Two oxides of a metal (M) contain 22.53% and 30.38% of oxygen. If the second oxide be `M_(2)O_(3)`, find the formula of first oxide.

Explanation

Let us-consider, atomic mass of M is a
Amount of oxygen in `M_(2)O_(3)=(48)/(2a+48)xx100=30.38%`
or, a=55
Let us consider, formula of the first oxide is `M_(2)O_(x)`
amount of oxygen in `M_(2)O_(x)=(16x)/(2a+16x)xx100=22.53%`
or, `(16x)/(110+16x)=0.2253%" "therefore x~~2`.
`therefore`Formula of the first oxide is `M_(2)O_(3)` or MO.