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Slide 2 New Way Chemistry for Hong Kong A-Level Book 11 Chapter 3 Chemical Equations and Stoichiometry 3.1Formulae of Compounds 3.2Derivation of Empirical Formulae 3.3Derivation of Molecular Formulae 3.4 Chemical Equations 3.5 Calculations Based on Equations 3.6 Simple Titrations Slide 3 New Way Chemistry for Hong Kong A-Level Book 12 Formulae of Compounds How can you describe the composition of compound X? 1st way = by chemical formula C?H?C?H? ratio of no. of atoms 3.1 Formulae of Compounds (SB p.54) Slide 4 New Way Chemistry for Hong Kong A-Level Book 13 How can you describe the composition of compound X? Compound X 2nd way = by percentage by mass Mass of carbon atoms inside = . g Mass of hydrogen atoms inside = . g carbon atoms hydrogen atoms 3.1 Formulae of Compounds (SB p.54) Slide 5 New Way Chemistry for Hong Kong A-Level Book 14 3.1 Formulae of Compounds (SB p.54) Answer Check Point 3-1 Give the empirical molecular and structural formula for the following compounds C 6 H 12 O 6 (d) Glucose C 2 H 5 OHC2H6OC2H6O(c) Ethanol HNO 3 (b) Nitric acid C3H6C3H6 CH 2 (a) Propene Structural formula Molecular formula Empirical formula Compound Slide 6 New Way Chemistry for Hong Kong A-Level Book 15 3.1 Formulae of Compounds (SB p.55) CompoundEmpirical formula Molecular formula Structural formula Carbon dioxide CO 2 O = C =O WaterH2OH2OH2OH2O MethaneCH 4 GlucoseCH 2 OC 6 H 12 O 6 Sodium fluoride NaFNot applicableNa + F - The different types of formulae of some compounds Slide 7 New Way Chemistry for Hong Kong A-Level Book 16 Example 3-1 A hydrogen was burnt completely in excess oxygen. It was found that 1.00 g of the hydrocarbon gives 2.93 g of carbon dioxide and 1.8 g of water. Find the empirical formula of the hydrocarbon.? (R.a.m.* : H = 1.0, C = 12.0, O = 16.0) Solution: The relative molecular mass of CO 2 = 12.0 + 2 x 16.0 = 44.0 Mass of carbon in 2.93 g of CO 2 = 2.93 g x 12.0/44.0 = 0.80 g The relative molecular mass of H 2 O = 2 x 1.0 + 16.0 = 18.0 Mass of hydrogen in 1.80 g of H 2 O = 1.80 g x 2.0/18.0 = 0.20 g Answer 3.2 Derivation of Empirical Formulae (SB p.56) Slide 8 New Way Chemistry for Hong Kong A-Level Book 17 3.2 Derivation of Empirical Formulae (SB p.57) Solution: (contd) Let the empirical formula of the hydrocarbon be C x H y. Mass of carbon in C x H y = Mass of carbon in CO 2 Mass of hydrogen in C x H y = Mass of hydrogen in H 2 O The simplest whole number ratio of x and y can be determined by the following the steps in the below table. Slide 9 New Way Chemistry for Hong Kong A-Level Book 18 3.2 Derivation of Empirical Formulae (SB p.57) CarbonHydrogen Mass (g)0.800.20 Number of moles (mol) 0.80/12.0 = 0.066 7 0.2/0.066 7 = 3 Relative number of moles 0.066 7/0.066 7=10.20/0.066 7 =3 Simplest mole ratio 13 Slide 10 New Way Chemistry for Hong Kong A-Level Book 19 Example 3-2 Compound X is known to contain carbon, hydrogen 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. (R.a.m.* : H = 1.0, C = 12.0, O = 16.0) Solution: Mass of compound X = 0.46g Mass of carbon in compound X = 0.88 g x 12.0/44.0 = 0.24 g Mass of hydrogen in compound X = 0.54 g x 2.0/18.0 = 0..06g Mass of oxygen in compound X = 0.46 g 0.24 g 0.06 g = 0.16 g Answer 3.2 Derivation of Empirical Formulae (SB p.57) Slide 11 New Way Chemistry for Hong Kong A-Level Book 110 3.2 Derivation of Empirical Formulae (SB p.57) Solution: (contd) Let the empirical formula of compound X be C x H y O z. Therefore, the empirical formula of compound X is C 2 H 6 O. CarbonHydrogenOxygen Mass (g)0.240.060.16 Number of moles (mol) 0.020.060.01 Relative number of moles 261 Simplest mole ratio 261 Slide 12 New Way Chemistry for Hong Kong A-Level Book 111 Check Point 3-2 (a) 5 g of sulphur forms 10 g of an oxide on burning.What is the empirical formula of the oxide? (R.a.m. : O = 16.0, S = 32.1) (b) 19.85 f of element M combines with 25.61 g of oxygen to form an oxide. If the relative atomic mass of M is 331.0, find the empirical formula of the oxide. (R.a.m. : O =16.0) (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 = 23.321g Mass of test tube + Mass of copper = 22.940g (R.a.m. : Cu = 63.5, O = 16.0) 3.1 Formulae of Compounds (SB p.58) Answer (a)Mass of sulphur = 5 g Mass of oxygen = (10 5) g The empirical formula of the sulphur oxide is SO 2. Sulphur Oxygen Mass (g)55 Number of moles (mol) 5/ 32.1 = 