Density

This lesson covers the concept of density, its definition, the equation linking density to mass and volume, and the practical methods used to measure the density of regular and irregular solids. This is a key topic in the Edexcel GCSE Combined Science specification (1SC0) Particle Model unit.

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What Is Density?

Density is a measure of how much mass is contained in a given volume. It tells us how tightly packed the particles are in a substance.

$\rho = \frac{m}{V}$ρ=Vm​

Symbol	Quantity	Unit
$\rho$ρ (rho)	Density	kilograms per cubic metre (kg/m³) or grams per cubic centimetre (g/cm³)
$m$m	Mass	kilograms (kg) or grams (g)
$V$V	Volume	cubic metres (m³) or cubic centimetres (cm³)

Exam Tip: The Greek letter $\rho$ρ (rho) is used for density — do not confuse it with the letter p. Make sure your handwriting clearly distinguishes the two.

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Rearranging the Density Equation

You must be able to rearrange the equation for any of the three quantities:

To find	Rearrangement
Density	$\rho = \frac{m}{V}$ρ=Vm​
Mass	$m = \rho \times V$m=ρ×V
Volume	$V = \frac{m}{\rho}$V=ρm​

The Formula Triangle

A useful memory aid is the formula triangle:

graph TD
    A["m"] --- B["ρ × V"]

Cover the quantity you want to find:

• Cover m → $\rho \times V$ρ×V
• Cover $\rho$ρ → $m \div V$m÷V
• Cover V → $m \div \rho$m÷ρ

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Typical Densities

Material	Density (kg/m³)	Density (g/cm³)
Air	1.2	0.0012
Water	1000	1.0
Ice	917	0.917
Aluminium	2700	2.7
Iron/Steel	7800	7.8
Gold	19 300	19.3

Exam Tip: You should know that the density of water is 1000 kg/m³ (or 1.0 g/cm³). This is a very commonly used value in calculations.

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Converting Units

To convert between the two common density units:

• g/cm³ → kg/m³: multiply by 1000
• kg/m³ → g/cm³: divide by 1000

Worked Example 1

A block of aluminium has a mass of 5.4 kg and a volume of 0.002 m³. Calculate its density.

$\rho = \frac{m}{V} = \frac{5.4}{0.002} = 2700 \text{ kg/m³}$ρ=Vm​=0.0025.4​=2700 kg/m³

Worked Example 2

A gold ring has a mass of 19.3 g and a volume of 1.0 cm³. Calculate its density.

$\rho = \frac{m}{V} = \frac{19.3}{1.0} = 19.3 \text{ g/cm³}$ρ=Vm​=1.019.3​=19.3 g/cm³

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Measuring Density of a Regular Solid

For a regularly shaped solid (cuboid, cylinder, sphere):

1. Measure the mass using a balance (in grams).
2. Measure the dimensions using a ruler or vernier calliper.
3. Calculate the volume using the appropriate formula:
   • Cuboid: $V = l \times w \times h$V=l×w×h
   • Cylinder: $V = \pi r^2 h$V=πr2h
   • Sphere: $V = \frac{4}{3} \pi r^3$V=34​πr3
4. Calculate density using $\rho = m / V$ρ=m/V.

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Measuring Density of an Irregular Solid

For an irregularly shaped solid (e.g. a stone, a key), the volume cannot be calculated from dimensions. Instead, use the displacement method:

1. Measure the mass of the object using a balance.
2. Fill a measuring cylinder with water and record the initial volume $V_1$V1​.
3. Carefully lower the object into the water.
4. Record the new volume $V_2$V2​.
5. Volume of object = $V_2 - V_1$V2​−V1​.
6. Calculate density using $\rho = m / V$ρ=m/V.

Alternatively, use a eureka can (displacement can):

1. Fill the can until water reaches the spout.
2. Place a measuring cylinder under the spout.
3. Lower the object into the can.
4. The displaced water collects in the measuring cylinder — this volume equals the volume of the object.

graph LR
    A["Object placed in eureka can"] --> B["Water displaced through spout"]
    B --> C["Volume measured in cylinder"]
    C --> D["ρ = m / V"]

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Floating and Sinking

Whether an object floats or sinks depends on its density compared with the density of the liquid:

Condition	Result
Object density < liquid density	Object floats
Object density > liquid density	Object sinks
Object density = liquid density	Object is neutrally buoyant (stays at any depth)

Why Does Ice Float?

