Question

when 12.3 g of an organic compound known to be 62.04% c, 10.4% h, and 27.55% o by mass is dissolved in 644.4 g of cyclohexane, the freezing point is 3.17 °c. the normal freezing point of cyclohexane is 6.59 °c. what is the molecular formula for the organic compound? assume that the organic compound is a molecular solid and does not ionize in water. $k_f$ values for various solvents are given in the colligative constants table.
molecular formula: cho

Explanation:

Step1: Calculate freezing point depression

$\Delta T_f = T_f^\circ - T_f = 6.59^\circ\text{C} - 3.17^\circ\text{C} = 3.42^\circ\text{C}$

Step2: Find $K_f$ for cyclohexane

From colligative constants, $K_f(\text{cyclohexane}) = 20.8^\circ\text{C}\cdot\text{kg/mol}$

Step3: Calculate molality of solution

$m = \frac{\Delta T_f}{K_f} = \frac{3.42^\circ\text{C}}{20.8^\circ\text{C}\cdot\text{kg/mol}} \approx 0.1644\text{ mol/kg}$

Step4: Calculate moles of solute

Mass of solvent = $644.4\text{ g} = 0.6444\text{ kg}$
$\text{Moles of solute} = m \times \text{mass of solvent} = 0.1644\text{ mol/kg} \times 0.6444\text{ kg} \approx 0.1059\text{ mol}$

Step5: Calculate molar mass of solute

$\text{Molar mass} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{12.3\text{ g}}{0.1059\text{ mol}} \approx 116.1\text{ g/mol}$

Step6: Find empirical formula

Assume 100g of compound:

• Moles of C: $\frac{62.04\text{ g}}{12.01\text{ g/mol}} \approx 5.166\text{ mol}$
• Moles of H: $\frac{10.4\text{ g}}{1.008\text{ g/mol}} \approx 10.32\text{ mol}$
• Moles of O: $\frac{27.55\text{ g}}{16.00\text{ g/mol}} \approx 1.722\text{ mol}$

Divide by smallest mole value (1.722):

• C: $\frac{5.166}{1.722} = 3$
• H: $\frac{10.32}{1.722} \approx 6$
• O: $\frac{1.722}{1.722} = 1$

Empirical formula: $\text{C}_3\text{H}_6\text{O}$, empirical mass = $3(12.01)+6(1.008)+16.00 = 58.08\text{ g/mol}$

Step7: Find molecular formula multiple

$n = \frac{\text{Molar mass}}{\text{Empirical mass}} = \frac{116.1\text{ g/mol}}{58.08\text{ g/mol}} \approx 2$
Multiply empirical formula by $n$: $(\text{C}_3\text{H}_6\text{O})_2$

Answer:

$\text{C}_6\text{H}_{12}\text{O}_2$