Two oxides of a metal contain 27.6% and 30% oxygen, respectively. If the formula of the first compound is M3O4, find the formula of the second compound.

Given,

In the case of the 1^{st}^{ }oxide, Mass of oxygen = 27.6

Therefore,

Mass of metal = 100 - 27.6 = 72.4

In the case of the 2^{nd} oxide, Mass of oxygen = 30

Therefore,

Mass of metal = 100 - 30 = 70

Also,

Formula of the 1^{st} oxide is M_{3}O_{4}

Therefore,

Number of atoms of metal in the 2^{nd} oxide = (3/72.4) x 70

= 2.9

Number of atoms of oxygen in the 2^{nd} oxide = (4/27.6) x 30

= 4.35

Thus,

Ratio of metal : oxygen in the 2^{nd} oxide = 2.9 : 4.35 or ≈ **2 : 3**

Hence,

The Formula of the 2^{nd} oxide is **M _{2}O_{3}.**

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