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I'd like to ask for an extension of Nicolau Saker Neto's answer here: What is the relative size of the (M+2) peak?

I'm trying to calculate isotopic abundances of molecules with more than 2 isotopes. I've been experimenting calculating relative abundances involving $\ce{^{16}O}, \ce{^{17}O}$ , and $\ce{^{18}O} $ across $\ce{O_3}$ and $\ce{O_4}$. I've been googling and looking through books and haven't found any direct examples of someone doing this.

Laid out in the same format, and using $m_{16}$ to indicate the mass of $\ce{^{16}O}$, etc.:

$$\begin{align} (0.9976m_{16} + 0.0004m_{17} + 0.002m_{18})^3 ={} & \ \ \ \ \ \binom {3} {0}(0.9976 m_{16})^3 \\[5pt] & + \binom {3} {1}(0.9976 m_{16})^2 (0.002 m_{18}) \\[5pt] & + \binom {3} {1}(0.9976 m_{16})^2 (0.0004 m_{17}) \\[5pt] & + \binom {3} {2}(0.9976 m_{16}) (0.002 m_{18})^2 \\[5pt] & + \binom {3} {2}(0.9976 m_{16}) (0.0004 m_{17})^2 \\[5pt] & + \binom {3} {2}(0.0004 m_{17}) (0.002 m_{18})^2 \\[5pt] & + \binom {3} {2}(0.002 m_{18}) (0.0004 m_{17})^2 \\[5pt] ... \\[5pt] & + \binom {?} {?}(0.9976 m_{16}) (0.0004 m_{17}) (0.002 m_{18}) \\ \end{align}$$

My real issue is with the factors involving the binomial for the triple-combination. Which I guess is a polynomial instead of a binomial. I know for sure the value multiplied by the product of % abundances should be 6. But I'm not understanding the logic behind why it's 6.

From what I understand, multiplying binomials is akin to to squaring them, but that would mean this would be 3 choose 1 squared, which is 9. Multiplying them could make sense if the second binomial had an $(n-1)$ for its upper value, this would give you 3*2. This makes sense as long as you approach the largest number first.

The above logic holds well until I test it on $\ce{O_4}$, where it seems to be too naive whenever all 3 isotopes are present.

I'm using a symbolic math solver to brute-force solve these to check my answers. The source of error is obviously coming from the binomial/polynomial term. Is there a proper way to calculate that for a molecule when $n$ different isotopes of the same element are present?

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  • $\begingroup$ There are programs which solved this problem (e.g., this one) or some of the sketchers (e.g., gchemcalc, however without numeric access to the probabilities on the GUI; however it's source may be accessible). Perhaps these may help you to compare your computed results quickly. Good luck! $\endgroup$
    – Buttonwood
    Oct 1, 2021 at 22:10

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I think it is much clearer to frame this as a question in combinatorics, rather than chemistry.

In an $\ce{O3}$ molecule, the product of natural abundances gives you the probability that the molecule contains those isotopes. So the probability of having one $\ce{^16O}$, one $\ce{^17O}$, and one $\ce{^18O}$ is given by

$$p_{16}p_{17}p_{18}$$

where $p_n$ is the natural abundance of $\ce{^nO}$. However, there are many different ways in which you could accomplish this: the first oxygen could be $16$, the second $17$, and the third $18$; or the first could be $17$, the second $16$, and the third $18$; or so on. In general you should have $3! = 6$ different possible permutations of oxygen isotopes, and therefore this term needs to be weighted by a coefficient of $6$.

(If the above is unclear, consider the analogous maths problem of: how many different ways are there to draw one red, one green, and one blue marble from a box of marbles? Answer: $6$. The same is true of oxygen isotopes. Of course, they need to be further multiplied by the natural abundance, but that's what the term $p_{16}p_{17}p_{18}$ is for.)

Consider now the case of $16/17/17$, which is the fifth line in your above analysis. The probability is $p_{16}p_{17}^2$, but now the weighting factor is no longer $6$, because the $17$'s are not distinguishable. So our previous factor of $3! = 6$ has overcounted by a factor of $2! = 2$, and the real coefficient is $6/2 = 3$. Notice that this is equivalent to your $3 \choose 2$.

In the $16/17/18$ case, we haven't overcounted; or you could say that we have overcounted by a factor of $1!$ for each isotope. Obviously, $1! = 1$, so this has no influence on the result. But this shows us the way to generalise the problem at hand, which I assert:

The number of ways of choosing $n_1$ objects of type $1$, $n_2$ objects of type $2$, $\ldots$, and $n_g$ objects of type $g$ (where $g$ denotes the total number of types) is given by

$$\frac{n!}{(n_1!)(n_2!)\cdots(n_g!)}$$

where $n = n_1 + n_2 + \cdots + n_g$.

These numbers are not called polynomial coefficients, but rather multinomial coefficients; they are denoted by

$${n \choose {n_1 n_2 \cdots n_g}}.$$

Finally, in the case where $g = 2$ this reduces to the familiar binomial coefficients.

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