Q. The complete combustion of $2.40 \,g$ of a compound of carbon, hydrogen and oxygen yielded $5.46 \,g$ of $CO _{2}$ and $2.23 \,g \,H _{2} O .$ When $8.69\, g$ of the compound was dissolved in $281 \,g$ of water, the freezing point depression was found to be $0.97^{\circ} C$. Thus, molecular formula of the organic compound is $\left[K_{f}\left( H _{2} O \right)=1.86^{\circ} mol ^{-1} \,kg \right]$
Solutions
Solution:
$\underset{12 g }{\text { Carbon }} \stackrel{[ O ]}{\longrightarrow } CO _{2}$
$C \%=\frac{12 \times 100 \times \text { mass of } CO _{2}}{44 \times mass \text { of substance }}$
$=\frac{12 \times 100 \times 5.46}{44 \times 2.40}$
$=62.05 \%$
$2 H \underset{ g }{\stackrel{[ O ]}{\longrightarrow }} \underset{18 g }{ H _{2} O }$
$H \%=\frac{2 \times \text { mass of } H _{2} O \times 100}{18 \times \text { mass of substance }}$
$=\frac{2 \times 2.23 \times 100}{18 \times 2.40}$
$=10.32 \%$
Oxygen (by difference) $=100-(62.05+10.32)$
$=27.63 \%$
Element
%
Moles
Ratio
C
62.05
5.17
3
H
10.32
10.32
6
O
27.63
1.73
1
Empirical formula $= C _{3} H _{6} O$
$\Delta T_{f}=\frac{1000 K_{f} w_{1}}{m_{1} w_{2}}$
$\therefore m_{1}=\frac{1000 K_{f} w_{1}}{w_{2} \Delta T_{f}}$
$m_{1}($ solute $)=\frac{1000 \times 1.86 \times 8.69}{281 \times 0.97}=59.3$
$m_{1}\left( C _{3} H _{6} O \right)=58$
Thus, solute is $C _{3} H _{6} O$.
| Element | % | Moles | Ratio |
|---|---|---|---|
| C | 62.05 | 5.17 | 3 |
| H | 10.32 | 10.32 | 6 |
| O | 27.63 | 1.73 | 1 |