The amount of cationic vacancies per mole of ${\mathrm{Fe}}_{0.88}\mathrm{O}$ is

In non-stoichiometric iron oxide, 3 mol of ${\mathrm{Fe}}^{2+}$ ions are replaced by 2 mol of ${\mathrm{Fe}}^{3+}$ + ions and 1 mol of cationic vacancies is created. If xis the amount of Fe(III) present in the given compound, then

$x(+3)+(0.88\mathrm{mol}-x)(+2)=\left(1\mathrm{mol}\right)\left(2\right)$

This gives x = 0.24 mol

Hence, the given compound is ${\mathrm{Fe}}_{0.24}^{\left(\mathrm{ul}\right)}{\mathrm{Fe}}_{0.64}^{\left(\mathrm{II}\right)}\mathrm{O}$. The cationic vacancies will be 0.24 mol/2 = 0.12 mol