Question

Complete combustion of 6.90 g of a hydrocarbon produced 22.2 g of CO2 and 7.57 g of H2O. What is the empirical formula for the hydrocarbon?

Answer 1

Conversion factor of CO2 to C = 12.01g/44.01g = 0.27289

Conversion factor of H2O to H = 2.02g/18.02g = 0.11210

mass of C in the sample = 0.27289 × 22.2g = 6.058g

mass of H in the sample = 0.11210 × 7.57g = 0.8486g

Number of moles of C in the sample = 6.058g/12.01g/mol = 0.5044

Number of moles of H in the sample = 0.8486g/1.01g/mol = 0.8402

divide the number of moles by lowest of them

C : 0.5044/0.5044 = 1

H : 0.8402/0.5044 = 1.667

converting H value to whole number by multiplying both with 3

C : 1× 3 = 3

H : 1.667× 3 = 5

Therefore

empirical formula of the compound is C3H5