0.156 5/16.0 = 0.313 Relative Number of moles 0.156/0.156=10.313/0.156 =2 Simplest mole ratio 12 Slide 13 New Way Chemistry for Hong Kong A-Level Book 112 3.1 Formulae of Compounds (SB p.58) (b) The empirical formula of the oxide is M 2 O 5. MO Mass (g)19.8525.61 Number of moles (mol) 19.85/31.0 = 0.64 25.61/16.0 = 1.6 Relative number of moles 0.64/0.64 = 1 1.6/0.64 = 2.5 Simplest mole ratio 25 Slide 14 New Way Chemistry for Hong Kong A-Level Book 113 3.1 Formulae of Compounds (SB p.58) (c) Mass of Cu = (22.940 - 21.430) g = 1.51g Mass of O = (23.321 - 22.940) = 0.381 g The empirical formula of the oxide is CuO. CuO Mass (g)1.510.381 Number of moles (mol) 1.51/63.5 = 0.0238 0.381/16.0 = 0.0238 Relative number of moles 0.0238/0.0238 =1 Simplest mole ratio 11 Slide 15 New Way Chemistry for Hong Kong A-Level Book 114 Determination of Empirical Formula Composition by mass Empirical formula 3.2 Derivation of Empirical Formulae (SB p.58) From Combustion by Mass Slide 16 New Way Chemistry for Hong Kong A-Level Book 115 Example3-3 Example3-3 Compound A contains carbon and hydrogen only. It is found that the compound contains 75% carbon by mass. Determine its empirical formula. (Relative atomic masses: C=12, H=1 ) 3.2 Derivation of Empirical Formulae (SB p.58) Solution: Let the empirical formula of the hydrocarbon be C x H y, and the mass of the compound be 100 g. Mass of carbon in the compound = 75 g Mass of hydrogen in the compound=(100 75) g = 25 g Therefore, the empirical formula of the hydrocarbon is CH 4. Answer CarbonHydrogen Mass (g)7525 Number of moles (mol) 75/12.0 = 6.2525/1.0 = 25 Relative number of moles 6.25/6.25 = 125/6.25 = 4 Simplest mole ratio 14 Slide 17 New Way Chemistry for Hong Kong A-Level Book 116 Example 3-4 The percentage by mass of phosphorus and chlorine in a sample of a phosphorus chloride are 22.55% and 77.45% respectively. Find the empirical formula of the chloride. (R.a.m. : P = 31.0, Cl = 35.5) Solution: Let the mass of phosphorus chloride be 100g. Then, Mass of phosphorus in the compound = 22.55g Mass of chloride in the compound = 77.45g Therefore, the empirical formula of the phosphorus chloride is PCl 3. Answer 3.2 Derivation of Empirical Formulae (SB p.59) PhosphorusChloride Mass (g)22.5577.45 Number of mole (mol)22.55/31.0 = 0.727 77.45/35.5 =2.182 Relative number of moles 0.727/0.727 = 1 2.182/0.727 = 3 Simplest mole ratio13 Slide 18 New Way Chemistry for Hong Kong A-Level Book 117 Check Point 3-3 (a)Find the empirical formula of vitamin C if it consists of 40.9% caarbon, 54.5% oxygen and 4.6% hydrogen by mass. ( R.a.m.: C = 12.0, H = 1.0, O = 16.0) (b)Each 325 mg tablet of aspirin consists of 195.0 mg carbon 14.6 mg hydrogen and 115.4mg oxygen. Determine the empirical formula of aspirin. (R.a.m. : C= 12.0, H = 1.0, O = 16.0) Answer (a)Let the mass of vitamin C analyzed be 100g. The empirical formula of vitamin C is C 3 H 4 O 3. 3.2 Derivation of Empirical Formulae (SB p.59) CarbonHydrogenOxygen Mass (g)40.94.654.5 Number of moles (mol) 40.9/12.0 = 3.41 4.6/1.0 =4.60 54.5/16.0 =3.41 Relative number of moles 3.41/3.41 =1 4.61/3.41 =1.35 3.41/3.41 =1 Simplest mole ratio343 Slide 19 New Way Chemistry for Hong Kong A-Level Book 118 3.2 Derivation of Empirical Formulae (SB p.59) (b) In order to facilitate calculation, the masses of the elements are multiplied by 1000 first. The empirical formula of aspirin is C 9 H 8 O 4. CarbonHydrogenOxygen Mass (g)195.014.6115.4 Number of moles (mol) 195.0/12.0 =16.25 14.6/7.21 =2.02 7.21/7.21 =1 Relative number of moles 16.25/7.21 = 2.25 14.6/7.21 = 2.02 7.21/7.21 =1 Simplest mole ratio 984 Slide 20 New Way Chemistry for Hong Kong A-Level Book 119 What is Molecular Formulae? Molecular formula = (Empirical formula) n ? 3.3 Derivation of Molecular Formulae (SB p.60) Slide 21 New Way Chemistry for Hong Kong A-Level Book 120 Empirical formulaMolecular mass Molecular formula 3.3 Derivation of Molecular Formulae (SB p.60) Slide 22 New Way Chemistry for Hong Kong A-Level Book 121 Example 3-5 A hydrogen was burnt completely in excess oxygen. It was found that 5.00 g of the hydrocarbon gives 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. hydrocarbon.? (R.a.m.