CHEM 1405

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CHEM 1405

• Class Meeting 15

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Assignments and Reminders

• Reading Assignment
• Chapter 7 by Tuesday
• Homework Problems due Thursday Mar 9th
• Chapter 6 Problems 1,6, 10, 12, 14,16, 18, 20,
  22, 24, 26, 28, 30, 32, 34, 36, 38
• Homework Problems due Thursday Mar 23rd
• Chapter 7 Problems 3, 4, 6, 8, 9,10, 11, 12,
  15,16,17,20,22
• EXAM 2 will be in test center March 7th-March
  11th
• Class website
  http//iws.ccccd.edu/jstankus/

Note Typo on HW6 solution set Chap5 Ques 4a Ca
is group 2A So Ru in CaRuO3 has oxidation state
of 4
Please use only one side of the page when
submitting Homework
3
Chemistry Help Resources

• My office hours
• Tuesdays after class in Lecture room
• Thursdays 1-2 in Math Lab
• Free Tutoring through college
• Students must submit a tutor request form in
  order to receive detailed information about the
  available tutoring services.  The form is
  available on Collin's website and in the
  following offices
• CPC room A108 (ask for Sonia Castillo)
• PRC room F109 (ask for Shontel Penny or Mary
  Eldridge)
• SCC rooms G200 and G141
• There are group tutoring services available for
  the following courses (SUBJECT TO CHANGE!)
• CHEM 1405, 1411, 1412, 2423,
•  
• Also available will be online tutoring in the
  following courses (SUBJECT TO CHANGE!)
• CHEM 1412 below

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Stoichiometry

• Relationship between reactants and products in a
  chemical reaction
• Converts between moles of different substances in
  a reaction

5
Energy in Chemical Reactions
Thermodynamics is the study of energy in chemical
reactions
6
Energy in Chemical Reactions
C O2 ? CO2
C O2
Net amount Of Energy Released
CO2
Exothermic
Reaction Diagram
7
Exothermic reactions
Burning of methane
DH -890.3 kJ
CH4 2O2 ? CO2 2H2O
DH is the enthalpy change ? heat of the
reaction Minus sign indicates that
this reaction is exothermic DH -890.3 kJ
means that 890.3 kJ of heat is
released when 1 mol of CH4 is burned
Exothermic describes a process or reaction in
which, in a nonisolated system, heat is given off
to the surroundings
8
Endothermic Reaction
2KClO3 ? 2KCl 3O2
2KCl 3O2
Net Amount of Energy Required
Activation Energy
2KClO3
9
Endothermic reactions
Decomposition of Mercury (II) Oxide
2HgO ? 2Hg O2
DH 181.66 kJ
DH is the enthalpy change ? heat of the
reaction Plus sign indicates that
this reaction is endothermic DH 181.66 kJ
means that 181.66 kJ of heat must be
added when 2 mol of HgO is decomposed
Endothermic describes a process or reaction in
which, in a nonisolated system, heat is absorbed
from the surroundings
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Forward and Reverse Reactions
Living cells burn glucose by
C6H12 6 O2 ? 6 CO2 6 H2O energy
Green Plants create glucose and oxygen by
photosynthesis
6 CO2 6 H2O energy ? C6H12 6 O2
Notice that they are the same reactions but in
opposite directions
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Forward and Reverse Reactions
C6H12 6 O2 6 CO2 6 H2O energy
12
Reaction Rates

• Kinetics is the study of reaction Rates
• Reaction rates are affected by
• Temperature
• Concentration
• Catalysts

