Quiz on empirical and molecular formula

  • #71
ghostanime2001
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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
ghostanime2001
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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 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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  • #76
ghostanime2001
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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
epenguin
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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
ghostanime2001
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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
epenguin
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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
ghostanime2001
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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
epenguin
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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
ghostanime2001
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finding x until C2H5Ox = 116.28 g/mol, C4H10Ox = 116.28 g/mol, C6H15Ox = 116.28 g/mol
 
  • #83
epenguin
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finding x until C2H5Ox = 116.28 g/mol, C4H10Ox = 116.28 g/mol, C6H15Ox = 116.28 g/mol

So find the x.
 
  • #84
ghostanime2001
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x=5.455 for C2H5Ox, x=3.6425 for C4H10Ox, x=1.83 for C6H15Ox
 
  • #85
epenguin
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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
ghostanime2001
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x=5.455=1091/200, x=3.6425=1457/400, x=1.83=183/100
 
  • #87
epenguin
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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
ghostanime2001
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no fractions, they can only be whole numbers
 
  • #89
epenguin
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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
ghostanime2001
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x=6.234 for C2H5Nx, x=4.163 for C4H10Nx, x=2.091 for C6H15Nx
 
  • #91
epenguin
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x=6.234 for C2H5Nx, x=4.163 for C4H10Nx, x=2.091 for C6H15Nx

OK, so what conclusion do you draw from that?
 
  • #92
ghostanime2001
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there is nitrogen in the compound C6H15N2 ?
 
  • #93
epenguin
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there is nitrogen in the compound C6H15N2 ?

Yes you could suspect that.

It is a slightly difficult one. The question is whether that 0.09 (I haven't checked) excess is significant.

As a source I believe I gave earlier http://www.chm.davidson.edu/vce/stoichiometry/ch.html says "Be aware that the elemental analysis is not perfectly accurate. The experimental error will generally produces atom ratios that are not perfect integers but are close to integers." But it doesn't give any details of accuracy to be expected from the combustion method. I'm think it is pretty accurate, and am fairly sure that nearly 5% deviation is enough to say that that is not a whole number. Yes it is surely more accurate than that and any problems would have to do with the sample. Comments by others invited.

So let us consider the remaining possibility that we have C, H, O and N and see if we get anything more convincing. Can you work out some possibilities?

Hint: actually I think you will find it easier to see where you are going if you use just the whole-number approximations to atomic masses for now.
 
  • #94
ghostanime2001
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so you mean like varying the number of N and O atoms until I get something close to 116.28 g/mol ?
 
  • #95
epenguin
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so you mean like varying the number of N and O atoms until I get something close to 116.28 g/mol ?

Yes. As I said, 116 would do and would be faster. There are not very many possibilities.
 
  • #96
ghostanime2001
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C2H5O2N4 = 117 g/mol
C4H10ON3 = 116.17 g/mol
C6H15ON = 117 g/mol
 
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  • #97
epenguin
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:approve: One of those is more convincing and is very close to the molar mass given in the problem, though not quite exact.
I can't be here continually so more comment later.
 
  • #98
epenguin
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Let me draw threads together and point out some issues.

First treating this as slightly more-than-average hard exercise routinely posed to students. What I do is I believe what I'm told, suppose accuracy of measurements good. Atomic masses are near whole numbers and it is faster, you will see, if I work with those at first. The analysis gives me a C/H ratio close to 2/5 - an empirical formula you could say C2H5. Then C8H20 fits the actual molar mass of the compound. But chemically no such compound is possible. We then looked for a smaller molecule with the same C/H ratio, but with O and/or N to make up the molar mass to 116. Initial candidates must be based on C2H5, C4H10 and C6H15. Using the whole-number atomic mass approximation I rapidly eliminate the first and third of these. These part-molecules having an odd number molar mass, combined with any number of O or N which have even number mass (16 and 14) cannot make the molar mass of 116 which is even. That narrows it down to C4H10 with approximate mass 58. From 116 we need to make up 116 - 58 = 58 with O and N. 58 is not a multiple of either 16 or 14, so only with both could we find an answer. From 58 subtract 16 once, twice, three times and see if any result is a multiple of 14. Turns out the only one corresponds to N3O. So we would conclude the molecule is C4H10N3O. When I calculate the exact molar mass of that I get 116.14. That would normally be considered excellent and conclusive agreement with the #1 question figure of 116.28 but more anon.

The didactic point is that this is just an example calculation of the kind students are required to perform, usual principles, though slightly more complex than average. I cannot see anything not straightforward and even obvious about it (at most you might miss some shortcut) and do not understand why it took so long. Maybe the student now realizes what point he was missing and needs to do some more exercises to make sure he is at ease with them.

There are however some dubious points. My calculated molar mass of 116.14 is extremely close to 116.28 but we are not really comparing theory with experiment which is approximate - we are comparing theory with theory. You would only work out a figure like that from a composition and the precisely known atomic masses, so I don't know what explains the discrepancy. It is possible that this was an invented school exercise, and the authors thought of C8H20 not noticing it was impossible. For that I calculate a MM of 116.24 which is nearer though still not quite the one of the problem. At this point I should say the atomic masses that are used in such calculations are updated from time to time. I am using
H 1.00794, C 12.010107, N 14.00674, O 15.9994 Perhaps someone will check, if there are other figures around.

Molecular masses as far as I know are these days mostly determined by mass spectroscopy. But then from the fragments you can determine the composition too, and for that matter the molecular structure from mass spectroscopy. But maybe one machine cannot do everything and I am not very familiar with the state of the art.

I said ‘believe the data’. My faith wavered a bit when the OP pointed out four formulae which were within about 1% of right. In fact my shortcut rather depends on these being ruled out beforehand. But actually you can weigh within 0.1% easily so I guess this is right (though accuracy downstream does nothing if there is a problem with the sample). But you see how critical the accuracy question is to the scope of the method. The didactic exercises like http://www.chm.davidson.edu/vce/stoichiometry/ch.html just give nice numbers and pass over this. The Addison-Wesley book cited surely devotes consideration to this question and maybe the student could re-read that bit with new appreciation. Accuracy, how to check this and what you think the experiment is telling you, problems encountered, precautions etc., essentially scientific and methodological questions enriching the bare academic exercise.

Finally I found it quite difficult to find a structure that corresponds to the formula C4H10N3O. But if we can find only one we can certainly find a number of others. Is there anything wrong with
NH2NHCCHOHCH2CH2NH OR NH2NHCCH2OCH2CH2NH?

The question looks like a mistake but it has given us some useful issues to examine.
 
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  • #99
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Is there anything wrong with
NH2NHCCHOHCH2CH2NH OR NH2NHCCH2OCH2CH2NH?
The last N in both molecules has just 2 bonds.
 
  • #100
sjb-2812
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I think you can tell by the nitrogen rule that an odd number of nitrogens leads to an odd number of hydrogens (or equivalents, like halogens), so I'm not convinced by this at all. Unless we have ions / radicals, which I think we discounted earlier?
 
  • #101
epenguin
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You're both right. It was a now you see it now you don't, that's it ah no sort of thing.

The other unconvincing thing was (I had thought of trying azo-compunds) I found a site that gives possible structures from molecular formula called ChemSpider which gave me

HN=N+=NCHCH3CHOHCH3 and variations.

We have a charged molecule. Charged molecules like quaternary amines are quite respectable. They have to be in salts though. And I thought things like this were very unstable, even explosive. Not likely to be in a student excercise.
 

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