GENERAL CHEMISTRY TOPICS

Dimensional analysis and unit conversions

Dimensional analysis. Equivalence relations for units and conversion factors. Exact and inexact conversion factors. Examples.

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Dimensional analysis, unit conversions

Dimensional analysis of a calculation simply means that we treat the units in a calculation algebraically - when we multiply or divide terms, we do the same with the units. In a ratio, common factors cancel. When adding or subtracting terms, those terms must be expressed in the same units, otherwise you are adding together apples and oranges, i.e. you can't sensibly add/subtract those numbers. When setting up and performing any calculation where numbers or measurements with units are involved, one should always check whether the dimensional analysis of the calculation yields the correct units. Arguments to and products of functions such as log and exp are formally dimensionless. Often, looking at the units can guide you in what calculation to perform. If the dimensional analysis of a calculation shows the correct units are produced, it does not guarantee the calculation is correct. However, if the dimensional analysis of a calculation shows that it yields the wrong units, the calculation is certainly wrong.

Always include units in writing out the calculations you perform. This will help you avoid many mistakes and help give your instructor at least the impression you know what you are doing!

Any definition of units of a quantity provides exact conversion factors. For example, the units of inches are defined in terms of centimeters (cm) as follows:

1 in = 2.54 cm exactly

is the conversion factor 2.54 cm/in. In this instance, the conversion is exact (no loss of significant figures). So to convert inches to centimeters, we need a conversion factor of cm/in to multiply the length quantity in inches.

(6.3360 × 10⁵ in ) (2.54 cm·in⁻¹) = 2.4945 × 10⁵ cm

For the opposite conversion, centimeters to inches, the inverse factor, in/cm, is applied as follows

30.50 cm (2.54 cm·in⁻¹)⁻¹ = 12.01 in

The decimal multipliers work similarly to exact conversion factors. If, for example, we needed to convert between meters (m) and nanometers (nm), we recognize that

(x nm ) (10⁻⁹ m·nm⁻¹) = y m
(y m ) (10⁹ nm·m⁻¹) = x nm

and the decimal multiplier conversion factor does not affect the number of significant figures.

Unit equivalence relations and conversion factors

Two reciprocal conversion factors can be constructed from a unit equivalence. For example, the unit equivalence relation between inches (in) and centimeters (cm) is

1 in = 2.54 cm

By dividing both sides of this relation by 1 in, we obtain

1 = 2.54 cm/in

the right-hand side of which is a conversion factor for in to cm. Conversely, if we divide both sides of the unit equivalence relation by 2.54 cm, we obtain

(1/2.54) in/cm = 1

where the left-hand side is a conversion factor for cm to in.

Some unit equivalence relations, such as those from unit prefixes and their decimal multipliers, are exact. These give rise to exact conversion factors that can be treated in calculations as having an infinite number of significant figures. Another example of an exact equivalence is the cm/in relation given above:

1 in = 2.54 cm   (exactly)

The inch in effect is defined as 2.54 cm exactly.

Inexact conversion factors and constants

Many conversion factors and fundamental physical constants that appear in chemical calculations can be introduced to varying degrees of accuracy (number of digits), and the value we choose depends on the requirements of the calculation. For example, we may have the equivalence relation 1 lb = 453.6 g given by one source (e.g. Ref. 2), and 1 lb = 453.59 g by another source (e.g. Ref. 1). Similarly, the speed of light is given (in Ref. 1, on page facing the back cover) as 2.99792458 × 10⁸ m/s. We may choose to use a rounded value with fewer digits for a particular calculation (though we are always free to use the most accurate value known!) which is fine where data that is not accurately or precisely known is also part of the calculation. The general situation is analogous to computing the area of a circle of a given radius roughly ( using π = 3), or more accurately (using π = 3.14159).

Tip: When using inexact conversion factors or constants in calculations, where possible use a sufficient number of significant figures so that the conversion factor or constant does not limit the accuracy of the calculation.

