A-Level Chemistry May/June 2025 Q2(d)(ii): Calculate the mass, in mg, of iron in one tablet. Assume that all the iron in the table…

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  The reaction stoichiometry is $5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}$. The molar ratio of $\text{MnO}_4^-$ to $\text{Fe}^{2+}$ is 1:5.

  Step 1: Moles of $\text{MnO}_4^-$ and $\text{Fe}^{2+}$ in 25.0 cm³ (Assuming data from part (d)(i): Titration of 25.0 cm³ aliquots of the iron tablet solution required a mean titre of 31.00 cm³ of 0.0300 mol dm⁻³ KMnO₄)

  Moles of $\text{MnO}_4^-$ in titre = concentration $\times$ volume $n(\text{MnO}_4^-) = 0.0300 \text{ mol dm}^{-3} \times \frac{31.00}{1000} \text{ dm}^3 = 9.30 \times 10^{-4} \text{ mol}$

  Moles of $\text{Fe}^{2+}$ in 25.0 cm³ aliquot = $5 \times n(\text{MnO}_4^-)$ $n(\text{Fe}^{2+}) = 5 \times 9.30 \times 10^{-4} \text{ mol} = 4.65 \times 10^{-3} \text{ mol}$

  Step 2: Moles of $\text{Fe}^{2+}$ in 150.0 cm³ (Assuming data from the method: Two tablets were dissolved in a 150.0 cm³ volumetric flask)

  Total moles of $\text{Fe}^{2+}$ = moles in aliquot $\times$ (total volume / aliquot volume) Total $n(\text{Fe}^{2+}) = 4.65 \times 10^{-3} \text{ mol} \times \frac{150.0}{25.0} = 2.79 \times 10^{-2} \text{ mol}$

  Step 3: Mass of Fe in two tablets Mass = moles $\times$ relative atomic mass ($A_r$ of Fe = 55.8) Mass of Fe = $2.79 \times 10^{-2} \text{ mol} \times 55.8 \text{ g mol}^{-1} = 1.55682 \text{ g}$

  Step 4: Mass of Fe in one tablet (in mg) Mass in one tablet = $\frac{1.55682 \text{ g}}{2} = 0.77841 \text{ g}$

  Mass in mg = $0.77841 \text{ g} \times 1000 = 778.41 \text{ mg}$

  Mass = 778.4 mg (to 4 s.f.)

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  How the marks are awarded

  • M1 — For correctly calculating the moles of permanganate ions used in the titration ($9.30 \times 10^{-4}$ mol) and then using the 1:5 stoichiometric ratio to find the moles of iron(II) ions in the 25.0 cm³ aliquot ($4.65 \times 10^{-3}$ mol).
  • M2 — For correctly scaling up the moles of iron(II) from the 25.0 cm³ sample to the total 150.0 cm³ solution, giving $2.79 \times 10^{-2}$ mol.
  • M3 — For converting the total moles of iron into a mass using the relative atomic mass of Fe, to find the mass of iron in two tablets (1.557 g or 1.55682 g).
  • M4 — For correctly calculating the mass of iron in a single tablet by dividing by two, and converting the final answer from grams to the required unit of milligrams, giving 778.4 mg.

  Common mistakes

  • Using an incorrect 1:1 molar ratio for the MnO₄⁻:Fe²⁺ reaction instead of the correct 1:5 ratio, leading to a significantly underestimated result.
  • Forgetting to scale up the moles from the 25.0 cm³ titrated aliquot to the 150.0 cm³ total solution, thereby calculating the mass in the aliquot only.
  • Forgetting to divide the total mass by two at the end to find the mass in a single tablet as the question asks.
  • Failing to convert the final answer from grams to milligrams, or making a power-of-ten error during the conversion.

  Examiner tip: For multi-step calculation questions, always map out the experimental process (e.g., sample preparation → dilution → titration) to ensure each calculation step logically follows the next.

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