Quantitative chemistry — GCSE Chemistry Revision
Everything you need to revise quantitative chemistry for GCSE Chemistry: clear notes, the key facts and terms to learn, the mistakes that cost students marks, and practice questions with answers.
Revision notes
Relative Atomic Mass and Relative Formula Mass
The relative atomic mass (Ar) of an element is the weighted average mass of its isotopes compared to 1/12th the mass of a carbon-12 atom. It's found on the periodic table. The relative formula mass (Mr) of a compound is the sum of the relative atomic masses of all the atoms shown in its chemical formula. For example, the Mr of H2O is (2 x Ar of H) + (1 x Ar of O).
The Mole and Avogadro's Constant
A mole (mol) is a unit for the amount of substance. One mole of any substance contains the same number of particles (atoms, molecules, or ions), known as Avogadro's constant (6.02 x 10^23 particles per mole). The mass of one mole of a substance in grams is numerically equal to its relative formula mass (Mr). This is often called the molar mass.
Calculating Moles, Mass, and Mr
The relationship between moles, mass, and relative formula mass (or relative atomic mass) is fundamental. The formula is: moles = mass (g) / Mr (g/mol). This can be rearranged to find mass (mass = moles x Mr) or Mr (Mr = mass / moles). Always ensure units are consistent, typically grams for mass and grams per mole for Mr when using this formula.
Limiting Reactants and Yield
In a chemical reaction, the limiting reactant is the reactant that is completely used up first, and it determines the maximum amount of product that can be formed. The theoretical yield is the maximum mass of product that could be formed. The percentage yield compares the actual yield (mass obtained) to the theoretical yield: % yield = (actual yield / theoretical yield) x 100%.
Key facts
- Relative atomic mass (Ar) is the weighted average mass of an element's isotopes.|Relative formula mass (Mr) is the sum of Ar values in a chemical formula.|A mole (mol) is 6.02 x 10^23 particles (Avogadro's constant).|Molar mass (g/mol) is numerically equal to Mr.|Moles = Mass / Mr.|The limiting reactant determines the maximum product.
Key terms
- Relative atomic mass (Ar)::The weighted average mass of isotopes of an element, compared to 1/12th the mass of a carbon-12 atom.|Relative formula mass (Mr)::The sum of the relative atomic masses of all the atoms in the chemical formula of a compound or molecule.|Mole (mol)::The amount of substance that contains 6.02 x 10^23 particles (Avogadro's constant).|Molar mass::The mass of one mole of a substance, expressed in grams per mole (g/mol).|Limiting reactant::The reactant in a chemical reaction that is completely used up, determining the maximum amount of product formed.|Percentage yield::The ratio of the actual yield to the theoretical yield, expressed as a percentage: (actual yield / theoretical yield) x 100%.
Common mistakes
- Forgetting to balance the chemical equation before using mole ratios in calculations.|Confusing relative atomic mass (Ar) with relative formula mass (Mr) or using the wrong one in calculations.|Not converting mass to grams before using the moles = mass / Mr formula.|Incorrectly calculating percentage yield by putting theoretical yield over actual yield.
Exam tips
- Always write down the balanced symbol equation first for any reaction calculation.|Show all your working steps clearly, even if you make a calculation error, you may get method marks.|Pay close attention to units in questions (e.g., kg vs g, cm³ vs dm³).|Practise rearranging the mole formula (moles = mass/Mr) to find different variables.
Quick quiz
1. What is the relative formula mass (Mr) of carbon dioxide (CO2)? (Ar: C=12, O=16)
- 28
- 32
- 44
- 48
Show answer
44 — The Mr of CO2 is (1 x Ar of C) + (2 x Ar of O) = (1 x 12) + (2 x 16) = 12 + 32 = 44.
2. How many moles are present in 120 g of calcium carbonate (CaCO3)? (Ar: Ca=40, C=12, O=16)
- 0.8 mol
- 1.2 mol
- 12 mol
- 120 mol
Show answer
1.2 mol — First, calculate Mr of CaCO3 = 40 + 12 + (3 x 16) = 100. Then, moles = mass / Mr = 120 g / 100 g/mol = 1.2 mol.
3. Which statement best describes a limiting reactant?
- The reactant with the largest mass.
- The reactant that is completely consumed during a reaction.
- The reactant that is left over after a reaction.
- The reactant that forms the most product.
Show answer
The reactant that is completely consumed during a reaction. — The limiting reactant is the one that gets used up first, stopping the reaction and determining the maximum product formed.
4. What is Avogadro's constant?
- 6.02 x 10^23 particles per gram
- 6.02 x 10^23 particles per mole
- 6.02 x 10^23 moles per particle
- 6.02 x 10^23 grams per mole
Show answer
6.02 x 10^23 particles per mole — Avogadro's constant specifies the number of particles (atoms, molecules, ions) in one mole of any substance.
5. If the actual yield of a reaction is 80g and the theoretical yield is 100g, what is the percentage yield?
- 20%
- 80%
- 120%
- 180%
Show answer
80% — Percentage yield = (actual yield / theoretical yield) x 100% = (80g / 100g) x 100% = 80%.
Exam-style questions
Magnesium reacts with oxygen to form magnesium oxide. 2Mg (s) + O2 (g) → 2MgO (s) A student reacts 6.0 g of magnesium with excess oxygen. Calculate the theoretical mass of magnesium oxide formed. (Ar: Mg=24, O=16) [3 marks]
Show mark scheme
- Moles of Mg = 6.0 g / 24 g/mol = 0.25 mol
- From equation, 2 mol Mg forms 2 mol MgO, so 0.25 mol Mg forms 0.25 mol MgO
- Mass of MgO = 0.25 mol x (24+16) g/mol = 0.25 mol x 40 g/mol = 10.0 g
A student carried out a reaction and obtained 15.0 g of product. The theoretical yield for this reaction was calculated to be 20.0 g. Calculate the percentage yield for this reaction. [2 marks]
Show mark scheme
- Percentage yield = (actual yield / theoretical yield) x 100%
- = (15.0 g / 20.0 g) x 100% = 75.0%
What is the concentration, in g/dm³, of a solution containing 5.85 g of sodium chloride (NaCl) dissolved in 250 cm³ of water? (Ar: Na=23, Cl=35.5) [3 marks]
Show mark scheme
- Convert volume to dm³: 250 cm³ = 0.250 dm³
- Concentration = mass / volume
- = 5.85 g / 0.250 dm³ = 23.4 g/dm³
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