Question

A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 30.00 gram sample of the alcohol produced 57.30 grams of CO2 and 35.22 grams of H2O. What is the empirical formula of the alcohol?

Answer 1

first we calculate the mass percent composition of three elements carbon hydrogen and oxygen.
Mass of CO₂ = 44.0096 g/mol

Mass of H₂O = 8.0153 g/mol

percent composition of carbon = (57.30 g CO2) / (44.0096 g CO2/mol) x (1/1) x (12.0108 g C/mol) / (30.00 g) = 0.521264 = 52.1264% C

Percent composition of hydrogen = (35.22 g H2O) / (18.0153 g H2O/mol) x (2/1) x (1.0079 g H/mol) / (30.00 g) = 0.131363 = 13.1363% H

percent composition of oxygen = 100% - 52.1264% C 13.1363% H = 34.7373% O

Now calculate the number of moles in the compound;
(52.1264 g C) / (12.0108 g C/mol) = 4.33996 mol C
(13.1363 g H) / (1.0079 g H/mol) = 13.0333 mol H
(34.7373 g O) / (15.9994 g O/mol) = 2.17116 mol O

Now divide by the smallest number of moles:
(4.33996 mol C) / (2.17116 mol) = 1.999 = 2
(13.0333 mol H) / (2.17116 mol) = 6.003 = 6
(2.17116 mol O)/ (2.17116 mol) = 1.000 = 1

So, the empirical formula is :
C₂H₆O