Physics · Mechanics

Density: mass and the chosen volume

Q: How do you calculate the density of a body?

Divide the mass by the volume: $\rho{} =\frac{ m}{V}$. Example: an aluminium cube with $m = 270 g$ and $V = 100 cm^{3}$ has $\rho{} =\frac{ 2.7 g}{cm^{3}}$, in SI units $\frac{2\,700 kg}{m^{3}}$. You find the mass with a scale; the volume of regular bodies by measuring, of irregular ones via water displacement in a measuring cylinder, where the rise of the water level equals the body volume. Unit discipline matters: work consistently in grams with cubic centimetres or kilograms with cubic metres, never mixed. For conversion, $\frac{1 g}{cm^{3}} =\frac{ 1\,000 kg}{m^{3}}$. With the computed density you can then look up in a reference table which substance it probably is.

Q: How do you rearrange $\rho{} =\frac{ m}{V}$ for mass or volume?

Solved for the mass: $m = \rho{}\cdot{}V$, which lets you estimate weights without a scale. A 200 litre aquarium holds $m =\frac{ 1\,000 kg}{m^{3}} \times{} 0.2 m^{3} = 200 kg$ of water, which is why you should never carry it filled. Solved for the volume: $V =\frac{ m}{\rho{}}$, which gives the space required: 7.9 kg of steel with $\rho{} =\frac{ 7\,900 kg}{m^{3}}$ fills exactly $V = 0.001 m^{3}$, one litre. Density is the conversion factor between the worlds of mass and volume; craftsmen and engineers use it constantly (how heavy will the concrete slab be, how much does the bucket of paint weigh?). After every rearrangement check the units: $\frac{kg}{m^{3}}$ times $m^{3}$ gives kg, then the formula is right.

Average density is mass per defined volume; compare material densities with temperature, pressure and material state specified.

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Formula

Calculate

ρ = \frac{m}{V}

ρ in kg/m³ (or g/cm³; 1 g/cm³ = 1,000 kg/m³) · m in kg · V in m³

Variables & units

Variables & units
Symbol	Meaning	Unit
$ρ$	Density (rho)	kg/m³
$m$	Mass	kg
$V$	Volume	m³

LaTeX

\rho = \frac{m}{V}

QUANTA STUDYquanta-study.de

Original schematic cube with side a=4 cm, labelled mass 172.8 g and volume 64 cm³. Below it, density ρ=m/V and 1 cm³=10⁻⁶ m³. — Original homogeneous model cube: a = 4 cm gives V = a³ = 64 cm³. With m = 172.8 g, ρ = 2.7 g/cm³ = 2700 kg/m³. The volume conversion factor is cubed.

Density calculator

Find density, mass or volume: enter two values and the third appears straight away; ρ is the mean density of the whole object.

Find

Mass m

Mass of the object from the scales, e.g. 172.8 g for an aluminium cube with 4 cm edges.

Volume V

Volume of the same object, e.g. 64 cm³ for a cube with 4 cm edges (4 · 4 · 4).

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for ρ, substitute, evaluate

1. Rearrange for ρ
$\rho=\frac mV$
2. Substitute values (in SI units)
- m = 172.8 g = 0.1728 kg
- V = 64 cm³ = 6.4 · 10⁻⁵ m³
$\rho=\frac{0.1728}{0.000064}=2700\,\mathrm{kg\,m^{-3}}$
3. Result
ρ = 2.7 g/cm³

Result: Density ρ

ρ = 2.7 g/cm³

With m = 172.8 g and V = 64 cm³, ρ = 2.7 g/cm³.

Quick tips

Formula triangle: m on top, ρ and V below. Cover the quantity you need and the rest is the calculation, e.g. V = m ÷ ρ.
Converting: 1 g/cm³ = 1 kg/L = 1000 kg/m³. Water has about 1 g/cm³; an object with a greater mean density sinks in it.

Assumptions and limits

Mass and explicitly selected volume refer to the same body/material. Positive volume, nonnegative mass. Mean density is not automatically bulk, material or pore-space density.

Derivation & background

Average density is $\rho{} =\frac{ m}{V}$ ; homogeneous material has uniform density under identical conditions. $\frac{1 g}{cm^{3}}$ equals $\frac{1000 kg}{m^{3}}$ . Pores and cavities change average body density relative to material density. Approximate values such as water near $\frac{1000 kg}{m^{3}}$ and aluminium near $\frac{2700 kg}{m^{3}}$ refer to particular state conditions. In the simple static buoyancy model, a sealed body floats freely when its average density is lower than that of the fluid.

Exam blueprint

Validity range

$\rho=\frac{m}{V}$ defines average density even for nonuniform bodies. Uniform material density requires a homogeneous sample and specified state conditions. State the chosen volume for pores and cavities.

Derivation steps

Average density is defined as mass divided by its corresponding volume; homogeneity makes it a uniform material density.

