Question

Two oxides of a metal contain $36.4\%$ and $53.4\%$ of oxygen by mass respectively. If the formula of the first oxide is $M_{2}O$, then that of the second is

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

Answer 1

Answer: (B)

For I oxide Oxygen $=36.4$%
Metal $=100-36.4=63.6$
Given, formula of oxide $=M_{2}O$
$\therefore 63.6\%$ of metal $=2$ atoms of metal
and $36.4\%$ of oxygen $=1$ atom of oxygen For II oxide Oxygen
$=53.4\%$
Metal $=100-53.4=46.6\%$
$\because 63.6\%$ of metal $=2$ atoms of metal
$\therefore$ $46.6\%$ of metal
$=\frac{2\times 46.6}{63.6}$
$=1.46$ atoms of metal Again
$\because 36.4\%$ of oxygen $=1$ atom of oxygen
$\therefore$ $53.4\%$ of oxygen
$=\frac{1\times 53.4}{36.4}$
$=1.46$ atoms of oxygen Ratio of metal and oxide
$=1.46:1.46$
$=1:1$
Hence, formula of metal oxide $=MO$