The Amazing World of Solutes, Solvents, and Solutions · Lesson 5 of 7
Finding the Volume and Density of Solids
“Choose a volume method for a solid’s shape, then combine that volume with measured mass to find density.”
• Calculate the volume of a cuboid using perpendicular dimensions. • Find an irregular solid’s volume from displaced water. • Explain why full immersion and suitable sample choice matter. • Combine mass and volume measurements to calculate density. • Check calculations using units and physical meaning.
Choose a Method That Fits the Shape
A notebook has dimensions that can be measured along straight edges. A stone has an uneven outline that is difficult to describe using length, width, and height alone. Both occupy volume, but their shapes suggest different measurement methods. Begin by deciding which method actually measures the space occupied by the object.
A cuboid has three perpendicular dimensions: length, width, and height. Measuring these with a ruler gives the information needed to calculate its volume. Use the same unit for all three dimensions. The product then gives cubic units because three lengths have been multiplied together.
Volume of a Cuboid
Picture a cuboid filled with small equal cubes. Its length and width determine how many cubes fit in one layer, and its height determines the number of layers. Multiplying all three dimensions therefore measures the total space inside the cuboid.
Problem
A notebook is 25 cm long, 18 cm wide, and 2 cm thick. Calculate its volume, treating its shape as a cuboid.
- 1.Use length, width, and height in the same unit: 25 cm, 18 cm, and 2 cm.
- 2.Calculate the base area: 25 × 18 = 450 cm².
- 3.Multiply by the thickness: 450 cm² × 2 cm = 900 cm³. The notebook’s approximate cuboid volume is 900 cm³.
Problem
A cube-shaped die has edge length 2 cm. Find its volume.
- 1.A cube is a cuboid with all three dimensions equal.
- 2.Substitute 2 cm for length, width, and height: 2 × 2 × 2 = 8.
- 3.The volume is 8 cm³. Multiplying only two edges would give area, not volume.
For the regular-object investigation, collect a notebook, a shoebox, or a cube-shaped object. Measure three perpendicular dimensions, record the units, and calculate each volume. Think carefully about what the result represents. The outside dimensions of a hollow shoebox give its outside volume, not the volume of cardboard alone; inside dimensions estimate its capacity. Do not use the outside volume as the material volume when calculating cardboard density.
A hollow object contains empty space. Multiplying its outside dimensions includes that space. For a material-density calculation, measure the volume occupied by the material itself. State clearly whether a measurement describes material, a whole object, or container capacity.
Irregular Solids and Water Displacement
An irregular stone still takes up space even though it has no convenient set of cuboid edges. When it is lowered completely into water, it occupies space that water previously could occupy. The displaced water causes the cylinder reading to rise, allowing us to measure the stone’s volume.
Choose a solid such as a stone or metal key that fits into the cylinder and does not dissolve, react noticeably with water, or soak up a significant amount during the investigation. Begin with enough water to cover the object when lowered. Record the initial reading at eye level. Tie the object with a thin thread and lower it gently until it is fully immersed. Record the final reading using the same convention.
The difference between final and initial readings is the volume of water displaced. For a completely immersed suitable solid, that equals the solid’s volume. Keep all the water in the cylinder, avoid trapped air bubbles, and do not include a finger or other support beneath the water. Otherwise the change in reading may not be due to the object alone.
| Object | Initial A (mL) | Final B (mL) | Displaced B − A (mL) | Object volume (cm³) |
|---|---|---|---|---|
| Stone | 50 | 55 | 5 | 5 |
| Metal key | Record reading | Record reading | Calculate B − A | Same numerical value |
| Another suitable solid | Record reading | Record reading | Calculate B − A | Same numerical value |
Problem
A key changes the cylinder reading from 30 mL to 34 mL when fully immersed. What is its volume?
- 1.Subtract initial from final volume: 34 mL − 30 mL = 4 mL.
- 2.The displaced water has volume 4 mL. Full immersion means this equals the key’s volume.
- 3.Use 1 mL = 1 cm³: the key occupies 4 cm³. The final reading of 34 mL includes the original water, so it is not the key’s volume.
Combine the Measurements to Find Density
You can now measure the two quantities needed for density. Measure the sample’s mass before immersing it so that water stuck to its surface does not add to the balance reading. Then find its volume using the method appropriate to its shape and divide mass by volume.
Problem
A dry stone has measured mass 16.400 g. Water rises from 50 mL to 55 mL when the stone is fully immersed. Find the stone’s density.
- 1.Find the stone volume: 55 − 50 = 5 mL = 5 cm³.
- 2.Use density = mass ÷ volume: 16.400 g ÷ 5 cm³ = 3.28 g/cm³.
- 3.Check the meaning: each cubic centimetre of the stone’s material has a mass of about 3.28 g. Do not divide by 55 mL, because that reading includes the original water.
Problem
A uniform cuboid has dimensions 4 cm × 3 cm × 2 cm and mass 72 g. Find its density.
- 1.Calculate volume: 4 × 3 × 2 = 24 cm³.
- 2.Divide the measured mass by this volume: 72 g ÷ 24 cm³ = 3 g/cm³.
- 3.The result is density, not total mass. If the same material were sampled in a different shape under unchanged conditions, its density would remain the same.
Check that the sample was dry for weighing, the container mass was excluded, the dimensions or cylinder readings use consistent units, and the object was fully immersed for displacement. A formula cannot correct an unsuitable measurement method.
Quiz
A cuboid is 5 cm × 2 cm × 3 cm. What is its volume?
A fully immersed object raises water from 42 mL to 49 mL. What is its volume?
Why must the object be fully immersed for this method?
A stone has mass 18 g and displacement volume 6 cm³. What is its density?
Which object is unsuitable for a simple water-displacement measurement of its original solid volume?
Practice Problems
- A rectangular solid measures 6 cm × 4 cm × 2 cm. Calculate its volume. Explain why the answer uses cubic units.
- An irregular solid raises a cylinder reading from 35 mL to 43 mL. Find its volume in mL and cm³.
- The solid in the previous question has dry mass 24 g. Calculate its density and show the correct units.
- Explain why a dissolving sugar lump, an absorbent sponge, and a partly immersed stone can give misleading displacement results.
- A stone has mass 16.400 g and changes the reading from 50 to 55 mL. Explain the error in calculating density as 16.400 ÷ 55.
- A hollow box has known outside dimensions. Explain why the product of those dimensions is not automatically the volume of the material used to make it.
- Plan a complete density measurement for a metal key, including mass measurement, volume measurement, and two checks that reduce error.
Key Takeaways
• For a cuboid, volume = length × width × height, with all dimensions in the same unit. • For a suitable fully immersed irregular solid, volume equals final reading minus initial reading. • A displacement of 1 mL represents 1 cm³ of solid volume. • Water loss, trapped air, dissolution, absorption, and partial immersion can spoil the measurement. • Find density by dividing sample mass by sample volume, not by the cylinder’s final total reading. • Distinguish a hollow object’s outside volume from the volume occupied by its material.