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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry>Unit 4

Lesson 4: Stoichiometry

# Stoichiometry

AP.Chem:
SPQ‑4 (EU)
,
SPQ‑4.A (LO)
,
SPQ‑4.A.1 (EK)
,
SPQ‑4.A.2 (EK)

## Introduction

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:
start text, F, e, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 3, end subscript, left parenthesis, s, right parenthesis, plus, start color #11accd, 2, end color #11accd, start text, A, l, end text, left parenthesis, s, right parenthesis, right arrow, start color #e84d39, 2, end color #e84d39, start text, F, e, end text, left parenthesis, l, right parenthesis, plus, start text, A, l, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 3, end subscript, left parenthesis, s, right parenthesis
The coefficients in the equation tell us that 1 mole of $\ce{Fe2O3}$ reacts with 2 moles of A, l, forming 2 moles of F, e and 1 mole of $\ce{Al2O3}$. We can write the relationship between the $\ce{Fe2O3}$ and the A, l as the following mole ratio:
1, start text, m, o, l, space, F, e, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 3, end subscript, colon, start color #11accd, 2, end color #11accd, start text, m, o, l, space, A, l, end text
Using this ratio, we could calculate how many moles of start text, A, l, end text are needed to fully react with a certain amount of start text, F, e, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 3, end subscript, 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:
$\ce{NaOH}(aq) + \ce{H2SO4}(aq) \rightarrow \ce{H2O}(l) + \ce{Na2SO4}(aq)$
How many grams of N, a, O, H are required to fully consume 3, point, 10 grams of $\ce{H2SO4}$?
First things first: we need to balance the equation! In this case, we have 1 N, a atom and 3 H atoms on the reactant side and 2 N, a atoms and 2 H atoms on the product side. We can balance the equation by placing a 2 in front of N, a, O, H (so that there are 2 start text, N, a, end text atoms on each side) and another 2 in front of $\ce{H2O}$ (so that there are 6 O atoms and 4 H atoms on each side). Doing so gives the following balanced equation:
$\ce{2NaOH}(aq) + \ce{H2SO4}(aq) \rightarrow \ce{2H2O}(l) + \ce{Na2SO4}(aq)$
Now that we have the balanced equation, let's get to problem solving. To review, we want to find the mass of N, a, O, H that is needed to completely react 3, point, 10 grams of $\ce{H2SO4}$. We can tackle this stoichiometry problem using the following steps:

### Step 1: Convert known reactant mass to moles

In order to relate the amounts $\ce{H2SO4}$ and N, a, O, H using a mole ratio, we first need to know the quantity of $\ce{H2SO4}$ in moles. We can convert the 3, point, 10 grams of $\ce{H2SO4}$ to moles using the molar mass of $\ce{H2SO4}$ (98, point, 08, space, g, slash, m, o, l):
3, point, 10, start cancel, start text, g, space, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, end cancel, times, start fraction, 1, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, divided by, 98, point, 08, start cancel, start text, g, space, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, end cancel, end fraction, equals, 3, point, 16, times, 10, start superscript, minus, 2, end superscript, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript

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

Now that we have the quantity of $\ce{H2SO4}$ in moles, let's convert from moles of $\ce{H2SO4}$ to moles of N, a, O, H using the appropriate mole ratio. According to the coefficients in the balanced chemical equation, 2 moles of N, a, O, H are required for every 1 mole of $\ce{H2SO4}$, so the mole ratio is
start fraction, 2, start text, m, o, l, space, N, a, O, H, end text, divided by, 1, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, end fraction
Multiplying the number of moles of $\ce{H2SO4}$ by this factor gives us the number of moles of N, a, O, H needed:
3, point, 16, times, 10, start superscript, minus, 2, end superscript, start cancel, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, end cancel, times, start fraction, 2, start text, m, o, l, space, N, a, O, H, end text, divided by, 1, start cancel, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, end cancel, end fraction, equals, 6, point, 32, times, 10, start superscript, minus, 2, end superscript, start text, m, o, l, space, N, a, O, H, end text
Notice how we wrote the mole ratio so that the moles of $\ce{H2SO4}$ cancel out, resulting in moles of N, a, O, H 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 N, a, O, H in grams, so our last step is to convert the 6, point, 32, times, 10, start superscript, minus, 2, end superscript moles of N, a, O, H to grams. We can do so using the molar mass of N, a, O, H (40, point, 00, space, g, slash, m, o, l):
6, point, 32, times, 10, start superscript, minus, 2, end superscript, start cancel, start text, m, o, l, space, N, a, O, H, end text, end cancel, times, start fraction, 40, point, 00, start text, g, space, N, a, O, H, end text, divided by, 1, start cancel, start text, m, o, l, space, N, a, O, H, end text, end cancel, end fraction, equals, 2, point, 53, start text, g, space, N, a, O, H, end text
So, 2, point, 53, space, g of N, a, O, H are required to fully consume 3, point, 10 grams of $\ce{H2SO4}$ in this reaction.
Shortcut: We could have combined all three steps into a single calculation, as shown in the following expression:
$\underbrace{3.10\; \cancel{\text{g H}_2\text{SO}_4} ~\times~ \dfrac{1\; \cancel{\text{mol H}_2\text{SO}_4}}{98.08\; \cancel{\text{g H}_2\text{SO}_4}}} ~~\times~~ \underbrace{\dfrac{2\; \cancel{\text{mol NaOH}}}{1\; \cancel{\text{mol H}_2\text{SO}_4}}} ~~\times~~ \underbrace{\dfrac{40.00\; \text{g NaOH}}{1\; \cancel{\text{mol NaOH}}}} ~=~ 2.53\; \text {g NaOH} \\ \kern5.6em \text{Step 1} \kern9.5em \text{Step 2} \kern5.6em \text{Step 3} \\ \kern3.1em \text{Find mol of H}_2\text{SO}_4 \kern4.8em \text{Use mole ratio} \kern2.1em \text{Find g of NaOH}$
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!).
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!

