Question

Complete combustion of 3.60 g of a hydrocarbon produced 11.6 g of CO2 and 3.95 g of H2O. What is the empirical formula for the hydrocarbon?

Answer 1

Mass of C in 11.6g CO2 = molar mass of C * mass of CO2/ Molar mass of CO2
                                                 = 12 *11.6 /44
                                                 = 3.164g
Mass of H in 3.95 g H2O molar mass of H* mass of H2O/ Molar mass of H2O
                                       =2 * 3.95 /18
                                      = 0.44 g

number of moles of C = 3.164/12 = 0.264
number of moles of H = 0.44/1 = 0.44
divide by smaller
C = 0.264/0.264 = 1
H = 0.44/0.264 = 1.67
Remove fraction - multiply by 3

C = 1*3 = 3
H = 1.67*3 = 5
Empirical formula = C3H5