Formula & equation sheet

✓ dimensionally homogeneous — n = m / M both reduce to amount, machine-verified.

Rearranged: $m = nM$$M = \dfrac{m}{n}$

The bridge between the mass you weigh and the moles the balanced equation counts. Because molar mass $M$ carries units of $\mathrm{g/mol}$, dividing a mass in grams by it cancels the grams and leaves moles — the machine checks that both sides are an amount.

Where it holds: Any pure substance whose molar mass is known; the definition of molar mass, so it always holds.

✓ dimensionally homogeneous — c = n / V both reduce to concentration, machine-verified.

Rearranged: $n = cV$$V = \dfrac{n}{c}$

Concentration is amount per volume — $\mathrm{mol/L}$, written $\mathrm{M}$. Dividing moles by litres gives the molarity; multiplying a molarity by a volume gives back moles, which is how a measured volume of a stock solution becomes an amount on the ledger.

Where it holds: Any solution; the definition of molar concentration. Volume is the total solution volume, not the solvent volume.

✓ dimensionally homogeneous — M₁V₁ = M₂V₂ both reduce to amount, machine-verified.

Rearranged: $M_2 = \dfrac{M_1 V_1}{V_2}$$V_2 = \dfrac{M_1 V_1}{M_2}$

Adding solvent does not add or remove solute — so the amount $M V$ (concentration times volume, an amount of moles) is the same before and after. Both sides of the equation carry the dimension of an amount, which is exactly the conserved quantity.

Where it holds: Diluting a solution with more solvent: the moles of solute are unchanged, only the volume grows. Not for mixing two different solutes.

✓ dimensionally homogeneous — N = n · N_A both reduce to dimensionless, machine-verified.

Rearranged: $n = \dfrac{N}{N_A}$

A mole is a count: $N_A = 6.02214076\times10^{23}$ particles per mole (exact since the 2019 SI redefinition). Multiplying an amount in moles by $N_A$ gives a pure number of particles — the moles cancel the per-mole of $N_A$, leaving a dimensionless count, which the machine confirms.

Where it holds: Counting discrete particles (atoms, molecules, ions, formula units). N_A is a fixed count per mole.

✓ dimensionally homogeneous — percent yield = (actual / theoretical) × 100 both reduce to dimensionless, machine-verified.

Percent yield is a ratio of two masses, so it is dimensionless — the grams cancel, and the $\times 100$ only rescales the fraction to a percent. The machine checks that both sides are dimensionless; the chemistry check (that actual never exceeds theoretical) lives in the lesson that emits it.

Where it holds: Comparing a measured (actual) product mass to the theoretical maximum from stoichiometry. Always ≤ 100% — you cannot collect more than forms.

✓ dimensionally homogeneous — PV = nRT both reduce to energy, machine-verified.

Rearranged: $V = \dfrac{nRT}{P}$$n = \dfrac{PV}{RT}$$P = \dfrac{nRT}{V}$$T = \dfrac{PV}{nR}$

The single relation tying a gas's pressure, volume, amount, and temperature. Each side, $PV$ and $nRT$, works out to an energy — the machine confirms the two are dimensionally the same, which is why the equation can hold. Solve it for any one variable given the other three; it is the gas-phase entry onto the species ledger (a volume of gas is an amount of moles).

Where it holds: Best when the gas is dilute — low pressure and temperature well above the boiling point. Real gases deviate near condensation (high P, low T), where intermolecular forces and molecular volume stop being negligible.

✓ dimensionally homogeneous — P₁V₁ / T₁ = P₂V₂ / T₂ both reduce to energy per temperature, machine-verified.

Rearranged: $P_2 = \dfrac{P_1 V_1 T_2}{T_1 V_2}$$V_2 = \dfrac{P_1 V_1 T_2}{T_1 P_2}$

For a sealed sample of gas, $PV/T$ is the same in any two states — because $nR$ is constant, $PV/T = nR$. Each side carries the dimension of energy-per-temperature; the machine checks they match. Set up the two states, cancel whatever is held constant, and solve for the missing quantity.

Where it holds: A fixed amount of gas taken from one state to another (Boyle's, Charles's, and Gay-Lussac's laws combined). Temperatures are absolute (kelvin).

✓ dimensionally homogeneous — q = m c ΔT both reduce to energy, machine-verified.

Rearranged: $c = \dfrac{q}{m\,\Delta T}$$\Delta T = \dfrac{q}{m\,c}$

The heat $q$ that flows into a sample raises its temperature in proportion to its mass and its specific heat capacity $c$. Multiplying mass (grams) by $c$ (energy per gram per degree) by the temperature change (degrees) leaves an energy — the machine confirms $q$ and $m c \Delta T$ share the dimension of energy. This is the first rung of the energy ledger.

Where it holds: Heating or cooling one substance with no phase change and no chemical reaction. ΔT is a temperature difference, so °C and K give the same number.

✓ dimensionally homogeneous — ΔH_rxn = Σ ν ΔH_f° (products − reactants) both reduce to molar energy, machine-verified.

Rearranged: $\Delta H_f^\circ(\text{unknown}) = \Delta H_{\text{rxn}} - \sum \nu\,\Delta H_f^\circ(\text{known species})$$\Delta H_{\text{reverse}} = -\,\Delta H_{\text{rxn}}$

The enthalpy of a reaction is the sum of its products' standard formation enthalpies minus its reactants', each weighted by its coefficient $\nu$ — because enthalpy is a state function, the path does not matter (Hess's law). Both sides carry the dimension of molar energy (kJ/mol), and the machine checks they match. A negative $\Delta H_{\text{rxn}}$ is exothermic. The heat an actual run delivers is $q = \Delta H_{\text{rxn}}\cdot\xi$, scaled by the extent the limiting reagent allows.

