Solve density and unit-conversion problems using dimensional analysis. Explain why units guide problem setup and help catch errors.
A lesson of Chemistry, in Matter, Measurement, and Chemical Thinking, about 35 minutes.
By the end of this lesson, you will be able to compute and rearrange density, convert between any units by tracking them, use density as a conversion factor, and let the units tell you when a setup is wrong.
Density Is Mass Per Volume
$$\rho = \frac{m}{V}$$
How much mass is packed into a given space. It is an intensive property, so it
identifies a substance regardless of how much you have: a gram of gold and a
kilogram of gold both come in at $19.3$ g/cm³.
Less dense floats on more dense. Ice at $0.92$ floats on water at $1.00$, and
that is close to unique: almost every substance is denser as a solid than as a
liquid, because cooling packs particles closer. Water expands when it freezes
because its molecules lock into an open hexagonal arrangement, and lakes
therefore freeze from the top down rather than solid from the bottom, which is
the reason fish survive winters.
$1$ cm³ $= 1$ mL exactly, by definition, so g/cm³ and g/mL are the same unit
written two ways.
Rearranging It
$$\rho = \frac{m}{V} \qquad m = \rho V \qquad V = \frac{m}{\rho}$$
One relationship, three forms. Do not memorize three formulas, rearrange the
one, which is safer because a misremembered rearrangement is invisible while a
misremembered formula is not.
And the units check the rearrangement for you. If you want a volume and your
setup produces g²/cm³, the arrangement is wrong and you know before computing
anything.
Example 1: Finding a Density
Here is every step, worked all the way through.
A metal block has mass $54.0$ g and measures $2.0 \times 3.0 \times 1.5$ cm.
What is the volume, in cubic centimeters? 9.0 cm^3
Divide the mass by the volume. What is the density, in g/cm^3? 6.0 g/cm^3
The measurements had 2 significant figures. How many should the answer have? 2
Comparing with the table, is this denser or less dense than aluminum? Type one word denser
Displacement: Volume of an Awkward Shape
A block can be measured with a ruler. A stone cannot.
Submerge it and read how much the water rises. The rise equals the object's
volume, because the object pushed exactly its own volume of water aside.
Water at $25.0$ mL. Add the stone: water reads $32.4$ mL.
$$V = 32.4 - 25.0 = 7.4 \text{ mL}$$
Subtract, do not read the final level. The final reading includes the water
that was already there.
It only works if the object sinks and does not dissolve or absorb water. A
floating object displaces only part of its volume, and a sponge gives a reading
that is meaningless.
Worked Example 1: Displacement
A stone of mass $58.2$ g raises water from $30.0$ mL to $52.0$ mL.
Dimensional Analysis: Let Units Do the Work
The most useful technique in all of quantitative chemistry, and it is one idea:
multiply by fractions that equal $1$, arranged so unwanted units cancel.
The km canceled because one was on top and one on the bottom.
Set it up so the unit you are removing sits opposite the one you have. If km
is on top in your quantity, it must be on the bottom in your conversion factor. If
the units do not cancel, the fraction is upside down, and that is the entire
error-detection system.
Same density, opposite orientation, chosen by what needed canceling. You never
have to decide whether to multiply or divide, which is the whole point: the
units decide, and they are visible on the page.
Worked Example 2: Chained Conversion
Convert $2.5$ g/cm³ into kg/m³.
Worked Example 3: Density Both Ways
Gold has density $19.3$ g/cm³.
Common Mistakes
Mistake 1: Reading the final water level as the volume
Subtract the starting level.
Mistake 2: A conversion factor upside down
If the units do not cancel, flip it. That is what the units are for.
Mistake 3: Not cubing a cubed unit
$100$ cm/m becomes $10^6$ cm³/m³.
Mistake 4: Dropping the units mid-calculation
Carry them through every line; they are the error check.
Mistake 5: Treating density as extensive
A gram and a kilogram of gold have the same density.
Where This Goes Next
Unit tracking is not a beginner's crutch. It is a genuine method that professional
scientists rely on, and it has a formal name.
Dimensional analysis treats every quantity as a combination of base
dimensions: mass, length, time. Any correct equation must have the same
dimensions on both sides, which is a real constraint on what an equation can
possibly be. Given that a pendulum's period might depend on its length, its mass
and gravity, dimensional analysis alone forces $T \propto \sqrt{L/g}$, because no
other combination has units of time. The mass cannot appear, and that surprising
prediction is correct.
Physicists use this to guess the form of results before deriving them, and to
catch errors instantly: an answer for an energy that has units of force is wrong,
no matter how careful the algebra looked.
The cost of not doing it is well documented. The Mars Climate Orbiter was lost in
1999 because one team worked in pound-seconds and another in newton-seconds. A
$$125$ million spacecraft flew into the atmosphere and broke up, and unit
tracking would have caught it.
Carry the units. They are the cheapest error-detection system available, and they
work even when you have no idea whether the answer is reasonable.
Check Your Understanding
Five of them, no instructions this time.
Summary
$$\rho = \frac{m}{V}$$
Density is intensive, so it identifies a substance at any amount, and less
dense floats on more dense.
Displacement gives the volume of an awkward shape, by subtraction.
Dimensional analysis means multiplying by fractions equal to $1$, arranged so
unwanted units cancel. If they do not cancel, the factor is upside down.
Density is a conversion factor between mass and volume, usable either way up.
Cubed units convert by the cube of the factor, and carrying units through is
the cheapest error check there is.