Question

Complete combustion of 4.80 g of a hydrocarbon produced 14.5 g of CO2 and 7.44 g of H2O. What is the empirical formula for the hydrocarbon?

Answer 1

The number of moles of CO2 in 14.5g of CO2 and number of moles of H2O in 7.44g of H2O can be calculated as follows:

Number of moles of CO2 = 14.5g of CO2 / (44.01g/mol)

Number of moles of CO2 = 0.3295

Number of moles of H2O = 7.44g of H2O / (18.015g/mol)

Number of moles of H2O = 0.4130

Thus, number of moles of carbon in given sample is same as number of moles of CO2 i.e. 0.3295 moles.

The number of moles of hydrogen (H) is two times the number of moles of H2O as H2O contains two H-atoms.

Thus, number of moles of H = 0.4130 x 2= 0.8260

Ratio of number of moles of C and H can be obtained by dividing both with lowest number i.e. 0.3295

Mole ratio of C = 0.3295 / 0.3295 = 1.00

Mole ratio of H = 0.8260 / 0.3295 = 2.50

Thus, ratio of C and H is 1.00 : 2.50. It can be rounded to nearest whole number by multiplying 2 on both sides.

Therefore, C:H is 2.00:5.00. Hence, empirical formula for the given hydrocarbon is C2H5