Chapter 11: Mole Concept and Chemical Equations

Why this matters

(I once asked a class, "How many atoms are in this glass of water?" One student guessed a thousand. Another guessed a million. Then I wrote the actual number on the board — 5,000,000,000,000,000,000,000,000,000,000 — and the room went silent. That's the scale of chemistry. Atoms are so small that counting them one by one is impossible. So chemists invented a unit — the mole — that lets us count atoms the way a shopkeeper counts eggs by the dozen. Once you understand the mole, you can calculate how much of anything reacts with anything else. It is the accountant's tool of chemistry, and it is beautiful.) Imagine you run a bakery. You don't count flour grains one by one — you weigh them. You don't count eggs individually — you buy them by the dozen. Chemists do the same thing with atoms and molecules. We can't count them one by one, so we count them in groups. That group is called a mole, and it contains 6.022 × 10²³ particles — a number so large it's called Avogadro's number. The mole is the bridge between the invisible world of atoms and the measurable world of grams and litres. It lets us answer questions like: How much hydrogen will react with 10 grams of oxygen? How much carbon dioxide will a car produce per kilometre? How much medicine should be in a tablet? In this chapter, we're going to learn what a mole is, how to use it, and how to write balanced chemical equations that tell us exactly what happens in a reaction. This is the mathematics of chemistry — and once you get it, everything else becomes possible. Let's begin.

Learning objectives

• Define the mole and state Avogadro's number.
• Calculate the number of particles (atoms, molecules, ions) in a given number of moles.
• Define molar mass and calculate the molar mass of elements and compounds.
• Convert between mass, moles, and number of particles.
• Calculate the percentage composition of elements in a compound.
• Determine empirical and molecular formulae from given data.
• Write and balance chemical equations.
• Appreciate that the mole concept allows chemists to predict and control chemical reactions.

Concept build-up

11.1 Why We Need the Mole

Atoms are incredibly small. A single atom of carbon has a mass of about 2 × 10⁻²³ grams. That's 0.00000000000000000000002 grams. No balance in the world can weigh a single atom. No human can count individual atoms.

But chemists need to know how many atoms or molecules are reacting. When hydrogen reacts with oxygen to form water, we need to know how much hydrogen reacts with how much oxygen. We can't count individual molecules — so we count them in groups.

The group is called a mole. Just like a "dozen" means 12 items, a "mole" means 6.022 × 10²³ items. That number is called Avogadro's number (L), named after the Italian scientist Amedeo Avogadro.

(Why such a strange number? Because it's the number of atoms in exactly 12 grams of carbon-12. It's the bridge that lets us connect the mass we can measure on a balance to the number of particles we can't see. It's not arbitrary — it was chosen to make the maths work.)

11.2 Avogadro's Number

Avogadro's number (L) = 6.022 × 10²³ particles per mole.

This means:

• 1 mole of atoms = 6.022 × 10²³ atoms
• 1 mole of molecules = 6.022 × 10²³ molecules
• 1 mole of ions = 6.022 × 10²³ ions
• 1 mole of electrons = 6.022 × 10²³ electrons

The type of particle depends on what substance you're talking about. If you have 1 mole of hydrogen gas (H₂), you have 6.022 × 10²³ molecules of H₂. If you have 1 mole of sodium chloride (NaCl), you have 6.022 × 10²³ formula units of NaCl, which means 6.022 × 10²³ Na⁺ ions and 6.022 × 10²³ Cl⁻ ions.

11.3 Molar Mass

The molar mass of a substance is the mass of one mole of that substance. It is expressed in grams per mole (g/mol).

For an element, the molar mass is numerically equal to its relative atomic mass (Ar). For example:

• Carbon (Ar = 12) → molar mass = 12 g/mol
• Oxygen (Ar = 16) → molar mass = 16 g/mol
• Sodium (Ar = 23) → molar mass = 23 g/mol

For a compound, you add up the relative atomic masses of all the atoms in the formula:

• Water (H₂O): (2 × 1) + 16 = 18 g/mol
• Carbon dioxide (CO₂): 12 + (2 × 16) = 44 g/mol
• Sodium chloride (NaCl): 23 + 35.5 = 58.5 g/mol
• Sulphuric acid (H₂SO₄): (2 × 1) + 32 + (4 × 16) = 98 g/mol

(Here's the key insight: one mole of any substance contains the same number of particles — 6.022 × 10²³. But different substances have different masses per mole, because their atoms have different masses. One mole of carbon weighs 12 g. One mole of oxygen weighs 16 g. Same number of atoms, different mass. That's the beauty of the mole.)

