Chemistry 2e · Stoichiometry of Chemical Reactions

Reaction Stoichiometry

9 min read

Numerical values (molar masses) are commonly taught reference values; verify against current sources before relying on them in assessments.

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
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In 30 seconds

is the quantitative study of the amounts of substances consumed and produced in a chemical reaction — the arithmetic of chemical recipes. Its foundation is the balanced equation from the first topic of this chapter: the coefficients are mole ratios. In

$N 2 (g) + 3 H 2 (g) \to 2 NH 3 (g)$

the coefficients say that 1 mol of N₂ reacts with 3 mol of H₂ to produce 2 mol of NH₃ — a fixed ratio that holds no matter how much material you start with. The stoichiometric plan uses the mole as the universal currency:

$mass A \to moles A \to moles B \to mass B$

with carrying you into moles and the balanced-equation ratio carrying you from A to B. If a reaction happens in solution, molarity and volume (Chapter 3) supply the moles instead of a balance. This topic develops that machinery with worked mass–mass calculations; the next topics extend it to limiting reactants, yields, and titration analysis.

Why this matters

Stoichiometry is where chemistry becomes predictive: given a recipe and a starting amount, you can compute exactly how much product will form — or how much reactant you must buy to make a target amount. Chemical industry runs on these numbers: ammonia plants scale the N₂:H₂ = 1:3 ratio to thousands of tons; airbag inflators use the stoichiometry of sodium azide decomposition to generate just the right volume of nitrogen gas. In environmental chemistry, stoichiometry converts pollutant masses to the oxygen demand a wastewater plant must supply. The same three-step logic applies everywhere: grams to moles (molar mass), moles to moles (balanced-equation ratio), moles back to grams (molar mass). Because this pattern recurs in nearly every quantitative chapter that follows — gases (Chapter 9), solutions (Chapter 11), equilibria, acids and bases, electrochemistry — mastering it now compounds like interest.

The college version

Core Concepts

Mole ratios: the language of the balanced equation

The coefficients of a balanced equation are stoichiometric factors — ratios converting moles of one substance to moles of another. From N₂ + 3H₂ → 2NH₃, valid factors include

$3 mol H 2 1 mol N 2 2 mol NH 3 1 mol N 2 2 mol NH 3 3 mol H 2$

Pick the factor that cancels the unit you have and leaves the unit you want — the same dimensional-analysis discipline used throughout this book. Coefficients are ratios of moles, never grams: 1 g of N₂ does not react with 3 g of H₂.

The stoichiometric pathway

For a reaction of pure substances, the universal plan is:

$mass of A (g) \div molar mass A ⟶ moles of A \times mole ratio ⟶ moles of B \times molar mass B ⟶ mass of B (g)$

Each arrow is a conversion factor, so the whole calculation can be strung together in one dimensional-analysis chain — write the formula first, then substitute numbers, then cancel units.

When reactions happen in solution

If a reactant or product is in solution, its amount is found from molarity and volume (Chapter 3): moles = M × V. The stoichiometric ratio then converts those moles as usual. This is the setup for titration problems in the final topic of this chapter — for example, using the volume of a standardized NaOH solution to find how many moles of acid were neutralized.

Beyond the balanced equation: what stoichiometry assumes

Stoichiometric predictions assume complete reaction of pure materials. In reality, reactions often stop short (equilibrium, Chapter 13), run at finite speed (kinetics, Chapter 12), or have a reactant that runs out first (the , next topic). Stoichiometry gives the theoretical maximum; the next topic shows how real yields compare.

How It Works / Step-by-Step Process

Write and balance the equation; identify the known substance (A) and target substance (B).
Convert the given mass of A to moles using A's molar mass.
Multiply by the from the balanced equation (moles B / moles A) to get moles of B.
Convert moles of B to grams using B's molar mass.
Check units at every step; verify the equation is balanced before trusting the ratio.

Common Confusions

Do Not Confuse	With	Difference
Mole ratios	Mass ratios	Coefficients are ratios of moles, not grams — 1 mol N₂ needs 3 mol H₂, not 3 g H₂
Moles of A	Mass of A	Convert with molar mass before applying the mole ratio; jumping from grams to grams skips the required mole step
Product mass > reactant mass	Violation of conservation of mass	Other reactants (like O₂) contribute atoms; total mass is still conserved
Mole ratio direction	Ratio flipped	Choose the factor that cancels the unit you have (e.g., mol A on the bottom)
Theoretical yield	Actual yield	Stoichiometry gives the maximum possible; real reactions yield less (next topic)
Balanced equation coefficients	Reaction rate	Coefficients tell how much, not how fast — kinetics (Chapter 12) handles speed

Eli explains

The same idea, in plain words

Explain it like I’m 10

A chemical equation is a recipe, and stoichiometry is the math of scaling that recipe. If one pancake recipe needs 1 cup of flour and 2 eggs, then 3 pancakes need 3 cups of flour and 6 eggs — the ratio stays the same. Chemists do exactly this with molecules: the balanced equation says "1 nitrogen plus 3 hydrogens make 2 ammonias," so if you start with 5 nitrogen molecules, you'll use 15 hydrogens and make 10 ammonias. Stoichiometry just counts molecules (in giant "dozen-like" groups called moles) to figure out how much of each ingredient you need or how much food you'll get.

