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3 1 Chemical Equations and Stoichiometry 3.1 Formulae of Compounds 3.2 Derivation of Empirical Formulae 3.3 Derivation of Molecular Formulae 3.4 Chemical Equations 3.5 Calculations Based on Chemical Equations 3.6 Simple Titrations Stoichiometry (化學計量學)p.19 Deals with quantitative relationships (a) among atoms, molecules and ions RAM / RMM / Formula Mass 2 Stoichiometry (化學計量學)p.19 Deals with quantitative relationships (b) among the constituent elements of a compound Empirical / Molecular Formulae 3 Stoichiometry (化學計量學)p.19 Deals with quantitative relationships (c) among the substances participating in chemical reaction Calculations involving chemical equation 4 3.1 5 Formulae of Compounds 3.1 Formulae of compounds (SB p.43) Empirical formula Shows the simplest whole number ratio of the atoms or ions present E.g. Methane, CH4 Sodium chloride, NaCl 6 3.1 Formulae of compounds (SB p.43) Molecular formula Shows the actual number of each kind of atoms present in one molecule E.g. CH4 methane Ionic compounds do not have molecular formulae 7 3.1 Formulae of compounds (SB p.43) Structural formula Shows the bonding order of atoms in one molecule H E.g. CH4 methane H C H 8 H 3.1 Formulae of compounds (SB p.44) Different types of formulae of some compounds Compound Empirical formula Molecular formula Structural formula Carbon dioxide Water CO2 CO2 O = C =O H2O H2O O Methane CH4 H H H CH4 H C H H Glucose CH2O C6H12O6 OH O H H H OH H HO OH H Sodium fluoride 9 NaF Not applicable OH Na+F- 3.2 10 Derivation of Empirical Formulae 3.2 Derivation of empirical formulae (SB p.45) From composition by mass Example 1 Mg + N2 MgxNy 0.450 g excess 0.623 g Mass of N used = (0.623-0.450)g = 0.173 g 11 3.2 Derivation of empirical formulae (SB p.45) From composition by mass Example 1 Mg + N2 MgxNy 0.450 g excess 0.623 g mMg nMg nN 12 MMg mN MN 0.450 1.5 3 24.31 0.173 1 2 14.01 Mg3N2 3.2 Derivation of empirical formulae (SB p.45) Water of Crystallization Derived from Composition by Mass Example 2 Q.14 Example 3-3C 13 Check Point 3-3A Q.14 CuSO4.xH2O heat 10.0 g CuSO4 + xH2O 6.4 g (10.0–6.4) g = 3.6 g 6.4 3.6 : 0.04 : 0.2 1 : 5 CuSO4 : H2O = 160 18 x = 5 14 From combustion data CxHyOz Vitamin C heat CxHyOz 0.2000 g 15 + O2 excess CO2(g) 0.2998 g + H2O(g) 0.0819 g CxHyOz + O2 CO2(g) + H2O(g) Mass of C in sample = mass of C in CO2 formed Mass of H in sample = mass of H in H2O formed Mass of O in sample = total mass of sample – mass of C – mass of H 16 12.0 mass of C mass of CO2 44.0 12.0 0.2998 0.320 g 44.0 2.0 mass of H mass of H2O 18.0 2.0 0.0819 0.0267 g 18.0 Mass of O in sample = (0.2000-0.0818-0.0091) g = 0.109 g 17 Mass (g) Number of moles (mol) Simplest whole no. ratio C H O 0.0818 0.0091 0.109 0.0091 9.1 10-3 1.0 0.109 6.8 10-3 16.0 4 3 0.0818 6.8 10-3 12.0 3 C3H4O3 18 3.3 19 Derivation of Molecular Formulae 3.3 Derivation of molecular formulae (SB p.49) What is molecular formulae? Molecular formula = (Empirical formula)n 20 3.3 Derivation of Molecular Formulae (SB p.49) From empirical formula and known relative molecular mass Empirical formula Molecular mass Example 3-3A Example 3-3B Molecular formula 21 Q.15 CnH2n+2(l) 473 K, 1.00 atm 12.0 g CnH2n+2(g) 3.28 dm3 3 1 1 12.0 g (0.082 atm dm K mol )(473K) mRT M 1.00 atm 3.28 dm3 PV = 142 g mol1 Relative molecular mass = 142 22 Q.15 CnH2n+2(l) 473 K, 1.00 atm 12.0 g CnH2n+2(g) 3.28 dm3 RMM = 12n + 2n+2 = 142 n = 10 The molecular formula is C10H22 23 Determination of Chemical Formulae C, H, O are present Qualitative Analysis Structural formula Ethanoic acid CH3COOH Quantitative Analysis Empirical formula CH2O 24 RMM = 60.0 IR, NMR, MS Molecular formula C2H4O2 H H C H O C O H Structural formula : bond-line structure 25 Calculate the % by mass of the constituent elements of soda alum. Na2SO4·Al2(SO4)3·24H2O 2 23.0 100% 5.02% Na : 916.4 S: 4 32.1 100% 14.0% 916.4 2 27.0 100% 5.89% Al : 916.4 26 Calculate the % by mass of the constituent elements of soda alum. Na2SO4·Al2(SO4)3·24H2O 40 16.0 100% 69.8% O: 916.4 H: 27 48 1.0 100% 5.2% 916.4 A certain compound was known to have a formula which could be represented as [PdCxHyNz](ClO4)2. Analysis showed that the compound contained 30.15% carbon and 5.06% hydrogen. When converted to the corresponding thiocyanate, [PdCxHyNz](SCN)2, the analysis was 40.46% carbon and 5.94% hydrogen. Calculate the values of x, y and z. (Relative atomic masses : C = 12.0, H = 1.0, N = 14.0, O = 16.0, Cl = 35.5, S = 32.0, Pd = 106.0) 28 Let M be the formula mass of [PdCxHyNz](ClO4)2 Then, the formula mass of [PdCxHyNz](SCN)2 = M + 2(32.0+12.0+14.0) – 2(35.5+416.0) = M – 83.0 % by mass of C in [PdCxHyNz](ClO4)2 12.0x 100% 30.15% M % by mass of C in [PdCxHyNz](SCN)2 12.0(x 2) 100% 40.46% M 83.0 29 12.0x 100% 30.15% M 12.0(x 2) 100% 40.46% M 83.0 (1) (2) Solving by simultaneous equations, x = 14, M = 557 % by mass of H in [PdCxHyNz](ClO4)2 1.0y 100% 5.06% 557 y = 28 557 = 106.0 + 12.014 + 1.028 + 14.0z + 2(35.5+416.0) z = 4 30 3.4 31 Chemical Equations 3.4 Chemical equations (SB p.53) Chemical equations aA+bB cC+dD a, b, c, d are stoichiometric coefficients na : nb : nc : nd a : b : c : d 32 3.5 Calculations Based on Equations (SB p.65) Calculations based on equations Calculations involving reacting masses Example 3-5A Example 3-5B Check Point 3-4 33 3.4 Chemical equations (SB p.53) Example (a) Excess oxygen 2Mg(s) 2.43 g + O2(g) 2MgO(s) excess ? 