Q. Complete combustion of $0.858\, g$ of compound $X$ gives $2.63\, g$ of $CO_2$ and $1.28 \,g$ of $H_2O$. The lowest molecular mass $X$ can have
Some Basic Concepts of Chemistry
Solution:
$\underset{\text{0.858 g}}{[CH]}+O_{2} \to \underset{\text{2.63}}{CO_{2}}+ \underset{\text{1.28}}{H_{2}O}$
% age $C=\frac{12}{44}\times \frac{2.63}{0.858}\times100$
$=83.59$
% age $H=\frac{2}{18}\times \frac{2.63}{0.858}\times100$
$=16.57$
Element
$\%$
At.
mass
$\%$ At. mass
Simplest
ratio
Simplest
whole
no. ratio
C
83.59
12
$\frac{83.59}{12}=6.97$
$\frac{6.97}{6.97}=1$
3
H
16.57
1
$=\frac{16.57}{1}=16.57$
$\frac{16.57}{6.97}=2.37$
7
$\therefore E.F. = C_{3}H_{7}$
$\therefore E.F.$ mass (lowest mol mass)
$=12\times 3+7\times 1=43$
| Element | $\%$ | At. mass | $\%$ At. mass | Simplest ratio | Simplest whole no. ratio |
|---|---|---|---|---|---|
| C | 83.59 | 12 | $\frac{83.59}{12}=6.97$ | $\frac{6.97}{6.97}=1$ | 3 |
| H | 16.57 | 1 | $=\frac{16.57}{1}=16.57$ | $\frac{16.57}{6.97}=2.37$ | 7 |