Stoichiometry and the Mole

5.2

The Mole

Everything in chapters 3 and 4 was written one atom or one molecule at a time. Nobody works that way. This section installs the unit that connects the invisible scale to the balance on your bench — and it is, at bottom, just a counting word.

So far, we have been talking about chemical substances in terms of individual atoms and molecules. Yet we don't typically deal with substances an atom or a molecule at a time; we work with millions, billions, and trillions of atoms and molecules at a time. What we need is a way to deal with macroscopic, rather than microscopic, amounts of matter. We need a unit of amount that relates quantities of substances on a scale that we can interact with.

Chemistry uses a unit called mole. A mole (mol) is a number of things equal to the number of atoms in exactly 12 g of carbon-12. Experimental measurements have determined that this number is very large:

Understand that a mole means a number of things, just like a dozen means a certain number of things—twelve, in the case of a dozen. But a mole is a much larger number of things. These things can be atoms, or molecules, or eggs; however, in chemistry, we usually use the mole to refer to the amounts of atoms or molecules. Although the number of things in a mole is known to eight decimal places, it is usually fine to use only two or three decimal places in calculations. The numerical value of things in a mole is often called Avogadro's number (N_A), which is also known as the Avogadro constant, after Amadeo Avogadro, an Italian chemist who first proposed its importance.

How big is a mole? It is very large. Suppose you had a mole of dollar bills that need to be counted. If everyone on earth (about 6 billion people) counted one bill per second, it would take about 3.2 million years to count all the bills. A mole of sand would fill a cube about 32 km on a side. A mole of pennies stacked on top of each other would have about the same diameter as our galaxy, the Milky Way. A mole is a lot of things—but atoms and molecules are very tiny. One mole of carbon atoms would make a cube that is 1.74 cm on a side, small enough to carry in your pocket.

Why is the mole unit so important? It represents the link between the microscopic and the macroscopic, especially in terms of mass. A mole of a substance has the same mass in grams as one unit (atom or molecules) has in atomic mass units. The mole unit allows us to express amounts of atoms and molecules in visible amounts that we can understand.

Previously we defined atomic and molecular masses as the number of atomic mass units per atom or molecule. Now we can do so in terms of grams. The atomic mass of an element is the number of grams in 1 mol of atoms of that element, while the molecular mass of a compound is the number of grams in 1 mol of molecules of that compound. Sometimes these masses are called molar masses to emphasize the fact that they are the mass for 1 mol of things. (The term molar is the adjective form of mole and has nothing to do with teeth.)

That sentence in italics is the hinge of the entire chapter, and it is easy to read past. It says the number you looked up in chapter 3 as an atomic mass — 18.015 for water — is the same number you weigh out in grams to have one mole. One table, two jobs. Everything below is bookkeeping on that fact.

Two things the builder is checking that a bare answer box cannot. First, that the factor you picked is the right way up — the unit you are carrying has to appear on the bottom of the next fraction or nothing cancels. Second, that you land on the unit you were asked for. A right number with the wrong units is a wrong answer, and the chain shows you which it was.

5.3

The Mole in Chemical Reactions

Chapter 4 balanced equations by counting atoms. This section shows that the same coefficients can be read as moles — which is what lets a balanced equation become a conversion factor, and that is the only new idea in the section.

Consider this balanced chemical equation:

We interpret this as “two molecules of hydrogen react with one molecule of oxygen to make two molecules of water.” The chemical equation is balanced as long as the coefficients are in the ratio 2:1:2. For instance, this chemical equation is also balanced:

This equation is not conventional—because convention says that we use the lowest ratio of coefficients—but it is balanced. So is this chemical equation:

Again, this is not conventional, but it is still balanced. Suppose we use a much larger number:

These coefficients are also in the ratio of 2:1:2. But these numbers are related to the number of things in a mole: the first and last numbers are two times Avogadro's number, while the second number is Avogadro's number. That means that the first and last numbers represent 2 mol, while the middle number is just 1 mol. Well, why not just use the number of moles in balancing the chemical equation?

is the same balanced chemical equation we started with! What this means is that chemical equations are not just balanced in terms of molecules; they are also balanced in terms of moles. We can just as easily read this chemical equation as “two moles of hydrogen react with one mole of oxygen to make two moles of water.” All balanced chemical reactions are balanced in terms of moles.

