Uploaded by Carolyn Campita

General Chemistry 1

advertisement
The Commission on Higher Education
in collaboration with the Philippine Normal University
Teaching Guide for Senior High School
GENERAL
CHEMISTRY 1
SPECIALIZED SUBJECT | ACADEMIC STEM
This Teaching Guide was collaboratively developed and reviewed by
educators from public and private schools, colleges, and universities. We
encourage teachers and other education stakeholders to email their
feedback, comments, and recommendations to the Commission on Higher
Education, K to 12 Transition Program Management Unit Senior High School Support Team at [email protected]
We value your feedback and recommendations.
Development Team
Team Leader: Wyona C. Patalinghug, Ph.D.
Writers: Wyona C. Patalinghug, Ph.D.,
Vic Marie I. Camacho, Ph.D.,
Fortunato B. Sevilla III, Ph.D.,
Maria Cristina D. Singson
Published by the Commission on Higher Education, 2016
Chairperson: Patricia B. Licuanan, Ph.D.
Commission on Higher Education
K to 12 Transition Program Management Unit
Office Address: 4th Floor, Commission on Higher Education,
C.P. Garcia Ave., Diliman, Quezon City
Telefax: (02) 441-0927 / E-mail Address: [email protected]
Technical Editors: Marlene B. Ferido, Ph.D., Janeth
M. Fuentes
Copy Reader: Patricia Marie W. Baun
Illustrator: Juan Miguel M. Razon, Rachelle Ann J.
Bantayan, Danielle Christine Quing
Cover Artists: Paolo Kurtis N. Tan, Renan U. Ortiz
Senior High School Support Team
CHED K to 12 Transition Program Management Unit
Program Director: Karol Mark R. Yee
Consultants
THIS PROJECT WAS DEVELOPED WITH THE PHILIPPINE NORMAL UNIVERSITY.
University President: Ester B. Ogena, Ph.D.
VP for Academics: Ma. Antoinette C. Montealegre, Ph.D.
VP for University Relations & Advancement: Rosemarievic V. Diaz, Ph.D.
Ma. Cynthia Rose B. Bautista, Ph.D., CHED
Bienvenido F. Nebres, S.J., Ph.D., Ateneo de Manila University
Carmela C. Oracion, Ph.D., Ateneo de Manila University
Minella C. Alarcon, Ph.D., CHED
Gareth Price, Sheffield Hallam University
Stuart Bevins, Ph.D., Sheffield Hallam University
Lead for Senior High School Support:
Gerson M. Abesamis
Lead for Policy Advocacy and Communications:
Averill M. Pizarro
Course Development Officers:
John Carlo P. Fernando, Danie Son D. Gonzalvo,
Stanley Ernest Yu
Teacher Training Officers:
Ma. Theresa C. Carlos, Mylene E. Dones
Monitoring and Evaluation Officer:
Robert Adrian N. Daulat
Administrative Officers:
Ma. Leana Paula B. Bato, Kevin Ross D. Nera,
Allison A. Danao, Ayhen Loisse B. Dalena
Printed in the Philippines by EC-TEC Commercial, No. 32 St.
Louis Compound 7, Baesa, Quezon City, [email protected]
This Teaching Guide by the
Commission on Higher Education is
licensed under a Creative
Commons AttributionNonCommercial-ShareAlike 4.0
International License. This means
you are free to:
Share — copy and redistribute the
material in any medium or format
Adapt — remix, transform, and
build upon the material.
The licensor, CHED, cannot revoke
these freedoms as long as you
follow the license terms. However,
under the following terms:
Attribution — You must give
appropriate credit, provide a link to
the license, and indicate if changes
were made. You may do so in any
reasonable manner, but not in any
way that suggests the licensor
endorses you or your use.
NonCommercial — You may not use
the material for commercial
purposes.
ShareAlike — If you remix,
transform, or build upon the
material, you must distribute your
contributions under the same license
as the original.
Table of Contents
DepEd Curriculum Guide
i
Lesson 18: Emission Spectrum of Hydrogen, and Dual
Lesson 1: Matter and Its Properties
1
Nature of Matter
Lesson 2: Matter and Its Various Forms
14
Lesson 19: Flame Test (Laboratory)
158
Lesson 3: Measurements
19
Lesson 20: Electronic Structure of the Atom
162
Lesson 4: Measurements (Laboratory)
25
Lesson 21: Electron Configuration
175
Lesson 5: Atoms, Molecules, and Ions (Lecture)
30
Lesson 22: Periodic Relationships among the Elements
190
Lesson 6: Atoms, Molecules, and Ions (Laboratory)
47
Lesson 23: Periodic Relationships of Main Group
203
Lesson 7: Atomic Mass
52
Elements (Laboratory)
Lesson 8: The Mole Concept and Molar Mass (Lecture)
60
Lesson 24: Ionic Bonds
208
Lecture 9: The Mole Concept and Molar Mass (Laboratory)
70
Lesson 25: Covalent Bonds and Lewis Structures
220
Lesson 10: Percent Composition and Chemical Formulas
76
Lesson 26: Geometry of Molecules and Polarity
236
Lesson 11: Chemical Reactions and Chemical Equations (Lecture)
82
of Compounds
Lesson 12: Chemical Reactions and Chemical Equations (Laboratory)
89
Lesson 27: Geometry of Molecules and Polarity
Lesson 13: Mass Relationships in Chemical Reactions (Lecture)
94
of Molecules (Laboratory)
Lesson 14: Mass Relationships in Chemical Reactions (Laboratory)
105
Lesson 28: Carbon Compounds
254
Lesson 15: Gases (Lecture)
110
Lesson 29: Polymers
284
Lesson 16: Gases (Laboratory)
128
Lesson 30: Biomolecules
297
Lesson 17: Electromagnetic Waves, Planck’s Quantum Theory, and
132
Biographical Notes
314
Additional Images
317
Photoelectric Effect
144
250
Introduction
As the Commission supports DepEd’s implementation of Senior High School (SHS), it upholds the vision
and mission of the K to 12 program, stated in Section 2 of Republic Act 10533, or the Enhanced Basic
Education Act of 2013, that “every graduate of basic education be an empowered individual, through a
program rooted on...the competence to engage in work and be productive, the ability to coexist in
fruitful harmony with local and global communities, the capability to engage in creative and critical
thinking, and the capacity and willingness to transform others and oneself.”
To accomplish this, the Commission partnered with the Philippine Normal University (PNU), the National
Center for Teacher Education, to develop Teaching Guides for Courses of SHS. Together with PNU, this
Teaching Guide was studied and reviewed by education and pedagogy experts, and was enhanced with
appropriate methodologies and strategies.
Furthermore, the Commission believes that teachers are the most important partners in attaining this
goal. Incorporated in this Teaching Guide is a framework that will guide them in creating lessons and
assessment tools, support them in facilitating activities and questions, and assist them towards deeper
content areas and competencies. Thus, the introduction of the SHS for SHS Framework.
SHS for SHS
Framework
The SHS for SHS Framework, which stands for “Saysay-Husay-Sarili for Senior High School,” is at the
core of this book. The lessons, which combine high-quality content with flexible elements to
accommodate diversity of teachers and environments, promote these three fundamental concepts:
SAYSAY: MEANING
HUSAY: MASTERY
SARILI: OWNERSHIP
Why is this important?
How will I deeply understand this?
What can I do with this?
Through this Teaching Guide,
teachers will be able to facilitate
an understanding of the value
of the lessons, for each learner
to fully engage in the content
on both the cognitive and
affective levels.
Given that developing mastery
goes beyond memorization,
teachers should also aim for
deep understanding of the
subject matter where they lead
learners to analyze and
synthesize knowledge.
When teachers empower
learners to take ownership of
their learning, they develop
independence and selfdirection, learning about both
the subject matter and
themselves.
Parts of the
Teaching Guide
This Teaching Guide is mapped and aligned to the DepEd SHS Curriculum, designed to be highly
usable for teachers. It contains classroom activities and pedagogical notes, and is integrated with
innovative pedagogies. All of these elements are presented in the following parts:
1. Introduction
• Highlight key concepts and identify the essential questions
• Show the big picture
• Connect and/or review prerequisite knowledge
• Clearly communicate learning competencies and objectives
• Motivate through applications and connections to real-life
2. Motivation
• Give local examples and applications
• Engage in a game or movement activity
• Provide a hands-on/laboratory activity
• Connect to a real-life problem
3. Instruction/Delivery
• Give a demonstration/lecture/simulation/hands-on activity
• Show step-by-step solutions to sample problems
• Give applications of the theory
• Connect to a real-life problem if applicable
4. Practice
• Discuss worked-out examples
• Provide easy-medium-hard questions
• Give time for hands-on unguided classroom work and discovery
• Use formative assessment to give feedback
5. Enrichment
• Provide additional examples and applications
• Introduce extensions or generalisations of concepts
• Engage in reflection questions
• Encourage analysis through higher order thinking prompts
6. Evaluation
• Supply a diverse question bank for written work and exercises
• Provide alternative formats for student work: written homework, journal, portfolio, group
individual projects, student-directed research project
iii
On DepEd Functional Skills and CHED College Readiness Standards
As Higher Education Institutions (HEIs) welcome the graduates of
the Senior High School program, it is of paramount importance to
align Functional Skills set by DepEd with the College Readiness
Standards stated by CHED.
On the other hand, the Commission declared the College
Readiness Standards that consist of the combination of knowledge,
skills, and reflective thinking necessary to participate and succeed without remediation - in entry-level undergraduate courses in
college.
The DepEd articulated a set of 21st century skills that should be
embedded in the SHS curriculum across various subjects and tracks.
These skills are desired outcomes that K to 12 graduates should
possess in order to proceed to either higher education,
employment, entrepreneurship, or middle-level skills development.
The alignment of both standards, shown below, is also presented in
this Teaching Guide - prepares Senior High School graduates to the
revised college curriculum which will initially be implemented by AY
2018-2019.
College Readiness Standards Foundational Skills
DepEd Functional Skills
Produce all forms of texts (written, oral, visual, digital) based on:
1.
2.
3.
4.
5.
Solid grounding on Philippine experience and culture;
An understanding of the self, community, and nation;
Visual and information literacies, media literacy, critical thinking
Application of critical and creative thinking and doing processes;
and problem solving skills, creativity, initiative and self-direction
Competency in formulating ideas/arguments logically, scientifically, and creatively; and
Clear appreciation of one’s responsibility as a citizen of a multicultural Philippines and a
diverse world;
Systematically apply knowledge, understanding, theory, and skills for the development of
the self, local, and global communities using prior learning, inquiry, and experimentation
Global awareness, scientific and economic literacy, curiosity,
critical thinking and problem solving skills, risk taking, flexibility
and adaptability, initiative and self-direction
Work comfortably with relevant technologies and develop adaptations and innovations for
significant use in local and global communities
Global awareness, media literacy, technological literacy,
creativity, flexibility and adaptability, productivity and
accountability
Communicate with local and global communities with proficiency, orally, in writing, and
through new technologies of communication
Global awareness, multicultural literacy, collaboration and
interpersonal skills, social and cross-cultural skills, leadership
and responsibility
Interact meaningfully in a social setting and contribute to the fulfilment of individual and
shared goals, respecting the fundamental humanity of all persons and the diversity of
groups and communities
Media literacy, multicultural literacy, global awareness,
collaboration and interpersonal skills, social and cross-cultural
skills, leadership and responsibility, ethical, moral, and spiritual
values
v
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
Grade: 11
Subject Title: General Chemistry 1 & 2
Semester: 1st and 2nd
No. of Hours/ Semester: 80 hours per semester
Subject Description: Composition, structure, and properties of matter; quantitative principles, kinetics, and energetics of transformations of matter; and fundamental
concepts of organic chemistry
CONTENT
Quarter 1 – General Chemistry
Matter and its properties
1. the particulate nature of
matter
2. states of matter
a. the macroscopic
b. microscopic view
3. Physical and chemical
properties
4. Extensive and intensive
properties
5. Ways of classifying matter
a. pure substances and
mixtures
b. elements and
compounds
c. homogeneous and
heterogeneous
mixtures
6. Methods of separating
mixtures into their
component substances
CONTENT STANDARD
PERFORMANCE
STANDARD
LEARNING COMPETENCIES
CODE
1
The learners
demonstrate an
understanding of:
the properties of matter
and its various forms
The learners:
design using multimedia,
demonstrations, or models,
a representation or
simulation of any of the
following:
a. atomic structure
b. gas behavior
c. mass relationships in
d. reactions
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
The learners:
1. recognize that substances are made up
of smaller particles
2. describe and/or make a representation of
the arrangement, relative spacing, and
relative motion of the particles in each of
the three phases of matter
3. distinguish between physical and
chemical properties and give examples
4. distinguish between extensive and
intensive properties and give examples
5. use properties of matter to identify
substances and to separate them
6. differentiate between pure substances
and mixtures
7. differentiate between elements and
compounds
8. differentiate between homogenous and
heterogenous mixtures
9. recognize the formulas of common
chemical substances
10. describe separation techniques for
mixtures and compounds
11. compare consumer products on the basis
of their components for use, safety,
quality and cost
12. (LAB) apply simple separation techniques
such as distillation, chromatography
STEM_GC11MP-Ia-b-1
STEM_GC11MP-Ia-b-2
STEM_GC11MP-Ia-b-3
STEM_GC11MP-Ia-b-4
STEM_GC11MP-Ia-b-5
STEM_GC11MP-Ia-b-6
STEM_GC11MP-Ia-b-7
STEM_GC11MP-Ia-b-8
STEM_GC11MP-Ia-b-9
STEM_GC11MP-Ia-b-10
STEM_GC11MP-Ia-b-11
STEM_GC11MP-Ia-b-12
Page 1 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
Measurements
1. Accuracy and precision
2. Significant figures in
calculations
3. Density measurement
1. the difference
between accuracy
and precision
2. different sources of
errors in
measurements
Atoms, Molecules, and Ions
1. Dalton’s atomic theory
2. Basic laws of matter
3. Atomic structure
4. Subatomic particles
(protons, electrons,
neutrons)
5. Molecules and Ions
6. Chemical Formulas
7. Naming Compounds
1. atomic structure
2. formulas and
names of
compounds
PERFORMANCE
STANDARD
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
LEARNING COMPETENCIES
CODE
1. differentiate between precision and
accuracy
STEM_GC11MT-Ib-13
2. (LAB) Determine the density of liquids &
solids
STEM_GC11MT-Ib-14
1. explain how the basic laws of matter (law
of conservation of mass, law of constant
composition, law of multiple proportion)
led to the formulation of Dalton’s Atomic
Theory
STEM_GC11AM-Ic-e-15
2. describe Dalton’s Atomic Theory
STEM_GC11AM-Ic-e-16
3. differentiate among atomic number, mass
number, and isotopes, and which of these
distinguishes one element from another
STEM_GC11AM-Ic-e-17
4. write isotopic symbols
STEM_GC11AM-Ic-e-18
5. recognize common isotopes and their
uses.
