Question

A combustion analysis of a 0.503 g sample of an unknown hydrocarbon yields 1.53 g CO2_2​2​​ and 0.756 g H2_2​2​​O. What is the empirical formula of the sample? Give your answer in the form C#H# where the number following the element’s symbol corresponds to the subscript in the formula. (Don’t include a 1 subscript explicitly). For example, the formula CH2_2​2​​O would be entered as CH2O.

Answer 1

Given,

Mass of sample = 0.503 g

Mass of CO₂ after combustion = 1.53 g

Mass of H₂O after combustion = 0.756 g

Now, calculating the number of moles of C and H from the given masses of CO₂ and H₂O,

= 1.53 g CO₂ x ( 1 mol CO₂ / 44.01 g CO₂) x (1 mol C / 1 mol CO₂)

= 0.0348 mol C

Similarly,

= 0.756 g H₂O x ( 1 mol H₂O / 18.02 g H₂O) x (2 mol H / 1 mol H₂O)

= 0.0839 mol H

Now, dividing each mole by least number of moles,

= 0.0348 mol C / 0.0348 = 1 mol C

also,

= 0.04195 mol H / 0.0348 = 2.4 mol H

Multiplying to each mole by 5 to get a simple whole number.

= 1 mol C x 5 = 5 mol C

Similarly,

= 2.4 mol H x 5 = 12 mol H

Thus, the empirical formula of hydrocarbon is C₅H₁₂