The mole, concentration and empirical formulaeEdexcel International A Level Chemistry: Revision notes
Section 1
The mole and the Avogadro constant
The mole (mol) is the unit of amount of substance. One mole contains 6.02 x 10^23 particles (atoms, molecules, ions or formula units) of a substance, a number called the Avogadro constant, L, with units mol-1. Number of particles = amount (mol) x L.
For masses use amount (mol) = mass (g) / molar mass (g mol-1). Example: 5.85 g of NaCl is 5.85 / 58.5 = 0.100 mol, which contains 0.100 x 6.02 x 10^23 = 6.02 x 10^22 formula units. Count the particle you are asked for: 0.100 mol of NaCl contains 0.100 mol of Na+ and 0.100 mol of Cl-, so 1.20 x 10^23 ions in total.
Mass and amount are different things: 'moles' is not a unit of mass. Always convert grams to moles with mass / molar mass.
Section 2
Concentration of a solution
Concentration is the amount of solute per unit volume of solution. In mol dm-3: concentration = amount (mol) / volume (dm3). In g dm-3: concentration = mass (g) / volume (dm3). To convert between them multiply or divide by the molar mass: g dm-3 = mol dm-3 x molar mass.
Always convert cm3 to dm3 first by dividing by 1000. Example: 5.85 g of NaCl in 250 cm3 is 0.100 mol in 0.250 dm3, so 0.400 mol dm-3 or 23.4 g dm-3.
Forgetting to convert cm3 to dm3 gives an answer 1000 times too big or too small.
Section 3
Empirical formulae from experimental data
To find an empirical formula: (1) write the mass (or percentage) of each element; (2) divide by its relative atomic mass to give moles; (3) divide all by the smallest to find the simplest ratio; (4) if a ratio is not whole, multiply to make whole numbers (1 : 1.5 becomes 2 : 3).
Example: 2.10 g Fe and 0.90 g O gives 0.0376 mol Fe and 0.0563 mol O, a ratio of 1 : 1.5, so Fe2O3. Percentages are treated as masses in 100 g. If one element is not measured, find its mass by difference.
Ratios like 1 : 1.5, 1 : 1.33 and 1 : 2.5 are common. Multiply by 2, 3 or 2 respectively rather than rounding down.
Section 4
Molecular formulae
The molecular formula is a whole-number multiple of the empirical formula, found from the relative molecular mass: n = relative molecular mass / empirical formula mass. Example: a hydrocarbon with 85.7% C and Mr = 70.0 has empirical formula CH2 (mass 14.0), so n = 5 and the molecular formula is C5H10.
Combustion data can also give empirical formulae. All carbon in a sample ends up in CO2 and all hydrogen in H2O. Find moles of C from CO2, moles of H from 2 x moles of H2O, then find oxygen by difference in mass.
Each water molecule contains two hydrogen atoms: n(H) = 2 x n(H2O).
Must Know
- 1 mol contains 6.02 x 10^23 particles (L, units mol-1)
- amount = mass / molar mass; number of particles = amount x L
- concentration = amount / volume in dm3; g dm-3 = mol dm-3 x molar mass
- empirical formula: mass or % to moles, divide by the smallest, make whole numbers
- molecular formula = n x empirical formula, n = Mr / empirical formula mass
That's the notes covered.
Carry on to the next subtopic.
Exam questions on The mole, concentration and empirical formulae
- A technician weighs 5.85 g of sodium chloride, NaCl (molar mass 58.5 g mol⁻¹), dissolves it in water and makes the solution up to 250 cm³. The Avogadro constant, L, is 6.02 × 10²³ mol⁻¹.Calculate the concentration of the solution in mol dm⁻³ and in g dm⁻³.2 marks
- A laboratory stock bottle holds 500 cm³ of sulfuric acid of concentration 0.250 mol dm⁻³. The molar mass of sulfuric acid, H₂SO₄, is 98.1 g mol⁻¹.Calculate the mass of sulfuric acid needed to make 2.50 dm³ of solution of the same concentration.2 marks
- A student analyses two compounds. Compound P, an oxide of iron, contains 2.10 g of iron and 0.90 g of oxygen. Compound Q is a hydrocarbon with relative molecular mass 70.0 that contains 85.7% carbon by mass. Relative atomic masses: H = 1.0, C = 12.0, O = 16.0, Fe = 55.8.Use the data to calculate the empirical formula of compound P.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).