Why Do Chemical Equations Need to Be Balanced?

A chemical reaction changes substances into new substances.

For example, hydrogen can react with oxygen to produce water:

hydrogen + oxygen → water

Using chemical formulas:

H₂ + O₂ → H₂O

However, this equation is not balanced.

Count the atoms:

Reactants:

• H = 2
• O = 2

Products:

• H = 2
• O = 1

One oxygen atom appears to have disappeared.

That cannot happen in an ordinary chemical reaction.

The equation must therefore be balanced.

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The Law of Conservation of Mass

The law of conservation of mass states:

Mass is neither created nor destroyed during an ordinary chemical reaction.

Atoms are rearranged during chemical reactions, but they are not created or destroyed.

Therefore:

total mass of reactants = total mass of products

This is the fundamental reason chemical equations must be balanced.

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Conservation of Mass at the Particle Level

Imagine a reaction involving several atoms.

Before the reaction, the atoms may be connected in one arrangement.

After the reaction, those same atoms may be connected differently.

The important idea is:

The atoms are rearranged, not replaced.

For example:

Before:

A–A + B–B

After:

A–B + A–B

There are still:

• 2 A atoms
• 2 B atoms

Only their arrangement has changed.

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Closed and Open Systems

Conservation of mass is easiest to observe in a closed system, where matter cannot enter or leave.

In a closed container:

mass before reaction = mass after reaction

Sometimes a reaction in an open container appears to lose mass because a gas escapes into the surroundings. The matter has not been destroyed; it has simply left the container.

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Reactants and Products

Every chemical equation has two main sides.

Reactants → Products

The reactants are the substances present at the beginning of the reaction.

The products are the new substances formed.

For example:

2Mg + O₂ → 2MgO

Reactants:

• magnesium, Mg
• oxygen, O₂

Product:

• magnesium oxide, MgO

The arrow means:

reacts to form or produces

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Reading Chemical Equations

Consider:

2H₂ + O₂ → 2H₂O

This can be read as:

Two molecules of hydrogen react with one molecule of oxygen to produce two molecules of water.

The equation communicates:

• which substances react
• which substances form
• the relative numbers of particles involved

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Chemical Formulas and Subscripts

A subscript tells us how many atoms of an element are present in one particle or formula unit.

For example:

H₂O

contains:

• 2 H atoms
• 1 O atom

CO₂

contains:

• 1 C atom
• 2 O atoms

CaCl₂

contains:

• 1 Ca atom
• 2 Cl atoms

Subscripts are part of the chemical formula.

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Coefficients

A coefficient is a number placed in front of a chemical formula.

For example:

3H₂O

The coefficient 3 means:

3 water molecules

Each water molecule contains:

• 2 H atoms
• 1 O atom

Therefore:

3H₂O

contains:

• 6 H atoms
• 3 O atoms

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Coefficients Multiply the Entire Formula

Consider:

4CO₂

One CO₂ molecule contains:

• 1 C
• 2 O

Four CO₂ molecules contain:

• 4 C
• 8 O

Therefore:

coefficient × subscript = total number of that atom

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Coefficients and Subscripts Are Different

This distinction is extremely important.

Consider:

2H₂O

The coefficient 2 tells us there are two water molecules.

The subscript 2 tells us each water molecule contains two hydrogen atoms.

Therefore:

2H₂O

contains:

4 H atoms and 2 O atoms

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Never Change Subscripts to Balance an Equation

Suppose we have:

H₂ + O₂ → H₂O

We need two oxygen atoms on the product side.

It may be tempting to change:

H₂O

to:

H₂O₂

But this changes the substance.

H₂O = water

H₂O₂ = hydrogen peroxide

They are different compounds.

When balancing equations:

Change coefficients, never chemical subscripts.

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Balancing the Formation of Water

Start:

H₂ + O₂ → H₂O

Count atoms.

Reactants:

• H = 2
• O = 2

Products:

• H = 2
• O = 1

Balance oxygen by placing 2 before H₂O:

H₂ + O₂ → 2H₂O

Now count again.

Products:

• H = 4
• O = 2

Oxygen is balanced, but hydrogen is not.

Place 2 before H₂:

2H₂ + O₂ → 2H₂O

Now:

Reactants:

• H = 4
• O = 2

Products:

• H = 4
• O = 2

Balanced.

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What Does the Balanced Equation Mean?

The balanced equation:

2H₂ + O₂ → 2H₂O

shows a particle ratio of:

2 : 1 : 2

This means:

2 hydrogen molecules react with 1 oxygen molecule to form 2 water molecules.

It could also represent:

4 hydrogen molecules + 2 oxygen molecules → 4 water molecules

because the same ratio is maintained.

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A Strategy for Balancing Equations

A reliable method is:

Step 1: Write the correct chemical formulas.

Step 2: Count each type of atom on both sides.

Step 3: Choose an element that is not balanced.

Step 4: Add a coefficient.

Step 5: Count the atoms again.

Step 6: Continue until every element is balanced.

Step 7: Reduce the coefficients to the smallest whole-number ratio if necessary.

Step 8: Perform a final atom count.

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Example 1: Magnesium and Oxygen

Start:

Mg + O₂ → MgO

Count:

Reactants:

• Mg = 1
• O = 2

Products:

• Mg = 1
• O = 1

Balance oxygen:

Mg + O₂ → 2MgO

Now products contain:

• Mg = 2
• O = 2

Balance magnesium:

2Mg + O₂ → 2MgO

Final count:

Reactants:

• Mg = 2
• O = 2

Products:

• Mg = 2
• O = 2

Balanced.

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Example 2: Sodium and Chlorine

Start:

Na + Cl₂ → NaCl

Count:

Reactants:

• Na = 1
• Cl = 2

Products:

• Na = 1
• Cl = 1

Balance chlorine:

Na + Cl₂ → 2NaCl

Now products contain:

• Na = 2
• Cl = 2

Balance sodium:

2Na + Cl₂ → 2NaCl

Balanced equation:

2Na + Cl₂ → 2NaCl

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Example 3: Formation of Ammonia

Start:

N₂ + H₂ → NH₃

Count nitrogen first.

Reactants:

N = 2

Products:

N = 1

Place 2 before NH₃:

N₂ + H₂ → 2NH₃

Now products contain:

H = 6

Place 3 before H₂:

N₂ + 3H₂ → 2NH₃

Final count:

Reactants:

• N = 2
• H = 6

Products:

• N = 2
• H = 6

Balanced.

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Example 4: Hydrogen Chloride

Start:

H₂ + Cl₂ → HCl

Reactants:

• H = 2
• Cl = 2

Products:

• H = 1
• Cl = 1

Place 2 before HCl:

H₂ + Cl₂ → 2HCl

Now:

Reactants:

• H = 2
• Cl = 2

Products:

• H = 2
• Cl = 2

Balanced.

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Example 5: Aluminium Oxide

Start:

Al + O₂ → Al₂O₃

This is more challenging.

Oxygen appears as:

2 atoms in O₂

and:

3 atoms in Al₂O₃

The smallest common multiple of 2 and 3 is:

6

Use:

3O₂

to give 6 oxygen atoms.

Use:

2Al₂O₃

to give 6 oxygen atoms.

Now:

Al + 3O₂ → 2Al₂O₃

The products contain:

4 Al atoms

So place 4 before Al:

4Al + 3O₂ → 2Al₂O₃

Balanced.

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Using Multiples When Balancing

The aluminium oxide example demonstrates an important strategy.

If one side contains oxygen in groups of 2 and the other in groups of 3:

2, 4, 6, 8...

and:

3, 6, 9, 12...

the first common value is:

6

This tells us useful coefficients are:

3O₂

and:

2Al₂O₃

Recognizing common multiples can make balancing much faster.

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Example 6: Iron and Oxygen

Start:

Fe + O₂ → Fe₂O₃

As before, oxygen appears in groups of 2 and 3.

Use 6 oxygen atoms:

Fe + 3O₂ → 2Fe₂O₃

Now the products contain:

4 Fe

So:

4Fe + 3O₂ → 2Fe₂O₃

Balanced.

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Example 7: Methane Combustion

Methane reacts with oxygen to produce carbon dioxide and water.

Start:

CH₄ + O₂ → CO₂ + H₂O

Balance carbon:

C is already balanced.

Balance hydrogen:

Reactants have:

4 H

Place 2 before water:

CH₄ + O₂ → CO₂ + 2H₂O

Now count oxygen on the product side:

CO₂ contains 2 O.

2H₂O contains 2 O.

Total:

4 O atoms

Therefore use:

2O₂

Balanced equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

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Example 8: Propane Combustion

Start:

C₃H₈ + O₂ → CO₂ + H₂O

Balance carbon:

C₃H₈ + O₂ → 3CO₂ + H₂O

Balance hydrogen:

C₃H₈ + O₂ → 3CO₂ + 4H₂O

Now count oxygen on the products:

3CO₂ gives:

6 O

4H₂O gives:

4 O

Total:

10 O

Therefore:

5O₂

Balanced equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

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A Useful Order for Balancing

There is no single order that works perfectly for every equation, but a useful strategy is:

• begin with elements appearing in only one compound on each side
• leave hydrogen and oxygen until later when possible
• balance unchanged polyatomic ions as groups when appropriate
• recount everything at the end

For combustion reactions involving hydrocarbons, a useful order is often:

carbon → hydrogen → oxygen

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Counting Atoms Carefully

Consider:

2Al₂O₃

The coefficient 2 multiplies the entire formula.

Aluminium:

2 × 2 = 4 Al

Oxygen:

2 × 3 = 6 O

So:

2Al₂O₃

contains:

• 4 aluminium atoms
• 6 oxygen atoms

This multiplication is essential when checking balanced equations.

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Particle Relationships

Balanced equations are not just bookkeeping.

They describe particle relationships.

Consider:

N₂ + 3H₂ → 2NH₃

At the particle level:

1 N₂ molecule

reacts with:

3 H₂ molecules

to produce:

2 NH₃ molecules

The coefficients give the relative numbers of particles.

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Ratios in Balanced Equations

Consider:

2H₂ + O₂ → 2H₂O

Coefficient ratio:

2 : 1 : 2

This means that if we double everything:

4 : 2 : 4

the reaction relationship is still correct.

Or multiply by 10:

20 : 10 : 20

The relative ratio remains:

2 : 1 : 2

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Coefficients Do Not Usually Represent Individual Atoms

Consider:

2Na + Cl₂ → 2NaCl

At a particle level, the equation describes the relative numbers of reacting particles or formula units.

For molecular substances, we can talk about molecules.

For ionic substances such as NaCl, we normally describe formula units, because solid sodium chloride forms a giant ionic lattice rather than existing as separate NaCl molecules.

This distinction becomes increasingly important in chemistry.

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Balanced Equations and Mass

Consider:

2H₂ + O₂ → 2H₂O

The equation conserves atoms.

Because atoms have mass, conserving the number and type of atoms also conserves total mass.

The atoms have simply changed their arrangement.

This connects the particle model directly to the law of conservation of mass.

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Why Mass May Appear to Change

Suppose a carbonate reacts with an acid in an open flask and produces carbon dioxide gas.

If the gas escapes, the measured mass of the flask and its contents decreases.

Does this violate conservation of mass?

No.

The carbon dioxide still exists. It has simply entered the surroundings.

If the entire reaction and gas were contained in a closed system, the total mass would remain constant.

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Reactions That Take In Gases

The opposite can also happen.

Suppose a metal reacts with oxygen from the air.

The solid product may have a greater mass than the original metal.

This does not mean mass was created.

The additional mass came from:

oxygen in the air

The total mass of the metal plus oxygen is conserved.

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State Symbols

Chemical equations sometimes include state symbols.

These show the physical state of each substance:

(s) = solid

(l) = liquid

(g) = gas

(aq) = aqueous, dissolved in water

For example:

2Mg(s) + O₂(g) → 2MgO(s)

State symbols provide additional information but do not affect whether the equation is balanced.

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Balancing with State Symbols

Consider:

H₂(g) + O₂(g) → H₂O(l)

First balance the formulas exactly as before:

2H₂(g) + O₂(g) → 2H₂O(l)

The state symbols remain attached to their substances.

Do not count state symbols as atoms.

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Word Equations and Symbol Equations

A word equation shows substance names:

magnesium + oxygen → magnesium oxide

A symbol equation uses chemical formulas:

Mg + O₂ → MgO

A balanced symbol equation shows correct formulas and conserved atoms:

2Mg + O₂ → 2MgO

These forms communicate increasingly detailed information.

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Checking Whether an Equation Is Balanced

Consider:

2Na + Cl₂ → 2NaCl

Count atoms.

Left:

• Na = 2
• Cl = 2

Right:

• Na = 2
• Cl = 2

Therefore:

balanced

Now consider:

Na + Cl₂ → NaCl

Left:

• Na = 1
• Cl = 2

Right:

• Na = 1
• Cl = 1

Therefore:

not balanced

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Worked Example 1

Balance:

H₂ + Br₂ → HBr

Count:

Left:

• H = 2
• Br = 2

Right:

• H = 1
• Br = 1

Add coefficient 2:

H₂ + Br₂ → 2HBr

Balanced.

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Worked Example 2

Balance:

K + O₂ → K₂O

Balance oxygen first:

K + O₂ → 2K₂O

Now the products contain:

4 K

Therefore:

4K + O₂ → 2K₂O

Balanced.

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Worked Example 3

Balance:

Ca + H₂O → Ca(OH)₂ + H₂

Start by examining Ca.

Ca is already balanced.

Ca(OH)₂ contains:

• 2 O
• 2 H in the hydroxide groups

Use 2H₂O:

Ca + 2H₂O → Ca(OH)₂ + H₂

Count:

Left:

• Ca = 1
• H = 4
• O = 2

Right:

• Ca = 1
• H = 4
• O = 2

Balanced.

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Worked Example 4

Balance:

Na + H₂O → NaOH + H₂

Start by balancing sodium and the water relationship:

2Na + 2H₂O → 2NaOH + H₂

Count:

Left:

• Na = 2
• H = 4
• O = 2

Right:

• Na = 2
• H = 4
• O = 2

Balanced.

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Worked Example 5

Balance:

CaCO₃ → CaO + CO₂

Count:

Left:

• Ca = 1
• C = 1
• O = 3

Right:

CaO gives:

• Ca = 1
• O = 1

CO₂ gives:

• C = 1
• O = 2

Total right-side oxygen:

3

The equation is already balanced:

CaCO₃ → CaO + CO₂

Not every equation needs additional coefficients.

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Worked Example 6

Balance:

Fe + HCl → FeCl₂ + H₂

Fe is already balanced.

The product contains:

2 Cl

Therefore use:

2HCl

Equation:

Fe + 2HCl → FeCl₂ + H₂

Count hydrogen:

Left = 2 H

Right = 2 H

Balanced.

