Unit 1: How can the diversity of materials be explained?

VICChemistrySyllabus dot point

How can knowledge of elements explain the properties of matter?

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

A focused VCE Chemistry Unit 1 answer on mass spectrometry as an analytical technique. Covers the four stages of a mass spectrometer (ionisation, acceleration, deflection, detection), interpreting a mass spectrum (m/z axis, base peak, molecular ion peak, isotope patterns) and an introduction to fragmentation for organic molecules.

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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+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+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+1M+1, M+2M+2, etc., caused by the natural abundance of heavier isotopes (mainly 13C^{13}C, 37Cl^{37}Cl, 81Br^{81}Br).

Isotope patterns to know

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

M+A++BM^{+} \rightarrow A^{+} + B^{\cdot}

Only A+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:

m/z Fragment From
15 IMATH_26 Loss of methyl
17 IMATH_27 (as a loss) Hydroxyl
29 IMATH_28 or IMATH_29 Ethyl or aldehyde
43 IMATH_30 or IMATH_31 Propyl or acetyl
45 IMATH_32 Carboxylic acid
77 IMATH_33 Phenyl

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 OHOH; m/z difference of 29 suggests loss of CHOCHO (aldehyde) or C2H5C_2H_5.

Worked example

A mass spectrum of an unknown compound shows M+=78M^{+} = 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+1M+1 ratio of 6.6% suggests about 6 C atoms (since each C contributes 1.1%).
  • A molecule with 6 C atoms and Mr=78M_r = 78 has 6 H atoms and no oxygen (since 6×12=726 \times 12 = 72, leaving 6 for hydrogens). That is C6H6C_6H_6.
  • IMATH_42 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+M^{+}. It often is not. The base peak is just the most abundant ion. For propan-1-ol the base peak is C3H7+C_3H_7^{+} at 43, not M+M^{+} at 60.

Forgetting that neutral fragments are invisible. A fragment of mass 17 lost as a neutral OHOH does not appear on the spectrum; only the surviving cation at M17M - 17 shows up.

Counting carbon atoms by the absolute height of M+1M+1. Use the ratio M+1/MM+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.

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

2025 VCE SAC-style6 marksA student records the mass spectrum of an unknown organic compound. The spectrum shows a small peak at m/z = 60, a base peak at m/z = 43, and a peak at m/z = 15. There is also a peak at m/z = 61 with about 3.3% of the height of the m/z = 60 peak. (a) Identify the molecular ion peak and explain what it tells you. (b) Explain why the base peak is at m/z = 43 even though the molecule has a higher mass. (c) The compound is suspected to be propan-1-ol ($CH_3CH_2CH_2OH$, $M_r = 60$). Propose a fragment for the m/z = 15 peak and a fragment for the m/z = 43 peak. (d) Explain the small m/z = 61 peak.
Show worked answer →

A 6-mark answer needs each peak identified and the role of fragmentation and isotopes named.

(a) Molecular ion peak: m/z = 60. This is the peak corresponding to the intact molecule that has lost one electron, M+M^{+}. Its m/z value gives the relative molecular mass of the compound (60), which is consistent with propan-1-ol C3H8OC_3H_8O.

(b) The base peak is the most abundant ion in the spectrum and is set to 100% relative intensity. It does not have to be the molecular ion. Here the base peak is at m/z = 43, meaning a fragment ion of mass 43 is produced more readily than the intact M+M^{+}. The molecular ion of propan-1-ol fragments easily on ionisation because the cation is not very stable; the fragments produced are more stable cations.

(c) m/z = 15 corresponds to CH3+CH_3^{+} (mass 12+3=1512 + 3 = 15), the methyl cation. m/z = 43 corresponds to loss of OHOH (Mr=17M_r = 17) from M+M^{+}, giving C3H7+C_3H_7^{+} (mass 6017=4360 - 17 = 43), the propyl cation.

(d) The small m/z = 61 peak is the **M+1M+1 isotope peak**. It comes from the small natural abundance of 13C^{13}C (about 1.1% per carbon). A 3-carbon molecule has about 3×1.1%=3.3%3 \times 1.1\% = 3.3\% chance of having one 13C^{13}C atom, which matches the observed intensity and confirms there are 3 C atoms in the molecule.

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