Mass-Mass Calculations

Many chemical problems begin with the mass of one substance and ask for the mass of another substance.

For example:

If 12.0 g of magnesium reacts completely with oxygen, what mass of magnesium oxide can form?

A balanced chemical equation gives relationships in moles, not directly in grams. Therefore, we cannot normally move directly from the mass of one substance to the mass of another.

Instead, we use the pathway:

mass A → moles A → moles B → mass B

This is called a mass-mass stoichiometric calculation.

 

 

 

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Why We Convert Through Moles

Consider:

2Mg + O₂ → 2MgO

The coefficients tell us:

2 mol Mg → 2 mol MgO

They do not tell us:

2 g Mg → 2 g MgO

Magnesium and magnesium oxide have different molar masses.

Using approximate values:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

Therefore:

2 mol Mg = 48.6 g

while:

2 mol MgO = 80.6 g

So the balanced equation represents:

48.6 g Mg + 32.0 g O₂ → 80.6 g MgO

The coefficients give a mole ratio. Molar masses allow us to convert that relationship into masses.

The Mass-Mass Roadmap

Nearly every basic mass-mass problem follows the same pathway:

MASS GIVEN

↓

MOLES GIVEN

↓

MOLE RATIO

↓

MOLES WANTED

↓

MASS WANTED

Or more simply:

g A → mol A → mol B → g B

There are three conversions:

Mass to moles

n = m/M

Moles of one substance to moles of another

Use the coefficients from the balanced equation.

Moles to mass

m = nM

This roadmap is worth remembering.

Step One: Balance the Equation

Always begin with a balanced chemical equation.

For example:

Mg + O₂ → MgO

is not balanced.

Correct:

2Mg + O₂ → 2MgO

The coefficients:

2 : 1 : 2

provide the mole ratios needed for the calculation.

If the equation is wrong, the mass calculation will also be wrong.

Step Two: Convert the Given Mass to Moles

Use:

n = m/M

where:

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

For example, how many moles are in 48.6 g Mg?

n = 48.6 / 24.3

n = 2.00 mol Mg

Step Three: Use the Mole Ratio

For:

2Mg + O₂ → 2MgO

the ratio Mg : MgO is:

2 : 2

Therefore:

2.00 mol Mg × (2 mol MgO / 2 mol Mg)

= 2.00 mol MgO

This is the step where we change from one chemical substance to another.

Step Four: Convert Moles to Mass

Use:

m = nM

For MgO:

m = 2.00 × 40.3

m = 80.6 g

Therefore:

48.6 g Mg → 80.6 g MgO

assuming sufficient oxygen is available.

The Complete Calculation

The entire calculation can be written as:

48.6 g Mg → 2.00 mol Mg → 2.00 mol MgO → 80.6 g MgO

This clearly shows the mass-mass pathway.

 

 

 

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Worked Example: Magnesium Oxide

How much MgO can form from 12.15 g Mg?

Equation:

2Mg + O₂ → 2MgO

Use:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

Convert Mg to moles

n = 12.15 / 24.3

= 0.500 mol Mg

Use the mole ratio

Mg : MgO = 2 : 2 = 1 : 1

Therefore:

0.500 mol MgO

Convert MgO to mass

m = 0.500 × 40.3

= 20.15 g

Answer

Approximately:

20.2 g MgO

Dimensional Analysis

Mass-mass calculations can also be written as one continuous calculation.

For the previous example:

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

Notice how the units cancel:

g Mg

↓

mol Mg

↓

mol MgO

↓

g MgO

The only unit remaining is:

g MgO

This is exactly the unit we want.

Why Unit Cancellation Is Useful

Suppose a student accidentally writes:

12.15 g Mg × (24.3 g Mg / 1 mol Mg)

The units become:

g²/mol

instead of moles.

That tells us immediately that the conversion factor has been written upside down.

Units are therefore not just labels. They help us check the mathematics.

Worked Example: Producing Water

Consider:

2H₂ + O₂ → 2H₂O

How much water can form from 10.0 g H₂, assuming sufficient oxygen?

Use:

M(H₂) = 2.0 g/mol

M(H₂O) = 18.0 g/mol

Convert H₂ to moles

10.0 ÷ 2.0 = 5.0 mol H₂

Use the mole ratio

H₂ : H₂O = 2 : 2 = 1 : 1

Therefore:

5.0 mol H₂O

Convert to mass

5.0 × 18.0 = 90 g

Answer

90 g H₂O

Why 10 g Can Produce 90 g

At first, this may seem impossible.

But hydrogen is not the only reactant.

The equation is:

2H₂ + O₂ → 2H₂O

The hydrogen combines with oxygen.

For 5 mol H₂:

5 mol H₂ = 10 g

The required oxygen is:

2.5 mol O₂

Mass of oxygen:

2.5 × 32 = 80 g

Therefore:

10 g H₂ + 80 g O₂ → 90 g H₂O

Mass has been conserved.

Conservation of Mass

The law of conservation of mass states that mass is not created or destroyed during an ordinary chemical reaction.

Therefore:

total mass of reactants = total mass of products

For:

2H₂ + O₂ → 2H₂O

using stoichiometric quantities:

4 g H₂ + 32 g O₂ → 36 g H₂O

Total before:

36 g

Total after:

36 g

Stoichiometric calculations are consistent with conservation of mass.

 

 

 

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Worked Example: Iron Oxide

Iron reacts with oxygen:

4Fe + 3O₂ → 2Fe₂O₃

What mass of Fe₂O₃ can form from 28.0 g Fe, assuming sufficient oxygen?

Use:

M(Fe) = 56.0 g/mol

M(Fe₂O₃) = 160 g/mol

Convert Fe to moles

28.0 ÷ 56.0 = 0.500 mol Fe

Apply the mole ratio

Fe : Fe₂O₃ = 4 : 2

0.500 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe)

= 0.250 mol Fe₂O₃

Convert to mass

0.250 × 160 = 40.0 g

Answer

40.0 g Fe₂O₃

Calculating the Mass of a Reactant

Mass-mass calculations can also work backward.

Consider:

4Fe + 3O₂ → 2Fe₂O₃

How much Fe is required to produce 80.0 g Fe₂O₃?

Convert Fe₂O₃ to moles

80.0 ÷ 160 = 0.500 mol Fe₂O₃

Use the mole ratio

Fe : Fe₂O₃ = 4 : 2

0.500 mol Fe₂O₃ × (4 mol Fe / 2 mol Fe₂O₃)

= 1.00 mol Fe

Convert to mass

1.00 × 56.0 = 56.0 g Fe

Answer

56.0 g Fe

So mass-mass calculations can determine:

reactant → product

or:

product → reactant

Worked Example: Methane Combustion

Methane burns completely according to:

CH₄ + 2O₂ → CO₂ + 2H₂O

How much CO₂ can form from 8.0 g CH₄?

Use:

M(CH₄) = 16.0 g/mol

M(CO₂) = 44.0 g/mol

Convert methane to moles

8.0 ÷ 16.0 = 0.500 mol CH₄

Use the mole ratio

CH₄ : CO₂ = 1 : 1

Therefore:

0.500 mol CO₂

Convert to mass

0.500 × 44.0 = 22.0 g

Answer

22.0 g CO₂

Predicting Water from the Same Reaction

Using:

CH₄ + 2O₂ → CO₂ + 2H₂O

How much water can form from 8.0 g CH₄?

We already know:

8.0 g CH₄ = 0.500 mol CH₄

Ratio:

1 mol CH₄ : 2 mol H₂O

Therefore:

0.500 × 2 = 1.00 mol H₂O

Use:

M(H₂O) = 18.0 g/mol

Mass:

1.00 × 18.0 = 18.0 g

Answer

18.0 g H₂O

 

 

 

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Checking Conservation of Mass in Combustion

We predicted from 8.0 g CH₄:

22.0 g CO₂

and:

18.0 g H₂O

Total product mass:

22.0 + 18.0 = 40.0 g

How much oxygen was required?

0.500 mol CH₄ requires:

1.00 mol O₂

Mass of oxygen:

1.00 × 32.0 = 32.0 g

Total reactant mass:

8.0 + 32.0 = 40.0 g

Therefore:

40.0 g reactants = 40.0 g products

Worked Example: Propane Combustion

Propane burns according to:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

What mass of CO₂ can form from 22 g C₃H₈?

Use:

M(C₃H₈) = 44 g/mol

M(CO₂) = 44 g/mol

Convert propane to moles

22 ÷ 44 = 0.50 mol C₃H₈

Use the mole ratio

C₃H₈ : CO₂ = 1 : 3

0.50 × 3 = 1.50 mol CO₂

Convert to mass

1.50 × 44 = 66 g

Answer

66 g CO₂

Worked Example: Decomposition

Calcium carbonate decomposes when strongly heated:

CaCO₃ → CaO + CO₂

What mass of CaO can form from 150 g CaCO₃?

