1. Chemical Reactions: Mole and Mass Relationships
Fundamentals of General, Organic, and Biological Chemistry
7th Edition
Chapter 6
Chemical Reactions:
Mole and Mass Relationships
2013 Pearson Education, Inc.

2. Chapter 6 Goals Topics and Goals for the Chapter:
What is the mole, and why is it useful in chemistry?
Be able to understand and utilize the multiple definitions of the mole.
Be able to understand and use Avogadro’s number.
How are molar quantities and mass quantities related?
Be able to recognize and perform gram-mole conversions, mole-mole
conversions and mole-gram conversions.
Be able to understand how mole and mass relationships are illustrated in the balanced chemical equations.
What is the limiting reagent and percent yield of a reaction?
Be able to define limiting reagent, percent yield, actual yield and
theoretical yield.
Be able to understand the equation for percent yield and use the
equation to solve problems.
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3. 6.1 The Mole and Avogadro’s Number
A balanced equation gives us the molecule and/or atom ratio for reactants and products.
CH4 (g) + 2O2 (g) → CO2 (g) H2O (ℓ)
# Molecules:
Molecules are very small. We can’t count out a certain # of molecules of reactants and react them.
Instead of counting out molecules – we weigh out a particular mass of a substance.
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4. How do we relate the mass of a substance to the number of molecules?
Elements
Atomic Weight – Average mass of the element’s atoms.
Molecules
Molecular Weight – The average mass of a substance’s molecules.
The sum of atomic weights of all atoms in a molecule.
Formula Weight - The sum of atomic masses of all atoms in one formula unit of any compound (ionic or covalent). Tend to use this for ionic compounds.
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5. Examples of Calculating Molecular Mass:
MW CH4 = Atomic weight C + 4(Atomic weight H)
= amu + 4( amu)
= amu
MW O2 = 2(Atomic weight O)
= 2( amu)
= amu
When comparing two substances the ratio of molecular weights is equal to a 1:2 molecular ratio.
amu CH4 : amu O2 = 1:2 ratio of CH4:O2 molecules
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6. What is the molecular weight of CHCl3?
Problem:
What is the molecular weight of CHCl3?
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7. Units can be anything – donuts, pencils, people, etc.
The Mole
When we are talking about the number of molecules or formula units that take part in a reaction, we use a counting unit called a mole.
Units can be anything – donuts, pencils, people, etc.
What are some familiar counting units?
Dozen = 12 units
Gross = 144 units
Pair = 2 units
Quartet = 4 units
What is a mole?
Mole = 6.022x1023 units
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8. Mass in grams = Molar Mass = the mass of a mole of pure matter
The Mole in Chemistry
The Mole (mol) – The amount of a substance (# of units) whose mass in grams is numerically equal to its molecular or formula weight in amu.
Mass in grams = Molar Mass = the mass of a mole of pure matter
1 mole = Avogadro’s number (NA) of molecules or formula units or atoms
NA = 6.022x1023
THE RELATIONSHIP
Mass of 1 molecule CH4 = amu
Mass of 1 mole CH4 = g = 6.022x1023 molecules of CH4
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9. Mass of 1 molecule CH4 = 16.043 amu
The Mole in Chemistry
Mass of 1 molecule CH4 = amu
Mass of 1 mole CH4 = g = 6.022x1023 molecules of CH4
Can we derive the above mathematically?
What is the mass (g) of a sample of x 1023 CH4 molecules?
Remember 1 amu = x g
Is it always the case that 1 mole of substance will have a mass in grams equivalent to the mass of one unit of substance in amu? YES
Mass of 1 mole of O2 = g O2
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Chapter 6

10. 1 mole = Avogadro’s number (NA) of molecules or formula units
NA = 6.022x1023
How big is Avogadro’s Number?
6.022 x Watermelon Seeds:
Would be found inside a watermelon slightly larger than the moon
6.022 x Donut Holes:
Would cover the earth and be 5 miles deep.
6.022 x Pennies:
Would make at least 7 stacks that would reach the moon.
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Chapter 6

11. The Mole
The Mole Song
Some helpful websites
Definitions of the mole:
Mole-gram conversions:
Also: Refer to Helpful Chemistry Websites handout
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Chapter 6

12. 6.2 Gram-Mole Conversions
Remember:
Molar Mass = mass 1 mole of substance
= mass of 6.022x1023 molecules of substance.
= molecular (formula) wt. of substance in grams.
Conversions
Molar mass serves as a conversion factor between numbers of moles and mass.
If you know how many moles you have, you can calculate their mass.
If you know the mass of a sample, you can calculate the number of moles.
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13. # moles ↔ mass Conversions
The molar mass of the substance is required to convert in either direction
MM = g substance/1 mol substance
1/MM = 1 mol substance/g substance
Converting # Moles → Mass in grams
# moles (MM) = Mass in grams
(# Moles of a substance) (g substance/mol substance) = g substance
Converting Mass in grams → # Moles
(Mass in grams) (1/MM) = moles of substance
(# g of a substance) (mol substance/g substance) = # moles substance
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14. Converting moles ↔ # molecules
1 mole = x 1023 units (molecules or formula units)
(# moles) (6.022x1023 molecules/1 mole) = # molecules
# molecules (1 mole/6.022x1023 molecules) = # moles
Now we have some convenient conversion factors.
For example we can come up with the following conversion factors for CH4:
g CH4 1 mol CH4 1 mol CH4______________ etc.
1 mol CH g CH x1023 molecules CH4
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Chapter 6

