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If a molecule contains four chlorine atoms, how many molecular ion peaks will this compound show in its mass spectrum?

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The major isotopes of chlorine are 35 and 37. If 4 chlorines are present in a molecule, then the following isotopic combinations of chlorine are possible

  • 35, 35, 35, 35
  • 35, 35, 35, 37
  • 35, 35, 37, 37
  • 35, 37, 37, 37
  • 37, 37, 37, 37

So there will be 5 molecular ion peaks. You could calculate their relative heights by factoring in the relative isotopic proportions, along with the number of permutations of each of the above combinations.

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If a molecule contains four chlorine atoms, how many molecular ion peaks will this compound show in its mass spectrum?

Limitations of existing answers

Ron's answer is good for molecules that consist solely of chlorine, i.e., a hypothetical $\ce{Cl4}$ molecule. But if there are other atoms [of elements that have more than one stable isotope] as well, then there will be more peaks.

An example using difluorotetrachlorobenzene

Suppose that the molecule is an asymmetric difluorotetrachlorobenzene, $\ce{C6Cl4F2}$. Then all of the fluorine atoms will be $\ce{^{19}F}$, but every one of the six carbon atoms could be either $\ce{^{13}C}$ or $\ce{^{12}C}$, and every one of the four chlorine atoms could be $\ce{^{35}Cl}$ or $\ce{^{37}Cl}$.

A complete list of all isotopologues

Since there are six carbon atoms, there are 7 possible values for the total mass of carbon in each molecule (0 $\ce{^{13}C}$ atoms, 1 $\ce{^{13}C}$ atom, ... up to 6 $\ce{^{13}C}$ atoms). Similarly there are 5 possible values for chlorine, so there are $7\times 5=~~$ 35 different molecular ions! Here is a complete list:

       mass log10(abund) 12C 13C 35Cl 37Cl 19F
1  249.8722   -0.5102721   6   0    4    0   2
2  251.8693   -0.4031194   6   0    3    1   2
3  253.8663   -0.7219355   6   0    2    2   2
4  255.8634   -1.3929341   6   0    1    3   2
5  257.8604   -2.4899015   6   0    0    4   2
6  250.8756   -1.6980651   5   1    4    0   2
7  252.8726   -1.5909124   5   1    3    1   2
8  254.8697   -1.9097285   5   1    2    2   2
9  256.8667   -2.5807271   5   1    1    3   2
10 258.8638   -3.6776945   5   1    0    4   2
11 251.8789   -3.2660693   4   2    4    0   2
12 253.8760   -3.1589166   4   2    3    1   2
13 255.8730   -3.4777327   4   2    2    2   2
14 257.8701   -4.1487313   4   2    1    3   2
15 259.8671   -5.2456987   4   2    0    4   2
16 252.8823   -5.1070748   3   3    4    0   2
17 254.8793   -4.9999221   3   3    3    1   2
18 256.8764   -5.3187382   3   3    2    2   2
19 258.8734   -5.9897368   3   3    1    3   2
20 260.8705   -7.0867042   3   3    0    4   2
21 253.8856   -7.1979577   2   4    4    0   2
22 255.8827   -7.0908051   2   4    3    1   2
23 257.8797   -7.4096212   2   4    2    2   2
24 259.8768   -8.0806198   2   4    1    3   2
25 261.8738   -9.1775871   2   4    0    4   2
26 254.8890   -9.5618420   1   5    4    0   2
27 256.8860   -9.4546893   1   5    3    1   2
28 258.8831   -9.7735054   1   5    2    2   2
29 260.8801  -10.4445040   1   5    1    3   2
30 262.8772  -11.5414714   1   5    0    4   2
31 255.8923  -12.3059456   0   6    4    0   2
32 257.8894  -12.1987929   0   6    3    1   2
33 259.8864  -12.5176090   0   6    2    2   2
34 261.8835  -13.1886076   0   6    1    3   2
35 263.8805  -14.2855750   0   6    0    4   2

Why the complete list isn't always observed

  1. Resolution limitations. Many types of mass spectrometers won't be able to resolve the very small mass differences between some of these ions. For example, $m/z=251.8693$ for $\ce{^{12}C6~^{37}Cl1~^{35}Cl3~F2}$ is only 0.0096 Daltons away in mass from $m/z=251.8789$ for $\ce{^{12}C4~^{13}C2~^{35}Cl4~F2}$. In many instruments such peaks would not be well-resolved.
  2. Sensitivity limitations. Additionally, many of the 35 distinct isotopologues are very rare in abundance, well below practical limits of detection. The abundance of $\ce{^13C6~^37Cl4~F2}$ is about fourteen orders of magnitude lower than the most abundant mass isotopologue, $\ce{^{12}C6~^{35}Cl3~^{37}Cl1~F2}$.

A theoretical spectrum at unit-mass resolution

The end result is that in most mass spectrometers, more than 5 but many fewer than 35 peaks will be observed for our example molecule of difluorotetrachlorobenzene.

Here are approximate results for an instrument with ~0.5 Da mass resolution, such as most quadrupole detectors.

Plot of peaks for difluorotetrachlorobenzene at 0.5 Da resolution

How I have been calculating these results

I am a big fan of an R package called ecipex. With that package these calculations are easy. Here is my code to generate the list above.

require(ecipex)
df <- ecipex('C6Cl4F2', id=T, limit=1e-16)[[1]]
dfordered <- df[order(df[,'12C', decreasing=True]), ]
dfordered$abundance <- log10(dfordered$abundance)
names(dfordered)[2] <- 'log10(abund)'
dfordered

And to generate the graph:

require(ggplot2)
ggplot(df, aes(x=mass, weights=abundance)) + 
  geom_bar(stat='bin', binwidth=0.5, color='black', fill='gray') + 
  theme_bw() + 
  xlab('m/z, Da') + 
  ylab('abundance') + 
  scale_x_continuous(breaks = 248:258, limits = c(248,258))
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    $\begingroup$ Both to shed complementary light on the lesser obvious limitation that comes along with first answer, as well as adding how to compute the expected pattern are appreciated. $\endgroup$ – Buttonwood Apr 26 '15 at 21:06
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This answer simply aims to add to the one by Curt F. that there is software available to do these calculations, too. Beside the ones with the mass spectrometers, and assuming Ali Naqvi participates in initial lab classes, the Chemical Calulator of the Gnome Chemistry utilities may be worth a trial to get familiar with typical isotope patterns of organic molecules. As a training, I found this beneficial to recognize them in a routine analysis with ease.

To return to the initial question, a $\ce{Cl4}$ would appear as this -- note there are five peaks. enter image description here

On contrast, Ron's example of C6Cl4F2 -- nine peaks -- were this one: enter image description here

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  • $\begingroup$ Its always nice to see two independent means of calculating something give the same results. Thanks for your answer Buttonwood. $\endgroup$ – Curt F. Apr 26 '15 at 23:33
  • $\begingroup$ $\ce{C6Cl4F2}$ was Curt F's example, not mine. $\endgroup$ – ron Mar 13 '16 at 16:25

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