A-Level Chemistry May/June 2025 Q5(b)(ii): Use your answer to (b)(i) to calculate the percentage by mass of vitamin C present in 1…

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  From (b)(i), the amount of I₂ that reacted is $1.8325 \times 10^{-4}$ mol.

  Since vitamin C reacts with I₂ in a 1:1 molar ratio, the amount of vitamin C in the titrated sample is also $1.8325 \times 10^{-4}$ mol.

  This sample was only a fraction of the total extract from the 150.0 g of lemon. We must scale this up to find the total amount.

  Step 1: Calculate the total mass of vitamin C in 150.0 g of lemon.

  The molar mass ($M_r$) of vitamin C (C₆H₈O₆) is $(6 \times 12.0) + (8 \times 1.0) + (6 \times 16.0) = 176.0\ \text{g mol}^{-1}$.

  The titrated sample was 1/4 of the total solution, so we multiply the moles by 4.

  Total moles of vitamin C = $1.8325 \times 10^{-4} \times 4 = 7.33 \times 10^{-4}$ mol

  Total mass of vitamin C = moles $\times M_r$

  Total mass of vitamin C = $(7.33 \times 10^{-4}) \times 176.0 = 0.129008$ g

  Step 2: Calculate the percentage by mass.

  Percentage by mass = $\frac{\text{mass of vitamin C}}{\text{total mass of lemon}} \times 100$

  Percentage by mass = $\frac{0.129008}{150.0} \times 100 = 0.086005... \%$

  Percentage by mass = $0.086\%$ (to 2 significant figures)

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  How the marks are awarded

  • M1 — Calculating the total mass of vitamin C in the 150.0 g of lemon by multiplying the moles from the titration by the scaling factor (4) and the molar mass (176.0). Showing the working (1.8325 x 10⁻⁴ x 4) x 176.0 or the resulting value 0.129 g earns this mark.
  • M2 — Correctly calculating the final percentage by dividing the mass from M1 by the total mass of the lemon (150.0 g) and multiplying by 100, leading to the final correct answer of 0.086%.

  Common mistakes

  • Forgetting to multiply by the scaling factor of 4, which accounts for the fact that only a fraction of the total lemon extract was titrated.
  • Dividing by the scaling factor of 4 instead of multiplying, or applying it at the wrong stage of the calculation.
  • Using an incorrect molar mass for vitamin C, often due to miscounting atoms or using incorrect atomic masses.
  • Rounding the final answer to an inappropriate number of significant figures, such as one significant figure (0.09%) or too many.

  Examiner tip: Always check if the analysed sample is only a fraction (an aliquot) of the original solution and remember to scale your result up accordingly before the final calculation.

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