Adapted by Nelson NuñezRodriguez
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Opening Essay

We have already established that quantities are important in science, especially in chemistry. It is important to make accurate measurements of a variety of quantities when performing experiments. However, it is also important to be able to relate one measured quantity to another, unmeasured quantity. In this chapter, we will consider how we manipulate quantities to relate them to each other.
Learning Objective 

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 H_{2}. If the chemical reaction follows the balanced chemical equation
2H_{2}(g) + O_{2}(g) → 2H_{2}O(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 stoichiometry. 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 1 "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
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,
$$\text{9}\overline{)\text{cpancakemix}}\times \frac{\text{1egg}}{\text{2}\overline{)\text{cpancakemix}}}=4.5\text{eggs}$$
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:
$$\text{9}\overline{)\text{cpancakemix}}\times \frac{\text{1/2cmilk}}{2\overline{)\text{cpancakemix}}}=2.25\text{cmilk}$$
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:
2H_{2}(g) + O_{2}(g) → 2H_{2}O(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:
2H_{2} molecules ⇔ 1O_{2} molecule ⇔ 2H_{2}O molecules
These equivalences allow us to construct conversion factors:
$$\begin{array}{ccc}\frac{{\text{2moleculesH}}_{\text{2}}}{{\text{1moleculeO}}_{\text{2}}}& \frac{{\text{2moleculesH}}_{\text{2}}}{{\text{2moleculesH}}_{2}\text{O}}& \frac{{\text{1moleculeO}}_{\text{2}}}{{\text{2moleculesH}}_{\text{2}}\text{O}}\end{array}$$
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:
$$16\overline{){\text{moleculesH}}_{\text{2}}}\times \frac{1{\text{moleculeO}}_{\text{2}}}{2\overline{){\text{moleculesH}}_{\text{2}}}}=8{\text{moleculesO}}_{\text{2}}$$
Note how the unit molecules H_{2} 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.
Example 1 
How many molecules of SO_{3} are needed to react with 144 molecules of Fe_{2}O_{3} given this balanced chemical equation? Fe_{2}O_{3}(s) + 3SO_{3}(g) → Fe_{2}(SO_{4})_{3} Solution We use the balanced chemical equation to construct a conversion factor between Fe_{2}O_{3} and SO_{3}. The number of molecules of Fe_{2}O_{3} goes on the bottom of our conversion factor so it cancels with our given amount, and the molecules of SO_{3} go on the top. Thus, the appropriate conversion factor is $$\frac{{\text{3moleculesSO}}_{\text{3}}}{{\text{1moleculeFe}}_{\text{2}}{\text{O}}_{\text{3}}}$$
$$144\overline{){\text{moleculesFe}}_{\text{2}}{\text{O}}_{\text{3}}}\times \frac{{\text{3moleculesSO}}_{\text{3}}}{\text{1}\overline{){\text{moleculeFe}}_{\text{2}}{\text{O}}_{\text{3}}}}=432{\text{moleculesSO}}_{\text{3}}$$ $$$$
Test Yourself How many molecules of H_{2} are needed to react with 29 molecules of N_{2} to make ammonia if the balanced chemical equation is N_{2} + 3H_{2} → 2NH_{3}? Answer 87 molecules 
Chemical equations also allow us to make conversions regarding the number of atoms in a chemical reaction because a chemical formula lists the number of atoms of each element in a compound. The formula H_{2}O indicates that there are two hydrogen atoms and one oxygen atom in each molecule, and these relationships can be used to make conversion factors:
$$\begin{array}{cc}\frac{\text{2atomsH}}{{\text{1moleculeH}}_{\text{2}}\text{O}}& \frac{{\text{1moleculeH}}_{\text{2}}\text{O}}{\text{1atomO}}\end{array}$$
Conversion factors like this can also be used in stoichiometry calculations.
Example 2 
How many molecules of NH_{3} can you make if you have 228 atoms of H_{2}? Solution From the formula, we know that one molecule of NH_{3} has three H atoms. Use that fact as a conversion factor:
Test Yourself How many molecules of Fe_{2}(SO_{4})_{3} can you make from 777 atoms of S? Answer 259 molecules 
Key Takeaway 

