Conversion Factors and Functions

Submitted by ChemPRIME Staff on Wed, 12/08/2010 - 23:36

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 massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object. of gold? Unity factors and conversion factors are conceptually different, and we'll see that the "dimensional analysisA technique in which the cancelling of units is used as a tool to check the correctness of a calculation." 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 gasA state of matter in which a substance occupies the full volume of its container and changes shape to match the shape of the container. In a gas the distance between particles is much greater than the diameters of the particles themselves; hence the distances between particles can change as necessary so that the matter uniformly occupies its container. (measured at normal atmospheric pressureForce per unit area; in gases arising from the force exerted by collisions of gas molecules with the wall of the container. and 15°C) are available. But none of these quantities tells us what we really want to know ― how much heatEnergy transferred as a result of a temperature difference; a form of energy stored in the movement of atomic-sized particles. energyA system's capacity to do work. 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.

Suppose we have a rectangular solidA state of matter having a specific shape and volume and in which the particles do not readily change their relative positions. 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 gramOne thousandth of a kilogram., 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

\text{V}\times \rho =\frac{\text{m}}{\text{V}}\times \text{V}=\text{m}

m = V × ρ or mass = volume × density       (1)

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

\text{Mass}=\text{m}=\text{V}\rho =\text{428 cm}^{3}\times \frac{\text{10}\text{.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. (1) are multiplied by 1/ρ, we have

  & \frac{\text{1}}{\rho }\times \text{m}=\text{V}\rho \times \frac{\text{1}}{\rho }=\text{V} \ 
 & \text{         V}=\text{m}\times \frac{\text{1}}{\rho } \ 
\end{align}       (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?

An Important Caveat

A mistake sometimes made by beginning students is to confuse density with concentrationA measure of the ratio of the quantity of a substance to the quantity of solvent, solution, or ore. Also, the process of making something more concentrated., which also may have units of g/cm3. By dimensional analysis, this looks perfectly fine. To see the error, we must understand the meaning of the function

C = \frac{m}{V}.

In this case, V refers to the volume of a solutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture., which contains both a soluteThe substance added to a solvent to make a solution. and solventThe substance to which a solute is added to make a solution..

Given a concentration of an alloyA solid that has metallic properties and is made up of two or more elements. is 10g gold in 100 cm3 of alloy, we see that it is wrong (although dimensionally correct as far as conversion factors go) to incorrectly calculate the volume of gold in 20 g of the alloy as follows:

20 g     x     \frac{100 cm^3}{10 g}     = 200 cm3

It is only possible to calculate the volume of gold if the density of the alloy is known, so that the volume of alloy represented by the 20 g could be calculated. This volume multiplied by the concentration gives the mass of gold, which then can be converted to a volume with the density function.

The bottom line is that using a simple unit cancellation method does not always lead to the expected results, unless the mathematical function on which the conversion factor is based is fully understood.


A solution of ethanol with a concentration of 0.1754 g / cm3 has a density of 0.96923 g / cm3 and a freezing pointThe temperature at which a liquid becomes a solid; also called melting point. of -9 ° F [1]. What is the volume of ethanol (D = 0.78522 g / cm3 at 25 °C) in 100 g of the solution?


The volume of 100 g of solution is

V = m / D = 100 g /0.96923 g cm-3 = 103.17 cm3.

The mass of ethanol in this volume is

m = V x C = 103.17 cm3 x 0.1754 g / cm3 = 18.097 g.

The volume of ethanol = m / D = 18.097 g / 0.78522 g / cm3 = 23.05 cm3.

Note that we cannot calculate the volume of ethanol by

 \frac {\frac{0.96923 g}{cm^3} x 100 cm^3}{\frac {0.78522 g}{cm^3}} = 123.4 cm3

even though this equation is dimensionally correct.

Note that this result required when to use the function C = m/V, and when to use the function D=m/V as conversion factors. Pure dimensional analysis could not reliably give the answer, since both functions have the same dimensions.

EXAMPLE 2 Find the volume occupied by a 4.73-g sample of benzene.

Solution The density of benzene is 0.880 g cm–3. Using Eq. (2),

\text{Volume = }V\text{ = }m\text{ }\times \text{ }\frac{\text{1}}{\rho }\text{ = 4}\text{.73 g }\times \text{ }\frac{\text{1 cm}^{\text{3}}}{\text{0}\text{.880 g}}\text{ = 5}\text{.38 cm}^{\text{3}}

(Note that taking the reciprocal of \tfrac{\text{0}\text{.880 g}}{\text{1 cm}^{3}} simply inverts the fraction ― 1 cm3 goes on top, and 0.880 g goes on the bottom.)

