Spectrometry detective
From the atomic masses recorded in a mass spectrometry analysis can
you deduce the possible form of these compounds?
Age
16 to 18
Challenge level
Problem
This problem involves mathematical reasoning concerning elements of chemistry. For an introduction to mass spectrometry read our article Inspect Your Gadgets.
A diatomic gas of an element with a single stable isotope is analysed in a mass spectrometer. How many peaks will there be? How many peaks will there be if the element forming a diatomic gas has $2$ or $3$ stable isotopes?
When water is analysed in a mass spectrometer there are peaks at relative atomic mass $17$ and $18$. What chemicals do these peaks correspond to? Why are there no peaks at $1$ and $16$?
A compound is analysed and has peaks at $35, 37, 70, 72$ and $74$. What is this compound?
Another compound has peaks at $12, 13, 14, 15, 16$. What might this be? What is it definitely not?
Another compound has peaks at $14, 15, 16, 17$. What might this be? What is it definitely not?
A mixture of two chemicals is analysed and has peaks at $35, 36, 37, 38$ and $40$. What might this be? What is it definitely not?
Extension: A final compound has peaks at (from tallest to smallest) $31, 45, 29, 27, 46, 43, 26, 30, 15, 42, 28, 19, 25, 14, 13, 41, 47, 44, 17, 24, 18, 33, 12$. Can you suggest a likely candidate for the compound? What could the various peaks correspond to?
Other mathematical chemistry problems can be found on the chemNRICH pages.
NOTES AND BACKGROUND
This problem gives the opportunity to practise analysing results from mass spectrometry experiments. In this important process compounds are heated and the relative mass and frequency of the resulting fragments are measured. From the mass-spectrum it is often possible to determine the chemical composition of the original compound. To do this requires a knowledge of the atomic mass and relative abundance of the isotopes of the elements. There is a strong experimental input to this process as more complex molecules, such as long hydrocarbons break into a wide variety of fragments which complicated the analysis.
Source of data:
Mass Spectra by NIST Mass Spec Data Center, S.E. Stein, director, in NIST Chemistry WebBook, NIST Standard Reference Database Number 69 , Eds. P.J. Linstrom and W.G. Mallard, National Institute of Standards and Technology, Gaithersburg MD, 20899, http://webbook.nist.gov, (retrieved May 22, 2009).
You can read more about mass spectrometry at http://en.wikipedia.org/wiki/Mass_spectrometry
Getting Started
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 $N^+$.
Look for key facts about the data: smallest numbers, largest numbers, conspicuous gaps in the sequence. What elements in individual form would give rise to some of these? From this you can probably rule out the existence of certain elements, which will allow you to make a start.
There will be an iteration of attempts as you rule out various possibilities.
Be sure to double-check that your final answer gives a convincing explanation for the whole spectrum.
Look for key facts about the data: smallest numbers, largest numbers, conspicuous gaps in the sequence. What elements in individual form would give rise to some of these? From this you can probably rule out the existence of certain elements, which will allow you to make a start.
There will be an iteration of attempts as you rule out various possibilities.
Be sure to double-check that your final answer gives a convincing explanation for the whole spectrum.
Student Solutions
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 =$C^+$
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 =$C^+$
Teachers' Resources
Why do this problem?
This problem gives practice in mathematical reasoning in a genuine scientific context. It involves reasonably basic mathematics of combinatorics but requires more advanced inductive reasoning. It might prove an interesting context for the prospective mathematician and will definitely be useful for those interested in studying chemistry at university.Possible approach
As this problem does not fit directly into the typical
chemistry or mathematics classroom it could be used as an end of
term activity in which the focus is to encourage some clear
mathematical thinking in a cross-curricular context.
Suggest that students interested in both chemistry and
mathematics try the problem, either individually or in small
groups. The basic chemical knowledge required is simply that of
isotopes and atomic mass. The mass spectrometer might be
unfamiliar, but should be simple to grasp in essence.
The focus throughout the problem should be on clarity in the
mathematical explanations, and there will be an element of
convincing others of the soundness of any resulting analysis.
When this problem was created it caused a great deal of
discussions amongst students. Hopefully some of this discussion
might be replicated amongst your students.
Finally, note that this problem is of an industrial, real
world sort. It gives a flavour of the types of real questions which
might be asked to professional scientists and mathematicians, where
errors in mathematical reasoning can be highly costly . It gives
great practice in such thinking.
Key questions
Have you understood the chemical terms?
Which numbers seem to stand out, when compared with the
periodic table?
Are you clear as to which aspects of you calculations are
mathematically certain?
Possible extension
Extension is built into this problem directly.
Possible support
To make this problem more straightforward, provide copies of
the periodic table. Suggest that students write down all the
diatomic gasses that they know and start from these.
For a more straightforward foray into combinatoric chemistry,
try the problem Heavy
Hydrocarbons.