4.5 Stoichiometry

Stoichiometry uses the coefficients in a balanced chemical equation as mole ratios to calculate how much product forms from a given amount of reactant, or how much reactant you need to make a target amount of product. The whole process runs on dimensional analysis: convert your known amount to moles, multiply by the right mole ratio, then convert to the unit the question wants. For AP Chemistry, keep units attached through every conversion.

AP Chem Stoichiometry Basics

AP Chem stoichiometry uses a balanced chemical equation to connect the amounts of reactants and products. Because atoms are conserved in a chemical process, coefficients in the balanced equation give mole ratios that you can use in calculations.

The exam move is almost always dimensional analysis. Convert the given amount to moles, use the coefficient ratio to switch substances, then convert to the unit the question asks for. Stoichiometry can also combine with the ideal gas law or molarity for gas and solution problems.

Why This Matters for the AP Chemistry Exam

Stoichiometry is one of the most reused skills in the entire course. You explain how the amounts of reactants and products change based on the balanced equation, and you do quantitative calculations that connect mass, moles, gas volume, and solution concentration.

This skill shows up across both multiple-choice and free-response work. You will combine stoichiometry with the ideal gas law (PV = nRT) and with molarity to study gases and solutions, and it sets up later topics like titration, limiting reactant analysis, and percent yield. If you get comfortable with mole ratios now, the math in later units feels much more familiar.

Key Takeaways

• Atoms are conserved in a chemical change, so you can calculate product amounts from known reactant amounts and vice versa.
• The coefficients in a balanced equation give the mole ratios that link any two substances in the reaction.
• Most stoichiometry problems follow the same path: convert the known to moles, multiply by a mole ratio, then convert to the target unit.
• Molar mass (g/mol) converts between grams and moles; Avogadro's number (6.022 x 10^23) converts between moles and particles.
• At STP, one mole of an ideal gas occupies 22.4 L, which lets you convert between moles and gas volume.
• Stoichiometry combines with the ideal gas law and molarity, so the same logic works for gases and solutions.

Core Concepts to Know

Stoichiometry is the quantitative side of chemical reactions. Earlier units covered moles, molar mass, molar volume, molarity, and Avogadro's number. Stoichiometry ties these together by using mole ratios from a balanced equation to convert from what you know to what you want.

A few definitions to keep straight:

1. Balanced chemical equation: A written representation of a reaction with reactants on the left and products on the right, where each element has equal atoms on both sides. This is required because mass and charge are conserved.
2. Mole: A unit for the amount of a substance. One mole equals Avogadro's number, 6.022 x 10^23 particles (atoms, molecules, ions, etc.).
3. Stoichiometric coefficients: The numbers in front of each formula in a balanced equation. They give the relative amounts of reactants and products, so they act as proportionality factors in calculations.
4. Stoichiometric calculations: Using the balanced equation and the mole concept to find required or produced amounts. These rest on conservation of mass and atoms.

Mole Ratios

A mole ratio compares the amounts of two substances in a reaction, in moles. The coefficients of the balanced equation give you these ratios directly. To use them well, know these conversions cold:

• At STP, one mole of an ideal gas fills 22.4 L. This is the molar volume, found on the AP Chemistry reference sheet.
• One mole = 6.022 x 10^23 particles. Avogadro's number is also on the reference sheet.
• Molar mass tells you the grams in one mole. Each element's molar mass is on the periodic table. Hydrogen, for example, is 1.008 grams per mole.

You read mole ratios straight from the balanced equation. For the combustion of ethanol below, 1 mole of C2H5OH and 3 moles of O2 produce 2 moles of CO2 and 3 moles of H2O. A sample mole ratio is 1 mole O2 to 2 moles CO2. Writing it as 2 moles CO2 to 1 mole O2 means the same thing, just flipped.

Worked Stoichiometry Problems

Question 1

How many moles of potassium metal are required to fully react with 11.6 moles of water?

Step 1. Write and balance the equation:

2K (s) + 2H₂O (l) → 2KOH (aq) + H₂ (g)

Step 2. Identify the known: 11.6 moles of water.

Step 3. The known is already in moles, so write the mole ratio that cancels moles of H₂O and brings in moles of K. From the equation, 2 moles H₂O react with 2 moles K:

$11.6 \text{ mol H}_2\text{O} \times \frac{2 \text{ mol K}}{2 \text{ mol H}_2\text{O}} = 11.6 \text{ mol K}$

Track your units the whole way. Writing out the units and the substance each value belongs to keeps you from flipping a ratio by mistake. Since the target unit is moles of K, every other unit must cancel.

Question 2

If you have 105.2 g of ethanol (C₂H₅OH), what is the maximum volume of carbon dioxide that can form at STP?

