# Stoichiometry

How to use mole ratios from a balanced reaction to calculate amounts of reactants.

## Introduction

Freshly baked chocolate chip cookies on a 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
What do cookies and chemistry have in common? Many things, it turns out! A balanced chemical equation is the recipe for a reaction: it contains a list of all the reactants (the ingredients) and products (the cookies) as well as their relative proportions.
Using a balanced chemical equation to calculate amounts of reactants and products is called stoichiometry. It is a super technical-sounding word that simply means using ratios from the balanced equation. In this article, we will discuss how to use mole ratios to calculate the amount of reactants needed for a reaction.

## Balanced reactions and mole ratios

The stoichiometric coefficients are the numbers we use to make sure our equation is balanced. We can make ratios using the stoichiometric coefficients, and the ratios will tell us about the relative proportions of the chemicals in our reaction. You might see this ratio called the mole ratio, the stoichiometric factor, or the stoichiometric ratio. The mole ratio can be used as a conversion factor between different quantities.
Problem solving tip: The first and most important step for all stoichiometry problems is the same no matter what you are solving for—make sure your equation is balanced! If the equation is not balanced, the mole ratios will be wrong, and the answers will not be correct.
For example, the stoichiometric coefficients for the following balanced equation tell us that 1 mole of F, e, start subscript, 2, end subscript, O, start subscript, 3, end subscript will react with 2 moles of A, l to yield 2 moles of F, e and 1 mole of A, l, start subscript, 2, end subscript, O, start subscript, 3, end subscript.
When no coefficient is written—such as for F, e, start subscript, 2, end subscript, O, start subscript, 3, end subscript and A, l, start subscript, 2, end subscript, O, start subscript, 3, end subscript—it means the stoichiometric coefficient is one.
F, e, start subscript, 2, end subscript, O, start subscript, 3, end subscript, left parenthesis, s, right parenthesis, plus, start color blueD, 2, end color blueD, A, l, left parenthesis, s, right parenthesis, right arrow, start color redD, 2, end color redD, F, e, left parenthesis, l, right parenthesis, plus, A, l, start subscript, 2, end subscript, O, start subscript, 3, end subscript, left parenthesis, s, right parenthesis
If we have a known mass of the reactant F, e, start subscript, 2, end subscript, O, start subscript, 3, end subscript, we can calculate how many moles of A, l we need to fully react with the F, e, start subscript, 2, end subscript, O, start subscript, 3, end subscript using the ratio of their coefficients:
M, o, l, e, space, r, a, t, i, o, space, b, e, t, w, e, e, n, space, A, l, space, a, n, d, space, F, e, start subscript, 2, end subscript, O, start subscript, 3, end subscript, equals, start fraction, start color blueD, 2, end color blueD, space, m, o, l, e, space, A, l, divided by, 1, space, m, o, l, e, space, F, e, start subscript, 2, end subscript, O, start subscript, 3, end subscript, end fraction

## Example: Using mole ratios to calculate mass of a reactant

For the following unbalanced reaction, how many grams of N, a, O, H will be required to fully react with 3.10 grams of H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript?
$\text {NaOH}(aq) + \text H_2 \text{SO}_4 (aq) \rightarrow \text H_2 \text O + \text {Na}_2 \text{SO}_4(aq)~~~~~~~\text{Not balanced!}$
For this reaction, we have 1 N, a and 3 H on the reactant side and 2 N, a and 2 H on the product side. We can balance our equation by multiplying N, a, O, H by two—so that there are 2 N, a on each side—and multiplying H, start subscript, 2, end subscript, O by two—so there are 6 O and 4 H on both sides. That gives the following balanced reaction:
$2 \text {NaOH}(aq) + \text H_2 \text{SO}_4 (aq) \rightarrow 2\text H_2 \text O + \text {Na}_2 \text{SO}_4(aq)~~~~ \text{Balanced, hooray!}$
Once we have the balanced equation, we can ask ourselves the following questions:
• For which reactant(s) do we already know the amount of the chemical?
• What are we trying to calculate?
In this example, we know the amount of H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript is 3.10 grams, and we would like to calculate the mass of N, a, O, H. Armed with the balanced equation and a clear sense of purpose—hopefully—we can use the following strategy to tackle this stoichiometry problem:

### Step 1: Convert known reactant amount to moles.

The known quantity in this problem is the mass of H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript. We can convert the mass of H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript to moles using the molecular weight. Given that the molecular weight of H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript is 98.09 g/mol, we can find the moles of H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript as follows:
The number 0.0316 can be written using scientific notation as 3.16 times 10start superscript, minus, 2, end superscript. Scientific notation is particularly useful for representing very small or very large numbers, which happens a lot in chemistry!
If you would like more details about writing numbers this way, you can watch this video on scientific notation.

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

We are interested in calculating the amount of N, a, O, H, so we can use the mole ratio between N, a, O, H and H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript. Based on our balanced chemical equation, we need 2 moles of NaOH for every 1 mole of H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript, which gives the following ratio:
M, o, l, e, space, r, a, t, i, o, space, b, e, t, w, e, e, n, space, N, a, O, H, space, a, n, d, space, H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript, equals, start fraction, 2, space, m, o, l, space, N, a, O, H, divided by, 1, space, m, o, l, space, H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript, end fraction
We can use the ratio to convert moles of H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript from step one to moles of N, a, O, H:
Notice that we can write the mole ratio in two ways:
$\dfrac{2\,\text{mol NaOH}}{1\, \text{mol H}_2 \text {SO}_4}~~ \greenD{\checkmark}~~~$or$~~~ \dfrac{1\, \text{mol H}_2 \text {SO}_4}{2\,\text{mol NaOH}}~~\redD{\mathsf X}$
Each format gives a different answer! However, only one ratio will allow the units of H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript to cancel out properly. The important message here is always check your units! For a video explaining how units can be treated as numbers for easier bookkeeping, you can watch this video on dimensional analysis.

### Step 3: Convert moles to mass.

We can convert the moles of N, a, O, H from Step 2 to mass in grams using the molecular weight of N, a, O, H:
We will need 2.53 grams of N, a, O, H to fully react with 3.10 grams of H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript in this reaction.
Shortcut: We could also combine all three steps into a single calculation, with the caveat that we should pay extra close attention to our units. In order to convert the mass of H, start subscript, 2, end subscript, S, O, start subscript, 4, end subscript to mass of N, a, O, H, we could solve the following expression:
If we look carefully at the expression, we can break it down into steps 1 to 3 above. The only difference is that instead of doing each conversion separately, we did them all at once.

## Summary

The coefficients from the balanced chemical reaction tell us the proportions of the reactants and products. We can use ratios of the coefficients to convert between amounts of reactants and products in our reaction.
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 by condesign on pixabay, CC0 1.0
For more information on other common stoichiometric calculations, check out this exciting sequel on limiting reagents and percent yield!