Answer to Question #97833 in General Chemistry for Archie

Question

Complete combustion of 3.70 g of a hydrocarbon produced 11.2 g of CO2 and 5.73 g of H2O. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary.

Answer 1

C_xH_y + (x+y/4)O₂ = xCO₂ + y/2H₂O

C → CO₂

12.011 g C → 44.01 g CO₂

C in 11.2 g of CO₂ = 11.2×12.01/44.01 = 3.056 g; %C = (3.056/3.70)×100 = 82.6%

2H → H₂O

2×1.008 g H → 18.015 g H₂O

Hence H in 5.73 g of H₂O =5.73×2×1.008/18.015 = 0.641 g; %H = (0.641/3.70)×100 = 17.3%

Atomic ratios C = 82.6/12.011 = 6.88; H = 17.3/1.008 = 17.19

divide through by smallest (6.88) C =1.0 H = 2.50; has to be integers so X by 2

= C₂H₅ this the required empirical formula but it is an unstable ethyl radical.

C₄H₁₀, butane, is stable