Calculate reacting masses in simple proportions. Calculations will not involve the mole concept

Stoichiometry – Formulae, Relative Masses & the Mole (IGCSE Chemistry 0620)

Learning Objectives (AO1‑AO3)

• Recall and use the definitions of relative atomic mass (Ar) and relative molecular mass (Mr).
• Write correct chemical formulas, word equations and **balanced** symbol equations (including state symbols s, l, g, aq).
• Apply the laws of definite and multiple proportions.
• Understand the mole, Avogadro’s constant and molar volume (24 dm³ mol⁻¹ at r.t.p.).
• Convert between mass, moles, number of particles and gas volume.
• Calculate masses, volumes or concentrations of reactants and products using:
  • simple proportional (no‑mole) method, and
  • the mole concept where required.
• Derive empirical and molecular formulas from percentage composition.

Key Concepts

Relative Atomic Mass (Ar)

Mass of an atom relative to 1/12 of a carbon‑12 atom. Values are dimensionless.

Relative Molecular Mass (Mr)

Sum of the Ar values of all atoms in a molecule. Important relationship: molar mass (M) in g mol⁻¹ = Mr.

The Mole

• 1 mol = 6.02 × 10²³ particles (Avogadro’s constant, N_A).
• Mass of 1 mol of a substance = its molar mass (M = Mr) in grams.
• Molar volume at r.t.p. (25 °C, 1 atm): 1 mol gas occupies 24 dm³.

Concentration (c)

Amount of solute per unit volume.

• Mass concentration: c (g dm⁻³) = m (g) ÷ V (dm³)
• Molar concentration: c (mol dm⁻³) = n (mol) ÷ V (dm³)

Laws of Proportion

• Law of definite proportions: A given compound always contains the same elements in the same mass ratio.
• Law of multiple proportions: When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole‑number ratio.

Relative Atomic Masses (Ar) – Elements 1 – 20

Element	Symbol	Ar
Hydrogen	H	1
Helium	He	4
Lithium	Li	7
Beryllium	Be	9
Boron	B	11
Carbon	C	12
Nitrogen	N	14
Oxygen	O	16
Fluorine	F	19
Neon	Ne	20
Sodium	Na	23
Magnesium	Mg	24
Aluminium	Al	27
Silicon	Si	28
Phosphorus	P	31
Sulphur	S	32
Chlorine	Cl	35.5
Argon	Ar	40
Potassium	K	39
Calcium	Ca	40

Calculating Relative Molecular Mass (Mr)

Add the Ar values of all atoms in the formula.

Example: Water, H₂O

Mr_H₂O = 2 × Ar_H + 1 × Ar_O = 2 × 1 + 16 = 18

Conversions Involving the Mole

• Mass ↔ Moles: n (mol) = m (g) ÷ M (g mol⁻¹)
• Moles ↔ Particles: N = n × N_A
• Gas volume ↔ Moles (r.t.p.): V (dm³) = n × 24

Simple Proportion (No‑Mole) Method

This method uses the mass ratios that arise directly from the balanced equation and the Mr values.

1. Write and **balance** the chemical equation (include state symbols).
2. Calculate the Mr of each reactant and product.
3. Set up a proportion using the **mass‑to‑Mr** ratios: \[ \frac{m_1}{\text{Mr}_1} = \frac{m_2}{\text{Mr}_2} = \dots \] where \(m_i\) is the mass of substance i.
4. Insert the known mass(es) and solve for the unknown mass.

Worked Example 1 – Combustion of Magnesium (no‑mole)

Balanced equation (including states): Mg(s) + ½ O₂(g) → MgO(s)

Substance	Formula	Mr
Mg	Mg	24
O (from ½ O₂)	O	16
MgO	MgO	40

Mass proportion:

\[ \frac{m_{\text{Mg}}}{24} = \frac{m_{\text{O}}}{16} = \frac{m_{\text{MgO}}}{40} \]

If 12 g Mg react:

• O required: \(m_{\text{O}} = 12 \text{g} \times \frac{16}{24} = 8 \text{g}\)
• MgO formed: \(m_{\text{MgO}} = 12 \text{g} \times \frac{40}{24} = 20 \text{g}\)

Worked Example 2 – Formation of Water (no‑mole)

Balanced equation (states): 2 H₂(g) + O₂(g) → 2 H₂O(l)

Substance	Mr
H₂	2
O₂	32
H₂O	18

Proportion (using the coefficient 2 for H₂ and H₂O):

\[ \frac{m_{\text{H₂}}}{2} = \frac{m_{\text{H₂O}}}{18} \]

Given 10 g H₂:

\[ m_{\text{H₂O}} = 10 \text{g} \times \frac{18}{2} = 90 \text{g} \]

Using the Mole Concept (when required)

Follow the standard three‑step route:

1. Convert the given mass to moles (using M = Mr).
2. Use the stoichiometric coefficients to relate moles of reactants and products.
3. Convert the required moles back to mass (or volume for gases).

Empirical & Molecular Formulas

These are derived from percentage composition and the given molar mass.

1. Assume a 100 g sample → percentages become masses in grams.
2. Convert each mass to moles: \(n = \frac{m}{\text{Ar}}\).
3. Divide all mole values by the smallest mole number.
4. Round to the nearest whole number → empirical formula.
5. Calculate empirical‑formula mass (E).
   Find the factor \(k = \dfrac{\text{molar mass}}{E}\).
   Multiply each subscript in the empirical formula by k → molecular formula.

Worked Example – C, H, O Compound

Given: 40 % C, 6.7 % H, 53.3 % O; molar mass = 180 g mol⁻¹.

1. Masses: C = 40 g, H = 6.7 g, O = 53.3 g.
2. Moles: C = 40 ÷ 12 = 3.33 mol; H = 6.7 ÷ 1 = 6.70 mol; O = 53.3 ÷ 16 = 3.33 mol.
3. Divide by smallest (3.33): C ≈ 1, H ≈ 2, O ≈ 1 → empirical formula CH₂O.
4. Empirical mass = 12 + 2 × 1 + 16 = 30 g mol⁻¹.
5. k = 180 ÷ 30 = 6 → molecular formula C₆H₁₂O₆.

Law of Multiple Proportions – Quick Check

Carbon and oxygen form CO and CO₂.

• Mass of O that combines with 12 g C in CO: 16 g.
• Mass of O that combines with 12 g C in CO₂: 32 g.

Ratio = 32 : 16 = 2 : 1 – a simple whole‑number ratio, confirming the law.

Practice Questions

1. Calcium reacts with chlorine gas to form calcium chloride.
   • Write the balanced equation (include states).
   • Calculate the mass of chlorine required when 40 g of calcium are used.
2. When 10 g of hydrogen reacts with excess oxygen to form water, what mass of water is produced?
3. Magnesium reacts with nitrogen to give magnesium nitride, Mg₃N₂. If 12 g of magnesium are used, how many grams of nitrogen are required?
4. A sample of a compound contains 52.14 % C, 34.73 % O and 13.13 % H. Its molar mass is 180 g mol⁻¹. Determine its empirical and molecular formulas.
5. At r.t.p., 0.75 mol of a gas occupies 18 dm³. What is the mass of the gas if its molar mass is 44 g mol⁻¹?
6. A solution contains 58 g of NaCl dissolved in 250 cm³ of water. Calculate its concentration in:
   • g dm⁻³
   • mol dm⁻³

Answers to Practice Questions

1. Balanced equation (states): 2 Ca(s) + Cl₂(g) → 2 CaCl₂(s)

   Mr values: Ca = 40, Cl₂ = 71, CaCl₂ = 111

   Proportion: \(\displaystyle \frac{m_{\text{Ca}}}{40} = \frac{m_{\text{Cl₂}}}{71}\)

   Mass of Cl₂ required: \(m_{\text{Cl₂}} = 40 \text{g} \times \frac{71}{40} = 71 \text{g}\).

2. Balanced equation (states): 2 H₂(g) + O₂(g) → 2 H₂O(l)

   Proportion: \(\displaystyle \frac{m_{\text{H₂}}}{2} = \frac{m_{\text{H₂O}}}{18}\)

   Mass of water: \(m_{\text{H₂O}} = 10 \text{g} \times \frac{18}{2} = 90 \text{g}\).

3. Balanced equation (states): 3 Mg(s) + N₂(g) → Mg₃N₂(s)

   Mr values: Mg = 24, N₂ = 28, Mg₃N₂ = 100

   Using the coefficient 3 for Mg:

   \[ \frac{m_{\text{Mg}}}{3 \times 24} = \frac{m_{\text{N₂}}}{28} \]

   Mass of N₂ required: \(m_{\text{N₂}} = 12 \text{g} \times \frac{28}{72} = 4.67 \text{g}\) (2 dp).

4. Assume 100 g sample: C = 52.14 g, O = 34.73 g, H = 13.13 g.

   Moles: C = 52.14 ÷ 12 = 4.345 mol; O = 34.73 ÷ 16 = 2.171 mol; H = 13.13 ÷ 1 = 13.13 mol.

   Divide by smallest (2.171): C ≈ 2.00, O ≈ 1.00, H ≈ 6.05 → empirical formula C₂H₆O.

   Empirical mass = 2 × 12 + 6 × 1 + 16 = 46 g mol⁻¹.

   k = 180 ÷ 46 ≈ 3.91 ≈ 4 (nearest whole number). Molecular formula = C₈H₂₄O₄.

5. Moles of gas: \(n = \dfrac{V}{24} = \dfrac{18 \text{dm³}}{24 \text{dm³ mol⁻¹}} = 0.75 \text{mol}\) (given).

   Mass = n × M = 0.75 mol × 44 g mol⁻¹ = 33 g.

6. Volume = 250 cm³ = 0.250 dm³.

   • Mass concentration: \(c = \dfrac{58 \text{g}}{0.250 \text{dm³}} = 232 \text{g dm⁻³}\).
   • Molar mass of NaCl ≈ 58.5 g mol⁻¹.
     Moles = 58 g ÷ 58.5 g mol⁻¹ = 0.992 mol.
     Molar concentration: \(c = \dfrac{0.992 \text{mol}}{0.250 \text{dm³}} = 3.97 \text{mol dm⁻³}\) (2 dp).