combustion analysis

Question

A 3.967 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 9.018 grams of CO₂ and 3.692 grams of H₂O are produced. 

In a separate experiment, the molar mass is found to be 116.2 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

1. Enter the elements in the order C, H, O
2. 
   

empirical formula =

1. 
   
2. 
   

molecular formula =

Answer 1

moles C = 9.018 g CO2 x 1 mol CO2 / 44 g x 1 mol C / mol CO2 = 0.205 moles C

moles H = 3.692 g H2O x 1 mol H2O/18 g x 2 mol H / mol H2O = 0.410 moles H

To find moles of O, we need to subtract mass of C and H from 3.967 g to get grams O and then convert to moles of O.

0.205 mols C x 12 g / mol = 2.46 g C

0.410 mols H x 1 g / mol = 0.410 g H

g O = 3.967 g - 2.46 g - 0.410 g = 1.097 g O

moles O = 1.097 g O x 1 mol / 16 g = 0.0686 mols O

To get whole numbers of moles, we divide all the moles by 0.0686 to get...

moles C = 0.205 / 0.0686 = 3

moles H = 0.410 / 0.0686 = 6

moles O = 0.0686 / 0.0686 = 1

Empirical formula = C₃H₆O

molar mass of C₃H₆O = 55 g / mol

116.2 g / mol ÷ 55 g / mol = 2

Molecular formula = C₆H₁₂O₂