Quiz on empirical and molecular formula

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Discussion Overview

The discussion revolves around determining the molecular formula of an unknown organic compound based on combustion analysis data. Participants explore the relationships between the masses of carbon dioxide and water produced, the molar mass of the compound, and the implications for the compound's composition, including potential oxygen content.

Discussion Character

  • Homework-related
  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a problem involving combustion analysis and expresses confusion over their previous answers, suggesting a misunderstanding of the chemical principles involved.
  • Another participant prompts the calculation of moles of carbon dioxide and water produced from the combustion.
  • Participants discuss the stoichiometry of the combustion reaction, noting the number of carbon and hydrogen atoms derived from the products.
  • There is a proposal that the empirical formula derived from the combustion data leads to a molecular formula that seems chemically impossible, raising questions about the assumptions made regarding the compound's composition.
  • Some participants suggest that the compound may contain oxygen, which complicates the determination of the molecular formula based solely on carbon and hydrogen data.
  • There is a discussion about the limitations of combustion analysis in determining the presence of heteroatoms like oxygen in the compound.
  • Participants express uncertainty about how to calculate the oxygen content and the implications of the given data on the molecular formula.
  • One participant suggests that the problem may not provide enough information to arrive at a definitive answer regarding the molecular formula.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the molecular formula of the compound. There are multiple competing views regarding the presence of oxygen and the implications for the empirical and molecular formulas derived from the combustion data.

Contextual Notes

Participants note the lack of information regarding the mass of the unknown sample, which is critical for determining the oxygen content. There are unresolved questions about the assumptions made regarding the composition of the organic compound and the applicability of combustion analysis in this context.

  • #61
ghostanime2001 said:
C4H10, C6H15

so C4H10 seems to be the most probable compound

Do you have a reason for saying that?
 
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  • #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
 
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  • #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.
 
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  • #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 ?
 
  • #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
Don't add halogens, please. Add oxygen and, if you are sure you did that right, add oxygen and nitrogen.
 
  • #73
mfb said:
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
ghostanime2001 said:
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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  • #76
You could try it yourself. I have attached my paper showing my method. I can send you the file, if it's hard to read on here. If I assume 100 g of the compound, then I can use subtraction to find the mass of oxygen and it turns out to be around 40-41 g
 

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  • #77
ghostanime2001 said:
You could try it yourself. I have attached my paper showing my method. I can send you the file, if it's hard to read on here. If I assume 100 g of the compound, then I can use subtraction to find the mass of oxygen and it turns out to be around 40-41 g

I can read it and I am sorry but you are repeating an error that was pointed out in #20, 27, 30 and elsewhere - you cannot find the original amount of oxygen in the compound from the amount in the combustion products. Some of it may have come from the compound and some or all of it from the air, we don't know how much of each.

Then you know how many moles of C and H there are in a mole of the compound - at least we have 3 different possibilities. So if all the rest in a mole of compoiund is O, as we are assuming, can't you work out how many moles that must be? 100g has nothing to do with anything.
 
  • #78
I don't know what 3 possibilities you are referring to. We can't be sure we have a mole of the compound, we not given any information concerning it's mass. If you are talking about something else concerning oxygen then please be more specific.
 
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  • #79
ghostanime2001 said:
I don't know what 3 possibilities you are referring to. We can't be sure we have a mole of the compound, we not given any information concerning it's mass. If you are talking about something else concerning oxygen then please be more specific.

Oh dear, the three possibilities it took a lot to lead you to and have been said in #63, 73. The molar mass of the compound has been stated repeatedly starting by you in#1!
 
  • #80
if you mean that adding oxygen atoms to C2H5, C4H10 and C6H15 until the correct molar mass is obtained then it either the molar mass is below the required molar mass or too high.
 
  • #81
ghostanime2001 said:
if you mean that adding oxygen atoms to C2H5, C4H10 and C6H15 until the correct molar mass is obtained then it either the molar mass is below the required molar mass or too high.

I think that's it. Could you be more specirfic, e.g. with numbers?
 
  • #82
finding x until C2H5Ox = 116.28 g/mol, C4H10Ox = 116.28 g/mol, C6H15Ox = 116.28 g/mol
 
  • #83
ghostanime2001 said:
finding x until C2H5Ox = 116.28 g/mol, C4H10Ox = 116.28 g/mol, C6H15Ox = 116.28 g/mol

So find the x.
 
  • #84
x=5.455 for C2H5Ox, x=3.6425 for C4H10Ox, x=1.83 for C6H15Ox
 
  • #85
ghostanime2001 said:
x=5.455 for C2H5Ox, x=3.6425 for C4H10Ox, x=1.83 for C6H15Ox

And can the numer of O atoms in the molecule be fractions like those?
 
  • #86
x=5.455=1091/200, x=3.6425=1457/400, x=1.83=183/100
 
  • #87
ghostanime2001 said:
x=5.455=1091/200, x=3.6425=1457/400, x=1.83=183/100

I don't know what that's about. The question was, can you have fractions of an atom?
 
  • #88
no fractions, they can only be whole numbers
 
  • #89
ghostanime2001 said:
no fractions, they can only be whole numbers

OK, so this was why some of us concluded time ago that the coompoud could not be made of C, H and O alone and there must be something else and we looked at N.

Now to make sure you understand how to do this, suppose we make the hypothesis that there is H, C and N alone, no O. Can you show us the calculation with this to find what possibilities there are for the molecule with C, H and N?
 
  • #90
x=6.234 for C2H5Nx, x=4.163 for C4H10Nx, x=2.091 for C6H15Nx
 

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