PLEASE HELP!
 

I need an answer to this quick, quick, quick. Well moreso, an explanation of how to solve it, because I have the answer.

From the following heats of combustion,
CH₃OH_(l) + 3/2O_2(g) --> CO_2(g) + 2H₂O_(l) enthalpy of reaction=-726.4 KJ/mol
C_(graphite) + O_2(g) --> CO_2(g) enthalpy of reaction=-393.5 KJ/mol
H_2(g) + 1/2O_2(g) --> H₂O_(l) enthalpy of reaction=-285.8 KJ/mol

calculate the enthalpy of formation of methanol (CH3OH) from it's elements:
C_(graphite) + 2H_2(g) + 1/2O_2(g) --> CH₃OH_(l)

How the heck do I find enthalpy of formation?
I KNOW this is in the wrong board, I know where this type of question belongs, but let's face it, it could take a whole day before anyone sees it.

Please leave this here long enough for someone to help me.

I went to a review session but the TA didn't know how to explain it, she just solved it. She solved it by multiplying the first reaction's enthalpy by (-1) and the third reaction's enthalpy by (2). Afterwards she added up all the enthalpies. The problem is that I don't understand <i>why</i> she did that and she couldn't explain it either >.<

Please help!

Check this out.
http://wine1.sb.fsu.edu/chm1045/notes/Energy/HeatForm/Energy05.htm
Hope it helps.

here is another link
http://dbhs.wvusd.k12.ca.us/webdocs/...Formation.html

They didn't help, but I do appreciate the effort. I *think* I got it now, though.

Well, I looked at an example in the book and I worked it through, and it turns out that it's not that hard. The reason why the TA couldn't explain it is because it's kind of a process of trial and error, and she seemed like she was in a hurry.

I apologize for the out-of-context-ness of this post, though. Feel free to move it to the appropriate board now.

Moogle on then my brother!! Moogle on.