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 opens Topic P1 (Matter) of OCR Gateway Science A by defining density, working through the equation in both directions, sorting out the units, and walking 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.

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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.

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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.

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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.

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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.

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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​.

An alternative is to part-fill a measuring cylinder with water, read the level, lower the object in, and read the new level: the rise in level is the object's volume. Either way, the object must be fully submerged and must sink for the method to work.

Worked Example 6 — Density of an irregular stone

A small stone has a mass of $54 \text{ g}$54 g. When lowered into a eureka can, it displaces $20 \text{ cm}^3$20 cm3 of water. Calculate the density of the stone.

Step 1 — the displaced water volume equals the stone's volume: $V = 20 \text{ cm}^3$V=20 cm3.

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

Step 3 — substitute and calculate: $\rho = \dfrac{54}{20} = 2.7 \text{ g/cm}^3$ρ=2054​=2.7 g/cm3.

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

Exam Tip: In displacement, the volume of water pushed out equals the volume of the object. Read the measuring cylinder at eye level, with your eye level with the bottom of the meniscus, to avoid a parallax error.

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Measuring the Density of a Liquid

To find the density of a liquid you measure a known volume of it and find that volume's mass.

Method (numbered):

1. Place an empty measuring cylinder on a balance and zero (tare) the balance so it reads $0 \text{ g}$0 g with the empty cylinder on it.
2. Pour a measured volume of the liquid — say $50 \text{ cm}^3$50 cm3 — into the cylinder, reading the volume at eye level from the bottom of the meniscus.
3. Read the mass of the liquid directly from the balance.
4. Calculate the density using $\rho = \dfrac{m}{V}$ρ=Vm​.

If your balance cannot be zeroed with the cylinder on it, weigh the empty cylinder first, then weigh it with the liquid, and subtract to find the mass of liquid alone.

Worked Example 7 — Density of a liquid

A measuring cylinder is zeroed, then $50 \text{ cm}^3$50 cm3 of cooking oil is poured in. The balance reads $46 \text{ g}$46 g. Calculate the density of the oil.

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

Step 2 — substitute: $\rho = \dfrac{46}{50}$ρ=5046​.

Step 3 — calculate: $\rho = 0.92 \text{ g/cm}^3$ρ=0.92 g/cm3.

Answer: the oil has a density of $0.92 \text{ g/cm}^3$0.92 g/cm3. Because this is less than water's $1.0 \text{ g/cm}^3$1.0 g/cm3, the oil floats on water.

Exam Tip: Remember to subtract the mass of the empty container (or zero the balance with it on) — forgetting this is the single most common mistake in the liquid-density practical, and it makes every density come out too high.

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Why Solids Are Usually Denser Than Gases: A Particle View

Density depends on how closely the particles are packed and on the mass of each particle. The same kind of particle (say, the molecules of a substance) can be arranged very differently in a solid, a liquid and a gas, and this is what makes the densities so different.

• In a solid, the particles are packed closely together, touching, in a regular arrangement. A large number of particles — and so a large mass — fits into a small volume, giving a high density.
• In a liquid, the particles are still close together and touching but arranged more randomly. The density is only slightly lower than the solid (usually a little less, though water is an exception).
• In a gas, the particles are far apart, with large empty spaces between them. Only a few particles — and so very little mass — occupy a given volume, giving a very low density, typically about a thousand times less than the solid or liquid.

So a substance is usually most dense as a solid and least dense as a gas, simply because the particles are squeezed together most tightly in the solid and spread out most widely in the gas. This is why a room full of air weighs only a little, while a same-sized block of metal would be impossibly heavy: the metal has its atoms packed tightly, while the air has them spread thinly through the space.

State	Particle spacing	Typical density
Solid	Very close, touching, regular	High
Liquid	Close, touching, random	Slightly lower than solid
Gas	Far apart, large gaps	Very low (≈ 1000× less)

Exam Tip: Link density to particle spacing: gases are far less dense than solids because their particles are much further apart, so far fewer particles (and so less mass) fit into the same volume.

