Instrumental Methods of Analysis
1. UV–visible, fluorescence and atomic absorption spectrophotometry
A UV–visible spectrophotometer has a continuum source (deuterium lamp for the UV, tungsten–halogen for the visible), a monochromator, the sample cell (quartz below about 350 nm, since glass absorbs) and a detector (photomultiplier or photodiode array). Quantitative work uses the Beer–Lambert law A = εbc through a calibration line. Apparent deviations come from chemistry (association, dissociation, pH-dependent equilibria), from stray light and from polychromatic radiation, and they are worst at high absorbance; measurements are most precise at A of roughly 0.2–0.8. Absorbances are additive, so a two-component mixture is solved from measurements at two wavelengths.
In fluorescence the sample is excited at one wavelength and emission is observed at a longer one, usually at 90° to the beam so that transmitted light is not seen. For dilute solutions the emitted intensity is I_F = kΦ_F I₀ εbc, linear in concentration, and because it is measured against a dark background rather than as a small difference between two large signals, fluorescence is typically 10³ times more sensitive than absorbance. At higher absorbance the inner-filter effect (the solution absorbing its own excitation and emission) bends the calibration over. Atomic absorption spectroscopy (AAS) measures elements: the sample is atomised in a flame or a graphite furnace (the furnace is far more sensitive), and a hollow-cathode lamp of the same element emits its sharp resonance lines, which the ground-state atoms absorb in proportion to their concentration. Almost all atoms are in the ground state at flame temperatures, which is why absorption, not emission, is the sensitive measurement. Chemical interferences (phosphate binding Ca) are removed with a releasing agent such as La³⁺, ionisation interference with an excess of an easily ionised element (K), and broad background absorption by deuterium-lamp or Zeeman correction.
2. FT-IR spectrophotometry and mass spectrometry
An FT-IR spectrometer replaces the monochromator with a Michelson interferometer: a moving mirror varies the path difference, the detector records an interferogram containing all wavelengths at once, and a Fourier transform converts it into the spectrum. The gains are the multiplex (Fellgett) advantage — every wavenumber is measured throughout the scan, so signal-to-noise improves for the same time — the throughput (Jacquinot) advantage of having no narrow slits, and the wavenumber accuracy (Connes) advantage from a reference laser. Attenuated total reflectance (ATR) lets solids and liquids be run without preparation. For metal carbonyls the number of IR-active ν(CO) bands follows from symmetry: trans-[M(CO)₄L₂] (D₄h) shows 1, cis-[M(CO)₄L₂] (C₂v) 4, fac-[M(CO)₃L₃] (C₃v) 2 and mer-[M(CO)₃L₃] (C₂v) 3, so the count distinguishes isomers.
A mass spectrometer ionises the sample (electron impact, EI, which fragments; the soft methods electrospray, ESI, and MALDI, which keep large and fragile molecules intact), separates ions by m/z in a magnetic sector, quadrupole or time-of-flight analyser, and counts them. Resolution R = m/Δm is the ability to separate neighbouring masses: telling N₂⁺ (28.0061) from CO⁺ (27.9949) needs R ≈ 2500. Isotope patterns identify elements: one Cl gives M and M + 2 in 3:1 (³⁵Cl:³⁷Cl), two Cl give 9:6:1, one Br 1:1; the nitrogen rule says an odd molecular mass means an odd number of nitrogens. Metal carbonyls fragment by successive loss of CO (28 u), which counts the carbonyls.
3. NMR and ESR spectroscopy of inorganic compounds
Inorganic NMR uses many nuclei — ¹H, ¹³C, ¹⁹F, ³¹P, ¹¹B, ¹⁹⁵Pt, ¹⁰³Rh. Coupling to n equivalent nuclei of spin I gives 2nI + 1 lines: the ³¹P signal of PF₅ is split by the fluorines, and because Berry pseudorotation makes all five F equivalent on the NMR timescale it is a 1:5:10:10:5:1 sextet, although the static molecule has axial and equatorial fluorines. Metal hydrides resonate far upfield (δ 0 to −30). A partly abundant spin-½ metal gives satellites: ¹⁹⁵Pt (33.8%) splits a ligand signal into a central line flanked by two satellites, each about one-sixth of the total.
