density and specific gravity are almost the same, so it’s common to simply use the density.
Tip: You may see density reported as g/cm3 or g/cc. These units are the same as g/mL. A cube measuring 1 centimeter on each edge (written as 1 cm3) has a volume of 1 milliliter (1 mL). Because 1 mL = 1 cm3, g/mL and g/cm3 are interchangeable. And because a cubic centimeter (cm3) is commonly abbreviated cc, g/cc also means the same thing. (You hear cc a lot in the medical profession. When you receive a 10 cc injection, you’re getting 10 milliliters of liquid. That’s a lot. You better believe we’re running the other way when we see a nurse coming with a 10 cc shot!)
Measuring density
Calculating density is pretty straightforward. You measure the mass of an object by using a balance or scale, determine the object’s volume, and then divide the mass by the volume.
Determining the volume of liquids is easy, but solids can be tricky. If the object is a regular solid, like a cube, you can measure its three dimensions and calculate the volume by multiplying the length by the width by the height (volume = l × w × h). But if the object is an irregular solid, like a rock, determining the volume is more difficult. With irregular solids, you can measure the volume by using something called Archimedes’ principle.
Archimedes’ principle states that the volume of a solid is equal to the volume of water it displaces. The Greek mathematician Archimedes discovered this concept in the third century BC, greatly simplifying the process for finding an object’s density. Say that you want to measure the volume of a small rock in order to determine its density. First, put some water into a graduated cylinder with markings for every mL and read the volume. (The example in Figure 2-1 shows 25 mL.) Next, put the rock in, making sure that it’s totally submerged, and read the volume again (29 mL in Figure 2-1). The difference in volume (4 mL) is the volume of the rock.
© John Wiley & Sons, Inc.
Figure 2-1: Determining the volume of an irregular solid: Archimedes’ principle.
Tip: Anything with a density lower than water’s floats when put into water, and anything with a density greater than 1 g/mL sinks.
For your pondering pleasure, Table 2-3 lists the density of some common materials.
Table 2-3 Densities of Typical Solids and Liquids
Note that gold has a pretty high density. One of those gold bars that are stored in Fort Knox weighs over 30 pounds. Remember that fact when you see those burglars on TV throwing a bunch of gold bars into a bag, throwing it over their shoulders, and carrying them away.
If you know the density of a substance and either its mass or volume, you can calculate the other.
Examples
Q. Suppose you have a 25.0 mL sample of mercury (d = 13.55 g/mL). What would be the mass of that sample of mercury?
A. The mass of the sample would be 339 grams. To solve, follow these easy steps:
1. Start with the density formula.
2. Switch it around so you’re solving for the mass.
3. Put in your density and volume and solve for the mass (m):
That’s about
Q. A physicist measures the density of a substance to be 20 kg/m3. His chemist colleague, appalled with the excessively large units, decides to change the units of the measurement to the more familiar grams per cubic centimeter. What is the new expression of the density?
A. The density is 0.02 g/cm3. A kilogram contains 1,000 (103) grams, so 20 kg equals 20,000 g. Well, 100 cm = 1 m, so (100 cm)3 = (1 m)3. In other words, there are 1003 (or 106) cubic centimeters in 1 cubic meter. Doing the division gives you 0.02 g/cm3. You can write out the conversion as follows:
Practice Questions
1. The pascal, a unit of pressure, is equivalent to 1 newton per square meter. If the newton, a unit of force, is equal to a kilogram-meter per second squared, what is the pascal expressed entirely in basic units?
2. A student measures the length, width, and height of a sample to be 10 mm, 15 mm, and 5 mm, respectively. If the sample has a mass of 0.9 dag, what is the sample’s density in grams per milliliter?
3. Another student takes some measurements of an unknown solution. She determines that the mass of the solution is 89.5 g and that the volume of the solution is 145.2 mL. What is the density of the unknown solution?
Practice Answers
1.
Now substitute newtons (expressed in fundamental units) into the equation for the pascal:
Simplify this equation to
2. 12 g/mL. Because a milliliter is equivalent to a cubic centimeter, the first thing to do is to convert all the length measurements to centimeters: 1 cm, 1.5 cm, and 0.5 cm. Then multiply the converted lengths to get the volume:
3. 0.616 g/mL. Simply plug the given values for volume and mass into the density formula and solve for density:
Using Conversion Factors
So what happens if you need to convert between one set of units and another, perhaps from a non-SI unit to an SI unit or between SI units? Well, first, you need to understand what a conversion factor is. Second, you need to know how to set up a conversion problem and solve it.
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