Single Step Conversions

In problem solving, proper care in setting up calculations is very important, and special attention should always be given to unit cancellation. If the units cancel properly, the problem should solve correctly. This method is called dimensional analysis and will be an important part of problem solving in any science course. In this section, we will be putting a lot of practice in learning to use the approach to solving chemical problems.

In order to use dimensional analysis, we must first talk about conversion factors. We have been using conversion factors throughout most of our lives without realizing it. By knowing how many dimes are in a dollar, we know that twenty dimes equals two dollars. Conversion factors allow us to convert from one unit (dimes) to another (dollars). The thing about setting up a conversion factor is to know the equivalence of the two units, that is, when the two units equal the same amount. For example, a dime isn’t the same amount as a dollar, but ten dimes equals the same amount of money as one dollar. Let’s take a closer look using this simple example to determine how many dollars equal 20 dimes. First, set up a conversion factor. In any problem or calculation involving conversions, we need to know the units involved, in this case the units are dimes and dollars. We state the equivalence as

10 dimes = 1 dollar

The equivalence can be written in following fractional forms called conversion factors. We can write two conversion factors for each equivalence. One of the conversion factors will be used for the calculation. Because the numerator and denominator are equal, the fractions are equal to 1.

\(\displaystyle\frac{10 \;dimes}{1 \;dollar}\qquad or \qquad\frac{1 \;dollar}{10 \;dimes}\)

For now we want to concentrate on setting up conversion factors, but as a preview to dimensional analysis, the following calculation shows how the conversion factor is used. In our example, we are asked how many dollars equal 20 dimes. To convert from dimes to dollars, the given (20 dimes) is multiplied by the conversion factor that cancels out the unit dimes.

\(\displaystyle\large\require{cancel}20 \cancel{dimes}\times\frac{1 \;dollar}{10\cancel{dimes}}\;=\;\mathbf{2\;dollars}\)

Notice how the dime units cancel out, leaving the dollar units in the answer. This is the basis for dimensional analysis. As complex as some chemical calculations seem, the dimensional analysis involved remains as simple as the preceding exercise. For now, let’s look at the following exercise that deals with setting up the conversion factors. More than one equivalence can be used for a conversion as you will see later. For now we will focus on single step conversions.

Let’s write conversion factors for the conversion of pounds to grams. First, we need an equivalence. We know that there are 454 g in one lb. The equivalence is written as

1 lb = 454 g
From the one equivalence we can write two conversion factors.

\(\displaystyle\frac{1 \;lb}{454 \;g}\qquad or \qquad\frac{454 \;g}{1 \;lb}\)
To convert from pounds to grams, we would use the second conversion factor.
Another equivalence for the conversion of pounds to grams is 0.00220 lb = 1 g. The two conversion factors are:

\(\displaystyle\frac{0.00220 \;lb}{1 \;g}\qquad or \qquad\frac{1 \;g}{0.00220 \;lb}\large\)

Again, the second conversion factor would be used to convert from pounds to grams.
Knowing how to set up conversion factors, we can now move into setting up calculations using dimensional analysis, which is also known as the factor-label method. When this simple method is used in a calculation, the correct answer is almost guaranteed. The basis for this method is keeping track of the units of the components in the calculations. To determine how many gallons in 24 quarts we first need to set up an equivalence.
The equivalence is written as

1 gallon = 4 quarts

The two possible conversion factors are

\(\displaystyle\large\frac{1 \;gallon}{4 \;quart}\qquad or \qquad\frac{4 \;quart}{1 \;gallon}\large\)

Next, we need to setup the calculation. For this part we need to know the two types of units in our calculation: a) Given Units are the units that have a given amount. In this calculation, the given units are quarts since we have 24 quarts and b) desired units, the units for which we are solving. In this calculation we are solving for gallons.
Having identified the units and determined the conversion factor, the calculation is set up as follows:

\(\displaystyle\;amount\;with \cancel{given \;units}\times\frac{desired \;units}{\cancel{given \;units}}\;=\;answer\; (in \;desired \;units)\)

Notice that the conversion factor used has the given units in the denominator which allows for proper cancellation of the units, that is, the given units cancel out, leaving only the desired units which will be in the answer.

