Question

The osmotic pressure of a solution containing 2.10 g of an unknown compound dissolved in 175.0 mL of solution at 25 ∘C is 1.93 atm . The combustion of 24.02 g of the unknown compound produced 28.16 g CO2 and 8.640 g H2O.

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

Express your answer as a chemical formula.

Answer 1

Concentration,
C = mass / (molar mass * V)
C = 2.1 / (M*0.175)
Let molar mass be M gram and V=175 mL = 0.175 L

T = 25 oC = (25+273)K = 298 K
use:
osmotic pressure = C*R*T
1.93 = 2.1 / (M*0.175) * 0.0821 * 298
M= 152 g/mol

mass of CO2 formed = 28.16 g
number of moles of CO2 = mass / molar mass = 28.16 / 44 = 0.64 mol
so, moles of C in original compound = 0.64 mol
Mass of C in original compound = number of moles * molar mass = 0.64 * 12 =7.68 g

mass of H2O formed = 8.64 g
number of moles of H2O = mass / molar mass = 8.64 / 18 = 0.48 mol
so, moles of H in original compound = 2*0.48= 0.96 mol (since 1 mol of H2O has 2 H)
Mass of H in original compound = number of moles * molar mass = 0.96 * 1 =0.96 g

Mass of O = 24.02 - 7.68 - 0.96 = 15.38 g
moles of O =mass/ molar mass = 15.38/16 = 0.96 mol

moles of C = 0.64
moles of H =0.96
moles of O =0.96

Get simplest ratio by dividing each by 0.64
C = 0.64/0.64=1
H = 0.96/0.64= 1.5
O = 0.96/0.64= 1.5

multiply each by 2 to get simplest ratio
C = 1*2 = 2
H = 1.5*2 =3
O = 1.5*2 =3

C2H3O3 is the empirical formula
empirical formula mass = 2*12 + 3*1 + 3*16 = 75 g/mol
molar mass = 152 g/mol
multiplying factor = 152/75 = 2 (approx)

So formula is
C4H6O6