Physics · 1 Motion, forces and energy

1.4 Density

Before you start

Mass is measured with a balance, in g or kg, and $1$ kg $= 1000$ g.
The volume of a cuboid is length × width × height, and volume is measured in cm³ or m³.
Reading the scale of a measuring cylinder, which is marked in cm³ or ml.
Rearranging a three-letter equation such as $F = m a$ to make any one letter the subject.
Standard form, e.g. $4.0 \times 10^{- 5}$ for $0.000 040$ .

Syllabus objectives covered (4)

Define density as mass per unit volume; recall and use the equation ρ = m / V
Describe how to determine the density of a liquid, of a regularly shaped solid and of an irregularly shaped solid which sinks in a liquid (volume by displacement), including appropriate calculations
Determine whether an object floats based on density data
Determine whether one liquid will float on another liquid based on density data given that the liquids do not mix

Part 1 of 7

Density is how much mass is packed into each unit of volume

A kilogram of steel and a kilogram of polystyrene have the same mass, but the polystyrene takes up far more space. Density is the quantity that captures that difference: how much mass is packed into a given volume.
Density is defined as mass per unit volume. 'Per unit volume' means 'for each cubic centimetre' or 'for each cubic metre'. If a question asks you to Define density, write those words, not just the symbols.
$ρ = \frac{m}{V}$
$ρ$ is the Greek letter rho. $m$ is the mass and $V$ is the volume. Paper 4 gives you no formula sheet, so you must know this equation by heart.
You also need it the other way round. Multiply both sides by $V$ to get the mass; then divide both sides of that by $ρ$ to get the volume. Each move keeps the equation balanced because you do the same thing to both sides.
$m = ρ V V = \frac{m}{ρ}$
Density belongs to the material, not to the size of the object. Cut a block in half and you halve its mass and you halve its volume, so mass ÷ volume stays the same. Two pieces cut from the same block have the same density, however different their sizes.

Part 2 of 7

Units: g/cm³ and kg/m³, and getting from one to the other

The unit of density comes straight from the equation: a mass unit divided by a volume unit. Mass in kg and volume in m³ give kg/m³, the SI unit. Mass in g and volume in cm³ give g/cm³.
Measuring cylinders are marked in cm³ or in ml. These are the same volume: $1$ ml $= 1$ cm³.
A cubic metre is a cube $100$ cm long on each side, so it holds $100 \times 100 \times 100$ cubic centimetres. A cubic metre is far bigger than it sounds.
$1 m^{3} = 1 000 000 {cm}^{3} = 10^{6} {cm}^{3}$
Now the conversion. $1$ g/cm³ means $1$ g in every cm³. A cubic metre holds $10^{6}$ cm³, so it holds $10^{6}$ g, which is $1000$ kg. To go from g/cm³ to kg/m³, multiply by $1000$ ; to go back, divide by $1000$ .
$1 {g/cm}^{3} = 1000 {kg/m}^{3}$
Water has a density of about $1.0$ g/cm³, which is $1000$ kg/m³. Keep this number in mind: it is the benchmark you compare against whenever something is put in water.
The trap is dividing a mass in one system by a volume in the other. A mass in kg divided by a volume in cm³ gives a number in neither g/cm³ nor kg/m³. Before you divide, make the pair match: g with cm³, or kg with m³.

Part 3 of 7

Finding the density of a liquid

For any density measurement you need two things: a mass and a volume. A balance gives the mass and a measuring cylinder gives the volume. The only question is how to get each one.
A liquid cannot sit on a balance without a container, and the container has mass too. So first measure the mass of the empty measuring cylinder.
Pour some of the liquid into the cylinder and read the volume. Water's surface curves where it meets the glass; this curve is the meniscus. Read the scale at the bottom of the meniscus, with your eye level with it, so that you avoid parallax error.
Measure the mass of the cylinder with the liquid in it. The mass of the liquid is the difference between the two balance readings.
$m_{liquid} = m_{cylinder + liquid} - m_{empty cylinder}$
Then divide: $ρ = m / V$ , using the mass of the liquid alone and the volume you read.
If the balance has a zero (tare) button, you can press it with the empty cylinder on the pan. The balance then reads the mass of the liquid directly. Either method is correct; in a written answer, say which one you used.

Part 4 of 7

Finding the density of a regularly shaped solid

The mass is easy: put the solid on a balance.
A regular shape, such as a cuboid, has a volume you can calculate from its measurements. Measure the length, width and height with a ruler and multiply them.
$V = length \times width \times height$
For other regular shapes, use the matching volume formula from maths. For a cylinder of radius $r$ and height $h$ , for example, $V = π r^{2} h$ .
Keep the units consistent from the start. Lengths in cm give a volume in cm³; lengths in m give m³. Mixing a length in mm with one in cm gives a volume in no standard unit.
Then $ρ = m / V$ . In a Describe answer, name both instruments (balance, ruler), say what you measure with each, and finish with the calculation.