For example, in a calculation of the time it takes light from the sun to reach the earth, using an average distance of 1.5 × 10¹¹ m, it is sufficient to use the value 2.998 × 10⁸ m/s for the speed of light,

(1.5 × 10¹¹ m) / (2.998 × 10⁸ m/s)  =  5.0 × 10² s

where we have rounded and reported the result with the correct number of significant figures. We would have obtained the same result using 2.99792458 × 10⁸ m/s for the speed of light.

Conversion factor for derived units from base unit conversion

If we are given a unit equivalence involving a base unit and need to perform a conversion involving units derived from that base unit, care should be taken in deriving the conversion factor. As an example, suppose we need to convert a volume expressed in cubic centimeters (cm³) to volume expressed in cubic meters (m³). The equivalence we know from the base unit of length is 1 m = 10² cm. To obtain the equivalence between cm³ and m³, be sure to treat the units and numbers the same:

1 m  =  10² cm   the equivalence relation involving the base unit meters (m)

(1 m)³  =  (10² cm)³      raise both sides of the relation to the third power

1³ m³  =  (10²)³ cm³      operate on numbers and units in the same way

1 m³  =  10⁶ cm³     1³ = 1 and (10²)³ = 10^2·3        this is the desired equivalence relation

The equivalence relation above results in the conversion factors

10⁶ cm³/m³ (for converting to cm³ from m³)   and   10⁻⁶ m³/cm³ (for converting to m³ from cm³)

A common mistake is to fail to treat the number in the same way as units, resulting in the incorrect conversion factors of 10² cm³/m³ (for converting to cm³ from m³) and 10⁻² m³/cm³ (for converting to m³ from cm).

Examples of unit conversions and dimensional analysis

Oftentimes problem solving in chemistry will require unit conversion. Dimensional analysis is a tool for application of proper conversion factors. At the same time, dimensional analysis an indispensable strategy to guide the construction or set-up of an equation that can yield the correct answer to a problem, once numerical values are substituted into the equation and the computation is performed.

Example: Calculate the mass in pounds of 1.00 gal of ethanol, which has a density of 0.789 g cm⁻³. 1 pound (lb) is equal to 453.59 grams (g), and 1 liter (L) is equal to 1.057 quarts (qt).

One way to interpret this problem is that we need a conversion factor for density that, when multiplied with density expressed in the units g cm⁻³, will give density expressed in the units lb gal ⁻¹.

( 0.789 g cm⁻³)( x lb cm³ g⁻¹ gal⁻¹ )  =  z lb gal ⁻¹.

We have the unit equivalence

1 L = 1.057 qt

Dividing this relation, in turn, by the factor on either side yields the two conversion factors

1 = 1.057 qt L⁻¹   and   (1/1.057) L qt⁻¹ =  1

Similarly, 1 lb = 453.59 g yields

1 = 453.59 g lb⁻¹   and   (1/453.59) lb g ⁻¹  =  1

Furthermore, for the units of volume

1000 mL  =  1000 cm³  =  1 L

there are the conversion factors

1 = 10³ cm³ L⁻¹   and   1 = 10⁻³ L cm⁻³

Our task is to choose the right combination of conversion factors so that all units cancel except for the correct units of the answer. As we found above, we need the units lb and cm³ in the numerator, and the units g and gal in the denominator. Therefore, we choose (1/453.59) lb g ⁻¹ as the mass conversion factor, and 10³ cm³ L ⁻¹ to help with volume conversion

( x lb cm³ g⁻¹ gal⁻¹ )  =  ( 1 lb / 453.59 g )( 10³ cm³ L⁻¹) ( y L gal⁻¹)

We also need the equivalence between L and gal, or use the fact that 1 gal = 4 qt (exact).

( y L gal⁻¹)  =  ( 4 qt gal⁻¹)( 1 L / 1.057 qt )  .

Finally, we can write the complete conversion of density from g cm⁻³ to lb gal ⁻¹ and compute the answer as

( 0.789 g cm⁻³)( 1 lb / 453.59 g )( 10³ cm³ L⁻¹)( 4 qt gal⁻¹)( 1 L / 1.057 qt )  =  6.58₂₆ lb gal⁻¹