If it is known that in Fe_0.96O, Fe is present in +2 and +3 oxidation state. What is the mole fraction of Fe²⁺ in the compound ?

Charge balance 2x +3 (0.96 - x)-2 =0,

2x +2.88 - 3x -2 =0

0.88 -x =0

x=0.88

Fraction of Fe⁺² =0.88 and Fe+3 =0.12
% of Fe⁺² =$\frac{0.88}{0.96}\times 100$ =91.6% =$\frac{11}{12}$