Electron spin resonance (ESR, EPR) detects unpaired electrons. In a field B the resonance condition is hν = gμ_B B; for a free electron g = 2.0023, and in metal complexes spin–orbit coupling shifts g and makes it anisotropic (g∥ and g⊥ in axial Cu(II) complexes). Coupling to nuclei gives hyperfine structure: n equivalent nuclei of spin I give 2nI + 1 lines. So Cu²⁺ (⁶³,⁶⁵Cu, I = 3/2) gives 4 lines, VO²⁺ (⁵¹V, I = 7/2) 8 lines, the methyl radical (3 H) 4 lines in 1:3:3:1, a nitroxide (¹⁴N, I = 1) 3 lines, and the naphthalene radical anion, with two sets of four equivalent protons, (4 + 1)(4 + 1) = 25 lines. Kramers’ theorem guarantees a signal for any odd-electron ion; integer-spin systems with large zero-field splitting may be ESR-silent.
4. Mössbauer spectroscopy (Fe, Sn) and X-ray crystallography
Mössbauer spectroscopy uses recoil-free emission and resonant absorption of γ-rays by nuclei bound in a solid. For iron the source is ⁵⁷Co, which decays to an excited ⁵⁷Fe emitting a 14.4 keV γ-ray; for tin, ¹¹⁹ᵐSn gives a 23.9 keV line. The lines are so sharp that the source is moved at a few mm/s to scan energies by the Doppler effect, ΔE = (v/c)E_γ — 1 mm/s shifts a 14.4 keV photon by only 4.8 × 10⁻⁸ eV. Three parameters are read. The isomer shift δ reflects the s-electron density at the nucleus and hence the oxidation and spin state: high-spin Fe(II) about 0.9–1.4 mm/s, high-spin Fe(III) about 0.3–0.5, low-spin iron lower still; for tin, Sn(II) and Sn(IV) are cleanly separated. The quadrupole splitting ΔE_Q arises because the I = 3/2 excited state of ⁵⁷Fe senses an electric-field gradient, giving a doublet: large for high-spin Fe(II) with its asymmetric d⁶ shell, small for high-spin Fe(III), whose half-filled d⁵ shell is spherical. Magnetic hyperfine splitting in an internal field gives six lines, as in α-Fe₂O₃ and magnetite. In Prussian blue the method separates the high-spin Fe(III) and low-spin Fe(II) sites directly.
Single-crystal X-ray crystallography measures the positions and intensities of thousands of Bragg reflections. The intensities give the amplitudes of the structure factors but not their phases (the phase problem), which are recovered by direct methods or, for compounds containing a heavy atom, by the Patterson (heavy-atom) method; the electron-density map is then refined until the calculated and observed intensities agree, measured by the R factor (a few per cent for a good structure). Systematic absences give the space group. Because X-rays scatter from electrons, heavy atoms dominate and hydrogen atoms are located poorly; neutron diffraction finds them. Powder diffraction identifies crystalline phases by their d-spacing fingerprint.
5. Electroanalytical methods: cyclic voltammetry and ion-selective electrodes
In cyclic voltammetry the potential of a working electrode in an unstirred solution is swept linearly to a switching potential and back, and the current is recorded. A reduction on the forward sweep gives a cathodic peak and re-oxidation on the reverse an anodic peak. For a reversible (Nernstian) couple the peak separation ΔE_p = E_pa − E_pc ≈ 59/n mV at 25 °C (strictly 57–59 mV for n = 1, depending on the switching potential) and independent of scan rate, the half-wave potential E½ ≈ (E_pa + E_pc)/2 approximates E°′, and i_pa/i_pc = 1. The Randles–Sevcik equation i_p = 2.69 × 10⁵ n^(3/2) A D^½ C ν^½ (at 25 °C) makes the peak current proportional to concentration and to the square root of scan rate, the signature of diffusion control. Larger, scan-rate-dependent separations indicate slow electron transfer (quasi-reversible or irreversible), and a missing return peak a chemical reaction following the electron transfer.
