Experimentally it was found that a metal oxide has formula M0.98O. Metal M is present as M2+ and M3+ in its oxide. Fraction of the metal which exists as M3+ would be:

Consider one mole of the oxide then, we have 1 mole of M0.98O and O2-. M0.98 have both M3+ and M2+ present in the 1 mole. Let moles of M3+ be x and moles of M2+ be 0.98 - x. Now, by balancing the charge, we get 3x + 2(0.98 - x) = 2 x = 0.04. Fraction of M3+ = x / 0.98 = 0.04 / 0.98 = 4.08%