Stoichiometry (article) | Chemical reactions

Stoichiometry (article) | Chemical reactions | Khan Academy (2025)

Introduction

A balanced chemical equation is analogous to a recipe for chocolate chip cookies. It shows what reactants (the ingredients) combine to form what products (the cookies). It also shows the numerical relationships between the reactants and products (such as how many cups of flour are required to make a single batch of cookies).

These numerical relationships are known as reaction stoichiometry, a term derived from the Ancient Greek words stoicheion ("element") and metron ("measure"). In this article, we'll look at how we can use the stoichiometric relationships contained in balanced chemical equations to determine amounts of substances consumed and produced in chemical reactions.

Balanced equations and mole ratios

A common type of stoichiometric relationship is the mole ratio, which relates the amounts in moles of any two substances in a chemical reaction. We can write a mole ratio for a pair of substances by looking at the coefficients in front of each species in the balanced chemical equation. For example, consider the equation for the reaction between iron(III) oxide and aluminum metal:

${Fe}_{2} O_{3} (s) + 2 Al (s) \to 2 Fe (l) + {Al}_{2} O_{3} (s)$ ‍

The coefficients in the equation tell us that $1$ ‍ mole of $Fe_{2} O_{3}$ ‍ reacts with $2$ ‍ moles of $Al$ ‍, forming $2$ ‍ moles of $Fe$ ‍ and $1$ ‍ mole of $Al_{2} O_{3}$ ‍. We can write the relationship between the $Fe_{2} O_{3}$ ‍ and the $Al$ ‍ as the following mole ratio:

$1 {mol Fe}_{2} O_{3} : 2 mol Al$ ‍

Using this ratio, we could calculate how many moles of $Al$ ‍ are needed to fully react with a certain amount of ${Fe}_{2} O_{3}$ ‍, or vice versa. In general, mole ratios can be used to convert between amounts of any two substances involved in a chemical reaction. To illustrate, let's walk through an example where we use a mole ratio to convert between amounts of reactants.

Example: Using mole ratios to calculate mass of a reactant

Consider the following unbalanced equation:

$NaOH (a q) + H_{2} SO_{4} (a q) \to H_{2} O (l) + Na_{2} SO_{4} (a q)$ ‍

How many grams of $NaOH$ ‍ are required to fully consume $3.10$ ‍ grams of $H_{2} SO_{4}$ ‍?

First things first: we need to balance the equation! In this case, we have $1$ ‍ $Na$ ‍ atom and $3$ ‍ $H$ ‍ atoms on the reactant side and $2$ ‍ $Na$ ‍ atoms and $2$ ‍ $H$ ‍ atoms on the product side. We can balance the equation by placing a $2$ ‍ in front of $NaOH$ ‍ (so that there are $2$ ‍ $Na$ ‍ atoms on each side) and another $2$ ‍ in front of $H_{2} O$ ‍ (so that there are $6$ ‍ $O$ ‍ atoms and $4$ ‍ $H$ ‍ atoms on each side). Doing so gives the following balanced equation:

$2 NaOH (a q) + H_{2} SO_{4} (a q) \to 2 H_{2} O (l) + Na_{2} SO_{4} (a q)$ ‍

Now that we have the balanced equation, let's get to problem solving. To review, we want to find the mass of $NaOH$ ‍ that is needed to completely react $3.10$ ‍ grams of $H_{2} SO_{4}$ ‍. We can tackle this stoichiometry problem using the following steps:

Step 1: Convert known reactant mass to moles

In order to relate the amounts $H_{2} SO_{4}$ ‍ and $NaOH$ ‍ using a mole ratio, we first need to know the quantity of $H_{2} SO_{4}$ ‍ in moles. We can convert the $3.10$ ‍ grams of $H_{2} SO_{4}$ ‍ to moles using the molar mass of $H_{2} SO_{4}$ ‍ ( $98.08 g / mol$ ‍):

