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There are physical characteristics of a substance that help identify the substance. One of these characteristics is density. Density (whose most common symbol is the lowercase letter d) is defined as mass per unit volume. Density is calculated by dividing the mass of an object by its volume. This is shown in equation form, as follows:

Density = mass ÷ volume

By the way, the lower-case Greek letter rho, ρ, is also used to symbolize density. Oftentimes, the rho shape that a textbook would use looks more like the lower-case letter p. However, a lower-case d is more often used in intoductory settings like the one you are currently reading.

We can calculate the density of a solid, liquid, or gas. The density of a gas will be dealt with in a later unit, because its density is very sensitive to temperature and pressure. Although the density of liquids and solids do change with temperature and pressure changes, the amount is fairly small. We will ignore these small amounts and act as if all our density problems are at the same temperature and pressure. Note the difference in units in the formulas of the density of a solid and liquid. The unit for cubic centimeters is cm^{3} and for milliliters is mL.

solids: d = grams ÷ cubic centimetersliquids: d = grams ÷ milliliters

Since one mL equals one cm^{3}, there is no functional difference between g/cm^{3} and g/mL.

This image will help you in figuring out how to solve density problems:

Simply cover up whichever value you need to calculate and the other two are shown in their proper placement, be it to multiply or to divide.

For example, cover up the M. This leave you with dV (ignore the fact that it is in the denominator). Density times volume will give you mass. You can also check it out by way of the units: (g / cm^{3}) x cm^{3} cancels out the volume unit leaving grams, the desired unit for mass.

1) A block of aluminum occupies a volume of 15.0 mL and weighs 40.5 g. What is its density?

2) Mercury metal is poured into a graduated cylinder that holds exactly 22.5 mL. The mercury used to fill the cylinder weighs 306.0 g. From this information, calculate the density of mercury.

3) What is the weight of the ethyl alcohol that exactly fills a 200.0 mL container? The density of ethyl alcohol is 0.789 g/mL.

4) A rectangular block of copper metal weighs 1896 g. The dimensions of the block are 8.4 cm by 5.5 cm by 4.6 cm. From this data, what is the density of copper?

5) A flask that weighs 345.8 g is filled with 225 mL of carbon tetrachloride. The weight of the flask and carbon tetrachloride is found to be 703.55 g. From this information, calculate the density of carbon tetrachloride.

6) Calculate the density of sulfuric acid if 35.4 mL of the acid weighs 65.14 g.

7) Find the mass of 250.0 mL of benzene. The density of benzene is 0.8786 g/mL.

8) A block of lead has dimensions of 4.50 cm by 5.20 cm by 6.00 cm. The block weighs 1591 g. From this information, calculate the density of lead.

9) 28.5 g of iron shot is added to a graduated cylinder containing 45.5 mL of water. The water level rises to the 49.1 mL mark, From this information, calculate the density of iron.

10) What volume of silver metal will weigh exactly 2500.0 g. The density of silver is 10.5 g/cm^{3}.

I'm going to show you the solution to a slightly different density problem. Here goes!

The SI unit for density is kg/m^{3}. Convert the density of ethanol (789 kg/m^{3}) to the more commonly-used unit of g/cm^{3}

**Solution:**

1) Convert kg/m^{3} to g/m^{3}:

(789 kg/m^{3}) (1000 g / 1 kg) = 789000 g/m^{3}

2) Convert g/m^{3} to g/cm^{3}

(21450000 kg/m^{3}) (1 m^{3}/ 100^{3}cm^{3}) = 0.789 g/cm^{3}Note the use of 100

^{3}. 1 m^{3}is a cube 100 cm on a side: 100 cm x 100 cm x 100 cm = 100^{3}cm^{3}.

3) Many teachers that teach dimensional analysis want the solution in one line of calculation steps:

(789 kg/m^{3}) (1000 g / 1 kg) (1 m^{3}/ 100^{3}cm^{3}) = 0.789 g/cm^{3}Note the interim values/units such as 789000 g/m

^{3}do not appear in a one-line dimensional analysis presentation.

Why did I do that? Because it is easy to predict a situation where a problem (test or homework) gives you the density in kg/m^{3}, but use of the g/cm^{3} value is required in the solution to the problem.

In the ChemTeam section on the metric system, I go into what I call 'two-unit conversions' and I have a problem similar to the one above in that section.

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