Question

Two oxides of a metal (M) have 20.12% and 11.19% oxygen respectively. The formula of the first oxide is MO. Determine the formula of the second oxide.

Answer 1

Let two oxides be MO and MO_x.
Also, let atomic mass of the metal be m.
For first oxide MO,
(16 + m) parts of oxide contain 16 parts of oxygen.
Parts of oxygen present in 100 parts of oxide = $\frac{16}{16+m}\times 100=20.12$
On solving for m, we get:
m = 63.52
Similarly, for second oxide MO_x,
[m + x(16)] parts of oxide contain 16x parts of oxygen.
Parts of oxygen present in 100 parts of oxide = $\frac{16x}{16x+m}\times 100=11.19$
Substituting the value of m in the above relation and solving for m, we get:
$$\begin{gathered} \frac{16x}{16x+63.52}\times 100=11.19 \\ x=0.498\approx 0.5 \end{gathered}$$

Or,
x = $\frac{1}{2}$
So, molecular formula for the oxide MO_x is $M{\mathrm{O}}_{\frac{1}{2}}$ or M₂O.