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I was introduced to this thing today, and saw a graph showing the detection of ions with different relative abundances.

Just as an example here, $\ce{Br2}$ is shown a detection chart. And at $160~\mathrm{m/z}$ the ion has the highest relative abundance, but why? Since the $\mathrm{m/z}$ ratio is just the mass, for $\mathrm{z}$ is always $+1$, the one which is supposed to have the highest peak should be at $162$?

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In $\ce{Br2}$, the possible combinations of isotopes are:

  • $\ce{^{79}Br-^{79}Br} : \mathrm{m/z}~158$
  • $\ce{^{79}Br-^{81}Br} : \mathrm{m/z}~160$
  • $\ce{^{81}Br-^{79}Br} : \mathrm{m/z}~160$
  • $\ce{^{81}Br-^{81}Br} : \mathrm{m/z}~162$

Bromine's isotopic distribution is essentially a $50:50$ ratio of mass $79$ and mass $81$. That means that there's a $25\%$ chance of $\mathrm{m/z}~158$ $(0.5 \cdot 0.5 = 0.25)$ arising from $\ce{^{79}Br-^{79}Br}$. There's a $50\%$ chance of $\mathrm{m/z}~160$ since there are two combinations ($\ce{^{79}Br-^{81}Br}$ and $\ce{^{81}Br-^{79}Br}$) that give that mass $(0.5\cdot0.5 + 0.5\cdot0.5 = 0.25)$. There's a $25\%$ chance of $\mathrm{m/z}~162$ $(0.5\cdot 0.5 = 0.25)$ arising from $\ce{^{81}Br-^{81}Br}$.

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  • $\begingroup$ Since they are all ions, how could ions be combined with ions to make new ions? And why can some large ions fragment? $\endgroup$ Sep 10 '14 at 2:29
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jerepierre's answer is correct, you should see a statistical distribution of 3 different peaks, but also keep in mind that the m/z is not always the same as the mass. Depending on the molecule you're looking at and the ionization technique, the charge can be much larger than 1. Electrospray ionization, commonly used for ionizing biomolecules, routinely produces multiply charged ions. +20 charge is not uncommon. Additionally, many types of ionization cause fragmentation of molecules. For electron impact ionization, it's often the case that a peak for the full molecule is not observed at all.

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