# Balancing Chemical Equations

> Count atoms, change only the big numbers, and balance any equation step by step.

From HSC Chemistry on WhizzNotes, free study notes for NSW students: https://whizznotes.com/chemistry/balancing-equations/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The sandwich shop
Mei works at a sandwich shop. One sandwich needs 2 slices of bread and 1 slice of cheese. A customer orders 3 sandwiches. Mei does not cut the bread thinner or invent new cheese. She simply takes 6 slices of bread and 3 slices of cheese.

$$\text{2 bread} + \text{1 cheese} \rightarrow \text{1 sandwich}$$
$$\text{6 bread} + \text{3 cheese} \rightarrow \text{3 sandwiches}$$

Every slice that goes in comes out inside a sandwich. Balancing equations is the same: you cannot change what a sandwich (a molecule) is made of. You can only change **how many** you use, so that every atom on the left turns up on the right.

**In short:** A balanced chemical equation has the same number of each kind of atom on both sides. Balance it by changing only the coefficients (the big numbers in front), never the subscripts, one element at a time, leaving hydrogen and oxygen until last.

## Why equations must balance

Atoms are not created or destroyed in a reaction (conservation of mass). So there must be the **same number of each type of atom** on both sides of the arrow.

This equation is **not** balanced:

$$\mathrm{H_2 + O_2 \rightarrow H_2O}$$

Left: 2 H, 2 O. Right: 2 H, 1 O. An oxygen atom has disappeared, which is impossible.

## Subscripts and coefficients

| | Subscript | Coefficient |
|---|---|---|
| What it looks like | Small number after a symbol: the 2 in $\mathrm{H_2O}$ | Big number in front: the 3 in $\mathrm{3H_2O}$ |
| What it means | Atoms **inside** one molecule | **How many** molecules |
| Can you change it to balance? | **Never** | **Yes** |

### Watch out: Never change a subscript
Changing $\mathrm{H_2O}$ to $\mathrm{H_2O_2}$ would "balance" the oxygen, but $\mathrm{H_2O_2}$ is hydrogen peroxide, a different chemical. Only change the coefficients.

## How to count atoms

Multiply the coefficient by each subscript.

- $\mathrm{3H_2O}$: H $= 3 \times 2 = 6$, O $= 3 \times 1 = 3$.
- $\mathrm{2Ca(OH)_2}$: Ca $= 2$, O $= 2 \times 2 = 4$, H $= 2 \times 2 = 4$. The subscript after a bracket multiplies everything inside.

## Step by step method

1. Write the **correct formulas** for every reactant and product.
2. **Count** the atoms of each element on both sides. A table helps.
3. Balance **one element at a time** by putting coefficients in front of formulas. Start with an element that appears in only one substance on each side.
4. Leave **hydrogen and oxygen until last**, and an element that is on its own (like $\mathrm{O_2}$ or Mg) to the very end.
5. **Recount** everything. Make sure coefficients are in the simplest whole number ratio.

### Worked example: Water
$$\mathrm{H_2 + O_2 \rightarrow H_2O}$$

O: 2 on the left, 1 on the right. Put 2 in front of $\mathrm{H_2O}$:

$$\mathrm{H_2 + O_2 \rightarrow 2H_2O}$$

Now H: 2 on the left, 4 on the right. Put 2 in front of $\mathrm{H_2}$:

$$\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$$

| Atom | Left | Right |
|---|---|---|
| H | 4 | 4 |
| O | 2 | 2 |

Balanced.

### Worked example: Magnesium and hydrochloric acid
$$\mathrm{Mg + HCl \rightarrow MgCl_2 + H_2}$$

Cl: 1 left, 2 right. Put 2 in front of HCl:

$$\mathrm{Mg + 2HCl \rightarrow MgCl_2 + H_2}$$

Check: Mg 1 and 1. H 2 and 2. Cl 2 and 2. Balanced.

### Worked example: Iron and oxygen
$$\mathrm{Fe + O_2 \rightarrow Fe_2O_3}$$

O: 2 left, 3 right. The lowest common multiple of 2 and 3 is 6. Put 3 in front of $\mathrm{O_2}$ and 2 in front of $\mathrm{Fe_2O_3}$:

$$\mathrm{Fe + 3O_2 \rightarrow 2Fe_2O_3}$$

Fe: 1 left, 4 right. Put 4 in front of Fe:

$$\mathrm{4Fe + 3O_2 \rightarrow 2Fe_2O_3}$$

### Exam tip: Keep polyatomic ions together
If an ion like $\mathrm{SO_4}$ or $\mathrm{NO_3}$ appears unchanged on both sides, count it as **one unit** instead of counting S and O separately. It makes balancing much faster.

$$\mathrm{H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O}$$

## Reading a balanced equation

$$\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$$

"Two molecules of hydrogen react with one molecule of oxygen to make two molecules of water." The coefficients give the **ratio** in which substances react.

