Mass Spectrometry: Isotope Effects
Mass Spectrometry: Isotope Effects
This blogpost focuses on the effects of isotopes on mass spectra. It also
addresses practical applications of isotope measurement using a mass
spectrometer.
Introduction
The
ability of a mass spectrometer to distinguish different isotopes is one
of the reasons why mass spectrometry is such a powerful technique.The
presence of isotopes – a presence that is ubiquitous in nature – gives
each fragment a characteristic series of peaks with different
intensities.These intensities can be predicted based on the abundance of
each isotope in nature, and the relative peak heights can also be used
to assist in the deduction of the empirical formula of the molecule
being analyzed.
Isotopes
An
‘isotope’ of any given element is an atom with the same number of
protons but a different number of neutrons, resulting in a different
overall mass. Almost all elements have a variety
of naturally occurring isotopes – some notable exceptions are fluorine,
phosphorous, sodium, and iodine. For example, Carbon has 6 protons. In its most common isotope, it also has 6 neutrons, with a total number of protons +neutrons of 12 (called the ‘Mass Number’). This carbon atom, with a mass number of 12, is written: 12C. Carbon has two other naturally occurring isotopes, 13C and 14C, which have 7 and 8 neutrons, respectively. Table 1 below gives the most abundant naturally occurring isotopes for a variety of important elements. The
abundance for secondary isotopes is reported in number of atoms of
secondary isotope for every 100 atoms of the most abundant isotope1.
Element
|
Most Abundant Isotope
|
Secondary Isotope
|
Abundance/100 atoms of Primary Isotope
|
Hydrogen
|
1H
|
2H
|
0.015
|
Carbon
|
12C
|
13C
|
1.080
|
Nitrogen
|
14N
|
15N
|
0.370
|
Oxygen
|
16O
|
17O
|
0.040
|
18O
|
0.200
| ||
Sulfur
|
32S
|
33S
|
0.800
|
34S
|
4.400
| ||
Chlorine
|
35Cl
|
37Cl
|
32.50
|
Bromine
|
79Br
|
81Br
|
98.00
|
Silicon
|
28Si
|
29Si
|
5.100
|
30Si
|
3.400
|
Table 1. Abundance of several naturally occurring biologically important isotopes1.
Qualitative Peak Analysis
Mass spectrometry distinguishes elements based on a mass to charge ratio, m/z. Because of this, isotopes play an important role in mass spectra. Each
isotope will show up as a separate line in any mass spectrum with good
enough resolution.The Y-axis on a mass spectrum is relative
intensity.Therefore, the height of each of the peaks will correspond to
the relative abundance of each isotope in the sample. As an example, examine Figure 1.This is a mass spectrum of a natural sample of atomic carbon. 12C, with a natural abundance of 98.89%, naturally has a very high peak. 13C, with a natural abundance of only 1.11%, has a very low peak. The radioactive 14C has such a low natural abundance that it is not even seen in relation to the other two carbon isotopes2. The 12C peak in this spectrum would be called the Base Peak, and would be labeled M+. The
base peak is the largest peak in a spectrum, and the intensity of every
other peak is reported in comparison to the base peak. The 13C peak, with an atomic mass of 1 greater than 12C, would be labeled (M+1)+. 14C, if it were present, would be (M+2)+.
Figure 1.
Mass spectrum for atomic carbon. Note how the natural abundance of the
carbon isotopes corresponds with peak height in the mass spectrum.
Picture adapted from the IN-VSEE.
Quantitative Peak Analysis
We
can see from our qualitative example and Figure 1 that the intensity of
each isotope peak is proportional to the abundance of the isotope in
the sample. Because of this, it is relatively easy to predict the relative heights of each isotope peak for any given molecule. We can use the natural abundance of each isotope to predict how large each isotope peak will be in comparison to the base peak. Looking at the Carbon atom example in 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.
References
- Skoog, DA. Holler, FJ. Crouch, SR. Principles of Instrumental Analysis, 6th Edition. Thomson Brooks/Cole (2007).
- Coursey, J. S., Schwab, D. J., Dragoset, R. A. (2001). Atomic Weights and Isotopic Compositions. National Institute of Standards and Technology, Gaithersburg, MD.
Outside Links
- http://en.wikipedia.org/wiki/Mass_spectrometer
- This wikipedia page is about the Mass Spectrometer instrument.
- http://en.wikipedia.org/wiki/Mass_spectrum_analysis
- This wikipedia page is more directly related to isotope effects, as it focuses on reading mass spectra.
- http://www.chem.uoa.gr/applets/AppletMS/Appl_Ms2.html
- This applett is fun to play with. It generates isotope peaks in a specified mass fragment.
MASS SPECTRA - THE M+2 PEAK
This page explains how the M+2 peak in a mass spectrum arises from
the presence of chlorine or bromine atoms in an organic compound. It
also deals briefly with the origin of the M+4 peak in compounds
containing two chlorine atoms. | |
| |
The effect of chlorine or bromine atoms on the mass spectrum of an organic compound
Compounds containing chlorine atoms One chlorine atom in a compound | |
Note: All the mass spectra on this page have been drawn using data from the Spectral Data Base System for Organic Compounds (SDBS) at the National Institute of Materials and Chemical Research in Japan. With one exception, they have been simplified by omitting all the minor lines with peak heights of 2% or less of the base peak (the tallest peak). | |
The molecular ion peaks (M+ and M+2) each contain one chlorine atom -
but the chlorine can be either of the two chlorine isotopes, 35Cl and 37Cl. The molecular ion containing the 35Cl isotope has a relative formula mass of 78. The one containing 37Cl has a relative formula mass of 80 - hence the two lines at m/z = 78 and m/z = 80. Notice that the peak heights are in the ratio of 3 : 1. That reflects the fact that chlorine contains 3 times as much of the 35Cl isotope as the 37Cl one. That means that there will be 3 times more molecules containing the lighter isotope than the heavier one. So . . . if you look at the molecular ion region, and find two peaks separated by 2 m/z units and with a ratio of 3 : 1 in the peak heights, that tells you that the molecule contains 1 chlorine atom. You might also have noticed the same pattern at m/z = 63 and m/z = 65 in the mass spectrum above. That pattern is due to fragment ions also containing one chlorine atom - which could either be 35Cl or 37Cl. The fragmentation that produced those ions was: | |
Note: If you aren't sure about fragmentation you might like to have a look at this link. | |
Two chlorine atoms in a compound | |
Note: This spectrum has been simplified by omitting all the minor lines with peak heights of less than 1% of the base peak (the tallest peak). This contains more minor lines than other mass spectra in this section. It was necessary because otherwise an important line in the molecular ion region would have been missing. | |
The lines in the molecular ion region (at m/z values of 98, 100 ands
102) arise because of the various combinations of chlorine isotopes that
are possible. The carbons and hydrogens add up to 28 - so the various
possible molecular ions could be:
So . . . if you have 3 lines in the molecular ion region (M+, M+2 and M+4) with gaps of 2 m/z units between them, and with peak heights in the ratio of 9:6:1, the compound contains 2 chlorine atoms. Compounds containing bromine atoms Bromine has two isotopes, 79Br and 81Br in an approximately 1:1 ratio (50.5 : 49.5 if you want to be fussy!). That means that a compound containing 1 bromine atom will have two peaks in the molecular ion region, depending on which bromine isotope the molecular ion contains. Unlike compounds containing chlorine, though, the two peaks will be very similar in height.
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