Converting ml -> grams... of methanol.

[Kirol]

Ok. So I was just wondering.
I've got 200ml of water that was heated up by 22°C, meaning that it'ld take 18480 Joules to get there. (cmΔT, remember.)
I've got several alcohols, but let's just work with methanol right now.
Methanol gives out 6147 kJ/mol of energy, right? (Including the O=O bonds used in combustion). Or am I really out there?
Anyway, tell me if any of this is sounding weired to you.
So, let's say that I weighed the Methanol before and after the combustion (which raised the water's (200ml) temperature by 22 degrees). The change in weight was 1.48 grams.
What calculations can I use to find out how much energy the methanol used?
Methanol burned = 1.48g.
Water used = 200ml.
Enthalpy change = 22.
Thanks for your time.

[Jay-qu]

I really dont understand, to me there are 2 ways of interpereting this 1. you are burning the methanol or 2. you are heating the methanol. If you are heating then I think you want the specific heat capacity of methanol.

[Kirol]

Originally Posted by Jay-qu

I really dont understand, to me there are 2 ways of interpereting this 1. you are burning the methanol or 2. you are heating the methanol. If you are heating then I think you want the specific heat capacity of methanol.

I was burning the methanol (ethanol, propanol, butanol, and pentanol,) in a spirit burner...
the flame was heating a calorimeter with 200ml of water.
The temperature of the water rose by 22 degrees, celcius.
I need to find out how much energy was given out by the methanol when I burned 1.42 grams or whatever I said in my last post.
From there, I can work out how much actually went into the water.
And from there, I can work out the efficienct of the experiment...

[Jay-qu]

ok I get you now. 1.48g of methanol is equal to 0.0463mol (1.48/32) now multiplying this by the energy released per mol, which I think is 726kJ/mol, you get: 33.6kJ
This is assuming pure methanol is used, and remember not all this energy will go into the water

[P-man]

I think Einstein never was wrong (reffering to the signature of Jay-qu and has nothing to do with this thread).

[Kirol]

Originally Posted by Jay-qu

ok I get you now. 1.48g of methanol is equal to 0.0463mol (1.48/32) now multiplying this by the energy released per mol, which I think is 726kJ/mol, you get: 33.6kJ
This is assuming pure methanol is used, and remember not all this energy will go into the water

OH! Of course.
I was looking for some really complicated way to do it, but this is much more simple. Thanks a load.
Also, the energy released per mole is equal to 6147kJ...
I think. o_O
After all, the bond energies broken give a total of 2809 kJ/mol, and those formed give -3338...
ΔH is products - reactants, which gives 6147 kJ...
6.147kJ / ml. Although it does seem kinda wrong.
Just 1 ml giving off all that much? It can't be right...
No, wait. I think I have the minuses the wrong way around, after all, it takes energy to break bonds (meaning that the reactant energy should be minus 2809kJ/mol)... lemme work this out again. Ugh. Now I get 6147 positive.
If I have the products as minus, I get a massive number. If I have them as positive, I get an endothermic reaction...
Methanol has 3 X C-H bonds each of which are equal to 413 kJ/mol, 1 C-O bond which is equal to 360 kJ/mol, and 1 O-H bond which is equal to 463 kJ/mol.
So in total, 1 mole of methanol has 2062 kJ of energy. ½O² adds 747kJ/mol to the Reactant’s Energy.
(Remember, CH³OH (l) + 1½O² (g) -> CO²(g) + 2H²O (l) )
In H²O, there are 2 H-O bonds being formed, each of which need 463 kJ/mol. So in H²O, there is 926 kJ/mol absorbed.
In CO², there are 2 C=O bonds being formed, each of which need 743 kJ/mol. So in CO², there is 1486 kJ/mol being absorbed.
Or am I way out, again?

[Kirol]

Anyone?

[Jay-qu]

I am unfamiliar with you working out - probably dont teach that at a high school level here in Australia

So I did a search and came up with this http://www.avogadro.co.uk/calculations/question.htm
method 4 looks like your methodm but has again a different delta H value, all the other methods turn up -726kJ/mol...

[Kirol]

Originally Posted by Jay-qu

I am unfamiliar with you working out - probably dont teach that at a high school level here in Australia

So I did a search and came up with this http://www.avogadro.co.uk/calculations/question.htm
method 4 looks like your methodm but has again a different delta H value, all the other methods turn up -726kJ/mol...

It's ok, I've gotten some realistic figures now. The only problem is that I don't know how I got them, exactally.
I'm not sure if it's displaying the enrgy of forming or breaking the bonds. ~_~
Ah, well.
Who cares when you have some results? xD

[Jay-qu]

I care a great deal about accuracy - doing titrations at school the teacher asks us to get 3 concordant results within .05 of a mL, but I kept going untill I got 3 exact results, well exact to within the error of the burrette