CxHy + O2 → CO2 + H2O
All of the C from the hydrocarbon is found in the CO2 while all of the H is in the water.
(12 g C/44 g CO2) x 15.3 g CO2 = 7.65 g C
(2 g H/18 g H2O) x 7.32 g H2O = 1.22 g H
Next, calculate the moles of C & H represented by these masses:
(1 n C/12 g C) x 7.65 g C = 0.637 n C
(1 n H/1.10 g H) x 1.22 g H = 1.109 n H
Dividing by the smaller value (i.e. 0.637) we get a 1:2 ratio. Therefore, the empirical formula would be CH2
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