Quiz on empirical and molecular formula

[epenguin]

x=6.234 for C₂H₅N_x, x=4.163 for C₄H₁₀N_x, x=2.091 for C₆H₁₅N_x

OK, so what conclusion do you draw from that?

 

[ghostanime2001]

there is nitrogen in the compound C₆H₁₅N₂ ?

 

[epenguin]

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.

 

[ghostanime2001]

so you mean like varying the number of N and O atoms until I get something close to 116.28 g/mol ?

 

[epenguin]

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.

 

[ghostanime2001]

C₂H₅O₂N₄ = 117 g/mol
C₄H₁₀ON₃ = 116.17 g/mol
C₆H₁₅ON = 117 g/mol

 

[epenguin]

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.

 

[epenguin]

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.

 

[mfb]

Is there anything wrong with
NH₂NHCCHOHCH₂CH₂NH OR NH₂NHCCH₂OCH₂CH₂NH?

The last N in both molecules has just 2 bonds.

 

[sjb-2812]

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?

 

[epenguin]

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 exercise.