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Worked Example 7

Balance:

P₄ + O₂ → P₂O₅

Balance phosphorus:

P₄ + O₂ → 2P₂O₅

Products now contain:

10 O

Therefore use:

5O₂

Final equation:

P₄ + 5O₂ → 2P₂O₅

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Worked Example 8

Balance:

C₂H₆ + O₂ → CO₂ + H₂O

Balance carbon:

C₂H₆ + O₂ → 2CO₂ + H₂O

Balance hydrogen:

C₂H₆ + O₂ → 2CO₂ + 3H₂O

Products contain:

4 + 3 = 7 oxygen atoms

This initially gives:

7/2 O₂

Fractions can be useful during working:

C₂H₆ + 7/2O₂ → 2CO₂ + 3H₂O

Multiply every coefficient by 2:

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Now all coefficients are whole numbers.

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Smallest Whole-Number Coefficients

Consider:

4H₂ + 2O₂ → 4H₂O

This equation is balanced.

However, all coefficients can be divided by 2:

2H₂ + O₂ → 2H₂O

Chemical equations are normally written using the smallest whole-number ratio.

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Particle Diagrams and Balanced Equations

A particle diagram should agree with its balanced equation.

For:

2H₂ + O₂ → 2H₂O

a correct particle model should show:

Before:

• 2 H₂ particles
• 1 O₂ particle

After:

• 2 H₂O particles

Count the atoms in the picture:

Before:

• 4 H
• 2 O

After:

• 4 H
• 2 O

The visual model confirms conservation of atoms.

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From Particle Diagram to Equation

Suppose a particle diagram shows:

Before:

• 1 N₂ molecule
• 3 H₂ molecules

After:

• 2 NH₃ molecules

The corresponding equation is:

N₂ + 3H₂ → 2NH₃

Particle diagrams can therefore be translated directly into coefficients.

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What Balanced Equations Do Not Tell Us

A balanced chemical equation provides important information, but it does not automatically tell us:

• how quickly the reaction occurs
• how much energy is released
• the reaction temperature
• the reaction mechanism
• whether the reaction will happen easily
• the actual amount used in a particular experiment

A balanced equation primarily describes:

which substances react and their relative particle relationships.

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Common Mistakes

Mistake 1: Changing subscripts

Incorrect:

H₂ + O₂ → H₂O₂

if the intended product is water.

Changing the subscript changes the substance.

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Mistake 2: Forgetting that coefficients multiply the whole formula

For:

3CO₂

there are:

3 C and 6 O

not 3 C and 2 O.

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Mistake 3: Balancing only one element

Every element must have the same number of atoms on both sides.

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Mistake 4: Forgetting diatomic elements

Some elements commonly appear as diatomic molecules, including:

H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂

For example, elemental oxygen is normally written:

O₂

not:

O

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Mistake 5: Not reducing coefficients

4H₂ + 2O₂ → 4H₂O

is balanced, but:

2H₂ + O₂ → 2H₂O

is the preferred simplest ratio.

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Error Analysis

A student balances:

Mg + O₂ → MgO

as:

Mg + O₂ → MgO₂

The student has changed the chemical formula.

That changes magnesium oxide into a different formula rather than balancing the original reaction.

Correct approach:

2Mg + O₂ → 2MgO

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Another Error Analysis

A student writes:

2H₂ + O₂ → 2H₂O

and counts:

Products:

H = 2

O = 1

This ignores the coefficient.

The coefficient multiplies the entire formula.

For:

2H₂O

Hydrogen:

2 × 2 = 4

Oxygen:

2 × 1 = 2

Therefore the equation is balanced.

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Why Balanced Equations Matter

Balanced equations are fundamental to chemistry because they allow chemists to:

• represent chemical reactions accurately
• demonstrate conservation of mass
• compare quantities of reactants and products
• predict particle relationships
• perform chemical calculations
• plan laboratory reactions
• calculate expected product quantities
• understand industrial chemical processes

Balanced equations provide the foundation for stoichiometry, where these particle ratios are used to calculate actual amounts of substances.

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A Final Balancing Checklist

Before deciding that an equation is balanced, check:

1. Are all chemical formulas correct?

2. Have only coefficients been changed?

3. Is every element present in equal numbers on both sides?

4. Have coefficients been applied to the entire formula?

5. Are the coefficients whole numbers?

6. Are they in the smallest whole-number ratio?

7. Does the equation make sense as a particle relationship?

If the answer to all seven is yes, the equation is properly balanced.

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Did You Know?

Chemical equations are a symbolic way of representing events happening on an enormous particle scale.

A balanced equation such as:

2H₂ + O₂ → 2H₂O

does not mean chemists normally react only two hydrogen molecules.

A laboratory sample contains enormous numbers of particles.

The equation tells us the ratio in which those particles react.

Whether we imagine:

2 : 1 : 2

or:

2,000 : 1,000 : 2,000

or enormously larger quantities, the same particle relationship applies.

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Key Terms

• Chemical reaction: Process in which substances are transformed into new substances.
• Chemical equation: Symbolic representation of a chemical reaction.
• Reactant: Starting substance in a chemical reaction.
• Product: Substance formed during a chemical reaction.
• Law of conservation of mass: Mass is not created or destroyed during an ordinary chemical reaction.
• Balanced equation: Chemical equation containing equal numbers of each type of atom on both sides.
• Coefficient: Number placed before a chemical formula showing the relative number of particles or formula units.
• Subscript: Small number in a chemical formula showing the number of atoms of an element within the formula.
• Molecule: Discrete group of covalently bonded atoms.
• Formula unit: Simplest whole-number ratio represented by an ionic compound's formula.
• Closed system: System in which matter cannot enter or leave.
• Open system: System in which matter can enter or leave.
• State symbol: Symbol showing whether a substance is solid, liquid, gas, or aqueous.
• Particle ratio: Relative numbers of particles represented by coefficients.
• Stoichiometry: Quantitative study of reactants and products using balanced chemical equations.

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Key Rules

Conservation of mass:

total mass of reactants = total mass of products

For every element:

number of atoms before reaction = number of atoms after reaction

When balancing equations:

Change coefficients only.

Never change subscripts.

Coefficients multiply:

the entire chemical formula

Balanced equations should normally use:

the smallest whole-number coefficients

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Key Takeaways

• Chemical reactions rearrange atoms into new combinations.
• Atoms are not created or destroyed during ordinary chemical reactions.
• The law of conservation of mass explains why chemical equations must be balanced.
• In a closed system, the total mass before and after a chemical reaction remains constant.
• Apparent mass loss can occur in an open system when a gaseous product escapes.
• Apparent mass gain can occur when a substance reacts with matter from the surroundings, such as oxygen.
• Reactants appear on the left side of a chemical equation.
• Products appear on the right side.
• The reaction arrow means "reacts to form" or "produces."
• Subscripts describe the composition of a chemical substance.
• Coefficients describe relative numbers of particles or formula units.
• A coefficient multiplies every atom in the formula following it.
• Chemical formulas must not be changed when balancing equations.
• Changing a subscript changes the identity of the substance.
• Equations are balanced by changing coefficients.
• Each element must have the same number of atoms on both sides of a balanced equation.
• Common multiples can help balance elements appearing in different numerical groups.
• Equations should normally be reduced to the smallest whole-number coefficient ratio.
• Particle diagrams provide a visual way to check conservation of atoms.
• Balanced equations describe particle relationships as ratios.
• Molecular substances can be interpreted in terms of molecules.
• Ionic substances are more appropriately described using formula units.
• State symbols provide information about physical state but do not affect atom balancing.
• A balanced equation does not automatically describe reaction rate, energy change, or reaction conditions.
• Balanced equations provide the foundation for quantitative chemical calculations and stoichiometry.
• A final atom count is one of the most reliable ways to check that an equation has been balanced correctly.

Mole Ratios

A mole ratio is a relationship between the amounts, in moles, of substances involved in a chemical reaction.

Mole ratios come directly from the coefficients in a balanced chemical equation.

For example:

2H₂ + O₂ → 2H₂O

The coefficients tell us that:

2 mol H₂ react with 1 mol O₂ to produce 2 mol H₂O

This gives several possible mole ratios:

H₂ : O₂ = 2 : 1

H₂ : H₂O = 2 : 2 = 1 : 1

O₂ : H₂O = 1 : 2

Mole ratios are one of the most important ideas in stoichiometry because they allow us to calculate how much reactant is required or how much product can be produced.

Balanced Equations Tell a Quantitative Story

A chemical equation tells us:

• which substances react
• which substances are produced
• the relative amounts of each substance involved

Consider:

N₂ + 3H₂ → 2NH₃

This equation describes the production of ammonia.

At the particle level:

1 molecule N₂ + 3 molecules H₂ → 2 molecules NH₃

At the mole level:

1 mol N₂ + 3 mol H₂ → 2 mol NH₃

Therefore:

N₂ : H₂ : NH₃ = 1 : 3 : 2

These numbers are not arbitrary. They are determined by the conservation of atoms.

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Why Equations Must Be Balanced

Chemical reactions obey the law of conservation of mass.

Atoms are rearranged during chemical reactions, but they are not created or destroyed.

Consider the unbalanced equation:

H₂ + O₂ → H₂O

Count the atoms.

Left side:

• H = 2
• O = 2

Right side:

• H = 2
• O = 1

The oxygen atoms are not balanced.

The balanced equation is:

2H₂ + O₂ → 2H₂O

Now:

Left side:

• H = 4
• O = 2

Right side:

• H = 4
• O = 2

Only after the equation is balanced can its coefficients be used correctly for mole ratios.

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What Do Coefficients Mean?

The large numbers written in front of chemical formulas are called coefficients.

Consider:

2CO + O₂ → 2CO₂

The coefficient of CO is:

2

The coefficient of O₂ is:

1

The coefficient of CO₂ is:

2

Remember that a coefficient of 1 is normally not written.

The equation therefore means:

2 mol CO + 1 mol O₂ → 2 mol CO₂

Coefficients vs. Subscripts

Do not confuse coefficients with subscripts.

Consider:

2H₂O

The 2 in front is the coefficient.

It means:

2 molecules of H₂O

or:

2 mol of H₂O

The small 2 in H₂O is a subscript.

It tells us that each water molecule contains:

2 hydrogen atoms

Therefore:

coefficient → amount of substance

subscript → composition of one particle

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Never Change Subscripts When Balancing

Suppose we want to balance:

H₂ + O₂ → H₂O

We cannot change H₂O into H₂O₂ simply to balance the oxygen.

H₂O and H₂O₂ are different substances.

H₂O = water

H₂O₂ = hydrogen peroxide

Instead, change the coefficients:

2H₂ + O₂ → 2H₂O

Changing coefficients changes the amount.

Changing subscripts changes the chemical substance.

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Mole Ratios

For any balanced equation, the coefficients provide the mole ratios.

Consider:

2Mg + O₂ → 2MgO

The coefficients are:

2 : 1 : 2

Therefore:

Mg : O₂ = 2 : 1

Mg : MgO = 2 : 2 = 1 : 1

O₂ : MgO = 1 : 2

We can write these as conversion factors.

For example:

2 mol Mg / 1 mol O₂

or:

1 mol O₂ / 2 mol Mg

Which form we use depends on what we are trying to calculate.

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Mole Ratios Are Conversion Factors

This is the key idea for calculations.

Suppose:

2H₂ + O₂ → 2H₂O

We want to convert moles of H₂ into moles of H₂O.

The mole ratio is:

2 mol H₂O / 2 mol H₂

Therefore:

moles H₂ × (2 mol H₂O / 2 mol H₂)

The units of mol H₂ cancel.

We are left with:

mol H₂O

This is why mole ratios work like conversion factors.

---

The Basic Stoichiometry Pattern

For mole-to-mole problems, use:

moles of known substance → mole ratio → moles of unknown substance

A useful general equation is:

moles wanted = moles given × (coefficient wanted / coefficient given)

This simple relationship solves many mole-ratio problems.

---

Worked Example: Hydrogen and Water

Consider:

2H₂ + O₂ → 2H₂O

How many moles of water can be produced from 6 mol H₂, assuming enough oxygen is available?

Identify the ratio

H₂ : H₂O

2 : 2

Therefore:

6 mol H₂ × (2 mol H₂O / 2 mol H₂)

The H₂ units cancel.

= 6 mol H₂O

Answer

6 mol H₂O

Because H₂ and H₂O have equal coefficients, their mole ratio is:

1 : 1

---

Worked Example: Hydrogen and Oxygen

Using:

2H₂ + O₂ → 2H₂O

How many moles of O₂ are required to react with 8 mol H₂?

Ratio:

2 mol H₂ : 1 mol O₂

Calculation:

8 mol H₂ × (1 mol O₂ / 2 mol H₂)

= 4 mol O₂

Answer

4 mol O₂

---

Think of the Equation as a Recipe

A balanced equation is similar to a recipe.

Suppose a fictional recipe says:

2 buns + 1 patty → 1 burger

If you have:

8 buns

then you need:

4 patties

and can make:

4 burgers

Chemical equations work similarly, except the quantities are measured in moles.

For:

2H₂ + O₂ → 2H₂O

the chemical "recipe" requires:

2 mol H₂ for every 1 mol O₂

Worked Example: Making Ammonia

Ammonia is produced according to:

N₂ + 3H₂ → 2NH₃

How many moles of ammonia can be produced from 6 mol H₂, assuming enough nitrogen is available?

Ratio:

3 mol H₂ : 2 mol NH₃

Calculation:

6 mol H₂ × (2 mol NH₃ / 3 mol H₂)

= 4 mol NH₃

Answer

4 mol NH₃

---

Another Ammonia Example

Using:

N₂ + 3H₂ → 2NH₃

How many moles of nitrogen are required to produce 10 mol NH₃?

Ratio:

1 mol N₂ : 2 mol NH₃

Calculation:

10 mol NH₃ × (1 mol N₂ / 2 mol NH₃)

= 5 mol N₂

Answer

5 mol N₂

---

Mole Ratios Can Be Used in Either Direction

Consider:

N₂ + 3H₂ → 2NH₃

To convert N₂ into NH₃:

2 mol NH₃ / 1 mol N₂

To convert NH₃ into N₂:

1 mol N₂ / 2 mol NH₃

These are reciprocal relationships.

The correct orientation is the one that allows the unwanted unit to cancel.

---

Unit Cancellation

Unit cancellation is an excellent way to check your calculation.

Suppose:

N₂ + 3H₂ → 2NH₃

We start with:

9 mol H₂

and want NH₃.

Use:

9 mol H₂ × (2 mol NH₃ / 3 mol H₂)

The unit:

mol H₂

appears on top and bottom, so it cancels.

We are left with:

mol NH₃

Calculation:

9 × 2/3 = 6

Therefore:

6 mol NH₃

If the units do not cancel correctly, the ratio has probably been placed upside down.

---

Worked Example: Formation of Magnesium Oxide

Magnesium burns in oxygen:

2Mg + O₂ → 2MgO

How many moles of MgO can form from 7 mol Mg, assuming excess oxygen?

