“Everything through moles”
How to start any numerical
Grams, litres and particle counts cannot be compared directly. Convert each to moles first, use the balanced equation, then convert back to whatever the question wants.
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Every chapter on one page, to read straight through or print. 10 chapters, 171 formulas with every symbol defined, 31 memory hooks, and the mistakes that cost the most marks.
The mole is the chapter everything else in chemistry leans on. Physical chemistry numericals almost always begin by converting what you were given into moles, so the few conversions here are worth more marks than their size suggests.
| What | Formula | Symbols and notes |
|---|---|---|
| Avogadro's number | NA = 6.022 × 10²³ mol−1 | particles in one mole |
| Moles from mass | n = given mass / molar mass | the first line of most numericals |
| Moles from particles | n = N / NA | N is the number of atoms or molecules |
| Moles of a gas at STP | n = volume / 22.4 L | at 273 K and 1 bar, for an ideal gas |
| Number of particles | N = n × NA | atoms, molecules or ions, so say which |
| What | Formula | Symbols and notes |
|---|---|---|
| Molarity | M = moles of solute / volume of solution in litres | changes with temperature, because volume does |
| Molality | m = moles of solute / mass of solvent in kg | independent of temperature |
| Mole fraction | x = ncomponent / ntotal | the fractions of all components add to 1 |
| Normality | N = gram equivalents / volume in litres | N = M × n-factor |
| Dilution | M₁V₁ = M₂V₂ | moles of solute do not change on adding water |
| Parts per million | ppm = (mass of solute / mass of solution) × 10⁶ | used for very dilute solutions |
| What | Formula | Symbols and notes |
|---|---|---|
| Empirical formula | the simplest whole number ratio of atoms | found from percentage composition |
| Molecular formula | molecular formula = n × empirical formula | n = molar mass / empirical formula mass |
| Percentage composition | % of element = (mass of element / molar mass) × 100 | per mole of compound |
| Percentage yield | (actual yield / theoretical yield) × 100 | theoretical yield comes from the limiting reagent |
| Limiting reagent | the reactant giving the fewest moles of product | divide moles by the coefficient and take the smallest |
How to remember this chapter
“Everything through moles”
How to start any numerical
Grams, litres and particle counts cannot be compared directly. Convert each to moles first, use the balanced equation, then convert back to whatever the question wants.
“Molarity has volume, molality has mass”
Telling the two apart
Molarity is per litre of solution, molality per kilogram of solvent. Only molality is safe when the temperature changes.
“Smallest share wins”
Finding the limiting reagent
Divide each reactant's moles by its coefficient in the balanced equation. The smallest answer is the limiting reagent, and it sets the yield.
The mistakes that cost the most
Atomic structure is where chemistry borrows from physics. The Bohr results and the quantum numbers carry the marks, and the rules for filling orbitals are asked in every board paper in one form or another.
| What | Formula | Symbols and notes |
|---|---|---|
| Wave relation | c = νλ | c = 3 × 10⁸ m s−1 |
| Photon energy | E = hν = hc/λ | h = 6.626 × 10−34 J s |
| Photoelectric effect | hν = hν₀ + ½mv² | ν₀ is the threshold frequency |
| de Broglie wavelength | λ = h/mv | every moving particle has one |
| Heisenberg uncertainty | Δx · Δp ≥ h/4π | position and momentum cannot both be exact |
| What | Formula | Symbols and notes |
|---|---|---|
| Energy of the nth level | En = −13.6Z²/n² eV | for hydrogen-like species only |
| Radius of the nth orbit | rn = 0.529n²/Z Å | grows as n² |
| Angular momentum | mvr = nh/2π | this quantisation is Bohr's postulate |
| Rydberg equation | 1/λ = R(1/n₁² − 1/n₂²) | R = 1.097 × 10⁷ m−1 |
| Number of spectral lines | n(n − 1)/2 | from the nth level down to the ground state |
| What | Formula | Symbols and notes |
|---|---|---|
| Principal, n | shell number, 1, 2, 3, … | decides size and most of the energy |
| Azimuthal, l | 0 to n − 1, shapes s, p, d, f | decides the shape of the orbital |
| Magnetic, m | −l to +l | gives 2l + 1 orbitals in a subshell |
| Spin, s | +½ or −½ | two electrons per orbital, opposite spins |
| Electrons in a shell | 2n² | 2, 8, 18, 32 for the first four shells |
| Electrons in a subshell | 2(2l + 1) | s 2, p 6, d 10, f 14 |
| Aufbau order by (n + l) | lower n + l fills first; on a tie, lower n | this is why 4s fills before 3d |
| Hund's rule | fill singly before pairing | all with parallel spins |
| Pauli exclusion | no two electrons share all four quantum numbers | hence two per orbital |
How to remember this chapter
“n plus l, then n”
The order in which orbitals fill
Lower n + l fills first. If two are equal, the one with the smaller n goes first. That single rule puts 4s before 3d without memorising the diagonal chart.
