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Franklin Ow

Franklin Ow

Stoichiometry I

Slide Duration:

Table of Contents

Section 1: Basic Concepts & Measurement of Chemistry
Basic Concepts of Chemistry

16m 26s

Intro
0:00
Lesson Overview
0:07
Introduction
0:56
What is Chemistry?
0:57
What is Matter?
1:16
Solids
1:43
General Characteristics
1:44
Particulate-level Drawing of Solids
2:34
Liquids
3:39
General Characteristics of Liquids
3:40
Particulate-level Drawing of Liquids
3:55
Gases
4:23
General Characteristics of Gases
4:24
Particulate-level Drawing Gases
5:05
Classification of Matter
5:27
Classification of Matter
5:26
Pure Substances
5:54
Pure Substances
5:55
Mixtures
7:06
Definition of Mixtures
7:07
Homogeneous Mixtures
7:11
Heterogeneous Mixtures
7:52
Physical and Chemical Changes/Properties
8:18
Physical Changes Retain Chemical Composition
8:19
Chemical Changes Alter Chemical Composition
9:32
Physical and Chemical Changes/Properties, cont'd
10:55
Physical Properties
10:56
Chemical Properties
11:42
Sample Problem 1: Chemical & Physical Change
12:22
Sample Problem 2: Element, Compound, or Mixture?
13:52
Sample Problem 3: Classify Each of the Following Properties as chemical or Physical
15:03
Tools in Quantitative Chemistry

29m 22s

Intro
0:00
Lesson Overview
0:07
Units of Measurement
1:23
The International System of Units (SI): Mass, Length, and Volume
1:39
Percent Error
2:17
Percent Error
2:18
Example: Calculate the Percent Error
2:56
Standard Deviation
3:48
Standard Deviation Formula
3:49
Standard Deviation cont'd
4:42
Example: Calculate Your Standard Deviation
4:43
Precisions vs. Accuracy
6:25
Precision
6:26
Accuracy
7:01
Significant Figures and Uncertainty
7:50
Consider the Following (2) Rulers
7:51
Consider the Following Graduated Cylinder
11:30
Identifying Significant Figures
12:43
The Rules of Sig Figs Overview
12:44
The Rules for Sig Figs: All Nonzero Digits Are Significant
13:21
The Rules for Sig Figs: A Zero is Significant When It is In-Between Nonzero Digits
13:28
The Rules for Sig Figs: A Zero is Significant When at the End of a Decimal Number
14:02
The Rules for Sig Figs: A Zero is not significant When Starting a Decimal Number
14:27
Using Sig Figs in Calculations
15:03
Using Sig Figs for Multiplication and Division
15:04
Using Sig Figs for Addition and Subtraction
15:48
Using Sig Figs for Mixed Operations
16:11
Dimensional Analysis
16:20
Dimensional Analysis Overview
16:21
General Format for Dimensional Analysis
16:39
Example: How Many Miles are in 17 Laps?
17:17
Example: How Many Grams are in 1.22 Pounds?
18:40
Dimensional Analysis cont'd
19:43
Example: How Much is Spent on Diapers in One Week?
19:44
Dimensional Analysis cont'd
21:03
SI Prefixes
21:04
Dimensional Analysis cont'd
22:03
500 mg → ? kg
22:04
34.1 cm → ? um
24:03
Summary
25:11
Sample Problem 1: Dimensional Analysis
26:09
Section 2: Atoms, Molecules, and Ions
Atoms, Molecules, and Ions

52m 18s

Intro
0:00
Lesson Overview
0:08
Introduction to Atomic Structure
1:03
Introduction to Atomic Structure
1:04
Plum Pudding Model
1:26
Introduction to Atomic Structure Cont'd
2:07
John Dalton's Atomic Theory: Number 1
2:22
John Dalton's Atomic Theory: Number 2
2:50
John Dalton's Atomic Theory: Number 3
3:07
John Dalton's Atomic Theory: Number 4
3:30
John Dalton's Atomic Theory: Number 5
3:58
Introduction to Atomic Structure Cont'd
5:21
Ernest Rutherford's Gold Foil Experiment
5:22
Introduction to Atomic Structure Cont'd
7:42
Implications of the Gold Foil Experiment
7:43
Relative Masses and Charges
8:18
Isotopes
9:02
Isotopes
9:03
Introduction to The Periodic Table
12:17
The Periodic Table of the Elements
12:18
Periodic Table, cont'd
13:56
Metals
13:57
Nonmetals
14:25
Semimetals
14:51
Periodic Table, cont'd
15:57
Group I: The Alkali Metals
15:58
Group II: The Alkali Earth Metals
16:25
Group VII: The Halogens
16:40
Group VIII: The Noble Gases
17:08
Ionic Compounds: Formulas, Names, Props.
17:35
Common Polyatomic Ions
17:36
Predicting Ionic Charge for Main Group Elements
18:52
Ionic Compounds: Formulas, Names, Props.
20:36
Naming Ionic Compounds: Rule 1
20:51
Naming Ionic Compounds: Rule 2
21:22
Naming Ionic Compounds: Rule 3
21:50
Naming Ionic Compounds: Rule 4
22:22
Ionic Compounds: Formulas, Names, Props.
22:50
Naming Ionic Compounds Example: Al₂O₃
22:51
Naming Ionic Compounds Example: FeCl₃
23:21
Naming Ionic Compounds Example: CuI₂ 3H₂O
24:00
Naming Ionic Compounds Example: Barium Phosphide
24:40
Naming Ionic Compounds Example: Ammonium Phosphate
25:55
Molecular Compounds: Formulas and Names
26:42
Molecular Compounds: Formulas and Names
26:43
The Mole
28:10
The Mole is 'A Chemist's Dozen'
28:11
It is a Central Unit, Connecting the Following Quantities
30:01
The Mole, cont'd
32:07
Atomic Masses
32:08
Example: How Many Moles are in 25.7 Grams of Sodium?
32:28
Example: How Many Atoms are in 1.2 Moles of Carbon?
33:17
The Mole, cont'd
34:25
Example: What is the Molar Mass of Carbon Dioxide?
34:26
Example: How Many Grams are in 1.2 Moles of Carbon Dioxide?
25:46
Percentage Composition
36:43
Example: How Many Grams of Carbon Contained in 65.1 Grams of Carbon Dioxide?
36:44
Empirical and Molecular Formulas
39:19
Empirical Formulas
39:20
Empirical Formula & Elemental Analysis
40:21
Empirical and Molecular Formulas, cont'd
41:24
Example: Determine Both the Empirical and Molecular Formulas - Step 1
41:25
Example: Determine Both the Empirical and Molecular Formulas - Step 2
43:18
Summary
46:22
Sample Problem 1: Determine the Empirical Formula of Lithium Fluoride
47:10
Sample Problem 2: How Many Atoms of Carbon are Present in 2.67 kg of C₆H₆?
49:21
Section 3: Chemical Reactions
Chemical Reactions

43m 24s

Intro
0:00
Lesson Overview
0:06
The Law of Conservation of Mass and Balancing Chemical Reactions
1:49
The Law of Conservation of Mass
1:50
Balancing Chemical Reactions
2:50
Balancing Chemical Reactions Cont'd
3:40
Balance: N₂ + H₂ → NH₃
3:41
Balance: CH₄ + O₂ → CO₂ + H₂O
7:20
Balancing Chemical Reactions Cont'd
9:49
Balance: C₂H₆ + O₂ → CO₂ + H₂O
9:50
Intro to Chemical Equilibrium
15:32
When an Ionic Compound Full Dissociates
15:33
When an Ionic Compound Incompletely Dissociates
16:14
Dynamic Equilibrium
17:12
Electrolytes and Nonelectrolytes
18:03
Electrolytes
18:04
Strong Electrolytes and Weak Electrolytes
18:55
Nonelectrolytes
19:23
Predicting the Product(s) of an Aqueous Reaction
20:02
Single-replacement
20:03
Example: Li (s) + CuCl₂ (aq) → 2 LiCl (aq) + Cu (s)
21:03
Example: Cu (s) + LiCl (aq) → NR
21:23
Example: Zn (s) + 2HCl (aq) → ZnCl₂ (aq) + H₂ (g)
22:32
Predicting the Product(s) of an Aqueous Reaction
23:37
Double-replacement
23:38
Net-ionic Equation
25:29
Predicting the Product(s) of an Aqueous Reaction
26:12
Solubility Rules for Ionic Compounds
26:13
Predicting the Product(s) of an Aqueous Reaction
28:10
Neutralization Reactions
28:11
Example: HCl (aq) + NaOH (aq) → ?
28:37
Example: H₂SO₄ (aq) + KOH (aq) → ?
29:25
Predicting the Product(s) of an Aqueous Reaction
30:20
Certain Aqueous Reactions can Produce Unstable Compounds
30:21
Example 1
30:52
Example 2
32:16
Example 3
32:54
Summary
33:54
Sample Problem 1
34:55
ZnCO₃ (aq) + H₂SO₄ (aq) → ?
35:09
NH₄Br (aq) + Pb(C₂H₃O₂)₂ (aq) → ?
36:02
KNO₃ (aq) + CuCl₂ (aq) → ?
37:07
Li₂SO₄ (aq) + AgNO₃ (aq) → ?
37:52
Sample Problem 2
39:09
Question 1
39:10
Question 2
40:36
Question 3
41:47
Chemical Reactions II

