# 8.3: The Mole in Chemical Reactions

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Learning Objectives

- Balance a chemical equation in terms of moles.
- Use the balanced equation to construct conversion factors in terms of moles.
- Calculate moles of one substance from moles of another substance using a balanced chemical equation.

Consider this balanced chemical equation:

\[\ce{2H2 + O2 → 2H2O}\nonumber \]

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:

\[\ce{100H2 + 50O2 → 100H2O}\nonumber \]

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:

\[\ce{5,000 H2 + 2,500 O2 → 5,000H2O}\nonumber \]

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

\[12.044 \times 10^{23} \ce{H2} + 6.022 \times 10^{23} \ce{O2} → 12.044 \times 10^{23} \ce{H2O}\nonumber \]

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?

\[\ce{2H2 + O2 → 2H2O}\nonumber \]

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 \(\PageIndex{1}\)

Interpret this balanced chemical equation in terms of moles.

\[\ce{P4 + 5O2 → P4O10}\nonumber \]

**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."

Exercise \(\PageIndex{1}\)

Interpret this balanced chemical equation in terms of moles.

\[\ce{N2 + 3H2 → 2NH3}\nonumber \]

**Answer**-
One mole of elemental nitrogen reacts with three moles of elemental hydrogen to produce two moles of ammonia.

In Section 4.1, we stated that a chemical equation is simply a recipe for a chemical reaction. As such, chemical equations also give us equivalents—equivalents between the reactants and the products. However, now we understand that *these equivalents are expressed in terms of moles*. Consider the chemical equation

\[\ce{2H2 + O2 → 2H2O} \nonumber\nonumber \]

This chemical reaction gives us the following equivalents:

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:

\[17.6\cancel{mol\, H_{2}}\times \frac{1\, mol\, O_{2}}{2\cancel{mol\, H_{2}}}=8.80\, mol\, O_{2}\nonumber \]

Note how the mol H_{2} unit cancels, and mol O_{2} 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!

Example \(\PageIndex{2}\)

For the balanced chemical equation

\[\ce{2C4H10(g) + 13O2 → 8CO2(g) + 10H2O(ℓ)}\nonumber \]

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:

\[154\cancel{mol\, O_{2}}\times \frac{8\, mol\, CO_{2}}{13\cancel{mol\, O_{2}}}=94.8\, mol\, CO_{2}\nonumber \]

The mol O_{2} unit is in the denominator of the conversion factor so it cancels. Both the 8 and the 13 are exact numbers, so they do not contribute to the number of significant figures in the final answer.

Exercise \(\PageIndex{2}\)

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.

## Summary

Balanced chemical reactions are balanced in terms of moles. A balanced chemical reaction gives equivalents in moles that allow stoichiometry calculations to be performed.