GCSE Chemistry • Topic 3

Quantitative Chemistry

Using numbers to predict, calculate, and explain chemical reactions.

📌 What You’ll Learn

  • Relative atomic mass, formula mass, and mole concept
  • Balancing equations correctly
  • Using mole ratios in calculations
  • Limiting reactants (Higher Tier)
  • Percentage yield and atom economy
  • Concentration of solutions

🧠 How to Study

  • Never skip units (g, mol, dm³)
  • Write the balanced equation before calculating
  • Use ratios from equations, not intuition
  • Show working clearly (marks are method-based)
1. Relative Masses (Foundation)

Relative Atomic Mass ($A_r$)

The weighted average mass of an atom compared to 1/12 of carbon-12. You find these on your Periodic Table.

Relative Formula Mass ($M_r$)

To find $M_r$, simply add up all the $A_r$ values of the atoms in the chemical formula.

Example: $H_2O$
$M_r = (2 \times 1) + 16 = 18$

⚠️ Exam Trap

Do NOT include units with $A_r$ or $M_r$. They are relative values and have no units.

Q1: Calculate the $M_r$ of Calcium Carbonate ($CaCO_3$). ($A_r$: $Ca=40, C=12, O=16$)

$$ M_r = 40 + 12 + (3 \times 16) = 100 $$

2. The Mole (Core Concept)

What is a Mole?

A mole is just a specific number of particles: $6.02 \times 10^{23}$ (Avogadro's constant). It's the standard unit for measuring the amount of a substance.

[Image of the mole triangle for mass, moles, and Mr]

$$ \text{moles} = \frac{\text{mass (g)}}{M_r} $$

Q1: How many moles are in 44g of $CO_2$? ($M_r$ of $CO_2 = 44$)

$$ 44 / 44 = 1 \, \text{mol} $$

3. Balancing Chemical Equations (Conservation)

Why Balance?

Atoms are never created or destroyed in a reaction (Law of Conservation of Mass). The number of atoms on the left must equal the number on the right.

[Image illustrating balancing chemical equations with coefficients vs subscripts]
⚠️ Exam Trap

Never change the small numbers (subscripts) in a formula. Only change the big numbers (coefficients) in front.

Q1: Balance this: $Mg + O_2 \rightarrow MgO$

$$ 2Mg + O_2 \rightarrow 2MgO $$

4. Using Mole Ratios (The 4-Step Method)

Step-by-Step Reacting Masses

To find the mass of a product from a given mass of reactant, follow this exact sequence:

[Image showing the steps for reacting mass calculations (Mole Ratio Method)]
  1. Write the balanced equation.
  2. Convert given mass to moles ($\text{mass} / M_r$).
  3. Use the mole ratio from the big numbers in the equation.
  4. Convert back to mass ($\text{moles} \times M_r$).

Q1: Why can't we use mass ratios directly from the coefficients?

Because different substances have different $M_r$ values; coefficients represent moles, not grams.

5. Limiting Reactants (Higher Only)

Which one runs out?

The limiting reactant is the one that is completely used up first. It limits the amount of product you can make.

Method: Calculate the moles of product each reactant could make. The reactant that makes the least amount of product is the limiting one.

Q1: If you have 10 moles of A and 5 moles of B, but the ratio is 1:1, which is limiting?

B is limiting because it will run out first.

6. Percentage Yield & Atom Economy (Efficiency)

Percentage Yield

Tells you how much product you actually got compared to what was theoretically possible.

$$ \% \, \text{Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 $$

Atom Economy (Higher Only)

Measures the amount of starting materials that end up as useful products.

$$ \text{Atom Economy} = \frac{M_r \text{ of Desired Product}}{\text{Total } M_r \text{ of Reactants}} \times 100 $$

Q1: Why is yield usually less than 100%?

Incomplete reactions, loss during transfer, or side reactions occurring.

7. Concentration of Solutions (Solutions)

Calculating Concentration

Concentration measures how much substance is dissolved in a certain volume of liquid.

[Image for unit conversion between cm3 and dm3]

$$ \text{conc (g/dm}^3\text{)} = \frac{\text{mass (g)}}{\text{volume (dm}^3\text{)}} $$

Crucial Conversion: $1000 \, \text{cm}^3 = 1 \, \text{dm}^3$

Q1: 50g of salt is dissolved in 500 $cm^3$ of water. What is the concentration in $g/dm^3$?

$$ 500 \, \text{cm}^3 = 0.5 \, \text{dm}^3. \text{ Conc} = 50 / 0.5 = 100 \, \text{g/dm}^3 $$

⚡ Quick Revision Checklist

Must-Memorise Equations:
  • $\text{moles} = \text{mass} \div M_r$
  • $\text{concentration} = \text{mass} \div \text{volume}$
  • $\% \, \text{yield}$ formula
  • Atom economy formula
Must-Use Concepts:
  • Mole ratios (coefficients)
  • Limiting reactant logic
  • Theoretical vs Actual yield
  • Unit conversion ($cm^3 \rightarrow dm^3$)

🚫 Brutal Exam Trap Summary

1. Don't calculate anything until you've checked if the equation is balanced.
2. Don't forget to convert $cm^3$ to $dm^3$ by dividing by 1000.
3. Don't confuse $M_r$ (whole formula) with $A_r$ (single atom).
4. Don't guess the limiting reactant based on mass; always calculate moles first.

← Topic 2 🏠 Home 📚 Index Next: Chemical Changes →