Tutorial on the Effects of Isotopes on Mass Spectral Patterns

     To understand how nuclear isotopes influence the mass spectral pattern, view the spectrum of methyl chloride while following the provided tutorial.

Activity 1:

Identify the most abundant ion and its isotope at m/z=52.


Q: Assuming the charge state of the ion is 1, what two ionic species principally contribute to the peaks at m/z=50 and m/z=52 ?



Measure the peak intensities of m/z=50 and m/z=52 (the y-axis values). Then use your measurements to calculate the ratio of 37Cl/35Cl in this sample.

/ =


      In a mass spectrum, the most abundant ion peak is interpreted as 100 percent relative abundance. Consequently, the other ion peaks are represented as a percentage relative to the most abundant ion. The experimental value you just obtained represents the ratio of  37Cl to  35Cl in the spectrum sample. If we are to know if our calculation represents a sample that is isotopically enriched (containing a percentage of isotopes not found naturally) we must check the reference values for the natural isotopic abundaces of Cl.



Now calculate 37Cl / 35Cl from the reference values below and compare it to your previously calculated experimental value.

Experimental Value:
    / =
Element/Isotope % Natural Abundance
35Cl 75.77 (5)
37Cl 24.23 (5)
clear gif
Values were obtained from the CRC Handbook of Chemistry and Physics. Vol. 72. D.R. Lide, editor-in-chief. CRC Press,Inc. (p. 11-33)






Q. According to your above results, is the spectrum sample of methyl chloride isotopically enriched?





Activity 2:

Look carefully at the region around m/z=15 on methyl chloride's mass spectrum.


Q. What is the ionic species of contributing the most to m/z=15? Hint: This peak represents a fragment of the parental molecule, (12C 1H335Cl)+.




The fragment at m/z=16 is much like the fragment at m/z=15 except it is only one mass unit away. It must contain either one 'heavy' carbon or one 'heavy' hydrogen (deuterium). Therefore we can expect that the ratio of m/z=16 and m/z=15 to be approximately equal to the natural isotopic abundance ratio of hydrogen or that of carbon.


Calculate the ratio of m/z=16 to m/z=15 from the spectrum.

/ =

Now calculate the ratio of  13C / 12C and  2H / 1H from the following table of natural isotopic abundances in order to compare it to your m/z=16 divided by m/z=15 ratio.


Element/Isotope % Natural Abundance
12C 98.90 (3)
13C 1.10 (3)
1H 99.985 (1)
2H 0.015 (1)
Experimental Value:

C ratio:       / =

H ratio: 3 x / =



We multiply the hydrogen ratio by 3 because each methyl ion has three hydrogen atoms. This means there is three times the probability of finding a single deuterium (2H).


Q. From your above results, what ion isotope is most likely the largest constituent of peak m/z=16?




Despite the enhanced probability for the hydrogen isotope, the abundance of 13C is still the greater contributor to the peak at m/z=16 and the isotopic abundances match fairly closely with the observed pattern. By using mathematical methods one can predict the expected distribution of isotopes and thus mass spectral peaks. This is an important method for showing consistency between the mass spectral data and the assigned molecular formula.