Question

An unknown compound contains only C, H, and O. Combustion of 6.80 g of this compund produced 16.0 g of CO₂ and 4.37 g of H₂O. What is the empirical formula of the unkown compound? C_xH_yO_z

Answer 1

1. 16 g CO₂ x (1 mol CO₂/ 44.0-g CO₂) = 0.36 moles CO₂ and 0.36 moles C x 12-g /mol = 4.36-g C
2. 4.37 g H₂O x ( 1 mol H₂O / 18.0-g H₂O) = 0.49 moles H₂O and 0.49 moles H x 2 g/mol = 0.98 H
3. 6.80 g - (4.36 - 0 .98) = 3.42-g O x 1mol O / 16.0-gO) = 0.124 mol O

4. C = 0.36 moles/0.124 moles = 2.9
...H = 0.49 moles/0.124 moles= 3.9
...O = 0.124 moles/0.124 moles= 1
5. Step 4 above didn't yield whole numbers, but if we multiply through by 2, we'll get whole numbers.
C = 3
H = 4
O = 1

So, the empirical formula is C3H4O