Question

The complete combustion of $2.40g$ of a compound of carbon, hydrogen and oxygen yielded $5.46g$ of $C{O}_2$ and $2.23g{H}_{2}O.$ When $8.69g$ of the compound was dissolved in $281g$ of water, the freezing point depression was found to be $0.97^{\circ }C$. Thus, molecular formula of the organic compound is $\left[K_f\left(H_{2}O\right)=1.86^{\circ }mo{l}^{-1}kg\right]$

• A. $C_2{H}_4{O}_2$
• B. $C_2{H}_{6}O$
• C. $C_6{H}_{12}{O}_6$
• D. $C_3{H}_{6}O$

Answer 1

Answer: (D)

$\underset{12g}{\mathrm{Carbon}}\stackrel{[O]}{⟶}C{O}_2$

$C\%=\frac{12\times 100\times \text{ mass of }C{O}_2}{44\times mass\text{ of substance }}$

$=\frac{12\times 100\times 5.46}{44\times 2.40}$

$=62.05\%$

$2H\underset{g}{\stackrel{[O]}{⟶}}\underset{18g}{H_{2}O}$

$H\%=\frac{2\times \text{ mass of }{H}_{2}O\times 100}{18\times \text{ mass of substance }}$

$=\frac{2\times 2.23\times 100}{18\times 2.40}$

$=10.32\%$

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 $=C_3{H}_{6}O$

$\Delta T_f=\frac{1000K_f{w}_1}{m_1{w}_2}$

$\therefore m_1=\frac{1000K_f{w}_1}{w_2\Delta T_f}$

$m_1($ 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 $C_3{H}_{6}O$.