Question

A 0.4987-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.9267 g of carbon dioxide and 0.1897 g of water. What is the empirical formula of the compound?

Answer 1

(0.9267 g CO2) / (44.00964 g CO2/mol) x (1 mol C / 1 mol CO2) x (12.01078 g C/mol) =
0.252908 g C

(0.1897 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) x
(1.007947 g H/mol) = 0.0212272 g H

Use the Law of Conservation of Mass:
(0.4987 g total) - (0.252908 g C) - (0.0212272 g H) = 0.224565 g O

Covert to moles:
(0.252908 g C) / (12.01078 g C/mol) = 0.0210568 mol C
(0.0212272 g H) / (1.007947 g H/mol) = 0.021060 mol H
(0.224565 g O) / (15.99943 g O/mol) = 0.0140358 mol O

Divide by the smallest number of moles:
(0.0210568 mol C) / 0.0140358 mol = 1.500
(0.021060 mol H) / 0.0140358 mol = 1.500
(0.0140358 mol O) / 0.0140358 mol = 1.000
To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers fo find the empirical formula:
C3H3O2