MASS SPECTRA - THE M+1 PEAK


This page explains how the M+1 peak in a mass spectrum can be used to estimate the number of carbon atoms in an organic compound.


Note:  This is a small corner of mass spectrometry. It would be a good idea not to attack this page unless you have a reasonable idea about how a mass spectrum is produced and the sort of information you can get from it. If you haven't already done so, explore the mass spectrometry menu before you go on.



What causes the M+1 peak?

What is an M+1 peak?

If you had a complete (rather than a simplified) mass spectrum, you will find a small line 1 m/z unit to the right of the main molecular ion peak. This small peak is called the M+1 peak.

In questions at this level (UK A level or its equivalent), the M+1 peak is often left out to avoid confusion - particularly if you were being asked to find the relative formula mass of the compound from the molecular ion peak.

The carbon-13 isotope

The M+1 peak is caused by the presence of the 13C isotope in the molecule. 13C is a stable isotope of carbon - don't confuse it with the 14C isotope which is radioactive. Carbon-13 makes up 1.11% of all carbon atoms.

If you had a simple compound like methane, CH4, approximately 1 in every 100 of these molecules will contain carbon-13 rather than the more common carbon-12. That means that 1 in every 100 of the molecules will have a mass of 17 (13 + 4) rather than 16 (12 + 4).

The mass spectrum will therefore have a line corresponding to the molecular ion [13CH4]+ as well as [12CH4]+.

The line at m/z = 17 will be much smaller than the line at m/z = 16 because the carbon-13 isotope is much less common. Statistically you will have a ratio of approximately 1 of the heavier ions to every 99 of the lighter ones. That's why the M+1 peak is much smaller than the M+ peak.


Using the M+1 peak

What happens when there is more than 1 carbon atom in the compound?

Imagine a compound containing 2 carbon atoms. Either of them has an approximately 1 in 100 chance of being 13C.

There's therefore a 2 in 100 chance of the molecule as a whole containing one 13C atom rather than a 12C atom - which leaves a 98 in 100 chance of both atoms being 12C.

That means that the ratio of the height of the M+1 peak to the M+ peak will be approximately 2 : 98. That's pretty close to having an M+1 peak approximately 2% of the height of the M+ peak.


Note:  You might wonder why both atoms can't be carbon-13, giving you an M+2 peak. They can - and do! But statistically the chance of both carbons being 13C is approximately 1 in 10,000. The M+2 peak will be so small that you couldn't observe it.


Using the relative peak heights to predict the number of carbon atoms

If there are small numbers of carbon atoms

If you measure the peak height of the M+1 peak as a percentage of the peak height of the M+ peak, that gives you the number of carbon atoms in the compound.

We've just seen that a compound with 2 carbons will have an M+1 peak approximately 2% of the height of the M+ peak.

Similarly, you could show that a compound with 3 carbons will have the M+1 peak at about 3% of the height of the M+ peak.

With larger numbers of carbon atoms

The approximations we are making won't hold with more than 2 or 3 carbons. The proportion of carbon atoms which are 13C isn't 1% - it's 1.11%. And the appoximation that a ratio of 2 : 98 is about 2% doesn't hold as the small number increases.

Consider a molecule with 5 carbons in it. You could work out that 5.55 (5 x 1.11) molecules will contain 1 13C to every 94.45 (100 - 5.55) which contain only 12C atoms. If you convert that to how tall the M+1 peak is as a percentage of the M+ peak, you get an answer of 5.9% (5.55/94.45 x 100). That's close enough to 6% that you might assume wrongly that there are 6 carbon atoms.

Above 3 carbon atoms, then, you shouldn't really be making the approximation that the height of the M+1 peak as a percentage of the height of the M+ peak tells you the number of carbons - you will need to do some fiddly sums!


Important!  Most syllabuses at this level (UK A level or its equivalent) don't ask for this. Before you get too bogged down in all this, check your syllabus and recent past papers to see whether you need to bother. This is a case where you really need to know what line your examiners are currently taking - for example, how far are they pushing the approximation.

If you are doing a UK-based exam, and haven't got a syllabus and past papers, follow this link to find out how to get them.




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© Jim Clark 2000