Why do objects float or sink, and how does density explain it?

An object floats in a liquid if its density is less than the liquid's density. Ice at about 0.92\ g/cm^3 floats on liquid water at 1.00\ g/cm^3; a steel bolt at about 7.9\ g/cm^3 sinks. Size does not matter — a huge log floats while a tiny pebble sinks, because floating depends on the density ratio, not on total mass.

Is g/mL the same as g/cm^3?

Yes. One milliliter is defined as exactly one cubic centimeter, so the two density units are numerically identical and interchangeable. Chemists tend to write g/mL for liquids and g/cm^3 for solids purely out of habit, but you can swap them in any calculation without a conversion factor.

How do I know whether to multiply or divide by the density?

Let the units decide instead of memorizing. Write the given quantity, then write density as a fraction oriented so the given unit sits in the denominator and cancels. Going from grams to volume, use 1\ mLD\ g; going from volume to grams, use D\ g1\ mL. If the leftover units are wrong, the factor is upside down.

Does density change with temperature?

Yes, for almost every substance. Heating usually expands a sample, so the same mass occupies more volume and the density drops. That is why reference tables list a temperature, and why a density measured on a hot sample may not match the room-temperature table value closely enough to identify it.

CHEM-1.4

Density & Dimensional Analysis

Learn to calculate density from mass and volume, use density to identify unknown substances, and cancel units reliably with the factor-label method.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Density & Dimensional Analysis, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Density & Dimensional Analysis free

What this lesson covers

A 1 kilogram block of lead and a 1 kilogram bag of feathers have the same mass, but nobody would confuse them. The difference is density — how much mass is packed into each unit of volume. Density is one of the most useful properties in chemistry because it is intensive: a chip of aluminum and a solid aluminum bar have exactly the same density, so a single measured number can tell you what an unknown sample is made of.

In this lesson you will calculate density from measured mass and volume, flip density around and use it as a conversion factor to go from grams to milliliters (or back), and set up chains of conversion factors so unwanted units cancel line by line. The factor-label method you learn here shows up in every later unit — stoichiometry, gas laws, and solution concentration are all dimensional analysis with chemistry-flavored conversion factors.

Density: Mass Packed Into Volume

Density is the ratio of a sample's mass to the volume it occupies:

D = \frac{m}{V}

Solids are usually reported in

\text{g/cm}^3

and liquids in

\text{g/mL}

. Those units are interchangeable because

1\ \text{cm}^3 = 1\ \text{mL}

exactly — a fact worth memorizing now, since it lets you slide between solid and liquid volume units without thinking. Gases are so spread out that chemists report their densities in

\text{g/L}

instead.

The formula rearranges two ways:

m = D \times V

and

V = \frac{m}{D}

. You do not have to memorize all three forms if you keep units straight, because the units themselves tell you whether to multiply or divide.

Mass comes from a balance. Volume comes from a graduated cylinder for liquids, from length times width times height for a regular solid, or from water displacement for an irregular solid: record the water level, drop the object in, record the new level, and subtract.

The most important conceptual point is that density is an intensive property. Cut an aluminum bar in half and each half has half the mass and half the volume, so the ratio is unchanged. Mass and volume are extensive — they depend on how much you have. That is exactly why density can identify a substance while mass alone cannot. A common error is assuming a heavier object is denser; a 500 gram foam block is far less dense than a 50 gram steel bolt.

Because density comes from measured values, apply the significant-figure rules from the previous lesson to your final answer rather than reporting every digit your calculator shows.

