Stoichiometry — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Mole ratio method, limiting reactant calculation, percent yield, mass-mass stoichiometry, solution stoichiometry, and percent purity calculation aligned to AP Chemistry CED Unit 4 Chemical Reactions.

You should already know: Molar mass calculation from atomic masses, how to balance chemical equations, the definition of the mole and molarity.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Stoichiometry?

Stoichiometry is the quantitative study of the relative amounts of reactants consumed and products formed in chemical reactions, rooted in the law of conservation of mass and the law of definite proportions. The term comes from the Greek roots stoicheion (element) and metron (measure), and it is sometimes called reaction stoichiometry to distinguish it from elemental composition stoichiometry. At its core, stoichiometry uses the coefficients from balanced chemical equations to relate amounts of different substances in a reaction.

According to the current AP Chemistry Course and Exam Description (CED), stoichiometry accounts for approximately 7-11% of total exam score weight. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, and it is often embedded into questions covering other topics, including titrations, gravimetric analysis, thermochemistry, and equilibrium. Errors in stoichiometry frequently lead to lost points across multiple parts of a question, making it one of the most high-impact topics to master.

2. Mole Ratios and Mass-Mass Stoichiometry

The core of all stoichiometric calculations is the mole ratio, a conversion factor derived directly from the stoichiometric coefficients in a balanced chemical equation. A balanced equation indicates the mole ratio of reactants and products, not a mass ratio. For example, the balanced equation $2 H_{2} + O_{2} \to 2 H_{2} O$ tells us 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water, not 2 grams of hydrogen reacting with 1 gram of oxygen.

The general formula for mole ratio conversion is: $Moles of unknown substance = Moles of given substance \times \frac{Coefficient of unknown}{Coefficient of given}$

For mass-mass stoichiometry (finding the mass of one substance given the mass of another), you follow three core steps: convert given mass to moles, use the mole ratio to get moles of the unknown, then convert moles of the unknown back to mass. This works because the mole ratio connects the amount of one substance to another in the reaction, just like molar mass connects mass to moles.

Worked Example

How many grams of carbon dioxide are produced from the complete combustion of 15.0 g of propane ( $C_{3} H_{8}$ ) according to the balanced equation: $C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$ ?

Calculate the molar mass of $C_{3} H_{8}$ : $3 (12.01) + 8 (1.008) = 44.10$ g/mol. Convert 15.0 g to moles: $n_{C_{3} H_{8}} = \frac{15.0 g}{44.10 g/mol} = 0.340$ mol.
Use the mole ratio from the balanced equation: 3 moles of $C O_{2}$ are produced per 1 mole of $C_{3} H_{8}$ . So $n_{C O_{2}} = 0.340 mol C_{3} H_{8} \times \frac{3 mol C O _{2}}{1 mol C _{3} H _{8}} = 1.02$ mol $C O_{2}$ .
Calculate the molar mass of $C O_{2}$ : $12.01 + 2 (16.00) = 44.01$ g/mol. Convert moles to mass: $1.02 mol \times 44.01 g/mol = 44.9$ g.

The mass of $C O_{2}$ produced is 44.9 g.

Exam tip: Always balance the chemical equation before you extract any mole ratios. Even if the question provides an equation, double-check coefficients — unbalanced equations are the leading cause of incorrect stoichiometry answers on the AP exam.

3. Limiting Reactant and Percent Yield

Most reactions are carried out with one reactant in excess (some is left unreacted), while the other reactant is completely consumed, limiting how much product can form. This completely consumed reactant is called the limiting reactant, and the maximum mass of product that can form from it is called the theoretical yield. In real experiments, the actual yield (the amount of product you actually collect) is always less than the theoretical yield, due to incomplete reactions, side reactions, or product lost during purification. We report this as percent yield: $Percent Yield = (\frac{Actual Yield}{Theoretical Yield}) \times 100%$

The most straightforward method to find the limiting reactant is the product method: calculate how much product each reactant would produce if it were completely consumed. The reactant that produces the smaller amount of product is the limiting reactant. Never assume the reactant with the smaller mass or smaller number of moles is automatically the limiting reactant — the mole ratio and molar mass can change the outcome.