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Reaction Rates in Exothermic Reactions
C O2 ? CO2
C O2
Energy Released Increase temperature And
increases reaction rate
CO2
Provides more Energy to overcome Activation
barrier
Reaction can be self-sustaining
14
Reaction Rates in Endothermic Reaction
2KClO3 ? 2KCl 3O2
2KCl 3O2
Energy Required
Activation Energy
May reduce temperature Decreasing reaction rates
2KClO3
Not self-sustaining
15
Concentration Effect on Reaction Rate
Low Concentration Less Frequent
Collisions Between Reactants Slower reaction rate
High Concentration More Frequent
Collisions Between Reactants Faster Reaction Rate
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Catalysts
High Barriers Can slow Reaction path
An alternate Pathway that Is lower Can speed
up The reaction
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Catalysts
2KClO3 ? 2KCl 3O2
High Activation Energy May slow reaction rates
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Catalysts
2KClO3 ? 2KCl 3O2
A catalyst enables a lower activation energy
pathway for the reaction
A catalyst is a substance that increases the rate
of a reaction without itself being consumed in
the reaction.
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Equilibrium
Saturated Salt Solution Much more salt than Can
dissolve is added Leaving solid salt in Bottom of
beaker
NaCl is continually going into solution and
coming out of solution
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Forward and Reverse Reactions

• Some reactions can proceed both forward and
  backwards

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Equilibrium in Chemical reactions
Rate of reactions Forward Rxn greatest No
reverse rxn Forward Rxn slows Reverse Rxn
starts Forward Rxn continues to slow Reverse Rxn
increases Forward rxn rate equals Reverse
reaction rate
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Le Chateliers Principle

• Le Châtelier's principle states that if a stress
  is applied to a system at equilibrium, the
  equilibrium shifts in the direction that will
  relieve the stress.

If we add heat to this equilibria it will be
driven to the left
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Le Chatelier Continued
Get 2 volumes NO2 for every N2O4 that does react
Higher pressure drives reaction to left to reduce
pressure
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Le Chatelier Continued
If we increase the nitrogen concentration the
equilibria Will be pushed towards the right
generating more ammonia
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Summary of views of redox reactions
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Oxidation Numbers

• Just the charge on a simple ion
• (rules for covalent compounds coming up next)
• Increase in oxidation number oxidation
• Decrease in oxidation number reduction

The oxidation number of an element is a means of
designating the number of electrons that its
atoms lose, gain, or share in forming compounds
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Oxidation Number Examples
Ionic compounds

• NaCl is an ionic compound made up of
• Na and Cl- ions
• Oxidation number of Na is 1
• Oxidation number of Cl- is -1
• CaCl2
• Ca2 and Cl- ions
• Oxidation number of Ca2 is 2
• Oxidation number of Cl- is -1

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Oxidation Numbers

• Fairly straightforward for ionic compounds
• For covalent compounds (remember sharing of
  electrons) need some rules
• Following rules listed by priority
• Earlier rules take precedence

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Oxidation numbers

• 1. The total of the oxidation numbers of all the
  atoms in a neutral molecule, an isolated atom, or
  a formula unit is 0.
• Examples
• The oxidation number of the Fe atom is 0.
• The sum of the oxidation numbers of all the atoms
  in each of the following molecules
• Cl2 , S8 , and C6H12O6  is 0,
• The sum of the oxidation numbers of the ions in
  MgBr2  is 0

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Oxidation numbers

• 2. In their compounds, the Group 1A metals all
  have an oxidation number of 1, and the Group 2A
  metals have an oxidation number of 2.
• Examples
• The oxidation number
• of Na in Na2SO4  is 1,
• of Ca in Ca3(PO4)2   is 2

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Oxidation numbers

• 3.In its compounds, hydrogen has an oxidation
  number of 1
• Examples
• The oxidation number of H is 1
• in HCl, H2O, NH3 , and CH4

The principal exception to rule 3 is when H is
bonded to an element that is less electronegative
than itself, as in metal hydrides.
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Oxidation numbers

• 4. In its compounds, oxygen has an oxidation
  number of -2.
• Examples
• The oxidation number of O is -2
• in CO, CH3OH, and C6H12O6

The principal exception to rule 4 is when oxygen
is bonded to itself, as in peroxides (for
example, H2O2).
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Oxidation numbers

