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.

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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.

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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

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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

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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

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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

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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.

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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.

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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

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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

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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

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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

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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

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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.

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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.

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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.

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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.

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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

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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

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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.

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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

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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

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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.

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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.

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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

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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.

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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

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Common Mistakes

Using an Unbalanced Equation

Mole ratios must come from a balanced equation.

Wrong:

H₂ + O₂ → H₂O

Correct:

2H₂ + O₂ → 2H₂O

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Using Subscripts Instead of Coefficients

For:

2H₂ + O₂ → 2H₂O

the H₂ : O₂ mole ratio is:

2 : 1

not:

2 : 2

Use coefficients.

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Changing Subscripts to Balance an Equation

Never change:

H₂O

into:

H₂O₂

just to balance an equation.

Change coefficients instead.

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Assuming the Coefficients Are Mass Ratios

For:

2H₂ + O₂ → 2H₂O

the ratio:

2 : 1 : 2

is a mole ratio, not a gram ratio.

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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.

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Forgetting an Invisible Coefficient

In:

CH₄ + 2O₂ → CO₂ + 2H₂O

CH₄ has coefficient:

1

CO₂ also has coefficient:

1

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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.

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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

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Using Every Coefficient in the Equation

Usually, you only need the coefficients of:

the substance given

and:

the substance wanted

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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.

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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

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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.