Analysis shows that a metal oxide has an empirical formula M0.96OM0.96O. Calculate the percentage of M2+M2+ and M3+M3+ ions in this crystal.
We know that the ratio of M2+M2+ and O2-O2- ion in the pure metal oxide sample has a ratio = 1:1 Let us assume that x ions of M2+M2+ are replaced with M3+M3+ ions. Number of M2+M2+ ions present will be = (0.96 – x) As we know that oxides are neutral in nature, the total charge on M atoms will be equal to the charge on oxygen atoms. This can be written as: 2(0.96 – x) +3 x = 2 x = 0.08 So, the percentage of M3+M3+ ions will be = 0.080.96 × 100 = 8.3 0.080.96 × 100 = 8.3 Now the percentage of M2+M2+ions will be = 100 – 8.3 = 91.7%
