Iron oxide has formula `Fe_"0.94"O.` What fraction of Fe exists as `Fe^(+3)` ?

Explanation

Let there are 100 oxide ions and 94 iron ions `Fe^(+x) : O^(-2) = 94 : 100`
Total negative charge on 100 `O^(2-)` ions=-200" It means as per the concept of charge neutrality, 94 iron ions should have total charge of +200. Let number of `Fe^(+3) = x `
` therefore ` Number of `Fe^(+2) = (94 -x)`
Total positive charge=3x+2(94-x)=200
On solving,x=12
` therefore % Fe^(+3) = 12/94 xx 100 = 12.8 %`