxygen consists, by definition, of the same number of atoms as carbon-12, but it has a mass of 15.999 grams. This mass difference due to the different number of Proton and Neutron In both atoms. Electron is not affecting mass of atom because it is too small. (b) Outline the different between empirical and molecular formula [3-4] Empirical describes the simplest ratio among atoms in the compound, while molecular describe the real ratio, and it is used to calculate molar mass of the compound. (c) Using your scientific knowledge, describe and explain the determination of magnesium oxide. Your level will be determined, be the detail in your response: explain to give a detailed account including reasons or causes. [5-6] and [7-8] A known mass of magnesium ribbon is heated and let it be exposed with oxygen until it’s burned and turned into white ash. The mass before and after heating then used to calculate mole ratio of Mg and O. The fire needs to contact the crucible so the temperature is high enough to burn Mg, the lid must be open a little to supply oxygen Section A: Strand One Question Two a. State two component of atom that determine the mass of atom [1-2] Proton and neutron b. Outline two types of chemical formula [3-4] Empirical formula is the simplest ratio Molecular formula is the actual ratio c. Describe the composition of carbon atom [5-6] Atom carbon has 6 neutron and proton in the nucleus/center and it has 12 electron moving around in two layers d. Explain why the relative mass does not has unit [7-8] Relative mass is mass of atom relative to 1/12 mass of carbon, since the unit of gram will be canceled out, there is no unit at the end. 1 π΄π = ππ£πππππ πππ π ππ ππ ππ‘ππ (ππππ) π₯ 12 π₯ πππ π ππ ππππππ (ππππ) Section B: Strand two Question Three In this question you must solve problems related "Mole calculation” (a) Calculate the number of mole of 300 g of iron [1-2] 1 πππ ππ 300 ππππ πΉπ π₯ = 5,357 πππ 56 ππππ ππ This Information below is for question B to D An organic compound A contains 62.0% by mass of carbon, 24.1% by mass of nitrogen, the remainder being hydrogen. (b) Determine the percentage by mass of hydrogen and the empirical formula of A. [3-4] %H = 100-(62+24.1) = 13.9 % C. : N. : H 62 24.1 13,9 βΆ βΆ 12 14 1 3 : 1 : 8 C3NH8 (c) The relative molecular mass of A is 116. Determine the molecular formula of A [5-6] (C3NH8)n= 116 ((12 x3)+14+(8x1))n=116 58n=116 n=2 C6N2H16 (d) How many number of hydrogen’s atom are present in 90 gram of this molecules. [7-8] = 90 ππππ ππππππ’πππ π₯ 23 75.33 x 10 Hydrogen CD.E FGHI JKLGMFNO CPP FGHI IMQNRSQNT π₯ C IMQ JULGMFNO C IMQN JULGMFNO π₯ V.PWW X CPYZ JULGMFNO C IMQN [ULGMFNO Section B: Strand two Question Four A solution containing NaCl 0.1M a. How many moles of NaCl present in 2 L of this solution [1-2] 0.1 πππ πππΆπ = 2 πππ‘ππ πππΆπ π₯ = 0.2 πππ 1 πππ‘ππ πππΆπ b. How many gram of NaCl present in 2 L of this solution [3-4] = 2 πππ‘ππ πππΆπ π₯ 0.1 πππ πππΆπ 58.5 ππππ πππΆπ π₯ = 11.7 ππππ 1 πππ‘ππ πππΆπ 1 πππ πππΆπ = c. Explain how to make 0,1 M NaCl [5-6] For one litter of solution, weigh 5,85 gram of NaCl. This is because in 0.1 M NaCl, it consist of 0,1 mol of NaCl or 5.85. Then dissolve it in distilled water. Note that for dissolving the NaCl the final volume must be 1 litter. d. If 100 ml of NaCl 0,1M is heated until the volume of solution become 50 ml because of evaporation of water, calculate the molarity after heating [7-8] Amount of NaCl is not changed after heating = 0.01 mol π π 0.01 ππππ π= = π. π π΄ 0.05 πππ‘π‘ππ π= Section C: Strand Three Question Five The table below list some data synthesis of a compound X from sulphur and oxygen, No Mass of Sulphur Mass of Oxygen Mass of compound Mass of reactant Formed remaining 1 6 6 10 2 2 6 7 10.7 1.3 3 6 8 13.3 0.7 4 6 9 15 0 5 6 10 15 1 a. How does the mass of compound form change as the mass of oxygen increases? [1-2] Increase then constant b. If you were to express the data above using visual format, what form would it take? You can sketch this below if you wish. [3-4] c. Einstein is asked to make conclusion from the data above. [5-6] [7-8] “I think some data are not valid. The data from 9 gram of oxygen and above showing a constant result of compound. These two data should be ignored because it’s not aligned with the other three data. So over all the conclusion from this experiment is the more oxygen mass added the more compound will be formed”. Do you agree with Einstein’s conclusion? You should use the idea you have learnt in class, and these results, to answer this question No, I disagree. Atoms in a compound have certain ratio. The ratio in this experiment is 2:3 . if the mass reacted does not follow the ratio, than there will be an excess reagent not reacted. In the experiment the mass compound increases as the mas oxygen increase but constant after 9 and 10 gram. The data of 9 and 10 gram of oxygen is still valid. The compound mass are constant because it has reached the maximum possible ratio.