In Chapter 4, Section 4.1, we stated that a chemical equation is simply a recipe for a chemical reaction. As such, chemical equations also give us equivalences—equivalences between the reactants and the products. However, now we understand that these equivalences are expressed in terms of moles. Consider the chemical equation 2H₂ + O₂ → 2H₂O. This chemical reaction gives us the following equivalences:

Any two of these quantities can be used to construct a conversion factor that lets us relate the number of moles of one substance to an equivalent number of moles of another substance. If, for example, we want to know how many moles of oxygen will react with 17.6 mol of hydrogen, we construct a conversion factor between 2 mol of H₂ and 1 mol of O₂ and use it to convert from moles of one substance to moles of another:

Note how the mol H₂ unit cancels, and mol O₂ is the new unit introduced. This is an example of a mole-mole calculation, when you start with moles of one substance and convert to moles of another substance by using the balanced chemical equation. The example may seem simple because the numbers are small, but numbers won't always be so simple!

5.4

Mole-Mass and Mass-Mass Calculations

Now the two halves join. 5.2 gave you the factor that moves between grams and moles; 5.3 gave you the factor that crosses the arrow. Put a molar mass on each end of a mole ratio and you can start at a balance reading and finish at a balance reading — which is what an actual laboratory question sounds like.

Mole-mole calculations are not the only type of calculations that can be performed using balanced chemical equations. Recall that the molar mass can be determined from a chemical formula and used as a conversion factor. We can add that conversion factor as another step in a calculation to make a mole-mass calculation, where we start with a given number of moles of a substance and calculate the mass of another substance involved in the chemical equation, or vice versa.

For example, suppose we have the balanced chemical equation 2Al + 3Cl₂ → 2AlCl₃. Suppose we know we have 123.2 g of Cl₂. How can we determine how many moles of AlCl₃ we will get when the reaction is complete? First and foremost, chemical equations are not balanced in terms of grams; they are balanced in terms of moles. So to use the balanced chemical equation to relate an amount of Cl₂ to an amount of AlCl₃, we need to convert the given amount of Cl₂ into moles. We know how to do this by simply using the molar mass of Cl₂ as a conversion factor. The molar mass of Cl₂ (which we get from the atomic mass of Cl from the periodic table) is 70.90 g/mol. We must invert this fraction so that the units cancel properly:

Now that we have the quantity in moles, we can use the balanced chemical equation to construct a conversion factor that relates the number of moles of Cl₂ to the number of moles of AlCl₃. The numbers in the conversion factor come from the coefficients in the balanced chemical equation:

Using this conversion factor with the molar quantity we calculated above, we get

So, we will get 1.159 mol of AlCl₃ if we react 123.2 g of Cl₂.

In this last example, we did the calculation in two steps. However, it is mathematically equivalent to perform the two calculations sequentially on one line:

The units still cancel appropriately, and we get the same numerical answer in the end.

It should be a trivial task now to extend the calculations to mass-mass calculations, in which we start with a mass of some substance and end with the mass of another substance in the chemical reaction. For this type of calculation, the molar masses of two different substances must be used—be sure to keep track of which is which.

5.6

Limiting Reagents

Everything so far assumed you have exactly enough of everything. Kitchens don't work that way and neither do reactions. This section is the chapter's payoff, and the method is simply the last section's calculation run twice — once from each reactant — and then a comparison.

One additional assumption we have made about chemical reactions—in addition to the assumption that reactions proceed all the way to completion—is that all the reactants are present in the proper quantities to react to products. This is not always the case.