Use:

M(CaCO₃) = 100 g/mol

M(CaO) = 56 g/mol

Convert CaCO₃ to moles

150 ÷ 100 = 1.50 mol CaCO₃

Use the mole ratio

CaCO₃ : CaO = 1 : 1

Therefore:

1.50 mol CaO

Convert to mass

1.50 × 56 = 84 g

Answer

84 g CaO

Finding the Other Product

For:

CaCO₃ → CaO + CO₂

What mass of CO₂ forms from the same 150 g CaCO₃?

Use:

M(CO₂) = 44 g/mol

We already know:

150 g CaCO₃ = 1.50 mol CaCO₃

Ratio:

CaCO₃ : CO₂ = 1 : 1

Therefore:

1.50 mol CO₂

Mass:

1.50 × 44 = 66 g

Answer

66 g CO₂

Now check:

84 g CaO + 66 g CO₂ = 150 g

This agrees exactly with conservation of mass.

Worked Example: Aluminum and Chlorine

Consider:

2Al + 3Cl₂ → 2AlCl₃

What mass of AlCl₃ can form from 13.5 g Al, assuming sufficient chlorine?

Use:

M(Al) = 27.0 g/mol

M(AlCl₃) = 133.5 g/mol

Convert aluminum to moles

13.5 ÷ 27.0 = 0.500 mol Al

Use the ratio

Al : AlCl₃ = 2 : 2 = 1 : 1

Therefore:

0.500 mol AlCl₃

Convert to mass

0.500 × 133.5 = 66.75 g

Answer

Approximately:

66.8 g AlCl₃

A More Challenging Mass Ratio

Consider:

2Al + 3Cl₂ → 2AlCl₃

How much Cl₂ is required to produce 53.4 g AlCl₃?

Use:

M(AlCl₃) = 133.5 g/mol

M(Cl₂) = 71.0 g/mol

Convert product to moles

53.4 ÷ 133.5 = 0.400 mol AlCl₃

Apply the mole ratio

Cl₂ : AlCl₃ = 3 : 2

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

= 0.600 mol Cl₂

Convert to mass

0.600 × 71.0 = 42.6 g

Answer

42.6 g Cl₂

Mass-Mass Calculations with Acid Reactions

Consider:

Mg + 2HCl → MgCl₂ + H₂

What mass of MgCl₂ can form from 4.86 g Mg, assuming sufficient HCl?

Use:

M(Mg) = 24.3 g/mol

M(MgCl₂) = 95.3 g/mol

Convert Mg to moles

4.86 ÷ 24.3 = 0.200 mol Mg

Apply the mole ratio

Mg : MgCl₂ = 1 : 1

Therefore:

0.200 mol MgCl₂

Convert to mass

0.200 × 95.3 = 19.06 g

Answer

Approximately:

19.1 g MgCl₂

Mass-Mass Calculations with Neutralization

Consider:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

How much NaOH is required to react completely with 9.8 g H₂SO₄?

Use:

M(H₂SO₄) = 98 g/mol

M(NaOH) = 40 g/mol

Convert sulfuric acid to moles

9.8 ÷ 98 = 0.100 mol H₂SO₄

Use the mole ratio

H₂SO₄ : NaOH = 1 : 2

Therefore:

0.100 × 2 = 0.200 mol NaOH

Convert to mass

0.200 × 40 = 8.0 g

Answer

8.0 g NaOH

Why We Cannot Simply Compare Masses

Consider:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

The coefficient ratio is:

1 mol H₂SO₄ : 2 mol NaOH

But the mass relationship is:

98 g H₂SO₄ : 80 g NaOH

not:

1 g : 2 g

The reason is that one mole of each substance has a different mass.

This is why:

moles are the bridge between masses

Deriving a Reacting Mass Ratio

Once a balanced equation is known, we can calculate a mass relationship.

Consider:

2Mg + O₂ → 2MgO

Molar masses:

Mg = 24.3 g/mol

O₂ = 32.0 g/mol

MgO = 40.3 g/mol

Multiply each molar mass by its coefficient:

2Mg = 2 × 24.3 = 48.6 g

O₂ = 1 × 32.0 = 32.0 g

2MgO = 2 × 40.3 = 80.6 g

Therefore:

48.6 g Mg + 32.0 g O₂ → 80.6 g MgO

This is the stoichiometric mass relationship for the reaction.

Scaling a Mass Relationship

Once the correct mass relationship has been established, it can be scaled.

If:

48.6 g Mg → 80.6 g MgO

then half as much magnesium gives:

24.3 g Mg → 40.3 g MgO

Double gives:

97.2 g Mg → 161.2 g MgO

The proportions remain constant.

This proportional approach can be useful, but the mole method is more flexible and works reliably for unfamiliar problems.

Practical Application: Manufacturing

Chemical manufacturers need to calculate how much raw material is required and how much product can theoretically be produced.

Mass-mass calculations help determine:

• raw material requirements
• expected product quantities
• storage needs
• transportation requirements
• production costs
• waste quantities

For large-scale production, even a small calculation error can represent a large amount of material.

 

 

 

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Practical Application: Pharmaceuticals

Pharmaceutical production requires carefully controlled quantities.

Chemists may need to determine:

mass of starting material → theoretical mass of pharmaceutical product

Accurate calculations help:

• reduce waste
• control costs
• plan purification
• maintain consistent production

Real pharmaceutical chemistry can involve many additional factors, but stoichiometry provides the basic quantitative framework.

Practical Application: Environmental Chemistry

Mass-mass calculations can determine how much chemical is needed to treat pollutants.

For example:

HCl + NaOH → NaCl + H₂O

If the mass of HCl in acidic waste is known, stoichiometry can determine the theoretical mass of NaOH needed for neutralization.

This helps avoid using:

• too little treatment chemical
• unnecessarily large excesses

Practical Application: Combustion and Emissions

Mass-mass calculations can also predict emissions.

For:

CH₄ + 2O₂ → CO₂ + 2H₂O

we found that:

16 g CH₄ → 44 g CO₂

during complete combustion.

Therefore, if the mass of methane burned is known, the theoretical mass of carbon dioxide produced can be calculated.

Similar methods can be used for other fuels.

Practical Application: Laboratory Planning

Before performing a reaction, a student or chemist can calculate the required masses.

For example, suppose an experiment requires approximately:

10 g product

Stoichiometry can be used backward:

mass product → mol product → mol reactant → mass reactant

This helps determine how much starting material should theoretically be required.

Mass-Mass Calculations and Theoretical Yield

A mass-mass calculation often predicts the theoretical yield.

Suppose stoichiometry predicts:

18.5 g product

Then:

theoretical yield = 18.5 g

If an experiment actually produces:

15.9 g

then:

actual yield = 15.9 g

The actual amount may be lower because:

• the reaction was incomplete
• product was lost
• side reactions occurred
• reactants contained impurities
• measurements had uncertainty

Checking Your Answer

After solving a mass-mass problem, ask:

Is the equation balanced?

If not, the mole ratio is wrong.

Did I convert mass to moles?

Remember:

g → mol

before changing substances.

Did I use the correct coefficients?

Use coefficients from the balanced equation.

Did I convert back to mass?

If the question asks for grams:

mol → g

Do my units cancel?

The final unit should be the unit requested.

Is the answer reasonable?

Estimate before accepting the result.

Checking with Conservation of Mass

Suppose a student calculates:

10 g A + 15 g B → 80 g C

in a closed system where C is the only product.

This cannot be correct.

Total reactant mass:

10 + 15 = 25 g

Therefore the product cannot have a mass of 80 g.

Conservation of mass is a powerful way to detect unreasonable answers.

Product Mass Can Exceed the Given Reactant Mass

This is not automatically an error.

Suppose:

24.3 g Mg → 40.3 g MgO

The product has more mass than the magnesium because oxygen also enters the product.

Always compare:

total reactant mass

with:

total product mass

not just one reactant with the product.

A General Mass-Mass Formula

Mass-mass calculations can be summarized as:

mass wanted = (mass given / molar mass given) × (coefficient wanted / coefficient given) × molar mass wanted

Or:

m(wanted) = [m(given) / M(given)] × [coefficient wanted / coefficient given] × M(wanted)

This combines the three steps into one expression.

However, writing the individual steps is often safer while learning stoichiometry.

Worked Example Using the Combined Method

Consider:

2Mg + O₂ → 2MgO

Given:

6.075 g Mg

Find:

mass MgO

Use:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

Calculation:

mass MgO = (6.075 / 24.3) × (2/2) × 40.3

= 0.250 × 1 × 40.3

= 10.075 g

Approximately:

10.1 g MgO

Multi-Step Example

Potassium chlorate decomposes:

2KClO₃ → 2KCl + 3O₂

What mass of O₂ can form from 24.5 g KClO₃?

