Chapter 5 Stoichiometry and the Mole

5.1 Stoichiometry

Consider a classic recipe for pound cake: 1 pound of eggs, 1 pound of butter, 1 pound of flour, and 1 pound of sugar. (That’s why it’s called “pound cake.”) If you have 4 pounds of butter, how many pounds of sugar, flour, and eggs do you need? You would need 4 pounds each of sugar, flour, and eggs.

Now suppose you have 1.00 g H₂. If the chemical reaction follows the balanced chemical equation

2H₂(g) + O₂(g) → 2H₂O(ℓ)

then what mass of oxygen do you need to make water?

Curiously, this chemical reaction question is very similar to the pound cake question. Both of them involve relating a quantity of one substance to a quantity of another substance or substances. The relating of one chemical substance to another using a balanced chemical reaction is called stoichiometryThe relating of one chemical substance to another using a balanced chemical reaction.. Using stoichiometry is a fundamental skill in chemistry; it greatly broadens your ability to predict what will occur and, more importantly, how much is produced.

Let us consider a more complicated example. A recipe for pancakes calls for 2 cups (c) of pancake mix, 1 egg, and 1/2 c of milk. We can write this in the form of a chemical equation:

2 c mix + 1 egg + 1/2 c milk → 1 batch of pancakes

If you have 9 c of pancake mix, how many eggs and how much milk do you need? It might take a little bit of work, but eventually you will find you need 4½ eggs and 2¼ c milk.

How can we formalize this? We can make a conversion factor using our original recipe and use that conversion factor to convert from a quantity of one substance to a quantity of another substance, similar to the way we constructed a conversion factor between feet and yards in Chapter 2 "Measurements". Because one recipe’s worth of pancakes requires 2 c of pancake mix, 1 egg, and 1/2 c of milk, we actually have the following mathematical relationships that relate these quantities:

2 c pancake mix ⇔ 1 egg ⇔ 1/2 c milk

where ⇔ is the mathematical symbol for “is equivalent to.” This does not mean that 2 c of pancake mix equal 1 egg. However, as far as this recipe is concerned, these are the equivalent quantities needed for a single recipe of pancakes. So, any possible quantities of two or more ingredients must have the same numerical ratio as the ratios in the equivalence.

We can deal with these equivalences in the same way we deal with equalities in unit conversions: we can make conversion factors that essentially equal 1. For example, to determine how many eggs we need for 9 c of pancake mix, we construct the conversion factor

$$\frac{\text{1 egg}}{\text{2 c pancake mix}}$$

This conversion factor is, in a strange way, equivalent to 1 because the recipe relates the two quantities. Starting with our initial quantity and multiplying by our conversion factor,

$$\text{9}\cancel{\text{ c pancake mix}}\times \frac{\text{1 egg}}{\text{2}\cancel{\text{ c pancake mix}}}=4.5\text{ eggs}$$

Note how the units cups pancake mix canceled, leaving us with units of eggs. This is the formal, mathematical way of getting our amounts to mix with 9 c of pancake mix. We can use a similar conversion factor for the amount of milk:

$$\text{9}\cancel{\text{ c pancake mix}}\times \frac{\text{1/2 c milk}}{2\cancel{\text{ c pancake mix}}}=2.25\text{ c milk}$$

Again, units cancel, and new units are introduced.

A balanced chemical equation is nothing more than a recipe for a chemical reaction. The difference is that a balanced chemical equation is written in terms of atoms and molecules, not cups, pounds, and eggs.

For example, consider the following chemical equation:

2H₂(g) + O₂(g) → 2H₂O(ℓ)

We can interpret this as, literally, “two hydrogen molecules react with one oxygen molecule to make two water molecules.” That interpretation leads us directly to some equivalences, just as our pancake recipe did:

