Difficulty: Intermediate. Prerequisites: significant figures, metric conversions, valence electrons, Lewis structures (see the first two sets of study notes).
Big picture: This is where chemistry becomes quantitative. You learn to balance equations, convert between grams and moles, identify which reactant runs out first, and work with solutions. Oxidation-reduction (redox), double displacement, and precipitation reactions are the main reaction types covered. If bonding is about how atoms stick together, stoichiometry and reactions are about how much of each substance is involved and what happens when they interact. This material dominates the calculation-heavy portion of most CHEM 111 exams.
Balanced equations give mole ratios. Use those ratios plus molar masses to convert between grams of one substance and grams of another. The limiting reactant is the one that runs out first and determines how much product forms. Solutions are described by molarity (moles of solute per litre of solution). Redox reactions track electron transfer through oxidation numbers. Double displacement reactions produce precipitates, and solubility rules tell you which products are insoluble.
Mole
The SI unit for the amount of a substance. One mole contains exactly 6.02214199 × 10²³ particles (atoms, molecules, ions, or formula units). Think of it as: a chemist's counting unit, the way a "dozen" means 12.
Avogadro's number
The fixed constant 6.022 × 10²³, defining the number of particles in one mole.
Molar mass
The mass of one mole of a substance, in grams per mole (g/mol). Numerically equal to the atomic or molecular mass in AMU. In simple terms, it is the bridge between grams on a balance and moles in an equation.
Molecular mass
The mass of a single molecule, measured in atomic mass units (AMU).
Formula mass
The mass of one formula unit of an ionic compound, measured in AMU.
Stoichiometric coefficients
The numbers in front of chemical formulas in a balanced equation. They give the mole ratio of reactants and products. Think of it as: the recipe ratios telling you how many moles of each ingredient you need.
Limiting reactant
The reactant that is completely consumed first in a reaction. It determines the maximum amount of product that can form. In simple terms, it is the ingredient you run out of first.
Molarity (M)
Concentration expressed as moles of solute per litre of solution.
Solvent
The largest component of a solution (usually the liquid that does the dissolving).
Solute
The substance dissolved in the solvent (the smaller component).
Electrolyte
A substance that dissociates into ions when dissolved in water, allowing the solution to conduct electricity. Strong electrolytes dissociate completely; weak electrolytes dissociate only partially.
Nonelectrolyte
A substance that dissolves in water but does not produce ions, so the solution does not conduct electricity.
Oxidation
Loss of electrons, gain of oxygen, or an increase in oxidation number. Think of it as: OIL (Oxidation Is Loss of electrons).
Reduction
Gain of electrons, loss of oxygen, or a decrease in oxidation number. Think of it as: RIG (Reduction Is Gain of electrons).
Oxidation number
A bookkeeping number assigned to each atom in a compound to track electron distribution. Rules: 0 for a free element, equal to charge for monoatomic ions, -2 for oxygen (except peroxides), +1 for hydrogen (except metal hydrides, where it is -1).
Percent composition
The percentage by mass of each element in a compound. Calculated as (mass of element in one mole of compound / molar mass of compound) × 100.
Double displacement reaction
A reaction where parts of two ionic compounds swap partners: AB + CD → AD + BC. In simple terms, two compounds trade ions to form two new compounds.
Precipitate
An insoluble solid that forms when two aqueous solutions are mixed.
Solubility
The maximum amount of solute that dissolves in a given amount of solvent at a specific temperature (typically 25 °C).
Spectator ions
Ions present in solution that do not participate in the reaction. They appear on both sides of the full ionic equation and are removed to write the net ionic equation.
A balanced equation has the same number of atoms of each element on both sides. The coefficients in front of formulas give mole ratios; the subscripts within formulas are fixed by the compound's identity and must not be changed.
Steps to balance:
Write the unbalanced equation with correct reactant and product formulas.
Balance atoms of one element at a time (save H and O for last if possible).
Balance the remaining elements.
Check that every element has equal counts on both sides.
The mole is the SI unit for amount of substance: exactly 6.02214199 × 10²³ particles. Avogadro's number is a fixed constant (redefined on 20 May 2019).
Molar mass (g/mol) is numerically equal to the atomic or formula mass in AMU. It converts between grams (what you weigh) and moles (what you use in equations).
