Determine the empirical formula of a compound which gives by combustion of $1.627\;g$ $3.254\;g$ of $CO2$ and $1.331\; g$ water, knowing that it contains only carbon, hydrogen and oxygen. Measuring the molar mass cryoscopically gives a lowering of $0.86°C$ of the freezing point of the aqueous solution realized by dissolving $4.07\;g$ in $100.0\;g$ water.
Let be $c$=compound and $e$= water:
$\mu_c$ $=$ $\frac{1000\cdot n_c}{m_e}$ $=$ $\frac{1000\cdot m_c}{M_c\cdot m_e}$
$\Delta $ $=$ $K_{fus}\mu_c$
so:
$M_c$ $=$ $\frac{K_{fus}\cdot1000\cdot m_c}{m_e\Delta T}$ $=$ $\frac{1000\cdot 1.86\cdot 4.07}{0.86\cdot 100 }$ $\approx$ $ 88.0\frac{g}{mol}$
Carbon in $1\;mole$ $=$ $88\;g$:
$m$ $=$ $\frac{3.254\cdot 12}{44}\frac{88.0}{1.627}$ $\approx$ $ 48.0\;g$ $=$ $4\;mol$
Hydrogen in $1\;mole$ $=$ $88\;g$:
$m=\frac{1.331\cdot 2}{18}\frac{88.0}{1.627}\approx 8\;g=8\;mol$
Oxygen in $1\;mole$ $=$ $88\;g$:
$88-48-8$ $=$ $32\;g$ $=$ $2\;mol$
Molecular formula:
$C_4H_8O_2$