Part 5 of 7

An irregular solid that sinks: volume by displacement

A stone has no formula for its volume. Instead you measure the volume of water it pushes aside. An object that is completely under water displaces a volume of water equal to its own volume.
Measure the mass of the stone on a balance first, while it is dry. Water left on it after it has been in the cylinder would add to the reading.
Partly fill a measuring cylinder with water and read the volume, $V_{1}$ . Tie the stone to a thread and lower it gently until it is completely under the water. Read the new volume, $V_{2}$ .
apparatus diagram — not drawn yet
Two measuring cylinders side by side. Left: water filled to the 35 cm³ mark, labelled V1 = 35 cm³. Right: the same cylinder with a stone hanging on a thread completely below the water surface, the water now at the 52 cm³ mark, labelled V2 = 52 cm³. An eye drawn level with the bottom of the meniscus on the right.
The rise in the water level, 52 − 35 = 17 cm³, is the volume of the stone.
The rise in the reading is the volume of the stone. Then $ρ = m / V$ as before.
$V_{stone} = V_{2} - V_{1}$
If the object is too big for a measuring cylinder, use a displacement can (an overflow can). Fill the can with water until it runs out of the spout, wait for the dripping to stop, then put an empty measuring cylinder under the spout. Lower the object in. The volume of water collected in the cylinder is the volume of the object.
Each precaution protects one of the two readings. Lower the object gently so that no water splashes out, which would make the volume too small. Check that no air bubbles cling to it, which would make the volume too large.
The objective says 'which sinks' for a reason. The method only works if the whole object is under the surface. A floating object sits partly above the water, so it displaces less water than its own volume, and the measured volume would be too small.

Part 6 of 7

Will it float? Compare two densities

To decide whether an object floats in a liquid, compare the density of the object with the density of the liquid. If the object is less dense than the liquid, it floats. If it is more dense, it sinks.
Mass on its own does not decide it. A steel ship floats, but a small steel ball bearing dropped into water sinks. What matters is mass compared with volume.
For a hollow object, use its average density: its total mass divided by its total volume, including the air inside. A ship encloses a large volume of air, so its average density is less than that of water, even though steel is denser than water.
$ρ_{average} = \frac{total mass}{total volume}$
Put both densities in the same unit before you compare them. $0.80$ g/cm³ and $800$ kg/m³ are the same density; $0.80$ and $1000$ are not comparable numbers until one is converted.
A full answer quotes both densities and the comparison, then the conclusion: 'the block's density, $0.65$ g/cm³, is less than the density of water, $1.0$ g/cm³, so the block floats.' A bare 'it floats' gives no reason.

Part 7 of 7

One liquid floating on another

Liquids that do not mix follow the same rule as solids: the less dense liquid floats on the denser one. Liquids that do not mix are called immiscible.
Oil and water do not mix. Pour them into a jar and the oil forms a layer on top of the water, because the oil is less dense.
With three or more immiscible liquids, they settle into layers in order of density: densest at the bottom, least dense at the top. Neither the order in which you pour them nor how much of each you pour changes the final order.
apparatus diagram — not drawn yet
A tall measuring cylinder holding three liquid layers labelled from bottom to top: R (1.60 g/cm³), Q (1.00 g/cm³), P (0.80 g/cm³). Four small solid objects: one floating on the top surface (0.70 g/cm³), one resting at the P/Q boundary (0.90 g/cm³), one resting at the Q/R boundary (1.20 g/cm³), one on the bottom of the cylinder (2.00 g/cm³).
Densest at the bottom. Each object sinks through the liquids less dense than itself and stops on the first denser one.
The condition in the objective matters. Liquids that mix, such as ethanol and water, form a single liquid instead of separate layers, so the layer rule does not apply to them.
A solid dropped into the layers sinks through every liquid that is less dense than itself, and floats on the first liquid that is denser. So a solid whose density lies between two layers comes to rest at the boundary between them.

Worked examples

Example 1. A metal block is a cuboid measuring $5.0$ cm × $4.0$ cm × $2.0$ cm. Its mass is $316$ g. Calculate its density in g/cm³ and in kg/m³.

