Question

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.

Answer 1

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