Question

Combustion analysis of 0.600 g of an unknown compound containing carbon, hydrogen, and oxygen produced 1.043 g of CO2 and 0.5670 g of H2O. The molar mass is 532.7 g/mol. What is the molecular formula of the compound?

Answer 1

The MM for CO₂ reported is 44 g/mol and for H₂O is 18 g/mol.

If we have 1.043 g of CO₂ means that we have:

44 g CO₂ --------> 12 g C

1.043 g ------------> X

X = 1.043 * 12/44 = 0.2845 g of C

For H:

18 g H₂O ----------> 2 g H

0.5670 g -----------> Y

Y = 0.5670 * 2/18 = 0.063 g

For O:

0.600 - 0.063 - 0.2845 = 0.2525 g of O

C = 0.2845 / 12 = 0.0237 / 0.0158 = 1.5

H = 0.063 / 1 = 0.063 / 0.0158 = 3.987

O = 0.2525 / 16 = 0.0158 / 0.0158 = 1

Empirical Formula = C₂H₄O

MM = 2*12 + 4*1 + 16 = 44 g/mol

n = 532.7 / 44 = 12.11

Molecular Formula = 12 * C₂H₄O

MF = C₂₄H₄₈O₁₂

Hope this helps