# Predicting isotope patterns in mass spectrometry

Imagine you have a mixture of $$\ce{^1H}$$, $$\ce{^2H}$$ and $$\ce{^3H}$$ (hydrogen, deuterium and tritium) present as diatomic molecules and that the numbers of the atoms of the three species are the same. Sketch the mass spectrum.

For peaks of mass= 2, 3, 4, 5, ratio of heights = 1:2:3:2:1.

I’m not getting why the answer is this?

Why there is no peaks of mass 1, is it because $$\ce{H2}$$ consisting of $$\ce{^1H}$$ can break in mass spectrometer?

• Try thinking of the simpler case, only hydrogen and deuterium. Every atom in every $\ce{H2}$ molecule has identical chances of being either hydrogen or deuterium. What would you expect?
– Jan
Oct 5 '15 at 18:29
• – user7951
Oct 5 '15 at 18:33
• why there is no peaks of mass 1, because H2 consisting of hydrogen-1 can break in mass spectrometer. Oct 5 '15 at 19:07
• This means they are doing it under the conditions when the molecules wouldn't break. Oct 5 '15 at 19:20
• I can't find where they mention that in the question? Oct 5 '15 at 19:21

The first step is to write all the possible combinations of H atoms: $$\mathrm{H^1H^1~~ H^1H^2~~ H^1H^3 ~~H^2H^1 ~~H^2H^2~~ H^2H^3~~ H^3H^1~~ H^3H^2 ~~H^3H^3}$$ Now it is important to note that each of these parings have an equal probability of forming and exist in equal quantities in the mixture. Their molar mass are respectively: $$2, 3, 4, 3, 4, 5, 4, 5, 6$$ So there is $1$ way to make $2$, $2$ ways to make $3$, $3$ ways to make $4$, $2$ ways to make $5$ and only $1$ way to make $6$. Since each way has an equal probability of occurring, then the ratio of the molar masses should be in the ratio - $~1:2:3:2:1$.
Hence for peaks of mass = $2, 3, 4, 5, 6$ ratio of heights = $1:2:3:2:1$
The reason why there is no peak for mass $1$, I am not too sure. However I think it is because if the hydrogen molecule was to break up, it would form a proton and a H radical. Both of these are extremely reactive and unstable and will probably just get lost in the machine and never actually get recorded. This could make sense as the mass spectra for $\ce{HCl}$ or $\ce{HBr}$ doesn't include a peak for mass $1$. Also in mass spectra for hydrocarbons, such as heptane, the smallest mass peak recorded is usually 29 which is the ethyl group. Therefore any thing lower than that (such as the methyl group) is probably too unstable.
• The question is not very good indeed. It's a 100% imaginary question and has nothing to do with real MS measurement. 1, The main purpose of Mass Spectroscopy(MS) is to identify compound by measuring the mass to charge ratios. It measures either $[M+H]^+$ or $[M-H]^-$, in positive mode or negative mode, respectively. M refers to the exact mass of the compound. For a organic molecule, it is usually several hundred Daltons. 2, Each isotope of an element has certain natural abundance percentage affecting peak intensity. Oct 6 '15 at 2:20
• @Dejian15 This is incorrect, it depends on the ionisation method. ESI is known to give association complexes such as $\ce{[M + H]+}$ while EI would give $\ce{[M]+}$.