“Bus seats”
Hund's rule
People take empty seats on a bus before sitting next to someone. Electrons fill each orbital of a subshell singly before any pairs up.
“Half and full are happiest”
Chromium and copper
Cr is 3d⁵4s¹ and Cu is 3d¹⁰4s¹, not the expected 3d⁴4s² and 3d⁹4s², because half-filled and fully filled subshells are extra stable.
The mistakes that cost the most
The gas laws are the most formula-dense part of Class 11 chemistry, and the most predictable. Almost every question is one of the named laws or the ideal gas equation, and the deviations at the end supply the theory mark.
| What | Formula | Symbols and notes |
|---|---|---|
| Boyle's law | p₁V₁ = p₂V₂ | at constant temperature and amount |
| Charles's law | V₁/T₁ = V₂/T₂ | at constant pressure; T in kelvin |
| Gay-Lussac's law | p₁/T₁ = p₂/T₂ | at constant volume |
| Combined gas law | p₁V₁/T₁ = p₂V₂/T₂ | the three above in one line |
| Avogadro's law | V ∝ n | equal volumes hold equal numbers of molecules |
| What | Formula | Symbols and notes |
|---|---|---|
| Ideal gas equation | pV = nRT | R = 8.314 J mol−1 K−1 = 0.0821 L atm mol−1 K−1 |
| In terms of density | pM = ρRT | a quick way to find molar mass from density |
| Dalton's law of partial pressures | ptotal = p₁ + p₂ + … | for non-reacting gases |
| Partial pressure from mole fraction | pi = xi × ptotal | the most used form of Dalton's law |
| Graham's law of diffusion | r₁/r₂ = √(M₂/M₁) | lighter gases diffuse faster |
| What | Formula | Symbols and notes |
|---|---|---|
| van der Waals equation | (p + an²/V²)(V − nb) = nRT | a corrects for attraction, b for molecular volume |
| Compressibility factor | Z = pV/nRT | Z = 1 for an ideal gas |
| Root mean square speed | urms = √(3RT/M) | M in kg per mole |
| Average speed | uav = √(8RT/πM) | slightly less than urms |
| Most probable speed | ump = √(2RT/M) | the smallest of the three |
| Critical temperature | above Tc a gas cannot be liquefied | however great the pressure |
How to remember this chapter
“Kelvin or nothing”
Every gas law
Every gas calculation needs absolute temperature. Using °C in Charles's law is the single most common error in the chapter.
“Light gases are quick”
Graham's law
Rate goes as one over the square root of molar mass. Hydrogen diffuses four times faster than oxygen, because 32 over 2 is 16 and its square root is 4.
“a for attraction, b for bulk”
The van der Waals constants
a corrects the pressure for intermolecular attraction; b corrects the volume for the space the molecules themselves take up.
The mistakes that cost the most
Thermodynamics in chemistry answers one question: will this reaction go by itself? Enthalpy, entropy and Gibbs energy together give the answer, and the sign conventions are the whole difficulty.
