Simple Whole Number Ratios

The reason why the chemical amount and its base unit, the mole, have been developed is to simplify stoichiometric calculations. They have been expressly designed so that they obey the following law:


We can easily describe this behavior as an algebraic equation of the form:

Here nA is the chemical amount of A consumed or produced in the reaction while nB is the chemical amount of B consumed or produced. Both and are simple whole numbers.

If A and B are elements then the compound they form has the formula:


If either A or B is a compound, then and correspond to the coefficients in the balanced chemical equation. For example, if the reaction is:

4 NH3 + 7 O2 = 4 NO2 + 6 H2O

then the chemical amounts of NH3 and O2 are in the ratio:


A ratio like this is often written in a form involving not only amounts but "decorated" amounts:

Strictly speaking we should not do this. Everyone will agree that 4 mol / 7 mol = 4/7. It is problematic, though, whether 4 mol NH3 / 7 mol O2 = 4/7 or not. This is especially the case if we later tell students not to cancel mol NH3 with mol O2.

In practice, decorating the mole like this makes it easier to avoid mistakes. Without any reminder,it is very easy to put down 7/4 instead of 4/7 for this ratio. Entering 4 mol NH3 / 7 mol O2 makes it much more difficult to invert this ratio.





The Stoichiometric Ratio

It is useful to give this ratio of amounts a name and a symbol. The author suggests the term "stoichiometric ratio" with the symbol S. If necessary, this symbol can be subscripted . In some texts the term "mole ratio" is used (without any symbol) for this quantity, but this term has the disadvantage of being unit-specific. By using the adjective "stoichiometric" we strongly suggest that this is not any old ratio of amounts, but the ratio when the reaction has gone exactly to completion, i.e. a ratio of simple whole numbers. Following this suggestion, for the reaction:

4 NH3 + 7 O2 = 4 NO2 + 6 H2O

we would give the stoichiometric ratio of NH3 to O2 the symbol:


The stoichiometric ratio enables us to calculate the chemical amount of one substance X that reacts with, or is produced by, a given chemical amount of a second substance Y. The two amounts are related by the formula:

For example, in the oxidation of ammonia, we can calculate the chemical amount of NH3 needed to oxidize 10.37 mol of O2 as follows:










Proportionality Factors

Both the density and the stoichiometric ratio are examples of what we have called a proportionality factor. We must now define and discuss this term. Suppose the two quantities A and B are proportional to each other, this relationship can be expressed algebraically as:


where k is the proportionality factor or proportionality constant relating A to B. This factor allows us to easily calculate quantity A from quantity B or vice versa by rearranging the above equation to read:

In words, we can find one quantity from the other, either by multiplying or dividing by the proportionality constant.

A proportionality relationship is thus a two-way street. The constant k enables us to find A from B or B from A. We can indicate the behavior diagrammatically in the form:


Our first example of a proportionality factor is the density. For any given substance the mass of a sample is proportional to its volume at a given temperature and pressure. The constant of this proportionality is called the density, symbol d. The density is a useful quantity to know. Since it relates the mass to the volume, we can use it to find the mass of a given volume of that substance or vice versa. The appropriate algebraic equations are:

The second and third of these equations tell us that, given either the volume or the mass of a substance, all we have to do is to multiply or divide by the density in order to find the other quantity.

The stoichiometric ratio behaves in an analogous way. With it, we can calculate the chemical amount of X from a given chemical amount of Y or vice versa. Only multiplication or division by the proportionality factor is needed:


The two-way character of these two proportionalities can again be described diagrammatically:

The Molar Mass

We are now in a position to introduce another proportionality factor which is important in stoichiometry. This is the molar mass, symbol MX. The molar mass of a substance X is defined as the mass per unit chemical amount of X.


It enables us to interconvert a mass and a chemical amount:

Note in particular that the molar mass is not defined as the mass of a mole of X. Like the density,the molar mass is an intensive property! Also note that, as usual, the molar mass is defined in unit-neutral terms. There is no necessity for describing it in the units g / mol. Quite often the units mg / mmol are more appropriate. By contrast, in calculating the root-mean square velocity of a gas, using the formula:

the most useful units to use are kg / mol.

