Chemistry notes, chapter 3
The Mole Concept, Chemical Formula and Equation
Textbook Chemistry Form 4 KSSM (DLP), pages 42–77
What this chapter covers
- Explain relative atomic mass, relative molecular mass and relative formula mass on the carbon-12 scale, and calculate them (3.1)
- Define the mole and the Avogadro constant, and interrelate moles, particles, mass and volume of gas using molar mass and molar volume (3.2)
- State the meaning of chemical, empirical and molecular formula and determine them from experimental data (3.3)
- Construct formulae and name ionic and simple molecular compounds (3.3)
- Write and balance chemical equations, interpret them and solve stoichiometry problems (3.4)
3.1Relative atomic mass and relative molecular mass
Atoms are far too small to count or weigh one by one, so chemists compare the mass of one atom with a standard atom. Since 1961 the standard is the carbon-12 atom, which is given a mass of exactly 12 units.
Relative atomic mass (RAM) of an element is the average mass of one atom of the element compared with 1⁄12 of the mass of one carbon-12 atom.
Relative molecular mass (RMM) of a substance is the average mass of one molecule compared with 1⁄12 of the mass of one carbon-12 atom.
RAM = 4 for helium means one helium atom is 4 times as heavy as 1⁄12 of a carbon-12 atom. Relative masses are comparisons, so they have no unit.
Calculating relative masses
- RMM (molecular substances) = sum of the RAMs of all atoms in the molecular formula.
- RFM, relative formula mass (ionic substances) = sum of the RAMs of all atoms in the formula.
Worked example
Aluminium sulphate, Al2(SO4)3 [Al = 27, S = 32, O = 16]
RFM = 2(27) + 3[32 + 4(16)] = 54 + 3(96) = 342
RFM = 2(27) + 3[32 + 4(16)] = 54 + 3(96) = 342
When a bracket has a subscript, multiply everything inside the bracket by that number: Mg(NO3)2 has 2 N and 6 O.
Calculate the RMM of sulphuric acid, H2SO4 [H = 1, S = 32, O = 16].
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2(1) + 32 + 4(16) = 2 + 32 + 64 = 98
3.2The mole concept
The mole (mol) is the SI unit of amount of substance. One mole contains 6.02 × 1023 particles. This number is the Avogadro constant, NA = 6.02 × 1023 mol−1.
| Type of substance | Particles in 1 mol | Example |
|---|---|---|
| Atomic (metals, noble gases) | atoms | 1 mol Cu = 6.02 × 1023 atoms |
| Molecular | molecules | 1 mol H2O = 6.02 × 1023 molecules |
| Ionic | formula units | 1 mol NaCl = 6.02 × 1023 units |
Three conversion factors
| Link | Conversion factor | Formula |
|---|---|---|
| Moles and particles | Avogadro constant, NA | n = number of particles ÷ NA |
| Moles and mass | Molar mass (g mol−1) | n = mass ÷ molar mass |
| Moles and gas volume | Molar volume | n = volume ÷ molar volume |
Molar mass is the mass of one mole of substance. Its value in g mol−1 equals the RAM, RMM or RFM.
Molar volume is the volume of one mole of any gas: 22.4 dm3 mol−1 at STP (0 °C, 1 atm) and 24 dm3 mol−1 at room conditions (25 °C, 1 atm).
Mole relationship chart: Chemistry Form 4 (DLP) textbook, Figure 3.9, p. 56.
Worked example: volume from mass
Volume of 26.4 g of CO2 at room conditions [C = 12, O = 16]
Molar mass = 12 + 2(16) = 44 g mol−1
n = 26.4 ÷ 44 = 0.6 mol
Volume = 0.6 × 24 = 14.4 dm3
Molar mass = 12 + 2(16) = 44 g mol−1
n = 26.4 ÷ 44 = 0.6 mol
Volume = 0.6 × 24 = 14.4 dm3
Worked example: particles from volume
Molecules in 672 cm3 of H2 at STP
672 cm3 = 0.672 dm3; n = 0.672 ÷ 22.4 = 0.03 mol
Molecules = 0.03 × 6.02 × 1023 = 1.806 × 1022
672 cm3 = 0.672 dm3; n = 0.672 ÷ 22.4 = 0.03 mol
Molecules = 0.03 × 6.02 × 1023 = 1.806 × 1022
Forgetting the unit conversion: 1 dm3 = 1000 cm3. Also, count atoms not molecules when asked: 1 mol of O2 has 2 mol of O atoms.
Always go through moles. Mass → moles → volume, never mass → volume directly. Determine the molar mass first.
How many moles are there in 4.7 g of K2O [K = 39, O = 16]?