Use:

M(KClO₃) = 122.5 g/mol

M(O₂) = 32.0 g/mol

Convert KClO₃ to moles

24.5 ÷ 122.5 = 0.200 mol KClO₃

Apply the mole ratio

KClO₃ : O₂ = 2 : 3

0.200 × (3/2) = 0.300 mol O₂

Convert to mass

0.300 × 32.0 = 9.60 g

Answer

9.60 g O₂

Another Multi-Step Example

Consider:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

What mass of CO₂ can form from 50.0 g CaCO₃?

Use:

M(CaCO₃) = 100 g/mol

M(CO₂) = 44 g/mol

Convert CaCO₃ to moles

50.0 ÷ 100 = 0.500 mol CaCO₃

Apply the ratio

CaCO₃ : CO₂ = 1 : 1

Therefore:

0.500 mol CO₂

Convert to mass

0.500 × 44 = 22.0 g

Answer

22.0 g CO₂

Common Mistakes

Using an Unbalanced Equation

Always balance first.

Going Directly from Grams to Grams Using Coefficients

Wrong approach:

10 g A × coefficient ratio = grams B

The coefficients describe moles.

Use:

g A → mol A → mol B → g B

Confusing Molar Mass with Coefficients

Molar mass comes from the chemical formula.

Mole ratios come from the balanced equation.

They serve different purposes.

Using the Wrong Molar Mass

For example:

M(O₂) = 32 g/mol

not:

16 g/mol

because oxygen gas contains two oxygen atoms.

Forgetting Parentheses

For:

Ca(OH)₂

there are:

• 1 Ca
• 2 O
• 2 H

Every atom must be included when calculating molar mass.

Using Subscripts as the Mole Ratio

The mole ratio comes from the numbers in front of formulas, not the numbers inside them.

Reversing the Conversion Factor

If converting Fe into Fe₂O₃:

4Fe + 3O₂ → 2Fe₂O₃

use:

2 mol Fe₂O₃ / 4 mol Fe

so that mol Fe cancels.

Forgetting the Final Mass Conversion

After calculating moles of the wanted substance, check the question.

If it asks for grams, multiply by molar mass.

Rounding Too Early

Keep several digits during intermediate steps.

Round the final answer.

Rejecting a Larger Product Mass

A product can have more mass than the starting reactant you were given because another reactant also contributes mass.

Key Terms

Mass-mass calculation — A stoichiometric calculation that converts the mass of one substance into the mass of another.

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

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

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

Mole ratio — The relationship between substances obtained from coefficients in a balanced chemical equation.

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

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

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 into another.

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

Stoichiometric mass relationship — The mass relationship between substances derived from a balanced equation and their molar masses.

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

Theoretical yield — The maximum amount of product predicted by stoichiometry under the stated assumptions.

Actual yield — The amount of product actually obtained experimentally.

Limiting reactant — The reactant that is consumed first and determines the maximum amount of product.

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

Key Takeaways

• Balanced chemical equations give relationships between substances in moles.
• Mass-mass calculations therefore require conversion through moles.
• The fundamental pathway is:

g A → mol A → mol B → g B

• Convert mass to moles using:

n = m/M

• Convert moles to mass using:

m = nM

• Convert between substances using the mole ratio from the balanced equation.
• Coefficients are mole ratios, not mass ratios.
• Different substances have different molar masses.
• A mass-mass calculation can determine product mass from reactant mass.
• It can also determine required reactant mass from product mass.
• Unit cancellation helps check whether conversion factors have been arranged correctly.
• Mass-mass stoichiometry obeys conservation of mass.
• A product may have more mass than one starting reactant because other reactants contribute mass.
• Total reactant mass must equal total product mass in a closed system.
• Mass-mass calculations are important in manufacturing, environmental chemistry, combustion, laboratory planning, and many other applications.
• Theoretical product masses may differ from experimentally obtained masses.
• Always check whether a calculated answer is chemically reasonable.

The central strategy is:

BALANCE → GRAMS TO MOLES → MOLE RATIO → MOLES TO GRAMS

Check Your Understanding

Use:

2Mg + O₂ → 2MgO

with:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

1. How many moles of Mg are present in 12.15 g Mg?

2. How many moles of MgO can form from this amount?

3. Calculate the mass of MgO produced.

4. Calculate the mass of MgO that can form from 48.6 g Mg.

5. Explain why the mass of MgO is greater than the mass of Mg.

Use:

2H₂ + O₂ → 2H₂O

with:

M(H₂) = 2.0 g/mol

M(O₂) = 32.0 g/mol

M(H₂O) = 18.0 g/mol

6. What mass of H₂O can form from 2.0 g H₂?

7. What mass of H₂O can form from 8.0 g H₂?

8. What mass of O₂ is required to react with 4.0 g H₂?

9. What mass of H₂ is required to produce 72 g H₂O?

10. Show that your answer to Question 6 agrees with conservation of mass.

Use:

4Fe + 3O₂ → 2Fe₂O₃

with:

M(Fe) = 56 g/mol

M(O₂) = 32 g/mol

M(Fe₂O₃) = 160 g/mol

11. Calculate the mass of Fe₂O₃ produced from 56 g Fe.

12. Calculate the mass of Fe₂O₃ produced from 112 g Fe.

13. Calculate the mass of O₂ required to react with 112 g Fe.

14. Use your answers to Questions 12 and 13 to demonstrate conservation of mass.

15. What mass of Fe is required to produce 320 g Fe₂O₃?

Use:

CH₄ + 2O₂ → CO₂ + 2H₂O

with:

M(CH₄) = 16 g/mol

M(CO₂) = 44 g/mol

M(H₂O) = 18 g/mol

16. What mass of CO₂ forms from 16 g CH₄?

17. What mass of H₂O forms from 16 g CH₄?

18. What mass of CO₂ forms from 40 g CH₄?

19. What mass of H₂O forms from 40 g CH₄?

20. Explain why the mass of CO₂ produced is greater than the mass of CH₄ burned.

Multi-Step Challenge

Use:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

with:

M(C₃H₈) = 44 g/mol

M(O₂) = 32 g/mol

M(CO₂) = 44 g/mol

M(H₂O) = 18 g/mol

21. Calculate the mass of CO₂ produced from 44 g C₃H₈.

22. Calculate the mass of H₂O produced from 44 g C₃H₈.

23. Calculate the mass of O₂ required to burn 44 g C₃H₈ completely.

24. Use Questions 21–23 to demonstrate conservation of mass.

25. Calculate the mass of propane required to produce 264 g CO₂.

Use:

2KClO₃ → 2KCl + 3O₂

with:

M(KClO₃) = 122.5 g/mol

M(KCl) = 74.5 g/mol

M(O₂) = 32.0 g/mol

26. Calculate the mass of O₂ produced from 122.5 g KClO₃.

27. Calculate the mass of KCl produced from 122.5 g KClO₃.

28. Add the masses from Questions 26 and 27. Compare the result with the starting mass.

29. Explain why the result demonstrates conservation of mass.

30. Calculate the mass of KClO₃ required to produce 48.0 g O₂.

Application and Reasoning

31. Explain why balanced-equation coefficients cannot normally be used directly as gram ratios.

32. Explain why moles act as the bridge in a mass-mass calculation.

33. Write the four-stage pathway used to convert the mass of reactant A into the mass of product B.

34. A student obtains an answer with units of mol when the question asks for grams. What step has probably been missed?

35. A student uses 16 g/mol as the molar mass of O₂. Explain the error.

36. A reaction uses 30 g of reactant A and 20 g of reactant B to form only one product. A student predicts 70 g of product. Explain why the answer cannot be correct.

37. A calculation predicts 45 g of product, but an experiment produces 39 g. Give three possible explanations.

38. Explain how mass-mass calculations could help a chemical factory reduce waste.

39. Explain how mass-mass calculations can be used to estimate carbon dioxide emissions from fuel combustion.

40. Explain why checking units and conservation of mass provides two independent ways of evaluating whether a stoichiometric answer is reasonable.

Limiting Reactant vs. Excess Reactant

These two concepts are closely connected.

Limiting Reactant

The reactant that:

• is completely consumed first
• stops the reaction when it runs out
• determines the maximum amount of product

Excess Reactant

The reactant that:

• is supplied in more than the required amount
• is not completely consumed
• remains after the reaction stops

If one reactant is limiting, another reactant is usually in excess.

Why Does an Excess Reactant Remain?

Chemical reactions occur according to specific particle ratios.

Consider:

N₂ + 3H₂ → 2NH₃

Every:

1 mol N₂

requires:

3 mol H₂

Suppose we have:

2 mol N₂

and:

9 mol H₂

The 2 mol N₂ require:

6 mol H₂

But:

9 mol H₂

are available.

Only 6 mol H₂ can react because all the nitrogen is then gone.

Therefore:

9 − 6 = 3 mol H₂

remain.

Hydrogen is the excess reactant.

The Basic Excess Reactant Calculation

The central calculation is:

amount remaining = amount initially present − amount consumed

The difficult part is usually determining the amount consumed.

To find it:

1. Identify the limiting reactant.
2. Use the limiting reactant and the mole ratio to determine how much excess reactant reacts.
3. Subtract that amount from the initial amount.

