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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: sundberg 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!

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+0mass, M+1 mass and so on.
so first we can have Cl35 or Cl37, thus M+0 and M+2. Then the chance of having Cl35 in your molecule is 3 times higher as Cl37.
Then you can have only F19 and in the last case we can have Br79 or Br81 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+1peak won't be seen, since there are no M+1isotopes > intensity = 0 (diagonal of the second numbers horizontal and vertical in the matrix
the M+2peak has intensity 1 + 3 = 4
In the same way, we find a M+3 and a M+4peakintensity 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 ;))

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,
^{35}Cl:^{37}Cl = 3:1 & ^{79}Br:^{81}Br = 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