Using dimensional analysis with derived units requires special care. When units are squared or cubed as with area or volume, the conversion factors themselves must also be squared or cubed. Two convenient volume units are the liter, which is equal to a cubic decimeter, and the milliliters, which is equal to a cubic centimeter. There are thus \(1000 \: \text{cm}^3\) in \(1 \: \text{dm}^3\), which is the same thing as saying there are \(1000 \: \text{mL}\) in \(1 \: \text{L}\). The conversion factor of \(1 \: \text{cm}^3 = 1 \: \text{mL}\) is a very useful conversion.
There are \(1000 \: \text{cm}^3\) in \(1 \: \text{dm}^3\). Since a \(\text{cm}^3\) is equal to a \(\text{mL}\), and a \(\text{dm}^3\) is equal to a \(\text{L}\), we can say that there are \(1000 \: \text{mL}\) in \(1 \: \text{L}\).
Example \(\PageIndex{2}\)
Convert \(3.6 \: \text{mm}^3\) to \(\text{mL}\).
Solution
Determine the units of the known value \(\left( \text{mm}^3 \right)\) and the units of the unknown value \(\left( \text{mL} \right)\). The starting and ending units will help guide the setup of the problem. Next, list any known conversion factors that might be helpful.
- \(1 \: \text{m} = 1000 \: \text{mm}\)
- \(1 \: \text{mL} = 1 \: \text{cm}^3\)
- \(1 \: \text{m} = 100 \: \text{cm}\)
Now, we can set up the problem to find the value in units of \(\text{mL}\). Once we know the starting units, we can then use the conversion factors to find the answer.
\[3.6 \: \text{mm}^3 \times \left( \dfrac{?}{?} \right) \nonumber\]
Continue to use the conversion factors between the units to set up the rest of the problem. Note that all of the units cancel except \(\text{mL}\), which are the requested units for the answer. Since the values in these conversion factors are exact numbers, they will not affect the number of significant figures in the answer. Only the original value (3.6) will be considered in determining significant figures.
\[3.6 \: \text{mm}^3 \times \left( \dfrac{1 \: \text{m}}{1000 \: \text{mm}} \right)^3 \times \left( \dfrac{?}{?} \right) \nonumber\]
Once you have solved the problem, always ask if the answer seems reasonable. Remember, a millimeter is very small and a cubic millimeter is also very small. Therefore, we would expect a small volume which means \(0.0036 \: \text{mL}\) is reasonable.
If you find that you forgot to cube numbers as well as units, you can setup the problem in an expanded form which is the equivalent to the previous method to cube the numerical values.