Find the volume of the object, not the final cylinder reading
Two water readings can reveal the volume of an irregular solid. The important step is subtraction: the rise in the reading belongs to the submerged object, while the final reading includes the original water as well. Once that distinction is secure, a density calculation becomes a short chain of measurements, subtraction and division.
These supplied-data problems concern fully submerged, non-porous solids that do not dissolve or react with the liquid. There are no trapped bubbles, no spilled liquid and no significant displacement from a support thread. The numbers describe fictional samples and are not identification certificates for real materials.
1. Keep three quantities separate
Mass tells us how much matter is represented by a balance reading. Volume is the space occupied by the object. Density compares them: density ρ = mass m ÷ volume V. A heavier object does not necessarily have a higher density; its volume may also be larger.
| Quantity | Symbol | Convenient units | SI units |
|---|---|---|---|
| Mass | m | g | kg |
| Object volume | V | cm3 or mL | m3 |
| Density | ρ | g/cm3 | kg/m3 |
The volume relationship is 1 mL = 1 cm3. A balance measures mass; a force meter reports weight in newtons. Do not put a weight reading directly into m/V. If weight is the supplied quantity, first use the stated gravitational field strength to obtain mass. No weight conversion is needed in the balance examples here.
2. Read a cylinder scale in intervals
First identify two labelled graduations, count the equal spaces between them, and divide the numerical difference by that number of spaces. Five spaces between 30 mL and 40 mL give 2 mL per space. Counting six lines instead would give the wrong interval.
For water with a concave meniscus, read the bottom of the curve at eye level. In a supplied sketch, use the stated meniscus reference rather than the top of the coloured fill. A camera or eye above the level can introduce parallax. In real measurements, the scale and instrument uncertainty limit the precision of the result; extra calculator digits do not improve the measurement.
Worked case 1: an irregular solid
A fictional dry sample has mass 67.2 g. The cylinder initially reads 34 mL and reads 58 mL with the sample fully submerged. The scale has 2 mL divisions. Find the volume and density, giving the density to two significant figures.
Native diagram alternative: Two identical cylinders have labelled marks at 20, 30, 40, 50, 60 and 70 mL, with 2 mL intermediate divisions. The left meniscus bottom is at 34 mL. The right meniscus bottom is at 58 mL above a fully submerged solid. A separate balance label states dry-object mass 67.2 g. The cylinder walls and solid outline are schematic; only the marked liquid readings supply the volume data.
Step 1, subtract: V = 58 − 34 = 24 mL = 24 cm3. The solid occupies the extra volume excluded to the liquid. The original 34 mL remains present, so it must not be counted as part of the solid.
Step 2, divide: ρ = 67.2 g ÷ 24 cm3 = 2.8 g/cm3. Step 3, convert: ρ = 2.8 × 1000 = 2800 kg/m3, or 2.8 × 103 kg/m3. The scientific notation makes the requested significant figures clear.
A mistaken division, 67.2 ÷ 58, produces about 1.16 g/cm3. That number is too small because its denominator includes water volume that does not belong to the object. The repair is to correct the volume, not to change the density formula.
3. Explain the factor of one thousand
One gram is 0.001 kg. One centimetre is 0.01 m, so one cubic centimetre is (0.01)3 = 0.000001 m3. Therefore 1 g/cm3 = 0.001 kg ÷ 0.000001 m3 = 1000 kg/m3. Cubing the length conversion is the essential step.
Choose one route and keep it consistent. Either calculate in grams and cubic centimetres, then convert the final density, or convert both mass and volume before dividing. Converting only the numerator and labelling the result kg/m3 mixes unit systems.
Worked case 2: remove the container mass
An empty weighing container has mass 31.6 g; container plus dry solid has mass 148.6 g. Cylinder readings are 42.0 mL before immersion and 57.0 mL afterward. Find the density of the solid.
Solution: The solid mass is 148.6 − 31.6 = 117.0 g. Its volume is 57.0 − 42.0 = 15.0 cm3. Hence ρ = 117.0 ÷ 15.0 = 7.80 g/cm3 = 7.80 × 103 kg/m3. The three-significant-figure result matches the stated volume precision.
Using 148.6 g would include the container in the numerator even though the container did not enter the cylinder. That error raises the calculated density. The correct mass and correct volume must refer to the same object, in the same relevant condition.
Worked case 3: cross-check with a regular block
A solid rectangular block measures 3.0 cm by 2.0 cm by 1.5 cm and has mass 72 g. A separate displacement record rises from 46 mL to 55 mL. Do the methods agree?
Solution: Geometrical volume = 3.0 × 2.0 × 1.5 = 9.0 cm3. Displacement volume = 55 − 46 = 9 cm3. The central values agree. Density from either is 72 ÷ 9 = 8 g/cm3. The dimensions support writing the geometrical result as 8.0 g/cm3; the integer cylinder record has a separate, unspecified reading uncertainty.
Agreement is a useful consistency check, not proof of perfect accuracy. Both methods can have measurement errors. If they disagree, check arithmetic, dimensions, immersion and bubbles before assuming that the material changed density.
Worked case 4: diagnose trapped air
A solid has a true volume of 18 cm3 and mass 54 g. In a faulty supplied record, an attached bubble excludes an additional 2 cm3 of liquid. Compare the correct and apparent densities.