Exercises 

Learning Objectives 

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 carbon12. Experimental measurements have determined that this number is very large:
1 mol = 6.02214179 × 10^{23} 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.
Example 3 
How many molecules are present in 2.76 mol of H_{2}O? How many atoms is this? Solution The definition of a mole is an equality that can be used to construct a conversion factor. Also, because we know that there are three atoms in each molecule of H_{2}O, we can also determine the number of atoms in the sample.
Test Yourself How many molecules are present in 4.61 × 10^{−2} mol of O_{2}? Answer 2.78 × 10^{22} molecules 
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 × 10^{23} atoms:
12 g C = 6.022 × 10^{23} 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 × 10^{25} atoms of carbon? This is a onestep conversion:
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 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.)
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, H_{2}. One molecule has a mass of 1.0079 + 1.0079 = 2.0158 u, while 1 mol H_{2} has a mass of 2.0158 g. A molecule of H_{2}O has a mass of about 18.01 u; 1 mol H_{2}O 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 × 10^{23} units: 6.022 × 10^{23} atoms of H, 6.022 × 10^{23} molecules of H_{2} and H_{2}O, 6.022 × 10^{23} units of NaCl ions. These relationships give us plenty of opportunities to construct conversion factors for simple calculations.
Example 4 

What is the molar mass of C_{6}H_{12}O_{6}? Solution To determine the molar mass, we simply add the atomic masses of the atoms in the molecular formula but express the total in grams per mole, not atomic mass units. The masses of the atoms can be taken from the periodic table:
Per convention, the unit grams per mole is written as a fraction. Test Yourself What is the molar mass of AgNO_{3}? Answer 169.87 g/mol 
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.
Example 5 
What is the mass of 3.56 mol of HgCl_{2}? The molar mass of HgCl_{2} is 271.49 g/mol. Solution
$$\text{3}\text{.56}\overline{){\text{molHgCl}}_{2}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{271.49{\text{gHgCl}}_{2}}{\overline{){\text{molHgCl}}_{2}}}=967{\text{gHgCl}}_{2}$$ Test Yourself
Answer

Example 6 
How many moles of H_{2}O are present in 240.0 g of water (about the mass of a cup of water)? Solution
$$240.0\overline{){\text{gH}}_{2}\text{O}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{{\text{1molH}}_{2}\text{O}}{18.015\overline{){\text{gH}}_{2}\text{O}}}=13.32{\text{molH}}_{2}\text{O}$$ Test Yourself How many moles are present in 35.6 g of H_{2}SO_{4} (molar mass = 98.08 g/mol)? Answer 0.363 mol 
Other conversion factors can be combined with the definition of mole—density, for example.
Example 7 
The density of ethanol is 0.789 g/mL. How many moles are in 100.0 mL of ethanol? The molar mass of ethanol is 46.08 g/mol. Solution Here, we use density to convert from volume to mass and then use the molar mass to determine the number of moles. $$100.0\overline{)\text{mL}}\text{ethanol}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{0.789\overline{)\text{g}}}{\overline{)\text{mL}}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{1\text{mol}}{46.08\overline{)\text{g}}}=1.71\text{molethanol}$$ Test Yourself If the density of benzene, C_{6}H_{6}, is 0.879 g/mL, how many moles are present in 17.9 mL of benzene? Answer 0.201 mol 
Key Takeaways 

Exercises 

Learning Objectives 

Consider this balanced chemical equation:
2H_{2} + O_{2} → 2H_{2}O
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:
100H_{2} + 50O_{2} → 100H_{2}O
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:
5,000H_{2} + 2,500O_{2} → 5,000H_{2}O
Again, this is not conventional, but it is still balanced. Suppose we use a much larger number:
12.044 × 10^{23} H_{2} + 6.022 × 10^{23} O_{2} → 12.044 × 10^{23} H_{2}O
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?
2H_{2} + O_{2} → 2H_{2}O
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.
Example 8 
Interpret this balanced chemical equation in terms of moles. P_{4} + 5O_{2} → P_{4}O_{10} Solution The coefficients represent the number of moles that react, not just molecules. We would speak of this equation as “one mole of molecular phosphorus reacts with five moles of elemental oxygen to make one mole of tetraphosphorus decoxide.”Test Yourself Interpret this balanced chemical equation in terms of moles. N_{2} + 3H_{2} → 2NH_{3} Answer One mole of elemental nitrogen reacts with three moles of elemental hydrogen to produce two moles of ammonia. 
In Chapter 5 "Chemical Reactions and Equations: 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
2H_{2} + O_{2} → 2H_{2}O
This chemical reaction gives us the following equivalences:
2 mol H_{2} ⇔ 1 mol O_{2} ⇔ 2 mol H_{2}O
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 H_{2} and 1 mol of O_{2} and use it to convert from moles of one substance to moles of another:
$$\text{17}\text{.6}\overline{){\text{molH}}_{2}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{{\text{1molO}}_{2}}{2\overline{){\text{molH}}_{2}}}=8.80{\text{molO}}_{2}$$
Note how the mol H_{2} unit cancels, and mol O_{2} is the new unit introduced. This is an example of a molemole 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!
Example 9 
For the balanced chemical equation 2C_{4}H_{10}(g) + 13O_{2} → 8CO_{2}(g) + 10H_{2}O(ℓ)if 154 mol of O_{2} are reacted, how many moles of CO_{2} are produced?
Solution We are relating an amount of oxygen to an amount of carbon dioxide, so we need the equivalence between these two substances. According to the balanced chemical equation, the equivalence is 13 mol O_{2} ⇔ 8 mol CO_{2}We can use this equivalence to construct the proper conversion factor. We start with what we are given and apply the conversion factor:
Test Yourself Using the above equation, how many moles of H_{2}O are produced when 154 mol of O_{2} react? Answer 118 mol 
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.
Key Takeaways 

