The composition of a sample of Wustite is `Fe_(0.93)O_(1.00)`. What percentage of the iron is present in the form of `Fe (III)`?

Explanation

The composition is ` Fe_(0-93)` interead of FeO because some ` Fe^(2+)` ions have been replaced by ` Fe^(3+)` ions. Let us first calcuate the number of ` Fe^(2+) and Fe^(3+)` ions present. The formula ` Fe_(0.93) O _(1.00)` implies that 93 Fe atoms are combined with 100 O - atoms. Out of 93 Fe atoms , suppose Fe atoms present as ` Fe^(3+)`= x , then `Fe^(2+) = 93 -x ` , as the compound is neutral, total charge on ` Fe^(2+) and Fe^(3+)` ions = total charge on ` O^(2-)` ions. Thus,
` 3 xx x + 2( 9 3-x) = 2xx 100 or 3 x +1 86 -2x = 200 or x=14 , i.e. Fe^(3+) =14`
Hence, ` Fe^(2+) = 93 - 14=79`
Thus, out of 93 Fe atoms, Fe presnet as ` Fe^(3+)=14 ` : % age of Fe presnet as Fe ( III) = ` 14/93 xx 100 = 15%`