Question

A 10.00 g sample of a compound containing only carbon, hydrogen, and oxygen forms 23.98 g CO2 and 4.91 g H2O upon complete combustion. What is the empirical formula of the compound?

Answer 1

(23.98 g CO2) / (44.00964 g CO2/mol) × (1 mol C / 1 mol CO2) = 0.544881 mol C
(0.544881 mol C) × (12.01078 g C/mol) = 6.54445 g C

(4.91 g H2O) / (18.01532 g H2O/mol) × (2 mol H / 1 mol H2O) = 0.54509 mol H
(0.54509 mol H) × (1.007947 g H/mol) = 0.54942 g H

10 g total - 6.54445 g C - 0.54942 g H = 2.90613 g O
(2.90613 g O) / (15.99943 g O/mol) = 0.18164 mol O

Divide by the smallest number of moles:
(0.544881 mol C) / 0.18164 mol = 2.9998
(0.54509 mol H) / 0.18164 mol = 3.0009
(0.18164 mol O) / 0.18164 mol = 1.0000
Round to the nearest whole number to find the empirical formula:
C3H3O