| What | Formula | Symbols and notes |
|---|---|---|
| First law | ΔU = q + w | q heat added to the system, w work done on the system |
| Work of expansion | w = −pextΔV | negative when the gas expands |
| Enthalpy | H = U + pV, ΔH = ΔU + ΔngRT | Δng is the change in moles of gas |
| At constant volume | qv = ΔU | no expansion work is done |
| At constant pressure | qp = ΔH | the usual laboratory condition |
| Heat capacity relation | Cp − Cv = R | per mole, for an ideal gas |
| What | Formula | Symbols and notes |
|---|---|---|
| Hess's law | ΔH is the same by any route | so thermochemical equations can be added |
| From formation enthalpies | ΔH = ΣΔHf(products) − ΣΔHf(reactants) | elements in their standard state have ΔHf = 0 |
| From bond enthalpies | ΔH = Σ(bonds broken) − Σ(bonds formed) | breaking costs energy, forming releases it |
| Exothermic | ΔH negative | heat is released, the beaker warms |
| Endothermic | ΔH positive | heat is absorbed, the beaker cools |
| What | Formula | Symbols and notes |
|---|---|---|
| Entropy change | ΔS = qrev/T | unit J K−1 mol−1 |
| Gibbs energy | ΔG = ΔH − TΔS | the test for spontaneity |
| Spontaneous | ΔG < 0 | at the given temperature |
| At equilibrium | ΔG = 0 | the forward and reverse tendencies balance |
| Gibbs energy and K | ΔG° = −2.303 RT log K | links thermodynamics to equilibrium |
| Second law | the entropy of the universe always increases | for any spontaneous change |
How to remember this chapter
“Negative means it goes”
Reading ΔG, ΔH and ΔS
Negative ΔG means spontaneous. Negative ΔH means heat released. Positive ΔS means more disorder. Only ΔG decides whether a reaction happens.
“Both negative, cold helps; both positive, heat helps”
When temperature decides
If ΔH and ΔS are both negative, low temperature makes ΔG negative. If both are positive, high temperature does. If signs differ, temperature cannot change the answer.
“Break costs, make pays”
Bond enthalpy calculations
Breaking bonds always absorbs energy and forming them always releases it. Bonds broken minus bonds formed gives ΔH, in that order.
The mistakes that cost the most
Equilibrium is two chapters treated as one: the equilibrium constant, and what happens in water. Between them they supply pH, buffers and solubility, which are among the most reliably asked numericals in the whole subject.
| What | Formula | Symbols and notes |
|---|---|---|
| In concentrations | Kc = [products]/[reactants], each raised to its coefficient | pure solids and liquids are left out |
| In partial pressures | Kp = Kc(RT)Δn | Δn is moles of gaseous products minus reactants |
| Reaction quotient | Q compared with K | Q < K goes forward, Q > K goes backward, Q = K is equilibrium |
| Reversed reaction | K' = 1/K | and for a reaction multiplied by n, K becomes Kn |
| Le Chatelier's principle | a system at equilibrium opposes any change imposed on it | pressure, concentration or temperature |
| What | Formula | Symbols and notes |
|---|---|---|
| Ionic product of water | Kw = [H⁺][OH⁻] = 10−14 | at 25 °C |
| pH | pH = −log[H⁺] | 7 is neutral at 25 °C |
| pOH and the relation | pOH = −log[OH⁻], pH + pOH = 14 | at 25 °C |
| Weak acid | [H⁺] = √(KaC) | C is the initial concentration |
| Degree of dissociation | α = √(Ka/C) | Ostwald's dilution law: dilution increases α |
| pK values | pKa = −log Ka | a smaller pKa means a stronger acid |
| What | Formula | Symbols and notes |
|---|---|---|
| Acidic buffer | pH = pKa + log([salt]/[acid]) | the Henderson-Hasselbalch equation |
| Basic buffer | pOH = pKb + log([salt]/[base]) | then pH = 14 − pOH |
| Solubility product, AB | Ksp = s² | for example AgCl |
| Solubility product, AB₂ | Ksp = 4s³ | for example CaF₂ |
| Solubility product, A₂B₃ | Ksp = 108s⁵ | for example Ca₃(PO₄)₂ type salts |
| Precipitation | precipitates when the ionic product exceeds Ksp | otherwise the salt stays dissolved |
| Common ion effect | adding a common ion suppresses dissociation | which is how buffers resist pH change |
How to remember this chapter
“Solids and liquids sit out”
Writing an equilibrium constant
Pure solids and pure liquids never appear in K. Their concentration does not change, so it is absorbed into the constant.
“Push it, it pushes back”
Le Chatelier's principle
Add a reactant and the reaction moves forward; raise the pressure and it shifts to the side with fewer gas molecules; heat an exothermic reaction and it goes backwards.
