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Density investigations

Key idea: To find density, measure a sample's mass and volume, then divide mass by volume. Use dimensions for a regular solid and the rise in water volume for a fully submerged irregular solid.

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  • SEC G3 Combined Science Physics component 2027
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Finding a density needs only a mass and a volume, but they must belong to the same sample, and the way you measure the volume depends on what the sample is. A liquid is poured into a measuring cylinder, a regular block is measured with a rule, and an odd-shaped stone is lowered into water to see how far the level rises. Each method has its own errors, and knowing which way they push the answer is part of the investigation.

Determine density

density = mass/volume

The mass and the volume must belong to the same sample. Measure mass with an electronic balance; the method for volume depends on the sample.

Two ways to find volumeA rectangular block has three perpendicular dimensions: length l, width w and height h, so V = l × w × h. For an irregular solid fully immersed in water, subtract the initial measuring-cylinder reading V₁ from the final reading V₂.Regular solidlhwV = l × w × hIrregular solidV₁V₂Object volume = V₂ − V₁
Two ways to find volumeA rectangular block has three perpendicular dimensions: length l, width w and height h, so V = l × w × h. For an irregular solid fully immersed in water, subtract the initial measuring-cylinder reading V₁ from the final reading V₂.Regular solidlhwV = l × w × hIrregular solidV₁V₂Object volume = V₂ − V₁
Use calculated volume for a regular solid and water displacement for an irregular solid that sinks.
  • Liquid: weigh an empty measuring cylinder, pour in the liquid, read the volume at eye level and weigh again. For a concave meniscus such as water’s, read its lowest point. The liquid’s mass is the difference.
  • Regular solid: measure its dimensions with a rule or digital calipers and calculate the volume. Repeat each dimension reading and check different positions. Small variations can be averaged for a shape close to regular; if it is appreciably irregular, use a suitable displacement method.
  • Irregular solid that sinks: weigh it dry first. Read the water level, lower the solid in on a thread, and read the new level. Keep it fully submerged without bubbles, and use a thin thread. Volume = final reading − initial reading. The solid must not dissolve or absorb appreciable water.

Worked example 1

Density of an irregular solid

Problem

A dry metal piece has a mass of 63.0 g. Fully submerging it without air bubbles raises the water level from 38.0 cm³ to 46.0 cm³. Assume it does not dissolve or absorb water. Find its density.

Show full solution

  1. Find the volume

    Method

    Subtract the initial reading from the final reading.

    Reason

    The fully submerged piece displaces its own volume of water. Assume it does not dissolve or absorb water and has no trapped air.

    Working

    Answer

    V = 46.0 - 38.0 = 8.0 cm³

  2. Divide and round

    Method

    Divide mass by volume and match the precision of the volume.

    Reason

    8.0 cm³ has two significant figures, so the density has two.

    Working

    Answer

    ρ = (63.0 g)/(8.0 cm³) = 7.875 ≈ 7.9 g/cm³

  3. State the density

    Working

    Answer

    The density is 7.9 g/cm³ to two significant figures.

Guided practice 2

Density of a liquid

About 4 min

Problem

An empty measuring cylinder has a mass of 52.40 g. With 40.0 cm³ of a liquid in it, the mass is 84.20 g. Find the density of the liquid.

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Unit: g
Show the steps Optional working
Unit: g
Unit: g/cm³

Hints

Hint 1

mass by difference

The balance reading includes the cylinder. Which mass belongs to the liquid alone?

Show solution step by step

  1. Mass of the liquid

    Method

    Subtract the empty cylinder’s mass.

    Reason

    Density needs the mass of the liquid only.

    Working

    Answer

    m = 84.20 - 52.40 = 31.80 g

  2. Density

    Method

    Divide by the volume read at the meniscus.

    Reason

    40.0 cm³ has three significant figures, so give three.

