Two oxides of a metal contain 50% and 40% metal M respectively. If the formula of the first oxide is MO₂, the
formula of the second oxide will be

First oxide               Second oxide

M    50%                   M = 40%

O    50%                    Formula = ?

Let the atomic weight of metal = x

$\therefore$ % of M in MO₂ = $\frac{1\times \mathrm{x}\times 100}{\mathrm{x}+32}=50 \left(\mathrm{given}\right)$

Hence x = 32

$\therefore$ Empirical formula of second oxide:

$$\begin{gathered} O=\frac{60}{16}=3.75 \\ M=\frac{40}{32}=1.25 \end{gathered}$$                $\left[\begin{array}{c}\frac{3.75}{1.25}=3\\ \frac{1.25}{1.25}=1\end{array}\right]$

Thus, the formula of second oxide = MO₃