“Buffer is pKa plus the log of the ratio”
Henderson-Hasselbalch
When salt and acid are equal the log is zero, so pH equals pKa. That is the best buffering point, and it is worth remembering on its own.
“s squared, 4s cubed, 108s to the five”
Solubility product expressions
AB gives s², AB₂ gives 4s³, A₂B₃ gives 108s⁵. The coefficients come from the stoichiometry, not from memory alone.
The mistakes that cost the most
Redox looks like bookkeeping and it is. Oxidation numbers, then balancing, then the language of oxidising and reducing agents. Every question in this chapter can be answered by assigning oxidation numbers correctly first.
| What | Formula | Symbols and notes |
|---|---|---|
| Free element | 0 | for example H₂, O₂, Na, S₈ |
| Hydrogen | +1, except −1 in metal hydrides | NaH has hydrogen at −1 |
| Oxygen | −2, except −1 in peroxides and +2 in OF₂ | H₂O₂ has oxygen at −1 |
| Group 1 and 2 metals | +1 and +2 | always, in compounds |
| Fluorine | −1 always | the most electronegative element |
| Sum rule | neutral compound sums to 0, an ion sums to its charge | this is how unknown states are found |
| What | Formula | Symbols and notes |
|---|---|---|
| Oxidation | loss of electrons, oxidation number rises | the substance oxidised is the reducing agent |
| Reduction | gain of electrons, oxidation number falls | the substance reduced is the oxidising agent |
| Oxidising agent | itself gets reduced | for example KMnO₄, K₂Cr₂O₇ |
| Reducing agent | itself gets oxidised | for example H₂, CO, metals |
| Disproportionation | one element both oxidised and reduced | for example 2H₂O₂ → 2H₂O + O₂ |
| What | Formula | Symbols and notes |
|---|---|---|
| Ion electron method | balance atoms, then O with H₂O, then H with H⁺, then charge with electrons | in basic medium add OH⁻ at the end |
| Oxidation number method | equalise the total rise and fall in oxidation number | then balance the rest by inspection |
| n-factor of an acid | number of replaceable H⁺ | H₂SO₄ has n-factor 2 |
| n-factor in redox | electrons gained or lost per formula unit | KMnO₄ in acid has n-factor 5 |
| Equivalent weight | molar mass / n-factor | used in titration calculations |
| Titration | N₁V₁ = N₂V₂ | normality times volume, on both sides |
How to remember this chapter
“OIL RIG”
Oxidation and reduction
Oxidation Is Loss of electrons, Reduction Is Gain. Say it once at the start of every redox question and the agents follow.
“The agent suffers the opposite”
Naming the agents
The oxidising agent is itself reduced; the reducing agent is itself oxidised. Whatever the agent does to the other, it has the reverse done to itself.
“KMnO₄ is 5 in acid, 3 in neutral”
n-factor of permanganate
In acidic medium manganese goes from +7 to +2, so five electrons. In neutral or faintly alkaline medium it goes to +4, so three.
The mistakes that cost the most
Colligative properties depend on how many particles are dissolved, not on what they are. That one sentence explains every formula here, and the van't Hoff factor is what turns it from theory into a numerical.
| What | Formula | Symbols and notes |
|---|---|---|
| Molarity | M = moles of solute / litres of solution | temperature dependent |
| Molality | m = moles of solute / kg of solvent | temperature independent, so used in these formulas |
| Mole fraction | x = ni/ntotal | adds to 1 over all components |
| Mass percentage | (mass of solute / mass of solution) × 100 | the everyday way of stating strength |
| What | Formula | Symbols and notes |
|---|---|---|
| Raoult's law | p = xsolvent × p° | p° is the vapour pressure of the pure solvent |
| Relative lowering | (p° − p)/p° = xsolute | equal to the mole fraction of the solute |
| Ideal solution | ΔHmix = 0, ΔVmix = 0 | obeys Raoult's law at all compositions |
| Positive deviation | weaker new interactions, higher vapour pressure | for example ethanol and acetone |
| Negative deviation | stronger new interactions, lower vapour pressure | for example chloroform and acetone |
| Henry's law | p = KHx | solubility of a gas rises with pressure |
| What | Formula | Symbols and notes |
|---|---|---|
| Boiling point elevation | ΔTb = iKbm | Kb for water is 0.52 K kg mol−1 |
| Freezing point depression | ΔTf = iKfm | Kf for water is 1.86 K kg mol−1 |
| Osmotic pressure | π = iCRT | C is molarity; the most sensitive of the four |
| Relative lowering of vapour pressure | (p° − p)/p° = ixsolute | the fourth colligative property |
| van't Hoff factor | i = observed value / calculated value | 2 for NaCl, 3 for CaCl₂, less than 1 for association |
| Molar mass from a colligative property | M = iwK/(ΔT × W) | w solute mass, W solvent mass in kg |
How to remember this chapter
“Count particles, not kind”
What colligative means
One mole of sugar and one mole of NaCl do not behave alike, because NaCl gives two particles. It is the number of particles in solution that matters, never their identity.
