Density–Mass–Volume (Rearrange)

\( \rho=\tfrac{m}{V},\quad m=\rho V,\quad V=\tfrac{m}{\rho} \)
Measures GCSE

\( A gold cube has density 19.3 g/cm^3 and side 2 cm. Find its mass. \)

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
\( Find volume: side^3 \)

Explanation

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Statement

The density–mass–volume relationship is a fundamental formula in physics and chemistry. It links three key properties of matter: density (\( \rho \)), mass (\( m \)), and volume (\( V \)). The core equation is:

\[\rho = \frac{m}{V}, \quad m = \rho V, \quad V = \frac{m}{\rho}\]

This means that density is defined as the mass of a substance per unit volume. By rearranging the equation, you can calculate mass if density and volume are known, or calculate volume if mass and density are given.

Why it’s true

  • Density measures how tightly matter is packed: more particles in the same space means higher density.
  • Rearranging formulas follows algebraic rules: multiplying or dividing both sides by the same term keeps the equation valid.
  • These rearrangements are simply different ways of expressing the same relationship.

Recipe (how to use it)

  1. Identify which quantities you know (density, mass, or volume).
  2. Select the correct rearranged form of the formula.
  3. Substitute the values, making sure units are consistent (e.g., grams with cm³, kilograms with m³).
  4. Calculate step by step, showing working clearly.

Spotting it

This formula is used whenever you need to connect mass, volume, and density. It appears in questions involving materials, fluids, or objects where size and weight are related.

Common pairings

  • Conversion of units (e.g., g/cm³ to kg/m³).
  • Calculations with regular solids (cubes, spheres, cylinders) where volume can be worked out first, then mass or density.
  • Archimedes’ principle or buoyancy problems, where density comparisons determine whether an object floats or sinks.

Mini examples

  1. Given: \( m = 200 \text{ g}, V = 50 \text{ cm}^3 \). Find: density. Answer: \( \rho = \frac{200}{50} = 4 \text{ g/cm}^3 \).
  2. Given: \( \rho = 8 \text{ g/cm}^3, V = 10 \text{ cm}^3 \). Find: mass. Answer: \( m = 8 \times 10 = 80 \text{ g} \).

Pitfalls

  • Forgetting to convert units (e.g., cm³ to m³).
  • Confusing which variable to rearrange for — always check the question carefully.
  • Misplacing numerator and denominator (e.g., writing \( V = \rho/m \) instead of \( V = m/\rho \)).
  • Leaving answers without units.

Exam strategy

  • Highlight known values in the question.
  • Write the correct version of the formula before substituting numbers.
  • Double-check unit consistency and convert if needed.
  • Box your final answer with the correct unit.

Summary

The density–mass–volume formula is one of the most useful relationships in science. By remembering that density equals mass divided by volume, you can quickly rearrange to find any missing variable. Consistent use of units, careful rearrangement, and clear working will ensure accuracy. This relationship underpins real-world applications such as material design, fluid science, and engineering.

Worked examples

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  1. \( A block has a mass of 600 g and a volume of 200 cm^3. Find its density. \)
    1. \( Use formula: density = mass / volume \)
    2. \( Density = 600 / 200 \)
    3. \( Density = 3 g/cm^3 \)
    Answer: \( 3 \text{ g/cm}^3 \)
  2. \( An object has density 7 g/cm^3 and volume 12 cm^3. Find its mass. \)
    1. \( Use formula: mass = density × volume \)
    2. \( Mass = 7 × 12 \)
    3. \( Mass = 84 g \)
    Answer: 84 g
  3. \( A rock weighs 2.5 kg and has a density of 2.5 g/cm^3. Find its volume in cm^3. \)
    1. \( Convert mass: 2.5 kg = 2500 g \)
    2. \( Use formula: V = m / ρ \)
    3. \( V = 2500 / 2.5 = 1000 cm^3 \)
    Answer: \( 1000 \text{ cm}^3 \)
  4. A cube of side 5 cm has a mass of 250 g. Find its density.
    1. \( Volume = side^3 = 5^3 = 125 cm^3 \)
    2. \( Density = 250 / 125 \)
    3. \( Density = 2 g/cm^3 \)
    Answer: \( 2 \text{ g/cm}^3 \)
  5. \( A liquid has density 1.2 g/cm^3. What mass is contained in 50 cm^3? \)
    1. \( m = ρ × V \)
    2. \( m = 1.2 × 50 \)
    3. \( m = 60 g \)
    Answer: 60 g
  6. \( An iron sphere has mass 1570 g and density 7.85 g/cm^3. Find its volume. \)
    1. \( V = m / ρ \)
    2. \( V = 1570 / 7.85 \)
    3. \( V = 200 cm^3 \)
    Answer: \( 200 \text{ cm}^3 \)
  7. A cylinder of radius 3 cm and height 10 cm has a mass of 850 g. Find its density.
    1. \( Volume = πr^2h = 3.1416×9×10 = 282.74 cm^3 \)
    2. \( Density = 850 / 282.74 \)
    3. \( Density ≈ 3.0 g/cm^3 \)
    Answer: \( 3.0 \text{ g/cm}^3 \)
  8. \( A car engine part has volume 200 cm^3 and mass 1.58 kg. Find its density in g/cm^3. \)
    1. \( Convert mass: 1.58 kg = 1580 g \)
    2. \( ρ = 1580 / 200 \)
    3. \( ρ = 7.9 g/cm^3 \)
    Answer: \( 7.9 \text{ g/cm}^3 \)
  9. \( A gold bar measures 10 cm × 5 cm × 2 cm and has a density of 19.3 g/cm^3. Find its mass. \)
    1. \( Volume = 10×5×2 = 100 cm^3 \)
    2. \( Mass = ρ × V = 19.3 × 100 \)
    3. \( Mass = 1930 g \)
    Answer: 1930 g
  10. \( A block of aluminium has a density of 2.7 g/cm^3 and a mass of 540 g. Find its volume. \)
    1. \( V = m / ρ \)
    2. \( V = 540 / 2.7 \)
    3. \( V = 200 cm^3 \)
    Answer: \( 200 \text{ cm}^3 \)