2H₂ molecules ⇔ 1O₂ molecule ⇔ 2H₂O molecules

These equivalences allow us to construct conversion factors:

$$\begin{array}{ccc}\frac{{\text{2 molecules H}}_{\text{2}}}{{\text{1 molecule O}}_{\text{2}}}& \frac{{\text{2 molecules H}}_{\text{2}}}{{\text{ 2 molecules H}}_2\text{O}}& \frac{{\text{1 molecule O}}_{\text{2}}}{{\text{2 molecules H}}_{\text{2}}\text{O}}\end{array}$$

and so forth. These conversions can be used to relate quantities of one substance to quantities of another. For example, suppose we need to know how many molecules of oxygen are needed to react with 16 molecules of H₂. As we did with converting units, we start with our given quantity and use the appropriate conversion factor:

$$16\cancel{{\text{ molecules H}}_{\text{2}}}\times \frac{1{\text{ molecule O}}_{\text{2}}}{2\cancel{{\text{ molecules H}}_{\text{2}}}}=8{\text{ molecules O}}_{\text{2}}$$

Note how the unit molecules H₂ cancels algebraically, just as any unit does in a conversion like this. The conversion factor came directly from the coefficients in the balanced chemical equation. This is another reason why a properly balanced chemical equation is important.

Chemical equations also allow us to make conversions regarding the number of atoms in a chemical reaction because a chemical formula lists the number of atoms of each element in a compound. The formula H₂O indicates that there are two hydrogen atoms and one oxygen atom in each molecule, and these relationships can be used to make conversion factors:

$$\begin{array}{cc}\frac{\text{2 atoms H}}{{\text{1 molecule H}}_{\text{2}}\text{O}}& \frac{{\text{1 molecule H}}_{\text{2}}\text{O}}{\text{1 atom O}}\end{array}$$

Conversion factors like this can also be used in stoichiometry calculations.

5.2 The Mole

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 moleThe number of things equal to the number of atoms in exactly 12 g of carbon-12; equals $6.022\times {10}^{23}$ things. (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:

1 mol = 6.02214179 × 10²³ things

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.

For example, we already know that, by definition, a mole of carbon has a mass of exactly 12 g. This means that exactly 12 g of C has 6.022 × 10²³ atoms:

12 g C = 6.022 × 10²³ atoms C

We can use this equality as a conversion factor between the number of atoms of carbon and the number of grams of carbon. How many grams are there, say, in 1.50 × 10²⁵ atoms of carbon? This is a one-step conversion:

$$1.50\times {10}^{25}\cancel{\text{ atoms C}}\times \frac{12.0000\text{ g C}}{6.022\times {10}^{23}\cancel{\text{ atoms C}}}=299\text{ g C}$$

But it also goes beyond carbon. 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 massesThe mass of 1 mol of a substance in grams. 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.)

Here are some examples. The mass of a hydrogen atom is 1.0079 u; the mass of 1 mol of hydrogen atoms is 1.0079 g. Elemental hydrogen exists as a diatomic molecule, H₂. One molecule has a mass of 1.0079 + 1.0079 = 2.0158 u, while 1 mol H₂ has a mass of 2.0158 g. A molecule of H₂O has a mass of about 18.01 u; 1 mol H₂O has a mass of 18.01 g. A single unit of NaCl has a mass of 58.45 u; NaCl has a molar mass of 58.45 g. In each of these moles of substances, there are 6.022 × 10²³ units: 6.022 × 10²³ atoms of H, 6.022 × 10²³ molecules of H₂ and H₂O, 6.022 × 10²³ units of NaCl ions. These relationships give us plenty of opportunities to construct conversion factors for simple calculations.

Knowing the molar mass of a substance, we can calculate the number of moles in a certain mass of a substance and vice versa, as these examples illustrate. The molar mass is used as the conversion factor.

Other conversion factors can be combined with the definition of mole—density, for example.

5.3 The Mole in Chemical Reactions

Consider this balanced chemical equation:

2H₂ + O₂ → 2H₂O

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:

100H₂ + 50O₂ → 100H₂O

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:

5,000H₂ + 2,500O₂ → 5,000H₂O

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

12.044 × 10²³ H₂ + 6.022 × 10²³ O₂ → 12.044 × 10²³ H₂O

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?

