Calculate moles, mass, Mr, concentration, reacting masses, percentage yield, and gas volumes with step-by-step solutions. Perfect for GCSE and A-Level Chemistry.
Moles, mass, and Mr relationship
Number of moles = mass ÷ relative formula mass
Sum of all atomic masses in the formula
Concentration = moles ÷ volume in dm³
Mass concentration = mass ÷ volume in dm³
Compares actual to theoretical yield
How much reactant becomes useful product
1 mole of gas = 24 dm³ at room temperature & pressure
Relates pressure, volume, moles, and temperature
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The mole is the SI unit for the amount of substance. One mole contains exactly 6.022 × 10²³ particles — this is called Avogadro's constant (Nₐ). These particles can be atoms, molecules, ions, or electrons.
Think of it like "a dozen" — a dozen always means 12, whether it's 12 eggs or 12 elephants. A mole always means 6.022 × 10²³, whether it's 6.022 × 10²³ hydrogen atoms or 6.022 × 10²³ water molecules. The mole lets chemists count particles by simply weighing them.
The mass of one mole of a substance (in grams) is numerically equal to its relative formula mass (Mr). For example, one mole of water (H₂O, Mr = 18) weighs exactly 18 grams. This relationship — n = m / Mr — is the foundation of quantitative chemistry.
The SI unit for amount of substance. 1 mol = 6.022 × 10²³ particles.
The average mass of one atom of an element relative to ¹/₁₂ of carbon-12.
The sum of all the Ar values in a chemical formula. Also called molar mass.
6.022 × 10²³ — the number of particles in one mole of substance.
The amount of solute dissolved per unit volume of solution. c = n / V.
At RTP (25°C, 1 atm), 1 mole of any gas occupies 24 dm³.
(Actual yield ÷ theoretical yield) × 100. Always ≤ 100% in practice.
The percentage of reactant atoms that become part of the desired product.
The moles triangle is one of the most important tools in chemistry. Cover the variable you want to find:
Worked example: 44g of CO₂ (Mr = 44) → n = 44 ÷ 44 = 1 mol
✗ Forgetting to convert cm³ to dm³
✓ Always divide by 1000: 250 cm³ = 0.25 dm³. This is the #1 error in concentration questions.
✗ Confusing Ar and Mr
✓ Ar is for individual elements (from the periodic table). Mr is for compounds (sum of all Ar values in the formula).
✗ Wrong mole ratio from unbalanced equation
✓ The mole ratio comes from the coefficients in a balanced equation. Always balance the equation first.
✗ Mixing up mol/dm³ and g/dm³
✓ mol/dm³ uses moles; g/dm³ uses mass in grams. To convert: g/dm³ = mol/dm³ × Mr.
✗ Forgetting units in final answer
✓ Always state units: mol for moles, g for mass, dm³ for volume, mol/dm³ for concentration.
✗ Getting percentage yield over 100%
✓ In practice, yield cannot exceed 100%. If it does, check for impurities, incomplete drying, or measurement error.
✗ Using 22.4 dm³ instead of 24 dm³
✓ 22.4 dm³/mol is at STP (0°C). At RTP (25°C), which GCSE uses, it's 24 dm³/mol.
✗ Not showing Mr working
✓ Examiners give marks for showing how you calculated Mr, even if your final answer is wrong. Always show element × count.
A mole is a unit of measurement for amount of substance. One mole contains exactly 6.022 × 10²³ particles (atoms, molecules, or ions). This is called Avogadro's constant. It allows chemists to count particles by weighing them.
Use the formula n = m / Mr, where n is the number of moles, m is the mass in grams, and Mr is the relative formula mass. For example, 36g of water (Mr = 18) = 36 ÷ 18 = 2 moles.
Relative formula mass (Mr) is the sum of all the relative atomic masses (Ar) of the atoms in a chemical formula. For example, H₂O has Mr = (2 × 1.0) + 16.0 = 18.0.
Use the formula c = n / V, where c is concentration in mol/dm³, n is the number of moles, and V is the volume in dm³. Remember to convert cm³ to dm³ by dividing by 1000.
Ar (relative atomic mass) is for individual elements. Mr (relative formula mass) is for compounds — the sum of all Ar values in the formula.
Percentage yield = (actual yield ÷ theoretical yield) × 100. It compares how much product you actually made to how much you could have made in theory.
Atom economy = (Mr of desired product ÷ sum of Mr of all products) × 100. It measures how efficiently atoms are used. A high atom economy means less waste.
At room temperature and pressure (25°C, 1 atmosphere), 1 mole of any gas occupies 24 dm³. This is called the molar gas volume.
The ideal gas law PV = nRT is used at A-Level when conditions are not at RTP. P is pressure in Pa, V is volume in m³, n is moles, R is 8.314 J/(mol·K), T is temperature in Kelvin.
The mole ratio comes directly from the coefficients in the balanced equation. For example, in 2Mg + O₂ → 2MgO, the ratio of Mg to MgO is 2:2, simplified to 1:1.
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