55m 40s

Intro
0:00
Lesson Overview
0:10
Arrhenius Definition
1:15
Arrhenius Acids
1:16
Arrhenius Bases
3:20
The Bronsted-Lowry Definition
4:48
Acids Dissolve In Water and Donate a Proton to Water: Example 1
4:49
Acids Dissolve In Water and Donate a Proton to Water: Example 2
6:54
Monoprotic Acids & Polyprotic Acids
7:58
Strong Acids
11:30
Bases Dissolve In Water and Accept a Proton From Water
12:41
Strong Bases
16:36
The Autoionization of Water
17:42
Amphiprotic
17:43
Water Reacts With Itself
18:24
Oxides of Metals and Nonmetals
20:08
Oxides of Metals and Nonmetals Overview
20:09
Oxides of Nonmetals: Acidic Oxides
21:23
Oxides of Metals: Basic Oxides
24:08
Oxidation-Reduction (Redox) Reactions
25:34
Redox Reaction Overview
25:35
Oxidizing and Reducing Agents
27:02
Redox Reaction: Transfer of Electrons
27:54
Oxidation-Reduction Reactions Cont'd
29:55
Oxidation Number Overview
29:56
Oxidation Number of Homonuclear Species
31:17
Oxidation Number of Monatomic Ions
32:58
Oxidation Number of Fluorine
33:27
Oxidation Number of Oxygen
34:00
Oxidation Number of Chlorine, Bromine, and Iodine
35:07
Oxidation Number of Hydrogen
35:30
Net Sum of All Oxidation Numbers In a Compound
36:21
Oxidation-Reduction Reactions Cont'd
38:19
Let's Practice Assigning Oxidation Number
38:20
Now Let's Apply This to a Chemical Reaction
41:07
Summary
44:19
Sample Problems
45:29
Sample Problem 1
45:30
Sample Problem 2: Determine the Oxidizing and Reducing Agents
48:48
Sample Problem 3: Determine the Oxidizing and Reducing Agents
50:43
Section 4: Stoichiometry
Stoichiometry I

42m 10s

Intro
0:00
Lesson Overview
0:23
Mole to Mole Ratios
1:32
Example 1: In 1 Mole of H₂O, How Many Moles Are There of Each Element?
1:53
Example 2: In 2.6 Moles of Water, How Many Moles Are There of Each Element?
2:24
Mole to Mole Ratios Cont'd
5:13
Balanced Chemical Reaction
5:14
Mole to Mole Ratios Cont'd
7:25
Example 3: How Many Moles of Ammonia Can Form If you Have 3.1 Moles of H₂?
7:26
Example 4: How Many Moles of Hydrogen Gas Are Required to React With 6.4 Moles of Nitrogen Gas?
9:08
Mass to mass Conversion
11:06
Mass to mass Conversion
11:07
Example 5: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂?
12:37
Example 6: How Many Grams of Hydrogen Gas Are Required to React With 6.4 Grams of Nitrogen Gas?
15:34
Example 7: How Man Milligrams of Ammonia Can Form If You Have 1.2 kg of H₂?
17:29
Limiting Reactants, Percent Yields
20:42
Limiting Reactants, Percent Yields
20:43
Example 8: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂ and 3.1 Grams of N₂
22:25
Percent Yield
25:30
Example 9: How Many Grams of The Excess Reactant Remains?
26:37
Summary
29:34
Sample Problem 1: How Many Grams of Carbon Are In 2.2 Kilograms of Carbon Dioxide?
30:47
Sample Problem 2: How Many Milligrams of Carbon Dioxide Can Form From 23.1 Kg of CH₄(g)?
33:06
Sample Problem 3: Part 1
36:10
Sample Problem 3: Part 2 - What Amount Of The Excess Reactant Will Remain?
40:53
Stoichiometry II

42m 38s

Intro
0:00
Lesson Overview
0:10
Molarity
1:14
Solute and Solvent
1:15
Molarity
2:01
Molarity Cont'd
2:59
Example 1: How Many Grams of KBr are Needed to Make 350 mL of a 0.67 M KBr Solution?
3:00
Example 2: How Many Moles of KBr are in 350 mL of a 0.67 M KBr Solution?
5:44
Example 3: What Volume of a 0.67 M KBr Solution Contains 250 mg of KBr?
7:46
Dilutions
10:01
Dilution: M₁V₂=M₁V₂
10:02
Example 5: Explain How to Make 250 mL of a 0.67 M KBr Solution Starting From a 1.2M Stock Solution
12:04
Stoichiometry and Double-Displacement Precipitation Reactions
14:41
Example 6: How Many grams of PbCl₂ Can Form From 250 mL of 0.32 M NaCl?
15:38
Stoichiometry and Double-Displacement Precipitation Reactions
18:05
Example 7: How Many grams of PbCl₂ Can Form When 250 mL of 0.32 M NaCl and 150 mL of 0.45 Pb(NO₃)₂ Mix?
18:06
Stoichiometry and Neutralization Reactions
21:01
Example 8: How Many Grams of NaOh are Required to Neutralize 4.5 Grams of HCl?
21:02
Stoichiometry and Neutralization Reactions
23:03
Example 9: How Many mL of 0.45 M NaOH are Required to Neutralize 250 mL of 0.89 M HCl?
23:04
Stoichiometry and Acid-Base Standardization
25:28
Introduction to Titration & Standardization
25:30
Acid-Base Titration
26:12
The Analyte & Titrant
26:24
The Experimental Setup
26:49
The Experimental Setup
26:50
Stoichiometry and Acid-Base Standardization
28:38
Example 9: Determine the Concentration of the Analyte
28:39
Summary
32:46
Sample Problem 1: Stoichiometry & Neutralization
35:24
Sample Problem 2: Stoichiometry
37:50
Section 5: Thermochemistry
Energy & Chemical Reactions

55m 28s

Intro
0:00
Lesson Overview
0:14
Introduction
1:22
Recall: Chemistry
1:23
Energy Can Be Expressed In Different Units
1:57
The First Law of Thermodynamics
2:43
Internal Energy
2:44
The First Law of Thermodynamics Cont'd
6:14
Ways to Transfer Internal Energy
6:15
Work Energy
8:13
Heat Energy
8:34
∆U = q + w
8:44
Calculating ∆U, Q, and W
8:58
Changes In Both Volume and Temperature of a System
8:59
Calculating ∆U, Q, and W Cont'd
11:01
The Work Equation
11:02
Example 1: Calculate ∆U For The Burning Fuel
11:45
Calculating ∆U, Q, and W Cont'd
14:09
The Heat Equation
14:10
Calculating ∆U, Q, and W Cont'd
16:03
Example 2: Calculate The Final Temperature
16:04
Constant-Volume Calorimetry
18:05
Bomb Calorimeter
18:06
The Effect of Constant Volume On The Equation For Internal Energy
22:11
Example 3: Calculate ∆U
23:12
Constant-Pressure Conditions
26:05
Constant-Pressure Conditions
26:06
Calculating Enthalpy: Phase Changes
27:29
Melting, Vaporization, and Sublimation
27:30
Freezing, Condensation and Deposition
28:25
Enthalpy Values For Phase Changes
28:40
Example 4: How Much Energy In The Form of heat is Required to Melt 1.36 Grams of Ice?
29:40
Calculating Enthalpy: Heats of Reaction
31:22
Example 5: Calculate The Heat In kJ Associated With The Complete Reaction of 155 g NH₃
31:23
Using Standard Enthalpies of Formation
33:53
Standard Enthalpies of Formation
33:54
Using Standard Enthalpies of Formation
36:12
Example 6: Calculate The Standard Enthalpies of Formation For The Following Reaction
36:13
Enthalpy From a Series of Reactions
39:58
Hess's Law
39:59
Coffee-Cup Calorimetry
42:43
Coffee-Cup Calorimetry
42:44
Example 7: Calculate ∆H° of Reaction
45:10
Summary
47:12
Sample Problem 1
48:58
Sample Problem 2
51:24
Section 6: Quantum Theory of Atoms
Structure of Atoms

42m 33s

Intro
0:00
Lesson Overview
0:07
Introduction
1:01
Rutherford's Gold Foil Experiment
1:02
Electromagnetic Radiation
2:31
Radiation
2:32
Three Parameters: Energy, Frequency, and Wavelength
2:52
Electromagnetic Radiation
5:18
The Electromagnetic Spectrum
5:19
Atomic Spectroscopy and The Bohr Model
7:46
Wavelengths of Light
7:47
Atomic Spectroscopy Cont'd
9:45
The Bohr Model
9:46
Atomic Spectroscopy Cont'd
12:21
The Balmer Series
12:22
Rydberg Equation For Predicting The Wavelengths of Light
13:04
The Wave Nature of Matter
15:11
The Wave Nature of Matter
15:12
The Wave Nature of Matter
19:10
New School of Thought
19:11
Einstein: Energy
19:49
Hertz and Planck: Photoelectric Effect
20:16
de Broglie: Wavelength of a Moving Particle
21:14
Quantum Mechanics and The Atom
22:15
Heisenberg: Uncertainty Principle
22:16
Schrodinger: Wavefunctions
23:08
Quantum Mechanics and The Atom
24:02
Principle Quantum Number
24:03
Angular Momentum Quantum Number
25:06
Magnetic Quantum Number
26:27
Spin Quantum Number
28:42
The Shapes of Atomic Orbitals
29:15
Radial Wave Function
29:16
Probability Distribution Function
32:08
The Shapes of Atomic Orbitals
34:02
3-Dimensional Space of Wavefunctions
34:03
Summary
35:57
Sample Problem 1
37:07
Sample Problem 2
40:23
Section 7: Electron Configurations and Periodicity
Periodic Trends

