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Topic 3.1.2: Amount of Substance - Exam Practice

AQA A-Level Chemistry • Physical Chemistry • Exam Practice

AQA A-LEVEL CHEMISTRY

Topic 3.1.2: Amount of Substance Exam Practice

Test your understanding of moles, the ideal gas equation, yields, atom economy, and titration calculations.

Amount of Substance › Exam Practice

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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]
Examiner tip: In part (a), do not forget that each H2O molecule contains 2 hydrogen atoms, so you must multiply the moles of H2O by 2 to get the moles of H. In part (b), the most common loss of marks is incorrect unit conversion for V (cm3 to m3 is divide by 1,000,000) and T (must be in Kelvin).

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]
Examiner tip: In part (b), when calculating total Mr of reactants, you must include the coefficients in the balanced equation (i.e. 4 * Mr of Na). The atom economy is low because a large mass of sodium chloride waste is produced.

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]
Examiner tip: The standard solution preparation question is extremely common. You must mention: using a weighing boat and weighing by difference, adding washings, and inverting the volumetric flask to ensure the solution is homogeneous.

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