Before You Begin: The Chemistry Your School Course Skipped

School chemistry teaches models as if they were facts. Lewis structures, VSEPR shapes, and the octet rule are useful approximations with domains of validity - and university chemistry begins by saying which approximation fails where and why. The transition needs three things: the arithmetic of the mole done without formula-sheet panic, enough quantum language to know what an orbital is, and the algebra of equilibria, which is logarithms and simultaneous equations.

This page rebuilds those, defining every term before use and proving every rule that can be proved. Take the self-test in §5; study what surprises you.

Chemistry is counting and energy. Counting: reactions happen in whole numbers of particles, and the mole is the exchange rate between the counting world and the weighing world. Energy: every reaction, phase change and bond is an energy ledger, and equilibrium is where the ledger’s bookkeeping; entropy, not order - balances. School chemistry states the ledger’s conclusions; university chemistry derives them, which requires the logarithm and the exponential to be friends of yours rather than buttons on a calculator.

Definition. A mole of entities is $N_A = 6.02214076 \times 10^{23}$ entities; a fixed count, exactly as a “dozen” is twelve.

Definition. The molar mass $M$ of a substance is the mass of one mole of its entities, in $\mathrm{g\,mol^{-1}}$. For an element, $M$ equals its relative atomic mass numerically.

Why the mole works. Atoms combine in small whole-number ratios because reactions conserve atoms. So a balanced equation is a counting statement: $2\mathrm{H_2} + \mathrm{O_2} \to 2\mathrm{H_2O}$ says 2 molecules react with 1. Multiplying the count by $N_A$ converts to moles; multiplying moles by $M$ converts to grams. The chain is

and every stoichiometry problem is a walk along this chain with a ratio from the balanced equation in the middle.

Worked example. What mass of $\mathrm{CO_2}$ forms from $24.0\,\mathrm{g}$ of carbon? $n(\mathrm{C}) = 24.0/12.0 = 2.00\,\mathrm{mol}$; the ratio C : CO₂ is $1 : 1$, so $n(\mathrm{CO_2}) = 2.00\,\mathrm{mol}$; mass $= 2.00 \times 44.0 = 88.0\,\mathrm{g}$.

Concentration. $c = n/V$ in $\mathrm{mol\,dm^{-3}}$ (school: “molarity”). Dilution conserves moles: $c_1V_1 = c_2V_2$; proved because $n = cV$ is the same $n$ before and after adding solvent.

Definition. An atomic orbital is a mathematical function $\psi$ whose square $|\psi|^2$ gives the probability density of finding an electron at a point. It is not a path.

This is where school and university part company. The school orbit; electrons in shells circling a nucleus; is wrong in a way that matters: an electron has no trajectory, and the orbital is a standing wave of probability. The “shells” are energy levels; the “subshells” $s, p, d$ are shapes of $|\psi|^2$; occupancy follows the Pauli exclusion principle (two electrons per orbital, opposite spins) and the Aufbau ordering by energy.

Why models are models. Lewis structures assign bonding to shared pairs and work well for main-group chemistry. They fail for electron-deficient species ($\mathrm{BF_3}$), for odd-electron species ($\mathrm{NO}$), and for anything where delocalisation matters (benzene). University chemistry keeps the model and states its domain; and when a model fails, that failure is data about the system, which is exactly the reasoning the mechanism sections of the tracks use.

Definition. For a reversible reaction at equilibrium, $a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}$, the equilibrium constant is

with $[\,]$ denoting equilibrium concentration. Only temperature changes $K_c$.

The algebra you need. Every equilibrium problem ends in a quadratic or a logarithm. Two examples of the entire technique:

Weak acid. For $\mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-}$ with $K_a$ and initial concentration $C$: the ICE table gives $K_a = \frac{x^2}{C - x}$. When $K_a \ll C^2$… more precisely when $x \ll C$, the approximation $C - x \approx C$ gives $x = \sqrt{K_a C}$; and the approximation is self-checking: compute $x/C$; if it exceeds about $5\%$, solve the quadratic instead. Knowing when an approximation is licensed is the skill.

pH. $\mathrm{pH} = -\log_{10}[\mathrm{H^+}]$, so $[\mathrm{H^+}] = 10^{-\mathrm{pH}}$; a definition, not a formula to derive. For a weak acid, $\mathrm{pH} = \tfrac12(\mathrm{p}K_a - \log C)$, obtained by taking $-\log$ of $x = \sqrt{K_aC}$. Logs convert the multiplication in $K_a$ into the addition in pH, which is why every equilibrium graph in the subject is drawn on log axes.

Hess’s law and energies. Enthalpy is a state function, so any route between the same states gives the same $\Delta H$; that is why Hess cycles work, and it is proved by the definition of a state function rather than by experiment.

• Units multiply and cancel like algebra. $\mathrm{mol\,dm^{-3}} \times \mathrm{dm^3} = \mathrm{mol}$. Writing units at every line is how errors are caught.
• Significant figures: a computed answer carries no more than the least precise input. Reporting $0.0666667\,\mathrm{M}$ from data good to two figures is a claim the data does not support.
• Ions carry charge as superscripts; states as subscripts $(s), (l), (g), (aq)$. A balanced equation conserves both atoms and charge - check both.

1. How many oxygen atoms are in $0.25\,\mathrm{mol}$ of $\mathrm{Al_2(SO_4)_3}$? (§1)
2. $25.0\,\mathrm{cm^3}$ of $0.100\,\mathrm{M\,NaOH}$ neutralises $20.0\,\mathrm{cm^3}$ of HCl. Find $c(\mathrm{HCl})$. (§1, §3)
3. Why is the Lewis model of bonding wrong for benzene, and what replaces it? (§2)
4. A $0.10\,\mathrm{M}$ solution of a weak acid has $\mathrm{pH} = 2.87$. Find $K_a$, stating the approximation you made. (§3)
5. Why does $\mathrm{pH} = -\log_{10}[\mathrm{H^+}]$ turn a product into a sum? (§3)

Answers. (1) $0.25 \times 12 \times N_A = 1.8 \times 10^{24}$ atoms ($3$ oxygens per formula unit $\times\, 4$ oxygens in each sulfate). (2) $c_1V_1 = c_2V_2$: $0.100 \times 25.0 = c \times 20.0$, so $c = 0.125\,\mathrm{M}$. (3) Benzene’s six $\pi$ electrons are delocalised over the ring; the Lewis model forces localised double bonds, predicting two distinct C–C bond lengths, while all six are equal; the delocalised $\pi$ system replaces it. (4) $[\mathrm{H^+}] = 10^{-2.87} = 1.35\times10^{-3}$; $K_a = x^2/(C - x) = (1.35\times10^{-3})^2/(0.10 - 0.00135) = 1.83\times10^{-5}$; the approximation $C - x \approx C$ was used and $x/C = 1.35\%$, well under $5\%$, so it is licensed. (5) Because $\log$ of a product is the sum of the logs: $-\log(ab) = -(\log a + \log b)$.

• Atomic Structure and Periodicity; orbitals done properly.
• Chemical Kinetics; rates, and the difference between equilibrium and speed.
• Thermodynamics; the energy ledger, derived.

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