Show the working

The block is a regular shape, so find its volume by multiplying the three lengths. They are all in cm, so the volume comes out in cm³.
$V = 5.0 \times 4.0 \times 2.0 = 40 {cm}^{3}$
Mass is in g and volume in cm³: a matching pair, so divide.
$ρ = \frac{m}{V} = \frac{316}{40} = 7.9 {g/cm}^{3}$
Convert to kg/m³ by multiplying by $1000$ , because $1$ g/cm³ $= 1000$ kg/m³.
$7.9 \times 1000 = 7900 {kg/m}^{3}$
Check it the long way, in SI units throughout: $316$ g $= 0.316$ kg and $40$ cm³ $= 40 \times 10^{- 6}$ m³ $= 4.0 \times 10^{- 5}$ m³.
$ρ = \frac{0.316}{4.0 \times 10^{- 5}} = 7900 {kg/m}^{3}$

Answer: $7.9$ g/cm³, which is $7900$ kg/m³.

Example 2. A stone has a mass of $45.9$ g. A measuring cylinder contains water up to the $35$ cm³ mark. When the stone is lowered in on a thread until it is completely under water, the reading rises to $52$ cm³. Calculate the density of the stone.

Show the working

The stone sinks and is completely under the water, so the volume of water it displaces equals its own volume. That is the rise in the reading, not the final reading.
$V = V_{2} - V_{1} = 52 - 35 = 17 {cm}^{3}$
Mass in g, volume in cm³, so the answer will be in g/cm³.
$ρ = \frac{m}{V} = \frac{45.9}{17} = 2.7 {g/cm}^{3}$
Sense check: $2.7$ g/cm³ is more than water's $1.0$ g/cm³, which agrees with the stone sinking. If you had divided by $52$ instead of $17$ you would get about $0.88$ g/cm³, a density that says the stone should float. A result that contradicts what you saw is a sign of a wrong subtraction.

Answer: $2.7$ g/cm³ (which is $2700$ kg/m³).

Example 3. An empty measuring cylinder has a mass of $84.0$ g. A liquid is poured in up to the $50.0$ cm³ mark, and the cylinder and liquid together have a mass of $130.0$ g. Calculate the density of the liquid in kg/m³.

Show the working

The balance weighed the cylinder too, so subtract the empty cylinder to get the mass of the liquid alone.
$m = 130.0 - 84.0 = 46.0 g$
Divide by the volume read from the cylinder.
$ρ = \frac{46.0}{50.0} = 0.92 {g/cm}^{3}$
The question asks for kg/m³, so multiply by $1000$ .
$0.92 \times 1000 = 920 {kg/m}^{3}$

Answer: $920$ kg/m³ (which is $0.92$ g/cm³).

Example 4. Three liquids that do not mix have densities P $0.80$ g/cm³, Q $1.00$ g/cm³ and R $1.60$ g/cm³. They are poured into a tall cylinder and left to settle. Four small solid objects, of densities $0.70$ , $0.90$ , $1.20$ and $2.00$ g/cm³, are then dropped in. Determine the order of the layers and where each object comes to rest.

Show the working

The liquids do not mix, so they form layers in order of density. Densest at the bottom: R ( $1.60$ ), then Q ( $1.00$ ), then P ( $0.80$ ) on top.
Now take each object in turn and ask, layer by layer from the top, 'is it denser than this liquid?' If yes, it sinks through; if no, it floats on it.
$0.70$ g/cm³: less dense than P, the top layer, so it floats on the top surface.
$0.90$ g/cm³: denser than P ( $0.80$ ), so it sinks through P; less dense than Q ( $1.00$ ), so it floats on Q. It rests at the P/Q boundary.
$1.20$ g/cm³: denser than P and Q, so it sinks through both; less dense than R ( $1.60$ ), so it rests at the Q/R boundary.
$2.00$ g/cm³: denser than all three liquids, so it sinks to the bottom of the cylinder.

Answer: Layers from the bottom: R, Q, P. The $0.70$ g/cm³ object floats on the surface of P; the $0.90$ object rests at the P/Q boundary; the $1.20$ object rests at the Q/R boundary; the $2.00$ object sinks to the bottom.

Where people go wrong

Mistake: Taking the final measuring cylinder reading (e.g. $52$ cm³) as the volume of the stone.

Instead: The volume of the stone is the rise in the reading: final minus initial, $V_{2} - V_{1}$ .

The final reading includes the water that was already in the cylinder. Only the extra volume was pushed up by the stone.

Mistake: Deciding that heavy things sink and light things float.

Instead: Compare the object's density with the liquid's density. Less dense floats; more dense sinks.

A steel ship has an enormous mass and floats; a steel ball bearing has a tiny mass and sinks. Mass on its own tells you nothing until you compare it with volume.

Mistake: Thinking a smaller piece of a material has a smaller density.