Ratio:

2 mol Mg : 2 mol MgO

Calculation:

7 mol Mg × (2 mol MgO / 2 mol Mg)

= 7 mol MgO

Answer

7 mol MgO

---

Worked Example: Oxygen Needed for Magnesium

Using:

2Mg + O₂ → 2MgO

How many moles of O₂ are required for 12 mol Mg?

Ratio:

2 mol Mg : 1 mol O₂

Calculation:

12 mol Mg × (1 mol O₂ / 2 mol Mg)

= 6 mol O₂

Answer

6 mol O₂

---

Worked Example: Decomposition

Mole ratios also work for decomposition reactions.

Consider:

2H₂O₂ → 2H₂O + O₂

Hydrogen peroxide decomposes into water and oxygen.

The ratio is:

2 mol H₂O₂ : 2 mol H₂O : 1 mol O₂

Suppose 8 mol H₂O₂ decomposes completely.

How many moles of O₂ form?

8 mol H₂O₂ × (1 mol O₂ / 2 mol H₂O₂)

= 4 mol O₂

Answer

4 mol O₂

---

Worked Example: Combustion

Methane burns according to:

CH₄ + 2O₂ → CO₂ + 2H₂O

The coefficient ratio is:

1 : 2 : 1 : 2

Therefore:

CH₄ : O₂ = 1 : 2

CH₄ : CO₂ = 1 : 1

CH₄ : H₂O = 1 : 2

O₂ : CO₂ = 2 : 1

O₂ : H₂O = 2 : 2 = 1 : 1

Combustion Calculation

Using:

CH₄ + 2O₂ → CO₂ + 2H₂O

How many moles of oxygen are required to burn 3 mol CH₄ completely?

Ratio:

1 mol CH₄ : 2 mol O₂

Calculation:

3 mol CH₄ × (2 mol O₂ / 1 mol CH₄)

= 6 mol O₂

Answer

6 mol O₂

---

Predicting Carbon Dioxide

Using the same reaction:

CH₄ + 2O₂ → CO₂ + 2H₂O

How many moles of CO₂ form when 7 mol CH₄ burns completely?

Ratio:

1 mol CH₄ : 1 mol CO₂

Calculation:

7 mol CH₄ × (1 mol CO₂ / 1 mol CH₄)

= 7 mol CO₂

Answer

7 mol CO₂

---

Ratios Do Not Represent Mass Ratios

This is extremely important.

Consider:

2H₂ + O₂ → 2H₂O

The mole ratio is:

2 : 1 : 2

This does not mean:

2 g H₂ + 1 g O₂ → 2 g H₂O

That would be incorrect.

Approximate molar masses are:

H₂ = 2 g/mol

O₂ = 32 g/mol

H₂O = 18 g/mol

Therefore:

2 mol H₂ = 4 g

1 mol O₂ = 32 g

2 mol H₂O = 36 g

So the mass relationship is:

4 g H₂ + 32 g O₂ → 36 g H₂O

Mass is conserved.

---

Mole Ratios vs. Particle Ratios

The coefficients can represent particle ratios and mole ratios.

For:

2H₂ + O₂ → 2H₂O

we can say:

2 molecules H₂ : 1 molecule O₂ : 2 molecules H₂O

or:

2 mol H₂ : 1 mol O₂ : 2 mol H₂O

Why?

Because one mole always represents the same number of particles:

6.022 × 10²³ particles

This is Avogadro's constant.

Scaling the particle ratio up to moles does not change the ratio.

---

Fractional Amounts Are Allowed

Suppose:

2H₂ + O₂ → 2H₂O

Could 1 mol H₂ react?

Yes.

The equation tells us the ratio:

2 : 1 : 2

Dividing everything by 2 gives:

1 mol H₂ : 0.5 mol O₂ : 1 mol H₂O

Coefficients in the balanced equation are normally written as the smallest whole-number ratio, but actual reacting amounts can include decimal values.

---

Scaling Chemical Equations

Consider:

N₂ + 3H₂ → 2NH₃

The basic ratio is:

1 : 3 : 2

Multiply everything by 2:

2 : 6 : 4

Multiply everything by 5:

5 : 15 : 10

Multiply everything by 10:

10 : 30 : 20

All represent the same chemical ratio.

This is why mole ratios allow us to scale reactions to different quantities.

---

A Three-Step Method

For most mole-ratio questions:

Step A: Write the balanced equation

Example:

2Al + 3Cl₂ → 2AlCl₃

Step B: Identify the required mole ratio

Suppose we want to convert Al into AlCl₃.

Ratio:

2 mol Al : 2 mol AlCl₃

Step C: Multiply by the conversion factor

If we have 5 mol Al:

5 mol Al × (2 mol AlCl₃ / 2 mol Al)

= 5 mol AlCl₃

---

Worked Example: Aluminum Chloride

Consider:

2Al + 3Cl₂ → 2AlCl₃

How many moles of chlorine gas are required to react with 8 mol Al?

Ratio:

2 mol Al : 3 mol Cl₂

Calculation:

8 mol Al × (3 mol Cl₂ / 2 mol Al)

= 12 mol Cl₂

Answer

12 mol Cl₂

---

Worked Example with a Decimal

Consider:

2Al + 3Cl₂ → 2AlCl₃

How many moles of AlCl₃ can be produced from 2.5 mol Cl₂, assuming enough aluminum is available?

Ratio:

3 mol Cl₂ : 2 mol AlCl₃

Calculation:

2.5 mol Cl₂ × (2 mol AlCl₃ / 3 mol Cl₂)

= 1.67 mol AlCl₃

Answer

Approximately:

1.67 mol AlCl₃

Mole-ratio calculations do not always produce whole numbers.

---

More Complex Coefficients

Consider combustion of propane:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

The ratio is:

1 : 5 : 3 : 4

This gives:

C₃H₈ : O₂ = 1 : 5

C₃H₈ : CO₂ = 1 : 3

C₃H₈ : H₂O = 1 : 4

O₂ : CO₂ = 5 : 3

CO₂ : H₂O = 3 : 4

Worked Example: Propane

How many moles of CO₂ are produced when 4 mol C₃H₈ burns completely?

Equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Ratio:

1 mol C₃H₈ : 3 mol CO₂

Calculation:

4 mol C₃H₈ × (3 mol CO₂ / 1 mol C₃H₈)

= 12 mol CO₂

Answer

12 mol CO₂

---

Another Propane Example

How many moles of O₂ are needed to produce 9 mol CO₂?

Equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Ratio:

5 mol O₂ : 3 mol CO₂

Calculation:

9 mol CO₂ × (5 mol O₂ / 3 mol CO₂)

= 15 mol O₂

Answer

15 mol O₂

---

Comparing Quantities

Mole ratios can also be used without a full calculation.

Consider:

4Fe + 3O₂ → 2Fe₂O₃

We can immediately say:

• 4 mol Fe react with 3 mol O₂
• 4 mol Fe produce 2 mol Fe₂O₃
• 3 mol O₂ produce 2 mol Fe₂O₃

Because:

Fe : Fe₂O₃ = 4 : 2 = 2 : 1

twice as many moles of Fe are required as moles of Fe₂O₃ produced.

---

Mole Ratios and Stoichiometry

Stoichiometry is the quantitative study of reactants and products in chemical reactions.

Mole ratios are at the centre of stoichiometry.

Many future calculations follow this pattern:

given quantity → moles → mole ratio → moles wanted → wanted quantity

For example:

mass → moles → mole ratio → moles → mass

or:

particles → moles → mole ratio → moles → particles

or:

solution volume → moles → mole ratio → moles → concentration

Understanding mole ratios now makes more advanced stoichiometry much easier later.

---

Why Chemists Use Moles

Individual atoms and molecules are far too small to count directly during ordinary laboratory work.

Chemists therefore use the mole to connect:

microscopic particles

with:

measurable laboratory quantities

One mole contains:

6.022 × 10²³ particles

A balanced equation therefore allows us to scale a reaction from individual particles to laboratory quantities.

Real-World Connection: Industrial Chemistry

Factories need precise amounts of reactants.

Using too much reactant can:

• waste money
• increase waste
• increase purification requirements
• increase environmental impact

Using too little can:

• reduce product yield
• leave another reactant unused
• reduce efficiency

Stoichiometric calculations help chemical engineers determine appropriate quantities for industrial reactions.

Applications include producing:

• fertilizers
• pharmaceuticals
• fuels
• plastics
• metals
• cleaning products

---

Real-World Connection: Ammonia Production

Ammonia is produced industrially using:

N₂ + 3H₂ ⇌ 2NH₃

The stoichiometric mole ratio is:

1 mol N₂ : 3 mol H₂

This means the balanced equation requires three times as many moles of hydrogen as nitrogen for the reaction ratio.

Ammonia is an important starting material for many nitrogen-containing products, especially fertilizers.

The balanced equation allows chemists and engineers to calculate the required quantities of raw materials.

---

Real-World Connection: Combustion

Fuel combustion also depends on chemical ratios.

For methane:

CH₄ + 2O₂ → CO₂ + 2H₂O

One mole of methane requires:

2 mol O₂

If insufficient oxygen is available, complete combustion cannot proceed exactly as represented by this equation.

Other products, including carbon monoxide or carbon, can form under oxygen-limited conditions.

Correct reactant ratios therefore matter in:

• engines
• furnaces
• boilers
• power generation

---

Common Mistakes

Using an Unbalanced Equation

Mole ratios must come from a balanced equation.

Wrong:

H₂ + O₂ → H₂O

Correct:

2H₂ + O₂ → 2H₂O

---

Using Subscripts Instead of Coefficients

For:

2H₂ + O₂ → 2H₂O

the H₂ : O₂ mole ratio is:

2 : 1

not:

2 : 2

Use coefficients.

---

Changing Subscripts to Balance an Equation

Never change:

H₂O

into:

H₂O₂

just to balance an equation.

Change coefficients instead.

---

Assuming the Coefficients Are Mass Ratios

For:

2H₂ + O₂ → 2H₂O

the ratio:

2 : 1 : 2

is a mole ratio, not a gram ratio.

---

Putting the Conversion Factor Upside Down

If converting H₂ into O₂:

2H₂ + O₂ → 2H₂O

use:

1 mol O₂ / 2 mol H₂

not:

2 mol H₂ / 1 mol O₂

Check that unwanted units cancel.

---

Forgetting an Invisible Coefficient

In:

CH₄ + 2O₂ → CO₂ + 2H₂O

CH₄ has coefficient:

1

CO₂ also has coefficient:

1

---

Assuming Mole Ratios Must Produce Whole Numbers

Actual amounts can be:

• 0.5 mol
• 1.25 mol
• 2.8 mol

Whole-number coefficients describe the simplest reaction ratio, not the only possible quantities.

---

Using Molar Masses Too Early

If a question gives moles and asks for moles, you usually do not need molar mass.

Simply use:

moles → mole ratio → moles

---

Using Every Coefficient in the Equation

Usually, you only need the coefficients of:

the substance given

and:

the substance wanted

---

Key Terms

Mole — The amount of substance containing 6.022 × 10²³ representative particles.

Avogadro's constant — 6.022 × 10²³ particles per mole.

Balanced chemical equation — An equation containing equal numbers of each type of atom on both sides.

Coefficient — A number placed before a chemical formula showing the relative amount of that substance in a balanced equation.

Subscript — A small number within a chemical formula showing the number of atoms of an element in one formula unit or molecule.

Mole ratio — The ratio between amounts in moles of substances in a balanced chemical equation.

Stoichiometry — The quantitative study of relationships between reactants and products in chemical reactions.

Reactant — A starting substance in a chemical reaction.

Product — A substance formed during a chemical reaction.

Conversion factor — A ratio used to convert one quantity or unit into another.

Unit cancellation — A calculation method in which matching units in the numerator and denominator cancel.

Molar mass — The mass of one mole of a substance, usually measured in g/mol.

Conservation of mass — The principle that matter is not created or destroyed during an ordinary chemical reaction.

Chemical equation — A symbolic representation of a chemical reaction.

Stoichiometric coefficient — The coefficient of a substance in a balanced chemical equation.

Stoichiometric ratio — The relative mole quantities specified by the balanced chemical equation.

Limiting reactant — The reactant that is consumed first and therefore limits the amount of product that can form.

Excess reactant — A reactant present in more than the stoichiometric amount required.

---

Key Takeaways

• Mole ratios come directly from balanced chemical equations.
• Chemical equations must be balanced before mole ratios are used.
• Coefficients represent relative numbers of particles and relative numbers of moles.
• A coefficient of 1 is usually not written.
• Subscripts describe the composition of a substance.
• Coefficients describe relative amounts of substances.
• Never change subscripts when balancing an equation.
• Mole ratios can compare reactant with reactant, reactant with product, or product with product.
• A mole ratio can be written in either direction.
• Choose the direction that allows unwanted units to cancel.
• The basic mole-to-mole calculation is:

moles wanted = moles given × (coefficient wanted / coefficient given)

• Mole ratios act as conversion factors.
• Unit cancellation helps identify whether the correct ratio has been used.
• Coefficients are not mass ratios.
• Whole-number coefficients do not mean actual reacting quantities must be whole numbers.
• Stoichiometry is based on quantitative relationships between reactants and products.
• Mole ratios are the central conversion step in most stoichiometry calculations.
• More advanced calculations often follow:

given quantity → moles → mole ratio → moles wanted → wanted quantity

The central idea is:

BALANCE THE EQUATION → READ THE COEFFICIENTS → BUILD THE MOLE RATIO → CONVERT THE MOLES

---

Check Your Understanding

1. What is a mole ratio?

2. Where do mole ratios come from?

3. Why must an equation be balanced before determining mole ratios?

4. What does a coefficient represent?

5. Explain the difference between a coefficient and a subscript.

For questions 6–10, use:

2H₂ + O₂ → 2H₂O

6. What is the mole ratio H₂ : O₂?

7. What is the mole ratio O₂ : H₂O?

8. How many moles of H₂O can form from 5 mol H₂?

9. How many moles of O₂ are required for 10 mol H₂?

10. How many moles of H₂O form from 3 mol O₂?

For questions 11–15, use:

N₂ + 3H₂ → 2NH₃

11. What is the mole ratio N₂ : H₂?

12. What is the mole ratio H₂ : NH₃?

13. How many moles of NH₃ can form from 12 mol H₂?

14. How many moles of N₂ are required to produce 8 mol NH₃?

15. How many moles of H₂ are required for 5 mol N₂?

For questions 16–20, use:

CH₄ + 2O₂ → CO₂ + 2H₂O

16. How many moles of O₂ are required for 4 mol CH₄?

17. How many moles of CO₂ form from 6 mol CH₄?

18. How many moles of H₂O form from 2.5 mol CH₄?

19. How many moles of CH₄ are needed to produce 12 mol H₂O?

20. How many moles of CO₂ form when 14 mol O₂ are completely consumed with sufficient methane?

For questions 21–25, use:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

21. Write the mole ratio C₃H₈ : O₂.

22. Write the mole ratio C₃H₈ : CO₂.

23. How many moles of O₂ are needed for 3 mol C₃H₈?

24. How many moles of CO₂ can form from 5 mol C₃H₈?

25. How many moles of H₂O can form from 10 mol O₂?

Challenge

Consider:

4NH₃ + 5O₂ → 4NO + 6H₂O

26. Determine the mole ratio NH₃ : O₂.

27. Determine the mole ratio O₂ : H₂O.

28. Calculate the moles of O₂ required to react with 12 mol NH₃.

29. Calculate the moles of NO produced from 7.5 mol NH₃.

30. Calculate the moles of H₂O produced from 15 mol O₂.

31. Calculate the moles of NH₃ required to produce 18 mol H₂O.

32. Explain why the ratio 4 : 5 : 4 : 6 is a mole ratio rather than a mass ratio.

33. Explain why the coefficients can also represent a ratio of molecules.

34. Explain why changing a subscript while balancing an equation is chemically incorrect.

35. Explain how unit cancellation can help you determine whether you have used the correct mole ratio.

Stoichiometric Calculations

Stoichiometry is the quantitative study of the amounts of reactants and products involved in chemical reactions.

