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)