Q. $1.25$ g of a metal $(M)$ reacts with oxygen completely to produce $1.68$ g of metal oxide. The empirical formula of the metal oxide is [molar mass of $M$ and $O$ are $69.7\, g\,mol^{-1}$ and $16.0\, g\, mol{-1}$, respectively]
KVPYKVPY 2017Organic Chemistry – Some Basic Principles and Techniques
Solution:
$M+O_{2}\rightarrow MO_{2}$
Percentage of $M = \frac{1.25}{1.68}\times 100$
$ = 74.4\%$
Percentage of oxygen in oxide
$= 100- 74.4\%= 25.6\%$
Element
% of element
At mass of element
Moles of element
Simplest molar ratio
Simplest whole no.
$M$
$74.4$
$69.7$
$\frac{74.4}{69.7} =1.06$
$\frac{1.06}{1.06} =1$
$1\times 2=2$
$O$
$25.6$
$16$
$\frac{25.6}{16} =1.6$
$\frac{1.6}{1.06} =1.50$
$1.50 \times 2=3$
$\therefore $ Empirical formula of metal oxide is $M_{2}O_{3}$
| Element | % of element | At mass of element | Moles of element | Simplest molar ratio | Simplest whole no. |
|---|---|---|---|---|---|
| $M$ | $74.4$ | $69.7$ | $\frac{74.4}{69.7} =1.06$ | $\frac{1.06}{1.06} =1$ | $1\times 2=2$ |
| $O$ | $25.6$ | $16$ | $\frac{25.6}{16} =1.6$ | $\frac{1.6}{1.06} =1.50$ | $1.50 \times 2=3$ |