Question

Combustion analysis of a hydrocarbon produced 33.01 g CO2 and 10.82 g H2O.

Part A Calculate the empirical formula of the hydrocarbon. Express your answer as a chemical formula.

Answer 1

Given,

Mass of CO₂ = 33.01 g

Mass of H₂O = 10.82 g

Calculating the number of moles of C and H from the given masses of CO₂ and H₂O respectively.

= 33.01 g CO₂ x ( 1 mol / 44.01 g) x ( 1 mol C / 1 mol CO₂)

= 0.750 mol C

Similarly,

= 10.82 g H₂O x ( 1 mol / 18.02 g) x ( 2 mol H / 1 mol H₂O)

= 1.20 mol H

Now, dividing each number of moles by the least number of moles,

= 0.750 mol C / 0.750 = 1 mol C

Similarly,

= 1.20 mol H / 0.750 = 1.6 mol H

Now, here, the number for moles of "H" is in fraction thus multiplying to each mole by 5 to get the whole number,

= 1 mol C x 5 = 5 mol C

Also,

= 1.6 mol H x 5 = 8 mol H

Thus, the empirical formula for the hydrocarbon in C₅H₈.