• 5.  In their binary (two-element) compounds with
  metals, Group 7A elements have an oxidation
  number of -1, Group 6A elements have an oxidation
  number of -2, and Group 5A elements have an
  oxidation number of -3.
• Examples
• The oxidation number
• Of Br in CaBr2 is -1   because Br is in group 7A
• Of S in Na2S is -2 because S is in group 6A
• of N in Mg3N2  is -3 because N is in group 5A

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Oxidation Number Examples

• I2
• Oxidation number of Iodine is 0 (rule 1)
• Cr2O3
• Oxidation number of Oxygen is -2 (rule 4)
• Oxidation number of Chromium is 3 (rule 1)
• AlCl3
• Oxidation number of chlorine is -1 (rule 5)
• Oxidation number of Aluminum is 3 (rule 1)

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Oxidation Number Examples

• Na2SO4
• Oxidation number of Na is 1 (Rule 2)
• Oxidation number of O is -2 (rule 4)
• Oxidation number of S is 6 (rule 1)
• CaH2
• Oxidation number of Ca is 2 (Rule 2)
• Oxidation number of H is -1 (rule 3 exception)

NEEDS WORK
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Oxidizing and Reducing Agents

• Oxidizing agent element or compound that gets
  reduced
• Causes oxidation of other substance
• Reducing agent element or compound that gets
  oxidized
• Causes reduction of other substance

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Identify the Oxidizing Agent and Reducing Agent

• 2 C O2 ? 2 CO
• Carbon gains oxygen and is
  oxidized
• therefore it must be Reducing agent
• Oxygen(O2) must be oxidizing agent
• N2 3 H2 ? 2 NH3
• Nitrogen gains hydrogen and is reduced
• So it is the oxidizing agent
• Therefore Hydrogen (H2) is the reducing agent

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Identify the Oxidizing Agent and Reducing Agent

• SnO H2 ? Sn H2O
• SnO loses oxygen and is
  reduced
• therefore it must be oxidizing agent
• Hydrogen(H2) must be reducing agent
• Mg Cl2 ? Mg2 2 Cl-
• Magnesium loses electrons and is oxidized
• So it is the reducing agent
• Therefore chlorine (Cl2) is the oxidizing agent

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Reducing Agents

• Production of metals
• SnO2 C ? Sn CO2
• Photography
• Used in process to develop film
• Antioxidants
• Inhibit damage by O2 to cells
• Some water soluble, some fat soluble

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Identifying Redox Reactions

• In an oxidation-reduction reaction,
• the oxidation number of one or more elements
  increasesan oxidation process
• and the oxidation number of one or more elements
  decreasesa reduction process.
• A redox reaction must always have both an
  oxidation and a reduction.

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Half-Reactions

• Can break redox reactions into separate oxidation
  and reduction reactions
• Oxidation Zn(s) ? Zn2(aq) 2
  e
• Reduction Cu2(aq) 2 e ? Cu(s)
• Overall Zn(s) Cu2(aq) ? Cu(s) Zn2(aq)

42
Half-Reaction Practice Problems

• Mg Cl2 ? Mg2 2 Cl-
• Half reactions
• Mg ? Mg2 2e-
• Cl2 2e- ? 2 Cl-
• Note that not only do the atoms balance but the
  electrons also balance

43
Density of Gases

• Application of Ideal gas law
• PV nRT
• Remember density Mass/Volume
• We can calculate Mass from
• n ? number of moles
• Mass n x molar mass
• We can calculate Volume from ideal gas law
• V nRT/P

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Simple Case

• At STP we know 1 mol occupies 22.4 L
• What is the density of hydrogen at STP?
• 1 mole of hydrogen has a mass of 2.016 g/mol

m V
d
d
0.090 g/L
45
Density of Gases

• Remember density Mass/Volume
• We can calculate Mass from
• n ? number of moles
• Mass n x molar mass
• We can calculate Volume from ideal gas law
• V nRT/P

46
Henrys Law? Effect of Pressure on Solubility

• Equilibrium
• CO2(g) ? CO2(aq)
• Application of Le Chateliers Principle

Henry's law states that the solubility of a gas
is directly proportional to the pressure
maintained in the gas above the solvent.
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The Effect of Pressure on Solubility

