1. Stoichiometry
Chapter 9

2. Stoichiometry – Chapter 9
Stoichiometry is the study of the quantities of materials consumed and produced in chemical reactions.
What do you need to know to be able to complete this chapter?
Chapter 7 – Chemical Quantities – moles/representative particles/molar mass/molar volume/MOLE ROADMAP
Chapter 8 – Balancing Chemical Equations
Chapter 3&4 – Dimensional Analysis

4. S’mores When you have a beach BBQ, you make s’mores, right?
When I make s’mores, here’s what my chemical reaction looks like:
2 graham crackers + 1 marshmallow + 1 piece chocolate
→ 1 s’more
Let’s say I wanted to make 8 s’mores
for my friends. How much of each
Ingredient would I need?
That’s stoichiometry!

5. S’mores
8 𝑠𝑚𝑜𝑟𝑒𝑠 𝑥 1 𝑝𝑖𝑒𝑐𝑒 𝑐ℎ𝑜𝑐𝑜𝑙𝑎𝑡𝑒 1 𝑠𝑚𝑜𝑟𝑒 = 8 𝑝𝑖𝑒𝑐𝑒𝑠 𝑐ℎ𝑜𝑐𝑜𝑙𝑎𝑡𝑒 8 𝑠𝑚𝑜𝑟𝑒𝑠 𝑥 2 𝑔𝑟𝑎ℎ𝑎𝑚 𝑐𝑟𝑎𝑐𝑘𝑒𝑟𝑠 1 𝑠𝑚𝑜𝑟𝑒 = 16 𝑔𝑟𝑎ℎ𝑎𝑚 𝑐𝑟𝑎𝑐𝑘𝑒𝑟𝑠 8 𝑠𝑚𝑜𝑟𝑒𝑠 𝑥 1 𝑚𝑎𝑟𝑠ℎ𝑚𝑎𝑙𝑙𝑜𝑤 1 𝑠𝑚𝑜𝑟𝑒 = 8 𝑚𝑎𝑟𝑠ℎ𝑚𝑎𝑙𝑙𝑜𝑤𝑠 That’s a conversion factor for making s’mores

6. Why stoichiometry?
Let’s say you want to relate the amount of product made in a chemical reaction to the amount of reactants you need to start off with.
N H2 → 2 NH3

7. So let’s go back to previous example:
N H2 → 2 NH →
2 atoms N atoms H → 2 atoms N and 6 atoms H
1 molecule N molecules H2 → 2 molecules NH3
10 molecule N molecules H2 → 20 molecules NH3
1(6.02E23) molecules N2 + 3(6.02E23) molecules H2 → 2(6.02E23) molecules NH3
1 mol N mol H2 → 2 mol NH3 (mol reactants ≠ mol products!)
(28.0) g N (2.0 g) H2 → 2 (17.0) g NH3
34 g reactants → 34 g products (conservation of mass/matter)
22.4 L N2 + 3(22.4) L H2 → 2(22.4) L NH (L reactants ≠ L products)

8. Mole-Mole Calculations
N2 (g) H2 (g) → 2 NH3 (g)
What would the six mole/mole ratios be for this reaction? These are the 6 possible conversion factors associated with this reaction.
Mole/mole ratios give the ratio of how much of two of the chemicals are needed to react together in this reaction.
1 𝑚𝑜𝑙𝑒 𝑁 2 3 𝑚𝑜𝑙 𝐻 𝑚𝑜𝑙 𝑁 2 2 𝑚𝑜𝑙 𝑁𝐻 𝑚𝑜𝑙 𝐻 2 2 𝑚𝑜𝑙 𝑁𝐻 3
3 𝑚𝑜𝑙𝑒 𝐻 2 1 𝑚𝑜𝑙 𝑁 𝑚𝑜𝑙 𝑁𝐻 3 1 𝑚𝑜𝑙 𝑁 𝑚𝑜𝑙 𝑁𝐻 3 3 𝑚𝑜𝑙 𝐻 2

9. Mole-Mole Calculations
So let’s say I started out with 0.4 mol of N2. How much ammonia (how many moles) could I produce using that reaction?
0.4 𝑚𝑜𝑙 𝑁 2 𝑥 2 𝑚𝑜𝑙 𝑁𝐻 3 1 𝑚𝑜𝑙 𝑁 2 =0.8 𝑚𝑜𝑙 𝑁𝐻 3
If I did make 0.8 mol of NH3, how much H2 would I have to supply to make that happen?
0.8 𝑚𝑜𝑙 𝑁𝐻 3 𝑥 3 𝑚𝑜𝑙 𝐻 2 2 𝑚𝑜𝑙 𝑁𝐻 3 =1.2 𝑚𝑜𝑙 𝐻 2

