Question

Complete combustion of a 0.30-mol sample of a hydrocarbon, CxHy, gives 1.20 mol of CO2 and 1.35 mol of H2O. The molecular formula of the original hydrocarbon is

Select one:

a. C4H9

b. C7H20

c. C3H5

d. C3H8

e. C6H6

Answer 1

C_xH_y + (x+${\frac {y} {4}}$)O₂ = xCO₂ + ${\frac {y} {2}}$H₂O

${\frac {n(CO_2)} {n(C_xH_y)}}=x={\frac {1.20} {0.30}}=4$

x = 4

${\frac {n(H_2O)} {n(C_xH_y)}}={\frac {y} {2}}={\frac {1.35} {0.30}}=4.5$

y = 9

Then the empirical formula of C_xH_y = C₄H₉, answer a

(Calculations are right, but i can't claim this answer as a correct one because hydrocarbon C₄H₉ doesn't exist)