Calculation of empirical formula - combustion analysis

Combustion analysis  of 0.18g of an organic compound produces  0.396 g of carbon dioxide  and 0.216 g of water . The empirical formula of the compound, given that it contains carbon, hydrogen and oxygen only is

A.  C₃H₈O

B.  C₄H₈O

C.  C₂H₈O₄

D.  C₆H₁₆O

Collate data ie

0.396 g carbon dioxide

0.216 g water

Empirical formula can only have  H O C

Equation for combustion    M   + O2     =   CO2   +  H2O

  multiply values by 1000   180g                396g        216g

Divide values by Relative Formula Mass (RFM) to obtain moles of carbon dioxide and moles of water

     396/44    =   9 moles of carbon dioxide,   216/18   =  12 moles of water 

Note: Some oxygen is present in M and some added during burning

9 moles of carbon dioxide contains 9x12 = 108 g of Carbon

12 moles of water   contains          12x2  = 24 g of   Hydrogen

Therefore, the mass of (C+H) present in 180 g of M = 10-8+24 = 132 g

The remainder is oxygen ie, 180 -132  = 48 g of oxygen

Summary

moles of  Carbon atoms       =  9

moles of  Hydrogen atoms   =  24

moles of  Oxygen  48/16      = 3

Divide by smallest value to get ratios  to give      9/3 = C 3,  24/3 = H   8,  3/3 =  O  1

Empirical Formula  C3 H8 1

Hope this was clear. Look out for other questions and answers.