* : H = 1.0, C = 12.0, O = 16.0) Solution: Let the empirical formula of the hydrocarbon be C x H y. Mass of carbon in the hydrocarbon = 14.6g x 12.0/44.0 = 4.0g Mass of hydrogen in the hydrocarbon = 9.0g x 2.0/18.0 = 1.0g 3.3 Derivation of Molecular Formulae (SB p.60) CarbonHydrogen Mass (g)4.01.0 Number of moles (mol) 4.0/12.0 = 0.333 1.0/1.0 = 1 Relative number of moles 0.333/0.333 = 11/0.333 = 3 Simplest mole ratio13 Answer Slide 23 New Way Chemistry for Hong Kong A-Level Book 122 3.3 Derivation of Molecular Formulae (SB p.60) Solution: (contd) Therefore, the empirical formula of the hydrocarbon is CH 3. The molecular formula of the hydrocarbon is (CH 3 ) n. Relative molecular mass of (CH 3 ) n = 30.0 n x (12.0 + 1.0 x 3) = 30.0 n= 2 Therefore, the molecular formula of the hydrocarbon is C 2 H 6. Slide 24 New Way Chemistry for Hong Kong A-Level Book 123 3.3 Derivation of Molecular Formulae (SB p.61) Example 3-6 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(R.a.m.* : H = 1.0, C = 12.0, O = 16.0) Solution: Let the empirical formula of the hydrocarbon be C x H y O z. Mass of carbon in the compound = 44.44g Mass of hydrogen in the compound = 6.18g Mass of oxygen in the compound = 49.38g Answer CarbonHydrogenOxygen Mass (g)44.446.1849.38 Number of moles (mol) 44.44/12.0 = 3.70 6.18/1.0 = 6.18 49.38/16.0 = 3.09 Relative number of moles 3.70/3.09 = 1.2 6.18/3.09 = 2 3.09/3.09 = 1 Simplest mole ratio 6105 Slide 25 New Way Chemistry for Hong Kong A-Level Book 124 Solution(contd) The empirical formula of compound X is C 6 H 10 O 5. The molecular formula of compound X is (C 6 H 10 O 5 ) n. Relative molecular mass of (C 6 H 10 O 5 ) n = 162.0 n x (12.0 x 6 + 1.0 x 10 + 16.0 x 5) = 162.0 n = 1 Therefore, the molecular formula of compound is C 6 H 10 O 5. 3.3 Derivation of Molecular Formulae (SB p.61) Slide 26 New Way Chemistry for Hong Kong A-Level Book 125 3.3 Derivation of Molecular Formulae (SB p.61) Water of Crystallization Derived from Composition by Mass Hydrated saltAnhydrous salt CuSO 4 5H 2 O Blue crystals Anhydrous CuSO 4 White powder Na 2 CO 3 10H 2 O Colourless crystals Anhydrous Na 2 CO 3 White powder CoCl 2 2H 2 O Pink crystals Anhydrous CoCl 2 Blue crystals Slide 27 New Way Chemistry for Hong Kong A-Level Book 126 Example 3-7 The chemical formula of hydrated copper(II) sulphate is known to be CuSO 4.xH 2 O. It is found that the percentage of water by mass in the compound is 36%. Find x. (R.a.m. : H=1.0, O=16.0, S=32.1, Cu=63.5) Solution: Let Relative molecular mass of CuSO 4 x H 2 O = 63.5 + 32.1 + 16.0 x 4 + (1.0x2 = 16.0) x = 159.6 + 18 x Relative molecular mass of water of crystallization =18 x 18 x /(159.6 + 18 x ) = 36/100 1800 x = 5745.6 + 648 x 1152 x = 5745.6 x = 4.99 5 Therefore, the chemical formula of hydrated copper(II) sulphate is CuSO 4 5H 2 O Answer 3.3 Derivation of Molecular Formulae (SB p.61) Slide 28 New Way Chemistry for Hong Kong A-Level Book 127 3.3 Derivation of Molecular Formulae (SB p.63) Check Point 3-4 (a)Find Compound Z is the major component 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, finds its molecular formula.(R.a.m. : C= 12.0, H = 1.0, O = 16.0) Answer (a) (i) Let the mass of compound Z be 100g. The empirical formula of compound Z is CH 2 O. CarbonHydrogenOxygen Mass (g)40.006.6753.33 Number of moles (mol) 40.00/12.0 = 3.33 6.67/1.0 = 6.67 53.33/16.0= 3.33 Relative number of moles 3.33/3.33 =1 6.67/3.33 =2 3.33/3.33 =1 Simplest mole ratio 121 Slide 29 New Way Chemistry for Hong Kong A-Level Book 128 3.3 Derivation of Molecular Formulae (SB p.63) (a) (ii) Let the molecular formula of compound Z be (CH 2 O) n. n x (12.0 = 1.0 x 2 = 16.0) = 180 30n = 180 n = 6 The molecular formula of Z is C 6 H 12 O 6. Slide 30 New Way Chemistry for Hong Kong A-Level Book 129 3.3 Derivation of Molecular Formulae (SB p.63) Check Point 3-4 (b) (NH 4 ) 2 S x contains 72.72% sulphur by mass is water. Find the value of x. (R.a.m.: H = 1.0, N = 14.0, O = 16.0) (c) In the compound MgSO 4 nH 2 O, 51.22% by mass is water. Find the value of n. (R.a.m.: H = 1.0, O = 16.0, Mg = 24.3, S = 32.1) Answer (b) Since the chemical formula of (NH 4 ) 2 S x is (NH 4 ) 2 S 3, the value of x is 3. (NH 4 ) unitS Mass (g)27.2872.72 Number of moles (mol) 27.28/18.0 = 1.52 72.72/32.1 =2.27 Relative number of moles 1.52/1.52 =1 2.27/1.52 =1.49 Simplest mole ratio 23 Slide 31 New Way Chemistry for Hong Kong A-Level Book 130 (c) Since the chemical formula of MgSO 4 nH 2 O is MgSO 4 7H 2 O, the value of x is 7. 