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?

  • $\begingroup$ 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
    Commented Oct 1, 2021 at 22:10

1 Answer 1


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


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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.