Figure 5.2 · Making Water. Here we are taking hydrogen atoms and oxygen atoms (left) to make water molecules (right). However, there are not enough oxygen atoms to use up all the hydrogen atoms. We run out of oxygen atoms and cannot make any more water molecules, so the process stops when we run out of oxygen atoms.

A similar situation exists for many chemical reactions: you usually run out of one reactant before all of the other reactant has reacted. The reactant you run out of is called the limiting reagent; the other reactant or reactants are considered to be in excess. A crucial skill in evaluating the conditions of a chemical process is to determine which reactant is the limiting reagent and which is in excess.

The key to recognizing which reactant is the limiting reagent is based on a mole-mass or mass-mass calculation: whichever reactant gives the lesser amount of product is the limiting reagent. What we need to do is determine an amount of one product (either moles or mass) assuming all of each reactant reacts. Whichever reactant gives the least amount of that particular product is the limiting reagent. It does not matter which product we use, as long as we use the same one each time. It does not matter whether we determine the number of moles or grams of that product; however, we will see shortly that knowing the final mass of product can be useful.

For example, consider this reaction:

Suppose we start a reaction with 50.0 g of As and 50.0 g of O₂. Which one is the limiting reagent? We need to perform two mole-mass calculations, each assuming that each reactant reacts completely. Then we compare the amount of the product produced by each and determine which is less. The calculations are as follows:

Comparing these two answers, it is clear that 0.334 mol of As₂O₃ is less than 1.04 mol of As₂O₃, so arsenic is the limiting reagent. If this reaction is performed under these initial conditions, the arsenic will run out before the oxygen runs out. We say that the oxygen is “in excess.”

Notice what the comparison is not. It is not which reactant you have less of in grams — both were 50.0 g here — and it is not which has fewer moles. It is which one produces less product once the mole ratio has had its say. That is why the calculation has to be run all the way to a product both times.

Identifying the limiting reagent, then, is straightforward. However, there are usually two associated questions: (1) what mass of product (or products) is then actually formed? and (2) what mass of what reactant is left over? The first question is straightforward to answer: simply perform a conversion from the number of moles of product formed to its mass, using its molar mass. For As₂O₃, the molar mass is 197.84 g/mol; knowing that we will form 0.334 mol of As₂O₃ under the given conditions, we will get

The second question is somewhat more convoluted to answer. First, we must do a mass-mass calculation relating the limiting reagent (here, As) to the other reagent (O₂). Once we determine the mass of O₂ that reacted, we subtract that from the original amount to determine the amount left over. According to the mass-mass calculation,

Because we reacted 16.0 g of our original O₂, we subtract that from the original amount, 50.0 g, to get the mass of O₂ remaining:

You must remember to perform this final subtraction to determine the amount remaining; a common error is to report the 16.0 g as the amount remaining.

Drills

The Games

Three short drills. They are short on purpose: the thing that builds the skill is repeatedly pulling the answer out of your own head, not reading another worked example. New numbers every round, so you cannot memorise your way through.

Why the page is built this way. The chain builder exists because Ball writes every problem as a chain with the units struck out — that notation is already an interface, so the page makes it operable, the way Aktiv Chemistry and ChemCollective handle dimensional analysis. It checks the chain rather than the number, because a right number reached by the wrong route teaches nothing.

Scaffolding is a button, not a rule. A 2021 study comparing hints with mandatory scaffolds in adult learners found that the scaffolded group requested significantly more correct answers, and that lower-prior-knowledge students were significantly less likely to finish. The authors recommend on-demand scaffolding that preserves the learner's agency. So every problem opens whole and "break it into steps" is yours to press.

The streak is honest decoration. A 2025 study of gamified feedback in adaptive retrieval practice found robust gains in felt competence, enjoyment and task value — and strong evidence against any effect on what was actually recalled days later. Points help you keep going; the retrieval is what teaches. Both are worth having, as long as nobody mistakes one for the other.