The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 atm . The combustion of 22.08 g of the unknown compound produced 36.26 g CO2 and 14.85 g H2O.

What is the molecular formula of the compound (which contains only carbon, hydrogen, and oxygen)?

The molecular formula of the  compound is C4H12O4.

π = icRT

π = osmotic pressure

c = concentration

R = molar constant

T = temperature

Now we know that; c = n/V

n = number of moles

V = volume

π = i(n/V)RT

π =inRT/V

n = πV/iRT

n = 2.13 atm × 0.175 L/1 × 0.082 atmLK-1mol-1 × 298 K

n = 0.3675/24.436

n = 0.015 moles

number of moles  = mass/molar mass

Molar mass = mass/ number of moles = 2.04 g /0.015 moles = 136 g/mol

From the combustion information;

mass of C = 36.26 g/44 × 12 = 9.9 g

Number of moles of C = 9.9 g/12 g/mol = 0.825 moles

Mass of H = 14.85 g/18 × 2 = 1.65 g

Number of moles of H = 1.65 g/ 1 g/mol = 1.65 moles

Mass of O = 22.08 g - (9.9 g + 1.65 g) = 10.53 g

Number of moles of oxygen = 10.53 g/16 g/mol = 0.658 moles

To obtain the empirical formula, we divide through by the least number of moles;

C - 0.825 moles/0.658 moles,  H - 1.65 moles/0.658 moles,  O - 0.658 moles/0.658 moles

C - 1, H - 3, O - 1

The empirical formula of the  compound is CH3O

The molecular formula of the compound is obtained thus;

[12 + 3 + 16]n = 136

n = 136/31

n = 4

The molecular formula is C4H12O4.

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