38m 50s

Intro
0:00
Lesson Overview
0:09
Introduction
0:36
Electron Configuration of Atoms
1:33
Electron Configuration & Atom's Electrons
1:34
Electron Configuration Format
1:56
Electron Configuration of Atoms Cont'd
3:01
Aufbau Principle
3:02
Electron Configuration of Atoms Cont'd
6:53
Electron Configuration Format 1: Li, O, and Cl
6:56
Electron Configuration Format 2: Li, O, and Cl
9:11
Electron Configuration of Atoms Cont'd
12:48
Orbital Box Diagrams
12:49
Pauli Exclusion Principle
13:11
Hund's Rule
13:36
Electron Configuration of Atoms Cont'd
17:35
Exceptions to The Aufbau Principle: Cr
17:36
Exceptions to The Aufbau Principle: Cu
18:15
Electron Configuration of Atoms Cont'd
20:22
Electron Configuration of Monatomic Ions: Al
20:23
Electron Configuration of Monatomic Ions: Al³⁺
20:46
Electron Configuration of Monatomic Ions: Cl
21:57
Electron Configuration of Monatomic Ions: Cl¹⁻
22:09
Electron Configuration Cont'd
24:31
Paramagnetism
24:32
Diamagnetism
25:00
Atomic Radii
26:08
Atomic Radii
26:09
In a Column of the Periodic Table
26:25
In a Row of the Periodic Table
26:46
Ionic Radii
27:30
Ionic Radii
27:31
Anions
27:42
Cations
27:57
Isoelectronic Species
28:12
Ionization Energy
29:00
Ionization Energy
29:01
Electron Affinity
31:37
Electron Affinity
31:37
Summary
33:43
Sample Problem 1: Ground State Configuration and Orbital Box Diagram
34:21
Fe
34:48
P
35:32
Sample Problem 2
36:38
Which Has The Larger Ionization Energy: Na or Li?
36:39
Which Has The Larger Atomic Size: O or N ?
37:23
Which Has The Larger Atomic Size: O²⁻ or N³⁻ ?
38:00
Section 8: Molecular Geometry & Bonding Theory
Bonding & Molecular Structure

52m 39s

Intro
0:00
Lesson Overview
0:08
Introduction
1:10
Types of Chemical Bonds
1:53
Ionic Bond
1:54
Molecular Bond
2:42
Electronegativity and Bond Polarity
3:26
Electronegativity (EN)
3:27
Periodic Trend
4:36
Electronegativity and Bond Polarity Cont'd
6:04
Bond Polarity: Polar Covalent Bond
6:05
Bond Polarity: Nonpolar Covalent Bond
8:53
Lewis Electron Dot Structure of Atoms
9:48
Lewis Electron Dot Structure of Atoms
9:49
Lewis Structures of Polyatomic Species
12:51
Single Bonds
12:52
Double Bonds
13:28
Nonbonding Electrons
13:59
Lewis Structures of Polyatomic Species Cont'd
14:45
Drawing Lewis Structures: Step 1
14:48
Drawing Lewis Structures: Step 2
15:16
Drawing Lewis Structures: Step 3
15:52
Drawing Lewis Structures: Step 4
17:31
Drawing Lewis Structures: Step 5
19:08
Drawing Lewis Structure Example: Carbonate
19:33
Resonance and Formal Charges (FC)
24:06
Resonance Structures
24:07
Formal Charge
25:20
Resonance and Formal Charges Cont'd
27:46
More On Formal Charge
27:47
Resonance and Formal Charges Cont'd
28:21
Good Resonance Structures
28:22
VSEPR Theory
31:08
VSEPR Theory Continue
31:09
VSEPR Theory Cont'd
32:53
VSEPR Geometries
32:54
Steric Number
33:04
Basic Geometry
33:50
Molecular Geometry
35:50
Molecular Polarity
37:51
Steps In Determining Molecular Polarity
37:52
Example 1: Polar
38:47
Example 2: Nonpolar
39:10
Example 3: Polar
39:36
Example 4: Polar
40:08
Bond Properties: Order, Length, and Energy
40:38
Bond Order
40:39
Bond Length
41:21
Bond Energy
41:55
Summary
43:09
Sample Problem 1
43:42
XeO₃
44:03
I₃⁻
47:02
SF₅
49:16
Advanced Bonding Theories

1h 11m 41s

Intro
0:00
Lesson Overview
0:09
Introduction
0:38
Valence Bond Theory
3:07
Valence Bond Theory
3:08
spᶟ Hybridized Carbon Atom
4:19
Valence Bond Theory Cont'd
6:24
spᶟ Hybridized
6:25
Hybrid Orbitals For Water
7:26
Valence Bond Theory Cont'd (spᶟ)
11:53
Example 1: NH₃
11:54
Valence Bond Theory Cont'd (sp²)
14:48
sp² Hybridization
14:49
Example 2: BF₃
16:44
Valence Bond Theory Cont'd (sp)
22:44
sp Hybridization
22:46
Example 3: HCN
23:38
Valence Bond Theory Cont'd (sp³d and sp³d²)
27:36
Valence Bond Theory: sp³d and sp³d²
27:37
Molecular Orbital Theory
29:10
Valence Bond Theory Doesn't Always Account For a Molecule's Magnetic Behavior
29:11
Molecular Orbital Theory Cont'd
30:37
Molecular Orbital Theory
30:38
Wavefunctions
31:04
How s-orbitals Can Interact
32:23
Bonding Nature of p-orbitals: Head-on
35:34
Bonding Nature of p-orbitals: Parallel
39:04
Interaction Between s and p-orbital
40:45
Molecular Orbital Diagram For Homonuclear Diatomics: H₂
42:21
Molecular Orbital Diagram For Homonuclear Diatomics: He₂
45:23
Molecular Orbital Diagram For Homonuclear Diatomic: Li₂
46:39
Molecular Orbital Diagram For Homonuclear Diatomic: Li₂⁺
47:42
Molecular Orbital Diagram For Homonuclear Diatomic: B₂
48:57
Molecular Orbital Diagram For Homonuclear Diatomic: N₂
54:04
Molecular Orbital Diagram: Molecular Oxygen
55:57
Molecular Orbital Diagram For Heteronuclear Diatomics: Hydrochloric Acid
1:02:16
Sample Problem 1: Determine the Atomic Hybridization
1:07:20
XeO₃
1:07:21
SF₆
1:07:49
I₃⁻
1:08:20
Sample Problem 2
1:09:04
Section 9: Gases, Solids, & Liquids
Gases

35m 6s

Intro
0:00
Lesson Overview
0:07
The Kinetic Molecular Theory of Gases
1:23
The Kinetic Molecular Theory of Gases
1:24
Parameters To Characterize Gases
3:35
Parameters To Characterize Gases: Pressure
3:37
Interpreting Pressure On a Particulate Level
4:43
Parameters Cont'd
6:08
Units For Expressing Pressure: Psi, Pascal
6:19
Units For Expressing Pressure: mm Hg
6:42
Units For Expressing Pressure: atm
6:58
Units For Expressing Pressure: torr
7:24
Parameters Cont'd
8:09
Parameters To Characterize Gases: Volume
8:10
Common Units of Volume
9:00
Parameters Cont'd
9:11
Parameters To Characterize Gases: Temperature
9:12
Particulate Level
9:36
Parameters To Characterize Gases: Moles
10:24
The Simple Gas Laws
10:43
Gas Laws Are Only Valid For…
10:44
Charles' Law
11:24
The Simple Gas Laws
13:13
Boyle's Law
13:14
The Simple Gas Laws
15:28
Gay-Lussac's Law
15:29
The Simple Gas Laws
17:11
Avogadro's Law
17:12
The Ideal Gas Law
18:43
The Ideal Gas Law: PV = nRT
18:44
Applications of the Ideal Gas Law
20:12
Standard Temperature and Pressure for Gases
20:13
Applications of the Ideal Gas Law
21:43
Ideal Gas Law & Gas Density
21:44
Gas Pressures and Partial Pressures
23:18
Dalton's Law of Partial Pressures
23:19
Gas Stoichiometry
24:15
Stoichiometry Problems Involving Gases
24:16
Using The Ideal Gas Law to Get to Moles
25:16
Using Molar Volume to Get to Moles
25:39
Gas Stoichiometry Cont'd
26:03
Example 1: How Many Liters of O₂ at STP are Needed to Form 10.5 g of Water Vapor?
26:04
Summary
28:33
Sample Problem 1: Calculate the Molar Mass of the Gas
29:28
Sample Problem 2: What Mass of Ag₂O is Required to Form 3888 mL of O₂ Gas When Measured at 734 mm Hg and 25°C?
31:59
Intermolecular Forces & Liquids

33m 47s

Intro
0:00
Lesson Overview
0:10
Introduction
0:46
Intermolecular Forces (IMF)
0:47
Intermolecular Forces of Polar Molecules
1:32
Ion-dipole Forces
1:33
Example: Salt Dissolved in Water
1:50
Coulomb's Law & the Force of Attraction Between Ions and/or Dipoles
3:06
IMF of Polar Molecules cont'd
4:36
Enthalpy of Solvation or Enthalpy of Hydration
4:37
IMF of Polar Molecules cont'd
6:01
Dipole-dipole Forces
6:02
IMF of Polar Molecules cont'd
7:22
Hydrogen Bonding
7:23
Example: Hydrogen Bonding of Water
8:06
IMF of Nonpolar Molecules
9:37
Dipole-induced Dipole Attraction
9:38
IMF of Nonpolar Molecules cont'd
11:34
Induced Dipole Attraction, London Dispersion Forces, or Vand der Waals Forces
11:35
Polarizability
13:46
IMF of Nonpolar Molecules cont'd
14:26
Intermolecular Forces (IMF) and Polarizability
14:31
Properties of Liquids
16:48
Standard Molar Enthalpy of Vaporization
16:49
Trends in Boiling Points of Representative Liquids: H₂O vs. H₂S
17:43
Properties of Liquids cont'd
18:36
Aliphatic Hydrocarbons
18:37
Branched Hydrocarbons
20:52
Properties of Liquids cont'd
22:10
Vapor Pressure
22:11
The Clausius-Clapeyron Equation
24:30
Properties of Liquids cont'd
25:52
Boiling Point
25:53
Properties of Liquids cont'd
27:07
Surface Tension
27:08
Viscosity
28:06
Summary
29:04
Sample Problem 1: Determine Which of the Following Liquids Will Have the Lower Vapor Pressure
30:21
Sample Problem 2: Determine Which of the Following Liquids Will Have the Largest Standard Molar Enthalpy of Vaporization
31:37
The Chemistry of Solids

