Question

1.25 g of a metal $\left(M\right)$ reacts with oxygen completely to produce 1.68 g of metal oxide. The empirical formula of the metal oxide is:
[Molar mass of $M$ and $O$ are 69.7 g ${mol}^{-1}$ and 16.0 g ${mol}^{-1}$, respectively]

Answer 1

$M+O_2⟶M_x{O}_y$

% of $M$ in oxide= $\frac{1.25}{1.68}=0.74\times 100=74$

% of $O$ in oxide= $1-0.74=0.26\times 100=26$

$M:\frac{74}{69.7}=1.06$

$O:\frac{26}{16}=1.625$

Multiply by $2$, we will get simplest whole number ratio, so, Empirical formula is $M_2{O}_3$ .