the principles of mass spectrometry as an analytical technique for identifying elements and compounds, including ionisation, acceleration, deflection and detection, the interpretation of a mass spectrum (m/z, base peak, molecular ion peak, isotope peaks) and an introduction to fragmentation

What this dot point is asking

VCAA wants you to describe mass spectrometry as an analytical technique, name the four stages of a mass spectrometer, read a mass spectrum (m/z axis, base peak, molecular ion peak, isotope peaks), and give an introductory explanation of fragmentation for organic molecules.

The answer

Why mass spectrometry matters

Mass spectrometry (MS) is one of the most powerful tools in analytical chemistry. It can:

• Determine the relative atomic mass of an element from the abundance of its isotopes.
• Determine the relative molecular mass of a pure organic or inorganic compound.
• Help identify an unknown compound from its fragmentation pattern.
• Detect trace amounts of substances (parts per billion) in forensic, environmental and biomedical samples.

Inside the instrument: four stages

1. Ionisation. A gaseous sample is bombarded with high-energy electrons (electron impact). Each collision knocks an electron off a sample molecule, producing a positive ion (the molecular ion $M^{+}$) and sometimes splitting it into smaller cations and neutral fragments.
2. Acceleration. A high voltage accelerates the cations to a common kinetic energy.
3. Deflection. A strong magnetic field deflects the ions; the radius of the curved path depends on the mass-to-charge ratio (m/z). Lighter ions and more highly charged ions deflect more.
4. Detection. Ions arriving at the detector produce a current proportional to their abundance. The instrument plots relative abundance against m/z.

Reading a mass spectrum

The output is a bar chart:

• x-axis: m/z (mass-to-charge ratio). For the singly-charged cations that dominate in this course, m/z equals the mass of the ion in atomic mass units.
• y-axis: relative abundance, with the largest peak (base peak) scaled to 100%.

Key features:

• Molecular ion peak ($M^{+}$): corresponds to the intact molecule that has lost a single electron. Its m/z value gives the molecular mass of the compound. Usually the rightmost significant peak in the spectrum.
• Base peak: the most abundant ion. Set to 100% relative intensity. Often a fragment, not the molecular ion.
• Fragment peaks: smaller cations produced when the molecular ion breaks apart inside the instrument.
• Isotope peaks: extra peaks at $M+1$, $M+2$, etc., caused by the natural abundance of heavier isotopes (mainly $^{13}C$, $^{37}Cl$, $^{81}Br$).

Isotope patterns to know

• Carbon ($^{13}C$, 1.1%): every C in a molecule contributes about 1.1% to the $M+1$ peak. So a molecule with $n$ carbons has an $M+1$ peak of roughly $1.1n\%$ of the $M^{+}$ height. This is a quick way to count carbon atoms in an unknown compound.
• Chlorine ($^{35}Cl$ : $^{37}Cl$, about 3:1): a molecule with one Cl shows an $M$ and $M+2$ pair of peaks in a 3:1 ratio. With two Cl atoms the pattern becomes 9:6:1 ($M$, $M+2$, $M+4$).
• Bromine ($^{79}Br$ : $^{81}Br$, about 1:1): a molecule with one Br shows an $M$ and $M+2$ pair of nearly equal height.

These patterns are diagnostic. Seeing a 3:1 doublet of peaks two units apart almost certainly means a chlorine atom is in the molecule.

Fragmentation

When a molecule is ionised, the molecular ion is excited and often falls apart along its weakest bond. The pieces include:

• A fragment cation (detected by the instrument; appears on the spectrum).
• A neutral fragment (radical or small molecule; not detected because it has no charge).

A typical fragmentation:

Only $A^{+}$ shows up on the spectrum, at m/z corresponding to its mass. Fragmentation patterns are reproducible for a given compound and are catalogued in spectral databases, so an unknown spectrum can often be matched to a known compound.

Common low-mass fragments for organic compounds:

Spotting a small molecular ion plus a base peak 15 lighter is a common signature of loss of a methyl group; m/z difference of 17 suggests loss of $OH$; m/z difference of 29 suggests loss of $CHO$ (aldehyde) or $C_2H_5$.

Worked example

A mass spectrum of an unknown compound shows $M^{+} = 78$, a base peak at m/z = 78, and a small peak at m/z = 79 about 6.6% of the base peak. Suggest a structure.