A balanced chemical equation acts like a chemical recipe. It tells us the relative amounts of substances that react and form.

For example:

2H₂ + O₂ → 2H₂O

This tells us:

2 mol H₂ + 1 mol O₂ → 2 mol H₂O

If we know the amount of one substance, we can use the balanced equation to calculate the amount of another.

This is the central idea of stoichiometry:

known amount → balanced equation → unknown amount

The Stoichiometric Relationship

Consider:

N₂ + 3H₂ → 2NH₃

The coefficients tell us:

1 mol N₂ : 3 mol H₂ : 2 mol NH₃

From this equation we can determine many relationships.

For every:

1 mol N₂

we need:

3 mol H₂

and theoretically produce:

2 mol NH₃

If the amount of one substance changes, the amounts of the others change proportionally.

---

The Mole Is the Bridge

The most important idea in stoichiometric calculations is that moles connect substances in a chemical equation.

A balanced equation directly relates:

moles ↔ moles

It does not directly relate grams to grams.

Therefore, if a question gives a quantity other than moles, we normally convert it into moles first.

The general pathway is:

given quantity → moles of given substance → mole ratio → moles of wanted substance → wanted quantity

This pathway is the foundation of most stoichiometry problems.

The Four Main Steps

A reliable approach is:

Balance the equation

Make sure the chemical equation is balanced.

Convert the given quantity to moles

If the question already gives moles, this step is unnecessary.

Use the mole ratio

Use coefficients from the balanced equation to convert between substances.

Convert to the requested quantity

If the answer is required in moles, stop.

If it is required in another quantity, perform the necessary conversion.

A useful summary is:

BALANCE → CONVERT TO MOLES → USE MOLE RATIO → CONVERT TO ANSWER

---

Mole-to-Mole Calculations

The simplest stoichiometric problems give one quantity in moles and ask for another quantity in moles.

Consider:

2H₂ + O₂ → 2H₂O

Suppose 5 mol H₂ react with sufficient oxygen.

How many moles of water can form?

Identify what is given

5 mol H₂

Identify what is wanted

mol H₂O

Find the mole ratio

From the equation:

2 mol H₂ : 2 mol H₂O

Calculate

5 mol H₂ × (2 mol H₂O / 2 mol H₂)

= 5 mol H₂O

Answer

5 mol H₂O

---

A Shortcut Formula

For mole-to-mole calculations:

moles wanted = moles given × (coefficient wanted / coefficient given)

For example:

N₂ + 3H₂ → 2NH₃

If we have 9 mol H₂:

moles NH₃ = 9 × (2/3)

= 6 mol NH₃

This formula is useful, but understanding the mole-ratio method is more important than memorizing the formula.

---

Worked Example: Ammonia

Consider:

N₂ + 3H₂ → 2NH₃

How many moles of NH₃ can form from 7.5 mol N₂?

Ratio:

1 mol N₂ : 2 mol NH₃

Calculation:

7.5 mol N₂ × (2 mol NH₃ / 1 mol N₂)

= 15 mol NH₃

Answer

15 mol NH₃

---

Worked Example: Finding a Reactant

Using:

N₂ + 3H₂ → 2NH₃

How many moles of H₂ are required to produce 12 mol NH₃?

Ratio:

3 mol H₂ : 2 mol NH₃

Calculation:

12 mol NH₃ × (3 mol H₂ / 2 mol NH₃)

= 18 mol H₂

Answer

18 mol H₂

Notice that stoichiometry can be used in either direction:

reactant → product

or:

product → reactant

---

Worked Example: Combustion

Methane burns according to:

CH₄ + 2O₂ → CO₂ + 2H₂O

How many moles of oxygen are required to burn 4.5 mol CH₄?

Ratio:

1 mol CH₄ : 2 mol O₂

Calculation:

4.5 mol CH₄ × (2 mol O₂ / 1 mol CH₄)

= 9 mol O₂

Answer

9 mol O₂

Multi-Step Stoichiometry

More advanced questions may require several conversions.

For example:

mass → moles → mole ratio → moles → mass

This is one of the most common stoichiometric pathways.

The key idea is that the mole ratio always connects the two different substances.

For example:

grams A → mol A → mol B → grams B

Notice where the chemical identity changes:

mol A → mol B

That is the mole-ratio step.

---

Mass and Moles

To convert between mass and moles:

n = m / M

where:

• n = amount in moles (mol)
• m = mass (g)
• M = molar mass (g/mol)

Rearranging:

m = nM

Therefore:

mass → moles

use:

n = m / M

and:

moles → mass

use:

m = nM

---

Worked Example: Mass to Moles to Moles

Consider:

2Mg + O₂ → 2MgO

Suppose 12.0 g Mg reacts with sufficient oxygen.

How many moles of MgO can form?

Use:

M(Mg) ≈ 24.3 g/mol

Convert Mg to moles

n = m / M

n = 12.0 / 24.3

n ≈ 0.494 mol Mg

Use the mole ratio

From:

2Mg + O₂ → 2MgO

Mg : MgO is:

2 : 2

or:

1 : 1

Therefore:

0.494 mol Mg × (2 mol MgO / 2 mol Mg)

= 0.494 mol MgO

Answer

0.494 mol MgO

---

Worked Example: Mass to Mass

Now suppose we want the mass of MgO produced.

Equation:

2Mg + O₂ → 2MgO

Given:

12.0 g Mg

Molar masses:

Mg ≈ 24.3 g/mol

MgO ≈ 40.3 g/mol

Convert Mg to moles

12.0 g ÷ 24.3 g/mol = 0.494 mol Mg

Apply the mole ratio

Mg : MgO = 1 : 1

Therefore:

0.494 mol MgO

Convert MgO to mass

m = nM

m = 0.494 × 40.3

m ≈ 19.9 g

Answer

19.9 g MgO

The complete pathway was:

12.0 g Mg → 0.494 mol Mg → 0.494 mol MgO → 19.9 g MgO

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Dimensional Analysis

The same calculation can be written as one continuous calculation:

12.0 g Mg × (1 mol Mg / 24.3 g Mg) × (2 mol MgO / 2 mol Mg) × (40.3 g MgO / 1 mol MgO)

Units cancel:

g Mg → mol Mg → mol MgO → g MgO

leaving:

19.9 g MgO

This method is called dimensional analysis.

It is extremely useful because the units show whether the calculation has been set up correctly.

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The Stoichiometry Road Map

Many problems can be understood using this structure:

mass of A

↓

moles of A

↓

MOLE RATIO

↓

moles of B

↓

mass of B

The middle step is always based on the balanced equation.

Worked Example: Producing Water

Consider:

2H₂ + O₂ → 2H₂O

How many grams of water can theoretically form from 8.0 g H₂, assuming sufficient oxygen?

Use approximate molar masses:

H₂ = 2.0 g/mol

H₂O = 18.0 g/mol

Convert H₂ to moles

8.0 g ÷ 2.0 g/mol = 4.0 mol H₂

Use the mole ratio

H₂ : H₂O = 2 : 2 = 1 : 1

Therefore:

4.0 mol H₂O

Convert water to mass

4.0 × 18.0 = 72 g

Answer

72 g H₂O

---

Worked Example: Oxygen Required

Using:

2H₂ + O₂ → 2H₂O

How many grams of oxygen are needed to react completely with 6.0 mol H₂?

Use the mole ratio

H₂ : O₂ = 2 : 1

6.0 mol H₂ × (1 mol O₂ / 2 mol H₂)

= 3.0 mol O₂

Molar mass:

O₂ = 32.0 g/mol

Convert to mass

m = nM

m = 3.0 × 32.0

= 96 g

Answer

96 g O₂

---

Worked Example: Propane Combustion

Propane burns according to:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

How many moles of CO₂ form when 2.5 mol C₃H₈ burns completely?

Mole ratio

C₃H₈ : CO₂ = 1 : 3

Calculate

2.5 mol C₃H₈ × (3 mol CO₂ / 1 mol C₃H₈)

= 7.5 mol CO₂

Answer

7.5 mol CO₂

---

Multi-Step Propane Problem

How many grams of CO₂ can form when 44 g C₃H₈ burns completely?

Equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Approximate molar masses:

C₃H₈ = 44 g/mol

CO₂ = 44 g/mol

Convert propane to moles

44 g ÷ 44 g/mol = 1 mol C₃H₈

Use the mole ratio

1 mol C₃H₈ : 3 mol CO₂

Therefore:

3 mol CO₂

Convert to mass

3 mol × 44 g/mol = 132 g

Answer

132 g CO₂

The pathway was:

44 g C₃H₈ → 1 mol C₃H₈ → 3 mol CO₂ → 132 g CO₂

---

Stoichiometry with Decomposition Reactions

Stoichiometry works with any correctly balanced reaction.

Consider the decomposition of calcium carbonate:

CaCO₃ → CaO + CO₂

The ratio is:

1 : 1 : 1

If 2.5 mol CaCO₃ decomposes completely:

2.5 mol CaCO₃ → 2.5 mol CaO + 2.5 mol CO₂

Worked Example: Calcium Carbonate

How many grams of CO₂ can form from 100 g CaCO₃?

Approximate molar masses:

CaCO₃ = 100 g/mol

CO₂ = 44 g/mol

Convert CaCO₃ to moles

100 g ÷ 100 g/mol = 1 mol CaCO₃

Apply the mole ratio

CaCO₃ : CO₂ = 1 : 1

Therefore:

1 mol CO₂

Convert to mass

1 × 44 = 44 g

Answer

44 g CO₂

---

Stoichiometry with Synthesis Reactions

Consider:

4Fe + 3O₂ → 2Fe₂O₃

Suppose 8 mol Fe reacts with sufficient oxygen.

How many moles of Fe₂O₃ can form?

Ratio:

4 mol Fe : 2 mol Fe₂O₃

Calculation:

8 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe)

= 4 mol Fe₂O₃

Answer

4 mol Fe₂O₃

---

A More Complex Mass Calculation

Consider:

4Fe + 3O₂ → 2Fe₂O₃

How many grams of Fe₂O₃ can theoretically form from 112 g Fe?

Use approximate molar masses:

Fe = 56 g/mol

Fe₂O₃ = 160 g/mol

Convert Fe to moles

112 ÷ 56 = 2 mol Fe

Use the mole ratio

2 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe)

= 1 mol Fe₂O₃

Convert to mass

1 × 160 = 160 g

Answer

160 g Fe₂O₃

---

Why the Product Can Have More Mass Than One Reactant

In the previous example:

112 g Fe → 160 g Fe₂O₃

Does this violate conservation of mass?

No.

The iron combines with oxygen from O₂.

The additional mass comes from oxygen.

The complete reaction conserves mass:

mass of all reactants = mass of all products

This is an important point when interpreting stoichiometric calculations.

---

Starting with the Product

Stoichiometry does not always move from reactant to product.

Suppose:

2KClO₃ → 2KCl + 3O₂

How many moles of KClO₃ are required to produce 9 mol O₂?

Ratio:

2 mol KClO₃ : 3 mol O₂

Calculation:

9 mol O₂ × (2 mol KClO₃ / 3 mol O₂)

= 6 mol KClO₃

Answer

6 mol KClO₃

The calculation can move backward through the equation.

---

Multi-Step Problems with Several Substances

Consider:

2Al + 3Cl₂ → 2AlCl₃

Suppose we want to determine how much chlorine is needed to produce 26.7 g AlCl₃.

Approximate molar mass:

AlCl₃ = 133.5 g/mol

Convert AlCl₃ to moles

26.7 ÷ 133.5 = 0.200 mol AlCl₃

Use the mole ratio

Cl₂ : AlCl₃ = 3 : 2

0.200 mol AlCl₃ × (3 mol Cl₂ / 2 mol AlCl₃)

= 0.300 mol Cl₂

If the question asks for moles, stop here.

Answer

0.300 mol Cl₂

If mass were requested, we would continue by multiplying by the molar mass of Cl₂.

---

Deciding Which Conversion to Use

Ask:

What unit do I have?

and:

What unit do I need?

If you have grams:

grams → moles

If you have moles:

you may be ready for the mole ratio.

If the answer requires grams:

moles → grams

This prevents unnecessary calculations.

---

The Mole Ratio Is the Chemical Bridge

Suppose substances A and B participate in a reaction.

You cannot normally jump directly from:

grams A → grams B

Instead:

grams A → mol A → mol B → grams B

The conversion:

mol A → mol B

comes from the balanced equation.

This is the chemical bridge between the two substances.

---

Why Coefficients Matter

Consider:

2Al + 3Cl₂ → 2AlCl₃

Suppose you have 6 mol Al.

A common mistake would be to assume you need 6 mol Cl₂.

But the equation says:

2 mol Al : 3 mol Cl₂

Therefore:

6 mol Al × (3 mol Cl₂ / 2 mol Al)

= 9 mol Cl₂

Stoichiometry depends on the actual coefficients, not simply on the number of substances present.

---

Multi-Step Calculation Strategy

When facing a longer problem, write this at the top of your page:

GIVEN → mol GIVEN → mol WANTED → WANTED

Then fill in the quantities.

For example:

24.3 g Mg → mol Mg → mol MgO → g MgO

This gives you a roadmap before you calculate anything.

---

Real-World Connection: Chemical Manufacturing

Industrial chemists use stoichiometry to calculate how much raw material is required to manufacture products.

For example, ammonia is produced using:

N₂ + 3H₂ ⇌ 2NH₃

The balanced equation gives a mole ratio of:

1 mol N₂ : 3 mol H₂ : 2 mol NH₃

Manufacturers need to know:

• how much nitrogen is required
• how much hydrogen is required
• how much ammonia could theoretically form
• how efficiently raw materials are being used

Real-World Connection: Pharmaceuticals

Pharmaceutical manufacturing also requires careful control of quantities.