1Identify the body and its mass m.
2Define whether V is material, envelope or total volume.
3Use consistent units: kg with $m^{3}$ or g with $cm^{3}$ .
4Form the ratio ρ̄ = m/V with $V > 0$ .
5For homogeneous material in the same state, $m = \rho{}V$ uses uniform ρ.
6For spatially variable density, $m = \int{}\rho{}(r)dV$ ; the overall ratio remains an average.
7Check temperature, pressure, pores and uncertainty before comparing material values.

Rearrangements

Mass from density and volume

m = \rho \cdot V

This is how you estimate weights without a scale, e.g. water in an aquarium.

Volume from mass and density

V = \frac{m}{\rho}

The basis of the Archimedes displacement method.

Task variant

What is the volume of 7.9 kg of steel $(\rho{} =\frac{ 7\,900 kg}{m^{3}})$ ?

$V=\frac{m}{\rho}=\frac{7.9\,\mathrm{kg}}{7900\,\mathrm{kg\,m^{-3}}}=0.001\,\mathrm{m^3}$ , or 1 litre.

A wooden block has $m = 0.3 kg$ and $V = 500 cm^{3}$ . Does it float on water?

$\rho=\frac{0.3\,\mathrm{kg}}{5\cdot10^{-4}\,\mathrm{m^3}}=600\,\mathrm{kg\,m^{-3}}<1000\,\mathrm{kg\,m^{-3}}$ , so it floats.

Common mistakes

Mixing $\frac{g}{cm^{3}}$ and $\frac{kg}{m^{3}}$ .

Factor 1,000: $\frac{1 g}{cm^{3}} =\frac{ 1\,000 kg}{m^{3}}$ .

Equating density with weight or mass.

Density is mass per volume; a small piece of gold is denser but lighter than a chunk of ice.

Combining volume in $cm^{3}$ with mass in kg.

Choose consistent units: kg with $m^{3}$ or g with $cm^{3}$ .

Exam context

Typical in buoyancy and material identification tasks: compute the average density and compare with the liquid.

Formula cluster

Material properties

Density links mechanics (buoyancy, pressure) with material data.

p = F/A

Pressure

pV = nRT

Ideal Gas Law

Q = m·c·ΔT

Heat Capacity: Q = m·c·ΔT

F_A = ρ·V·g

Buoyant Force (Archimedes)

Worked example

An aluminium cube has $m = 270 g = 0.27 kg$ and $V = 100 cm^{3} = 10^{-4} m^{3}$ . Approach: $\rho=\frac mV$ . Substitution: $\rho=\frac{0.27\,\mathrm{kg}}{10^{-4}\,\mathrm{m^3}}=2700\,\mathrm{kg\,m^{-3}}=2.7\,\mathrm{g\,cm^{-3}}$ .

Applications

Materials testing, buoyancy and shipbuilding, density determination in the lab (pycnometer), meteorology (air layering)

How do you calculate with Density?

Here is how to work through a typical Density (ρ = m/V) task step by step:

1
Task
What is the volume of 7.9 kg of steel $(\rho{} =\frac{ 7\,900 kg}{m^{3}})$ ?
Solution path
$V=\frac{m}{\rho}=\frac{7.9\,\mathrm{kg}}{7900\,\mathrm{kg\,m^{-3}}}=0.001\,\mathrm{m^3}$ , or 1 litre.
2
Task
A wooden block has $m = 0.3 kg$ and $V = 500 cm^{3}$ . Does it float on water?
Solution path
$\rho=\frac{0.3\,\mathrm{kg}}{5\cdot10^{-4}\,\mathrm{m^3}}=600\,\mathrm{kg\,m^{-3}}<1000\,\mathrm{kg\,m^{-3}}$ , so it floats.

Understand and apply the equation

A large object need not be denser than a small one. Density describes how much mass occupies a particular volume. Learn to calculate this comparison, measure volume reliably and choose the appropriate model for cavities, heating and flotation.

What is the density formula, and which volume counts?

Two cubes of equal size can show very different masses on a balance. Comparing equal volumes reveals a material property; mass alone does not. Conversely, a large piece of light foam can have more mass than a small metal object. Start by comparing mass per volume, then establish which volume belongs to the system under study.[1]

$\overline\rho=\frac mV$ defines average mass density for the chosen mass and volume $V>0$ . For homogeneous material under identical conditions, differently sized samples have the same ratio. For a composite body, it may be only an overall average. Density is neither mass nor weight.[1]

\overline\rho=\frac{m}{V},\qquad m=\rho V,\qquad V=\frac{m}{\rho}

Using a single material density in m = ρV assumes homogeneity; the final rearrangement requires ρ > 0.

A small dense metal body can have less mass than a large low-density foam body. In an original comparison, 10 cm³ at 8 g/cm³ has mass 80 g, while 1000 cm³ at 0.1 g/cm³ has mass 100 g. The denser object is lighter here.

