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

Answer: 85

Let the formula of the sample be ${\left(F{e}^{2+}\right)}_x{\left(F{e}^{3+}\right)}_{y}O$
On looking at the given formula of the compound
$x+y=0.93\dots \text{ (i) }$
Total positive charge on ferrous and ferric ions should balance the two units of negative charge on oxygen.
Therefore, $2x+3y=2$...(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 $F{e}^{2+}$ 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.