Isotopic profiling [closed]








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If a molecule contains 12 carbon atoms and 2 nitrogens, what's the probability that it contains precisely 1 carbon-13 and precisely 1 nitrogen-15?


my working: 1.1(12)(0.36)(2) = 9.5 %


where 1.1 is the natural abundance (%) of carbon-13 and 0.36 is the natural abundance (%) of nitrogen-15


This seems too high. What is the correct probability calculation and why?




Yes, 9.5% is much to high a value. 


Let's break the problem into two parts. 


* $p\_\mathrm{C}$ will be the probability of getting exactly one $\ce{^{13}C}$ atom in a molecule with 12 carbon atoms.
* $p\_\mathrm{N}$ will be the probability of getting exactly one $\ce{^{15}N}$ atom in a molecule with 2 nitrogen atoms.
* $p$, the probability sought will be $p = p\_\mathrm{C}\times p\_\mathrm{N}$


**For carbon**


The probability of getting one $\ce{^{13}C}$ atom when selecting the first carbon is 1.1% or 0.011.


The probability of not getting any $\ce{^{13}C}$ atom when selecting eleven additional carbon atoms is $(1-0.011)^{11} = 0.8854$. 


So the probability of having the first carbon atom be a $\ce{^{13}C}$ atom and the next eleven atoms not being a $\ce{^{13}C}$ atom is $0.011\times 0.8854 = 0.009740$


But the $\ce{^{13}C}$ atom could have been chosen in any one of the 12 positions. So


$p\_\mathrm{C} = 12\times 0.009740 = 0.1169 $ 


**For nitrogen**


The probability of getting one $\ce{^{15}N}$ atom when selecting the first nitrogen is 0.36 % or 0.0036.


The probability of not getting a $\ce{^{15}N}$ atom when selecting one additional carbon atoms is (1−0.0036)=0.9964


So the probability of having the first nitrogen atom be a $\ce{^{15}N}$
atom and the next atom not being a $\ce{^{15}N}$ atom is:


$0.0036\times 0.9964 = ‭0.003587$


But the $\ce{^{15}N}$ atom could be the first or second so the value needs to be multiplied by 2. So


$p\_N =2 \times ‭0.003587=0.007174$


Finally $p = p\_C \times p\_N = 0.1169 \times 0.007174 = 0.0008386 = 0.084\%$




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**BONUS**
Figuring out the number of combinations with just one $\ce{^{13}C}$ atom and the others being non-$\ce{^{13}C}$ atoms was easy. However listing all the carbon possibilities when there are four $\ce{^{13}C}$ atoms and six non-$\ce{^{13}C}$ atoms can get tricky. Fortunately there a mathematical method call Combinatorics which allows the answer to be found easily. 


$\large\binom{n}{m} = \small\dfrac{n!}{(n-m)!m!}$ is read as "n chose m."


So for for four $\ce{^{13}C}$ atoms and six non-$\ce{^{13}C}$ atoms we'd have:


$\large\binom{10}{4} = \small\dfrac{10!}{(10-4)!4!} = 210$ possible arrangements.


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