Question

Two oxides of a metal contain $27.6$% and $30$% of oxygen respectively. If the formula of the first oxide is $M_3{O}_4$, then formula for the second oxide is:

• A. $MO$
• B. $M{O}_2$
• C. $M_2{O}_3$
• D. $M_{2}O$

Answer 1

The correct option is C $M_2{O}_3$

First oxide of a metal $M_3{O}_4$ contain $27.6$% oxygen.

The atomic weight of $O=16g/mol$

$\frac{4\times 16}{3M+4\times 16}\times 100=27.6$%

$M=55.96g/mol$

Second oxide of metal contains $30$ % oxygen.

The proportion of $M$ and $O$ in oxide $M_x{O}_y$ would be

$\frac{56x}{16y}=\frac{70}{30}$

$\frac{x}{y}=\frac{2}{3}$

Hence, the formula of the second oxide is $M_2{O}_3.$