📋 Structured Questions
Complete each question on paper, then check your answers against the mark scheme.
Question 1: Empirical Formula and Gas Calculations
8 marks(a) A sample of compound X contains only carbon, hydrogen, and oxygen. On complete combustion, 1.20 g of X produced 2.64 g of carbon dioxide and 1.44 g of water. Show by calculation that the empirical formula of X is C3H8O. [4]
(b) In another experiment, 0.352 g of a volatile liquid Y was vaporised at 98.0 °C and 100 kPa. The volume of gas produced was 114 cm3. Calculate the relative molecular mass (Mr) of Y. Give your answer to 3 significant figures. (R = 8.314 J K^-1 mol^-1). [4]
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(a)
- Moles of C = Moles of CO2 = 2.64 / 44.0 = 0.0600 mol. Mass of C = 0.0600 * 12.0 = 0.720 g. [1]
- Moles of H = 2 * Moles of H2O = 2 * (1.44 / 18.0) = 0.160 mol. Mass of H = 0.160 * 1.0 = 0.160 g. [1]
- Mass of O = 1.20 - 0.720 - 0.160 = 0.320 g. Moles of O = 0.320 / 16.0 = 0.0200 mol. [1]
- Ratio of C : H : O = 0.0600 : 0.160 : 0.0200 = 3 : 8 : 1, giving empirical formula C3H8O. [1]
(b)
- Convert units: p = 100,000 Pa; V = 114 x 10^-6 m3 = 1.14 x 10^-4 m3; T = 98.0 + 273 = 371 K [1]
- n = pV / RT = (100,000 * 1.14 x 10^-4) / (8.314 * 371) [1]
- n = 11.4 / 3084.5 = 0.003696 mol [1]
- Mr = mass / n = 0.352 / 0.003696 = 95.2 [1]
Question 2: Yield and Atom Economy
5 marks(a) Titanium is manufactured by the reaction of titanium(IV) chloride with sodium: TiCl4(l) + 4Na(s) -> Ti(s) + 4NaCl(s). In a batch, 50.0 kg of TiCl4 reacted with excess sodium to produce 11.5 kg of titanium. Calculate the percentage yield of titanium. Give your answer to 3 significant figures. [3]
(b) Calculate the percentage atom economy for the production of titanium in this reaction. Give your answer to 3 significant figures. [2]
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(a)
- Moles of TiCl4 used = 50,000 g / 189.9 = 263.3 mol [1]
- Theoretical mass of Ti = 263.3 * 47.9 = 12,612 g = 12.61 kg [1]
- Percentage yield = (11.5 / 12.61) * 100 = 91.2% [1]
(b)
- Atom economy = (Mr of Ti / (Mr of TiCl4 + 4 * Mr of Na)) * 100 [1]
- Atom economy = (47.9 / (189.9 + 4 * 23.0)) * 100 = (47.9 / 281.9) * 100 = 17.0% [1]
Question 3: Standard Solutions & Titration Practical
7 marks(a) Describe the experimental procedure to prepare 250 cm3 of a standard solution of sodium hydrogensulfate (NaHSO4) starting from a solid. [5]
(b) 25.0 cm3 of the prepared 0.100 mol dm^-3 NaHSO4 solution was neutralised by exactly 22.4 cm3 of sodium hydroxide solution: NaHSO4(aq) + NaOH(aq) -> Na2SO4(aq) + H2O(l). Calculate the concentration of the sodium hydroxide solution. Give your answer to 3 significant figures. [2]
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(a)
- Weigh the solid in a weighing boat on a 2-decimal place balance, record the mass, transfer to a beaker and re-weigh the weighing boat (difference method). [1]
- Add distilled water and stir with a glass rod until all the solid dissolves. [1]
- Transfer the solution to a 250 cm3 volumetric flask. [1]
- Rinse the beaker, glass rod, and funnel with distilled water and add all washings to the flask. [1]
- Make up to the graduation mark with distilled water (until the bottom of the meniscus is on the line), then invert the flask multiple times to mix. [1]
(b)
- Moles of NaHSO4 = 0.0250 * 0.100 = 0.00250 mol. Moles of NaOH = 0.00250 mol (1:1 ratio). [1]
- Concentration of NaOH = moles / volume = 0.00250 / 0.0224 = 0.112 mol dm^-3. [1]
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