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

1. Skoog, DA. Holler, FJ. Crouch, SR. Principles of Instrumental Analysis, 6^th Edition. Thomson Brooks/Cole (2007).
2. 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.
  
   
  
   
   
  

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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: 28 + 35 + 35 = 98 28 + 35 + 37 = 100 28 + 37 + 37 = 102 If you have the necessary maths, you could show that the chances of these arrangements occurring are in the ratio of 9:6:1 - and this is the ratio of the peak heights. If you don't know the right bit of maths, just learn this ratio! 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. The carbons and hydrogens add up to 29. The M+ and M+2 peaks are therefore at m/z values given by: 29 + 79 = 108 29 + 81 = 110 So . . . if you have two lines in the molecular ion region with a gap of 2 m/z units between them and with almost equal heights, this shows the presence of a bromine atom in the molecule.