$\begin{array}{1 1}\large\frac{9}{8}\\\large\frac{5}{8}\\\large\frac{7}{8}\\\large\frac{3}{8}\end{array} $

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Let there be 1 mol of iron atom.

Amount of Fe(III)=$\large\frac{1}{4}$ mol

Amount of Fe(II)=$\large\frac{3}{4}$ mol

Total positive charges =$\large\frac{1}{4}$$(3)+\large\frac{3}{4}$$(2)$

$\Rightarrow \large\frac{9}{4}$ mol

Let $x$ be the amount of oxygen atoms,we have

Total negative charges =$x(2)$

To satisfy electrical neutrality,we have

Total positive charges =Total negative charges

$\large\frac{9}{4}$ mol = $2x$

$x=\large\frac{9}{8}$ mol

Hence composition is $FeO_{9/8}$ or $Fe_8O_9$

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