1 Mole Concept Stoichiometry Tutorial Ans

SH1 H2 Chemistry 2023 The Mole Concept and Stoichiometry Discussion Questions Solutions Calculation of Relative Molecular Mass 1 Bromine consists of two isotopes, 79Br and 81Br, in the relative abundance ratio of 1:1. Bromine exists as diatomic molecule, Br2, at room temperature and pressure. Calculate the possible relative molecular masses, M r, of Br 2 molecules formed by these two isotopes and their relative abundance ratio. [1:2:1] Molecular Formula 79Br79Br 79Br81Br or 81Br79Br 81Br81Br Mr 158 160 162 Relative Ratio 4 1 2 1 2 1 =     2 2 1 2 1 2 1 =     4 1 2 1 2 1 =     Abundance Ratio 1 2 1 Calculation involving Parts Per Million 2 Gardeners sometimes fumigate their greenhouses to rid them of pests and moulds by burning a sulfur ‘candle’. A gaseous concentration of sulfur dioxide of 50 ppm (parts per million) by volume is effective. Calculate how many grams of sulfur a gardener need s to burn in order to produce a concentration of 50 ppm of SO2 in a greenhouse that measures 2 m × 3 m × 4 m. Assume room conditions. [1.61 g] Equation for reaction: S(s) + O2(g) → SO2(g) Volume of greenhouse = (2)(3)(4) = 24 m3 = 24000 dm3 [Recall: 1 m3 = 103 dm3] Given that concentration of SO2 = 50 ppm  In 106 dm3 of air, there is 50 dm3 of SO2,  In 24000 dm3 of air, there is 𝟓𝟎 𝟏𝟎𝟔 x 24000 = 1.20 dm3 of SO2 Amt of SO2 = 1.20 ÷ 24.0 = 0.0500 mol [Recall room conditions = r.t.p = 24.0 dm3 mol-1] Mass of S = 0.0500 x 32.1 = 1.605 = 1.61 g ( to 3 sf) Calculation using Mole Concept 3 Phosgene, COCl2, was once used as a war gas. It is poisonous because when inhaled, it reacts with water in the lungs to produce carbon dioxide gas and hydrochloric acid which causes severe lung damage, leading to death ultimately. (a) Calculate the percentage by mass of chlorine in COCl2. [71.7%] ( ) ( ) 2 2 35.5% by mass of C in COC 100% 12.0 16.0 2 35.5 =  = ++ 71.7 %ll (b) Write a balanced equation for the reaction between COCl2 and H2O. COCl2 + H2O → CO2 + 2HCl 𝒙 𝟐𝟒𝟎𝟎𝟎 𝐱 𝟏𝟎𝟔 = 𝟓𝟎 𝒙 = 1.20 dm3

SH1 H2 Chemistry 2023 (c) Calculate the amount of HCl that will be produced by the complete reaction of 0.430 mol of COCl2. [0.860] Since COCl2 : HCl= 1 : 2, l =  =amount of HC produced 0.430 0.860 mol2 (d) Identify the limiting reagent when 0.200 mol of COCl2 is mixed with 6.20 g of H2O. Hence, calculate the amount of HCl produced at the end of the reaction. [0.400] ( ) 2 2 6.20amount of H O used 2 1.0 16.0 amount of COC used 0.200 mol == + = 0.344 mol l 0.200 mol of COCl2 will require 0.200 mol of H2O, Since 0.344 mol of H2O is present (excess), COCl2 is the limiting reagent. Since COCl2 : HCl is 1 : 2, l =  =amount of HC produced 0.200 0.400 mol2 Calculation involving % purity & % yield 4 The mineral phosphorite, Ca3(PO4)2, exists as phosphate rock in its impure form. Elemental phosphorus can be prepared from phosphate rock by reduction using carbon in the presence of sand, SiO2. The reduction of phosphorite also produces solid A and carbon monoxide. (a) A has the following composition by mass: Ca: 34.2% Si: 24.5% O: 41.3% Calculate the empirical formula for solid A. [CaSiO3] elements Ca Si O % by mass 34.2 24.5 41.3 amount / mol 34.2 40.1 = 0.853 24.5 28.1 = 0.872 41.3 16.0 = 2.58 Simplest whole number mole ratio 0.853 0.853 = 1 0.872 0.853  1 2.58 0.853  3  The empirical formula of solid A is CaSiO3. (b) Write a balanced equation for the reaction. Ca3(PO4)2 + 5C + 3SiO2 → 2P + 3CaSiO3 + 5CO (c) A 30.0 g sample of phosphate rock was subjected to the above reaction and produced 5.3 g of phosphorus. Calculate the percentage purity of phosphorite in the rock sample. [88.4%] amount of phosphorus = = 0.171 mol Since Ca3(PO4)2 : P is 1 : 2 amount of Ca3(PO4)2 = ½ × 0.171 = 0.0855 mol mass of Ca3(PO4)2 = 0.0855 × [3(40.1) + 2(31.0) + 8(16.0)] = 26.53 g % purity of Ca3(PO4)2 in the rock sample= 26.53 30.0 ×100% = 88.4% 5.3 31.0