2H₂ + O₂ → 2H₂O

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 "Chemical Reactions and Equations", Section 4.1 "The Chemical Equation", 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:

2 mol H₂ ⇔ 1 mol O₂ ⇔ 2 mol H₂O

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:

$$\text{17}\text{.6}\cancel{{\text{ mol H}}_2}\times \frac{{\text{1 mol O}}_2}{2\cancel{{\text{ mol H}}_2}}=8.80{\text{ mol O}}_2$$

Note how the mol H₂ unit cancels, and mol O₂ is the new unit introduced. This is an example of a mole-mole calculationA stoichiometry calculation when one starts with moles of one substance and convert to moles of another substance using the balanced chemical equation., 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!

It is important to reiterate that balanced chemical equations are balanced in terms of moles. Not grams, kilograms, or liters—but moles. Any stoichiometry problem will likely need to work through the mole unit at some point, especially if you are working with a balanced chemical reaction.

5.4 Mole-Mass and Mass-Mass Calculations

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 calculationA calculation in which you 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., 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:

$$\text{123}\text{.2}\cancel{{\text{ g Cl}}_2}\times \frac{1{\text{ mol Cl}}_2}{70.90\cancel{{\text{ g Cl}}_2}}=1.738{\text{ mol Cl}}_2$$

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:

$$\frac{{\text{2 mol AlCl}}_3}{{\text{3 mol Cl}}_2}$$

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

$$1.738\cancel{{\text{ mol Cl}}_2}\times \frac{2{\text{ mol AlCl}}_3}{\text{3}\cancel{{\text{ mol Cl}}_2}}=1.159{\text{ mol AlCl}}_3$$

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:

$$\text{123}\text{.2}\cancel{{\text{ g Cl}}_2}\times \frac{1\cancel{{\text{ mol Cl}}_2}}{70.90\cancel{{\text{ g Cl}}_2}}\times \frac{2{\text{ mol AlCl}}_3}{3\cancel{{\text{ mol Cl}}_2}}=1.159{\text{ mol AlCl}}_3$$

The units still cancel appropriately, and we get the same numerical answer in the end. Sometimes the answer may be slightly different from doing it one step at a time because of rounding of the intermediate answers, but the final answers should be effectively the same.

A variation of the mole-mass calculation is to start with an amount in moles and then determine an amount of another substance in grams. The steps are the same but are performed in reverse order.

It should be a trivial task now to extend the calculations to mass-mass calculationsA calculation in which you start with a given mass of a substance and calculate the mass of another substance involved in the chemical equation., 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. Again, however, it is important to emphasize that before the balanced chemical reaction is used, the mass quantity must first be converted to moles. Then the coefficients of the balanced chemical reaction can be used to convert to moles of another substance, which can then be converted to a mass.

For example, let us determine the number of grams of SO₃ that can be produced by the reaction of 45.3 g of SO₂ and O₂:

2SO₂(g) + O₂(g) → 2SO₃(g)

First, we convert the given amount, 45.3 g of SO₂, to moles of SO₂ using its molar mass (64.06 g/mol):

$$\text{45}\text{.3}\cancel{{\text{ g SO}}_2}\times \frac{{\text{1 mol SO}}_2}{\text{64}\text{.06}\cancel{{\text{ g SO}}_2}}=0.707{\text{ mol SO}}_2$$

Second, we use the balanced chemical reaction to convert from moles of SO₂ to moles of SO₃:

$$\text{0}\text{.707}\cancel{{\text{ mol SO}}_2}\times \frac{{\text{2 mol SO}}_3}{\text{2}\cancel{{\text{ mol SO}}_2}}=\text{0}{\text{.707 mol SO}}_3$$

Finally, we use the molar mass of SO₃ (80.06 g/mol) to convert to the mass of SO₃:

$$\text{0}\text{.707}\cancel{{\text{ mol SO}}_3}\times \frac{\text{80}{\text{.06 g SO}}_3}{1\cancel{{\text{ mol SO}}_3}}=\text{56}{\text{.6 g SO}}_3$$

We can also perform all three steps sequentially, writing them on one line as

$$\text{45}\text{.3}\cancel{{\text{ g SO}}_2}\times \frac{\text{1}\cancel{{\text{ mol SO}}_2}}{\text{64}\text{.06 }\cancel{{\text{ g SO}}_2}}\times \frac{\text{2}\cancel{{\text{ mol SO}}_3}}{\text{2}\cancel{{\text{ mol SO}}_2}}\times \frac{\text{80}{\text{.06 g SO}}_3}{1\cancel{{\text{ mol SO}}_3}}=\text{56}{\text{.6 g SO}}_3$$

We get the same answer. Note how the initial and all the intermediate units cancel, leaving grams of SO₃, which is what we are looking for, as our final answer.

