Question

complete combustion of 3.70 g of a hydrocarbon produced 11.9 g of CO2 and 4.06 g of H20. What is the empirical formula for the hydrocarbon?

Answer 1

mass of CO2 = 11.9 g

moles of CO2 = 11.9 / 44.01 = 0.27

moles of C = 0.2704 mol

mass of C = 3.245 g

moles of H2O = 4.06 / 18.02 = 0.225 mol

moles of H = 0.4506

mass of H = 0.4506 g

ratio :         C      H

               0.2704      0.4506

                 1.00         1.67

                     1             5/3

                      3              5

Emperical formula = C3H5