What percent, by mass, of the mixture was FeSO4?

Question

A mixture consists of FeSO4 and Fe2(SO4)3. A 0.1223 g sample of this mixture was dissolved in water and titrated with a 0.00420 M KMnO4 solution. The titration required 18.85 mL of the KMnO4 solution. What percent, by mass, of the mixture was FeSO4?

Answer 1

First, we have to look at the redox reaction between FeSO4 and KMnO4.

10FeSO₄ + 2KMnO₄ + 8H₂SO₄ → 5Fe₂(SO₄)₃ + K₂SO₄ + 2MnSO₄ + 8H₂O

(note: this is a redox reaction in acid. Instead of 8H₂SO₄, we could have just used 16H⁺, etc.)

From the information provided, we can calculate the moles of KMnO₄ used. We can then find the moles of FeSO₄ originally present. We then convert moles FeSO₄ to grams FeSO₄. We divided this by the original mass to find percent FeSO₄.

moles KMnO₄ used = 18.85 ml x 1 L / 1000 ml x 0.00420 mol/L = 7.917x10^-5 moles

moles FeSO₄ present = 7.917x10^-5 moles KMnO₄ x 10 mols FeSO₄ / 2 mol KMnO₄ = 3.959x10^-4 mols FeSO₄

molar mass FeSO4 = 151.9 g / mol

grams FeSO₄ present = 3.959x10^-4 mols FeSO₄ x 151.9 g/mol = 0.06013 g FeSO₄

% FeSO₄ = 0.06014 g / 0.1223 g (x100%) = 49.17% by mass