Exercises 

Learning Objectives 

Molemole 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 molemass 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 + 3Cl_{2} → 2AlCl_{3}
Suppose we know we have 123.2 g of Cl_{2}. How can we determine how many moles of AlCl_{3} 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 Cl_{2} to an amount of AlCl_{3}, we need to convert the given amount of Cl_{2} into moles. We know how to do this by simply using the molar mass of Cl_{2} as a conversion factor. The molar mass of Cl_{2} (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 Cl_{2} to the number of moles of AlCl_{3}. The numbers in the conversion factor come from the coefficients in the balanced chemical equation:
$$\frac{{\text{2molAlCl}}_{3}}{{\text{3molCl}}_{2}}$$
Using this conversion factor with the molar quantity we calculated above, we get
$$1.738\overline{){\text{molCl}}_{2}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{2{\text{molAlCl}}_{3}}{\text{3}\overline{){\text{molCl}}_{2}}}=1.159{\text{molAlCl}}_{3}$$
So, we will get 1.159 mol of AlCl_{3} if we react 123.2 g of Cl_{2}.
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:
$$\text{123}\text{.2}\overline{){\text{gCl}}_{2}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{1\overline{){\text{molCl}}_{2}}}{70.90\overline{){\text{gCl}}_{2}}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{2{\text{molAlCl}}_{3}}{3\overline{){\text{molCl}}_{2}}}=1.159{\text{molAlCl}}_{3}$$
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.
Example 10 
How many moles of HCl will be produced when 249 g of AlCl_{3} are reacted according to this chemical equation? 2AlCl_{3} + 3H_{2}O(ℓ) → Al_{2}O_{3} + 6HCl(g)
Solution We will do this in two steps: convert the mass of AlCl_{3} to moles and then use the balanced chemical equation to find the number of moles of HCl formed. The molar mass of AlCl_{3} is 133.33 g/mol, which we have to invert to get the appropriate conversion factor: $$\text{249}\overline{){\text{gAlCl}}_{3}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{1{\text{molAlCl}}_{3}}{\text{133}\text{.33}\overline{){\text{gAlCl}}_{3}}}=1.87{\text{molAlCl}}_{3}$$
$$\text{1}\text{.87}\overline{){\text{molAlCl}}_{3}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{\text{6molHCl}}{\text{2}\overline{){\text{molAlCl}}_{3}}}=5.61\text{molHCl}$$
Test Yourself How many moles of Al_{2}O_{3} will be produced when 23.9 g of H_{2}O are reacted according to this chemical equation? 2AlCl_{3} + 3H_{2}O(ℓ) → Al_{2}O_{3} + 6HCl(g) Answer 0.442 mol 
A variation of the molemass 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.
Example 11 
How many grams of NH_{3} will be produced when 33.9 mol of H_{2} are reacted according to this chemical equation? N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)
Solution The conversions are the same, but they are applied in a different order. Start by using the balanced chemical equation to convert to moles of another substance and then use its molar mass to determine the mass of the final substance. In two steps, we have
Test Yourself How many grams of N_{2} are needed to produce 2.17 mol of NH_{3} when reacted according to this chemical equation? N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)Answer 30.4 g (Note: here we go from a product to a reactant, showing that molemass problems can begin and end with any substance in the chemical equation.) 
It should be a trivial task now to extend the calculations to massmass 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. 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 SO_{3} that can be produced by the reaction of 45.3 g of SO_{2} and O_{2}:
2SO_{2}(g) + O_{2}(g) → 2SO_{3}(g)
First, we convert the given amount, 45.3 g of SO_{2}, to moles of SO_{2} using its molar mass (64.06 g/mol):
Second, we use the balanced chemical reaction to convert from moles of SO_{2} to moles of SO_{3}:
$$\text{0}\text{.707}\overline{){\text{molSO}}_{2}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{{\text{2molSO}}_{3}}{\text{2}\overline{){\text{molSO}}_{2}}}=\text{0}{\text{.707molSO}}_{3}$$
Finally, we use the molar mass of SO3 (80.06 g/mol) to convert to the mass of SO3:
$$\text{0}\text{.707}\overline{){\text{molSO}}_{3}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{\text{80}{\text{.06gSO}}_{3}}{1\overline{){\text{molSO}}_{3}}}=\text{56}{\text{.6gSO}}_{3}$$
We can also perform all three steps sequentially, writing them on one line as
$$\text{45}\text{.3}\overline{){\text{gSO}}_{2}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{\text{1}\overline{){\text{molSO}}_{2}}}{\text{64}\text{.06}\overline{){\text{gSO}}_{2}}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{\text{2}\overline{){\text{molSO}}_{3}}}{\text{2}\overline{){\text{molSO}}_{2}}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{\text{80}{\text{.06gSO}}_{3}}{1\overline{){\text{molSO}}_{3}}}=\text{56}{\text{.6gSO}}_{3}$$
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.
Example 12 
What mass of Mg will be produced when 86.4 g of K are reacted? MgCl_{2}(s) + 2K(s) → Mg(s) + 2KCl(s) Solution We will simply follow the steps mass K → mol K → mol Mg → mass MgIn addition to the balanced chemical equation, we need the molar masses of K (39.09 g/mol) and Mg (24.31 g/mol). In one line, Test Yourself What mass of H_{2} will be produced when 122 g of Zn are reacted? Zn(s) + 2HCl(aq) → ZnCl_{2}(aq) + H_{2}(g)Answer 3.77 g 
Key Takeaways 

