Video Summary - IGCSE CHEMISTRY REVISION [Syllabus 4]

Main ideas / lessons

Stoichiometry is made easier by mastering foundations: understand valency, how to write correct chemical formulae, how to balance equations, then use mole-based conversions.
Valency drives formula writing and ion charges:
- Valency is how many electrons atoms lose or gain to reach a full outer shell.
- It depends on the group number on the periodic table (typical IGCSE guidance: skip transition metals).
Chemical equations must be balanced so that the number of each type of atom is equal on both sides.
- You do not change subscripts in formulas; instead you add coefficients (numbers in front).
- Those coefficients correspond to mole ratios.
Stoichiometric calculations rely on moles and three major calculation areas:
1. Mass calculations
2. Gas (volume) calculations
3. Solution (concentration/volume) calculations

Methodology / instruction-style content (detailed)

1) Constructing the formula of a compound (using valency)

Step A: Determine valency from the periodic table

Group 1 → valency 1 (e.g., lithium has 1 outer-shell electron)
Group 2 → valency 2 (e.g., beryllium has 2)
Group 6 → valency 2 (e.g., oxygen “needs two” more electrons)
Group 7 → valency 1 (non-metal tendency to gain 1)

Step B: Write the compound formula

Swap the valency numbers (cation valency and anion valency).
Cancel if possible (reduce to simplest whole-number ratio).

Examples

Potassium oxide: K (1) and O (2) → swap → K₂O
Magnesium oxide: Mg (2) and O (2) → swap → Mg₂O₂ → cancel → MgO

2) Writing and balancing chemical equations (critical rule)

Rule: Atom conservation

Total number of each type of atom on the left must equal the right.

Key constraint

Do not alter subscripts in chemical formulas to “fix” imbalance.

Fixing imbalance

Add coefficients in front of compounds.
The coefficients create the balanced mole ratio.

Example concept

Magnesium + oxygen → magnesium oxide
- If oxygen doesn’t match (e.g., 2 O on left but 1 O on right), multiply the appropriate compound (e.g., put a 2 in front of magnesium oxide) to balance.

3) Core definitions needed for mole calculations

Relative atomic mass (Ar)
- Average mass of naturally occurring atoms, using a relative scale (carbon = 12).
Relative formula mass (Mr)
- Sum of Ar values of all atoms in a compound.
- Example: MgO → 24 + 16 = 40
Mole and Avogadro’s constant
- 1 mole contains 6 × 10²³ particles (atoms/ions/molecules).
- Mass of 1 mole = relative formula mass (in grams).
- Example: 1 mole of MgO has mass 40 g.
- Example: O₂ has relative formula mass 32 → 1 mole weighs 32 g.

Calculation procedures

A) Mass calculations (using moles + Mr)

Three-step approach used

Write/confirm balanced equation.
Convert given mass to moles for the starting substance:
- [ \text{moles}=\frac{\text{mass}}{\text{relative formula mass}} ]
Use the mole ratio (from balanced equation coefficients) to find moles of the required product.
Convert moles of product to mass:
- [ \text{mass}=\text{moles}\times \text{relative formula mass} ]

Examples

Conceptual
- 1 mole of water (H₂O): Mr = 18 → mass = 18 g
Worked method
- Question: mass of MgO formed when 3 g Mg reacts with excess oxygen.
- Step outcomes described:
  - moles Mg = 3 ÷ 24 = 0.125 mol
  - From balanced ratio: Mg : MgO = 1 : 1
  - moles MgO = 0.125 mol
  - mass MgO = 0.125 × 40 = 5 g

B) Gas calculations (1 mole gas volume at RTP)

Key concept

At room temperature and pressure (RTP), 1 mole of any gas occupies:
- 24 dm³

Useful equation (rearranged)

[ \text{moles of gas}=\frac{\text{volume}}{24} ]
[ \text{volume}=\text{moles}\times 24 ]

Unit reminder

1 dm³ = 1000 cm³
If volume is in cm³, convert to dm³ before using the 24 dm³ rule.

Example described

Find volume of O₂ needed to burn 1.4 g of butane.
Steps described:
1. Compute moles butane:
  - moles butane = 1.4 ÷ Mr(butane)
2. Use mole ratio:
  - butane : oxygen = 1 : 6
  - moles O₂ = 6 × moles butane
3. Convert moles O₂ to volume:
  - volume O₂ = moles O₂ × 24
Final stated result: 3.6 dm³ of O₂

C) Solution calculations (concentration × volume)

Formula used

[ \text{moles}=\text{concentration}\times \text{volume} ]

Unit conversion needed

Often volume is given in cm³, so convert:
- [ \text{dm}^3=\frac{\text{cm}^3}{1000} ]

Stoichiometric mole ratio

Use the balanced equation to relate moles.
- For sulfuric acid vs sodium hydroxide: ratio 1 : 2
  - 1 mol H₂SO₄ requires 2 mol NaOH

When asked for volume of solute

Rearrange:
- [ \text{volume}=\frac{\text{moles}}{\text{concentration}} ]