1. NEW Office hours as of Mon. Sept. 26
Chemistry 140 Fall 2002
Change in Office Rooms
NEW Office hours as of Mon. Sept. 26
Mondays 9:00-11:30 Chemistry Conference Room (Room 273 Essex Hall).
Fridays 2:00-4:00 SRC Essex Hall.
Often do not specify Z when writing. For example 14C, C specifies Z = 12.
Special names for some isotopes. For example hydrogen, deuterium, tritium.
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

2. Chapter 4: Chemical Reactions
Chemistry 140 Fall 2002
General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chapter 4: Chemical Reactions
Philip Dutton
University of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
(modified 2003 by Dr. Paul Root and 2005 by Dr. David Tramontozzi)
Iron rusts
Natural gas burns
Represent chemical reactions by chemical equations.
Relationship between reactants and products is STOICHIOMETRY
Many reactions occur in solution-concentration uses the mole concept to describe solutions
General Chemistry: Chapter 4
Prentice-Hall © 2002

3. Contents 4-1 Chemical Reactions and Chemical Equations
Chemistry 140 Fall 2002
Contents
4-1 Chemical Reactions and Chemical Equations
4-2 Chemical Equations and Stoichiometry
4-3 Chemical Reactions in Solution
4-4 Determining the Limiting reagent
4-5 Other Practical Matters in Reaction Stoichiometry
Focus on Industrial Chemistry
Chemical reactions are the central concern for the entire science of chemistry
Learn how to express chemical reactions as equations and establish a quantitative relationship among the reactants and products
Because most reactions occurs in solution, Chapter 4 introduces the concept of molarity – a method of describing the composition of a solution
Finally, the concept of the mole will resurface in new types of problem solving
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

4. 4-1 Chemical Reactions and Chemical Equations
Chemistry 140 Fall 2002
4-1 Chemical Reactions and Chemical Equations
As reactants are converted to products we observe:
Color change [KCrO4(yellow)  AgCrO4(red-brown)]
Precipitate formation
Gas evolution (penny and nitric acid)
Heat absorption or evolution (cooling packs)
A chemical reaction by definition is a process which one set of substances (reactants) is converted into a new set of substances (products)
In other words – a chemical reaction is the process in which chemical change occurs
Chemical reactions are typically evident by physical evidence such as:
Colour change – ie. iron or copper oxidation
Precipitation formation – ie.
Gas evolution – ie. Baking powder releases CO2 while baking to make cakes/breads rise
Heat absorption or evolution – ie. Combustion, propane BBQ’s
Chemical evidence may be necessary. (i.e.-a detailed chemical analysis of products)
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

5. Nitrogen monoxide + oxygen → nitrogen dioxide
Chemistry 140 Fall 2002
Chemical Reaction
Nitrogen monoxide + oxygen → nitrogen dioxide
Step 1: Write the reaction using chemical symbols.
Step 2: Balance the chemical equation.
Consider the reaction of colorless nitrogen monoxide with oxygen to give brown nitrogen dioxide.
The arrow represents a reaction proceeding
Replace the words with chemical symbols
Step 1
Explain the chemical symbols – one nitrogen, one oxygen = nitrogen monoxide
- oxygen is diatomic, existing as O2 instead of O
-one nitrogen, two oxygen = nitrogen dioxide
Explain the reasoning for balancing chemical equations – Conservation of Mass
Step 2
- Coefficients required to balance a chemical equation are called stoichiometric coefficients
these are important for a variety of calculations that will be evident later in this chapter and throughout the course 2
NO + O2 → NO2 1 2
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

6. Molecular Representation
Chemistry 140 Fall 2002
Molecular Representation
Same reaction of NO + O2  NO2 using a space filling model to visualize reactants and the products
Note all atoms are balanced
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

7. Balancing Equations NO + O2 → NO2 + O NO + O2 → NO3
Chemistry 140 Fall 2002
Balancing Equations
Never introduce extraneous atoms to balance.
NO + O2 → NO2 + O
Never change a formula for the purpose of balancing an equation.
Rules for balancing equations
No new atoms introduced
No changing formulas
NO + O2 → NO3
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

8. Balancing Equation Strategy
Chemistry 140 Fall 2002
Balancing Equation Strategy
Balance elements that occur in only one compound on each side first.
Balance free elements last.
Balance unchanged polyatomics as groups.
Fractional coefficients are acceptable and can be cleared at the end by multiplication.
4 step strategy
1. Atoms that are present in one compound in the reactant side (left side) and one compound on the product side (right side)
2. Free elements are balanced last – have the least impact on any other element in the equation
3. Polyatomics as groups is easier than attempting to balance each individual atoms. Polyatomics must be unchanged following the reacation
4. Fractions are find until the final step which eliminates them
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

9. Always exist as diatomics, never as an individual atom.
Unusual Free Elements
Oxygen O2
Nitrogen N2
Hydrogen H2
Chlorine Cl2
Fluorine F2
Bromine Br2
Phosphorus P4
Always exist as diatomics, never as an individual atom.
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

10. Chemistry 140 Fall 2002
Example 4-2
Writing and Balancing an Equation: The Combustion of a Carbon-Hydrogen-Oxygen Compound.
Liquid triethylene glycol, C6H14O4, is used a a solvent and plasticizer for vinyl and polyurethane plastics. Write a balanced chemical equation for its complete combustion.
Any combustions reaction involving hydrocarbons requires:
Hydrocarbon + O2  CO2 + H2O
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

