Chemistry
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AP Chemistry — Cheatsheet

Formulas, exam-day tips, and key terms on one page.

Formulas & relationships

Avogadro's number
1 mole = 6.022 × 10²³ particles
Particles can be atoms, molecules, ions, or electrons — always state what you are counting.
Average atomic mass
avg mass = Σ (isotope mass × fractional abundance)
Fractional abundance is the percentage written as a decimal (e.g. 75% → 0.75).
Aufbau filling order
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s
Note the swap: 4s fills before 3d because it is slightly lower in energy.
Trend summary (top-right is the extreme)
→ across & ↑ up: radius ↓, ionization energy ↑, electronegativity ↑
Formal charge
formal charge = valence − nonbonding − ½ bonding
Count the atom's own valence electrons, subtract its lone-pair (nonbonding) electrons, then subtract half of its shared (bonding) electrons.
Bond type from electronegativity difference (ΔEN)
ΔEN ≳ 1.7 → ionic · 0.4–1.7 → polar covalent · < 0.4 → nonpolar covalent
These cutoffs are approximate guides, not hard walls — bonding is a continuum from pure sharing to full transfer.
Electron-domain geometry by domain count
2 → linear (180°) · 3 → trigonal planar (120°) · 4 → tetrahedral (109.5°)
This is the arrangement of ALL domains. The shape named by the atoms (molecular geometry) can differ when some domains are lone pairs.
Hybridization from electron domains
2 domains → sp (linear) · 3 → sp² (trigonal planar) · 4 → sp³ (tetrahedral)
Superscript = number of p orbitals mixed with the one s orbital; total hybrids = domains.
IMF strength ranking (per interaction)
ion–dipole > hydrogen bonding > dipole–dipole > London dispersion
A caution: in large molecules, many LDFs can add up to outweigh a single dipole–dipole attraction — compare total forces, not just the type.
Boiling condition
liquid boils when: vapor pressure = external pressure
Lower the external pressure and the boiling point drops; raise it (a pressure cooker) and the boiling point climbs.
Ideal gas law
PV = nRT
R = 0.08206 L·atm·mol⁻¹·K⁻¹. Always convert T to kelvin and match your pressure/volume units to R before plugging in.
Partial pressure from mole fraction
Pᵢ = Xᵢ × P_total, where Xᵢ = nᵢ ÷ n_total
Molarity
M = mol solute ÷ L solution
It is liters of *solution*, not liters of solvent — you dissolve the solute and then fill to the mark.
Building a net ionic equation
molecular → split strong electrolytes into ions → cancel spectators → net ionic
Only strong electrolytes split. Precipitates (s), gases (g), water, and weak acids stay written as whole formulas.
Grams ↔ moles
moles = mass ÷ molar mass
The universal on-ramp and off-ramp: grams → (÷ molar mass) → moles → (× mole ratio) → moles of target → (× molar mass) → grams.
Percent yield
percent yield = (actual yield ÷ theoretical yield) × 100%
A value above 100% signals an error (or impurity/wet product) — you cannot make more than theory allows.
Oxidation-number sum rule
Σ (oxidation numbers) = overall charge of the species
= 0 for a neutral compound; = the ion charge for a polyatomic ion. Solve for the unknown atom.
Molarity
M = mol ÷ L (so mol = M × V)
Volume must be in liters. 25.0 mL = 0.0250 L.
Titration relationships
mol titrant = M × V → mol unknown = mol titrant × (mole ratio) → M unknown = mol ÷ V
For a 1:1 reaction only, this simplifies to MₐVₐ = M_bV_b. When the ratio is not 1:1, you must apply it explicitly.
Average reaction rate
rate = −Δ[reactant] / Δt = +Δ[product] / Δt
For a A → b B, divide each term by its coefficient so all species give the same rate: −(1/a)Δ[A]/Δt = (1/b)Δ[B]/Δt.
Factors that speed a reaction
↑ concentration, ↑ temperature, ↑ surface area, add catalyst → ↑ rate
Concentration, temperature, and surface area raise the collision rate; temperature and a catalyst also change the fraction of collisions that succeed.