• The $M+1$ ratio of 6.6% suggests about 6 C atoms (since each C contributes 1.1%).
• A molecule with 6 C atoms and $M_r = 78$ has 6 H atoms and no oxygen (since $6 \times 12 = 72$, leaving 6 for hydrogens). That is $C_6H_6$.
• $C_6H_6$ is benzene, and the strong base peak at the molecular ion (little fragmentation) is consistent with the very stable aromatic ring.

Common traps

Calling the y-axis the mass

The y-axis is relative abundance. The mass-to-charge ratio is the x-axis.

Assuming the base peak is always $M^{+}$

It often is not. The base peak is just the most abundant ion. For propan-1-ol the base peak is $C_3H_7^{+}$ at 43, not $M^{+}$ at 60.

Forgetting that neutral fragments are invisible

A fragment of mass 17 lost as a neutral $OH$ does not appear on the spectrum; only the surviving cation at $M - 17$ shows up.

Counting carbon atoms by the absolute height of $M+1$

Use the ratio $M+1 / M$, not the raw height.

Confusing m/z with mass for multiply charged ions

At Unit 1 level almost all ions are +1, so m/z equals the mass; for +2 ions m/z would be half the mass. Worth noting but rare.

In one sentence

Mass spectrometry ionises a gaseous sample, accelerates the cations, deflects them according to their mass-to-charge ratio and detects the abundance at each m/z, producing a spectrum whose molecular ion peak gives the relative molecular mass, whose base peak is the most abundant ion, whose isotope peaks reveal the elements present (especially Cl, Br and the number of carbons), and whose fragmentation pattern identifies the structure of the unknown compound.

Examples in context

Example 1. Doping control at the Australian Sports Anti-Doping Laboratory. Forensic chemists at the Australian Sports Drug Testing Laboratory in Pymble screen athlete urine samples taken at events such as the Australian Open. They use gas chromatography coupled to mass spectrometry. After a sample is ionised, the molecular ion $M^{+}$ of an anabolic steroid such as nandrolone appears at m/z $= 274$, with a characteristic loss of $\text{CH}_3$ (15) giving a fragment at 259 and a loss of the side chain giving 213. Quantification is by comparing peak heights to a deuterated internal standard. A confirmed identification needs at least three diagnostic fragment ions within $\pm 0.1$ m/z and ratios matching a reference spectrum.

Example 2. Detecting chlorinated pesticides in Murray-Darling water. Researchers at the Australian Centre for Research on Water and Wetlands use mass spectrometry to identify trace pesticides in irrigation runoff. Atrazine has $M_r = 215$, and its molecular ion at m/z $= 215$ shows a small $M+2$ peak about $32\%$ of $M$ because of the single $^{37}\text{Cl}$ atom (the natural Cl-35 to Cl-37 ratio is roughly $3:1$). Fragmentation cleaves the isopropyl side chain to give a strong peak at m/z $= 200$ (loss of 15). A detection threshold near $10^{-9}$ g per litre allows monitoring against the Australian Drinking Water Guideline of $40 \, \mu\text{g}$ per litre.

Try this

Q1. Outline the four stages of operation of a mass spectrometer. [4 marks]

• Cue. Ionisation (electron bombardment knocks an electron off, forming cation); acceleration (electric field); deflection (magnetic field, m/z dependent); detection (count ions, plot abundance vs m/z).

Q2. A compound shows a molecular ion at m/z $= 88$ with an $M+1$ peak that is $4.4\%$ of $M$, and no significant $M+2$. (a) Determine the number of carbon atoms. (b) Suggest a molecular formula. [3 marks]

• Cue. (a) $4.4 / 1.1 = 4$ carbons. (b) $C_4 H_8 O_2$ (mass $= 48 + 8 + 32 = 88$); for example butanoic acid or ethyl ethanoate.

Q3. A mass spectrum of an unknown organic compound shows $M^{+}$ at m/z $= 78$ as the base peak with no isotope cluster suggesting Cl or Br. (a) Calculate the relative molecular mass. (b) Suggest a formula consistent with the very stable base peak. (c) Explain why little fragmentation is observed. [2+2+2 marks]

• Cue. (a) $M_r = 78$. (b) $C_6 H_6$ (benzene). (c) Delocalised aromatic ring makes the cation unusually stable, so fragmentation is energetically unfavourable.