Jump to content

Complete Combustion of Butanol


Recommended Posts

i have got these bond energies:

Butanol 5580 (total)

Water 928

Carbon Dioxide 1210

oxygen 498.3


When i work out the total bond energies on both sides, it shows me that the complete combustion of Butanol is ENDOTHERMIC!!!! Whats wrong with my calculation


C4H9OH + 6O2 ----------> 4CO2 + 5H2O


14149.8 11080



Link to comment
Share on other sites

well, honestly to tell you, i havent a clue what you are talking about!!! But all i know, is that the butanol we are using has a formula of C4H9OH


P.S i read ur question on the smallest movement that is detected and i have a question, Why do u ask impossible-to-answer questions :) lol

Link to comment
Share on other sites

LOL, well in reverse order, it`s not an impossible question, there has to be some threshold for this movement idea.


Next, lets have a look at your formula....

C4 H9 OH well we can re-shuffle those a little as there`s 2 piles with H in it :)


C4 H10 O1


and we have as many Oxygens as we can use in our air too :)


we can make 4 lots of CO2 using 8 O from the air


and 5 lots of H2O using 4 lots of O from the air and remaining Oxygen in our fuel.


total air usage to combust 1 molecule of butanol would be 8 oxygen from the Air.


leaving 4 parts CO2 and 5 parts water

and completely combusted... can you work from there?

Link to comment
Share on other sites

aommaster said in post #5 :

so is the balanced equation:


c4h9oh + 8o2 ------------> 4co2 + 5h20




4x CO2 (that`s 4xO2)

5x H2O (that`s 5xO... or 2xO2 and just a single O)


don`t forget the fuel contains ONE of the O atoms :)


I count 13 O ATOMS from the air, or 6 O2 molecules and 1 ATOM.

and one used from the fuel making it 7O2 :)


Link to comment
Share on other sites

there`s 2 schools of thought about that in all honesty. for instance Phosphourous Pentoxide is often writen as P2O5 when in fact the molecule is P4O10.

so that`s entirely up to you, whichever works best and remains factual at the same time, is great :)

Link to comment
Share on other sites

ok, so then, i'll divide it by 2 since my chemistry teacher tells be to simplify it. Thanx, and in case u were wondering y i was a total idiot, it was because im only in year 11, and balancing formulas as complex as this, i have not practiced





Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.