General rate law
rate = k[A]ᵐ[B]ⁿ
m is the order in A, n the order in B, and (m + n) the overall order. Orders are usually 0, 1, or 2 and come only from data.
Integrated rate laws & linear plots
Zero: [A] = [A]₀ − kt (plot [A] vs t) | First: ln[A] = ln[A]₀ − kt (plot ln[A] vs t) | Second: 1/[A] = 1/[A]₀ + kt (plot 1/[A] vs t)
Whichever plot is a straight line reveals the order; the slope gives k (−k for zero/first order, +k for second order).
First-order half-life
t½ = 0.693 / k
0.693 is ln 2. Because [A]₀ cancels out, first-order half-life is independent of starting concentration. (Zero order: t½ = [A]₀/2k; second order: t½ = 1/(k[A]₀) — these do depend on [A]₀.)
Spotting each species in a mechanism
Intermediate: appears first as a product, then a reactant. Catalyst: appears first as a reactant, then a product.
Both cancel out of the overall equation — the order in which they appear (made-then-used vs. used-then-remade) is what tells them apart.
Sign convention for enthalpy
ΔH = H(products) − H(reactants); exothermic ΔH < 0, endothermic ΔH > 0
Negative means energy leaves the system (released); positive means energy enters the system (absorbed).
Heat and temperature change
q = m·c·ΔT
q = heat (J), m = mass (g), c = specific heat capacity (J·g⁻¹·°C⁻¹), ΔT = T(final) − T(initial). For water, c = 4.18 J·g⁻¹·°C⁻¹.
Enthalpy from formation enthalpies
ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants)
Each ΔH°f is multiplied by its coefficient. The ΔH°f of any element in its standard state is exactly 0.
Gibbs free energy
ΔG = ΔH − T·ΔS
T is absolute temperature in kelvin. Watch units: ΔH is usually kJ while ΔS is usually J·K⁻¹ — convert ΔS to kJ·K⁻¹ (divide by 1000) before subtracting.
Equilibrium-constant expression
For aA + bB ⇌ cC + dD: Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)
Products on top, reactants on the bottom, each raised to its coefficient. Omit any pure solid (s) or pure liquid (l).
Kp for gaseous equilibria
Kp = Kc (RT)^Δn, where Δn = (moles gas product) − (moles gas reactant)
Kp uses partial pressures instead of concentrations. If the number of gas moles is unchanged (Δn = 0), then Kp = Kc.
Reaction quotient Q vs. K
Q < K → shift right (forward); Q = K → at equilibrium; Q > K → shift left (reverse)
Q has the exact same form as K but uses the current (not necessarily equilibrium) concentrations. Comparing Q to K predicts the direction of change.
ICE relationship
Equilibrium concentration = Initial + Change (Change = ± coefficient × x)
Reactants are consumed (−), products are formed (+). The coefficient in front of x must match the balanced equation.
Temperature and K (heat-as-reagent rule)
Exothermic (ΔH < 0): heat is a product → raising T shifts left, K decreases. Endothermic (ΔH > 0): heat is a reactant → raising T shifts right, K increases.
Only a temperature change alters the numerical value of K. Concentration, volume, pressure, and catalyst changes never do.
Solubility product
For AₘBₙ(s) ⇌ m Aⁿ⁺(aq) + n Bᵐ⁻(aq): Ksp = [Aⁿ⁺]ᵐ [Bᵐ⁻]ⁿ
Only the dissolved ions appear, each raised to its coefficient. A larger Ksp (for the same ion-count stoichiometry) means a more soluble salt.
Ksp ↔ molar solubility
MX type: Ksp = s². MX₂ type: Ksp = 4s³. M₂X type: Ksp = 4s³.
Always rebuild the relationship from the balanced dissolution equation; do not assume Ksp = s² for every salt.
Definition of pH
pH = −log[H⁺]
The minus sign makes pH positive for typical solutions. Each whole pH unit is a factor of 10 in [H⁺]: pH 3 is ten times more acidic than pH 4.
The pH–pOH relationship
pH + pOH = 14 (at 25 °C, since Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴)
pOH = −log[OH⁻]. Once you know any one of pH, pOH, [H⁺], or [OH⁻], you can reach the other three.