25m 13s

Intro
0:00
Lesson Overview
0:07
Introduction
0:46
General Characteristics
0:47
Particulate-level Drawing
1:09
The Basic Structure of Solids: Crystal Lattices
1:37
The Unit Cell Defined
1:38
Primitive Cubic
2:50
Crystal Lattices cont'd
3:58
Body-centered Cubic
3:59
Face-centered Cubic
5:02
Lattice Enthalpy and Trends
6:27
Introduction to Lattice Enthalpy
6:28
Equation to Calculate Lattice Enthalpy
7:21
Different Types of Crystalline Solids
9:35
Molecular Solids
9:36
Network Solids
10:25
Phase Changes Involving Solids
11:03
Melting & Thermodynamic Value
11:04
Freezing & Thermodynamic Value
11:49
Phase Changes cont'd
12:40
Sublimation & Thermodynamic Value
12:41
Depositions & Thermodynamic Value
13:13
Phase Diagrams
13:40
Introduction to Phase Diagrams
13:41
Phase Diagram of H₂O: Melting Point
14:12
Phase Diagram of H₂O: Normal Boiling Point
14:50
Phase Diagram of H₂O: Sublimation Point
15:02
Phase Diagram of H₂O: Point C ( Supercritical Point)
15:32
Phase Diagrams cont'd
16:31
Phase Diagram of Dry Ice
16:32
Summary
18:15
Sample Problem 1, Part A: Of the Group I Fluorides, Which Should Have the Highest Lattice Enthalpy?
19:01
Sample Problem 1, Part B: Of the Lithium Halides, Which Should Have the Lowest Lattice Enthalpy?
19:54
Sample Problem 2: How Many Joules of Energy is Required to Melt 546 mg of Ice at Standard Pressure?
20:55
Sample Problem 3: Phase Diagram of Helium
22:42
Section 10: Solutions, Rates of Reaction, & Equilibrium
Solutions & Their Behavior

38m 6s

Intro
0:00
Lesson Overview
0:10
Units of Concentration
1:40
Molarity
1:41
Molality
3:30
Weight Percent
4:26
ppm
5:16
Like Dissolves Like
6:28
Like Dissolves Like
6:29
Factors Affecting Solubility
9:35
The Effect of Pressure: Henry's Law
9:36
The Effect of Temperature on Gas Solubility
12:16
The Effect of Temperature on Solid Solubility
14:28
Colligative Properties
16:48
Colligative Properties
16:49
Changes in Vapor Pressure: Raoult's Law
17:19
Colligative Properties cont'd
19:53
Boiling Point Elevation and Freezing Point Depression
19:54
Colligative Properties cont'd
26:13
Definition of Osmosis
26:14
Osmotic Pressure Example
27:11
Summary
31:11
Sample Problem 1: Calculating Vapor Pressure
32:53
Sample Problem 2: Calculating Molality
36:29
Chemical Kinetics

37m 45s

Intro
0:00
Lesson Overview
0:06
Introduction
1:09
Chemical Kinetics and the Rate of a Reaction
1:10
Factors Influencing Rate
1:19
Introduction cont'd
2:27
How a Reaction Progresses Through Time
2:28
Rate of Change Equation
6:02
Rate Laws
7:06
Definition of Rate Laws
7:07
General Form of Rate Laws
7:37
Rate Laws cont'd
11:07
Rate Orders With Respect to Reactant and Concentration
11:08
Methods of Initial Rates
13:38
Methods of Initial Rates
13:39
Integrated Rate Laws
17:57
Integrated Rate Laws
17:58
Graphically Determine the Rate Constant k
18:52
Reaction Mechanisms
21:05
Step 1: Reversible
21:18
Step 2: Rate-limiting Step
21:44
Rate Law for the Reaction
23:28
Reaction Rates and Temperatures
26:16
Reaction Rates and Temperatures
26:17
The Arrhenius Equation
29:06
Catalysis
30:31
Catalyst
30:32
Summary
32:02
Sample Problem 1: Calculate the Rate Constant and the Time Required for the Reaction to be Completed
32:54
Sample Problem 2: Calculate the Energy of Activation and the Order of the Reaction
35:24
Principles of Chemical Equilibrium

34m 9s

Intro
0:00
Lesson Overview
0:08
Introduction
1:02
The Equilibrium Constant
3:08
The Equilibrium Constant
3:09
The Equilibrium Constant cont'd
5:50
The Equilibrium Concentration and Constant for Solutions
5:51
The Equilibrium Partial Pressure and Constant for Gases
7:01
Relationship of Kc and Kp
7:30
Heterogeneous Equilibria
8:23
Heterogeneous Equilibria
8:24
Manipulating K
9:57
First Way of Manipulating K
9:58
Second Way of Manipulating K
11:48
Manipulating K cont'd
12:31
Third Way of Manipulating K
12:32
The Reaction Quotient Q
14:42
The Reaction Quotient Q
14:43
Q > K
16:16
Q < K
16:30
Q = K
16:43
Le Chatlier's Principle
17:32
Restoring Equilibrium When It is Disturbed
17:33
Disturbing a Chemical System at Equilibrium
18:35
Problem-Solving with ICE Tables
19:05
Determining a Reaction's Equilibrium Constant With ICE Table
19:06
Problem-Solving with ICE Tables cont'd
21:03
Example 1: Calculate O₂(g) at Equilibrium
21:04
Problem-Solving with ICE Tables cont'd
22:53
Example 2: Calculate the Equilibrium Constant
22:54
Summary
25:24
Sample Problem 1: Calculate the Equilibrium Constant
27:59
Sample Problem 2: Calculate The Equilibrium Concentration
30:30
Section 11: Acids & Bases Chemistry
Acid-Base Chemistry

43m 44s

Intro
0:00
Lesson Overview
0:06
Introduction
0:55
Bronsted-Lowry Acid & Bronsted -Lowry Base
0:56
Water is an Amphiprotic Molecule
2:40
Water Reacting With Itself
2:58
Introduction cont'd
4:04
Strong Acids
4:05
Strong Bases
5:18
Introduction cont'd
6:16
Weak Acids and Bases
6:17
Quantifying Acid-Base Strength
7:35
The pH Scale
7:36
Quantifying Acid-Base Strength cont'd
9:55
The Acid-ionization Constant Ka and pKa
9:56
Quantifying Acid-Base Strength cont'd
12:13
Example: Calculate the pH of a 1.2M Solution of Acetic Acid
12:14
Quantifying Acid-Base Strength
15:06
Calculating the pH of Weak Base Solutions
15:07
Writing Out Acid-Base Equilibria
17:45
Writing Out Acid-Base Equilibria
17:46
Writing Out Acid-Base Equilibria cont'd
19:47
Consider the Following Equilibrium
19:48
Conjugate Base and Conjugate Acid
21:18
Salts Solutions
22:00
Salts That Produce Acidic Aqueous Solutions
22:01
Salts That Produce Basic Aqueous Solutions
23:15
Neutral Salt Solutions
24:05
Diprotic and Polyprotic Acids
24:44
Example: Calculate the pH of a 1.2 M Solution of H₂SO₃
24:43
Diprotic and Polyprotic Acids cont'd
27:18
Calculate the pH of a 1.2 M Solution of Na₂SO₃
27:19
Lewis Acids and Bases
29:13
Lewis Acids
29:14
Lewis Bases
30:10
Example: Lewis Acids and Bases
31:04
Molecular Structure and Acidity
32:03
The Effect of Charge
32:04
Within a Period/Row
33:07
Molecular Structure and Acidity cont'd
34:17
Within a Group/Column
34:18
Oxoacids
35:58
Molecular Structure and Acidity cont'd
37:54
Carboxylic Acids
37:55
Hydrated Metal Cations
39:23
Summary
40:39
Sample Problem 1: Calculate the pH of a 1.2 M Solution of NH₃
41:20
Sample Problem 2: Predict If The Following Slat Solutions are Acidic, Basic, or Neutral
42:37
Applications of Aqueous Equilibria

55m 26s

Intro
0:00
Lesson Overview
0:07
Calculating pH of an Acid-Base Mixture
0:53
Equilibria Involving Direct Reaction With Water
0:54
When a Bronsted-Lowry Acid and Base React
1:12
After Neutralization Occurs
2:05
Calculating pH of an Acid-Base Mixture cont'd
2:51
Example: Calculating pH of an Acid-Base Mixture, Step 1 - Neutralization
2:52
Example: Calculating pH of an Acid-Base Mixture, Step 2 - React With H₂O
5:24
Buffers
7:45
Introduction to Buffers
7:46
When Acid is Added to a Buffer
8:50
When Base is Added to a Buffer
9:54
Buffers cont'd
10:41
Calculating the pH
10:42
Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer
14:03
Buffers cont'd
14:10
Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer: Step 1 -Neutralization
14:11
Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer: Step 2- ICE Table
15:22
Buffer Preparation and Capacity
16:38
Example: Calculating the pH of a Buffer Solution
16:42
Effective Buffer
18:40
Acid-Base Titrations
19:33
Acid-Base Titrations: Basic Setup
19:34
Acid-Base Titrations cont'd
22:12
Example: Calculate the pH at the Equivalence Point When 0.250 L of 0.0350 M HClO is Titrated With 1.00 M KOH
22:13
Acid-Base Titrations cont'd
25:38
Titration Curve
25:39
Solubility Equilibria
33:07
Solubility of Salts
33:08
Solubility Product Constant: Ksp
34:14
Solubility Equilibria cont'd
34:58
Q < Ksp
34:59
Q > Ksp
35:34
Solubility Equilibria cont'd
36:03
Common-ion Effect
36:04
Example: Calculate the Solubility of PbCl₂ in 0.55 M NaCl
36:30
Solubility Equilibria cont'd
39:02
When a Solid Salt Contains the Conjugate of a Weak Acid
39:03
Temperature and Solubility
40:41
Complexation Equilibria
41:10
Complex Ion
41:11
Complex Ion Formation Constant: Kf
42:26
Summary
43:35
Sample Problem 1: Question
44:23
Sample Problem 1: Part a) Calculate the pH at the Beginning of the Titration
45:48
Sample Problem 1: Part b) Calculate the pH at the Midpoint or Half-way Point
48:04
Sample Problem 1: Part c) Calculate the pH at the Equivalence Point
48:32
Sample Problem 1: Part d) Calculate the pH After 27.50 mL of the Acid was Added
53:00
Section 12: Thermodynamics & Electrochemistry
Entropy & Free Energy

