1. Quantities in
Chemical Reactions:
Stoichiometry

2. Quantities in Chemical Reactions
Balanced Chem Equation gives MOLE ratio
In a chemical reaction, amount of every substance used or made is related to the amounts of all the other substances
Law of Conservation of Mass
Reaction Stoichiometry: the study of the numerical relationship between chemical quantities in a chemical reaction

3. Chapter Objectives
***Use the balanced equation to solve mole-to-mole and mass-to-mass stoichiometry problems
**Understand limiting reagent/reactant vs. excess reactant
Understand theoretical yield and percent yield

4. About Bunsen Burner Robert Bunsen: the Burner, the spectroscopy
Recall how you would adjust the flame in a Bunsen Burner:
Small flame: more gas, more air
Yellow “fluffy”: more air
Noisy unsteady: less air
Conclusion: Chemical reaction follows certain ratio

5. Recipe: Making Pancakes
The number of pancakes you can make depends on the amount of the Ingredients you use
1 cup Flour + 2 Eggs + ½ tsp Baking Powder  5 Pancakes
Mathematically (as Conversion Factor)
1 cu flour  2 eggs  ½ tsp baking powder  5 pancakes

6. More Pancakes? More Everything!
1 cu flour  2 eggs  ½ tsp baking powder  5 pancakes
Make more than 5 pancakes?
The number of eggs, and baking power too, determines the number of pancakes you can make
Conversion Factor, again!
Example: To make 20 pancakes, how many eggs we need? assuming you have enough everything else
20 pancakes x = eggs

7. Balanced Equation: Mole-to-Mole Conversions
Balanced equation = “recipe” for a chemical reaction, providing Conversion Factors for calculation.
Equation 3 H2(g) + N2(g)  2 NH3(g) means:
___ molecules of H2 react with ____ molecule of N2, make ____ molecules of NH3 or
__ molecules H2  __ molecule N2  __ molecules NH3
since we count molecules by moles
___ moles H2  ___ mole N2  ___ moles NH3

8. Mole Ratio in Chemical reaction: N2 + 3H2  2NH3
Every 3 molecules of H2 react with exactly 1 molecule of N2 to produce exactly 2 molecules of NH3

9. Example: Mole-to-Mole Conversions

10. Find: ? moles water Write down the quantity to find and/or its units.
Example: How many moles of H2O result from the complete combustion of 3.4 mol of CH4 in the reaction below? CH4 + 2O2 ® CO2 + 2H2O
Write down the quantity to find and/or its units.
Find: ? moles water
Collect Needed Conversion Factors:
1 mole CH4  2 moles H2O
Start with non-Converion factor:
3.4 mol CH4 x = mol H2O
6.8 moles H2O

11. Group Practice – For the following equation, find how many mole of 2nd product will be produced when 1.23 mole of 1st reactant is consumed.
2HBr(aq) + K2SO3(aq) ® H2O(l) + SO2(g) + 2 KBr(aq)
3K2CO3(aq) + 2(NH4)3PO4(aq) ® 3(NH4)2CO3(aq) + 2K3PO4(aq)
Ba(OH)2(aq) + H2SO4(aq) ® BaSO4(s) + 2H2O(l)
Stoichiometry: 
Stoichiometry: 
Tro: Chemistry: A Molecular Approach, 2/e 11

12. Stoichiometry: Mass-to-Mass Conversions
Relationship between Mass and Number of Moles of a chemical: Molar Mass
1 mole = Molar Mass in grams
Molar Mass of the chemicals in the reaction + Balanced chemical equation : convert from the Mass of any chemical in the reaction to the Mass of any other

13. Example: Mass-to-Mass Conversions
Balanced equation:
aA + bB  cC + dD
Mass A  _____ A  ____ B  Mass B
Mass B  _____ B  ____ D  Mass D

14. Collect Needed Conversion Factors:
Example: How many grams of glucose (C6H12O6) can be synthesized from 58.5 g of CO2 in the photosynthesis? 6 CO2(g) + 6 H2O(l)  6 O2(g) + C6H12O6(aq)
Information
Given: 55.4 g CO2
Find: g C6H12O6
Collect Needed Conversion Factors:
Molar Mass C6H12O6 = g/mol
Molar Mass CO2 = g/mol
From the equation:
1 mole C6H12O6  ______ mol CO2

15. 58.5 g CO2 x ----------------- x ---------------- x ------------------
Example: How many grams of glucose can be synthesized from 58.5 g of CO2 in the reaction? 6 CO2(g) + 6 H2O(l)  6 O2(g) + C6H12O6(aq)
CF: 1 mol C6H12O6 = g
1 mol CO2 = g
1 mol C6H12O6  6 mol CO2 g
CO2
mol
CO2
mol
C6H12O6 g
C6H12O6
58.5 g CO2 x x x
= g C6H12O6
(___ sig figs)
39.9 g C6H12O6

