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.