Molecular mass is the mass of one molecule in AMU. Formula mass is the same concept applied to one formula unit of an ionic compound.
Mole-to-mole: Start with a balanced equation. Write down the known moles of one substance, apply the mole ratio from the coefficients, and calculate moles of the target substance.
Mass-to-mass: A four-step bridge:
Start with grams of substance A.
Convert to moles of A using its molar mass (÷ molar mass).
Convert to moles of B using the mole ratio from the balanced equation.
Convert to grams of B using its molar mass (× molar mass).
In most reactions, one reactant is used up before the others. This is the limiting reactant, and it determines the maximum amount of product. The other reactant(s) are in excess.
To identify the limiting reactant: convert each reactant's mass to moles, use the mole ratio to see which one produces the least product, and that one is limiting.
Percent composition tells you the mass percentage of each element in a compound. Calculate it as (mass of that element in one mole of compound / molar mass of compound) × 100.
Oxidation is loss of electrons (increase in oxidation number). Reduction is gain of electrons (decrease in oxidation number). The mnemonic OIL RIG helps: Oxidation Is Loss, Reduction Is Gain.
Oxidation number rules:
Free element: 0
Monoatomic ion: equal to the ion's charge
Oxygen: -2 in most compounds (exception: -1 in peroxides)
Hydrogen: +1 in most compounds (exception: -1 in metal hydrides)
The sum of oxidation numbers in a compound equals zero; in a polyatomic ion, it equals the ion's charge
Example: In CO₃²⁻, carbon has an oxidation number of +4 (since 3 × (-2) for oxygen = -6, and the ion's charge is -2, so carbon must be +4 to make the sum -2).
A solution has a solvent (the largest component) and a solute (the dissolved substance). Both can be gas, liquid, or solid.
Electrolytes and nonelectrolytes:
Strong electrolytes dissociate completely in water (e.g. NaCl, HCl) and conduct electricity well.
Weak electrolytes dissociate only partially (e.g. acetic acid) and conduct poorly.
Nonelectrolytes do not produce ions and do not conduct.
Molarity = moles of solute / litres of solution.
Dilution formula: M₁V₁ = M₂V₂ (initial molarity × initial volume = final molarity × final volume).
In a double displacement reaction, two ionic compounds swap partners: AB + CD → AD + BC. The driving force is the formation of a precipitate (an insoluble product), water, or a gas.
Solubility rules (simplified):
All alkali metal compounds are soluble.
All nitrates and acetates are soluble.
All ammonium compounds are soluble.
Most sulfates are soluble (exceptions include BaSO₄, PbSO₄).
Most phosphates, carbonates, chromates, and sulfides are insoluble.
Most hydroxides are insoluble (exceptions: NaOH, KOH, Ba(OH)₂).
Write the balanced molecular equation.
Split all soluble ionic compounds (those marked "aq") into their ions.
Identify spectator ions (those appearing identically on both sides).
Remove spectator ions. What remains is the net ionic equation.
When light passes through a solution, some wavelengths are absorbed. The colour we see is the wavelength reflected or transmitted.
A UV-vis spectrometer works by sending light through a sample. A monochromator selects specific wavelengths; a detector measures how much light comes through.
Beer's law: more concentrated solutions absorb more light (less light is transmitted). This relationship is used to determine unknown concentrations experimentally.
Stoichiometry is how chemical engineers scale reactions from the lab bench to the factory floor. Limiting reactant analysis ensures raw materials are used efficiently. Molarity and dilution calculations are routine in any laboratory setting, from clinical blood tests to wastewater treatment. Redox chemistry underpins batteries, corrosion, and electroplating.
Molar mass: numerically equal to atomic/molecular mass in AMU, expressed in g/mol
Moles from mass: n = mass (g) / molar mass (g/mol)
Molarity: M = moles of solute / litres of solution
Dilution: M₁V₁ = M₂V₂
Percent composition: (mass of element in 1 mol of compound / molar mass of compound) × 100
Oxidation number sum: all oxidation numbers in a neutral compound sum to 0; in a polyatomic ion, they sum to the ion's charge
Students confuse subscripts with coefficients. Subscripts are fixed by the compound's formula and must not be changed when balancing. Only coefficients (the numbers in front) can be adjusted.