A Useful Roadmap

For excess-reactant problems:

BALANCE THE EQUATION

↓

CONVERT REACTANTS TO MOLES

↓

IDENTIFY THE LIMITING REACTANT

↓

CALCULATE EXCESS REACTANT CONSUMED

↓

INITIAL EXCESS − CONSUMED EXCESS

↓

EXCESS REACTANT REMAINING

This is closely related to limiting-reactant calculations, but the final goal is different.

Worked Example: Hydrogen and Oxygen

Consider:

2H₂ + O₂ → 2H₂O

Suppose:

10 mol H₂

and:

3 mol O₂

are available.

Identify the limiting reactant

Three moles of O₂ require:

3 × 2 = 6 mol H₂

We have:

10 mol H₂

Therefore:

O₂ is limiting

and:

H₂ is in excess

Determine how much H₂ reacts

From the equation:

1 mol O₂ requires 2 mol H₂

Therefore:

3 mol O₂ require 6 mol H₂

Calculate the amount remaining

Initial H₂:

10 mol

Consumed:

6 mol

Remaining:

10 − 6 = 4 mol

Answer

4 mol H₂ remain after the reaction.

Calculate the Product Too

Using the same reaction:

2H₂ + O₂ → 2H₂O

and:

10 mol H₂ + 3 mol O₂

we determined that O₂ is limiting.

The ratio is:

1 mol O₂ : 2 mol H₂O

Therefore:

3 mol O₂ → 6 mol H₂O

At the end:

• H₂O formed = 6 mol
• O₂ remaining = 0 mol
• H₂ remaining = 4 mol

This gives us a complete picture of the reaction.

Worked Example: Ammonia

Consider:

N₂ + 3H₂ → 2NH₃

Suppose:

5 mol N₂

and:

12 mol H₂

are available.

Identify the limiting reactant

Five moles N₂ would require:

15 mol H₂

But only:

12 mol H₂

are available.

Therefore:

H₂ is limiting

and:

N₂ is in excess

Calculate N₂ consumed

Ratio:

3 mol H₂ : 1 mol N₂

Therefore:

12 mol H₂ × (1 mol N₂ / 3 mol H₂)

= 4 mol N₂

Calculate N₂ remaining

Initial:

5 mol N₂

Consumed:

4 mol N₂

Remaining:

5 − 4 = 1 mol N₂

Answer

1 mol N₂ remains in excess.

Worked Example: Another Ammonia Problem

Suppose:

4 mol N₂

and:

9 mol H₂

are available.

Equation:

N₂ + 3H₂ → 2NH₃

Nine moles H₂ require:

9 × (1/3) = 3 mol N₂

We have:

4 mol N₂

Therefore:

H₂ is limiting

and:

N₂ is in excess

N₂ remaining:

4 − 3 = 1 mol N₂

Product:

9 mol H₂ × (2 mol NH₃ / 3 mol H₂)

= 6 mol NH₃

At the end:

• H₂ = 0 mol
• N₂ = 1 mol
• NH₃ = 6 mol

When Masses Are Given

Excess-reactant problems often provide masses rather than moles.

Because balanced equations describe mole ratios, first convert the reactants to moles.

The pathway becomes:

grams → moles → identify limiting/excess reactants → calculate excess consumed → calculate excess remaining

 

 

 

5

Worked Example: Magnesium and Oxygen

Consider:

2Mg + O₂ → 2MgO

Suppose:

36.45 g Mg

react with:

16.0 g O₂

Use:

M(Mg) = 24.3 g/mol

M(O₂) = 32.0 g/mol

Convert magnesium to moles

36.45 ÷ 24.3 = 1.50 mol Mg

Convert oxygen to moles

16.0 ÷ 32.0 = 0.500 mol O₂

Identify the limiting reactant

The equation requires:

2 mol Mg : 1 mol O₂

Therefore:

0.500 mol O₂

requires:

1.00 mol Mg

We have:

1.50 mol Mg

Therefore:

O₂ is limiting

and:

Mg is in excess

Calculate the Excess Magnesium Remaining

Initial Mg:

1.50 mol

Mg consumed:

1.00 mol

Therefore:

1.50 − 1.00 = 0.50 mol Mg

Convert to mass:

m = nM

m = 0.50 × 24.3

= 12.15 g

Answer

12.15 g Mg remain after the reaction.

Check the Entire Reaction

The limiting O₂ produces MgO.

From:

2Mg + O₂ → 2MgO

0.500 mol O₂ → 1.00 mol MgO

Use:

M(MgO) = 40.3 g/mol

Therefore:

40.3 g MgO

are produced.

Initial mass:

36.45 + 16.0 = 52.45 g

Final mass:

40.3 + 12.15 = 52.45 g

Therefore:

initial mass = final mass

The leftover excess reactant must be included when checking conservation of mass.

Why the Excess Reactant Cannot Keep Reacting

Suppose magnesium remains after all the oxygen has been consumed.

You might ask:

Why doesn't the remaining magnesium continue reacting?

Because the reaction requires oxygen.

Once there are no O₂ molecules left, the remaining Mg atoms have nothing to react with.

The reaction stops even though magnesium is still present.

Adding more oxygen would allow the reaction to continue.

Worked Example: Iron and Oxygen

Consider:

4Fe + 3O₂ → 2Fe₂O₃

Suppose:

112 g Fe

react with:

64 g O₂

Use:

M(Fe) = 56 g/mol

M(O₂) = 32 g/mol

Convert to moles

Fe:

112 ÷ 56 = 2.00 mol

O₂:

64 ÷ 32 = 2.00 mol

Equal numbers of moles do not mean equal stoichiometric quantities.

The equation requires:

4 mol Fe : 3 mol O₂

Identify the Excess Reactant

Two moles Fe require:

2 × (3/4) = 1.50 mol O₂

But:

2.00 mol O₂

are available.

Therefore:

Fe is limiting

and:

O₂ is in excess

Calculate Oxygen Remaining

Initial O₂:

2.00 mol

Consumed:

1.50 mol

Remaining:

2.00 − 1.50 = 0.50 mol O₂

Convert to mass:

0.50 × 32 = 16 g

Answer

16 g O₂ remain.

Check the Product

Two moles Fe produce:

1 mol Fe₂O₃

Use:

M(Fe₂O₃) = 160 g/mol

Product:

160 g Fe₂O₃

Initial mass:

112 + 64 = 176 g

Final mass:

160 + 16 = 176 g

Again:

mass is conserved

Worked Example: Aluminum and Chlorine

Consider:

2Al + 3Cl₂ → 2AlCl₃

Suppose:

54.0 g Al

react with:

142 g Cl₂

Use:

M(Al) = 27.0 g/mol

M(Cl₂) = 71.0 g/mol

Convert to moles

Al:

54.0 ÷ 27.0 = 2.00 mol Al

Cl₂:

142 ÷ 71.0 = 2.00 mol Cl₂

Required ratio:

2 mol Al : 3 mol Cl₂

Two moles Cl₂ require:

2 × (2/3) = 1.33 mol Al

We have:

2.00 mol Al

Therefore:

Cl₂ is limiting

and:

Al is in excess

Calculate Aluminum Remaining

Al consumed:

2.00 mol Cl₂ × (2 mol Al / 3 mol Cl₂)

= 1.33 mol Al

Initial Al:

2.00 mol

Remaining:

2.00 − 1.33 = 0.67 mol Al

Convert to mass:

0.67 × 27.0 ≈ 18.0 g Al

Answer

Approximately:

18.0 g Al remain.

Using Product Amount to Find Excess Consumed

Sometimes you already know how much product formed.

Consider:

2H₂ + O₂ → 2H₂O

Suppose the reaction produces:

8 mol H₂O

How much O₂ was consumed?

Ratio:

1 mol O₂ : 2 mol H₂O

Therefore:

8 mol H₂O × (1 mol O₂ / 2 mol H₂O)

= 4 mol O₂

If initially there were:

6 mol O₂

then:

6 − 4 = 2 mol O₂

remain.

This is another way to calculate leftover reactant.

Calculating Percent Excess

In more advanced stoichiometry, chemists may describe how much extra reactant has been supplied using percent excess.

First determine how much reactant is actually required.

Then:

excess amount = actual amount − required amount

and:

percent excess = (excess amount / required amount) × 100%

Worked Example: Percent Excess

Suppose a reaction requires:

20 g of reactant B

but:

25 g

are supplied.

Excess:

25 − 20 = 5 g

Percent excess:

(5 / 20) × 100% = 25%

Therefore:

B was supplied at 25% excess.

This does not mean that 25% of the original amount necessarily remains in every situation; it describes the extra amount relative to the stoichiometric requirement.

Why Use an Excess Reactant?

Using an excess reactant may sound wasteful, but it can be useful.

An excess reactant may help:

• ensure the limiting reactant reacts completely
• increase conversion of an expensive reactant
• improve production efficiency
• drive some reactions toward greater product formation
• compensate for practical losses
• maintain desired reaction conditions

The choice of which substance to use in excess can be economically important.