15. # Moles # Molecules Mass (g)
Example Using CH4
MM CH4 = g
# Moles
# Molecules
Mass (g) 5
30.11 x 1023
80.215 2
x 1023
32.086 1
6.022 x 1023
16.043
0.5 (1/2)
3.011 x 1023
8.0215
0.2 (1/5)
x 1023
3.2086
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16. How many moles of NaCl do we have?
Example:
What if we have 12.5 g of NaCl.
How many moles of NaCl do we have?
How many formula units do we have?
Plan your strategy for problem-solving:
What information do you have?
What information do you need?
Where can you obtain the needed information?
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17. A bottle contains 2.5 moles of CO2
Another Example:
A bottle contains 2.5 moles of CO2
How many molecules of CO2 do we have?
How many grams of CO2 do we have?
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18. 6.3 Mole Relationships and Chemical Equations
The coefficients in a balanced chemical equation tell how many molecules, and thus how many moles, of each reactant are needed and how many molecules, and thus moles, of each product are formed.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (ℓ)
For the above reaction we can set-up mole ratios and use those ratios as conversion factors.
1 mole CH4 1 mol CH4 2 mol O2___ etc.
2 mol O2 1 mole CO2 1 mole CO2
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19. Moles CH4 Moles O2 CO2 Moles H2O
What are the mole ratios for the reaction below:
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (ℓ)
Moles CH4
Moles O2
CO2
Moles H2O 1 2
0.5
0.25
0.2 20
How are you determining the values for the table?
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20. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (ℓ)
Example:
How many moles of H2O will be formed by the complete oxidation reaction of 1.7 moles of CH4 (g)?
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (ℓ)
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Chapter 6
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21. 6.4 Mass Relationships and Chemical Reactions
The coefficients in a balanced equation give mole → mole relationship.
In the lab we obtain quantities of substances by mass.
Need to develop mass → mass relationships
Reaction: A → B
If: Moles A → Moles B
Then: Mass A → Mass B
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Chapter 6
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22. We know how to perform the following types of conversions:
A → B
We know how to perform the following types of conversions:
Moles A ↔ Moles B → Using coefficient ratio from balanced equation
Moles A ↔ Mass A → Using molar mass (MM) of A
Moles A ↔ # of molecules of A → Using Avogadro's number
Moles B ↔ Mass B → Using molar mass (MM) of B
Moles B ↔ # of molecules of B → Using Avogadro's number
And, we can combine a series of conversions.
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23. Convert using the series below: Mass A ↔ Moles A ↔ Moles B ↔ Mass B
Question:
How do we combine our knowledge of chemical equations and all of the relationships to go from mass reactants ↔ mass products
A → B
Mass A → Mass B
Answer:
Convert using the series below:
Mass A ↔ Moles A ↔ Moles B ↔ Mass B
By the way – we are not limited to Mass reactants ↔ Mass Products
We can also convert
Mass reactant #1 ↔ Mass reactant #2
and
Mass product #1 ↔ Mass product #2
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24. Chapter 6
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25. Write the balanced chemical equation.
Four Step Method for Determining The Mass Relationship Between Reactants and Products
Write the balanced chemical equation.
Choose molar masses and mole ratios to convert known information
into needed information.
Set up the factor-label expression, and calculate the answer.
Estimate or check the answer using a ballpark solution.
Hint: It is very helpful to ORGANIZE your information.
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Chapter 6
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26. Given 2NaN3 (s) → 2Na (s) + 3N2 (g)
Example:
Given 2NaN3 (s) → 2Na (s) + 3N2 (g)
How many grams of NaN3 are required to produce 18.0 g of N2?
How do we approach the problem?
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27. Given 2NaN3 (s) → 2Na (s) + 3N2 (g)
On Your Own:
Given 2NaN3 (s) → 2Na (s) + 3N2 (g)
How many grams of Na (s) will be formed if 18.0 g of N2 are formed?
Is the Law of Conservation of Mass obeyed? Explain your answer.
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28. 6.5 Limiting Reagent and Percent Yield
Yield From Chemical Reactions
In theory – All of reactants should react to make product.
In practice – It is rare that a reaction produces 100% of the expected product.
Theory – A recipe states that is produces 36 chocolate chip cookies.
Practice – You made 34 cookies, not 36.
Why is this?
Many reactions take several steps – We may lose some of our reactants, intermediates or products at each step through processing or handling.
May lose molecules to “side-reactions” – Reactions other than the reaction
of interest.
Yield is less than 100% for every reaction because of the above types of losses.
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29. Percent Yield = (Actual yield ÷ Theoretical yield) x 100%
The actual yield is found by weighing the product obtained.
The theoretical yield is found by a mass-to-mass calculation.
For our cookie recipe problem – What is the percent yield?
Limiting Reagent
If we run the reaction without the exact amounts of reagents to allow all of them to react completely, the reactant exhausted first is called the limiting reagent.
Example of limiting reagent: Grilled cheese sandwiches
2 Bread + 1 Cheese = 1 Grilled Cheese Sandwich
How many sandwiches can we make with 11 bread and 4 cheese?
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30. Suppose we react 16.0 g CH4 and form 15.0 g CO2. What is the % Yield?
% Yield Problem:
Suppose we react 16.0 g CH4 and form 15.0 g CO2.
What is the % Yield?
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (ℓ)
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Chapter 6

31. Putting It All Together
Chapter 6
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32. Chapter Summary
A mole refers to Avogadro’s number of formula units of a substance. One mole of any substance has a mass equal to its formula weight in grams.
Molar masses act as conversion factors between numbers of molecules and masses in grams.
The coefficients in a balanced chemical equation represent the numbers of moles of reactants and products in a reaction.
Mole ratios relate amounts of reactants and/or products. Using molar masses and mole ratios in factor-label calculations relates unknown masses to known masses or molar amounts.
The yield is the amount of product obtained.
The percent yield is the amount of product obtained divided by the amount theoretically possible and multiplied by 100%.
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