3.3 Derivation of Molecular Formulae (SB p.63) MgSO 4 H2OH2O Mass (g)48.7851.22 Number of moles (mol) 48.78/120.4 =0.405 51.22/18.0 =2.846 Relative number of moles 0.405/0.405 =1 2.846/0.405 =7 Simplest mole ratio 17 Slide 32 New Way Chemistry for Hong Kong A-Level Book 131 Example 3-8 The chemical formula of ethanoic acid is CH 3 COOH. Calculate the percentages by mass of carbon, hydrogen and oxygen by mass respectively. (R.a.m. : C=12.0, H=1.0, O=16.0 ) Solution: Relative molecular mass of CH 3 COOH = 12.0 x 2 + 1.0 x 4 + 16.0 x 2 = 60.0 % by mass of C = 12.0 x 2/ 60.0 x 100%= 40.00% % by mass of H = 1.0 x 4 /60.0 x 100% = 6.67% % by mass of O = 16.0 x 2/60.0 x 100% = 53.33% The percentage by mass of carbon, hydrogen and oxygen are 40.00%, 6.67% and 53.33% respectively. Answer 3.3 Derivation of Molecular Formulae (SB p.63) Slide 33 New Way Chemistry for Hong Kong A-Level Book 132 Example 3-9 Calculate the mass of iron metal in a sample of 20g of hydrated iron (II) sulphate, FeSO 4 7H 2 O. (R.a.m. : Fe = 55.8, H=1.0, O=16.0 ) Solution: Relative molecular mass of FeSO 4 7H 2 O = 55.8 + 32.1 + 16.0 x 4 + (1.0x2+16.0) x 7=277.9 % by mass of Fe = 55.8/277.9 x 100% = 20.08% Mass of Fe = 20g x 20.08% = 4.02g Answer 3.3 Derivation of Molecular Formulae (SB p.63) Slide 34 New Way Chemistry for Hong Kong A-Level Book 133 3.3 Derivation of Molecular Formulae (SB p.63) Check Point 3-5 (a) Calculate percentages by mass of potassium, chromium and oxygen in potassium chromate (VI), K 2 Cr 2 O 7.(R.a.m. : K = 39.1. Cr = 52.0, O = 16.0) (b) Find the mass of metal and water of crystallization in (i)100 g of Na 2 SO 4 10H 2 O; (ii)70g of Fe 2 O 3 8H 2 O. (R.a.m.: H = 1.0, O = 16.0, Na = 23, S = 32.1, Fe = 55.8) Answer (a)Molar mass of K 2 Cr 2 O 7 = (39.1x2+52.0+16.0x7) g mol -1 = 294.2 g mol -1 % by mass of K = 39.1 x 2 g mol -1 /294.2 g mol -1 x 100% = 26.58% % by mass of Cr = 52.0 x 2 g mol -1 /294.2g mol -1 x 100% =35.25% % by mass of O = 16.0 x 7 g mol -1 /294.2g mol -1 x 100% = 38.07% Slide 35 New Way Chemistry for Hong Kong A-Level Book 134 3.3 Derivation of Molecular Formulae (SB p.63) (b)( i) Molar mass of Na 2 SO 4 10H 2 O= 322.1 g mol -1 Mass of Na = 23.0 x 2 g mol -1 / 322.1 g mol -1 x 100g = 14.28 g Mass of H 2 O = 18.0 x 10 g mol -1 / 322.1 g mol -1 x 100g = 14.28 g (ii) Molar mass of Fe 2 O 3 8H 2 O= 303.6 g mol -1 Mass of Fe = 55.8 x 2 g mol -1 /303.6g mol -1 x 70g = 25.73 g Mass of H 2 O = 18.0 x 8 g mol -1 /303.6g mol -1 x 70g = 33.20 g Slide 36 New Way Chemistry for Hong Kong A-Level Book 135 Chemical Equations a A + b B c C + d D mole ratios (can also be volume ratios for gases) Stoichiometry = relative no. of moles of substances involved in a chemical reaction. 3.4 Chemical Equations (SB p.64) Slide 37 New Way Chemistry for Hong Kong A-Level Book 136 Answer 3.4 Chemical Equations (SB p.64) Check Point 3-6 Give the chemical equations for the following reactions: (a) Zinc + steam zinc oxide + hydrogen (b) Magnesium + silver nitrate silver + magnesium nitrate (c) Butane + oxygen carbon dioxide + water (a) Zn(s) + H 2 O(g) ZnO(s) + H 2 (g) (b) Mg(s) + 2 AgNO 3 (aq) 2Ag(s) + Mg(NO 3 ) 2 (aq) (c) 2C 4 H 10 (g) + 13O 2 (g) 8CO 2 (g) + 10H 2 O(l) Slide 38 New Way Chemistry for Hong Kong A-Level Book 137 Calculations Based on Equations 3.5 Calculations Based on Equations (SB p.65) Calculations involving Reacting Masses Slide 39 New Way Chemistry for Hong Kong A-Level Book 138 Example 3-10 Calculate the mass of copper formed when 12.45g of copper(II) oxide is completely reduced by hydrogen. (R.a.m. : H=1.0, O=16.0, Cu = 63.5 ) Answer 3.5 Calculations Based on Equations (SB p.65) Solution: CuO(s) + H 2 (g) Cu(s) + H 2 O(l) As the mole ratio of Cu : CuO is 1 : 1, the number of moles of Cu formed is the same as the number of moles of CuO reduced. Number of moles of CuO reduced = 12.45/ (63.5 + 13.0) g mol -1 = 0.157 mol Number of mole of Cu formed = 0.157 mol Mass of Cu / 63.5 g mol -1 = 0.157 Mass of Cu = 0.157 mol x 63.5 g mol -1 = 9.97g Therefore, the mass of copper formed in the reaction is 9.97g. Slide 40 New Way Chemistry for Hong Kong A-Level Book 139 3.5 Calculations Based on Equations (SB p.65) Answer Example 3-11 Sodium hydrogencarbonate decomposes according to the following equation. 