The older term for molar mass is 'molecular weight.' One should avoid using this term because it has as many as four meanings. In freshman texts, for example, the term molecular weight of carbon dioxide can be found to have the following meanings:

      1. The relative molecular weight, 44.01
      2. The mass of a CO2 atom, 44.01 amu = 7.308 10-29 kg
      3. The mass of one mole of CO2, 44.01 g
      4. The molar mass of CO2, 44.01 g mol-1

Next to the stoichiometric ratio, the molar mass is the most frequently used proportionality factor in chemical calculations. There is a simple reason for this. The mass of a sample of substance is easy to measure if it is a liquid or a solid. We simply place the sample In a weighed container on a digital balance. Bycontrast, there is no corresponding convenient device for measuring the chemical amount of a substance. Because of this situation, we tend to think of "how much" of a substance we have in terms of its mass rather than in terms of its chemical amount. When it comes to chemical reactions though, the chemical amount of a substance is a distinctly more convenient measure of "how much" is consumed or produced. This is because the chemical amount of one substance involved in a reaction is simply related to the chemical amount of any other substance involved by a simple whole number ratio. The ease of measurement on the one hand, and the ease of calculation on the other, means that we find ourselves constantly switching from mass to chemical amount and vice versa when we do chemical calculations.





Road Mapping

Perhaps the most common type of stoichiometric calculation is the task of finding the mass of substance X produced or consumed chemically by a given mass of a second substance Y. We will now consider an example of this kind of calculation with the object of showing how the logic of the calculation can be deduced without mentioning units at all! At the same time a technique for succinctly describing this logic will be developed.

The problem to be explored in this way reads:

Find the mass of H2O produced when 100 g of NH3 is burned in excess oxygen according to the equation:

4 NH3 + 7 O2 = 4 NO2 + 6 H2O

This problem tells us "how much" NH3 is consumed and asks us "how much" H2O will be produced as a result. There is only one way to go. There is only one bridge which connects "how much" of one substance with "how much" of another substance involved with it chemically. This bridge is the stoichiometric ratio. We must use the chemical amount as a measure of "how much". This is because the chemical amounts involved will then be in a simple whole number ratio, in this case four to six. In terms of diagramming the logic we need to involve the step:


In this step we either multiply or divide the chemical amount of NH3 by the stoichiometric ratio in order to find the chemical amount of H2O.

Of course we don't know the chemical amount of NH3 at this stage. We only know the mass of ammonia. How do we find the chemical amount from the mass? We must multiply or divide by the molar mass of NH3. We can diagram this step as follows:


Finally, we need to find the mass of H2O from its chemical amount. This step has the reverse of the logic used in the step just calculated. We divide or multiply the chemical amount of H2O by the molar mass of H2O:

We can now combine these three steps into a 'road-map' of the logic for the total calculation in the form:

We have now essentially solved this problem, and we have solved it without any recourse to units. Our argument has been purely about quantities. All we need to do now is to substitute in the values for and the three proportionality constants. Only at this point do the units become useful. The answer is given by:

Here the units tell us if we have gotten the proportionality factors the correct way round or not. They tell us that we must divide by the molar mass of NH3 so as to cancel out the gram units, and that we should multiply by the molar mass of H2O so as to end up with grams. Likewise, they tell us that 4 mol NH3 should go below the line in the stoichiometric ratio.

The final answer given above has been written in a form closely resembling what would be obtained using 'dimensional analysis.' This is not the only way of proceeding. We could also have obtained the answer one step at a time, beginning with:


followed by:

and finally by:

Here, we have deliberately refrained from decorating the mole. Instead, we have used more algebra than previously. The subscripts on the algebraic symbols used serve much the same purpose as 'decorations' of the mole and other units.

