Hydrocarbons are organic compounds composed entirely of hydrogen and carbon. A 0.1647 g sample of a pure hydrocarbon was burned in a combustion apparatus to produce 0.4931 g of CO2 and 0.2691 g of water. Determine the empirical formula; enter as C#H# (for example: C1H1, write "1" if appropriate) Through another experiment it was determined that the molecular weight of this hydrocarbon is approximately 132 amu. What is the molecular formula of this compound? Enter as C#H# (for example C2H6, write "1" if appropriate)

Empirical Formula = > C₂H₅

Molecular Formula = > C₉H₂₄

Explanation:

0.1647g CₓHᵪ + Excess Oxy = > 0.4931g CO₂ + 0.2691g H₂O

If %C in CO₂ is 27.27% by weight and %H in H₂O is 11% by weight then the masses of C & H consumed from CₓHᵪ are ...

C = 0.2727 (0.4931g) = 0.1345g C

H = 0.1100 (0.2391g) = 0.0296g H

Therefore ...

Wt%C in CₓHᵪ = (0.1345g/0.1647g) 100% = 81.17% (w/w)

Wt%H in CₓHᵪ = 100% - 81.17% = 17.83% (w/w)

Empirical Ratio is calculated following the sequence ...

% (per 100wt) = > g (per 100wt) = > moles = > mole ratio = > reduce mole ratio = > empirical ratio

C = 81.17% = > 81.17g = > (81.17g/12g/mol) = 6.764 mole C

H = 17.83% = > 17.83g = > (17.83g/1g/mol) = 17.83 mole H

Empirical Ratio = > C:H = 6.764/6.764:17.83/6.764 = 1:2.6

Adjusted to smallest whole no. ratio = > 2 (1:2.6) ~ 2:5 = > Empirical Formula = C₂H₅

Molecular Formula Wt = N (Empirical Formula Wt); N = whole number multiple of empirical ratio.

N = 132/29 = 4.5 = > Molecular Formula = (C₂H₅) ₄.₅ ≈ C₉H₂₄*

*Need add 1.5H to match formula weight of 132amu.