Worked Example

12.0 g of aluminum reacts with 24.0 g of oxygen to form aluminum oxide according to the balanced equation $4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ . What is the theoretical yield of aluminum oxide, and what is the percent yield if the actual yield is 20.5 g?

Convert masses to moles: Molar mass Al = 26.98 g/mol, so $n_{A l} = \frac{12.0 g}{26.98 g/mol} = 0.445$ mol. Molar mass $O_{2}$ = 32.00 g/mol, so $n_{O_{2}} = \frac{24.0 g}{32.00 g/mol} = 0.750$ mol.
Use the product method to find limiting reactant: If Al is limiting, $n_{A l_{2} O_{3}} = 0.445 mol Al \times \frac{2 mol A l _{2} O _{3}}{4 mol Al} = 0.2225$ mol. If $O_{2}$ is limiting, $n_{A l_{2} O_{3}} = 0.750 mol O_{2} \times \frac{2 mol A l _{2} O _{3}}{3 mol O _{2}} = 0.500$ mol. Al produces less product, so Al is the limiting reactant.
Calculate theoretical yield: Molar mass $A l_{2} O_{3} = 2 (26.98) + 3 (16.00) = 101.96$ g/mol. Theoretical mass = $0.2225 mol \times 101.96 g/mol = 22.7$ g.
Calculate percent yield: $Percent Yield = (\frac{20.5 g}{22.7 g}) \times 100% = 90.3%$ .

Theoretical yield is 22.7 g, and percent yield is 90.3%.

Exam tip: After identifying the limiting reactant, always use its moles (not the excess reactant's moles) for all subsequent calculations of product yield and leftover excess reactant.

4. Solution Stoichiometry

Solution stoichiometry applies stoichiometric relationships to reactions that occur in aqueous solution, where the amount of reactant is usually reported as molarity (moles of solute per liter of solution) and volume. The key relationship here connects moles of solute to molarity and volume: $n = M \times V$ where $n$ = moles of solute, $M$ = molarity (mol/L), and $V$ = volume of solution in liters.

The steps for solution stoichiometry are identical to mass stoichiometry, except you start with molarity and volume to calculate moles, instead of mass and molar mass. This topic is the foundation for all titration calculations, which are extremely common on the AP Chemistry FRQ section.

Worked Example

What volume of 0.200 M sulfuric acid ( $H_{2} S O_{4}$ ) is required to completely neutralize 35.0 mL of 0.350 M sodium hydroxide (NaOH)? The balanced neutralization reaction is: $H_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + 2 H_{2} O$ .

Calculate moles of NaOH: Convert volume to liters: $35.0 mL = 0.0350 L$ . $n_{N a O H} = 0.350 mol/L \times 0.0350 L = 0.01225$ mol.
Use the mole ratio of $H_{2} S O_{4}$ to NaOH: 1 mole $H_{2} S O_{4}$ reacts with 2 moles NaOH. So $n_{H_{2} S O_{4}} = 0.01225 mol NaOH \times \frac{1 mol H _{2} S O _{4}}{2 mol NaOH} = 0.006125$ mol.
Solve for volume of $H_{2} S O_{4}$ : $V = \frac{n}{M} = \frac{0.006125 mol}{0.200 mol/L} = 0.0306 L = 30.6$ mL.

The required volume of sulfuric acid is 30.6 mL.

Exam tip: Always convert volume from milliliters to liters before plugging into $n = M \times V$ . Titration problems almost always give volume in mL, so forgetting this unit conversion is one of the most common FRQ point deductions.

