# 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?

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1. 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?

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2. I got C4H5. And it says that is wrong. Can someone help me on this one?

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3. 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?

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4. If 10.0 mol of butane burn completely, wgat thermal energy would be produced?

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