2.43 g nMg reacted 0.100 mol 1 24.3 g mol nMg 2 1 nMgO 2 nMgO produced 0.100 mol 34 3.4 Chemical equations (SB p.53) Example (a) Excess oxygen 2Mg(s) 2.43 g + O2(g) 2MgO(s) excess ? mMgO 0.100 mol (24.3 16.0) g mol 1 = 4.03 g 35 3.4 Chemical equations (SB p.53) Example(b) limiting reagent to be determined 2Mg(s) 2.43 g + O2(g) 2MgO(s) 1.28g ? nMg 0.100mol nO2 36 1.28 g 0.040 mol 1 32.0 g mol 3.4 Chemical equations (SB p.53) Example(b) limiting reagent to be determined 2Mg(s) 0.100 mol nMg nO2 + O2(g) 2MgO(s) 0.040 mol 0.100 mol 2 2.5 0.040 mol 1 Mg is in excess, O2 is the limiting reagent 37 ? 3.4 Chemical equations (SB p.53) Example(b) limiting reagent to be determined 2Mg(s) + O2(g) 2MgO(s) 0.040 mol 0.080 mol mMgO 0.080 mol (24.3 16.0) g mol = 3.22 g 1 Q.16, 17 Check Point 3-4 38 Q.16 P4 (s) 4.00 g + 5O2(g) 2P2O5(s) 6.00 g 4.00 g nP4 0.0323 mol 1 31.0 4 g mol nO2 39 6.00 g 0.188 mol 1 32.0 g mol limiting reactant P4(s) 0.0323 mol + excess 5O2(g) 0.188 mol nO2 0.188 mol 5.82 5 nP4 0.0323 mol O2 is in excess 40 2P2O5(s) Q.16 P4(s) + 5O2(g) 2P2O5(s) 20.0323 mol 0.0323 mol mP2O5 2 0.0323 mol 142 g mol = 9.17 g 41 1 Q.17 0.27 g nAl reacted 0.01 mol 1 27 g mol 0.96 g nCu produced 0.015 mol 1 64 g mol nAl : nCu 1 : 1.5 = 2 : 3 2Al(s) + 3Cu2+(aq) 2Al3+(aq) + 3Cu(s) 42 3.5 Calculations Based on Equations (SB p.66) Calculations involving volumes of gases 1.Gases not at the same conditions 2.Gases at the same conditions - Gay Lussac’s Law 43 CH4(g) + 2O2(g) CO2(g) + 2H2O(g) 2.8 dm3 25C 1.65 atm 35.0 dm3 31C 1.25 atm ? dm3 125C 2.50 atm nCH4 PV 1.65 atm 2.8 dm3 0.189 mol 3 1 1 RT 0.082 atm dm K mol 298 K nO2 PV 1.25 atm 35.0 dm3 1.75 mol 3 1 1 RT 0.082 atm dm K mol 304 K nO2 nCH4 44 1.75 2 0.189 1 O2 is in excess and CH4 is the limiting reactant CH4(g) + 2O2(g) CO2(g) + 2H2O(g) 0.189 mol VCO2 nCO2 RT 0.189 mol ? dm3 125C 2.50 atm P 0.189 mol 0.082 atm dm3 K 1 mol 1 (273 125) K 2.50 atm = 2.47 dm3 45 Gay Lussac’s law : When gases reacts, they do so in volumes which bears a simple whole number ratio to one another, and to the volumes of gaseous products, all volumes being measured under the same conditions of temperature and pressure. Watch video 46 39.5 cm3 Ammonia Fe as catalyst 20 cm3 H2(g) + CuO nitrogen + hydrogen 10 cm3 heat 29.5 cm3 Cu(s) + H2O(l) 2NH3(g) 1N2(g) + 3H2(g) VNH3 : VN2 : VH2 20 : 10 : 29.5 2 : 1 : 3 47 Gay Lussac’s law is an application of the Avogadro’s law. a A(g) + b B(g) c C(g) + d D(g) na : nb : nc : nd a : b : c : d At fixed T & P, nV Va : Vb : Vc : Vd a : b : c : d 48 Gay Lussac’s law a A(g) + b B(g) c C(g) + d D(g) At fixed T & P Va : Vb : Vc : Vd a : b : c : d 49 Determination of Molecular Formula from Reacting Volumes of Gases For CxHy y y CxHy(g) + x O2(g) xCO2(g) + H2O(?) 4 2 VC x H y : VO2 : VCO2 50 y 1: x : x 4 Determination of Molecular Formula from Reacting Volumes of Gases For CxHyOz y y z CxHyOz(g) + x O2(g) xCO2(g) + H2O(?) 2 4 2 VC x H y Oz : VO2 : VCO2 51 y z 1: x : x 4 2 Q.18 y y CxHy(g) + x O2(g) xCO2(g) + H2O(l) 4 2 15 cm3 75 cm3 45 cm3 Volume of CO2 formed = volume of gas absorbed by NaOH = (70 – 25) cm3 = 45 cm3 52 Before the reaction, VO2 100cm3 After the reaction, VO2 25cm3 VO2used = 75 cm3 Q.18 y y CxHy(g) + x O2(g) xCO2(g) + H2O(l) 4 2 15 cm3 VCO2 VCxHy VO2 75 cm3 45 cm3 x 45 cm x 3 3 1 15 cm 3 y 4 x 75 cm 5 3 VCxHy 1 15 cm y (3 ) 5 y 8 4 53 3 C3H8 Q.19 y y z CxHyOz(g) + x O2(g) xCO2(g) + H2O(g) 2 4 2 50 cm3 VCO2 VCxHyOz VH2O VCxHyOz VO2 VCxHyOz 54 100 cm3 x 100 cm3 x 2 3 1 50 cm y 3 100 cm 2 y4 3 1 50 cm x 4y 2z 100 cm3 2 3 1 50 cm 100 cm3 100 cm3 C2H4O2 4 z (2 ) 2 z 2 4 2 3.6 55 Simple Titrations 3.6 Simple titrations (SB p.58) Simple titrations Simple titrations Acid-base titrations 56 Redox titrations 3.6 Simple titrations (SB p.58) Simple titrations Acid-base titrations Acid-base titrations with indicators Acid-base titrations without indicators Refer to Notes on ‘Detection of Equivalence Point in Acid-base Titration 57 End point : The point at which the indicator undergoes a sharp colour change in a titration Equivalence point : The point at which the reaction is just complete without excess of any reactant. 