Using incorrect proportions can:

• waste expensive reactants
• reduce product formation
• produce unwanted by-products
• make purification more difficult

Stoichiometry allows chemists to predict how much starting material is required to produce a desired amount of product.

---

Real-World Connection: Environmental Chemistry

Stoichiometry can be used to determine quantities needed to:

• neutralize acidic waste
• remove pollutants
• treat wastewater
• calculate combustion emissions
• analyze atmospheric reactions

For example:

HCl + NaOH → NaCl + H₂O

The mole ratio between HCl and NaOH is:

1 : 1

Therefore, 1 mol NaOH is stoichiometrically required to neutralize 1 mol HCl.

---

Real-World Connection: Combustion and Emissions

Consider:

CH₄ + 2O₂ → CO₂ + 2H₂O

The equation predicts:

1 mol CH₄ → 1 mol CO₂

Therefore, knowing how much methane is burned allows us to calculate the theoretical amount of carbon dioxide produced.

Stoichiometry is therefore important when estimating emissions from chemical processes and fuels.

---

Theoretical Amounts vs. Actual Amounts

Stoichiometric calculations predict what should happen according to the balanced equation.

These calculated quantities are theoretical.

Real experiments may produce less product because:

• reactions may not go to completion
• material may be lost during transfer
• competing reactions may occur
• products may be lost during purification
• experimental measurements have uncertainty

Later, these ideas lead to concepts such as:

theoretical yield

and:

percentage yield

---

Assumptions in Basic Stoichiometry

Simple stoichiometric calculations often assume:

• the equation is correct and balanced
• reactants are pure
• the reaction proceeds as written
• the required reactants are available
• the reaction proceeds completely
• there are no significant competing reactions
• no product is lost

These assumptions create an idealized calculation.

Actual laboratory results may differ.

---

Common Mistakes

Not Balancing the Equation First

Wrong:

H₂ + O₂ → H₂O

Correct:

2H₂ + O₂ → 2H₂O

The mole ratio must come from the balanced equation.

---

Using Subscripts as Mole Ratios

For:

2H₂ + O₂ → 2H₂O

the H₂ : O₂ ratio is:

2 : 1

Use coefficients, not subscripts.

---

Skipping the Mole Conversion

If a question gives grams, do not usually apply the coefficients directly to the masses.

Use:

grams → moles → mole ratio

---

Treating Coefficients as Gram Ratios

For:

2H₂ + O₂ → 2H₂O

the coefficients do not mean:

2 g H₂ + 1 g O₂ → 2 g H₂O

They represent relative numbers of moles.

---

Using the Mole Ratio Backwards

Suppose:

N₂ + 3H₂ → 2NH₃

and we are converting H₂ into NH₃.

Use:

2 mol NH₃ / 3 mol H₂

because mol H₂ must cancel.

---

Using the Wrong Molar Mass

For O₂:

M = 32.0 g/mol

not 16.0 g/mol.

For CO₂:

M ≈ 44.0 g/mol

not 12.0 g/mol.

Always calculate molar mass for the complete chemical formula.

---

Rounding Too Early

Keep several digits during intermediate calculations.

Round appropriately at the end.

Early rounding can make the final answer less accurate.

---

Forgetting Units

Always include units.

For example:

0.50 mol

18.0 g

44.0 g/mol

Units help reveal calculation errors.

---

Doing Unnecessary Steps

If the question gives moles and asks for moles:

moles → mole ratio → moles

There is no reason to calculate mass first.

---

Key Terms

Stoichiometry — The quantitative study of the relationships between reactants and products in chemical reactions.

Stoichiometric calculation — A calculation using a balanced chemical equation to determine quantities of substances.

Balanced chemical equation — An equation containing equal numbers of each type of atom on both sides.

Coefficient — A number before a chemical formula showing the relative amount of that substance in a reaction.

Mole — The amount of substance containing 6.022 × 10²³ representative particles.

Mole ratio — A ratio between amounts of substances obtained from the coefficients of a balanced equation.

Molar mass — The mass of one mole of a substance, usually expressed in g/mol.

Reactant — A starting substance in a chemical reaction.

Product — A substance produced by a chemical reaction.

Conversion factor — A ratio used to convert one quantity into another.

Dimensional analysis — A calculation method that uses conversion factors and unit cancellation.

Unit cancellation — The cancellation of identical units appearing in the numerator and denominator of conversion factors.

Conservation of mass — The principle that total mass is conserved during a chemical reaction.

Theoretical amount — The quantity predicted from a balanced chemical equation under ideal conditions.

Theoretical yield — The maximum amount of product predicted by stoichiometric calculations.

Limiting reactant — The reactant that is consumed first and limits how much product can form.

Excess reactant — A reactant present in more than the amount required by the reaction ratio.

Yield — The amount of product obtained from a chemical reaction.

---

Key Takeaways

• Stoichiometry connects quantities of different substances in chemical reactions.
• Every stoichiometric calculation begins with a balanced chemical equation.
• Coefficients provide mole ratios.
• The mole is the central unit connecting different substances.
• A balanced equation directly relates moles, not grams.
• If a problem gives mass, convert mass to moles before using the mole ratio.
• Use n = m/M to convert mass into moles.
• Use m = nM to convert moles into mass.
• Mole-to-mole problems require only the mole ratio.
• Mass-to-mass problems usually require three conversions.
• The standard mass-to-mass pathway is:

mass A → mol A → mol B → mass B

• The conversion between mol A and mol B comes from the balanced equation.
• Unit cancellation can be used to check whether a calculation has been arranged correctly.
• Stoichiometry can calculate reactants from products or products from reactants.
• Multi-step calculations become easier when a conversion pathway is written before calculating.
• Stoichiometric calculations predict theoretical quantities.
• Actual experimental results may differ from theoretical predictions.
• Stoichiometry is used in manufacturing, medicine, environmental science, energy production, and laboratory chemistry.

The most useful roadmap is:

GIVEN QUANTITY → MOLES GIVEN → MOLE RATIO → MOLES WANTED → WANTED QUANTITY

---

Check Your Understanding

For questions 1–5, use:

2H₂ + O₂ → 2H₂O

1. How many moles of H₂O can form from 8 mol H₂?

2. How many moles of O₂ are required for 14 mol H₂?

3. How many moles of H₂O can form from 4.5 mol O₂?

4. How many grams of H₂O can form from 2 mol H₂?

5. How many grams of O₂ are required for 5 mol H₂?

For questions 6–10, use:

N₂ + 3H₂ → 2NH₃

6. How many moles of NH₃ can form from 6 mol N₂?

7. How many moles of H₂ are required to produce 10 mol NH₃?

8. How many moles of N₂ are required to produce 16 mol NH₃?

9. How many grams of NH₃ can form from 3 mol N₂? Use M(NH₃) = 17 g/mol.

10. How many grams of H₂ are required to produce 34 g NH₃? Use M(H₂) = 2 g/mol.

For questions 11–15, use:

CH₄ + 2O₂ → CO₂ + 2H₂O

11. How many moles of CO₂ form from 3.5 mol CH₄?

12. How many moles of O₂ are required for 7 mol CH₄?

13. How many grams of CO₂ form from 2 mol CH₄?

14. How many grams of H₂O form from 1.5 mol CH₄?

15. How many moles of CH₄ must burn to produce 88 g CO₂?

For questions 16–20, use:

2Mg + O₂ → 2MgO

Use:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

16. How many moles of MgO form from 4 mol Mg?

17. How many moles of O₂ are required for 10 mol Mg?

18. How many moles of Mg are present in 48.6 g Mg?

19. How many grams of MgO can theoretically form from 48.6 g Mg?

20. How many grams of Mg are required to produce 80.6 g MgO?

Multi-Step Challenge

Use:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Approximate molar masses:

C₃H₈ = 44 g/mol

O₂ = 32 g/mol

CO₂ = 44 g/mol

H₂O = 18 g/mol

21. How many moles of O₂ are required to burn 2 mol C₃H₈?

22. How many moles of CO₂ form from 5 mol C₃H₈?

23. How many grams of CO₂ form from 44 g C₃H₈?

24. How many grams of H₂O form from 88 g C₃H₈?

25. How many grams of O₂ are required to burn 132 g C₃H₈ completely?

26. How many grams of C₃H₈ must burn to produce 264 g CO₂?

27. Write the complete conversion pathway for converting grams of C₃H₈ into grams of H₂O.

28. Explain why the mole ratio must be taken from a balanced equation.

29. Explain why grams of one substance cannot normally be converted directly into grams of another using the coefficients.

30. A student calculates that 112 g of iron can produce 160 g of iron oxide and claims that mass has been created. Explain why this conclusion is incorrect.

Reacting Quantities

Chemical reactions occur in specific quantitative proportions. A balanced chemical equation tells us not only which substances react, but also the relative amounts required.

For example:

2H₂ + O₂ → 2H₂O

This means:

2 mol H₂ react with 1 mol O₂

Therefore, if we know how much hydrogen is available, we can determine exactly how much oxygen is required.

This is the main idea behind reacting quantities:

balanced equation → mole ratio → required amount

These calculations are extremely useful because chemists rarely want to mix reactants randomly. They want to know how much of each substance is needed for the reaction.

Reactants Must Be Present in the Correct Ratio

Consider:

2H₂ + O₂ → 2H₂O

The required ratio is:

H₂ : O₂ = 2 : 1

So:

• 2 mol H₂ require 1 mol O₂
• 4 mol H₂ require 2 mol O₂
• 6 mol H₂ require 3 mol O₂
• 10 mol H₂ require 5 mol O₂

The quantities change, but the ratio remains:

2 : 1

This is called the stoichiometric ratio.

---

Why Balanced Equations Matter

The reacting quantities must come from a balanced chemical equation.

Consider:

Mg + O₂ → MgO

This equation is not balanced.

The balanced equation is:

2Mg + O₂ → 2MgO

Therefore:

2 mol Mg react with 1 mol O₂

If we used the unbalanced equation, we might incorrectly assume:

1 mol Mg reacts with 1 mol O₂

That would give the wrong reacting quantities.

Always:

BALANCE FIRST → CALCULATE SECOND

---

Coefficients Give Mole Ratios

Consider:

N₂ + 3H₂ → 2NH₃

The coefficients are:

1 : 3 : 2

Therefore:

1 mol N₂ reacts with 3 mol H₂

and theoretically produces:

2 mol NH₃

For reacting-quantity questions, we often focus on the two reactants:

N₂ : H₂ = 1 : 3

The coefficients provide the conversion factor between them.

---

Reacting Quantities in Moles

The simplest reacting-quantity problems give one reactant in moles and ask how many moles of another reactant are required.

The general calculation is:

moles wanted = moles given × (coefficient wanted / coefficient given)

For example:

2H₂ + O₂ → 2H₂O

How many moles of O₂ are required for 8 mol H₂?

8 mol H₂ × (1 mol O₂ / 2 mol H₂)

= 4 mol O₂

Therefore:

8 mol H₂ require 4 mol O₂

---

A Simple Calculation Method

For most reacting-quantity problems:

Balance the equation

Make sure the equation is correct.

Identify the known reactant

What quantity has been given?

Identify the required reactant

What quantity must be calculated?

Convert to moles if necessary

If mass is given:

n = m / M

Apply the mole ratio

Use the coefficients from the balanced equation.

Convert to the required unit

If mass is required:

m = nM

The overall pathway is:

KNOWN REACTANT → MOLES → MOLE RATIO → MOLES OF REQUIRED REACTANT → REQUIRED QUANTITY

Worked Example: Hydrogen and Oxygen

Consider:

2H₂ + O₂ → 2H₂O

How many moles of O₂ are required to react completely with 7 mol H₂?

Identify the ratio

H₂ : O₂ = 2 : 1

Calculate

7 mol H₂ × (1 mol O₂ / 2 mol H₂)

= 3.5 mol O₂

Answer

3.5 mol O₂

Notice that reacting quantities do not have to be whole numbers.

---

Worked Example: Making Ammonia

Consider:

N₂ + 3H₂ → 2NH₃

How many moles of H₂ are required to react completely with 4 mol N₂?

Ratio:

1 mol N₂ : 3 mol H₂

Calculation:

4 mol N₂ × (3 mol H₂ / 1 mol N₂)

= 12 mol H₂

Answer

12 mol H₂

---

Worked Example: Working Backwards

Using:

N₂ + 3H₂ → 2NH₃

How many moles of N₂ are required to react with 15 mol H₂?

Ratio:

1 mol N₂ : 3 mol H₂

Calculation:

15 mol H₂ × (1 mol N₂ / 3 mol H₂)

= 5 mol N₂

Answer

5 mol N₂

Mole ratios can be used in either direction.

---

Reacting Quantities Involving Mass

Laboratory chemicals are often measured by mass, not by counting moles directly.

Therefore, many practical questions follow:

mass A → moles A → moles B → mass B

This is one of the most important calculation pathways in chemistry.

Remember:

n = m / M

and:

m = nM

where:

• n = amount in mol
• m = mass in g
• M = molar mass in g/mol

---

Worked Example: Magnesium and Oxygen

Magnesium reacts with oxygen:

2Mg + O₂ → 2MgO

How many grams of O₂ are required to react completely with 24.3 g Mg?

Molar masses:

Mg = 24.3 g/mol

O₂ = 32.0 g/mol

Convert Mg to moles

n = m / M

n = 24.3 / 24.3

= 1.00 mol Mg

Use the mole ratio

From:

2Mg + O₂ → 2MgO

2 mol Mg : 1 mol O₂

Therefore:

1.00 mol Mg × (1 mol O₂ / 2 mol Mg)

= 0.500 mol O₂

Convert O₂ to mass

m = nM

m = 0.500 × 32.0

= 16.0 g

Answer

16.0 g O₂

The complete pathway was:

24.3 g Mg → 1.00 mol Mg → 0.500 mol O₂ → 16.0 g O₂

---

Why Mass Ratios Are Different from Mole Ratios

Consider again:

2Mg + O₂ → 2MgO

The mole ratio is:

2 mol Mg : 1 mol O₂

But this does not mean:

2 g Mg : 1 g O₂

Convert the amounts into mass:

2 mol Mg:

2 × 24.3 = 48.6 g

1 mol O₂:

1 × 32.0 = 32.0 g

Therefore, the reacting mass relationship is:

48.6 g Mg : 32.0 g O₂

Mole ratios and mass ratios are not generally the same.

---

Worked Example: Finding the Reacting Mass

Aluminum reacts with chlorine:

2Al + 3Cl₂ → 2AlCl₃

How many grams of chlorine gas are required to react completely with 5.40 g Al?

