Revision notes: The Mole Concept, Chemical Formula and Equation

This chapter is the calculation engine of Form 4 Chemistry, and almost every question comes down to one skill: converting a quantity into moles, using a ratio, and converting back. Work through the standards below in order, and keep a periodic table beside you, because the relative atomic masses (Ar) you need are read from it in the exam.

3.1 Relative atomic mass and relative molecular mass

The relative atomic mass (Ar) of an element is the average mass of one atom compared with one-twelfth of the mass of a carbon-12 atom. It has no unit because it is a ratio. The relative molecular mass (Mr) of a molecule is the sum of the relative atomic masses of all the atoms in its formula; for an ionic compound we call the same total the relative formula mass.

Method. To find Mr, multiply each element’s Ar by how many atoms of it the formula shows, then add. For water, H₂O = 2(1) + 16 = 18. For carbon dioxide, CO₂ = 12 + 2(16) = 44. For calcium hydroxide, Ca(OH)₂ = 40 + 2(16 + 1) = 74. For hydrated copper(II) sulfate, CuSO₄·5H₂O = 64 + 32 + 4(16) + 5(18) = 250, the five water molecules must be included.

3.2 The mole and the Avogadro constant

A mole is the amount of substance that contains as many particles as there are atoms in exactly 12 g of carbon-12. That number is the Avogadro constant, 6.02 x 10²³ mol⁻¹. So one mole of any substance contains 6.02 x 10²³ mol⁻¹ particles, atoms, molecules or ions, depending on the substance.

The link between moles and number of particles is: number of particles = number of moles × the Avogadro constant. For example, 0.5 mol of carbon dioxide contains 0.5 × 6.02 x 10²³ mol⁻¹ = 3.01 x 10²³ molecules. Remember that one CO₂ molecule contains three atoms, so the same sample contains 3 × 3.01 x 10²³ = 9.03 x 10²³ atoms.

3.3 Mole, mass and number of particles

The molar mass (M) of a substance is its Ar or Mr expressed in grams per mole. The central relationship of the whole chapter is:

number of moles = mass ÷ molar mass, or n = m ÷ M.

Rearranged, mass = moles × molar mass, and molar mass = mass ÷ moles. To find the number of moles in 8 g of sodium hydroxide, first find M(NaOH) = 23 + 16 + 1 = 40, then n = 8 ÷ 40 = 0.2 mol. To find the mass of 0.25 mol of calcium carbonate, M(CaCO₃) = 40 + 12 + 3(16) = 100, so mass = 0.25 × 100 = 25 g. Chaining these two steps, mass to moles, then moles to particles, answers a very common style of question.

3.4 Mole and the volume of gas

Equal volumes of all gases, measured at the same temperature and pressure, contain the same number of molecules. The volume occupied by one mole of any gas is the molar volume. At room conditions this is 24 dm³ mol⁻¹; at standard temperature and pressure (STP) it is 22.4 dm³ mol⁻¹. Read the question to see which conditions apply.

Method. number of moles of gas = volume of gas ÷ molar volume. So 6 dm³ of oxygen at room conditions is 6 ÷ 24 = 0.25 mol; and 2 mol of any gas at room conditions occupies 2 × 24 = 48 dm³. Watch your units: 1 dm³ = 1000 cm³, so convert cm³ to dm³ before dividing by a molar volume given in dm³ mol⁻¹.

3.5 Empirical formula and molecular formula

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 one molecule; it is the empirical formula, or a whole-number multiple of it.

Empirical formula method. (1) Write the mass or percentage of each element. (2) Divide each by its Ar to get moles of atoms. (3) Divide every result by the smallest of them to get a ratio. (4) If needed, multiply to reach whole numbers. For a compound that is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen: C = 40.0 ÷ 12 = 3.33; H = 6.7 ÷ 1 = 6.7; O = 53.3 ÷ 16 = 3.33. Dividing by 3.33 gives 1 : 2 : 1, so the empirical formula is CH₂O.

To reach the molecular formula, use the relative molecular mass. If the compound above has Mr = 60, the empirical formula mass of CH₂O is 12 + 2 + 16 = 30, so n = 60 ÷ 30 = 2, and the molecular formula is C₂H₄O₂.

3.6 Chemical equations and stoichiometry

A chemical equation shows reactants and products with correct formulae, balanced so that the number of atoms of each element is equal on both sides, with state symbols (s, l, g, aq). The balancing numbers (coefficients) give the mole ratio in which substances react.

Stoichiometry method. (1) Write and balance the equation. (2) Convert the given quantity to moles. (3) Use the mole ratio from the coefficients to find moles of the substance you want. (4) Convert those moles to the quantity asked for. Example: what mass of magnesium oxide forms when 6 g of magnesium burns? The equation is 2Mg + O₂ → 2MgO. n(Mg) = 6 ÷ 24 = 0.25 mol; the ratio Mg : MgO is 2 : 2 = 1 : 1, so n(MgO) = 0.25 mol; M(MgO) = 40, so mass = 0.25 × 40 = 10 g.

How to revise this chapter

Do not read this chapter, practise it. Redraw the conversion map from memory: mass ⇄ moles ⇄ particles (through the Avogadro constant) and moles ⇄ volume of gas (through the molar volume). For every calculation, write the formula, substitute the numbers with units, then evaluate, so an examiner can award method marks even if an arithmetic slip creeps in. A one-to-one teacher can watch you set out this working and catch the exact step where marks are usually dropped, most often a wrong molar mass or a mole ratio used upside down. Because these conversions reappear in acids and bases, electrochemistry and thermochemistry across SPM Chemistry, the time you spend making them automatic now is repaid many times in Form 4 and Form 5.

Quick self-check

Cover the answers and try these. Mr of ammonium sulfate, (NH₄)₂SO₄? [2(14 + 4) + 32 + 4(16) = 132.] Moles in 22 g of CO₂? [22 ÷ 44 = 0.5 mol.] Volume of 0.1 mol of hydrogen at room conditions? [0.1 × 24 = 2.4 dm³.] Empirical formula of a compound with a C : H atom ratio of 1 : 2 and Mr = 42? [Empirical CH₂, mass 14, n = 42 ÷ 14 = 3, molecular formula C₃H₆.] If you can produce the working, not just the answer, you are ready for the practice questions.