Dimensional Analysis

This math trick is based on the identity property of multiplication: the product of any number and one is equal to that number. Conversion factors are written as ratios. While the numerals are not equal to one, the physical quantities are the same, so the ratio has a dimension of one.

Dimensional analysis (also called the unit factor method) is a method for converting from one system of measure to another. In this technique, all measurements are written with their units and conversion factors are written as ratios such that units cancel, top to bottom, like common factors in fractions. For example, to convert my height from English inches to metric meters, I need a conversion factor for English to metric length. The only one I know is 1 inch = 2.54 cm. But I also know that 100 cm = 1m. My height is 67 inches, so I start with that. Then I choose a conversion factor for inches, and write it as a ratio with the inch unit in the denominator. That gives centimeters in the numerator, so I choose another conversion factor with centimeters, and write it as a ratio with centimeters in the denominator. The inches and centimeters cancel out leaving me with the meters as the units. Multiply the numerators and divide by the denominators.

 

67 in (2.54 cm/1in )(1m/100 cm) = 1.7 m

 

In this case, the answer is rounded to two significant figures because the original measurement had two significant figures. The other values are exact numbers. The conversion between inches and centimeters is an unusual example of a definition between two totally different measurement systems.

 

After the first chapter review, chemistry students will seldom convert English to metric system. However, some topics require students to convert metric and SI derived units. For example, the there are two values for the gas law constant, R: 0.08206 L atm/mol K and 8.314 J/mol K. These two values are the same quantity but use different units. We can show this to be the case using unit analysis and some information about the derived units from the previous section. We know that a joule can be reduced to SI base units. That means we need to convert liters and atmospheres into base units; the moles and degrees Kelvin are fine. One milliliter is the same as a cubic centimeter. A cubic meter is a cube with each side 100 centimeters long, or 1×106 cm3. Setting up conversion factors as ratios such that the units cancel top to bottom gives us:

 

1 L (1000 mL/1 L) (1 cubic cm/1 mL) (1 cubic m/1×106 cubic cm) = 0.001 cubic m

 

So we have a new conversion factor: 1L = 0.001 m3. If we convert atmospheres to pascals, we can find SI base units. Using the conversion for atmospheres to kilopascals, knowing that there are 1000 pascals in a kilopascal, and using the definitions of the pascal and the joule, we get:

 

1 atm (101.325 kPa/1atm) (1000 Pa/1kPa) (1 N per sq m/1Pa) (1 J/1 Nm) = 1.01325×105 J/cubic m

 

So, 1 atm = 1.01325×105 J/m3. Note that the meters units do not cancel. We have cubic meters in the denominator. We can use these conversions to convert the gas law constant:

 

0.08206 L atm/mol K (0.001 cubic m/1L) (1.01325×105 J/cubic m) = 8.315 J/mol K

 

This is close enough to the accepted value to prove the point. Now we need to think a bit about the significant figures. The original constant, 0.08206, has four significant figures. The atmosphere to kilopascals conversion has six significant figures. All of the other conversions are exact, because they are based on metric system definitions rather than measurements. The result is rounded to four significant figures by the multiplication/division rule.