11. Example 4-2 Chemical Equation: 15 2 6 7 C6H14O4 + O2 → CO2 + H2O 6 7 6
Chemistry 140 Fall 2002
Example 4-2
Chemical Equation: 15 2
C6H14O O2 → CO H2O 6
1. Balance C.
2. Balance H.
3. Balance O.
4. Multiply by two
Step 1 – C occurs in one one compound on either side of the reaction
6 C on reactant side requires “6” CO2
Step 2 – balance free atoms last, so next we balance H
14 H on reactant side, requires “7” H2O
Step 3 – balance O2
product side has 19 O atoms
reactant side has 4 O from hydrocarbon, requires 15 more O; 15/2 O2 since oxygen is diatomic
Step 4 – eliminate fraction by multiplying each co-efficient by 2
2 C6H14O O2 → 12 CO H2O
and check all elements.
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

12. 4-2 Chemical Equations and Stoichiometry
Chemistry 140 Fall 2002
4-2 Chemical Equations and Stoichiometry
Stoichiometry (means to measure the elements) includes all the quantitative relationships involving:
atomic and formula masses
chemical formulas.
Stoichiometry literally means “to measure elements”, in more practical terms, it includes all the quantitative relationships involving atomic and formula masses, chemical formulas, and the chemical equation
The mole ratio (also called a stoichiometric factor) is based on balanced chemical equationsallows one to relate any two substances involved in a chemical reaction
Mole ratio is a central conversion factor This is the conversion from moles of A to moles of B.
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

13. Example 4-3 Relating the Numbers of Moles of Reactant and Product.
How many moles of H2O are produced by burning 2.72 mol H2 in an excess of O2?
H O2 → H2O
Write the Chemical Equation:
Balance the Chemical Equation: 2
Use the stoichiometric factor or mole ratio in an equation:
nH2O = 2.72 mol H2 × = 2.72 mol H2O
2 mol H2O
2 mol H2
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

14. Chemistry 140 Fall 2002
Example 4-6
Additional Conversion Factors in a Stoichiometric Calculation: Volume, Density, and Percent Composition.
An alloy used in aircraft structures consists of 93.7% Al and 6.3% Cu by mass. The alloy has a density of 2.85 g/cm3. A cm3 piece of the alloy reacts with an excess of HCl(aq). If we assume that all the Al but none of the Cu reacts with HCl(aq), what is the mass of H2 obtained?
For HCl(aq) ; the (aq) denotes the compound is in an aqueous (water) solution
- could be (g) gas, (l) liquid, or (s) solid
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

15. Example 4-6 Al + HCl → AlCl3 + H2 Write the Chemical Equation:
Balance the Chemical Equation: 2 6 3
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

16. Example 4-6 2 Al + 6 HCl → 2 AlCl3 + 3 H2 Plan the strategy:
Chemistry 140 Fall 2002
Example 4-6
2 Al + 6 HCl → 2 AlCl H2
Plan the strategy:
cm3 alloy → g alloy → g Al → mole Al → mol H2 → g H2
We need 5 conversion factors!
× ×
Write the Equation
mH2 = cm3 alloy × × ×
2.85 g alloy
1 cm3
93.7 g Al
100 g alloy
1 mol Al
26.98 g Al
3 mol H2
2 mol Al
2.016 g H2
1 mol H2
= g H2
and Calculate:
This solutions is shown as a one step solution
it may be easier to break it into separate steps
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

17. 4-3 Chemical Reactions in Solution
Close contact between atoms, ions and molecules necessary for a reaction to occur.
Solvent
We will usually use aqueous (aq) solution.
Solute
A material dissolved by the solvent.
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

18. This volume is total solution volume.
Molarity
Molarity (M) =
Volume of solution (L)
Amount of solute (mol solute)
If mol of urea is dissolved in enough water to make L of solution the concentration is:
curea =
1.000 L
0.444 mol urea
= M CO(NH2)2
This volume is total solution volume.
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

19. Preparation of a Solution
Chemistry 140 Fall 2002
Preparation of a Solution
Weigh the solid sample.
Dissolve it in a volumetric flask partially filled with solvent.
Carefully fill to the mark.
In a practical sense, how do you make up a solution.
2 ways: 1 if your compound is a solid you need to weight out a certain mass and dissolve it in a specific volume
2 if your compound is in solution, you need to know either the concentration or density and dilute it in a specific volume
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

20. Example 4-9
Calculating the Mass of Solute in a Solution of Known Molarity.
We want to prepare exactly L (250 mL) of a M K2CrO4 solution in water. What mass of K2CrO4 should we use?
Plan strategy:
Volume → moles → mass
We need 2 conversion factors!
Write equation and calculate:
mK2CrO4 = L × × = 12.1 g
0.250 mol
1.00 L g
1.00 mol
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

21. Solution Dilution Mi × Vi = ni = nf = Mf × Vf M = n V Mi × Vi Mf = Vf
Chemistry 140 Fall 2002
Solution Dilution
Mi × Vi = ni
Mi × Vi
Mf × Vf
= nf = Mf × Vf
M = n V
All of the solute taken from the initial concentrated solution appears in the final diluted solution
This is new here !!!!! For today’s lecture
mols = concentration x volume
When performing a dilution the number of moles of substance is conserved.
Mi × Vi
Mf = Vf
= Mi Vi Vf
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

22. Example 4-10 Preparing a solution by dilution.
Chemistry 140 Fall 2002
Example 4-10
Preparing a solution by dilution.
A particular analytical chemistry procedure requires M K2CrO4. What volume of M K2CrO4 should we use to prepare L of M K2CrO4?
Plan strategy:
Mf = Mi Vi Vf
Vi = Vf Mf Mi
Calculate:
VK2CrO4 = L × × = L
mol
1.00 L
1.000 L
0.250 mol
State what this means:
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002

23. Preparation of a Solution
VK2CrO4 = L × × = L
mol
1.00 L
1.000 L
0.250 mol Mi Vi Vf Mf
Chemistry 140 Fall 2002 Dutton
Prentice-Hall © 2002