Use:

Al = 27.0 g/mol

Cl₂ = 71.0 g/mol

Convert aluminum to moles

5.40 ÷ 27.0 = 0.200 mol Al

Apply the mole ratio

Al : Cl₂ = 2 : 3

0.200 mol Al × (3 mol Cl₂ / 2 mol Al)

= 0.300 mol Cl₂

Convert chlorine to mass

0.300 × 71.0 = 21.3 g

Answer

21.3 g Cl₂

---

Writing the Calculation as One Line

The same calculation can be written using dimensional analysis:

5.40 g Al × (1 mol Al / 27.0 g Al) × (3 mol Cl₂ / 2 mol Al) × (71.0 g Cl₂ / 1 mol Cl₂)

The units cancel:

g Al → mol Al → mol Cl₂ → g Cl₂

leaving:

21.3 g Cl₂

This method can make complicated stoichiometric calculations easier to organize.

---

Worked Example: Iron and Oxygen

Iron reacts with oxygen:

4Fe + 3O₂ → 2Fe₂O₃

How many grams of O₂ are required to react completely with 112 g Fe?

Use:

Fe = 56 g/mol

O₂ = 32 g/mol

Convert Fe to moles

112 ÷ 56 = 2 mol Fe

Use the mole ratio

Fe : O₂ = 4 : 3

2 mol Fe × (3 mol O₂ / 4 mol Fe)

= 1.5 mol O₂

Convert to mass

1.5 × 32 = 48 g

Answer

48 g O₂

Therefore:

112 g Fe reacts with 48 g O₂

If the reaction forms only Fe₂O₃, conservation of mass predicts:

112 g + 48 g = 160 g Fe₂O₃

---

Conservation of Mass

Reacting quantities must obey the law of conservation of mass.

For:

4Fe + 3O₂ → 2Fe₂O₃

we found:

112 g Fe + 48 g O₂ → 160 g Fe₂O₃

Total reactant mass:

160 g

Total product mass:

160 g

Mass has been conserved.

Worked Example: Combustion of Methane

Methane burns according to:

CH₄ + 2O₂ → CO₂ + 2H₂O

How many grams of O₂ are required to burn 16 g CH₄ completely?

Use:

CH₄ = 16 g/mol

O₂ = 32 g/mol

Convert methane to moles

16 ÷ 16 = 1 mol CH₄

Use the mole ratio

CH₄ : O₂ = 1 : 2

Therefore:

1 mol CH₄ requires 2 mol O₂

Convert oxygen to mass

2 × 32 = 64 g

Answer

64 g O₂

Therefore:

16 g CH₄ requires 64 g O₂

for complete combustion according to this equation.

---

Worked Example: Propane Combustion

Propane burns according to:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

How many grams of oxygen are required to burn 44 g propane?

Use:

C₃H₈ = 44 g/mol

O₂ = 32 g/mol

Convert propane to moles

44 ÷ 44 = 1 mol C₃H₈

Use the ratio

1 mol C₃H₈ : 5 mol O₂

Therefore:

5 mol O₂

Convert to mass

5 × 32 = 160 g

Answer

160 g O₂

So:

44 g propane requires 160 g oxygen

for complete combustion.

Why Combustion Requires So Much Oxygen

Students are sometimes surprised that a relatively small mass of fuel can require a much larger mass of oxygen.

For example:

44 g propane requires 160 g O₂

This happens because combustion combines the fuel with oxygen from the surrounding air.

The mass of the combustion products therefore includes:

mass from the fuel + mass from oxygen

This is why combustion products can have a greater total mass than the original fuel alone.

---

Practical Situation: Acid Neutralization

Reacting quantities are important in neutralization.

Consider:

HCl + NaOH → NaCl + H₂O

The mole ratio is:

1 mol HCl : 1 mol NaOH

Therefore:

0.50 mol HCl requires 0.50 mol NaOH

If the molar mass of NaOH is:

40.0 g/mol

then:

m = nM

m = 0.50 × 40.0

= 20.0 g NaOH

Therefore:

0.50 mol HCl requires 20.0 g NaOH

according to the balanced equation.

---

Practical Situation: Acid and Carbonate

Consider:

2HCl + CaCO₃ → CaCl₂ + H₂O + CO₂

The reacting mole ratio is:

2 mol HCl : 1 mol CaCO₃

Suppose we have:

0.40 mol HCl

How much CaCO₃ is required?

0.40 mol HCl × (1 mol CaCO₃ / 2 mol HCl)

= 0.20 mol CaCO₃

Molar mass of CaCO₃:

100 g/mol

Therefore:

0.20 × 100 = 20 g

Answer

20 g CaCO₃

---

Why This Matters in Neutralization

If too little carbonate is used:

some acid remains

If exactly the stoichiometric amount is used:

the reactants are present in the proportion required by the equation.

If more carbonate is added than required:

carbonate remains in excess

This introduces two important ideas:

limiting reactant

and:

excess reactant

---

Exact Stoichiometric Quantities

Suppose:

2H₂ + O₂ → 2H₂O

We mix:

4 mol H₂

and:

2 mol O₂

The required ratio is:

2 : 1

Our ratio is:

4 : 2

which simplifies to:

2 : 1

Therefore, the reactants are present in the exact stoichiometric proportion.

If the reaction proceeds completely as written:

• all H₂ can be consumed
• all O₂ can be consumed
• neither is left in excess

---

What Happens If the Ratio Is Wrong?

Suppose instead we mix:

4 mol H₂

with:

5 mol O₂

But only:

2 mol O₂

are required for 4 mol H₂.

Therefore:

3 mol O₂ remain

after all the hydrogen has reacted, assuming the reaction proceeds completely.

Hydrogen is the:

limiting reactant

Oxygen is the:

excess reactant

A later topic may examine limiting reactants in more detail, but reacting quantities provide the foundation.

---

Another Example of Excess Reactant

Consider:

N₂ + 3H₂ → 2NH₃

Suppose we have:

2 mol N₂

How much H₂ is required?

Ratio:

1 : 3

Therefore:

2 mol N₂ require 6 mol H₂

If we actually supply:

10 mol H₂

then:

6 mol H₂ are required

and:

4 mol H₂ are extra

Hydrogen is present in excess.

---

Reacting Quantities and Laboratory Planning

Before performing an experiment, chemists can calculate the required reactant quantities.

Suppose a student wants to react:

0.10 mol Mg

with hydrochloric acid.

Equation:

Mg + 2HCl → MgCl₂ + H₂

The ratio is:

1 mol Mg : 2 mol HCl

Therefore:

0.10 mol Mg requires 0.20 mol HCl

This calculation can be performed before the experiment begins.

Worked Example: Magnesium and Hydrochloric Acid

How many grams of HCl are required to react completely with 4.86 g Mg?

Equation:

Mg + 2HCl → MgCl₂ + H₂

Use:

Mg = 24.3 g/mol

HCl = 36.5 g/mol

Convert Mg to moles

4.86 ÷ 24.3 = 0.200 mol Mg

Apply the mole ratio

Mg : HCl = 1 : 2

0.200 mol Mg × (2 mol HCl / 1 mol Mg)

= 0.400 mol HCl

Convert HCl to mass

0.400 × 36.5 = 14.6 g

Answer

14.6 g HCl

---

Practical Chemistry and Safety

Reacting-quantity calculations can also improve laboratory safety.

If chemists calculate quantities before an experiment, they can avoid using unnecessarily large amounts of chemicals.

This can reduce:

• chemical waste
• cost
• exposure to hazardous substances
• quantities requiring disposal
• severity of possible spills

Good stoichiometry therefore supports both:

efficient chemistry

and:

safer chemistry

---

Reacting Quantities and Green Chemistry

Using excessive quantities of reactants can create unnecessary waste.

Suppose a reaction requires:

1 mol A : 2 mol B

Using much more B than necessary may:

• waste raw material
• require additional separation
• increase disposal requirements
• increase production costs

Industrial chemists therefore carefully control reacting quantities.

This connects stoichiometry with:

green chemistry and sustainability

---

Real-World Application: Fertilizer Production

Ammonia is an important raw material for fertilizer production.

It can be produced using:

N₂ + 3H₂ ⇌ 2NH₃

The stoichiometric relationship is:

1 mol N₂ : 3 mol H₂

Large chemical plants must carefully control the amounts of gases entering industrial processes.

Even though real industrial systems involve additional complications such as equilibrium, recycling, temperature, pressure, and conversion efficiency, the balanced equation provides the basic quantitative relationship.

---

Real-World Application: Combustion

Engines, furnaces, boilers, and burners require appropriate amounts of fuel and oxygen.

Too little oxygen can cause:

incomplete combustion

For hydrocarbons, incomplete combustion may produce substances including:

• carbon monoxide
• carbon
• unburned hydrocarbons

Correct reacting quantities therefore have implications for:

• efficiency
• pollution
• fuel consumption
• safety

---

Real-World Application: Environmental Treatment

Stoichiometric calculations can help determine how much chemical is required to:

• neutralize acidic waste
• treat alkaline waste
• remove contaminants
• precipitate dissolved substances
• control water chemistry

Using too little treatment chemical may leave contaminants untreated.

Using excessive amounts may:

• waste chemicals
• increase costs
• create additional environmental problems

---

Comparing Mole and Mass Relationships

Consider:

2H₂ + O₂ → 2H₂O

Mole relationship

2 mol H₂ : 1 mol O₂

Using molar masses:

H₂ = 2 g/mol

O₂ = 32 g/mol

Mass relationship

2 mol H₂:

2 × 2 = 4 g

1 mol O₂:

1 × 32 = 32 g

Therefore:

4 g H₂ reacts with 32 g O₂

Notice:

mole ratio = 2 : 1

but:

mass ratio = 4 : 32 = 1 : 8

These ratios are very different.

---

Scaling Reacting Quantities

Once we know the correct reacting quantities, we can scale them.

For:

4 g H₂ : 32 g O₂

divide both by 4:

1 g H₂ : 8 g O₂

Multiply by 10:

10 g H₂ : 80 g O₂

Multiply by 25:

25 g H₂ : 200 g O₂

The mass ratio remains constant because it comes from the stoichiometric mole relationship and the substances' molar masses.

---

Worked Example: Scaling by Mass

Suppose:

2H₂ + O₂ → 2H₂O

We know:

4 g H₂ requires 32 g O₂

How much oxygen is required for:

12 g H₂?

12 g is three times 4 g.

Therefore:

32 × 3 = 96 g O₂

Answer

96 g O₂

This proportional method works when the reacting mass relationship is already known.

For unfamiliar reactions, the mole method is generally safer.

---

A Reliable Problem-Solving Checklist

Before calculating, ask:

Is the equation balanced?

Then identify:

What substance do I know?

What substance do I need?

What unit was I given?

What unit is required?

Then write the pathway:

given → mol given → mol wanted → wanted unit

Finally ask:

Does my answer make chemical sense?

---

Worked Example: Full Multi-Step Problem

Calcium reacts with water:

Ca + 2H₂O → Ca(OH)₂ + H₂

How many grams of water are required to react completely with 20.0 g Ca?

Use:

Ca = 40.0 g/mol

H₂O = 18.0 g/mol

Convert calcium to moles

20.0 ÷ 40.0 = 0.500 mol Ca

Apply the mole ratio

Ca : H₂O = 1 : 2

0.500 mol Ca × (2 mol H₂O / 1 mol Ca)

= 1.00 mol H₂O

Convert water to mass

1.00 × 18.0 = 18.0 g

Answer

18.0 g H₂O

The pathway was:

20.0 g Ca → 0.500 mol Ca → 1.00 mol H₂O → 18.0 g H₂O

---

Worked Example: A More Challenging Reaction

Consider:

2Al + 3CuCl₂ → 2AlCl₃ + 3Cu

How many grams of CuCl₂ are required to react completely with 5.40 g Al?

Use:

Al = 27.0 g/mol

CuCl₂ = 134.5 g/mol

Convert Al to moles

5.40 ÷ 27.0 = 0.200 mol Al

Apply the ratio

Al : CuCl₂ = 2 : 3

0.200 mol Al × (3 mol CuCl₂ / 2 mol Al)

= 0.300 mol CuCl₂

Convert to mass

0.300 × 134.5 = 40.35 g

Answer

Approximately:

40.4 g CuCl₂

---

Checking the Answer

After completing a calculation, check:

Equation

Was it balanced?

Mole Ratio

Did you use coefficients rather than subscripts?

Direction

Did the given substance cancel?

Molar Mass

Did you calculate the complete formula correctly?

Units

Does your final answer have the requested unit?

Magnitude

Does the answer seem reasonable?

These checks catch many common errors.

---

Common Mistakes

Using an Unbalanced Equation

Mole ratios only work correctly with balanced equations.

---

Treating Mole Ratios as Mass Ratios

For:

2H₂ + O₂ → 2H₂O

2 : 1 is a mole ratio, not a gram ratio.

---

Forgetting to Convert Mass to Moles

If mass is given, usually begin:

mass → moles

before using the coefficients.

---

Using Subscripts Instead of Coefficients

For:

2Mg + O₂ → 2MgO

Mg : O₂ = 2 : 1

Do not use the subscripts in the formulas to create the mole ratio.

---

Reversing the Mole Ratio

If converting Mg into O₂:

2Mg + O₂ → 2MgO

use:

1 mol O₂ / 2 mol Mg

so that mol Mg cancels.

---

Using Atomic Mass Instead of Molecular Molar Mass

Oxygen gas is:

O₂

Therefore:

M(O₂) = 32.0 g/mol

not 16.0 g/mol.

Similarly:

Cl₂ ≈ 71.0 g/mol

not 35.5 g/mol.

---

Assuming Equal Masses React

Equal numbers of moles do not necessarily have equal masses.

Different substances have different molar masses.

---

Assuming More Reactant Is Always Better

Excess reactant may:

• waste material
• increase cost
• require separation
• increase waste

The correct quantity depends on the reaction and purpose.

---

Forgetting Conservation of Mass

If a product has more mass than one reactant, that does not mean mass was created.

Other reactants contributed mass.

---

Rounding Too Early

Keep extra digits during intermediate calculations and round at the end.

---

Key Terms

Reacting quantity — The amount of a substance required or involved in a chemical reaction.

Stoichiometry — The quantitative study of relationships between reactants and products.

Stoichiometric ratio — The mole relationship between substances specified by a balanced equation.

Stoichiometric amount — The amount of a substance required according to the balanced chemical equation.

Balanced chemical equation — An equation with equal numbers of each type of atom on both sides.

Coefficient — A number before a chemical formula indicating its relative amount in the reaction.

Mole ratio — A ratio between amounts in moles obtained from coefficients in a balanced equation.

Mole — An amount of substance containing 6.022 × 10²³ representative particles.

Molar mass — The mass of one mole of a substance, usually expressed in g/mol.

Reactant — A starting substance in a chemical reaction.

Product — A substance formed during a chemical reaction.

Conversion factor — A ratio used to convert between quantities.

Dimensional analysis — A method of calculation using conversion factors and unit cancellation.

Conservation of mass — The principle that total mass remains constant during a chemical reaction.

Limiting reactant — The reactant that is consumed first and therefore limits product formation.

Excess reactant — A reactant present in more than the stoichiometric amount required.

Complete combustion — Combustion in sufficient oxygen that, for a hydrocarbon, ideally produces carbon dioxide and water.

