# What is the relative size of the (M+2) peak?

The $(M+1)$ peak is often considered in the high-resolution mass spectra of organic molecules as it reveals the number of carbon atoms in the sample. In general, it is known that the ratio of the size of the $M$ to $(M+1)$ peaks is $98.9 : 1.1 \times n$ since the relative abundance in nature of $^{13}$C is $1.1$% for the mass spectrum of an organic molecule containing $n$ carbon atoms and no heteroatoms. It is mentioned that the $(M+2)$ peak is statistically insignificant on this site. However, I believe that only applies for organic molecules with a relatively small number of carbon atoms and this peak would become significant when considering the mass spectra of larger organics. Using simple mathematics, I derived that the ratio of the $M$ to $(M+2)$ peak is $98.9^{2} : 1.1^{2} \times _nC _2$. I would like to verify if this is correct. If it is not, could someone then correct it by posting an answer?

You are correct on all accounts.

To a very good approximation, molecules can be thought as made of elements (with their respective isotope distributions) combining completely independently. You can think of it like rolling multiple die at once. This means that a simple multinomial distribution will describe this problem mathematically.

Let's start with something easy and consider the hypothetical molecule $\ce{C_5}$. Furthermore, let us consider that the only carbon isotopes with significant natural occurrence are $\ce{^{12}C}$ (98.9%) and $\ce{^{13}C}$ (1.1%). We can find all of the isotopic peaks and their relative abundances by then expanding the binomial $(0.989\times m[^{12}C] + 0.011\times m[^{13}C])^5$, where $m[^{12}C]$ and $m[^{13}C]$ denote the exact masses of the carbon-12 and carbon-13 isotopes, respectively. Expanding the binomial yields:

\begin{aligned} (0.989\times m[^{12}C] + 0.011\times m[^{13}C])^5 ={} & \ \ \ \ \ \binom {5} {0}(0.989\times m[^{12}C])^5 \\ & + \binom {5} {1}(0.989\times m[^{12}C])^4 \times (0.011\times m[^{13}C]) \\ & + \binom {5} {2}(0.989\times m[^{12}C])^3 \times (0.011\times m[^{13}C])^2 \\ & + \binom {5} {3}(0.989\times m[^{12}C])^2 \times (0.011\times m[^{13}C])^3 \\ & + \binom {5} {4}(0.989\times m[^{12}C]) \times (0.011\times m[^{13}C])^4 \\ & + \binom {5} {5} (0.011\times m[^{13}C])^5 \\ \end{aligned}

Calculating the coefficients in each term:

\begin{aligned} (0.989\times m[^{12}C] + 0.011\times (m[^{13}C])^5 ={} & \ \ \ \ \ 0.946 \times (m[^{12}C])^5 \\ & + 0.0526\times (m[^{12}C])^4 \times m[^{13}C] \\ & + 0.00117 \times (m[^{12}C])^3 \times (m[^{13}C])^2 \\ & + 0.0000130 \times (m[^{12}C])^2 \times (m[^{13}C])^3 \\ & + 7.24×10^{-8}\times (m[^{12}C]) \times ( m[^{13}C])^4 \\ & + 1.61×10^{-10} \times m[^{13}C])^5 \\ \end{aligned}

From the expansion, we see that 94.6% of all $\ce{C_5}$ molecules contain only carbon-12 (the lowest possible mass for the molecule), and almost all of the rest (5.3% out of the remaining 5.4%) is accounted for by molecules that contain a single carbon-13 atom. Only about 0.1% of $\ce{C_5}$ molecules contain two or more carbon-13 atoms.

But what happens in very large molecules? Intuitively, if there are many atoms, you would expect a higher chance of there being at least one less common isotope in the mix. Let's see the first few terms for the molecule $\ce{C_100}$:

\begin{aligned} (0.989\times m[^{12}C] + 0.011\times m[^{13}C])^{100} ={} & \ \ \ \ \ \binom {100} {0}(0.989\times m[^{12}C])^{100} \\ & + \binom {100} {1}(0.989\times m[^{12}C])^{99} \times (0.011\times m[^{13}C]) \\ & + \binom {100} {2}(0.989\times m[^{12}C])^{98} \times (0.011\times m[^{13}C])^2 \\ & + \ ...\\ \end{aligned}

Calculating the coefficients:

\begin{aligned} (0.989\times m[^{12}C] + 0.011\times m[^{13}C])^{100} ={} & \ \ \ \ \ 0.331 \times (m[^{12}C])^{100} \\ & + 0.368 \times (m[^{12}C])^{99} \times (m[^{13}C]) \\ & + 0.203 \times (m[^{12}C])^{98} \times (m[^{13}C])^2 \\ & +\ ...\\ \end{aligned}

Well that's interesting. Now only 33.1% of the molecules contain only carbon-12 atoms, and in fact more molecules contain exactly one carbon-13 atom, at 36.8% of the total. Even molecules with two carbon-13 atoms are quite abundant, at 20.3%.

