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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?

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    $\begingroup$ You can see the issue with your approach by considering what probability you would get of having exactly 1 carbon-13 in a chain of 100 carbon atoms. With you method, you would get over 100%, which tells you that your missing a piece that will keep your probability bounded between 0 and 1. $\endgroup$ – Tyberius Apr 14 at 22:47
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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 $p = p_C \times p_N = 0.1169 \times 0.007174 = 0.0008386 = 0.084\%$


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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    $\begingroup$ V clear answer, but you seem to have chosen 10 C atoms instead of the 12 in the question, $\endgroup$ – porphyrin Apr 14 at 7:54
  • $\begingroup$ @porphyrin - Thanks for point out my stupid mistake! $\endgroup$ – MaxW Apr 14 at 15:20
  • $\begingroup$ no problem I make them all the time :) $\endgroup$ – porphyrin Apr 14 at 16:08
  • $\begingroup$ I should vote to close this question as off-topic because it's really about mathematics :-) $\endgroup$ – Buck Thorn Apr 14 at 20:01
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    $\begingroup$ @BuckThorn , maybe so, but a number of non-mathematical chemists will come to chemistry stackexchange and find this answer very useful as it's not only a common problem encountered in chemistry, but also explains the maths at a level that is easily understood by a chemist. maths stackexchange would likely either dismiss this as 'trivial' or give an overly complicated answer. $\endgroup$ – TSA Apr 15 at 16:05

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