[hoppingma-d]

Hey guys...

I'm doing a high-school prac including testing a series of fuels for the energy released during combustion...

We make a bomb calorimeter of some kind, (which means, we light the fuel with a wick and put a beaker of water over the flame and test the temperature change) of course, nothing I made was completely accurate... I know how to explain all that in the discussion though.

The fuels we tested were methanol, ethanol, butan-1-ol, and pentan-1-ol. (The series being the number of carbon atom's in each fuel increases).

I was hoping for help in the sense that, can any of you tell me much about the enthalpy of these fuels in relation to each other? I've got numbers and stuff from wikipedia but I can't really make any sense from the whole thing...

For example:
Methanol has 1 carbon atoms, so the chemical formula is CH3OH but it's density is 0.7918 g cm-3
While Ethanol has 2 carbon atoms, the chemical formula is C2H5OH and it's density 0.789 g cm-3 so it's less than methanol.

Yet butanol and pentanol both have higher density's in the 0.8's ?? Surely if the molecular weight increases the density increases (slightly at least). So why not for Ethanol???

And what does the density and molecular weight have to do with the energy released (or the enthalpy) when tested in the bomb calirometer??? Can someone please help me understand a bit??

Much Appreciated

[csrscience.com]

At what temperature? Are you sure that you are comparing both methanol and ethanol at the same temp?

[hoppingma-d]

At what temperature? Are you sure that you are comparing both methanol and ethanol at the same temp?

They were both room temperature, we didn't heat nor cool either fuel before the experiment... but I didn't find those numbers from the test... those were the set densities (given on Wikipedia which also = ? for accuracy)
actually the numbers I found were quite different now that I think about it... I measured 50mL in a beaker (accuracy = ?) and weighed them...

mm dear...

The weights I had for 50mL
methanol - 40.64g - 0.8184
ethanol - 40.07g - 0.8014
butanol - 39.76g - 0.7752
pentanol - 39.45g - 0.7748

Surely THAT makes no sense?

Would all this matter for the enthalpy or energy released?

Would you know where I could find the official set density's of these fuels vs. temperature online?

What should I be getting

[UG]

What are you actually trying to find? If you are doing what I think you're doing all you need the change in temperature and the mass of the water which was heated and that's about all, none of this density business.

[hoppingma-d]

Sounds about right...

ACTUALLY trying to find... to find the energy released from fuels in a series...
what we did was light the fuel up... waited until the water rose 5 degree's (we actually timed that for extra depth) and weighed the fuel after.

But doesn't the enthalpy depend on the density? depend on the molecular weight? and depend on.. WHAT? I just really don't understand...

[UG]

Okay, I see.
So let's say you had 10 grams of methanol to start off with and after the temperature of your water rose by 5^oC, there was 7 grams left, meaning that 3 grams was consumed. The enthalpy change = mass (of water) x c (specific heat capacity of water) x  :delta:T (change in temperature), c= 4.18 so you'll have, 50 x 4.18 x 5 = 1045 Joules of energy released. Then to calculate the enthalpy change per mole of ethanol, you work out the number of moles = 3/32 = 0.09375 moles, so the enthalpy change per mole is 1045 x 10.66 = 11140 J

[hoppingma-d]

Okay, I see.
So let's say you had 10 grams of methanol to start off with and after the temperature of your water rose by 5^oC, there was 7 grams left, meaning that 3 grams was consumed. The enthalpy change = mass (of water) x c (specific heat capacity of water) x  :delta:T (change in temperature), c= 4.18 so you'll have, 50 x 4.18 x 5 = 1045 Joules of energy released. Then to calculate the enthalpy change per mole of ethanol, you work out the number of moles = 3/32 = 0.09375 moles, so the enthalpy change per mole is 1045 x 10.66 = 11140 J

Well that makes THAT easier 

Am I shot if I ask though...

What makes say... butanol have a higher enthalpy change than methanol?
and is there a place I can find all the official enthalpy changes so I can compare my results to them?

Thanks for your help so far!!

[UG]

Note what you are looking for is the standard enthalpy of combustion of methanol, ethanol etc... not just standard enthalpy of ethanol because the word enthalpy is involved in alot of other things such as the standard enthalpy change of formation, standard enthalpy change of fusion etc...

Methanol: –715.0 kJ/mol

http://en.wikipedia.org/wiki/Methanol_(data_page)

Ethanol: –1370.7 kJ/mol

http://en.wikipedia.org/wiki/Methanol_(data_page)

You get the idea...

[hoppingma-d]

Yes! Of combustion... I guess that's important haha... thanks the references on those wiki pages are great!