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Common Misconceptions

Misconception	The correct idea
"A bigger object always has a higher density"	Density is mass per unit volume; a large and a small block of the same material have the same density
"Heavy objects are always dense"	A heavy object can have low density if it is also large (e.g. a wardrobe); density compares mass to volume, not mass alone
"To convert $\text{g/cm}^3$g/cm3 to $\text{kg/m}^3$kg/m3 you divide by 1000"	You multiply by 1000; water is $1 \text{ g/cm}^3 = 1000 \text{ kg/m}^3$1 g/cm3=1000 kg/m3
"Density is volume divided by mass"	Density is mass divided by volume, $\rho = \dfrac{m}{V}$ρ=Vm​
"The displaced water volume must be weighed"	The displaced water's volume equals the object's volume; you do not weigh the water
"You can find an irregular object's volume with a ruler"	Irregular shapes have no simple formula — use displacement instead

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Model Answer — Measuring the Density of an Irregular Object

Question (6 marks): Describe how you would measure the density of a small irregular pebble accurately. Include the apparatus you would use, the measurements you would take, and how you would calculate the density.

Mid-band response: "Weigh the pebble on a balance to find its mass. Put it in water and see how much the water goes up — that is the volume. Then do density = mass ÷ volume."

Examiner-style commentary: The three essential ideas (mass by balance, volume by displacement, $\rho = m/V$ρ=m/V) are present but the method is vague. To climb a band, name the apparatus (measuring cylinder or eureka can), say the volume of water displaced equals the volume of the pebble, and mention reading at eye level.

Stronger response: "Measure the mass of the pebble using a balance. Part-fill a measuring cylinder with water and read the volume. Lower the pebble in on a thread until it is fully submerged, and read the new volume. The increase in volume is the volume of the pebble. Then calculate the density using $\rho = \dfrac{m}{V}$ρ=Vm​."

Examiner-style commentary: A clear, correct method that names the apparatus and explains how the volume is found. To reach the top band, add a precision detail (read the meniscus at eye level), state the units, and note that the pebble must sink and be fully submerged.

Top-band response: "First, measure the mass of the pebble with a digital balance in grams, zeroing the balance first. To find the volume, use displacement: part-fill a measuring cylinder with water and record the level at eye level from the bottom of the meniscus. Lower the pebble in gently on a thread until it is fully submerged, and record the new level. The rise in water level equals the volume of the pebble in $\text{cm}^3$cm3, because the pebble displaces its own volume of water. Finally, calculate the density from $\rho = \dfrac{m}{V}$ρ=Vm​, giving the answer in $\text{g/cm}^3$g/cm3. Reading at eye level avoids a parallax error, and the pebble must sink for the method to work."

Examiner-style commentary: Full marks. It names every piece of apparatus, distinguishes the mass and volume measurements, explains why displacement gives the volume, gives the equation with units, and adds the precision and validity points examiners reward.

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Exam Tips

• Density is mass per unit volume: $\rho = \dfrac{m}{V}$ρ=Vm​; rearrange to $m = \rho V$m=ρV and $V = \dfrac{m}{\rho}$V=ρm​.
• Convert $\text{g/cm}^3 \to \text{kg/m}^3$g/cm3→kg/m3 by multiplying by 1000; convert back by dividing by 1000.
• For a regular solid, find volume from its dimensions; for an irregular solid, use water displacement.
• For a liquid, zero the balance with the empty cylinder, then read the mass of a known volume.
• Always show equation, substitution, answer with unit, and read scales at eye level.
• Solids are denser than gases because their particles are packed far more closely together.

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Summary

• Density is the mass per unit volume of a material, $\rho = \dfrac{m}{V}$ρ=Vm​, measured in $\text{kg/m}^3$kg/m3 or $\text{g/cm}^3$g/cm3.
• The equation rearranges to $m = \rho V$m=ρV and $V = \dfrac{m}{\rho}$V=ρm​.
• $1 \text{ g/cm}^3 = 1000 \text{ kg/m}^3$1 g/cm3=1000 kg/m3: multiply by 1000 going from $\text{g/cm}^3$g/cm3 to $\text{kg/m}^3$kg/m3.
• A regular solid's volume comes from its dimensions; an irregular solid's from displacement; a liquid's density from a known volume weighed on a balance.
• Solids are denser than gases because the particles are packed closely in a solid but far apart in a gas.

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This content is aligned with OCR Gateway Science A GCSE Physics (J249), Topic P1 Matter (density; the density equation; measuring the density of solids and liquids — required practical). Refer to the official OCR specification document for the exact wording.