[kimi85]

The two main isotopes , ³⁵Cl and ³⁷Cl, have relative abundances of approximately 75% and 25% respectively.  Consider now the mass spectrum of p-dichlorobenzene.  If the peak at m/e 146 has intensity of 100 units, what is the expected intensity for the m/e 148 peak?

The answer is 66.7% but I don't know how. Thank you

[sjb]

Not sure if this is quite the right way to look at it, but it does give the right answer.

146 is the m/e peak corresponding to the bis ³⁵Cl substituted benzene.

For every 3 ³⁵Cl atoms you have 1 ³⁷Cl.

As there are 2 chlorines in the molecule, you'd expect 2 x 1/3 of the molecules to be mono ³⁷Cl substituted, which equates to the 66.7% observed.

Follow-up, assuming that chlorine isotopes are the only ones with signficant isotope splitting (fair assumption, as D only exists in nature with < 0.1 %, and ¹³-C is about 1%), what intensity do you anticipate for the m/e peak at 150, again based on 146 having 100% intensity?

S

[sdekivit]

i do not agree with 1/3 part. the way i used to solve these problem is to create a matrix. we have 2 isotopes: Cl-35 (call mass  = M) and Cl-37 (mass = M+2) with relative abundance = 3:1 (so you'd expect 3/4 to be Cl-35 and 1/4 to be Cl-37)

Now create a matrix with the ratios of the isotopes: M - M+1 - M+2 horizontal and vertical wit M = 3 and M+2 = 1.

now the result will be M : M+1 : M+2 : M+3 : M+4 = 9 : 0 : 6 : 0 : 1

--> so if the peak intensity of M = 100, the intensity of the peak generated by M+2 would be (6 * 100) / 9 = 66.7

or without matrices:

Cl-35 / Cl-35 would have a chance 3/4 * 3/4 = 9/16
Cl-35 / Cl-37 would have a chance 2 * (1/4 * 3/4) = 6/16

again this would result in a peak intesity 600/9 = 66.7

(thus the peak intensity at m/z = 146 is not 100% but 9/16 = 56.25%)

[sjb]

...(so you'd expect 1/4 to be Cl-35 and 3/4 to be Cl-37)..

The other way round, surely? 3/4 ³⁵Cl and 1/4 ³⁷Cl..?

(thus the peak intensity at m/z = 146 is not 100% but 9/16 = 56.25%)

56.25% of the molecular ion's total intensities, yes, but we've rescaled so that this has relative intensity 100%. I think we're agreed that 6/9 is the same as 2/3....

Otherwise, yes your way is probably slightly better - always got a bit confused with matrices

S

[sdekivit]

yes it's the other way --> modified it

indeed the ratio between the peask intensity 146 : 148 = 9:6 and so the peak intensity of m/z = 148 = 2/3 of the peak intensity of m/z = 146.

but i can't follow your calculation 2 * 1/3 because this is not the way to calculate the problem (it's not mathematically correct when talking about probabilities)

[macman104]

The way I always calculated it was, you can have the following combinations:

35, 35 = 146   (3/4 * 3/4) = 9/16
35, 37 = 148   (3/4 * 1/4) = 3/16
37, 35 = 148   (1/4 * 3/4) = 3/16
37, 37 = 150   (1/4 * 1/4) = 1/16

The two 148s combine for a ratio of 6/16

Normalizing....
9/16 over 9/16 = 1
6/16 over 9/16 = 2/3
1/16 over 9/16 = 1/9

Relative heights of 9x, 6x and 1x for 146, 148, and 150 respectively.

That is how I've always solved these problems.

[sdekivit]

that's the same method as i used in my second possibility to get to the answer.

[sjb]

but i can't follow your calculation 2 * 1/3 because this is not the way to calculate the problem (it's not mathematically correct when talking about probabilities)

I've put together a little sketchet to try and help. This only really works for two atoms with isotopes, gets a bit more complicated with 3 or more, which is where your method is better!

In this region a has area a third that of the region marked "main", as does region b, which is where my 2 x 1/3 is coming from.

If my method confuses you, ignore it

S

[kimi85]

thank you very much!

God bless you all.

[sdekivit]

sjb, now i understand your method and of course is completely correct

[AWK]

I've put together a little sketchet to try and help. This only really works for two atoms with isotopes, gets a bit more complicated with 3 or more, which is where your method is better!

For 3 isotopes a cube works in analogous way

[sjb]

I've put together a little sketchet to try and help. This only really works for two atoms with isotopes, gets a bit more complicated with 3 or more, which is where your method is better!

For 3 isotopes a cube works in analogous way

For 3, yes a cube is just about useful, I hesitate in imagining a tesseract for 4 isotopes and above though . My method is also not that amenable in its basic form for different isotopic abundances, for instance C₆H₆ClBr - but can be made to work eventually....

S