Exercises 

Learning Objective 

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 carboncontaining compounds, such as methane, some CO is produced as well as CO_{2}. 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 12 "Chemical Equilibrium"). When we calculate an amount of product assuming that all the reactant reacts, we calculate the theoretical yield, 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 yield. 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 yield is a comparison between the actual yield and the theoretical yield and is defined as
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.
Example 13 
A worker reacts 30.5 g of Zn with nitric acid and evaporates the remaining water to obtain 65.2 g of Zn(NO_{3})_{2}. What are the theoretical yield, the actual yield, and the percent yield? Zn(s) + 2HNO_{3}(aq) → Zn(NO_{3})_{2}(aq) + H_{2}(g) Solution A massmass calculation can be performed to determine the theoretical yield. We need the molar masses of Zn (65.39 g/mol) and Zn(NO_{3})_{2} (189.41 g/mol). In three steps, the massmass calculation is $$30.5\overline{)\text{gZn}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{\text{1}\overline{)\text{molZn}}}{65.39\overline{)\text{gZn}}}\times \frac{\text{1}\overline{){\text{molZn(NO}}_{3}{)}_{2}}}{1\overline{)\text{molZn}}}\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\frac{189.41{\text{gZn(NO}}_{3}{)}_{2}}{1\overline{){\text{molZn(NO}}_{3}{)}_{2}}}=88.3{\text{gZn(NO}}_{3}{)}_{2}$$
Thus, the theoretical yield is 88.3 g of Zn(NO_{3})_{2}. The actual yield is the amount that was actually made, which was 65.2 g of Zn(NO_{3})_{2}. To calculate the percent yield, we take the actual yield and divide it by the theoretical yield and multiply by 100:
The worker achieved almost threefourths of the possible yield. Test Yourself A synthesis produced 2.05 g of NH_{3} from 16.5 g of N_{2}. What is the theoretical yield and the percent yield? N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g) Answer $\text{}$theoretical yield = 20.1 g; percent yield = 10.2% 
Chemistry Is Everywhere: Actual Yields in Drug Synthesis and Purification 

Many drugs are the product of several steps of chemical synthesis. Each step typically occurs with less than 100% yield, so the overall percent yield might be very small. The general rule is that the overall percent yield is the product of the percent yields of the individual synthesis steps. For a drug synthesis that has many steps, the overall percent yield can be very tiny, which is one factor in the huge cost of some drugs. For example, if a 10step synthesis has a percent yield of 90% for each step, the overall yield for the entire synthesis is only 35%. Many scientists work every day trying to improve percent yields of the steps in the synthesis to decrease costs, improve profits, and minimize waste. Even purifications of complex molecules into drugquality purity are subject to percent yields. Consider the purification of impure albuterol. Albuterol (C_{13}H_{21}NO_{2}; accompanying figure) is an inhaled drug used to treat asthma, bronchitis, and other obstructive pulmonary diseases. It is synthesized from norepinephrine, a naturally occurring hormone and neurotransmitter. Its initial synthesis makes very impure albuterol that is purified in five chemical steps. The details of the steps do not concern us; only the percent yields do:

Key Takeaways 

Exercises 

Additional Exercises 
