# 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?

• Why the down votes? This is actually a decent question. – Zhe Dec 3 '16 at 20:53
• @Zhe in its initial form it was very difficult to understand. I retracted my -1. – orthocresol Dec 4 '16 at 0:01

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.

• (p+(1−p))^n, why it is there? – Ahmad Dec 4 '16 at 10:35
• The is the probability distribution for $n$ carbons. Notice that this value is 1 because $p + 1 - p = 1$, and $1^{n}=1$. But this expansion tells us the exact distribution for all combinations of $i$ out of the $n$ carbons being carbon-13 where $i$ is the exponent on the $(1-p)$ term. – Zhe Dec 4 '16 at 14:53
• Why can't the formula be 1.1*n/100=abundance of M1 in percent? – Ahmad Apr 10 '17 at 15:55
• Try $n=1000$ and that clearly doesn't work. – Zhe Apr 10 '17 at 17:27
• Actually, shouldn't it be 98.9, instead of 100? – Tan Yong Boon Dec 20 '17 at 8:20