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Class 11 Chemistry: Essential Formulas & Concepts (PDF) — All Important Formulas Free

Free class 11 chemistry: essential formulas & concepts — all 64 key formulas on one page, downloadable as PDF for fast revision before CBSE board exams, JEE & NEET. No signup.

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TL;DR: Free class 11 chemistry: essential formulas & concepts — all 64 key formulas on one page, downloadable as PDF for fast revision before CBSE board exam…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Master the foundational concepts of chemistry: atomic structure, mole theory, gas laws, thermodynamics, equilibrium, and redox reactions. All formulas in one reference sheet for quick problem-solving.

Mole Concept & Stoichiometry

FormulaExpression
Number of moles from massn = given mass (g) / molar mass (g/mol) — n is always in moles; divide grams by atomic/molecular weight
Number of moles from particlesn = N / Nₐ — N = number of particles, Nₐ = Avogadro's number = 6.022 × 10²³
Molar mass from densityM = dRT / P — d = density (g/L), R = 0.0821 L·atm/(mol·K), T in Kelvin, P in atm
Percentage composition% element = (atomic mass × no. of atoms / molar mass) × 100 — Sum of all percentages = 100%
Empirical formula massn = molar mass / empirical formula mass — Molecular formula = (empirical formula)ₙ
Stoichiometric ratiomol A / mol B = coefficient A / coefficient B — From balanced chemical equation
Limiting reagentCompare moles of reactant / stoichiometric coefficient — Smallest ratio is limiting; controls theoretical yield
Percent yield% yield = (actual yield / theoretical yield) × 100 — Actual from experiment; theoretical from stoichiometry

Concentration Terms

FormulaExpression
MolarityM = n / V(L) = moles of solute / litres of solution — Temperature-dependent; liters of final solution, not solvent
Molalitym = n / W(kg) = moles of solute / kg of solvent — Temperature-independent; uses mass of solvent only
NormalityN = n × equivalents / V(L) — Equivalents = n × valency for acids/bases/salts
Mole fractionχₐ = nₐ / (nₐ + nᵦ + ...) — Sum of all mole fractions = 1; dimensionless
Mass percent (w/w)w/w % = (mass of solute / mass of solution) × 100 — Solution = solute + solvent
Volume percent (v/v)v/v % = (volume of solute / volume of solution) × 100 — For liquids; may not be additive
Relation: Molarity & Molalitym = M / (1 - M × M / 1000) — M = molar mass of solute; holds for dilute solutions
Dilution formulaM₁V₁ = M₂V₂ — Moles solute conserved; works for any molarity scale

Gas Laws & Ideal Gas Equation

FormulaExpression
Ideal gas equationPV = nRT — P (Pa or atm), V (m³ or L), n (mol), R = 8.314 J/(mol·K) or 0.0821 L·atm/(mol·K), T (K)
Boyle's LawP₁V₁ = P₂V₂ — At constant T and n; inverse relationship
Charles' LawV₁/T₁ = V₂/T₂ — At constant P and n; direct relationship
Gay-Lussac's LawP₁/T₁ = P₂/T₂ — At constant V and n; direct relationship
Combined gas lawP₁V₁/T₁ = P₂V₂/T₂ — No mole change needed; compare two states
Dalton's law of partial pressuresPₜₒₜₐₗ = P₁ + P₂ + P₃ + ... — Each gas contributes its own pressure independently
Henry's LawPgas = Kₕ × χsolute (in solution) — Kₕ = Henry's law constant; Pgas = partial pressure
Graham's law of diffusionr₁/r₂ = √(M₂/M₁) — r = rate of diffusion; M = molar mass; lighter gas diffuses faster
Density of gas at STPd = PM / RT — d (g/L); relates density to molar mass

Atomic Structure

FormulaExpression
Wavelength-frequency relationc = λν — c = 3 × 10⁸ m/s, λ = wavelength (m), ν = frequency (Hz)
Photon energyE = hν = hc/λ — h = 6.626 × 10⁻³⁴ J·s; higher frequency = higher energy
Rydberg equation (hydrogen)1/λ = R∞(1/n₁² − 1/n₂²) — R∞ = 1.097 × 10⁷ m⁻¹; n₁ < n₂ for emission; n₁ = 1 is Lyman, 2 is Balmer
Energy levels (Bohr model)Eₙ = −13.6/n² eV — n = 1, 2, 3...; negative energy means bound state
de Broglie wavelengthλ = h / p = h / mv — All particles have wavelike behavior; h = Planck constant
Uncertainty principle (Heisenberg)Δx × Δp ≥ h / 4π — Cannot know position and momentum simultaneously with certainty
Ionization energy (hydrogen-like)IE = 13.6 × Z² / n² eV — Z = atomic number, n = principal quantum number
Electron in orbital probabilityψ² = probability density at a point in space — ψ = wave function; ψ² gives electron density

Thermodynamics

FormulaExpression
Heat absorbed/releasedq = mcΔT — m = mass (g), c = specific heat capacity (J/g·°C), ΔT = temperature change (K or °C)
Enthalpy changeΔH = q_p (at constant pressure) — ΔH > 0 endothermic, ΔH < 0 exothermic
Hess's LawΔH_reaction = Σ ΔH_products − Σ ΔH_reactants — Enthalpy is a state function; independent of reaction path
Standard enthalpy of formationΔH°f = enthalpy change for 1 mol pure compound from elements — ΔH°f(element) = 0; sum of ΔH°f gives ΔH°rxn
Entropy changeΔS = q_rev / T — S > 0 disorder increases; state function
Gibbs free energyΔG = ΔH − TΔS — ΔG < 0 spontaneous, ΔG > 0 non-spontaneous at that T
Gibbs energy and equilibriumΔG° = −RT ln Kₑq — R = 8.314 J/(mol·K); Kₑq = equilibrium constant
Calorimetry (coffee cup)q_solution + q_calorimeter + q_reaction = 0 — Heat lost = heat gained (closed system)