Equal volume, different mass

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Two uniform model materials have densities of 1 and 2.7 g/cm³. At 100 cm³ each, their masses are 100 and 270 g. Slope in the mass-volume graph is density; the lines compare different sample sizes under matching state conditions.[1]

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A definition rather than a proof of a physical law

Density is defined as a ratio, not proved from a motion law. With uniform ρ and material conditions, multiplying by a subvolume gives its mass. Doubling mass and volume together leaves the ratio unchanged. Compressing a fixed mass changes volume and thus density.

\rho(\vec r)=\frac{\mathrm dm}{\mathrm dV},\qquad m=\int_V\rho(\vec r)\,\mathrm dV

Local continuum density may vary in space; the integral sums masses of small volume elements.

For a more detailed description, imagine dividing a body into small volume elements. Each should still contain enough particles for its average density to be meaningful. Density may then differ between locations, for instance within a wall containing cavities. For the whole body, $\overline\rho=\frac mV$ remains average density. It need not equal density at every location. Treating density as a smooth field is called a continuum model.

Calculating mass and volume: converting g/cm³ to kg/m³

1\,\mathrm{cm^3}=(10^{-2}\,\mathrm m)^3=10^{-6}\,\mathrm{m^3},\qquad1\,\mathrm{g/cm^3}=1000\,\mathrm{kg/m^3}

Each of the three spatial dimensions contributes a factor 10⁻².

1 mL equals 1 cm³; 1 L equals 1000 cm³ or 10⁻³ m³. Use either grams with cubic centimetres or kilograms with cubic metres. Do not relabel kg/cm³ as kg/m³ without conversion. 2.5 g/cm³ equals 2500 kg/m³, not 0.0025 kg/m³.

A balance measures mass; dimensions can provide volume. For a cube, $V=a^3$ : edge-length error therefore contributes three times its relative size to volume error in the linear approximation. Many decimal places in a density value do not replace precise measurement.

How do you measure density using dimensions or displacement?

Density of a geometric body

Given: Cube edge 4.0 cm; mass 172.8 g. All measurements refer to the same body.
Find: Find average density in g/cm³ and kg/m³.

Approach: Use cube geometry for total volume and then determine the same body’s average density.
$\begin{aligned}V&=a^3\\\rho&=\frac{m}{V}\end{aligned}$
Calculate volume from edge length.
$V=4^3=64\,\mathrm{cm^3}=6.4\cdot10^{-5}\,\mathrm{m^3}$
Divide mass by volume.
$\rho=\frac{172.8}{64}=2.7\,\mathrm{g/cm^3}$
Convert to SI units.
$\rho=2700\,\mathrm{kg/m^3}$

Result: Average density in this calculation is 2.7 g/cm³ or 2700 kg/m³.

Check: Back-substitution: 2700·6.4·10⁻⁵ kg = 0.1728 kg. The value is consistent with aluminium but does not uniquely identify material.

Measuring an irregular body by displacement

Given: Body mass 500 g; liquid readings 350 mL before and 550 mL after immersion. The body is fully submerged, insoluble and nonabsorbing, with no trapped air bubbles.
Find: Find volume and average density.

Approach: For complete bubble-free immersion without absorption, the volume-reading difference equals body volume; divide mass by that volume.
$\begin{aligned}V&=V_{\mathrm{after}}-V_{\mathrm{before}}\\\rho&=\frac{m}{V}\end{aligned}$
Subtract liquid readings.
$V=550-350=200\,\mathrm{mL}=200\,\mathrm{cm^3}$
Divide mass by displaced volume.
$\rho=\frac{500}{200}=2.5\,\mathrm{g/cm^3}$
Find SI density.
$\rho=2500\,\mathrm{kg/m^3}$

Result: Displaced volume is 200 cm³; average density is 2500 kg/m³.

Check: Back-substitution: 2500·0.0002 = 0.5 kg. An assumed volume error of ±2 cm³ with unchanged mass gives a ratio between about 2475 and 2525 kg/m³.

What density do hollow bodies and mixtures have?

For a sealed hollow object, external volume includes the enclosed cavity. Average density may be far below wall-material density. For open pores, displacement depends on whether liquid enters them. Material volume, envelope volume and interparticle spaces in bulk material must not be confused.

Original hollow-body comparison: 60 g of wall material enclosed in an external volume of 100 cm³ gives body density 0.60 g/cm³. That does not establish wall-material density, which requires actual material volume.

\rho_{\mathrm{mix}}=\frac{m_1+m_2}{V_{\mathrm{mix}}}

Use measured total volume; it need not equal V₁ + V₂.

Only additive component volumes give $\rho_{\mathrm{mix}}=\frac{\rho_1V_1+\rho_2V_2}{V_1+V_2}$ . Even then, the arithmetic density mean applies only to equal component volumes. Equal masses require different weighting. Mixing liquids can change volume; use actual mixture volume in that case.

How are density, temperature and pressure related?

