When 19.8 g of an organic compound known to be 54.53% C , 9.2% H , and 36.32% O by mass is dissolved in 711.5 g of cyclohexane, the freezing point is 0.02 ∘C

Question

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. 𝐾fvalues for various solvents are given in the colligative constants table.

Answer 1

∆T = imK

∆T = change in freezing point = 6.59º - 0.02º = -6.57º

i = van't Hoff factor = 1 (since it's an organic compound we assume no dissociation or ionization)

m = molality = moles compound / kg cyclohexane = ?

K = freezing constant = -20.2 (found on Wikipedia, could not access the link provided)

Solving for m we have...

m = -6.57 / (1)(-20.2)

m = 0.325 moles/kg

Since we have 0.7115 kg cyclohexane, we can find moles...

0.325 mol/kg x 0.7115 kg = 0.231 moles of unknown

molar mass = 19.8 g /0.231 moles = 85.7 g/mole

We can find the molecular formula from the empirical formula and the molar mass. So first, we need the empirical formula.

C: 54.53 g x 1 mol/12 g = 4.5 moles

H: 9.2 g x 1 mol/1 g = 9.2 moles

O: 36.32 g x 1 mol/16 g = 2.27

This gives us C₂H₄O as the empirical formula with a molar mass of 44 g/mol

Since molar mass of the unknown is 86 g/mol, the molecular formula of the unknown is C₄H₈O₂

You could also have arrived at this same molecular formula by using the calculated molar mass and % composition as follows:

54.53% x 86 g x 1 mol/12 g = 3.9 moles C

9.2% x 86 g x 1 mol/1 g = 7.9 moles H

36.32% x 86 g x 1 mol/16 g = 1.95 moles O

These become C₄H₈O₂ as above