Density of Materials

Hold a small block of iron in one hand and a same-sized block of expanded polystyrene in the other, and the iron feels enormously heavier — even though the two blocks take up exactly the same amount of space. The property that captures this difference is density: it tells you how much mass is packed into each unit of volume of a material. Density explains why some objects float and others sink, why a kilogram of feathers fills a much bigger bag than a kilogram of lead, and how engineers choose materials for everything from aircraft wings to ship hulls. This lesson, part of Topic P1 (Matter) of OCR Gateway Combined Science A, defines density, works through the equation in both directions, sorts out the units, and walks through the required practical in which you measure the density of solids and a liquid.

By the end of this lesson you should be able to define density, use and rearrange the density equation, convert between $\text{kg/m}^3$kg/m3 and $\text{g/cm}^3$g/cm3, describe how to measure the density of a regular solid, an irregular solid and a liquid, and explain in terms of particles why a solid is usually much denser than a gas.

This lesson is AO1 for defining density and the required-practical methods, strongly AO2 for applying and rearranging $\rho = \dfrac{m}{V}$ρ=Vm​ and converting units, and AO3 when you evaluate measured data — for example judging which object is denser from mass and volume readings, or spotting a source of error in the displacement method.

---

What Is Density?

Density is the mass per unit volume of a material — in plain words, how much "stuff" (mass) is squeezed into a given amount of space (volume). A dense material has a lot of mass packed into a small volume; a material of low density has only a little mass in the same volume.

The symbol for density is the Greek letter rho, written $\rho$ρ. It is defined by the equation:

$\rho = \dfrac{m}{V}$ρ=Vm​

where $\rho$ρ is the density, $m$m is the mass and $V$V is the volume. Because density is a property of the material and not of the size of the sample, a tiny gold ring and a huge gold bar have exactly the same density — the bar simply has more mass and proportionally more volume, so the ratio $m/V$m/V is unchanged.

Exam Tip: Density is mass per unit volume, so the equation is always $\rho = \dfrac{m}{V}$ρ=Vm​ — mass divided by volume, never the other way round. If you ever get a silly answer, check you have not accidentally divided volume by mass.

---

The Density Equation and Rearranging It

The density equation can be rearranged to make mass or volume the subject, depending on what the question asks for:

$\rho = \dfrac{m}{V} \qquad m = \rho V \qquad V = \dfrac{m}{\rho}$ρ=Vm​m=ρVV=ρm​

A formula triangle is a handy memory aid: cover the quantity you want and the triangle shows you the calculation.

Mass sits at the top, so $m = \rho V$m=ρV; density and volume sit side by side at the bottom, so $\rho = \dfrac{m}{V}$ρ=Vm​ and $V = \dfrac{m}{\rho}$V=ρm​.

Worked Example 1 — Finding density from mass and volume

A block of aluminium has a mass of $540 \text{ g}$540 g and a volume of $200 \text{ cm}^3$200 cm3. Calculate its density.

Step 1 — write the equation: $\rho = \dfrac{m}{V}$ρ=Vm​.

Step 2 — substitute the values: $\rho = \dfrac{540}{200}$ρ=200540​.

Step 3 — calculate: $\rho = 2.7 \text{ g/cm}^3$ρ=2.7 g/cm3.

Answer: the density of aluminium is $2.7 \text{ g/cm}^3$2.7 g/cm3.

Worked Example 2 — Rearranging for mass

A copper pipe has a volume of $30 \text{ cm}^3$30 cm3. Copper has a density of $8.9 \text{ g/cm}^3$8.9 g/cm3. Calculate the mass of the pipe.

Step 1 — rearrange to make mass the subject: $m = \rho V$m=ρV.

Step 2 — substitute: $m = 8.9 \times 30$m=8.9×30.

Step 3 — calculate: $m = 267 \text{ g}$m=267 g.

Answer: the pipe has a mass of $267 \text{ g}$267 g.

Worked Example 3 — Rearranging for volume

A quantity of mercury has a mass of $1360 \text{ g}$1360 g. The density of mercury is $13.6 \text{ g/cm}^3$13.6 g/cm3. Calculate its volume.

Step 1 — rearrange to make volume the subject: $V = \dfrac{m}{\rho}$V=ρm​.

Step 2 — substitute: $V = \dfrac{1360}{13.6}$V=13.61360​.

Step 3 — calculate: $V = 100 \text{ cm}^3$V=100 cm3.

Answer: the mercury occupies $100 \text{ cm}^3$100 cm3.

Exam Tip: Always lay calculations out in three lines — equation, substitution, answer with unit. Examiners award marks for each stage, so even if you slip on the arithmetic you can still pick up the method marks if the equation and substitution are shown.

---

Units of Density and Converting Between Them

Density is a mass divided by a volume, so its unit is a mass unit divided by a volume unit. The two you must know are:

• $\text{kg/m}^3$kg/m3 (kilograms per cubic metre) — the standard SI unit, used on the OCR data sheet;
• $\text{g/cm}^3$g/cm3 (grams per cubic centimetre) — convenient for small laboratory samples.

