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We use the mass and volume of a substance to determine its density. Thus, the units of density are defined by the base units of mass and length.

The density of a substance is the ratio of the mass of a sample of the substance to its volume. The SI unit for density is the kilogram per cubic meter (kg/m3). For many situations, however, this as an inconvenient unit, and we often use grams per cubic centimeter (g/cm3) for the densities of solids and liquids, and grams per liter (g/L) for gases. Although there are exceptions, most liquids and solids have densities that range from about 0.7 g/cm3 (the density of gasoline) to 19 g/cm3 (the density of gold). The density of air is about 1.2 g/L. Table 1 shows the densities of some common substances.

Densities of Common Substances
Solids Liquids Gases (at 25 °C and 1 atm)
ice (at 0 °C) 0.92 g/cm3 water 1.0 g/cm3 dry air 1.20 g/L
oak (wood) 0.60–0.90 g/cm3 ethanol 0.79 g/cm3 oxygen 1.31 g/L
iron 7.9 g/cm3 acetone 0.79 g/cm3 nitrogen 1.14 g/L
copper 9.0 g/cm3 glycerin 1.26 g/cm3 carbon dioxide 1.80 g/L
lead 11.3 g/cm3 olive oil 0.92 g/cm3 helium 0.16 g/L
silver 10.5 g/cm3 gasoline 0.70–0.77 g/cm3 neon 0.83 g/L
gold 19.3 g/cm3 mercury 13.6 g/cm3 radon 9.1 g/L

While there are many ways to determine the density of an object, perhaps the most straightforward method involves separately finding the mass and volume of the object, and then dividing the mass of the sample by its volume.

    \[ \text{density} = \frac{\text{mass}}{\text{volume}} \]

In the following example, the mass is found directly by weighing, but the volume is found indirectly through length measurements.

Calculation of Density Gold—in bricks, bars, and coins—has been a form of currency for centuries. In order to swindle people into paying for a brick of gold without actually investing in a brick of gold, people have considered filling the centers of hollow gold bricks with lead to fool buyers into thinking that the entire brick is gold. It does not work: Lead is a dense substance, but its density is not as great as that of gold, 19.3 g/cm3.

What is the density of lead if a cube of lead has an edge length of 2.00 cm and a mass of 90.7 g?

Solution The density of a substance can be calculated by dividing its mass by its volume. The volume of a cube is calculated by cubing the edge length.

    \[\text{volume of lead cube}=2.00 \si{cm} \times 2.00 \si{cm} \times 2.00 \si{cm}=8.00 \si{cm^3} \] \[\text{density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{90.7 g}}{8.00 \si{cm^3}}=\frac{11.3 \si{g}}{1.00 \si{cm^3}}=11.3 \si{g/cm^3} \]

(We will discuss the reason for rounding to the first decimal place in the next section.)

Check Your Learning

(a) To three decimal places, what is the volume of a cube (cm3) with an edge length of 0.843 cm?

(b) If the cube in part (a) is copper and has a mass of 5.34 g, what is the density of copper to two decimal places?

 

ANSWER:  (a) 0.599 cm3; (b) 8.91 g/cm3

Using Displacement of Water to Determine Density

This PhET simulation illustrates another way to determine density, using displacement of water. Determine the density of the red and yellow blocks.

Solution

When you open the density simulation and select Same Mass, you can choose from several 5.00-kg colored blocks that you can drop into a tank containing 100.00 L water. The yellow block floats (it is less dense than water), and the water level rises to 105.00 L. While floating, the yellow block displaces 5.00 L water, an amount equal to the weight of the block. The red block sinks (it is more dense than water, which has density = 1.00 kg/L), and the water level rises to 101.25 L.

The red block therefore displaces 1.25 L water, an amount equal to the volume of the block. The density of the red block is:

    \[\text{density} = \frac{\text{mass}}{\text{volume}} = \frac{\SI{5.00}{kg} }{\SI{1.25}{L}} = \SI{4.00}{kg/L} \]

Note that since the yellow block is not completely submerged, you cannot determine its density from this information. But if you hold the yellow block on the bottom of the tank, the water level rises to 110.00 L, which means that it now displaces 10.00 L water, and its density can be found:

    \[\text{density} = \frac{\text{mass}}{\text{volume}} = \frac{\SI{5.00}{kg} }{\SI{10.00}{L}} = \SI{0.500}{kg/L} \]

Check Your Learning
Remove all of the blocks from the water and add the green block to the tank of water, placing it approximately in the middle of the tank. Determine the density of the green block.
ANSWER:  2.00 kg/L

Experimental Determination of Density Using Water Displacement

A piece of rebar is weighed and then submerged in a graduated cylinder partially filled with water, with results as shown.

This diagram shows the initial volume of water in a graduated cylinder as 13.5 milliliters. A 69.658 gram piece of metal rebar is added to the graduated cylinder, causing the water to reach a final volume of 22.4 milliliters

(a) Use these values to determine the density of this piece of rebar.

(b) Rebar is mostly iron. Does your result in (a) support this statement? How?

Solution

The volume of the piece of rebar is equal to the volume of the water displaced:

\text{volume} = \SI{22.4}{mL} - \SI{13.5}{mL} = \SI{8.9}{mL} = \SI{8.9}{cm^3}

(rounded to the nearest 0.1 mL, per the rule for addition and subtraction)

The density is the mass-to-volume ratio:

\text{density} = \frac{\text{mass}}{\text{volume}} = \frac{\SI{69.658}{g}}{\SI{8.9}{cm^3}} = \SI{7.8}{g/cm^3}

(rounded to two significant figures, per the rule for multiplication and division)

From [link], the density of iron is 7.9 g/cm3, very close to that of rebar, which lends some support to the fact that rebar is mostly iron.

Check Your Learning

An irregularly shaped piece of a shiny yellowish material is weighed and then submerged in a graduated cylinder, with results as shown.

This diagram shows the initial volume of water in a graduated cylinder as 17.1 milliliters. A 51.842 gram gold colored rock is added to the graduated cylinder, causing the water to reach a final volume of 19.8 milliliters

(a) Use these values to determine the density of this material.

(b) Do you have any reasonable guesses as to the identity of this material? Explain your reasoning.

 

ANSWER:  (a) 19 g/cm3; (b) It is likely gold; the right appearance for gold and very close to the density given for gold in [link].

Key Concepts and Summary

Measurements provide quantitative information that is critical in studying and practicing chemistry. Each measurement has an amount, a unit for comparison, and an uncertainty. Measurements can be represented in either decimal or scientific notation. Scientists primarily use the SI (International System) or metric systems. We use base SI units such as meters, seconds, and kilograms, as well as derived units, such as liters (for volume) and g/cm3 (for density). In many cases, we find it convenient to use unit prefixes that yield fractional and multiple units, such as microseconds (10−6 seconds) and megahertz (106 hertz), respectively.

[1]

Key Equations

  •     \[ \text{density} = \frac{\text{mass}}{\text{volume}} \]

  1. OpenStax Chemistry: Atom First, Chemistry: Atoms First. OpenStax CNX. Nov 13, 2018. Download for free at http://cnx.org/contents/4539ae23-1ccc-421e-9b25-843acbb6c4b0@9.17. ↵

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