Using Density as a Conversion Factor

Any equality can be written as a fraction equal to 1, and a density is an equality in disguise. Saying that mercury has a density of

13.6\ \text{g/mL}

is saying that

13.6\ \text{g}

of mercury and

1\ \text{mL}

of mercury are the same sample. So you may write either

\frac{13.6\ \text{g}}{1\ \text{mL}} \qquad \text{or} \qquad \frac{1\ \text{mL}}{13.6\ \text{g}}

You pick the orientation that puts the unit you want to cancel in the denominator. Converting 40.8 g of mercury to a volume, grams must cancel, so grams goes on the bottom:

40.8\ \text{g} \times \frac{1\ \text{mL}}{13.6\ \text{g}} = 3.00\ \text{mL}

Had you multiplied by

13.6\ \text{g}/1\ \text{mL}

instead, the answer would carry units of

\text{g}^2/\text{mL}

— nonsense, which is precisely the warning system dimensional analysis gives you.

Identification works in reverse. Measure a sample's density, then compare to a reference table.

Substance	Density ( $\text{g/cm}^3$ at room temperature)
Ethanol	0.789
Water	1.00
Aluminum	2.70
Iron	7.87
Copper	8.96
Lead	11.3
Mercury	13.6
Gold	19.3

If your calculated density is

2.7\ \text{g/cm}^3

, the sample is consistent with aluminum. Say "consistent with" rather than "is" — other materials can share a density, and density alone is not proof of identity. It is strong evidence that must agree with other properties such as color, melting point, and reactivity. Density also changes with temperature, so reference tables always state the temperature they apply to.

The Factor-Label Method, Step by Step

The factor-label method converts units by multiplying by fractions that equal 1. Because multiplying by 1 never changes the amount, only the labels change.

Step	What you do	Example: 2.50 kg to milligrams
1	Write the given quantity with its unit	$2.50\ \text{kg}$
2	Write the target unit you need	mg
3	List equalities that bridge them	$1\ \text{kg} = 1000\ \text{g}$ ; $1\ \text{g} = 1000\ \text{mg}$
4	Build fractions so old units cancel	$\frac{1000\ \text{g}}{1\ \text{kg}}$ , then $\frac{1000\ \text{mg}}{1\ \text{g}}$
5	Multiply across, divide, check units	$2.50 \times 1000 \times 1000 = 2.50 \times 10^{6}\ \text{mg}$

Written as one chain:

2.50\ \text{kg} \times \frac{1000\ \text{g}}{1\ \text{kg}} \times \frac{1000\ \text{mg}}{1\ \text{g}} = 2.50 \times 10^{6}\ \text{mg}

Cross out each canceled unit with your pencil as you go. If the only unit left standing is the one you wanted, the setup is right — even before you touch a calculator. This is the real power of the method: it checks your reasoning, not just your arithmetic.

Two practical notes. First, metric definitions such as

1\ \text{kg} = 1000\ \text{g}

are exact and never limit significant figures; measured values in the problem do. Second, do the whole chain in one calculator pass instead of rounding at each intermediate step, since repeated rounding drifts the answer.

When the given quantity is a ratio like a speed or a density, convert the top and the bottom separately in the same chain. Nothing about the procedure changes; you just have two units to chase instead of one.

Squared, Cubed, and Ratio Units

Volume conversions trip up more students than any other part of this topic. The equality

1\ \text{m} = 100\ \text{cm}

applies to length. To convert cubic meters to cubic centimeters you must cube the entire conversion factor, parentheses and all:

\left(\frac{100\ \text{cm}}{1\ \text{m}}\right)^{3} = \frac{10^{6}\ \text{cm}^3}{1\ \text{m}^3}

2.5\ \text{m}^3 = 2.5 \times 10^{6}\ \text{cm}^3

, not

250\ \text{cm}^3

. The number 100 gets cubed too, which is why the factor is a million and not a hundred. The same idea applies to area with an exponent of 2. A quick sanity check: a cubic meter is a box one meter on each side, and you could clearly fit far more than 250 sugar-cube-sized centimeter cubes inside it.