The two calculations just done show that density is a conversion factor which changes volume to mass, and the reciprocal of density is a conversion factor changing mass into volume. This can be done because the mathematical formula defining density relates it to mass and volume. Algebraic manipulation of this formula gave us expressions for mass and for volume [Eq. (1) and (2)], and we used them to solve our problems. If we understand the function D = m/V and heed the caveat above, we can devise appropriate converstion factors by unit cancellation, as the following example shows:

EXAMPLE 3 A student weighs 98.0 g of mercury. If the density of mercury is 13.6 g/cm3, what volume does the sample occupy?


We know that volume is related to mass through density.


V = m × conversion factor

Since the mass is in grams, we need to get rid of these units and replace them with volume units. This can be done if the reciprocal of the density is used as a conversion factor. This puts grams in the denominator so that these units cancel:

V=m\times \frac{\text{1}}{\rho }=\text{98}\text{.0 g}\times \frac{\text{1 cm}^{3}}{\text{13}\text{.6 g}}=\text{7}\text{.21 cm}^{3}

If we had multiplied by the density instead of its reciprocal, the units of the result would immediately show our error:

V=\text{98}\text{.0 g}\times \frac{\text{13,6 }g}{\text{1 cm}^{3}}=\text{1}\text{.333}{\text{g}^{2}}/{\text{cm}^{3}}\; (no cancellation!)

It is clear that square grams per cubic centimeter are not the units we want.

Using a conversion factor is very similar to using a unity factor — we know the conversion factor is correct when units cancel appropriately. A conversion factor is not unity, however. Rather it is a physical quantity (or the reciprocal of a physical quantity) which is related to the two other quantities we are interconverting. The conversion factor works because of the relationship [ie. the definition of density as defined by Eqs. (1) and (2) includes the relationships between density, mass, and volume], not because it is has a value of one. Once we have established that a relationship exists, it is no longer necessary to memorize a mathematical formula. The units tell us whether to use the conversion factor or its reciprocal. Without such a relationship, however, mere cancellation of units does not guarantee that we are doing the right thing.

A simple way to remember relationships among quantities and conversion factors is a “road map“of the type shown below:

\text{Mass }\overset{density}{\longleftrightarrow}\text{ volume   or   }m\overset{\rho }{\longleftrightarrow}V\text{ }

This indicates that the mass of a particular sample of matterAnything that occupies space and has mass; contrasted with energy. is related to its volume (and the volume to its mass) through the conversion factor, density. The double arrow indicates that a conversion may be made in either direction, provided the units of the conversion factor cancel those of the quantity which was known initially. In general the road map can be written

\text{First quantity }\overset{\text{conversion factor}}{\longleftrightarrow}\text{ second quantity}

As we come to more complicated problems, where several steps are required to obtain a final result, such road maps will become more useful in charting a path to the solution.

EXAMPLE 4 Black ironwood has a density of 67.24 lb/ft3. If you had a sample whose volume was 47.3 ml, how many grams would it weigh? (1 lb = 454 g; 1 ft = 30.5 cm).

Solution The road map

V\xrightarrow{\rho }m\text{ }

tells us that the mass of the sample may be obtained from its volume using the conversion factor, density. Since milliliters and cubic centimeters are the same, we use the SI unitsThe international system of units (Système International d'Unité) based on seven fundamental units: meter, kilogram, second, ampere, kelvin, candela, mole. for our calculation:

Mass = m = 47.3 cm3 × \frac{\text{67}\text{.24 lb}}{\text{1 ft}^{3}}

Since the volume units are different, we need a unity factor to get them to cancel:

m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\left( \frac{\text{1 ft}}{\text{30}\text{.5 cm}} \right)^{\text{3}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}

We now have the mass in pounds, but we want it in grams, so another unity factor is needed:

m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ }\times \text{ }\frac{\text{454 g}}{\text{ 1 lb}}\text{ = 50}\text{.9 g}

In subsequent chapters we will establish a number of relationships among physical quantities. Formulas will be given which define these relationships, but we do not advocate slavish memorization and manipulation of those formulas. Instead we recommend that you remember that a relationship exists, perhaps in terms of a road map, and then adjust the quantities involved so that the units cancel appropriately. Such an approach has the advantage that you can solve a wide variety of problems by using the same technique.

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