 

 

 

5

Excess Reactants in Industry

Suppose reactant A is very expensive and reactant B is inexpensive.

A manufacturer may deliberately use excess B.

This helps ensure that as much of A as possible reacts.

Afterward, unused B may sometimes be:

• separated
• purified
• recycled
• returned to the reactor

This can reduce both costs and waste.

Excess Oxygen in Combustion

Combustion provides a familiar example.

For complete methane combustion:

CH₄ + 2O₂ → CO₂ + 2H₂O

The stoichiometric ratio requires:

1 mol CH₄ : 2 mol O₂

In practical combustion systems, oxygen may be supplied in excess to help ensure more complete combustion of the fuel.

Insufficient oxygen can contribute to incomplete combustion and the formation of products such as carbon monoxide.

Worked Combustion Example

Suppose:

2 mol CH₄

react with:

6 mol O₂

Equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

Two moles CH₄ require:

4 mol O₂

Available:

6 mol O₂

Therefore:

CH₄ is limiting

and:

O₂ is in excess

O₂ remaining:

6 − 4 = 2 mol O₂

Products:

2 mol CO₂

and:

4 mol H₂O

When There Is No Excess Reactant

Not every reaction mixture has an excess reactant.

For:

2H₂ + O₂ → 2H₂O

suppose:

4 mol H₂

and:

2 mol O₂

are available.

The ratio is exactly:

2 : 1

Both reactants are completely consumed.

Therefore:

H₂ remaining = 0

O₂ remaining = 0

The reactants were present in stoichiometric proportions.

Excess Reactants and Conservation of Mass

Excess reactants are especially important when accounting for mass.

Suppose:

70 g

of reactants are initially present.

After the reaction:

12 g

of an excess reactant remain.

If there is only one product:

product mass = 70 − 12

= 58 g

The excess reactant remains part of the system.

It cannot simply be ignored.

A Complete Mass Balance

Suppose a reaction begins with:

25 g A

and:

40 g B

Total initial mass:

65 g

Suppose A is limiting and:

15 g B

remain afterward.

Mass consumed:

25 g A + 25 g B = 50 g

If there is one product:

product mass = 50 g

Final mass:

50 g product + 15 g B = 65 g

Therefore:

initial mass = final mass

Practical Laboratory Example

Imagine mixing two solutions to form a precipitate.

If one dissolved reactant is supplied in excess:

• all of the limiting reactant may be consumed
• the solid product forms
• some excess reactant remains dissolved in the solution

The excess reactant has not disappeared simply because it cannot be seen.

It may remain as dissolved ions in the solution.

This is important when interpreting laboratory results.

 

 

 

6

Choosing Which Reactant Should Be in Excess

Chemists may consider several factors:

• cost
• availability
• safety
• toxicity
• ease of separation
• environmental impact
• ability to recycle unused material
• desired reaction efficiency

For example, it may make sense to use an inexpensive and easily removed substance in excess rather than an expensive or hazardous one.

A Full Problem-Solving Example

Consider:

2Na + Cl₂ → 2NaCl

Suppose:

69 g Na

react with:

71 g Cl₂

Use:

M(Na) = 23 g/mol

M(Cl₂) = 71 g/mol

M(NaCl) = 58.5 g/mol

Convert to moles

Na:

69 ÷ 23 = 3 mol Na

Cl₂:

71 ÷ 71 = 1 mol Cl₂

Determine the limiting reactant

One mole Cl₂ requires:

2 mol Na

We have:

3 mol Na

Therefore:

Cl₂ is limiting

and:

Na is in excess

Calculate Na consumed

1 mol Cl₂ × (2 mol Na / 1 mol Cl₂)

= 2 mol Na

Calculate Na remaining

Initial:

3 mol Na

Consumed:

2 mol Na

Remaining:

1 mol Na

Mass remaining:

1 × 23 = 23 g Na

Calculate product

One mole Cl₂ produces:

2 mol NaCl

Mass:

2 × 58.5 = 117 g NaCl

Check mass conservation

Initial:

69 + 71 = 140 g

Final:

117 + 23 = 140 g

Everything is accounted for.

Common Mistakes

Confusing Excess with Limiting

The limiting reactant runs out.

The excess reactant remains.

Assuming the Larger Mass Is Excess

A substance is not excess simply because more grams are present.

Molar masses and stoichiometric ratios must be considered.

Assuming the Larger Number of Moles Is Excess

The balanced equation may require different mole quantities.

For:

N₂ + 3H₂ → 2NH₃

having more H₂ moles than N₂ does not automatically mean H₂ is in excess.

Three times as much H₂ is required.

Subtracting the Limiting Reactant from the Excess Reactant

Do not calculate:

initial excess − limiting amount

unless the mole ratio happens to be 1 : 1.

First use the balanced equation to determine the amount of excess reactant consumed.

Subtracting Different Units

Do not calculate something such as:

10 g − 0.2 mol

Both quantities must be expressed in compatible units.

Forgetting to Convert Back to Grams

If the question asks for the mass remaining, convert leftover moles to mass.

Using Initial Excess to Calculate Product

The product is controlled by the limiting reactant, not by the total amount of excess reactant supplied.

Forgetting Leftover Material in Mass Conservation

If excess reactant remains, it must be included in the final mass.

Rounding Too Early

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

Key Terms

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

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

Leftover reactant — The portion of an excess reactant remaining after the reaction stops.

Excess consumed — The amount of excess reactant that participates in the reaction.

Excess remaining — The amount of excess reactant left after the limiting reactant has been consumed.

Stoichiometric ratio — The quantitative relationship between substances given by a balanced equation.

Stoichiometric proportions — Reactants present in exactly the required mole ratio.

Mole ratio — A relationship between quantities of substances based on balanced-equation coefficients.

Percent excess — The amount supplied beyond the stoichiometric requirement, expressed as a percentage of the required amount.

Theoretical yield — The maximum amount of product predicted from the limiting reactant.

Mass balance — Accounting for all mass entering, leaving, reacting, and remaining in a chemical system.

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

Key Takeaways

• An excess reactant is present in more than the amount required by the balanced equation.
• The excess reactant is not completely consumed.
• The limiting reactant runs out first.
• The limiting reactant determines how much product forms.
• The excess reactant determines how much material may remain afterward.
• Limiting and excess reactants must be identified using stoichiometric ratios.
• Neither mass nor number of moles alone reliably identifies the excess reactant.
• When masses are given, convert them to moles before comparing reactants.
• To find leftover reactant:

amount remaining = amount initially present − amount consumed

• Use the limiting reactant to calculate how much of the excess reactant is consumed.
• If the question asks for leftover mass, convert the remaining moles back to grams.
• Sometimes reactants are present in exact stoichiometric proportions and neither remains in excess.
• Excess reactants may be deliberately used in laboratories and industrial processes.
• Excess reactants can help ensure that a more valuable reactant is consumed as completely as possible.
• Unused excess material may sometimes be recovered and recycled.
• Leftover reactant must be included when checking conservation of mass.

The main pathway is:

BALANCE → CONVERT TO MOLES → IDENTIFY LIMITING REACTANT → IDENTIFY EXCESS REACTANT → CALCULATE EXCESS CONSUMED → SUBTRACT → FIND EXCESS REMAINING

Check Your Understanding

Use:

2H₂ + O₂ → 2H₂O

1. If 8 mol H₂ react with 3 mol O₂, identify the excess reactant.

2. How many moles of the excess reactant are consumed?

3. How many moles of the excess reactant remain?

4. How many moles of H₂O form?

5. If 6 mol H₂ react with 3 mol O₂, is either reactant in excess? Explain.

Use:

N₂ + 3H₂ → 2NH₃

6. If 4 mol N₂ react with 9 mol H₂, identify the excess reactant.

7. Calculate the amount of N₂ consumed.

8. Calculate the amount of N₂ remaining.

9. Calculate the amount of NH₃ produced.

10. If 3 mol N₂ react with 12 mol H₂, calculate the amount of excess reactant remaining.

Mass Problems

Use:

2Mg + O₂ → 2MgO

with:

M(Mg) = 24.3 g/mol

M(O₂) = 32.0 g/mol

M(MgO) = 40.3 g/mol

11. If 48.6 g Mg react with 16.0 g O₂, identify the excess reactant.

12. Calculate the mass of excess reactant consumed.

13. Calculate the mass of excess reactant remaining.

14. Calculate the mass of MgO formed.

15. Show that the initial and final masses are equal.

Use:

4Fe + 3O₂ → 2Fe₂O₃

with:

M(Fe) = 56 g/mol

M(O₂) = 32 g/mol

M(Fe₂O₃) = 160 g/mol

16. If 112 g Fe react with 96 g O₂, identify the excess reactant.

17. Calculate the amount of excess reactant consumed.

18. Calculate the mass of excess reactant remaining.

19. Calculate the mass of Fe₂O₃ produced.

20. Check your answer using conservation of mass.

Challenge Problems

Use:

