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Adam R. Johnson, Harvey Mudd College
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deuterium labeling in CHN analysis

My student sent a labeled compound for CHN analysis.  Her data came back all wrong.  I remembered (in the dim recesses...) that when doing CHN analyses for deuterated compounds, you have to do something different.  But I have no idea what it is.  But, technology has advanced in the intervening years, so I had chemdraw automatically calculate numbers.

Chemdraw calcd for C14H15D6NO2:  %C:  69.67, %H:  11.27, %N:  5.80

Chemdraw calcd for C14H21NO2:  %C:  71.46, %H:  8.99, %N:  5.95

Found:  %C:  69.50, %H:  8.42, %N:  5.46

 

we know the compound is pure, so were are basically trying to remember how to massage the data so that it is right.  Chemdraw apparently does NOT do it right?

Adam 

 

 

 

Nancy Scott Burke Williams, Scripps College, Pitzer College, Claremont McKenna College
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This is a SWAG (Scientific Wild-Assed Guess).

It certainly used to be that CHN was determined by finding the mass of water emitted by burning the compound in an O2 flame (I think by absorbing the H2O on CaCl2 and weighing it or some damn fool thing like that).  I'm not sure that this is how it's actually done at whichever lab you used.  For example, if they're determining moles of hydrogen by AA or ICP (I have no idea whether these are practical for determining H), and dividing by the mass of sample, then you won'i get a mass corresponding to the deuteria, and your % H s/b 8.76% H (in other words, actually lower than the per-protio compound!).

Jeffrey Bodwin, Minnesota State University Moorhead
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I've always understood CHN the same way, the amount of combustion products are determined by mass after being adsorbed/attached to some sort of substrate.  Doing a quick web search, I'm not so sure that's always true.  There's a little footnote on the following link:

http://web.chemistry.gatech.edu/~marder/website/resources/eapricelist2005.pdf

This seems to imply that their CHN analyzer does not determine the mass of combustion products, but actually counts the combustion product molecules coming off the sample.  {Hmm, maybe they're not even combustion products in this method....}  So it would appear that the "correct" way to analyze your results depends upon the type of analyzer used.  I'd contact the lab that did the analysis, if they can't answer your question, then I think we're all in trouble.  Be sure to let us know the results.

Adam R. Johnson, Harvey Mudd College
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IIRC, you calculate your mass %H by assuming all the D's are H's and that the lab doesn't actually weigh anything, but determines moles.  But, they did weigh the sample originally, and report "moles H + D"*(1.00794)/(mass sample) which would be 8.77% (as Scott mentioned). 

I contacted the lab...

Dear Adam Johnson,

Thank you for your message.

Your communication is important to us and you will be contacted soon by a Columbia Analytical representative.

Thank you,

Columbia Analytical Services
1.800.695.7222
www.caslab.com

I'll let you know... 

Adam R. Johnson, Harvey Mudd College
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    first response from Columbia.  Based on that, I am guessing that Scott was right and they are determining moles of H2O/D2O, and the correct %H is 8.77.


The CHN analyses performed on your samples is based on the Dumas method.  Durring the combustion of the sample in the presences of oxygen, carbon is converted to CO2, hydrogen to H2O, and Nitrogen to NOx then reduced to N.  These gases are separated by a packed column and their intensity is measured by thermo-conductivity.  The intensity is proportional to mass of the gases formed and is converted and reported on a percent by weight basis of C, H, and N.

Deuterium is going to react the same as hydrogen in the instrument.  I will need to do some research on how deutrium impacts the way our instrument calculates final results.  I will give you an answer tomorrow.

Nancy Scott Burke Williams, Scripps College, Pitzer College, Claremont McKenna College
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Sounds right.  Since the Dumas method measures moles of gas by PV=nRT (sheesh-they do this with *water*? No wonder my EAs are off!), they're measuring moles, not mass.

Graeme McAlister, University of York, UK
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Hi, new member here.

I know this topics was a while ago (?), but does anyone know if there was ever an answer?

I'm running samples of completely deuterated material, and wondered how the results could be adjusted to account for the heavier D.

The analyser uses a similar method of detection as that in answer #5, ie thermal conductivity before and after chemical removal of the water/D2O produced during combustion

 

Many thanks

Kyle Grice, DePaul University
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Hi Graeme,

Welcome! What's the analyzer you are using?

If its going by TCD, the signal would be attenuated by the difference in thermal conductivity... 

Doing a quick search, i found this paper:

https://aip.scitation.org/doi/abs/10.1063/1.1725357

It looks like it depends on the temp you are using...I'd be able to help more if I knew what instrument you are using.

Cheers,

Kyle 

Graeme McAlister, University of York, UK
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Hi Kyle Thanks for getting back to me. It's an Exeter Analytical CE440 that I use. Cheers Graeme
Kyle Grice, DePaul University
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Hi Graeme,

Have you emailed the company? It looks like they'd be responsive/helpful based on their internet presence. 

Another option would be to put a deuterated NMR solvent in (if you can do liquid samples) and see what it tells you. 

Let us know what you find out

Kyle 

Graeme McAlister, University of York, UK
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Hi

I contacted the US branch of the suppliers, and they outlined a method of converting H to D using a conversion factor based on the relative masses of H and D. In addition, that paper you cited, said that the themal conductivities of H2O and D2O were the same at my instrument operating temperature

Long story short, I ran the sample today and the results look good:

C12D17Br

Theory: %C, 55.80  %D, 13.26  %Rest, 30.94

Found: %C, 55.58  %H, 6.64  %Rest, 37.78

However, the analyser doesn't differentiate H and D and since the TC's are the same, it just calculates %H

Applying the conversion factor (1.9985) we get:

Found: %C, 55.58  %D, 13.27  %Rest, 31.15

:-)

Thanks for all your help!

Graeme

Kyle Grice, DePaul University
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Hi Graeme,

Cool, glad it was figured out.

Kyle