Honors Chemistry, Chapter 9

1
Chapter 9 Stoichiometry
2
Reaction Stoichiometry

• Reaction Stoichiometry deals with the mass
  relationships between reactants and products in a
  chemical reaction.
• There are four basic types of stoichiometry
  problems.

3
Stoichiometry Problem Type 1

• Amount of Given Amount of Unknown
• Substance ? Substance in moles
• in moles

4
Stoichiometry Problem Type 2

• Amount of Given Amount of Unknown
• Substance ? Substance in moles ?
• in moles
• Mass of Unknown
• Substance
• in grams

5
Stoichiometry Problem Type 3

• Mass of Given Amount of Given
• Substance ? Substance in moles ?
• in grams
• Amount of Unknown
• Substance
• in moles

6
Stoichiometry Problem Type 4

• Mass of Given Amount of Given
• Substance ? Substance in moles ?
• in grams
• Amount of Unknown Mass of Unknown
• Substance ? Substance
• in moles In grams

7
Mole Ratio

• Mole Ratio is a conversion factor that relates
  the amounts in moles of any two substances
  involved in a chemical reaction.
• Example
• 2Al2O3(l) ? 4Al(s) 3O2(g)

8
Example Possible Ratios

• 2 mol Al2O3 4 mol Al_
  
• 4 mol Al or 2 mol Al2O3
• 2 mol Al2O3 3 mol O2__
• 3 mol O2 or 2 mol Al2O3
• _4 mol Al 3 mol O2_
  
• 3 mol O2 or 4 mol Al

9
Formula Mass

• The formula mass for a compound is the sum of the
  masses of each of the atoms the compound
• Example formula mass of Al2O3
• Al 26.98 g/mol x 2 53.96 g/mol
• O 16.00 g/mol x 3 48.00 g/mol
• Total 101.96
  g/mol
• (Remember Formula Mass Molar Mass)

10
Molar Mass

• Molar mass is the conversion factor that relates
  the mass of a substance to the amount in moles of
  that substance.

11
Example Molar Mass Conversion Factors
101.96 g Al2O3 mol
Al2O3_ mol Al2O3 or
101.96 g Al2O3 26.98 g Al
mol Al __ mol Al
or 26.98 g Al _32.00 g O2
mol O2 _ mol O2
or 32.00 g O2
12
Example Calculation

• Balanced Reaction
• 2Al2O3(l) ? 4Al(s) 3O2(g)
• Given 13 moles of Al2O3
• Calculate
• a. Amount in moles of Al
• b. Mass in grams of Al

13
Example Calculations

• 13 moles Al2O3 x 4 mol Al 26.0 mol Al
• 2 mol Al2O3
• 26.0 mol Al x 26.98 g Al 701 g of Al
• mol Al

14
Conversions of Moles to Moles

• Conversion Factor mol A
• mol B
• mol B given x mol A mol A
• mol B

15
Chapter 9, Section 1 Review

• Define stoichiometry.
• Describe the importance of the mole ratio in
  stoichiometric calculations.
• Write a mole ratio relating two substances in a
  chemical equation.

16
Example Spacecraft CO2 Absorber

• CO2(g) 2LiOH(s) ? Li2CO3(s) H2O(l)
• How many moles of LiOH are needed to absorb 20
  mol of CO2, exhaled by each astronaut per day?
• Conversion Factor ?
• Conversion Factor 2 mol LiOH
• 1 mol CO2

17
CO2 Absorber Calculation

• 20 mol CO2 x 2 mol LiOH 40 mol LiOH
• 1 mol CO2
• How many moles of Li2CO3?
• How many moles of H2O?
• Why not Na, K, Cs, or RbOH?

18
Conversions of Moles to Mass

• Conversion Factors
• Mol Unknown and Unknown Molar Mass
• mol given g/mol
• mol given x mol Unknown x mass Unk. g.
• mol given mol
  Unknown
• mass Unknown in grams

19
Example Photosynthesis

• 6CO2(g) 6H2O(l) ? C6H12O6(s) 6O2(g)
• 3 mols of water produces how many grams of
  glucose?
• How many grams of CO2 are needed to react with 3
  moles of H2O?

20
Photosynthesis Example Conversion Factors
6CO2(g) 6H2O(l) ? C6H12O6(s) 6O2(g) 3 mols
of water produces how many grams of glucose? 1
mole glucose and 180.18 g glucose 6 mol
water mol glucose
21
Photosynthesis Calculation
3 mol H2O x 1 mol glucose x 180.18 g gluc.
6 mol H2O mol glucose
90.1 g
glucose How many moles of O2 formed? How many
moles of CO2 consumed?
22
Photosynthesis Example Conversion Factors
6CO2(g) 6H2O(l) ? C6H12O6(s) 6O2(g) How many
grams of CO2 are needed to react with 3 moles of
H2O? 6 mole CO2 and 44.01 g CO2_
6 mol water mol CO2
23
Photosynthesis Calculation
3 mol H2O x 6 mol CO2 x 44.01 g CO2
6 mol H2O mol CO2
132 g CO2 What is the
reverse reaction called?
24
Conversions of Mass to Moles
Conversion Factors Mol given and
moles Unknown gram given
moles given gram given x mol given x moles
Unk. gram given mol
given moles Unknown
25
Example Ammonia Oxidation

• NH3(g) O2(g) ? NO(g) H2O(g)
• 
  (Unbalanced)
• 4NH3(g) 5O2(g) ? 4NO(g) 6H2O(g)
• Given 824 g of ammonia
• How many moles of NO are produced?

