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.

Answer

`M^(2+): 91.67%; M^(3+) : 8.33`

Explanation

`M_(0.96)O_(1.00)` is a non-stoichiometric compound and contains mixture of `M^(2+)` and `M^(3+)` ions.
Let x of `M^(3+)` ions be present in the compound. This means that x `M^(2+)` ions have been replaced by `M^(3+)` ions.
`:.` No. of `M^(2+)` ions present in the oxide = 0.96-x
For electrical neutrality,
Total positive charge on cations = Total charge on anions
`:.` 2 (0.96-x)+3x=2 or 1.92+x=2 or x=0.08`
`:.` Percentage of `M^(3+)` ions =`(0.08)/(0.96) xx 100 = 8.33`
Percentage of `M^(2+)` ions = `(0.96-0.08)/(0.96)xx 100 = 91.67`