Question

The osmotic pressure of a solution containing 1.02 g of an unknown compound dissolved in 175.0 mL of solution at 25 ∘C is 1.93 atm. The combustion of 23.40 g of the unknown compound produced 41.70 gCO2 and 17.07 gH2O. What is the molecular formula of the compound (which contains only carbon, hydrogen, and oxygen)?

Answer 1

We know that,

Osmotic Pressure = CRT

=> 1.93 = C x 0.0821 x 298

=> C = 0.079 M

Molarity = Moles / Volume

=> 0.079 = Moles of solute / 0.175 (Volume of solution in liters)

=> Moles of Solute = 0.079 x 0.175 = 0.0138 moles

Given,

Mass of solute = 1.02 g

=> Molar Mass of Solute = 1.02 / 0.0138 = 73.9 g / mol

All the C and H atoms in the product are from the solute

Moles of C in the product = 41.7 / 44 = 0.948

=> Mass of C in solute = 0.948 x 12 = 11.376 g

Moles of H in product = 2 x 17.07 / 18 = 1.9

=> Mass of H in solute = 1.9 x1 = 1.9 g

and Mass of O = 23.4 - (1.9 + 11.376) = 10.124 g

Moles of O = 10.124 / 16 = 0.633

Summary:

Moles of C = 0.948

Moles of H = 1.9

Moles of O = 0.633

Dividing each of the above by 0.633

Moles of C / 0.633 = 0.948 / 0.633 = 1.5

Moles of H / 0.633 = 1.9 / 0.633 = 3

Moles of O / 0.633 = 1

Therefore the emperical and molecular formula of the compound is,

C3H6O2