# How to find polynomial factor of relative (M+n) peak when factoring more than 2 isotopes

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?

• 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! Oct 1, 2021 at 22:10

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.