STEM_GC11AM-Ic-e-19
6. differentiate among atoms, molecules,
ions and give examples
STEM_GC11AM-Ic-e-20
7. represent compounds using chemical
formulas, structural formulas and models
STEM_GC11AM-Ic-e-21
8. give the similarities and differences
between the empirical formula and
molecular formula of a compound
STEM_GC11AM-Ic-e-22
Page 2 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
LEARNING COMPETENCIES
CODE
9. name compounds given their formula and
write formula given the name of the
compound
STEM_GC11AM-Ic-e-23
10. (LAB) Practice chemical nomenclature:
writing the chemical formulas of ionic
compounds; naming ionic compounds
from formulas
Stoichiometry
1. Atomic mass
2. Avogadro’s number
3. The mole concept
4. Percent composition and
chemical formulas
1. the mole concept in
relation to
Avogadro’s number
and mass
2. the relationship of
percent composition
and chemical
formula
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
STEM_GC11AM-Ic-e-24
1. explain relative atomic mass and average
atomic mass
STEM_GC11S-Ie-25
2. define a mole
STEM_GC11S-Ie-26
3. illustrate Avogadro’s number with
examples
STEM_GC11S-Ie-27
4. determine the molar mass of elements
and compounds
STEM_GC11S-Ie-28
5. calculate the mass of a given number of
moles of an element or compound or vice
versa
STEM_GC11S-Ie-29
6. calculate the mass of a given number of
particles of an element or compound or
vice versa
STEM_GC11S-Ie-30
1. calculate the percent composition of a
compound from its formula
STEM_GC11PC-If-31
2. calculate the empirical formula from the
percent composition of a compound
STEM_GC11PC-If-32
Page 3 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
LEARNING COMPETENCIES
3. calculate molecular formula given molar
mass
5. Chemical reactions and
chemical equations
6. Types of chemical
reactions in aqueous
solutions
3. the use of chemical
formulas to
represent chemical
reactions
4. write equations for chemical reactions and
balance the equations
5. interpret the meaning of a balanced
chemical reaction in terms of the law of
conservation of mass
6. describe evidences that a chemical
reaction has occurred
7. (LAB) Perform exercises on writing and
balancing chemical equations
7. Mass relationships in
chemical reactions
Gases
1. Pressure of a gas
a. Units of pressure
2. The Gas laws
4. the quantitative
relationship of
reactants and
products in a
chemical reaction
5. the mathematical
relationship between
pressure, volume,
and temperature of
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
1. construct mole or mass ratios for a
reaction in order to calculate the amount
of reactant needed or amount of product
formed in terms of moles or mass
2. Calculate percent yield and theoretical
yield of the reaction
3. explain the concept of limiting reagent in
a chemical reaction; identify the excess
reagent(s)
4. calculate reaction yield when a limiting
reagent is present
5. (LAB) Determine mass relationship in a
chemical reaction
1. define pressure and give the common
units of pressure
2.
express the gas laws in equation form
CODE
STEM_GC11PC-If-33
STEM_GC11CR-If-g-34
STEM_GC11CR-If-g-35
STEM_GC11CR-If-g-36
STEM_GC11CR-If-g-37
STEM_GC11MR-Ig-h-38
STEM_GC11MR-Ig-h-39
STEM_GC11MR-Ig-h-40
STEM_GC11MR-Ig-h-41
STEM_GC11MR-Ig-h-42
STEM_GC11G-Ih-i-43
STEM_GC11G-Ih-i-44
Page 4 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
a. Boyle’s Law
b. Charles’ Law
c. Avogadro’s Law
3. Ideal Gas Equation
a gas
4. Dalton’s Law of partial
pressures
6. the partial pressures
of gases in a mixture
5. Gas stoichiometry
7. quantitative
relationships of
reactants and
products in a
gaseous reaction
6. Kinetic molecular theory of
gases
PERFORMANCE
STANDARD
LEARNING COMPETENCIES
use the gas laws to determine pressure,
volume, or temperature of a gas under
certain conditions of change
4. use the ideal gas equation to calculate
pressure, volume, temperature, or
number of moles of a gas
5. use Dalton’s law of partial pressures to
relate mole fraction and partial pressure of
gases in a mixture
CODE
3.
8. the behavior and
properties of gases
at the molecular
level
STEM_GC11G-Ih-i-45
STEM_GC11G-Ih-i-46
STEM_GC11DL-Ii-47
6. apply the principles of stoichiometry to
determine the amounts (volume, number
of moles, or mass) of gaseous reactants
and products
STEM_GC11GS-Ii-j-48
7. explain the gas laws in terms of the
kinetic molecular theory of gases
STEM_GC11KMT-Ij-49
8. relate the rate of gas effusion with molar
mass
STEM_GC11KMT-Ij-50
9. (LAB) Demonstrate Graham’s law of
effusion in an experiment
STEM_GC11KMT-Ij-51
Quarter 2 – General Chemistry 1
Electronic Structure of
Atoms
1. Quantum mechanical
description of the atom
2. Schrodinger’s model of the
hydrogen atom and wave
functions
3. Main energy levels, sublevels
and orbitals
the quantum mechanical
description of the atom
and its electronic
structure
illustrate the reactions at
the molecular level in any
of the following:
1. enzyme action
2. protein denaturation
3. separation of
components in
coconut milk
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
1. describe the quantum mechanical model
of the atom
2. describe the electronic structure of atoms
in terms of main energy levels, sublevels,
and orbitals, and relate this to energy
3. use quantum numbers to describe an
electron in an atom
4. (LAB) Perform exercises on quantum
numbers
STEM_GC11ES-IIa-b-52
STEM_GC11ES-IIa-b-53
STEM_GC11ES-IIa-b-54
STEM_GC11ES-IIa-b-55
Page 5 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
4. Quantum numbers
5. Electron Configuration
a. Aufbau Principle
b. Pauli Exclusion Principle
c. Hund’s Rule
d. Diamagnetism and
Paramagnetism
e. Orbital diagrams
Electronic Structure and
Periodicity
1. The Electron Configuration
and the Periodic Table
2. Periodic Variation in Atomic
Properties
a. Atomic Radius and
effective nuclear charge;
the shielding effect in
many-electron atoms
b. Ionic radius
c. Ionization energy
d. Electron affinity
Chemical Bonding
Ionic Bonds
1. The stability of noble gases
2. Forming ions
3. Ionic bonding
4. Ionic compounds
5. Formulas
6. Structure
7. Properties
the arrangement of
elements in the periodic
table and trends in the
properties of the
elements in terms of
electronic structure
1. ionic bond formation
in terms of atomic
properties
2. the properties of ionic
compounds in relation
to their structure
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
LEARNING COMPETENCIES
CODE
5. write the electronic configuration of atoms
STEM_GC11ES-IIa-b-56
6. determine the magnetic property of the
atom based on its electronic configuration
7. draw an orbital diagram to represent the
electronic configuration of atoms
8. (LAB) Perform exercises on writing
electronic configuration
1. explain the periodic recurrence of similar
properties among elements in the periodic
table in terms of electronic structure
2. relate the number of valence electrons of
elements to their group number in the
periodic table
3. compare the properties of families of
elements
4. predict the properties of individual
elements based on their position in the
periodic table
5. describe and explain the trends in atomic
properties in the periodic table
6. (LAB) Investigate reactions of ions and
apply these in qualitative analysis
7. (LAB) Determine periodic properties of
the main group elements
1. relate the stability of noble gases to their
electron configuration
STEM_GC11ES-IIa-b-57
STEM_GC11ES-IIa-b-58
STEM_GC11ES-IIa-b-59
STEM_GC11ESP-IIc-d-60
STEM_GC11ESP-IIc-d-61
STEM_GC11ESP-IIc-d-62
STEM_GC11ESP-IIc-d-63
STEM_GC11ESP-IIc-d-64
STEM_GC11ESP-IIc-d-65
STEM_GC11ESP-IIc-d-66
STEM_GC11CB-IId-g-67
2. state the octet rule
STEM_GC11CB-IId-g-68
3. determine the charge of the ions formed
by the representative elements and relate
this to their ionization energy or electron
affinity, valence electron configuration
and position in the periodic table
STEM_GC11CB-IId-g-69
Page 6 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
Covalent Bonds
1. Formation of covalent bonds
2. Formulas of molecular
compounds
3. Lewis structure of molecules
4. Molecules of elements
5. Molecules of compounds
6. Structure and properties of
molecular compounds
7. Strength of covalent bonds
8. Electronegativity and bond
polarity
9. Geometry of molecules
10. Polarity of compounds
CONTENT STANDARD
PERFORMANCE
STANDARD
1. covalent bond
formation in terms of
atomic properties
2. the properties of
molecular covalent
compounds in
relation to their
structure
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
LEARNING COMPETENCIES
CODE
4. draw the Lewis structure of ions
STEM_GC11CB-IId-g-70
5. predict the formula of the ionic compound
formed by a metal and non-metal among
the representative elements
STEM_GC11CB-IId-g-71
6. Lewis structure of ionic compounds
STEM_GC11CB-IId-g-72
7. list the properties of ionic compounds and
explain these properties in terms of their
structure
8. (LAB) Perform exercises on writing Lewis
structures of ions/ionic compounds and
molecules
9. describe covalent bonding in terms of
electron sharing
10. apply the octet rule in the formation of
molecular covalent compounds
11. write the formula of molecular compounds
formed by the nonmetallic elements of the
representative block
12. draw Lewis structure of molecular
covalent compounds
13. explain the properties of covalent
molecular compounds in terms of their
structure.
14. determine the polarity of a bond based on
the electronegativities of the atoms
forming the bond
15. describe the geometry of simple
compounds
STEM_GC11CB-IId-g-73
STEM_GC11CB-IId-g-74
STEM_GC11CB-IId-g-75
STEM_GC11CB-IId-g-76
STEM_GC11CB-IId-g-77
STEM_GC11CB-IId-g-78
STEM_GC11CB-IId-g-79
STEM_GC11CB-IId-g-80
STEM_GC11CB-IId-g-81
16. determine the polarity of simple molecules
STEM_GC11CB-IId-g-82
17. (LAB) Determine and/or observe
evidence of molecular polarity
STEM_GC11CB-IId-g-83
Page 7 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
Organic compounds
1. The carbon atom
2. Bonding patterns in
hydrocarbons
3. Properties and reactivities
of common functional
groups
4. Polymers
5. Biomolecules
CONTENT STANDARD
PERFORMANCE
STANDARD
the properties of organic
compounds and
polymers in terms of
their structure
LEARNING COMPETENCIES
1.
describe the special nature of carbon
2.
list general characteristics of organic
compounds
describe the bonding in ethane,
ethene(ethylene) and
ethyne(acetylene) and explain their
geometry in terms of hybridization and
σ and ¶ carbon-carbon bonds
describe the different functional
groups
cite uses of representative examples of
compounds bearing the different
functional groups
describe structural isomerism; give
examples
describe some simple reactions of
organic compounds: combustion of
organic fuels, addition, condensation,
and saponification of fats
describe the formation and structure of
polymers
3.
4.
5.
6.
7.
8.
9.
STEM_GC11OC-IIg-j-84
STEM_GC11OC-IIg-j-85
STEM_GC11OC-IIg-j-86
STEM_GC11OC-IIg-j-87
STEM_GC11OC-IIg-j-88
STEM_GC11OC-IIg-j-89
STEM_GC11OC-IIg-j-90
STEM_GC11OC-IIg-j-91
give examples of polymers
STEM_GC11OC-IIg-j-92
10.
explain the properties of some
polymers in terms of their structure
STEM_GC11OC-IIg-j-93
11.
describe some biomolecules: proteins,
nucleic acids, lipids, and carbohydrates
describe the structure of proteins,
nucleic acids, lipids, and
carbohydrates, and relate them to
their function
12.
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
CODE
STEM_GC11OC-IIg-j-94
STEM_GC11OC-IIg-j-95
Page 8 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
LEARNING COMPETENCIES
13.
14.
15.
Third Quarter – General Chemistry 2
Intermolecular Forces and
1. the properties of
Liquids and Solids
liquids and solids to
1. Kinetic molecular model of
the nature of forces
liquids and solids
between particles
2. Intermolecular Forces
2. phase changes in
3. Dipole-dipole forces
terms of the
4. Ion-dipole forces
accompanying
5. Dispersion forces
changes in energy
6. Hydrogen bonds
and forces between
7. Properties of liquids and
particles
IMF
8. Surface Tension
9. Viscosity
10. Vapour pressure, boiling
point
11. Molar heat of vaporization
12. Structure and Properties of
Water
13. Types and properties of
solids
14. Crystalline and amorphous
solids
15. Types of Crystals – ionic,
design a simple
investigation to determine
the effect on boiling point
or freezing point when a
solid is dissolved in water
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
(LAB) Perform exercises on the
structure of organic compounds using
of models
(LAB) Prepare selected organic
compound and describe their
properties
(LAB) Perform laboratory activities on
enzyme action, protein denaturation,
separation of components in coconut
milk
CODE
STEM_GC11OC-IIg-j-96
STEM_GC11OC-IIg-j-97
STEM_GC11OC-IIg-j-98
1.
use the kinetic molecular model to
explain properties of liquids and solids
STEM_GC11IMF-IIIa-c-99
2.
describe and differentiate the types of
intermolecular forces
STEM_GC11IMF-IIIa-c100
3.
predict the intermolecular forces
possible for a molecule
STEM_GC11IMF-IIIa-c101
4.
describe the following properties of
liquids, and explain the effect of
intermolecular forces on these
properties: surface tension, viscosity,
vapor pressure, boiling point, and
molar heat of vaporization
explain the properties of water with its
molecular structure and intermolecular
forces
5.
STEM_GC11IMF-IIIa-c102
STEM_GC11IMF-IIIa-c103
6.
describe the difference in structure of
crystalline and amorphous solids
STEM_GC11IMF-IIIa-c104
7.
describe the different types of crystals
and their properties: ionic, covalent,
molecular, and metallic.
STEM_GC11IMF-IIIa-c105
Page 9 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
covalent, molecular,
metallic
16. Phase Changes
- phase diagrams of water
and carbon dioxide
LEARNING COMPETENCIES
8.
9.
10.
11.