SH1 H2 Chemistry 2023 5 In the Solvay process, ammonia is recovered by the reaction: 2NH4Cl (aq) + CaO (s) → CaCl2 (aq) + H2O (l) + 2NH3 (g) (a) What is the maximum volume of ammonia that can be recovered, at s.t.p., from 20.0 g of NH4Cl and 4.50 g of CaO? State the limiting reagent, if any, and assume that any impurity present is unreactive. [3.64 dm3] Amount of NH4Cl = 53.5 20.0 = 0.3738 mol Amount of CaO = 56.1 4.50 = 0.08021 mol 0.08021 mol of CaO will require 2(0.08021) = 0.1604 mol of NH4Cl Since 0.3738 mol of NH4Cl is present (excess), hence CaO is the limiting reagent. Since CaO:NH3 is 1 : 2, Amount of NH3 = 2 × 0.08021 = 0.1604 mol Vol of NH3 = 0.1604 × 22.7 = 3.64 dm3 (b) What is the percentage yield if only 3.03 dm3 of NH3 is obtained experimentally at s.t.p.? [83.2%] % yield = 3.03 3.64 × 100 = 83.2 % Calculation involving % composition by mass 6 In small quantities, nicotine in tobacco is addictive. In large quantities, it is a deadly poison. Determine the molecular formula of nicotine, CxHyNz, if 4.38 mg of nicotine burns to form 11.9 mg of carbon dioxide and 3.41 mg of water. [Mr of nicotine = 162.0] [C10H14N2] Mass of C in nicotine = 11.9 × 44.0 12.0 = 3.245 mg Mass of H in nicotine = 3.41 × 18.0 2.0 = 0.3789 mg Mass of N in nicotine = 4.38 – 3.245 – 0.3789 = 0.7561 mg Element C H N Mass / mg 3.245 0.3789 0.7561 Amount / mol 0.0002704 0.0003789 5.401 x 10-5 Simplest whole number mole ratio 5 7 1 ∴ Empirical formula of nicotine is C5H7N. Let the molecular formula be (C5H7N)n Mr of (C5H7N)n = 162.0 {(5(12.0) + 7(1.0) + 14.0)}n = 162.0 n = 2 Molecular formula of nicotine is C10H14N2.

SH1 H2 Chemistry 2023 Calculation involving Volumes of gas 7 Buckminsterfullerene, C60, is a molecule with 60 carbon atoms arranged in pentagons and hexagons that are similar to those in a football. C 60 reacts with hydrogen to form hydrofullerenes with the molecular formula C60Hn. When 60 g of C 60 is reacted with hydrogen gas at 273 K and 1 bar, the volume of hydrogen gas is decreased by 34 dm3. Find the value of n in the formula of hydrofullerene, C60Hn. [36] Amount of C60 = 60 60 x 12.0 = 0.08333 mol Amount of H2 = 34 22.7 = 1.498 mol Mole ratio of C60 : H2 = 0.08333 : 1.498 1 : 18 Mole ratio of C60 to H = 1 : 36 Thus, molecular formula of the hydrofullerene formed is C60H36. n= 36 Calculation using Volumes of gas in Combustion reactions 8 20 cm3 of a gaseous hydrocarbon A, CxHy, was exploded with an excess of oxygen. Upon cooling to room temperature, there was a contraction in volume of 50 cm 3. When the products were treated with excess potassium hydroxide, there was a further contraction of 60 cm3. Deduce the molecular formula of A. All volumes were measured at r.t.p. [C3H6] Before reaction After reaction & cooled to r.t.p After reaction with KOH(aq) 2C H O addedVV + xy decrease in vol. of 22CO O VV + unreacted further decrease of 2O V unreacted 20 cm3 + excess O2 50 cm3 60 cm3 Vol of CO2 = 60 cm3 (2nd Contraction) Initial vol – Final vol = 50 cm3 (1st Contraction) (Vol CxHy + Total Vol O2) – (Vol CO2 + Vol unreacted O2) = 50 cm3 (20 + Total Vol O2) – (