• Why does a soda fizz when you open it?
• Carbonated water is a saturated solution of
  carbon dioxide in water under pressure
• When opened the pressure over the soda drops to 1
  atm causing the solubility of carbon dioxide to
  be reduced (Henrys Law)
• The excess carbon dioxide comes out of solution

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Gas Mixtures

• A lot of gases we deal with are not pure
• Air (78 nitrogen, 21 oxygen, 1 other)
• How do we deal with these?
• Remember from the Kinetic theory of gases
  molecules of a gas are non-interacting (unless
  the collide) discrete particles
• They act independently

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Partial Pressures
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Partial Pressures

• Using ideal gas law
• PVnRT
• Partial Pressure is
• where PA is partial pressure of A
• and nA is number of moles of A

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Partial Pressures

• Total pressure is the sum of
• Ptot PA PB PC ..
• Total moles of material
• ntot nA nB nC ..

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Partial Pressure Examples

• What is the total pressure in a container that
  contains
• methane at a partial pressure of 0.75 atm,
• hydrogen at a partial pressure of 0.40 atm
• propane at a partial pressure of 0.50 atm?
• Ptot Pmethane Phydrogen Ppropane
• Ptot 0.75 atm 0.40 atm 0.50 atm
• Ptot 1.65 atm
• Remember Significant Figures rules for addition

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Partial Pressure Examples

• What is the Partial Pressure of Nitrogen in air
  at STP?
• First what do we know?
• The pressure at STP is 1 atm
• Nitrogen makes up 78 by volume of air
• Since 78 of the molecules in air are nitrogen it
  contributes 78 of the total pressure
• Partial pressure of nitrogen in air at STP is
  0.78 atm

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Density of Gases (more complex problems)
In a hot air balloon the air temperature may be
100CWhat is the density of air in the balloon at
100C and in surrounding air at 20C?
M n x molar mass
V nRT/P
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Molecular Shape

• Determine reactive, chemical and physical
  properties
• Very important in biological systems
• Must draw electron-dot diagram to predict shape
  of molecule
• Look at central atom

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Valence Shell Electron Pair Repulsion (VSEPR)
Theory

• Electron pairs arrange themselves to minimize
  repulsion

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Shape of Molecules

• Draw electron-dot structure
• Count number of bonding and nonbonding pairs of
  electrons
• Determine number of electron sets
• Sketch shape

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Basic Shapes

• Electron pairs arrange themselves to minimize
  repulsion

Linear 180
2 Sets of electrons
Trigonal Planar 120 in plane of page
3 Sets of electrons
Tetrahedral 109.5
4 Sets of electrons
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Nonbonding Pairs Effect on Shape
4 Sets of electrons
4 Bonding Pairs 0 Lone Pairs (LP)
Tetrahedral
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Lone Pairs Effect on Shape
4 Sets of electrons
3 Bonding Pairs 1 Lone Pairs (LP)
Pyramidal
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Lone Pairs Effect on Shape
4 Sets of electrons
2 Bonding Pairs 2 Lone Pairs (LP)
Bent
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Possible Shapes
65
Electronegativity

• Fluorine most electronegative element

Electronegativity Increases
Electronegativity Increases
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Categorizing Bonds by Electronegativity Difference
Increasingly Polar
Nonpolar
0
1.7 -1.8
0.4
Ionic
Covalent
Example H-Cl Cl 3.0
H 2.1 Difference 0.9 Polar Covalent Bond
Elecronegativity Difference
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Shapes and Properties
d-
d-
d
d-
d
d-
d
Net Polarity
d
2.6
2.6
3.5
2.2
CO2
H2S
Bond Polarity adds up Polar Molecule
Bond Polarity Cancels out Nonpolar Molecule
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Shapes and Properties

• Polar molecule has separate centers of positive
  and negative charge

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Formula Weight of Molecules

• Formula weight is simply the sum of the masses of
  the component elements
• Water H2O
• Formula wt of H2O
• 2 x Atomic wt of Hydrogen Atomic
  wt of Oxygen
• 2 x (1.008 amu) 16.00 amu
• 18.02 amu