The author's experience suggests that doing stoichiometric calculations one step at a time in this way makes more sense to the average student than the unit-juggling inherent in the "dimensional analysis" approach. Each step involves the calculation of one quantity from another using a proportionality factor. Each quantity and factor has a name and a symbol. With this exact language, it is easy to explain the rationale for each step both mathematically and in words. Once students have acquired sufficient facility with each type of step, so that they can do them "without thinking",then the one-fell-swoop approach may be more appropriate. Whatever approach is adopted a logical road-map should be mandated.







Unity Factors

We have already encounterd the term "unity factor" earlier when discussing Example I:


As we saw then, a factor like 1 lb / 453.6 g can be considered algebraically equivalent to unity since 1 lb and 453.6 g are different descriptions of the same quantity . We also saw that when a unity factor operates on a quantity it only changes the units without changing the quantity itself. In the first stage of Example I for instance, we have :

Here the unity factor 1 lb / 453.6 g converts 2.70 g / cm3 to 0.00595 lb./ cm3. However, no new quantity is obtained. Both are alternative descriptions of the density of aluminum. The density of aluminum, symbolized by dAl, is a constant fixed property, the same for all samples of aluminum, and is not changed by our choice of units to describe it. If we change units from g /cm3 to g /ft.3, the number accompanying the units changes from 2.70 to 0.00595 but the density of aluminum remains unaltered. The density of this metal is determined by the size and mass of the Al atoms and also by the closeness of their packing. It is not changed by the words and numbers used to describe it. There is nothing strange about this. The words homo, anthropos, and muntu all have the same meaning as the English word "man".

It is very important to be able to distinguish, almost at sight, whether a given factor is a proportionality factor or a unity factor. The major clues to telling the difference are the dimensions of the numerator and the denominator. If these two dimensions are different, the factor must be a proportionality factor. In the case of unity factors, not only must the numerator and the denominator have the same dimensions, they must also be alternative descriptions of the same quantity.

Decide which of the following factors is a unity factor, and which is a proportionality factor. If it is a proportionality factor, give its name and algebraic symbol.


a) Although the denominator looks like a length, the units mmHg are actually units of pressure as are kilopascals. Both 100 kPa and 750 mmHg are equal to the standard pressure of 1 bar. (The atmosphere is no longer a standard pressure.) Accordingly, this is a unity factor.

b) The numerator is a mass, while the denominator is an amount. Accordingly, this factor must be a proportionality factor. It is, of course the molar mass of O2, with the symbol .

c) Both liters and cubic centimeters are units of volume. Strictly speaking, these two units are not exactly equal to each other, since 1000.00 cm3 = 1.00003 L. Unless we are dealing with very accurate measurements, we can ignore this difference.Accordingly, we can regard this factor as a unity factor.

d) This is obviously a stoichiometric ratio, with the symbol S or, more accurately, SH2O/NH3. Note that both numerator and denominator are chemical amounts, but not the same chemical amount. Thus the factor must be a proportionality factor.

An important property of unity factors is that they are " algebraically silent". When we use a unity factor to adjust units, all that we are doing is multiplying by one. No convenient algebraic way exists for describing this operation. In Example I we changed the units describing the density of aluminum from g/cm3 to lb/ft3. The best we can do to describe these manipulations algebraically is to write:


Though unity factors are algebraically silent, the same is, of course, not true of proportionality factors. All the most useful proportionality factors have both names and algebraic symbols. Their proportionality relationships to other quantities are easily expressed algebraically. For example:

Almost no introductory texts make this distinction between unity factors and proportionality factors. Consequently, these books do not explain which factors can be represented by an algebraic symbol and which cannot. Equally, no attempt is made to point out which factors produce a new quantity and which do not. Stoichiometric calculations tend to be treated purely as exercises in unit-replacement with little or no reference to what each factor in each step is actually achieving.

When the differences between proportionality factors and unity factors are not made clear, it becomes difficult to distinguish whether two quantities are actually equal to each other, or merely proportional to each other. Alas! textbooks often compound this difficulty by equating two quantities that are manifestly different. Some egregious examples are: