# Formulae and Chemical Equations

> Chemistry · CIE IGCSE 0620 (2026-2028)
> Source: https://www.owlsprep.com/study/cie-0620-u3-formulae-and-chemical-equations/

This guide covers writing chemical formulae, balancing full equations with state symbols, and Extended-only mass deduction skills, aligned to CIE IGCSE Chemistry 0620 Unit 3 Stoichiometry syllabus point 3.1.

**Prerequisites:** [Knowledge of element names, symbols and periodic table group numbers](https://www.owlsprep.com/study/cie-0620-u1-periodic-table/); [Basic understanding of ionic and covalent bonding](https://www.owlsprep.com/study/cie-0620-u2-bonding-basics/)

## Learning objectives

- Write correct chemical formulae of ionic and covalent compounds using valency rules
- Balance full chemical equations with required state symbols for Core tier
- Deduce balanced equations from reactant/product mass data for Extended tier
- Apply IUPAC naming conventions for common inorganic compounds
- Avoid common formula and equation errors tested in CIE IGCSE 0620 exams

## Writing Chemical Formulae of Common Compounds (Core)

**Valency** — The combining power of an atom or ion, equal to the charge of the ion for ionic compounds, or the number of covalent bonds formed for covalent compounds.

*Example:* Sodium (Group 1) has valency +1, oxygen (Group 16) has valency -2

For all neutral compounds, the sum of positive valencies equals the sum of negative valencies. For ionic compounds, swap the valencies of the two ions to get the subscript count for each ion, and bracket polyatomic ions if you have more than one copy of the ion.

**Worked example:** Write the correct chemical formula of calcium nitrate.

1. Step 1: Identify valencies of the ions: Ca²⁺ = +2, NO₃⁻ = -1
2. Step 2: Find the lowest common multiple of 2 and 1 = 2
3. Step 3: Calculate number of each ion needed: 1 Ca²⁺ (2/2 =1), 2 NO₃⁻ (2/1=2)
4. Step 4: Write the formula, bracketing the polyatomic nitrate ion: Ca(NO₃)₂

> **tip**
>
> Never write the number 1 as a subscript in chemical formulae, it is implied if no number is written.

*Calculator:* forbidden

## Balancing Chemical Equations (Core)

**Balanced Chemical Equation** — An equation where the number of atoms of every element is identical on the reactant (left) and product (right) sides, complying with the law of conservation of mass.

*Example:* 2H₂(g) + O₂(g) → 2H₂O(l): 4 H atoms and 2 O atoms on both sides

First write the unbalanced equation with correct formulae for all reactants and products. Balance elements that appear in only one reactant and one product first, then balance elements that appear in multiple substances. Only adjust whole number coefficients in front of formulae, never change subscripts (this changes the identity of the compound). Add state symbols last.

**Worked example:** Balance the equation for the reaction of magnesium with hydrochloric acid, including state symbols: Mg + HCl → MgCl₂ + H₂

1. Step 1: Count atoms on each side: Left = 1 Mg, 1 H, 1 Cl; Right = 1 Mg, 2 H, 2 Cl
2. Step 2: Balance Cl first: add coefficient 2 in front of HCl: left Cl = 2, left H = 2
3. Step 3: Verify H is now balanced (2 on both sides), Mg is already balanced
4. Step 4: Add state symbols: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
5. Step 5: Final check: all atom counts are equal on both sides, equation is correct

**Exam command terms**

Common command terms for this topic:

- **Write the formula of** — Provide the neutral chemical formula of the compound, no coefficients required *(Write the formula of copper(II) oxide: CuO)*

- **Balance the following equation** — Add whole number coefficients to the given unbalanced equation to equalise atom counts *(Balance H₂ + O₂ → H₂O: 2H₂ + O₂ → 2H₂O)*

*Calculator:* forbidden

## Extended Only: Deducing Balanced Equations from Mass Data

For Extended tier candidates, you may be given masses of reactants and products to deduce the balanced equation. Use the mole formula to calculate the moles of each substance, then find the simplest whole number ratio of moles to get the coefficients for the balanced equation.

$$Moles = \frac{Mass}{A_r \text{ (for atoms)} \text{ or } M_r \text{ (for molecules/ions)}}$$

**Worked example:** 8.0g of copper(II) oxide reacts with 0.2g of hydrogen to form 6.4g of copper and 1.8g of water. Deduce the balanced equation for this reaction. A_r values: Cu=64, O=16, H=1.

