Quiz on empirical and molecular formula

  • #26
I understand that, I can give you the masses of carbon and hydrogen contained in the original sample. But there is no information if the compound has oxygen or not. Are you saying that there is oxygen in the sample too ? The problem statement just said organic substance, so I assume it has ONLY carbon & hydrogen.
 
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  • #27
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EDIT: Can you now tell me why we cannot determine the O content in the manner which you proposed?

The information given is that it is an organic compound. This means it can be made up of C, H, O, N etc. Combustion analysis, in a typical question like this, does not allow one to know if or how many hetero-atoms (except O) are present. Given that they tell you the products are water and carbon dioxide, you can assume that the compound has C, H and maybe O. Otherwise you'd get other combustion products of N, S etc. So we are left to decide if we have only a hydrocarbon or an O containing hydrocarbon. Seeing as the assumption that we have only a hydrocarbon, without O, gives us a molecular formula which is impossible we are left to assume that the compound should contain O atoms. Then we are left with the task of determining the relative amount of O content in the compound, but I don't think there is a way to figure that out based on the information provided. Typically the O content is calculated by using conservation of mass but that requires knowing the mass of the sample that was combusted.

I hope I didn't give you the false impression that I know how to solve the question exactly, I was only trying to get you to understand the thought process. I'm inclined now to believe that there is not enough information in the problem to get an answer. There may be some trick, or what not, which may give an answer. Guess and check might work. You know the relative amounts of C and H in your compound and you know the molar mass of the compound. It may be possible to try different molecular formulas where you keep the C and H ratio but vary the O content until you get the correct molecular weight. It is extremely annoying but maybe possible.

Are you sure that this was the question exactly as written?
 
  • #28
yes, I have typed the question exactly as it was written on my test paper. I was thinking that too, I am not given the mass of the unknown sample, so I can't find the mass of oxygen. The only way I know to find the mass of oxygen is to add the masses of oxygen in carbon dioxide and water together.

vary the O content until you get the correct molecular weight
what you mean by that ?
 
  • #29
The mole ratios I come up with if I include oxygen in addition to carbon & hydrogen is C : H : O = 1 : 2.5 : 3.25 so if I multiply each by 4 to get whole numbers, I get 4 : 10 : 13 (C4H10O13), which is just absurd!

There are more oxygen atoms than carbon and hydrogen atoms !! :(
 
  • #30
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That is because you are CANNOT assume that ALL the Oxygen atoms in your products are from the unknown compound.

Before we proceed further I need you to understand why you can't count the O atoms in the products (that is the water and carbon dioxide) as coming only from your organic compound. I'd like you to read the link I posted, or a Gen Chem textbook or any other resource you'd like and explain to me why you cannot assume the Oxygen atoms in the water and carbon dioxide are ALL from the unknown organic compound. This is the first step that you need to get straight in your head.

Before I help you any further I want you to explain to me the basic principles behind combustion analysis. You are still blindly throwing around procedures without thinking about what you are doing.
 
  • #31
ok I understand why now, because dry oxygen gas was used to burn the unknown sample in the analyzer
 
  • #32
epenguin
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Not so obvious. I agree the C/H proportions give you C8H20 which also fits the molar mass nicely. But as the OP said there are no C8 compounds that can fit that ratio. There is too much H.

But even if you have oxygen in the compound, it doesn't help as far as I can see. If you replace any C-H with C-O-H you don't increase the H/C ratio. So i see just two ways to increase the H/C. One is by water of crystallisation. But then to be plausible you have to incorporate O into the molecule as well. Considering the molar mass I think this needs to be a C4 molecule in which case you could fit the C/H ratio even without the water of crystallisation. There is an outside possibility with a C6 molecule and water, but I can't see anything that adds up in either case.

I am taking it that any molar mass measurement that works at all will give you the MMass of compound without the water, and that we can take the accuracy of the measurements given as good.
 
  • #33
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ok I understand why now, because dry oxygen gas was used to burn the unknown sample in the analyzer

That is good that you get this because at least you got something from this situation.

Not so obvious.

I am glad that I'm not the only one who is stumped. I haven't been able to figure it out myself.

Nothing that I've tried has seemed to work. Anything beyond C6 doesn't fit the 2n+2 criteria. Trying C4H10 gets ugly numbers.

Maybe there is supposed to be some kind of other hetero-atom, but that seems like a silly possibility.
 
  • #34
looks like we are all stumped on this question. Where do we go from here ?
 
  • #35
epenguin
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Maybe I've got it! Maybe we were meant to bring in something that was mentioned and then forgotten - nitrogen. When I include the possibility of N and use the well-known approximate whole number atomic masses there is a solution that I think is unique.

The only trouble is when I calculate it using accurate atomic masses I have disagreement with the given 116.26 by about 0.1 - a whole 1 in 1,000 discrepancy!:blushing: Perhaps the OP could try the calculation with this idea, or someone else check the figures, or maybe I've overlooked some other solution?

That incredibly accurate :surprised molar mass figure given makes the problem look to me a contrived one. Surely a figure that accurate could not be result of experimental measurement; one would only calculate that figure once one knew the answer?
 
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  • #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.
 
  • #37
epenguin
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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...
 
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  • #39
epenguin
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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 realised 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 ?
 
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  • #41
epenguin
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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
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@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
epenguin
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@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
epenguin
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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?
 
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  • #46
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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 highschool, radicals in quantitative problems was not in the curriculum. Let's not try to get off topic guys please...
 
  • #48
epenguin
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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
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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.

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.
 

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