The ratio of ${\mathrm{Fe}}^{3+}\text{ and }{\mathrm{Fe}}^{2+}\text{ ions in }{\mathrm{Fe}}_{0.9}{ \mathrm{S}}_{1.0}\text{ is }$ 

Fe³⁺=x 
Fe²⁺=0.9−x 
Using charge balance 
Fe_0.9​S_1.0​ =Charge on neutral compound =0 
(Fe²⁺& Fe³⁺) 
⇒3x+2(0.9−x)=1×2 
3x+1.8−2x=2 
x=2−1.8 
x=0.2 
Thus the no. of Fe⁺² ion =0.93−x 
=0.9−0.2 
=0.7 
Ratio Fe³⁺​/Fe²⁺=0.2​ /0.7
=0.28