An ion-selective electrode develops a potential across a membrane that responds to one ion: E = constant + (2.303RT/zF) log aᵢ, so the ideal slope is 59.2/z mV per decade at 25 °C — 59.2 mV for H⁺ or F⁻ (with the sign reversed for anions), 29.6 mV for Ca²⁺. The glass electrode for pH uses a hydrated silicate membrane; the fluoride electrode a single crystal of LaF₃ doped with EuF₂; others use ion-exchangers or ionophores such as valinomycin (K⁺) in a PVC membrane. Interference by other ions is described by the Nikolsky–Eisenman equation through a selectivity coefficient; the glass electrode’s "alkaline error" at high pH is its response to Na⁺.
6. Thermo-analytical methods: TGA, DTA and DSC
| Method | What is measured against temperature | What it detects |
|---|---|---|
| TGA (thermogravimetry) | sample mass | dehydration, decomposition, oxidation — any change of mass, quantitatively |
| DTA (differential thermal analysis) | temperature difference ΔT between sample and inert reference | endothermic and exothermic events, with or without mass change (melting, phase transitions, crystallisation) |
| DSC (differential scanning calorimetry) | heat flow needed to keep sample and reference at the same temperature | enthalpies of transitions quantitatively, heat capacity, the glass transition, purity |
The classic TGA curve is calcium oxalate monohydrate, CaC₂O₄·H₂O (M = 146.12 g/mol), which loses mass in three clean steps: water near 100–200 °C (to CaC₂O₄, 87.7% of the original mass), carbon monoxide near 400–500 °C (to CaCO₃, 68.5%) and carbon dioxide near 700–850 °C (to CaO, 38.4%). Each plateau’s mass identifies the intermediate. DTA and DSC on the same sample show the first and last steps as endotherms; in air the CO step appears exothermic because CO burns. DSC also measures the glass transition of polymers, a step in heat capacity with no latent heat, which neither TGA nor a melting-point apparatus can see.
Key takeaways
- UV–vis quantifies through A = εbc (most precise at A ≈ 0.2–0.8); fluorescence is ~10³ times more sensitive; AAS uses an element-specific hollow-cathode lamp and ground-state atoms.
- FT-IR has multiplex, throughput and wavenumber-accuracy advantages; ν(CO) band counts separate cis/trans and fac/mer carbonyls; MS isotope patterns (Cl 3:1, Br 1:1) and R = m/Δm.
- NMR and ESR: n equivalent nuclei of spin I give 2nI + 1 lines — Cu²⁺ 4, VO²⁺ 8, CH₃• 4, naphthalene⁻ 25; hν = gμ_B B.
- Mössbauer: isomer shift gives oxidation and spin state, quadrupole doublet the field gradient (large for HS Fe²⁺), six lines a magnetic field; X-ray crystallography solves the phase problem and reports an R factor.
- Reversible CV: ΔE_p ≈ 59/n mV, i_p ∝ √ν; ISE slope 59.2/z mV per decade; TGA weighs, DTA measures ΔT, DSC measures heat flow — CaC₂O₄·H₂O ends as 38.4% CaO.
Practice questions (18)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Calcium oxalate monohydrate, CaC₂O₄·H₂O (M = 146.12 g/mol), is heated to 1000 °C in a thermobalance. What percentage of the original mass remains, to one decimal place? (M(CaO) = 56.08 g/mol)
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Answer: 38.4
By 1000 °C the sample has lost H₂O, CO and CO₂ and is CaO: 56.08/146.12 × 100 = 38.4%. Stopping at CaCO₃ gives 68.5%, and at anhydrous CaC₂O₄ 87.7%; each plateau identifies its intermediate.Which statements about thermal methods are correct?