Density is not an immutable material number under all conditions. At fixed mass, increasing volume lowers density. With $V_2\approx V_1(1+\beta\Delta T)$ , density becomes $\rho_2\approx\frac{\rho_1}{1+\beta\Delta T}$ within the thermal volume-expansion approximation. Phase changes or large temperature intervals require more. Linear thermal expansion concerns a length first; for freely expanding isotropic material and small changes, approximately $\beta=3\alpha$ . Do not equate the length and volume coefficients unchanged.[2]

Water is a counterexample to the claim that heating always lowers density: between roughly 0 and 4 °C, liquid-water density increases under ordinary pressure conditions. Read tabulated densities with temperature, pressure and, where relevant, material state.[2]

\rho=\frac{pM}{RT}

For an ideal gas this follows from pV = nRT and m = nM. p is absolute pressure in Pa, M molar mass in kg/mol and R the molar gas constant in J/(mol·K); use kelvins for T.[3]

At unchanged pressure and composition, doubling absolute temperature halves ideal-gas density. At fixed volume and particle number, density stays constant while pressure rises. Boundary conditions, not temperature alone, decide. The ideal gas law supplies the state relation used here, with absolute pressure and temperature in kelvins.[3]

When does a body float, and how much volume is submerged?

For free flotation in static equilibrium without other vertical forces, buoyancy balances weight. In homogeneous liquid, $\rho_{\mathrm{fluid}}gV_{\mathrm{sub}}=mg$ . With body average $\rho_{\mathrm{body}}=\frac m{V_{\mathrm{outer}}}$ , submerged volume fraction is $\frac{\rho_{\mathrm{body}}}{\rho_{\mathrm{fluid}}}$ when this ratio is below one.[4]

The original sealed 60 g body of external volume 100 cm³ would be 60% submerged by volume in water modelled at 1 g/cm³. That is not automatically 60% of its height; shape determines the volume–height relationship.

Density alone does not settle cases involving surface tension, external support, strongly stratified fluid or a body filling with liquid. Material identification also needs more than one density number: alloy composition, pores, temperature and uncertainty can yield similar results.

Five problems with solutions

Calculate · Foundation

A sample has mass 84 g and volume 12 cm³. Find density in both common units.

Show hint

First calculate in g/cm³.

Learning goal: Calculate mass density and convert units.

Show solution and calculation

Approach: Average density is mass divided by the corresponding volume.
Grams and cubic centimetres may be used together here; then multiply by 1000 for SI.
Result: 7 g/cm³ or 7000 kg/m³.

Check: Density times 12 cm³ reproduces the original 84 g.

\begin{aligned}\rho&=\frac mV=\frac{84\,\mathrm g}{12\,\mathrm{cm^3}}=7\,\mathrm{g\,cm^{-3}}\\&=7000\,\mathrm{kg\,m^{-3}}\\\rho V&=7\cdot12\,\mathrm g=84\,\mathrm g\end{aligned}

[1]

Rearrange · Foundation

Find mass for 2.5 L at density 800 kg/m³.

Show hint

Convert litres to cubic metres.

Learning goal: Find mass from density and volume.

Show solution and calculation

Approach: Rearrange for mass and convert volume to cubic metres before substitution.
Result: 2 kg.

Check: Two kilograms in 2.5 litres correspond to 0.8 kg per litre, matching the given density. Conversion changes units, not physical mass.

\begin{aligned}m&=\rho V\\V&=2.5\cdot10^{-3}\,\mathrm{m^3}\\m&=800\cdot0.0025\,\mathrm{kg}=2\,\mathrm{kg}\end{aligned}

[1]

Transfer · Application

Two incompressible components with additive volumes each have 100 cm³ and densities 2 and 4 g/cm³. Find overall average density.

Show hint

Add masses and divide by total volume.

Learning goal: Find composite density from additive masses and volumes.

Show solution and calculation

Approach: Calculate each component’s mass, then divide total mass by the additive total volume.
Result: 3 g/cm³.

Check: This lies between both component densities. It equals their arithmetic mean here because component volumes are equal; equal masses would not generally justify that average.

\begin{aligned}m_i&=\rho_iV_i\\m_1&=200\,\mathrm g,\qquad m_2=400\,\mathrm g\\\overline\rho&=\frac{m_1+m_2}{V_1+V_2}=\frac{600\,\mathrm g}{200\,\mathrm{cm^3}}=3\,\mathrm{g\,cm^{-3}}\end{aligned}

[1]

Interpret · Application

A closed body expands by 5% in volume while mass stays constant. Initial density is 1000 kg/m³. Find its new average density. Is an exact 5% reduction correct?

Show hint

New density is inversely proportional to new volume.

Learning goal: Interpret density change during mass-conserving expansion.

Show solution and calculation

Approach: V₂=1.05V₁ with m₂=m₁, giving ρ₂= $\frac{\rho_1}{1.05}$ ≈952.381 kg/m³.
The relative decrease is about 4.7619%, not exactly 5%.

