[interminable]

So I have the following data:

m/z (RI)
720 (100)
721 (65.9)
722 (20.
723 (4.7)

And I know that it's C₆₀. And I can figure out how relative intensities of M+1 and M+2 peaks are predicted. But I can't figure out how you determine M+3.

I have a question to do where I'm given data like this, and I'm supposed to determine the molecular formula. I've figured out an answer that's consistent with my M+, M+1 and M+2 peaks in that case, but I can't figure out how to predict an intensity for M+3 because I want to see if it's consistent with the provided value.

So essentially I'm trying to work backwards from this example, figuring that if I can understand how to predict a theoretical M+3 for C60, I could do it for my actual problem.

I thought it had something to do with the probability of getting 3 carbon-13's...but I can't seem to figure out how they got 4.7

[AWK]

What method of ionization did you use?

[interminable]

EI-MS

[MOTOBALL]

You have to calculate the probability of 1 x C13, 2 x C13, 3 x C13 atoms..
This is the "how many ways to pick 1 from 60, 2 from 60, 3 from 60 "; the general form is

       _nC_r = n!/r!(n - r)!               where n = 60 and r = 1, 2, 3 etc

₆₀C₁ = 60!/1!(60 -1)! = 60 x 0.0107/0.9893 x 100 = 64.9%

₆₀C₂ = 60!/2!(60 - 2)! = (60 x 59)/2 x [(0.0107/0.9893)]² x 100 = 20.7%

₆₀C₃ = 60!/3!(60 - 3)! = (60 x 59 x 58)/6 x [(0.0107/0.9893)]³ x 100 = 4.33%

where 0.9893 is relative abundance of 12C; 0.0107 is rel. abund. of 13C

Your take-home quiz is to calculate the relative intensity of m/z 724

Good Luck,

Motoball

[MOTOBALL]

And spot my (non-deliberate) mistakes in first line..

not 1 x C13, 2 x C13 but 1 x ¹³C and 2 x ¹³C etc

Motoball