Question

The osmotic pressure of a solution containing 2.16 g of an unknown compound dissolved in 175.0 mL of solution at 25 ºC is 2.65 atm. The combustion of 25.05 g of the unknown compound produced 58.02 g CO2 and 19.80 g H2O. The compound contains only carbon, hydrogen, and oxygen.

What is the molecular formula of the compound?

Express your answer as a chemical formula

Answer 1

First work out the empirical formula of the compound:
Mass of C in 58.02 g CO₂ = (12/44) 58.02 = 15.824 g
Mass of H in 19.80 g H₂O = (2/18) 19.80 = 2.20 g
Mass of O in sample = 25.05 - (15.824 + 2.20) = 7.026 g

Divide each mass by respective atomic mass =
C = 15.824/12 = 1.319
H = 2.20/1 = 2.20
O = 7.026/16 = 0.439

Divide through by smallest value
C = 1.319/0.439 = 3
H = 2.20/0.439 = 5
O = 0.439/0.439 = 1

Empirical formula = C₃H₅O

Use the other data to calculate the molarity of the solution - use osmotic pressure equation:

Osmotic pressure, = iMRT
where
i is the dimensionless van 't Hoff factor, M is the molarity, R = 0.0821 LatmK^-1mol^-1 is the gas constant, T is the thermodynamic (absolute) temperature = 25^oC = 298 K

Substituting above,
2.65 = 1 M 0.0821 298
M = 2.65/(0.0821 298)
M = 2.65/24.46
M = 0.108 M

You have 2.16 g solute in 175 mL solution,

Molarity = (mass/molar mass) (1/volume in litres)

0.108 = (2.16/molar mass) (1/0.175)

Molar Mass = 114.28 g/mol

Formula mass of empirical formula = C₃H₅O = 57 g
So, 114/57 = 2

There are two formula units per molecule = molecular formula = (C₃H₅O)₂ = C₆H₁₀O₂ which is the molecular formula.