The mole and molar massEdexcel A-Level Chemistry: Revision notes
Section 1
The mole and the Avogadro constant
The mole (mol) is the unit for amount of substance. One mole contains the Avogadro constant, L, of particles: L = 6.02 × 10²³ mol⁻¹. The particles may be atoms, molecules, ions or electrons, so always say which.
Number of particles N = n × L, so n = N ÷ L.
Worked example: 0.250 mol of CO₂ contains 0.250 × 6.02 × 10²³ = 1.51 × 10²³ molecules, and twice as many oxygen atoms, 3.01 × 10²³.
Counting molecules when the question asks for atoms. Multiply by the number of atoms of that element in each formula unit.
Section 2
Molar mass and amount of substance
Molar mass, M, is the mass per mole of a substance, in g mol⁻¹. It equals the relative formula mass in grams.
n = m ÷ M, so m = n × M and M = m ÷ n.
Worked example: 22.0 g of CO₂ (M = 44.0 g mol⁻¹) is 22.0 ÷ 44.0 = 0.500 mol.
For an ionic compound use the formula unit. For hydrates include the water, e.g. M(MgSO₄·7H₂O) = 120.4 + 7 × 18.0 = 246.4 g mol⁻¹.
Write the formula n = m ÷ M with units on every line, and rearrange before substituting.
Section 3
Empirical and molecular formulae
The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula is the actual number of atoms of each element in a molecule, and is a whole-number multiple of the empirical formula.
Glucose: molecular formula C₆H₁₂O₆, empirical formula CH₂O.
Molecular formula = empirical formula × (M ÷ empirical formula mass).
Section 4
Calculating empirical formulae from data
From percentage composition or masses:
- Divide each mass (or percentage) by the molar mass of the element to get moles.
- Divide all by the smallest to get a ratio.
- Multiply up to whole numbers if needed (e.g. 1 : 1.5 becomes 2 : 3).
Example: 54.5% C, 9.1% H, 36.4% O gives 4.54 : 9.1 : 2.28 mol, so 2 : 4 : 1 and the empirical formula is C₂H₄O.
Combustion data: all carbon becomes CO₂, so n(C) = n(CO₂), and all hydrogen becomes H₂O, so n(H) = 2 × n(H₂O). Oxygen is found by difference. For a hydrate, the mass lost on heating is water, and x = n(H₂O) ÷ n(anhydrous salt).
Rounding 1.5 to 1 or 2. Ratios such as 1 : 1.5 must be multiplied up to 2 : 3.
Section 5
Using pV = nRT to find molar mass
For a gas or a volatile liquid vaporised at a known temperature, pV = nRT gives the amount of gas, n.
- p in pascals (1 kPa = 10³ Pa)
- V in m³ (1 cm³ = 10⁻⁶ m³; 1 dm³ = 10⁻³ m³)
- T in kelvin (K = °C + 273)
- R = 8.31 J K⁻¹ mol⁻¹
Then M = m ÷ n, or directly M = mRT ÷ pV. Compare M with the empirical formula mass to find the molecular formula.
Example: 0.320 g of vapour at 373 K and 101 kPa occupies 112 cm³: n = (101 000 × 1.12 × 10⁻⁴) ÷ (8.31 × 373) = 3.65 × 10⁻³ mol, so M = 87.7 g mol⁻¹.
Convert units before you start. The most common error is leaving V in cm³ or p in kPa.
Must know
- n = m ÷ M; N = n × L with L = 6.02 × 10²³ mol⁻¹
- Empirical formula = simplest ratio; molecular formula = multiple of it
- Empirical formula: divide by Ar, divide by smallest, make whole numbers
- pV = nRT with p in Pa, V in m³, T in K
- M = mRT ÷ pV, then molecular formula = empirical × (M ÷ empirical mass)
That's the notes covered.
Carry on to the next subtopic.
Exam questions on The mole and molar mass
- A technician uses a 22.0 g block of dry ice (solid carbon dioxide, CO₂) for a school demonstration. Relative atomic masses: C 12.0, O 16.0. The Avogadro constant, L, is 6.02 × 10²³ mol⁻¹ and the gas constant, R, is 8.31 J K⁻¹ mol⁻¹.The block turns completely into gas in a large bag at 298 K and 101 kPa. Calculate the volume of the gas in dm³, assuming it behaves as an ideal gas.2 marks
- An organic liquid contains only carbon, hydrogen and oxygen. Analysis shows that it contains 54.5% carbon, 9.1% hydrogen and 36.4% oxygen by mass. Relative atomic masses: H 1.0, C 12.0, O 16.0. The gas constant, R, is 8.31 J K⁻¹ mol⁻¹.A 0.320 g sample of the liquid is vaporised and occupies 112 cm³ at 373 K and 101 kPa. Calculate its molar mass and deduce its molecular formula.2 marks
- A student heats 2.46 g of hydrated magnesium sulfate, MgSO₄·xH₂O, until it reaches constant mass. The anhydrous magnesium sulfate left has a mass of 1.20 g. M(MgSO₄) = 120.4 g mol⁻¹ and M(H₂O) = 18.0 g mol⁻¹. The Avogadro constant, L, is 6.02 × 10²³ mol⁻¹.Calculate the amount (mol) of anhydrous magnesium sulfate and of water driven off, and hence find the value of x.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).