58 3.6 Simple titrations (SB p.62) Titration without an indicator By following the change (a) in pH value (pH titration) (b) in temperature (thermometric titration) (c) in electrical conductivity (conductometric titration) in the course of the reaction 59 pH Titration Curves Phenolphthalein Equivalence point, pH 7.00 Methyl orange 60 pH Titration Curves Equivalence point, pH 5.28 Methyl orange 61 pH Titration Curves Equivalence point, pH 8.72 Phenolphthalein 62 pH Titration Curves Slow change in pH around equivalence point Equivalence point, ~ pH 7 Most indicators are not suitable 63 Thermometric Titration Temperature / K Equivalence point Volume of acid added / cm3 64 H+(aq) + OH(aq) H2O(l) + heat cooling by excess acid Temperature / K Equivalence point Volume of acid added / cm3 65 Conductometric Titration – NaOH vs HCl Electrical Conductivity Equivalence point Volume of acid added /cm3 66 Conductometric Titration – NaOH vs HCl more mobile + + Electrical Conductivity H (aq) + Cl (aq) + Na (aq) + OH (aq) less mobile H2O(l) + Na+(aq) + Cl(aq) Volume of acid added /cm3 67 Conductometric Titration – NaOH vs HCl Electrical Conductivity Beyond the equivalence point, conductivity sharply due to excess H+ (most mobile) & Cl Steeper slope Equivalence point Volume of acid added /cm3 68 At the equivalence point Electrical Conductivity H+(aq) + Cl(aq) + Na+(aq) + OH(aq) H2O(l) + Na+(aq) + Cl(aq) conductivity > 0 Volume of acid added /cm3 69 Conductometric Titration – NH3 vs HCl Electrical Conductivity Equivalence point Volume of acid added /cm3 70 Initial conductivity is low NH3(aq) + H2O(l) NH4+(aq) + OH(aq) Electrical Conductivity Equivalence point Volume of acid added /cm3 71 H+(aq) + Cl(aq) + NH3(aq) Electrical Conductivity NH4+(aq) + Cl(aq) less conducting more conducting Volume of acid added /cm3 72 Electrical Conductivity Beyond the equivalence point, conductivity sharply due to excess H+ (most mobile) & Cl steeper slope Equivalence point Volume of acid added /cm3 73 Conductometric Titration – NaOH vs CH3COOH Electrical Conductivity Equivalence point Volume of acid added /cm3 74 CH3COOH(aq) + Na+(aq) + OH(aq) more conducting Electrical Conductivity less conducting H2O(l) + Na+(aq) + CH3COO(aq) Volume of acid added /cm3 75 Electrical Conductivity Beyond the equivalence point, conductivity slowly because the excess acid is weak smaller slope Equivalence point Volume of acid added /cm3 76 Conductometric Titration – NH3 vs CH3COOH Electrical Conductivity Equivalence point Volume of acid added /cm3 77 Initial conductivity is low Electrical Conductivity NH3(aq) + H2O(l) NH4+(aq) + OH(aq) Equivalence point Volume of acid added /cm3 78 CH3COOH (aq) + NH3(aq) Electrical Conductivity CH3COO(aq) + NH4+(aq) more conducting Volume of acid added /cm3 79 Electrical Conductivity Beyond the equivalence point, conductivity slowly because the excess acid is weak smaller slope Equivalence point Volume of acid added /cm3 80 Conductometric Titration – Ba(OH)2 vs H2SO4 Electrical Conductivity At equivalence point, conductivity 0 because BaSO4 is insoluble in water Equivalence point Volume of acid added / cm3 81 3.6 Simple titrations (SB p.58) Simple titrations 6B Simple titrations Acid-base titrations 82 Redox titrations 3.6 Simple Titrations (SB p.65) Redox titrations 1. Iodometric titration I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O62-(aq) brown colourless S2O3 2-(aq) I2 brown 83 Very few I2 yellow During the course of the titration 3.6 Simple Titrations (SB p.65) Redox titrations 1. Iodometric titration I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O62-(aq) brown colourless Starch Very few I2 yellow 84 solution Complex of starch & I2 dark blue During the course of the titration 3.6 Simple Titrations (SB p.65) Redox titrations Example 3-6E 1. Iodometric titration I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O62-(aq) brown colourless At the end point S2O32-(aq) 2I-(aq) + Complex of starch & I2 dark blue 85 S4O62-(aq) colourless 3.6 Simple Titrations (SB p.66) 2. Titrations involving potassium manganate(VII) MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + H2O(aq) + 5Fe3+(aq) purple pale green colourless yellow H+(aq)/MnO4(aq) Fe2+ & Fe2+ Fe3+ During the course of the titration 86 3.6 Simple Titrations (SB p.66) 2. Titrations involving potassium manganate(VII) MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + H2O(aq) + 5Fe3+(aq) purple pale green colourless yellow H+(aq)/MnO4(aq) Fe2+ & Fe3+ At the end point Example 3-6F 87 Fe3+ & MnO4 Check Point 3-6 NaOH + HCl NaCl + H2O Na2CO3 + HCl NaHCO3 + NaCl Red in phenolphthalein colourless in phenolphthalein Double indicator titration 88 NaHCO3 + HCl NaCl + H2O + CO2 yellow in methyl orange red in methyl orange Double indicator titration 89 The END 90 3.1 Formulae of compounds (SB p.45) Back Give the empirical,Empirical molecular and structural formulae for the Compound Molecular Structural formula formula following compounds: formula (a) Propene (a) Propene (b) Nitric acid (b) Nitric (c) Ethanol acid (c) Ethanol (d) Glucose (d) Glucose CH2 C3H6 H H H H C C C H HNO3 HNO3 O H O N O C2H6O C6H12O6 C2H5OH C6H12O6 H H H C C H H Answer OH OH O H H H OH H H OH HO 91 H OH 3.2 Derivation of empirical formulae (SB p.46) A hydrocarbon was burnt completely in excess oxygen. It was found that 1.00 g of the hydrocarbon gives 2.93 g of carbon dioxide and 1.80 g of water. Find the empirical formula of the hydrocarbon. Answer 92 3.2 Derivation of empirical formulae (SB p.46) The relative molecular mass of CO2 = 12.0 + 16.0 2 = 44.0 Mass of carbon in 2.93 g of CO2 = 2.93 g 12.0 = 0.80 g 44.0 The relative molecular mass of H2O = 1.0 2 + 16.0 = 18.0 2 .0 Mass of hydrogen in 1.80 g of H2O = 1.80 g = 0.20 g 18.0 Let the empirical formula of the hydrocarbon be CxHy. Mass of carbon in CxHy = Mass of carbon in CO2 Mass of hydrogen in CxHy = Mass of hydrogen in H2O 93 3.2 Derivation of empirical formulae (SB p.46) Back Carbon Hydrogen Mass (g) 0.80 0.20 No. of moles (mol) 0.80 0.0667 12.0 0.20 0.20 1 .0 Relative no. of moles 0.0667 1 0.0667 0.20 3 0.0667 Simplest mole ratio 1 3 Therefore, the empirical formula of the hydrocarbon is CH3. 