Physical Properties of
Solutions
1. Types of Solutions
2. Energy of solution
formation
3. Concentration Units and
comparison of
concentration units
a. percent by mass, by
volume
b. mole fraction
c. molality
d. molarity
e. percent by volume,
percent by mass, ppm
4. Solution stoichiometry
5. Factors affecting Solubility
6. Colligative Properties of
Nonelectrolyte and
electrolyte solutions
properties of solutions,
solubility, and the
stoichiometry of
reactions in solutions
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
CODE
describe the nature of the following
phase changes in terms of energy
change and the increase or decrease in
molecular order: solid-liquid, liquidvapor, and solid-vapor
STEM_GC11IMF-IIIa-c106
interpret the phase diagram of water
and carbon dioxide
STEM_GC11IMF-IIIa-c107
(LAB) Measure and explain the
difference in the viscosity of some
liquids
(LAB) Determine and explain the
heating and cooling curve of a
substance
STEM_GC11IMF-IIIa-c108
STEM_GC11IMF-IIIa-c109
1.
describe the different types of
solutions
STEM_GC11PP-IIId-f-110
2.
use different ways of expressing
concentration of solutions: percent by
mass, mole fraction, molarity, molality,
percent by volume, percent by mass,
ppm
STEM_GC11PP-IIId-f-111
3.
perform stoichiometric calculations for
reactions in solution
STEM_GC11PP-IIId-f-112
4.
explain the effect of temperature on
the solubility of a solid and of a gas
STEM_GC11PP-IIId-f-113
5.
explain the effect of pressure on the
solubility of a gas
STEM_GC11PP-IIId-f-114
6.
describe the effect of concentration on
the colligative properties of solutions
STEM_GC11PP-IIId-f-115
7.
differentiate the colligative properties
of nonelectrolyte solutions and of
electrolyte solutions
STEM_GC11PP-IIId-f-116
Page 10 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
LEARNING COMPETENCIES
8.
Calculate boiling point elevation and
freezing point depression from the
concentration of a solute in a solution
STEM_GC11PP-IIId-f-117
9.
calculate molar mass from colligative
property data
STEM_GC11PP-IIId-f-118
10.
(LAB) Perform acid-base titration to
determine concentration of solutions
STEM_GC11PP-IIId-f-119
11.
(LAB) Determine the solubility of a
solid in a given amount of water at
different temperatures
(LAB) Determine the molar mass of a
solid from the change of melting point
or boiling point of a solution
explain the energy changes during
chemical reactions
distinguish between exothermic and
endothermic processes
explain the first law of
thermodynamics
12.
Thermochemistry
1. Energy Changes in
Chemical Reactions:
exothermic and
endothermic processes
2. First Law of
Thermodynamics
3. Enthalpy of a Chemical
Reaction
- thermochemical equations
4. Calorimetry
5. Standard Enthalpy of
Formation and Reaction
Hess’ Law
energy changes in
chemical reactions
1.
2.
3.
4.
explain enthalpy of a reaction.
5.
Write the thermochemical equation for
a chemical reaction
Calculate the change in enthalpy of a
given reaction using Hess Law
(LAB) Do exercises on
thermochemical calculations
(LAB)Determine the heat of
neutralization of an acid
describe how various factors influence
the rate of a reaction
write the mathematical relationship
between the rate of a reaction, rate
constant, and concentration of the
6.
7.
8.
Chemical Kinetics
1. The rate of a
1. The Rate of a Reaction
reaction and the
2. Factors that influence
various factors that
reaction rate
influence it
3. The Rate Law and its
2. the collision theory
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
CODE
1.
2.
STEM_GC11PP-IIId-f-120
STEM_GC11PP-IIId-f-121
STEM_GC11TC-IIIg-i-122
STEM_GC11TC-IIIg-i-123
STEM_GC11TC-IIIg-i-124
STEM_GC11TC-IIIg-i-125
STEM_GC11TC-IIIg-i-126
STEM_GC11TC-IIIg-i-127
STEM_GC11TC-IIIg-i-128
STEM_GC11TC-IIIg-i-129
STEM_GC11CK-IIIi-j-130
STEM_GC11CK-IIIi-j-131
Page 11 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
4.
5.
CONTENT STANDARD
PERFORMANCE
STANDARD
components
Collision theory
Catalysis
LEARNING COMPETENCIES
3.
4.
5.
6.
7.
8.
9.
10.
Fourth Quarter – General Chemistry 2
Chemical Thermodynamics
spontaneous change,
1. Spontaneous processes
entropy, and free energy
2. Entropy
3. The Second Law of
Thermodynamics
4. Gibbs Free Energy and
Chemical Equilibrium
Chemical Equilibrium
1. The equilibrium condition
Chemical equilibrium
and Le Chatelier’s
prepare a poster on a
specific application of one
of the following:
a. Acid-base
equilibrium
b. Electrochemistry
Include in the poster the
concepts, principles, and
chemical reactions involved,
and diagrams of processes
and other relevant
materials
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
reactants
differentiate zero, first-, and secondorder reactions
write the rate law for first-order
reaction
discuss the effect of reactant
concentration on the half-time of a
first-order reaction
explain the effect of temperature on
the rate of a reaction
explain reactions qualitatively in terms
of molecular collisions
explain activation energy and how a
catalyst affects the reaction rate
cite and differentiate the types of
catalysts
(LAB)Determine the effect of various
factors on the rate of a reaction
CODE
STEM_GC11CK-IIIi-j-132
STEM_GC11CK-IIIi-j-133
STEM_GC11CK-IIIi-j-134
STEM_GC11CK-IIIi-j-135
STEM_GC11CK-IIIi-j-136
STEM_GC11CK-IIIi-j-137
STEM_GC11CK-IIIi-j-138
STEM_GC11CK-IIIi-j-139
1. predict the spontaneity of a process based
on entropy
STEM_GC11CT-IVa-b-140
2. determine whether entropy increases or
decreases if the following are changed:
temperature, phase, number of particles
STEM_GC11CT-IVa-b-141
3. explain the second law of
thermodynamics and its significance
STEM_GC11CT-IVa-b-142
4. use Gibbs’ free energy to determine the
direction of a reaction
STEM_GC11CT-IVa-b-143
1. describe reversible reactions
STEM_GC11CE-IVb-e-144
Page 12 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
2. Writing the reaction
quotient/equilibrium
constant expression
3. Predicting the direction of a
reaction
4. Significance of the
equilibrium constant
5. Le Chatelier’s Principle
CONTENT STANDARD
PERFORMANCE
STANDARD
Principle
LEARNING COMPETENCIES
CODE
2. explain chemical equilibrium in terms of
the reaction rates of the forward and the
reverse reaction
STEM_GC11CE-IVb-e-145
3. write expressions for the reaction
quotient/equilibrium constants
STEM_GC11CE-IVb-e-146
4. explain the significance of the value of the
equilibrium constant.
STEM_GC11CE-IVb-e-147
5. calculate equilibrium constant and the
pressure or concentration of reactants or
products in an equilibrium mixture
6. state the Le Chatelier’s principle and apply
it qualitatively to describe the effect of
changes in pressure, concentration and
temperature on a system at equilibrium
Acid-Base Equilibria and Salt
Equilibria
1. Bronsted acids and bases
2. The acid-base properties of
water
3. pH- a measure of acidity
4. Strength of acids and bases
5. Weak acids/weak bases and
1. acid-base equilibrium
and its applications
to the pH of
solutions and the
use of buffer
solutions
2. solubility equilibrium
and its applications
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
STEM_GC11CE-IVb-e-148
STEM_GC11CE-IVb-e-149
7. (LAB) Describe the behavior of reversible
reactions
STEM_GC11CE-IVb-e-150
8. (LAB) Describe the behavior of a reaction
mixture when the following takes place:
a. change in concentration of reactants
or products
b. change in temperature
STEM_GC11CE-IVb-e-151
9. (LAB) Perform calculations involving
equilibrium of gaseous reactions
STEM_GC11CE-IVb-e-152
1. define Bronsted acids and bases
STEM_GC11AB-IVf-g-153
2. discuss the acid-base property of water
STEM_GC11AB-IVf-g-154
3. define pH
STEM_GC11AB-IVf-g-155
4. calculate pH from the concentration of
hydrogen ion or hydroxide ions in
aqueous solutions
STEM_GC11AB-IVf-g-156
Page 13 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
ionization constants
6. Relationship between the
ionization constants of acids
and their conjugate bases
7. The Common Ion Effect
8. Buffer solutions
9. Solubility equilibria
Electrochemistry
1. Redox reactions
2. Galvanic cells
3. Standard reduction
potentials
4. Spontaneity of redox
reactions
5. Batteries
6. Corrosion
7. Electrolysis
LEARNING COMPETENCIES
5. determine the relative strength of an acid
or a base, from the value of the ionization
constant of a weak acid or base
6. determine the pH of a solution of weak
acid or weak base
7. explain the Common Ion Effect
Redox reactions as
applied to galvanic and
electrolytic cells
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
8. describe how a buffer solution maintains
its pH
9. calculate the pH of a buffer solution using
the Henderson-Hasselbalch equation
10. explain and apply the solubility product
constant to predict the solubility of salts
11. describe the common ion effect on the
solubility of a precipitate
12. explain the effect of pH on the solubility
of a precipitate
13. (LAB) Determine the pH of solutions of a
weak acid at different concentrations and
in the presence of its salt
14. (LAB)Determine the behavior of the pH
of buffered solutions upon the addition of
a small amount of acid and base
1. define oxidation and reduction reactions
CODE
STEM_GC11AB-IVf-g-157
STEM_GC11AB-IVf-g-158
STEM_GC11AB-IVf-g-159
STEM_GC11AB-IVf-g-160
STEM_GC11AB-IVf-g-161
STEM_GC11AB-IVf-g-164
STEM_GC11AB-IVf-g-165
STEM_GC11AB-IVf-g-166
STEM_GC11AB-IVf-g-167
STEM_GC11AB-IVf-g-168
STEM_GC11AB-IVf-g-169
2. balance redox reactions using the change
in oxidation number method
STEM_GC11AB-IVf-g-170
3. draw the structure of a galvanic cell and
label the parts
STEM_GC11AB-IVf-g-171
4. identify the reaction occurring in the
different parts of the cell
STEM_GC11AB-IVf-g-172
Page 14 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
LEARNING COMPETENCIES
CODE
5. write the half-equations for the reactions
occurring in the electrodes
STEM_GC11AB-IVf-g-173
6. write the balanced overall cell reaction
7. give different examples of galvanic cell
8. define reduction potential, oxidation
potential, and cell potential
9. describe the standard hydrogen electrode
10. calculate the standard cell potential
11. relate the value of the cell potential to the
feasibility of using the cell to generate an
electric current
12. describe the electrochemistry involved in
some common batteries:
a. leclanche dry cell
b. button batteries
c. fuel cells
d. lead storage battery
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
STEM_GC11AB-IVf-g-174
STEM_GC11AB-IVf-g-175
STEM_GC11AB-IVf-g-176
STEM_GC11AB-IVf-g-177
STEM_GC11AB-IVf-g-178
STEM_GC11AB-IVf-g-179
STEM_GC11AB-IVf-g-180
13. apply electrochemical principles to explain
corrosion
STEM_GC11AB-IVf-g-181
14. explain the electrode reactions during
electrolysis
STEM_GC11AB-IVf-g-182
15. describe the reactions in some
STEM_GC11AB-IVf-g-183
Page 15 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
LEARNING COMPETENCIES
CODE
commercial electrolytic processes
16. (LAB) Determine the potential and
predict the cell reaction of some
assembled electrochemical cells
17. (LAB) Describe the reactions at the
electrodes during the electrolysis of
water; cite the evidence for your
conclusion
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
STEM_GC11AB-IVf-g-184
STEM_GC11AB-IVf-g-185
Page 16 of 17
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
Code Book Legend
Sample: STEM_GC11AB-IVf-g-183
DOMAIN/ COMPONENT
LEGEND
SAMPLE
Learning Area and
Strand/ Subject or
Specialization
Science, Technology,
Engineering and Mathematics
General Chemistry
Uppercase
Letter/s
Domain/Content/
Component/ Topic
Grade 11
STEM_GC11AB
Acid-Base Equilibria and Salt
Equilibria
Roman Numeral
*Zero if no specific
quarter
Quarter
Fourth Quarter
IV
Lowercase
Letter/s
*Put a hyphen (-) in
between letters to
indicate more than a
specific week
Week
Weeks six to seven
f-g
Arabic Number
Competency
describe the reactions in
some commercial electrolytic
processes
Matter and Its Properties
MP
Measurements
MT
Atoms, Molecules and Ions
AM
S
Stoichiometry
First Entry
Grade Level
CODE
183
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 December 2013
Percent Composition and Chemical Formulas
PC
Mass Relationships in Chemical Reactions
MR
Chemical reactions and chemical equations
CR
Gases
G
Dalton’s Law of partial pressures
DL
Gas stoichiometry
GS
KMT
Kinetic molecular theory of gases
Electronic Structure of Atoms
ES
Electronic Structure and Periodicity
ESP
Chemical Bonding
CB
Organic compounds
OC
Intermolecular Forces and Liquids and Solids
MF
Physical Properties of Solutions
PP
Thermochemistry
TC
Chemical Kinetics
CK
Chemical Thermodynamics
CT
Chemical Equilibrium
CE
Acid-Base Equilibria and Salt Equilibria
AB
Page 17 of 17
General Chemistry 1
120 MINS
Lesson 1: Matter and its properties
Content Standard
Lesson Outline
The learners demonstrate an understanding of the properties of
matter and its various forms.
Introduction
Presentation of Learning Objectives and
Important Keywords
Performance Standards
Motivation
Application of the Particulate State of Matter
through Syringe Test
15
Instruction
Matter and its Properties
60
Enrichment
Demonstration on the Visualization of Matter
30
Evaluation
Written Task
10
The learners shall be able to:
1. Make a representation of the particulate nature of the three
phases of matter;
2. Discuss the difference between:
a. Pure substances and mixtures
b. Elements and compound
c. Homogeneous and heterogeneous mixtures;
3. Classify the properties of matter as:
a. Physical or chemical
b. Intensive or extensive; and
4. Perform simple separation procedures.
5
Materials
Projector, Computer, Flip charts
Resources
(1) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York: McGraw-Hill.
Learning Competencies
At the end of the lesson, the learners:
1. Recognize that substances are made of smaller particles
(STEM_GGC11-MP-Ia-b-1);
2. Describe and make a representation of the arrangements,
relative spacing, and relative motion of the particles in the three
phases of matter (STEM_GGC11-MP-Ia-b-2);
3. Distinguish between physical and chemical properties and give
examples (STEM_GGC11-MP-Ia-b-3);
4. Distinguish between extensive and intensive properties and give
examples (STEM_GGC11-MP-Ia-b-4);
5. Use properties of matter to identify substances and to separate
them (STEM_GGC11-MP-Ia-b-5);
6. Differentiate between pure substances and mixtures
(STEM_GGC11-MP-Ia-b-6);
7. Differentiate between elements and compounds
(STEM_GGC11-MP-Ia-b-7);
8. Differentiate between homogenous and heterogeneous
mixtures (STEM_GGC11-MP-Ia-b-8);
9. Recognize the formula of common chemical substances
(STEM_GGC11-MP-Ia-b-9);
10. Describe separation techniques for mixtures and compounds
(STEM_GGC11-MP-Ia-b-10); and
11. Compare consumer products on the basis of their components
for use, safety, quality, and cost (STEM_GGC11-MP-Ia-b-11).