“i is how many it breaks into”
The van't Hoff factor
NaCl gives 2, CaCl₂ gives 3, glucose gives 1. If the solute associates instead, as benzoic acid does in benzene, i falls below 1.
“Salt on ice, sugar in the pot”
Which way each effect goes
Adding solute lowers the freezing point and raises the boiling point. Salt melts ice on a road; sugar syrup boils above 100 °C.
The mistakes that cost the most
Electrochemistry joins redox to energy. The Nernst equation and the link between cell potential and Gibbs energy carry the numericals, and conductance supplies the rest. Sign conventions decide most of the marks.
| What | Formula | Symbols and notes |
|---|---|---|
| Cell potential | E°cell = E°cathode − E°anode | both as reduction potentials |
| Galvanic cell | spontaneous, E°cell positive | anode is negative, cathode is positive |
| Electrolytic cell | non-spontaneous, needs an external supply | anode is positive, cathode is negative |
| Nernst equation | E = E° − (0.0591/n)log([products]/[reactants]) | at 298 K |
| At equilibrium | E = 0 and E° = (0.0591/n)log K | which links potential to the equilibrium constant |
| Gibbs energy | ΔG° = −nFE°cell | F = 96500 C mol−1 |
| What | Formula | Symbols and notes |
|---|---|---|
| Conductance | G = 1/R | unit siemens |
| Conductivity | κ = G × (l/A) | l/A is the cell constant, unit m−1 |
| Molar conductivity | Λm = κ × 1000/C | C in mol per litre |
| On dilution | Λm rises, κ falls | more ions are free but they are further apart |
| Kohlrausch's law | Λ°m = λ°+ + λ°− | limiting molar conductivity adds from the ions |
| Degree of dissociation | α = Λm/Λ°m | for a weak electrolyte |
| What | Formula | Symbols and notes |
|---|---|---|
| Faraday's first law | m = ZIt | Z is the electrochemical equivalent |
| In terms of equivalents | m = (E × I × t)/96500 | E is the equivalent weight |
| Faraday's second law | m₁/m₂ = E₁/E₂ | for the same charge passed |
| One faraday | 96500 C deposits one gram equivalent | one mole of electrons |
How to remember this chapter
“Red Cat, An Ox”
What happens at each electrode
REDuction at the CAThode, ANode is OXidation. True in both galvanic and electrolytic cells, even though the signs of the electrodes swap.
“Cathode minus anode”
Calculating cell potential
E°cell = E°cathode − E°anode, with both written as reduction potentials. Never reverse a sign and subtract as well; do one or the other.
“0.0591 over n”
The Nernst equation at room temperature
At 298 K the factor 2.303RT/F is 0.0591 volts. Memorise the number and the equation becomes one line.
The mistakes that cost the most
Kinetics asks how fast, not how far. Order, rate constant and half-life are the whole chapter, and the first order results are asked so often that they are worth memorising as a block.