5.5 Yields

In all the previous calculations we have performed involving balanced chemical equations, we made two assumptions: (1) the reaction goes exactly as written, and (2) the reaction proceeds completely. In reality, such things as side reactions occur that make some chemical reactions rather messy. For example, in the actual combustion of some carbon-containing compounds, such as methane, some CO is produced as well as CO₂. However, we will continue to ignore side reactions, unless otherwise noted.

The second assumption, that the reaction proceeds completely, is more troublesome. Many chemical reactions do not proceed to completion as written, for a variety of reasons (some of which we will consider in Chapter 13 "Chemical Equilibrium"). When we calculate an amount of product assuming that all the reactant reacts, we calculate the theoretical yieldAn amount that is theoretically produced as calculated using the balanced chemical reaction., an amount that is theoretically produced as calculated using the balanced chemical reaction.

In many cases, however, this is not what really happens. In many cases, less—sometimes much less—of a product is made during the course of a chemical reaction. The amount that is actually produced in a reaction is called the actual yieldThe amount that is actually produced in a chemical reaction.. By definition, the actual yield is less than or equal to the theoretical yield. If it is not, then an error has been made.

Both theoretical yields and actual yields are expressed in units of moles or grams. It is also common to see something called a percent yield. The percent yieldActual yield divided by theoretical yield times 100% to give a percentage between 0% and 100%. is a comparison between the actual yield and the theoretical yield and is defined as

percent yield=actual yieldtheoretical yield×100%

It does not matter whether the actual and theoretical yields are expressed in moles or grams, as long as they are expressed in the same units. However, the percent yield always has units of percent. Proper percent yields are between 0% and 100%—again, if percent yield is greater than 100%, an error has been made.

5.6 Limiting Reagents

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.

Consider 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 reagentThe reactant that runs out first.; 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:

4As(s) + 3O₂(g) → 2As₂O₃(s)

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:

$$\text{50}\text{.0}\cancel{\text{ g As}}\times \frac{\text{1}\cancel{\text{ mol As}}}{\text{74}\text{.92}\cancel{\text{ g As}}}\times \frac{{\text{2 mol As}}_2{\text{O}}_3}{\text{4}\cancel{\text{ mol As}}}=0.334{\text{ mol As}}_2{\text{O}}_3$$ $$\text{50}\text{.0}\cancel{{\text{ g O}}_2}\times \frac{\text{1}\cancel{{\text{ mol O}}_2}}{\text{ 32}\text{.00}\cancel{{\text{ g O}}_2}}\times \frac{2{\text{ mol As}}_2{\text{O}}_3}{\text{3}\cancel{{\text{ mol O}}_2}}=1.04{\text{ mol As}}_2{\text{O}}_3$$

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.”

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

$$\text{0}\text{.334}\cancel{{\text{ mol As}}_2{\text{O}}_3}\times \frac{\text{197}{\text{.84 g As}}_2{\text{O}}_3}{1\cancel{{\text{ mol As}}_2{\text{O}}_3}}=66.1{\text{ g As}}_2{\text{O}}_3$$

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,

$$\text{50}\text{.0}\cancel{\text{ g As}}\times \frac{\text{1}\cancel{\text{ mol As}}}{74.92\cancel{\text{ g As}}}\times \frac{\text{3}\cancel{{\text{ mol O}}_2}}{\text{4}\cancel{\text{ mol As}}}\times \frac{32.00{\text{ g O}}_2}{\text{1}\cancel{{\text{ mol O}}_2}}=16.0{\text{ g O}}_2\text{ reacted}$$

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:

50.0 g O₂ − 16.0 g O₂ reacted = 34.0 g O₂ left over

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.