10. Let’s try another example
First we have to balance the following equation ___ C5H12 + ____O2 → ____CO2 + ____H2O Step 1: C 1 C5H12 + ____O2 → 5 CO2 + ____H2O Step 2: H 1 C5H12 + ____O2 → 5 CO2 + 6 H2O Step 3: O 1 C5H O2 → 5 CO2 + 6 H2O Total of 5 C, 12 H and 16 O to balance it.

11. Mole-Mole Calculations
We can now relate moles of product to the moles of reactant (or any species involved in the reaction, really) in mathematical equations.
Mole ratio: A conversion factor derived from the coefficients of a balanced equation.
You have to start with a balanced equation.
Use a mole ratio when you want to relate one chemical’s quantity in the reaction to another chemical’s quantity.

12. Mole Ratios 1 C5H12 + 8 O2 → 5 CO2 + 6 H2O
Let’s try a few mole ratios using the balanced equation from a few pages ago.
How many moles of water are produced when 4 moles of oxygen are reacted?
Start with writing the given 4 moles oxygen on the left…
Then write what is wanted on the right… moles of water
Then use an appropriate mole-mole conversion factor
4 𝑚𝑜𝑙 𝑂 2 x 6 𝑚𝑜𝑙 𝐻 2 𝑂 8 𝑚𝑜𝑙 𝑂 2 =3 mol 𝐻 2 𝑂

13. Mole ratios 1 C5H12 + 8 O2 → 5 CO2 + 6 H2O
How many moles of pentane are needed to produce 12 moles of carbon dioxide?
Start by writing the given on the left (12 moles CO2)
Next write the thing you want to find on the right (moles of pentane)
Then use an appropriate conversion factor
12 𝑚𝑜𝑙 𝐶𝑂 2 𝑥 1 𝑚𝑜𝑙 𝐶 5 𝐻 𝑚𝑜𝑙 𝐶𝑂 2 = 2.4 𝑚𝑜𝑙𝑒𝑠 𝐶 5 𝐻 12

14. You try it on whiteboards
Given the equation 4 Al (s) + 3 O2 (g) → 2 Al2O3 (s)
Write the six possible mole ratios on the white board.
How many moles of Al are needed to form 3.7 moles of Al2O3 ? Show the calculation on the white board.
How many moles of O2 are required to react with 14.8 moles of Al?
How many moles of Al2O3 are formed when 0.78 moles of O2 react with excess Al?

15. Now recall the mole roadmap

16. Stoichiometry Concept Map All roads lead to the balanced equation and mole ratios!

17. Problems not given in moles:
Example: How many moles of carbon dioxide will be produced when 10 grams of pentane react with excess oxygen gas?
Note how the question is not moles to moles - it is moles and grams, so we need to convert.
Strategy:
Convert the given quantity in the problem to moles. If it is already given in moles, skip this step.
Use the mole ratio. Hint: you will never skip this step.
Convert the value to the unit asked for in the problem.

18. Let’s try an example 1 C5H12 + 8 O2 → 5 CO2 + 6 H2O
How many moles of carbon dioxide will be produced when 20.5 grams of pentane reacts with excess oxygen?
What goes on the left side of the paper?
What goes on the right side of the paper?
What is the molar mass of pentane?
5(12.0) + 12(1.0) = 72 g/mole pentane
20.5 𝑔 𝐶 5 𝐻 12 x 1 𝑚𝑜𝑙 𝐶 5 𝐻 𝑔 𝐶 5 𝐻 12 𝑥 5 𝑚𝑜𝑙 𝐶𝑂 2 1 𝑚𝑜𝑙 𝐶 5 𝐻 12 =1.42 𝑚𝑜𝑙 𝐶𝑂 2
molar mass pentane mole/mole conversion factor

19. You try this one on your whiteboard
1 C5H O2 → CO H2O
How many grams of oxygen gas are needed to react if 12.8 moles of water are formed?
What do you write on the left side of the board?
12.8 𝑚𝑜𝑙 𝐻 2 𝑂
What do you write on the right side of the board?
= 𝑔 𝑂 2
What do you need the molar mass of?
𝑜𝑥𝑦𝑔𝑒𝑛 𝑔𝑎𝑠
You should get an answer of about 546 g of O2 – did you?

