Question

An unknown compound contains only C, H, and O. Combustion of 8.20 g of this compound produced 16.4 g of CO2 and 6.71 g of H2O.

What is the empirical formula of the unknown compound?

Answer 1

Number of moles of CO2 = mass of CO2 / molar mass CO2

= 16.4/44

= 0.3727

Number of moles of H2O = mass of H2O / molar mass H2O

= 6.71/18

= 0.3728

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.3727

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.3728 = 0.7456

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 8.2 - 0.3727*12 - 0.7456*1

= 2.9817

number of mol of O = mass of O / molar mass of O

= 2.9817/16.0

= 0.1864

Divide by smallest to get simplest whole number ratio:

C: 0.3727/0.1864 = 2

H: 0.7456/0.1864 = 4

O: 0.1864/0.1864 = 1

So empirical formula is:C₂H₄O