Question

An unknown compound contains only C , H , and O . Combustion of 4.10 g of this compound produced 10.0 g CO2 and 4.10 g H2O . What is the empirical formula of the unknown compound? Insert subscripts as needed.

Answer 1

Moles of CO2 = (10 g) / (44.10 g/mol) = 0.2267 mol

Each CO2 have 1 Carbon in it

So, moles of C = 0.2267 mol

Weight of C = (0.2267 mol) * (12.01 g/mol) = 2.72 g

Moles of H2O = (4.10 g) / (18.02 g/mol) = 0.227 mol

Each H2O have 2 hydrogens in it

Moles of H = (0.227 mol) * 2 = 0.455 mol

Weight of H = (0.455 mol) * (1.01 g/mol) = 0.46 g

Weight of oxygen = (4.1 g) - (2.72 g + 0.46 g) = 0.92 g

Moles of O = (0.92 g) / (16 g/mol) = 0.057 mol

Simplest mol ratio

For C = (0.2267 mol) / (0.057 mol) = 4

For H = (0.455 mol) / (0.057 mol) = 8

For O = (0.057 mol) / (0.057 mol) = 1

Empirical formula is: C4H8O