2Al + 3Cl₂ → 2AlCl₃

with:

M(Al) = 27.0 g/mol

M(Cl₂) = 71.0 g/mol

M(AlCl₃) = 133.5 g/mol

21. If 81.0 g Al react with 213 g Cl₂, determine whether either reactant is in excess.

22. If 81.0 g Al react with 142 g Cl₂, identify the excess reactant.

23. Calculate the mass of excess reactant remaining in Question 22.

24. Calculate the mass of AlCl₃ produced.

25. Demonstrate that mass is conserved.

Use:

CH₄ + 2O₂ → CO₂ + 2H₂O

with:

M(CH₄) = 16 g/mol

M(O₂) = 32 g/mol

26. If 32 g CH₄ react with 160 g O₂, identify the excess reactant.

27. Calculate the mass of excess reactant consumed.

28. Calculate the mass of excess reactant remaining.

29. Calculate the moles of CO₂ and H₂O produced.

30. Explain why the reaction stops even though one reactant remains.

Reasoning and Application

31. Define an excess reactant in your own words.

32. Explain the relationship between limiting and excess reactants.

33. Why can't the reactant with the larger mass automatically be identified as excess?

34. Why must the balanced equation be used when calculating leftover reactant?

35. A reaction begins with 80 g of total reactants and leaves 12 g of excess reactant. If only one product forms, calculate the mass of product.

36. Explain how Question 35 demonstrates conservation of mass.

37. A reaction requires 40 g of reactant B, but 50 g are supplied. Calculate the mass supplied in excess.

38. Calculate the percent excess in Question 37.

39. Explain why an industrial process might deliberately use one reactant in excess.

40. Describe the complete procedure for determining the mass of an excess reactant remaining when the initial masses of two reactants are known.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant, according to the balanced chemical equation.

It is called theoretical because it represents what should be produced under ideal conditions.

The calculation assumes that:

• the reaction goes completely to products
• no product is lost
• no unwanted side reactions occur
• the reactants are pure
• the chemical equation accurately represents the reaction

In a real experiment, the amount of product collected is often lower than the theoretical yield.

 

 

 

5

Theoretical Yield and Stoichiometry

Theoretical yield is calculated using stoichiometry.

A balanced chemical equation tells us the mole relationship between reactants and products.

For example:

2Mg + O₂ → 2MgO

This tells us:

2 mol Mg → 2 mol MgO

or:

1 mol Mg → 1 mol MgO

If we know how much magnesium reacts, we can calculate the maximum amount of magnesium oxide that could theoretically form.

The basic pathway is:

amount of reactant → moles of reactant → moles of product → theoretical yield

If the answer is required in grams:

g reactant → mol reactant → mol product → g product

A Simple Example

Consider:

2H₂ + O₂ → 2H₂O

Suppose:

4 mol H₂

react with sufficient oxygen.

The mole ratio is:

2 mol H₂ : 2 mol H₂O

Therefore:

4 mol H₂ → 4 mol H₂O

So the theoretical yield is:

4 mol H₂O

If we want the answer in grams:

M(H₂O) = 18 g/mol

Therefore:

m = nM

m = 4 × 18

= 72 g

Theoretical yield

72 g H₂O

Theoretical Yield Is a Maximum

The word maximum is important.

If stoichiometry predicts:

72 g H₂O

then the theoretical yield is:

72 g

Under the assumptions of the calculation, the reaction cannot produce more product from the stated amount of limiting reactant.

In a laboratory, however, we might collect:

68 g

or:

61 g

or some other amount below the theoretical value.

The amount actually obtained is called the actual yield.

Theoretical Yield vs. Actual Yield

Theoretical Yield

The maximum amount predicted by stoichiometry.

It is calculated.

Actual Yield

The amount actually obtained during an experiment.

It is usually measured.

For example:

Theoretical yield:

25.0 g

Actual yield:

21.3 g

The difference indicates that not all of the theoretically possible product was successfully obtained.

Where Theoretical Yield Comes From

Theoretical yield comes from three pieces of information:

The Balanced Equation

This provides the mole ratio.

The Amount of Reactant

This tells us how much material is available.

The Molar Masses

These allow us to convert between grams and moles.

Together, these allow us to predict the maximum product.

The Basic Calculation Method

When the mass of one reactant is given:

Balance the equation.

Convert the given mass to moles.

Use:

n = m/M

Use the mole ratio.

Convert:

mol reactant → mol product

Convert product moles to mass.

Use:

m = nM

The resulting mass is the theoretical yield.

Worked Example: Magnesium Oxide

Consider:

2Mg + O₂ → 2MgO

What is the theoretical yield of MgO when 12.15 g Mg reacts with excess oxygen?

Use:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

Convert Mg to moles

n = 12.15 / 24.3

= 0.500 mol Mg

Use the mole ratio

From:

2Mg → 2MgO

the ratio is:

1 : 1

Therefore:

0.500 mol Mg → 0.500 mol MgO

Convert MgO to mass

m = 0.500 × 40.3

= 20.15 g

Answer

Theoretical yield = 20.15 g MgO

Approximately:

20.2 g MgO

 

 

 

5

Worked Example: Formation of Water

Consider:

2H₂ + O₂ → 2H₂O

What is the theoretical yield of water from 10.0 g H₂, assuming excess oxygen?

Use:

M(H₂) = 2.0 g/mol

M(H₂O) = 18.0 g/mol

Convert H₂ to moles

10.0 ÷ 2.0 = 5.0 mol H₂

Use the mole ratio

H₂ : H₂O = 2 : 2

Therefore:

5.0 mol H₂ → 5.0 mol H₂O

Convert to mass

5.0 × 18.0 = 90 g

Answer

Theoretical yield = 90 g H₂O

Why the Product Can Have More Mass

In the previous example:

10 g H₂

can theoretically produce:

90 g H₂O

This does not violate conservation of mass.

Oxygen also contributes mass to the product.

The reaction requires:

5 mol H₂

and:

2.5 mol O₂

Mass of O₂:

2.5 × 32 = 80 g

Therefore:

10 g H₂ + 80 g O₂ → 90 g H₂O

Mass is conserved.

Worked Example: Calcium Carbonate

Calcium carbonate decomposes when heated:

CaCO₃ → CaO + CO₂

What is the theoretical yield of CaO from 250 g CaCO₃?

Use:

M(CaCO₃) = 100 g/mol

M(CaO) = 56 g/mol

Convert CaCO₃ to moles

250 ÷ 100 = 2.50 mol CaCO₃

Use the mole ratio

CaCO₃ : CaO = 1 : 1

Therefore:

2.50 mol CaO

Convert to mass

2.50 × 56 = 140 g

Answer

Theoretical yield = 140 g CaO

Predicting the Other Product

For:

CaCO₃ → CaO + CO₂

what is the theoretical yield of CO₂ from the same 250 g CaCO₃?

Use:

M(CO₂) = 44 g/mol

We already know:

250 g CaCO₃ = 2.50 mol CaCO₃

The ratio is:

1 mol CaCO₃ : 1 mol CO₂

Therefore:

2.50 mol CO₂

Mass:

2.50 × 44 = 110 g

Answer

Theoretical yield = 110 g CO₂

Check:

140 g CaO + 110 g CO₂ = 250 g

Conservation of mass is satisfied.

Theoretical Yield with Different Coefficients

Consider:

2KClO₃ → 2KCl + 3O₂

What is the theoretical yield of oxygen from 49.0 g KClO₃?

Use:

M(KClO₃) = 122.5 g/mol

M(O₂) = 32.0 g/mol

Convert KClO₃ to moles

49.0 ÷ 122.5 = 0.400 mol KClO₃

Use the mole ratio

KClO₃ : O₂ = 2 : 3

Therefore:

0.400 × (3/2)

= 0.600 mol O₂

Convert to mass

0.600 × 32.0 = 19.2 g

Answer

Theoretical yield = 19.2 g O₂

Theoretical Yield and Limiting Reactants

When two or more reactant quantities are given, you cannot simply choose one reactant to calculate theoretical yield.

You must first identify the limiting reactant.

The limiting reactant determines the theoretical yield because it is consumed first.

Once it runs out, no additional product can form.

The pathway becomes:

reactant amounts → identify limiting reactant → calculate product from limiting reactant → theoretical yield

Worked Example with Two Reactants

Consider:

2H₂ + O₂ → 2H₂O

Suppose:

10 g H₂

react with:

64 g O₂

Use:

M(H₂) = 2 g/mol

M(O₂) = 32 g/mol

Convert H₂ to moles

10 ÷ 2 = 5 mol H₂

Convert O₂ to moles

64 ÷ 32 = 2 mol O₂

The equation requires:

2 mol H₂ : 1 mol O₂

Two moles O₂ require:

4 mol H₂

We have:

5 mol H₂

Therefore:

O₂ is limiting

and H₂ is in excess.

