Question

The osmotic pressure of a solution containing 1.44 g of an unknown compound dissolved in 175.0 mL of solution at 25 ∘C is 1.93 atm. The combustion of 32.88 g of the unknown compound produced 55.60 gCO2 and 22.76 gH2O. What is the molecular formula of the compound (which contains only carbon, hydrogen, and oxygen)? Express your answer as a chemical formula.

Answer 1

moles of unknown = 1.44 g

volume = 175 ml = 0.175 L

temperature = 298 K

osmatic pressure = 1.93 atm

P = M R T

1.93 = M x 0.0821 x 298

Molarity = 0.0789 M

molarity = moles / volume

0.0789 = moles / 0.175

moles = 0.0138

moles = mass / molar mass

0.0138 = 1.44 / molar mass

molar mass of unknown compound = 104.31 g/mol

moles of CO2 = 55.60 / 44 = 1.264

moles of C = 1.264

mass of C = 1.264 x 12 = 15.17 g

moles of H2O = 22.76 / 18 = 1.264

moles of H = 2 x 1.264 = 2.529

mass of H = 2.529 g

mass of O = 32.88 - (15.17 + 2.529) = 15.18 g

moles of O = 0.949

C          H             O

4/3       8/3                1

4            8                3

emperical formula = C4H8O3

mass of emperical formula = 104 g/mol

mass of unknown = 104 g/mol

n = 104 / 104 = 1

molecualer formula = n x emperical formula

                                 = C4H8O3