# Thermochemistry : % Mass Question

1. Feb 24, 2009

### ElementUser

1. The problem statement, all variables and given/known data

When a 10.0g sample of a mixture of CH4 and C2H6 is burned in excess oxygen, exactly 525 KJ of heat is produced. What is the percentage by mass of CH4 in the original mixture?

Given: CH4(g) +2 O2(g) -> CO2(g) + 2 H2O(l) $$\Delta$$H= -890.4 KJ

C2H6(g) + $$\frac{7}{2}$$ O2(g) -> 2 CO2(g) + 3 H2O(l) $$\Delta$$H= -1560.0 KJ

2. Relevant equations

N/A

3. The attempt at a solution

I take the ratio of the enthalpy of CH4 to C2H6. To do this, I do -890.4/-1560.0. This gives me approx. 0.57 (note that I keep as many digits as possible when doing the actual calculations). Therefore, for every one mol of C2H6 burned, 0.57 mol of CH4 is burned.

I then find out how much of the % of the mixture is burned for the CH4.

$$\frac{0.57}{1.57}$$=36.3%

I proceed to multiply the 525 KJ by this percentage and then convert to grams of CH4 to get 3.44 g of CH4. Finally, I find the % mass by dividing my 3.44 g of CH4 by 10.0g of the original mixture, multiply it by 100% to obtain 34.4% of the original mixture as my final solution.

However, the answer appears to be 17.7%; precisely half of the % mass I acquired. Can someone help point out on what I did wrong?

Any help is appreciated. Thanks in advance~

2. Feb 26, 2009

### Staff: Mentor

17.7% doesn't look as a correct answer to me.

3. Feb 26, 2009

### ElementUser

Hm ok, I was just wondering if there was another method to solve this or if the question's answer was just wrong.