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

Step I : Calculation of the atomic mass of the metal
In the first oxide of the metal,
Percentage of oxygen = 27.6
Percentage of metal `=100-27.6 = 72.4`
Formula of the metal oxide `= M_(3)O_(4)`
Let the atomic mass of the metal = x
Percentage by weight of the metal in the oxide `M_(3)O_(4)=(3x xx100)/((3x+64))=72.4` (given)
`:. (3x)/(3x+64)=0.724or 3x-2.17x=46.366`
`:.0.828x=46.336orx=(46.336)/(0.828)=56`
The atomic mass of the metal = 56
Step II. Calculation of the formula of the second oxide
`{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("M",70.0,56,(70.0)/(56)=1.25,(1.25)/(1.25)=1,2),("O",30.0,16,(30.0)/(16)=1.88,(1.88)/(1.25)=1.5,3):}`
The simplest ratio of the elements are : `M:O : : 2 :3`
`:.` Empirical formula of metal oxide `= M_(2)O_(3)`.