# Ratios and Rearranging Equations

> Ratios, proportion, percentages, density, and how to rearrange any formula to find the unknown.

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

### Story time: Mixing cordial
Jack's grandma makes cordial with a rule: **1 part syrup to 4 parts water**. For one glass she uses 50 mL of syrup and 200 mL of water.

At Jack's party there are twenty glasses to fill. Jack does not guess. He keeps the ratio the same: 1 L of syrup and 4 L of water. Every glass tastes exactly like grandma's.

Chemistry works the same way. Water is always 2 hydrogen atoms to 1 oxygen atom. Reactions always use chemicals in fixed ratios. If you can handle ratios and rearrange a formula, most chemistry maths becomes easy.

**In short:** A ratio compares amounts, and you share in a ratio by adding the parts and finding the size of one part. To rearrange a formula, do the same thing to both sides: density $\rho = \dfrac{m}{V}$ gives $m = \rho V$ and $V = \dfrac{m}{\rho}$.

## Ratios

A **ratio** compares two amounts. It is written with a colon: syrup : water = 1 : 4.

**Simplify** a ratio by dividing both sides by the same number, just like a fraction.

$$12 : 8 = 3 : 2 \quad (\text{divide both by } 4)$$

**Scale** a ratio by multiplying both sides by the same number.

$$1 : 4 \;\rightarrow\; 5 : 20 \quad (\text{multiply both by } 5)$$

### Worked example: Ratios in a formula
In water, $\mathrm{H_2O}$, the ratio of hydrogen atoms to oxygen atoms is 2 : 1.

How many hydrogen atoms are in 30 water molecules? Each molecule has 2, so $30 \times 2 = 60$ hydrogen atoms.

## Sharing in a ratio

To split an amount in a ratio:

1. Add the parts.
2. Divide the total by the number of parts to find one part.
3. Multiply.

### Worked example: Bronze
Bronze is copper and tin in the ratio 9 : 1 by mass. How much copper is in 500 g of bronze?

Parts: $9 + 1 = 10$. One part: $500 \div 10 = 50 \text{ g}$. Copper: $9 \times 50 = 450 \text{ g}$.

## Percentages

$$\text{percentage} = \frac{\text{part}}{\text{whole}} \times 100\%$$

### Worked example: Percentage of an element
A 20 g sample of a compound contains 8 g of carbon. What percentage is carbon?

$$\frac{8}{20} \times 100\% = 40\%$$

## Proportion

Two quantities are **directly proportional** when doubling one doubles the other. If 2 g of magnesium makes 3.3 g of magnesium oxide, then 4 g makes 6.6 g.

The **unitary method** finds the amount for 1 first:

- 2 g makes 3.3 g, so 1 g makes $3.3 \div 2 = 1.65$ g.
- 5 g makes $5 \times 1.65 = 8.25$ g.

## Rearranging equations

Many chemistry formulas have three quantities. For example, **density**:

$$\rho = \frac{m}{V}$$

where $\rho$ (the Greek letter rho) is density, $m$ is mass and $V$ is volume.

To find a different quantity, **rearrange**. The golden rule: **do the same thing to both sides**.

To find $m$: multiply both sides by $V$.

$$m = \rho V$$

To find $V$: start with $m = \rho V$ and divide both sides by $\rho$.

$$V = \frac{m}{\rho}$$

### Key idea: The formula triangle
For any formula like $a = \dfrac{b}{c}$, put $b$ on top and $a$ and $c$ side by side underneath. Cover the one you want:
- Cover $b$: you see $a \times c$.
- Cover $a$: you see $b \div c$.
- Cover $c$: you see $b \div a$.

*Diagram: Formula triangle with m on top, rho and V on the bottom*
*Cover the quantity you want to find.*

### Worked example: Density of a metal block
A block of aluminium has a mass of 54 g and a volume of 20 cm³.

$$\rho = \frac{m}{V} = \frac{54}{20} = 2.7 \text{ g/cm}^3$$

What volume would 135 g of aluminium take up?

$$V = \frac{m}{\rho} = \frac{135}{2.7} = 50 \text{ cm}^3$$

### Exam tip: Three steps for any calculation
1. Write the formula.
2. Rearrange it **before** putting numbers in.
3. Substitute with units, then calculate.

### Watch out: Converting before calculating
If density is in g/cm³, the mass must be in g and volume in cm³. Convert first, then use the formula.

**Mind map**

```text
Chemistry maths
  Ratios
    Simplify: divide both sides
    Scale: multiply both sides
    Share: add parts first
  Percentages
    part ÷ whole × 100%
  Proportion
    Unitary method: find 1 first
  Rearranging
    Same to both sides
    Formula triangle
  Density
    ρ = m / V
```

### Year 11: The mole and molar mass
Chemists count particles in **moles**. One mole is $6.02 \times 10^{23}$ particles (Avogadro's number).

The **molar mass** $M$ is the mass of one mole, in g/mol. It equals the relative atomic or formula mass from the periodic table. For water, $M = 2(1.0) + 16.0 = 18.0 \text{ g/mol}$.

$$n = \frac{m}{M} \qquad m = nM \qquad M = \frac{m}{n}$$

How many moles in 36 g of water? $n = \frac{36}{18.0} = 2.0 \text{ mol}$.

It is the same rearranging skill. Only the letters changed.

### Year 12: Mole ratios and concentration
A balanced equation is a ratio recipe. In $\mathrm{2H_2} + \mathrm{O_2} \rightarrow \mathrm{2H_2O}$ the mole ratio is $\mathrm{H_2 : O_2 : H_2O} = 2 : 1 : 2$.

If 5.0 mol of $\mathrm{O_2}$ reacts, $\mathrm{H_2O}$ formed $= 5.0 \times 2 = 10$ mol.

**Concentration** is moles per litre:

$$c = \frac{n}{V} \qquad n = cV \qquad V = \frac{n}{c}$$

with $c$ in mol/L and $V$ in **litres**. How many moles of salt are in 250 mL of a 0.40 mol/L solution? $V = 0.250 \text{ L}$, so $n = 0.40 \times 0.250 = 0.10 \text{ mol}$.

## Quick summary

- A ratio compares amounts. Simplify or scale by doing the same to both sides.
- To share in a ratio: add the parts, find one part, multiply.
- Percentage $= \dfrac{\text{part}}{\text{whole}} \times 100\%$.
- Rearranging: do the same thing to both sides, or use the formula triangle.
- Density: $\rho = \dfrac{m}{V}$, so $m = \rho V$ and $V = \dfrac{m}{\rho}$.
- Convert units before you substitute.

## Quiz

Try these questions from the chapter quiz.

1. Simplify the ratio 15 : 10.
   - A. 1 : 2
   - B. 3 : 2
   - C. 5 : 3
   - D. 2 : 3
   - Answer: B. 3 : 2. Divide both sides by 5.

2. In carbon dioxide, $\mathrm{CO_2}$, the ratio of carbon atoms to oxygen atoms is:
   - A. 1 : 2
   - B. 2 : 1
   - C. 1 : 1
   - D. 1 : 3
   - Answer: A. 1 : 2. One C and two O in each molecule.

3. How many oxygen atoms are in 10 molecules of $\mathrm{CO_2}$?
   - A. 10
   - B. 20
   - C. 30
   - D. 12
   - Answer: B. 20. Each molecule has 2 oxygen atoms. 10 × 2 = 20.

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