## Want to join the conversation?

• "1 mole of Fe2O3" Can i say 1 molecule ? Because im new at this amu/mole thing
• No, because a mole isn't a direct measurement. Mole is a term like dozen - a dozen eggs, a dozen cows, no matter what you use dozen with, it always means twelve of whatever the dozen is of. So a mole is like that, except with particles. There are always 6.022*10^23 atoms in a mole, no matter if that mole is of iron, or hydrogen, or helium.
• How did you manage to get [2]molNaOH/1molH2SO4.
How will you know if you're suppose to place 3 there? Or 4?
• Go back to the balanced equation.
2 NaOH + H2SO4 -> 2 H2O + Na2SO4
Look at the left side (the reactants). You have 2 NaOH's, and 1 H2SO4's. The ratio of NaOH to H2SO4 is 2:1. So you get 2 moles of NaOH for every 1 mole of H2SO4. I hope that answered your question!
• Is mol a version of mole? I just see this a lot on the board when my chem teacher is talking about moles.
• Mole is the SI unit for "amount of substance", just like kilogram is, for "mass". And like kilograms are represented by the symbol 'kg', moles are represented by the symbol 'mol'. That's it! :)
• Where did you get the value of the molecular weight of 98.09 g/mol for H2SO4?? Are we suppose to know that?
• To get the molecular weight of H2SO4 you have to add the atomic mass of the constituent elements with the appropriate coefficients.
Here the molecular weight of H2SO4 = (2 * atomic mass of H) + (atomic mass of S) + (4 * atomic mass of O)
Hope that helped!
• Can someone tell me what did we do in step 1? why did we multiply the given mass of HeSO4 by 1mol H2SO4/ 98.09 g HeSO4?
• That is converting the grams of H2SO4 given to moles of H2SO4. When we do these calculations we always need to work in moles.
Why moles? Because 1 gram of hydrogen has more atoms than 1 gram of sulfur, for example. But 1 mole of hydrogen has exactly the same number of atoms as 1 mole of sulfur.
• hi!
i am new to this stoichiometry,i am a bit confused about the the problem solving tip you gave in the article.
• What it means is make sure that the number of atoms of each element on the left side of the equation is exactly equal to the numbers on the right side. The equation is then balanced.

If the numbers aren't the same, left and right, then the stoichiometric coefficients need to be adjusted until the equation is balanced - earlier videos showed how this was done.

When counting up numbers of atoms, you need to take account of both the atom subscripts and the stoichiometric coefficients. For example, Fe2O3 contains two iron atoms and three oxygen atoms. However, if it was 2Fe2O3, then this would be four iron atoms and six oxygen atoms, because the stoichiometric coefficient of 2 multiplies everything.
• What is the relative molecular mass for Na?
• 58.5g is the molecular mass of na
• Can someone explain step 2 please why do you use the ratio?
• We use the ratio to find the number of moles of NaOH that will be used. If the ratio of 2 compounds of a reaction is given and the mass of one of them is given, then we can use the ratio to find the mass of the other compound.
For eg. in the oxidation of magnesium (Mg+O2 -> 2MgO), we get that O2 and MgO are in the ratio 1:2. So, if 3.16 (completely random number) moles of oxygen is involved, we know that 6.32 (2 x 3.16) moles of MgO will be formed. This info can be used to tell how much of MgO will be formed, in terms of mass.
Hope this helps.