Acid ionization constant
Ka = [H⁺][A⁻] / [HA] pKa = −log(Ka)
Larger Ka ↔ smaller pKa ↔ stronger acid. For a conjugate pair, Ka × Kb = Kw = 1.0 × 10⁻¹⁴.
Henderson–Hasselbalch equation
pH = pKa + log([A⁻] / [HA])
When [A⁻] = [HA] the log term is log(1) = 0, so pH = pKa. More conjugate base raises pH; more acid lowers it.
Half-equivalence point
at half-equivalence: [HA] = [A⁻] ⇒ pH = pKa
This is Henderson–Hasselbalch with log([A⁻]/[HA]) = log(1) = 0. It is the single most useful point for extracting pKa from a curve.
Gibbs free energy
ΔG = ΔH − TΔS
ΔG < 0 favored, ΔG > 0 not favored, ΔG = 0 at equilibrium. T is in kelvin, so the TΔS term always grows with temperature.
Free energy and the equilibrium constant
ΔG° = −RT ln K
R = 8.314 J·mol⁻¹·K⁻¹, T in kelvin. K > 1 ⇒ ln K > 0 ⇒ ΔG° < 0 (products favored). K < 1 ⇒ ΔG° > 0 (reactants favored).
Standard cell potential
E°cell = E°cathode − E°anode
Both values are standard *reduction* potentials read straight from the table. Do NOT multiply a potential by the number of electrons or by a balancing coefficient — potential is an intensive property.
Free energy from cell potential
ΔG° = −nFE°
n = moles of electrons transferred, F = 96 485 C·mol⁻¹ (Faraday constant). E° > 0 ⇒ ΔG° < 0 ⇒ spontaneous; E° < 0 ⇒ ΔG° > 0 ⇒ nonspontaneous.
Faraday's laws of electrolysis
moles e⁻ = Q ÷ F = (I·t) ÷ 96 485
Then scale by the half-reaction: e.g. Ag⁺ + e⁻ → Ag needs 1 mol e⁻ per mol Ag, while Al³⁺ + 3e⁻ → Al needs 3 mol e⁻ per mol Al.
The Nernst equation (at 298 K)
E = E° − (0.0592 / n) log Q
General form: E = E° − (RT/nF) ln Q. At 25 °C the constants collapse to 0.0592/n with base-10 log. Q < 1 raises E above E°; Q > 1 lowers it.
The PES energy balance
binding energy = photon energy − kinetic energy of ejected electron
Because every subshell holds its electrons with a characteristic tightness, each subshell produces its own peak at a fixed binding energy.
Coulomb's law (attraction)
F ∝ q₁q₂ / r²
q₁ is the effective nuclear charge (Zₑff), q₂ the electron’s charge, and r the distance between them. Doubling the distance quarters the force — the inverse-square term dominates.
Formal charge
formal charge = valence electrons − nonbonding electrons − ½(bonding electrons)
Count the atom's group valence, subtract all of its lone-pair electrons, then subtract half of the electrons it shares in bonds. The formal charges of a species must sum to its overall charge.
Average bond order (resonance hybrid)
bond order = (total bonds shared among the equivalent positions) ÷ (number of equivalent positions)
Equivalently, average the bond (single = 1, double = 2, triple = 3) for one linkage across all resonance structures. For NO₃⁻: 4 bonds over 3 equal N–O positions → 4/3 ≈ 1.33.
Sigma and pi bonds per bond type
single = 1σ · double = 1σ + 1π · triple = 1σ + 2π
To total a molecule: number of σ bonds = number of bonded atom pairs (every bonded connection has exactly one σ, including every X–H bond); number of π bonds = (double bonds) + 2 × (triple bonds).
Graham’s law of effusion
rate₁ / rate₂ = √(M₂ / M₁)
Lighter gas is faster. The heavier gas goes in the numerator under the root, so the lighter gas’s rate comes out larger.
Dalton’s law (partial pressures)
P_total = P₁ + P₂ + … , Pᵢ = Xᵢ × P_total , P_gas = P_total − P_water
The last form is for a gas collected over water: subtract the temperature-dependent vapor pressure of water to get the dry-gas pressure.
