Quiz on empirical and molecular formula

  • #51
epenguin
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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
epenguin
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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?
 
  • #55
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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
epenguin
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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.
 
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  • #58
empirical formula = C2H5
 
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  • #59
epenguin
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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
 
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  • #61
epenguin
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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
epenguin
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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
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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
 
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  • #67
epenguin
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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 realised 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.
 
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  • #69
epenguin
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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 realised 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 realise 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 ?
 
  • #71
Yanik, now I understand what you mean lol....

C2H5 is possible if combined with another element. Some hydrocarbons include halogens. But, which halogen or non-halogen element is present in the unknown compound ? How can we find out the atomic number of the unknown element ? Now the question becomes finding out the number of atoms of an unknown element before solving to get the molecular formula (C2H5Ax)
 
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  • #72
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Don't add halogens, please. Add oxygen and, if you are sure you did that right, add oxygen and nitrogen.
 
  • #73
epenguin
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Don't add halogens, please. Add oxygen and, if you are sure you did that right, add oxygen and nitrogen.

I agree. If there had been anything else the questioners would have told you, and there are too many possibilities to go through. And there are regular chemical analysis methods for most of them. But they might not have told you about oxygen or even nitrogen whose analyses happen to be more difficult.

It doesn't really matter what this damned substance is (I am now convinced that the question contains a mistake). It does matter that the OP who appears to be revising chemistry, can do elementary reasoning and calculations about molecules. If he looks back he will see there was nothing difficult about the reasoning that got us from the data to the conclusion that the molecule must contain one of C2H5, C4H10 or C6H15. If you additionally assume the only other element is O it should equally be a matter of a minute to see what possible molecular formula, if any, is possible. Another kick yourself afterwards question.

If necessary, about combustion analysis I found this http://www.chem.ucalgary.ca/courses/351/Carey5th/useful/cdea.html Though personally I'd find it easier to do the calculation requested off my head than wade through it.
 
  • #74
The empirical formula with oxygen is C4H10O13. I don't know how I would add nitrogen since, I don't have the mass of the unknown compound so subtraction from 100% would not work.
 
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  • #75
epenguin
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The empirical formula with oxygen is C4H10O13. I don't know how I would add nitrogen since, I don't have the mass of the unknown compound so subtraction would not work and there are no compounds containing oxygen.

You don't give any reasoning how you get that formula so hard to help. However with 13 O atoms that empirical formula is also a minimum molecular formula, and it should hit you in the eye that that gives you way more than the molecular weight of 116. But more serious I can't see any way you could possibly get that 13, there must be some mistake. Are you using the molar mass?
 
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