These two units are not the same size, and a very common exam task is to convert between them. The key fact is:

$1 \text{ g/cm}^3 = 1000 \text{ kg/m}^3$1 g/cm3=1000 kg/m3

So to change a density from $\text{g/cm}^3$g/cm3 into $\text{kg/m}^3$kg/m3 you multiply by 1000, and to go the other way you divide by 1000. The reason for the factor of 1000 is worth understanding rather than just memorising: although $1 \text{ kg} = 1000 \text{ g}$1 kg=1000 g (which on its own would divide the number by 1000), there are $1\,000\,000 \text{ cm}^3$1000000 cm3 in $1 \text{ m}^3$1 m3, and the volume conversion is the larger effect, so overall the number is multiplied by 1000.

Density in $\text{g/cm}^3$g/cm3	Density in $\text{kg/m}^3$kg/m3
Water — 1.0	1000
Aluminium — 2.7	2700
Iron — 7.9	7900
Copper — 8.9	8900
Gold — 19.3	19 300

Worked Example 4 — Converting units

The density of ice is $0.92 \text{ g/cm}^3$0.92 g/cm3. Express this in $\text{kg/m}^3$kg/m3.

Step 1 — recall the conversion: multiply $\text{g/cm}^3$g/cm3 by 1000 to get $\text{kg/m}^3$kg/m3.

Step 2 — calculate: $0.92 \times 1000 = 920 \text{ kg/m}^3$0.92×1000=920 kg/m3.

Answer: ice has a density of $920 \text{ kg/m}^3$920 kg/m3. Notice this is less than the $1000 \text{ kg/m}^3$1000 kg/m3 of liquid water, which is exactly why ice floats.

Exam Tip: Going from $\text{g/cm}^3$g/cm3 to $\text{kg/m}^3$kg/m3 the number gets bigger (×1000); going the other way it gets smaller (÷1000). A quick sanity check: water is $1 \text{ g/cm}^3$1 g/cm3 and $1000 \text{ kg/m}^3$1000 kg/m3 — if your converted figure does not sit sensibly relative to water, you have multiplied when you should have divided.

---

Measuring the Density of a Regular Solid

This is the first part of the P1 required practical on density. A regular solid is one with a simple shape — a cube, a cuboid (rectangular block) or a cylinder — whose volume can be calculated from measurements of its dimensions.

Method (numbered):

1. Use a balance to measure the mass of the object in grams. Place the balance on a flat surface and zero (tare) it first.
2. Use a ruler, or better a vernier caliper or micrometer for greater precision, to measure the lengths needed to find the volume. For a cuboid, measure the length, width and height; for a cylinder, measure the diameter and length.
3. Calculate the volume from the dimensions. For a cuboid, $V = \text{length} \times \text{width} \times \text{height}$V=length×width×height. For a cylinder, $V = \pi r^2 h$V=πr2h.
4. Calculate the density using $\rho = \dfrac{m}{V}$ρ=Vm​.

Worked Example 5 — Density of a metal cuboid

A steel block measures $4 \text{ cm} \times 3 \text{ cm} \times 2 \text{ cm}$4 cm×3 cm×2 cm and has a mass of $189.6 \text{ g}$189.6 g. Calculate its density.

Step 1 — find the volume: $V = 4 \times 3 \times 2 = 24 \text{ cm}^3$V=4×3×2=24 cm3.

Step 2 — write the density equation: $\rho = \dfrac{m}{V}$ρ=Vm​.

Step 3 — substitute and calculate: $\rho = \dfrac{189.6}{24} = 7.9 \text{ g/cm}^3$ρ=24189.6​=7.9 g/cm3.

Answer: the density is $7.9 \text{ g/cm}^3$7.9 g/cm3, which matches the density of steel.

Exam Tip: For a regular solid, measure mass with a balance and volume from the dimensions — there is no need for water displacement. Using a caliper rather than a ruler reduces the percentage uncertainty in the small lengths, improving the precision of the final density.

---

Measuring the Density of an Irregular Solid (Displacement)

An irregular solid — a stone, a crown, an oddly-shaped piece of metal — has no simple formula for its volume, so you find the volume by displacement of water. The object pushes aside (displaces) a volume of water exactly equal to its own volume.

Method using a eureka (displacement) can:

1. Measure the mass of the object on a balance.
2. Fill a eureka can (a beaker with a spout) with water until water just runs out of the spout, then let it stop dripping.
3. Place an empty measuring cylinder under the spout.
4. Gently lower the object into the can on a thread. The water it displaces runs out of the spout and is collected in the measuring cylinder.
5. Read the volume of water collected — this equals the volume of the object (since $1 \text{ cm}^3$1 cm3 of water has a volume of $1 \text{ cm}^3$1 cm3).
6. Calculate the density using $\rho = \dfrac{m}{V}$ρ=Vm​.