    Working

    Answer

    ρ = 31.80 ÷ 40.0 = 0.795 g/cm³

  3. State the liquid density

    Working

    Answer

    The liquid’s density is 0.795 g/cm³. Using the mass of the filled cylinder would include the container and overestimate the density.

Subtract the container's mass before calculating liquid densityThe same measuring cylinder is weighed twice in this illustrative example. Empty, the balance reads 52.40 grams. Filled with 40.0 cubic centimetres of liquid, it reads 84.20 grams. The liquid mass is the difference, 31.80 grams. The volume scale is shown alongside a concave meniscus, read at its lowest point at eye level. The diagram illustrates the supplied readings, not an actual measurement or a statement of measurement uncertainty.Mass of liquid = filled − emptyEmpty cylinder204060cm³52.40 gCylinder + liquid204060cm³84.20 g40.0 cm³lowest pointof meniscus
Subtract the container's mass before calculating liquid densityThe same measuring cylinder is weighed twice in this illustrative example. Empty, the balance reads 52.40 grams. Filled with 40.0 cubic centimetres of liquid, it reads 84.20 grams. The liquid mass is the difference, 31.80 grams. The volume scale is shown alongside a concave meniscus, read at its lowest point at eye level. The diagram illustrates the supplied readings, not an actual measurement or a statement of measurement uncertainty.Mass of liquid = filled − emptyEmpty cylinder204060cm³52.40 gCylinder + liquid204060cm³84.20 g40.0 cm³lowest pointof meniscus
Use the same cylinder for both weighings. The liquid's mass is 84.20 − 52.40 = 31.80 g; its volume is 40.0 cm³. For a concave meniscus, read the lowest point at eye level. These are illustrative readings.

Limitations and improvements

SourceEffectImprovement
Air bubbles stuck to the solidVolume too large, so density too smallLower the solid slowly on a thread and tap the cylinder to free the bubbles.
Solid weighed after it was in the waterIf water remains on the solid, the mass is too large, so density is too largeWeigh the solid dry, before measuring its volume.
A small volume change read on a cylinder with 1 cm³ divisionsReading ambiguity affects both levels; their difference can have a large fractional uncertainty for a small volume changeUse a suitable cylinder with finer divisions, or a larger sample that still fits fully below the water.
Liquid left on the cylinder’s wall above the meniscusIt is weighed but not in the volume reading, so density too largePour carefully down the inside and let the liquid drain before reading.

The timing method and gradient examples are in Free-fall investigation.

The trolley method and friction example are in Balanced and unbalanced forces investigation.

Check your understanding 1 An offset and a bubble are not the same error

A 48 g solid raises a cylinder reading from 30 to 36 cm³. Both readings have the same +2 cm³ zero error. A trapped bubble adds 1 cm³ to the displacement. Find the solid’s volume and density. Which error cancels when the readings are subtracted?

Show the steps Optional working
Unit: cm³
Unit: cm³
Show the steps Optional working
Unit: cm³
Unit: g/cm³
Error that cancels
Show answer

The recorded difference is 36.0 − 30.0 = 6.0 cm³. The common additive offset cancels: (36.0 − 2.0) − (30.0 − 2.0) = 6.0 cm³. But the piece and bubble both displace water, so the piece alone occupies 6.0 − 1.0 = 5.0 cm³ and its density is 48.0/5.0 = 9.6 g/cm³. Do not subtract the offset only once from the difference. Weigh the dry piece, submerge it completely on a thin thread, remove attached bubbles and repeat the volume readings at eye level. In a real experiment the bubble’s volume would usually be unknown, so remove it rather than inventing a correction.

Common mistakes

  • Using the mass of the filled cylinder instead of subtracting the empty cylinder’s mass.
  • Leaving air bubbles on an irregular solid, which makes the volume too large and the density too small.
  • Weighing a solid after it has been in the water.
  • Mixing grams with m³, or kilograms with cm³, in one calculation.
  • Assuming a measured density alone identifies the material.

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