Ratio units need a factor for the numerator and another for the denominator. To express a density of

8.96\ \text{g/cm}^3

\text{kg/m}^3

\frac{8.96\ \text{g}}{1\ \text{cm}^3} \times \frac{1\ \text{kg}}{1000\ \text{g}} \times \frac{10^{6}\ \text{cm}^3}{1\ \text{m}^3} = 8960\ \frac{\text{kg}}{\text{m}^3}

Grams cancel against the top of the first factor; cubic centimeters in the bottom cancel against the cubic centimeters in the numerator of the second.

Where students actually go wrong: forgetting the exponent on the factor, canceling

\text{cm}

against

\text{cm}^3

as though they were the same unit, and inverting a density because they grabbed the number without asking which unit needed to disappear. Slow down at the setup stage and let the canceling decide the arrangement for you — that habit carries directly into mole conversions later in the course.

Key terms

Density.: The ratio of a substance's mass to its volume, $D = m/V$ , commonly reported in $\text{g/cm}^3$ for solids, $\text{g/mL}$ for liquids, and $\text{g/L}$ for gases.
Intensive property.: A property whose value does not depend on the amount of substance present, such as density, melting point, or color.
Extensive property.: A property that changes with sample size, such as mass, volume, or length.
Conversion factor.: A fraction built from an equality that equals 1, used to change units without changing the quantity, for example $\frac{1000\ \text{mg}}{1\ \text{g}}$ .
Factor-label method (dimensional analysis).: A problem-solving technique in which a quantity is multiplied by conversion factors arranged so unwanted units cancel and only the target unit remains.
Water displacement.: A method for finding the volume of an irregular solid by measuring how much the liquid level rises when the object is submerged.
Exact number.: A defined or counted value, such as $1\ \text{kg} = 1000\ \text{g}$ , that is treated as having unlimited significant figures.

Worked example

A student masses an irregular chunk of metal and finds 66.3 g. A graduated cylinder holds 25.0 mL of water; after the metal is fully submerged, the level reads 32.4 mL. (a) Find the density of the metal. (b) Identify the metal using the reference table. (c) Express that density in

\text{kg/m}^3

Part (a). First get the volume by displacement. The metal pushed the water level up, and the amount of rise equals the metal's volume:

V = 32.4\ \text{mL} - 25.0\ \text{mL} = 7.4\ \text{mL}

Both readings have one decimal place, so the difference keeps one decimal place: 7.4 mL, which is two significant figures.

Now divide mass by volume:

D = \frac{66.3\ \text{g}}{7.4\ \text{mL}} = 8.959...\ \text{g/mL} \approx 9.0\ \text{g/mL}

The volume limits the answer to two significant figures, so report

9.0\ \text{g/mL}

.

Part (b). Since

1\ \text{mL} = 1\ \text{cm}^3

, this is

9.0\ \text{g/cm}^3

. Scanning the table, copper is

8.96\ \text{g/cm}^3

— the only close match. Iron at 7.87 and lead at 11.3 are both far outside the measurement's range. Conclusion: the sample is consistent with copper. A complete answer says "consistent with" and notes that a second property, such as its reddish color, supports the identification.

Part (c). Convert the numerator and denominator in one chain, using the unrounded value

8.96\ \text{g/cm}^3

for the metal:

\frac{8.96\ \text{g}}{1\ \text{cm}^3} \times \frac{1\ \text{kg}}{1000\ \text{g}} \times \left(\frac{100\ \text{cm}}{1\ \text{m}}\right)^{3} = \frac{8.96\ \text{g}}{1\ \text{cm}^3} \times \frac{1\ \text{kg}}{1000\ \text{g}} \times \frac{10^{6}\ \text{cm}^3}{1\ \text{m}^3}

Grams cancel, cubic centimeters cancel, and

8.96 \times 10^{6}/1000 = 8960

. The result is

8.96 \times 10^{3}\ \text{kg/m}^3

. Notice that the numerical value jumped by a factor of 1000 — a useful shortcut to remember for converting any

\text{g/cm}^3

value to

\text{kg/m}^3

Practice questions

Ethanol has a density of

0.789\ \text{g/mL}

. What volume of ethanol has a mass of 155 g?

