Experimentally, it was found that a metal oxide has formula ${\mathrm{M}}_{0.98}\mathrm{O}$. Metal M is present as M²⁺ and M³⁺ in its oxide. Fraction of the metal which exists as M³⁺ would be

${\mathrm{M}}_{0.98}\mathrm{O}$
It represents stoichiometric defect.If number of moles of M²⁺= x
Then M³⁺ =(0.98 - x)Thus, by charge balance of ${\mathrm{M}}_{0.98}\mathrm{O}$
$\begin{array}{l}(\mathrm{x}\times 2)+(0.98-\mathrm{x})\times 3-(1\times 2)=0-\mathrm{x}=2-0.98\times 3\mathrm{x}=0.94\\ {\mathrm{M}}^{3+}=0.98-0.94=0.04\text{ Percentage of }{\mathrm{M}}^{3+}=\frac{0.04}{0.98}\times 100=4.08\mathrm{\%}\end{array}$