How do logs stay afloat in water?
After trees are cut, logging companies often move the raw material down a river to a sawmill where it can be shaped into building materials or other products. The logs float on the water because they are less dense than the water they are in. Knowledge of density is important in the characterization and separation of materials. Information about density allows us to make predictions about the behavior of matter.
Density
A golf ball and a table tennis ball are about the same size. However, the golf ball is much heavier than the table tennis ball. Now imagine a similar size ball made out of lead. That would be very heavy indeed! What are we comparing? By comparing the mass of an object relative to its size, we are studying a property called density. Density is the ratio of the mass of an object to its volume.
\[\text{Density} = \frac{\text{mass}}{\text{volume}}\nonumber \]
Density is an intensive property, meaning that it does not depend on the amount of material present in the sample. Water has a density of \(1.0 \: \text{g/mL}\). That density is the same whether you have a small glass of water or a swimming pool full of water. Density is a property that is constant for the particular identity of the matter being studied.
The SI unit of density is kilograms per cubic meter \(\left( \text{kg/m}^3 \right)\), since the \(\text{kg}\) and the \(\text{m}\) are the SI units for mass and length respectively. In everyday usage in a laboratory, this unit is awkwardly large. Most solids and liquids have densities that are conveniently expressed in grams per cubic centimeter \(\left( \text{g/cm}^3 \right)\). Since a cubic centimeter is equal to a milliliter, density units can also be expressed as \(\text{g/mL}\). Gases are much less dense than solids and liquids, so their densities are often reported in \(\text{g/L}\). Densities of some common substances at \(20^\text{o} \text{C}\) are listed in the table below.
Table \(\PageIndex{1}\): Densities of Some Common Substances
Liquids and Solids 
Density at \(20^\text{o} \text{C} \: \left( \text{g/mL} \right)\) 
Gases 
Density at \(20^\text{o} \text{C} \: \left( \text{g/L} \right)\) 
Ethanol 
0.79 
Hydrogen 
0.084 
Ice \(\left( 0^\text{o} \text{C} \right)\) 
0.917 
Helium 
0.166 
Corn oil 
0.922 
Air 
1.20 
Water 
0.998 
Oxygen 
1.33 
Water \(\left( 4^\text{o} \text{C} \right)\) 
1.000 
Carbon dioxide 
1.83 
Corn syrup 
1.36 
Radon 
9.23 
Aluminum 
2.70 


Copper 
8.92 


Lead 
11.35 


Mercury 
13.6 


Gold 
19.3 


Since most materials expand as temperature increases, the density of a substance is temperature dependent, and usually decreases as temperature increases.
Ice floats in water, and it can be seen from the table that ice is less dense. Corn syrup, being more dense, would sink if placed in water.
Example \(\PageIndex{1}\)
An \(18.2 \: \text{g}\) sample of zinc metal has a volume of \(2.55 \: \text{cm}^3\). Calculate the density of zinc.
Solution
Step 1: List the known quantities and plan the problem.
Known
 Mass \(= 18.2 \: \text{g}\)
 Volume \(= 2.55 \: \text{cm}^3\)
Unknown
Use the equation for density, \(D = \frac{m}{V}\), to solve the problem.
Step 2: Calculate.
\[D = \frac{m}{V} = \frac{18.2 \: \text{g}}{2.55 \: \text{cm}^3} = 7.14 \: \text{g/cm}^3\nonumber \]
Step 3: Think about your result.
If \(1 \: \text{cm}^3\) of zinc has a mass of about 7 grams, then 2 and a half \(\text{cm}^3\) will have a mass about 2 and a half times as great. Metals are expected to have a density greater than that of water, and zinc's density falls within the range of the other metals listed above.
Since density values are known for many substances, density can be used to determine an unknown mass or an unknown volume. Dimensional analysis will be used to ensure that units cancel appropriately.
Example \(\PageIndex{2}\)
 What is the mass of \(2.49 \: \text{cm}^3\) of aluminum?
 What is the volume of \(50.0 \: \text{g}\) of aluminum?
Solution
Step 1: List the known quantities and plan the problem.
Known
 Density \(= 2.70 \: \text{g/cm}^3\)
 1. Volume \(= 2.49 \: \text{cm}^3\)
 2. Mass \(= 50.0 \: \text{g}\)
Unknown
 1. Mass \(= ? \: \text{g}\)
 2. Volume \(= ? \: \text{cm}^3\)
Use the equation for density, \(D = \frac{m}{V}\), and dimensional analysis to solve each problem.
Step 2: Calculate.
\[1. \: \: 2.49 \: \text{cm}^3 \times \frac{2.70 \: \text{g}}{1 \: \text{cm}^3} = 6.72 \: \text{g}\nonumber \]
\[2. \: \: 50.0 \: \text{g} \times \frac{1 \: \text{cm}^3}{2.70 \: \text{g}} = 18.5 \: \text{cm}^3\nonumber \]
In problem 1, the mass is equal to the density multiplied by the volume. In problem 2, the volume is equal to the mass divided by the density.
Step 3: Think about your results.
Because a mass of \(1 \: \text{cm}^3\) of aluminum is \(2.70 \: \text{g}\), the mass of about \(2.5 \: \text{cm}^3\) should be about 2.5 times larger. The \(50 \: \text{g}\) of aluminum is substantially more than its density, so that amount should occupy a relatively large volume.