2NaHCO 3 (s) Na 2 CO 3 (s) + CO 2 (g) + H 2 O(l) In order to obtain 240 cm 3 of CO 2 at room temperature and pressure, what is the minimum amount of sodium hydrogencarbonate required? (R.a.m. : H = 1.0, C =12.0, O = 16.0, Na = 23.0; molar volume of gas at R.T.P. = 24.0 dm 3 mol -1 ) Solution: Number of moles of CO 2 formed = 240cm 3 / 24000cm 3 mol -1 = 0.01 mol From the equation, 2 moles of NaHCO 3 (s) will form 1 mole of CO 2 (g). Number of moles of NaHCO 3 required = 0.01 x 2 = 0.02 mol Mass of NaHCO 3 required = 0.02 mol x(23.0 + 1.0 + 16.0 x 3) g mol -1 = 0.02 mol x 84.0g mol -1 = 1.68 g Therefore, the minimum amount of sodium hydrogencarbonate required is 1.68g. Slide 41 New Way Chemistry for Hong Kong A-Level Book 140 Calculations Based on Equations 3.5 Calculations Based on Equations (SB p.66) Calculations involving Volumes of Gases Slide 42 New Way Chemistry for Hong Kong A-Level Book 141 Answer 3.5 Calculations Based on Equations (SB p.66) Example 3-12 Calculate the volume of carbon dioxide formed when 20 cm 3 of ethane and 70 cm 3 of oxygen are exploded, assuming all volumes are measured at room temperature and pressure. Solution: Number of moles of CO 2 formed C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(l) 2 mol : 7 mol : 4 mol : 6 mol (from equation) 2 volumes: 7 volumes : 4 volumes : - (by Avogadros law) It can be judged from the equation that the mole ratio of CO 2 : C 2 H 6 is 4 :2, and the volume ratio of CO 2 : C 2 H 6 should also be 4:2. Let x be the volume of CO 2 (g) formed x /20cm 3 = 4/2 x = 40 cm 3 Therefore, the volume of CO 2 formed is 40 cm 3. Slide 43 New Way Chemistry for Hong Kong A-Level Book 142 Answer 3.5 Calculations Based on Equations (SB p.67) Example 3-13 10 cm 3 of a gaseous hydrocarbon was mixed with 80cm 3 of oxygen which was in excess. The mixture was exploded and then cooled. The volume left was 70cm 3. Upon passing the resulting gaseous mixture through concentrated sodium hydroxide solution ( to absorb carbon dioxide), the volume of the residual gas became 50 cm 3. Find the molecular formula of the hydrocarbon. Solution: Let the molecular formula of the hydrocarbon be C x H y. Volume of hydrogen reacted = 10 cm 3 Volume of O 2 (g) unreacted = 50 cm 3 Volume of O 2 (g) reacted = 30 cm 3 Volume of CO 2 (g) formed = 20 cm 3 C x H y + (x + y/4) O 2 CO 2 + y/2 H 2 O 1 volume : (x + y/4) volumes : x volumes Volume of CO 2 (g)/ volume of C x H y (g) = 20 cm 3 / 10cm 3 = 2 X =2 Volume of O 2 (g) / volume of C x H y (g) =(x + y/4) / 1= 30/ 10 (x + y/4)= 3 Y = 2 Molecular formula is C 2 H 4 Slide 44 New Way Chemistry for Hong Kong A-Level Book 143 3.5 Calculations Based on Equations (SB p.68) Answer Check Point 3-7 (a)Find the volume of hydrogen produced at R.T.P. when 2.43g of magnesium reacts with excess hydrochloric acid. (R.a.m. : Mg = 24.3; molar volume of gas at R.T.P. = 24.0 dm 3 mol -1. (b)Find the minimum mass of chlorine required to produced 100 g of phosphorus trichloride ( PCl 3 ). (c)20 cm3 of a gaseous hydrocarbon and 150 cm 3 of oxygen were exploded in a closed vessel. After cooling, 110 cm 3 of gases remained. After passing through a solution of concentrated sodium hydroxide, the volume left was 50 cm 3. Determine the molecular of the hydrocarbon. (d)Calculate the volume of carbon dioxide formed when 5cm 3 of methane burns in excess oxygen, assuming all volumes are measured at room temperature and pressure. (a) No. of moles of H 2 = No. of moles of Mg Volume of H 2 / 24.0 dm 3 mol -1 = 2.43 g / 24.3 g mol -1 Volume of H 2 = 2.4 dm 3 (b) 1/3 x no. of moles of Cl 2 = 1/2x no. of moles of PCl 3 1/3 x mass of Cl 2 / (35.5 x 2) g mol -1 = 1/2 x 100g / (31.0 + 35.5 + 3 ) g mol -1 Mass of Cl 2 = 77.45g Slide 45 New Way Chemistry for Hong Kong A-Level Book 144 3.5 Calculations Based on Equations (SB p.68) (c)Volume of C x H y used = 20 cm 3 Volume of CO 2 formed = 60 cm 3 Volume of O 2 used = 100 cm 3 Volume of C x H y : volume of CO 2 = 1 : x = 20 : 60 x = 3 Volume of C x H y : volume of O 2 = 1 : x + y/4 = 20 : 100 x + y/4 = 5 3 + y/4 = 5 y = 8 Slide 46 New Way Chemistry for Hong Kong A-Level Book 145 3.5 Calculations Based on Equations (SB p.68) (d)Volume of C x H y used = 20 cm 3 It can be judged from the equation that the mole ratio of CO 2 : CH 4 is 1:1, the volume ratio of CO 2 : CH 4 should also be 1:1. x / 5 = 1/1 x = 5 The volume of carbon dioxide gas is 5 cm 3. Slide 47 New Way Chemistry for Hong Kong A-Level Book 146 Simple Titrations Acid-Base Titrations Acid-Base Titrations with Indicators Acid-Base Titrations without Indicators (to be discussed in later chapters) 3.6 Simple Titrations (SB p.68) Slide 48 New Way Chemistry for Hong Kong A-Level Book 147 Finding the concentration of a solution + solute solvent solution Copper(II) sulphate Water Copper(II) sulphate solution 3.6 Simple Titrations (SB p.69) Slide 49 New Way Chemistry for Hong Kong A-Level Book 148 Finding the concentration of a solution ~50 cm 3 3.6 Simple Titrations (SB p.69) Slide 50 New Way Chemistry for Hong Kong A-Level Book 149 ~50 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 51 New Way Chemistry for Hong Kong A-Level Book 150 ~50 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 52 New Way Chemistry for Hong Kong A-Level Book 151 ~50 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 53 New Way Chemistry for Hong Kong A-Level Book 152 ~50 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 54 New Way Chemistry for Hong Kong A-Level Book 153 ~50 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 55 New Way Chemistry for Hong Kong A-Level Book 154 50 cm 3 Solution A 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 56 New Way Chemistry for Hong Kong A-Level Book 155 ~50 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 57 New Way Chemistry for Hong Kong A-Level Book 156 ~50 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 58 New Way Chemistry for Hong Kong A-Level Book 157 ~50 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 59 New Way Chemistry for Hong Kong A-Level Book 158 ~50 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 60 New Way Chemistry for Hong Kong A-Level Book 159 ~50 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 61 New Way Chemistry for Hong Kong A-Level Book 160 ~50 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 62 New Way Chemistry for Hong Kong A-Level Book 161 50 cm 3 Solution B 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 63 New Way Chemistry for Hong Kong A-Level Book 162 ~100 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 64 New Way Chemistry for Hong Kong A-Level Book 163 ~100 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 65 New Way Chemistry for Hong Kong A-Level Book 164 ~100 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 66 New Way Chemistry for Hong Kong A-Level Book 165 ~100 cm 3 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Slide 67 New Way Chemistry for Hong Kong A-Level Book 166 Finding the concentration of a solution ~100 cm 3 3.6 Simple Titrations (SB p.69) Slide 68 New Way Chemistry for Hong Kong A-Level Book 167 Finding the concentration of a solution ~100 cm 3 3.6 Simple Titrations (SB p.69) Slide 69 New Way Chemistry for Hong Kong A-Level Book 168 Finding the concentration of a solution Solution C 100 cm 3 3.6 Simple Titrations (SB p.69) Slide 70 New Way Chemistry for Hong Kong A-Level Book 169 Comment on the Concentrations of Solutions A, B and C ! contain the same amount of solute (same concentration) 2 x the amount of solute Concentration of solution B is 2 times that of the concentrations of solutions A & B. 3.6 Simple Titrations (SB p.69) Concentration is the amount of solute in a unit volume of solution. Slide 71 New Way Chemistry for Hong Kong A-Level Book 170 Class Practice Suppose the right-handed side figure shows the number of solute particles in solution D. Draw similar particle models for Solutions A, B and C. 3.6 Simple Titrations (SB p.69) Slide 72 New Way Chemistry for Hong Kong A-Level Book 171 Class Practice Answers 3.6 Simple Titrations (SB p.69) Slide 73 New Way Chemistry for Hong Kong A-Level Book 172 Comment on the Concentrations of Solutions A, B and C ! Concentration is the amount of solute in a unit volume of solution. no. of spoons mass no. of moles 3.6 Simple Titrations (SB p.69) Slide 74 New Way Chemistry for Hong Kong A-Level Book 173 Molarity Molarity is the number of moles of solute dissolved in 1 dm 3 (1000 cm 3 ) of solution. A way of expressing concentrations Unit: moles/dm 3 (M) 3.6 Simple Titrations (SB p.69) Slide 75 New Way Chemistry for Hong Kong A-Level Book 174 What does this mean? 1 dm 3 contains 2 moles of HCl In every 1 dm 3 of the solution, 2 moles of HCl is dissolved. 3.6 Simple Titrations (SB p.69) Slide 76 New Way Chemistry for Hong Kong A-Level Book 175 Answer Example 3-14 25.0cm 3 of sodium hydroxide solution was titrated against 0.067 M of sulphuric(VI) acid using methyl orange as indicator. The indicator changed colour from yellow to red when 22.5 cm 3 of sulphuric(VI) acid had benn added. Calculate the molarity of the sodium hydroxide solution. 