Back to the calculation

\(\displaystyle 24 \cancel{quarts}\times\frac{1 \;gallon}{4\cancel{quarts}}\;=\;\mathbf{6\;gallons}\)

In the example we converted 24 quarts to gallons. Most of us could have performed the calculation without setting up equivalences and conversion factors. Many chemistry problems require unit conversions and this is a good method to use regardless of the type of problem encountered. It is important to identify the given and the desired quantities in any problem. This is good practice for the many problems you will encounter in this and future chemistry and science courses.

Multi Step Conversions and Road Maps

The preceding discussion was based on simple single step conversions. The following problems will require multistep conversions in the calculations, that means more than one conversion factor and a road map. Road maps are very handy to use in doing calculations. Don’t ever think that this approach is beneath you. Taking the time to “sketch out” the calculation will ensure the correct answer. In the following example, we’ll show how to use a road map in the calculation. Keep in mind that each type of problem can be done with as many or as few conversion factors as you can write. The number of conversion factors used for each problem will depend on the types and number of equivalences that you use.

Problem: A student needs 2,361 μL of ethyl alcohol for an experiment. He will use a graduated cylinder that reads in milliliter gradations. How many milliliters of ethyl alcohol will he measure? Identify the given units and the desired units:

Given Units: 2,361 μL
Desired Units: mL

If it’s not a single step calculation, develop a road map. Although there is a way to develop a conversion factor which will give us a one-step calculation, for the sake of this example, let’s proceed with a two-step method. We can state the following two relationships:

1 μL = 10^-6 L
1 mL = 10^-3 L
Looking at the two equivalences, we can see the common unit of Liters, L, between them. Converting from microliters (μL) to Liters,(L), is the first step in the calculation:

\(\displaystyle μL\;\rightarrow\;L\)

This is the first part of the road map. Now convert from liters (L) to milliliter(mL), which will be the second step of the calculation.

\(\large m\;\rightarrow\;mL\)

In terms of the road map, it would look like this

\( μL\;\rightarrow\;L\rightarrow\;mL\) In this two-step method, we will covert as follows:
microliters to liters and liters to milliliters

Write an equivalence and conversion factors for the conversion microliters to liters
1 μL = 10^-6 L

\(\displaystyle\frac{1\mu{L}}{10^{-6}L}\qquad\;or\qquad\frac{10^{-6}L}{1\mu{L}}\)
Write an equivalence and conversion factors for liters to milliliters
1 mL = 10^-3 L

\(\displaystyle\frac{1\;mL}{10^{-3}L}\qquad\;\;or\qquad\frac{10^{-3}L}{1\;mL}\)

Notice that one equivalence and one set of conversion factors is written for each arrow in the roadmap. Now, we can set up the calculation. Start with the given, 2,361 μL.

\(\large2,361\cancel{\mu{L}}\times\frac{10^{-6}\cancel{L}}{1\cancel{\mu{L}}}\times\frac{1\;mL}{10^{-3}\cancel{L}}\;=\;\mathbf{2.361\;mL}\;(4\;significant\;figures)\)

There are many ways to solve these problems which will depend on what equivalences you remember. We could have solved the problem using 1 equivalence, 10³μL = 1 mL. Using this equivalence we have:

\(\large2,361\;\cancel{\mu{L}}\times\frac{1\;mL}{10^3\cancel{\mu{L}}}\;=\;\mathbf{2.361\;mL}\)

Sometimes, you might have to use 3, 4, 5 or more equivalences to get the desired unit. Again, it will depend on the equivalences that you remember. In the meantime, you will need to practice, practice, and more practice. Watch the following videos. Work the following exercises!! If you have a question, please ask in the comment section.

Worksheet: Conversion Factors and Roadmaps
Worksheet: Conversions