36m 13s

Intro
0:00
Lesson Overview
0:08
Introduction
0:53
Introduction to Entropy
1:37
Introduction to Entropy
1:38
Entropy and Heat Flow
6:31
Recall Thermodynamics
6:32
Entropy is a State Function
6:54
∆S and Heat Flow
7:28
Entropy and Heat Flow cont'd
8:18
Entropy and Heat Flow: Equations
8:19
Endothermic Processes: ∆S > 0
8:44
The Second Law of Thermodynamics
10:04
Total ∆S = ∆S of System + ∆S of Surrounding
10:05
Nature Favors Processes Where The Amount of Entropy Increases
10:22
The Third Law of Thermodynamics
11:55
The Third Law of Thermodynamics & Zero Entropy
11:56
Problem-Solving involving Entropy
12:36
Endothermic Process and ∆S
12:37
Exothermic Process and ∆S
13:19
Problem-Solving cont'd
13:46
Change in Physical States: From Solid to Liquid to Gas
13:47
Change in Physical States: All Gases
15:02
Problem-Solving cont'd
15:56
Calculating the ∆S for the System, Surrounding, and Total
15:57
Example: Calculating the Total ∆S
16:17
Problem-Solving cont'd
18:36
Problems Involving Standard Molar Entropies of Formation
18:37
Introduction to Gibb's Free Energy
20:09
Definition of Free Energy ∆G
20:10
Spontaneous Process and ∆G
20:19
Gibb's Free Energy cont'd
22:28
Standard Molar Free Energies of Formation
22:29
The Free Energies of Formation are Zero for All Compounds in the Standard State
22:42
Gibb's Free Energy cont'd
23:31
∆G° of the System = ∆H° of the System - T∆S° of the System
23:32
Predicting Spontaneous Reaction Based on the Sign of ∆G° of the System
24:24
Gibb's Free Energy cont'd
26:32
Effect of reactant and Product Concentration on the Sign of Free Energy
26:33
∆G° of Reaction = -RT ln K
27:18
Summary
28:12
Sample Problem 1: Calculate ∆S° of Reaction
28:48
Sample Problem 2: Calculate the Temperature at Which the Reaction Becomes Spontaneous
31:18
Sample Problem 3: Calculate Kp
33:47
Electrochemistry

41m 16s

Intro
0:00
Lesson Overview
0:08
Introduction
0:53
Redox Reactions
1:42
Oxidation-Reduction Reaction Overview
1:43
Redox Reactions cont'd
2:37
Which Reactant is Being Oxidized and Which is Being Reduced?
2:38
Redox Reactions cont'd
6:34
Balance Redox Reaction In Neutral Solutions
6:35
Redox Reactions cont'd
10:37
Balance Redox Reaction In Acidic and Basic Solutions: Step 1
10:38
Balance Redox Reaction In Acidic and Basic Solutions: Step 2 - Balance Each Half-Reaction
11:22
Redox Reactions cont'd
12:19
Balance Redox Reaction In Acidic and Basic Solutions: Step 2 - Balance Hydrogen
12:20
Redox Reactions cont'd
14:30
Balance Redox Reaction In Acidic and Basic Solutions: Step 3
14:34
Balance Redox Reaction In Acidic and Basic Solutions: Step 4
15:38
Voltaic Cells
17:01
Voltaic Cell or Galvanic Cell
17:02
Cell Notation
22:03
Electrochemical Potentials
25:22
Electrochemical Potentials
25:23
Electrochemical Potentials cont'd
26:07
Table of Standard Reduction Potentials
26:08
The Nernst Equation
30:41
The Nernst Equation
30:42
It Can Be Shown That At Equilibrium E =0.00
32:15
Gibb's Free Energy and Electrochemistry
32:46
Gibbs Free Energy is Relatively Small if the Potential is Relatively High
32:47
When E° is Very Large
33:39
Charge, Current and Time
33:56
A Battery Has Three Main Parameters
33:57
A Simple Equation Relates All of These Parameters
34:09
Summary
34:50
Sample Problem 1: Redox Reaction
35:26
Sample Problem 2: Battery
38:00
Section 13: Transition Elements & Coordination Compounds
The Chemistry of The Transition Metals

39m 3s

Intro
0:00
Lesson Overview
0:11
Coordination Compounds
1:20
Coordination Compounds
1:21
Nomenclature of Coordination Compounds
2:48
Rule 1
3:01
Rule 2
3:12
Rule 3
4:07
Nomenclature cont'd
4:58
Rule 4
4:59
Rule 5
5:13
Rule 6
5:35
Rule 7
6:19
Rule 8
6:46
Nomenclature cont'd
7:39
Rule 9
7:40
Rule 10
7:45
Rule 11
8:00
Nomenclature of Coordination Compounds: NH₄[PtCl₃NH₃]
8:11
Nomenclature of Coordination Compounds: [Cr(NH₃)₄(OH)₂]Br
9:31
Structures of Coordination Compounds
10:54
Coordination Number or Steric Number
10:55
Commonly Observed Coordination Numbers and Geometries: 4
11:14
Commonly Observed Coordination Numbers and Geometries: 6
12:00
Isomers of Coordination Compounds
13:13
Isomers of Coordination Compounds
13:14
Geometrical Isomers of CN = 6 Include: ML₄L₂'
13:30
Geometrical Isomers of CN = 6 Include: ML₃L₃'
15:07
Isomers cont'd
17:00
Structural Isomers Overview
17:01
Structural Isomers: Ionization
18:06
Structural Isomers: Hydrate
19:25
Structural Isomers: Linkage
20:11
Structural Isomers: Coordination Isomers
21:05
Electronic Structure
22:25
Crystal Field Theory
22:26
Octahedral and Tetrahedral Field
22:54
Electronic Structure cont'd
25:43
Vanadium (II) Ion in an Octahedral Field
25:44
Chromium(III) Ion in an Octahedral Field
26:37
Electronic Structure cont'd
28:47
Strong-Field Ligands and Weak-Field Ligands
28:48
Implications of Electronic Structure
30:08
Compare the Magnetic Properties of: [Fe(OH₂)₆]²⁺ vs. [Fe(CN)₆]⁴⁻
30:09
Discussion on Color
31:57
Summary
34:41
Sample Problem 1: Name the Following Compound [Fe(OH)(OH₂)₅]Cl₂
35:08
Sample Problem 1: Name the Following Compound [Co(NH₃)₃(OH₂)₃]₂(SO₄)₃
36:24
Sample Problem 2: Change in Magnetic Properties
37:30
Section 14: Nuclear Chemistry
Nuclear Chemistry

16m 39s

Intro
0:00
Lesson Overview
0:06
Introduction
0:40
Introduction to Nuclear Reactions
0:41
Types of Radioactive Decay
2:10
Alpha Decay
2:11
Beta Decay
3:27
Gamma Decay
4:40
Other Types of Particles of Varying Energy
5:40
Nuclear Equations
6:47
Nuclear Equations
6:48
Nuclear Decay
9:28
Nuclear Decay and the First-Order Kinetics
9:29
Summary
11:31
Sample Problem 1: Complete the Following Nuclear Equations
12:13
Sample Problem 2: How Old is the Rock?
14:21
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Lecture Comments (30)

0 answers

Post by duganbrandon225 on March 1 at 05:29:15 PM

Why do we use 10(3) power when doing the problems? I do not know where you got that.

0 answers

Post by Mirriel Akoto on July 18, 2017

For sample problem 1 on stochiometryI. I do not think the calculation is right.

0 answers

Post by Evan Wang on May 18, 2017

Can you give us harder examples?
I feel these are too simple.

0 answers

Post by jacob featehrstone on November 19, 2015

just noticed a mistake in the video.. 2.6 moles of H2O has 5.2 moles of H not 5.6

1 answer

Last reply by: Peter Ke
Sat Sep 12, 2015 8:30 PM

Post by Peter Ke on September 12, 2015

I am so confused in Sample Problem 3: Part 1. Please help.

Didn't the mg for CH4 converted to CO2 was 6.3 x 10^7 mg CO2 that you got in Sample Problem 2? But you put 10^7.

1 answer

Last reply by: Peter Ke
Sat Sep 12, 2015 5:14 PM

Post by Peter Ke on September 12, 2015

I was wondering when using dimensional analysis, when you get a answer with decimals don't you have to round it? Because the answer for 17.29 is 6.75 and you didnt round it up. Why is that?

