You may also like

problem icon

Approximating Pi

By inscribing a circle in a square and then a square in a circle find an approximation to pi. By using a hexagon, can you improve on the approximation?

problem icon

Road Maker

Which of these roads will satisfy a Munchkin builder?

problem icon

Twin Equivalent Sudoku

This Sudoku problem consists of a pair of linked standard Suduko puzzles each with some starting digits

Spectrometry Detective

Stage: 5 Challenge Level: Challenge Level:1
This solution was constructed by one of our undergraduate summer students in chemistry. Whilst not all students would be expected to create such a detailed solution, it is artfully clear and well worth reading for the way in which quantitative reasoning is applied to chemistry. 


A diatomic gas with a single stable isotope will give two peaks in the mass spectrometer. One of these peaks will be the molecular ion. The other peak will be seen at half the m/z of the molecular ion, and will be due to a fragment of the diatomic gas. For example, Nitrogen gas would give a molecular ion peak due to $N_2^+$ and a fragment peak due to $N^+$.

If the diatomic gas has 2 stable isotopes, five peaks will be seen in the mass spectrometer, of which two will be due to fragments.
i.e. Denoting $X$ as the element in question, with stable isotopes $^{\alpha}X$ and $^{\beta}X$, gives the five peaks due to:

$^{\alpha}X^+$
$^{\beta}X^+$
$[^{\alpha}X - ^{\alpha}X]^+$
$[^{\alpha}X - ^{\beta}X]^+$
$[^{\beta}X - ^{\beta}X]^+$

If the diatomic gas has 3 stable isotopes, nine peaks will be observed, of which three will be due to fragments. Denoting $^{\gamma}X$ as the third of these isotopes:

$^{\alpha}X^+$
$^{\beta}X^+$
$^{\gamma}X^+$
$[^{\alpha}X - ^{\alpha}X]^+$
$[^{\beta}X - ^{\beta}X]^+$
$[^{\gamma}X - ^{\gamma}X]^+$
$[^{\alpha}X - ^{\beta}X]^+$
$[^{\alpha}X - ^{\gamma}X]^+$
$[^{\beta}X - ^{\gamma}X]^+$

It is additionally possible in all of the above cases that double ionisation may occur, in which case a smaller m/z will be observed, and so more signals will be present. However, the amount of double ionisation that occurs is small, and so these will not be notable.


When water is analysed in a mass spectrometer, a peak at 18 is observed due to $H_2O^+$ and at 17 due to $HO^+$. No peak is observed at 1 since any $H^+$ will combine with $H^-$ that is present, to form a neutral $H_2$ molecule. This will not be observed in the mass spectrometer. Additionally, no 16 peak is observed since it is too energetically unfavourable for $O^+$ to be formed.


The compound with peaks at 35, 37, 70, 72 and 74 is $\mathbf{Cl_2}$ (Chlorine gas). The peaks are assigned as follows:

$^{35}Cl^+$ fragment gives the 35 peak
$^{37}Cl^+$ fragment gives the 37 peak
$[^{35}Cl - ^{35}Cl]^+$ gives 70 peak
$[^{35}Cl - ^{37}Cl]^+$ gives 72 peak
$[^{37}Cl - ^{37}Cl]^+$ gives 74 peak

The compound with peaks at 12, 13, 14, 15, 16 is $CH_4$ (Methane). The peaks are assigned as follows:

$CH_4^+$ gives the 16 peak
$CH_3^+$ gives the 15 peak
$CH_2^+$ gives the 14 peak
$CH^+$ gives the 13 peak
$C^+$ gives the 12 peak

[Note that the $^{13}C$ isotope is in low enough abundance that it does not give a significant peak at 17]


The compound with peaks at 14, 15 ,16, 17 is $NH_3$ (Ammonia). The peaks are assigned as follows:

$NH_3^+$ gives the 17 peak
$NH_2^+$ gives the 16 peak
$NH^+$ gives the 15 peak
$N^+$ gives the 14 peak

This compound is definitely not methane since the $^{13}C$ isotope is in low enough abundance to give no significant peak at 17. Additionally, peaks at 13 and 12 would also be observed.


The mixture of two chemicals is $HCl$ (Hydrogen Chloride) and $Ar$ (Argon). One way of initially analysing the data is to notice that the 35, 36, 37, and 38 peaks are sequential, but that there is no 39 peak before the 40 peak. This might indicate that the 40 peak is due to one of the chemicals, and the other peaks due to the others. This is in fact the case, and the peaks are assigned below:

$Ar$ gives the 40 peak
$[H\ ^{37}Cl]^+$ gives the 38 peak
$^{37}Cl^+$ gives the 37 peak
$[H\ ^{35}Cl]^+$ gives the 36 peak
$^{35}Cl^+$ gives the 35 peak


The final compound is Ethanol ($\mathbf{C_2H_5OH}$). The data given is initially overwhelming, so it is crucial to identify key peaks, which aid in ascertaining what the molecule is. One of the tallest peaks (45) also has one of the greatest sizes, and so it is quite plausible that this corresponds to the molecular ion or to the molecular ion which has lost a hydrogen. It is reassuring to see that there is a smaller 46 and even smaller 47 peak, as these could quite likely be due to heavier, but less abundant, isotopes in the molecular ion. In fact, the 46 peak is the molecular ion peak, with the 45 peak due to the loss of a hydrogen. The 47 peak is largely due to the possibility of $^{13}$C.

The tallest given peak is at 31, which corresponds to a fragment. It is likely that given the abundance of this that it is due to only a single fragmentation. It is 15 less than the molecular ion, and so quite plausibly is due to the loss of a methyl group.

Another large fragment is at 29, which corresponds to a loss of 17. This is likely to be due to the loss of an -OH group.

So far, it is has been hypothesised that the compound contains a methyl group and an -OH group. Initial calculations reveal that methanol, the simplest alcohol, an RMM of 32, which is too small to be the compound in question. However, the next largest alcohol, ethanol, fits all of the data, and so is a suitable candidate to be the molecule in question.

All the peaks can be assigned as follows [note, not all possibilities shown!]:

31 = $ [CH_2OH]^+$
45 = $[CH_2CH_2OH]^+$
29 = $[CH_3CH_2]^+$
27 = $[CH_3C]^+$
46 = $[CH_3CH_2OH]^+$
43 = $[CCH_2OH]^+$
26 = $[CH_2C]^+$
30 = $[CHOH]^+$
15 = $[CH_3]^+$
42 = $[CCHOH]^+$
28 = $[CH_3CH]^+$
19 = $[^{18}OH]^+$
25 = $[CCH]^+$
14 = $[CH_2]^+$
13 = $[CH]^+$
41 = $[CCOH]^+$
47 = $[^{13}CH_3CH_2OH]^+$
44 = $[CHCH_2OH]^+$
17 = $[OH]^+$
24 = $[CC]^+$
18 = $[HOH]^+$
33 =$[CH_3OH]^+$
12 =$C^+$