1
INTRODUCTION (5 minutes)
Teacher Tip
Display the objectives prominently on the
board, so that the learners can track the
progress of their learning.
2. Present the keywords for the concepts to be learned:
a. Atoms
b. Chemical properties
c. Compounds
d. Distillation
e. Elements
f. Extensive properties
g. Filtration
h. Gas
i. Heterogeneous mixtures
j. Homogeneous mixtures
k. Intensive properties
l. Ions
m. Liquid
n. Magnetic separation
o. Mixtures
p. Molecules
q. Physical properties
r. Pure substances
s. Solid
Teacher Tip
List these keywords on the board or through
PowerPoint slides. Alternatively, you can write
them on flip charts. The learners will be asked
to complete a concept map using words from
this list.
1. Introduce the learning objectives by using the suggested protocol (Read-aloud):
a. I will be able to describe the particulate nature of the different forms of matter
b. I will be able to classify the properties of matter
c. I will be able to differentiate pure substance and mixtures; elements and compounds;
homogeneous and heterogeneous mixtures
d. I will be able to recognize the formulas of some common substances
e. I will be able to discuss methods to separate the components of a mixtures
f. I will be able to recognize chemical substances present in some consumer products
Another approach is to write these keywords
in meta cards of different colors.
2
MOTIVATION (15 minutes)
1. Present two 60-mL plastic syringes with the needle removed and replaced by a seal. One syringe
contains a small block of wood, while the other contains entrapped air. The plunger is set to touch
the wood block, as shown below:
2. Ask them what will happen if the plunger will be pushed down the syringe.
3. Make one learner push the plunger in the two syringes, and check if they have predicted the
behavior of the plunger in the two syringes correctly.
4. Ask them to answer the question: Why is it easier to compress the entrapped air than the wood
block?
5. Highlight that a particulate model for matter is very useful in explaining the properties of matter.
Point out that some basic concepts on matter that have been introduced in junior high school will
be reviewed in this lesson.
INSTRUCTION (60 minutes)
1. Construct the following block diagram and make the learners fill it up using the keywords listed in
the board.
Teacher Tip
This demonstration is meant to make them
realize the usefulness of visualizing matter
being made up of particles.
It is likely that their answers will be based on
what they will recall from experience and from
what they learned from junior high school:
that the plunger can be moved more easily in
the syringe containing an entrapped gas than
in the other syringe containing a solid.
A gas is made up of particles that are far
•
apart from each other, which can be
pushed closer towards each other;
•
A solid is made up of particles which are
compact, so that it is no longer possible
to push these particles closer to each
other.
Let them recognize that the keywords to be
encountered in the lesson are commonly used
to describe the things around them.
Teacher Tip
Make them take turns in filling up each box
with the correct keyword (or in placing the
proper meta cards). The block diagram can
be presented through PowerPoint slides
projected on a white board. Alternatively, it
can be prepared on flip charts or on manila
paper.
Answer for Number 1
The keywords to be placed are: atoms; ions;
molecules.
2. Ask them to answer the question: How do the following particles differ from each other?
a. Atoms
b. Molecules
c. Ions
Answer for Number 2
a. Atoms – the smallest particle
b. Molecules – composed of atoms
c. Ions – particles with charges
In case they fail to recall the differences, a
short discussion might be necessary. Also,
refer them to read Chapter 1 of the resource
book (Chang, R. & Goldsby, K., Chemistry).
3
3. Construct the following block diagram and make them fill it up using the keywords listed in the
board.
Answer for Number 3
The block diagram can be presented through
PowerPoint slides projected on a white board.
Alternatively, it can be prepared on flip charts
or on manila paper.
The keywords to be placed are:
solid; liquid; gas
Answer for Number 4
The arrangement of the particles for solid,
liquid, and gas, respectively are:
4. For the bottom layer of boxes, ask them to illustrate how the particles are distributed or arranged
in each state of matter using circles.
5. Ask them to answer the following questions:
a. How separated are the particles in each state of matter?
b. How free are the particles to move in each state of matter?
6. Ask them to classify the following substances according to the three states of matter:
a. Iron nail
b. Sugar
c. Syrup
d. Air
e. Ice
f. Alcohol
4
Answer for Number 5
•
Solid: closely packed; restricted motion
•
Liquid: far apart; free movement
•
Gas: very far apart; very free (chaotic)
movement
Answer for Number 6
a. Solid
b. Solid
c. Liquid
d. Gas
e. Solid
f. Liquid
7. Construct the following block diagram and make them fill it up using the keywords listed in the
board.
Answer for Number 7
The block diagram can be presented through
PowerPoint slides projected on a white board.
Alternatively, it can be prepared on flip charts
or on manila paper.
The keywords to be placed are: physical
properties and chemical properties (left
cluster); and extensive properties and
intensive properties (right cluster).
8. Ask them to answer the following questions:
a. What is the difference between physical properties and chemical properties?
b. How do the extensive properties differ from the intensive properties?
9. Ask them to classify the following examples as physical or chemical properties:
a. Melting of ice
b. Evaporation of water
c. Rusting
d. Digestion
5
Answer for Number 8
a. In physical properties, no change in
composition takes place during the
determination or measurement of these
properties. On the other hand, in
chemical properties, a change in
composition occurs during the
determination or measurement of these
properties.
b. Extensive properties change their value
when the amount of matter or substance
is changed. Meanwhile, intensive
properties do not change their value
when the amount of matter is changed.
In case they fail to recall the differences, a
short discussion might be necessary. Also,
refer them to read Chapter 1 of the resource
book (Chang, R. & Goldsby, K., Chemistry).
Answer for Number 9
a. Physical property
b. Physical property
c. Chemical property
d. Chemical property
10. Ask them to classify the following examples as intensive or extensive properties:
a. Boiling point
b. Weight
c. Volume
Answer for Number 10
a. Intensive property
b. Extensive property
c. Extensive property
d. Intensive property
d. Density
11. Construct the following block diagram and make them fill it up using the keywords listed in the
board.
Answer for Number 11
The block diagram can be presented through
PowerPoint slides projected on a white board.
Alternatively, it can be prepared on flip charts
or on manila paper.
The keywords to be placed are: pure
substances and mixtures (top cluster);
elements and compounds (bottom left
cluster); homogeneous mixtures and
heterogeneous mixture (bottom right
cluster).
12. Ask them to answer the question: How do pure substances differ from mixtures?
Answer for Number 12
Pure substances are composed of only one
component, while mixtures are composed of
several components.
In case they fail to recall the differences, a
short discussion might be necessary. Also,
refer them to read Chapter 1 of the resource
book (Chang, R. & Goldsby, K., Chemistry).
6
13. Present the following substances (or pictures of these substances), and ask them to answer the
question: Which of the following are pure substances and which are mixtures?
a. Table sugar
b. Table salt
c. Iodized salt
d. Brown sugar
e. Distilled water
f.
Answer for Number 13
a. Pure substance
b. Pure substance
c. Mixture
d. Mixture
e. Pure substance
f. Mixture
g. Pure substance
h. Mixture
Soft drinks
g. Oxygen gas (in tank)
h. Human breath
14. Ask them to answer the question: What is the difference between elements and compounds?
Give examples of each.
15. Ask them to answer the question: What is the difference between homogeneous and
heterogeneous mixtures?
7
Answer for Number 14
Elements are pure substances that are
•
made up of only one kind of atoms.
Possible examples: iron; gold; mercury
Compounds are pure substances made
•
up of two or more kinds of atoms.
Possible examples: salt; sugar; water
In case they fail to recall the differences, a
•
short discussion might be necessary.
Also, refer them to read Chapter 1 of the
resource book (Chang, R. & Goldsby, K.,
Chemistry).
Answer for Number 15
A homogeneous mixture has a uniform
•
composition and exhibits the same
properties in different parts of the
mixture.
A heterogeneous mixture has a non•
uniform composition and its properties
vary in different parts of the mixture.
In case they fail to recall the differences, a
•
short discussion might be necessary.
Also, refer them to read Chapter 1 of the
resource book (Chang, R. & Goldsby, K.,
Chemistry).
16. Present the following mixtures (or pictures of these mixtures), and ask them the question: Which of
the following are homogeneous mixtures? Which are heterogeneous mixtures?
a. Rubbing alcohol
b. Mixture of water and oil
c. Mixture of salt and pepper
Answer for Number 16
a. Homogeneous mixture
b. Heterogeneous mixture
c. Heterogeneous mixture
d. Homogeneous mixture
e. Homogeneous mixture
d. Carbonated soft drink
e. Human breath
17. Construct the following block diagram and make them fill it up using the keywords listed in the
board.
18. Learner prompt: Look at this diagram. Give three common ways to separate the components of
a mixture?
19. Ask them to answer the question: When can each method be used in separating the components
of a mixture?
20. Ask them to answer the question: How can the following components of the following mixtures
be separated?
a. Salt from salt water
b. Salt from a mixture of iron and salt
8
Answer for Number 17
The block diagram can be presented through
PowerPoint slides projected on a white board.
Alternatively, it can be prepared on flip charts
or on manila paper.
Answer for Number 18
Some keywords that can placed are filtration;
distillation; magnetic separation;
decantation; sublimation.
Answer for Number 19
Filtration: to separate a solid from a liquid
•
in a heterogeneous mixture using a
filtering membrane, like paper or cloth
Distillation: to separate a liquid in a
•
homogeneous mixture
Magnetic separation: to separate a
•
magnetic solid from a heterogeneous
mixture
Decantation: to separate a solid from a
•
liquid in a heterogeneous mixture based
on gravity
Sublimation: to separate a volatile solid
•
from a non-volatile solid
Answer for Number 20
a. Heating to evaporate the water
b. By adding water to dissolve the salt, and
filter or decant to separate the iron.
ENRICHMENT (30 minutes)
1. Present a demonstration for the visualization of matter. This will reinforce the concept on the
differences between pure substances, mixtures, elements, and compounds. See attached sheet.
Teacher Tip
This activity can be done at the end of the
lecture session.
2. Then conduct the learner’s activity on Visualization and Classification of Matter. See the teacher’s
guide and learner’s worksheet.
EVALUATION (10 minutes)
1. Make them do an activity wherein they will apply the visualization of matter to classify pure
substances, mixtures, elements, and compounds. See attached sheet.
2. Written task (assignment): Classify some substances found in the kitchen and in the bathroom as
pure substances or mixtures; elements or compounds; and homogeneous or heterogeneous
mixture.
Teacher Tip
This activity can be done at the end of the
lecture session. In case there is no longer
enough time, it can be done during the
laboratory session.
EVALUATION
EXCEEDS EXPECTATIONS
MEETS EXPECTATIONS
The learner classified six or more
substances in Part 1 correctly, and
ten or more substances in their
list in Part 2.
The learner classified four to five
substances in Part 1 correctly, and
six to nine substances in their list
for Part 2.
NEEDS IMPROVEMENT
NOT VISIBLE
The learner classified less than
The learner did not do the
four substances in Part 1 correctly, assigned tasks.
and less than five substances in
their list for Part 2.
9
VISUALIZATION AND CLASSIFICATION OF MATTER
Introduction
In this activity, physical models, such as balls or beads, will be used to illustrate that matter is made up of particles. A ball or a bead will
represent an atom of an element, and a combination of balls or bead will represent a compound. A collection of single balls and/or combined
balls will be used to show the difference between pure substances and mixtures.
This activity was adapted from Chemistry with Charisma, published by Terrific Science Press.
Materials
a. A set of balls or beads of two or more colors
b. Zip lock bags
Procedure
1. Assemble the following sets of balls and place them in
unlabelled zip lock bags.
4. Ask them the following questions:
a. Are the balls the same or different?
b. Do the balls represent a pure substance of a mixture?
c. Do the balls represent an element or a compound?
5. Repeat Step 3 with the mixture set.
6. Repeat Step 3 with the compound set.
7. Repeat Step 3 with the diatomic set. Highlight and discuss the
answer to the last question.
8. Introduce the concept of formulas. Each ball of a certain color
will be assigned a letter (e.g. A for the white, B for the black,
and C for another color).
9. Ask them for a possible formula for:
a. the monoatomic element (Answer: A)
b. the diatomic element (Answer: A2)
c. the compound (Answer: AB)
2. Show the bags to the learners and tell them that their task is to
find out if the bag contains a pure substance or a mixture. If the
content is a pure substance, they have to determine if it is a
monoatomic element, a diatomic element, or a compound.
3. Show them the bag with monoatomic elements, and ask them if
it is a pure substance or a mixture. Take out the contents from
the bag one by one, and show them to the learners.
10
LEARNER WORKSHEET
BAG LABEL
PURE SUBSTANCE or
MIXTURE
ELEMENT(S) or
COMPOUND(S)
A
B
C
D
11
FORMULA FOR EACH
SUBSTANCE IN THE BAG
LEARNER’S ACTIVITY: TEACHER’S GUIDE
Introduction
In this activity, physical models, such as balls or beads, will be used to illustrate that matter is made up of particles. A ball or a bead will
represent an atom of an element, and a combination of balls or bead will represent a compound. A collection of single balls and/or combined
balls will be used to show the difference between pure substances and mixtures.
This activity was adapted from Chemistry with Charisma, published by Terrific Science Press.
Materials
•
A set of balls or beads of two or more colors
•
Zip lock bags
Procedure
1. Assemble the following sets of balls and place them in zip lock
bags labeled only with the letters.
2. Distribute the set of bags and ask the learners to fill up the
provided worksheet (see below) using the bags labeled A to H.
3. Ask them to compare their results.
4. For the bags containing models of compounds, ask them to
write the formula of the compound represented by the model.
12
LEARNER’S ACTIVITY: LEARNER WORKSHEET:
VISUALIZATION AND CLASSIFICATION OF MATTER
Introduction
In this activity, physical models, such as balls or beads, will be used to illustrate that matter is made up of particles. A ball or a bead will
represent an atom of an element, and a combination of balls or bead will represent a compound. A collection of single balls and/or combined
balls will be used to show the difference between pure substances and mixtures. This activity was adapted from Chemistry with Charisma,
published by Terrific Science Press.
Materials
1. A set of balls or beads of two or more colors
2. Zip lock bags
Procedure
1. Obtain a set of bags with physical models of the particles of different substances from your teacher.
2. Examine the particles in each bag and classify them as pure substances or mixtures, monoatomic elements, or diatomic elements. Fill up
the worksheet provided below using the bags labeled A to H.
