Question

Two metallic oxides contain $27.6$ % and $30.0$ % oxygen respectively. If the formula of the first oxide is $M_3{O}_4$, then that of the second will be

• A. $M_2{O}_5$
• B. $M_2{O}_3$
• C. $MO$
• D. $M{O}_2$

Answer 1

The correct option is B

$M_2{O}_3$

As first oxide is $M_2{O}_4$,
$72.4$ parts of $M=3$ atoms of $M$
and $27.6$ parts of $O=4$ atoms of $O$
$\therefore 70\text{parts of}M=\frac{3}{72.4}\times 70\text{atoms of}M=2.9\text{atoms of}M$
$30\text{parts of}O=\frac{4}{27.6}\times 30\text{atoms of}O=4.35\text{atoms of}M$
$\therefore$ ratio of $M:O$ in the second oxide
$=2.9:4.35=1:1.5$ or $2:3$
Hence, the formula is $M_2{O}_3$