Beer–Lambert law
A = εbc
A is absorbance (unitless); ε is molar absorptivity (L·mol⁻¹·cm⁻¹); b is path length (cm); c is molar concentration (mol·L⁻¹). Rearranged, c = A ÷ (εb).
Dilution relationship
M₁V₁ = M₂V₂
Moles of solute are conserved on dilution. Subscript 1 is the concentrated stock, 2 is the diluted solution; any consistent volume unit works since it cancels.
Half-reaction balancing checklist
atoms (≠ O,H) → O with H₂O → H with H⁺ → charge with e⁻ → scale & add → (basic: + OH⁻ both sides, combine to H₂O, cancel)
Electrons always land on the more-positive side of a half-reaction: the reactant side for reduction, the product side for oxidation.
Yield and back-titration relationships
percent yield = (actual ÷ theoretical) × 100% | mol reacted with analyte = mol added − mol left over
Route every problem through moles: grams ÷ molar mass, or molarity × volume(L). Convert back to grams or molarity only at the very end.
Integrated rate laws as straight lines
Zero: [A] = −kt + [A]₀ (plot [A] vs t) | First: ln[A] = −kt + ln[A]₀ (plot ln[A] vs t) | Second: 1/[A] = kt + 1/[A]₀ (plot 1/[A] vs t)
In each case y = mx + b: the y-intercept is the starting value and the slope is ±k. Zero and first order give slope = −k; second order gives slope = +k.
Half-life by order
Zero: t½ = [A]₀ / (2k) First: t½ = 0.693 / k Second: t½ = 1 / (k[A]₀)
Only first-order t½ is independent of [A]₀ (0.693 = ln 2). If successive half-lives are equal → first order; if they get shorter → zero order; if they get longer → second order.
Arrhenius equation
k = A·e^(−Eₐ/RT)
R = 8.314 J·mol⁻¹·K⁻¹ (use Eₐ in J/mol). Two-temperature form: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂). A plot of ln k vs 1/T is linear with slope = −Eₐ/R.
Energy relationships on the diagram
Eₐ(forward) = E(transition state) − E(reactants) Eₐ(reverse) = E(transition state) − E(products) ΔH = Eₐ(forward) − Eₐ(reverse)
ΔH also equals E(products) − E(reactants). If Eₐ(forward) < Eₐ(reverse) the reaction is exothermic (ΔH < 0); if Eₐ(forward) > Eₐ(reverse) it is endothermic (ΔH > 0).
Heat and temperature change
q = m·c·ΔT with q(hot) + q(cold) = 0 (insulated)
q = heat (J), m = mass (g), c = specific heat (J·g⁻¹·°C⁻¹), ΔT = T(final) − T(initial). The hot object has ΔT < 0 (q < 0); the cold object has ΔT > 0 (q > 0).
Two cross-checks for ΔH°rxn
ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants); ΔH ≈ Σ(bonds broken) − Σ(bonds formed)
Each ΔH°f is weighted by its coefficient; ΔH°f of an element in its standard state is 0. The bond-enthalpy estimate is gas-phase only and approximate (tabulated bonds are averages).
Entropy change of a reaction
ΔS°rxn = Σ S°(products) − Σ S°(reactants)
S° values are in J·mol⁻¹·K⁻¹ and are always positive (even for elements). More moles of gas on the product side → ΔS > 0.
Gibbs free energy
ΔG = ΔH − T·ΔS
T is absolute temperature (kelvin). Convert ΔS from J·K⁻¹ to kJ·K⁻¹ before subtracting. ΔG < 0 spontaneous, ΔG > 0 nonspontaneous, ΔG = 0 at equilibrium.
Crossover (equilibrium) temperature
ΔG = 0 ⟹ T = ΔH / ΔS
Use ΔH in joules to match ΔS in J·K⁻¹ (or ΔH in kJ with ΔS in kJ·K⁻¹). For (+, +) the reaction turns spontaneous above T; for (−, −) it turns nonspontaneous above T.