Since there are 2 hydrogen atoms in H2O
For Molecules, Formula weight is also known as
Molecular weight
70
Mole Concept

• Atoms are very small
• Practically, we must deal with a large number of
  atoms at a time
• We could try keeping track of individual
  molecules
• But thats a bit unwieldy
• To make things easier we define a unit, the Mole,
  that represents a very large number of atoms or
  molecules
• 1 mole 6.023 x 1023 items
• (think of this similarly to 1 dozen 12 items)

6.023 x 1023 molecules CH4
1 mole of CH4
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MolesA mole is just a quantity

• 1 mole of helium atoms ? 6.023 x 1023 He atoms
• 1 mole of apples ? 6.023 x 1023 apples
• 1 mole of water ? 6.023 x 1023 H2O molecules

This is the chemists equivalent of saying a
dozen doughnuts or a gross of pencils
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Avogadros Numberor why 1 mole 6.023 x 1023
items?

• To make things simple
• It was defined that 1 mole of 12C would have a
  mass of 12.000 grams
• 6.023 x 1023 is the number of items in a mole
  such that 1 mole of items weighing 1 amu weigh 1
  gram
• In other words there are 6.023 x 1023 amu per
  gram

Atomic wt of 12C is 12.000 amu
73
Molar Weight Examples

• Water H2O Molecular wt 18.011 amu
• Mass of 1 mole of H2O is 18.011 grams
• Propane C3H8 Molecular wt 44.097 amu
• Mass of 1 mole of C3H8 is 44.097 grams
• Aspirin C9H8O4 Molecular wt 180.163 amu
• Mass of 1 mole of Aspirin is 180.163 grams
• Ammonium orthophosphate (NH4)2HPO4 Molecular wt
  132.0559 amu
• Mass of 1 mole of Ammonium Orthophosphate is
  132.0559 grams

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Chemical Equations

• A notation to describe a chemical reaction
• Components of a chemical equation
• Reactants (the substances that react)
• Products (the substances that are produced)
• The relative amounts of the substances

Reactants ? Products Example Combustion (burning)
of methane CH4 2O2 ? CO2 2H2O Methane reacts
with two Oxygen molecules to yield Carbon Dioxide
and two water molecules
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A closer look at the example
Reactants ? Products CH4 2O2 ? CO2 2H2O
Two Oxygen Molecules
One Methane Molecule
Reactants
Two Water Molecules
Carbon Dioxide Molecule
Products

• Relative Amounts
• 2 molecules of Water are produced for every 1
  molecule of Carbon dioxide produced

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Mole to Mass Relationships

• Balanced chemical reaction gives molar ratios

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Suggestions for Balancing Chemical Equations

• If element occurs once on each side, balance
  first
• Balance any reactants or products that exist as
  free elements last
• CANNOT change subscripts
• CANNOT add/delete products or reactants

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Balanced Equation

• Since Matter is conserved both sides have to have
  the same number of atoms

Reactants ? Products CH4 2O2 ? CO2 2H2O
79
Balanced Equation

• Since Matter is conserved both sides have to have
  the same number of atoms

Reactants ? Products CH4 2O2 ? CO2 2H2O
1 Carbon
1 Carbon
Balanced
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Balanced Equation

• Since Matter is conserved both sides have to have
  the same number of atoms

Reactants ? Products CH4 2O2 ? CO2 2H2O
1 C
1 C
Carbon Balanced
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Balanced Equation

• Since Matter is conserved both sides have to have
  the same number of atoms

Reactants ? Products CH4 2O2 ? CO2 2H2O
1 C
1 C
Carbon Balanced
2 Water molecules with 2 Hydrogen Atoms each
4 Hydrogen Atoms
4 Hydrogen Atoms
Balanced
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Balanced Equation