2 answers

Last reply by: Parsa Abadi
Sun Apr 23, 2017 12:14 PM

Post by Muhammad Ziad on October 27, 2014

Shouldn't the answer to sample 1 be approximately 600.5 grams? I did the same work as you but my answer was different. By the way, your lecture was very helpful. Thank you so much.

2 answers

Last reply by: Professor Franklin Ow
Mon Nov 3, 2014 11:03 PM

Post by Okwudili Ezeh on October 27, 2014

The answer to sample problem 1 should be 600gms of carbon.

1 answer

Last reply by: Professor Franklin Ow
Fri Oct 17, 2014 5:34 PM

Post by Virginia Vizconde on October 17, 2014

Why 2.6 mole of H20 is 2 mole H? Thank you

1 answer

Last reply by: Professor Franklin Ow
Sun Oct 12, 2014 11:58 AM

Post by Saadman Elman on October 12, 2014

You didn't answer my Question. Can you please let me know if i am right or wrong?

Thanks.

1 answer

Last reply by: Parsa Abadi
Sun Apr 23, 2017 12:36 PM

Post by Saadman Elman on September 20, 2014

The answer of sample 3 is wrong. The question was how many of excess (CH4) will remain. You said the answer is approximately 5000 g. But the answer is 17325 gram of CH4 will remain.

1 answer

Last reply by: Professor Franklin Ow
Fri Jul 18, 2014 5:49 AM

Post by Kurt Kamena on July 15, 2014

Shouldn't the answer to Sample Question 3 (Part 2) be roughly 17,000 grams?

1 answer

Last reply by: David Millaud
Mon Jun 23, 2014 2:46 PM

Post by David Millaud on June 23, 2014

is that math correct on your conversion using the molar mass of hydrogen i thought you would only calculate 1 gram seeing how hydrogen is one gram/mole.  Or is it because the element is diatomic due to the subscript is why you use 2 grams for one mole?

2 answers

Last reply by: Professor Franklin Ow
Sat Apr 26, 2014 5:19 PM

Post by Jia Cheong on March 31, 2014

2.6 moles of H2O x 2 moles of H = 5.2 not 5.6 moles of H2O...

0 answers

Post by Shane Lynch on March 31, 2014

Is that last answer about the excess CH4 correct? I got 7100g...

1 answer

Last reply by: Professor Franklin Ow
Sun Mar 16, 2014 12:21 AM

Post by Meredith Roach on March 9, 2014

Is the math correct in Sample problem #1 (33:01)?

Stoichiometry I

  • Stoichiometry uses coefficients from a balanced chemical equation as a conversion factor to relate any (2) reactants and/or products.
  • Mole to mole ratios are central to any stoichiometry problem.
  • The limiting reagent or reactant dictates how much product is expected to form (known as the theoretical yield).
  • Molarity expresses solution concentration, and can be used as a conversion factor.
  • Acid-base titrations are used to determine (or standardize) the concentration of an unknown solution.

Stoichiometry I

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Lesson Overview 0:23
  • Mole to Mole Ratios 1:32
    • Example 1: In 1 Mole of H₂O, How Many Moles Are There of Each Element?
    • Example 2: In 2.6 Moles of Water, How Many Moles Are There of Each Element?
  • Mole to Mole Ratios Cont'd 5:13
    • Balanced Chemical Reaction
  • Mole to Mole Ratios Cont'd 7:25
    • Example 3: How Many Moles of Ammonia Can Form If you Have 3.1 Moles of H₂?
    • Example 4: How Many Moles of Hydrogen Gas Are Required to React With 6.4 Moles of Nitrogen Gas?
  • Mass to mass Conversion 11:06
    • Mass to mass Conversion
    • Example 5: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂?
    • Example 6: How Many Grams of Hydrogen Gas Are Required to React With 6.4 Grams of Nitrogen Gas?
    • Example 7: How Man Milligrams of Ammonia Can Form If You Have 1.2 kg of H₂?
  • Limiting Reactants, Percent Yields 20:42
    • Limiting Reactants, Percent Yields
    • Example 8: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂ and 3.1 Grams of N₂
    • Percent Yield
    • Example 9: How Many Grams of The Excess Reactant Remains?
  • Summary 29:34
  • Sample Problem 1: How Many Grams of Carbon Are In 2.2 Kilograms of Carbon Dioxide? 30:47
  • Sample Problem 2: How Many Milligrams of Carbon Dioxide Can Form From 23.1 Kg of CH₄(g)? 33:06
  • Sample Problem 3: Part 1 36:10
  • Sample Problem 3: Part 2 - What Amount Of The Excess Reactant Will Remain? 40:53

Transcription: Stoichiometry I

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is going to be another...0003

It is one of those central topics that you are going to be using for the rest of your general chemistry career.0007

That is going to be the concept of stoichiometry.0014

This is going to be the first of two lectures on stoichiometry.0017

Let's go over the lesson overview.0024

What is very core to stoichiometry problems is what is called a mole to mole ratio.0027

This is what we are going to introduce right from the start.0033

After we introduce the mole to mole ratio, we are going to go through0038

some traditional types of problems and sample calculations you can encounter.0041

Moreover the mole to mole ratio can be used to solve0048

a very common type of problem which is called the mass to mass conversion.0053

Right after the mass to mass conversion, we are going to encounter0060

another unique type of problem which is called a limiting reactant.0064

In addition to that, we are going to be discussing what is called a percent yield.0069

I am going to not only show you how to do it.0073

But more importantly, how do you know when you are dealing with a limiting reactant problem?0075

Because it involves a different strategy of its own.0079

We are going to finish up with a brief summary as always.0083

Then we will go ahead and tackle some sample problems together.0087

What exactly is a mole to mole ratio?0094

In chemical formulas, that is the first place where we are going to look.0098

In chemical formulas, the subscripts can actually be interpreted as the following.0102

The moles of that element for every one mole of compound.0108

For example, in one mole of water, there are two moles of hydrogen and there is going to be one mole of oxygen.0113

We literally can translate the subscripts in terms of moles.0128

That is going to make calculations a lot easier to interpret later on.0135

We not only have to deal with one mole of a compound.0141

For example, let's go ahead and look at the next question.0144

It is the same question: how many moles are there of each element?0147

But not in one mole of water but this time in two moles of water.0150

Remember the concept of dimensional analysis.0156

We saw that dimensional analysis used what was called a conversion factor.0160

In this whole lesson, conversion factors are key.0166

We are going to be doing dimensional analysis so much in this chapter.0175

But the nice news about this is that you already know how to do dimensional analysis.0182

You already know how to use conversion factors.0186

It is the same mathematical tools that we were introduced to several lectures ago.0189

Basically 2.6 moles of water is given to me; 2.6 moles of water.0195

Let's go ahead and get the moles of hydrogen.0204

2.6 moles of water times something over something, that is going to give me my moles of hydrogen.0208

Let's go ahead and enter the conversion factor.0218

I want moles of water to cancel; so moles of water goes downstairs.0220

I want moles of hydrogen to go upstairs to get carried through to the final answer.0224

Where do I get the numbers for the conversion factor?0231

You get it from the subscripts in the chemical formula.0233

For every one mole of water, there is going to be two moles of hydrogen.0237

We get an answer of 5.6 moles of hydrogen; that is it; it is that simple.0242

Let's go ahead and finish up the problem and now calculate the moles of oxygen that are contained in 2.6 moles of water.0250

2.6 moles of water times something over something equals to the moles of oxygen.0258

We are going to be using the same tools as before.0272

Moles of water is going to get cancelled; that goes downstairs.0276

Moles of oxygen now is going to the numerator.0280

That is going to get carried through to the final answer.0286

My mole to mole ratio is just 1:1, very conveniently.0289

In 2.6 moles of water, there are 2.6 moles of oxygen.0294

What we just did was we used and we came up with mole to mole ratios from a chemical formula.0300

Mole to mole ratios from a chemical formula.0311

The nice thing about mole to mole ratios is that they easily carry over from the chemical formula to a balanced chemical equation.0317

For example, in the following equation, nitrogen plus three hydrogens goes on to form two ammonias.0329

The coefficients in the balanced chemical equation, they are actually interpreted and treated as moles; we treat the coefficients as moles.0341

For example, let's go ahead and translate this.0354

One mole of N2 requires three moles of H2 for this reaction to occur.0357

Next line, we can expect two moles of ammonia to form from one mole of N2.0366

Finally we can expect two moles of ammonia to form from three moles of H2; from three moles of H2.0372

That should be H2; my apologies.0380

It is that simple; again we use the coefficients in terms of moles now.0386

This now leads us to our definition of stoichiometry.0394

If you look at everything in this phrase here, every sentence is relating one compound from the chemical reaction to another one.0399

Anytime you relate any two elements or molecules in a chemical reaction, this is what is called stoichiometry.0410

Once again this is what is called stoichiometry.0419

Another name to be specific for these coefficients, sometimes you will hear this, these are also called stoichiometric coefficients.0423

Again it is just a fancy name for what you already know how to do.0435

That is to use whole numbers to balance a chemical reaction.0439

Let's go ahead and now apply this to some problems.0447

We are going to take the following, the same reaction, the ammonia formation.0450

The question is the following: how many moles of ammonia can form if you have 3.1 moles of H2?0456

I want you to get into habit; what type of problem is this?0465

This is stoichiometry because we are asked to relate one compound of the chemical reaction to another.0467

In other words, we are asked to relate ammonia with H2.0475

We are going to use dimensional analysis again to do our problem.0481

3.1 moles of H2 times something over something is going to give me the moles of ammonia that we can expect to form.0485

The moles of hydrogen goes downstairs to get cancelled.0500

The moles NH3 goes upstairs to get carried through to the final answer.0508

What is going to make the numbers for the conversion factor?--the coefficients; that is it.0515