## From words to a balanced equation

Exam questions often give a reaction in words. Turn it into a balanced equation in three steps.

1. **Word equation**: names only.
2. **Skeleton equation**: correct formulas and state symbols, but **not yet balanced**.
3. **Balanced equation**: add coefficients.

### Worked example: Iron and fluorine
"Solid iron reacts with fluorine gas to form solid iron(III) fluoride."

1. Word: iron + fluorine → iron(III) fluoride
2. Skeleton: $\mathrm{Fe(s)} + \mathrm{F_2(g)} \rightarrow \mathrm{FeF_3(s)}$. Iron(III) is $\mathrm{Fe^{3+}}$, so it needs three $\mathrm{F^-}$. Fluorine is diatomic, so it is $\mathrm{F_2}$.
3. Balance F: 2 on the left, 3 on the right. The lowest common multiple is 6, so use $\mathrm{3F_2}$ and $\mathrm{2FeF_3}$. Then Fe: use $\mathrm{2Fe}$.

$$\mathrm{2Fe(s)} + \mathrm{3F_2(g)} \rightarrow \mathrm{2FeF_3(s)}$$

### Watch out: Remember the diatomic elements
Hydrogen, nitrogen, oxygen and the halogens are written as $\mathrm{H_2}$, $\mathrm{N_2}$, $\mathrm{O_2}$, $\mathrm{F_2}$, $\mathrm{Cl_2}$, $\mathrm{Br_2}$ and $\mathrm{I_2}$ in the skeleton equation when they are on their own. Writing "O" for oxygen gas is a very common mistake.

**Mind map**

```text
Balancing equations
  Why
    Atoms conserved
    Same atoms each side
  Numbers
    Subscript: never change
    Coefficient: change this
    Bracket multiplies inside
  Method
    Correct formulas first
    Count atoms in a table
    One element at a time
    H and O last
    Recount
  Shortcuts
    Polyatomic ions as one unit
    Lowest common multiple
```

### Year 11: Balancing harder equations
**Combustion of hydrocarbons**: balance C, then H, then O last.

$$\mathrm{C_3H_8 + O_2 \rightarrow CO_2 + H_2O}$$

C: 3, so $\mathrm{3CO_2}$. H: 8, so $\mathrm{4H_2O}$. O on the right: $3 \times 2 + 4 \times 1 = 10$, so $\mathrm{5O_2}$.

$$\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$$

**Odd oxygen trick**: if you need an odd number of O atoms from $\mathrm{O_2}$, use a fraction, then double everything.

$$\mathrm{C_2H_6 + \tfrac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O} \quad\Rightarrow\quad \mathrm{2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O}$$

### Year 12: Using a balanced equation to calculate masses
The coefficients give the **mole ratio**, which lets you calculate amounts.

What mass of water forms when 8.0 g of hydrogen burns? $\mathrm{2H_2} + \mathrm{O_2} \rightarrow \mathrm{2H_2O}$

1. Moles of $\mathrm{H_2}$: $n = \frac{m}{M} = \frac{8.0}{2.0} = 4.0 \text{ mol}$.
2. Mole ratio $\mathrm{H_2 : H_2O} = 2 : 2$, so $n(\mathrm{H_2O}) = 4.0 \text{ mol}$.
3. Mass of water: $m = nM = 4.0 \times 18.0 = 72 \text{ g}$.

Check with conservation of mass: 4.0 mol $\mathrm{H_2}$ needs 2.0 mol $\mathrm{O_2}$ = 64 g. $8.0 + 64 = 72$ g. It matches.

## Quick summary

- A balanced equation has the **same number of each atom** on both sides.
- Change only **coefficients** (big numbers in front), never **subscripts**.
- Count atoms by multiplying the coefficient by each subscript. A bracket subscript multiplies everything inside.
- Balance one element at a time, leave H and O until last, then recount.
- Treat unchanged polyatomic ions as one unit.

## Quiz

Try these questions from the chapter quiz.

1. Why must chemical equations be balanced?
   - A. To make them look neat
   - B. Because atoms are not created or destroyed in a reaction
   - C. Because molecules are created
   - D. Because energy is conserved
   - Answer: B. Because atoms are not created or destroyed in a reaction. Every atom in the reactants must appear in the products.

2. When balancing an equation, you may change:
   - A. The coefficients only
   - B. The subscripts only
   - C. Both coefficients and subscripts
   - D. The products to different chemicals
   - Answer: A. The coefficients only. Changing a subscript changes the chemical itself.

3. How many oxygen atoms are in $\mathrm{3CO_2}$?
   - A. 2
   - B. 3
   - C. 5
   - D. 6
   - Answer: D. 6. 3 × 2 = 6.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/chemistry/balancing-equations/#quiz