Worked example

Example 1: Mass of product from mass of reactant

Problem: How many grams of CO₂ form when 10.0 g of CH₄ (molar mass 16.04 g/mol) burns completely in oxygen? The balanced equation is

$CH 4 (g) + 2 O 2 (g) \to CO 2 (g) + 2 H 2 O (g)$

Step 1 — Convert mass of CH₄ to moles:

$10.0 g CH 4 \times 1 mol CH 4 16.04 g CH 4 = 0.623 mol CH 4$

Step 2 — Apply the mole ratio (1 mol CH₄ : 1 mol CO₂):

$0.623 mol CH 4 \times 1 mol CO 2 1 mol CH 4 = 0.623 mol CO 2$

Step 3 — Convert moles of CO₂ to grams (molar mass 44.01 g/mol):

$0.623 mol CO 2 \times 44.01 g CO 2 1 mol CO 2 = 27.4 g CO 2$

Answer: 10.0 g of methane produces 27.4 g of CO₂ — more than the starting mass, because the oxygen contributes most of the CO₂'s mass. That is not a violation of conservation of mass; it reflects the oxygen consumed.

Example 2: One-chain dimensional analysis (mass → mass)

Problem: How many grams of water form when 5.00 g of H₂ (molar mass 2.016 g/mol) reacts with excess O₂? Balanced equation:

$2 H 2 (g) + O 2 (g) \to 2 H 2 O (l)$

String all three conversions into one chain, writing the formula before substituting:

$5.00 g H 2 \times 1 mol H 2 2.016 g H 2 \times 2 mol H 2 O 2 mol H 2 \times 18.02 g H 2 O 1 mol H 2 O = 44.7 g H 2 O$

Check units: g H₂ cancels, mol H₂ cancels, mol H₂O cancels, leaving g H₂O. Answer: 44.7 g of water. The mole ratio 2:2 (i.e., 1:1) came straight from the coefficients.

Example 3: Mole ratio with different coefficients

Problem: How many grams of NH₃ (molar mass 17.03 g/mol) can be made from 28.0 g of N₂ (molar mass 28.02 g/mol) with excess H₂?

$N 2 (g) + 3 H 2 (g) \to 2 NH 3 (g)$

Step 1 — Moles of N₂:

$28.0 g N 2 \times 1 mol N 2 28.02 g N 2 = 0.999 mol N 2 \approx 1.00 mol N 2$

Step 2 — Mole ratio 2 mol NH₃ per 1 mol N₂:

$1.00 mol N 2 \times 2 mol NH 3 1 mol N 2 = 2.00 mol NH 3$

Step 3 — Grams of NH₃:

$2.00 mol NH 3 \times 17.03 g NH 3 1 mol NH 3 = 34.1 g NH 3$

Answer: 34.1 g of NH₃ — the 1:2 coefficient ratio doubled the mole count, exactly what the balanced equation demands.

Key takeaways

Balanced-equation coefficients are mole ratios — the conversion factors between substances (never grams directly).
The universal plan: mass A → moles A (÷ molar mass) → moles B (× mole ratio) → mass B (× molar mass).
Dimensional analysis: set up conversion factors so units cancel; write the formula, then substitute numbers.
In solution, moles = M × V; the same mole-ratio logic then applies.
Stoichiometry predicts theoretical amounts assuming complete reaction of pure materials; real reactions may be limited by a reactant or fall short of completion.
Check your answer: more product mass than reactant mass is possible (extra atoms come from the other reactant), but every atom count must balance.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

Why must the equation be balanced before any stoichiometric calculation?
Show answer
Only a balanced equation gives correct mole ratios; an unbalanced equation's coefficients are meaningless as conversion factors.
In 2H₂ + O₂ → 2H₂O, how many moles of H₂O form from 4 mol of H₂?
Show answer
4 mol H₂ × (2 mol H₂O / 2 mol H₂) = 4 mol H₂O.
How many grams of H₂O form from 5.00 g of H₂ with excess O₂? (Molar masses: H₂ = 2.016, H₂O = 18.02.)
Show answer
5.00 g H₂ × (1 mol / 2.016 g) × (2 mol H₂O / 2 mol H₂) × (18.02 g / 1 mol) = 44.7 g H₂O.
What are the three conversion steps in a mass → mass problem?
Show answer
Mass A → moles A (÷ molar mass of A) → moles B (× mole ratio) → mass B (× molar mass of B).
If a product's mass exceeds the mass of the single reactant you started with, is something wrong? Explain.
Show answer
No — other reactants (e.g., O₂ in combustion) contribute atoms to the product; total mass of all reactants equals total mass of all products.
How would you find moles of a reactant if it is provided as a solution concentration and volume?
Show answer
Moles = molarity × volume (n = M × V), then use the mole ratio as usual.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Stoichiometry: The quantitative relationships between amounts of reactants and products
Mole ratio: Ratio of coefficients from the balanced equation
Molar mass: Grams per mole of a substance (g/mol)
Dimensional analysis: Unit-canceling calculation method
Theoretical amount: Amount predicted by stoichiometry for complete reaction
Limiting reactant: The reactant that runs out first, stopping the reaction
Aqueous solution: A substance dissolved in water

Sources & references

openstax.org — Chemistry 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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In 30 seconds

Why this matters

The college version

Core Concepts

Mole ratios: the language of the balanced equation

The stoichiometric pathway

When reactions happen in solution

Beyond the balanced equation: what stoichiometry assumes

How It Works / Step-by-Step Process

Common Confusions

The same idea, in plain words

Worked example

Example 1: Mass of product from mass of reactant

Example 2: One-chain dimensional analysis (mass → mass)

Example 3: Mole ratio with different coefficients

Key takeaways

Check yourself

Keep learning

Key vocabulary

Related lessons

Sources & references