When I look here, it is basically a 3:2 ratio of hydrogen to ammonia.0523

We put the coefficient in front of ammonia there and you put the coefficient in front of H2; just like that.0533

When all is said and done, we are going to get an answer of 2.1 moles of ammonia.0542

Again all we are doing is we are using the coefficients in a balanced chemical equation directly into our conversion factor.0550

It is that simple; let's go on to another illustration of this.0557

Same reaction, now how many moles of hydrogen gas are required to react with 6.4 moles of nitrogen gas?0563

Once again this is stoichiometry all the way through because we are asked to relate one compound of the balanced chemical equation with another.0572

6.4 moles of nitrogen times something over something is going to give me the moles of H2 required.0583

Moles of N2 is going to go downstairs to get cancelled.0600

Moles of H2 is going to go upstairs to get carried through to the final answer.0606

Once again where do we get the numbers from?--we get it from the coefficients.0611

When I go ahead and look at the balanced chemical equation... that should be a three; this should be a two.0618

It is going to be three moles of H2 for every one mole of N2.0629

3 goes in front of the moles of H2 and 1 goes in front of the moles of N2.0636

When all is said and done, we are going to get 19.2 moles of H2 required.0644

Once again what we just did was we were able to derive a conversion factor from the coefficients in the balanced chemical equation.0655

It is always more practical to work not in terms of moles but in grams.0668

Because when you go into the laboratory, that is what you are actually weighing.0672

So what we are going to do now is a mass to mass conversion.0675

What we previously did was we converted from moles of one compound to moles of another.0680

All we are going to do next is now convert from mass of one compound to mass of another.0686

That is usually of course going to be grams.0692

So we need a conversion factor that relates moles to grams; but we already know that.0700

Do you remember what the conversion factor is called that relates moles of one entity to grams of the same entity?0705

It is called molar mass; once again we are using nothing new.0715

We are using material that we have learned previously.0720

Remember, let's just go ahead and do a refresher.0724

If we start with the grams of one compound and we want to go to moles of that compound.0727

Grams goes downstairs and 1 mole goes upstairs; so we are essentially going to divide by molar mass.0736

If I start with moles of the compound and I want to go to grams of the compound.0744

1 mole goes downstairs and grams goes upstairs; I am essentially multiplying by molar mass.0750

Let's go ahead and apply this to a question now.0759

I am going to now work with the same identical question.0764

Except that you notice that I changed the units of moles to grams now.0768

How many grams of ammonia can form if you have 3.1 grams of H2?0773

Before what we did was we did moles of A to moles of B.0781

But now we want to go from grams of A to grams of B.0790

All we are going to do, we are going to retain this very important step.0794

All we are going to do is put two steps on the outside now.0799

We are going to start with grams of A and then go to moles of A.0803

Once we are in moles of A, we are going to go to moles of B.0808

Then once we are in moles of B, we are going to finish up the problem and go on to grams of B.0811

We already know how to do this.0820

This is just a mole to mole ratio from the balanced chemical equation; that is what we just did.0822

If I want to go from grams of A to moles of A, I am going to divide by molar mass of A.0829

Here if I want to go from moles of B to grams of B, I am going to multiply by molar mass of B.0838

We are looking essentially at a minimum of three steps, three conversion factors.0844

Let's go ahead and do this; 3.1 grams of H2, the first step is to go to moles of H2.0850

Times 1 mole of H2 divided by roughly 2.016 grams of H2; that gets me as the moles of H2.0861

Now from moles of H2, I am going to go on to the next step to go to moles of ammonia.0870

That is going to be moles of NH3 on top and moles of H2 on the bottom.0876

That is going to be a 2:3 ratio; finally this gets me into moles of NH3 after everything cancels.0883

Now I want to go to grams of NH3; times something over something gives me grams of ammonia.0891

1 mole of ammonia goes downstairs and grams of ammonia goes upstairs.0898

This is going to be roughly 17 grams for the molar mass.0904

After following this three step process, we are going to get an answer of 17.4 grams of ammonia.0910

What we just did again is called a mass to mass conversion.0924

Grams of one compound to grams of another; it is a basic three step process.0929

Let's go ahead and do one more problem.0936

How many grams of hydrogen gas are required to react with 6.4 grams of nitrogen gas?0939

It is the same repetitive machinery; we are just going to go through the work now.0949

6.4 grams of N2, the first step is to go from grams of A to moles of A.0956

Times something over something; grams of N2 goes downstairs; 1 mole of N2 goes upstairs.0965

The molar mass of molecular nitrogen is going to be roughly 28 grams of N2.0973

After I am in moles of A, I then go on to moles of B.0980

Times something over something; moles of H2 on top; moles of N2 on the bottom.0983

Remember I get this mole to mole ratio from the balanced chemical equation, the coefficient.0992

That is going to be 3 moles of H2 on top and 1 mole of N2 on the bottom.0996

I am now in units of moles of B; let's go ahead and finish up now and go to grams of B.1000

Times something over something is going to give me grams of H2 required.1005

1 mole of H2 on the bottom; grams of H2 on top.1012

The molar mass of molecular hydrogen is roughly 2.016.1016

When all is said and done, we are going to get an answer of 1.4 grams of hydrogen gas that is required.1021

Remember it is a basic three step process.1033

You go from grams of A to moles of A; moles of A to moles of B.1036

Then on to moles of B to grams of B; this is called a mass to mass conversion.1043

Because we are working with mass, that does not mean we have to end in grams.1053

Any of the mass units, you should be able to get.1058

Let me go ahead and redraw what we just did.1062

What we did was we went from grams of A to moles of A.1065

Then to moles of B; then to grams of B.1071

But remember we know the prefixes for a multiplier, such as kilo, such as milli; this is always important.1075

From grams of A, you can also work with milligrams, kilograms, etc.1086

Similarly for the mass of the product, you can also work with milligrams, kilograms, etc.1094

This is a nice cumulative problem; it really requires you to know those prefixes1102

that you are undoubtedly going to have to memorize; especially milli and kilo.1107

Let's go ahead and do an example.1113

How many milligrams of ammonia can form if you have 1.2 kilograms of H2?1116

We are just going to follow this basic flow chart.1121

1.2 kilograms of H2, we want to go to grams of A or grams of H2.1128

Times grams of H2 for every 1 kilogram of H2.1136

Remember what the prefix multiplier is for kilo?--it is 103.1140

Remember you always put the multiplier with the prefix-less unit; times.1144

Now I want to go from grams of A to moles of A.1152

Times moles of H2 for every gram of H2.1156

That is going to be roughly 2.016 for the molar mass of H2.1162

Now I am in moles of H2, moles of A.1167

Now I want to go to moles of B which is going to be moles of ammonia.1170

Times moles of ammonia for every mole of H2.1174

Remember I get the mole to mole ratio from the balanced chemical equation.1180

That is going to be 2 here and 3 there.1183

Now I can go from moles of ammonia to grams of ammonia.1187

Times roughly 17 grams of NH3 for every one mole of NH3.1191

The question is asking for milligrams; so now I am going to multiply this by 1 mg over 10-3 grams.1198

When all is said and done, I am going to get 6.7 times 106 milligrams of ammonia.1212

That is expected to form from 1.2 kilograms of H2.1219

Just be on the lookout for that again.1225

Just because the most common unit of mass we work with is grams, it doesn't mean we have to stop there or start there.1229

It is any of the mass units we can work with using stoichiometry and mole to mole ratios.1235

We now move on to a very specific type of stoichiometry problem.1244

This is what is called the limiting reactant and percent yields.1248

What we have done is the following before.1253

How many grams of ammonia can form if you have 3.1 grams of H2?1256

You notice one thing is that only one reactant amount is specified; that is of hydrogen gas.1260

When you encounter these types of problems where only one reactant amount is specified, you are making several assumptions.1268

First, you have enough of the other reactant.1276

Second, there are no side reactions or unexpected occurrences.1280

Finally, third, say you are doing this in a lab.1286

You are assuming that everything goes perfect; there are no experimental errors at all.1290

If all goes well, you expect the maximum amount possible of product to form.1297

This is what we call the theoretical yield.1306

In practice, you never ever get the theoretical yield because errors happen all the time.1312

There are going to be inefficiencies in the system; you may get side reactions, etc.1318

Again the theoretical yield is never obtained; in practice never obtained.1324

Remember there is no such thing as a 100 percent efficient process.1335

Let's look at the following question.1346

How many grams of ammonia can form if you have 3.1 grams of H2 and 3.1 grams of N2?1348

This is the first time we have encounter this type of problem, when both reactant amounts is specified.1355

This is what we call a limiting reactant problem.1361

Because one of these reactant amounts is going to dictate, is going to limit how much product we can get.1364

To solve a limiting reactant problem, we are going to be using what is called the smaller amount reacted.1371

Let me go ahead and demonstrate.1377

We are going to first determine the amount of product that can form from each of the reactants; let's go ahead and do both.1380

3.1 grams of H2 times 1 mole of H2 divided by 2.016 grams of H2.1389

Times 2 moles of ammonia for every 3 moles of H2.1401

That is going to give me 1.03 moles of ammonia expected; remember that is the key word, expected.1415

What we are going to do now, we are going to repeat the process.1426

But for the other reactant amount, the 3.1 grams of nitrogen gas.1428

3.1 grams of N2 times 1 mole of N2 divided by roughly 28 grams of N2.1432

Now times 2 moles of ammonia for every 1 mole of N2.1441

That is going to give me 0.22 moles of NH3 expected.1449

This is again called the smaller amount method for the following reason.1459

The route that forms the smaller amount of product is going to be what is called the limiting reactant.1464

