Analytical Methods
Analytical chemistry answers two questions: "What is it?" (qualitative analysis) and "How much is there?" (quantitative analysis). The methods range from classical wet chemistry — gravimetry, titration, and separation — to instrumental techniques that exploit the interaction of matter with electromagnetic radiation, electric fields, and crystal lattices. This skill covers the major techniques with worked examples of data interpretation.
Agent affinity: hodgkin (analytical/structural chemistry, primary)
Concept IDs: chem-mixtures-pure-substances, chem-physical-chemical-properties, chem-density
Mixtures vs. Pure Substances
Understanding what you are analyzing starts with classification:
| Category | Subcategory | Characteristics | Example |
|---|---|---|---|
| Pure substance | Element | Cannot be decomposed further | Gold (Au), oxygen (O2) |
| Pure substance | Compound | Fixed composition, decomposable | Water (H2O), NaCl |
| Mixture | Homogeneous (solution) | Uniform composition, single phase | Saltwater, air, brass |
| Mixture | Heterogeneous | Non-uniform, multiple phases | Granite, oil-and-water |
Analytical methods separate mixtures into components and identify/quantify each.
Physical and Chemical Properties in Analysis
Physical properties are observed without changing composition: density, melting point, boiling point, color, refractive index, solubility. These are the basis of physical methods of separation and identification.
Chemical properties describe how a substance reacts: flammability, reactivity with acids, oxidation tendency. Chemical tests (flame tests, precipitation reactions, pH indicators) exploit these.
Intensive properties (independent of amount: density, melting point, refractive index) are especially useful for identification because they are characteristic of the substance itself.
Density and Its Analytical Uses
Density (rho) = mass / volume. Units: g/mL or g/cm^3 for liquids and solids; g/L for gases.
Worked example. A mineral sample has mass ite ite ite 15.6 g and displaces 5.20 mL of water. Identify it.
rho = 15.6 g / 5.20 mL = 3.00 g/mL.
Consulting a density table: calcite = 2.71, fluorite = 3.18, apatite = 3.19. The value 3.00 does not match common minerals exactly — further analysis (XRD, elemental) is needed. But the density narrows the field dramatically. This illustrates why density alone is a screening tool, not definitive identification.
Worked example. Determine whether a gold ring (mass 19.3 g, volume 1.50 mL) is pure gold.
rho = 19.3 / 1.50 = 12.9 g/mL. Pure gold = 19.3 g/mL. This ring is NOT pure gold — likely an alloy. The density is too low by 33%.
Separation Techniques
Filtration
Separates an insoluble solid from a liquid. Gravity filtration for routine work; vacuum filtration for faster throughput and drier precipitates.
Distillation
Separates liquids by boiling point differences. Simple distillation works when boiling points differ by more than 25 C. Fractional distillation (using a fractionating column) resolves closer boiling points by providing multiple vaporization-condensation cycles.
Extraction
Separates compounds by differential solubility in two immiscible solvents (typically water and an organic solvent). The partition coefficient K = [solute in organic layer] / [solute in aqueous layer] governs the distribution.
Worked example. Caffeine has K = 4.6 between dichloromethane and water. If 100 mg of caffeine is dissolved in 100 mL water, how much is extracted by a single 100 mL portion of DCM?
Mass in DCM = K x mass-in-water. Total mass = mass-in-DCM + mass-in-water. Let x = mass in water after extraction. mass-in-DCM = 100 - x. K = (100 - x)/100 / (x/100) = (100 - x) / x = 4.6. 100 - x = 4.6x. 100 = 5.6x. x = 17.9 mg in water.
Extracted: 100 - 17.9 = 82.1 mg (82.1% recovery in one extraction).
Multiple smaller extractions are more efficient. Three 33.3 mL portions recover more than one 100 mL portion — a fundamental principle of liquid-liquid extraction.
