Two oxides of a metal contain `27.6%` and `30.0%` of Oxygen, respecttively. If the formula of the first be `M_(3) O_(4)`. Find that of the second.

Explanation

`{:("First oxide","Second oxide"),("Oxygen"=27.6%, "Oxygen"=30%),("Metal"=72.4%,"Metal"=70%):}`
Formula of first oxide `=M_(3)O_(4)`
Suppose the atomic weight of metal =x
Percentage of metal in the compound `M_(3)O_(4)`
`=(3x)/(3x + 64) xx 100`
`therefore (3x)/(3x + 64) xx 100 = 72.4`
or `300 x = 217.2 + 4633.6`
Now in the second oxide, metal and oxygen are 70% and 30%. Therefore, their atomic ratio will be
M:O
`70/56 : 30/16`
`1.25 : 1.875`
or `1:1.5`
or `2:3`
Therefore, formula of the compound `=M_(2)O_(3)`.