Question

Actual formula of a sample of ferrous oxide is $F_{e_{0.93}} O_{1.00}$. What is the percentage fraction of $F e^{2+}$ ions among all the iron ions in this sample? Give answer after rounding off to the nearest integer value.

Answer 1

Let the formula of the sample be $\left( Fe ^{2+}\right)_x\left( Fe ^{3+}\right)_y O$
On looking at the given formula of the compound
$x+y=0.93 \ldots \text { (i) }$
Total positive charge on ferrous and ferric ions should balance the two units of negative charge on oxygen.
Therefore, $2 x+3 y=2$...(ii)
$\Rightarrow x+\frac{3}{2} y=1 \text {...(iii) }$
On subtracting equation (i) from equation (iii) we have
$\frac{3}{2} y-y=1-0 \cdot 93$
$\Rightarrow \frac{1}{2} y=0 \cdot 07 \Rightarrow y=0 \cdot 14$
On putting the value of $y$ in equation (i), we get
$x+0.14=0.93 $
$\Rightarrow x=0.93-0.14 \Rightarrow x=0.79$
Fraction of $Fe ^{2+}$ ions present in the sample $=\frac{0.79}{0.93}=0 \cdot 849 \cong 85 \%$
Metal deficiency defect is present in the sample because iron is less in
amount than that required for stoichiometric composition.