3. For the bags with models of compounds, write the formula of the compound represented by the model.
BAG LABEL
PURE SUBSTANCE or MIXTURE
ELEMENT(S) or COMPOUND(S)
A
B
C
D
E
F
G
H
13
FORMULA FOR EACH SUBSTANCE
IN THE BAG
General Chemistry 1
120 MINS
Lesson 2: Matter and its Various Forms
Content Standard
The learners demonstrate an understanding of the properties of matter and its
various forms.
Lesson Outline
Introduction
Pre-Laboratory Work
Motivation
Inquiry
The learners shall be able to
Instruction
Experiment
90
1. Perform simple separation procedures.
Enrichment
Discussion of Alternative Procedures for
the Separation
15
Evaluation
Submission of the Report on the
Experiment
Performance Standard
Learning Competency
At the end of the lesson, the learners:
1. Apply simple separation techniques such as distillation, chromatography
(STEM_GGC11-MP-Ia-b-12).
10
5
Materials
Laboratory glassware or alternative containers
Resources
(1) Separation of a mixture [PDF file]. Retrieved from Princeton High
School web site: http://phs.princetonk12.org/teachers/jgiammanco/
Chem%201/Labs/C2-SepMixtureLab.pdf
(2) Solar still challenge [PDF file]. Retrieved from American Chemical
Society web site: http://www.acs.org/content/dam/acsorg/global/
iyc2011/global-water-experiment-purification.pdf
14
INTRODUCTION (10 minutes)
1. This introduction can serve as a pre-laboratory discussion prior to the experiment proper.
2. Ask the learners to recall how to differentiate a pure substance from a mixture.
3. Point out that mixtures are common and that in some situations, it is necessary to separate the
components or to isolate one component of a mixture.
Teacher Tip
A laboratory experiment sheet has to be prepared
and distributed to the learners.
The experiment described in Annex 1 could be
adopted or revised to suit the available facilities.
4. State the objective of the experiment they will be performing.
MOTIVATION (5 minutes)
1. Ask them how table salt is obtained from seawater.
2. As an alternative, you can ask how drinking water is obtained from seawater.
INSTRUCTION (90 minutes)
1. Provide each group with a prepared mixture of salt, sand, and iron filings.
2. Ask them to follow the procedure in the experiment sheet.
Teacher Tip
This could be given as an assignment before the
laboratory session. They will be asked to search
the internet on how these processes are actually
carried out.
Teacher Tip
Low-cost (or zero-cost materials) can be used in
place of the materials described in the experiment
sheet:
a. A vial can be used instead of the evaporating
dish, and the watch glass can be omitted. A
moistened filter paper can be used to cover
the vial.
b. A vial or a small bottle can be used in place of
a beaker.
c. A plastic funnel used at home can be a
substitute for the glass funnel.
They can be asked to make a flow diagram of the
procedure.
If desired, the experiment can be performed as a
quantitative procedure wherein the isolated
substances will be dried and weighed.
15
ENRICHMENT (15 minutes)
1. In the post-laboratory discussion, ask them what properties of the components were used to
separate each from the other.
2. Discuss possible alternative procedures for the separation.
3. They can be asked to perform the Solar Still Challenge, as described in the following internet
webpage: http://www.acs.org/content/dam/acsorg/global/iyc2011/global-water-experimentpurification.pdf
EVALUATION
1. Ask them to submit a report on the experiment.
2. They could be provided with a worksheet that they have to fill up, which could include some
questions.
Teacher Tip
The volatile nature of naphthalene enabled its
sublimation. Point out that the odor of
naphthalene is caused by the vapor it produces.
The difference in the solubility of sodium chloride
and sand (or silicon dioxide) in water was used in
separating the two components.
An alternative procedure could involve the
differences in the solubility of the components in
alcohol and in water.
a. Naphthalene dissolves in ethanol but not in
water.
b. Sodium chloride dissolves in water, but not in
alcohol.
c. Silica does not dissolve in alcohol and in
water.
This experiment was conducted as part of the
Global Experiment for the International Year of
Chemistry in 2011. It could be done to motivate
their innovative skills.
EVALUATION
EXCEEDS EXPECTATIONS
The learner:
i. performed the experiment
correctly;
ii. described the results
correctly;
iii. discussed the results of the
experiment very well; and
iv. performed the Solar Still
Challenge.
MEETS EXPECTATIONS
NEEDS IMPROVEMENT
The learner:
The learner:
i. performed the experiment
i. performed the experiment
correctly;
correctly;
ii. described the results correctly; ii. described the results correctly;
iii. discussed the results of the
but
iii. did not discuss the results of
experiment well, but
iv. did not perform the Solar Still
the experiment, and
iv. did not perform the Solar Still
Challenge.
Challenge.
16
NOT VISIBLE
The learner:
i. did not do the assigned
task.
SEPARATION OF THE COMPONENTS IN A MIXTURE
Introduction
Several components, which retain their identity and characteristic properties, are present in a mixture. No chemical reactions occur between
the components of a mixture. Many of the materials surrounding us are mixtures, such as soil, cement, soft drinks, and pharmaceuticals. In this
experiment, the components of a mixture will be separated from each other. The techniques applied for this separation does not involve a
chemical reaction, so that the isolated components will retain their identity.
Materials
1. A mixture containing the following:
a. Sodium chloride, NaCl
b. Naphthalene
c. Silicon dioxide, SiO2 (sand)
2. Digital balance
3. Beaker
4.
5.
6.
7.
8.
9.
Funnel
Watch glass
Masking tape
Evaporating dish
Filter paper
Hot plate
Procedure
1. Weigh 0.50 to 0.60 g of the mixture on the digital balance.
2. Place the mixture on an evaporating dish and cover it with the pre-weighed watch glass.
3. Seal the sides with masking tape.
4. Place a moist tissue paper over the watch glass, and gently heat the evaporating dish until white vapors are emitted.
5. Cool the setup and carefully remove the watch glass. Describe the solid adhering to the watch glass.
6. Pour distilled water into the mixture remaining in the evaporating dish and stir it carefully.
7. Filter the mixture and collect the filtrate in the pre-weighed beaker. Wash the residual solid in the filter paper with a small amount of water,
combining the washing with the filtrate.
8. Gently heat to evaporate the water in the filtrate.
9. Cool the beaker. Describe the solid remaining in it
10. Dry the filter paper with the sand in an oven at 100oC. Describe the solid remaining in the filter paper.
17
Treatment of Results
1. Record the description of the substances isolated in the experiment. Tabulate your data below:
DESCRIPTION
Solid adhering to the watch glass
Solid remaining in the beaker
Solid remaining in the filter paper
2. Knowing the substances present in the mixture, identify the isolated solids.
IDENTITY
Solid adhering to the watch glass
Solid remaining in the beaker
Solid remaining in the filter paper
3. Devise another procedure to separate the components of the mixture used in the experiment.
18
General Chemistry 1
60 MINS
Lesson 3: Measurements
Content Standard
The learners demonstrate an understanding of measurement and the
difference between accuracy and precision.
Lesson Outline
Introduction
Communicating Learning Objectives
3
Motivation
Why is Measurement Important?
7
1. Discuss the need and describe the result of a measurement, in general;
Instruction
Demonstration
30
2. Differentiate between the accuracy and precision of a measurement;
Enrichment
Laboratory Experiment
15
Evaluation
Take-home Activity
Performance Standards
The learners shall be able to:
3. Point out possible sources of errors in a measurement; and
4. Carry out a measurement and report the results correctly.
Materials
Projector, Computer, Flip charts
Learning Competencies
Resources
(1) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York:
McGraw-Hill.
At the end of the lesson, the learners:
1. Explain the need for measurements;
2. Describe how to carry out measurements of length, mass, and volume; and
3. Dfferentiate between precision and accuracy (STEM_GC11MT-Ib-13).
19
5
INTRODUCTION (3 minutes)
1. Introduce the following learning objectives using the suggested protocol (Read-aloud):
a. I will be able to describe the need for measurement
Teacher Tip
The lesson is essentially a review of some concepts
presented and used in junior high school.
b. I will be able to carry out simple measurements of length, volume, and mass
c. I will be able to differentiate the accuracy and the precision of a measurement
Teacher Tip
List these keywords on the board or through
PowerPoint slides. Alternatively, you can write
them on flip charts.
2. Present the keywords for the concepts to be learned:
a. Measurements
b. Units of measurement
c. Accuracy
d. Precision
e. Significant figures
f.
Errors
MOTIVATION (7 minutes)
1. Present to two plastic bottles containing different amounts of water, and ask the learners to
describe and differentiate the two objects. Make them realize the need to use a number (the
volume of the water content or the weight of the bottles and their contents) to describe the
objects more clearly and to differentiate them.
2. Make them realize the need for a quantitative or a numerical description of some properties of
matter, and how this is applied in their daily lives. Ask them to cite some situations in daily life
where a measurement is important.
INSTRUCTION (30 minutes)
1. After the motivation, they will see the importance of a quantitative description of some
parameters, such as length, mass, and volume.
2. Call two learners separately. Ask each one to measure the length of a table without using a
ruler, meter stick, or tape measure. Make them write their measurements on the board
(number, unit: e.g., 3 hand spans).
3. Ask the class to compare the results and explain for differences or similarities. Ask them to
answer the question: Why is there a need to use a common unit for measurement?
20
Teacher Tip
Alternatively, a small and a big ball of the same
color and material can be used. Another option is
to use a small and a long plastic ruler.
Sample Responses
a. Measuring the ingredients during cooking (or
baking)
b. Measuring the weight of salt being purchased
Teacher Tip
It is expected that the learners will use the span of
their fingers, hands, or arms for the measurement.
4. Introduce the concept of unit of measurement, which is a means for a quantitative description
of a property. Highlight the need for a common or universally accepted unit of measurement.
5. Point out that for scientific measurements, a common system has been agreed upon and is
used by all scientists anywhere and all the time. Ask them to answer the question: What is the
measurement system adopted in scientific measurements?
6. Post this table on the board and ask them to supply the unit for each property.
PROPERTY
Teacher Tip
The results of the measurements will be different
because of the difference in the length of their
finger, hand, or arms.
Make them recall from their Science class in junior
high school that the International System (or SI) of
Measurement is being used in measurements in
science.
SI Units
Teacher Tip
The table can be presented through PowerPoint
slides projected on a white board. Alternatively, it
can be prepared on flip charts or on manila paper.
Length
Mass
It is expected that they will be able to fill up the
table, recalling what they have learned from junior
high school.
Volume
Time
Temperature
PROPERTY
SI Units
Length
7. Ask them to cite some examples where these units of measurements are used in real life.
Mass
Sample responses:
Volume
•
Length – in measuring the height of a person; distances; the size of cloths
•
Mass – in measuring the weight of a person; the amount of salt or sugar being bought
Time
•
Volume – in measuring the amount of a liquid (e.g. soft drinks)
•
Time – in measuring the duration of an event (e.g. to run through a distance)
•
Temperature – in measuring the body temperature of a person or of the atmosphere.
8. Ask them to group into pairs. Tell them to measure the length, width, and thickness of a
book, and record their results on the following table (to be shown on the board).
Temperature
In case they fail to recall the correct units of
measurement, a short discussion might be
necessary. Also, refer them to read Chapter 1 of
the resource book (Chang, R. & Goldsby, K.,
Chemistry).
All pairs should measure the same book.
21
MEASUREMENTS
TRIAL 1
TRIAL 2
TRIAL 3
Length
Mass
Volume
Teacher Tip
The correct results will include two decimal units.
Time
Temperature
9. When the pairs have completed the measurements and recorded their results on the table,
ask them to answer the question: How many significant figures did you use in reporting your
measurements?
10. Explain that based on the calibration of the ruler, the measurement is certain until the first
decimal unit and that the result can include one insignificant or uncertain figure.
!
The concept of significant figures has been
presented in junior high school, but it might not
have been fully understood. Therefore, reviewing
it would be worthwhile. For the guidelines for
using significant figures, see Chapter 1 of the
resource book (Chang, R. & Goldsby, K.,
Chemistry).
Let them examine the ruler they used.
At the end of this short activity, you should
address misconceptions that they have on the
concepts presented.
The concepts of accuracy and precision have been
presented in junior high school. It would be worth
reviewing these concepts.
11. Ask them to examine the results of the three measurements that they made on the length,
width, and thickness of the book.
Ask them to answer the following questions:
a. Are the results of each measurement (length, width or thickness) close to each other?
b. Were the measurements accurate or precise?
12. Write the actual length, width, and thickness of the book on the board, and ask them to
compare their results with this value.
22
Point out that the closeness of the results of a
measurement to each other is expressed by its
precision.
It is not suggested that they should be made to
quantify precision in terms of standard deviation.
This will be done in their course on Mathematics
(or Statistics).
Ask them to answer the following questions:
a. Are the results of each measurement (length, width, or thickness) close to the true value?
b. Were the measurements accurate or precise?
13. Let them recall the difference between accuracy and precision. Then, state the definitions of
accuracy and precision as used in measurement.
Answer Key
It is likely that:
a. The results will be close to the true value;
b. The measurements were accurate
Point out that the closeness of the results of a
measurement to the true value is expressed by its
accuracy.
14. Evaluating the accuracy of a measurement will require the true value. However, the true value
for the dimensions of the book is not available. Point out that if twenty or more
measurements were done, the mean value can be taken as the true value. This is an
assumption in statistics.
15. Draw the following dot plots on the board, and explain that each dot is the result of a
measurement whose value is indicated in the horizontal (or x-) axis. Tell them that the plot
presents the results of six measurements of the weight of a pebble whose true weight is 8.0 g.
Ask them to determine whether each measurement is accurate or inaccurate, and precise or
imprecise.
Teacher Tip
The dot plot can be drawn on a manila paper
before class, or presented through a PowerPoint
slide.
Answer Key
(A) Accurate and precise
(B) Accurate and imprecise
(C) Inaccurate and imprecise
(D) Inaccurate and precise
23
16. Highlight that the measurement they made could have errors, which could:
I.
Cause the result to be far from the true value (low accuracy). These errors are known as
systematic errors.
II. Cause the results to be different from each other (low precision). These errors are known
as random errors.
Ask them to answer the question: What possible errors did the person who made the
measurements commit to lower the accuracy of the results? To lower the precision of the results?
ENRICHMENT (15 minutes)
Make the learners perform a laboratory experiment on the determination of density. This activity
will reinforce the concept of measurements, the units used, and the concept of significant figures.
EVALUATION (5 minutes)
1. Assign them to read the labels of some canned or bottled goods in the kitchen, and report
the mass or volume of the contents.
2. Let them classify the following measurement data as high precision or low precision:
a. Volume of a liquid: 11.0 cm3, 11.3 cm3, 10.9 cm3, 11.1 cm3
b. Mass of a solid: 25.0 g, 23.0 g, 20.0 g, 28.0 g
24
Teacher Tip
Point out that the errors could be due to the
measuring instrument or due to the person doing
the measurement
Teacher Tip
Refer to the Teacher’s Guide for this laboratory
activity.
