Combustion of 28.30 g of a compound containing only carbon, hydrogen, and oxygen produces 32.75 gco2 and 13.41 gh2o. what is the empirical formula of the compound?

The empirical formula is C₂H₄O₃.

We must calculate the masses of C, H, and O from the masses given.

Mass of C = 32.75 g CO₂ × (12.01 g C/44.01 g CO₂) = 8.9372 g C

Mass of H = 13.4 g H₂O × (2.016 H/18.02 g H₂O) = 1.5067 g H

Mass of O = Mass of compound - Mass of C - Mass of H

                  = (28.30 – 8.9372 – 1.5067) g = 17.856 g

Now, we must convert these masses to moles and find their ratios. If the ratios are not integers, we multiply them by a number to make them close to integers.

From here on, I like to summarize the calculations in a table.

Element       m/g       n/mol     Ratio        ×2      Integers  

     C           8.9372   0.74415   1               2                  2

     H           1.5067    1.4948    2.0086    4.0174          4

     O        17.856       1.1160      1.4997     2.9994         3

The empirical formula is C₂H₄O₃.

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