# 1.14: Density

Learning Objectives
• Define density.
• Apply an equation to calculate a density.

Density (d) is an intrinsic physical property that is calculated by dividing the mass of an object by its volume.  The equation for calculating a density, based on its definition, can be written as

$$\text{Density} = \dfrac{\text{Mass}}{\text{Volume}}$$

or, using variables, as

$$\text{d} = \dfrac{\text{m}}{\text{V}}$$

Since the quantities in the numerator and denominator must have different units, unit cancelation cannot occur.  Therefore, the unit for density must include both a mass unit and a volume unit, expressed in a fractional format.  Masses of objects are typically measured in grams (g), and volumes can be reported in either milliliters (mL) or cubic centimeters (cm3).  As a result, density can be calculated in both metric (grams per milliliter, g/mL) and SI (grams per cubic centimeter, g/cm3) units.

An intrinsic property is defined as a quantity that does not change, regardless of the amount of substance that is present.  Density is a property of the material itself, so each unique substance has its own corresponding density, which can vary slightly with temperature.  For example, liquid water has a density of 1.00 g/mL (or 1.00 g/cm3), regardless of the quantity of water being studied.  (Since water is such a common and versatile chemical, most chemistry instructors encourage students to memorize as many facts about water as possible.  The density of liquid water is the first such quantity that you should commit to memory.)  The densities of four common liquids are shown below in Table $$\PageIndex{1}$$.

Table $$\PageIndex{1}$$: Densities of Four Common Liquids
Substance Density at 25°C (g/cm3)
corn oil 0.922
liquid water 1.00
honey 1.42
mercury 13.6

Intrinsic properties are non-intuitive, as most people associate having a greater amount of a substance with larger numerical values.  Quantities that do scale with the amount of substance that is present are called extrinsic properties

Example $$\PageIndex{1}$$

A 30.2 milliliter sample of ethanol has a mass of 23.71 grams.  Calculate the density of ethanol.

Solution

The equation

$$\text{d} = \dfrac{\text{m}}{\text{V}}$$

must be applied to complete this calculation.  Since both the mass, 23.71 grams, and the volume, 30.2 milliliters, are provided in units that are appropriate for a density calculation, these values can be directly inserted into the density equation, as shown below.

$$\text{d} = \dfrac{\text{23.71 g}}{\text{30.2 mL}} = {\text {0.7850993... g/mL}}$$

Since density is a calculated quantity, significant figures must be applied.  The quantity in the numerator has 4 significant figures, and the quantity in the denominator has 3 significant figures.  Since the answer must be limited to the lesser count of significant figures, the denominator is the limiting quantity, and the final answer must have 3 significant figures.  Since the number of digits that the answer actually has is more than the number of digits that it should have, the calculated answer must be rounded to 0.785 g/mL.

Exercise $$\PageIndex{1}$$

A 2.130 milliliter sample of acetic acid has a mass of 0.002234 kilograms.  Calculate the density of acetic acid.

The equation

$$\text{d} = \dfrac{\text{m}}{\text{V}}$$

must again be applied to complete this calculation.  However, the mass, 0.002234 kilograms, of the sample is not reported in a unit that is typically appropriate for a density calculation.  Therefore, this mass must first be converted to grams.

$${0.002234 \; \cancel{\rm{kg}}} \times$$ $$\dfrac{1,000 \; \rm{g}}{\cancel{\rm{kg}}}$$ = $${\text {2.234 g}}$$

The density equation can then be used, as shown below.

$$\text{d} = \dfrac{\text{2.234 g}}{\text{2.130 mL}} = {\text {1.04882629... g/mL}}$$

After applying the correct number of significant figures, the density of acetic acid would be reported as 1.049 g/mL.