Amounts of Substances in Equations | Chemistry

📖 In-Depth Theory

Molar Ratios from Balanced Equations

A BALANCED equation gives the MOLAR RATIO of reactants and products — the coefficients show the ratio of moles.

Examples:

2Mg + O₂ → 2MgO

Means: 2 mol Mg reacts with 1 mol O₂ to produce 2 mol MgO.

Molar ratio Mg : O₂ : MgO = 2 : 1 : 2.

Mg + 2HCl → MgCl₂ + H₂

Means: 1 mol Mg reacts with 2 mol HCl to produce 1 mol MgCl₂ and 1 mol H₂.

N₂ + 3H₂ → 2NH₃

Means: 1 mol N₂ reacts with 3 mol H₂ to produce 2 mol NH₃.

Key principle: the molar ratio in the balanced equation is always maintained, whatever the actual quantities used.

Calculating Masses Using Moles — The Method

STANDARD METHOD for calculating masses in reactions:

Step 1: Write the balanced equation.

Step 2: Write the molar ratio from the equation.

Step 3: Convert given mass to moles: n = mass ÷ Mr.

Step 4: Use the molar ratio to find moles of the unknown substance.

Step 5: Convert moles back to mass: mass = n × Mr.

EXAMPLE:

How much magnesium oxide forms when 4.8 g of magnesium burns?

2Mg + O₂ → 2MgO

Step 1: 2 mol Mg → 2 mol MgO (ratio 1:1)

Step 2: n(Mg) = 4.8 ÷ 24 = 0.2 mol

Step 3: n(MgO) = 0.2 mol (same ratio)

Step 4: mass(MgO) = 0.2 × 40 = 8.0 g

Percentage Yield

THEORETICAL YIELD: the maximum mass of product calculated from the equation.

ACTUAL YIELD: the mass actually obtained in the experiment.

Actual yield is always LESS than or equal to theoretical yield.

Why actual yield < theoretical yield:

Reaction may not go to completion (reversible reactions).

Side reactions produce unwanted products.

Product lost during transfer/purification.

Reactants may be impure.

PERCENTAGE YIELD:

% yield = (actual yield ÷ theoretical yield) × 100

EXAMPLE:

Theoretical yield of MgO = 8.0 g. Actual yield = 7.2 g.

% yield = (7.2 ÷ 8.0) × 100 = 90%

ATOM ECONOMY:

Atom economy = (Mr of desired products ÷ Mr of ALL products) × 100

High atom economy = less waste, more sustainable, more cost-effective.

Addition reactions have 100% atom economy (one product only).

⚠️ Common Mistake

Always use the MOLAR RATIO from the balanced equation when converting between moles of different substances. If 2 mol of A produces 1 mol of B, then 0.4 mol of A produces 0.2 mol of B — not 0.4 mol of B. The ratio comes from the COEFFICIENTS in the balanced equation.

📐 Variables

nAmount in moles (n) is measured in mol (mol)

mMass (m) is measured in grams (g)

MrRelative formula mass (Mr) is measured in ()

📐 Key Equations

n = m ÷ Mr

% yield = (actual yield ÷ theoretical yield) × 100

atom economy = (Mr desired products ÷ Mr all products) × 100

📌 Key Note

Balanced equation coefficients = molar ratios. Method: mass → moles (÷Mr) → use ratio → moles → mass (×Mr). % yield = (actual ÷ theoretical) × 100. Atom economy = (Mr desired ÷ Mr all products) × 100. High atom economy = less waste.

🎯 Matching Activity — Molar Ratios from Equations

Match each equation to the correct molar ratio statement. — drag the symbols on the right to match the component names on the left.

2Mg + O₂ → 2MgO

Drop here

N₂ + 3H₂ → 2NH₃

Drop here

Mg + 2HCl → MgCl₂ + H₂

Drop here

CaCO₃ → CaO + CO₂

Drop here

2 mol Mg reacts with 1 mol O₂ — Mg:MgO ratio is 1:1

1 mol CaCO₃ produces 1 mol CaO and 1 mol CO₂

1 mol Mg reacts with 2 mol HCl to produce 1 mol MgCl₂ and 1 mol H₂

1 mol N₂ reacts with 3 mol H₂ — N₂:NH₃ ratio is 1:2

⚽ FIFA Worked Examples

Mass Calculation from Equation

Calculate the mass of hydrogen produced when 1.2 g of magnesium reacts with excess HCl. Mg + 2HCl → MgCl₂ + H₂. Ar: Mg=24, H=1.

F

n = mass ÷ Mr; then use molar ratio; then mass = n × Mr

I

n(Mg) = 1.2 ÷ 24 = 0.05 mol. Ratio Mg:H₂ = 1:1, so n(H₂) = 0.05 mol. Mr(H₂) = 2

F

mass(H₂) = 0.05 × 2 = 0.1

A

0.1 g of hydrogen produced

🎯 Test Yourself

Question 1 of 2

1. N₂ + 3H₂ → 2NH₃. How many moles of NH₃ are produced from 0.6 mol of H₂?

2. The theoretical yield of a product is 25 g but only 20 g is obtained. What is the percentage yield?

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🧪 Amounts of Substances in Equations

Molar Ratios from Balanced Equations

Calculating Masses Using Moles — The Method

Percentage Yield

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