94 3.2 Derivation of empirical formulae (SB p.46) Compound X is known to contain carbon, hdyrogen and oxygen only. When it is burnt completely in excess oxygen, carbon dioxide and water are given out as the only products. It is found that 0.46 g of compound X gives 0.88 g of carbon dioxide and 0.54 g of water. Find the empirical formula of compound X. Answer 95 3.2 Derivation of empirical formulae (SB p.47) Mass of compound X = 0.46 g 12.0 Mass of carbon in compound X = 0.88 g = 0.24 g 44.0 Mass of hydrogen in compound X = 0.54 g 2.0 = 0.06 g 18.0 Mass of oxygen in compound X = 0.46 g – 0.24 g – 0.06 g = 0.16 g Let the empirical formula of compound X be CxHyOz. 96 3.2 Derivation of empirical formulae (SB p.47) Back Carbon Hydrogen Oxygen Mass (g) 0.24 0.06 0.16 No. of moles (mol) 0.80 0.0667 12.0 0.06 0.06 1 .0 Relative no. of moles 0.02 2 0.01 0.06 6 0.01 0.01 1 0.01 Simplest mole ratio 2 6 1 0.16 0.01 16.0 Therefore, the empirical formula of compound X is C2H6O. 97 3.2 Derivation of empirical formulae (SB p.47) (a) 5 g of sulphur forms 10 g of an oxide on complete combustion. What is the empirical formula of the oxide? Answer (a) Mass of sulphur = 5 g Mass of oxygen = (10 – 5) g = 5 g Sulphur Oxygen Mass (g) 5 5 No. of moles (mol) 5 0.156 32.1 0.156 1 0.156 1 5 0.313 16.0 0.313 2 0.156 2 Relative no. of moles Simplest mole ratio 98 3.2 Derivation of empirical formulae (SB p.47) (b) 19.85 g of element M combines with 25.61 g of oxygen to form an oxide. If the relative atomic mass of M is 31.0, find the empirical formula of the oxide. Answer (b) M O Mass (g) 19.85 25.61 No. of moles (mol) 19.85 0.64 31.0 0.64 1 0.64 2 Relative no. of moles Simplest mole ratio The empirical formula of the oxide is M2O5. 99 25.61 1 .6 16.0 1 .6 2.5 0.64 5 3.2 Derivation of empirical formulae (SB p.47) (c) Determine the empirical formula of copper(II) oxide using the following results. Experimental results: Mass of test tube = 21.430 g Mass of test tube + Mass of copper(II) oxide = 23.321 g Mass of test tube + Mass of copper = 22.940 g Answer 100 3.2 Derivation of empirical formulae (SB p.47) Back (c) Mass of Cu = (22.940 – 21.430) g = 1.51 g Mass of O = (23.321 – 22.940) g = 0.381 g Copper Oxygen Mass (g) 1.51 0.381 No. of moles (mol) 1.51 0.0238 63.5 0.381 0.0238 16.0 Relative no. of moles 0.0238 1 0.0238 0.0238 1 0.0238 Simplest mole ratio 1 1 Therefore, the empirical formula of copper(II) oxide is CuO. 101 3.2 Derivation of empirical formulae (SB p.48) Compound A contains carbon and hydrogen atoms only. It is found that the compound contains 75 % carbon by mass. Determine its empirical formula. Answer 102 3.2 Derivation of empirical formulae (SB p.48) Back Let the empirical formula of compound A be CxHy, and the mass of the compound be 100 g. Then, mass of carbon in the compound = 75 g Mass of hydrogen in the compound = (100 – 75) g = 25 g Carbon Hydrogen Mass (g) 75 25 No. of moles (mol) 75 6.25 12.0 25 25 1 .0 Relative no. of moles 6.25 1 6.25 25 4 6.25 Simplest mole ratio 1 Therefore, the empirical formula of compound A is CH4. 103 4 3.2 Derivation of empirical formulae (SB p.48) The percentages by mass of phosphorus and chlorine in a sample of phosphorus chloride are 22.55 % and 77.45 % respectively. Find the empirical formula of the phosphorus chloride. Answer 104 3.2 Derivation of empirical formulae (SB p.48) Back Let the mass of phosphorus chloride be 100 g. Then, Mass of phosphorus in the compound = 22.55 g Mass of chlorine in the compound = 77.45 g Mass (g) Phosphorus Chlorine 22.55 77.45 No. of moles (mol) 22.55 0.727 31.0 77.45 2.182 35.5 Relative no. of moles 0.727 1 0.727 2.182 3 0.727 Simplest mole ratio 1 3 Therefore, the empirical formula of the phosphorus chloride is PCl3. 105 3.2 Derivation of empirical formulae (SB p.49) (a) Find the empirical formula of vitamin C if it consists of 40.9 % carbon, 54.5 % oxygen and 4.6 % hydrogen by mass. (a) Let the mass of vitamin C analyzed be 100 g. Carbon Hydrogen Oxygen Mass (g) 40.9 4.6 54.5 No. of moles (mol) 40.9 3.41 12.0 3.41 1 3.41 3 4 .6 4.60 1 .0 4 .6 1.35 3.41 4 Relative no. of moles Simplest mole ratio The empirical formula of vitamin C is C3H4O3. 