20. Mass-Mass conversions
Now what if both the “given” and the desired answer are in grams rather than moles?
How many grams of water will be formed when grams of pentane are reacted in the presence of oxygen?
1 C5H O2 → CO H2O
552.3 𝑔 𝐶 5 𝐻 12 𝑥 1 𝑚𝑜𝑙 𝐶 5 𝐻 𝑔 𝐶 5 𝐻 12 𝑥 6 𝑚𝑜𝑙 𝐻 2 𝑂 1 𝑚𝑜𝑙 𝐶 5 𝐻 12 𝑥 18.0 𝑔 𝐻 2 𝑂 1 𝑚𝑜𝑙 𝐻 2 𝑂 =
882.5 g H2O
Here, because both given and answer were in grams, you had to use two molar mass conversion factors, plus the mole ratio conversion factor (that you always use).

21. Representative particle practice
Using the reaction: 2 KClO3 → 2 KCl + 3 O2
How many molecules of O2 are produced by the decomposition of 6.54 grams of KClO3 ?
First think about what goes on far left and far right.
Then figure out what mole-mole conversion factor to use.
Then you need one molar mass KClO3 = (16.0)=122.5
Then you need Avogadro’s number conversion factor
6.54 𝑔 𝐾𝐶𝑙𝑂 3 𝑥 1 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂 𝑔 𝐾𝐶𝑙𝑂 3 𝑥 3 𝑚𝑜𝑙 𝑂 2 2 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂 3 𝑥 𝑥 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝑂 2 1 𝑚𝑜𝑙 𝑂 2
=4.82 𝑥 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑂 2

22. Volume of a gas practice
Given the reaction: 2 CO + 1 O2 → 2 CO2
How many liters of O2 are required to burn 3.86 liters of CO?
Recall the conversion factor for gases at STP that liters = 1 mole.
What do you start with on the left? On the right?
3.86 𝐿 𝐶𝑂 𝑥 1 𝑚𝑜𝑙 𝐶𝑂 22.4 𝐿 𝐶𝑂 𝑥 1 𝑚𝑜𝑙 𝑂 2 2 𝑚𝑜𝑙 𝐶𝑂 𝑥 𝐿 𝑂 2 1 𝑚𝑜𝑙 𝑂 2 =1.93 𝐿 𝑂 2
Do you see that the 2nd and 4th term look like they cancel each other out? Careful, your units won’t work out if you skip them and you won’t get credit on the test if you skip them.

23. Try one on your whiteboard
Given the reaction: H2O → 2 H2 + O2
How many molecules of oxygen are produced when a sample of 29.2 g of water is decomposed by electrolysis according to the equation above?

24. Try one on your whiteboard
Given the reaction: H2O → 2 H2 + O2
How many molecules of oxygen are produced when a sample of 29.2 g of water is decomposed by electrolysis according to the equation above?
29.2 𝑔 𝐻 2 𝑂 𝑥 1 𝑚𝑜𝑙 𝐻 2 𝑂 18.0 𝑔 𝐻 2 𝑂 𝑥 1 𝑚𝑜𝑙 𝑂 2 2 𝑚𝑜𝑙 𝐻 2 𝑂 𝑥 𝑥 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑂 2 1 𝑚𝑜𝑙 𝑂 2
= 4.88 x molecules of O2

25. Last one on your whiteboard
Given the reaction: 2 SO2 (g) + O2 (g) → SO3 (g)
First of all, is that balanced?
Once you have it balanced, then assuming STP (standard temperature and pressure), how many liters of oxygen are needed to produce 19.8 L SO3 ?
What goes on the left?
What goes on the right?
Are you going to need any molar masses?

26. Last one on your whiteboard
Given the reaction: 2 SO2 (g) + O2 (g) → 2 SO3 (g)
How many liters of oxygen are needed to produce L SO3 ?
19.8 𝐿 𝑆𝑂 3 𝑥 1 𝑚𝑜𝑙 𝑆𝑂 𝐿 𝑆𝑂 3 𝑥 1 𝑚𝑜𝑙 𝑂 2 2 𝑚𝑜𝑙 𝑆𝑂 3 𝑥 𝐿 𝑂 2 1 𝑚𝑜𝑙 𝑂 2 =9.90 L 𝑂 2

27. Section 9.3 Limiting Reagents
Let’s go back to our discussion of s’mores.
Let’s say I want to make 40 s’mores.
Each s’more is made up of
2 graham crackers
1 marshmallow
1 square of chocolate
If I have 70 graham crackers, 42 marshmallows and 45 squares of chocolate, how many s’mores can I make?
The limiting factor is the graham crackers. I can only make 35 s’mores from the graham crackers I have. That is the limiting reagent. I have enough marshmallows and chocolate to make 40, so they are NOT limiting reagents.

