Density

1. Definition

A. Density

Density, ρ, is the mass per unit volume of a substance.

ρ = m/V

• ρ = density (kg m⁻³)
• m = mass (kg)
• V = volume (m³)
• Density is a scalar quantity.

Density is the gradient of a mass–volume graph

A straight-line graph of mass against volume for a single material. The gradient equals density.

Scroll across the graph to read all labels.

Open full-size graph

View figure data

Values for Density is the gradient of a mass–volume graph

Volume (cm³)	Example material (ρ = 2.7 g cm⁻³)
0	0
2	5.4
4	10.8
6	16.2
8	21.6
10	27
12	32.4

Values for Density is the gradient of a mass–volume graph

Volume (cm³)	Example material (ρ = 2.7 g cm⁻³)
0	0
2	5.4
4	10.8
6	16.2
8	21.6
10	27
12	32.4

2. Key Ideas

• Rearrangements:
  • m = ρ V
  • V = m/ρ
• Use consistent units:
  • kg with m³ → kg m⁻³
  • g with cm³ → g cm⁻³
• Useful conversions:
  • 1 g = 1 × 10⁻³ kg
  • 1 cm³ = 1 × 10⁻⁶ m³
  • 1 g cm⁻³ = 1000 kg m⁻³
• Density of the same material is the same anywhere at the same temperature, because mass and volume do not depend on gravitational field strength.

3. Detailed Explanations

A. What density tells you

Density compares how “packed” the matter is:

• higher density → more mass in the same volume
• lower density → less mass in the same volume

B. Volume (how you measure it)

Volume, V, is the amount of space occupied by a three-dimensional object.

• SI unit of volume is m³.
• Other common units: cm³ and mL (where 1 cm³ = 1 mL).

How volume is found in exams:

• regular solid: use dimensions (e.g. V = lwh)
• irregular solid: use water displacement
• liquid: read the volume in a measuring cylinder

C. Using the density equation (and rearranging)

Density is given by:

ρ = m/V

Rearranging gives:

• m = ρ V (use this when density and volume are known)
• V = m/ρ (use this when density and mass are known)

D. Unit conversions you should memorise

1. Converting between cm³ and m³:

2. Converting between g cm⁻³ and kg m⁻³:

1 g cm⁻³ = (10⁻³ kg)/(10⁻⁶ m³) = 10³ kg m⁻³

E. Densities of common substances (typical values)

F. Link to pressure in liquids (next lesson)

Density appears directly in hydrostatic pressure:

p = hρ g

See: Hydrostatic Pressure.

4. Common Mistakes

• Using weight (N) instead of mass (kg) in ρ = m/V.
• Mixing units (e.g. mass in g but volume in m³) without converting.
• Forgetting that volume conversions are cubed:
  • 1 cm³ = 10⁻⁶ m³ (not 10⁻²).
• For displacement questions: using the final reading as the volume instead of (final - initial).

5. Exam Tips

• Start with the formula you need ( ρ = m/V, m = ρ V, or V = m/ρ).
• Write units beside your values before you calculate.
• State the final density with the correct unit (kg m⁻³ or g cm⁻³).
• In practical-style questions, describe measurements clearly: “measure mass with a balance” and “find volume by displacement”.

6. Worked Examples

Modelled example 1

Mass from density and dimensions (SI units)

Core

Problem

A block of concrete has dimensions 0.40 m × 0.30 m × 0.10 m and density 2500 kg m⁻³. Find its mass.

Study the worked solution

1. Find the block's volume

   Method

   Multiply the three perpendicular dimensions.

   Reason

   A rectangular block’s volume is V = lwh, and all dimensions are already in metres.

   Working

   V = (0.40)(0.30)(0.10) = 0.012 m³

2. Choose the density rearrangement

   Reason

   Density and volume are known, so rearrange ρ = m/V to make mass the subject.

   Working

   m = ρ V

3. Calculate the mass

   Reason

   The density and volume use compatible SI units, so no conversion is needed.

   Working

   m = (2500)(0.012) = 30 kg

Guided practice 2

Density of an irregular solid (displacement)

About 5 min

Problem

An irregular stone has mass 120 g. Water in a measuring cylinder rises from 50 cm³ to 95 cm³ when the stone is fully immersed. Find the density of the stone in g cm⁻³.

Complete the displacement calculation

Volume of the stone

cm³Unit: cm³

Density of the stone

g cm⁻³Unit: g cm⁻³

Check my attempt

Hints

Hint 1: find the displaced volume

Subtract the initial water reading from the final reading.

Hint 2: choose the density equation

Once the stone’s volume is known, use ρ = m/V with grams and cubic centimetres.

View solution step by step

1. Find the stone's volume

   Method

   Use the volume of water displaced by the fully immersed stone.

   Reason

   The stone is irregular, so its volume is the rise in the measuring-cylinder reading.

   Working

   V = 95-50 = 45 cm³

2. Calculate the density

   Reason

   The mass is in grams and volume in cubic centimetres, so the result is directly in g cm⁻³.

   Working

   ρ = m/V = 120/45 = 2.67 g cm⁻³ ≈ 2.7 g cm⁻³

Common misconception 3

Convert density units

Find and correct the mistake

Learner response

Aluminium has density 2.7 g cm⁻³. A student starts converting it to kg m⁻³ by writing:

Because 1 cm = 10⁻² m, use 1 cm³ = 10⁻² m³.

Locate the first error and give the correct density in kg m⁻³.

Diagnose and correct the conversion

Where is the first error?

The centimetre-to-metre factor was not cubed for volume.Density cannot be represented by rho.Aluminium cannot have a density.

Correct density

kg m⁻³Unit: kg m⁻³

Check my attempt

View solution step by step

1. Locate the volume-conversion error

   Method

   Cube the linear conversion factor.

   Reason

   Volume has three length dimensions, so converting centimetres to metres requires (10⁻²)³.

   Working

   1 cm³ = 10⁻⁶ m³

2. Combine the mass and volume conversions

   Reason

   One gram contributes 10⁻³ kg while one cubic centimetre contributes 10⁻⁶ m³.

   Working

   1 g cm⁻³ = (10⁻³ kg)/(10⁻⁶ m³) = 10³ kg m⁻³

3. Convert the aluminium density

   Working

   2.7 g cm⁻³ = 2.7 × 10³ kg m⁻³

Examiner practice 4

Mass and volume from density (copper)

4 marks

Examination question

Copper has density 9.0 g cm⁻³. Calculate:

1. the mass of 5.0 cm³ of copper;
2. the volume occupied by 63 g of copper. [4 marks]

Show both rearrangements before viewing the mark scheme

Show the equations, substitutions and units for both parts

View solution step by step

1. Choose the mass relationship

   1 mark

   Method

   Rearrange density to make mass the subject.

   Reason

   Density and volume are known in compatible g–cm³ units.

   Working

   m = ρ V

2. Calculate the mass

   1 mark

   Reason

   Multiplying g cm⁻³ by cm³ leaves grams.

   Working

   m = (9.0)(5.0) = 45 g

3. Choose the volume relationship

   1 mark

   Method

   Rearrange density to make volume the subject.

   Reason

   Mass and density are known for the second sample.

   Working

   V = m/ρ

4. Calculate the volume

   1 mark

   Reason

   Dividing grams by g cm⁻³ leaves cubic centimetres.

   Working

   V = 63/9.0 = 7.0 cm³

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Self-mark each rearrangement and answer separately.

Mark points

1. 1 mark: Uses m = rho V for the first sample.
2. 1 mark: Obtains 45 g.My response includes the correct unit.
3. 1 mark: Uses V = m/rho for the second sample.
4. 1 mark: Obtains 7.0 cm³.My response includes the correct unit.

Calculate my mark

Challenge 5

Identify a sheet material from mass and volume

Minimal support

Data transfer

An unlabelled metal sheet has mass 200 g and volume 73 cm³.

1. Calculate its density.
2. Compare the result with the table in section 3E and identify the most likely material.
3. Explain why density is more useful for this identification than the sheet’s mass alone.

Calculate, identify and justify

Density of the sheet

g cm⁻³Unit: g cm⁻³

Most likely material

AluminiumCopperLead

Check my attempt

Explain why density is more useful than mass alone

Hints

Hint 1: calculate the comparison value

Use ρ = m/V before consulting the reference table.

Hint 2: compare a property rather than sample size

A sample’s mass changes with its size; density characterises the material at a stated temperature.

View solution step by step

1. Calculate the sheet's density

   Method

   Divide its mass by its volume.

   Reason

   The quantities are already in grams and cubic centimetres, matching the reference table’s density unit.

   Working

   ρ = 200/73 = 2.74 g cm⁻³

2. Identify the closest reference material

   Reason

   The calculated value is closest to aluminium’s listed density of 2.7 g cm⁻³.

   Working

   The most likely material is aluminium.

3. Justify using density

   Reason

   Mass alone depends on how large the sample is, while a uniform material has the same mass-to-volume ratio at the same temperature.

   Working

   A small and a large aluminium sheet can have different masses but the same density.

Next step

Try Mind Stretcher 2 without using the worked-solution structure. That separate response is practice;

7. Mind Stretchers

Mind stretcher 1: Does density change when you cut an object?Extension

A cube has density 8.0 g cm⁻³. It is cut into 8 identical smaller cubes. What is the density of each smaller cube? Explain.

Show Answer

Density stays the same.

When you cut the cube, both the mass and the volume of each piece become 1/8 of the original, so the ratio ρ = m/V is unchanged.

Mind stretcher 2: Mixed units (volume in cm³)Extension

A piece of wood has mass 0.60 kg and density 800 kg m⁻³. Find its volume in cm³.

Show Answer

Volume in m³: V = m/ρ = 0.60/800 = 7.5 × 10⁻⁴ m³

Convert to cm³ (since 1 m³ = 10⁶ cm³): V = (7.5 × 10⁻⁴)(10⁶) = 7.5 × 10² = 750 cm³

8. Practice and next step

Complete the Pressure quiz, then continue to What is pressure? to use force and area before studying liquid columns.