Q. Two oxides of a metal contain $50 \%$ and $40 \%$ metal ( $M$ ) respectively. If formula of first oxide is $M O _{2}$, the formula of second oxide will be
Some Basic Concepts of Chemistry
Solution:
First oxide is $M O _{2}$ containing $50 \%$ of oxygen $(32 \,g )$
$\therefore $ Metal contained is also $32 \,g$
Atomic mass of metal $=\frac{\text { Weight }}{\text { Number of moles }}=\frac{32}{1}=32$
Second oxide:
Percentage
Mole ratio
Simple ratio
M
40
$\frac{40}{32}=1.25$
1
O
60
$\frac{60}{16}=3.75$
3
Empirical formula $= MO_{3}$
| Percentage | Mole ratio | Simple ratio | |
|---|---|---|---|
| M | 40 | $\frac{40}{32}=1.25$ | 1 |
| O | 60 | $\frac{60}{16}=3.75$ | 3 |