3.6 Simple Titrations (SB p.71) Solution: Number of moles of NaOH(aq) 2 Number of moles of H 2 SO 4 (aq) = 1 x Number of moles of NaOH(aq) = Number of moles of H 2 SO 4 (aq) = 0.067 mol dm -3 x 22.5 x 10 -3 dm 3 = 1.508 x 10 -3 mol Number of moles of NaOH(aq) = 2 x 1.508 x 10 -3 mol = 3.016 x 10 -3 mol Molarity of NaOH(aq) = 3.012 x 10 -3 mol / 25.0 x 10 -3 mol = 0.1221 mol dm -3 The molarity of NaOH is 0.121M Slide 77 New Way Chemistry for Hong Kong A-Level Book 176 Answer Example 3-15 2.52 g of a pure dibasic acid with formula mass of 126.0 was dissolved in water and made up to 250.0 cm 3 in a volumetric flask 25.0 cm 3 of this solution was found to neutralize 28.5 cm 3 of sodium hydroxide solution. (a)Calculate the molarity of the acid solution. (b)If the dibasic acid is represented by H 2 X, write an equation for the reaction between the acid and sodium hydroxide. (c)Calculate the molarity of the sodium hydroxide solution. 3.6 Simple Titrations (SB p.71) Solution: (a)Number of moles of acid = 2.52 g/ 126.0 g mol -1 = 0.02 mol Molarity of acid solution = 0.02 mol / 250 x 10 -3 dm 3 = 0.08M (b) H 2 X(aq) + 2NaOH(aq) Na 2 X(aq) + 2NaOH(l) (c) Number of moles of H 2 X = x number of moles of NaOH Molarity of NaOH = 0.14M Slide 78 New Way Chemistry for Hong Kong A-Level Book 177 Answer Example 3-16 0.186g of sample of hydrate sodium carbonate, NaCO 2 nH 2 O, was dissolved in 100 cm 3 of distilled water in conical flask. 0.10 M by hydrochloric acid was added from a burette, 2 cm 3 at a time. The pH value of the solution was measured by a pH meter. The result was recorded and shown in the following figure. Calculate the value of n in NaCO 2 nH 2 O. 3.6 Simple Titrations (SB p.72) Solution: (a)Number of moles of acid = 2.52 g / 126.0 g mol -1 There is a sudden drop in the pH value of the solution (from pH 3 to pH 8) with the end point at 30.0 cm 3. Na 2 CO 3 nH 2 O(s) + 2 HCl(aq) 2NaCl(aq) + CO 2 (g) + (n+1)H2O(l) Number of moles of Na 2 CO 3 nH 2 O = x 0.1 mol dm -3 x 30 x 10 -3 dm 3 106.0 + 18.0n = 124.0 n = 1 The formula is Na 2 CO 3 H 2 O Slide 79 New Way Chemistry for Hong Kong A-Level Book 178 Answer 3.6 Simple Titrations (SB p.73) Example 3-17 5 cm 3 of 0.5M sulphuric(VI) acid was added to 25.0 cm 3 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: (a) Plot the graph of temperature against volume of sulphuric(VI) acid added. (b) Calculate the molarity of the potassium hydroxide solution. (c) Explain why the temperature rose to a maximum and the fell. Solution: (a) Slide 80 New Way Chemistry for Hong Kong A-Level Book 179 3.6 Simple Titrations (SB p.74) Solution: (contd) (b) From the graph, it is found that the end point of the titration is reached when 20 cm 3 of H 2 SO 4 is added. Number of moles of = 0.5 mol dm -3 x 20/1000 dm 3 = 0.01 mol 2KOH(aq) + H 2 SO 4 (aq) K 2 SO 4 (aq) + 2H 2 O(l) 2 mol 1 mol Mole of KOH(aq) : H2SO4 = 2 : 1 Number of moles of KOH(aq) = 2 x 0.01 mol = 0.02 mol Molarity of KOH(aq) = 0.02 mol / (25 x 10 -3 dm 3 ) = 0.8M Slide 81 New Way Chemistry for Hong Kong A-Level Book 180 3.6 Simple Titrations (SB p.74) Solution: (contd) (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 cooled down the reacting solution, causing the temperature to drop. Slide 82 New Way Chemistry for Hong Kong A-Level Book 181 Iodometric Titrations Some Examples I 2 (aq) + 2S 2 O 3 2- (aq) 2I - (aq) + S 4 O 6 2- (aq) browncolourless in conical flask in burette 3.6 Simple Titrations (SB p.76) Redox Titrations Slide 83 New Way Chemistry for Hong Kong A-Level Book 182 Iodometric Titrations Some Examples I 2 (aq) + 2S 2 O 3 2- (aq) 2I - (aq) + S 4 O 6 2- (aq) browncolourless Add starch During titration : brown yellow in conical flaskin burette 3.6 Simple Titrations (SB p.76) Redox Titrations Slide 84 New Way Chemistry for Hong Kong A-Level Book 183 Iodometric Titrations Some Examples I 2 (aq) + 2S 2 O 3 2- (aq) 2I - (aq) + S 4 O 6 2- (aq) browncolourless During titration : brown yellow 3.6 Simple Titrations (SB p.76) Redox Titrations Slide 85 New Way Chemistry for Hong Kong A-Level Book 184 Iodometric Titrations Some Examples I 2 (aq) + 2S 2 O 3 2- (aq) 2I - (aq) + S 4 O 6 2- (aq) browncolourless End point : blue black colourless (after addition of starch indicator) During titration : brown yellow 3.6 Simple Titrations (SB p.76) Redox Titrations Slide 86 New Way Chemistry for Hong Kong A-Level Book 185 Answer Example 3-18 When excess potassium iodide solution (KI) is added to 25.0 cm 3 of acidified potassium iodate solution (KIO 3 ) of unknown concentration, the solution turns brown. This brown solution requires 22.0 cm 3 of 0.05 M sodium thiosulphate solution to react completely with the iodine formed, using starch solution as indicator. Find the molarity of the potassium iodate solution. 