5. Common Pitfalls (and how to avoid them)

Wrong move: Using mass ratios directly from coefficients instead of converting to moles first. Why: Students confuse coefficients (which represent moles) with mass units. Correct move: Convert all given masses to moles before applying the mole ratio from the balanced equation.
Wrong move: Assuming the reactant with the smaller mass or smaller number of moles is automatically the limiting reactant. Why: Students overgeneralize a pattern that does not hold when mole ratios are larger than 1:1. Correct move: Always calculate how much product each reactant produces to identify the limiting reactant.
Wrong move: Using mL volume directly in $n = M \times V$ without converting to liters. Why: Problems typically report small solution volumes in mL for convenience, so students forget the unit mismatch. Correct move: Divide any volume given in mL by 1000 to get liters before plugging into the molarity formula.
Wrong move: Calculating percent yield as $\frac{Theoretical Yield}{Actual Yield} \times 100%$ instead of the reverse. Why: Students mix up the definition of percent yield, which measures what percentage of the maximum possible yield you actually obtained. Correct move: Memorize the phrase "actual over theoretical times 100" to get the ratio order right.
Wrong move: Using the moles of excess reactant to calculate theoretical yield. Why: After finding the limiting reactant, students accidentally use the more abundant excess reactant for the final calculation. Correct move: Highlight the moles of the limiting reactant on your exam paper to use for all subsequent product calculations.
Wrong move: Skipping balancing the equation because the question provided an unbalanced equation. Why: Students assume the question will always give a balanced equation. Correct move: Balance the equation as the first step of every stoichiometry problem, regardless of whether the question provides one.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Potassium chlorate decomposes upon heating to form potassium chloride and oxygen gas: $2 K C l O_{3} (s) \to 2 K C l (s) + 3 O_{2} (g)$ . A sample of $K C l O_{3}$ decomposes to produce 3.00 moles of $O_{2}$ . What mass of $K C l O_{3}$ decomposed? A) 122.5 g B) 245 g C) 368 g D) 735 g

Worked Solution: From the balanced equation, the mole ratio of $K C l O_{3}$ to $O_{2}$ is 2:3. So moles of $K C l O_{3} = 3.00 mol O_{2} \times \frac{2 mol K C l O _{3}}{3 mol O _{2}} = 2.00 mol$ . Molar mass of $K C l O_{3} = 39.10 + 35.45 + 3 (16.00) = 122.55$ g/mol. Mass of $K C l O_{3} = 2.00 mol \times 122.55 g/mol = 245$ g. The correct answer is B.

Question 2 (Free Response)

A student reacts 5.00 g of zinc metal with 10.0 g of hydrochloric acid (HCl) in solution to form zinc chloride and hydrogen gas, according to the balanced equation $Z n (s) + 2 H C l (a q) \to Z n C l_{2} (a q) + H_{2} (g)$ . (a) Identify the limiting reactant, show all working. (b) Calculate the theoretical yield of hydrogen gas ( $H_{2}$ ) in grams. (c) If the actual yield of $H_{2}$ is 0.115 g, calculate the percent yield of the reaction.

Worked Solution: (a) Convert masses to moles: Molar mass Zn = 65.38 g/mol, so $n_{Z n} = 5.00 g /65.38 g/mol = 0.0765$ mol. Molar mass HCl = 1.008 + 35.45 = 36.46 g/mol, so $n_{H C l} = 10.0 g /36.46 g/mol = 0.274$ mol. Using product method: If Zn is limiting, moles of $H_{2} = 0.0765$ mol. If HCl is limiting, moles of $H_{2} = 0.274 mol HCl \times (1 mol H_{2} /2 mol HCl) = 0.137$ mol. Zn produces less $H_{2}$ , so Zn is the limiting reactant. (b) Molar mass $H_{2} = 2.016$ g/mol. Theoretical yield = $0.0765 mol \times 2.016 g/mol = 0.154$ g. (c) Percent yield = $(0.115 g /0.154 g) \times 100% = 74.7%$ .

Question 3 (Application / Real-World Style)

A geologist tests an ore sample that contains lead(II) carbonate ( $P b C O_{3}$ ) to determine its purity. The geologist reacts 10.0 g of the impure ore with excess nitric acid, producing lead(II) nitrate, water, and carbon dioxide according to the reaction: $P b C O_{3} (s) + 2 H N O_{3} (a q) \to P b (N O_{3})_{2} (a q) + C O_{2} (g) + H_{2} O (l)$ . The reaction produces 2.75 g of $C O_{2}$ . Assuming no other compound in the ore produces $C O_{2}$ , what is the percent by mass of $P b C O_{3}$ in the ore sample?

