# How to calculate the composition of a mixture of iron oxides containing inert impurities with titration data?

An $$\pu{8 g}$$ sample of $$\ce{Fe3O4}$$ and $$\ce{Fe2O3}$$ containing some inert impurity was treated with excess of aqueous $$\ce{KI}$$ in acidic medium, which converted all iron to $$\ce{Fe^2+}$$. The resulting solution was then diluted to $$\pu{50 mL}$$.

$$\pu{10 mL}$$ of it was taken and the liberated iodine required $$\pu{7.2 mL}$$ of $$\pu{1 M}$$ sodium thiosulphate to reduce all iodine.

Another $$\pu{25 mL}$$ was taken and the iodine was removed. The remaining solution required $$\pu{4.2 mL}$$ of $$\pu{1 M}$$ $$\ce{KMnO4}$$ to oxidize all $$\ce{Fe^2+}$$.

Calculate the percentage of composition of the mixture.

The initial solution already contains $$\ce{Fe^2+}$$ from $$\ce{FeO}$$, which does not react until the second titration.

Working backwards, milliequivalents (m.eq) of $$\ce{Fe^2+}$$ is $$5\times 4.2\times 1 = 21$$ in $$\pu{25 mL}$$ solution. So in $$\pu{50 mL}$$, the amount of $$\ce{Fe^2+}$$ is $$42$$ m.eq.

Similarly, the m.eq of iodine from titration with thiosulphate is $$36$$.

From the first reaction (with $$\ce{KI}$$), $$\text{m.eq of iodine liberated = m.eq of \ce{Fe^2+} formed.}$$ The excess $$\ce{Fe^2+}$$ is from $$\ce{FeO}$$ which is present in $$\ce{Fe3O4}$$.

The amount of $$\ce{FeO}$$ is $$6$$ m.eq. Hence its weight is $$\frac{6}{1000}\times\frac{72}{2}=\pu{0.216 g}$$ since $$n$$-factor is $$\mathrm{2}$$.

Since $$\pu{1 mol}$$ of $$\ce{Fe3O4}$$ contains $$\pu{72 g}$$ of $$\ce{FeO}$$, the mass of $$\ce{Fe3O4}$$ is $$\pu{0.6945 g}$$. But this answer is wrong. Is there anything wrong in my procedure?