16. More Example: When using gasoline in the car engine, CO2 gas is produced as part of the emission gas. How many grams of CO2 will be produced from the complete combustion of 2.83  103 g of C8H18 (about 1.0 gallon)? 2C8H O2 ® 16CO2 + 18H2O You will see how we are contributing to the greenhouse gas in the environment? g
C8O18

17. Group Practice – For the following equation, how many grams of 2nd product can be produced when 1.23 g of 1st reactant is consumed?
2HBr(aq) + K2SO3(aq) ® H2O(l) + SO2(g) + 2 KBr(aq)
3K2CO3(aq) + 2(NH4)3PO4(aq) ® 3(NH4)2CO3(aq) + 2K3PO4(aq)
Ba(OH)2(aq) + H2SO4(aq) ® BaSO4(s) + 2H2O(l)
Stoichiometry: 
Stoichiometry: 
Tro: Chemistry: A Molecular Approach, 2/e 17

18. Making Pancakes: What if ….?
Same receipe:
1 cup Flour + 2 Eggs + ½ tsp Baking Powder  5 Pancakes
How many pancakes we can make if we had 3
cups of flour, 10 eggs, and 4 tsp of baking powder?

19. Recipes: 1 cup Flour + 2 Eggs + ½ tsp Baking Powder  5 Pancakes
3 cups flour + 10 eggs, 4 tsp baking powder = ? pancakes

20. Given certain amount of each ingredient, The actual number of pancakes is the ______ amount of pancakes as calculated below.
3 cups flour + 10 eggs, 4 tsp baking powder
let’s find out #pancakes from each ingredient:

21. What we learned from making pancakes?
Each ingredient could potentially make a different number of pancakes
but all the ingredients have to work together, following the right ratio!
The maximal amount of pancakes depends on the ingredient that gives the least amount of pancakes. Here as flour
Once we make 15 pancakes, the flour runs out, no more pancakes can be made!
Chemical reaction is very similar in terms of ratio

22. From making pancakes to Chemical Reaction
According to the balanced equation, all the reactants have to work together, following the right MOLE ratio!
The amount of products depends on the reactant that runs out the first, since if one reactant runs out, no more products can be made.
The reactant that rans out the first is the Limiting Reactant (also known as “Limiting Reagent”)

23. Ratio in Chemical reaction: N2 + 3H2  2NH3
Hydrogen gas would be consumed first, therefore it would be the Limiting Reagent

24. How to find the Limiting Reactant (L.R.)
When the amount of all reactants are given (either as MOLE or as Mass), the actual amount of products depends on the L.R.
To find the L.R., calculate the amount of ONE product from each reactant (assuming there are enough OTHER reactants)
The REACTANT that gives the LEAST amount of product is the L.R.
Theoretical yield: The maximum amount of products we can make from the reaction. It depends on the limiting reactant.

25. Example: Limiting Reagent from Moles of Reactants
In the internal combustion engine, 10.0 moles of oxygen gas is mixed with 10.1 moles of octane to produce carbon dioxide and water. Calculate the maximal amount of carbon dioxide may be produced from the reaction.
2C8H O2 ® 16CO2 + 18H2O
The plan:
mol
C8H18
CO2
mol O2
CO2

26. In the internal combustion engine, 10
In the internal combustion engine, 10.0 moles of oxygen gas is mixed with 10.1 moles of octane to produce carbon dioxide and water. The reaction is as follows: 2C8H O2 ® 16CO2 + 18H2O Calculate the maximal amount of carbon dioxide may be produced from the reaction.
10.1 mol C8H18 x = _________ mol CO2
10.0 mol O2 x = _________ mol CO2
From octane: 80.8 mol. From oxygen, 6.40 mol. Oxygen as L.R. Answer: 6.40 mol CO2

27. Example: Limiting Reagent from Mass of Reactants given
In the internal combustion engine, 100. g of oxygen gas is mixed with 101. g of octane to produce carbon dioxide and water. Calculate the theoretical yield (in grams) of carbon dioxide may be produced from the reaction. 2C8H O2 ® 16CO2 + 18H2O
The Plan: g
C8H18
mol
CO2 g O2
mol
CO2

28. In the internal combustion engine, 100
In the internal combustion engine, 100. g of oxygen gas is mixed with 101. g of octane to produce carbon dioxide and water: 2C8H O2 ® 16CO2 + 18H2O Calculate the theoretical yield (in grams) of carbon dioxide from the reaction.
Oxygen as L.R mol CO2. Answer: 88.0 g CO2