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Answer: A — TGA records mass as a function of temperature; B — DSC can measure the enthalpy of a phase transition quantitatively; C — DTA detects melting even though no mass is lost
TGA weighs the sample; DSC measures the heat flow, so it gives ΔH; DTA sees the temperature lag of an endotherm such as melting. The glass transition involves no mass change, so TGA is blind to it — it is a DSC measurement.For a reversible, diffusion-controlled couple, the cathodic peak current in cyclic voltammetry is 10 μA at a scan rate of 50 mV/s. What is it at 200 mV/s, in μA?
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Answer: 20
By Randles–Sevcik i_p ∝ √ν: the scan rate rises fourfold, so i_p rises by √4 = 2, to 20 μA. A current proportional to ν itself (40 μA) would indicate an adsorbed species rather than diffusion.A cyclic voltammogram shows E_pc = +0.215 V and E_pa = +0.245 V, with i_pa/i_pc = 1, independent of scan rate. What does this indicate?
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Answer: A — A reversible two-electron couple with E°′ ≈ +0.230 V
ΔE_p = 30 mV ≈ 59/n with n = 2, constant with scan rate, and equal peak currents mark a reversible two-electron process; E°′ ≈ (0.215 + 0.245)/2 = 0.230 V. A one-electron couple would need about 59 mV; a following chemical reaction would shrink the return peak.What is the ideal Nernstian slope of a Ca²⁺ ion-selective electrode at 25 °C, in mV per tenfold change in Ca²⁺ activity, to one decimal place? (2.303RT/F = 59.16 mV)
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Answer: 29.6
Slope = 2.303RT/(zF) = 59.16/2 = 29.6 mV per decade for the divalent ion. A monovalent ion such as K⁺ or F⁻ gives 59.2 mV; using that value for Ca²⁺ overstates the response twofold.The sensing membrane of the fluoride ion-selective electrode is
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Answer: A — a single crystal of LaF₃ doped with EuF₂
F⁻ ions move through vacancies in the LaF₃ lattice created by the Eu²⁺ dopant, so the crystal responds almost only to fluoride (OH⁻ is the main interferent). Glass is the pH electrode and valinomycin the K⁺ ionophore; platinum is an inert redox electrode.How many hyperfine lines are expected in the ESR spectrum of a Cu²⁺ complex, taking copper as a single nucleus of spin I = 3/2?
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Answer: 4
2nI + 1 with n = 1 and I = 3/2 gives 4 lines. Coupling to ligand nitrogens adds superhyperfine structure on each line; taking I = 1/2 gives 2.How many hyperfine lines appear in the ESR spectrum of the vanadyl ion VO²⁺ (⁵¹V, I = 7/2, essentially 100% abundant)?
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Answer: 8
V(IV) is d¹ with one unpaired electron coupled to one ⁵¹V nucleus: 2 × 1 × 7/2 + 1 = 8 equally intense lines, a textbook fingerprint of vanadyl. Counting I + 1 = 4.5 is meaningless; the rule is 2I + 1.The naphthalene radical anion has two sets of four equivalent protons with different coupling constants. How many lines does its ESR spectrum show?