General Chemistry 1
120 MINS
Lesson 4: Measurements (Laboratory)
Content Standard
Lesson Outline
The learners demonstrate understanding of basic measurement skills.
Introduction
State the Objectives of the Experiment
5
The learners shall be able to:
Motivation
Application of Density Data
5
1. Carry out a measurement and report correctly the results.
Instruction
Experiment
90
Enrichment
Discussion of the Interpretation of the
Graph
20
Evaluation
Report
Performance Standard
Learning Competency
At the end of the lesson, the learners:
1. determine the density of a liquid (STEM_GC11MT-Ib-14).
Materials
Simple laboratory glassware or low-cost alternatives
Resources
(1) Laboratory experiment in Annex 1
25
INTRODUCTION (5 minutes)
1. State the objective of the experiment that the learners will be performing.
2. Ask them to recall the definition of density and the formula for calculating it.
3. Review the methods for measuring weight and volume.
Teacher Tip
A laboratory experiment sheet has to be prepared
and distributed to the learners. The experiment
found in the Annex makes use of low-cost
materials.
Density is used as a means to obtain the
concentration of a solution.
MOTIVATION (5 minutes)
1. Point out some application of density data in industry.
INSTRUCTION (90 minutes)
Each group should be provided with different
concentrations so that the relationship between
density and concentration can be shown. Sugar
solution can be used instead of salt solution.
ENRICHMENT (20 minutes)
This relationship can be used as a means to
determine the concentration of a solution.
1. Provide each group with a salt solution of a given concentration.
2. Ask them to follow the procedure in the experiment sheet.
1. Discuss the interpretation of the graph between density and the concentration of the solution.
2. Assign them internet research on the density of the following:
a. Regular soda in can
b. Light soda in can
c. Soda with aspartame in can
3. Ask them to explain the difference in density of these soft drinks.
EVALUATION
Point out that this relationship is used in industry
to monitor the concentration of some solutions.
The different drinks contain different
concentrations of sugar, so their density will vary.
They could be provided with a worksheet that they
have to fill up. It could include some questions.
1. Ask them to submit a report on the experiment.
EXCEEDS EXPECTATIONS
MEETS EXPECTATIONS
NEEDS IMPROVEMENT
The learner:
The learner:
The learner:
i.
i.
i.
performed the experiment
correctly;
performed the experiment
correctly;
performed the experiment
correctly;
ii. described the results
correctly; and
ii. described the results correctly;
and
ii. described the results correctly;
but
iii. discussed the results of the
experiment very well.
iii. discussed the results of the
experiment well.
iii. did not discuss the results of
the experiment.
26
NOT VISIBLE
The learner:
i. did not do the assigned
task.
DENSITY OF AN AQUEOUS SOLUTION
Introduction
Density is an important property of matter. It expresses the weight of a unit volume of a substance, is used to characterize substances, and can
provide a means for the identification of a solid, a liquid, or a gas.
In this experiment, the density of an aqueous solution will be determined by measuring the weight of different volumes of these solutions.
Several solutions containing different concentration of a solute will be assigned to different groups, and the variation of the density of the
solutions with the solute concentration will be studied. The behavior that you will observe has important applications in industrial and in health
monitoring.
Materials
1. NaCl solution, in 5%, 10%, 15%, and 20% concentrations
2. Digital balance
3. Syringe, 1 mL
4. Plastic mini tray
Procedure
1. Place the plastic mini tray on the stage of the digital balance and measure its weight.
2. Measure 1 mL of the test solution into the syringe, making sure that no air bubbles are trapped.
3. Slowly transfer the liquid in the syringe onto the mini tray. Measure the weight of the tray with the solution in it.
4. Repeat Steps 1 to 3 to provide a duplicate measurement. This will be used to check the repeatability of the results.
5. Repeat the whole procedure using 2 mL and 3 mL of the solution.
27
Treatment of results
1. Record the weight of the mini tray at the beginning of the experiment. Record the weight after each addition of 1 mL, 2 mL, and 3 mL of
the sample solution.
MEASUREMENTS
TRIAL 1
TRIAL 2
Weight of empty container
Weight of empty container + 1 mL solution
Weight of empty container
Weight of empty container + 2 mL solution
Weight of empty container
Weight of empty container + 3 mL solution
2. From the data above, calculate the weight of each of the different volumes that you have added to the plastic mini tray by subtracting the
weight before the addition from the weight after the addition. Calculate the average value of the measured weights.
MEASUREMENTS
TRIAL 1
Weight of 1 mL solution
Weight of 2 mL solution
Weight of 3 mL solution
28
TRIAL 2
3. From the data in the previous table, calculate the density of the solution. Calculate the average value of the density.
MEASUREMENTS
DENSITY OF SOLUTION
Based on 1 mL solution
Based on 2 mL solution
Based on 3 mL solution
AVERAGE
4. Obtain the results from the other groups who used different concentrations of the solution. Tabulate the density of the various solutions
studied.
CONCENTRATION
5%
10%
15%
20%%
Density, g/mL
5. Plot the concentration of the solution (in the x-axis) against its density (in the y-axis). Infer how the density varies based on the concentration
of the solution.
29
General Chemistry 1
160 MINS
Lesson 5: Atoms, Molecules, and Ions
(Lecture)
Lesson Outline
Content Standard
The learners demonstrate understanding of the structure of an atom and the
formula and the name of compounds.
Introduction
Presentation of Learning Objectives and
Important Keywords
5
Performance Standards
Motivation
The Particles that Make Up an Atom
5
Instruction
The Laws of Chemical Changes
Enrichment
Laboratory Session
10
Evaluation
Check Up Quiz
20
The learners shall be able to:
1. Describe the structure of an atom of an element;
2. Recognize and differentiate atoms, molecules, and ions; and
3. Write the formula and give the name of simple compounds.
Learning Competencies
120
Materials
Projector, Computer, Flip charts
At the end of the lesson, the learners:
1. Explain how the basic laws of matter (Law of Conservation of
Mass, Law of Constant Composition, and Law of Multiple
Proportion) led to the formulation of Dalton’s Atomic Theory
(STEM_GC11AM-Ic-e-15);
2. Describe Dalton’s Atomic Theory (STEM_GC11AM-Ic-e-16);
3. Differentiate among atomic number, mass number, and
isotopes, and which of these distinguishes one element from
another (STEM_GC11AM-Ic-e-17);
4. Write isotopic symbols (STEM_GC11AM-Ic-e-18);
5. Recognize common isotopes and their uses (STEM_GC11AMIc-e-19);
6. Differentiate among atoms, molecules, ions, and give examples
(STEM_GC11AM-Ic-e-20);
Resources
(1) Chang, R. & Goldsby, K. (2016). Chemistry (12th ed.). New York:
McGraw-Hill.
7. Represent compounds using chemical formulas, structural
formulas, and models (STEM_GC11AM-Ic-e-21);
8. Give the similarities and differences between the empirical
formula and molecular formula of a compound
(STEM_GC11AM-Ic-e-22); and
9. Name compounds given their formula and write formulas given
the name of the compound (STEM_GC11AM-Ic-e-23).
30
INTRODUCTION (5 minutes)
1. Introduce the following learning objectives using the suggested protocol (Read-aloud):
a. I will be able to describe and discuss the basic laws of chemical change
Teacher Tip
Display the objectives prominently on the board,
so that the learners can track the progress of their
learning.
b. I will be able to discuss how Dalton’s Atomic Theory could explain the basic laws of
chemical changes
c. I will be able to give the information provided by the atomic number and mass number of
an atom and its isotopes
d. I will be able to differentiate atoms, molecules, and ions
e. I will be able to write the chemical formula of some molecules
f.
I will be able to differentiate a molecular formula and an empirical formula
g. I will be able to give the name of a compound, given its chemical formula
Teacher Tip
List these keywords on the board. They will be
asked to complete a concept map based on words
on this list.
2. Present the keywords for the concepts to be learned:
a. Law of Conservation of Matter
b. Law of Definite Proportion
c. Law of Multiple Proportion
d. Dalton’s Atomic Theory
e. Atomic number
f.
Mass number
g. Isotope
h. Atom
i.
Molecule
j.
Ion
k. Chemical formula
l.
Molecular formula
m. Empirical formula
31
MOTIVATION (5 minutes)
1. Call one of the learners to the front and give him/her a piece of paper. Ask him/her to cut the
paper in half, and then cut one of the halves again in half, and again and again. Let him/her
proceed as long as s/he can cut a piece into half.
2. Ask him/her the question: Can you go on cutting the paper into half?
3. Tell him/her that though the cutting can go on and on mentally, there is a physical limit to this
process. It is impossible to cut the paper into half forever. There is a limit – a point where the
piece can no longer be divided.
4. Highlight that the limit is an indivisible piece, which was called by the Greek philosopher
Democritus as the atom.
5. However, beginning in the late 1800s, experiments have indicated that atoms are made up of
smaller particles.
6. Ask them the question: What are these particles that make up the atom?
7. Point out that the science of chemistry is based on the concept of the atom and molecules.
Knowledge of the atoms and molecules in the environment and in biological systems has
provided an understanding of the changes occurring in them. It has also allowed the
prediction of their behavior and the solution to any problem observed in their behavior.
INSTRUCTION (120 minutes)
1. Present the laws of chemical changes. These laws were inferred from several experiments
conducted during the 18th century using a balance for the measurements:
a. Law of Conservation of Mass
b. Law of Definite Proportion
c. Law of Multiple Proportion
2. Introduce the Law of Conservation of Mass: In a chemical reaction, no change in mass takes
place. The total mass of the products is equal to the total mass of the reactant.
3. Antoine Lavoisier, a brilliant French chemist, formulated this law by describing one of his
experiments involving mercuric oxide. He placed a small amount of mercuric oxide, a red
solid, inside a retort and sealed the vessel tightly.
32
Teacher Tip
The law might have been presented in the Science
course in junior high school. In this case, ask a
learner to state the law.A PowerPoint slide can be
prepared for this part.
He weighed the system, and then subjected it to high temperature. During the heating, the red
solid turned into a silvery liquid. This observation indicated that a chemical reaction took place.
After which, the setup was cooled and then weighed. The weight of the system was found to be
the same as before heating.
Illustrate an application of this law through the following problems. Ask them to solve the
problems in their seats, and ask one learner to write his/her solution on the board:
a. How many grams of water will be formed if 1.00 g hydrogen gas reacts with 8.00 g
oxygen? The reaction can be represented by the following word equation:
hydrogen + oxygen ! water
b. 5.58 g iron reacted with 3.21 g sulfur. How many grams of iron (II) sulfide were produced?
The reaction involved was:
iron + sulfur ! iron(II) sulfide
c. Magnesium burns in air to form magnesium oxide, as represented by the following word
equation:
magnesium + oxygen ! magnesium oxide
When 2.43 g magnesium was burned, 4.03 g magnesium oxide was produced. How many
grams of oxygen reacted with the magnesium?
d. Ammonia is produced by the reaction of nitrogen with hydrogen:
nitrogen + hydrogen ! ammonia
33
Teacher Tip
The law might have been presented in the Science
course in junior high school. In this case, ask a
learner to state the law.A PowerPoint slide can be
prepared for this part.
How many grams of nitrogen combined with 50.0 g hydrogen is needed to yield 283.3 g
ammonia?
4. State the Law of Definite Proportion: A compound always contains the same constituent
elements in a fixed or definite proportion by mass.
If water samples coming from different sources are analyzed, all the samples will contain the
same ratio by mass of hydrogen to oxygen.
Teacher Tip
The law might have been presented in the Science
course in junior high school. In this case, ask a
learner to state the law.
This experiment can be best described using a
PowerPoint slide. A picture of the burning
magnesium can be included in the slide.
5. Illustrate the application of this law using the previous example of magnesium reacting with
oxygen:
a. Describe an experiment wherein different amounts of magnesium powder are heated
in air.
b. Magnesium burns brightly in air and reacts with oxygen. During the reaction, the gray
powder turns into a white substance. The reaction causes the weight of the solid to
increase.
c. The following data were collected:
WEIGHTS OF
MAGNESIUM
WEIGHT OF
PRODUCT
WEIGHT OF
OXYGEN
COMBINED WITH
MAGNESIUM
Length
Mass
Volume
Time
Temperature
34
RATIO OF MASS
OF OXYGEN TO
MASS OF
MAGNESIUM
Magnesium
Product
Oxygen
Ratio
3.00
7.56
4.56
1.52
5.00
12.60
7.60
1.52
7.00
17.64
10.64
1.52
d. Ask them to complete the third column by applying the Law of Conservation of Mass.
e. Ask them to fill up the fourth column by dividing the mass of oxygen (third column) by
the mass of the magnesium (first column).
6. Ask them to solve the following problems:
a. In the first problem given earlier, it was given that 1.00 g hydrogen combines with 8.00 g
oxygen. How many grams of hydrogen will react with 10.00 g oxygen?
b. In the previous set of problem, it was seen that 5.58 g iron reacted with 3.21 g sulfur.
Based on this information, calculate how many grams of iron will combine with 80.0 g
sulfur.
Teacher Tip
Ask them to solve the problem in their seats. Call
one learner to write his/her solution on the board.
Answer Key
1. 1.25 g
Solution:
2.
139 g
Solution:
7. Present the Law of Multiple Proportions: If two elements can combine to form more than
one compound, the masses of one element that will combine with a fixed mass of the other
element are in a ratio of small whole numbers.
8. Illustrate the application of this law using the example of carbon which reacts with oxygen to
form carbon monoxide and carbon dioxide.
a. In carbon monoxide, 1.00 g carbon combines with 1.33 g oxygen; whereas, in carbon
dioxide, 1.00 g carbon combines with 2.66 g oxygen.
b. It can be seen that the ratio is 1:2.
9. Remind them that laws are derived from experimental results. A theory is formulated to
provide an explanation to the laws.
35
The law might have been presented in the Science
course in junior high school. In this case, ask a
learner to state the law.
Pictures or meta cards with chemical formulas may
be posted on the board and used to facilitate
discussion. It is highly encouraged to use pictures
of actual substances.
Dalton’s Atomic Theory, proposed by John Dalton, can be used to explain the laws of chemical
change. This theory is based on the following set of postulates:
1. Elements are made up of very small particles known as atoms.
2. All the atoms of an element are identical in mass and size, and are different from the atoms
of another element. Dalton used the different shapes or figures to represent different
elements, as follows:
Teacher Tip
Draw atoms to clarify each postulate, particularly
Postulates 2, 3, and 4.
Drawing the Dalton symbols for the element will
facilitate the understanding of Postulates 2 and 3.
•
Oxygen
Hydrogen
Carbon
Nitrogen
Phosphorus
Sulfur
3. Compounds are composed of atoms of more than one element, combined in definite ratios
with whole number values.
