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I have been reading about resolution in mass spectrometry and there are a few things I do not understand.

The resolving power is determined by $m/(m_2-m_1)$ at full width half height of a peak. The higher the value obtained the better. This is where I struggle to form a link between the resolution of the mass spectrometer and the resolving power. In image (a) below the resolution of the mass spectrometer is 1000 and would have a very low mass resolution value, but in image (b) the resolution is 5000 and the nominal peaks masses are separated, therefore providing a high mass resolution value.

Why can a mass spectrometer with a resolution of 5000 resolve the nominal masses and the mass spectrometer with a resolution of 1000 cannot? And how does the mass resolution come into play? Also, what do the values of 1000 and 5000 mean?

enter image description here

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I'm not sure I completely understand what exactly your problem is, but:

You want to resolve to signals at approx. m/z = 1060.2 and m/z = 1061.2. So $\Delta m = 1$ The required resolution is: $\frac{m}{\Delta m} = \frac{1061}{1} \approx 1060$: The spectrometer in (a) has only 1000 which is too low. The spectrometer in (b) has 5000 which is quite more than needed.

Maybe it is easier to think of the reciprocal: $\frac{\Delta m}{m}$ which tells you how far two signal need to be apart so your spectrometer can resolve them.

In (a) this is $\frac{\Delta m}{m} = \frac{1}{1000} = \frac{1.06}{1060}$ two signals can be resolved if they are at least 1.06 m/z apart at m = 1060. I.e. if they were slightly further apart than they are.

The spectrometer in (b) can resolve signals that are at least $\Delta m/z = 0.21$ apart at $m/z = 1060$.

FWHM and resolution

The FWHM is basically part of the resolution. The $\Delta m$ is the m/z difference between two signals that are just resolved.

Have a look at the IUPAC's definitions of resolution in mass spec. There are several approaches to specify what "just resolved" means.

The FWHM can be used directly as the $\Delta m$ of the "peak width definition". If instrument A has $5\times$ the FWHM than instrument B, then resolution of A $\approx \frac{1}{5}$ resolution of B.

I'm more used to the "10 % valley" which roughly equals the full with at 5 % height: you specify how much signal is allowed at the minimum between the two peaks, e.g. 10 %.

It is important to realize that these two definitions differ by almost a factor 2!

Here are some results of "simulated" peaks at different resolutions and $\Delta m/z$ (all gaussian, which of course doesn't need to be the case in reality). I use the full width at 5% = 10 % valley definitions for the resolution.

The first row is resolution 5000 with the, second row is resolution 1000. You see that the two signals are much better resolved than in your example picture above. Which means that above another resolution definition was used, possibly $\Delta m$ = FWHM. In any case, a resolution of 1000 should be almost sufficient to resolve m/z 1060.2 from 1061.2.

The third row shows two signals of equal height with $\Delta m/z $ = 1 = FWHM. According to the resolution definitions above, this is a resolution of ca. 520. Note that while you can say that there are at least two peaks, you cannot take the two maxima as the m/z of the two underlying signals (grey).

The fourth row is almost the same, but I multiplied the second signal by 56% to account for the frequency of having bradykinin with one $^{13}$C. Looks quite similar to your picture.

example calculation for mass spec resolution

Also, have a look at M. P. Balogh: Debating Resolution and Mass Accuracy, LC•GC Europe, 17(3), 152–159 (2004).

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  • $\begingroup$ Firstly thank you for the answer it has cleared up some issues. But in the second paragraph you looked at the distance between the two peaks, when does the FWHM/FWHH come into this? Also, in the fourth paragraph where were the values 1.06 and 1060 obtained? $\endgroup$ – Harpal May 10 '12 at 11:48
  • $\begingroup$ @Harpal: the 1060 is approximately the m/z you're looking at. FWHM: I add this to answer. $\endgroup$ – cbeleites supports Monica May 12 '12 at 14:01
  • $\begingroup$ I didn't think properly when I wrote the first answer - of course $\Delta m/z = 1$. I got confused because I'm more familiar with the 10%-valley definition of the resolution, and the R = 1000 example definitively didn't look like "almost" resolved according to that... It is corrected now and I added a discussion of the two resolution concepts. $\endgroup$ – cbeleites supports Monica May 12 '12 at 15:55
  • $\begingroup$ Resolution has more than one definition - take a look at Harris' textbook, it has been of great help to me. Especially the subsection "Oh Mass Spectrum, Speak to me!" $\endgroup$ – CHM May 12 '12 at 16:30
  • $\begingroup$ @CHM: don't have the Harris here, and I didn't knowt the web site, thx. $\endgroup$ – cbeleites supports Monica May 12 '12 at 17:30

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