The ratio of Fe³⁺ and Fe²⁺ ions in Fe_0.9S_1.0 is

$\mathrm{Formula}: {\mathrm{Fe}}_{0.9}{\mathrm{O}}_{1.0} \cong {\mathrm{Fe}}_{90}{\mathrm{O}}_{100}$

Let,

Number of moles of Fe⁺² ions =X 

Then, number of moles of Fe⁺³ = [ 90-X ]

we now that,

Sum of the charges of all the atoms in a compound = 0

$$\begin{gathered} \mathrm{X}\left[+2\right]+\left[90-\mathrm{X}\right]+3+100\left[-2\right]=0 \\ 2\mathrm{X}-3\mathrm{X}+270-200=0 \\ 70-\mathrm{X}=0 \\ \mathrm{X}\left[\mathrm{Moles} \mathrm{of} {\mathrm{Fe}}^{+2} \mathrm{ions}\right]=70 \\ \mathrm{Moles} \mathrm{of} {\mathrm{Fe}}^{+3} \mathrm{ions}=90-\mathrm{X}=90-70=20 \\ \frac{{\mathrm{Fe}}^{+3}}{{\mathrm{Fe}}^{+2}}=\frac{20}{70}=0.28 \end{gathered}$$