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Introduction
The M+1, M+2 peaks (and beyond) can contain contributions from any of the heavy isotopes present in a molecule. For instance, 13C will contribute to the M+1 peak, but where two atoms of 13C are present in a molecule, that molecule will appear in the M+2 peak, indistinguishable from a molecule containing no heavy carbon, but a single heavy oxygen 18O atom. The size of the M+j peak will be the sum of all the contributions that each individual isotope could make to that peak. Therefore this page shows merely how to calculate the contribution of a single isotope to a single M+j peak. To get a full isotope pattern for a particular empirical formula (for instance C6H12O6) you would need to do a whole set of these calculations and sum the results. note: in theory the difference in mass between 13C and 12C is not exactly the same as that between 2H and 1H, and neither is exactly 1, so to talk of an "M+1" peak is an approximation. But it is an excellent approximation and mass specs are not precise enough to detect a cluster of peaks at approximately +1 caused by different heavy isotopes. How it works - the M+1 peak If you pick an atom at random, let the probability of it being heavy be a. Therefore the probability of it not being heavy is (1-a) Now imagine a molecule with a chain of atoms that might be heavy (shading indicates a heavy atom): The probability of finding this is given by: where n is the number of atoms of the interesting element in this molecule. The probability of finding the totally-light version is given by Therefore the relative intensity of these two things is But if we expressed the chance of finding a heavy isotope relative to that of finding the light isotope, rather than as a probability in total, then this would also be a/(1-a). Therefore the relative intensity of the heavy peak caused by molecules looking like A above is actually just the relative intensity of the heavy isotope itself, which we'll call α However, there are n atoms of the element in the chain, which means that there are n ways to draw a molecule containing a single heavy atom. For instance, we could equally well have drawn Therefore the final relative intensity of the M+1 peak is actually nα How it works - the (M+2) peak We are now interested in the probability of finding The probability of finding exactly this combination of heavy and light atoms is given by: So its relative intensity, relative to the entirely-light molecule (B) is: You will be spotting a pattern here: the probability of any number of heavies is simply α to the power of the number of heavies you want. However, there are now n(n-1) ways to arrange the heavy atoms (you can put the first one in any of the n positions, but the second one will now have its choice restricted to the remaining n-1 positions). Of these n(n-1) arrangements, half are indistinguishable. It doesn't matter if you put the first heavy in position 1 and the second in position 2, or the first in position 2 and the second in position 1. Therefore the number of M+2 combinations is given by And the final relative intensity of the M+2 peak is given by Some books quote an approximate method of calculating the size of the M+2 peak for a typical organic molecule based on
oxygen and carbon. You can very easily derive this approximate formula from the information above, bearing in mind that we
have worked entirely in fractions rather than percentages (so our α ranges
from 0 to 1, but can be expressed as 0 to 100%). The approximate formula also assumes that at high carbon content,
(n-1) is nearly equal to n, so the top half of the fraction can be replaced simply by n2. I personally feel this is a pointless approximation, because it saves virtually no work but yields an inaccurate answer: you might as well ignore carbon completely at low carbon content if you are going to go this way. How it works - M+j peaks As we saw above, the intensity of a molecule with a particular pattern of labelling, relative to the totally-light
version, is always αj where j is the number of heavy atoms. Therefore all that is needed to work out the relative intensity of the M+j peak is to find the number of ways the heavy atoms can be arranged. The number of ways we can place the heavy atoms is simply because there is one less choice available each time we place a new atom. This series goes on until we've placed j atoms. We can write this as a series: or we can write it more conveniently as: Actually if you were calculating this and wanted to do it quickly, you would probably use the series method to avoid wasted calculations, but if you're using a pocket calculator, then the factorial version is quite handy. Given j heavy atoms, they can be arranged in j! indistinguishable ways, so the final number of heavy arrangements we have to deal with is given by Therefore the relative intensity of the M+j peak is given by: (I am not a mathematician, but this method agrees with results calculated using the isotope viewer of the Excalibur software from Finnigan, which is probably a very good sign.)