• Ice has a density of about 917 kg/m³, which is less than water at 1000 kg/m³.
• The water molecules in ice form a rigid crystalline structure with gaps, making ice less dense than liquid water.
• This is unusual — most substances are denser as a solid than as a liquid.

Exam Tip: A common exam question asks you to explain why ice floats on water. Always state that ice is less dense than water, and link this to the particle arrangement.

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Worked Example 3

A student measures the mass of a stone as 150 g. She places it in a measuring cylinder containing 40 cm³ of water. The water level rises to 98 cm³. Calculate the density of the stone.

Step 1 — Find the volume:

$V = V_2 - V_1 = 98 - 40 = 58 \text{ cm}^3$V=V2​−V1​=98−40=58 cm3

Step 2 — Calculate density:

$\rho = \frac{m}{V} = \frac{150}{58} = 2.59 \text{ g/cm}^3 \text{ (to 3 s.f.)}$ρ=Vm​=58150​=2.59 g/cm3 (to 3 s.f.)

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Common Misconceptions

Misconception	Correction
Heavy objects always sink	It depends on density, not just mass — a large ship floats because it has a low average density
Density and mass are the same thing	Density is mass per unit volume — two objects can have the same mass but very different densities
All metals sink in water	Some metals (e.g. lithium, sodium, potassium) are less dense than water and float

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Summary

• Density = mass / volume: $\rho = m / V$ρ=m/V.
• Units: kg/m³ or g/cm³.
• Regular solids: measure dimensions and calculate volume.
• Irregular solids: use the displacement method (measuring cylinder or eureka can).
• An object floats if its density is less than the liquid it is placed in; it sinks if its density is greater.
• Ice floats because it is less dense than liquid water due to its open crystalline structure.

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Extended Worked Examples

Worked Example 3 — Mixed units

A copper pipe has a mass of 850 g and a volume of 95 cm³. Calculate its density in kg/m³.

Step 1 — convert mass to kg: $m = 0.850 \text{ kg}$m=0.850 kg.

Step 2 — convert volume to m³: $V = 95 \times 10^{-6} = 9.5 \times 10^{-5} \text{ m³}$V=95×10−6=9.5×10−5 m³.

Step 3 — apply $\rho = m / V$ρ=m/V:

$\rho = \frac{0.850}{9.5 \times 10^{-5}} \approx 8947 \text{ kg/m³}$ρ=9.5×10−50.850​≈8947 kg/m³

The density of pure copper is around 8960 kg/m³, so the answer is sensible.

Exam Tip: When a question mixes grams and cubic centimetres, you can work in g/cm³ throughout and convert only at the end. Converting early is a common source of arithmetic error.

Worked Example 4 — Floating and sinking

Seawater has a density of 1030 kg/m³. Pine wood has a density of 510 kg/m³, and granite has a density of 2750 kg/m³. Predict whether each floats in seawater.

Material	$\rho$ρ (kg/m³)	Floats in seawater?
Pine wood	510	Yes — less dense than seawater
Granite	2750	No — more dense than seawater

Worked Example 5 — Finding mass of a lead sinker

A cylindrical lead weight has a radius of 1.5 cm and a height of 6.0 cm. Lead has a density of 11 340 kg/m³. Calculate its mass in grams.

Step 1 — volume of cylinder: $V = \pi r^2 h = \pi \times 1.5^2 \times 6.0 \approx 42.4 \text{ cm³}$V=πr2h=π×1.52×6.0≈42.4 cm³.

Step 2 — convert density to g/cm³: $11 340 \div 1000 = 11.34 \text{ g/cm³}$11340÷1000=11.34 g/cm³.

Step 3 — rearrange $m = \rho V = 11.34 \times 42.4 \approx 481 \text{ g}$m=ρV=11.34×42.4≈481 g.

