Density of fluids

Pick up a bottle of honey and a bottle of water of the same size, and the honey feels heavier in your hand. Both are liquids, both fill the same volume, yet one packs more mass into that space than the other. That packed-in-ness is density, and for fluids — liquids and, in spirit, gases — it is measured the same careful way every time: weigh the fluid, know its volume exactly, and divide. This article extends the density idea you already know for solids to liquids, introduces the neat trick of a density bottle for getting an exact volume without reading a scale, and shows how comparing any liquid's density to water's gives a unit-less number called relative density that tells you, at a glance, whether it will float or sink.

The core insight

The density of a fluid is its mass divided by its volume, $\rho = m/V$ , and comparing that to water's density gives the unit-less relative density that predicts floating and sinking.

The core idea

Density is defined identically for solids, liquids and gases — it never changes meaning, only the method of measuring $m$ and $V$ changes with the state of matter.

\rho = \frac{m}{V}

where: $\rho$ is density, $m$ is mass, and $V$ is volume. For a liquid, mass is found by weighing (weigh the container, then weigh container-plus-liquid, and subtract), while volume can be read off a graduated measuring cylinder at eye level, at the bottom of the meniscus. A density bottle gives a far more precise volume than reading a cylinder scale. It is a small flask of exactly known fixed volume (often $50\ \text{cm}^3$ ), closed with a stopper that has a fine capillary hole through it. Fill the bottle to the brim, push the stopper in, and any excess liquid squeezes out through the capillary — leaving exactly the bottle's fixed volume inside, every single time, with no scale-reading error at all. To compare fluids without worrying about units, we use relative density (RD), the ratio of a substance's density to that of water at $4$ °C:

\text{RD} = \frac{\text{Density of substance}}{\text{Density of water}}

where: both densities are in the same units, so RD is a pure number with no units at all. Because water's density is exactly $1\ \text{g/cm}^3$ ( $1000\ \text{kg/m}^3$ ), a liquid's RD is numerically identical to its density measured in g/cm³ — a convenient shortcut. If RD is greater than $1$ , the liquid is denser than water and sinks in it (glycerine, RD about $1.26$ ); if RD is less than $1$ , it floats on water (kerosene, RD about $0.8$ ). Gases fit the same formula, $\rho = m/V$ , but their particles are spread far apart compared with liquids, so a given volume of gas holds far less mass — gas densities are typically hundreds of times smaller than liquid densities. Precise gas-density measurement uses special apparatus and is not required at this level; the formula and the comparison idea are what matter here.

Figure

Two identical measuring cylinders side by side, one holding kerosene which floats a wooden cube and one holding glycerine in which the same cube sinks, illustrating relative density less than one and greater than one

Derivation AGetting relative density from a density bottle

The goal is to turn two simple weighings and one known fixed volume into a unit-less relative density, showing exactly why the volume never needs to be read off any scale and why the water reference is what cancels the units away in the final ratio. Take an empty density bottle of fixed volume $V$ , and suppose it is later filled first with the unknown liquid and separately with water, each time reading only a mass on a balance.

Step 1. Weigh the empty density bottle to get its own mass, $m_{\text{bottle}}$ . This value is needed only so it can be subtracted off later; it plays no role in the final answer beyond that.

Step 2. Fill the bottle completely with the unknown liquid, using the capillary-hole stopper so the volume is exactly the bottle's fixed value $V$ with no excess left inside. Weigh the full bottle to get $m_{\text{bottle+liquid}}$ , and subtract to isolate the liquid's mass: $m_{\text{liquid}} = m_{\text{bottle+liquid}} - m_{\text{bottle}}$ .

Step 3. Compute the liquid's density using the bottle's known fixed volume: $\rho_{\text{liquid}} = m_{\text{liquid}} / V$ . Because $V$ is fixed and certified by the bottle's manufacture, this density is far more trustworthy than one obtained by reading a graduated cylinder.

