Quiz on empirical and molecular formula

In summary, the conversation discusses a problem in which an unknown organic compound was analyzed using a carbon-hydrogen analyzer. The combustion of the compound produced 35.2 g of carbon dioxide and 18.0 g of water, and the molar mass of the compound was found to be 116.28 g/mol. The participants try to determine the molecular formula of the compound, with one student initially suggesting C2O4.H2O but later realizing this is incorrect. They then discuss the maximum number of hydrogens in alkanes, alkenes, and alkynes and try to determine the empirical formula of the compound. The conversation also explores the possibility of the compound containing oxygen and the challenges in determining the molecular formula from
  • #36
The question said carbon-hydrogen analyzer, and reading from the textbook "Chemistry 11" 2003 from Addison-Wesley it doesn't say anything about nitrogen concerning the bulbs that absorb carbon dioxide or water.
 
Physics news on Phys.org
  • #37
ghostanime2001 said:
The question said carbon-hydrogen analyzer, and reading from the textbook "Chemistry 11" 2003 from Addison-Wesley it doesn't say anything about nitrogen concerning the bulbs that absorb carbon dioxide or water.

All you have is the C/H ratio (you are not given the weight of original sample) and the molar mass of the compound. You have to try to home in on an answer just from those.

Experimental.
Notoriously there are no good universal methods of classical chemical type for measuring total nitrogen. (You can fairly easily measure certain types of N such as primary amines.) I do not have the book you mentioned but it is probably something like here http://www.chm.davidson.edu/vce/stoichiometry/ch.html where it says "any nitrogen in the sample is reduced to N2, which is unreactive" i.e. will leave no trace in measurements. I seem to vaguely remember doing an experiment like that at school; the CO2 was measured gravimetically as an insoluble BaCO3 precipitate I think.

Theoretical So you have a C/H ratio which you can take as accurate. You have the total molar mass. Although there is O in the CO2 and H2O you don't know how much of it was in the original molecules, so the measurements tell you nothing about this. And you don't know how much N there was.

But you also have atomic masses. And you have chemical principles. And from these you were already able to exclude one possiblity. And the fact that no one was able to find a solution when, I think, they considered only C, H and O is positive - it means the data are very constraining! So try and come up with something in as systematic a reasoning as possible (try and spell it out) involving C, H, O and N even if there remain some doubts like I have.
A hint is I think the fact the approximate whole number atomic masses of O and N are both even helps rapid elimination of some possibilities.
 
  • #38
I have read that webpage that you posted. How would I determine the mass of the unknown element in the unknown compound ? Do I assume 100 g ? so then I can subtract the mass of carbon & hydrogen from 100 g to get the mass of the unknown element. But I don't know the molar mass of the compound :( this sucks...
 
Last edited:
  • #39
ghostanime2001 said:
to be very honest, I am stumped on this one, from head to toe.

Most of the progress so far you made yourself so don't give up. Stumped by negative results is different from stumped by not trying. You got C8H20. But you realized that's is chemically impossible by itself. You cannot get that proportion of H to C with C and H alone. Not with C20. But you can with a smaller molecule. Not just anything smaller, the ratios limit you to few possibilities. List them. But then a smaller molecule doesn't have mole mass around 116. But we are free to, and we have to, make up that mass with O and/or N atoms.

(I could also have got to the same conclusion saying adding O to my C20 doesn't increase the possible H/C as explained previously. I can increase the ratio if I have N in the molecule. But that will make the MM greater than 116 too, unless the number of C is less.)

So try some possibilities and report even results that seem negative or we can't help you.
 
  • #40
Assuming 100 g of the compound and including nitrogen in the sample, doesn't exactly make anything easier. By the way, what calculation made you believe that this is a chemically possible organic compound ?
 
Last edited:
  • #41
ghostanime2001 said:
I have read that webpage that you posted. How would I determine the mass of the unknown element in the unknown compound ? Do I assume 100 g ? so then I can subtract the mass of carbon & hydrogen from 100 g to get the mass of the unknown element. But I don't know the molar mass of the compound :( this sucks...