Chemical Equilibrium

FormulaExpression
Equilibrium constant (Kc)Kc = [C]^c[D]^d / [A]^a[B]^b — For aA + bB ⇌ cC + dD; square brackets = molar concentration at equilibrium
Equilibrium constant (Kp)Kp = P_C^c × P_D^d / P_A^a × P_B^b — P = partial pressure (atm or Pa)
Relation between Kp and KcKp = Kc(RT)^Δn — Δn = (c + d) − (a + b) = change in moles of gas
Ion product of waterKw = [H⁺][OH⁻] = 10⁻¹⁴ at 25°C — Always 10⁻¹⁴ at 25°C regardless of acid/base
pH definitionpH = −log[H⁺]; pOH = −log[OH⁻] — pH + pOH = 14 at 25°C; pH < 7 acidic, pH > 7 basic
Acid dissociation constantKa = [H⁺][A⁻] / [HA] — Larger Ka = stronger acid; pKa = −log Ka
Base dissociation constantKb = [BH⁺][OH⁻] / [B] — Ka × Kb = Kw for conjugate acid-base pair
Henderson-Hasselbalch equationpH = pKa + log([A⁻]/[HA]) — For buffer solutions; [A⁻] = conjugate base, [HA] = weak acid
Buffer capacityBuffer resists pH change when small amounts of acid/base are added — Best when pH ≈ pKa; ratio [A⁻]/[HA] ≈ 1
Common ion effectAdding common ion shifts equilibrium left, decreases solubility — E.g. adding NaCl to NaCl(aq) ⇌ Na⁺ + Cl⁻

Redox Reactions & Electrochemistry

FormulaExpression
Oxidation number rulesElement = 0; monatomic ion = charge; O usually −2; H usually +1; F always −1 — Sum of oxidation numbers = 0 (neutral) or charge (ion)
Oxidation and reductionOxidation = loss of e⁻ (O.N. increases); Reduction = gain of e⁻ (O.N. decreases) — OIL RIG; redox pair involved in every redox reaction
Balancing redox (half-reaction method)Balance atoms except O & H → balance O with H₂O → balance H with H⁺ or OH⁻ → balance charge with e⁻ — Multiply half-reactions to equalize electrons lost/gained
Equivalent massEq.M = molar mass / valency change per atom — For redox, valency change = no. of e⁻ transferred per formula unit
Normality for redoxN = n / (Eq.M × V) × 1000; n_e⁻ = N × V — Electrons gained = electrons lost in balanced equation
Electrochemical cell notationAnode (−) | salt bridge | Cathode (+); read −→ + for conventional current — Oxidation at anode (left), reduction at cathode (right)
Standard cell potentialE°cell = E°cathode − E°anode — E° > 0 spontaneous, E° < 0 non-spontaneous; look up in tables
Gibbs energy and cell potentialΔG° = −nFE°cell — n = moles of e⁻, F = Faraday constant = 96,500 C/mol

Quick Key Concepts Review

FormulaExpression
Avogadro's numberNₐ = 6.022 × 10²³ particles/mol — Defines the mole; used to count atoms, molecules, ions
Molar volume at STPVm = 22.4 L/mol (at 0°C, 1 atm) — Useful for gas stoichiometry; nearly 24 L/mol at ~25°C, 1 atm
Gas constant RR = 8.314 J/(mol·K) = 0.0821 L·atm/(mol·K) = 2 cal/(mol·K) — Use first form for SI; second for gas law calculations
Faraday's constantF = 96,500 C/mol e⁻ — Charge of 1 mole of electrons; coulomb = amp × second
Atomic mass unit1 u = 1.66 × 10⁻²⁷ kg — Mass of one nucleon (proton or neutron); ¹²C standard = 12.000 u

FAQs

How do I find the empirical formula from percentage composition?

Assume 100 g sample. Convert % to grams, divide by atomic masses to get moles, divide by smallest to get ratio, multiply by a whole number if needed. Example: C 80%, H 20% → 100 g sample → 80/12 = 6.67 mol C, 20/1 = 20 mol H → divide by 6.67 → 1:3 ratio → empirical formula CH₃.

When should I use Kc vs Kp?

Use Kc when concentrations (mol/L) are given; Kp when partial pressures (atm/Pa) are given. For gases, they relate by Kp = Kc(RT)^Δn where Δn is the change in moles of gas. Both describe the same equilibrium.

What is the difference between an exothermic and endothermic reaction?

Exothermic releases heat (ΔH < 0, q > 0); reactants have more energy than products. Endothermic absorbs heat (ΔH > 0, q < 0); products have more energy than reactants. Combustion is always exothermic; melting ice is endothermic.

More Class 11 Chemistry Formula Sheets

  • Class 11 Chemistry: Equilibrium Formulas
  • Class 11 Chemistry: Some Basic Concepts of Chemistry Formulas
  • Class 11 Chemistry: Structure of Atom Formulas
  • Class 11 Chemistry: Thermodynamics (Chemistry) Formulas

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