• Since Matter is conserved both sides have to have
  the same number of atoms

Reactants ? Products CH4 2O2 ? CO2 2H2O
1 C
1 C
Carbon Balanced
Hydrogen Balanced
4 H
4 H
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Balanced Equation

• Since Matter is conserved both sides have to have
  the same number of atoms

Reactants ? Products CH4 2O2 ? CO2 2H2O
1 C
1 C
Carbon Balanced
Hydrogen Balanced
4 H
4 H
2 Oxygen molecules with 2 Oxygen Atoms each
2 Water molecules with 1 Oxygen Atom each
2 Oxygen Atoms
4 Oxygen atoms
4 Oxygen atoms
Balanced
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Balanced Equation

• Since Matter is conserved both sides have to have
  the same number of atoms

Reactants ? Products CH4 2O2 ? CO2 2H2O
1 C
1 C
Carbon Balanced
Hydrogen Balanced
4 H
4 H
Oxygen Balanced
4 O
4 O
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Balanced Equation

• Since Matter is conserved both sides have to have
  the same number of atoms

Reactants ? Products CH4 2O2 ? CO2 2H2O
1 C
1 C
Carbon Balanced
Hydrogen Balanced
4 H
4 H
Oxygen Balanced
4 O
4 O
All elements are in equal amounts on both sides
of the equation Therefore this is a balanced
equation
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How to balance equations
Combustion of hydrogen to form water
H2 O2 ? H2O
2 H
2 H
Not Balanced
2 O
1 O
In order to account for both Oxygen atoms two
water molecules must be formed
H2 O2 ? 2H2O
Not Balanced
2 H
4 H
2 O
2 O
Two Hydrogen molecules are needed to produce two
water molecules
2H2 O2 ? 2H2O
Balanced
4 H
4 H
2 O
2 O
87
Equation Balancing Example
Look at Cl first
Al HCl ? AlCl3 H2 Reactants Products 1
- Al - 1 1 - H - 2 1 - Cl - 3 Unbalanced Equation
Al 3HCl ? AlCl3 H2 Reactants Products 1 -
Al - 1 3 - H - 2 3 - Cl - 3 Unbalanced Equation
Examine H
Examine remaining Al
Al 6HCl ? 2AlCl3 3H2 Reactants Products 1
- Al - 2 6 - H - 6 6 - Cl - 6 Unbalanced Equation
2Al 6HCl ? 2AlCl3 3H2 Reactants Products 2
- Al - 2 6 - H - 6 6 - Cl - 6 Balanced Equation
88
Calculations based on Chemical Equations

• Chemical equations can be used to calculate the
  amount of substances involved in chemical
  reactions

CH4 2O2 ? CO2 2H2O
Molecular level
1 Molecule Methane
2 Molecules Oxygen
1 Molecule Carbon Dioxide
2 Molecules Water
Macroscopic level
1 Mole of Methane
2 Moles Oxygen
1 Mole of Carbon Dioxide
2 Moles Water
89
Calculations based on Chemical Equations
CH4 2O2 ? CO2 2H2O
1 Mole of Methane
2 Moles Oxygen
1 Mole of Carbon Dioxide
2 Moles Water
Molar wt of Methane is 16.0 g
Molar wt of Oxygen molecule is 32.0 g
Molar wt of Carbon Dioxide is 44.0 g
Molar wt of Water is 18.0 g
Mass
16.0 g
64.0 g
44.0 g
36.0 g
80 grams Reactants
80 grams Products
90
Percent Yields from Chemical Reactions
C6H6 HNO3 ? C6H5NO2 H2O
1 Mole of Benzene
1 Mole Nitric Acid
1 Mole of Nitrobenzene
1 Mole Water
78.1 g
63.0 g
123.1 g
18.0 g
Molar wt
In a lab experiment reacting 1 mole benzene with
1 mole nitric acid only 90 g of Nitrobenzene was
produced
Actual yield
90 g
x 100 73 Yield
Yield x 100
Theoretical yield
123.1 g
91
Calculations with real amounts
C6H6 HNO3 ? C6H5NO2 H2O
78.1 g
63.0 g
123.1 g
18.0 g
Molar wt
Experiment 19.5 g of benzene is reacted with
15.8 g of nitric acid to form 21.2 g
of nitrobenzene - Calculate yield
19.5 g benzene
Moles of benzene
0.25 moles of benzene
78.1 g/mol benzene
15.8 g nitric acid
Moles of nitric acid
0.25 moles of nitric acid
63.0 g/mol nitric acid
From the equation we know 1 mol of nitrobenzene
produced for each mole of the reactants we
theoretically expect 0.25 moles of nitrobenzene
produced theoretical mass(nitrobenzene) 0.25
mol x 123.1g/mol nitrobenzene 30.8 g
Actual yield
21.2 g
x 100 69 Yield
Yield x 100
Theoretical yield
30.8 g
92
Limiting Reagent Concept

