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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ContinueFinding 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
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.
- 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.
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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 cm³Divide 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/cm³State the density
Working
Answer
The density is 7.9 g/cm³ to two significant figures.
Quick check
Guided practice 2
Density of a liquid
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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Hints
Hint 1
mass by difference
The balance reading includes the cylinder. Which mass belongs to the liquid alone?
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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 gDensity
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
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.
Quick check
A density below 1.00 g/cm³ means the liquid is less dense than water, which is sensible for an oil or alcohol.
Limitations and improvements
| Source | Effect | Improvement |
|---|---|---|
| Air bubbles stuck to the solid | Volume too large, so density too small | Lower the solid slowly on a thread and tap the cylinder to free the bubbles. |
| Solid weighed after it was in the water | If water remains on the solid, the mass is too large, so density is too large | Weigh the solid dry, before measuring its volume. |
| A small volume change read on a cylinder with 1 cm³ divisions | Reading ambiguity affects both levels; their difference can have a large fractional uncertainty for a small volume change | Use 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 meniscus | It is weighed but not in the volume reading, so density too large | Pour 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
Show the steps Optional working
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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Syllabus and review details
- SEC G2 Science Physics component 2027 · 2027
Physics section, PDF pages 8–23
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