Reacting Masses

This lesson covers reacting mass calculations as required by the AQA GCSE Chemistry specification (4.3.2). You need to be able to use balanced equations and relative formula masses to calculate the masses of reactants and products in chemical reactions. These calculations are essential for predicting how much product you can make from a given amount of reactant, or how much reactant you need.

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The Principle of Reacting Masses

In a balanced chemical equation, the coefficients (large numbers in front of the formulae) tell you the ratio of moles in which the substances react and are produced. Because we know the relationship between moles and mass (moles = mass / Mr), we can convert between moles and grams to find unknown masses.

Step	What to Do
1	Write the balanced symbol equation
2	Identify the substance you know and the substance you want to find
3	Calculate the Mr of both substances
4	Use the balanced equation to find the mole ratio
5	Calculate the unknown mass

Exam Tip: You must always start with a balanced equation. If the equation is not given in the question, you will need to balance it yourself before doing any calculations.

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Method 1: The Three-Line Method

The three-line method is a systematic approach that works for all reacting mass calculations. It involves writing down the mole ratio, the Mr values, and then the masses.

Worked Example 1

What mass of magnesium oxide is produced when 6 g of magnesium burns in excess oxygen?

Balanced equation: 2Mg + O2 --> 2MgO

Line 1 — Mole ratio from equation:

• 2Mg : 2MgO = 1 : 1

Line 2 — Mr values:

• Mr of Mg = 24
• Mr of MgO = 24 + 16 = 40

Line 3 — Calculation:

• moles of Mg = mass / Mr = 6 / 24 = 0.25 mol
• From the equation, moles of MgO = 0.25 mol (1:1 ratio)
• mass of MgO = moles x Mr = 0.25 x 40 = 10 g

Worked Example 2

What mass of carbon dioxide is produced when 25 g of calcium carbonate decomposes?

Balanced equation: CaCO3 --> CaO + CO2

Step 1: Mr of CaCO3 = 40 + 12 + (3 x 16) = 100 Mr of CO2 = 12 + (2 x 16) = 44

Step 2: moles of CaCO3 = 25 / 100 = 0.25 mol

Step 3: From the equation, the ratio CaCO3 : CO2 = 1 : 1 So moles of CO2 = 0.25 mol

Step 4: mass of CO2 = 0.25 x 44 = 11 g

Exam Tip: Always state the mole ratio you are using. Writing "from the equation, the ratio is 1:1" (or whatever it is) shows the examiner that you understand the method, and earns you a method mark even if you make an arithmetic error later.

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Method 2: The Proportion Method

For Foundation Tier or simpler problems, you can use a direct proportion approach without explicitly calculating moles.

Worked Example 3

What mass of water is produced when 4 g of hydrogen reacts with excess oxygen?

Balanced equation: 2H2 + O2 --> 2H2O

From the equation:

• 2 moles of H2 produce 2 moles of H2O
• Mr of H2 = 2, so 2 moles of H2 = 2 x 2 = 4 g
• Mr of H2O = 18, so 2 moles of H2O = 2 x 18 = 36 g

Therefore, 4 g of H2 produces 36 g of H2O.

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Reacting Mass Calculation Flow

graph TD
    A["Start: Write balanced equation"] --> B["Identify known and unknown substances"]
    B --> C["Calculate Mr for both substances"]
    C --> D["Calculate moles of known substance"]
    D --> E["Use mole ratio to find moles of unknown"]
    E --> F["Convert moles to mass: mass = moles x Mr"]
    F --> G["State answer with correct units (g)"]

    style A fill:#2c3e50,color:#fff
    style B fill:#34495e,color:#fff
    style C fill:#2980b9,color:#fff
    style D fill:#3498db,color:#fff
    style E fill:#27ae60,color:#fff
    style F fill:#2ecc71,color:#fff
    style G fill:#e67e22,color:#fff

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More Worked Examples

Worked Example 4: Finding Mass of a Reactant

What mass of iron is needed to react with 16 g of sulfur to make iron sulfide?

Balanced equation: Fe + S --> FeS

Step 1: Mr of S = 32, Mr of Fe = 56

Step 2: moles of S = 16 / 32 = 0.5 mol

Step 3: Ratio Fe : S = 1 : 1, so moles of Fe = 0.5 mol

Step 4: mass of Fe = 0.5 x 56 = 28 g

Worked Example 5: Non-1:1 Ratio

What mass of aluminium oxide is produced when 54 g of aluminium reacts with excess oxygen?

Balanced equation: 4Al + 3O2 --> 2Al2O3

Step 1: Mr of Al = 27, Mr of Al2O3 = (2 x 27) + (3 x 16) = 54 + 48 = 102

Step 2: moles of Al = 54 / 27 = 2 mol

Step 3: From the equation, ratio Al : Al2O3 = 4 : 2 = 2 : 1 So moles of Al2O3 = 2 / 2 = 1 mol

Step 4: mass of Al2O3 = 1 x 102 = 102 g

Exam Tip: When the mole ratio is not 1:1, be very careful with the ratio step. Write out the full ratio from the equation (e.g. 4:2), simplify it if needed, and then apply it. This is where most students make errors.

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Excess and Limiting Reactants

In many reactions, one reactant is present in excess — there is more of it than is needed. The other reactant is the limiting reactant — it is completely used up and determines the amount of product formed.

When a question says a reactant is "in excess", it means you should base your calculation on the other reactant (the one that is not in excess), because that is the limiting reactant.

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Practice Problems

Equation	Given	Find	Answer
2Na + Cl2 --> 2NaCl	4.6 g Na	mass of NaCl	11.7 g
CaCO3 --> CaO + CO2	50 g CaCO3	mass of CaO	28 g
Mg + 2HCl --> MgCl2 + H2	12 g Mg	mass of H2	1 g
2Fe + 3Cl2 --> 2FeCl3	11.2 g Fe	mass of FeCl3	32.5 g

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Common Mistakes in Reacting Mass Calculations