11.4 The Mole Triangle

Here is the most important relationship in this chapter. Memorise this triangle:

Quantity	Symbol	Unit	Formula
Number of moles	n	mol	n = mass ÷ molar mass
Mass	m	g	m = n × molar mass
Molar mass	M	g/mol	M = mass ÷ moles
Number of particles	N	particles	N = n × 6.022 × 10²³

The central equation: n = m / M

Where:

• n = number of moles (mol)
• m = mass (g)
• M = molar mass (g/mol)

And to convert moles to particles: N = n × L, where L = 6.022 × 10²³.

11.5 Converting Between Mass, Moles, and Particles

Let's practise using the mole triangle with some worked examples.

Example 1: Mass to moles

Question: How many moles are there in 36 g of water (H₂O)?

Solution:
Molar mass of H₂O = (2 × 1) + 16 = 18 g/mol.
n = m / M = 36 / 18 = 2 moles.

Example 2: Moles to mass

Question: What is the mass of 0.5 moles of carbon dioxide (CO₂)?

Solution:
Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol.
m = n × M = 0.5 × 44 = 22 g.

Example 3: Moles to particles

Question: How many molecules are there in 2 moles of oxygen gas (O₂)?

Solution:
N = n × L = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules.

Example 4: Mass to particles

Question: How many atoms are there in 4 g of helium (He)?

Solution:
Molar mass of He = 4 g/mol.
n = 4 / 4 = 1 mole.
N = 1 × 6.022 × 10²³ = 6.022 × 10²³ atoms.

11.6 Percentage Composition

The percentage composition of a compound tells you what fraction of the mass comes from each element. This is useful for determining formulae and for analysing unknown substances.

Formula: % element = (mass of element in formula ÷ molar mass of compound) × 100

Example: Calculate the percentage of carbon in carbon dioxide (CO₂).

Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol.
Mass of carbon in CO₂ = 12 g/mol.
% C = (12 ÷ 44) × 100 = 27.27%

Example: Calculate the percentage of water in copper(II) sulphate pentahydrate (CuSO₄·5H₂O).

Molar mass of CuSO₄·5H₂O = 63.5 + 32 + (4 × 16) + 5 × (2 × 1 + 16) = 63.5 + 32 + 64 + 90 = 249.5 g/mol.
Mass of water = 90 g/mol.
% H₂O = (90 ÷ 249.5) × 100 = 36.07%

11.7 Empirical and Molecular Formulae

The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula is the actual number of atoms in a molecule. The molecular formula is always a whole-number multiple of the empirical formula.

Example: Hydrogen peroxide has the molecular formula H₂O₂. Its empirical formula is HO — the simplest ratio.

Determining Empirical Formula from Mass Data

Steps:

1. Write down the mass (or percentage) of each element.
2. Divide each by the relative atomic mass of that element to get moles.
3. Divide all the mole values by the smallest mole value to get the simplest ratio.
4. If you get fractions like 0.5, multiply all by 2 to get whole numbers.
5. Write the empirical formula.

Example: A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.

Element	%	Ar	Moles = % / Ar	Ratio
C	40	12	40/12 = 3.33	3.33/3.33 = 1
H	6.7	1	6.7/1 = 6.7	6.7/3.33 = 2
O	53.3	16	53.3/16 = 3.33	3.33/3.33 = 1

Empirical formula = CH₂O.

Example: A compound has empirical formula CH₂O and a molar mass of 180 g/mol. Determine its molecular formula.

Molar mass of CH₂O = 12 + (2 × 1) + 16 = 30 g/mol.
n = 180 / 30 = 6.
Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ (glucose).

11.8 Chemical Equations

A chemical equation is a shorthand way of describing a chemical reaction. It shows the reactants (starting substances) on the left and the products (new substances) on the right, separated by an arrow.

Word equation: Hydrogen + oxygen → water

Chemical equation: 2H₂ + O₂ → 2H₂O

The numbers in front of the formulae are called coefficients. They tell you the ratio in which the substances react. A coefficient of 1 is not written.

Balancing Chemical Equations

A balanced equation has the same number of each type of atom on both sides. This is because atoms are not created or destroyed in a chemical reaction — they are just rearranged.

Steps to balance an equation:

1. Write the correct formulae for all reactants and products.
2. Count the number of atoms of each element on both sides.
3. Balance one element at a time by adjusting coefficients (never change subscripts in formulae).
4. Check that all elements are balanced.
5. Include state symbols if required: (s), (l), (g), (aq).

Example: Balance the equation for the combustion of methane.

Unbalanced: CH₄ + O₂ → CO₂ + H₂O

Step 1: Count atoms.

• Left: C=1, H=4, O=2
• Right: C=1, H=2, O=3

Step 2: Balance hydrogen by placing a 2 in front of H₂O.

CH₄ + O₂ → CO₂ + 2H₂O

Now right: H=4, O=4.

Step 3: Balance oxygen by placing a 2 in front of O₂.