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RFM = 2(39) + 16 = 94; n = 4.7 ÷ 94 = 0.05 mol
3.3Chemical formula
A chemical formula uses symbols and subscript numbers to show the atoms in a substance. Two types are used for compounds.
| Formula | Shows | Glucose |
|---|---|---|
| Molecular formula | the actual number of atoms of each element in one molecule | C6H12O6 |
| Empirical formula | the simplest whole-number ratio of atoms | CH2O |
Molecular formula = (empirical formula)n, where n is a whole number, so n = RMM ÷ mass of the empirical formula.
Finding an empirical formula
- Write the mass (or % by mass) of each element.
- Divide by its RAM to get moles of atoms.
- Divide all by the smallest number.
- If needed, multiply by 2 or 3 to get whole numbers.
Worked example
1.35 g of Al combines with 1.2 g of O [Al = 27, O = 16]
Moles: Al 1.35 ÷ 27 = 0.05; O 1.2 ÷ 16 = 0.075
Ratio: 1 : 1.5 → ×2 → 2 : 3. Empirical formula = Al2O3
Moles: Al 1.35 ÷ 27 = 0.05; O 1.2 ÷ 16 = 0.075
Ratio: 1 : 1.5 → ×2 → 2 : 3. Empirical formula = Al2O3
| Oxide | Method | Why |
|---|---|---|
| Magnesium oxide | Heat magnesium ribbon in a crucible with a lid; weigh until the mass is constant | Mg is reactive and burns in air |
| Copper(II) oxide | Heat the oxide in a stream of hydrogen; mass lost = mass of oxygen | Cu is not reactive enough to burn easily; hydrogen removes the oxygen |
Heat and weigh again until the mass is constant, so you know the reaction is complete. For copper(II) oxide, hydrogen must flow before heating and until cool, to stop copper re-oxidising and to avoid an explosion.
Formulae of ionic compounds
Use the cross-change method: swap the charge numbers to give the number of ions, then check that the total charge is zero.
Worked example
Aluminium oxide: Al3+ and O2− → 2 Al, 3 O → Al2O3 (2 × +3 = +6; 3 × −2 = −6)
Naming compounds
- Ionic: cation first, then anion (sodium chloride). Use Roman numerals for metals with more than one ion: Fe2+ iron(II), Fe3+ iron(III).
- Simple molecular: the first element keeps its name, the second ends in ‘-ide’, and Greek prefixes give the numbers: mono (1), di (2), tri (3), tetra (4), penta (5), hexa (6). CO is carbon monoxide; SO3 is sulphur trioxide.
The empirical formula of a hydrocarbon is CH2 and its RMM is 56. Find its molecular formula [C = 12, H = 1].
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Mass of CH2 = 14; n = 56 ÷ 14 = 4; molecular formula = C4H8
3.4Chemical equation
Reactants are on the left, products on the right, and the arrow means ‘produces’. State symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous solution. A Δ over the arrow means heating.
Writing and balancing
- Write the word equation.
- Replace names with correct formulae.
- Balance by changing the coefficients (numbers in front), never the subscripts.
- Add state symbols.
Worked example
Hydrogen + oxygen → water
H2 + O2 → H2O (not balanced: 2 O on the left, 1 on the right)
2H2(g) + O2(g) → 2H2O(l)
H2 + O2 → H2O (not balanced: 2 O on the left, 1 on the right)
2H2(g) + O2(g) → 2H2O(l)
Balancing is needed because of the law of conservation of mass: atoms are neither created nor destroyed.
Stoichiometry
The coefficients give the mole ratio. 2Na + Cl2 → 2NaCl means 2 mol Na react with 1 mol Cl2 to give 2 mol NaCl.
Worked example
4Al + 3O2 → 2Al2O3. Mass of Al2O3 from 5.4 g Al [Al = 27, O = 16]
n(Al) = 5.4 ÷ 27 = 0.2 mol
Ratio 4 : 2, so n(Al2O3) = 0.1 mol
Mass = 0.1 × 102 = 10.2 g
n(Al) = 5.4 ÷ 27 = 0.2 mol
Ratio 4 : 2, so n(Al2O3) = 0.1 mol
Mass = 0.1 × 102 = 10.2 g
| Step | What to do |
|---|---|
| 1 | Balanced equation |
| 2 | Convert the given quantity to moles |
| 3 | Use the mole ratio to get moles of the wanted substance |
| 4 | Convert moles to the quantity asked (mass, volume or particles) |
Using the mass ratio from the equation. The ratio in the coefficients is a mole ratio, not a ratio of masses.
Zn + 2HCl → ZnCl2 + H2. What mass of zinc gives 0.5 mol of H2 [Zn = 65]?
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Ratio 1 : 1, so 0.5 mol Zn; mass = 0.5 × 65 = 32.5 g