Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, Maqqwertd ygoijpk[l
‗?
Introductory Chemistry, 3rd Edition
Nivaldo Tro
Chapter 8
Quantities in
Chemical Reactions
2009, Prentice Hall
Car an octane and oxygen
molecules
and carbon dioxide and
water
Outline
• 8.1 Global Warming : Too much Carbon Dioxide
• 8.2 Making Pancakes: Relationships between Ingredients
• 8.3 Making Molecules: Mole to Mole Conversions
• 8.4 Making Molecule: Mass to Mass Conversions
• 8.5 More Pancakes: Limiting Reactant, Theoretical Yield,
and Percent Yield
• 8.6 Limiting Reactant, Theoretical Yield, and Percent
Yield form Initial Reactants
• 8.7 Enthalpy: Measure of the Heat Evolved or Absorbed
during a Chemical Reaction
2Tro's ―Introductory Chemistry‖,
Chapter 8
Burning of Fossil
Fuels
Internal combustion engines
produe Carbon dioxide by
burning fuels.
For example octane, C8H18,
in gasoline.
2C8H18 (l) + 25O2(g) 16 CO2 (g) + 18H2O(g)
Car and molecules
from chapter opening
5Tro's ―Introductory Chemistry‖,
Chapter 8
Global Warming• Average 0.6 °C rise in atmospheric temperature since
1860.
• In the same period atmospheric CO2 levels have risen 25%.
• Are the two related?
Figure 8.1 and Figure 8.2
Showing global tempertature
Rise and greenhouse effect.
6Tro's ―Introductory Chemistry‖,
Chapter 8
Increased CO2
• The primary source of the increased CO2
levels is combustion of fossil fuels
1860 = Industrial Revolution in the U.S. and Europe.
(g)(g)(g)(g) OH 2 CO O 2 CH 2224
Methane in natural gas
7Tro's ―Introductory Chemistry‖,
Chapter 8
Quantities in Chemical Reactions
• All substances in a chemical reaction are related.
• The numerical relationships between compounds in a chemical reaction is called stoichiometry.
9Tro's ―Introductory Chemistry‖,
Chapter 8
Making Pancakes
• Pancakes are made with specific amounts of
ingredients. Your book forgot buttermilk!
1 cup flour + 2 eggs + ½ tsp baking powder +3/4 cup buttermilk
5 pancakes
Pancake recipe figure on page 244
10Tro's ―Introductory Chemistry‖,
Chapter 8
Making Pancakes, Continued• You can scale the recipe up of down.
• Notice that for every 2 eggs 5 pancakes
• What if you have 8 eggs ? pancakes
pancakes 20 eggs 2
pancakes 5eggs 8
Graphics illustrating the above equation
12Tro's ―Introductory Chemistry‖,
Chapter 8
Making Molecules
Mole-to-Mole Conversions
• The balanced equation = chemical ―recipe‖
• 3 H2(g) + N2(g) 2 NH3(g)
3 molecules + 1 molecule 2 molcules
H2 N2 NH3
H-H H H
H-H + N-N H-N-H H-N-H
H-H
Molecule to Molecule and
Mole to Mole conversions• 3 H2(g) + N2(g) 2 NH3(g)
• This equation may be scaled up if the ratio of H2:N2:NH3 is kept
3:1:2
• Multiply by 4
• 12 H2(g) + 4N2(g) 8 NH3(g)
• Multiply by 6.022 x 1023
• 3x(6.022 x1023) H2(g) + 1x(6.022 x1023)N2(g) 2x(6.022 x1023)NH3(g)
• This number of molecules can be expressed as moles.
• 3 mol H2(g) + 1 mol N2(g) 2 mol NH3(g)
13Tro's ―Introductory Chemistry‖,
Chapter 8
The coefficients in a balanced equation give the number of molecules as well as the number of moles
of each substance
We can use the ratios of the coefficients to convert
between moles of substances in a chemical reaction
Molar Ratios
•3 H2(g) + N2(g) 2 NH3(g)
•This equation has 6 molar ratios relating the
reactant and products.
•For example H2 and N2 are related by
1 mole N2 3 mol H2
or
3 mol H2 1 mole N2
Mole to Mole Conversions
• 3 H2(g) + N2(g) 2 NH3(g)
• 12 moles of H2 requires how many mole of
N2
• The ratio must be H2:N2 is 3:1, so so 12:4 is
the same so 4 mole of N2 is needed.
• How many moles of N2 are needed for 1.74
mol of H2
17Tro's ―Introductory Chemistry‖,
Chapter 8
Mole to Mole Conversion• 3 H2(g) + N2(g) 2 NH3(g)
• How many moles of N2 are needed for 1.74 mol of H2?
• We need to use the molar ratio.
• mol H2 mole N2
1 mole N2
1.74 mol H2 x = 0.580 mole N2
3 mol H2
18Tro's ―Introductory Chemistry‖,
Chapter 8
21
Making Molecules
Mass-to-Mass Conversions• We learned previously to convert between moles and grams
using the molar mass.