Posts overlapped. You do have the molar mass of the compound, see #1.
 
  • #42
@epenguin: While I can find a C,H,N,O combination with the approximate molar mass, how would that look like as a molecule? I think you get 1/2 of a bond somewhere, or a missing/additional hydrogen.
 
  • #43
You're right
Posts overlapped. You do have the molar mass of the compound, see #1.
but how would I use that to determine the mass or molar mass of the unknown element ? and therefore the atomic mass of the unknown element ?
 
  • #44
mfb said:
@epenguin: While I can find a C,H,N,O combination with the approximate molar mass, how would that look like as a molecule? I think you get 1/2 of a bond somewhere, or a missing/additional hydrogen.

That's running ahead! :smile:

I'd think with the number of atoms we have to play with we will be able to think of many molecules within the rules.
 
  • #45
ghostanime2001 said:
You're right
but how would I use that to determine the mass or molar mass of the unknown element ? and therefore the atomic mass of the unknown element ?

Just start with the fact that it can't be C8H20. But the ratio of number of atoms C/H is 8/20. So what are the possibilities for the number of C and H atoms in the molecule?
 
Last edited:
  • #46
epenguin said:
That's running ahead! :smile:
How could others follow (your advice) if you are not running ahead?

I'd think with the number of atoms we have to play with we will be able to think of many molecules within the rules.
I did not find a proper solution. Sure, you can make ions or radicals, but I doubt this is an intended solution.
I just have some doubts that this N addition will lead to something useful.
 
  • #47
This was high school, radicals in quantitative problems was not in the curriculum. Let's not try to get off topic guys please...
 
  • #48
The OP seems to have abandoned, though the next step is very easy - we have established that there are less than 8 carbon atoms and I think most people know all the numbers that are less than 8.

That seems to me to swiftly lead to a unique answer that is anywhere near; it is as I said very near but not quite. Assuming this is the molecular formula I do after all find it difficult to think of a molecular one. Well I did find one that is notoriously semi-stable. :wink: I say one, but you are always going to get several for the price of one as you can move groups about. These searches could be a bit formalised, I am halfway doing that. Keep getting near, but there is a conspiracy giving me one H to many or too few - when you change anything you usually gain or lose two H, but I have not concluded.
 
  • #49
searching for a compound with fewer than 8 carbons seems like painful work. There are thousands of organic compounds and several compounds could have similar molar mass. I don't think this question was meant to be more confusing than it's meant to be.
 
  • #50
epenguin said:
Keep getting near, but there is a conspiracy giving me one H to many or too few - when you change anything you usually gain or lose two H, but I have not concluded.
That is not just bad luck, it is a fundamental problem and the basis of my post #42.
You can calculate the number of hydrogen atoms needed if there are no rings and no double bindings - for the number of C, N O you assumed, this is an odd number. Every ring and every double binding reduces this number by 2 - you always keep an odd number, this does not fit to the even number you need to satisfy the C/H ratio together with the molecular mass.

That's why I asked if you can imagine a molecule with that. Otherwise, this direction leads nowhere.

ghostanime2001 said:
searching for a compound with fewer than 8 carbons seems like painful work. There are thousands of organic compounds and several compounds could have similar molar mass. I don't think this question was meant to be more confusing than it's meant to be.
With just C N O H, a known C/H ratio and the total (small) molecular mass, it is not so hard to find all possible sum formulas.
 
  • #51
ghostanime2001 said:
searching for a compound with fewer than 8 carbons seems like painful work. There are thousands of organic compounds and several compounds could have similar molar mass. I don't think this question was meant to be more confusing than it's meant to be.

Right now we are trying to establish a molecular formula, the numbers of atoms in the molecule, on the basis of information we have. This is really quite easy, certainly the next step. There are not thousands of numbers less than 7, there are only 7 of them! You know the ratio of H atoms to C atoms. On that basis what number of C atoms are possible?

Frankly after you get that you'll kick yourself. I think you'll do well then to ask yourself what was stopping you get it, it may be useful to you since you seem to be voluntarily going through old papers for revision or something. I would be interested to understand myself.

