Which of the following molecular formulas is consistent with the analysis?

Question

Combustion analysis of a hydrocarbon produces 0.396 g of CO2 and 0.163 g of H2O.
A) CH4
B) C2H6
C) C2H2
D) C2H4
E) C3H8

Answer 1

Find the number of moles of CO2 released:
.396g / 44(grams/mole) = .009

Find the number of moles of H2O released:
.163g / 18(grams/mole) = .009

In a combustion, O is in excess, so we disregard it and make a ratio of the C and the H.
.009/ 2*.009 (we multiply H times two since water has two Hydrogens)
so C/H is equal to the ratio 1/2, which would be consistent with answer D.

Answer 2

First you should calculate the number of moles per element:
CO2 = 0.396/ 44 = 0.009
H2O = 0.163 / 18 = 0.00906

since the resulting products has the same number of moles then D is the answer... it will give off thesame molecular ratio of product assuming it is a complete combustion.

Answer 3

None of your proposed 'answers' contain oxygen. Remember that matter can be neither created nor destroyed. Further, the basic definition of a hydrocarbon contains carbon and hydrogen.
If you have Oxygen in the products, you have to have oxygen in the reactants. With this said, none of your answers are correct.

Answer 4

Use the percent composition of C in CO2 and multiply by 0.396g to calculate the grams of carbon

Do the same for Hydrogen in H2O times 0.163g

Now you have the grams of C and H, so you can calculate the empirical formula and solve your question

Answer 5

I believe it is C2H4. It has been so long since I did this kind of stuff, but that is what I think it is.