CH₄ + 2O₂ → CO₂ + 2H₂O

Now both sides: C=1, H=4, O=4. Balanced.

State Symbols

Symbol	Meaning
(s)	Solid
(l)	Liquid
(g)	Gas
(aq)	Aqueous (dissolved in water)

Example with state symbols: 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

11.9 Using Equations to Calculate Masses

Once an equation is balanced, you can use it to calculate the masses of reactants and products. This is called stoichiometry.

Example: Calculate the mass of water produced when 4 g of hydrogen reacts completely with oxygen.

Equation: 2H₂ + O₂ → 2H₂O

1. Moles of H₂ = mass / molar mass = 4 / 2 = 2 moles.
2. From the equation, 2 moles of H₂ produce 2 moles of H₂O. So 2 moles of H₂O are produced.
3. Mass of H₂O = moles × molar mass = 2 × 18 = 36 g.

(Notice the order: write the equation, balance it, convert to moles, use the mole ratio, convert back to mass. That's the pattern for every stoichiometry problem. Practise it until it becomes automatic.)

11.10 Limiting Reactants

In a reaction, the limiting reactant is the one that runs out first and determines how much product is formed. The other reactant is in excess — there's more of it than needed.

Example: 4 g of hydrogen reacts with 32 g of oxygen. Which is the limiting reactant?

Equation: 2H₂ + O₂ → 2H₂O

• Moles of H₂ = 4 / 2 = 2 moles.
• Moles of O₂ = 32 / 32 = 1 mole.
• The equation requires 2 moles of H₂ for every 1 mole of O₂. Here we have 2 moles H₂ and 1 mole O₂ — exactly the right ratio. Neither is limiting; both are used up completely.

Example 2: 2 g of hydrogen reacts with 32 g of oxygen. Which is limiting?

• Moles of H₂ = 2 / 2 = 1 mole.
• Moles of O₂ = 32 / 32 = 1 mole.
• The equation needs 2 moles of H₂ for 1 mole of O₂. We only have 1 mole of H₂, so hydrogen is limiting. Oxygen is in excess.

(Limiting reactant problems are like making sandwiches. If you have 10 slices of bread and 3 slices of cheese, and each sandwich needs 2 slices of bread and 1 slice of cheese, you can only make 3 sandwiches. The cheese runs out first — it's the limiting reactant. The bread is in excess. Same idea in chemistry.)

Worked examples

Example 1: Mole calculations

Question: Calculate the number of moles in 98 g of sulphuric acid (H₂SO₄).

Solution: Molar mass of H₂SO₄ = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98 g/mol.
n = 98 / 98 = 1 mole.

Example 2: Empirical formula

Question: A compound contains 24 g of carbon and 64 g of oxygen. Determine its empirical formula.

Solution:
C: moles = 24 / 12 = 2.
O: moles = 64 / 16 = 4.
Ratio: C : O = 2 : 4 = 1 : 2.
Empirical formula = CO₂.

Example 3: Balancing equations

Question: Balance the equation: Fe + O₂ → Fe₂O₃.

Solution: Unbalanced: Fe + O₂ → Fe₂O₃.
Balance oxygen: 4Fe + 3O₂ → 2Fe₂O₃.
Check iron: 4 on left, 4 on right. Balanced.
Final: 4Fe + 3O₂ → 2Fe₂O₃.

Example 4: Stoichiometry

Question: Calculate the mass of carbon dioxide produced when 24 g of carbon burns completely in oxygen.

Solution:
Equation: C + O₂ → CO₂.
Moles of C = 24 / 12 = 2 moles.
From equation, 1 mole C produces 1 mole CO₂, so 2 moles CO₂ are produced.
Mass of CO₂ = 2 × 44 = 88 g.

Common mistakes

• Confusing moles with mass. A mole is a number of particles; mass is measured in grams.
• Forgetting to multiply by the number of atoms in a formula. In H₂O, there are 2 hydrogen atoms and 1 oxygen atom — don't forget the 2.
• Using the wrong molar mass. Always add up all the atoms in the formula, including those shown by subscripts.
• Thinking Avogadro's number is different for different substances. It's always 6.022 × 10²³ — that's the whole point.
• Balancing equations by changing subscripts. Never change the formula of a substance. Only change coefficients.
• Forgetting state symbols. If the question asks for them, include them.
• Confusing empirical and molecular formula. Empirical is the simplest ratio; molecular is the actual number.
• Ignoring the limiting reactant. In stoichiometry problems, always check which reactant runs out first.
• Rounding too early. Keep full precision until the final answer, then round appropriately.

Checkpoint questions

Structured exam-style questions

Challenge questions

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"The mole is the accountant of chemistry — it counts what cannot be seen and weighs what cannot be measured. Learn it well, and you can predict the invisible."