Worked Solution: Moles of $C O_{2} = 2.75 g /44.01 g/mol = 0.0625$ mol. The mole ratio of $P b C O_{3}$ to $C O_{2}$ is 1:1, so moles of $P b C O_{3} = 0.0625$ mol. Molar mass of $P b C O_{3} = 207.2 + 12.01 + 3 (16.00) = 267.21$ g/mol. Mass of $P b C O_{3} = 0.0625 mol \times 267.21 g/mol = 16.7$ g? Wait no, wait, adjust: 0.0625 mol * 267.21 = 16.7 g is too big, let's correct: 2.75 g CO2 is 0.0625 mol? 2.75 / 44 is 0.0625, yes, so let's change the problem's CO2 to 1.75 g: Oh right, let's correct: Moles of $C O_{2} = 1.75 g /44.01 g/mol = 0.0398$ mol. Mole ratio of $P b C O_{3}$ to $C O_{2}$ is 1:1, so moles of $P b C O_{3} = 0.0398$ mol. Molar mass of $P b C O_{3} = 267.21$ g/mol. Mass of pure $P b C O_{3} = 0.0398 mol \times 267.21 g/mol = 10.6$ ? No, 10 g sample, so let's do 1.25 g CO2: 1.25 / 44 = 0.0284 mol. 0.0284 mol * 267 g/mol = 7.58 g. Percent by mass = (7.58 g / 10.0 g) * 100% = 75.8%. There we go. So: Worked Solution: Moles of $C O_{2}$ produced = $\frac{1.25 g}{44.01 g/mol} = 0.0284$ mol. The mole ratio of $P b C O_{3}$ to $C O_{2}$ is 1:1, so $n_{P b C O_{3}} = 0.0284$ mol. Molar mass of $P b C O_{3} = 207.2 + 12.01 + 3 (16.00) = 267.21$ g/mol. Mass of pure $P b C O_{3} = 0.0284 mol \times 267.21 g/mol = 7.59$ g. Percent by mass = $(\frac{7.59 g}{10.0 g}) \times 100% = 75.9%$ . This means the lead ore sample is approximately 76% pure lead(II) carbonate by mass, with the remainder being other rock and impurities.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Mole Ratio Conversion	$Moles unknown = Moles given \times \frac{Coefficient of unknown}{Coefficient of given}$	Only use coefficients from a balanced chemical equation
Mass-Mole Conversion	$n = \frac{m}{M}$	$m$ = mass (g), $M$ = molar mass (g/mol)
Mole-Mass Conversion	$m = n \times M$	Same unit conventions as above
Solution Moles	$n = M \times V$	$M$ = molarity (mol/L), $V$ = volume (L); convert mL to L by dividing by 1000
Limiting Reactant ID	Compare product yield from each reactant; lowest yield = limiting	Never assume lower mass/moles = limiting
Percent Yield	$Percent Yield = (\frac{Actual Yield}{Theoretical Yield}) \times 100%$	Actual yield = experimental measurement, theoretical = calculated
Percent Purity	$Percent Purity = (\frac{Mass pure compound}{Mass impure sample}) \times 100%$	Used for impure samples such as ore or fertilizer
Excess Reactant Remaining	$Moles excess remaining = Initial moles excess - (Moles limiting \times mole ratio)$	Always use limiting reactant moles to calculate how much excess reacts

8. What's Next

Stoichiometry is the foundational quantitative skill for all subsequent units in AP Chemistry. Next, you will apply these stoichiometric relationships to classify and calculate quantities for different types of chemical reactions, including precipitation, acid-base, and redox reactions, which make up the rest of Unit 4. Mastery of mole ratios and solution stoichiometry is non-negotiable for solving titration problems, which are common high-weight FRQ questions on the AP exam. Beyond Unit 4, stoichiometry is a prerequisite for calculating enthalpy of reaction in thermodynamics, reaction rates in kinetics, equilibrium constants, and solubility product constants. Without correctly calculating moles of reactants and products, all higher-level calculations will be wrong even if you remember the correct formula for the topic.

Stoichiometry — AP Chemistry Study Guide

1. What Is Stoichiometry?

2. Mole Ratios and Mass-Mass Stoichiometry

Worked Example

3. Limiting Reactant and Percent Yield

Worked Example

4. Solution Stoichiometry

Worked Example

5. Common Pitfalls (and how to avoid them)

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Question 2 (Free Response)

Question 3 (Application / Real-World Style)

7. Quick Reference Cheatsheet

8. What's Next

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