Question

9. Complete combustion of a 0.20 mole sample of a hydrocarbon, CxHy, yields 0.80 mol of CO2 and 1.0 mol of H2O. What is the empirical formula for the hydrocarbon?
A. C2H5
B. C4H5
C. C4H8
D. C4H10
E. C3H8


I have C4H8 Correct?
I don't know why that didn't post...

Complete combustion of a 0.20 mole sample of a hydrocarbon, CxHy, yields 0.80 mol of CO2 and 1.0 mol of H2O. What is the empirical formula for the hydrocarbon?
A. C2H5
B. C4H5
C. C4H8
D. C4H10
E. C3H8

I got an answer of C4H5. Is that correct?
I got C4H5. And it says that is wrong. Can someone help me on this one?
Check my thinking on this.
CxHy + O2 ==> CO2 + H2O
CxHy = 0.20 mol
CO2 = 0.80 mol
H2O = 1.0 mol

Convert 0.20 mol CxHy to 1.0 mol by multiplying by 5. Therefore, we should get 5 x 0.80 mol CO2 (4.0 mols) and 5 x 1.0 mol H2O (5.0 mols). The equation must be something like this:
CxHy + 13/2 O2 ==> 4CO2 + 5H2O which means CxHy must be C4H10?
If 10.0 mol of butane burn completely, wgat thermal energy would be produced?

Related Questions

complete combustion of a 0.0100mol sample of hydrocarbon, CxHy, gives 1.344 L of CO2 at STP and 0.72... Unsure how to solve this without a formula... I think its combustion analysis? CxHy + O2 --> CO... A combustion analysis of a 0.00134g sample (CxHy) of a hydrocarbon yields 0.00375g of CO2 and 0.0016... Complete combustion of a 0.40-mol sample of a hydrocarbon, CxHy, gives 2.40 mol of CO2 and 1.60 mol...