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

M:O : Let number of `M^(+2)=x`
96:100 `therefore ` Number of `M^(+3)`=(96-x)
As per the principle of charge neutrality 2x + 3(96-x)=200
`therefore` x=88
`therefore` % of `M^(+2)=(88xx100)/96`=91.6%
`therefore `% of `M^(+3)`=100-91.6=8.4 %