The output data of mass spectrometers is relative intensity (y axis) over m/z (x axis). How is the distance between m/z values on the x-axis called (in case of centroided data and profile/raw data respectively)?

  • 4
    $\begingroup$ I think using "distance between $m/z$" or "distance between peaks" is already fine. Alternatively, maybe use $Δ(m/z)$. $\endgroup$ – andselisk Oct 28 '19 at 15:18
  • $\begingroup$ Probably the question was not clear. I mean the distance between m/z values on the x-axis, not the distance between m/z values of actual samples. $\endgroup$ – thinwybk Oct 28 '19 at 15:49
  • $\begingroup$ I still stand by my answer. The term 'neutral loss' is only applicable inside a single fragmentation series. When you have a larger molecule, you are ought to have parallel, independent fragmentation series, and he distance between two random peaks is then generally not due to a loss of a neutral species. The resolution, also mentioned above, is only defined between two peaks that are just slightly separated by your MS instrument. In general, the distance between two arbitrary peaks in a complicated spectrum has no name. $\endgroup$ – Ezze Oct 29 '19 at 14:28
  • $\begingroup$ @Ezza After discussion with a collegue of mine the answer is still not completely clear. Pls let me some more time for more internal discussions. $\endgroup$ – thinwybk Oct 29 '19 at 16:09

I wanted to back up my comment and make sure it's justified, so I decided to flip through Gross'Mass spectrometry: a textbook [1] and it looks like there are indeed basically two possible variations, including distance between $m/z$ values, as you suggested:

  • {Distance | difference | mass difference} between {peaks | signals | peak tips}
  • {Difference in $m/z$ | $m/z$ difference | $Δ(m/z)$}

Both refer to describing isotopic distribution as well as to reading and interpreting the spectra.

Selected parts touching standardization (emphasis mine):

From [1, p. 11]:

The distance between peaks on that axis has the meaning of a neutral loss from the ion at higher $m/z$ to produce the fragment ion at lower $m/z.$ Therefore, the amount of this neutral loss is given as “$x~\pu{u}$”, where the symbol $\pu{u}$ stands for unified atomic mass. It is important to notice that the mass of the neutral is only reflected by the difference between the corresponding $m/z$ values, i.e., $Δ(m/z).$ This is because the mass spectrometer detects only charged species, i.e., the charge-retaining group of a fragmenting ion.

Further summary from [1, p. 20]:

1.7.1 Basic Terminology in Describing Mass Spectra


  1. Ranges in spectra or ranges set in operating a mass spectrometer are to be referred to in the form of “$m/z~10–100$” or “$m/z~10$ to $m/z~100$”.


  1. Neutral losses (rarely called dark matter […]) are exclusively recognized from the distance between peaks expressed in terms of the difference $Δ(m/z).$ The mass of the corresponding neutral is then given in units of $\mathrm{u}.$


  1. Gross, J. H. Mass Spectrometry: A Textbook, 3rd ed.; Springer International Publishing: Cham, Switzerland, 2017. ISBN 978-3-319-54397-0.
  • $\begingroup$ Right, but this neutral loss is only meaningful if the smaller peak is produced from the larger peak, which is not true for all peaks in a spectrum. But this actually might be the definition that the poster is looking for. $\endgroup$ – Ezze Oct 28 '19 at 16:02
  • $\begingroup$ @Ezze I suspect this is just an example where using atomic mass units makes sense, but I guess the terminology can be used beyond this case. $\endgroup$ – andselisk Oct 28 '19 at 16:05

Hmm, I haven't heard about such a term. After a quick search in the IUPAC suggested terminilogy I haven't found anything either. Perhaps there simply is no name for this quantity. I guess you can simply call it separation of the peaks.

  • $\begingroup$ I've looked into IUPAC as well without success. I don't think "separation of peaks" is suitable cause not every intensity-m/z sample needs to be a peak. The distance depends on the mass spectrometer specs (mass resolution, mass accuracy, etc.), means it's performance for sure. However I don't know to which spec it relates and how. $\endgroup$ – thinwybk Oct 28 '19 at 15:29
  • $\begingroup$ @thinwybk I do not believe that the separation actually depends on the instrument as if one ion has an m/z value of 23 and a different one has an m/z value of 90 then their separation is 67 m/z on all instruments, right? Unless you are considering peaks that are really close to each other, in which case, the resolution comes into play - but this does have an accurate IUPAC definition. $\endgroup$ – Ezze Oct 28 '19 at 15:37
  • $\begingroup$ The question was not clearly stated enough. I'm interested in the difference of the m/z values on the x-axis, not the distance between samples. I'm also considering high resolution data. As far as I know the resolution does influence the minimal distance between distinguishable peaks, not the "granularity" of the m/z axis values. Which IUPAC definition do you mean? $\endgroup$ – thinwybk Oct 28 '19 at 15:57
  • $\begingroup$ But the distance on the x axis really depends on the size of the paper you print it on, doesn't it? And as for the definition of resolution: goldbook.iupac.org/terms/view/R05318 $\endgroup$ – Ezze Oct 28 '19 at 16:00

The distance between two peaks is characterized not by $\Delta m$ but rather by mass resolution. The distance between the peaks is often referred to as $\Delta m$ and assumes that the $z = 1$, but this is not always the case.

IUPAC addresses this in detail: Resolution, DOI: 10.1351/goldbook.R05318

(10 per cent valley definition): Let two peaks of equal height in a mass spectrum at masses m and m−Δm be separated by a valley which at its lowest point is just 10 per cent of the height of either peak. For similar peaks at a mass exceeding m, let the height of the valley at its lowest point be more (by any amount) than ten per cent of either peak height. Then the resolution (10 per cent valley definition) is m/Δm. It is usually a function of m. The ratio m/Δm should be given for a number of values of m.

  • $\begingroup$ Since it can be expected that the UI of the Gold Book may change (like it did last year, which broke quite a few links), I've switched to the issued DOI and encourage you to do the same. Please also note, that I have wrapped MathJax around the whole expression, as this ensures the post doesn't break on other devices than your own. $\endgroup$ – Martin - マーチン Oct 29 '19 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.