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Answer: 25
Each set of four protons gives 2 × 4 × ½ + 1 = 5 lines, and the two independent sets multiply: 5 × 5 = 25 lines (a quintet of quintets). Treating all eight protons as equivalent gives 9, which is wrong because the α and β positions couple differently.In ⁵⁷Fe Mössbauer spectroscopy (E_γ = 14.4 keV), by how much is the γ-ray energy shifted when the source moves at 1.0 mm/s? Give the answer in units of 10⁻⁸ eV, to one decimal place. (c = 2.998 × 10⁸ m/s)
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Answer: 4.8
ΔE = (v/c)E_γ = (1.0 × 10⁻³/2.998 × 10⁸) × 14.4 × 10³ eV = 4.8 × 10⁻⁸ eV. So a velocity scan of a few mm/s covers the whole hyperfine pattern, which shows how extraordinarily narrow the resonance is. Forgetting to convert keV to eV gives 4.8 × 10⁻¹¹.How many lines does a ⁵⁷Fe Mössbauer spectrum show when the nucleus experiences an internal magnetic field (magnetic hyperfine splitting)?
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Answer: 6
The ground state (I = 1/2) splits into 2 levels and the excited state (I = 3/2) into 4; the selection rule Δm_I = 0, ±1 allows 6 transitions, the sextet of α-Fe or haematite. A quadrupole interaction alone gives a doublet.Why does high-spin Fe(III) usually show a much smaller Mössbauer quadrupole splitting than high-spin Fe(II)?
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Answer: A — Its half-filled d⁵ shell is spherically symmetric and adds no field gradient
⁶A₁ high-spin d⁵ puts one electron in each d orbital, a spherical charge distribution, so only the more distant ligands create a field gradient; high-spin d⁶ has one extra, asymmetric electron and a large ΔE_Q. Nuclear spin belongs to the ⁵⁷Fe nucleus, whatever the oxidation state.How many IR-active C–O stretching bands are expected for mer-[Mo(CO)₃(PR₃)₃] (C₂v)?
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Answer: 3
The three CO stretches of the mer isomer span 2A₁ + B₁ (or B₂) in C₂v, and all three are IR active. The fac isomer (C₃v) gives A₁ + E, only 2 bands; the count distinguishes the two isomers.In the mass spectrum of a compound containing two chlorine atoms (³⁵Cl:³⁷Cl = 3:1), what is the intensity of the M + 2 peak relative to the M peak, to two decimal places?
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Answer: 0.67
(3 + 1)² expands to 9 : 6 : 1 for M : M + 2 : M + 4, so M + 2/M = 6/9 = 0.67. A single chlorine gives 1/3 = 0.33, and a single bromine about 1.0.What mass-resolving power R = m/Δm is needed to separate N₂⁺ (m = 28.0061) from CO⁺ (m = 27.9949)? Give the nearest whole number, taking m = 28.0.
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Answer: 2500
Δm = 28.0061 − 27.9949 = 0.0112, so R = 28.0/0.0112 = 2500. A low-resolution instrument (R of a few hundred) shows both as one peak at m/z 28.The multiplex (Fellgett) advantage of an FT-IR spectrometer arises because
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Answer: A — all wavenumbers are measured simultaneously throughout the scan
The interferometer sends every wavelength to the detector at once, so each spectral element is observed for the whole scan time and the signal-to-noise ratio improves as √N. The absence of slits is the separate Jacquinot advantage; a grating is what FT-IR replaces.In AAS a 2.0 ppm standard gives an absorbance of 0.25. A sample gives 0.40 under the same conditions. Assuming a linear calibration through the origin, what is the sample concentration, in ppm, to one decimal place?
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Answer: 3.2
A is proportional to concentration, so c = 2.0 × 0.40/0.25 = 3.2 ppm. Inverting the ratio gives 1.25 ppm; in practice several standards bracketing the sample are used rather than a single point.Why does atomic absorption spectroscopy use a hollow-cathode lamp of the analyte element rather than a continuum source?
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Answer: A — Atomic absorption lines are far narrower than a monochromator bandpass, so only a line source of the same element gives measurable, selective absorption
Atomic lines are about 0.002 nm wide, so with a continuum source and a 0.2 nm bandpass the atoms would absorb a negligible fraction of the light. The lamp emits the element’s own resonance lines, even narrower than the absorption lines, giving sensitivity and selectivity. The flame or furnace atomises; deuterium lamps provide UV continua.