29. Group Practice: Limiting Reactant A. when 1
Group Practice: Limiting Reactant A. when 1.23 g each of reactant are mixed, calculate the theoretical yield (in grams) of 2nd product. B. Which reagent is in excess?
2HBr(aq) + K2SO3(aq) ® H2O(l) + SO2(g) + 2 KBr(aq)
3K2CO3(aq) + 2(NH4)3PO4(aq) ® 3(NH4)2CO3(aq) + 2K3PO4(aq)
Ba(OH)2(aq) + H2SO4(aq) ® BaSO4(s) + 2H2O(l) 29

30. Actual Yield vs. Percent Yield
Actual yield: The actual amount of product made in a chemical reaction (due to Controllable/Uncontrollable factors that results in less product. Such as during making pancakes, spill some of the batter, burn a pancake, drop one on the floor or else so that finally 11 pancakes are made out of 15.
Actual yield can NOT be calculated, only from experiment.
Percent yield : the efficiency of production during chemical reactions.

31. Theoretical and Actual Yield
To determine the theoretical yield,
Use reaction stoichiometry to determine the amount of product EACH reactants could make.
The theoretical yield will always be the least possible amount of product.
The theoretical yield will always come from the _______ reactant.
Because of both controllable and uncontrollable factors, Actual yield < Theoretical yield, so percent yield is no more than 100%.

32. Example: Finding Theoretical Yield and Percent Yield

33. Example: Most people believe that exercise helps “burning” carbohydrates and prevents weight gain.
When a person drinks two cans of soda (100.0 g of sugar, C12H22O11) per day. Meanwhile, he didn’t work out, neither did he eat any other carbohydrates, nor protein or fat. Measurement showed he released g CO2 during this day through this reaction:
C12H22O O2  12CO2 + 11H2O
Question:
What is the Theoretical Yield of CO2 from complete “burning sugar” in his body?
How much of the sugar was “burned” into CO2?
What is the Percent Yield of CO2?

34. Collect Needed Conversion Factors:
Example: When g of C12H22O11 reacts with O2, g of CO2 are produced. Find the Theoretical Yield and Percent Yield. C12H22O O2  12CO2 + 11H2O
Information
Given: g C12H22O11,
144.0 g of CO2 produced
Find: Theor. Yld., % Yld.
Collect Needed Conversion Factors:
Molar Mass C12H22O11 = g/mol
Molar Mass CO2 = g/mol
1 mole C12H22O11  12 mol CO2 (from the chem. equation)

35. Example: When 100. 0 g of C12H22O11 reacts with O2, 144
Example: When g of C12H22O11 reacts with O2, g of CO2 are produced. Find the Theoretical Yield and Percent Yield. C12H22O O2  12CO2 + 11H2O
Information
Given: 100.0g C12H22O11,
144.0 g of CO2 produced
Find: Theoretical Yld., % Yld.
Write a Solution Map: g
C12H22O11
mol
C12H22O11
mol
CO2 g
CO2

36. Apply the Solution Map:
Example: When g of C12H22O11 reacts with O2, g of CO2 are produced. Find the Theoretical Yield and Percent Yield. C12H22O O2  12CO2 + 11H2O
Apply the Solution Map:
Theoretical yield of CO2 (“total burn out of carbs”)
Percent yield = 144.0g/154.3g x 100% = 93.32%

37. Apply the Solution Map:
Example: How many grams of glucose can be synthesized from 58.5 g of CO2 in the reaction? 6 CO2(g) + 6 H2O(l)  6 O2(g) + C6H12O6(aq)
Information
Given: 58.5 g CO2
Find: g C6H12O6
CF: 1 mol C6H12O6 = g
1 mol CO2 = g
1 mol C6H12O6  6 mol CO2
SM: g CO2  mol CO2 
mol C6H12O6  g C6H12O6
Apply the Solution Map:
= g C6H12O6
Sig. Figs. & Round:
= 39.9 g C6H12O6

38. How to find the Limiting Reactant (L.R.): An alternative method
Starting from the mole of 1st reactant in the equation, calculate the mole of 2nd reactant required for complete reaction (use mole ratio);
If calculated mol 2nd reactant is smaller than actual (given), 2nd reactant is more than enough, thus 1st reactant is L.R.;
If calculated mol 2nd reactant is larger than given, 2nd reactant is not enough, thus 2nd reactant is L.R.
If more than two reactants in the equation, decide “L.R.” (pseudoLR) between 1st and 2nd reactants first; then compare pseudoLR with 3rd reactant.
If pseudoLR is the