Question

The composition of a sample is $F{e}_{0.93}{O}_{1.0}$. What percentage of iron is present in the form of Fe(III)?

• A. 20%
• B. 15%
• C. 5%
• D. 10%

Answer 1

The correct option is B 15%

$F{e}_{0.93}{O}_{1.0}$ is a non stoichiometric compound. It is a mixture of $F{e}^{2+}$ and $F{e}^{3+}$. Let x atoms of $F{e}^{3+}$ ions be present in the compound for every $O^{2-}$
So, (0.93 - x) atoms of $F{e}^{2+}$ ions are present in the compound.
Total positive charge + total negative charge = 0
$\therefore x\times 3+(0.93-x)\times 2+(-2)=0$
$3x+0.93\times 2-2x=2$
$x+0.93\times 2=2$
$x=0.14$
$\therefore$ % of $F{e}^{3+}$ ions present = $\frac{0.14}{0.93}\times 100\approx 15\text{\%}$