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Potassium Permanganate

The chromium ...

Question

The chromium in a $1.0 g$ sample of chromite $(F e C r_{2} O_{4})$ was oxidized to $C r^{+ 6}$ state by fusion with $N a_{2} O_{2}$ . The fused mixture was treated with excess of water and boiled to destroy the excess of peroxide. After acidification, the sample was treated with $50 mL$ of $0.16 M$ $F e S O_{4}$ solution.
$6.67 mL$ of $0.05 M$ $C r_{2} O_{7}^{2 -}$ solution was required to oxidize the $F e^{2 +}$ ion left unreacted. Determine the mass percentage of chromite in the original sample $[C r = 52, F e = 56]$

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Solution

Miliequivalents of $F e^{2 +}$ taken $= 50 \times 0.16 = 8$
${6 + C r}_{2} O_{7}^{2 -} + F e^{2 +} \to 0 C r + F e_{2} O_{3}$
Miliequivalents of $F e^{2 +}$ ion left $= 6.67 \times 0.05 \times 6 = 2$
Miliequivalents of $C r^{+ 6}$ in solution $= 8 - 2 = 6$
Milimoles of $C r^{+ 6}$ in solution $= 6 / 3 = 2$
Milimoles of chromite $= 2 / 2 = 1$
Molecular mass of $F e C r_{2} O_{4} = 56 + 104 + 64 = 224$
Mass of chromite in original sample $= \frac{1 \times 224}{1000} = 0.224$ g
Percentage of chromite in the original sample $= 22.4 %$

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