28. Limiting reagents N2 + 3 H2 → 2 NH3 This is the balanced reaction.
Let’s say I start out with 2 moles of N2 and 3 moles of H2 in a container and I want to make ammonia.
N H2 → 2 NH3 This is the balanced reaction.
Which reactant will be the limiting reagent?
Let’s look at the hydrogen first:
3 moles of hydrogen require 1 mole of nitrogen to make 2 moles of ammonia gas.
Do I have enough nitrogen to do that?
Yes, I have 2 moles of nitrogen and I only need 1 mole (there is excess nitrogen).
Now let’s look at the nitrogen instead:
2 moles of nitrogen would mix with 6 moles of hydrogen to give 4 moles of NH3 gas (refer to the balanced equation to see that).
Do I have enough hydrogen to do that?
No, I need 6 moles of hydrogen but I only have 3 moles of it. So hydrogen is the limiting reagent.

29. Limiting reagents
Looking at it graphically, which reactant was limited?
(which reactant had excess, and which got all used up?)
Before reaction After reaction
N H2 → 2 NH N H2 → 2 NH3

30. Limiting reagents example 9-9
2 Cu + S → Cu2S
What is the limiting reagent when 80.0 g of Cu reacts with 25.0 g S ?
Need to know how many moles that is for Cu and for S.
80.0 𝑔 𝐶𝑢 𝑥 1 𝑚𝑜𝑙 𝐶𝑢 63.5 𝑔 𝐶𝑢 =1.26 𝑚𝑜𝑙 𝐶𝑢
25.0 𝑔 𝑆 𝑥 1 𝑚𝑜𝑙 𝑆 32.1 𝑔 𝑆 =0.779 𝑚𝑜𝑙 𝑆
1.26 𝑚𝑜𝑙 𝐶𝑢 𝑥 1 𝑚𝑜𝑙 𝑆 2 𝑚𝑜𝑙 𝐶𝑢 =0.630 𝑚𝑜𝑙 𝑆 𝑛𝑒𝑒𝑑𝑒𝑑.
If we compare the amount of S needed (0.630 mole) to the amount of S we have (0.779 mole), we have enough S so it is not the limiting reagent, Cu is the limiting reagent.
This is how much of each reactant we HAVE

31. Section 9.3: Percent yield
If you do an experiment and mix two reactants and expect a certain amount of products, do you always get that much?
If there is experimental error (and there will be), then you will never get exactly that much.
The percent yield for your experiment is :
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑦𝑖𝑒𝑙𝑑= 𝑎𝑐𝑡𝑢𝑎𝑙 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑜𝑟 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑦𝑖𝑒𝑙𝑑 𝑥 100%

32. Percent yield – example 9-10
Calcium carbonate is decomposed by heating:
CaCO3 → CaO(s) + CO2
Find the theoretical yield and the percent yield if 13.1 g of CaO is produced.
Given: the starting mass of calcium carbonate is 24.8 g.
Molar mass of CaCO3 is g/mol
Molar mass of CaO is 56.1 g/mol
Mole ratio of CaO to CaCO3 is 1:1
24.8 𝑔 𝐶𝑎𝐶 𝑂 3 𝑥 1 𝑚𝑜𝑙 𝐶𝑎𝐶 𝑂 𝑔 𝐶𝑎𝐶 𝑂 3 𝑥 1 𝑚𝑜𝑙 𝐶𝑎𝑂 1 𝑚𝑜 𝐶𝑎𝐶 𝑂 3 𝑥 56.1 𝑔 𝐶𝑎𝑂 1 𝑚𝑜𝑙 𝐶𝑎𝑂 =13.9 𝑔 𝐶𝑎𝑂
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑦𝑖𝑒𝑙𝑑= 𝑔 𝐶𝑎𝑂 13.9 𝑔 𝐶𝑎𝑂 𝑥100%=94.2%