Question

Hi, I need help with a chemistry virtual lab and the questions that go with it? I really don't understand so if someone could help me get to answer then that would be awesome.

Answer 1

Post the problems.

Answer 2

1. Calculate \[DeltaT \] for the water in the benzoic acid combustion by \[DeltaT = |T _{f}-T _{i}|\]

Calculate the moles of benzoic acid (MW = 122.13 g/mol) . n = mass sample/molecular weight

Answer 3

That is the first problem. How would I do this?

Answer 4

Not enough information.

Answer 5

What information is needed? I don't know how to do this. I have a table also but I don't know what this question means or what information to use. What is this question asking?

Answer 6

@primeralph Are you still here? Can you help me?

Answer 7

You should provide all the information given in the problem. I have to go though.
Good luck.

Answer 8

Hi,

You would need to obtain the final and initial temperatures of the benzoic acid from the chemistry virtual lab (this is a practical session done virtually i.e. in your teacher's simulated chemistry lab environment, which would possibly be hosted online). You can then input these values for Tf & Ti and use them to calculate Delta(T). The same goes for the benzoic acid - you will need to record the mass of benzoic acid which you have used for the combustion (in your chemistry virtual lab session) and use this to deduce the number of moles of benzoic acid used.

Answer 9

benzoic acid sucrose
mass of sample 0.9488 0.9953
initial temperature 25.001 25.009
final temperature 30.411 28.264

That is the information that I recorded in the virtual lab. Where do I go from here?

Answer 10

You can use the information in my previous post and in the original question to deduce the values. Try it & post your answers here again.

Answer 11

I don't know how. How do I do this? Sorry, I'm just super confused and have no clue how to even do this problem or what is what.

Answer 12

Delta(T) = |Tf - Ti| = |30.411 - 25.001| = 5.41 degrees
Number of moles of benzoic acid = m/MW = 0.9488/122.13 = 0.007769 mol = 7.769mmol

Answer 13

Oh. OH. God, I just realized the Tf is the final temperature and Ti is the initial temperature, huh? God I feel so stupid. God, how am I not completely failing? Wow, I feel dumb.

Answer 14

2. \[DeltaH _{comb}\] for benzoic acid can be calculated by \[DeltaH = C _{system}DeltaT/n\] , where n is the moles of benzoic acid in the sample and \[C _{system}\] is the heat capacity of the calorimetric system. If the accepted value for the heat of combustion for benzoic acid is 3226 kJ/mol, calculate the heat capacity \[(C _{system})\] of the calorimetric system.

What does this mean? Do you understand this and know what I should do?

Answer 15

Try it. You already know Delta(T) from the previous part. DeltaH is the heat of combustion for benzoic acid.

And at your previous post: all of us make slips at times. Is this a new topic for you? if so, it is highly acceptable that you would be unlikely to infer that Ti refers to initial temperature and Tf refers to final temperature.

Answer 16

Alright, I'll try.
Yes, this is a new topic. It's super confusing to me.

Answer 17

Give it a go - you'll see how easy it gets when you do it. Also, please post your answer and method here when you've done it.

Answer 18

So Delta(H) is 3226kJ/mol
Delta(T) is 5.41
N is 7.769
right?

Answer 19

Take care of the units - they must all be in SI units. So Delta(H) should be in joules and n should be in moles.

Answer 20

How do I do that?

Answer 21

Convert DeltaH to Joules (converting from kJ to J is the same as converting from kg to grams - how do you do this?)

In my previous post, I gave the value of n in moles (abbreviated as mol) and in millimoles (mmol). You'd need to use the value represented in mol, NOT the one in mmol.

Answer 22

Oh okay so n is 0.007769 mol

Answer 23

Yes that's right ... can you work out what is DeltaH (in J/mol)?

Answer 24

Don't you like divide or multiply by 1000 or something?

Answer 25

Yes, but do you divide or multiply? Remember, kJ is a bigger unit than J, so a number which appears large when represented in J will be reduced when converted to kJ - that's the reason why we use these units anyway ...

Answer 26

Are you able to figure out the answer to this question? If so, please post your answer here. Thanks.

Answer 27

So DeltaH = 3226000 J/mol ...

Answer 28

Sorry, things came up and I haven't been able to continue until now. Sorry.

So to go from kJ to J you multiply by 1000. So 3226 * 1000 = 3226000
Then it'd be DeltaH = (Csystem * DeltaT)/n which is
3226000 = (Csystem * 5.41)/0.007769 right?
I know there's like some way to figure out Csystem using like an algebraic equation or something but I don't know how to rearrange things to solve for it. How would I figure it out using the information that I have? @shiraz14

Answer 29

@shiraz14
would it be:
(n * DeltaH)/DeltaT = Csystem
Making it:
(0.007769 * 3226000)/5.41 = Csystem
Csystem = 4632.679
Is that right or wrong? That's what someone else said that they thought would be the equation to get Csystem but since they aren't sure I was wondering if you knew.

Answer 30

Or does anyone else know that can help? Thanks

Answer 31

@primeralph
Do you know if this is correct?

Answer 32

@JJ_Stark : You're correct. So, the heat capacity of the system (Csystem) is 4.632679 kJ/mol.

Answer 33

Oh yay! Thank you! Can you check my work for the next set of problems? I think I got the answers but I want to make sure I did them right.

Answer 34

@JJ_Stark : Please upload them in a separate thread. Thanks.

Answer 35

Even if they're related to this question?