# The heat released when 1.00 g of ethanol (Mr = 46.0) undergoes complete combustion is

I have attemoted this question 3 times and still do not understand how I do it i was thinking to work out the moles of ethanol times it by 29.8KJ but before convert it into J and then times this by avg const. but im not getting any answer that is close? , am i miss an imoortant steP?

Sorry the full question is here: The heat released when 1.00 g of ethanol (Mr = 46.0) undergoes complete
combustion is 29.8 kJ
What is the heat released by each molecule, in joules, when ethanol undergoes
complete combustion?
(the Avogadro constant L = 6.022 × 1023 mol–1)
(edited 7 months ago)
Original post by Hersoul
I have attemoted this question 3 times and still do not understand how I do it i was thinking to work out the moles of ethanol times it by 29.8KJ but before convert it into J and then times this by avg const. but im not getting any answer that is close? , am i miss an imoortant steP?

Sorry the full question is here: The heat released when 1.00 g of ethanol (Mr = 46.0) undergoes complete
combustion is 29.8 kJ
What is the heat released by each molecule, in joules, when ethanol undergoes
complete combustion?
(the Avogadro constant L = 6.022 × 1023 mol–1)

Have you tried following essentially the same method, except try dividing by Avogadro’s constant instead of multiplying.

Edit: apologies, I had misread your initial method a little and now can see where the error is.
(edited 7 months ago)
Original post by TypicalNerd
Have you tried following essentially the same method, except try dividing by Avogadro’s constant instead of multiplying.

I have done it this way as well unfortunately it gives the answer 1.07 x 10^19 when the correct answer is meant to be 2.28 x 10^-18 J
Original post by Hersoul
I have done it this way as well unfortunately it gives the answer 1.07 x 10^19 when the correct answer is meant to be 2.28 x 10^-18 J

I just tried it by the method I had in mind and it worked. Let’s try a slightly different way.

Start by calculating the number of moles of ethanol in 1 gram
Original post by TypicalNerd
I just tried it by the method I had in mind and it worked. Let’s try a slightly different way.

Start by calculating the number of moles of ethanol in 1 gram

moles=mass/mr 1g/46(mr)= 0.0217 moles, then to work out the number of moles its 0.0217 x 6.022x10^23= 1.039x10^22 but then do we times by 0.0298J
Original post by Hersoul
moles=mass/mr 1g/46(mr)= 0.0217 moles, then to work out the number of moles its 0.0217 x 6.022x10^23= 1.039x10^22 but then do we times by 0.0298J

You pre-empted my next step… you are meant to work out how many molecules of ethanol are in the sample, but you do not multiply by 0.0298 J.

You then divide 29.8 kJ by the number of molecules in 1 gram, since 29.8 kJ is how much energy is released when 1 gram of ethanol is burnt.
Original post by TypicalNerd
You pre-empted my next step… you are meant to work out how many molecules of ethanol are in the sample, but you do not multiply by 0.0298 J.

You then divide 29.8 kJ by the number of molecules in 1 gram, since 29.8 kJ is how much energy is released when 1 gram of ethanol is burnt.

the ans I got for this step is 4.39 x 10^20
Original post by Hersoul
the ans I got for this step is 4.39 x 10^20

Are you sure that (29.8 kJ)/(1.309 x 10^22) equals that?

Also, apologies for the late response - I'm now a full time uni student.
(edited 7 months ago)
Original post by TypicalNerd
Have you tried following essentially the same method, except try dividing by Avogadro’s constant instead of multiplying.

Edit: apologies, I had misread your initial method a little and now can see where the error is.

First you have to time the Mr (46) with the converted combustion ( 29.8 x 10^3). Then divide it by the Avogadro constant L(6.022x10^23) and it gives you 2.276x10^-18 J by doing 3 significant figures 2.28x10^-18 J
I did it like that and It gave me the correct solution. I jope this can help other people
(edited 4 months ago)
Original post by garoi001.309
First you have to time the Mr (46) with the converted combustion ( 29.8 x 10^3). Then divide it by the Avogadro constant L(6.022x10^23) and it gives you 2.276x10^-18 J by doing 3 significant figures 2.28x10^-18 J
I did it like that and It gave me the correct solution. I jope this can help other people

That is a correct solution. Nicely done