Analytical techniques: TCE Chemistry (Tasmania)

Syllabus dot point

“Select and interpret spectroscopic and chromatographic techniques to determine structure and concentration”

10 min readUpdated 2 Jun 2026

Quick answer

Chemists identify and measure substances using techniques such as mass spectrometry (molar mass), infrared spectroscopy (functional groups), NMR (the carbon-hydrogen framework), chromatography (separation of mixtures) and titration or UV-visible spectroscopy (concentration). Match each technique to the information it provides, and combine several to build a complete structure.

What this dot point is asking

Analytical chemistry answers two questions: what is in a sample (qualitative analysis) and how much is present (quantitative analysis). Different instruments probe different aspects of a molecule, so chemists often combine several techniques. TASC expects you to match a technique to the information it provides and interpret the data it generates.

Mass spectrometry

Mass spectrometry measures the mass-to-charge ratio ( $m/z$ ) of ions. A sample is vaporised and ionised, often breaking into fragments, and the ions are separated by mass. The peak at the highest $m/z$ , the molecular ion peak, gives the molar mass of the original molecule. The pattern of smaller fragment peaks gives structural clues, since particular fragments correspond to particular groups. Mass spectrometry is also used to determine relative isotopic abundances and hence relative atomic mass.

Molecular ion peak

The peak in a mass spectrum produced by the unfragmented ionised molecule. Its $m/z$ value gives the molar mass of the compound, the key starting point for working out a molecular formula.

Infrared spectroscopy

Infrared (IR) spectroscopy identifies functional groups. Bonds absorb infrared radiation at characteristic frequencies that make them vibrate, so the wavenumbers at which absorptions occur reveal which bonds are present. A broad absorption around $3200$ to $3550\ \text{cm}^{-1}$ suggests an $\text{O}-\text{H}$ group, while a strong absorption near $1700\ \text{cm}^{-1}$ indicates a carbonyl ( $\text{C}=\text{O}$ ) group. The complex fingerprint region is unique to each compound and confirms identity by comparison with reference spectra.

UV-visible spectroscopy

UV-visible spectroscopy is mainly used for quantitative analysis of coloured solutions. The absorbance at a particular wavelength is proportional to concentration (the Beer-Lambert relationship). By measuring the absorbance of standards of known concentration, you build a calibration curve and read off the concentration of an unknown.

NMR spectroscopy

Nuclear magnetic resonance (NMR) gives detailed structural information about the carbon-hydrogen framework. Proton ( $^1\text{H}$ ) NMR reveals the different chemical environments of hydrogen atoms: the number of peaks shows how many environments there are, the area under each peak (integration) shows the relative number of hydrogens in each environment, and the chemical shift indicates the type of environment. Carbon-13 NMR similarly shows the number of distinct carbon environments.

Interpreting an infrared spectrum

Read the molar mass

An unknown compound has molar mass $60\ \text{g mol}^{-1}$ from mass spectrometry. Its IR spectrum shows a broad absorption near $3000\ \text{cm}^{-1}$ and a strong absorption near $1710\ \text{cm}^{-1}$ .

Identify the functional groups

The strong absorption near $1710\ \text{cm}^{-1}$ indicates a carbonyl ( $\text{C}=\text{O}$ ). The broad absorption around $3000\ \text{cm}^{-1}$ (overlapping the $\text{O}-\text{H}$ region) is consistent with the $\text{O}-\text{H}$ of a carboxylic acid.

Match to a formula

A carboxylic acid of molar mass $60$ is ethanoic acid, $\text{CH}_3\text{COOH}$ , so the compound is most likely ethanoic (acetic) acid.

Chromatography

Chromatography separates the components of a mixture by their different affinities for a stationary phase and a mobile phase; components that interact more strongly with the stationary phase move more slowly. In thin-layer and paper chromatography, the retention factor $R_f = \dfrac{\text{distance moved by spot}}{\text{distance moved by solvent front}}$ helps identify components by comparison with standards. High-performance liquid chromatography and gas chromatography give precise separations and can be coupled to a mass spectrometer for both separation and identification.

Volumetric analysis

Titration is a classic quantitative method: a standard solution of known concentration is reacted with a measured volume of the unknown until an indicator or instrument shows the equivalence point. From the balanced equation, volumes and concentrations, you calculate the unknown concentration using mole ratios.

Confusing what each technique tells you

A frequent error is to expect IR to give molar mass or NMR to identify functional groups directly. Mass spectrometry gives molar mass, IR identifies functional groups, and NMR maps the hydrogen and carbon environments. Match the technique to the question: mass spectrometry for mass, IR for functional groups, NMR for the carbon-hydrogen skeleton, and titration or UV-visible for concentration.

In exams, justify your choice of technique by stating exactly what information it provides, and when interpreting spectra, link each piece of data to a specific structural feature before drawing a conclusion.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

An unknown organic compound has a molecular ion peak at

m/z = 46

in its mass spectrum. Its infrared spectrum shows a broad absorption at

3300\ \text{cm}^{-1}

and no strong absorption near

1700\ \text{cm}^{-1}

. Its

^1\text{H}

NMR spectrum shows three peaks with integration ratio

3:2:1

. Deduce the structure of the compound, justifying each conclusion.

Show worked answer →

The molecular ion at $m/z = 46$ gives a molar mass of $46\ \text{g mol}^{-1}$ . (1 mark)

The broad IR absorption at $3300\ \text{cm}^{-1}$ indicates an $\text{O}-\text{H}$ group, and the absence of a band near $1700\ \text{cm}^{-1}$ rules out a carbonyl ( $\text{C}=\text{O}$ ), so it is an alcohol rather than a carboxylic acid. (1 mark)

Three $^1\text{H}$ environments in a $3:2:1$ ratio match $\text{CH}_3$ (3 H), $\text{CH}_2$ (2 H) and $\text{OH}$ (1 H). (1 mark)

A $\text{C}_2\text{H}_6\text{O}$ alcohol of molar mass $46$ is ethanol, $\text{CH}_3\text{CH}_2\text{OH}$ . (1 mark)

Practice3 marks

A student measured the absorbance of five standard

\text{Cu}^{2+}

solutions at

600\ \text{nm}

and plotted a calibration line, then measured an unknown solution. (a) State the technique used and why it suits a coloured solution. (b) Explain how the calibration line is used to find the unknown concentration, and state the relationship absorbance has with concentration.

Show worked answer →

(a) UV-visible spectroscopy (colorimetry). It suits coloured solutions because the $\text{Cu}^{2+}$ ion absorbs visible light, and the amount absorbed depends on concentration. (1 mark)

(b) Absorbance is directly proportional to concentration (the Beer-Lambert relationship), so the calibration line is a straight line through the origin. Measure the unknown's absorbance, then read the corresponding concentration off the calibration line (or substitute into its equation). (2 marks)

Practise this

Sources & how we know this

Chemistry Level 4 (CHM415115) course document (2015 onwards) — TASC