# 9.1 Stoichiometry Basics

## Learning Objectives

By the end of this section, you will be able to:

• Explain the concept of stoichiometry as it pertains to chemical reactions
• Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products

A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction’s stoichiometry, a term derived from the Greek words stoicheion (meaning “element”) and metron (meaning “measure”). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.

The general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Food preparation, for example, offers an appropriate comparison. A recipe for making eight pancakes calls for 1 cup pancake mix, $\frac{3}{4}$ cup milk, and one egg. The “equation” representing the preparation of pancakes per this recipe is

$1 \;\text{cup mix} + \frac{3}{4} \;\text{cup milk} + 1 \;\text{egg} \longrightarrow 8 \;\text{pancakes}$

If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is

$24 \;\rule[0.5ex]{4em}{0.1ex}\hspace{-4em}\text{pancakes} \times \frac{1 \;\text{egg}}{8 \;\rule[0.25ex]{3em}{0.1ex}\hspace{-3em}\text{pancakes}} = 3 \;\text{eggs}$

Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive stoichiometric factors that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:

$\text{N}_2(g) + 3\text{H}_2(g) \longrightarrow 2\text{NH}_3(g)$

This equation shows ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit:

$\frac{2 \;\text{NH}_3 \;\text{molecules}}{3 \;\text{H}_2 \;\text{molecules}} \;\text{or} \;\frac{2 \;\text{doz NH}_3 \;\text{molecules}}{3 \;\text{doz H}_2 \;\text{molecules}} \;\text{or} \;\frac{2 \;\text{mol NH}_3 \;\text{molecules}}{3 \;\text{mol H}_2 \;\text{molecules}}$

These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.

Watch Determining the Mole Ratio (6 mins)

Video Source: SmarterTeacher (2013, January 30). Determining the mole ratio [Video]. YouTube.

### Example 9.1a

#### Moles of Reactant Required in a Reaction

How many moles of I2 are required to react with 0.429 mol of Al according to the following equation (see Figure 9.1a)?

$2\text{Al} + 3\text{I}_2 \longrightarrow 2\text{AlI}_3$

#### Solution

Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is $\frac{3 \;\text{mol I}_2}{2 \;\text{mol Al}}$. The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:

$\begin{array}{r @{{}={}} l} & 0.429 \;\rule[0.5ex]{3.25em}{0.1ex}\hspace{-3.25em}\text{mol Al} \times \frac{3 \;\text{mol I}_2}{2 \;\rule[0.25ex]{2em}{0.1ex}\hspace{-2em}\text{mol Al}} \\[1em] & = 0.644 \;\text{mol I}_2 \end{array}$

### Exercise 9.1a

How many moles of Ca(OH)2 are required to react with 1.36 mol of H3PO4 to produce Ca3(PO4)2 according to the equation:

$3\text{Ca(OH)}_2 + 2\text{H}_3 \text{PO}_4 \longrightarrow \text{Ca}_3 \text{(PO}_4)_2 + 6\text{H}_2 \text{O}$?

These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related.

### Exercise 9.1b

Practice using the following PhET simulation: Reactants, Products, and Leftovers

Activity source: Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC-BY-4.0

### Link to Interactive Learning Tools

Practice Mole-to-mole ratios from eCampusOntario H5P Studio.

Practice Stoichiometry – Relationships (Apprentice Difficulty Level ONLY) from The Physics Classroom.