Instead: A piece cut from a block has the same density as the whole block.

Cutting it reduces the mass and the volume by the same factor, so mass ÷ volume is unchanged.

Mistake: Dividing a mass in kg by a volume in cm³, or a mass in g by a volume in m³.

Instead: Convert first so the units match: g with cm³ gives g/cm³; kg with m³ gives kg/m³.

$1$ m³ is $10^{6}$ cm³, so a mismatched pair gives an answer that is wrong by a factor of $1000$ or more, and in no recognised unit.

Mistake: Thinking the liquid poured in first, or the one there is most of, ends up at the bottom.

Instead: Immiscible liquids settle in order of density alone: densest at the bottom, least dense at the top.

Whether a liquid floats on another depends on comparing densities, and density does not depend on the amount of liquid or the order of pouring.

Check yourself

Define density and give the equation that links density, mass and volume.
Answer
Density is mass per unit volume: $ρ = m / V$ .
A concrete slab has a volume of $0.60$ m³ and a mass of $1440$ kg. Calculate its density.
Answer
$ρ = m / V = 1440 / 0.60 = 2400$ kg/m³.
An oil has a density of $0.92$ g/cm³. Calculate the mass of $250$ cm³ of the oil.
Answer
$m = ρ V = 0.92 \times 250 = 230$ g.
Convert $2.7$ g/cm³ into kg/m³, and $13 600$ kg/m³ into g/cm³.
Answer
$2.7 \times 1000 = 2700$ kg/m³. $13 600 \div 1000 = 13.6$ g/cm³.
A measuring cylinder reads $60.0$ cm³. Six identical steel ball bearings with a total mass of $62.4$ g are dropped in, and the reading rises to $68.0$ cm³. Calculate the density of the steel, and suggest why six ball bearings are used rather than one.
Answer
$V = 68.0 - 60.0 = 8.0$ cm³, so $ρ = 62.4 / 8.0 = 7.8$ g/cm³. One ball bearing would raise the level by only about $1.3$ cm³, which is hard to read accurately on the scale; six give a larger change, so the reading is a more precise measure of the volume.
Explain why the displacement method in a measuring cylinder cannot be used directly to find the volume of a piece of cork that floats in water.
Answer
The cork floats with part of it above the surface, so it does not displace its whole volume. The rise in the water level would be smaller than the volume of the cork.
A sealed hollow steel container has a mass of $3.0$ kg and an external volume of $0.0040$ m³. Determine whether it floats in water of density $1000$ kg/m³.
Answer
Average density $= 3.0 / 0.0040 = 750$ kg/m³. This is less than $1000$ kg/m³, so the container floats.
Two liquids that do not mix have densities $1.15$ g/cm³ and $0.87$ g/cm³. Which forms the upper layer? Where does a bead of density $1.05$ g/cm³ come to rest?
Answer
The $0.87$ g/cm³ liquid is less dense, so it forms the upper layer. The bead is denser than $0.87$ so it sinks through the upper layer, but less dense than $1.15$ so it floats on the lower one: it rests at the boundary between the two liquids.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A student wants to find the density of a small stone. The stone sinks in water. She has a balance, a measuring cylinder, water and some thread.

(a) Define density. 1 mark No calculator

Mark scheme

B1 mass per unit volume (accept mass ÷ volume)

(b) Describe how she can use the measuring cylinder and water to determine the volume of the stone. 3 marks No calculator

Mark scheme

B1 partly fill the measuring cylinder with water and read the volume
B1 lower the stone (on the thread) until it is completely under the water and read the new volume
B1 volume of stone = new reading − first reading, owtte

(c) The reading on the measuring cylinder rises from $42$ cm³ to $57$ cm³. The mass of the stone is $39$ g. Calculate the density of the stone. 3 marks

Mark scheme

C1 volume = 57 − 42 = 15 cm³
C1 ρ = m / V or 39 ÷ 15 seen, ecf for the volume
A1 2.6 g/cm³ (accept 2600 kg/m³); unit required

(d) Suggest one precaution the student should take when reading the volume on the measuring cylinder. 1 mark No calculator

Mark scheme

B1 any one from: eye level with the water surface (to avoid parallax error) / read the bottom of the meniscus / cylinder standing on a level surface

Question 2. An empty measuring cylinder has a mass of $95.0$ g. A liquid is poured in up to the $120$ cm³ mark. The cylinder and liquid now have a mass of $203.0$ g. The liquid does not mix with water. The density of water is $1000$ kg/m³.