So try and answer the above question. If having tried you still can't say anything about that come back for more hints.
 
  • #52
I know the reduced ratio of C:H is 2:5 from 8:20 ratio but I don't undestand what you said there about something having only 7 compounds with the above mentioned ratio.. to be quite honest, I don't really know if this question was meant to be this complicated. I guess I do need some more hints.
 
  • #53
ghostanime2001 said:
I know the reduced ratio of C:H is 2:5 from 8:20 ratio but I don't undestand what you said there about something having only 7 compounds with the above mentioned ratio.. to be quite honest, I don't really know if this question was meant to be this complicated. I guess I do need some more hints.

I said essentially, several times now, the compound can have only 7, 6, 5, 4, 3, 2 or 1 C atoms.

Which of those are commpatible with your 2:5 C:H ratio?
 
  • #54
2 carbon compound ?
 
  • #55
ghostanime2001 said:
2 carbon compound ?

Yes, that appears compatible. Why? Are any others compatible? Which are incompatible?
 
  • #56
Any others would not be compatible because the molecular formula is a integer multiple of empirical formula ?
 
  • #57
ghostanime2001 said:
Any others would not be compatible because the molecular formula is a integer multiple of empirical formula ?

A true statement but I don't see any logic as an answer. But maybe we can do it this way.

What is the empirical formula of this compound? I mean just the C and H part of it.
 
Last edited:
  • #58
empirical formula = C2H5
 
Last edited:
  • #59
ghostanime2001 said:
empirical formula was C2H5

What then are possible molecular formulae in our case?
 
  • #60
The possible molecular formulas may be:

1. C4H10 n=2
2. C6H15 n=3
3. C8H20 n=4
 
Last edited:
  • #61
ghostanime2001 said:
C4H10, C6H15

so C4H10 seems to be the most probable compound

Do you have a reason for saying that?
 
  • #62
C4H10 has the correct maximum number of hydrogens, 2n+2
 
  • #63
ghostanime2001 said:
C4H10 has the correct maximum number of hydrogens, 2n+2

That is rather dodgy. Something like that might be true if this were an alkane, but don't we already know the compound is not C4H10 and there must also be some other kind of atom there?

So for the moment the only possibilities are C2H5, C4H10 and C6H15.

Now suppose we think there is only one other type of atom there, oxygen. Can you see how to go about finding what number of O atoms there could be in the above three cases?
 
  • #64
Why do you say
but don't we already know the compound is not C4H10

get the total number of oxygen atoms from CO2 and H2O then multiply that by 2 or 3 (molecular formula integer multiple in C4H10 and C6H15)
 
  • #65
ghostanime2001 said:
Why do you say

get the total number of oxygen atoms from CO2 and H2O then multiply that by 2 or 3 (molecular formula integer multiple in C4H10 and C6H15)

We've already said we do not know where these O atoms come from. Whatever number there are in the compound won't affect how much CO2 and H2O we get, which will just depend on the C and H.

Another hint. A key in school chemistry (and math) and not just school is "use all the information you are given". Usually questions see to it you have just the information you need (real life problems are not so easy and you have to know yourself what you need). But I have to keep reminding you of information (little) you were given! Refer to #1 again.
 
  • #66
let me just ask, do you already know the answer ? I have re-read post #1 and I still don't know other information missing. We won't get into the matter of oxygen in the compound, as you've already established that. How would I know there exists another unknown element without knowing the mass of the unknown compound. I have used all the information I'm given, I don't see any other information that I have missed. Maybe it said only "carbon-hydrogen analyzer" so I assume there is carbon and hydrogen ONLY ? Maybe I need to assume 100 g of the compound so I can determine mass percent ? .. since mass percentage remain the same regardless of the amount of the compound. I think you mean something like this maybe? http://chemwiki.ucdavis.edu/Analyti...ve_Analysis/Virtual:_Carbon_Hydrogen_Analysis
 
Last edited:
  • #67
ghostanime2001 said:
let me just ask, do you already know the answer ? I have read it over 3 times and I still do not know what other information I'm missing. We won't get into the matter of oxygen in the compound, as you've already established that. How would I know there exists another unknown element without knowing the mass of the unknown compound. I have used all the information I'm given, I don't see any other information that I have missed. Maybe it said only "carbon-hydrogen analyzer" so I assume there is carbon and hydrogen ONLY ?