It is going to be what dictates the amount of product formed.1476

We are not going to get the 1.03 moles of ammonia expected.1481

But instead we expect the smaller amount, the 0.22 moles of NH3.1484

Because of this, we conclude that N2 is a limiting reactant in this example.1489

Also H2 therefore is something we have plenty of; we have more than enough.1504

The nitrogen gas, the limiting reactant, we don't have enough of it; that is why it is limiting.1511

That is why it is going to dictate how much product we are going to get.1516

We say that H2 is in excess; we have too much of it and not enough of N2.1519

The percent yield, the percent yield is the following.1528

This is what is called the actual yield divided by the theoretical yield times 100 percent.1534

The actual yield is what you physically get in the lab; this is from practice.1547

You are usually told this in the question; given.1558

The theoretical yield is what you always calculate from a mole to mole ratio.1562

In other words, it is the 0.22 moles; from a calculation.1568

In practice of course we want the percent yield to be as close to 100 percent as possible.1577

But that is just never ever going to happen.1581

What your percent yield tells you is pretty much how well of an experiment1586

you were able to pull off doing this type of synthesis problem.1593

What we just did was called the smaller amount method.1602

It is used to solve when two different reactants are given.1604

Another common type of question you can encounter is the following.1611

How many grams of the excess reactant is going to remain?1614

In other words, how much H2 is going to be left over?1620

This is another type of common question you can be asked.1630

I have seen it asked many many times which is why I want to bring it up.1633

Basically what we are going to do is we are going to once again use stoichiometry mole to mole ratios.1637

Let's start with what we know for sure; we know for sure we are going to consume all of the limiting reactant.1646

We don't have enough of it, the 3.1 grams of N2.1650

Basically we are going to go from 3.1 grams of N2 to how much grams of H2 required.1655

Once we get the grams of H2 required, we simply subtract that from the amount that we have to get the amount left over.1667

Grams of H2 initially have minus grams of H2 required is going to give you the grams of H2 remaining.1675

Let's go ahead and demonstrate; 3.1 grams of N2 times 1 mole of N2 divided by roughly 28 grams of N2.1693

Now we go to moles of H2; 3 moles of H2 for every 1 moles of N2.1707

Now we go to grams of H2; 2.016 grams of H2 for every 1 mole of H2.1715

That is going to give us 0.67 grams of H2.1723

What this number is, this 0.67 grams, this is the amount of H21727

that is going to react with the 3.1 grams of N2... will react.1731

But the question is asking for now how much is going to react or how much is going to remain.1738

Now 3.1 grams of H2 minus 0.67 grams of H2.1743

That is going to say 2.43 grams of H2 left over.1750

Once again we are not reinventing the wheel at all today.1757

We are simply using tools that we know already, dimensional analysis.1760

We are really using the concept of the mole to mole ratio incredibly heavily.1765

Let's go ahead and summarize before we get into the sample problems.1776

The mole to mole ratio is incredibly fundamental; I cannot underscore this enough.1780

We derive the mole to mole ratio from two sources today.1786

Number one was the chemical formula; number two was from a balanced chemical equation.1789

That is why you always have to make sure to balance your chemical equation.1794

Always make sure it is balanced if not done so for you already.1798

After introducing the mole to mole ratio, we then define stoichiometry.1802

Once again stoichiometry refers to relating the amounts of one compound to that of another in a chemical formula or a reaction.1807

Finally we introduced a very specific commonly asked stoichiometry problem, the limiting reactant.1821

How do you know you are doing a limiting reactant?1828

Because you are going to be specified two reactant amounts.1830

That is our summary for this first presentation of stoichiometry.1836

Let's now spend some time on doing some calculations and representative problems.1841

How many grams of carbon are in 2.2 kilograms of carbon dioxide?1848

How many grams of carbon are in so much carbon dioxide?1854

You don't see any chemical equation or there is no mention of any chemical equation.1858

This is using a mole to mole ratio from a chemical formula... mole to mole ratio from a chemical formula.1863

For carbon dioxide, it is basically 1 mole of carbon dioxide contains 1 mole of carbon.1878

And 1 mole of carbon dioxide contains 2 moles of oxygen.1886

So 1 mole of CO2 for every 1 mole of carbon.1890

And 1 mole of CO2 for every 2 moles of oxygen.1894

Remember we use the subscripts directly as moles from a chemical formula.1900

Now that we have our conversion factors that we can possibly use, let's go ahead and solve using dimensional analysis.1906

2.2 kilograms of carbon dioxide, remember we want to get this to grams first.1912

Times 103 grams of CO2 for every 1 kilogram of CO2.1919

Now that we are in grams of CO2... remember grams of A.1927

We now go to moles of A or moles of CO2; times 1 mole of CO2.1930

The molar mass of carbon dioxide is roughly 44 grams of CO2.1937

Now we are in moles of A, moles of carbon dioxide.1943

We want to go to moles of B which is going to be the moles of carbon.1950

That is going to be 1 mole of carbon for every 1 mole of CO2.1954

Finally moles of B goes to grams of B; times 12.01 grams of carbon for every 1 mole of carbon.1960

When all is said and done, we are going to get 2.6 times 104 grams of carbon.1971

Again this is a representative problem of using a mole to mole ratio directly from a chemical formula.1979

Right away, sample problem two, you see a chemical equation immediately.1989

You know this is going to be using a mole to mole ratio from the coefficients of the balanced chemical equation.1994

The very first thing you want to do is to balance or make sure it is balanced.2001

This is an ordinary combustion problem that we have looked at before.2005

We are going to need in the end two waters and two oxygens.2009

Now the question: how many milligrams of carbon dioxide can form from so many kilograms of CH4 gas?2017

You see milligrams and you see kilograms; therefore what type of problem is this?2027

That is right; this is a mass to mass conversion.2032

In addition, do you see amounts of both reactants specified?2039

We don't; we only see the amount of one reactant specified, the CH4.2045

So we know that this is not limiting reactant at all... not limiting reactant.2052

Let's go ahead and jump into the problem now.2063

23.1 kilograms of CH4, again we want to get to grams first.2068

Times 103 grams of CH4 for every 1 kilogram of CH4; that gets us into grams of A.2074

Now from grams of A, I want to go to moles of A.2083

Times 1 mole of CH4 for every... molar mass of CH4 is roughly 16 grams of CH4.2086

That gives me moles of A; now from moles of A, I want to go to moles of B.2096

That is going to be carbon dioxide; it is conveniently a 1:1 ratio.2102

1 mole of CO2 for every 1 mole of CH4; moles of B.2111

Now from moles of B, I want to go to grams of B.2120

Carbon dioxide is roughly 44 grams for every 1 mole.2125

The question is asking for milligrams of carbon dioxide; times 1 milligram divided by 10-3 grams.2129

After we do this calculation, we are going to get 6.3 times 107 milligrams of CO2.2143

That is expected to form from 23.1 kilograms of CH4.2150

Again this is a mole to mole ratio from the balanced equation and not a limiting reactant problem.2162

Sample problem number three; once you see a chemical equation, make sure it is balanced.2174

Again it is going to be two of these and two of those.2183

How many milligrams of carbon dioxide can form when 23.1 kilograms of CH4 are combined with 23.1 kilograms of oxygen gas?2189

First of all, you see that both of the reactant amounts are specified.2205

You know right away, 100 percent, there is no question, it is limiting reactant.2211

Because it is a limiting reactant problem, we are going to use our method2219

that we call the smaller quantity approach or the smaller quantity method... smaller quantity method.2223

Remember we are going to now determine the theoretical amount of product that can form from each reactant and compare.2235

The smaller one is going to be the answer.2242

Now the 23.1 kilograms of CH4; we already did that.2246

We got roughly 107 milligrams of CO2; that is just directly from the previous example.2259

Now we need to determine how much CO2 is going to form from 23.1 kilograms of oxygen gas.2271

23.1 kilograms of oxygen, I want to get to grams first.2280

Times 103 grams for every 1 kilogram of oxygen; that is grams of A.2288

Now from grams of A, I go to moles of A.2296

Times 1 mole of oxygen for every... the molar mass of molecular oxygen is going to be 32 grams of O2.2298

That is moles of A; now from moles of A, I go to moles of B.2314

That is going to be 1 mole of CO2 for every 2 moles of oxygen.2318

Remember I am getting this from the balanced chemical equation.2324

Now from this I want to go ahead and get the grams of CO2.2329

Again I want to get the grams of CO2.2342

Now I am going to multiply this by 44 grams of carbon dioxide for every 1 mole of CO2.2345

When all is said and done, we are going to get approximately 16,000 grams of CO2.2355

I apologize; I don't have that number in front of me.2363

But it is going to be roughly that number; roughly 16,000 grams of CO2.2365

What is important is that this is going to be smaller than the 107 milligrams of CO2 gotten from the other answer.2374

Just go ahead and compare; 16,000 grams of CO2 is going to be...2390

That is going to be 16,000,000 milligrams of CO2 or 1.6 times 107 milligrams of CO2.2396

From the previous answer, this was I believe 6.3 times 107 milligrams.2409

The 1.6 times 107 milligrams of CO2 is going to be our actual answer.2418

We conclude that molecular oxygen is the limiting reactant.2429

CH4 in this case is going to be in excess.2440

We don't have enough oxygen; that is why it is limiting.2448

But we have plenty of CH4.2450

What that means is that we can calculate the amount of CH4 that is going to remain.2454

We started with 23.1 kilograms initially of CH4.2461

We are going to be using roughly 16,000 grams of CH4 required; roughly.2471

That means 23,100 grams minus 16,000 grams tells me approximately 5,000 grams of CH4, the excess amount, will remain unused.2489

What we just did again for sample problem three was a limiting reactant problem.2511

We not only determined the limiting reactant.2516

We also determined how much of the excess reactant is going to remain.2519

Thank you for using Educator.com; it was great to see you guys again.2525

I will see you all next time.2529

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