A 27.0 g sample of an unknown hydrocarbon was burned in excess oxygen to form 88.0 g of carbon dioxide an?

Question

A 27.0 g sample of an unknown hydrocarbon was burned in excess oxygen to form 88.0 g of carbon dioxide and 27.0 grams of water. What is a possible molecular formula for the hydrocarbon?

A. CH4
B. C2H2
C. C4H3
D. C4H6
E. C4H10

Answer 1

88g/44g = 2 mol CO2
27g/18g= 1.5 mol H2O

Find number of Carbon and Hydrogen from the mols. Verify it matches with the 27g. I will finish the problem later if no one else does. I have to run.

Answer 2

Based on the previous answer, there are 2 mols of C (from 2 mols of CO2) and 3 mols of H (from 1.5 mols of H2O) that must have come from the hydrocarbon. Since matter cannot be destroyed, only converted, this 2:3 element ratio must exist in the hydrocarbon 's formula.

The only compound in the above list that meets this qualification is C4H6 (Answer D)......4 C's/6 H's = 2 C's/3 H's

You can verify that this is the right answer by writing out the combustion reaction and balancing it:

2 C4H6 + 11 O2 ----------> 8 CO2 + 6 H2O

(1 mol C4H6 / 4 mols CO2) * 2 mols CO2

= 0.5 mols C4H6 consumed in reaction

MW of C4H6 = 4(12) + 6(1) = 54 g/mol

27.0 g of C4H6 = 27/54 = 0.5 mols