A common iron ore, fool’s gold, contains the mineral iron pyrite, FeS₂ - VCE - SSCE Chemistry - Question 2 - 2016 - Paper 1

To find the percentage by mass of FeS₂ in the mineral sample, we first need to determine the mass of BaSO₄ produced:

1. Calculate the mass of BaSO₄:

   (Mass of BaSO₄) = (Mass of crucible, filter paper and dry BaSO₄) - (Mass of crucible and filter paper)

   (Mass of BaSO₄) = 174.99 g - 123.40 g = 51.59 g

2. Calculate the number of moles of BaSO₄ produced:

   n(BaSO₄) = \frac{mass}{molar mass} = \frac{51.59 g}{233.4 g/mol} \approx 0.221 mol

3. The reaction shows that 1 mole of FeS₂ produces 1 mole of BaSO₄. Therefore, moles of FeS₂ is also 0.221 mol.

4. Calculate the mass of FeS₂:

   mass(FeS₂) = n(FeS₂) \times M(FeS₂) = 0.221 mol \times 120.0 g/mol \approx 26.52 g

5. Finally, use the mass of FeS₂ to determine the percentage by mass:

   percentage = \frac{mass(FeS₂)}{mass(sample)} \times 100 = \frac{26.52 g}{14.3 g} \times 100 \approx 185.5%

Given that this seems impossibly high, a calculation error likely occurred earlier or assumptions regarding the purity of reactants may need to be reviewed.