106 Answer 54.5 3.41 16.0 3.41 1 3.41 3 3.2 Derivation of empirical formulae (SB p.49) Back (b) Each 325 mg tablet of aspirin consists of 195.0 mg carbon, 14.6 mg hydrogen and 115.4 mg oxygen. Determine the empirical formula of aspirin. Answer (b) The masses of the elements are multiplied by 1000 first. Mass (g) Carbon Hydrogen Oxygen 195.0 14.6 115.4 No. of moles 195.0 16.25 (mol) 12.0 Relative no. 16.25 2.25 of moles 7.21 Simplest 9 mole ratio 14.6 14.6 1. 0 14.6 2.02 7.21 8 The empirical formula of aspirin is C9H8O4. 107 115.4 7.21 16.0 7.21 1 7.21 4 3.3 Derivation of molecular formulae (SB p.50) A hydrocarbon was burnt completely in excess oxygen. It was found that 5.0 g of the hydrocarbon gave 14.6 g of carbon dioxide and 9.0 g of water. Given that the relative molecular mass of the hydrocarbon is 30.0, determine its molecular formula. Answer 108 3.3 Derivation of molecular formulae (SB p.50) Let the empirical formula of the hydrocarbon be CxHy. Mass of carbon in the hydrocarbon = 14.6 g 12.0 = 4.0 g 44.0 2 .0 Mass of hydrogen in the hydrocarbon = 9.0 g 18.0 = 1.0 g Carbon Hydrogen Mass (g) 4.0 1.0 No. of moles (mol) 4.0 0.333 12.0 1 .0 1 1 .0 Relative no. of moles 0.333 1 0.333 1 3 0.333 Simplest mole ratio 109 1 3 3.3 Derivation of molecular formulae (SB p.50) Back Therefore, the empirical formula of the hydrocarbon is CH3. Let the molecular formula of the hydrocarbon be (CH3)n. Relative molecular mass of (CH3)n = 30.0 n (12.0 + 1.0 3) = 30.0 n=2 Therefore, the molecular formula of the hydrocarbon is C2H6. 110 3.3 Derivation of molecular formulae (SB p.50) Compound X is known to contain 44.44 % carbon, 6.18 % hydrogen and 49.38 % oxygen by mass. A typical analysis shows that it has a relative molecular mass of 162.0. Find its molecular formula. Answer 111 3.3 Derivation of molecular formulae (SB p.50) Let the empirical formula of compound X is CxHyOz, and the mass of the compound be 100 g. Then, Mass of carbon in the compound = 44.44 g Mass of hydrogen in the compound = 6.18 g Mass of oxygen in the compound = 49.38 g Carbon Hydrogen Oxygen Mass (g) 44.44 6.18 49.38 No. of moles (mol) 44.44 3.70 12.0 3.70 1 .2 3.09 6 6.18 6.18 1.0 6.18 2 3.09 10 49.38 3.09 16.0 3.09 1 3.09 5 Relative no. of moles Simplest mole ratio 112 empirical formula of compound X is C6H10O5. The 3.3 Derivation of molecular formulae (SB p.50) Back Let the molecular formula of compound X be (C6H10O5)n. Relative molecular mass of (C6H10O5)n = 162.0 n (12.0 6 + 1.0 10 + 16.0 5) = 162.0 n=1 Therefore, the molecular formula of compound X is C6H10O5. 113 3.3 Derivation of molecular formulae (SB p.51) The chemical formula of hydrated copper(II) sulphate is known to be CuSO4 · xH2O. It is found that the percentage of water of crystallization by mass in the compound is 36 %. Find the value of x. Answer 114 3.3 Derivation of molecular formulae (SB p.51) Relative formula mass of CuSO4 · xH2O Back = 63.5 + 32.1 + 16.0 4 + (1.0 2 + 16.0)x = 159.6 + 18x Relative molecular mass of water of crystallization = 18x 18 x 36 159.6 18 x 100 1800x = 5745.6 + 648x 1152x = 5745.6 x = 4.99 5 Therefore, the chemical formula of the hydrated copper(II) sulphate is CuSO4 · 5H2O. 115 3.3 Derivation of molecular formulae (SB p.52) (a) Compound Z is the major ingredient of a healthy drink. It contains 40.00 % carbon, 6.67 % hydrogen and 53.33 % oxygen. (i) Find the empirical formula of compound Z. (ii) If the relative molecular mass of compound Z is 180, find its molecular formula. Answer 116 3.3 Derivation of molecular formulae (SB p.52) (a) (i) Let the mass of compound Z be 100 g. Carbon Hydrogen Oxygen Mass (g) 40.00 6.67 53.33 No. of moles (mol) 40.00 3.33 12.0 6.67 6.67 1.0 53.33 3.33 16.0 Relative no. of moles 3.33 1 3.33 6.67 2 3.33 3.33 1 3.33 Simplest mole ratio 1 2 1 Therefore, the empirical formula of compound Z is CH2O. 117 3.3 Derivation of molecular formulae (SB p.52) (ii)Let the empirical formula of compound Z be (CH2O)n. n (12.0 + 1.0 2 + 16.0) = 180 30n = 180 n =6 Therefore, the molecular formula of compound Z is C6H12O6. 118 3.3 Derivation of molecular formulae (SB p.52) (b) (NH4)2Sx contains 72.72 % sulphur by mass. Find the value of x. Answer (b) (NH4) unit S Mass (g) 27.28 72.72 No. of moles (mol) 27.28 1.52 18.0 72.72 2.27 32.1 Relative no. of moles Simplest mole ratio 1.52 1 1.52 2 2.27 1.49 1.52 3 Since the chemical formula of (NH4)Sx is (NH4)2S3, the value of x is 3. 119 3.3 Derivation of molecular formulae (SB p.52) Back (c) In the compound MgSO4 · nH2O, 51.22 % by mass is water. Find the value of n. Answer (c) MgSO4 H 2O Mass (g) 48.78 51.22 No. of moles (mol) 48.78 0.405 120.4 51.22 2.846 18.0 Relative no. of moles Simplest mole ratio 0.405 1 0.405 1 2.846 7 0.405 7 Since the chemical formula of MgSO4 · nH2O is MgSO4 · 7H2O, the value of n is 7. 120 3.3 Derivation of molecular formulae (SB p.52) The chemical formula of ethanoic acid is CH3COOH. Calculate the percentage of mass of carbon, hydrogen and oxygen respectively. Answer 121 3.3 Derivation of molecular formulae (SB p.52) Relative molecular mass of CH3COOH Back = 12.0 2 + 1.0 4 + 16.0 2 = 60.0 12.0 2 100% 60.0 = 40.00 % % by mass of H = 1.0 4 100% 60.0 = 6.67 % 16.0 2 100% % by mass of O = 60.0 = 53.33 % % by mass of C = The percentage by mass of carbon, hydrogen and oxygen are 40.00 %, 6.67 % and 53.33 % respectively. 