• Ham sandwich analogy
• Suppose you have 20 slices of ham
• and 36 slices of bread
• How many ham sandwiches (1 slice of ham and
  2 slices of bread) can you make?

18 sandwiches can be made before you run out of
bread
The bread is the limiting reagent
2 slices of ham will be leftover this is the
excess reagent
93
Limiting Reagent
C6H6 HNO3 ? C6H5NO2 H2O
78.1 g
63.0 g
123.1 g
18.0 g
Molar wt
Experiment 39.1 g of benzene is reacted with 75
g of nitric acid calculate
theoretical yield of nitrobenzene
39.1 g benzene
Moles of benzene
0.50 moles of benzene
78.1 g/mol benzene
Limiting Reagent
75 g nitric acid
Moles of nitric acid
1.2 moles of nitric acid
63.0 g/mol nitric acid
Since benzene is the limiting reagent and 1 mol
of nitrobenzene produced for each mole benzene we
theoretically expect 0.50 moles of nitrobenzene
to be produced theoretical mass(nitrobenzene)
0.50 mol x 123.1g/mol nitrobenzene 61.6 g
94
Calculations based on Chemical Equations

• Chemical equations can be used to calculate the
  amount of substances involved in chemical
  reactions

CH4 2O2 ? CO2 2H2O
Molecular level
1 Molecule Methane
2 Molecules Oxygen
1 Molecule Carbon Dioxide
2 Molecules Water
Macroscopic level
1 Mole of Methane
2 Moles Oxygen
1 Mole of Carbon Dioxide
2 Moles Water
95
Calculations based on Chemical Equations
CH4 2O2 ? CO2 2H2O
1 Mole of Methane
2 Moles Oxygen
1 Mole of Carbon Dioxide
2 Moles Water
Molar wt of Methane is 16.0 g
Molar wt of Oxygen molecule is 32.0 g
Molar wt of Carbon Dioxide is 44.0 g
Molar wt of Water is 18.0 g
Mass
16.0 g
64.0 g
44.0 g
36.0 g
80 grams Reactants
80 grams Products
Law of Conservation of Mass
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Stoichiometry
For the balanced equation a A b B ? c C d D
Grams of A
Use Molar mass as Conversion Factor
Moles of A
Use Coefficients in balanced equation to find
Mole ratios
Moles of B
Use Molar mass as Conversion Factor
Grams of B
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Calculations Examples
Usually dont have exactly a mole of reactants
How many Grams of water (steam) are produced
from 81.55 g of methane?
CH4 2O2 ? CO2 2H2O
81.55 g CH4
Use Molar mass as Conversion Factor
5.10 Moles CH4
Use Coefficients in balanced equation to find
Mole ratios
10.2 Moles H2O
Use Molar mass as Conversion Factor
184 g water
98
Calculations Examples
How many Grams of NaOH are needed to react
with 25.0 g of Cl2
2NaOH(aq) Cl2 ? NaOCl(aq) 2H2O
25.0 g of Cl2
Use Molar mass as Conversion Factor
0.353 Moles Cl2
Use Coefficients in balanced equation to find
Mole ratios
0.706 Moles NaOH
Use Molar mass as Conversion Factor
28.24 g NaOH
99
Percent Yields from Chemical Reactions
Not all reactions go to Completion
C6H6 HNO3 ? C6H5NO2 H2O
1 Mole of Benzene
1 Mole Nitric Acid
1 Mole of Nitrobenzene
1 Mole Water
78.1 g
63.0 g
123.1 g
18.0 g
Molar wt
In a lab experiment reacting 1 mole benzene with
1 mole nitric acid only 90 g of Nitrobenzene was
produced
Actual yield
90 g
73 Yield
x 100
Yield x 100
Theoretical yield
123.1 g
100
Calculations with real amounts
C6H6 HNO3 ? C6H5NO2 H2O
78.1 g
63.0 g
123.1 g
18.0 g
Molar wt
Experiment 19.5 g of benzene is reacted with
15.8 g of nitric acid to form 21.2 g
of nitrobenzene - Calculate yield
19.5 g benzene
Moles of benzene
0.25 moles of benzene
78.1 g/mol benzene
15.8 g nitric acid
Moles of nitric acid
0.25 moles of nitric acid
63.0 g/mol nitric acid
From the equation we know 1 mol of nitrobenzene
produced for each mole of the reactants we
theoretically expect 0.25 moles of nitrobenzene
produced theoretical mass(nitrobenzene) 0.25
mol x 123.1g/mol nitrobenzene 30.8 g
Actual yield
21.2 g
x 100 69 Yield
Yield x 100
Theoretical yield
30.8 g
101
Limiting Reagent Concept

