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Analytical Techniques: Question 6

Syllabus 22.2

Multiple choice AS 1 mark

The mass spectrum of an organic compound, X, containing only carbon and hydrogen, shows a molecular ion peak at m/z=128m/z = 128 with relative abundance 100, and a smaller peak at m/z=129m/z = 129 (the [M+1]+[\text{M}+1]^+ peak) with relative abundance 9.9. No peak of significant height appears at m/z=130m/z = 130.

Given that carbon-13 makes up approximately 1.1% of all naturally occurring carbon atoms, which of the following is the correct interpretation of this data?

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Worked solution

Reading the M+1 peak

The molecular ion, M+\text{M}^+, appears at m/z=128m/z = 128 with relative abundance 100. A second peak, only one mass unit heavier at m/z=129m/z = 129, has a much smaller relative abundance of 9.9. A peak just one mass unit above M+\text{M}^+ is not the [M+2]+[\text{M}+2]^+ pattern used to detect chlorine or bromine (which sits two mass units above M+\text{M}^+); instead it is the [M+1]+[\text{M}+1]^+ peak, produced by the small fraction of molecules that happen to contain one atom of the heavier carbon isotope, 13C^{13}\text{C}, in place of the common 12C^{12}\text{C}.

Calculating the number of carbon atoms

Because 13C^{13}\text{C} makes up only about 1.1% of all naturally occurring carbon atoms, the relative height of the [M+1]+[\text{M}+1]^+ peak (compared with M+\text{M}^+) is approximately proportional to the number of carbon atoms present, nn, each contributing 1.1% of the molecular ion abundance. Rearranging this relationship:

n=100×abundance of [M+1]+1.1×abundance of M+n = \frac{100 \times \text{abundance of } [\text{M}+1]^+}{1.1 \times \text{abundance of } \text{M}^+}

n=100×9.91.1×100=990110=9n = \frac{100 \times 9.9}{1.1 \times 100} = \frac{990}{110} = 9

So compound X contains 9 carbon atoms. With Mr=128M_r = 128 and only carbon and hydrogen present, this is consistent with the saturated hydrocarbon C9H20\text{C}_9\text{H}_{20} (nonane): 9(12)+20(1)=108+20=1289(12) + 20(1) = 108 + 20 = 128.

Why the absence of a peak at m/z = 130 matters

No significant peak appears at m/z=130m/z = 130, i.e. no [M+2]+[\text{M}+2]^+ peak. If X contained one chlorine atom, a [M+2]+[\text{M}+2]^+ peak roughly a third the height of M+\text{M}^+ would be expected at m/z=130m/z = 130; one bromine atom would give a [M+2]+[\text{M}+2]^+ peak almost as tall as M+\text{M}^+. Neither is observed, so compound X contains no chlorine or bromine. Confirming that the m/z=129m/z = 129 peak is the much smaller carbon-13 [M+1]+[\text{M}+1]^+ peak, not a halogen isotope peak.

Why the other options are wrong

  • B applies the right general idea, that a small peak above M+\text{M}^+ relates to composition, but skips the essential 1.1% correction factor for the natural abundance of 13C^{13}\text{C}; treating 9.9% directly as “10 carbons” overstates nn by rounding the raw percentage instead of dividing it by 1.1.
  • C mistakes the [M+1]+[\text{M}+1]^+ peak (one mass unit above M+\text{M}^+, from carbon-13) for the [M+2]+[\text{M}+2]^+ pattern used to detect chlorine or bromine (two mass units above M+\text{M}^+); the two effects have different origins and appear at different m/zm/z values.
  • D wrongly dismisses the m/z=129m/z = 129 peak as noise. A [M+1]+[\text{M}+1]^+ peak arising from natural carbon-13 abundance is a real, reproducible, and diagnostically useful feature of every organic mass spectrum, not measurement error.

Final answer

A. Compound X contains 9 carbon atoms.