122 3.3 Derivation of molecular formulae (SB p.53) Back Calculate the mass of iron in a sample of 20 g of hydrated iron(II) sulphate, FeSO4 · 7H2O. Answer Relative formula mass of FeSO4 · 7H2O = 55.8 + 32.1 + 16.0 4 + (1.0 2 + 16.0) 7 = 277.9 55.8 100% % by mass of Fe = 277.9 = 20.08 % Mass of Fe = 20 g 20.08 % = 4.02 g 123 3.3 Derivation of molecular formulae (SB p.53) (a) Calculate the percentages by mass of potassium , chromium and oxygen in potassium dichromate(VI), K2Cr2O7. Answer -1 (a) Molar mass of K2Cr2O7 = (39.1 2 + 52.0 2 + 16.0 7) g mol = 294.2 g mol-1 (39.1 2)gmol 1 100% % by mass of K = 1 294.2gmol = 26.58 % (52.0 2)gmol 1 100% % by mass of Cr = 1 294.2gmol = 35.35 % (16.0 7)gmol 1 100% % by mass of O = 1 294.2gmol = 38.07 % 124 3.3 Derivation of molecular formulae (SB p.53) (b) Find the mass of metal and water of crystallization in (i) 100 g of Na2SO4 · 10H2O (ii) 70 g of Fe2O3 · 8H2O Answer 125 3.3 Derivation of molecular formulae (SB p.53) Back (b) (i) Molar mass of Na2SO4 · 10H2O = 322.1 g mol-1 (23.0 2)gmol 1 Mass of Na = 100g 1 322.1gmol = 14.28 g (18.0 10)gmol 1 Mass of H2O = 100g 1 322.1gmol = 55.88 g (ii) Molar mass of Fe2O3 · 8H2O = 303.6 g mol-1 (55.8 2)gmol 1 Mass of Fe = 70g 1 303.6gmol = 25.73 g (18.0 8)gmol 1 70g Mass of H2O = 303.6gmol 1 = 33.20 g 126 3.4 Chemical equations (SB p.54) Back Give the chemical equations for the following reactions: • Zinc + steam zinc oxide + hydrogen (a) Zn(s) + H2O(g) ZnO(s) + H2(g) (b) Magnesium + silver nitrate silver + magnesium nitrate (b) Mg(s) + 2AgNO3(aq) 2Ag(s) + Mg(NO3)2(aq) (c) Butane + oxygen carbon dioxide + water (c) 2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(l) Answer 127 3.5 Calculations based on chemical equations (SB p.55) Calculate the mass of copper formed when 12.45 g of copper(II) oxide is completely reduced by hydrogen. Answer 128 3.5 Calculations based on chemical equations (SB p.55) Back CuO(s) + H2(g) Cu(s) + H2O(l) As the mole ratio of CuO : Cu is 1 : 1, the number of moles of Cu formed is the same as the number of moles of CuO reduced. 12.45g Number of moles of CuO reduced = (63.5 16.0)gmol 1 = 0.157 mol Number of moles of Cu formed = 0.157 mol Mass of Cu = 0.157 mol 63.5 g mol 1 Mass of Cu = 0.157 mol 63.5 g mol-1 = 9.97 g Therefore, the mass of copper formed in the reaction is 9.97 g. 129 3.5 Calculations based on chemical equations (SB p.55) Sodium hydrogencarbonate decomposes according to the following chemial equation: 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(l) In order to obtain 240 cm3 of CO2 at room temperature and pressure, what is the minimum amount of sodium hydrogencarbonate required? (Molar volume of gas at R.T.P. = 24.0 dm3 mol-1) 130 Answer 3.5 Calculations based on chemical equations (SB p.55) Back Number of moles of CO2 required 240 cm3 = = 0.01 mol 3 1 24000 cm mol From the chemical equation, 2 moles of NaHCO3(s) give 1 mole of CO2(g). Number of moles of NaHCO3 required = 0.01 2 = 0.02 mol Mass of NaHCO3 required = 0.02 mol (23.0 + 1.0 + 12.0 + 16.0 3) g mol-1 = 0.02 mol 84.0 g mol-1 = 1.68 g Therefore, the minimum amount of sodium hydrogencarbonate required is 1.68 g. 131 3.5 Calculations based on chemical equations (SB p.56) Calculate the volume of carbon dioxide formed when 20 cm3 of ethane and 70 cm3 of oxygen are exploded, assuming all volumes of gases are measured at room temperature and pressure. Answer 132 3.5 Calculations based on chemical equations (SB p.56) Back 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l) 2 mol 7 mol : 4 mol : 6 mol (from equation) 7 volumes : 4 volumes : - (by Avogadro’s law) : 2 volumes : It can be judged from the chemical equation that the mole ratio of CO2 : C2H6 is 4 : 2, and the volume ratio of CO2 : C2H6 should also be 4 : 2 according to the Avogadro’s law. Let x be the volume of CO2(g) formed. Number of moles of CO2(g) formed : number of moles of C2H6(g) used =4:2 Volume of CO2(g) : volume of C2H6(g) = 4 : 2 x : 20 cm3 = 4 : 2 x = 40 cm3 Therefore, the volume of CO2(g) formed is 40 cm3. 133 3.5 Calculations based on chemical equations (SB p.57) 10 cm3 of a gaseous hydrocarbon was mixed with 80 cm3 of oxygen which was in excess. The mixture was exploded and then cooled. The volume left was 70 cm3. Upon passing the resulting gaseous mixture through concentrated sodium hydroxide solution (to absorb carbon dioxide), the volume of the residual gas became 50 cm3. Find the molecular formula of the hydrocarbon. Answer 134 3.5 Calculations based on chemical equations (SB p.57) Let the molecular formula of the hydrocarbon be CxHy. Volume of hydrocarbon reacted = 10 cm3 Volume of O2(g) unreacted = 50 cm3 (the residual gas after reaction) Volume of O2(g) reacted = (80 – 50) cm3 = 30 cm3 Volume of CO2(g) formed = (70 – 50) cm3 = 20 cm3 CxHy + O2 xCO2 1 mol : mol : x mol volumes : x volumes 1 volume : 135 + H2 O 3.5 Calculations based on chemical equations (SB p.57) Back Volume of CO2 (g) = x Volume of C xHy (g) 1 20 cm3 = x 10 cm3 1 x=2 y x Volume of O 2 (g) 4 = Volume of C xHy (g) 1y 3 x 30 cm = 4 3 10 cm 1 y x =3 4 y 2 =3 As x = 2, 4 y=4 Therefore, the molecular formula of the hydrocarbon is C2H4. 