Step 1. Write and balance the equation:

C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (l)

Step 2. Identify the known: 105.2 g of ethanol.

Step 3. The known is in grams, so convert to moles using molar mass. For C₂H₅OH:

2(12.01 g) + 6(1.008 g) + 16.00 g = 46.07 g/mol

$105.2 \text{ g C}_2\text{H}_5\text{OH} \times \frac{1 \text{ mol C}_2\text{H}_5\text{OH}}{46.07 \text{ g C}_2\text{H}_5\text{OH}} = 2.284 \text{ mol C}_2\text{H}_5\text{OH}$

Step 4. You want CO₂, not ethanol, so use the mole ratio. From the equation, 1 mole C₂H₅OH makes 2 moles CO₂:

$2.284 \text{ mol C}_2\text{H}_5\text{OH} \times \frac{2 \text{ mol CO}_2}{1 \text{ mol C}_2\text{H}_5\text{OH}} = 4.568 \text{ mol CO}_2$

Step 5. You have moles of CO₂ but want volume. Since the reaction is at STP, use the molar volume:

$4.568 \text{ mol CO}_2 \times \frac{22.4 \text{ L}}{1 \text{ mol CO}_2} = 102 \text{ L CO}_2$

With 105.2 g of ethanol, you can make about 102 L of CO₂ at STP.

Stoichiometry also pairs with the ideal gas law (PV = nRT) and with molarity (moles per liter of solution, written M). Those let you handle non-STP gases and solution problems with the same mole-ratio logic.

General Steps

A reliable routine for any stoichiometry problem:

1. Write and balance the chemical equation if it is not given.
2. Identify the known measurement.
3. If the known is in grams, convert it to moles using molar mass.
4. Multiply by a mole ratio using the coefficients from the equation.
5. Repeat steps 3 and 4 as needed to reach the target substance.
6. Convert to the final unit, then check that your units cancel to leave exactly what the question asks.

Try It Yourself

The following reaction occurs at STP:

Br₂ (g) + 5F₂ (g) → 2BrF₅ (g)

How many particles of BrF₅ are produced from 160.0 g of Br₂?

Answer

Step 1. Convert grams of Br₂ to moles using its molar mass (Br₂ = 2 × 79.90 = 159.8 g/mol):

$160.0 \text{ g Br}_2 \times \frac{1 \text{ mol Br}_2}{159.8 \text{ g Br}_2} = 1.001 \text{ mol Br}_2$

Step 2. Use the mole ratio of Br₂ to BrF₅ from the coefficients:

$1.001 \text{ mol Br}_2 \times \frac{2 \text{ mol BrF}_5}{1 \text{ mol Br}_2} = 2.002 \text{ mol BrF}_5$

Step 3. Use Avogadro's number to convert moles to particles:

$2.002 \text{ mol BrF}_5 \times \frac{6.022 \times 10^{23} \text{ particles}}{1 \text{ mol BrF}_5} = 1.206 \times 10^{24} \text{ particles}$

How to Use This on the AP Chemistry Exam

Problem Solving

• Always start from a balanced equation. An unbalanced equation gives wrong mole ratios and a wrong answer.
• Build the conversion as a chain of fractions so every unit cancels except the one you want. This is dimensional analysis, and it catches setup errors fast.
• Keep extra digits during the calculation and round only at the end to avoid rounding error.

Free Response

• Show each conversion step clearly, including the mole ratio you used. Partial credit follows clear setup.
• Label units on every quantity. Graders look for correct units, not just a number.
• When a problem gives gas conditions other than STP, reach for PV = nRT instead of the 22.4 L molar volume.

Common Trap

• Comparing mass to mass instead of converting to moles first. Coefficients are mole ratios, not mass ratios, so you must be in moles before you apply them.

Common Misconceptions

• Coefficients are mass ratios. They are mole ratios. A 2:1 coefficient ratio does not mean a 2:1 mass ratio, because different substances have different molar masses.
• You can skip balancing. Mole ratios come from the balanced coefficients. If the equation is not balanced, the ratios are wrong.
• 22.4 L works for any gas problem. That molar volume only applies to an ideal gas at STP. For other temperatures or pressures, use PV = nRT.
• Avogadro's number converts grams to particles directly. It converts moles to particles. To go from grams to particles, convert grams to moles with molar mass first, then use Avogadro's number.
• More starting reactant always means more product without limits. The amount of product is set by the reactant amounts and the mole ratios, which is the idea you build on when you study limiting reactants.

Related AP Chemistry Guides

• Unit 4 Overview: Chemical Reactions
• 4.1 Introduction to Reactions
• 4.2 Net Ionic Equations
• 4.6 Introduction to Titration
• 4.3 Representations of Reactions
• 4.7 Types of Chemical Reactions