Question

An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percentages of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of 25.208C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g of water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Short answer

The empirical formula of the compound is ${\text{C}}_2{\text{H}}_4{\text{O}}_3$, the molar mass of the compound will be 151g whereas the molecular formula of the compound ${\text{C}}_4{\text{H}}_8{\text{O}}_6$.

Step-by-step solution

Empirical formula

According to the question, it is given that

The mass percent of carbon = 31.57%

The mass percent of hydrogen = 5.30%

So, the mass percent of oxygen = $100\text{-}(31.57+5.30)$ = 63.13

The empirical formula of the compound will be:

Atom	No. of Moles	Ratio	Whole ratio
C	31.57/12 = 2.63	2.63/2.63 = 1	2
H	5.30/1 = 5.30	5.30/2.63 = 2	4
O	63.13/16 = 3.94	3.94/2.63 = 1.5	3

Therefore, according to the ratio, the empirical formula of the compound will be ${\text{C}}_2{\text{H}}_4{\text{O}}_3$

Molar mass

It is given that

$\Delta \text{T = 5.20}°\text{C}$

${\text{k}}_{\text{f}}\text{= 1.86}°\text{C/m}$

$\text{Molality =}\frac{\text{Mass of solute}\times \text{1000}}{\text{Molar mass}\times \text{Mass of solvent}}$

Thus, the relation between freezing point and molar mass will be given by:

$\Delta \text{T =}\frac{{\text{k}}_{\text{f}}\times \text{Mass of solute}\times \text{1000}}{\text{Molar mass}\times \text{Mass of solvent}}$

$\text{5.20 =}\frac{\text{1.86}\times \text{10.56}\times \text{1000}}{\text{Molar mass}\times \text{25}}$

Molar mass == 151g

Molecular formula

The relation between empirical and molecular formula is given by

$\text{n =}\frac{\text{Molecular formula}}{\text{Empirical formula}}$

$\text{n =}\frac{151}{76} \text{= 1.98}~\text{2}$

Therefore, the molecular formula will be $\text{2}\times {\text{C}}_2{\text{H}}_4{\text{O}}_3$= ${\text{C}}_4{\text{H}}_8{\text{O}}_6$