Calculate Theoretical Yield from the Limiting Reactant

Equation:

2H₂ + O₂ → 2H₂O

From:

2 mol O₂

we obtain:

4 mol H₂O

Mass:

4 × 18 = 72 g

Answer

Theoretical yield = 72 g H₂O

We must use the limiting reactant because it determines the maximum amount of product.

What If We Used the Wrong Reactant?

Suppose we incorrectly calculated the product from all 5 mol H₂.

The 1 : 1 ratio between H₂ and H₂O would predict:

5 mol H₂O

or:

90 g H₂O

But only enough oxygen exists to produce:

72 g H₂O

Therefore:

90 g is impossible under the stated conditions.

This demonstrates why identifying the limiting reactant is essential.

Worked Example: Iron Oxide

Consider:

4Fe + 3O₂ → 2Fe₂O₃

Suppose:

112 g Fe

react with:

64 g O₂

Use:

M(Fe) = 56 g/mol

M(O₂) = 32 g/mol

M(Fe₂O₃) = 160 g/mol

Convert reactants to moles

Fe:

112 ÷ 56 = 2 mol Fe

O₂:

64 ÷ 32 = 2 mol O₂

Identify the limiting reactant

Two moles Fe require:

2 × (3/4) = 1.5 mol O₂

We have:

2 mol O₂

Therefore:

Fe is limiting

and O₂ is in excess.

Calculate product

Ratio:

4 mol Fe : 2 mol Fe₂O₃

Therefore:

2 mol Fe → 1 mol Fe₂O₃

Mass:

1 × 160 = 160 g

Answer

Theoretical yield = 160 g Fe₂O₃

Worked Example: Aluminum Chloride

Consider:

2Al + 3Cl₂ → 2AlCl₃

Suppose:

27.0 g Al

react with:

71.0 g Cl₂

Use:

M(Al) = 27.0 g/mol

M(Cl₂) = 71.0 g/mol

M(AlCl₃) = 133.5 g/mol

Convert to moles

Al:

27.0 ÷ 27.0 = 1.00 mol

Cl₂:

71.0 ÷ 71.0 = 1.00 mol

The equation requires:

2 mol Al : 3 mol Cl₂

Divide by coefficients:

Al:

1.00 ÷ 2 = 0.500

Cl₂:

1.00 ÷ 3 = 0.333

Therefore:

Cl₂ is limiting

Calculate theoretical yield

Ratio:

3 mol Cl₂ : 2 mol AlCl₃

Therefore:

1.00 mol Cl₂ × (2/3)

= 0.667 mol AlCl₃

Mass:

0.667 × 133.5 ≈ 89.0 g

Answer

Theoretical yield ≈ 89.0 g AlCl₃

Theoretical Yield and Actual Experiments

Real chemical reactions rarely behave perfectly.

Suppose stoichiometry predicts:

50.0 g product

but the experiment produces:

43.2 g product

Then:

theoretical yield = 50.0 g

actual yield = 43.2 g

The theoretical calculation has not necessarily been wrong.

Instead, practical factors may have prevented all of the theoretical product from being collected.

 

 

 

8

Why Theoretical Yield Is Often Not Achieved

There are many possible reasons.

The Reaction May Not Go to Completion

Some reactants may remain unreacted.

If not all of the limiting reactant becomes product, the actual yield will be lower.

Product May Be Lost During Transfer

Some product may remain:

• inside a beaker
• on a stirring rod
• on filter paper
• inside a funnel
• in another piece of apparatus

Small losses can occur every time material is transferred.

Product May Be Lost During Filtration

A precipitate may:

• pass through the filter
• remain dissolved
• stick to glassware
• be spilled

This reduces the amount collected.

Side Reactions May Occur

Reactants may undergo unwanted reactions that form other products.

This means some reactant is used without forming the desired product.

Reactants May Contain Impurities

Suppose a sample has a mass of:

10.0 g

but only:

8.5 g

is actually the desired reactant.

Using the full 10.0 g in a theoretical calculation would overestimate how much product can form.

Some Product May Remain Dissolved

When a solid product forms in solution, some may remain dissolved rather than being collected.

Gas May Escape

If the desired product is a gas, some may escape before it is collected or measured.

Reversible Reactions May Not Go to Completion

Some reactions reach equilibrium rather than converting all reactants into products.

This can reduce the amount of desired product.

Experimental Error and Theoretical Yield

Measurement uncertainty can also affect the comparison between theoretical and actual yield.

Possible sources include:

• balance uncertainty
• inaccurate volume measurements
• incomplete drying
• loss during heating
• contamination
• incomplete collection

The theoretical yield represents an ideal prediction, while the actual yield reflects the real experimental process.

Can Actual Yield Be Greater Than Theoretical Yield?

In a correctly performed and interpreted experiment, the actual amount of pure desired product should not exceed the theoretical yield calculated from the true limiting reactant.

However, an experiment may appear to produce more than 100% of the theoretical yield.

For example:

Theoretical yield:

10.0 g

Measured product:

11.2 g

Possible explanations include:

• the product was wet
• impurities were present
• unreacted reactant remained with the product
• another substance contaminated the sample
• the theoretical calculation was incorrect
• the limiting reactant was identified incorrectly

So a measured mass greater than theoretical yield is usually evidence that something needs to be investigated.

Theoretical Yield and Conservation of Mass

Theoretical yield must be consistent with conservation of mass.

Consider:

2Mg + O₂ → 2MgO

Suppose:

24.3 g Mg

react completely.

This requires:

16.0 g O₂

Total reacting mass:

24.3 + 16.0 = 40.3 g

Therefore:

theoretical yield = 40.3 g MgO

The predicted product mass exactly matches the total mass of reactants consumed.

Theoretical Yield and Excess Reactants

An excess reactant does not increase the theoretical yield once the limiting reactant has been completely consumed.

Suppose:

2H₂ + O₂ → 2H₂O

You have:

4 mol H₂

and:

10 mol O₂

Only:

2 mol O₂

are needed to react with the 4 mol H₂.

Adding even more oxygen cannot produce more water because the hydrogen has already been completely consumed.

Therefore:

H₂ is limiting

and:

theoretical yield = 4 mol H₂O

The remaining oxygen is simply excess reactant.

Increasing Theoretical Yield

To increase the theoretical yield, you generally need to increase the amount of the limiting reactant.

Adding more excess reactant will not increase the maximum product.

For example:

2H₂ + O₂ → 2H₂O

Suppose:

2 mol H₂ + 10 mol O₂

Hydrogen is limiting.

Adding another 5 mol O₂ changes nothing.

But increasing H₂ can increase the theoretical yield.

This is an important idea in industrial chemistry.

Theoretical Yield in Manufacturing

Chemical manufacturers use theoretical yield calculations to predict how much product should be possible from their raw materials.

These calculations help determine:

• how much reactant to purchase
• expected production levels
• production costs
• equipment requirements
• waste quantities
• process efficiency

A factory may compare its actual production with theoretical yield to determine how effectively the process is operating.

 

 

 

5

Theoretical Yield in Pharmaceutical Chemistry

Pharmaceutical manufacturing requires careful control of chemical quantities.

Suppose a synthesis theoretically produces:

100 kg

of a pharmaceutical compound.

If the process consistently produces only:

65 kg

chemists may investigate:

• incomplete reactions
• side reactions
• purification losses
• inefficient separation
• decomposition of the product

Improving the process can increase actual production without necessarily increasing the amount of starting material.

Theoretical Yield and Green Chemistry

Higher actual yields can often mean that fewer resources are wasted.

Low yields may result in:

• wasted reactants
• additional solvent use
• greater energy consumption
• more waste requiring disposal
• higher production costs

For this reason, improving reaction efficiency is an important goal of green chemistry.

However, yield is only one measure of sustainability. A high-yield reaction can still create significant waste or require hazardous materials.

A Complete Problem-Solving Strategy

For a theoretical-yield problem:

Balance the chemical equation.

Never perform stoichiometry using an unbalanced equation.

Determine what information is given.

Are you given:

• moles?
• mass?
• quantities of two reactants?

Convert to moles if necessary.

Use:

n = m/M

Identify the limiting reactant if necessary.

If quantities of multiple reactants are provided, determine which runs out first.

Use the mole ratio.

Convert:

mol limiting reactant → mol product

Convert product to the requested unit.

For mass:

m = nM

State the theoretical yield clearly.

Include:

• numerical value
• unit
• substance

For example:

The theoretical yield is 35.6 g CaO.

One-Line Stoichiometric Method

A theoretical-yield calculation can also be written as one continuous calculation.

Consider:

2Mg + O₂ → 2MgO

Starting with:

12.15 g Mg

Calculation:

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

= 20.15 g MgO

The units cancel:

g Mg → mol Mg → mol MgO → g MgO

Therefore:

theoretical yield = 20.15 g MgO

Worked Example: Methane Combustion

Consider:

CH₄ + 2O₂ → CO₂ + 2H₂O

What is the theoretical yield of CO₂ from 32 g CH₄, assuming excess oxygen?