Worked Example 6 — Displacement method

An irregular stone is dropped into a measuring cylinder containing 60 cm³ of water. The new reading is 83 cm³. The stone has a mass of 67 g. Calculate its density.

• Volume of stone = $83 - 60 = 23 \text{ cm³}$83−60=23 cm³.
• $\rho = 67 \div 23 \approx 2.9 \text{ g/cm³}$ρ=67÷23≈2.9 g/cm³.

The density is close to that of a typical rock such as basalt.

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Comparing Materials by Density

Category	Typical material	Density (kg/m³)
Gases	Hydrogen	0.09
Gases	Air	1.2
Liquids	Petrol	740
Liquids	Water	1000
Liquids	Seawater	1030
Liquids	Mercury	13 600
Solids	Cork	240
Solids	Ice	917
Solids	Aluminium	2700
Solids	Iron	7800
Solids	Lead	11 340
Solids	Gold	19 300
Solids	Osmium	22 590

Exam Tip: Solids and liquids have similar densities because particles are in contact. Gases are about 1000 times less dense because particles are far apart.

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Density, Particle Spacing and the Particle Model

graph LR
    S["Solid: particles touching, regular lattice, high density"] --> L["Liquid: particles touching, disordered, similar density"]
    L --> G["Gas: particles far apart, very low density"]

The particle model explains why most substances become less dense when they melt or boil. When a solid melts, the spacing between particles increases slightly (density falls slightly). When a liquid boils into a gas, the spacing increases enormously, so the density falls by a factor of around a thousand. Water is an anomaly: ice is less dense than liquid water because water molecules form an open hexagonal lattice when they freeze.

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Common Mistakes Callout

Common Mistake: Writing the units of density as "kg/m" or "g/cm". Density is mass per cubic metre or per cubic centimetre. Always check the cube.

Common Mistake: Forgetting to subtract the initial water level in a displacement experiment — the answer is the change in volume, not the final reading.

Common Mistake: Quoting density to too many significant figures. If your mass is given to 2 s.f., the final density should also be given to 2 s.f.

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Answering at different grade levels

Grade band	What a strong answer looks like
Grades 1–3	Recall $\rho = m/V$ρ=m/V; state units kg/m³ or g/cm³; know that water has density 1000 kg/m³; identify if something floats or sinks using a comparison with water.
Grades 4–5	Rearrange $\rho = m/V$ρ=m/V for mass or volume; describe the displacement method for an irregular solid; explain that objects less dense than water float; convert between g/cm³ and kg/m³.
Grades 6–7	Give full method for measuring density of a regular or irregular solid including apparatus, measurements and uncertainty; relate differences in density to differences in particle spacing; explain why gases are around 1000× less dense than solids or liquids.
Grades 8–9	Analyse experimental data with percentage error and random/systematic uncertainty; justify why ice is less dense than water using hydrogen bonding and the open hexagonal lattice; combine $\rho = m/V$ρ=m/V with volume formulas (cylinder, sphere) to solve multi-step problems; use floating/sinking arguments quantitatively (e.g. fraction submerged = $\rho_{\text{object}} / \rho_{\text{liquid}}$ρobject​/ρliquid​).

Error Analysis in a Density Experiment

In the Core Practical on density, the main sources of error are:

Source	Type	Effect
Water clinging to the object when removed	Systematic	Mass too high → density too high
Parallax error reading meniscus	Random	Slight scatter
Object partially absorbing water	Systematic	Mass too high after immersion
Air bubbles on the object	Systematic	Volume too high → density too low
Uneven container walls	Systematic	Volume of regular solid miscalculated

Students should repeat readings and take a mean, reject obvious anomalies, and quote final density values to appropriate significant figures.

Exam Tip: If asked to suggest an improvement, aim for something specific such as "dry the object with filter paper between the mass and volume measurements" rather than a vague "be more careful".

Edexcel alignment: This content is aligned with Edexcel GCSE Combined Science (1SC0) Physics Topic 14 Particle model / Topic 6 Radioactivity — specifically CP15 Particle model (Core Practical — investigate the densities of solids and liquids), CP16 Forces and matter, and specification points on density, particle arrangement and the particle model of matter. Assessed on Physics Papers 1 and 2.