Quadratic formula (for the exact ICE solve)
ax² + bx + c = 0 → x = (−b ± √(b² − 4ac)) / 2a
Use when the small-x approximation fails the 5% test. Discard the root that gives a negative concentration or a value exceeding the initial amount.
Kp from Kc
Kp = Kc (RT)^Δn, Δn = (moles of gaseous product) − (moles of gaseous reactant)
R = 0.0821 L·atm·mol⁻¹·K⁻¹ and T is in kelvin. If Δn = 0 the factor is 1, so Kp = Kc. Only gas-phase species count toward Δn.
Solubility product and molar solubility
AₘBₙ(s) ⇌ m Aⁿ⁺ + n Bᵐ⁻: Ksp = [Aⁿ⁺]ᵐ[Bᵐ⁻]ⁿ. MX: Ksp = s². MX₂ or M₂X: Ksp = 4s³. MX₃: Ksp = 27s⁴.
The numerical prefix (4, 27, …) is (coefficient)^(coefficient) summed over the ions. To get s from Ksp for a 1:2 salt, take the cube root of Ksp/4.
Precipitation criterion
Q > Ksp → precipitate forms; Q = Ksp → just saturated; Q < Ksp → no precipitate
Q uses concentrations in the combined solution (after mixing dilutes them). Same Q-vs-K logic as reaction equilibria, applied to dissolving.
Conjugate relationship
Kb = Kw / Ka (and Ka · Kb = Kw = 1.0 × 10⁻¹⁴ at 25 °C)
To find how basic a conjugate base A⁻ is, divide Kw by the Ka of its parent acid HA. A weaker acid (small Ka) has a stronger conjugate base (large Kb).
Henderson–Hasselbalch (solved for the ratio)
pH = pKa + log([A⁻] / [HA]) ⇒ [A⁻] / [HA] = 10^(pH − pKa)
The exponent pH − pKa drives the design: it is 0 (ratio 1) at pH = pKa, +1 (ratio 10) one unit above, and −1 (ratio 0.1) one unit below.
Free energy from cell potential
ΔG° = −nFE°
n = moles of electrons transferred in the balanced reaction, F = 96 485 C·mol⁻¹. E° > 0 ⇒ ΔG° < 0 ⇒ spontaneous. E° is intensive — never scale it by n or by balancing coefficients; the n out front is what accounts for reaction size.
Free energy from the equilibrium constant
ΔG° = −RT ln K
R = 8.314 J·mol⁻¹·K⁻¹, T in kelvin. K > 1 ⇒ ΔG° < 0 (products favored); K < 1 ⇒ ΔG° > 0 (reactants favored).
Voltage and equilibrium (298 K shortcut)
log K = nE° / 0.0592
Derived by equating ΔG° = −nFE° with ΔG° = −RT ln K at 25 °C. A modest E° of a few tenths of a volt produces an enormous K — voltage and equilibrium are exponentially related.
Faraday's laws of electrolysis
mol e⁻ = Q ÷ F = (I·t) ÷ 96 485
Then divide by the electrons-per-ion from the half-reaction: Cu²⁺ + 2e⁻ → Cu needs 2 mol e⁻ per mol Cu; Ag⁺ + e⁻ → Ag needs 1.
The Nernst equation (298 K)
E = E° − (0.0592 / n) log Q
General form E = E° − (RT/nF) ln Q; at 25 °C the constants collapse to 0.0592/n with base-10 log. Q < 1 raises E above E°; Q > 1 lowers it; Q = 1 recovers E = E°.
Concentration cell (identical electrodes, 298 K)
E = −(0.0592 / n) log Q, with E° = 0
Q = [ion]dilute ÷ [ion]concentrated < 1, so log Q < 0 and E > 0. The concentrated half-cell is always the cathode; electrons flow to shrink the gradient.

On the exam

How to get a 5

Key terms

Ideal gas law equation and R valuePV = nRT; R = 0.0821 L·atm·mol⁻¹·K⁻¹ (or 8.314 J·mol⁻¹·K⁻¹).
Definition of a bufferA solution of a weak acid and its conjugate base (or weak base + conjugate acid) that resists pH change when small amounts of acid or base are added.
