4.3.2.1MolesHigher tier
Chemical amounts are measured in moles. The symbol for the unit mole is mol.
The mass of one mole of a substance in grams is numerically equal to its relative formula mass. Carbon has an Ar of 12, so one mole of carbon atoms has a mass of 12 g. Water has an Mr of 18, so one mole of water has a mass of 18 g.
One mole of a substance contains the same number of the stated particles — atoms, molecules or ions — as one mole of any other substance. The number of atoms, molecules or ions in a mole of a given substance is the Avogadro constant, and its value is 6.02 × 10²³ per mole.
You should understand that the measurement of amounts in moles can apply to atoms, molecules, ions, electrons, formulae and equations. For example, in one mole of carbon (C) the number of atoms is the same as the number of molecules in one mole of carbon dioxide (CO₂) — both are 6.02 × 10²³.
Be careful about which particle is being counted, though. One mole of carbon dioxide contains 6.02 × 10²³ molecules, but three times that many atoms, because each molecule contains three atoms.
You should be able to use the relative formula mass of a substance to calculate the number of moles in a given mass of that substance, and vice versa — that is, both directions of moles = mass ÷ Mr.
Worked example
Calculate the number of moles in 8.8 g of carbon dioxide, CO₂, and the number of molecules this contains. (Ar: C = 12, O = 16.)
- Relative formula mass: 12 + (2 × 16) = 44.
- moles = mass ÷ Mr = 8.8 ÷ 44 = 0.2 mol.
- Number of molecules = moles × Avogadro constant.
- 0.2 × 6.02 × 10²³ = 1.204 × 10²³.
Answer: 0.2 mol, containing 1.20 × 10²³ molecules (to 3 significant figures).
Common mistake
Give your answer to an appropriate number of significant figures, and keep standard form tidy. Writing 1.204 × 10²³ when the data has two significant figures overstates what the measurement supports; writing 12.04 × 10²² is not standard form at all.
- Mole
- the unit of chemical amount, symbol mol; the mass of one mole of a substance in grams is numerically equal to its relative formula mass
- Avogadro constant
- the number of atoms, molecules or ions in one mole of a substance, 6.02 × 10²³ per mole
4.3.2.2Amounts of substances in equationsHigher tier
The masses of reactants and products can be calculated from balanced symbol equations.
Chemical equations can be interpreted in terms of moles. For example, Mg + 2HCl → MgCl₂ + H₂ shows that one mole of magnesium reacts with two moles of hydrochloric acid to produce one mole of magnesium chloride and one mole of hydrogen gas.
That is the whole method in one sentence: the big numbers in a balanced equation are a ratio of moles, so converting a mass into moles lets you use the equation, and converting back gives the mass you were asked for.
You should be able to calculate the masses of substances shown in a balanced symbol equation, and to calculate the masses of reactants and products from the balanced symbol equation and the mass of a given reactant or product.
You should also be able to change the subject of a mathematical equation, and to substitute numerical values into algebraic equations using appropriate units for physical quantities. In practice that means being fluent in all three arrangements of moles = mass ÷ Mr.
Worked example
Calculate the mass of magnesium oxide made when 4.8 g of magnesium burns. 2Mg + O₂ → 2MgO. (Ar: Mg = 24, O = 16.)
- Moles of magnesium = mass ÷ Ar = 4.8 ÷ 24 = 0.2 mol.
- Read the ratio from the equation: 2 Mg gives 2 MgO, so the ratio is 1 : 1.
- Moles of magnesium oxide = 0.2 mol.
- Mr of MgO = 24 + 16 = 40.
- Mass = moles × Mr = 0.2 × 40.
Answer: 8.0 g of magnesium oxide
Common mistake
The mole ratio comes from the balanced equation, not from the formulae. In 2Mg + O₂ → 2MgO the magnesium to oxygen ratio is 2 : 1, so 0.2 mol of magnesium needs only 0.1 mol of oxygen molecules.
4.3.2.3Using moles to balance equationsHigher tier
The balancing numbers in a symbol equation can be calculated from the masses of reactants and products by converting the masses in grams to amounts in moles, and converting the numbers of moles to simple whole number ratios.
You should be able to balance an equation given the masses of reactants and products. This is the previous method run backwards: instead of using the ratio to find a mass, you use the masses to find the ratio.
Worked example
4.6 g of sodium reacts completely with 1.6 g of oxygen to make 6.2 g of sodium oxide. Work out the balanced equation. (Ar: Na = 23, O = 16.)
- Moles of sodium = 4.6 ÷ 23 = 0.2 mol.
- Moles of oxygen molecules O₂ = 1.6 ÷ 32 = 0.05 mol.
- Moles of sodium oxide Na₂O = 6.2 ÷ 62 = 0.1 mol.
- Write the ratio: 0.2 : 0.05 : 0.1.
- Divide through by the smallest, 0.05: 4 : 1 : 2.
