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Freshly baked chocolate chip cookies on a wire cooling rack.
You might use stoichiometry skills to double a cookie recipe! Image credit: "Chocolate Chip Cookies" by Kimberley Vardeman on Wikimedia Commons, CC BY 2.0.
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:
The coefficients in the equation tell us that 1 mole of FeA2OA3 reacts with 2 moles of Al, forming 2 moles of Fe and 1 mole of AlA2OA3. We can write the relationship between the FeA2OA3 and the Al as the following mole ratio:
1mol Fe2O3:2mol Al
Using this ratio, we could calculate how many moles of Al are needed to fully react with a certain amount of Fe2O3, 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:
How many grams of NaOH are required to fully consume 3.10 grams of HA2SOA4?
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 HA2O (so that there are 6 O atoms and 4 H atoms on each side). Doing so gives the following balanced equation:
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 HA2SOA4. We can tackle this stoichiometry problem using the following steps:

Step 1: Convert known reactant mass to moles

In order to relate the amounts HA2SOA4 and NaOH using a mole ratio, we first need to know the quantity of HA2SOA4 in moles. We can convert the 3.10 grams of HA2SOA4 to moles using the molar mass of HA2SOA4 (98.08 g/mol):
3.10g H2SO4×1mol H2SO498.08g H2SO4=3.16×102mol H2SO4

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

Now that we have the quantity of HA2SOA4 in moles, let's convert from moles of HA2SOA4 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 HA2SOA4, so the mole ratio is
2mol NaOH1mol H2SO4
Multiplying the number of moles of HA2SOA4 by this factor gives us the number of moles of NaOH needed:
3.16×102mol H2SO4×2mol NaOH1mol H2SO4=6.32×102mol NaOH
Notice how we wrote the mole ratio so that the moles of HA2SOA4 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 6.32×102 moles of NaOH to grams. We can do so using the molar mass of NaOH (40.00 g/mol):
6.32×102mol NaOH×40.00g NaOH1mol NaOH=2.53g NaOH
So, 2.53 g of NaOH are required to fully consume 3.10 grams of HA2SOA4 in this reaction.
Shortcut: We could have combined all three steps into a single calculation, as shown in the following expression:
3.10g H2SO4 × 1mol H2SO498.08g H2SO4  ×  2mol NaOH1mol H2SO4  ×  40.00g NaOH1mol NaOH = 2.53g NaOHStep 1Step 2Step 3Find mol of H2SO4Use mole ratioFind g of NaOH
Be sure to pay extra close attention to the units if you take this approach, though!


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!).
Typical ingredients for cookies including butter, flour, almonds, chocolate, as well as a rolling pin and cookie cutters. Everything is scattered over a wooden table.
Running a chemical reaction is like making cookies. Hopefully your lab bench is cleaner than this kitchen counter, though! Image credit: Congerdesign on Pixabay, Pixabay License.
To learn about other common stoichiometric calculations, check out this exciting sequel on limiting reactants and percent yield!

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