Question

Two oxides of metal contain 30.0 and 27.6 per cent of oxygen respectively. If the formula of the first oxide is $M_2{O}_3$, then find the formula of the second oxide.

• A. $M{O}_4$
• B. $M_5{O}_4$
• C. $M_{3}O$
• D. $M_3{O}_4$

Answer 1

The correct option is D $M_3{O}_4$

% oxygen = $\frac{\text{no. of oxygen atom in molecule}\times 16}{molarmass}\times 100$

30 =$\frac{3\times 16}{2M^{\prime }+48}\times$ 100

2M' +48 = 160

M'=Atomic mass of M =56

For 2nd oxide M${}_x{O}_y$

27.6=$\frac{16y}{56x+16y}\times 100$

1545.5 x = 1158.4y

$\frac{x}{y}$ = $\frac{3}{4}$

2nd oxide is M${}_3{O}_4$.

Hence, option $D$ is correct.