Complete combustion of 7.40 g of a hydrocarbon produced 23.8 g of co2 and 8.11 g of h2o. what is the empirical formula for the hydrocarbon?

You have to assume that the hydrocarbon contains only C, H and, possibly, O.

Then, all the C in CO2 and the H in H2O come from the hydrocarbon.

a) Number of moles of C in 23.8 g of CO2.

molar mass of CO2 = 12.0 g + 2 * 16.0 g = 44.0 g / mol

number of moles of C = mass in grams / molar mass = 23.8 g / 44.0 g/mol = 0.5409 mol

b) number of moles of H in 8.11 g of H2O

number of moles = mass in grams / molar mass = 8.11 grams / 18.0 g/mol = 0.4506 mol

c) mass of O in the hydrocarbon, m of O:

m of O = 7.40 g - mass of C - mass of H

mass of C = number of moles * atomic mass = 0.5409 mol * 12 g/mol = 6.49 g

m of H = 0.4506 mol * 1 g / mol = 0.45 g

=> grams of O = 7.40 g - 6.49 g - 0.45 g = 0.46 g of O.

d) number of moles of O = mass / atomic mass

=> number of moles of O = 0.46 g / 16 g / mol = 0.02875 mol O.

e) Empirical formula

C: 0.5409 / 0.02875 = 18.8 ≈ 19

H: 0.4506 / 0.02875 = 15.7 ≈ 16

O: 0.02875 / 0.02875 = 1

=> Empirical formula = C19 H16 O