Chemical reactions transform substances by breaking and forming bonds. Stoichiometry is the quantitative bookkeeping — tracking atoms, moles, and energy through these transformations. This skill covers equation balancing, reaction classification, mole-based calculations, acid-base chemistry, redox reactions, and thermochemistry.
Agent affinity: lavoisier (chair, reactions and conservation of mass, primary)
Lavoisier's Law. In a chemical reaction, matter is neither created nor destroyed. Every atom present in the reactants must appear in the products.
Balancing procedure:
Write the unbalanced equation with correct formulas.
Balance one element at a time, starting with the most complex molecule.
Balance hydrogen and oxygen last (they appear in many compounds).
Use the smallest whole-number coefficients.
Verify: count every element on both sides.
Worked Example: Combustion of Propane
Unbalanced: C3H8 + O2 -> CO2 + H2O
Step 1. Balance C: 3 carbons on left, so 3 CO2 on right.
C3H8 + O2 -> 3 CO2 + H2O
Step 2. Balance H: 8 hydrogens on left, so 4 H2O on right.
C3H8 + O2 -> 3 CO2 + 4 H2O
Step 3. Balance O: Right side has 3(2) + 4(1) = 10 oxygens. Left needs 10/2 = 5 O2.
C3H8 + 5 O2 -> 3 CO2 + 4 H2O
Verify: C: 3 = 3. H: 8 = 8. O: 10 = 10. Balanced.
Worked Example: Balancing a More Complex Equation
Unbalanced: Fe2O3 + CO -> Fe + CO2
Step 1. Balance Fe: 2 Fe on left, so 2 Fe on right.
Fe2O3 + CO -> 2 Fe + CO2
Step 2. Balance O: Left has 3 (from Fe2O3) + 1 (from CO) = 4 if 1 CO. Right has 2 from CO2. Try: 3 CO on left gives 3 + 3 = 6 oxygens total on left... Systematic approach: Fe2O3 + 3 CO -> 2 Fe + 3 CO2.
Percent yield = (actual yield / theoretical yield) x 100%. If the experiment produced 10.5 g: (10.5 / 11.3) x 100% = 92.9%.
Reaction Types
The Five Classical Types
Type
Pattern
Example
Synthesis (combination)
A + B -> AB
2 Na + Cl2 -> 2 NaCl
Decomposition
AB -> A + B
2 HgO -> 2 Hg + O2
Single replacement
A + BC -> AC + B
Zn + CuSO4 -> ZnSO4 + Cu
Double replacement (metathesis)
AB + CD -> AD + CB
AgNO3 + NaCl -> AgCl + NaNO3
Combustion
CxHy + O2 -> CO2 + H2O
CH4 + 2 O2 -> CO2 + 2 H2O
Activity series for single replacement. A metal replaces another in solution only if it is more active (higher on the activity series). Li > K > Ba > Ca > Na > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Ag > Pt > Au. Zinc replaces copper; copper does not replace zinc.
Precipitation Reactions
A double replacement reaction where an insoluble product (precipitate) forms. Use solubility rules:
Soluble: All Na+, K+, NH4+ salts. All nitrates. Most chlorides (except AgCl, PbCl2).
Insoluble: Most carbonates, phosphates, sulfides (except Group 1 and NH4+).
Worked example.Write the net ionic equation for mixing AgNO3(aq) and NaCl(aq).
Full molecular: AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)
Net ionic (cancel spectators Na+ and NO3-): Ag+(aq) + Cl-(aq) -> AgCl(s)
The net ionic equation captures the chemistry — silver and chloride ions combine to form the insoluble precipitate.
Acids and Bases
Three Definitions
Theory
Acid
Base
Arrhenius
Produces H+ in water
Produces OH- in water
Bronsted-Lowry
Proton (H+) donor
Proton (H+) acceptor
Lewis
Electron pair acceptor
Electron pair donor
Each definition is progressively more general. Bronsted-Lowry is the workhorse for aqueous chemistry. Lewis acid-base theory extends to non-aqueous and coordination chemistry.
Conjugate Pairs
Every Bronsted-Lowry acid has a conjugate base (what remains after donating H+), and every base has a conjugate acid (what forms after accepting H+).
Problem. Calculate the pH of 0.10 M acetic acid (Ka = 1.8 x 10^-5).
CH3COOH <=> CH3COO- + H+
Let x = [H+] at equilibrium. Ka = x^2 / (0.10 - x). Since Ka is small, assume 0.10 - x is approximately 0.10.
x^2 = 1.8 x 10^-5 x 0.10 = 1.8 x 10^-6.
x = 1.34 x 10^-3 M. Check assumption: 1.34 x 10^-3 / 0.10 = 1.3% < 5%. Valid.
pH = -log(1.34 x 10^-3) = 2.87.
Titration and Equivalence Point
At the equivalence point, moles of acid = moles of base. For a strong acid + strong base titration, the equivalence point pH is 7.00. For a weak acid + strong base, the equivalence point pH is above 7 (conjugate base in solution is basic).
Buffer solutions. Mixtures of a weak acid and its conjugate base (or weak base and conjugate acid) resist pH changes. Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]).
Oxidation-Reduction (Redox)
Oxidation States
Rules for assigning oxidation states (priority order):
Free elements: 0 (Na, O2, P4 all have oxidation state 0)
Oxidation = increase in oxidation state (loss of electrons). Reduction = decrease in oxidation state (gain of electrons). Mnemonic: OIL RIG (Oxidation Is Loss, Reduction Is Gain).
Half-Reaction Method (Acidic Solution)
Worked example.Balance: MnO4- + Fe^2+ -> Mn^2+ + Fe^3+ in acidic solution.
Standard reduction potentials (E-zero) predict spontaneous redox reactions. A positive cell potential (E-zero-cell = E-zero-cathode - E-zero-anode) means the reaction is spontaneous. This connects stoichiometry to electrical energy — the basis of batteries and electrolysis.
If a reaction can be expressed as the sum of two or more steps, the overall delta-H is the sum of the delta-H values of the steps. Enthalpy is a state function — only initial and final states matter, not the path.
