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The presence of sulfur in a mass spec is 96% and 4%. I believe this to be because the natural occurrence of sulfur isotopes are within these proportions.

However, chlorine behaves in a 66%/33% ratio in mass spec, but the most abundance isotopes of chlorine are 75%/25%. Why is this so?

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    $\begingroup$ See also chemguide mentioning 3:1 atomic ions and 9:6:1 molecular ions, based on webbook.nist.gov $\endgroup$
    – Poutnik
    Commented Sep 3, 2020 at 13:40
  • $\begingroup$ What sort of compounds are you using ? $\endgroup$
    – Maurice
    Commented Sep 3, 2020 at 14:03
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    $\begingroup$ Or in more general terms: ratio (66:33 relative to each other) vs percent (75% and 25% relative to their sum). $\endgroup$ Commented Sep 3, 2020 at 15:29
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    $\begingroup$ Side note: isotopic distributions may vary both along time and place. This is why IUPAC's atomic weight for Li seems to be less and less precise today and compared to other elements. Isotopic fingerprints are used in geology not only to tell age, but origin of the samples (example) e.g.for the analysis of gemstones as well as in forensics (example). $\endgroup$
    – Buttonwood
    Commented Sep 3, 2020 at 21:35

1 Answer 1

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You are probably mixing natural abundance (NA) and relative abundance (RA). In mass spectrometry RA is a more valuable parameter as it can be directly obtained as the $y$-coordinate of a plotted mass spectra: the most abundant ion (isotope) corresponds to the base peak, which is always $100\%.$ In other words, RA reflect isotope ratio, not NA. For the isotopes of the elements RAs can easily be derived from NAs via normalization; however, the problem of finding RAs of the various isotopic molecular species is a bit less trivial [1].

The following table contains compiled data for NAs [2, p. 1-12] and RAs [3, p. 89] for both elements you've mentioned:

$$ \newcommand{\d}[2]{#1.&\hspace{-1em}#2} \begin{array}{lllrlrlr} \hline Z & \text{Isotope} & & &\text{Mass}/\pu{u} & \text{NA}&\hspace{-1em}/\% & \text{RA}&\hspace{-1em}/\%\\ \hline 16 & \ce{^{32}S} & \ce{[E]} & \d{31}{9720711744(14)} & \d{94}{99(26)} & \d{100}{000} &\hspace{-1em} \\ & \ce{^{33}S} & \ce{[E + 1]} & \d{32}{9714589098(15)} & \d{0}{75(2)} & \d{0}{789} \\ & \ce{^{34}S} & \ce{[E + 2]} & \d{33}{96786700(5)} & \d{4}{25(24)} & \d{4}{433} \\ \hline 17 & \ce{^{35}Cl} & \ce{[E]} & \d{34}{96885268(4)} & \d{75}{76(10)} & \d{100}{000} \\ & \ce{^{37}Cl} & \ce{[E + 2]} & \d{36}{96590260(6)} & \d{24}{24(10)} & \d{32}{399} \\ \hline \end{array} $$

References

  1. Margrave, J. L.; Polansky, R. B. Relative Abundance Calculations for Isotopic Molecular Species. J. Chem. Educ. 1962, 39 (7), 335. DOI: 10.1021/ed039p335.
  2. Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.
  3. Gross, J. H. Mass Spectrometry: A Textbook, 3rd ed.; Springer International Publishing: Cham, Switzerland, 2017. ISBN 978-3-319-54397-0.
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    $\begingroup$ So 66% / 33% is wrong, it should be 100% / 33% or 75% / 25% if you round a bit. For sulfur with the low abundance of S-34, ratio or fraction are similar, at about 4%. $\endgroup$
    – Karsten
    Commented Sep 4, 2020 at 1:17

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