Carbon
monoxide
Carbon
dioxide
Nitric
oxide
4. During a chemical reaction, atoms combine, separate, or rearrange. No atoms are created
and no atoms disappear.
+""""2
Carbon
Oxygen
Carbon
dioxide
5. Ask them which postulate could provide an explanation for the:
Answer Key
a. Postulate 4
b. Postulate 3
a. Law of Conservation of Mass
b. Law of Definite Proportion
6. Remind them that during the time of Dalton, the atom was believed to be the smallest
particle comprising substances. However, before the end of the 19th century, experiments
provided proof of the existence of smaller particles within the atom.
36
7. Ask them to recall the particles contained in an atom (or the subatomic particles) and
differentiate the particles in terms of location, charge, and relative mass by filling up the
following table:
PARTICLE
LOCATION
CHARGE
RELATIVE MASS
Ask them to recall the information about the composition of an atom provided by the
following:
a. Atomic number
b. Mass number
Teacher Tip
This has been presented in the Science course in
junior high school.
PARTICLE
LOCATION
CHARGE
RELATIVE
MASS
PROTON
Nucleus
+1
1
ELECTRON
Outside
nucleus
-1
0.0006
NEUTRON
Nucleus
0
1
As enrichment, assign them to read and make a
report on the discovery of the existence of the
electron, proton, and nucleus.
The concepts of atomic number and mass number
have been presented in the Science course in
junior high school.
Confirm that the above numbers are defined by the following equations:
a. Atomic number = number of protons = number of electrons in a neutral atom
b. Mass number = number of protons + number of neutrons
The table can be presented through PowerPoint
slides projected on a white board. Alternatively, it
can be prepared on a flip chart or on manila paper.
8. To apply these concepts, ask them to fill up the following table:
ATOMIC
NUMBER
MASS
NUMBER
4
9
14
28
NUMBER OF
PROTONS
NUMBER OF
ELECTRONS
8
NUMBER OF
NEUTRON
9
11
12
52
24
19
20
37
Atomic
Number
Mass
number
Number of
Protons
Number of
electrons
Number of
neutrons
4
9
4
4
5
14
28
14
14
14
8
7
8
8
9
11
23
11
11
12
24
52
24
24
28
19
39
19
19
20
9. Introduce the concept of isotopes – atoms of an element having the same atomic number
but different mass number. The existence of isotopes was shown by mass spectroscopy
experiments, wherein elements were found to be composed of several types of atoms, each
with different masses.
a. The atomic number identifies an element. The atoms of isotopes of an element have
the same number of protons and electrons.
b. The atoms of isotopes of an element differ in the number of neutrons.
10. To apply the concept of isotopes, ask them to complete the following table containing
information about the isotopes of hydrogen:
PROTIUM
(Hydrogen)
DEUTERIUM
TRITIUM
Atomic Number
1
1
1
Mass number
1
2
3
ISOTOPE
Number of protons
Number of electrons
Number of neutrons
The common hydrogen atom is protium, while deuterium is found in heavy water.
Ask them to recall the difference between the following particles:
a. Atom
b. Molecule
c. Ion
38
Teacher Tip
For better understanding of the concept of
isotopes, they can be assigned to read about mass
spectroscopy. Make them refer to General
Chemistry books instead of the internet, because
the latter might lead them to complicated
description of this technique.
Ask them to answer the following questions
afterwards:
1. What does a mass spectrometer do?
2. How does the mass spectro-meter separate
isotopes of different masses?
The table can be presented in PowerPoint slides
projected on a white board. Alternatively, it can
be prepared in flip charts or on manila paper.
ISOTOPE
PROTIUM
DEUTERIUM
TRITIUM
Atomic
Number
1
1
1
Mass
number
1
2
3
Number of
protons
1
1
0
Number of
electrons
1
1
1
Number of
neutrons
1
1
2
Let them complete the following concept map showing the relationship of these particles:
Atoms
Gain of
electrons
?
Loss of
electrons
Answer Key:
?
11. Emphasize that each element has a characteristic atom.
a. Dalton differentiated the elements and their atoms through drawings.
b. However, in present day, elements are differentiated and represented through symbols.
i.
Assign them to find information from the internet
on useful isotopes. These concepts might have
been presented in the Science course in junior
high school.
The concepts of characteristic atoms and ions
might have been presented in the Science course
in junior high school.
Many symbols are abbreviations derived from the name of the element.
ii. Some symbols are derived from their Latin names.
Atoms
Gain of
electrons
Call five or more learners to write some elements and their names and symbol on the board.
Make them recall that the difference between an ion and an atom is the presence of charges.
The simple ions are derived from atoms through the gain or loss of an electron.
Let them complete the following concept map showing the relationship of these particles:
Ions can be made up of only one atom (monoatomic) or more than one type of atom
(polyatomic).
39
CATIONS
(Positive Ions)
An alternative diagram could be:
Loss of
electrons
ANIONS
(Negative Ions)
12. Monoatomic ions are named based on the element.
a. For cations, the name of the element is unchanged. If an element can form two ions of
different charges, the name, which is usually derived from its Latin name, is modified by
the suffix –ic for the ion with the higher charge, and –ous for that with the lower charge.
Teacher Tip
The naming of the compound or molecule will be
discussed later.
b. For anions, the name of the element is modified by the suffix –ide.
Answers for Number 13
a. Zn2+ – zinc ion
b. Mg2+ – magnesium ion
c. K+ – potassium ion
d. Fe2+ – ferrous ion or iron (II) ion
e. Fe3+ – ferric ion or iron(III) ion
13. Ask them to name the following cations:
a. Zn2+
b. Mg2+
c. K+
d. Fe2+
e.
Answers for Number 14
a. Br- – bromide ion
b. S2- – sulfide ion
c. O2- – oxide ion
d. I- – iodide
Fe3+
14. Ask them to name the following anions:
a. Brb.
Teacher Tip
Provide them with a list of the common anions,
together with their names.
S2-
c. O2d. ISeveral anions are polyatomic and are named based on the atomic constituents and the
suffix – ide.
15. The most common examples are:
a. OH- – hydroxide ion
b. CN- – cyanide ion
40
16. A number of polyatomic anions containing oxygen atoms are named based on the root word
of the central (or non-oxygen) atom and the suffix –ate for the one with more oxygen atoms
and –ite for the one with less oxygen atom.
a. NO3- – nitrate ion
b. NO2- – nitrite ion
c. SO32- – sulfite ion
d. SO42- – sulfate ion
e. PO43- – phosphate ion
17. Some anions have common names ending with the suffix –ate.
a. C2H3O2- – acetate ion
b. C2O42- – oxalate ion
Point out that the composition of a molecule or an ion can be represented by a chemical
formula. The formula consists of the symbols of the atoms making up the molecule. If there
is more than one atom present, a numerical subscript is used. Examples are the following:
a. O2
– oxygen gas
b. H2O
– water
c. NaOH – sodium hydroxide (liquid Sosa)
d. HCl
– hydrochloric acid (muriatic acid)
18. Discuss that there are two types of chemical formulas:
a. Molecular formula – gives the composition of the molecule, in terms of the actual
number of atoms present. Examples are the following:
i. C6H12O6
ii. K3PO4
iii. Na2C2O4
41
Teacher Tip
They might be able to recall some compounds that
have been presented in the Science course in
junior high school, such as sodium chloride and
carbon dioxide.
b. Empirical formula – gives the composition of the molecule, in terms of the smallest ratio
of the number of atoms present. Examples are the following:
i. CH2O
ii. NaCO2
19. The naming of compounds follows a set of rules.
Start the lesson with the rule of naming of binary compounds. Binary compounds – made
up of two elements. Discuss the rules for naming in two groups of binary compounds:
Answers for Number 19
– sodium iodide
i. NaI
ii. MgCl2 – magnesium chloride
– iron (II) sulfide
iii. FeS
iv. K2O – potassium oxide
a. Ionic compounds – made up of a cation and an anion. They are named by giving the
name of the cation first, followed by the name of the anion. Ask them to name the
following compounds:
i. NaI
ii. MgCl2
iii. FeS
iv. K2O
b. Molecular compounds – made up of two non-metals. They are named by giving the
name of the first nonmetal and then that of the second nonmetal modified by the ending ide. Molecular compounds are usually gases. Ask them to name the following compounds:
i. HCl
ii. CO2
iii. SO3
20. After they have learned how to name binary compounds, discuss the rules for naming
ternary compounds – made up of three elements. The naming of ternary compounds
follows the same rule as that of the binary ionic compound: the name of the cation is given
first, followed by the name of the anion.
42
Answers for Number 20
i. HCl – hydrogen chloride
ii. CO2 – carbon dioxide
iii. SO3 – sulfur trioxide
Answers for Number 20
i. NaNO3 – sodium nitrate
ii. BaCrO4 – barium chromate
iii. K2SO4 – potassium sulfate
Ask them to name the following compounds:
i. NaNO3
ii. BaCrO4
iii. K2SO4
21. Discuss next the naming of acids. Acids – yield hydrogen ions in aqueous solutions.
a. Binary acids – composed of hydrogen and another element, usually a nonmetal. The first
part of the name starts with the prefix hydro- followed by the name of the element,
modified by the ending –ic. The second part consists of the word ‘acid’. Name = hydro(root name of element) -ic + acid
Answers for Number 21.a
i. HCl – hydrochloric acid
ii. H2S – hydrosulfuric acid
iii. HI – hydroiodic acid
Ask them to name the following binary acids:
i. HCl
ii. H2S
iii. HI
b. Ternary acids – made up of hydrogen and an anion, usually containing oxygen. The first
part of the name consists of the root word of the name of the element, modified by the
ending –ic. The second part consists of the word ‘acid’. If there is another acid with the
same atoms, the suffix –ous is used to denote the one with less number of atoms. Name =
(root name of element) -ic (or –ous) + acid
Ask them to name the following ternary acids:
i. HNO3
ii. HNO2
iii. H2SO4
iv. H2SO3
v. H3PO4
43
Answers for Number 21.b
i. HNO3 – nitric acid
ii. HNO2 – nitrous acid
iii. H2SO4 – sulfuric acid
iv. H2SO3 – sulfurous acid
v. H3PO4 – phosphoric acid
22. After they have become familiar with the naming of compounds, it would be easy to write the
formula of the compound. Emphasize that in writing the formula, the total positive charges
of the cations should be equal to the total of the negative charges of the anion. The net
charge should be zero.
Answers for Number 22
– AgNO3
i. Silver nitrate
ii. Potassium iodide – KI
iii. Nitrogen dioxide – NO2
iv. Barium chloride – BaCl2
v. Hydrobromic acid – HBr
Ask them to write the formula of the following compounds, given the name of the
compound:
i. Silver nitrate
ii. Potassium iodide
iii. Nitrogen dioxide
iv. Barium chloride
v. Hydrobromic acid
ENRICHMENT
1. Conduct a laboratory session on the naming of compounds and on formula writing.
EVALUATION (20 minutes)
Check-up Quiz
Choose the best answer from among the choices given:
1. In one experiment, 0.558 g element X was found to react with 0.320 g element Y to form
only one product, compound Z. How many grams of compound Z were formed?
A. 0.238 g
C. 0.558 g
B. 0.320 g
D. 0.878 g
2. When 24.3 g magnesium reacts completely with 16.0 g oxygen, exactly 40.3 g magnesium
oxide is formed. Which of the following laws is illustrated by this observation?
A. Law of Definite Proportion
C. Law of Conservation of Mass
B. Law of Multiple Proportion
D. Law of Conservation of Energy
44
Teacher Tip
Refer to the laboratory teaching guide of this
lesson as well as the Formula Writing and Naming
of Compounds data table.
3. Which of the following statements is consistent with Dalton’s Atomic Theory?
A. The atoms of element A are identical with the atoms of another element D.
B. The atoms of element A have the same mass as the atoms of another element D.
C. The atoms of element A are different from the atoms of another element D.
D. The atoms of element A have the same properties as the atoms of another element D.
4. According to Dalton’s atomic theory, which of the following is involved in a chemical
reaction?
A. The conversion of one atom into another
C. The formation of a new atom
B. The combination of atoms
D. The disappearance of an atom
5. Which of the following subatomic particles has the smallest mass?
A. Electron
C. Proton
B. Neutron
D. Nucleus
6. In which of the following quantities will two isotopes of an element have different values?
A. Atomic number
C. Number of protons
B. Mass number
D. Number of electrons
7. Which of the following information on the number of protons (p), electrons (e) and neutrons
(n) is correct for 92 U238?
A. 92 p, 92 n, 92 e
C. 238 p, 146 n, 238 e
B. 92 p, 146 n, 92 e
D. 146 p, 82 n, 92 e
8. What is the mass number of an atom which has 11 protons, 11 electrons, and 12 neutrons?
A. 11
C. 22
B. 12
D. 23
45
9. Which of the following data is correct for the Mg2+ ion (atomic number = 12)?
A. 12 protons and 13 electrons
C. 14 protons and 12 electrons
B. 12 protons and 10 electrons
D. 12 protons and 14 electron
10. Which of the following symbols corresponds to the element tin?
A. Ti
C. Pb
B. Zn
D. Sn
11. Which of the following takes place when a monovalent cation is formed from an atom?
A. One electron is gained.
C. Two electrons are gained.
B. One electron is lost.
D. Two electrons are shared.
12. Which of the following anions is polyatomic?
A. Iodide
C. Sulfide
B. Nitrite
D. Bromide
13. Which of the following is the correct formula of copper (II) nitrate?
A. CuNO3
C. Cu(NO3)2
B. Cu2NO3
D. Cu2(NO3)2
14. Which of the following is a binary compound?
A. Sodium nitrate
C. Sodium hydroxide
B. Sodium oxide
D. Sodium carbonate
46
General Chemistry 1
90 MINS
Lesson 6: Atoms, Molecules,
and Ions (Laboratory)
Content Standard
The learners demonstrate an understanding of the formula and the name of
compounds.
Performance Standard
The learners shall be able to:
Lesson Outline
Introduction
Review
Motivation
Names and Formulas of Compounds
Practice
Activity
Enrichment
Discussion of Answers
Materials
Exercise sheets
1. Write the formula and give the name of simple compounds.
Learning Competency
Resources
(1) Chang, R. & Goldsby, K. (2016). Chemistry (12th ed.). New York:
McGraw-Hill.
At the end of the lesson, the learners:
1. Write the chemical formulas of ionic compounds and name ionic
compounds from their formulas (STEM_GC11AM-Ic-e-24).
47
15
5
70
INTRODUCTION (15 minutes)
1. Reiterate to the learners the importance of the names and formulas of compounds. Make
them recall the basic rules involved in formula writing and chemical nomenclature.
2. Review the symbols of the common elements encountered in compound.
3. State the objective of the exercise that they will work on for the laboratory period.
MOTIVATION (5 minutes)
1. Point out that the names and formulas of compounds will be needed in the succeeding
lessons, particularly in writing chemical equations.
