Course: CHEM 101, Michigan State University | Source: Compounds and Stoichiometry lecture notes
Difficulty: Introductory | Prerequisites: Molecules, Moles, and Molecular Weight notes (Part 1 of this set)
Once you can count atoms and convert between grams and moles, the next question is: what happens when substances react? This set of notes covers how compounds are represented (empirical vs. molecular formulas), the major categories of chemical reactions, how to balance equations, and how to use stoichiometry to predict quantities. These skills underpin nearly every problem set and exam in general chemistry.
Chemical formulas come in two flavours: empirical (simplest ratio) and molecular (actual atom count). Reactions fall into a handful of types (combination, decomposition, combustion, single-displacement, double-displacement, neutralisation). Balancing equations and using mole ratios lets you calculate exactly how much product forms, identify the limiting reagent, and find percent yield.
Law of Constant Composition (Law of Definite Proportions)
Any pure sample of a given compound always contains the same elements in the same mass ratio. In simple terms, water is always 11.2% hydrogen and 88.8% oxygen by mass, no matter where you find it.
Empirical formula
The simplest whole-number ratio of atoms in a compound. Think of it as the reduced version of the molecular formula.
Molecular formula
The actual number of atoms of each element in one molecule. It is always a whole-number multiple of the empirical formula.
Percent composition
The percentage by mass of each element in a compound.
Combination reaction (synthesis)
Two or more reactants combine to form a single product. In simple terms, small pieces join into one bigger piece.
Decomposition reaction
A single compound breaks apart into two or more products. The reverse of a combination reaction.
Combustion reaction
A substance (usually a hydrocarbon) reacts with O₂ to produce CO₂ and H₂O. Think of it as burning fuel.
Single-displacement reaction
An atom or ion in a compound is replaced by an atom or ion of a different element. Also called a redox reaction in many contexts.
Double-displacement (metathesis) reaction
Ions from two compounds swap partners to form two new compounds. Driven by formation of a precipitate, a gas, or a weak electrolyte.
Neutralisation reaction
A specific type of double-displacement reaction where an acid reacts with a base to form a salt and (often) water.
Limiting reagent (limiting reactant)
The reactant that runs out first, determining the maximum amount of product that can form.
Percent yield
The ratio of actual product obtained to the theoretical maximum, expressed as a percentage.
The Law of Constant Composition guarantees that a compound's elemental ratio is fixed. Every sample of water has two H atoms for every O atom.
Empirical formula gives the simplest ratio. For glucose the empirical formula is CH₂O.
Molecular formula gives the true count. Glucose's molecular formula is C₆H₁₂O₆ (six times the empirical formula).
Ionic compounds only have empirical formulas, because they form extended lattices rather than discrete molecules.
Percent composition = (mass of element in one formula unit / molar mass of compound) x 100%.
Combination (synthesis)
Two or more reactants form one product.
Example: 2 Mg + O₂ --> 2 MgO
Decomposition
One compound breaks into two or more products. The reverse of combination.
Example: 2 H₂O --> 2 H₂ + O₂
Combustion
A fuel (typically a hydrocarbon) reacts with O₂ to produce CO₂ and H₂O.
Example: CH₄ + 2 O₂ --> CO₂ + 2 H₂O
Single-displacement
One element replaces another in a compound. These are redox reactions.
Example: Cu + AgNO₃ --> Ag + CuNO₃
Double-displacement (metathesis)
Two compounds exchange ions to form two new compounds.
Occurs when a product precipitates, forms a gas, or is a weak electrolyte.
Example: CaCl₂ + 2 AgNO₃ --> Ca(NO₃)₂ + 2 AgCl
Neutralisation
A special case of double-displacement: acid + base --> salt + water.
Example: HCl + NaOH --> NaCl + H₂O
Atoms must be conserved: the same number of each element appears on both sides.
Balance one element at a time, leaving H and O for last.
Adjust coefficients (the numbers in front of formulas), never subscripts.
Use the balanced equation's coefficients as mole ratios to convert between substances.
At STP, 1 mol of any ideal gas occupies 22.4 L.