(a) Calculate the density of the liquid in kg/m³. 3 marks

Mark scheme

C1 mass of liquid = 203.0 − 95.0 = 108 g
C1 108 ÷ 120 or 0.90 g/cm³ seen, or 0.108 ÷ (1.20 × 10⁻⁴) seen, ecf for the mass
A1 900 kg/m³; unit required

(b) Some of the liquid is gently poured into a beaker of water and left to settle. Predict what is seen, and give a reason. 2 marks No calculator

Mark scheme

B1 the liquid forms a layer on top of the water / floats on the water
B1 because the density of the liquid is less than the density of water, ecf from (a), ORA

(c) A small plastic bead of density $950$ kg/m³ is dropped into the beaker. Deduce where the bead comes to rest. 2 marks No calculator

Mark scheme

B1 it sinks through the (top) layer of liquid because 950 kg/m³ is greater than the density of the liquid, ecf from (a)
B1 it floats on the water / rests at the boundary between the two liquids because 950 kg/m³ is less than 1000 kg/m³

Question 3. A sealed hollow metal box has a mass of $2.4$ kg and an external volume of $3000$ cm³. The metal it is made from has a density of $7900$ kg/m³. The density of water is $1000$ kg/m³.

(a) Calculate the average density of the box in kg/m³. 3 marks

Mark scheme

C1 3000 cm³ = 0.0030 m³ (3.0 × 10⁻³ m³), or 2.4 kg = 2400 g seen
C1 ρ = m / V, or 2.4 ÷ 0.0030 or 2400 ÷ 3000 or 0.80 g/cm³ seen
A1 800 kg/m³; unit required (an answer left as 0.80 g/cm³ does not score this mark, because the question asks for kg/m³)

(b) Explain why the box floats in water even though the metal is denser than water. 2 marks No calculator

Mark scheme

B1 the box is hollow / contains air, so its total volume is much larger than the volume of the metal alone
B1 so its average density is less than the density of water (800 < 1000 kg/m³, ecf from (a)), and an object less dense than water floats

How this is examined

Describing how to find the volume of an irregular solid by displacement, then its density: a three- or four-mark Describe about a stone or similar object that sinks, which can be followed by a Calculate using readings from the experiment Mass measured on a balance; measuring cylinder partly filled with water and the volume read; object lowered in until completely under the water and the new volume read; volume of object = new reading − first reading; density = mass ÷ volume. Write each step as its own sentence so that each one can be credited.
Calculating a density from two readings that have to be subtracted first: a two- or three-mark Calculate from balance readings (cylinder empty and full) or measuring cylinder readings (before and after) The subtraction written out (mass of liquid alone, or rise in volume); $ρ = m / V$ written with numbers in; the final answer with its unit, g/cm³ or kg/m³. The unit is part of the final mark.
Converting between g/cm³ and kg/m³, or cm³ and m³: a Calculate whose data are in g and cm³ but which asks for the answer in kg/m³, or whose volume is in cm³ and mass in kg Both quantities in a matching pair before dividing; then × $1000$ from g/cm³ to kg/m³, or $1$ cm³ $= 10^{- 6}$ m³ used. A correct value in the wrong unit does not score the final mark.
Deciding whether an object floats from density data: a one- or two-mark Determine, Deduce or Explain with a table of densities, or after you have calculated a density Both densities quoted in the same unit, the comparison stated ('less than' / 'greater than' the liquid's density), and the conclusion floats or sinks. For a hollow object, the average density compared with the liquid's.
Which immiscible liquid ends up on top: a one- or two-mark Predict or Determine about two or more liquids that do not mix, which can include a solid object dropped in 'The liquid with the lower density forms the top layer', with the two densities quoted. For an object, the boundary it rests at, with the reason that it is denser than the liquid above and less dense than the liquid below.
The definition of density: a one-mark Define, which can open a longer density question The words 'mass per unit volume'.

Words to know

Density: Mass per unit volume, $ρ = m / V$ . Measured in kg/m³ (SI) or g/cm³; $1$ g/cm³ $= 1000$ kg/m³.
Volume: The amount of space an object or a sample of liquid takes up, measured in cm³ (the same as ml) or m³.
Displacement: Pushing water aside. An object completely under water displaces a volume of water equal to its own volume.
Displacement can: A can with a spout near the top. Water pushed out by an object lowered into it runs out of the spout and can be collected in a measuring cylinder.
Meniscus: The curved surface of a liquid in a narrow container. Read a water level at the bottom of the curve, with your eye level with it.
Parallax error: A reading error caused by looking at a scale from an angle instead of straight on.
Average density: The total mass of an object divided by its total volume, including any air spaces inside it.
Immiscible: Describes liquids that do not mix. Immiscible liquids settle into separate layers, densest at the bottom.

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