Isotope Patterns in Mass Spectrometry

Type a molecular formula. Each atom in a real sample is one of its element's isotopes, chosen at natural abundance, so the molecules come in several masses. The page computes how many sit at each whole mass, M, M+1, M+2 and on, draws the peaks a mass spectrometer would show, and works out the first two by hand.

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Lesson

The theory — Isotope Patterns in Mass Spectrometry

M is the molecule built entirely from each element's lightest isotope; M+k is every isotopologue k whole mass units heavier. For n atoms of an element with a light isotope of abundance a₀ and a heavy one of abundance a, the share with j heavy atoms is the binomial C(n, j) aʲ a₀ⁿ⁻ʲ. A formula's whole cluster is the product of one such distribution per element, collected by total mass shift, which is a convolution. The first two ratios have a closed form: M+1 / M = Σ nₑ a₊₁/a₀, since only one atom can be swapped for a +1 isotope.

What each symbol means

M
the peak at the molecule's lightest possible mass, every atom its lightest isotope
a₀, a₊₁, a₊₂
the abundances of an element's lightest isotope and of the isotopes one and two mass units heavier
nₑ
the number of atoms of element e in the formula
Assumes
Natural abundances as the CIAAW last gave them as single values: ³⁵Cl 75.76%, ⁷⁹Br 50.69%, ¹³C 1.07%, ³⁴S 4.25%, with ²H, ¹⁵N, ¹⁷O and ¹⁸O at theirs. Since 2013 the CIAAW gives these elements as ranges, because real samples vary: carbon-13 runs from 0.96% to 1.16% in terrestrial material, which moves C₆₀'s M+1 from 58% to 70% of M. Peaks are grouped by whole mass units, the molecule carries one charge, and it does not fragment or pick up a hydrogen on the way in.
Breaks when
At high resolution, where ¹³C and ²H no longer land in the same M+1 peak (they differ by 0.003 u), and where the mass defect separates the nominal mass from the true one: insulin's M is nominally 5,801 and actually 5,803.6377. For ions carrying several charges, peaks sit 1/z apart on the m/z axis. And in enriched or labelled samples, whose abundances are not natural.

Three peaks two units apart mean two halogens, and their heights say which 🖖

Chlorine is 75.76% ³⁵Cl and 24.24% ³⁷Cl, so one chlorine gives M and M+2 in heights 100 : 32.00, about 3 : 1. Bromine is 50.69% ⁷⁹Br and 49.31% ⁸¹Br, so one bromine gives 100 : 97.28, nearly 1 : 1. Two halogens give three main peaks whichever two they are. Press CH₂Cl₂: 100 : 63.99 : 10.24, falling away. Press CH₂Br₂: 51.40 : 100 : 48.64, the middle peak tallest. Type CH2BrCl: 77.36 : 100 : 24.08, which is neither. The number of peaks counts the halogens and their heights say which ones, because each is a row of the binomial with that element's own odds, and with odds this close to even bromine's row is Pascal's 1 : 2 : 1.

The tallest peak leaves M at two bromines, four chlorines or 93 carbons 🖖

A peak with one heavy atom is n × (heavy / light) times as tall as M, because any of the n atoms can be the heavy one. Once that product passes 1, M stops being the tallest. For bromine the ratio is 0.9728, so two bromines already do it. For chlorine it is 0.3200, so it takes four: carbon tetrachloride's M is 78.14% of its M+2. For carbon it is 0.0108, so it takes 93. Sixty carbons sound like plenty, since 60 × 1.1% is 66%, but that is still short of M: press C₆₀ and M+1 is 64.89% of M. The two cross at 93 carbons, and in larger molecules the tallest peak keeps moving up the cluster. Insulin's is M+3.

Only 2.90% of insulin molecules have insulin's monoisotopic mass 🖖

Press Insulin. Its 257 carbons alone make the M+1 peak 2.8 times the height of M, and hydrogen, nitrogen, oxygen and six sulfurs add to it. The tallest peak is M+3; M sits at 15.54 on that scale; and 97.10% of the molecules contain at least one heavy isotope. The monoisotopic mass, 5,803.6377 u, belongs to a molecule that is rare in its own sample. A spectrometer that resolves single mass units shows insulin as an envelope of peaks, and the tallest of them is three units heavier.

Problem solved in full

  1. Naming a compound from its cluster 4 steps

    An unknown gives a molecular-ion cluster at m/z 118, 120, 122 and 124 with heights 100 : 96 : 31 : 3.3, and a small peak at 119 about 1.1% of the one at 118. Nothing else in that cluster rises above 1%. How many halogen atoms, which halogen, how many carbons, and what is it?

    Open this problem in the tool

    1. Four peaks two units apart mean three atoms that each have a +2 isotope. Among common elements that is chlorine or bromine.

    2. For three chlorines the heights are 1 : 3r : 3r² : r³ with r = 0.2424 / 0.7576. That matches. Three bromines would put M+2 above M, and so would every mixture containing bromine.

    3. M+1 counts carbon: each carbon adds 1.0816% of M, each hydrogen 0.0115%. 1.1% is one carbon.

    4. What is left of the mass after three ³⁵Cl and one ¹²C is hydrogen.

    Answer

    Three chlorines, one carbon and one hydrogen: CHCl₃, chloroform. Press Chloroform and the page gives 100 : 95.99 : 30.71 : 3.28, with M+1 at 1.09%. The mixed candidates miss badly: two chlorines and a bromine give 62.0 : 100 : 44.9 : 6.2, one and two give 44.1 : 100 : 69.2 : 13.4, three bromines 34.3 : 100 : 97.3 : 31.5.

References (3)