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

To determine the percentage of iron in the original sample, we first find the molar mass of iron(III) oxide (Fe₂O₃):

• Molar mass of Fe₂O₃ = 159.70 g mol⁻¹

Then, we calculate the moles of Fe₂O₃ produced:

• Moles of Fe₂O₃ = ( \frac{4.21 \text{ g}}{159.70 \text{ g mol}^{-1}} = 0.02632 \text{ mol} )

Next, we calculate the moles of iron in Fe₂O₃:

• Moles of Fe = 2 × moles of Fe₂O₃ = 2 × 0.02632 = 0.05274 mol

Now, we find the mass of iron:

• Mass of Fe = 0.05274 mol × 55.85 g mol⁻¹ = 2.9446 g

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

• Percentage of iron = ( \frac{2.9446 \text{ g}}{4.32 \text{ g}} \times 100 = 68.2% )