The mole is the SI unit of amount of substance. 1 mole = $6.022\times10^{23}$ (Avogadro constant $N_A$) particles (atoms, molecules, ions, electrons — any entity). The molar mass in grams per mole equals the relative atomic/molecular mass in amu. Key relationships: $n = m/M$ (moles = mass/molar mass). $N = n \times N_A$ (number of particles = moles × Avogadro number). $V = n \times 22.4$ L at STP for ideal gases (0°C, 1 atm). $V = n \times 24.0$ L at NTP (25°C, 1 atm). Importance: the mole concept links the macroscopic world (grams, litres) to the microscopic world (atoms, molecules). Every stoichiometric calculation uses the mole as the currency of conversion.

$N_A = 6.022 \times 10^{23}$ mol$^{-1}$. Historical determination: from electrolysis (Faraday constant $F = N_A e$), X-ray crystal diffraction, Brownian motion (Perrin), sedimentation equilibrium. Modern determination: silicon sphere with precisely measured lattice parameter. The large magnitude of $N_A$ reflects the tiny size of atoms. 1 mole of pennies stacked would reach Sun and back $3\times10^{10}$ times. 1 mole of water molecules fills ~18 mL (molar volume = 18 mL/mol for liquid water since M=18 g/mol, density=1 g/mL). $N_A$ connects: atomic mass units (1 amu = 1.66×10⁻²⁴ g = 1/N_A grams) to grams. Coulombs to electron charge (F = N_A × e = 96485 C/mol). Boltzmann constant (k_B = R/N_A = 1.38×10⁻²³ J/K).

Empirical formula: simplest whole number ratio of atoms. Molecular formula: actual number of atoms. $n = M_{molecular}/M_{empirical}$. Steps to determine empirical formula: (1) Find % by mass of each element. (2) Divide by atomic mass to get mole ratio. (3) Divide by smallest to get simplest ratio. (4) If non-integer, multiply all by suitable factor. Example: 40% C, 6.67% H, 53.33% O. Mole ratio: C=40/12=3.33, H=6.67/1=6.67, O=53.33/16=3.33. Ratio C:H:O = 1:2:1. Empirical formula = CH2O (formaldehyde or glucose etc). If M=180, n=180/30=6, molecular = C6H12O6 (glucose). Combustion analysis: burning organic compound in excess O2, measuring CO2 and H2O to determine C and H content. If S, N, halogens present: special analyses needed.

Chemical equations: mole ratios of reactants and products. Stoichiometric calculations: (1) Balance equation. (2) Convert given masses to moles. (3) Use mole ratios from equation. (4) Convert moles back to mass/volume. Limiting reagent: the reactant that is completely consumed first; determines maximum product. Excess reagent: remains after reaction. Theoretical yield: maximum calculated from limiting reagent. Actual yield: experimentally obtained. Percentage yield = (actual/theoretical)×100. Example: $N_2 + 3H_2 \to 2NH_3$. If 28g N2 and 6g H2: N2 = 1 mol, H2 = 3 mol. Need 3 mol H2 per mol N2 → exactly stoichiometric! → 2 mol NH3 = 34g. If 28g N2 and 3g H2: H2 = 1.5 mol. H2 is limiting (need 3 mol for 1 mol N2). Produces 1 mol NH3 = 17g.

Molarity (M) = moles of solute / litres of solution. Most common lab unit. Changes with temperature (volume changes). Molality (m) = moles of solute / kg of solvent. Temperature-independent. Used in colligative properties. Normality (N) = equivalents / litre. N = M × n-factor. Mole fraction ($x_i$) = moles of i / total moles. $\sum x_i = 1$. Parts per million (ppm) = mg/kg or mg/L (for dilute aqueous solutions). Used for trace contaminants. Mass percent = (mass solute/mass solution)×100. Volume percent = (volume solute/volume solution)×100. Converting between: M = (ρ × % × 10)/M_r where ρ = density in g/mL, % = mass percent, Mr = molar mass. Dilution: $M_1V_1 = M_2V_2$ (moles preserved on dilution).

Equivalent weight = molar mass / n-factor. n-factor depends on reaction type: Acid-base: n = number of H⁺ or OH⁻ transferred. H2SO4: n=2, HCl: n=1, H3PO4: n=1,2,3 depending on reaction. Redox: n = change in oxidation number per formula unit. KMnO4 in acid: Mn goes from +7 to +2, n=5. In neutral: +7 to +4, n=3. In basic: +7 to +6, n=1. K2Cr2O7: Cr goes from +6 to +3, 2 Cr atoms, n=6. Salts: n = valency × cation count. The milliequivalent (meq) concept is used in medicine for electrolyte concentrations (Na⁺, K⁺, Ca²⁺, Cl⁻ etc.) in blood plasma.

% by mass of element X = (mass of X in 1 mol compound/molar mass of compound)×100. Example: Na2CO3 (M=106): %Na = 46/106×100 = 43.4%, %C = 12/106×100 = 11.3%, %O = 48/106×100 = 45.3%. Uses: quality control (verify compound identity), environmental analysis (% pollutant), nutritional labelling (% protein, fat, carbohydrate). Combustion analysis for organic compounds: weigh sample → burn in O2 → collect CO2 (from C) and H2O (from H) → mass difference gives O content (if only C,H,O present). Modern elemental analysers automate this. CHNS analysers determine C, H, N, S simultaneously.

At STP (0°C, 1 atm): 1 mol any ideal gas = 22.4 L. At NTP (25°C, 1 atm): 22.7 L. Ideal gas law: $PV = nRT$ where $R = 0.0821$ L·atm/(mol·K) = 8.314 J/(mol·K). Number of moles from gas: $n = PV/RT$. Gay-Lussac law of combining volumes: gases combine in simple whole number volume ratios (at same T, P). $H_2 + Cl_2 \to 2HCl$: 1L + 1L → 2L. Dalton law of partial pressures: $P_{total} = \sum P_i$. Graham law of diffusion/effusion: rate ∝ 1/√M. Lighter gases diffuse faster. Separation of isotopes: $^{235}UF_6$ from $^{238}UF_6$ by gaseous diffusion (used in uranium enrichment for nuclear fuel/weapons).