Relative masses and the moleAQA A-Level Chemistry: Revision notes
Section 1
Relative atomic mass and relative molecular mass
Masses of atoms are compared with the carbon-12 atom, which is the standard.
Relative atomic mass (Ar) is the weighted mean mass of an atom of an element compared with 1/12 of the mass of an atom of ¹²C. It is a weighted mean because most elements have several isotopes, so Ar is often not a whole number.
Relative molecular mass (Mr) is the weighted mean mass of a molecule compared with 1/12 of the mass of an atom of ¹²C. It is found by adding the relative atomic masses of all the atoms in the formula.
For ionic compounds, which have no molecules, the term relative formula mass is used, calculated in the same way. All of these are ratios, so they have no units.
Do not forget the multiplier outside a bracket: Mg(NO₃)₂ contains 2 N and 6 O. Relative masses have no units, unlike molar mass (g mol⁻¹).
Section 2
The mole and the Avogadro constant
The mole is the unit of amount of substance. The Avogadro constant, L = 6.02 × 10²³ mol⁻¹, is the number of particles in one mole. The value need not be recalled, as it is given in the data.
The particles can be atoms, molecules, ions, electrons or formula units, and the mole also applies to formulas and equations: in 2H₂ + O₂ → 2H₂O, the coefficients give the ratio of moles.
Number of particles = amount (mol) × 6.02 × 10²³
Take care to say which particles you are counting. One mole of H₂O has 6.02 × 10²³ molecules but 1.20 × 10²⁴ hydrogen atoms.
Do not mix up molecules and atoms. Multiply by the number of atoms of each element in the formula.
Section 3
Amount, mass and Mr
The mass of one mole is the relative formula mass in grams (molar mass, g mol⁻¹).
amount (mol) = mass (g) ÷ Mr
Worked example. What is the amount in 5.85 g of NaCl? Mr = 23.0 + 35.5 = 58.5.
Amount = 5.85 ÷ 58.5 = 0.100 mol
The formula can be rearranged to mass = amount × Mr, or Mr = mass ÷ amount.
Number of atoms: 0.100 mol of NaCl contains 0.100 × 6.02 × 10²³ = 6.02 × 10²² formula units, or 1.20 × 10²³ ions in total.
Section 4
Concentration of a solution
The concentration of a solution is the amount of solute per dm³ of solution, in mol dm⁻³.
amount (mol) = concentration (mol dm⁻³) × volume (dm³), that is n = cV
Convert volumes in cm³ to dm³ by dividing by 1000.
Worked example. What is the concentration of 2.12 g of Na₂CO₃ (Mr 106.0) in 250 cm³ of solution?
- n = 2.12 ÷ 106.0 = 0.0200 mol
- V = 250 ÷ 1000 = 0.250 dm³
- c = 0.0200 ÷ 0.250 = 0.0800 mol dm⁻³
For ionic solutes, the ion concentration depends on the formula: 0.0800 mol dm⁻³ Na₂CO₃ contains 0.160 mol dm⁻³ Na⁺.
Write the units at each step of the calculation. Convert cm³ to dm³ before using n = cV.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Relative masses and the mole
- Magnesium nitrate, Mg(NO₃)₂, is used in some fireworks as an oxidising agent. A technician needs to convert between masses and amounts of this compound, using relative atomic masses Mg 24.3, N 14.0 and O 16.0.Calculate the amount, in mol, of magnesium nitrate in 7.42 g of Mg(NO₃)₂. Give your answer to 3 significant figures.2 marks
- A student has 3.60 g of water, H₂O, in a beaker. Use relative atomic masses H 1.0 and O 16.0, and the Avogadro constant, 6.02 × 10²³ mol⁻¹.Calculate the mass, in grams, of one molecule of water. Give your answer in standard form to 3 significant figures.2 marks
- A technician prepares a solution of sodium carbonate, Na₂CO₃, by dissolving 6.36 g of the solid in water and making the volume up to exactly 250 cm³ in a volumetric flask. Use relative atomic masses Na 23.0, C 12.0 and O 16.0, and the Avogadro constant, 6.02 × 10²³ mol⁻¹.Calculate the concentration of the sodium carbonate solution in mol dm⁻³.3 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).