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Topic: Mass spectrometry and molar ion patterns  (Read 4594 times)

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Offline sundberg

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Mass spectrometry and molar ion patterns
« on: October 23, 2005, 09:52:15 AM »
I'm trying to figure out how to predict molar ion patterns of halogenated organic compounds.
I have no problems when it's just one halogen involved, using the formula (a+b)^n. Where a+b is the isotopic ratio. But how do you solve for example CHFClBr? Flourine has a isotopic ratio of 1, chlorine 3:1 and bromine 1:1. But how should I apply the formula then, using it one time for each halogen doesn't work obviosly.

Thanks!

Offline sdekivit

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Re:Mass spectrometry and molar ion patterns
« Reply #1 on: October 23, 2005, 10:46:06 AM »
it's just a probability problem. Compare with balls in a vase with a different color and you can pick 3 out the vase. You can make a matrix from this with the M+0-mass, M+1 mass and so on.

so first we can have Cl-35 or Cl-37, thus M+0 and M+2. Then the chance of having Cl-35 in your molecule is 3 times higher as Cl-37.

Then you can have only F-19 and in the last case we can have Br-79 or Br-81 with a ratio 1:1.

Now we can make a matrix with Cl horizontal and Br verical in order to masses:

Cl --> 3 : 0 : 1
Br     |    
1     |     3   0   1
0     |     0   0   0
1     |     3   0   1

In this matrix we can see our isotopepattern with the peakintensity in our mass spectrum:

the M+0 peak has the mass of the original molecule is intensity 3.
Then the M+1-peak won't be seen, since there are no M+1-isotopes --> intensity = 0 (diagonal of the second numbers horizontal and vertical in the matrix
the M+2-peak has intensity 1 + 3 = 4

In the same way, we find a M+3 and a M+4-peakintensity of 0 respectively 1.

Thus in our spectrum we will see a M+0 to M+4 peak in the ratio:

3 : 0 : 4 : 0 : 1

In the same way it's for example also possible to calculate the peakpattern for example for ZnCl2 (you should try it ;))
« Last Edit: October 23, 2005, 10:48:20 AM by sdekivit »

Offline Winga

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Re:Mass spectrometry and molar ion patterns
« Reply #2 on: October 24, 2005, 04:20:39 AM »
I think the formula (a+b)^n doesn't work in heteronuclear cases, because it is difficult to find out the ratio of natural abundance of different elements.

As you know,
35Cl:37Cl = 3:1 & 79Br:81Br = 1:1

Cl   3 : 1
------------
Br  |3:1
1:1|    3:1
------------
      |3:4:1

or

Br   1 : 1
------------
Cl  |3:3
3:1|    1:1
------------
      |3:4:1

M : M+2 : M+4 = 3:4:1
« Last Edit: October 24, 2005, 04:27:32 AM by Winga »

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