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

Let `Mn^(2+)` ions in the crystal be x so that `M^(3+)` ions in the crystal will be `0.96-x`. Total charge on the compound must be zero, so that
`+2xx x+3(0.96-x)-2=0`
`2x-2.88-3x-2=0`
`x=0.88`
`%" of M"^(2+)=(0.88)/(0.96)xx100=91.67`
`%" of "M^(3+)=100-91.67=8.33`.