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. $ MO_{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 $= MO$