The oxidation number of carbon in CO₃²⁻ is often miscalculated. Students sometimes forget that the overall charge of the ion (-2) must be accounted for: C = +4 (not +6, which would be the case if the ion were neutral).
"Limiting reactant" does not mean the reactant present in the smallest amount by mass. It is the reactant that produces the least product based on its mole ratio. A heavier reactant can still be in excess.
Students frequently forget that molarity uses litres, not millilitres, in the denominator. Always convert to litres before dividing.
⚠️ Mass-to-mass conversions are the most tested calculation type. Practise the four-step bridge (grams → moles → mole ratio → grams) until it is automatic.
⚠️ Limiting reactant problems appear on nearly every exam. The most common error is skipping the mole-ratio comparison and assuming the lighter reactant is limiting.
⚠️ Solubility rules must be memorised. Expect a question asking you to predict whether a precipitate forms when two solutions are mixed.
⚠️ Net ionic equation questions test whether you can identify and remove spectator ions. Practise writing all three forms (molecular, full ionic, net ionic) for double displacement reactions.
⚠️ Oxidation number assignments are commonly tested in isolation and as part of redox balancing.
True or False: In a balanced equation, you may change subscripts to balance atoms.
Fill in the blank: One mole of any substance contains ______ particles.
True or False: The limiting reactant is always the one present in the smallest mass.
Fill in the blank: Molarity equals moles of solute divided by ______ of solution.
True or False: Spectator ions appear in the net ionic equation.
Answers: 1. False (only coefficients may be changed). 2. 6.022 × 10²³. 3. False (it depends on mole ratios, not mass). 4. Litres. 5. False (they are removed).
Q: You have 10.0 g of hydrogen (H₂) and 80.0 g of oxygen (O₂). Which is the limiting reactant for the formation of water (2H₂ + O₂ → 2H₂O)?
A: Moles of H₂ = 10.0 / 2.016 = 4.96 mol. Moles of O₂ = 80.0 / 32.00 = 2.50 mol. The equation requires 2 mol H₂ per 1 mol O₂, so 4.96 mol H₂ would need 2.48 mol O₂. You have 2.50 mol O₂, which is just enough. H₂ is the limiting reactant (it runs out first relative to the ratio).
Q: What is the oxidation number of sulfur in SO₄²⁻?
A: Each oxygen is -2, so four oxygens contribute -8. The ion's charge is -2. Therefore S + (-8) = -2, so S = +6.
Q: You dissolve 5.85 g of NaCl (molar mass 58.44 g/mol) in enough water to make 500 mL of solution. What is the molarity?
A: Moles of NaCl = 5.85 / 58.44 = 0.100 mol. Volume = 500 mL = 0.500 L. Molarity = 0.100 / 0.500 = 0.200 M.
Q: When solutions of silver nitrate (AgNO₃) and sodium chloride (NaCl) are mixed, what is the net ionic equation?
A: Ag⁺(aq) + Cl⁻(aq) → AgCl(s). The sodium and nitrate ions are spectators and are removed.
Q: If you dilute 25.0 mL of 6.00 M HCl to a final volume of 500 mL, what is the final molarity?
A: M₁V₁ = M₂V₂. (6.00)(0.0250) = M₂(0.500). M₂ = 0.300 M.
Balancing equations and mole ratios are the arithmetic backbone of every quantitative topic that follows in general chemistry, including thermochemistry, gas laws, and equilibrium. Redox chemistry connects to electrochemistry (batteries and galvanic cells) later in the course. Solution chemistry and molarity reappear in acid-base chemistry and equilibrium constant expressions. Beer's law connects to analytical chemistry and any lab where you measure concentration by absorbance.
Stoichiometry, mole, Avogadro's number, molar mass, molecular mass, formula mass, AMU, balancing equations, stoichiometric coefficients, mole ratio, mass-to-mass conversion, limiting reactant, excess reactant, percent composition, oxidation, reduction, redox, OIL RIG, oxidation number, molarity, concentration, dilution, M1V1 M2V2, solvent, solute, electrolyte, nonelectrolyte, strong electrolyte, weak electrolyte, double displacement, precipitation, precipitate, solubility rules, net ionic equation, spectator ions, Beer's law, UV-vis, absorbance, CHEM 111, general chemistry, Purdue