1 mol of any substance = 6.02 x 10²³ particles.
Convert each reactant's given quantity to moles.
Use the mole ratio to find which reactant produces the least product. That reactant is the limiting reagent.
The other reactant is in excess.
Theoretical yield is the maximum product predicted by stoichiometry (based on the limiting reagent).
Actual yield is what you collect in the lab.
% yield = (actual / theoretical) x 100%
\text{Percent composition of element} = \frac{\text{mass of element in 1 mol of compound}}{\text{molar mass of compound}} \times 100\%\text{Molecular formula} = n \times \text{Empirical formula}, \quad n = \frac{\text{molar mass}}{\text{empirical formula mass}}\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%\text{Volume of gas at STP} = n \times 22.4 \text{ L/mol}Combustion stoichiometry is how automotive engineers tune air-fuel ratios in engines: too little oxygen means incomplete combustion and wasted fuel; too much means excess emissions. Percent yield matters in pharmaceutical manufacturing, where even a small improvement in yield can save millions at production scale.
Students often confuse empirical and molecular formulas. The empirical formula CH₂O could represent formaldehyde (CH₂O), acetic acid (C₂H₄O₂), or glucose (C₆H₁₂O₆). You need the molar mass to determine which.
Balancing equations by changing subscripts is wrong. Subscripts define the compound. Only coefficients change.
Students sometimes assume the reactant present in the smallest mass is the limiting reagent. It is the reactant that produces the fewest moles of product, which requires converting to moles first.
A percent yield over 100% does not mean you created matter. It means there is an experimental error (impure product, incomplete drying, or a measurement mistake).
⚠️ Identifying reaction types is a guaranteed exam topic. Be able to classify a reaction from its equation alone.
⚠️ Limiting reagent problems appear on nearly every general chemistry exam. Practise the full workflow: grams to moles, apply ratio, compare.
⚠️ Balancing equations is a prerequisite skill, not an optional one. If you cannot balance, you cannot do stoichiometry.
⚠️ Percent composition and empirical formula determination are common calculation problems. Know how to go from mass percentages to the empirical formula and back.
True or False: The molecular formula is always a whole-number multiple of the empirical formula. (True)
Fill in the blank: In a combustion reaction, a hydrocarbon reacts with ______ to produce CO₂ and H₂O. (O₂)
True or False: In a single-displacement reaction, two compounds exchange ions. (False, that is double-displacement)
Fill in the blank: The limiting reagent is the reactant that produces the ______ amount of product. (least)
True or False: Percent yield can never exceed 100% in a correctly performed experiment. (True)
Q: What is the empirical formula of a compound that is 40.0% C, 6.7% H, and 53.3% O by mass?
A: Assume 100 g. C: 40.0/12.01 = 3.33 mol; H: 6.7/1.008 = 6.65 mol; O: 53.3/16.00 = 3.33 mol. Ratio = 1:2:1. Empirical formula: CH₂O.
Q: Balance the equation: Fe + O₂ --> Fe₂O₃
A: 4 Fe + 3 O₂ --> 2 Fe₂O₃.
Q: If 10.0 g of hydrogen reacts with 10.0 g of oxygen to form water, which is the limiting reagent?
A: 2 H₂ + O₂ --> 2 H₂O. Moles H₂ = 10.0/2.016 = 4.96. Moles O₂ = 10.0/32.00 = 0.3125. H₂ needs 4.96/2 = 2.48 mol O₂, but only 0.3125 is available. O₂ is the limiting reagent.
Q: A reaction has a theoretical yield of 25.0 g and an actual yield of 18.7 g. What is the percent yield?
A: (18.7 / 25.0) x 100% = 74.8%.
Q: Classify: 2 KClO₃ --> 2 KCl + 3 O₂
A: Decomposition. One reactant breaks into two products.
Stoichiometry is the bridge between this chapter and solution chemistry: once you can find moles from a balanced equation, calculating molarity and performing dilution problems follows directly. Reaction types reappear in electrochemistry (redox) and acid-base equilibrium (neutralisation). Combustion analysis is a standard method for determining empirical formulas of organic compounds, linking this material to organic chemistry.
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