Two oxides of a metal contain 30.0 and 27.6 per cent of oxygen respectively. If the formula of the first oxide be `M_(2)O_(3)`. Find that of the second oxide.

Explanation

Per cent of oxygen in first oxide=30.0
Per cent of metal in first oxide =70.0
Let the atomic mass of the emtal be=x g
Molecular mass of first oxide `M_(2)O_(3)=2xx x+16xx3`
`=(2x+48)g`
Percentage of oxygen by mass in the first oxide.
`=(48)/((2x+48))xx100`
`[(48xx100)/((2x+48))]=30`
So, x=56g
Determination of the formula of second oxide.

The formula of the second oxide is `M_(3)O_(4)`.