Incomplete combustion — Combustion occurring with insufficient oxygen, potentially producing carbon monoxide, carbon, and other products.

Neutralization — A reaction in which an acid and base react, typically producing a salt and water.

---

Key Takeaways

• Chemical reactions require reactants in specific proportions.
• These proportions come from balanced chemical equations.
• Coefficients provide the mole ratios between reactants.
• Reacting quantities are fundamentally based on moles.
• A balanced equation must be used before any stoichiometric calculation.
• If one reacting quantity is known, the required quantity of another reactant can be calculated.
• For mole-to-mole problems:

moles wanted = moles given × (coefficient wanted / coefficient given)

• For mass-to-mass reacting-quantity problems:

mass A → mol A → mol B → mass B

• Convert mass to moles using:

n = m/M

• Convert moles to mass using:

m = nM

• Mole ratios are not usually the same as mass ratios.
• Different substances have different molar masses.
• Unit cancellation helps verify that calculations are arranged correctly.
• Reacting quantities can be scaled while maintaining the same stoichiometric proportions.
• If reactants are supplied in exactly the required ratio, neither should remain in excess after complete reaction as written.
• If one reactant is supplied in excess, another reactant limits how far the reaction can proceed.
• Conservation of mass applies to all reacting quantities.
• Practical stoichiometry helps reduce waste, control costs, and improve laboratory safety.
• Reacting-quantity calculations are used in combustion, neutralization, manufacturing, environmental treatment, and many other chemical processes.

The central strategy is:

BALANCE → CONVERT TO MOLES → USE THE REACTANT MOLE RATIO → CONVERT TO THE REQUIRED QUANTITY

---

Check Your Understanding

For questions 1–5, use:

2H₂ + O₂ → 2H₂O

1. How many moles of O₂ are required for 6 mol H₂?

2. How many moles of H₂ are required for 4 mol O₂?

3. How many grams of O₂ are required for 4 mol H₂?

4. How many grams of H₂ are required to react with 64 g O₂? Use M(H₂) = 2.0 g/mol.

5. Explain why the mole ratio 2 : 1 is not the same as the reacting mass ratio.

For questions 6–10, use:

N₂ + 3H₂ → 2NH₃

6. How many moles of H₂ are required for 5 mol N₂?

7. How many moles of N₂ are required for 21 mol H₂?

8. How many grams of H₂ are required for 2 mol N₂?

9. How many moles of H₂ are required for 28 g N₂? Use M(N₂) = 28 g/mol.

10. How many grams of N₂ are required to react with 12 g H₂?

For questions 11–15, use:

2Mg + O₂ → 2MgO

Use:

M(Mg) = 24.3 g/mol

M(O₂) = 32.0 g/mol

11. How many moles of O₂ are required for 6 mol Mg?

12. How many grams of O₂ are required for 48.6 g Mg?

13. How many grams of Mg are required to react with 16.0 g O₂?

14. A student mixes 4 mol Mg with 2 mol O₂. Are the reactants in the correct stoichiometric proportion? Explain.

15. A student mixes 4 mol Mg with 5 mol O₂. Which substance is present in excess?

For questions 16–20, use:

CH₄ + 2O₂ → CO₂ + 2H₂O

Use:

M(CH₄) = 16 g/mol

M(O₂) = 32 g/mol

16. How many moles of O₂ are required for 3 mol CH₄?

17. How many grams of O₂ are required to burn 16 g CH₄?

18. How many grams of CH₄ can react completely with 128 g O₂?

19. Explain why 16 g CH₄ requires a much greater mass of oxygen.

20. Why can insufficient oxygen change the products formed during combustion?

Multi-Step Challenge

Use:

2Al + 3Cl₂ → 2AlCl₃

Molar masses:

Al = 27.0 g/mol

Cl₂ = 71.0 g/mol

21. How many moles of Cl₂ are required for 4 mol Al?

22. How many grams of Cl₂ are required for 2 mol Al?

23. How many moles of Al are required for 6 mol Cl₂?

24. How many grams of Cl₂ are required to react with 10.8 g Al?

25. How many grams of Al are required to react with 35.5 g Cl₂?

26. Write the complete conversion pathway for calculating grams of Cl₂ required from grams of Al.

27. Explain why coefficients rather than subscripts determine reacting quantities.

28. Explain why chemists calculate reacting quantities before performing laboratory experiments.

29. Explain how accurate reacting-quantity calculations can reduce chemical waste.

30. A factory uses much more of one reactant than the balanced equation requires. Explain two possible disadvantages of doing this.

Extended Challenge

Calcium carbonate reacts with hydrochloric acid:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Use:

M(CaCO₃) = 100 g/mol

M(HCl) = 36.5 g/mol

31. How many moles of HCl are required for 1 mol CaCO₃?

32. How many moles of HCl are required for 2.5 mol CaCO₃?

33. How many grams of HCl are required for 100 g CaCO₃?

34. How many grams of CaCO₃ are required to react with 73 g HCl?

35. A student has 50 g CaCO₃. Calculate the mass of HCl required for complete reaction.

36. Explain what would happen if the student used less HCl than the calculated amount.

37. Explain what would happen if considerably more HCl were added than required.

38. Identify which reactant would be in excess in Question 37.

39. Explain how this reaction demonstrates the connection between mole ratios and practical reacting quantities.

40. Explain why the pathway

mass CaCO₃ → mol CaCO₃ → mol HCl → mass HCl

is more reliable than simply comparing the masses of CaCO₃ and HCl directly.

Predicting Product Amounts

One of the most useful applications of stoichiometry is predicting how much product can form from a known amount of reactant.

A balanced chemical equation provides the relationship between reactants and products.

For example:

2H₂ + O₂ → 2H₂O

This tells us:

2 mol H₂ → 2 mol H₂O

Therefore, if enough oxygen is available:

5 mol H₂ → 5 mol H₂O

The general idea is:

known reactant → balanced equation → predicted product

The calculated product amount represents the amount expected from the chemical equation under the assumptions given in the problem.

Balanced Equations Predict Product Quantities

Consider:

N₂ + 3H₂ → 2NH₃

The coefficients tell us:

1 mol N₂ + 3 mol H₂ → 2 mol NH₃

Therefore:

• 1 mol N₂ can produce 2 mol NH₃
• 2 mol N₂ can produce 4 mol NH₃
• 5 mol N₂ can produce 10 mol NH₃
• 10 mol N₂ can produce 20 mol NH₃

These predictions assume that enough hydrogen is available.

The equation provides the stoichiometric relationship between the reactant and product.

---

The Main Calculation Pathway

When predicting product amounts, use:

KNOWN REACTANT → MOLES OF REACTANT → MOLE RATIO → MOLES OF PRODUCT → REQUIRED PRODUCT UNIT

If both quantities are measured in moles:

mol reactant → mol product

If the reactant is given in grams and the product is required in grams:

g reactant → mol reactant → mol product → g product

This second pathway is one of the most important calculations in stoichiometry.

---

Predicting Product in Moles

Consider:

2Mg + O₂ → 2MgO

How many moles of MgO can form from 7 mol Mg, assuming sufficient oxygen?

Identify the mole ratio

Mg : MgO = 2 : 2

This simplifies to:

1 : 1

Calculate

7 mol Mg × (2 mol MgO / 2 mol Mg)

= 7 mol MgO

Answer

7 mol MgO

---

A Useful Formula

For mole-to-mole product calculations:

moles product = moles reactant × (coefficient product / coefficient reactant)

For example:

4Fe + 3O₂ → 2Fe₂O₃

If we begin with 8 mol Fe:

moles Fe₂O₃ = 8 × (2/4)

= 4 mol Fe₂O₃

This formula works only when the equation is balanced and the selected reactant is available to react as assumed.

---

Worked Example: Ammonia

Consider:

N₂ + 3H₂ → 2NH₃

How many moles of ammonia can theoretically form from 6 mol N₂, assuming sufficient hydrogen?

Mole ratio

N₂ : NH₃ = 1 : 2

Calculate

6 mol N₂ × (2 mol NH₃ / 1 mol N₂)

= 12 mol NH₃

Answer

12 mol NH₃

---

Worked Example: Starting with Hydrogen

Using:

N₂ + 3H₂ → 2NH₃

How many moles of NH₃ can form from 9 mol H₂, assuming sufficient nitrogen?

Ratio:

3 mol H₂ : 2 mol NH₃

Calculation:

9 mol H₂ × (2 mol NH₃ / 3 mol H₂)

= 6 mol NH₃

Answer

6 mol NH₃

The product prediction depends on which reactant quantity is given.

---

Predicting Product Mass

Laboratory quantities are often measured in grams.

To calculate product mass:

mass reactant → moles reactant → moles product → mass product

Use:

n = m/M

to convert mass to moles.

Then use the mole ratio.

Finally use:

m = nM

to convert product moles to product mass.

Worked Example: Magnesium Oxide

Magnesium burns in oxygen:

2Mg + O₂ → 2MgO

What mass of MgO can form from 24.3 g Mg, assuming sufficient oxygen?

Use:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

Convert magnesium to moles

n = m/M

n = 24.3 / 24.3

= 1.00 mol Mg

Convert Mg to MgO

Mg : MgO = 1 : 1

Therefore:

1.00 mol MgO

Convert MgO to mass

m = nM

m = 1.00 × 40.3

= 40.3 g

Answer

40.3 g MgO

The pathway was:

24.3 g Mg → 1.00 mol Mg → 1.00 mol MgO → 40.3 g MgO

---

Why the Product Has More Mass

In the previous example:

24.3 g Mg → 40.3 g MgO

It may appear that mass has been created.

It has not.

Magnesium combines with oxygen:

2Mg + O₂ → 2MgO

The additional mass comes from oxygen.

For 1 mol Mg:

24.3 g Mg + 16.0 g O → 40.3 g MgO

The total mass is conserved.

---

Worked Example: Predicting Water

Hydrogen burns according to:

2H₂ + O₂ → 2H₂O

What mass of water can form from 6.0 g H₂, assuming sufficient oxygen?

Use:

M(H₂) = 2.0 g/mol

M(H₂O) = 18.0 g/mol

Convert hydrogen to moles

6.0 ÷ 2.0 = 3.0 mol H₂

Apply the mole ratio

H₂ : H₂O = 2 : 2 = 1 : 1

Therefore:

3.0 mol H₂O

Convert water to mass

3.0 × 18.0 = 54 g

Answer

54 g H₂O

---

Worked Example: Iron Oxide

Iron reacts with oxygen:

4Fe + 3O₂ → 2Fe₂O₃

What mass of Fe₂O₃ can form from 56 g Fe, assuming sufficient oxygen?

Use:

M(Fe) = 56 g/mol

M(Fe₂O₃) = 160 g/mol

Convert iron to moles

56 ÷ 56 = 1 mol Fe

Use the mole ratio

Fe : Fe₂O₃ = 4 : 2

1 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe)

= 0.5 mol Fe₂O₃

Convert to mass

0.5 × 160 = 80 g

Answer

80 g Fe₂O₃

---

Dimensional Analysis

The iron oxide calculation can also be written as one continuous calculation:

56 g Fe × (1 mol Fe / 56 g Fe) × (2 mol Fe₂O₃ / 4 mol Fe) × (160 g Fe₂O₃ / 1 mol Fe₂O₃)

Units cancel:

g Fe → mol Fe → mol Fe₂O₃ → g Fe₂O₃

leaving:

80 g Fe₂O₃

This method is useful for checking that each conversion has been arranged correctly.

---

Predicting Products from Combustion

Methane undergoes complete combustion:

CH₄ + 2O₂ → CO₂ + 2H₂O

From the equation:

1 mol CH₄ → 1 mol CO₂

and:

1 mol CH₄ → 2 mol H₂O

Therefore, burning one substance can produce different mole quantities of different products.

Worked Example: Carbon Dioxide from Methane

How many grams of CO₂ can form when 32 g CH₄ burns completely?

Use:

M(CH₄) = 16 g/mol

M(CO₂) = 44 g/mol

Convert methane to moles

32 ÷ 16 = 2 mol CH₄

Apply the mole ratio

CH₄ : CO₂ = 1 : 1

Therefore:

2 mol CO₂

Convert to mass

2 × 44 = 88 g

Answer

88 g CO₂

---

Worked Example: Water from Methane

Using the same reaction:

CH₄ + 2O₂ → CO₂ + 2H₂O

How many grams of water can form from 32 g CH₄?

We already know:

32 g CH₄ = 2 mol CH₄

Ratio:

1 mol CH₄ : 2 mol H₂O

Therefore:

2 mol CH₄ × (2 mol H₂O / 1 mol CH₄)

= 4 mol H₂O

Convert to mass:

4 × 18 = 72 g

Answer

72 g H₂O

So complete combustion of 32 g methane theoretically produces:

88 g CO₂

and:

72 g H₂O

assuming sufficient oxygen.

---

Checking Conservation of Mass

For:

CH₄ + 2O₂ → CO₂ + 2H₂O

Suppose:

32 g CH₄

reacts.

This is:

2 mol CH₄

It requires:

4 mol O₂

Mass of oxygen:

4 × 32 = 128 g

Total reactant mass:

32 + 128 = 160 g

Predicted products:

88 g CO₂ + 72 g H₂O = 160 g

Therefore:

mass of reactants = mass of products

The prediction agrees with conservation of mass.

---

Predicting Multiple Products

Some reactions produce more than one product.

Consider:

CaCO₃ → CaO + CO₂

One mole of calcium carbonate produces:

1 mol CaO

and:

1 mol CO₂

Suppose 2 mol CaCO₃ decompose completely.

We predict:

2 mol CaO

and:

2 mol CO₂

Each product can then be converted into mass if required.

---

Worked Example: Thermal Decomposition

How much CO₂ can form from 250 g CaCO₃?

Equation:

CaCO₃ → CaO + CO₂

Use:

M(CaCO₃) = 100 g/mol

M(CO₂) = 44 g/mol

Convert CaCO₃ to moles

250 ÷ 100 = 2.5 mol CaCO₃

Apply the mole ratio

CaCO₃ : CO₂ = 1 : 1

Therefore:

2.5 mol CO₂

Convert to mass

2.5 × 44 = 110 g

Answer

110 g CO₂

Predicting the Other Product

Using:

CaCO₃ → CaO + CO₂

What mass of CaO forms from the same 250 g CaCO₃?

We already know:

250 g CaCO₃ = 2.5 mol CaCO₃

The ratio CaCO₃ : CaO is:

1 : 1

Therefore:

2.5 mol CaO

Use:

M(CaO) = 56 g/mol

Mass:

2.5 × 56 = 140 g

Answer

140 g CaO

Now check:

140 g CaO + 110 g CO₂ = 250 g

The predicted products equal the original reactant mass.

---

Predicting Product from an Acid Reaction

Magnesium reacts with hydrochloric acid:

Mg + 2HCl → MgCl₂ + H₂

How many moles of hydrogen gas can form from 0.30 mol Mg, assuming sufficient HCl?

