When 11.1 g of an organic compound known to be 55.81% C , 7.0% H , and 37.17% O by mass is dissolved in 909.6 g of benzene, the freezing point is 5.01 ∘C .

Question

The normal freezing point is 5.49 ∘C . What is the molecular formula for the organic compound?

Answer 1

Use ∆T = imK

∆T = change in freezing point = 5.49º - 5.01º = 0.48º

i = van't Hoff factor = 1 for an organic compound that does not dissociate or ionize (non electrolyte)

m = molality = mols solute / kg solvent = see below

K = freezing constant for benzene = 5.12º/m (I looked this up. It should have been provided)

Solving for m, we have...

m = ∆T / (i)(K) = 0.48º / (1)(5.12º/m)

m = 0.0938 mols/kg

To find moles, we have...

0.0938 mols / kg unknown x 0.9096 kg =0.0853 moles of unknown

molar mass = grams / moles = 11.1 g / 0.-0853 moles

molar mass = 130 g / mole

To find the molecular formula, we must first find the empirical formula, and then determine how many empirical formulas "fit into" the molar mass.

If we have11.1 g of unknown, we have...

11.1 g x 55.81% C = 6.19 g C x 1 mol C / 12 g = 0.516 mols C

11.1 g x 7.0% H = 0.777 g H x 1 mo H / 1 g = 0.777 mols H

11.1 g x 37.17% O = 4.126 g O x 1 mol O / 16 g = 0.258 mols O

Dividing all by 0.258 to try to get whole number, we have...

0.516 / 0.258 = 2 mols C

0.777 / 0.258 = 3 mols H

0.258 / 0.258 = 1 mol O

Empirical formula = C₂H₃O

molar mass = 2x12 + 3x1 + 1x16 = 43 g / mol

To find molecular formula, divide this into the molar mass: 130 / 43 = 3

Molecular formula = C₆H₉O₃