analytical-methods

# 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 =