26
Ammonia Oxidation Example Conversion Factors
4NH3(g) 5O2(g) ? 4NO(g) 6H2O(g) 824 grams
of ammonia produces how many moles of NO? 1
mole NH3 and 4 moles of NO 17.04 g NH3
4 moles of NH3
27
Ammonia Oxidation Calculation
824 g NH3 x 1 mol NH3 x 4 mol NO
17.04 g NH3 4 mol NH3
48.4 mol NO
28
Example Ammonia Oxidation
4NH3(g) 5O2(g) ? 4NO(g) 6H2O(g) Given 824
g of ammonia How many moles of H2O are produced?
29
Ammonia Oxidation Example Conversion Factors
4NH3(g) 5O2(g) ? 4NO(g) 6H2O(g) 824 grams
of ammonia produces how many moles of H2O? 1
mole NH3 and 6 moles of H2O 17.04 g NH3
4 moles of NH3
30
Ammonia Oxidation Calculation
824 g NH3 x 1 mol NH3 x 6 mol H2O
17.04 g NH3 4 mol NH3
72.6 mol H2O
31
Conversions of Mass to Mass
gram given x mol given x moles Unk.
gram given mol given x
Unknown g/mol mass Unknown in g
32
Example Stannous Fluoride Production
Sn(s) 2HF ? SnF2 H2(g) Given 30 grams of
HF. How many grams of SnF2 are produced?
33
Stannous Fluoride Production Example Conversion
Factors
Sn(s) 2HF ? SnF2 H2(g) 1 mol HF
1 mol SnF2 156.71 g SnF2 20.01 g HF
2 mol HF mol SnF2
34
Stannous Fluoride Production Calculation
30.00 g HF x 1 mol HF x 1mole SnF2
20.01 g HF 2 mol HF 156.71 g
SnF2 117.5 gram SnF2 1 mol SnF2
35
Chapter 9, Section 2 Review

• Calculate the amount in moles of a reactant or
  product from the amount in moles of a different
  reactant or product.
• Calculate the mass of a reactant or product from
  the amount in moles of a different reactant or
  product.

36
Chapter 9, Section 2 Review

• Calculate the amount in moles of a reactant or
  product from the mass of a different of a
  different reactant or product.
• Calculate the mass of a reactant or product from
  the mass of a different reactant or product.

37
Limiting Reactants

• The limiting reactant is the reactant that limits
  the amount of the other reactants that can
  combine and the amount of the product that can
  form in a chemical reaction.
• The excess reactant is the substance that is not
  used up completely in a reaction.

38
Limiting Reactant Example

• SiO2(s) 4HF ? SiF4(g) 2H2O(l)
• If 2.0 mol of HF are reacted with 4.5 mol of
  SiO2, which is the limiting reactant?

39
Limiting Reactant Example Calculations

• SiO2(s) 4HF ? SiF4(g) 2H2O(l)
• 2 mol HF x 1 mol SiO2 0.5 mol SiO2
• 4 mol HF
• Or
• 4.5 mol SiO2 x 4 mol HF 18 mol of HF
• 1 mol SiO2

40
Black Iron Oxide Example

• 3Fe(s) 4H2O(g) ? Fe3O4(s) 4H2(g)
• Given 36 g H2O
• 167 g Fe
• What is limiting?
• What is mass of black iron oxide produced?
• What mass of excess reactant is left?

41
Black Iron Oxide Calculations

• Convert weights to moles
• 36 g of H2O x 1mol H2O 2.00 mol H2O
• 18.02 g H2O
• 167 g Fe x 1 mol Fe 2.99 mol Fe
• 55.85 g Fe

42
Black Iron Oxide Calculations

• 3Fe(s) 4H2O(g) ? Fe3O4(s) 4H2(g)
• Given 2.00 mol H2O
• 2.99 mol Fe
• Water is limiting.
• What is mass of black iron oxide produced?
• What mass of excess reactant is left?

43
Black Iron Oxide Produced

• 2.00 mol H2O x 1 mol Fe3O4_
• 4 mol H2O
• x 231.55 g Fe3O4_
• 1 mol Fe3O4 116 g Fe3O4
• 
  produced

44
Iron Remaining

• 2.00 mol H2O x 3 mol Fe_
• 4 mol H2O
• x 55.85 g Fe_
• 1 mol Fe 83.8 g Fe consumed
• 167 g Fe 83.8 g 83.2 g of Fe remaining

45
Percent Yield

• Theoretical yield is the maximum amount of
  product that can be produced from a given amount
  of reactant
• Actual yield is the measured amount of product
  from a reaction.
• Percent yield actual yield x 100
• theoretical yield

46
Percent Yield Example

• C6H6(l) Cl2(g) ? C6H5Cl(s) HCl(g)
• Given 36.8 g of C6H6
• Actual yield is 38.8 g.
• What is percent yield?

47
Percent Yield Calculations

• 36.8 g C6H6 x 1 mol C6H6 x 1 mol C6H5Cl
• 78.12 g C6H6 1 mol
  C6H6
• x 112.56 g C6H5Cl 53.0 g of C6H5Cl
• mol C6H5Cl
• Percent Yield 38.8 g x 100 73.2
• 53.0 g

48
Chapter 9, Section 3 Review

• Describe a method for determining which of two
  reactants is a limiting reactant.
• Calculate the amount in moles or mass in grams of
  a product, given the amounts in moles or masses
  in grams of two reactants, one of which is in
  excess.

49
Chapter 9, Section 3 Review

• Distinguish between theoretical yield, actual
  yield, and percent yield.
• Calculate percent yield, given the actual yield
  and the quantity of a reactant.