Check: ρ₂V₂=ρ₁V₁ conserves mass. A linear 5% decrease is only approximate; no expansion coefficient is needed because volume change is supplied.

\begin{aligned}\frac{\rho_2}{\rho_1}&=\frac1{1.05}\\\frac{\rho_1-\rho_2}{\rho_1}&=1-\frac1{1.05}\approx0.047619\end{aligned}

[1][2]

Error analysis · Application

Two components each have mass 1.00 kg and densities 1000 and 2000 kg/m³. Their volumes add. A solution averages densities to 1500 kg/m³. Check and correct it.

Show hint

Equal masses imply unequal volumes here.

Learning goal: Use correct mass and volume weights for average density.

Show solution and calculation

Approach: first calculate component volumes $\frac m\rho$ : 0.001 and 0.0005 m³.
Dividing total mass 2 kg by 0.0015 m³ gives 1333.33 kg/m³.
The arithmetic density mean would fit equal component volumes, which are not supplied here.

Check: overall density lies between both component densities and reconstructs their combined volume.

\overline\rho=\frac{m_1+m_2}{\frac{m_1}{\rho_1}+\frac{m_2}{\rho_2}}=\frac{2}{0.0015}\,\mathrm{kg\,m^{-3}}\approx1333.33\,\mathrm{kg\,m^{-3}}

[1]

Key points

Density relates the mass and volume of the same object. Convert both quantities to consistent units and distinguish mean from local density. For composite objects, total mass and total volume determine mean density; simply averaging material densities is insufficient.

Sources for this explanation

Quanta · Updated October 1, 2026

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This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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Change history

October 1, 2026Made the density definition, rearrangements, volume choice, measurement and flotation accessible through precise headings; retained hollow-body and measurement examples and linked the gas law, limited thermal-expansion model and buoyancy in context.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Revised the learning introduction and solution paths; independently checked model assumptions and calculations against primary sources.

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Quanta. Density. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/dichte

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Frequently asked questions about Density

How do you calculate the density of a body?+

Divide the mass by the volume: $\rho{} =\frac{ m}{V}$ . Example: an aluminium cube with $m = 270 g$ and $V = 100 cm^{3}$ has $\rho{} =\frac{ 2.7 g}{cm^{3}}$ , in SI units $\frac{2\,700 kg}{m^{3}}$ . You find the mass with a scale; the volume of regular bodies by measuring, of irregular ones via water displacement in a measuring cylinder, where the rise of the water level equals the body volume. Unit discipline matters: work consistently in grams with cubic centimetres or kilograms with cubic metres, never mixed. For conversion, $\frac{1 g}{cm^{3}} =\frac{ 1\,000 kg}{m^{3}}$ . With the computed density you can then look up in a reference table which substance it probably is.

Why does ice float on water?+

Under ordinary conditions, ice is less dense than liquid water. In the simple static model without other vertical forces, displaced water must have the same mass as the ice body. Using approximate values $\frac{917 kg}{m^{3}}$ for ice and $\frac{1000 kg}{m^{3}}$ for water gives a submerged volume fraction of $\frac{917}{1000} = 0.917$ , about 92%. This is a volume fraction, not generally the same fraction of height. The values depend on temperature, pressure and composition; seawater has a different density. Surface tension, trapped air or snow can further affect real cases. Liquid water also has a density maximum near 4 °C.

How do you rearrange

\rho{} =\frac{ m}{V}

for mass or volume?+

Solved for the mass: $m = \rho{}\cdot{}V$ , which lets you estimate weights without a scale. A 200 litre aquarium holds $m =\frac{ 1\,000 kg}{m^{3}} \times{} 0.2 m^{3} = 200 kg$ of water, which is why you should never carry it filled. Solved for the volume: $V =\frac{ m}{\rho{}}$ , which gives the space required: 7.9 kg of steel with $\rho{} =\frac{ 7\,900 kg}{m^{3}}$ fills exactly $V = 0.001 m^{3}$ , one litre. Density is the conversion factor between the worlds of mass and volume; craftsmen and engineers use it constantly (how heavy will the concrete slab be, how much does the bucket of paint weigh?). After every rearrangement check the units: $\frac{kg}{m^{3}}$ times $m^{3}$ gives kg, then the formula is right.

How do you determine the density of an irregularly shaped body?+

Weigh the body and measure its volume by fully immersing it in a suitable liquid. The difference between volume readings gives displaced volume; 1 mL equals $1 cm^{3}$ . For 500 g and readings changing from 350 to 550 mL, $\rho{} =\frac{ 500}{200} =\frac{ 2.5 g}{cm^{3}}$ . The body must not dissolve, absorb liquid or react with it; attached air bubbles and supporting equipment can distort the measurement. Open pores require clarifying which volume is measured. A density value alone does not identify material uniquely. Comparing reference values requires matching temperature and material state, together with realistic measurement uncertainty.

