The iron content of an impure sample (4.32 g) was determined by the process shown in the flow chart - HSC - SSCE Chemistry - Question 28 - 2022 - Paper 1

First, calculate the molar mass of iron(III) oxide (Fe₂O₃):

$\text{Molar mass of Fe}_2\text{O}_3 = 2 \times 55.85 \text{ g mol}^{-1} + 3 \times 16.00 \text{ g mol}^{-1} = 159.70 \text{ g mol}^{-1}$

Next, find the number of moles of iron(III) oxide produced:

$\text{Amount of Fe}_2\text{O}_3 = \frac{4.21 \text{ g}}{159.70 \text{ g mol}^{-1}} = 0.02632\text{ mol}$

Iron content in moles:

$\text{Amount of Fe} = 2 \times \text{moles of Fe}_2\text{O}_3 = 2 \times 0.02632 = 0.05274\text{ mol}$

Calculating the mass of iron:

$\text{Mass of Fe} = 0.05274\text{ mol} \times 55.85\text{ g mol}^{-1} = 2.9446\text{ g}$

Finally, calculate the percentage of iron in the original sample:

$\text{Percentage of Fe in original impure sample} = \left( \frac{2.9446\text{ g}}{4.32\text{ g}} \right) \times 100 = 68.2\%$