Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produced 36.86g CO2 and 10.06g H2O. The molar mass of estriol is 288.38g/mol.

Find the molecular formula of the unknown compound

Answer: molecular formula = C12H16O8

Explanation:

NB Mm CO2 = 44g/mol

Mm H2O = 18g/mol

Moles of CO2 = 36.86/44=0.84mol

0.84mole of CO2 has 0.84 mol of C

Moles of H2O = 10.06/18 = 0.56mol

1mol of H20 contains 1mol of O and 2 mol H,

Hence there are 0.56mol O and (0.56*2) mol H

Hence the compound contains

C = 0.84 mol H = 1.12mol O=0.56mol

Divide through by smallest number

C = 0.83/0.56 = 1.5mol

H = 1.12/0.55 = 2mol

O = 0.56/0.56 = 1mol

Multiply all by 2 to have whole number of moles = 3:4:2

Hence empirical formula = C3H4O2

(C3H4O2) n = 288.38

[ (12*3) + 4 + (16*2) ]n = 288.38

72n=288.38

n = 4

:. Molecular formula = (C3H4O2) 4 = C12H16O8