Ratio:

1 mol Mg : 1 mol H₂

Therefore:

0.30 mol Mg → 0.30 mol H₂

Answer

0.30 mol H₂

---

Predicting Product Mass from an Acid Reaction

Suppose 12.15 g Mg reacts with sufficient hydrochloric acid.

Equation:

Mg + 2HCl → MgCl₂ + H₂

Use:

M(Mg) = 24.3 g/mol

M(MgCl₂) ≈ 95.3 g/mol

Convert magnesium to moles

12.15 ÷ 24.3 = 0.500 mol Mg

Apply the mole ratio

Mg : MgCl₂ = 1 : 1

Therefore:

0.500 mol MgCl₂

Convert to mass

0.500 × 95.3 = 47.65 g

Answer

Approximately:

47.7 g MgCl₂

The additional mass comes from chlorine supplied by hydrochloric acid.

---

Predicting Product from a Precipitation Reaction

Consider:

AgNO₃ + NaCl → AgCl + NaNO₃

Silver chloride, AgCl, forms as a solid precipitate.

The mole ratio is:

1 mol AgNO₃ : 1 mol AgCl

If 0.25 mol AgNO₃ reacts with sufficient NaCl:

0.25 mol AgCl

is predicted to form.

If:

M(AgCl) ≈ 143.5 g/mol

then:

m = 0.25 × 143.5

≈ 35.9 g AgCl

Predicted amount

35.9 g AgCl

---

Product Predictions Are Theoretical

Stoichiometric calculations tell us how much product should form according to the balanced equation and assumptions of the problem.

This is sometimes called the theoretical yield.

For example, a calculation may predict:

25.0 g product

But an experiment might actually produce:

21.8 g product

The calculation is not necessarily wrong.

Real reactions are not always perfectly efficient.

---

Why Actual Product Amounts May Be Lower

The actual amount of product may be lower because:

• the reaction does not go to completion
• product is lost during transfer
• product remains in laboratory equipment
• competing reactions occur
• reactants contain impurities
• some product is lost during filtration
• some product is lost during heating or purification
• measurement uncertainty affects results

This means:

predicted amount ≠ always actual amount

---

Theoretical Yield

The theoretical yield is the maximum amount of product predicted by stoichiometry from the available reactant under the stated assumptions.

For example:

Calculation predicts:

12.5 g Cu

Therefore:

theoretical yield = 12.5 g Cu

If only 10.8 g is collected experimentally:

actual yield = 10.8 g Cu

These values can later be used to calculate percentage yield.

---

Is a Prediction Reasonable?

A calculation should never end with simply writing a number.

Ask whether the answer is reasonable.

Useful checks include:

• Is the equation balanced?
• Did I use the correct mole ratio?
• Did my units cancel correctly?
• Did I use the correct molar masses?
• Does the size of the answer make sense?
• Does the answer agree with conservation of mass?
• Did I accidentally use coefficients as mass ratios?
• Did I round too early?

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Reasonableness Check: Mole Ratio

Consider:

2Al + 3Cl₂ → 2AlCl₃

If we start with:

4 mol Al

we should produce:

4 mol AlCl₃

because Al : AlCl₃ is:

2 : 2 = 1 : 1

If someone calculates:

12 mol AlCl₃

we should immediately question the result.

The balanced equation provides a quick estimate before detailed calculations begin.

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Reasonableness Check: Conservation of Mass

Suppose a reaction has:

20 g of reactant A

and:

30 g of reactant B

and both react completely to form one product.

The product cannot have a mass of:

80 g

because only:

50 g

of reactants were present.

If no matter enters or leaves the system:

maximum total product mass = 50 g

Conservation of mass provides an important check.

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Reasonableness Check: Order of Magnitude

Suppose:

1 mol reactant → 1 mol product

and both substances have similar molar masses.

If you begin with approximately:

10 g reactant

but calculate:

10,000 g product

something is probably wrong.

Possible errors include:

• incorrect molar mass
• incorrect units
• inverted conversion factor
• calculator entry error

Estimating before calculating helps identify these mistakes.

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The Reactant Used for the Prediction Matters

Consider:

2H₂ + O₂ → 2H₂O

Suppose we have:

10 mol H₂

and:

2 mol O₂

Using hydrogen alone would predict:

10 mol H₂O

But using oxygen predicts:

4 mol H₂O

Both cannot be produced.

Why?

There is not enough oxygen to react with all the hydrogen.

The reactant that runs out first controls the maximum product amount.

This reactant is called the limiting reactant.

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A Preview of Limiting Reactants

For:

2H₂ + O₂ → 2H₂O

Given:

10 mol H₂

and:

2 mol O₂

The oxygen can react with:

4 mol H₂

and produce:

4 mol H₂O

Therefore:

O₂ is the limiting reactant

and:

H₂ is in excess

The maximum product is:

4 mol H₂O

This is why product predictions require careful attention when quantities of both reactants are provided.

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Worked Example: Product Prediction with Two Reactants

Consider:

N₂ + 3H₂ → 2NH₃

Suppose:

2 mol N₂

and:

9 mol H₂

are available.

For 2 mol N₂, the required H₂ is:

2 × 3 = 6 mol H₂

But 9 mol H₂ are available.

Therefore, hydrogen is available in excess.

The 2 mol N₂ determine the product amount.

Ratio:

1 mol N₂ : 2 mol NH₃

Therefore:

2 mol N₂ → 4 mol NH₃

Maximum predicted product

4 mol NH₃

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Industrial Product Predictions

Chemical manufacturers must predict how much product can be made from available raw materials.

Stoichiometric predictions help determine:

• required reactant quantities
• expected production
• raw-material costs
• equipment requirements
• storage requirements
• waste production
• process efficiency

Large industrial processes may involve thousands or millions of kilograms of material, making accurate calculations extremely important.

Application: Ammonia Production

Ammonia can be produced using:

N₂ + 3H₂ ⇌ 2NH₃

Suppose an idealized calculation begins with:

100 mol N₂

and sufficient hydrogen.

The ratio is:

1 mol N₂ : 2 mol NH₃

Therefore:

100 mol N₂ → 200 mol NH₃

The stoichiometric prediction is:

200 mol NH₃

Real industrial production is more complicated because the reaction is reversible and does not simply convert every molecule in a single pass, but the balanced equation still provides the fundamental quantitative relationship.

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Application: Environmental Chemistry

Product predictions can help estimate quantities of substances released into the environment.

For example:

CH₄ + 2O₂ → CO₂ + 2H₂O

The equation predicts:

1 mol CH₄ → 1 mol CO₂

Therefore, the amount of methane burned can be used to estimate the theoretical amount of carbon dioxide produced during complete combustion.

Similar calculations can be applied to:

• fuel combustion
• industrial emissions
• waste treatment
• neutralization
• water treatment

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Application: Laboratory Planning

Suppose a laboratory investigation requires approximately:

5.0 g of a product

Before performing the experiment, a chemist can work backward using stoichiometry to determine how much reactant should theoretically be required.

This helps:

• reduce waste
• control costs
• choose appropriate equipment
• improve safety
• plan experiments efficiently

Stoichiometry therefore allows us to predict both:

reactant → product

and:

desired product → required reactant

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Common Mistakes

Using an Unbalanced Equation

Always balance before calculating.

Wrong:

H₂ + O₂ → H₂O

Correct:

2H₂ + O₂ → 2H₂O

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Using Subscripts Instead of Coefficients

Mole ratios come from coefficients.

For:

2Mg + O₂ → 2MgO

Mg : MgO is:

2 : 2

not a ratio taken from the subscripts.

---

Treating Coefficients as Mass Ratios

For:

2H₂ + O₂ → 2H₂O

2 mol H₂ produce 2 mol H₂O.

But:

4 g H₂ → 36 g H₂O

not:

2 g H₂ → 2 g H₂O

---

Skipping the Mole Step

For mass-to-mass calculations, use:

mass reactant → mol reactant → mol product → mass product

Do not normally jump directly from one mass to another using coefficients.

---

Using the Wrong Molar Mass

For example:

CO₂ = 12 + (2 × 16) = 44 g/mol

Make sure every atom in the formula is included.

---

Reversing the Mole Ratio

If converting Fe into Fe₂O₃:

4Fe + 3O₂ → 2Fe₂O₃

use:

2 mol Fe₂O₃ / 4 mol Fe

so mol Fe cancels.

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Assuming Predicted Yield Equals Actual Yield

Stoichiometry gives a theoretical prediction.

Actual laboratory yield may be lower.

---

Ignoring the Other Reactant

If quantities of two reactants are given, do not automatically calculate product from whichever appears first.

One may be the limiting reactant.

---

Rejecting an Answer Because Product Mass Is Larger

A product can have more mass than one individual reactant because mass from another reactant has been added.

Always consider the total mass of all reactants.

---

Rounding Too Early

Keep extra digits during intermediate calculations.

Round the final answer appropriately.

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Key Terms

Product — A substance formed during a chemical reaction.

Predicted product amount — The quantity of product calculated from stoichiometry.

Stoichiometry — The quantitative study of relationships between reactants and products.

Balanced chemical equation — An equation containing equal numbers of each type of atom on both sides.

Coefficient — A number placed before a chemical formula indicating relative amounts in a reaction.

Mole ratio — A ratio between amounts of substances obtained from a balanced equation.

Mole — An amount of substance containing 6.022 × 10²³ representative particles.

Molar mass — The mass of one mole of a substance, usually expressed in g/mol.

Reactant — A starting substance in a chemical reaction.

Conversion factor — A ratio used to convert one quantity into another.

Dimensional analysis — A method that uses conversion factors and unit cancellation.

Conservation of mass — The principle that total mass remains constant during a chemical reaction.

Theoretical yield — The maximum quantity of product predicted by stoichiometric calculations under the stated assumptions.

Actual yield — The quantity of product actually obtained experimentally.

Limiting reactant — The reactant consumed first, which determines the maximum amount of product that can form.

Excess reactant — A reactant present in more than the amount required to react with the limiting reactant.

Complete reaction — A reaction in which the relevant reactant is assumed to react as fully as the problem specifies.

Reasonableness check — An evaluation of whether a calculated result is consistent with chemical principles, units, ratios, and expected magnitude.

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Key Takeaways

• Balanced equations can be used to predict quantities of products.
• Product calculations are based on mole ratios.
• Mole ratios come from coefficients in balanced equations.
• For mole-to-mole calculations:

mol reactant → mol product

• For mass-to-mass calculations:

g reactant → mol reactant → mol product → g product

• Convert mass to moles using:

n = m/M

• Convert moles to mass using:

m = nM

• Product mass can be greater than the mass of one reactant because other reactants contribute mass.
• Total mass must still obey conservation of mass.
• Predicted product quantities represent theoretical results under the assumptions of the calculation.
• Actual experimental amounts may be lower than predicted amounts.
• Product predictions should always be checked for reasonableness.
• Unit cancellation is useful for checking calculations.
• Conservation of mass provides another useful check.
• When quantities of multiple reactants are given, the limiting reactant determines the maximum product amount.
• Stoichiometric predictions are important in laboratories, manufacturing, environmental science, energy production, and many other applications.

The main pathway to remember is:

KNOWN REACTANT → MOLES REACTANT → MOLE RATIO → MOLES PRODUCT → PRODUCT QUANTITY

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Check Your Understanding

For questions 1–5, use:

2H₂ + O₂ → 2H₂O

1. How many moles of H₂O can form from 6 mol H₂?

2. How many moles of H₂O can form from 3 mol O₂?

3. How many grams of H₂O can form from 4 mol H₂?

4. How many grams of H₂O can form from 32 g O₂?

5. Explain why the mass of water produced can be greater than the mass of hydrogen used.

For questions 6–10, use:

N₂ + 3H₂ → 2NH₃

6. How many moles of NH₃ can form from 4 mol N₂?

7. How many moles of NH₃ can form from 12 mol H₂?

8. How many grams of NH₃ can form from 2 mol N₂? Use M(NH₃) = 17 g/mol.

9. How many grams of NH₃ can theoretically form from 28 g N₂?

10. Explain why these calculations represent theoretical predictions.

For questions 11–15, use:

2Mg + O₂ → 2MgO

Use:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

11. How many moles of MgO can form from 5 mol Mg?

12. How many grams of MgO can form from 2 mol Mg?

13. How many grams of MgO can theoretically form from 48.6 g Mg?

14. A student predicts 40.3 g MgO from 24.3 g Mg. Explain why this does not violate conservation of mass.

15. Identify the additional reactant contributing mass to MgO.

For questions 16–20, use:

CH₄ + 2O₂ → CO₂ + 2H₂O

16. How many moles of CO₂ form from 5 mol CH₄?

17. How many moles of H₂O form from 5 mol CH₄?

18. How many grams of CO₂ form from 32 g CH₄?

19. How many grams of H₂O form from 16 g CH₄?

20. Why are the masses of CO₂ and H₂O produced not equal even though they come from the same reaction?

Multi-Step Challenge

Use:

4Fe + 3O₂ → 2Fe₂O₃

Use:

M(Fe) = 56 g/mol

M(Fe₂O₃) = 160 g/mol

21. How many moles of Fe₂O₃ can form from 8 mol Fe?

22. How many grams of Fe₂O₃ can form from 4 mol Fe?

23. How many grams of Fe₂O₃ can theoretically form from 112 g Fe?

24. Write the complete conversion pathway for Question 23.

25. Explain why product mass is greater than the mass of iron used.

Use:

CaCO₃ → CaO + CO₂

with:

M(CaCO₃) = 100 g/mol

M(CaO) = 56 g/mol

M(CO₂) = 44 g/mol

26. How many moles of CO₂ form from 3 mol CaCO₃?

27. How many grams of CaO form from 200 g CaCO₃?

28. How many grams of CO₂ form from 200 g CaCO₃?

29. Add your answers to Questions 27 and 28. How does the result compare with the original mass of CaCO₃?

30. Explain how Question 29 demonstrates conservation of mass.

Reasonableness Challenge

31. A student calculates that 1 mol Mg produces 10 mol MgO from:

2Mg + O₂ → 2MgO

Explain why the answer cannot be correct.

32. A calculation predicts 5000 g of product from a total of 50 g of reactants in a closed system. Explain why the prediction is unreasonable.

33. A student predicts 36 g H₂O from 4 g H₂ reacting with sufficient O₂. Explain why the product can have a greater mass than the hydrogen.

34. A reaction is predicted to produce 25.0 g of product, but only 21.2 g is collected. Give three possible reasons for the difference.

35. Explain the difference between a predicted theoretical amount and an actual experimental amount.

36. Why should a chemist estimate the approximate size of an answer before completing a stoichiometric calculation?

37. Explain how units can help identify an incorrectly arranged calculation.

38. Explain why the balanced equation is essential for predicting product amounts.

39. Describe how product predictions could help a chemical manufacturer plan production.

40. Explain why identifying the limiting reactant becomes important when quantities of two or more reactants are provided.