Use:

M(CH₄) = 16 g/mol

M(CO₂) = 44 g/mol

Convert CH₄ to moles

32 ÷ 16 = 2 mol CH₄

Use the mole ratio

CH₄ : CO₂ = 1 : 1

Therefore:

2 mol CO₂

Convert to mass

2 × 44 = 88 g

Answer

Theoretical yield = 88 g CO₂

Worked Example: Sodium Chloride

Consider:

2Na + Cl₂ → 2NaCl

Suppose:

46 g Na

react with excess chlorine.

Use:

M(Na) = 23 g/mol

M(NaCl) = 58.5 g/mol

Convert Na to moles

46 ÷ 23 = 2 mol Na

Use the ratio

2 mol Na → 2 mol NaCl

Therefore:

2 mol NaCl

Convert to mass

2 × 58.5 = 117 g

Answer

Theoretical yield = 117 g NaCl

More Challenging Example

Consider:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Suppose:

22 g C₃H₈

react with excess oxygen.

Use:

M(C₃H₈) = 44 g/mol

M(CO₂) = 44 g/mol

Find the theoretical yield of CO₂.

Convert propane to moles

22 ÷ 44 = 0.500 mol C₃H₈

Use the mole ratio

C₃H₈ : CO₂ = 1 : 3

Therefore:

0.500 × 3 = 1.50 mol CO₂

Convert to mass

1.50 × 44 = 66 g

Answer

Theoretical yield = 66 g CO₂

Common Mistakes

Forgetting to Balance the Equation

The mole ratio comes from the balanced equation.

An incorrect equation produces an incorrect theoretical yield.

Using Grams Directly with the Coefficients

Coefficients represent mole ratios, not mass ratios.

Use:

grams → moles → mole ratio → grams

Ignoring the Limiting Reactant

If two reactant amounts are given, theoretical yield must be based on the limiting reactant.

Using the excess reactant will overestimate the yield.

Choosing the Smaller Mass as Limiting

The reactant with fewer grams is not automatically limiting.

Convert to moles and compare using the balanced equation.

Choosing the Smaller Number of Moles as Limiting

The balanced equation may require unequal numbers of moles.

Always consider the coefficients.

Confusing Theoretical and Actual Yield

Theoretical yield is calculated.

Actual yield is measured experimentally.

Assuming Theoretical Yield Is Always Obtained

Theoretical yield represents ideal conditions.

Real experiments usually involve some loss or inefficiency.

Thinking a Product Cannot Have More Mass Than One Reactant

Other reactants also contribute mass to the product.

Compare total reacting mass, not just one reactant.

Accepting More Than 100% Without Investigation

An apparent yield greater than the theoretical amount usually suggests:

• contamination
• incomplete drying
• measurement error
• incorrect calculations

Rounding Too Early

Keep several digits during intermediate calculations and round the final result appropriately.

Key Terms

Theoretical yield — The maximum amount of product predicted by stoichiometry from the available limiting reactant.

Actual yield — The amount of product actually obtained experimentally.

Stoichiometry — The quantitative relationship between reactants and products in chemical reactions.

Limiting reactant — The reactant consumed first and therefore responsible for determining theoretical yield.

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

Mole ratio — The ratio between substances given by coefficients in a balanced equation.

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

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

Maximum yield — Another way of describing the greatest quantity of product theoretically possible.

Side reaction — An unwanted reaction that consumes reactants or products and forms substances other than the desired product.

Reaction completion — The extent to which the available limiting reactant has been converted into products.

Product loss — Desired product that forms but is not successfully collected or measured.

Experimental error — Measurement or procedural uncertainty that affects experimental results.

Purity — The proportion of a sample consisting of the desired substance rather than impurities.

Key Takeaways

• Theoretical yield is the maximum amount of product predicted by stoichiometry.
• It is calculated from a balanced chemical equation.
• The calculation assumes ideal reaction conditions.
• When one reactant quantity is given and other reactants are in excess, use the given reactant to calculate theoretical yield.
• When multiple reactant quantities are given, first identify the limiting reactant.
• The limiting reactant determines theoretical yield.
• Excess reactant cannot produce additional product after the limiting reactant has been consumed.
• The basic mass pathway is:

g reactant → mol reactant → mol product → g product

• Molar mass converts between grams and moles.
• Balanced-equation coefficients provide the mole ratio.
• Actual yield is the amount obtained experimentally.
• Actual yield is often lower than theoretical yield.
• Product can be lost through transfers, filtration, purification, heating, or other procedures.
• Side reactions, impurities, incomplete reactions, and equilibrium can also reduce actual yield.
• An apparent actual yield greater than theoretical yield should be investigated.
• Increasing the excess reactant alone does not increase theoretical yield.
• Increasing the limiting reactant can increase theoretical yield.
• Theoretical-yield calculations are important in laboratory chemistry, manufacturing, pharmaceuticals, environmental chemistry, and process design.

The main strategy is:

BALANCE → MOLES → IDENTIFY LIMITING REACTANT → MOLE RATIO → PRODUCT → THEORETICAL YIELD

Check Your Understanding

Use:

2Mg + O₂ → 2MgO

with:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

1. Define theoretical yield.

2. Calculate the theoretical yield of MgO from 24.3 g Mg, assuming excess oxygen.

3. Calculate the theoretical yield of MgO from 48.6 g Mg.

4. Calculate the theoretical yield of MgO from 6.075 g Mg.

5. Explain why the mass of MgO can be greater than the initial mass of Mg.

Use:

CaCO₃ → CaO + CO₂

with:

M(CaCO₃) = 100 g/mol

M(CaO) = 56 g/mol

M(CO₂) = 44 g/mol

6. Calculate the theoretical yield of CaO from 100 g CaCO₃.

7. Calculate the theoretical yield of CO₂ from 100 g CaCO₃.

8. Calculate both theoretical yields from 350 g CaCO₃.

9. Add the masses of the two products from Question 8. What do you notice?

10. Explain how your answer demonstrates conservation of mass.

Limiting Reactant Problems

Use:

2H₂ + O₂ → 2H₂O

with:

M(H₂) = 2 g/mol

M(O₂) = 32 g/mol

M(H₂O) = 18 g/mol

11. If 8 g H₂ react with 32 g O₂, identify the limiting reactant.

12. Calculate the theoretical yield of H₂O.

13. Calculate the mass of excess reactant remaining.

14. If 4 g H₂ react with 64 g O₂, calculate the theoretical yield.

15. Explain why adding even more excess reactant would not increase the theoretical yield.

Use:

4Fe + 3O₂ → 2Fe₂O₃

with:

M(Fe) = 56 g/mol

M(O₂) = 32 g/mol

M(Fe₂O₃) = 160 g/mol

16. If 112 g Fe react with 96 g O₂, identify the limiting reactant.

17. Calculate the theoretical yield of Fe₂O₃.

18. Calculate the mass of excess reactant remaining.

19. If 224 g Fe react with 96 g O₂, calculate the theoretical yield of Fe₂O₃.

20. Determine whether any reactant remains after Question 19.

More Challenging Problems

Use:

2Al + 3Cl₂ → 2AlCl₃

with:

M(Al) = 27.0 g/mol

M(Cl₂) = 71.0 g/mol

M(AlCl₃) = 133.5 g/mol

21. If 54.0 g Al react with 213 g Cl₂, calculate the theoretical yield of AlCl₃.

22. If 54.0 g Al react with 142 g Cl₂, identify the limiting reactant.

23. Calculate the theoretical yield for Question 22.

24. Calculate the mass of excess reactant remaining.

25. Explain why the limiting reactant, rather than the excess reactant, must be used to determine theoretical yield.

Application and Reasoning

26. Explain the difference between theoretical yield and actual yield.

27. Give three reasons why actual yield may be lower than theoretical yield.

28. Explain how product lost during filtration affects actual yield.

29. Explain how a side reaction can reduce the amount of desired product.

30. Explain how impure reactants can affect a theoretical-yield calculation.

31. A calculation predicts a theoretical yield of 50.0 g. An experiment produces 43.0 g. Which value is the theoretical yield and which is the actual yield?

32. A reaction has a theoretical yield of 20.0 g, but a student measures 21.8 g of product. Give two possible explanations.

33. Explain why an apparent actual yield greater than theoretical yield should be investigated.

34. A manufacturer doubles the amount of an excess reactant but keeps the limiting reactant unchanged. What happens to the theoretical yield? Explain.

35. How could a manufacturer increase the theoretical yield of a process?

36. Explain why theoretical-yield calculations are useful before conducting a laboratory experiment.

37. Explain why theoretical-yield calculations are important in industrial chemistry.

38. Describe the complete procedure for calculating theoretical yield when the mass of one reactant is given and all other reactants are in excess.

39. Describe how the procedure changes when the masses of two reactants are given.

40. Explain why theoretical yield represents an ideal maximum rather than a guarantee of how much product will actually be collected.