• Ham sandwich analogy
• Suppose you have 20 slices of ham
• and 36 slices of bread
• How many ham sandwiches (1 slice of ham and
  2 slices of bread) can you make?

18 sandwiches can be made before you run out of
bread
The bread is the limiting reagent
2 slices of ham will be leftover this is the
excess reagent
102
Limiting Reactant Example
Lithium Oxide is used to remove water from air
supply of space shuttle
Li2O(s) H2O(g) ? 2 LiOH(s)
If 80.0 kg of water is to be removed and 65 kg of
Li2O is available Which reactant is limiting?
8.00 x 104 g H2O
4.44 x 103 moles H2O
Excess Reactant
18.0 g/mol
6.5 x 104 g Li2O
2.2 x 103 moles Li2O
Limiting Reactant
29.88 g/mol
From balanced equation 11 ratio of moles Li2O
to H2O
So which will we run out of first?
103
Limiting Reactant Example
Lithium Oxide is used to remove water from air
supply of space shuttle
Li2O(s) H2O(g) ? 2 LiOH(s)
If 80.0 kg of water is to be removed and 65 kg of
Li2O is available How much excess reactant is
remaining?
4.44 x 103 moles H2O
Excess Reactant
2.2 x 103 moles Li2O
Limiting Reactant
From balanced equation 11 ratio of moles Li2O
to H2O 2.2 x 103 moles Li2O will react with
2.2 x 103 moles H2O 4.44 x 103 moles H2O
available - 2.2 x 103 moles H2O Leaving
2.2 x 103 moles of water unreacted
2.2 x 103 moles H2O x 18.0 g/mol 4.0 x 105 g
water or 40. kg
104
Limiting Reagent
C6H6 HNO3 ? C6H5NO2 H2O
78.1 g
63.0 g
123.1 g
18.0 g
Molar wt
Experiment 39.1 g of benzene is reacted with 75
g of nitric acid calculate
theoretical yield of nitrobenzene
39.1 g benzene
Moles of benzene
0.50 moles of benzene
78.1 g/mol benzene
Limiting Reagent
75 g nitric acid
Moles of nitric acid
1.2 moles of nitric acid
63.0 g/mol nitric acid
Since benzene is the limiting reagent and 1 mol
of nitrobenzene produced for each mole benzene we
theoretically expect 0.50 moles of nitrobenzene
to be produced theoretical mass(nitrobenzene)
0.50 mol x 123.1g/mol nitrobenzene 61.6 g