$3.10 {g H}_{2} {SO}_{4} \times \frac{1 {mol H}_{2} {SO}_{4}}{98.08 {g H}_{2} {SO}_{4}} = 3.16 \times 10^{- 2} {mol H}_{2} {SO}_{4}$ ‍

Step 2: Use the mole ratio to find moles of other reactant

Now that we have the quantity of $H_{2} SO_{4}$ ‍ in moles, let's convert from moles of $H_{2} SO_{4}$ ‍ to moles of $NaOH$ ‍ using the appropriate mole ratio. According to the coefficients in the balanced chemical equation, $2$ ‍ moles of $NaOH$ ‍ are required for every $1$ ‍ mole of $H_{2} SO_{4}$ ‍, so the mole ratio is

$\frac{2 mol NaOH}{1 {mol H}_{2} {SO}_{4}}$ ‍

Multiplying the number of moles of $H_{2} SO_{4}$ ‍ by this factor gives us the number of moles of $NaOH$ ‍ needed:

$3.16 \times 10^{- 2} {mol H}_{2} {SO}_{4} \times \frac{2 mol NaOH}{1 {mol H}_{2} {SO}_{4}} = 6.32 \times 10^{- 2} mol NaOH$ ‍

Notice how we wrote the mole ratio so that the moles of $H_{2} SO_{4}$ ‍ cancel out, resulting in moles of $NaOH$ ‍ as the final units. To learn how units can be treated as numbers for easier bookkeeping in problems like this, check out this video on dimensional analysis.

Step 3: Convert moles of other reactant to mass

We were asked for the mass of $NaOH$ ‍ in grams, so our last step is to convert the $6.32 \times 10^{- 2}$ ‍ moles of $NaOH$ ‍ to grams. We can do so using the molar mass of $NaOH$ ‍ ( $40.00 g / mol$ ‍):

$6.32 \times 10^{- 2} mol NaOH \times \frac{40.00 g NaOH}{1 mol NaOH} = 2.53 g NaOH$ ‍

So, $2.53 g$ ‍ of $NaOH$ ‍ are required to fully consume $3.10$ ‍ grams of $H_{2} SO_{4}$ ‍ in this reaction.

Shortcut: We could have combined all three steps into a single calculation, as shown in the following expression:

$\begin{aligned} \underset{⏟}{3.10 {g H}_{2} {SO}_{4} \times \frac{1 {mol H}_{2} {SO}_{4}}{98.08 {g H}_{2} {SO}_{4}}} \times \underset{⏟}{\frac{2 mol NaOH}{1 {mol H}_{2} {SO}_{4}}} \times \underset{⏟}{\frac{40.00 g NaOH}{1 mol NaOH}} = 2.53 g NaOH \\ Step 1 Step 2 Step 3 \\ {Find mol of H}_{2} {SO}_{4} Use mole ratio Find g of NaOH \end{aligned}$ ‍

Be sure to pay extra close attention to the units if you take this approach, though!

Summary

A balanced chemical equation shows us the numerical relationships between each of the species involved in the chemical change. We can use these numerical relationships to write mole ratios, which allow us to convert between amounts of reactants and/or products (and thus solve stoichiometry problems!).

To learn about other common stoichiometric calculations, check out this exciting sequel on limiting reactants and percent yield!

This article was adapted from the following articles:

"Chemical Reactions" from UC Davis ChemWiki, CC BY-NC-SA 3.0
"Stoichiometry and Balancing Reactions" from UC Davis ChemWiki, CC BY-NC-SA 3.0
"Reaction Stoichiometry" from Boundless Chemistry, CC BY-SA 4.0

The modified article is licensed under a CC BY-NC-SA 4.0 license.

Additional references:

Zumdahl, S. S., and Zumdahl, S. A. (2014). Stoichiometric Calculations: Amounts of Reactants and Products. In Chemistry (9th ed., pp. 108–114). Belmont, CA: Brooks/Cole.