# The enthalpy of combustion of alcohols, help with experiment?

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In chemistry we're supposed to find the enthalpy of combustion of propanol. In order to do that, we burned the propanol and heated up the water with it. Now we have:

time: 300s

mass of water: 50.0g

change in temperature of water: 24.2K

change in mass of propanol: -1.5g

specific heat capacity of water: 4.18J/g/K

molar mass of propanol: 60.11g/mol

So change in enthalpy of water is = cmΔT = 5057.8J

This is also change in enthalpy of 1.5g of propanol

We want the change of 1 mole so we calculate:

(5057.8 x -1.5)/60.11 = 202683J = 202.68kJ

However, the enthalpy of combustion of 1 mole of propanol is 2026.8kJ. It seems that I'm off by one decimal place, but where? I can't find my mistake!! I'm desperate please help me? Thank you!

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I think your temperature change is completely unrealistic, either your calorimetry was sort of missed up so so much energy was lost, or you didn't calculate it well. I think you should repeat the lab for a better lab.

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I think your temperature change is completely unrealistic, either your calorimetry was sort of missed up so so much energy was lost, or you didn't calculate it well. I think you should repeat the lab for a better lab.

It's change in temperature, not temperature, so it's fine.

We want the change of 1 mole so we calculate:

(5057.8 x -1.5)/60.11 = 202683J = 202.68kJ

Are you only off by one decimal place? I did that on the calculator, and that answer is wrong. How can you get a bigger number from that?

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I think your temperature change is completely unrealistic, either your calorimetry was sort of missed up so so much energy was lost, or you didn't calculate it well. I think you should repeat the lab for a better lab.

It's change in temperature, not temperature, so it's fine.

We want the change of 1 mole so we calculate:

(5057.8 x -1.5)/60.11 = 202683J = 202.68kJ

Are you only off by one decimal place? I did that on the calculator, and that answer is wrong. How can you get a bigger number from that?

Yeah i know that it is the change in temperature and it is very very low,

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I think your temperature change is completely unrealistic, either your calorimetry was sort of missed up so so much energy was lost, or you didn't calculate it well. I think you should repeat the lab for a better lab.

It's change in temperature, not temperature, so it's fine.

We want the change of 1 mole so we calculate:

(5057.8 x -1.5)/60.11 = 202683J = 202.68kJ

Are you only off by one decimal place? I did that on the calculator, and that answer is wrong. How can you get a bigger number from that?

Oh I actually meant (5057.8*60.11)/(-1.5)

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I think your temperature change is completely unrealistic, either your calorimetry was sort of missed up so so much energy was lost, or you didn't calculate it well. I think you should repeat the lab for a better lab.

But even if my temperature change was wrong, then the right one would have to be ten times bigger in order for me to get the right results...the mistake has to be somewhere else.

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dH for water is 5057.8J

1.5/60.11=~0.02495mol of propanol

5057.8/0.02495=202.7kJ. Different from your answer since I rounded. But I'm off by a decimal too...

The temp change seems find, a dT of 24.2K is the same as a dT of 24.2C, seems fine to me.

Tried it a different way and got the same answer. One of your numbers is wrong somewhere...and idk where

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It's not a calculation that's wrong, it's a systematic error. You're losing so much heat to the environment that your dH values are 1/10th of the literature values. I suspect that if you tried burning other fuels in the same manner you would experience the same error.

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It's not a calculation that's wrong, it's a systematic error. You're losing so much heat to the environment that your dH values are 1/10th of the literature values. I suspect that if you tried burning other fuels in the same manner you would experience the same error.

^Forgot about how terrible calorimeters are at being accurate. I think he might have found your problem

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It's not a calculation that's wrong, it's a systematic error. You're losing so much heat to the environment that your dH values are 1/10th of the literature values. I suspect that if you tried burning other fuels in the same manner you would experience the same error.

It's not a calculation that's wrong, it's a systematic error. You're losing so much heat to the environment that your dH values are 1/10th of the literature values. I suspect that if you tried burning other fuels in the same manner you would experience the same error.

Wow, only one tenth? hmm well if nobody can spot any calculation errors, I guess that is the problem :/

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It depends on what type of material you used. Styrofoam tends to work well but there is still huge amounts of error. And yes, 1/10th can happen. The heat is going into the air, outside of your container, into the container itself, into the gases made from burning the stuff, etc.

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