Why does the density of gases depend so strongly on pressure and temperature?+

Because gas molecules, unlike those in solids and liquids, are far apart and easily compressed or expanded. If you raise the pressure, the particles move closer: same gas, smaller volume, higher density. If you warm the gas at constant pressure, it expands and the density falls, which is why warm air rises, the principle of the hot-air balloon. Quantitatively this sits in the ideal gas law: $\rho{} =\frac{ p\cdot{}M}{(R\cdot{}T)}$ with molar mass M. Air at 0 °C and normal pressure has about $\frac{1.29 kg}{m^{3}}$ , at 20 °C only $\frac{1.20 kg}{m^{3}}$ . That is why gas density figures always come with the conditions (pressure and temperature), whereas for solids the density is nearly constant.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 1, §14.1 Fluids, Density, and Pressure

Related formulas

\Delta L = \alpha L_0 \Delta T

Linear thermal expansion

More formulas: Mechanics

\sum \vec F_{\mathrm{ext}} = \vec F_{\mathrm{res}} = m \vec a

Newton's Second Law

F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}

Newton's Law of Universal Gravitation

E_{kin} = \frac{1}{2} m v^2

Kinetic Energy

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

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Density: mass and the chosen volume

Average density is mass per defined volume; compare material densities with temperature, pressure and material state specified.

BasicExam-relevantLower secondaryUpper secondary

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Symbol

Meaning

Unit

ρ

Density (rho)

kg/m³

m

Mass

V

Volume

m³

Derivation & background

Exam blueprint

Validity range

Derivation steps

Average density is defined as mass divided by its corresponding volume; homogeneity makes it a uniform material density.

1Identify the body and its mass m.
2Define whether V is material, envelope or total volume.
3Use consistent units: kg with $m^{3}$ or g with $cm^{3}$ .
4Form the ratio ρ̄ = m/V with $V > 0$ .
5For homogeneous material in the same state, $m = \rho{}V$ uses uniform ρ.
6For spatially variable density, $m = \int{}\rho{}(r)dV$ ; the overall ratio remains an average.
7Check temperature, pressure, pores and uncertainty before comparing material values.

How do you calculate with Density?

Here is how to work through a typical Density (ρ = m/V) task step by step:

Task

What is the volume of 7.9 kg of steel $(\rho{} =\frac{ 7\,900 kg}{m^{3}})$ ?

Solution path

$V=\frac{m}{\rho}=\frac{7.9\,\mathrm{kg}}{7900\,\mathrm{kg\,m^{-3}}}=0.001\,\mathrm{m^3}$ , or 1 litre.

Task

A wooden block has $m = 0.3 kg$ and $V = 500 cm^{3}$ . Does it float on water?

Solution path

$\rho=\frac{0.3\,\mathrm{kg}}{5\cdot10^{-4}\,\mathrm{m^3}}=600\,\mathrm{kg\,m^{-3}}<1000\,\mathrm{kg\,m^{-3}}$ , so it floats.

What is the density formula, and which volume counts?

\overline\rho=\frac{m}{V},\qquad m=\rho V,\qquad V=\frac{m}{\rho}

Using a single material density in m = ρV assumes homogeneity; the final rearrangement requires ρ > 0.

Equal volume, different mass

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A definition rather than a proof of a physical law

\rho(\vec r)=\frac{\mathrm dm}{\mathrm dV},\qquad m=\int_V\rho(\vec r)\,\mathrm dV

Local continuum density may vary in space; the integral sums masses of small volume elements.

Calculating mass and volume: converting g/cm³ to kg/m³

1\,\mathrm{cm^3}=(10^{-2}\,\mathrm m)^3=10^{-6}\,\mathrm{m^3},\qquad1\,\mathrm{g/cm^3}=1000\,\mathrm{kg/m^3}

Each of the three spatial dimensions contributes a factor 10⁻².

How do you measure density using dimensions or displacement?

Density of a geometric body

Given: Cube edge 4.0 cm; mass 172.8 g. All measurements refer to the same body.
Find: Find average density in g/cm³ and kg/m³.

Approach: Use cube geometry for total volume and then determine the same body’s average density.
$\begin{aligned}V&=a^3\\\rho&=\frac{m}{V}\end{aligned}$
Calculate volume from edge length.
$V=4^3=64\,\mathrm{cm^3}=6.4\cdot10^{-5}\,\mathrm{m^3}$
Divide mass by volume.
$\rho=\frac{172.8}{64}=2.7\,\mathrm{g/cm^3}$
Convert to SI units.
$\rho=2700\,\mathrm{kg/m^3}$

Result: Average density in this calculation is 2.7 g/cm³ or 2700 kg/m³.

Check: Back-substitution: 2700·6.4·10⁻⁵ kg = 0.1728 kg. The value is consistent with aluminium but does not uniquely identify material.

Measuring an irregular body by displacement

Given: Body mass 500 g; liquid readings 350 mL before and 550 mL after immersion. The body is fully submerged, insoluble and nonabsorbing, with no trapped air bubbles.
Find: Find volume and average density.

