Explain why the mass of carbon dioxide produced when excess hydrochloric acid reacts with calcium carbonate can be used to determine the percentage yield of the reaction, and outline ONE limitation of using mass loss to measure the amount of carbon dioxide produced.

IB DP Chemistry Standard Level (2023 syllabus) — R2.1 How much? The amount of chemical change · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

A student reacts a weighed sample of impure calcium carbonate with excess hydrochloric acid in an open conical flask placed on a balance. The balanced equation for the reaction is: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

Model answer (4 marks)

The reaction produces one mole of CO₂ for every mole of CaCO₃ that reacts, so the moles of CO₂ lost from the flask are directly proportional to the moles of CaCO₃ that have been converted. By measuring the mass loss of the flask and the sample, the mass of CO₂ that has escaped can be calculated. Converting this mass to moles (using 44.01 g mol⁻¹) gives the actual moles of CaCO₃ that reacted. The theoretical yield is found from the mass of pure CaCO₃ in the weighed sample (using 100 g mol⁻¹), giving the maximum possible moles of CaCO₃ that could react. The percentage yield is then (actual yield ÷ theoretical yield) × 100. One limitation of using mass loss is that CO₂ may escape from the open flask before the balance reading is taken; this loss of gas before measurement will make the recorded mass loss larger than the true amount produced during the timed period, leading to an over‑estimate of the yield.

Examiner tips

  • State the 1:1 stoichiometry between CaCO₃ and CO₂
  • Show the conversion from mass loss to moles of CO₂
  • Explain how theoretical yield is calculated from the initial CaCO₃ mass
  • Mention the specific limitation of CO₂ escape

Common mistakes

  • Confusing the mass of CO₂ with the mass of CaCO₃ in the calculation
  • Using the wrong molar mass for CaCO₃ (e.g. 100 g mol⁻¹ instead of 100.09 g mol⁻¹)
  • Failing to note that the flask is open and CO₂ can escape before the balance is read

Mark scheme (4 marks)

  1. The moles of CO₂ produced are stoichiometrically equal to the moles of CaCO₃ that reacted (1:1 molar ratio from the equation), so the actual yield of CaCO₃ converted can be calculated from the mass of CO₂ lost.
  2. The theoretical yield is calculated from the moles of pure CaCO₃ in the sample (using molar mass 100 g mol⁻¹), and percentage yield is actual yield divided by theoretical yield multiplied by 100.
  3. A limitation is that CO₂ may escape before the reaction is complete or before the balance reading is taken, causing the recorded mass loss to be greater than the true amount produced during the timed measurement, leading to an overestimate.
  4. Another limitation is that water vapour (or HCl vapour / spray) may also be lost from the open flask, contributing to the recorded mass loss and causing the mass of CO₂ to be overestimated.

Key terms in this question

percentage yield · mass loss

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