Chromatography
All chromatographic methods separate components based on differential interaction with a stationary phase and a mobile phase.
| Method | Stationary phase | Mobile phase | Analytes |
|---|---|---|---|
| Paper chromatography | Cellulose paper | Solvent (water/organic mix) | Dyes, amino acids |
| Thin-layer (TLC) | Silica/alumina on plate | Organic solvent | Quick screening |
| Column chromatography | Silica/alumina in column | Organic solvent | Preparative separations |
| Gas chromatography (GC) | Coated capillary column | Carrier gas (He, N2) | Volatile organics |
| HPLC | Packed column (C18, silica) | Liquid solvent gradient | Non-volatile organics, biologics |
| Ion chromatography | Ion-exchange resin | Buffer solution | Ions in water |
Retention factor (Rf) for TLC: Rf = distance traveled by spot / distance traveled by solvent front. Each compound has a characteristic Rf under given conditions.
Worked example. A TLC plate shows three spots at distances 2.1, 3.5, and 4.8 cm. The solvent front traveled 6.0 cm. Calculate Rf values.
Rf1 = 2.1 / 6.0 = 0.35. Rf2 = 3.5 / 6.0 = 0.58. Rf3 = 4.8 / 6.0 = 0.80.
The spot at Rf = 0.80 is least polar (traveled farthest in a normal-phase system where silica is polar and the mobile phase is organic).
Spectroscopic Methods
UV-Visible Spectroscopy
Principle. Molecules absorb UV or visible light, promoting electrons from bonding/nonbonding orbitals to antibonding orbitals. Conjugated systems (alternating single and double bonds) absorb at longer wavelengths (lower energy).
Beer-Lambert Law: A = epsilon x b x c, where A is absorbance, epsilon is the molar absorptivity (L/mol-cm), b is path length (cm), and c is concentration (mol/L).
Worked example. A solution of KMnO4 has an absorbance of 0.750 at 525 nm in a 1.00 cm cell. If epsilon = 2455 L/mol-cm, what is the concentration?
c = A / (epsilon x b) = 0.750 / (2455 x 1.00) = 3.06 x 10^-4 mol/L = 0.306 mM.
Applications. Quantitative analysis of colored solutions, enzyme kinetics (following absorbance change over time), determining concentration of DNA/proteins (A260 and A280 nm).
Infrared (IR) Spectroscopy
Principle. Molecules absorb IR radiation, causing bonds to vibrate (stretch, bend). Each functional group absorbs at characteristic frequencies.
Key absorptions (wavenumber in cm^-1):
| Functional group | Absorption range | Appearance |
|---|---|---|
| O-H (alcohol) | 3200-3550 | Broad |
| O-H (carboxylic acid) | 2500-3300 | Very broad |
| N-H | 3300-3500 | Medium, 1 or 2 peaks |
| C-H (sp3) | 2850-2960 | Strong |
| C=O (carbonyl) | 1650-1750 | Strong, sharp |
| C=C | 1600-1680 | Medium |
| C-O | 1000-1260 | Strong |
Worked example. An IR spectrum shows a strong broad absorption at 3300 cm^-1 and a strong sharp peak at 1710 cm^-1. What functional groups are present?
3300 broad = O-H stretch (carboxylic acid pattern, given the breadth). 1710 = C=O stretch. Together: carboxylic acid (-COOH). The broad O-H from 2500-3300 overlapping with C-H is the signature "acid O-H."
Nuclear Magnetic Resonance (NMR)
Principle. Nuclei with spin (1H, 13C) in a magnetic field absorb radio-frequency radiation. The absorption frequency depends on the electronic environment — shielded nuclei absorb at different frequencies than deshielded ones.
1H NMR key concepts:
- Chemical shift (delta, ppm): Position on the spectrum. TMS = 0 ppm reference. Alkyl H near 0-2 ppm; adjacent to electronegative groups 2-5 ppm; aromatic H 6-8 ppm; aldehyde H near 9-10 ppm; carboxylic acid H near 10-12 ppm.
- Integration: Area under each peak is proportional to the number of equivalent H atoms.
- Splitting (multiplicity): n+1 rule — a signal split by n equivalent neighboring H atoms appears as n+1 peaks (singlet, doublet, triplet, quartet, etc.).
Worked example. *A compound