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 (\(M_r\)) of Y. Give your answer to 3 significant figures. (\(R = 8.314\text{ J K}^{-1}\text{mol}^{-1}\)). [4]
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(a)
- Moles of C = Moles of CO₂ = \(\frac{2.64}{44.0} = 0.0600\text{ mol}\). Mass of C = 0.0600 × 12.0 = 0.720 g. [1]
- Moles of H = 2× Moles of \(\text{H}_2\text{O} = 2 \times \frac{1.44}{18.0} = 0.160\text{ mol}\). 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 \(\text{O} = \frac{0.320}{16.0} = 0.0200\text{ mol}\). [1]
- Ratio of C:H:O = 0.0600:0.160:0.0200 = 3:8:1, giving empirical formula C₃H₈O. [1]
(b)
- Convert units: \(p = 100,000\text{ Pa}\); \(V = 114 \times 10^{-6}\text{ m}^3 = 1.14 \times 10^{-4}\text{ m}^3\); \(T = 98.0 + 273 = 371\text{ K}\) [1]
- \(n = \frac{pV}{RT} = \frac{100,000 \times 1.14 \times 10^{-4}}{8.314 \times 371}\) [1]
- \(n = \frac{11.4}{3084.5} = 0.003696\text{ mol}\) [1]
- \(M_r = \frac{\text{mass}}{n} = \frac{0.352}{0.003696} = 95.2\) [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 = \(\frac{50,000\text{ g}}{189.9} = 263.3\text{ mol}\) [1]
- Theoretical mass of Ti = 263.3 × 47.9 = 12,612 g = 12.61 kg [1]
- Percentage yield = \(\frac{11.5}{12.61} \times 100 = 91.2\%\) [1]
(b)
- Atom economy = \(\frac{M_r(\text{Ti})}{M_r(\text{TiCl}_4) + 4 \times M_r(\text{Na})} \times 100\) [1]
- Atom economy = \(\frac{47.9}{189.9 + 4 \times 23.0} \times 100 = \frac{47.9}{281.9} \times 100 = 17.0\%\) [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 \(\text{NaOH} = \frac{\text{moles}}{\text{volume}} = \frac{0.00250}{0.0224} = 0.112\text{ mol dm}^{-3}\). [1]
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