| What | Formula | Symbols and notes |
|---|---|---|
| Rate of reaction | rate = −Δ[R]/Δt = +Δ[P]/Δt | divided by the stoichiometric coefficient |
| Rate law | rate = k[A]x[B]y | x and y come from experiment, not from the equation |
| Order | order = x + y | can be zero, fractional or negative |
| Molecularity | the number of species colliding in one step | always a whole number, never zero |
| Units of k, zero order | mol L−1 s−1 | the rate itself |
| Units of k, first order | s−1 | independent of concentration units |
| Units of k, second order | L mol−1 s−1 | general form: (mol L−1)1−n s−1 |
| What | Formula | Symbols and notes |
|---|---|---|
| Zero order | [R] = [R]₀ − kt | a straight line of concentration against time |
| Zero order half-life | t1/2 = [R]₀/2k | proportional to the starting concentration |
| First order | k = (2.303/t)log([R]₀/[R]) | the most used equation in the chapter |
| First order half-life | t1/2 = 0.693/k | independent of the starting concentration |
| First order graph | log[R] against t is a straight line | slope = −k/2.303 |
| After n half-lives | [R] = [R]₀/2n | the quick route for whole numbers of half-lives |
| What | Formula | Symbols and notes |
|---|---|---|
| Arrhenius equation | k = Ae−Ea/RT | A is the frequency factor |
| Logarithmic form | log k = log A − Ea/2.303RT | a straight line of log k against 1/T |
| Two temperatures | log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂) | the standard numerical |
| Rule of thumb | rate roughly doubles for every 10 K rise | near room temperature |
| Catalyst | lowers Ea, speeds both directions equally | it does not change ΔH or the position of equilibrium |
How to remember this chapter
“Order from experiment, molecularity from the step”
Telling them apart
Order is measured and can be zero or fractional. Molecularity is counted from a single step and is always a whole number.
“First order forgets where it started”
Half-life
For first order, t1/2 = 0.693/k, whatever the initial concentration. For zero order the half-life depends on it.
“A catalyst lowers the hill”
What a catalyst does
It provides a path with lower activation energy. Both forward and reverse rates rise by the same factor, so the equilibrium position does not move.
The mistakes that cost the most
Solid state is geometry with chemistry attached. Unit cells, packing and density make up nearly all of it, and every numerical uses the same density formula, so it is one of the fastest chapters to secure.
| What | Formula | Symbols and notes |
|---|---|---|
| Simple cubic | atoms per cell Z = 1 | corners only, 8 × ⅛ |
| Body centred cubic | Z = 2 | corners plus one at the centre |
| Face centred cubic | Z = 4 | corners plus six face centres, each shared by two |
| Edge and radius, simple cubic | a = 2r | atoms touch along the edge |
| Body centred cubic | a√3 = 4r | atoms touch along the body diagonal |
| Face centred cubic | a√2 = 4r | atoms touch along the face diagonal |
| What | Formula | Symbols and notes |
|---|---|---|
| Packing efficiency, simple cubic | 52.4% | the most wasteful arrangement |
| Packing efficiency, bcc | 68% | better, and common in alkali metals |
| Packing efficiency, fcc and hcp | 74% | the closest packing possible for equal spheres |
| Coordination number | simple cubic 6, bcc 8, fcc and hcp 12 | nearest neighbours touching an atom |
| Density of a unit cell | ρ = ZM/(a³NA) | a in metres gives density in kg m−3 |
| Voids in close packing | 2 tetrahedral and 1 octahedral per atom | for N atoms, 2N and N respectively |
| Radius ratio | r+/r− decides the geometry | 0.225 to 0.414 tetrahedral, 0.414 to 0.732 octahedral |
| What | Formula | Symbols and notes |
|---|---|---|
| Schottky defect | a cation and an anion both missing | density falls; seen in NaCl, KCl |
| Frenkel defect | an ion moves to an interstitial site | density unchanged; seen in AgCl, ZnS |
| Metal excess defect | gives colour, as F-centres | for example NaCl heated in sodium vapour turns yellow |
| n-type semiconductor | doped with a group 15 element | extra electrons carry the current |
| p-type semiconductor | doped with a group 13 element | holes carry the current |
How to remember this chapter
“1, 2, 4”
Atoms per unit cell
Simple cubic 1, body centred 2, face centred 4. Almost every solid state numerical starts by picking the right Z.
“Edge, face, body: 2, √2, √3”
Relating a and r
Simple cubic touches along the edge, fcc along the face diagonal, bcc along the body diagonal. a = 2r, a√2 = 4r, a√3 = 4r.
“Schottky slims, Frenkel shuffles”
The two defects
Schottky removes a pair of ions, so density falls. Frenkel just moves an ion inside, so density is unchanged.
The mistakes that cost the most
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