Question

what is the empirical formula of a hydrocarbon which produces 2.703 g of CO2 and 1.108 g of H2O when combusted?

Answer 1

Let the empirical formula of the compound may be CxHy.

First calculate the moles of CO2 in 2.703 g of CO2 as follows:

2.703 g of CO2 / molar mass of CO2 = 2.703 g / 44.01 g /mol = 0.0614 mol

Now Number of carbon atom in 0.0614 mol CO2 :

Number of carbon atom = 0.0614 mol CO2* 1 C mol / 1 mol CO2 = 0.0614 mol

second calculate the moles of H2O in 1.108 g of H2O as follows:

1.108 g of H2O / molar mass of H2O

= 1.108 g of H2O / 18.02 g /mol = 0.0615 mol

Now Number of H atom in 0.0615 mol H2O :

Number of H atom = 0.0615 mol H2O * 2 H mol / 1 mol H2O = 0.123 mol

let's simplify those ratios by dividing by the smallest.

C = 0.0614 moles / 0.0614 mol = 1.00
H = 0.123 moles / 0.0614 mol = 2.00

and our empirical formula is CH2.