A white solid is a mixture of sodium ethanoate (Na2C2O4), ethanoic acid dihydrate (H2C2O4·2H2O) and an inert solid - AQA - A-Level Chemistry - Question 11 - 2017 - Paper 1

From the balanced equation, it can be seen that 1 mole of MnO₄⁻ reacts with 5 moles of C₂O₄²⁻. Therefore, the moles of C₂O₄²⁻ (sodium ethanoate) are:

$5.30 \times 10^{-4} \text{ mol MnO₄⁻} \times 5 = 2.65 \times 10^{-3} \text{ mol C₂O₄²⁻}$

Using the second titration, where the moles of H₂C₂O₄ can be found. The moles of Na₂C₂O₄ produced in moles from the hydroxide reaction is:

$\text{Moles of } H₂C₂O₄ = 0.100 \text{ mol/dm}^3 \times \frac{10.45 \text{ cm}^3}{1000} = 1.045 \times 10^{-3} \text{ mol}$

As per the stoichiometry of the reaction, 1 mole of H₂C₂O₄ produces 1 mole of C₂O₄²⁻. Thus:

$\text{Moles of } C₂O₄^{2-} = 1.045 \times 10^{-3} \text{ mol}$

Total moles of C₂O₄²⁻ in the solid:

$2.65 \times 10^{-3} + 1.045 \times 10^{-3} = 3.695 \times 10^{-3} \text{ mol}$

The molar mass of sodium ethanoate (Na₂C₂O₄) is approximately 82.03 g/mol. Therefore, the mass can be calculated as:

$\text{Mass} = 3.695 \times 10^{-3} \text{ mol} \times 82.03 \text{ g/mol} = 303.68 \text{ g}$

The total mass of the solid sample is 1.90 g. The percentage is calculated as:

$\text{Percentage by mass} = \left( \frac{303.68 \text{ g}}{1.90 \text{ g}} \right) \times 100 = 58.68 \% \approx 58.7\% \, \text{to 3 significant figures}$