Where it holds: Assembling a reaction's enthalpy from tabulated standard enthalpies of formation. Each ΔH_f° is weighted by its stoichiometric coefficient ν; an element in its standard state has ΔH_f° = 0. The result is per mole of reaction, as the equation is balanced.

✓ dimensionally homogeneous — K_a = [H⁺][A⁻] / [HA] both reduce to dimensionless, machine-verified.

An equilibrium constant is a ratio of activities, and an activity $a_X = [X]/c^\circ$ is a concentration measured against the $c^\circ = 1\ \mathrm{M}$ standard state — so it is dimensionless. That is why $K_a$ carries no units even though it is written with concentrations: every bracket is really an activity, the $\mathrm{mol/L}$ cancels against $c^\circ$, and the machine checks that both sides reduce to the same (zero) dimension. What $K_a$ equals numerically is the sourced datum (`data/ionization-constants.toml`); what is machine-checked here is that the relation is dimensionally consistent. The equilibrium position it encodes is the disclosed model.

Where it holds: For a weak monoprotic acid at equilibrium, $\mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-}$. A larger $K_a$ means a stronger (more ionized) weak acid. For a polyprotic acid each proton has its own $K_a$ ($K_{a1} \gg K_{a2} \gg \dots$).

✓ dimensionally homogeneous — K_b = [BH⁺][OH⁻] / [B] both reduce to dimensionless, machine-verified.

Like $K_a$, $K_b$ is a ratio of activities ($a_X = [X]/c^\circ$ against the $1\ \mathrm{M}$ standard state), so it is dimensionless — the machine checks that both sides reduce to the zero dimension. The solvent water is a pure liquid (activity 1) and is excluded, exactly as the pure solid is excluded from $K_{sp}$. A base and its conjugate acid are tied together by $K_a K_b = K_w$ (see that entry); its numeric value is the sourced datum.

Where it holds: For a weak base reacting with water at equilibrium, $\mathrm{B} + \mathrm{H_2O} \rightleftharpoons \mathrm{BH^+} + \mathrm{OH^-}$. Water is the pure solvent (activity 1), so it does not appear. A larger $K_b$ means a stronger weak base.

✓ dimensionally homogeneous — K_w = [H⁺][OH⁻] both reduce to dimensionless, machine-verified.

$K_w$ is the product of the hydrogen-ion and hydroxide-ion activities ($a_X = [X]/c^\circ$), so it is dimensionless — the machine checks that both sides reduce to the zero dimension. Because it always holds, it ties $[\mathrm{H^+}]$ and $[\mathrm{OH^-}]$ together: a weak base fixes $[\mathrm{OH^-}]$, and $K_w$ converts that to $[\mathrm{H^+}]$ and thus a pH. Its value ($1.0\times10^{-14}$ at 25 °C) is the sourced datum; what is checked here is dimensional consistency.

Where it holds: The autoionization of water, $\mathrm{H_2O} \rightleftharpoons \mathrm{H^+} + \mathrm{OH^-}$, at every point in any aqueous solution. At 25 °C, $K_w = 1.0\times10^{-14}$. It is the bridge that turns an $[\mathrm{OH^-}]$ into an $[\mathrm{H^+}]$ (and pOH into pH): $[\mathrm{H^+}] = K_w/[\mathrm{OH^-}]$.

✓ dimensionally homogeneous — K_a K_b = K_w both reduce to dimensionless, machine-verified.

Each of $K_a$, $K_b$, $K_w$ is a dimensionless activity constant, so their relation is a dimensionless identity — the machine checks that both sides reduce to the zero dimension. It is why you never need a separate table of $K_b$ values: given $K_a$ and $K_w$, $K_b = K_w/K_a$. This is exactly the step a weak-acid/strong-base titration uses at the equivalence point, where the acid has all become its conjugate base.

Where it holds: For any conjugate acid-base pair in water: the acid's $K_a$ and its conjugate base's $K_b$ multiply to the water ion-product $K_w$. So a strong weak-acid has a weak conjugate base and vice versa, and $K_b = K_w/K_a$ — the identity that turns a titration's equivalence point (all conjugate base) into a weak-base solve.

✓ dimensionally homogeneous — K_sp = [Mⁿ⁺]ᵃ [Xᵐ⁻]ᵇ both reduce to dimensionless, machine-verified.

$K_{sp}$ is a product of ion activities ($a_X = [X]/c^\circ$ against the $1\ \mathrm{M}$ standard state), each dimensionless — so $K_{sp}$ is dimensionless whatever the stoichiometric coefficients $a, b$ are (a dimensionless quantity raised to any power stays dimensionless). The machine checks that both sides reduce to the zero dimension. The pure solid is excluded (activity 1), the same move that lets water drop out of $K_b$; its value is the sourced datum (`data/solubility-products.toml`).

Where it holds: For a sparingly soluble salt dissolving at saturation, $\mathrm{M}_a\mathrm{X}_b(s) \rightleftharpoons a\,\mathrm{M}^{n+} + b\,\mathrm{X}^{m-}$. The dissolving solid is a pure phase (activity 1), so it does not appear. Comparing the reaction quotient $Q$ to $K_{sp}$ predicts whether a precipitate forms ($Q > K_{sp}$).