Why (abundance of M+1)/(abundance of M)=(number of carbon atoms)*(1.1/100)?

In mass spectrometry, why is this equation true?


$\frac{A\_{\mathrm{M1}}}{A\_{\mathrm{M}0}} = n\_\ce{C}\;\;\times\;\;
\frac{1.1}{100}$


Here, $A\_{\mathrm{M1}}$ is the abundance of the M+1 peak and $A\_{\mathrm{M}0}$ is the abundance of M.


I do not understand the logic behind the equation. Let's say that in mass spectrometer I got $A\_{\mathrm{M1}}=1\%$ and $A\_{\mathrm{M}0}=99\%$ for methane.
Now, if I use the equation I get, $(1/99)=1 \times(1.1)/100$. The right hand side does not exactly match the left hand side, but maybe the answer is correct enough if I round it up?


Can anyone explain to me how the equation derived and the logic behind it?




Let's look at where these numbers come from. For carbon, the main isotopes are carbon-12 and carbon-13. Carbon-13 has an abundance of 1.109%.


Consider a molecule with $n$ carbons in it. The ratio of the $M\_{1}$ peak to the $M\_{0}$ peak is related to this above fraction, but you have to account for the fact that there are $n$ ways to create this molecule with one carbon as carbon-13 and the rest as carbon-12. This is basically one of the binomial cofficients:


$$\binom{n}{1} = n$$


You can take the total distribution of $p + (1-p)$, where $p$ is abundance of carbon-13. If you have $n$ carbons, we have:


$$\left(p + (1-p)\right)^{n}=(1-p)^{n} + \binom{n}{1}(1-p)^{n-1}p+\ldots$$


Each of $n+1$ terms corresponds to the relative height of one of your MS peaks. The ratio between peak 0 and peak 1 is:


$$\frac{\binom{n}{1}(1-p)^{n-1}p}{(1-p)^{n}}=\frac{np}{1-p}$$


Since $p=0.01109$, this expression comes to approximately:


$$n\cdot\frac{1.1}{100}$$


This means if you have $n$, you can estimate the height of the first peak relatively to the zero-th order peak. Technically, given the method I provided, you can estimate an higher order peak.


Order, if you have the base peak and the first peak, you can estimate $n$. This isn't supposed tobe perfect if you're using 2 significant figures, but your measurement of the MS peak intensity is probably similar in terms of error, but you should find that even in your example, $1/99$ is a lot closer to $1.1/100$ than $n\cdot (1.1/100)$ for values of $n$ other than 1.


 1