We've already been through that! With only C and H you got C8H20 - this fits the molar mass but you realized it was chemically impossible. So we conclude it must be from a smaller molecule with the given C/H ratios, and have now narrowed the possibilities to three. But with the numbers of C and H atoms in those cases we don't make the molar mass which is near 116. For that reason we decided there must be another element there and the most obvious was O.

Now it doesn't matter whether you agree about this, or whether (as I am beginning to think) the guys who set the problem made a mistake or you miscopied it, if from the data you can't reason something out about what the formula of a substance containing C, H, and O must be using the given information, then you do have a problem with elementary chemistry that you need to address. So I propose we continue the reasoning to see the answer we can get.

I mentioned the need to use all information. It is a help that there is so little. But anyway you know that to get a molecular formula from the empirical formula you need the molar mass, so you know what to look for.

Do you know what you need now to get a molecular formula in the C4 compound and the other possibles if the molecule, contains only C, H and O?
 
  • #68
I have not copied it wrong, I can scan the test paper to show you if you think I miscopied it. Also, I have done exactly what is said in the textbook. The textbook I used is Addison-Wesley Chemistry 11 ISBN: 0201708124 http://toronto.canadianlisted.com/b...11-ontario-edition-25-north-york_1749785.html. If you think I have a problem with elementary chemistry, why don't you tell me what you think is the answer then ? if you don't, then you're just as lost as me.
 
Last edited:
  • #69
ghostanime2001 said:
let me just ask, do you already know the answer ? I have read it over 3 times and I still do not know what other information I'm missing. We won't get into the matter of oxygen in the compound, as you've already established that. How would I know there exists another unknown element without knowing the mass of the unknown compound. I have used all the information I'm given, I don't see any other information that I have missed. Maybe it said only "carbon-hydrogen analyzer" so I assume there is carbon and hydrogen ONLY ?

We've already been through that! With only C and H you got C8H20 - this fits the molar mass but you realized it was chemically impossible. So we concluded it must be from a smaller molecule with the given C/H ratios, and have now narrowed the possibilities to three. But with the numbers of C and H atoms in those cases we don't make the molar mass which is near 116. For that reason we decided there must be another element there and the most obvious was O.

Now it doesn't matter whether you agree about this, or whether (as I am beginning to think) the guys who set the problem made a mistake or you miscopied it, if from the data you can't reason something out about what the formula of a substance containing C, H, and O must be using the given information, then you do have a problem with elementary chemistry that you need to address. So I propose we continue the reasoning to see the answer we can get.

I mentioned the need to use all information. It is a help that there is so little. But anyway you know that to get a molecular formula from the empirical formula you need the molar mass, so you know what to look for.

Do you know what you need now to get a molecular formula in the C4 compound and the other possibles if the molecule, contains only C, H and O, without having measured the O content?

OK as this is to slow and I will have to continue tomorrow, I hope you realize you need the approximate atomic mass of O , which probably you remember is 16. So what conclusions can you make using that?
 
  • #70
add atomic mass of oxygen to empirical formula mass ?
 

Similar threads

  • Biology and Chemistry Homework Help
Replies
4
Views
1K
  • Biology and Chemistry Homework Help
Replies
1
Views
2K
  • Biology and Chemistry Homework Help
Replies
8
Views
3K
  • Biology and Chemistry Homework Help
Replies
13
Views
3K
  • Biology and Chemistry Homework Help
Replies
10
Views
2K
  • Biology and Chemistry Homework Help
Replies
28
Views
4K
  • Biology and Chemistry Homework Help
Replies
4
Views
2K
  • Biology and Chemistry Homework Help
Replies
2
Views
1K
  • Biology and Chemistry Homework Help
Replies
6
Views
5K
  • Biology and Chemistry Homework Help
Replies
6
Views
8K
Back
Top