136 3.5 Calculations based on chemical equations (SB p.58) (a) Find the volume of hydrogen produced at R.T.P. when 2.43 g of magnesium reacts with excess hydrochloric acid. (Molar volume of gas at R.T.P. = 24.0 dm3 mol-1) (a) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) No. of moles of H2 = No. of moles of Mg 2.43 g Volume of H2 = 24.3 g mol -1 24.0 dm3 mol -1 Volume of H2 = 2.4 dm3 137 Answer 3.5 Calculations based on chemical equations (SB p.58) (b) Find the minimum mass of chlorine required to produce 100 g of phosphorus trichloride (PCl3). Answer 3 35.5 mCl 100 g 77.5 g 3 35.5 31.0 138 3.5 Calculations based on chemical equations (SB p.58) (c) 20 cm3 of a gaseous hydrocarbon and 150 cm3 of oxygen (which was in excess) were exploded in a closed vessel. After cooling, 110 cm3 of gases remained. After passing the resulting gaseous mixture through concentrated sodium hydroxide solution, the volume of the residual gas became 50 cm3. Determine the molecular formula of the hydrocarbon. Answer 139 3.5 Calculations based on chemical equations (SB p.58) y y ( x ) O2(g) xCO2(g) + H2O(l) 4 2 3 Volume of CxHy used = 20 cm (c) CxHy(g) + Volume of CO2 formed = (110 – 50) cm3 = 60 cm3 Volume of O2 used = (150 – 50) cm3 = 100 cm3 Volume of CxHy : Volume of CO2 = 1 : x = 20 : 60 x=3 y 4 = 20 : 100 Volume of CxHy : Volume of O2 = 1 : x =5 140 3.5 Calculations based on chemical equations (SB p.58) (c) As x = 3, 3 y =5 4 y =2 4 y=8 The molecular formula of the hydrocarbon is C3H8. 141 3.5 Calculations based on chemical equations (SB p.58) Back (d) Calculate the volume of carbon dioxide formed when 5 cm3 of methane was burnt completely in excess oxygen, assuming all volumes of gases are measured at room temperature and pressure. (d) CH4(g) 1 mol 2O2(g) CO2(g) + 2H2O(l) : 2 mol : : 2 mol (from equation) 1 volume : 5 cm3 Answer + 2 volumes : 1 mol 1 volume: - (from Avogadro’s law) x cm3 It can be judged from the equation that the mole ratio of CO2 : CH4 is 1 : 1, the volume ratio of CO2 : CH4 should also be 1 : 1. x 1 = 5 cm3 1 x = 5 cm3 The volume of CO2(g) formed is 5 cm3. 142 3.6 Simple titrations (SB p.61) 25.0 cm3 of sodium hydroxide solution was titrated against 0.067 M sulphuric(VI) acid using methyl orange as an indicator. The indicator changed colour from yellow to red when 22.5 cm3 of sulphuric(VI) acid had been added. Calculate the molarity of the sodium hydroxide solution. Answer 143 3.6 Simple titrations (SB p.61) Back 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) Number of moles of NaOH(aq) 2 Number of moles of H2 SO 4 (aq) = 1 1 Number of moles of NaOH(aq) = Number of moles of H2SO4(aq) 2 Number of moles of H2SO4(aq) = 0.067 mol dm-3 22.5 10-3 dm3 = 1.508 10-3 mol Number of moles of NaOH(aq) = 2 1.508 10-3 mol = 3.016 10-3 mol 3.016 10 -3 mol Molarity of NaOH(aq) = 25.0 10 - 3 dm3 = 0.121 mol dm-3 Therefore, the molarity of the sodium hydroxide solution was 0.121 M. 144 3.6 Simple titrations (SB p.61) 2.52 g of a pure dibasic acid with formula mass of 126.0 was dissolved in water and made up to 250.0 cm3 in a volumetric flask. 25.0 cm3 of this solution was found to neutralize 28.5 cm3 of sodium hydroxide solution. (a)Calculate the molarity of the acid solution. (a) Number of moles of acid = Molarity of acid solution = 145 2.52 g = 0.02 mol -1 126.0 g mol 0.02 mol = 0.08 M 3 3 250 10 dm Answer 3.6 Simple titrations (SB p.61) (b) If the dibasic acid is represented by H2X, write an equation for the reaction between the acid and sodium hydroxide. Answer (b) H2X(aq) + 2NaOH(aq) Na2X(aq) + 2H2O(l) 146 3.6 Simple titrations (SB p.61) Back (c) Calculate the molarity of the sodium hydroxide solution. Answer (c) Number of moles of H2X = 1 Number of moles of NaOH 2 3 -3 -3 0.08 mol dm 25.0 10 dm 1 = Molarity of NaOH 28.5 10-3 dm3 2 Molarity of NaOH = 0.14 M Therefore, the molarity of the sodium hydroxide solution was 0.14 M. 147 3.6 Simple titrations (SB p.62) 0.186 g of a sample of hydrated sodium carbonate, Na2CO3 · nH2O, was dissolved in 100 cm3 of distilled water in a conical flask. 