Chemistry help!

Question

A solution contains 10.00 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -5.99 ∘C. The mass percent composition of the compound is 38.70% C, 9.74% H, and the rest is O.

 

What is the molecular formula of the compound?

Answer 1

To get molecular formula for the unknown, you need the following

1. molar mass in gram/mole.  You were given the mass, so you need the number of moles. The number of moles can be calculated from the freezing point information.

2. The empirical formula of the unknown which can be obtained from the given mass percent composition. 

 

From the freezing point of the solution formula ΔT = k_f m where kf = -1.86°C/m and ΔT = -5.99°C, the molality of the solution is

m = - 5.99/-1.86 = 3.22 moles of solute/kg of solvent (water).

 

The number of moles of the unknown compound in the solution is

n = (3.22 mol/kg)(0.05 kg)= 0.161 mole

 

So the molar mass of the unknown compound is

M = mass/n = 10.0/0.161 = 62.1 g

 

For the empirical formula, for every 100g of the solution, there are  38.7 g of C, 9.74 g of H and 100-38.7 - 9.74 = 51.6 g of O in the solution.

So the number of moles of C, H and O are

C: 38.7/12 = 3.225

H: 9.74/1 = 9.74

O: 51.6/16 = 3.225

 

Since 9.74/3.225 ≅ 3, the empirical formula for the unknown is CH₃O, and the corresponding empirical mass is 31 g.

 

Since molar mass/empirical mass = 62g/31g = 2, the molecular formula of the unknown is

 

C₂H₆O₂