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

In the first oxide, oxygen = 27.6, metal = 100-27.6=72.4 parts by mass.
As the formula of the oxide is `M_(3)O_(4)`, this means
72.4 parts by mass of metal = 3 atoms of metal and 4 atoms of oxygen = 27.6 parts by mass.
In the second oxide, oxygen = 30.0 parts by mass and metal = `100-30=70` parts by mass.
But 72.4 parts by mass of metal = 3 atoms of metal
`therefore" 70 parts by mass of metal"=(3)/(72.4)xx70" atoms of metal"=2.90" atoms of metal"`
Also, 27.6 parts by mass of oxygen = 4 atoms of oxygen
`therefore" 30 parts by mass of oxygen "=(4)/(27.6)xx30` atoms of oxygen = 4.5 atoms of oxygen
Hence, ratio of `M:O` in the second oxide = `2.90:4.35=1:1.5=2:3`
`therefore" Formula of the metal oxide is "M_(2)O_(3)`.