An ore of iron wustite has formula Fe(0.93)O(1). The molefraction of total iron present in the form of Fe(2+) is ?

Oxidation no of O = -2

Oxidation no of Fe were = +2

composition would have been FeO

Fe is in the ratio Fe0.93O1.00

Thus , Fe has some atoms in +2 oxidation state and some in +3 

Let the amount of Fe(II) be x and Fe(III) be y
then
x +y =0.93 ...........( given composition)
2x + 3y = 2 ..........(Charge of Fe (2+) and charge of Fe (+3)  equal to charge of the compound }
solving them we get x = 0.79 and y = 0.14
now,
% of Fe(III) = 0.14/0.93 * 100 = 15%
 :)15%

Regards

Arun (askIItians forum expert)