Answer: 4Na + O₂ → 2Na₂O
Common mistake
Use the Mr of the molecule as it exists, not of a single atom. Oxygen gas is O₂ with an Mr of 32; dividing by 16 gives twice as many moles and a ratio that will not simplify to whole numbers.
4.3.2.4Limiting reactantsHigher tier
In a chemical reaction involving two reactants, it is common to use an excess of one of the reactants to ensure that all of the other reactant is used. The reactant that is completely used up is called the limiting reactant, because it limits the amount of products.
You should be able to explain the effect of a limiting quantity of a reactant on the amount of products it is possible to obtain, in terms of amounts in moles or masses in grams.
Worked example
2.4 g of magnesium is added to 3.65 g of hydrogen chloride in solution. Mg + 2HCl → MgCl₂ + H₂. Which reactant is limiting, and what mass of hydrogen is made? (Ar: Mg = 24, H = 1, Cl = 35.5.)
- Moles of magnesium = 2.4 ÷ 24 = 0.1 mol.
- Moles of HCl = 3.65 ÷ 36.5 = 0.1 mol.
- The equation needs 2 mol of HCl for every 1 mol of Mg, so 0.1 mol of Mg would need 0.2 mol of HCl — but only 0.1 mol is available.
- So the hydrogen chloride runs out first and is the limiting reactant; the magnesium is in excess.
- Work from the limiting reactant: 0.1 mol HCl gives 0.05 mol H₂, because the ratio is 2 : 1.
- Mass of H₂ = 0.05 × 2 = 0.1 g.
Answer: Hydrogen chloride is limiting, and 0.1 g of hydrogen is produced. Only half the magnesium reacts.
Common mistake
The limiting reactant is not simply the one with the smaller mass or the fewer moles — it is the one that runs out first once the equation's ratio is taken into account. Always compare against the ratio, as in step 3 above.
- Limiting reactant
- the reactant that is completely used up in a reaction, because it limits the amount of products that can be formed
4.3.2.5Concentration of solutions
Many chemical reactions take place in solutions. The concentration of a solution can be measured in mass per given volume of solution, for example grams per dm³ (g/dm³).
concentration in g/dm³ = mass of solute in g ÷ volume of solution in dm³
One dm³ is 1000 cm³, so a volume given in cm³ must be divided by 1000 before it is used.
You should be able to calculate the mass of solute in a given volume of solution of known concentration, in terms of mass per given volume of solution.
Worked example
Calculate the mass of sodium chloride in 250 cm³ of a solution of concentration 20 g/dm³.
- Convert the volume to dm³: 250 ÷ 1000 = 0.25 dm³.
- Rearrange the relationship: mass = concentration × volume.
- Substitute: 20 × 0.25.
- Check the units: g/dm³ × dm³ gives g, which is what was asked for.
Answer: 5 g of sodium chloride
Common mistake
Convert cm³ to dm³ before doing anything else. Forgetting the factor of 1000 is the single most common error in these questions, and it makes the answer wrong by three orders of magnitude.
- Concentration
- the amount of solute in a given volume of solution, measured in g/dm³ as mass of solute divided by volume of solution in dm³
4.3.2.5Explaining concentrationHigher tier
Higher tier students should also be able to explain how the mass of a solute and the volume of a solution is related to the concentration of the solution.
The explanation, rather than the calculation: concentration measures how much solute is packed into each unit of volume. Adding more solute to the same volume increases the concentration, because the same volume now holds more. Adding more solvent to the same mass of solute decreases the concentration, because that solute is spread through a greater volume — even though the mass of solute has not changed at all.
So concentration is directly proportional to the mass of solute and inversely proportional to the volume of solution. Doubling the mass doubles the concentration; doubling the volume halves it.
Worked example
A student takes 100 cm³ of a 40 g/dm³ solution and adds water until the volume is 200 cm³. What is the new concentration, and why?
- The mass of solute has not changed — only water was added.
- Mass of solute = concentration × volume = 40 × 0.100 = 4 g.
- The new volume is 0.200 dm³.
- New concentration = 4 ÷ 0.200 = 20 g/dm³.
- The volume doubled and the mass stayed the same, so the concentration halved.
Answer: 20 g/dm³ — the same solute is spread through twice the volume, so the concentration is inversely proportional to the volume.
Before you practise
- Sections 4.3.2.1 to 4.3.2.4 are Higher tier only; concentration in g/dm³ is for both tiers.
- The mass of one mole of a substance in grams is numerically equal to its relative formula mass.
- The Avogadro constant is 6.02 × 10²³ per mole — the number of particles in one mole of any substance.
- moles = mass ÷ Mr, and mass = moles × Mr.
- The big numbers in a balanced equation are a mole ratio: convert to moles, apply the ratio, convert back.
- To balance from masses: convert each mass to moles, then simplify to the smallest whole number ratio.
- The limiting reactant is the one that runs out first once the equation's ratio is applied; it decides how much product forms.
- concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³), and 1 dm³ = 1000 cm³.