# 1.10: Conversion Factors and Functions


Earlier we showed how unity factors can be used to express quantities in different units of the same parameter. For example, a density can be expressed in g/cm3 or lb/ft3. Now we will see how conversion factors representing mathematical functions, like D = m/v, can be used to transform quantities into different parameters. For example, what is the volume of a given mass of gold? Unity factors and conversion factors are conceptually different, and we'll see that the "dimensional analysis" we develop for unit conversion problems must be used with care in the case of functions.

When we are referring to the same object or sample of material, it is often useful to be able to convert one parameter into another. For example, in our discussion of fossil-fuel reserves we find that 318 Pg (3.18 × 1017 g) of coal, 28.6 km3 (2.68 × 1010 m3) of petroleum, and 2.83 × 103 km3 (2.83 × 1013 m3) of natural gas (measured at normal atmospheric pressure and 15°C) are available. But none of these quantities tells us what we really want to know ― how much heat energy could be released by burning each of these reserves? Only by converting the mass of coal and the volumes of petroleum and natural gas into their equivalent energies can we make a valid comparison. When this is done, we find that the coal could release 7.2 × 1021 J, , the petroleum 1.1 × 1021 J, and the gas 1.1 × 1021 J of heat energy. Thus the reserves of coal are more than three times those of the other two fuels combined. It is for this reason that more attention is being paid to the development of new ways for using coal resources than to oil or gas. Conversion of one kind of quantity into another is usually done with what can be called a conversion factor, but the conversion factor is based on a mathematical function (D = m / V) or mathematical equation that relates parameters. Since we have not yet discussed energy or the units (joules) in which it is measured, an example involving the more familiar quantities mass and volume will be used to illustrate the way conversion factors are employed. The same principles apply to finding how much energy would be released by burning a fuel, and that problem will be encountered later.

For helpful context about the above discussion, check out the following Crash Course Chemistry video: https://youtu.be/hQpQ0hxVNTg

Suppose we have a rectangular solid sample of gold which measures 3.04 cm × 8.14 cm × 17.3 cm. We can easily calculate that its volume is 428 cm3 but how much is it worth? The price of gold is about 5 dollars per gram, and so we need to know the mass rather than the volume. It is unlikely that we would have available a scale or balance which could weigh accurately such a large, heavy sample, and so we would have to determine the mass of gold equivalent to a volume of 428 cm3. This can be done by manipulating the equation which defines density, ρ = m / V. If we multiply both sides by V, we obtain

$V \times \rho =\dfrac{m}{V}\times V = m\label{1}$

$m = V \times \rho$

or

$mass = \text{volume} \times \text{ density }$

Taking the density of gold from a reference table, we can now calculate

$\text{Mass}= m =V \rho =\text{428 cm}^{3}\times \dfrac{\text{10}\text{0.32 g}}{\text{1 cm}^{3}}=8.27\times \text{10}^{3}\text{g}=\text{8}\text{.27 kg}$

This is more than 18 lb of gold. At the price quoted above, it would be worth over 40 000 dollars!

The formula which defines density can also be used to convert the mass of a sample to the corresponding volume. If both sides of Eq. $$\ref{1}$$ are multiplied by 1/ρ, we have

$\dfrac{\text{1}}{\rho }\times m=V \rho \times \dfrac{\text{1}}{\rho }=V$

$V=m \times \dfrac{\text{1}}{\rho }\label{2}$

Notice that we used the mathematical function D = m/V to convert parameters from mass to volume or vice versa in these examples. How does this differ from the use of unity factors to change units of one parameter?