Question

A combustion analysis of a 0.44g sample of an unknown compound yields 0.88 g CO2 and 0.36 g H2O. If the sample a molar mass of 132 g/mol, what is the molecular formula of the sample?

Answer 1

moles of CO2 = 0.88 / 44 = 0.02

moles of C = 0.02

mass of Carbon = 12 x 0.02 = 0.24 g

moles of H2O = 0.36 / 18 = 0.02

moles of H = 2 x 0.02 = 0.04

mass of hydroegen = 0.04 g

Oxygen mass = 0.44 - (0.24 + 0.04 ) = 0.16 g

moles of O = 0.16 /16 = 0.01

C               H    O           

0.02         0.04        0.01

2                 4                   1

C2H4O -----------------------> empirical formula

empirical formula mass = 24 +4 + 16 = 44

n = molar mass / empirical formula mass

    = 132 / 44

     = 3

molecular formula = n x empirical formula

                                = C6H12O3