Henderson–Hasselbalch equationpH = pKa + log([A⁻]/[HA]). pH = pKa when [A⁻] = [HA].
Relationship: Kc and KpKp = Kc(RT)^Δn, where Δn = moles gaseous products − moles gaseous reactants.
Sign of ΔG for spontaneityΔG < 0 spontaneous; ΔG > 0 nonspontaneous; ΔG = 0 at equilibrium. ΔG = ΔH − TΔS.
ΔG° relationship to KΔG° = −RT ln K. K > 1 → ΔG° < 0; K < 1 → ΔG° > 0.
What does a catalyst do?Lowers activation energy via an alternate pathway; speeds forward and reverse rates equally; does NOT change ΔH, ΔG, or K.
Photoelectron spectroscopy (PES): peak position and heightPeak position (binding energy) indicates the subshell/energy level; peak height (area) indicates the number of electrons in that subshell.
Periodic trend: atomic radiusDecreases across a period (increasing Zeff), increases down a group. Cations smaller than parent atom; anions larger.
Strong acids to memorizeHCl, HBr, HI, HNO₃, H₂SO₄, HClO₄ (and HClO₃) — they ionize completely in water.
Oxidation vs reduction (LEO/GER)Oxidation = Loss of Electrons (oxidation number increases, at the anode); Reduction = Gain of Electrons (number decreases).
Hess's Law / ΔH°rxn from formationΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants). ΔH°f of an element in its standard state = 0.
Average atomic mass from isotopic abundancesAverage mass = Σ(fractional abundance × isotopic mass). It is a weighted mean, so the answer always lies closer to the more abundant isotope, never at the simple midpoint.
How to read a photoelectron spectrum (PES)Peak position gives binding energy (further left/higher = closer to the nucleus, so 1s is at highest energy). Peak height is proportional to the number of electrons in that subshell. Add the heights to get the total electron count and identify the element.
Empirical vs molecular formulaPercent composition gives only the empirical (simplest whole-number) formula. To get the molecular formula you also need the molar mass: divide it by the empirical formula mass to find the multiplier.
Limiting-reactant procedureConvert each reactant mass to moles, divide by its stoichiometric coefficient, and the smallest quotient is the limiting reactant. Compute all yields from the limiting reactant only — the excess reactant is a distractor.
Dalton and gas collected over waterP(total) = ΣP(partial), and P(gas) = X(gas) × P(total). For a gas collected over water, subtract the water vapor pressure at that temperature: P(dry gas) = P(total) − P(H₂O). Forgetting this overestimates the moles of product.
Ideal gas law and its R valuesPV = nRT. R = 0.08206 L·atm·mol⁻¹·K⁻¹ (with P in atm, V in L) or 8.314 J·mol⁻¹·K⁻¹ (for energy calculations). Temperature must always be in kelvin.
First-order integrated rate law and half-lifeln[A]t = −kt + ln[A]₀ (a straight line for ln[A] vs t) and t½ = 0.693/k, independent of initial concentration. Second order: 1/[A] is linear in t and t½ = 1/(k[A]₀), which lengthens as the reaction proceeds.
Deriving a rate law from a mechanismThe rate law comes from the slow (rate-determining) step, using its reactants and their coefficients as orders. Species that appear only in fast steps after the bottleneck do not appear in the rate law, even though they are in the overall equation.
What a catalyst does and does not changeLowers Ea by providing an alternate pathway, so it speeds both forward and reverse reactions equally. It does NOT change ΔH, ΔG, K, or the equilibrium position — it only shortens the time needed to reach equilibrium.
Q versus KQ < K: too few products, net forward reaction. Q > K: too many products, net reverse reaction. Q = K: at equilibrium. Only temperature changes the value of K; concentration and volume changes move Q instead.
Ksp and molar solubilityFor AB: Ksp = s². For AB₂ or A₂B: Ksp = 4s³. Solve for s by taking the appropriate root. Adding a common ion decreases solubility but leaves Ksp unchanged.
Le Châtelier and temperature for an exothermic reactionTreat heat as a product when ΔH < 0. Raising T shifts toward reactants and DECREASES K; lowering T increases K and yield. This is the only stress that actually changes the value of K.