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.

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.

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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.

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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.

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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

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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.

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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.

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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.

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

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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.

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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.

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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

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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.

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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.

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

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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

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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

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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

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

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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.

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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

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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

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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

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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

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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

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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.

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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.

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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.

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.

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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

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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.

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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.

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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

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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.

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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.

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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.

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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.

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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

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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

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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

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Common Mistakes

Using an Unbalanced Equation

Always balance first.

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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

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Confusing Molar Mass with Coefficients

Molar mass comes from the chemical formula.

Mole ratios come from the balanced equation.

They serve different purposes.

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Using the Wrong Molar Mass

For example:

M(O₂) = 32 g/mol

not:

16 g/mol

because oxygen gas contains two oxygen atoms.

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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.

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Reversing the Conversion Factor

If converting Fe into Fe₂O₃:

4Fe + 3O₂ → 2Fe₂O₃

use:

2 mol Fe₂O₃ / 4 mol Fe

so that mol Fe cancels.

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Forgetting the Final Mass Conversion

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

If it asks for grams, multiply by molar mass.

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Rounding Too Early

Keep several digits during intermediate steps.

Round the final answer.

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Rejecting a Larger Product Mass

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

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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.

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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

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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.