Indeed, peaks containing rarer isotopes eventually dominate. For the huge molecule $\ce{C_10000}$, the strongest mass spectrum signal would come from molecules contaning 110 carbon-13 atoms, corresponding to 3.8% of the total, while a measly $9.2\times 10^{-47}\%$ of molecules contain only carbon-12. This happens because when $n$ is large, the term $\binom {n} {k}$ grows very quickly as $k$ rises from zero, overwhelming the increase in the exponent of the rarer isotope. You can see this behaviour quite nicely in this sequence of mass spectra of molecules with increasing size.

To calculate the specific $M/M+2$ ratio for a molecule containing only $n$ carbon atoms, all you need is to get the ratio for first and third terms in the binomial:

\begin{aligned} (0.989\times m[^{12}C] + 0.011\times m[^{13}C])^n ={} & \ \ \ \ \ \color{#0000ff}{ \binom {n} {0}(0.989\times m[^{12}C])^n} \\ & + \binom {n} {1}(0.989\times m[^{12}C])^{n-1} \times (0.011\times m[^{13}C]) \\ & + \color{#0000ff}{\binom {n} {2}(0.989\times m[^{12}C])^{n-2} \times (0.011\times m[^{13}C])^2} \\ & +\ ...\\ \end{aligned}

The ratio is then:

$$\frac{\binom {n} {0}0.989^n}{\binom {n} {2}0.989^{n-2} \times 0.011^2}=\frac{2\times 0.989^2}{n(n-1)\times 0.011^2}$$

Technically this only holds if there are no other elements which contain multiple isotopes, though it will hold approximately if the other elements only have very rare alternate isotopes, such as hydrogen (99.98% hydrogen-1, 0.02% hydrogen-2).

As a last curiosity, all of the above extends to analysing more complicated molecules. For example, glucose ($\ce{C6H12O6}$) will have a mass spectrum exactly described by the expression:

$$(0.989\times m[^{12}C] + 0.011\times m[^{13}C])^6 \times (0.9998\times m[^{1}H] + 0.002\times m[^{2}H])^{12} \times (0.9976\times m[^{16}O] + 0.004\times m[^{17}O] + 0.020\times m[^{18}O])^6$$

Happy expanding!

Let’s assume your compound is $\ce{C_nH_xO_y}$. Thankfully, both hydrogen and oxygen are elements that only have one significant naturally occuring isotope. Therefore, we can treat the entire contribution of $\ce{H_xO_y}$ as a constant $c$. All we need to answer is how large the $M+1$ and $M+2$ peaks are given the number of carbons, $n$. (This treatment is not entirely correct as highlighted by orthocresol, but it is close enough for my purposes.)

Assuming you do not have any isotope enrichments, each carbon will independently have a chance of either being $\ce{^12C}$ or $\ce{^13C}$ (again, we will ignore all other isotopes such as $\ce{^14C}$). The independence is the big key word here. We can use general principles of stochastics to calculate the result.

The probability of all carbon atoms of one molecule of your compound are $\ce{^12C}$ is: $$P(\ce{^12C_nH_xO_y}) = 0.989^n$$ The $M+1$ peak is represented by one single atom being $\ce{^13C}$. Again, the principles of stochastics apply: $$P(\ce{^12C_{n-1}^13C1H_xO_y}) = \left ( n\atop 1\right) \times 0.989^{n-1} \times 0.011$$ And finally for the $M+2$ peak: $$P(\ce{^12C_{n-2}^13C2H_xO_y}) = \left ( n \atop 2\right) \times 0.989^{n-2} \times 0.011^2$$

Using this, we can plot what the probability of each peak is – and use that as their relative heights.

We see that from $n=90$ the $M+1$ peak is actually larger than the $M$ peak. From $n=128$, even the $M+2$ peak will be larger than the $M$ peak. And from $n=181$ the $M+2$ peak becomes the largest of these three.

Of course, further peaks will also start appearing, meaning that the mass spectra of very large molecules will become difficult to analyse.

Now the numbers from which $M$ is no longer te principal peak are quite alrge. Unless you are synthesising maitotoxin, you will probably not need to resort to any analysis that does not only involve $M+1$.

• Oxygen-18 has 0.2% abundance. In a molecule with six carbons (for example) the probability of having two 13C is 0.17%. Even if the molecule only has one oxygen, the probability of having one 18O is comparable and not negligible. (Obviously the significance of 18O will depend on the C/O ratio.) Now, I’m thankful that HRMS exists :) Commented Dec 20, 2017 at 11:50
• Thanks for showing the graphical analysis and the trend. They certainly helped me to understand it a lot better, which is why I'm awarding you the bounty. Thanks! However, I accepted the other answer as I preferred something more mathematically rigorous. Commented Jan 17, 2018 at 9:06