Step 4. Repeat with water in the very same bottle to get $m_{\text{water}} = m_{\text{bottle+water}} - m_{\text{bottle}}$ , so $\rho_{\text{water}} = m_{\text{water}}/V$ . Divide the two densities: $\text{RD} = \rho_{\text{liquid}}/\rho_{\text{water}} = m_{\text{liquid}}/m_{\text{water}}$ , since the identical volume $V$ appears in both and cancels exactly. This is the elegant payoff: relative density can be found from two mass readings alone, without ever calculating either density separately, because the same bottle guarantees the same volume both times.

Worked example

A density bottle of fixed volume $V = 50\ \text{cm}^3$ is used to find the relative density of a liquid, with density of water taken as $1\ \text{g/cm}^3$ . Mass of empty density bottle: $20\ \text{g}$ . Mass of density bottle completely filled with the liquid: $65\ \text{g}$ .

Step 1. Find the mass of the liquid alone by subtracting the bottle's own mass. $m = 65 - 20 = 45\ \text{g}$ .

Step 2. Find the density of the liquid using the bottle's known fixed volume. $\rho = \dfrac{m}{V} = \dfrac{45}{50} = 0.9\ \text{g/cm}^3$ .

Step 3. Find the relative density by dividing by water's density. $\text{RD} = \dfrac{\rho_{\text{liquid}}}{\rho_{\text{water}}} = \dfrac{0.9}{1} = 0.9$ . Since $\text{RD} = 0.9 < 1$ , this liquid is less dense than water and floats on it — consistent with a light oil. The liquid's density is $0.9\ \text{g/cm}^3$ and its relative density is $0.9$ .

Competitive stretch

Suppose the same bottle, filled instead with milk, weighs $140\ \text{g}$ (bottle $40\ \text{g}$ + milk $100\ \text{g}$ found earlier by a measuring cylinder), giving $\rho = 100/50 = 2.0\ \text{g/cm}^3$ if the bottle's volume were $50\ \text{cm}^3$ — but real milk sits close to $1\ \text{g/cm}^3$ , so this mismatch is a quick way to sanity-check that the volume you used actually matches the container that produced the mass reading; mixing up a cylinder reading with a different-sized density bottle is a classic source of wildly wrong answers.

Common traps

Forgetting to subtract the container's mass. The liquid's mass alone is (container + liquid) minus (empty container), never the total weighing by itself.
Reading the measuring cylinder scale incorrectly. Always read the volume at eye level, at the bottom of the curved liquid surface (meniscus).
Thinking relative density has units. RD is a pure ratio of two densities, so it is always unit-less — never write RD in g/cm³ or kg/m³.
Assuming "denser" always means "sinks." A liquid with RD greater than $1$ sinks below water (like glycerine); RD less than $1$ floats above it (like kerosene) — the direction depends on the comparison, not just the word "dense."

Ask the tutor: Want more practice turning mass-and-volume readings into density and relative density, or a walkthrough of how RD connects to floating and sinking objects?

Why does a density bottle give a better answer than a measuring cylinder?

A measuring cylinder relies on you reading a printed scale at the liquid's meniscus, and even a careful reading has some error from parallax or the coarseness of the markings. A density bottle sidesteps all of that: its capillary-hole stopper forces out any extra liquid the instant it is inserted, so the volume inside is always exactly the bottle's certified fixed value, no matter how carefully or carelessly you poured. The only readings you take are masses on a balance, which are far more precise than a volume read by eye.

If relative density has no units, how is it useful at all?

Precisely because it has no units, RD lets you compare any two liquids directly without worrying about whether one was measured in g/cm³ and another in kg/m³ — the units cancel in the ratio. It also gives an instant floating-or-sinking prediction: since water is the reference at RD $=1$ , any liquid with RD above $1$ sinks in water and any liquid with RD below $1$ floats on it, which is exactly the kind of quick, practical judgement RD was designed for.

Does relative density change if I use a bigger density bottle?

No. Relative density is $\rho_{\text{liquid}}/\rho_{\text{water}}$ , and both densities are mass divided by the same volume $V$ , so $V$ cancels out of the ratio entirely — $\text{RD} = m_{\text{liquid}}/m_{\text{water}}$ for the same bottle. A bigger bottle simply gives bigger mass readings for both liquid and water, but their ratio, and hence the RD, comes out identical.

Density of fluids

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads