Question

Complete combustion of 4.80 g of a hydrocarbon produced 14.8 g of CO2 and 6.81 g of H2O. What is the empirical formula for the hydrocarbon?

Answer 1

mass of CO2 = 14.8 g

moles of CO2 = 14.8 / 44 = 0.336

moles of C = 0.336

mass of C = 4.036 g

moles of H2O = 6.81 / 18.02 = 0.378

moles ofH = 0.756

mass of H = 0.756 g

   C             H

0.336       0.756        

1             9/4

4              9

Emperical formula = C₄H₉