But your still not explaining what makes methanol –715.0 kJ/mol when ethanol is Ethanol: –1370.7 kJ/mol

The difference being the extra carbon and hydrogen atoms... but... at a molecular level what is actually happening that ethanol is releasing that much more heat energy than methanol?

(feel free to leave the perfectionist at this point since you've already helped her heaps haha)

[UG]

Okay, miss perfectionist, tell me what you think of this explanation.
I am going to answer this by talking about bond enthalpies, I hope you know what they are, if you don't you'll find out soon enough I reckon...
Okay, sooo...bond breaking requires energy, whereas bond forming releases energy so they are endo- and exo- thermic reactions respectively. If we look at the combustion of methanol, CH₃OH + 3/2 O2  CO₂ + 2H₂O, we can calculate the enthalpy change by looking at the bonds broken and the bonds formed,  ΔH = bonds broken - bonds formed. Bonds broken = 3 C-H, 1 C-O, 1 O-H and 3/2 O-O. Bonds formed = 2 C-O and 4 H-O, taking some date out of an old text book, we have ΔH = (3x413 + 358 + 464 + 3/2x498) - (2x805 + 4x464) = -658 kJ/mol 
And for ethanol, C₂H₅OH + 3O₂  2CO₂ + 3H₂O bonds broken = 5 C-H, C-O, O-H and 3 O-O. Bonds formed = 4 C-O and 6 O-H.  ΔH =  -1623 kJ/mol. As you can see the C=O and O-H bonds formed are a lot stronger than the bonds that are broken, so more energy is released for ethanol because more of these stronger bonds are being formed than in methanol. I hope I have explained it properly...     

[hoppingma-d]

Okay, miss perfectionist, tell me what you think of this explanation.
I am going to answer this by talking about bond enthalpies, I hope you know what they are, if you don't you'll find out soon enough I reckon...
Okay, sooo...bond breaking requires energy, whereas bond forming releases energy so they are endo- and exo- thermic reactions respectively. If we look at the combustion of methanol, CH₃OH + 3/2 O2  CO₂ + 2H₂O, we can calculate the enthalpy change by looking at the bonds broken and the bonds formed,  ΔH = bonds broken - bonds formed. Bonds broken = 3 C-H, 1 C-O, 1 O-H and 3/2 O-O. Bonds formed = 2 C-O and 4 H-O, taking some date out of an old text book, we have ΔH = (3x413 + 358 + 464 + 3/2x498) - (2x805 + 4x464) = -658 kJ/mol 
And for ethanol, C₂H₅OH + 3O₂  2CO₂ + 3H₂O bonds broken = 5 C-H, C-O, O-H and 3 O-O. Bonds formed = 4 C-O and 6 O-H.  ΔH =  -1623 kJ/mol. As you can see the C=O and O-H bonds formed are a lot stronger than the bonds that are broken, so more energy is released for ethanol because more of these stronger bonds are being formed than in methanol. I hope I have explained it properly...     

yeah! Sounds about right... I'll find out the energy requried to break C-O, C-H and O-O and I'll be able to use that for all of them! Thanks so much... I'm so glad I don't have to explain why the densities don't aren't in sequence... haha Now all I have to figure out is why the Time's are so out of whack... Leave that for another time!

Thanks for your help and time! Appreciated 

[UG]

Just remember that the bond enthalpies are all in the gaseous phase whereas the standard ethalpy of combustion requires the substances to be in their standard states so that's probably why the numbers I got are not close to the ones on Wiki. Energy will be gained and released for converting liquid into gas and vice versa.

[hoppingma-d]

Okay, I see.
So let's say you had 10 grams of methanol to start off with and after the temperature of your water rose by 5^oC, there was 7 grams left, meaning that 3 grams was consumed. The enthalpy change = mass (of water) x c (specific heat capacity of water) x  :delta:T (change in temperature), c= 4.18 so you'll have, 50 x 4.18 x 5 = 1045 Joules of energy released. Then to calculate the enthalpy change per mole of ethanol, you work out the number of moles = 3/32 = 0.09375 moles, so the enthalpy change per mole is 1045 x 10.66 = 11140 J

quick question... I get all of it up to 1045 x 10.66 = 11140J

where's 10.66 coming from?

[UG]

0.09375 moles is 3/32 of a mole. So to calculate the enthalpy change per mole you multiply 1045 by 32/3, which is 10.66

[hoppingma-d]

So you multiply 0.9375 by 1045^2 = 11140J or 11.14 Kj/mol??