I am wondering why mathane is burned to CO$_2$ for isotopic analysis, e.g with a mass spec? I have looked into several text books, they all describe how to do it but not why. Is it because of a interference on mass 16? Analysing methane it it is would give you also information about the D/H ratio. So why?


1 Answer 1


There are different types of mass spectrometers with different modes of sample introduction, analyte ionization and ion selection/focusing. You cannot generalize "with a mass spec."

In many cases of isotope ratio mass spectrometry (IRMS), compound specific isotopic analysis (CSIA), or Gas Chromatography Combustion Isotope Ratio Mass Spectrometry organic compounds (not just methane) are burned and oxidized to carbon dioxide and water and analyzed as such to determine the total ratio of carbon-13/carbon-12 and hydrogen-2/hydrogen-1. However, combustion is not the only way to introduce compounds in the mass spectrometer.

Another secondary issue is that there if often more noise at low mass ranges. Methane is a small molecule (m/z =16) and carbon dioxide is relatively large (m/z=44).

  • $\begingroup$ Sorry for being not that precise. Let's consider ambient air. So methane is extracted and afterwards burned to CO2 because of noise. With noise you mean mass interferences like H2O? $\endgroup$
    – Zorg
    Commented Nov 11, 2020 at 13:54
  • $\begingroup$ Noise, means noise overall (from the instrument) not interferences. $\endgroup$
    – ACR
    Commented Nov 11, 2020 at 14:23
  • 1
    $\begingroup$ You are using improper notations for isotopes. For example, "C-13" should be either carbon-13 or $\ce{^{13}C}.$ $\endgroup$
    – andselisk
    Commented Nov 11, 2020 at 14:35
  • $\begingroup$ I have edited it. I know you like to have perfection in symbolism but one cannot be too orthodox or rigid in notations. Too much rigidity makes it easy to break. Do a Google Book search of "C-13" NMR and you will find book titles with C-13 NMR, and of course plenty of publications. $\endgroup$
    – ACR
    Commented Nov 11, 2020 at 14:46
  • 2
    $\begingroup$ @andselisk - I think you are too rigid about formatting on this site. This site isn't an ACS publication. I certainly understood what C-13 meant. It sure is a lot easier to input than $\ce{^{13}C}$. $\endgroup$
    – MaxW
    Commented Nov 11, 2020 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.