Solution: Correct density = 54 ÷ 18 = 3.0 g/cm3. Apparent volume = 18 + 2 = 20 cm3, so apparent density = 54 ÷ 20 = 2.7 g/cm3. An extra positive contribution to volume makes the calculated density smaller when mass is unchanged. This direction-of-error explanation is more useful than simply saying “bubbles cause error”.
4. Know when the simple model fails
If an object is only partly immersed, the displaced liquid represents only its submerged portion. If it absorbs liquid or dissolves, the final reading may not represent an unchanged solid plus the initial liquid. If liquid is spilled, the rise is reduced. These situations need additional information or a different method; do not silently apply the fully submerged model.
A sealed hollow object also needs careful language. Its mass divided by the outer displaced volume gives its average density, including the enclosed space. That is not automatically the density of the wall material. The solid samples in the main examples avoid this distinction by being non-porous and without a hollow cavity.
Three tasks with complete solutions
Task 1: direct displacement
A 91.2 g sample raises the reading from 26.0 mL to 38.0 mL. Find its volume and density in both unit systems.
Solution: V = 38.0 − 26.0 = 12.0 cm3. Density = 91.2 ÷ 12.0 = 7.60 g/cm3 = 7.60 × 103 kg/m3. Check by multiplication: 7.60 × 12.0 returns 91.2 g.
Task 2: predict the final reading
A non-porous solid has mass 0.156 kg and density 2600 kg/m3. Initially the cylinder reads 45 mL. Assuming it fits and is fully submerged, what final reading does the model predict?
Solution: V = m/ρ = 0.156 ÷ 2600 = 0.000060 m3 = 60 cm3 = 60 mL. Final reading = 45 + 60 = 105 mL. A cylinder with capacity below 105 mL would be unsuitable for this proposed record; the arithmetic does not guarantee instrument capacity.
Task 3: repair a result
A 64 g sample raises the reading from 35 mL to 43 mL. A learner reports 64/43 = 1.49 g/cm3. Explain the error and give the corrected central value.
Solution: The denominator is the final reading instead of the increase. V = 43 − 35 = 8 cm3, giving ρ = 64/8 = 8 g/cm3. State measurement precision separately; the extra digits in 1.49 did not make the incorrect method reliable.
Relative density: compare with a stated reference
Relative density is the ratio ρsample/ρreference. Use matching density units, so they cancel. The result has no unit. For this classroom comparison, use water at the stated reference temperature of 4 °C and the supplied rounded reference density 1.00 g/cm3 = 1000 kg/m3. Treat all sample densities as values at the fixed measurement condition in the question. Water density is not exactly constant at every temperature.
Worked comparison using the retained displacement record
The first irregular solid above has mass 67.2 g and displacement volume 58 − 34 = 24 cm3. Its density is 2.8 g/cm3. Relative to the supplied water reference, d = (2.8 g/cm3)/(1.00 g/cm3) = 2.8. Equivalently, 2800/1000 = 2.8 using kg/m3.
The meaning is an equal-volume comparison: 24 cm3 of the reference water has mass 24 g under the rounded model, so the mass ratio is 67.2/24 = 2.8. It does not mean the sample has a mass of 2.8 g. Writing “2.8 g/cm3” answers a density question, not a relative-density question.
Independent calculation and interpretation
A fictional material has density 0.800 g/cm3 at its supplied measurement condition. Find its relative density against the same water reference. What mass does a 50.0 cm3 sample have? A second sample has relative density 1.25 against this reference: find its density in kg/m3.
Checked answers: First relative density = 0.800/1.00 = 0.800, dimensionless. Its mass is ρV = 0.800 × 50.0 = 40.0 g; the same reference-water volume has mass 50.0 g. The second density is 1.25 × 1000 = 1250 kg/m3. A ratio below one means lower density than the specified reference. Predicting an actual object’s floating behaviour additionally needs its average density, the actual surrounding fluid and any support or other forces; a ratio alone is not a material-identification certificate.
Definition check: IUPAC Gold Book, relative density. Mass–volume units and the water reference are consistent with OpenStax, density.
Frequently asked questions
Why subtract the initial cylinder reading?
The final reading includes the original liquid. Subtracting the initial reading isolates the volume excluded to the liquid by the fully submerged object.
Are millilitres and cubic centimetres equivalent?
Yes. One millilitre equals one cubic centimetre, so a 24 mL rise represents 24 cm³ under the stated displacement assumptions.
How do I convert g/cm³ to kg/m³?
Multiply by 1000. To convert the other way, divide by 1000. This combines the gram-to-kilogram conversion with the cubed centimetre-to-metre conversion.
What do attached bubbles do to the result?
They add to the apparent excluded volume. With the same measured object mass, that makes the calculated density lower than its true value in the stated model.
Can I use the method for an absorbing solid?
Not with the simple unchanged-object model. Absorption, dissolution or reaction can invalidate the assumption that the rise equals the original object volume.
Is density the same as weight?
No. Density is mass per volume. Weight is a gravitational force measured in newtons. A weight reading needs conversion to mass before it can be used in m/V.
Is relative density measured in g/cm³?
No. Relative density is a ratio to a stated reference density and has no unit. Density itself retains mass-per-volume units.
Continue revision
Browse the Physics topical revision collection; individual resources may require membership. The OpenStax density reference supports the mass-per-volume definition and unit relationship. All datasets and drawings here are original.
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