• Combining this with moles to moles conversions allows us to
related grams of one substrance in an equation to grams of
another substance.
g of A mol of A mol of B g of B
Molar
mass of AMolar
mass of B
Coefficients
of the Balanced
Equation
22Tro's ―Introductory Chemistry‖,
Chapter 8
Example 8.2—How Many Grams of Glucose Can
Be Synthesized from 58.5 g of CO2 in
Photosynthesis? • Photosynthesis:
6 CO2(g) + 6 H2O(g) C6H12O6(s) + 6 O2(g)
Molar masses needed (sum masses of all atoms)
CO2 = 44.01 g/mol
Glucose = 180.16 g/mol
g of CO2 mol of CO2 mol of C6H12O6 g of C6H12O6
Grams moles moles grams
6 CO2(g) + 6 H2O(g) C6H12O6(s) + 6 O2(g)58.5 g CO2 1 mol CO2
x = 1.33 mol CO2
44.01 g CO2
1.33 mol g CO2 1 mol C6H12O6
x = 0.222 mol C6H12O6
6 mol CO2
0.222 mol g CO2 180.16 g C6H12O6
x = 40.0 g C6H12O6
1 mol C6H12O6
Another stoichiometry exampleHow many g of Al2(SO4)3 are required to completely
react with 24.7 g of Ba(NO3)2?
Al2(SO4)3(aq) + 3Ba(NO3)2(aq) 3BaSO4(s) + 2Al(NO3)3(aq)
g of mol of mol of g of
Ba(NO3)2 Ba(NO3)2 Al2(SO4)3 Al2(SO4)3
molar masses needed
Ba(NO3)2 = 261.34g/mol
Al2(SO4)3 = 342.15 g/mol
Grams moles moles grams
Al2(SO4)3(aq) + 3Ba(NO3)2(aq) 3BaSO4(s) +Al(NO3)3(aq)
24.7 g Ba(NO3)2 1 mol Ba(NO3)2
x = 0.0945 mol Ba(NO3)2
261.34 g Ba(NO3)2
0.0945 mol Ba(NO3)2 1 mol Al2(SO4)3
x = 0.0315 mol Al2(SO4)3
3 mol Ba(NO3)2
0.0315 mol g Al2(SO4)3 342.15 g Al2(SO4)3
x = 10.8 g Al2(SO4)3
1 mol Al2(SO4)3
8.5 More Pancakes: Limiting
Reactant, Theoretical Yield, and
Percent Yield
8.6 Limiting Reactant,
Theoretical Yield, and Percent
Yield form Initial Reactants
26Tro's ―Introductory Chemistry‖,
Chapter 8
Note in these two sections, we are only going to
cover percent yield
Actual; yield
% Yield = x 100
Theoretical yield
Actual yield – amount actually obtained
Theoretical yield – amount obtained if
reaction goes to 100% completion.27Tro's ―Introductory Chemistry‖,
Chapter 8
% yield example
A is ran with 100 g of Hg reacting with
oxygen. 75.3 grams of HgO is isoloted.
What is the % yield?
Hg(s) + O2(g) HgO(s)
Molar masses:
Hg = 200.59 g/mol
HgO = 216.59 g/mol28
% yield example (cont)
• First calculate the theoretical yield.
• g Hg mol Hg mol HgO g HgO
• 100 g Hg 1 mol Hg
x = 0.499 mol Hg
200.59 g Hg
• 0.499 mol Hg 1 mol HgO
x = 0.499 mol HgO
1 mol Hg
• 0.499 mol Hg 216.59 g HgO
x = 108 g HgO (theoretical
1 mol HgO yield)
29
% yield example
30Tro's ―Introductory Chemistry‖,
Chapter 8
75.0 g HgO
% Yield = x 100
108 g HgO
= 69.4 % Yield of HgO
.
31Tro's ―Introductory Chemistry‖,
Chapter 8
Enthalpy Change
• We previously described processes as
exothermic if they released heat, or
endothermic if they absorbed heat.
• The enthalpy of reaction is the amount of
thermal energy that flows through a process.
At constant pressure.
DHrxn
32
Sign of Enthalpy Change• For exothermic reactions, the sign of the enthalpy
change is negative when:
Thermal energy is produced by the reaction.
The surroundings get hotter.
DH = ─
For the reaction CH4(s) + 2 O2(g) CO2(g) + 2 H2O(l), the DHrxn = −802.3 kJ per mol of CH4.
• For endothermic reactions, the sign of the enthalpy change is positive when:
Thermal energy is absorbed by the reaction.
The surroundings get colder.
DH = +
For the reaction N2(s) + O2(g) 2 NO(g), the DHrxn = +182.6 kJ per mol of N2.
33Tro's ―Introductory Chemistry‖,
Chapter 8
Enthalpy and Stoichiometry
• The amount of energy change in a reaction depends on the amount of reactants.
You get twice as much heat out when you burn twice as much CH4.
• Writing a reaction implies that amount of energy changes for the stoichiometric amount given in the equation.
For the reaction C3H8(l) + 5 O2(g) 3 CO2(g) + 4 H2O(g) DHrxn = −2044 kJ
So 1 mol C3H8 5 mol O2 3 mol CO2 4 mol H2O
−2044 kJ.
Enthalpy Calculation• C3H8(l) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
DHrxn = −2044 kJ
• How much heat is released when 53.8 g of O2
completely reacts with propane?
• g O2 mol O2 heat (J)
• 53.8 g of O2 1 mol O2
x = 1.68 mol O2
32.00 g O2
• 1.68 mol of O2 -2044 J
x = 687 J of heat
5 mol O234
Example 8.7—How Much Heat Is Associated with the
Complete Combustion of 11.8 x 103 g of C3H8(g)?
1 mol C3H8 = -2044 kJ, Molar mass = 44.11 g/mol
The sign is correct and the value is reasonable.
11.8 x 103 g C3H8,
heat, kJ
Check:
Solution:
Solution Map:
Relationships:
Given:
Find:
g 09.44
HC mol 1 83
kJ 1047.5HC mol 1
kJ 2044-
HC g 44.11
HC mol 1 HC g 1011.8 5
8383
8383
3
83HC mol 1
kJ 2044-
mol C3H8 kJg C3H8