Question

# When 3.036 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 10.26 grams...

When 3.036 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 10.26 grams of CO2 and 2.101 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 26.04 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Empirical formula is the least coefficient formula, that is

CxHy

we must find x,y,z via gravimetry.

Typically we do this relating to moles of C,H,O

mol of CO2 = mass of CO2/MW of CO2 = (10.26)/(44) = 0.2331 mol of CO2

1 mol of CO2 = 1 mol of C ---> 0.2331 mol of CO2 = 0.2331 mol of C

mol of H2O = mass of H2O/MW of H2O= (2.101)/(18) = 0.1167 mol of H2O

1 mol of H2O= 2 mol of H ---> 0.1167 mol of H2O= 0.1167*2 = 0.2334 mol of H

ratio

H:C =0.2334 / 0.2331 = 1

CH is the empirical formula

for overall formula

MW empirical = C+H = 12+1 = 13 g/mol

Ratio

MW real / MW empirical = 26/13 = 2x

so

C2H4 ---> molecular formula