3.6 Simple Titrations (SB p.76) Solution: IO 3- (aq) + 5I - + 6H + (aq) 3I 2 (aq) + 3H 2 O(l) (1) I 2 (aq) + 2S 2 O 3 2- (aq) 2I - (aq) + S 4 O 6 2- (aq) (2) From (1), Number of moles of IO 3- (aq) =1/3 x number of moles of I 2 (aq) From(2), Number of moles of I 2 (aq) =1/2 x number of moles of S 2 O 3 2- (aq) Number of moles of IO 3- (aq) = 1/6 x number of moles of S 2 O 3 2- (aq) Molority of IO 3- (aq) x 25.0/1000 dm 3 =1/6 x 0.05 mol dm -3 x 22.0/1000 dm 3 Molarity of IO 3- (aq) = 7.33 x 10 -3 M Slide 87 New Way Chemistry for Hong Kong A-Level Book 186 Titrations Involving Potassium Permanganate Some Examples MnO 4 - (aq) + 8H + (aq) + 5Fe 2+ (aq) Mn 2+ (aq) + H 2 O(aq) + 5Fe 3+ (aq) purplecolourless In buretteIn conical flask 3.6 Simple Titrations (SB p.76) Redox Titrations Slide 88 New Way Chemistry for Hong Kong A-Level Book 187 Titrations Involving Potassium Permanganate Some Examples MnO 4 - (aq) + 8H + (aq) + 5Fe 2+ (aq) Mn 2+ (aq) + H 2 O(aq) + 5Fe 3+ (aq) purple colourless During titration : pale green yellow In buretteIn conical flask 3.6 Simple Titrations (SB p.76) Redox Titrations Slide 89 New Way Chemistry for Hong Kong A-Level Book 188 Titrations Involving Potassium Permanganate Some Examples MnO 4 - (aq) + 8H + (aq) + 5Fe 2+ (aq) Mn 2+ (aq) + H 2 O(aq) + 5Fe 3+ (aq) purple colourless End point : yellow light purple During titration : pale green yellow In buretteIn conical flask 3.6 Simple Titrations (SB p.76) Redox Titrations Slide 90 New Way Chemistry for Hong Kong A-Level Book 189 Answer Example 3-19 A piece of impure iron wire weighs 0.22g. When it is dissolved in hydrochloric acid, it is oxidized to iron(II) ions. The solution requires 36.5 cm 3 of 0.02 M acidified potassium manganate(VII) for complete reaction to form iron(III) ions. What is the percentage purity of the iron wire? 3.6 Simple Titrations (SB p.77) Solution: MnO 4- (aq) + 8H + (aq) + 5Fe 2+ (aq) Mn 2+ (aq) + 4H 2 O(l) + 5Fe 3+ (aq) Number of moles of Fe 2+ (aq) = 5 x number of moles of MnO 4- (aq) = 5 x 0.02 mol dm -3 x 36.5 x 10 -3 dm 3 = 3.65 x 10 -3 mol Number of moles of Fe dissolved = number of mole of Fe 2+ formed = 3.65 x 10 -3 mol Mass of Fe = 3.65 x 10 -3 mol x 55.8 g mol -1 = 0.204g Percentage purity of Fe = 0.204g/0.22g x 100% = 92.73% Slide 91 New Way Chemistry for Hong Kong A-Level Book 190 Answer Check Point 3-8 (a)5g of anhydrous sodium carbonate is added to 100 cm 3 of 2 M hydrochloric acid. What is the volume of gas evolved at room temperature and pressure? (R.a.m. : C = 12.0, O = 16.0, Na = 23.0; molar volume of gas at R.T.P. = 24.0 dm 3 mol -1 ) (b) 8.54g of impure hydrated iron(II) sulphate (formula mass of 392.14) was dissolved in water and made up to 250 cm 3. 25cm 3 of this solution required 20.76cm 3 of 0.0203M acidified potassium manganate(VII) solution for complete reaction. Determine the percentage purity of the hydrated iron(II) sulphate. 3.6 Simple Titrations (SB p.78) (a) Na 2 CO 3 (s) + 2 HCl (aq) 2NaCl(aq) + H 2 O(l) + CO 2 (g) No. of moles of Na 2 CO 3 used = 5g / (23.0x 2 +12 +16.0 x 3)g mol -1 = 0.0472 mol No. of moles of HCl used = 2M x 100/ 1000 dm 3 = 0.2 mol Since HCl is in excess, Na 2 CO 3 is the limiting agent. No. of moles of CO 2 produced =no. of moles of Na 2 CO 3 used = 0.0472 mol Volume of CO 2 produced = 0.0472 mol x 24.0 dm 3 mol -1 = 1.133 dm 3 Slide 92 New Way Chemistry for Hong Kong A-Level Book 191 3.6 Simple Titrations (SB p.78) (b)No. of moles of MnO 4- = 0.0203M x 20.76/1000 dm 3 = 4.214 x 10 -4 mol No. of moles of Fe 2+ = 5 x no. of moles of MnO 4- = 2.107 x 10 -3 mol No. of mole of Fe 2+ in 25.0 cm 3 solution = 2.107 x 10 -3 mol No. of mole of Fe 2+ in 250.0 cm 3 solution = 0.021 07 mol Molar mass of hydrated FeSO 4 = 392.14 g mol 1 Mass of hydrated FeSO 4 = 0.021 07 mol x 392.14 g mol 1 = 8.26g 5 purity of FeSO 4 = 8.26g/8.54g x 100% = 96.72% Slide 93 New Way Chemistry for Hong Kong A-Level Book 192 The END