0.10 M hydrochloric acid was added from a burette, 2 cm3 at a time. The pH value of the reaction mixture was measured with a pH meter. The results were recorded and shown in the following figure. Calculate the value of n in Na2CO3 · nH2O. Answer 148 3.6 Simple titrations (SB p.63) Back There is a sudden drop in the pH value of the solution (from pH 8 to pH 3) with the equivalence point at 30.0 cm3. Na2CO3 · nH2O(s) + 2HCl(aq) 2NaCl(aq) + CO2(g) + (n + 1)H2O(l) 1 Number of moles of Na2CO3 · nH2O = Number of moles of HCl 2 0.186 g -1 (23.0 2 12.0 16.0 3 18.0n) g mol1 = 0.10 mol dm-3 30.0 10-3 dm3 2 106.0 + 18.0n = 124.0 n=1 Therefore, the chemical formula of the hydrated sodium carbonate is Na2CO3 · H2O. 149 3.6 Simple titrations (SB p.63) 5 cm3 of 0.5 M sulphuric(VI) acid was added to 25.0 cm3 of potassium hydroxide solution. The mixture was then stirred and the highest temperature was recorded. The experiment was repeated with different volumes of the sulphuric(VI) acid. The laboratory set-up and the results were as follows: 150 Volume of H2SO4 added (cm3) Temperature (oC) 0 20.0 5 21.8 10 23.4 15 25.0 20 26.5 25 25.2 30 24.0 3.6 Simple titrations (SB p.63) (a) Plot a graph of temperature against volume of sulphuric(VI) acid added. Answer 151 3.6 Simple titrations (SB p.63) (b) Calculate the molarity of the potassium hydroxide solution. (b) From the graph, it is found that the equivalence point of the titration is Answer reached when 20 cm3 of H2SO4 is added. Number of moles of H2SO4 = 0.5 mol dm-3 20 10-3 dm3 = 0.01 mol 2KOH(aq) + H2SO4(aq) K2SO4(aq) + 2H2O(l) 2 mol : 1 mol From the equation, mole ratio of KOH(aq) : H2SO4(aq) = 2 : 1 Number of moles of KOH(aq) = 2 0.01 mol = 0.02 mol 0.02 mol Molarity of KOH(aq) = = 0.8 M 3 3 25.0 10 dm Therefore, the molarity of potassium hydroxide solution was 0.8 M. 152 3.6 Simple titrations (SB p.63) Back (c) Explain why the temperature rose to a maximum and then fell. Answer (c) Neutralization is an exothermic reaction. When more and more sulphuric(VI) acid was added and reacted with potassium hydroxide, the temperature rose. The temperature rose to a maximum value at which the equivalence point of the reaction was reached. After that, any excess sulphuric(VI) acid added would cool down the reacting solution, causing the temperature to drop. 153 3.6 Simple titrations (SB p.66) When excess potassium iodide solution (KI) is added to 25.0 cm3 of acidified potassium iodate solution (KIO3) of unknown concentration, the solution turns brown. This brown solution requires 22.0 cm3 of 0.05 M sodium thiosulphate solution to react completely with the iodine formed, using starch solution as an indicator. Find the molarity of the acidified potassium iodate solution. Answer 154 3.6 Simple titrations (SB p.66) Back IO3-(aq) + 5I-(aq) + 6H+(aq) 3I2(aq) + 3H2O(l) … … (1) I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O62-(aq) … … (2) From (1), Number of moles of IO3-(aq) = 1 Number of moles of I2(aq) 3 1 From (2), Number of moles of I2(aq) = Number of moles of S2O32-(aq) 2 1 Number of moles of S2O32-(aq) 6 1 -3 3 Molarity of IO3 (aq) 25.0 10 dm = 0.05 mol dm-3 22.0 10-3 dm3 6 Number of moles of IO3-(aq) = Molarity of IO3-(aq) = 7.33 10-3 M Therefore, the molarity of the acidified potassium iodate solution is 7.33 10-3 M. 155 3.6 Simple titrations (SB p.67) A piece of impure iron wire weighs 0.22 g. When it is dissolved in hydrochloric acid, it is oxidized to iron(II) ions. The solution requires 36.5 cm3 of 0.02 M acidified potassium manganate(VII) solution for complete reaction to form iron(III) ions. What is the percentage purity of the iron wire? Answer 156 3.6 Simple titrations (SB p.67) Back MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) Number of moles of MnO4-(aq) : Number of moles of Fe2+(aq) = 1 : 5 Number of moles of Fe2+(aq) = 5 Number of moles of MnO4-(aq) = 5 0.02 mol dm-3 36.5 10-3 dm3 = 3.65 10-3 mol Number of moles of Fe dissolved = Number of moles of Fe2+ formed = 3.65 10-3 mol Mass of Fe = 3.65 10-3 mol 55.8 g mol-1 = 0.204 g 0.204 g Percentage purity of Fe = 100 % = 92.73 % 0.22 g Therefore, the percentage purity of the iron wire is 92.73 %. 157 3.6 Simple titrations (SB p.67) (a) 5 g of anhydrous sodium carbonate is added to 100 cm3 of 2 M hydrochloric acid. What is the volume of gas evolved at room temperature and pressure? (Molar volume of gas at R.T.P. = 24.0 dm3 mol-1) Answer 158 3.6 Simple titrations (SB p.67) Na2CO3(s) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g) 5g No. of moles of Na2CO3 used = (23.0 2 12.0 16.0 3) g mol -1 = 0.0472 mol 100 dm3 No. of moles of HCl used = 2 M 1000 = 0.2 mol Since HCl is in excess, Na2CO3 is the limiting agent. No. of moles of CO2 produced = No. of moles of Na2CO3 used = 0.0472 mol Volume of CO2 produced = 0.0472 mol 24.0 dm3 mol-1 = 1.133 dm3 159 3.6 Simple titrations (SB p.67) (b) 8.54 g of impure hydrated iron(II) sulphate (formula mass of 392.14) was dissolved in water and made up to 250.0 cm3. 25.0 cm3 of this solution required 20.76 cm3 of 0.020 3 M acidified potassium manganate(VII) solution for complete reaction. Determine the percentage purity of the hydrated iron(II) sulphate. Answer 160 3.6 Simple titrations (SB p.67) Back (b) MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) 20.76 dm 3 1000 -4 = 4.214 10 mol No. of moles of MnO4- ions = 0.0203 M No. of moles of Fe2+ ions = 5 No. of moles of MnO4- ions = 2.107 10-3 mol No. of moles of Fe2+ ions in 25.0 cm3 solution = 2.107 10-3 mol No. of moles of Fe2+ ions in 250.0 cm3 solution = 0.02107 mol Molar mass of hydrated FeSO4 = 392.14 g mol-1 Mass of hydrated FeSO4 = 0.02107 mol 392.14 g mol-1 = 8.26 g 8.26 g 100% = 96.72 % % purity of FeSO4 = 8.54 g 161