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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.

The book's answer is

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?

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    $\begingroup$ 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? $\endgroup$
    – Jan
    Commented Oct 5, 2015 at 18:29
  • $\begingroup$ Related: Mass Spectrum and Molecular Ion peaks $\endgroup$
    – user7951
    Commented Oct 5, 2015 at 18:33
  • $\begingroup$ why there is no peaks of mass 1, because H2 consisting of hydrogen-1 can break in mass spectrometer. $\endgroup$
    – Ahmad
    Commented Oct 5, 2015 at 19:07
  • $\begingroup$ This means they are doing it under the conditions when the molecules wouldn't break. $\endgroup$
    – Ivan Neretin
    Commented Oct 5, 2015 at 19:20
  • $\begingroup$ I can't find where they mention that in the question? $\endgroup$
    – Ahmad
    Commented Oct 5, 2015 at 19:21

1 Answer 1

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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.

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    $\begingroup$ The question is not very good because (i) most mass spectrometers that chemists are going to come across don't measure below 5 or 10 Daltons and thus are not useful for measuring hydrogen; (ii) tritium in equimolar abundance to deuterium or protium would produce a dangerously radioactive material, unsafe to handle; and (iii) mass spectrometry actually measures mass to charge ratios, not mass, and it is unclear what ions would be derived from the hydrogen. All that said, your answer to this not so great question is very good. $\endgroup$
    – Curt F.
    Commented Oct 5, 2015 at 23:14
  • $\begingroup$ 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. $\endgroup$
    – Dejian15
    Commented Oct 6, 2015 at 2:20
  • $\begingroup$ @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 $\ce{[M]+}$. $\endgroup$
    – Jan
    Commented Oct 6, 2015 at 11:25
  • $\begingroup$ @Jan Thanks for the comment. I should have pointed it out. This is a great website. $\endgroup$
    – Dejian15
    Commented Oct 6, 2015 at 16:49

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