Approach: For complete bubble-free immersion without absorption, the volume-reading difference equals body volume; divide mass by that volume.
$\begin{aligned}V&=V_{\mathrm{after}}-V_{\mathrm{before}}\\\rho&=\frac{m}{V}\end{aligned}$
Subtract liquid readings.
$V=550-350=200\,\mathrm{mL}=200\,\mathrm{cm^3}$
Divide mass by displaced volume.
$\rho=\frac{500}{200}=2.5\,\mathrm{g/cm^3}$
Find SI density.
$\rho=2500\,\mathrm{kg/m^3}$

Result: Displaced volume is 200 cm³; average density is 2500 kg/m³.

Check: Back-substitution: 2500·0.0002 = 0.5 kg. An assumed volume error of ±2 cm³ with unchanged mass gives a ratio between about 2475 and 2525 kg/m³.

What density do hollow bodies and mixtures have?

\rho_{\mathrm{mix}}=\frac{m_1+m_2}{V_{\mathrm{mix}}}

Use measured total volume; it need not equal V₁ + V₂.

How are density, temperature and pressure related?

\rho=\frac{pM}{RT}

For an ideal gas this follows from pV = nRT and m = nM. p is absolute pressure in Pa, M molar mass in kg/mol and R the molar gas constant in J/(mol·K); use kelvins for T.[3]

When does a body float, and how much volume is submerged?

Five problems with solutions

Calculate · Foundation

A sample has mass 84 g and volume 12 cm³. Find density in both common units.

Show hint

First calculate in g/cm³.

Learning goal: Calculate mass density and convert units.

Show solution and calculation

Approach: Average density is mass divided by the corresponding volume.
Grams and cubic centimetres may be used together here; then multiply by 1000 for SI.
Result: 7 g/cm³ or 7000 kg/m³.

Check: Density times 12 cm³ reproduces the original 84 g.

\begin{aligned}\rho&=\frac mV=\frac{84\,\mathrm g}{12\,\mathrm{cm^3}}=7\,\mathrm{g\,cm^{-3}}\\&=7000\,\mathrm{kg\,m^{-3}}\\\rho V&=7\cdot12\,\mathrm g=84\,\mathrm g\end{aligned}

[1]

Rearrange · Foundation

Find mass for 2.5 L at density 800 kg/m³.

Show hint

Convert litres to cubic metres.

Learning goal: Find mass from density and volume.

Show solution and calculation

Approach: Rearrange for mass and convert volume to cubic metres before substitution.
Result: 2 kg.

Check: Two kilograms in 2.5 litres correspond to 0.8 kg per litre, matching the given density. Conversion changes units, not physical mass.

\begin{aligned}m&=\rho V\\V&=2.5\cdot10^{-3}\,\mathrm{m^3}\\m&=800\cdot0.0025\,\mathrm{kg}=2\,\mathrm{kg}\end{aligned}

[1]

Transfer · Application

Two incompressible components with additive volumes each have 100 cm³ and densities 2 and 4 g/cm³. Find overall average density.

Show hint

Add masses and divide by total volume.

Learning goal: Find composite density from additive masses and volumes.

Show solution and calculation

Approach: Calculate each component’s mass, then divide total mass by the additive total volume.
Result: 3 g/cm³.

Check: This lies between both component densities. It equals their arithmetic mean here because component volumes are equal; equal masses would not generally justify that average.

\begin{aligned}m_i&=\rho_iV_i\\m_1&=200\,\mathrm g,\qquad m_2=400\,\mathrm g\\\overline\rho&=\frac{m_1+m_2}{V_1+V_2}=\frac{600\,\mathrm g}{200\,\mathrm{cm^3}}=3\,\mathrm{g\,cm^{-3}}\end{aligned}

[1]

Interpret · Application

A closed body expands by 5% in volume while mass stays constant. Initial density is 1000 kg/m³. Find its new average density. Is an exact 5% reduction correct?

Show hint

New density is inversely proportional to new volume.

Learning goal: Interpret density change during mass-conserving expansion.

Show solution and calculation

Approach: V₂=1.05V₁ with m₂=m₁, giving ρ₂= $\frac{\rho_1}{1.05}$ ≈952.381 kg/m³.
The relative decrease is about 4.7619%, not exactly 5%.

Check: ρ₂V₂=ρ₁V₁ conserves mass. A linear 5% decrease is only approximate; no expansion coefficient is needed because volume change is supplied.

\begin{aligned}\frac{\rho_2}{\rho_1}&=\frac1{1.05}\\\frac{\rho_1-\rho_2}{\rho_1}&=1-\frac1{1.05}\approx0.047619\end{aligned}

[1][2]

Error analysis · Application

Two components each have mass 1.00 kg and densities 1000 and 2000 kg/m³. Their volumes add. A solution averages densities to 1500 kg/m³. Check and correct it.

Show hint

Equal masses imply unequal volumes here.

Learning goal: Use correct mass and volume weights for average density.