1. Step 1: Calculate M_r values for compounds: CuO = 64+16=80, H₂=2, H₂O=18
2. Step 2: Calculate moles of each substance:
3. $$Moles \text{ } CuO = \frac{8.0}{80} = 0.1 \text{ mol}$$
4. $$Moles \text{ } H_2 = \frac{0.2}{2} = 0.1 \text{ mol}$$
5. $$Moles \text{ } Cu = \frac{6.4}{64} = 0.1 \text{ mol}$$
6. $$Moles \text{ } H_2O = \frac{1.8}{18} = 0.1 \text{ mol}$$
7. Step 3: Divide all mole values by the smallest value (0.1) to get a 1:1:1:1 ratio
8. Step 4: Write the balanced equation with state symbols: CuO(s) + H₂(g) → Cu(s) + H₂O(l)
9. Step 5: Verify atom counts are equal on both sides: 1 Cu, 1 O, 2 H each side, correct

> **Exam tip**
>
> Always use the relative molecular mass for diatomic elements (H₂, O₂, Cl₂, etc.) when calculating moles of the element in its standard state, not the relative atomic mass.

*Calculator:* allowed

## IUPAC Naming Rules for Common Inorganic Compounds

- Ionic compounds: Name the positive ion first, then the negative ion. Negative monatomic ions end in -ide, e.g. sodium chloride, magnesium oxide
- Polyatomic ions with oxygen end in -ate (more oxygen atoms) or -ite (fewer oxygen atoms), e.g. potassium sulfate, sodium sulfite
- Binary covalent compounds: Use prefixes (di-, tri-, tetra-) to show the number of each atom, omit mono- for the first element, e.g. carbon dioxide, dinitrogen trioxide

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Writing OH₂ instead of (OH)₂ for two hydroxide ions in calcium hydroxide formula
  - Why it fails: Incorrect notation implies 2 hydrogen atoms attached to one oxygen, rather than two separate hydroxide groups, and will lose marks in exams
  - Correct: Always bracket polyatomic ions if you have more than one copy, so write Ca(OH)₂
- **Wrong:** Changing subscripts in formulae when balancing equations, e.g. changing H₂O to H₂O₂ to balance oxygen
  - Why it fails: Subscripts define the identity of the compound, changing them creates a different substance that is not part of the reaction
  - Correct: Only adjust whole number coefficients in front of formulae to balance atom counts
- **Wrong:** Forgetting to add state symbols to equations even when not explicitly asked
  - Why it fails: CIE IGCSE 0620 awards 1 mark for correct state symbols in nearly all equation questions, regardless of explicit instruction
  - Correct: Add state symbols (s), (l), (g), (aq) to every balanced equation unless told otherwise
- **Wrong:** Writing the formula of sodium sulfate as NaSO₄
  - Why it fails: Sodium has valency +1, sulfate has valency -2, so the sum of charges is -1, making the formula not neutral
  - Correct: Use 2 sodium ions per sulfate ion to get neutral formula Na₂SO₄
- **Wrong:** Using A_r of O=16 instead of M_r of O₂=32 when calculating moles of oxygen gas (Extended only)
  - Why it fails: Oxygen exists as diatomic molecules in its standard state, so 1 mole of O₂ has a mass of 32g, not 16g
  - Correct: Use relative molecular mass for all diatomic elements (H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂) when calculating moles of the standard state element

## Cheatsheet

| Task | Rule | Example |
| --- | --- | --- |
| Write ionic formula | Sum of positive valencies = sum of negative valencies; bracket polyatomic ions if >1 | Aluminium sulfate: Al₂(SO₄)₃ |
| Balance equation | Only change coefficients, never subscripts; add state symbols last | Combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) |
| Extended: Deduce equation from mass | Calculate moles of all substances, divide by smallest mole value to get coefficient ratio | Mole ratio 1:2:1:2 → Zn + 2HCl → ZnCl₂ + H₂ |
| Name ionic compound | Positive ion first, negative ion ends in -ide/-ate/-ite | Fe(NO₃)₃ = iron(III) nitrate |

## What's next

Now that you can write and balance chemical equations, you have the foundational skill required for all stoichiometry calculations in CIE IGCSE Chemistry 0620 Unit 3. Incorrect formulae or unbalanced equations will make all subsequent calculation answers invalid, so mastering this subtopic eliminates a major source of mark loss in both Core and Extended tier papers. Next, you will learn to use balanced equations to calculate reacting masses, gas volumes, and solution concentrations, which make up ~15-20% of the total exam marks. Extended tier candidates will also use these equation writing skills for ionic half equations in redox reactions later in the course.

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