Let's say the exchange fraction of 0.000475 would have been incorrectly determined (for whatever reasons). Let's also say the actual exchange fraction is 0.05 (as it is more or less for raw cotton; see RCM 2014, 28, 545-552).
With a d2H(non-ex) of -177.8 the choice of equilibration waters would have yielded d2H(total) of -178.7 and -168.2 for water A (-196.1) and water B (+14.5) respectively. The difference between d2H(A) and d2(B) for equilibrated samples would thus be 10.5.
From: Stable Isotope Geochemistry [mailto:[log in to unmask]] On Behalf Of WASSENAAR, Leonard Irwin
Sent: 20 December 2018 08:06
To: [log in to unmask]
Subject: Re: [ISOGEOCHEM] help interpreting 2H exchange exp't
Hi Andrew - the devil is always in the prep and analysis details. Something seems wrong to me with such a low exch-H fraction. What is the residual moisture content (gravimetrically)? How did you treat / dry the sample, etc. It's also possible - by a fluke - that the H isotope fractionations of the 2 waters you used landed you right on the identical value of the non-exch H. I would go far outside the natural range - i.e. very positive on one of the waters (like +500). Len
From: Stable Isotope Geochemistry <[log in to unmask]> On Behalf Of A.J. Tanentzap
Sent: Wednesday, 19 December 2018 20:35
To: [log in to unmask]
Subject: [ISOGEOCHEM] help interpreting 2H exchange exp't
I've run an exchange experiment on a solid organic as described in Meier-Augenstein et al. 2011. Rapid Commun Mass Spectrom.
I estimated the non-exchangeable 2H ratio to be -178. The estimated fraction of exchangeable H in the sample was 0.0005.
When I re-run the sample in a bulk analysis, the total (i.e.
exchangeable + non-exchangeable) ratio was -190.
Would someone be so kind as to explain how this is possible with a near-zero fraction of exchangeable H?
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