122 mL
155 mL
196 mL
0.00509 mL

Answer: 196 mL

Grams must cancel, so density goes in the flipped orientation:

155\ \text{g} \times \frac{1\ \text{mL}}{0.789\ \text{g}} = 196\ \text{mL}

. The choice 122 mL comes from multiplying by 0.789 instead of dividing — a setup that would leave units of

\text{g}^2/\text{mL}

. A quick reasoning check catches the error too: ethanol is less dense than water, so 155 g of it must take up more than 155 mL, and only 196 mL fits that expectation.

A solid cube measures 2.00 cm on each edge and has a mass of 21.6 g. Calculate its density, identify a likely substance, and explain why a classmate's claim that "a bigger piece of the same metal would have a higher density" is incorrect.

Answer: Density is $2.70\ \text{g/cm}^3$ , consistent with aluminum; density is intensive, so a bigger piece has the same density.

Volume of a cube is edge cubed:

(2.00\ \text{cm})^3 = 8.00\ \text{cm}^3

— remember to cube the number as well as the unit. Then

D = 21.6\ \text{g} \div 8.00\ \text{cm}^3 = 2.70\ \text{g/cm}^3

, which matches aluminum in the reference table. The classmate is confusing mass with density. Doubling the size of the sample doubles both the mass and the volume, so the ratio

m/V

is unchanged. Density is intensive; mass and volume are extensive. That invariance is exactly what makes density usable as an identifying property.

How many cubic centimeters are in

0.45\ \text{m}^3

45 $\text{cm}^3$
$4.5 \times 10^{3}\ \text{cm}^3$
$4.5 \times 10^{5}\ \text{cm}^3$
$4.5 \times 10^{7}\ \text{cm}^3$

Answer: $4.5 \times 10^{5}\ \text{cm}^3$

Cube the whole length factor:

\left(\frac{100\ \text{cm}}{1\ \text{m}}\right)^3 = \frac{10^{6}\ \text{cm}^3}{1\ \text{m}^3}

. Then

0.45\ \text{m}^3 \times 10^{6}\ \text{cm}^3/\text{m}^3 = 4.5 \times 10^{5}\ \text{cm}^3

. The answer 45 comes from using 100 instead of

10^6

, the most frequent mistake with cubed units. Sanity check: a cube 1 m on a side holds a million centimeter cubes, so a fraction of a cubic meter should still be in the hundreds of thousands.

FAQ

Why do objects float or sink, and how does density explain it?: An object floats in a liquid if its density is less than the liquid's density. Ice at about $0.92\ \text{g/cm}^3$ floats on liquid water at $1.00\ \text{g/cm}^3$ ; a steel bolt at about $7.9\ \text{g/cm}^3$ sinks. Size does not matter — a huge log floats while a tiny pebble sinks, because floating depends on the density ratio, not on total mass.
Is $\text{g/mL}$ the same as $\text{g/cm}^3$ ?: Yes. One milliliter is defined as exactly one cubic centimeter, so the two density units are numerically identical and interchangeable. Chemists tend to write $\text{g/mL}$ for liquids and $\text{g/cm}^3$ for solids purely out of habit, but you can swap them in any calculation without a conversion factor.
How do I know whether to multiply or divide by the density?: Let the units decide instead of memorizing. Write the given quantity, then write density as a fraction oriented so the given unit sits in the denominator and cancels. Going from grams to volume, use $\frac{1\ \text{mL}}{D\ \text{g}}$ ; going from volume to grams, use $\frac{D\ \text{g}}{1\ \text{mL}}$ . If the leftover units are wrong, the factor is upside down.
Does density change with temperature?: Yes, for almost every substance. Heating usually expands a sample, so the same mass occupies more volume and the density drops. That is why reference tables list a temperature, and why a density measured on a hot sample may not match the room-temperature table value closely enough to identify it.

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