Ch 3 · The Mole and the Avogadro Constant · O Level Chemistry 5070

Topic 3 of 3 in Stoichiometry — O Level Chemistry 5070 — 56 lessons.

Introduction

The central bridge of all chemical calculations is the mole. Whether you are given a mass in grams, a solution volume in cm^3 or a gas volume at room temperature and pressure (r.t.p.), your first step is always the same: convert to moles. Then use the balanced equation to reach any other substance, and convert back to the unit required.

Key ideas — The mole and the three core equations

the mole is included. Everything in this lesson is examined.

🧑‍🏫 Teacher's explanation

Think of a 'mole' just like a 'dozen'. If I tell you I have a dozen eggs, you know I have 12 eggs. In chemistry, atoms are so tiny that counting them one by one is impossible. Instead, we use a 'chemist's dozen' called the mole. One mole is simply a package containing 6.02 × 10^23 particles.

📖 Definitions

🔍 Worked example — Easy

Calculate the number of moles in 44 g of carbon dioxide, CO_2. (A_r: C=12, O=16)

🔍 Worked example — Standard

A solution has a concentration of 0.5 mol/dm^3. Calculate the number of moles in 250 cm^3 of this solution.

🔍 Worked example — Exam-style

Calculate the volume of 0.2 moles of oxygen gas (O_2) at room temperature and pressure (r.t.p.).

⚠️ Common errors

🧠 Quick check 1

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🎯 Exam question

Key ideas — Reacting masses, gas volumes and titrations

For any mass, volume or concentration of a product or reactant, use three steps.

🧑‍🏫 Teacher's explanation

Stoichiometry is essentially a recipe for chemistry. Just as a recipe for pancakes tells you that 2 eggs and 500g of flour makes 10 pancakes, a balanced chemical equation tells you exactly how many 'units' (moles) of each substance are needed to react completely.

📖 Definitions

🔍 Worked example — Easy

Calculate the mass of magnesium oxide (MgO) produced when 4.8 g of magnesium (Mg) reacts completely with oxygen. (A_r: Mg = 24, O = 16)

🔍 Worked example — Exam-style

A titration is performed between 25.0 cm^3 of 0.150 mol/dm^3 NaOH and HCl. If 20.0 cm^3 of HCl is required for neutralization, calculate the concentration of the HCl. The reaction is NaOH + HCl arrow NaCl + H_2O

⚠️ Common errors

🧠 Quick check 1

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🎯 Exam question

Key ideas — Limiting reactants

When the amounts of both reactants are given, one will run out first. That limiting reactant controls how much product forms; the other is in excess. Convert both to moles, divide each by its coefficient in the equation, and the smaller result is the limiting reactant. Product calculations must start from it, never from the reactant in excess.

🧑‍🏫 Teacher's explanation

Imagine you are making sandwiches. Each sandwich requires 2 slices of bread and 1 slice of cheese. If you have 10 slices of bread and 2 slices of cheese, you can only make 2 sandwiches. Even though you have plenty of bread left over, the cheese 'limits' how many sandwiches you can make. In chemistry, the 'bread' and 'cheese' are your reactants, and the 'sandwiches' are your products.

📖 Definitions

🔍 Worked example — Easy

Calculate the limiting reactant when 10.0 g of H_2 reacts with 10.0 g of O_2 to form H_2O. Equation: 2H_2 + O_2 arrow 2H_2O.

🔍 Worked example — Standard

How many grams of H_2O are produced when 5.0 g of Mg reacts with 5.0 g of HCl? Equation: Mg + 2HCl arrow MgCl_2 + H_2.

🔍 Worked example — Exam-style

A student mixes 0.50 mol of NaOH with 0.20 mol of H_2SO_4. Calculate the mass of Na_2SO_4 produced. Equation: 2NaOH + H_2SO_4 arrow Na_2SO_4 + 2H_2O.

⚠️ Common errors

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🎯 Exam question

Key ideas — Empirical and molecular formulae

To find an empirical formula from masses or percentages by mass: divide each mass by the A_r of that element to get moles, divide every result by the smallest, and round to whole numbers (multiply through if you get 0.5 or 0.33). Heating 2.4 g of magnesium gives 4.0 g of oxide, so 1.6 g of oxygen combined:

🧑‍🏫 Teacher's explanation

Think of a recipe for cookies. The 'empirical formula' is like the simplest ratio of ingredients—for example, 1 cup of flour to 2 cups of sugar. Even if you make a massive batch using 10 cups of flour and 20 cups of sugar, the basic ratio remains 1:2. The 'molecular formula' is the actual amount used in one specific batch, like 10:20.

📖 Definitions

🔍 Worked example — Easy

A compound contains 4.0 g of Calcium and 1.2 g of Oxygen. Determine its empirical formula. (A_r: Ca = 40, O = 16)

🔍 Worked example — Standard

A hydrocarbon has an empirical formula of CH_2 and a relative molecular mass (M_r) of 56. Determine its molecular formula.

🔍 Worked example — Exam-style

A sample of an oxide of iron contains 70.0% Iron and 30.0% Oxygen by mass. Determine the empirical formula. (A_r: Fe = 56, O = 16)

⚠️ Common errors

🧠 Quick check 1

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🎯 Exam question

Key ideas — Percentage yield and percentage purity

The theoretical yield is the mass calculated from the equation. Real reactions lose product through side reactions, incomplete reaction and transfer losses, so

🧑‍🏫 Teacher's explanation

Imagine you are baking cookies. The recipe says you should get exactly 24 cookies (this is your 'theoretical yield'). However, you accidentally drop 2 cookies on the floor, and one batch burns slightly, so you only end up with 18 cookies (this is your 'actual yield'). The percentage yield tells you how efficient you were at following the recipe without making mistakes.

Ch 3 · The Mole and the Avogadro Constant · O Level Chemistry 5070

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