Show solution and calculation

Approach: first calculate component volumes $\frac m\rho$ : 0.001 and 0.0005 m³.
Dividing total mass 2 kg by 0.0015 m³ gives 1333.33 kg/m³.
The arithmetic density mean would fit equal component volumes, which are not supplied here.

Check: overall density lies between both component densities and reconstructs their combined volume.

\overline\rho=\frac{m_1+m_2}{\frac{m_1}{\rho_1}+\frac{m_2}{\rho_2}}=\frac{2}{0.0015}\,\mathrm{kg\,m^{-3}}\approx1333.33\,\mathrm{kg\,m^{-3}}

[1]

Key points

Sources for this explanation

Quanta · Updated October 1, 2026

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Made the density definition, rearrangements, volume choice, measurement and flotation accessible through precise headings; retained hollow-body and measurement examples and linked the gas law, limited thermal-expansion model and buoyancy in context.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Revised the learning introduction and solution paths; independently checked model assumptions and calculations against primary sources.

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Quanta. Density. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/dichte

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Frequently asked questions about Density

How do you calculate the density of a body?+

Why does ice float on water?+

How do you rearrange

\rho{} =\frac{ m}{V}

for mass or volume?+

How do you determine the density of an irregularly shaped body?+

Why does the density of gases depend so strongly on pressure and temperature?+

Density: mass and the chosen volume

Variables & units

Density calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Material properties

Worked example

Applications

How do you calculate with Density?

What is the density formula, and which volume counts?

Equal volume, different mass

A definition rather than a proof of a physical law

Calculating mass and volume: converting g/cm3 to kg/m3

How do you measure density using dimensions or displacement?

Density of a geometric body

Measuring an irregular body by displacement

What density do hollow bodies and mixtures have?

How are density, temperature and pressure related?

When does a body float, and how much volume is submerged?

Five problems with solutions

A sample has mass 84 g and volume 12 cm3. Find density in both common units.

Find mass for 2.5 L at density 800 kg/m3.

Two incompressible components with additive volumes each have 100 cm3 and densities 2 and 4 g/cm3. Find overall average density.

A closed body expands by 5% in volume while mass stays constant. Initial density is 1000 kg/m3. Find its new average density. Is an exact 5% reduction correct?

Two components each have mass 1.00 kg and densities 1000 and 2000 kg/m3. Their volumes add. A solution averages densities to 1500 kg/m3. Check and correct it.

Key points

Sources for this explanation

Frequently asked questions about Density

Subject references and further reading

Related formulas

More formulas: Mechanics

Retain Density for exams

Density: mass and the chosen volume

Variables & units

Density calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Material properties

Worked example

Applications

How do you calculate with Density?

What is the density formula, and which volume counts?

Equal volume, different mass

A definition rather than a proof of a physical law

Calculating mass and volume: converting g/cm3 to kg/m3

How do you measure density using dimensions or displacement?

Density of a geometric body

Measuring an irregular body by displacement

What density do hollow bodies and mixtures have?

How are density, temperature and pressure related?

When does a body float, and how much volume is submerged?

Five problems with solutions

A sample has mass 84 g and volume 12 cm3. Find density in both common units.

Find mass for 2.5 L at density 800 kg/m3.

Two incompressible components with additive volumes each have 100 cm3 and densities 2 and 4 g/cm3. Find overall average density.

A closed body expands by 5% in volume while mass stays constant. Initial density is 1000 kg/m3. Find its new average density. Is an exact 5% reduction correct?

Two components each have mass 1.00 kg and densities 1000 and 2000 kg/m3. Their volumes add. A solution averages densities to 1500 kg/m3. Check and correct it.

Key points

Sources for this explanation

Frequently asked questions about Density

Subject references and further reading

Related formulas

More formulas: Mechanics

Retain Density for exams

Calculating mass and volume: converting g/cm³ to kg/m³

A sample has mass 84 g and volume 12 cm³. Find density in both common units.

Find mass for 2.5 L at density 800 kg/m³.

Two incompressible components with additive volumes each have 100 cm³ and densities 2 and 4 g/cm³. Find overall average density.

A closed body expands by 5% in volume while mass stays constant. Initial density is 1000 kg/m³. Find its new average density. Is an exact 5% reduction correct?

Two components each have mass 1.00 kg and densities 1000 and 2000 kg/m³. Their volumes add. A solution averages densities to 1500 kg/m³. Check and correct it.

Calculating mass and volume: converting g/cm³ to kg/m³

A sample has mass 84 g and volume 12 cm³. Find density in both common units.

Find mass for 2.5 L at density 800 kg/m³.

Two incompressible components with additive volumes each have 100 cm³ and densities 2 and 4 g/cm³. Find overall average density.

A closed body expands by 5% in volume while mass stays constant. Initial density is 1000 kg/m³. Find its new average density. Is an exact 5% reduction correct?

Two components each have mass 1.00 kg and densities 1000 and 2000 kg/m³. Their volumes add. A solution averages densities to 1500 kg/m³. Check and correct it.