Analysis shows that a metal oxide has the empirical formula `M_(0.96)O_(1.00)`. Calculate the percentage of `M^(2+)` and `M^(3+)` ions in the sample.

Explanation

The ratio of `M^(2+)` and `O^(2-)` ions in the pure sample of metal oxide = `1:1`. Let x ions `M^(2+)` be replaced by `M^(3+)` ions in the given sample.
No. of `M^(2+)` ions present = `(0.96-x)`
Since the oxide is neutral in nature,
Total charge on M atoms = Charge on oxygen atoms
`2(0.96-x)+3x=2`
`1.92-2x+3x=2`
`x=2-1.92=0.08`
`%"of M"^(3+)"ions in the metal oxide"=("No. if M"^(3+)"ions")/("Total no. of M atoms")xx100=(0.08)/(0.96)xx100=8.33%`
`%"of M"^(2+)"ions in the metal oxide"=(100-8.33)=91.67%`