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12 (6 marks) Mole Ratio Whole Ratio m c = 30.1 % x 100 g = 30.1 g n c = 30.1 g /12.01 g/mol = 2.51 mol C 2.51/1.88 = 1.34 x3 = 4 m H = 3.16 % x 100 g = 3.16 g n H = 3.16/1.01 g/mol = 3.13 mol H 3.13/1.88 = 1.66 x3 = 5 n Cl = 66.7 % x 100 g = 66.7 g n Cl = 66.7/35.45 g/mol = /1.88 = 1 x3 = 3 The empirical formula is C 4 H 5 Cl 3. If the molecular formula is twice the empirical then the molecular formula must be C 8 H 10 Cl 6. (2 marks) MM C8H10Cl6 = g/mol n C8H10Cl6 = 38.0 g / g/mol = mol C 8 H 10 Cl 6 (2 marks) % yield = (actual yield/theoretical yield) x 100% theoretical yield = actual yield / (% yield / 100%) = mol / (68% / 100%) = mol (2 marks) OR % yield = (actual yield/theoretical yield) x 100% theoretical yield = actual yield / (% yield / 100%) 38.0 g / (68% / 100%) = 55.9 g n C 8 H 10 Cl 6 = 55.9 g / g/mol = mol n A = 3 mol A/ 2 mol C x mol C = mol A (2 marks) MM A = g/mol m B = n x MM = mol x g/mol = 18.6 g A (1 mark) BONUS: Since we know that compound B doesn’t contain any chlorine, compound A must, by inspection we can see that the molar mass for A corresponds to Cl 2. With this value we can see that our chemical equation would be: 3 Cl 2 + B -> C 8 H 10 Cl 6 In order to balance this equation B must be C 8 H 10 (2 marks)

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