Skip to main content
Chemistry LibreTexts

Mass Spectrometry: Isotope Effects

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Figure 1 and the isotope abundance data from Table 1, we can see that there are 1.08 13C atoms for every 100 12C atoms, and the 13C peak will be 1.08% as large as the 12C peak.

    The example above is simple, but the same methods can be applied to determine isotope peaks in more complicated molecules as well. The molecule C4Br1O2H5 has several isotope effects: 13C, 2H, 81Br, 17O, and 18O all must be taken into account. First we will look at the (M+1)+ peak in comparison with the M+ peak. Only isotopes that will increase the value of M by 1 must be taken into consideration here – since 81Br and 18O would both increase M by 2, they can be ignored (the most abundant isotopes for Br and O are 79Br and 16O). Like the previous example, there are 1.08 13C atoms for every 100 12C atoms. However, there are 4 carbon atoms in our molecule, and any one of them being a 13C atom would result in a molecule with mass (M+1). So it is necessary to multiply the probability of an atom being a 13C atom by the number of C atoms in the molecule. Therefore, we have:

    4C * 1.08 = 4.32 = molecules with a 13C atom per 100 molecules

    We can repeat this analysis for 2H and 17O:

    5H * 0.015 = 0.075 = molecules with a 2H atom per 100 molecules

    2O * 0.04 = 0.08 = molecules with a 17O atom per 100 molecules

    Any of the three isotopes, 13C, 2H, or 17O occurring in our molecule would result in an (M+1)+ peak. To get the ratio of (M+1)+/M+, we need to add all three probabilities:

    4.32 + 0.075 + 0.08 = 4.475 = (M+ 1)+ molecules per 100 M+ molecules

    We can say then that the (M+1)+ peak is 4.475% as high as the M+ peak.

    A similar analysis can be easily repeated for (M+2)+:

    1Br * 98 = 98 = molecules with an 81Br molecule per 100 molecules

    2O * 0.2 = 0.4 = molecules with an 18O molecule per 100 molecules

    98 + 0.4 = 98.4 = (M+2)+ molecules per 100 M+ molecules

    The (M + 2)+ peak is therefore 98.4% as tall as the M+ peak.

    This method is useful because using isotopic differences, it is possible to differentiate two molecules of identical mass numbers.

    Contributors and Attributions

    • Morgan Kelley (UCD)

    Mass Spectrometry: Isotope Effects is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?