IB Bridging Programme_Chem : Stoichiometric relationships 1FORMULAS Chemical formulas provide us with a lot of information about chemical substances, such as (a) kinds of atoms, their (b) relative number and (c) the way the atoms are arranged (chemical bonding and structure). There are several types of formulas to represent the chemical substances, namely empirical formula, molecular formula and structural formula. 1. Empirical formula It shows the simplest whole number ratio of the atoms or ions present. It is applicable to all kinds of compounds. 2. Molecular formula It shows the actual number of each kind of atoms in one molecule of the substance. It is applicable to molecular compounds and the elements which contain molecules. 3. Structural formula It shows how the constituent atoms are joined together in a molecule of the substance. It is applicable to molecular compounds and the elements which contain molecules. To determine the structural formula of a molecule, more analyses are required such as chemical tests, mass spectroscopy, infrared spectroscopy and nuclear magnetic spectroscopy etc. (These spectroscopic methods will be discussed in Grade 12.) Example 1 Sodium chloride Empirical formula : Molecular formula : Example 2 NaCl N/A Structural formula : N/A CH3 Molecular formula : C2H6 Ethane Empirical formula : Structural formula : 1 H H H C C H H H What is a chemical formula? Definition, types and examples. Retrieved from http://study.com/academy/lesson/what- is-a-chemical-formula-definition-types-examples.html -1- IB Bridging Programme_Chem : Stoichiometric relationships MOLE CONCEPT As the size of atoms, ions or even molecules in substances are very small; there are no suitable common units to describe their quantity. Scientists then use a special unit to show the quantity of the chemical species of a substance. This unit is known as mole. 2One mole of a species, represented by a formula, is the amount containing the same number of species as the number of atoms in exactly 12.000 00 g of carbon-12. In one mole of carbon-12, i.e. 12.000 00 g, there are 6.02 1023 carbon-12 atoms. This number is known as Avogadro’s constant, L. No. of chemical species = No. of moles of the formula units Avogadro’s constant Example 3 Calculate the number of CO2 molecules in 0.25 mole of CO2(g) ………………………………………………………………………………………… Example 4 Calculate the number of moles of H2O when a sample contains 3.01 10 24 H2O molecules ………………………………………………………………………………………… Carbon-12 atom is used as the standard atom. Exactly 12.000 00 g of carbon-12 has been chosen as the reference unit for the mole. This means that 12.000 00 g of carbon-12 contains one mole, or 6.02 1023 of carbon-12 atoms. Mass, in grams, of one mole of a substance is known as molar mass (unit : g mol-1). Molar mass of a chemical species = sum of relative atomic mass of each atom in the species (with unit) Example 5 Calculate the molar mass of the following substances. (a) Sodium chloride, NaCl ………………………………………………………… 2 (b) Water, H2O ………………………………………………………… (c) Alcohol, C2H5OH ………………………………………………………… (d) Carbon dioxide, CO2 ………………………………………………………… (a) Atomic mass. Retrieved from http://chemwiki.ucdavis.edu/Physical_Chemistry/Atomic_Theory/Atomic_Mass (b) Shmoop Editorial Team, "Atomic Mass - Shmoop Chemistry," Shmoop University, Inc., Last modified November 11, 2008, http://www.shmoop.com/stoichiometry/atomic-mass.html. -2- IB Bridging Programme_Chem : Stoichiometric relationships In other words, if the number of moles of a substance is larger, then its mass will be greater. i.e. Mass of a substance = No. of moles of the substance Molar mass of the substance It is common for the composition data to be given in the form of percentage by mass, and we use the figures to deduce the ratio of atoms present. An understanding of percentage by mass data helps us evaluate information that we commonly see on some commercial products such as food, drinks, medicine and household cleaners. % by mass of element A in a compound = relative atomic mass of A no. of atoms of A in the formula formula mass of the compound Example 6 Determine the percentage by mass of each element in the following substances. (a) Water, H2O H : ……………………………….… O : …………………………………. (b) Alcohol, C2H5OH C : ……………………………….… H : …………………………………. O : ……………………………….… (c) Carbon dioxide, CO2 C : ……………………………….… O : …………………………………. In fact, the percentage by mass of a species can help us to determine the mass of the species in the sample. Mass of the species in the sample = % by mass of the species in the compound mass of the sample Example 7 Calculate the mass of carbon in 5.350 g of CO2(g). ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… Example 8 Calculate the mass of lead in 15.550 g of PbCl2(s). ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… -3- IB Bridging Programme_Chem : Stoichiometric relationships Determination of formula When a new chemical substance is synthesized in a laboratory or isolated from a natural source, chemists need to determine its elemental composition, empirical formula, molecular formula and structure to understand its properties. To determine its formulas, we can conduct combustion analysis. This analysis requires us to burn a substance in a large excess of oxygen to convert the constituent elements of the substance into their corresponding oxide forms. The combustion products will then be collected and weighed separately. Once all the relevant information is collected, we can finally determine its formula. Depending on the type of substance, these two absorbers will not be required. Guidelines : Step 1 : Step 2 : Step 3 : Step 4 : Step 5 : Find the mass of various constituent elements (in “atomic form”). Find the number of moles of each constituent element. Find the relative number of moles. Multiple the relative number of moles by a constant, if necessary. Substitute the numbers found into the “proposed” empirical formula of the compound. -4- IB Bridging Programme_Chem : Stoichiometric relationships Part I : Determination of formula of ionic compounds Example 9 Determine the formula of oxide of magnesium Mass of crucible and the lid 21.510 g Mass of crucible with Mg ribbon and the lid 23.860 g Mass of crucible with the product and the lid 25.450 g Crucible with magnesium ribbon Bunsen burner Let MgxOy be the formula of the oxide of magnesium Step Mg O Step 1 Mass of various constituent elements Step 2 Number of moles of each constituent element Step 3 Relative number of moles Step 4 Multiple the relative number of moles by a constant, if necessary. Step 5 : Substitute the numbers found into the “proposed” empirical formula of the compound. ……………………………………………………………………………………………… -5- IB Bridging Programme_Chem : Stoichiometric relationships Example 10 Determine the formula of oxide of sodium Mass of crucible and the lid 21.510 g Mass of crucible with Na and the lid 25.325 g Mass of crucible with oxide of sodium and the lid 30.655 g Let NaxOy be the formula of the oxide of sodium Step Na O Step 1 Mass of various constituent elements Step 2 Number of moles of each constituent element Step 3 Relative number of moles Step 4 Multiple the relative number of moles by a constant, if necessary. Step 5 : The formula of the compound is …………………………………………………………... -6- IB Bridging Programme_Chem : Stoichiometric relationships Example 11 54.00 g of metal X is allowed to mix with chlorine gas in an experiment to produce 267.00 g of the chloride of metal X. Determine the formula of the chloride of metal X (Given : The relative atomic mass of metal X is 27.0.) ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… Example 12 2.760 g of iron is allowed to mix with oxygen gas in an experiment to produce 3.550 g of the oxide of iron. Determine the formula of the oxide of iron ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… -7- IB Bridging Programme_Chem : Stoichiometric relationships Part II : Determination of formula of covalent compounds The covalent compounds which are made up of carbon and hydrogen are called hydrocarbons. Their formula can be determined in school laboratory using the following set-up. Absorb H2O Absorb CO2 Example 13 Determine the empirical formula of a hydrocarbon X (a) The above experiment was done on a hydrocarbon X. The following table shows the experimental results. Mass of hydrocarbon X used Mass of the tube with P2O5(s) Mass of the tube with NaOH(s) 1.00 g Before combustion 51.30 g After combustion 53.10 g Before combustion 52.65 g After combustion 55.58 g Step C H Step 1 : Mass of element / g Step 2 : No. of moles of element / mol Step 3 : Relative no. of moles Step 4 : Multiple the relative number of moles by a constant, if necessary. Step 5 : The formula of the hydrocarbon X is ……….………………………………………… (b) If the molecular mass of hydrocarbon X is about 45. Determine the molecular formula of X. …………………………………………………………………………………………………… …………………………………………………………………………………………………… -8- IB Bridging Programme_Chem : Stoichiometric relationships Example 14 Determine the molecular formula of a hydrocarbon Y (a) A hydrocarbon Y is completely burnt with excess oxygen to yield 7.873 g of carbon dioxide and 4.030 g of water. Find the empirical formula for the hydrocarbon Y. …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… (b) If the molecular mass of hydrocarbon Y is about 88. Determine the molecular formula of Y. …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… Example 15 Determine the molecular formula of a hydrocarbon P (a) A compound P is made up of carbon and hydrogen. Its composition by mass of C and H is 85.1% and 14.9% respectively. (You can assume the mass of the hydrocarbon P is 100.0 g) …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… (b) If the molecular mass of hydrocarbon P is about 57. Determine the molecular formula of P. …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… -9- IB Bridging Programme_Chem : Stoichiometric relationships Example 16 Determine the molecular formula of a compound Q (a) Compound Q is made up of carbon, hydrogen and oxygen. It is burnt completely in air to form carbon dioxide and water as the only products. It is found that 2.430 g of compound Q gives 3.96 g of carbon dioxide and 1.35 g of water. Find the empirical formula of compound Q. …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… (b) If the molecular mass of hydrocarbon Q is about 162. Determine the molecular formula of Q. …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… Example 17 Determine the number of water of crystallization in hydrated Na2CO3 When 2.48 g of Na2CO3 nH2O is strongly heated, 2.12 g of the anhydrous salt is left. Determine the value of n. ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… - 10 - IB Bridging Programme_Chem : Stoichiometric relationships Example 18 Determine the number of water of crystallization in hydrated MgSO4 The formula of a hydrated magnesium sulphate is MgSO4 nH2O. If the percentage by mass of water of crystallization is 51.22%, what is the value of n? ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… Example 19 Determine the number of water of crystallization in a hydrated salt Z When 0.166 mole of the hydrated salt Z is strongly heated, it gives 17.940 g of water. Determine the number of moles of water of crystallization in one mole of the hydrated salt Z. ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… - 11 - IB Bridging Programme_Chem : Stoichiometric relationships Calculation based on equation Mole concepts can also be applied to predict the amount of products formed or the amount of reactants required for giving certain amount of products. In this, we need a balanced chemical equation and apply the mole concepts to it. Before this, there are two important concepts required discussion. 3 Limiting reactant is the reactant in a chemical reaction that limits the amount of product formed. The reaction will stop when all of the limiting reactant is consumed. Excess reactant is the reactant that remains when a reaction stops. Reaction between A and B : A + B C For example Before reaction has started Reactant A After reaction is completed Reactant B Product C To measure the efficiency of a reaction, percentage yield (or amount) of a product can be determined. It is the ratio between the actual yield and the theoretical yield of the product. The actual yield is the amount of the product obtained in a chemical reaction and the theoretical yield is that calculated according to the amount of limiting reagent and the stoichiometric equation. Percentage yield Guidelines : actual yield 100% theoretica l yield Step 1 : Convert the amounts of given substances into number of moles. Step 2 : Identify the limiting reactant of the process. Step 3 : Find the mole ratio between the limiting reactant and the required substance based on the balanced equation and hence the number of moles of the required substance. Step 4 : Convert the number of moles of the required substance into the quantity required. 3 (a) Stoichiometry : Limiting reactant. Retrieved from https://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm (b) Limiting reagents. Retrieved from http://www.sparknotes.com/testprep/books/sat2/chemistry/chapter7section4.rhtml - 12 - IB Bridging Programme_Chem : Stoichiometric relationships Example 20 Determine the mass of zinc oxide formed Zinc When 6.538 g of zinc powder is heated in air completely, how many grams of zinc oxide are formed? 2Zn(s) + O2(g) 2ZnO(s) Step 1 : Convert the amounts of given substances into number of moles Number of moles of Zn = …………………………………………… …………………………………………… Step 2 : Identify the limiting reactant of the process Limiting reactant : …………………………………………………………………………………… Step 3 : Find the mole ratio between the limiting reactant and the required substance based on the equation and hence the number of moles of the required substance ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… Step 4 : Convert the number of moles of the required substance into the quantity required. Mass of oxide = ………………………………………………………………………………………. Example 21 Iron can be extracted from its ore, haematite. At high temperature, carbon reacts with iron(III) oxide in haematite. In the process, carbon removes oxygen from iron(III) oxide to form iron and carbon dioxide. If 500.0 g of iron(III) oxide is heated with excess carbon powder, how many grams of iron can be produced? 2Fe2O3(s) + 3C(s) 4Fe(s) + 3CO2(g) - 13 - IB Bridging Programme_Chem : Stoichiometric relationships Example 22 (a) A piece of sodium metal Na(s) of 3.500 g is put into a container of chlorine gas Cl2(g) of 36.00 g. The reaction is triggered by adding a drop of water into the container and sodium chloride NaCl(s) is then formed. Calculate the amount, in g, of NaCl(s) obtained in the reaction. 2Na(s) + Cl2(g) 2NaCl(s) (b) If 6.853 g of NaCl(s) is obtained from the experiment, calculate its percentage yield. Example 23 (a) A piece of sodium granule Na(s) of 1.515 g is put into a container of oxygen O2(g) of 3.080 g. The reaction starts when the Na(s) is ignited. Calculate the mass of sodium oxide Na2O(s) obtained in the reaction. 4Na(s) + O2(g) 2Na2O(s) (b) If 1.377 g of Na2O(s) is obtained from the experiment, calculate its percentage yield. - 14 - IB Bridging Programme_Chem : Stoichiometric relationships Example 24 Magnesium nitride Mg3N2(s) is a yellow solid, which can speed up some industrial processes. A student wants to prepare the compound in a school laboratory. What is the minimum amount, in grams, of magnesium Mg(s) and nitrogen N2(g) required to give 5.500 g of magnesium nitride? 3Mg(s) + N2(g) Mg3N2(s) Example 25 (a) When hydrogen sulphide H2S(g) reacts with chlorine Cl2(g), hydrogen chloride HCl(g) and sulphur S(s) are formed. H2S(g) + Cl2(g) 2HCl(g) + S(s) Find the mass of HCl(g) formed if 10.00 g of H2S(g) and 10.00 g of Cl2(g) are allowed to react. (b) If 5.193 of HCl(g) is obtained from the experiment, calculate its percentage yield. - 15 -
