Conversion Factors

Referring to the SI units and prefixes described in Section 1.6 “The International System of Units”,  1 cm is 1/100th of a meter.

[latex]1\text{ cm}=\frac{1}{100\text{ m}}[/latex] 
or
100 cm = 1 m

Suppose we divide both sides of the equation by 1 m (both the number and the unit):

[latex]\frac{100\text{ cm}}{1\text{ m}}=\frac{1\text{ m}}{1\text{ m}}[/latex] 

As long as we perform the same operation on both sides of the equals sign, the expression remains an equality. Look at the right side of the equation; it now has the same quantity in the numerator (the top) as it has in the denominator (the bottom). Any fraction that has the same quantity in the numerator and the denominator has a value of 1:

We know that 100 cm is 1 m, so we have the same quantity on the top and the bottom of our fraction, although it is expressed in different units. A fraction that has equivalent quantities in the numerator and the denominator but expressed in different units is called a conversion factor.

Here is a simple example. How many centimeters are there in 3.55 m? Perhaps you can determine the answer in your head. If there are 100 cm in every meter, then 3.55 m equals 355 cm. To solve the problem more formally with a conversion factor, we first write the quantity we are given, 3.55 m. Then we multiply this quantity by a conversion factor, which is the same as multiplying it by 1. We can write 1 as 100 cm1 m and multiply:

[latex]3.55\text{ m}\times\frac{100\text{ cm}}{1\text{ m}}[/latex] 

The 3.55 m can be thought of as a fraction with a 1 in the denominator. Because m, the abbreviation for meters, occurs in both the numerator and the denominator of our expression, they cancel out:

[latex]\frac{3.55\cancel{\text{ m}}}{1}\times\frac{100\text{ cm}}{1\cancel{\text{ m}}}[/latex] 

The final step is to perform the calculation that remains once the units have been canceled:

[latex]\frac{3.55}{1}\times\frac{100\text{ cm}}{1}=355\text{ cm}[/latex] 

In the final answer, we omit the 1 in the denominator. Thus, by a more formal procedure, we find that 3.55 m equals 355 cm. A generalized description of this process is as follows:

quantity (in old units) × conversion factor = quantity (in new units)

You may be wondering why we use a seemingly complicated procedure for a straightforward conversion. In later studies, the conversion problems you encounter will not always be so simple. By mastering the technique of applying conversion factors, you will be able to solve a large variety of problems.

In the previous example, we used the fraction $$\frac{100\text{ cm}}{1\text{ m}}$$ as a conversion factor. Does the conversion factor  $$\frac{1\text{ m}}{100\text{ cm}}$$ also equal 1? Yes, it does; it has the same quantity in the numerator as in the denominator, except that they are expressed in different units. Why did we not use that conversion factor? If we had used the second conversion factor, the original unit would not have canceled, and the result would have been meaningless. Here is what we would have gotten:

[latex]3.55\text{ m} \times \frac{1\text{ m}}{100\text{ cm}}=0.0355  \frac{\text{m}^2}{\text{cm}}[/latex]

For the answer to be meaningful, we have to use the conversion factor in a form that causes the original unit to cancel out. Figure 1.10 “A Concept Map for Conversions” shows a concept map for constructing a proper conversion. The steps for doing a unit conversion problem are summarized in the margin.

Figure 1.10 A Concept Map for Conversions.This is how you construct a conversion factor to convert from one unit to another.

Significant Figures in Conversions

How do you determine the correct number of sig figs after a calculation involving conversion factors? The initial number is a measured number for which you can determine the number of sig figs. The numbers in metric-to-metric conversion factors and in counted numbers are exact, so they have infinite sig figs and do not limit the number of sig figs in the final answer.  However, conversion factors that come from measurements, such as density, percent, and  most relationships between metric and US customary units have a limited number of significant figures.  Since the use of conversion factors amounts to multiplication and division, the final answer should be rounded to the smallest number of sig figs based on the original measurement and and conversion factors that came from measurements.  Again, numbers from definitions or counting do not limit the number of sign figs.

13.6 g mercury = 1 mL mercury

This relationship can be used to construct two conversion factors:

[latex]\frac{13.6\text{ g}}{1\text{ mL}}\text{ and }\frac{1\text{ mL}}{13.6\text{ g}}[/latex]

Which one do we use? It depends, as usual, on the units we need to cancel and introduce. For example, suppose we want to know the mass of 16 mL of mercury. We would use the conversion factor that has milliliters on the bottom (so that the milliliter unit cancels) and grams on top so that our final answer has a unit of mass:

[latex]16\cancel{\text{ mL}}\times \frac{13.6\text{ g}}{1\cancel{\text{ mL}}}=217.6\text{ g}=220\text{ g}[/latex]

In the last step, we limit our final answer to two significant figures because the given volume quantity has only two significant figures; we inserted the 1 written with the mL in the density conversion factor as an interpretation, so it does not affect the number of significant figures.

The other conversion factor would be useful if we were given a mass and asked to find volume, as the following example illustrates.

Suppose you wanted to know how many grams of NaCl are in 235 g of solution.  Which conversion factor would you use?

[latex]235\cancel{\text{ g solution}} \times \frac{0.9 \text{ g NaCl}}{100\cancel{\text{ g solution}}}[/latex]  is the correct set up so that units cancel.

But if you needed to know how many grams of solution would  supply 15 g of NaCl, you would choose the other version of the conversion factor.

[latex]15\cancel{\text{ g NaCl}} \times \frac{100 \text{ g solution}}{0.9\cancel{\text{ g NaCl}}}[/latex]

To do a stepwise conversion, we first convert the given amount to the base unit. In this example, the base unit is meters. We know that there are 1,000 m in 1 km:

[latex]54.7\cancel{\text{ km}}\times \frac{1,000\text{ m}}{1\cancel{\text{ km}}}=54,700\text{ m}[/latex]

Then we take the result (54,700 m) and convert it to millimeters, remembering that there are 1,000 mm for every 1 m:

[latex]54,700\cancel{\text{ m}}\times \frac{1,000\text{ m}}{1\cancel{\text{ m}}}=54,700,000\text{ mm}=5.47\times10^7\text{ mm}[/latex]

We have expressed the final answer in scientific notation.

As a shortcut, both steps in the conversion can be combined into a single, multistep expression:

[latex]54.7\cancel{\text{ km}}\times \frac{1,000\cancel{\text{ m}}}{1\cancel{\text{ km}}}\times\frac{1,000\text{ mm}}{1\cancel{\text{ m}}}\times=54,700,000\text{ mm}=5.47\times10^7\text{ mm}[/latex]

Either method—one step at a time or all the steps together—is acceptable. If you do all the steps together, the restriction for the proper number of significant figures should be done after the last step. As long as the math is performed correctly, you should get the same answer no matter which method you use.