PRACTICE (70 minutes)
1. Provide each of them a copy of the exercise worksheet, and ask them to answer the exercise.
ENRICHMENT
Teacher Tip
Point out that the formula gives qualitative and
quantitative information about the composition of
a compound. It shows what elements make up the
compound (qualitative information) and the mole
ratio of the elements (quantitative information).
Call the learners one by one and ask him/her to
give the symbol of an element which you will
name.
Teacher Tip
The exercise worksheet given in Annex 1 could be
adopted or revised.
Each learner will work independently. It might be
best to keep the exercise as a closed-book activity,
and discourage them from consulting one another.
At the end of the exercise, let them check the
answers of their fellow learners who are seated
away from them.
1. Ask them to identify where they committed mistakes. Discuss the correct answers.
EVALUATION
EXCEEDS EXPECTATIONS
MEETS EXPECTATIONS
The learner answered more than
90% of the items correctly.
The learner answered 70% to 89%
of the items correctly.
NEEDS IMPROVEMENT
The learner answered less than
70% of the items correctly.
48
NOT VISIBLE
The learner did not answer any
item correctly.
FORMULA WRITING AND NAMING OF COMPOUNDS
Section 1: Ion names
Section 2: Ions from formulas
Complete the table by writing the name or formula of the ionic
species.
Complete the chart by writing the formula of the ions and of the
compounds.
ION
NAME
COMPOUND
Na+
KCl
Ca2+
Ba(NO3)2
magnesium ion
FeSO4
manganese (II) ion
Li2CO3
Fe3+
Na2O
chromium (III) ion
(NH4)2SO4
Ba2+
Al(OH)3
ClNO3phosphate ion
OHchromate ion
C2O42permanganate ion
49
POSITIVE ION
NEGATIVE ION
Section 3: Writing formulas from chemical names
Section 4: Chemical names from formulas
Write the formula of the ions expected from the following
compounds.
Write the chemical name of the ions expected from the following
compounds.
COMPOUND
POSITIVE
ION
NEGATIVE
ION
FORMULA
FORMULA
Calcium
sulfate
ZnCl2
K3PO4
Potassium
chloride
Cu(NO3)2
Na2CrO4
Tin (IV) oxide
Ni(OH)2
Lead iodide
BaO
Bismuth
nitrate
(NH4)2C2O4
Sodium
carbonate
Strontium
chromate
50
POSITIVE
ION
NEGATIVE
ION
NAME
Section 5: Binary covalent compounds
Section 6: Acids and bases
Complete the table below by filling up the missing formula or
chemical name.
Complete the table below by filling up the missing formula or
chemical name.
FORMULA
NAME
FORMULA
NO2
NAME
hydroiodic acid
phosphorus trichloride
potassium hydroxide
carbon monoxide
HClO
SbBr5
H2S
sulfur tetraiodide
perchloric acid
hydrogen peroxide
Zn(OH)2
P2O5
H3PO4
silicon dioxide
nickel(II) hydroxide
nitrogen trifluoride
sulfuric acid
CI4
HNO2
Mg(OH)2
carbonic acid
51
General Chemistry 1
60 MINS
Lesson 7: Atomic Mass
Content Standard
Lesson Outline
The learners demonstrate an understanding of the mole concept in relation to
Avogadro’s number and mass.
Introduction
Communicating Learning Objectives
Performance Standards
Motivation
Activity: Counting by Weighing
10
The learners shall be able to design, using multimedia, demonstrations, or
models, a representation or simulation of any of the following:
Instruction
Average Atomic Mass
35
1. Atomic structure
Enrichment
Vitamins and Minerals
2. Mass relationships in reactions
Evaluation
Check Up Quiz
Learning Competency
At the end of the lesson, the learners:
1. Explain relative atomic mass and average atomic mass (STEM_GC11SIe-25).
Specific Learning Outcomes
At the end of the lesson, the learners shall be able to:
1. Define atomic mass unit;
2. Calculate the average atomic mass of elements;
3. Determine the average molecular mass of molecules; and
4. Determine the average formula mass of ionic compounds.
52
5
10
Resources
(1) Burdge, J & Overby, J. (2012). Chemistry: Atoms first. New York:
McGraw-Hill.
(2) Chang, R. &Goldsby, K. (2016). Chemistry. (12th ed.). New York:
McGraw-Hill.
(3) Isotopes and atomic mass [Simulation]. Retrieved from Phet Interactive
Simulations website: https://phet.colorado.edu/en/simulation/
isotopes-and-atomic-mass
(4) Moore, J.W., Stanitski, C.L. &Jurs, P.C. (2012). Chemistry: The
molecular science (4th ed.). Belmont, CA: Brooks Cole/Cengage
Learning.
(5) Zumdahl, SS. & Zumdahl, S.A. (2012). Chemistry: An atoms first
approach. Belmont, CA: Brooks/Cole Cengage Learning.
INTRODUCTION (5 minutes)
1. Introduce the following learning objectives using any of the suggested protocol (Verbatim,
Own Words, or Read-aloud):
a. Define atomic mass unit
b. Calculate the average atomic mass of elements
c. Determine the average molecular mass of molecules
d. Determine the average formula mass of ionic compounds
2. Present the keywords for the concepts to be learned:
a. Atomic mass unit (amu)
b. Average atomic mass
c. Molecular mass
d. Formula mass
e. Avogadro’s number
f.
Mole
3. Review isotopes
a. What are isotopes?
b. Give examples of isotopes.
c. What is the similarity between Mg-24 and Mg-25? What is their difference?
MOTIVATION (10 minutes)
Activity: Counting by Weighing
1. Ask the learners if they can count objects by weighing them. Present to them this situation:
Ms. Lilia sells shelled peanuts in a store. But she meets customers asking for 150 peanuts,
another for 750 peanuts, and another for 2,000 peanuts. Obviously, it will take Ms. Lilia a very
long time to count the peanuts. What would be another way to count them?
53
Answers for Number 3
a. Isotopes are atoms that have the same
number of protons but different number of
neutrons.)
b. Here are some examples of isotopes:
i. U-235 and U-238
ii. O-17 and O-18
iii. Kr-80, Kr-82, Kr-83
c. Mg-24 and Mg-25 both have 12 protons.
However, Mg-24 has 12 neutrons while Mg-25
has 13 neutrons.
Ms. Lilia takes 20 peanuts and weighs them. She finds out that 20 peanuts weigh 32 g. How
much then will each peanut weigh?
Teacher Tip
Display the question clearly. Present the problem
to the class.
This activity can be done as a class or in groups.
Give them about five minutes to reflect on the
problem. Then, guide them to the process and
the answer using the example given.
Hence the weight of 150 peanuts would be:
Take note and emphasize that not all the peanuts
will have the same mass of 1.6 g. They are not all
identical. Some will be heavier while some will be
lighter. What was done was to get the average
mass of the peanut and 1.6 g is the average mass
of a peanut. However, for purposes of counting,
what is needed is only the average mass.
It will be easier to weigh the peanuts than to count them.
Now, 960 g is appropriately how many peanuts?
This method of counting by weighing is useful for
counting very small objects, e.g. small candies,
beans, etc.
Ask them to draw a conclusion. Is it possible to count objects by weighing? Summarize the
procedure done with the peanuts. This can be done with other objects like mongo beans,
marbles, etc.
54
Answer Key
The procedure is as follows:
1. Count a given number of peanuts and weigh
them.
2. Get the average mass of a peanut. This
assumes that the objects are identical.
3. Divide the mass of a sample of peanuts by the
average mass to get the number of peanuts in
the sample.
INSTRUCTION (35 minutes)
What is the Atomic Mass and the Atomic Mass Unit?
Relate the exercise on counting peanuts by weighing to counting atoms. Ask them if it is possible
to use the same procedure to count atoms. Why or why not?
Whether it is peanuts or mongo beans or candies or atoms, the
procedure should be the same. The problem, however, is
atoms are very, very small and it is not possible to see them and
count them individually to get the average mass. We need to
look for another way to get the average mass of the atom.
Teacher Tip
Ask them to check the meaning of the word
relative when used as an adjective. Ask them to
provide their source of information. Then, let
them express the meaning in English and in
Filipino.
Briefly, relative, when used as an adjective, means
‘compared to something or to someone’.
Emphasize that they should always use the
appropriate unit in calculations. For atomic mass,
the unit is amu.
Additional information:
A mass spectrometer is used to experimentally
compare and determine the masses of atoms to a
very high degree of accuracy.
Experiments have shown that atoms have different masses
relative to one another.
For example, a Mg atom is
experimentally reported to be twice as heavy as a carbon atom;
a silicon atom is twice the mass of a nitrogen atom. It is
possible to make a relative scale if one atom is chosen as the
reference or standard atom against which the masses of the
other atoms are measured.
Answer key
6.410 x 12 amu = 76.92 amu
By international agreement, the reference atom chosen is the C-12 isotope which contains six
protons and six neutrons. By definition, one atom of C-12 has a mass of exactly 12 atomic mass
units (amu). One amu, therefore, is one-twelfth (1/12) the mass of a C-12 atom.
The atomic mass of Cu-63 is 62.93 amu. This means that relative to C-12, one atom of Cu-63 is
62.93/12 or 5.244 times the mass of a C-12 atom.
Ask them to answer this example:
One atom of Se-77 is 6.410 times as heavy as an atom of C-12. What is the atomic mass of
Se-77?
55
Average Atomic Mass
Now, ask them to look up the atomic mass for carbon in the periodic table. The expected answer
is 12.01 amu. Then, proceed to explaining the average atomic mass.
Teacher Tip
They should all have the same version of the
periodic table so that average atomic masses are
reported with the same number of significant
figures.
If C has six protons and six neutrons, why is the relative atomic mass of carbon given as 12.01
amu and not 12 amu? There are no individual atoms of carbon with a mass of 12.01 amu.
The periodic table provides the average atomic mass which takes into account the different
isotopes of an element and their relative abundances. It is not a simple average that is taken but
a weighted average.
Ask them to look up the atomic masses of other
elements to familiarize them with using the
periodic table.
Illustrate a weighted average using final grade calculation:
For the class in Chem 345, the teacher informs the class that the final grade will be based on
Exam 1 (15%), Exam 2 (15%), Problem Sets (30%), and Final Exam (40%). To pass the course, the
learner must get a final grade of 75% or higher. Calculate the final grade of learner Ms. Julita if
she got the following scores:
COMPONENTS OF FINAL GRADE
WEIGHT
SCORES OF
MS. JULITA
Exam 1
15.0%
83%
Exam 2
15.0%
95%
Problem Sets
30.0%
65%
Final Exam
40.0%
88%
The final grade will be computed as follows:
(.150 x .83) + (.150 x .95%) + (.300 x .65) + (.400 x .88) = 81%
Therefore, Ms. Julita passes the course!
56
You may want to show the difference between
simple average and weighted average using the
same values in the example given.
Always observe the use of significant figures in
calculations.
Isotopes of elements occur in different abundances. Some are more abundant than others.
Chlorine has two isotopes. The natural abundance of Cl-35 is 75% while that of Cl-37 is 25%.
This means that if you have 100 atoms of chlorine, 75 of them will be Cl-35 and 25 of them will be
Cl-37. Magnesium, on the other hand, has three isotopes with varying abundances: Mg-24,
Mg-25, and Mg-26, 11.01 have 78.99%, 10.00%, and 11.01% abundance, respectively.
Teacher Tip
Note that the atomic mass of C-12 is exactly 12
amu. In calculations, this is treated as an exact
number.
For carbon, the natural abundance of C-12 is 98.90% while that of C-13 is 1.10%. The atomic
mass of C-13 has been determined to be 13.00335 amu while that of C-12 is exactly 12 amu.
Now, we calculate the average atomic mass of carbon:
Misconception
They may think that there is a carbon atom with a
mass of 12.01 amu. There is none. There are only
atoms of C-12 and C-13. The value 12.01 amu is
an average atomic mass.
Review how exact numbers are treated in
calculations.
= (atomic mass of C-12) (% abundance of C-12) + (atomic mass of C-13) (% abundance of C-13)
= (12.0000 amu) (.9890) + (13.00335 amu) (.0110)
= 12.01 amu
Ask them to answer these practice exercises:
1. From the periodic table, look up the average atomic mass of the following elements: Co, Be,
Al, Zn.
2. Copper has two stable isotopes with the following masses and % abundances: Cu-63 (62.93
amu, 69.09% abundance) and Cu-65 (64.9278 amu, 30.91% abundance). Calculate the
average atomic mass of copper.
3. An element consists of an isotope with mass of 10.0129 amu and 19.91% abundance, and
another isotope with mass of 11.0093 amu and 80.09% abundance. Calculate the average
atomic mass of this element. Refer to the periodic table and identify the element.
57
Answer Key
1. Co (58.93 amu), Be (9.012 amu), Al (26.98
amu), Zn (65.39 amu)
2. 63.55 amu
3. 10.81 amu; the element is boron
Ensure that they observe the proper use of
significant figures in all their calculations.
Average Molecular Mass (also referred to as molecular mass)
The molecular mass is the sum of the average atomic masses of the atoms in the molecule.
Ask them to answer the following examples:
1. What is the molecular mass of carbon dioxide, CO2?
2. Determine the molecular mass of the following molecules:
Teacher Tip
Note the difference between molecular mass and
formula mass. Molecular mass is used for covalent
compounds while formula mass is used for ionic
compounds.
For brevity, many books refer to the average
molecular mass as simply molecular mass.
a. Water, H2O
b. Methane, CH4
Answer Key
1. Molecular mass of CO2
= atomic mass of C + 2 (atomic mass of O)
= 12.01 amu + 2 (16.00 amu)
= 44.01 amu
2.
Average Formula Mass (also referred to as formula mass)
The formula mass is the sum of the atomic masses of the atoms in the ionic compound.
What is the formula mass of sodium chloride, NaCl?
2.
What is the formula mass of magnesium chloride, MgCl2?
Teacher Tip
Keep the examples simple. It is the concept that
needs to be introduced. This will be taken up
again in the next lesson.
Answer key
1. Formula mass of NaCl
= atomic mass of Na + atomic mass of Cl)
= 22.99 amu + 35.45 amu = 58.44 amu
2. Formula mass of MgCl2 = 95.21 amu
Ask them to answer the following examples:
1.
a. molecular mass of water = 18.02 amu
b. molecular mass of methane
= 16.04 amu
58
Vitamins and minerals
1. Vitamins and minerals are nutrients for the body. An example of a vitamin is Vitamin C. Look
up the molecular formula of Vitamin C and determine its average molecular mass. What is
another common name for Vitamin C? Give at least one important use of Vitamin C in the
body.
Teacher Tip
This can be given as an assignment.
2. Minerals include potassium, calcium, iron, and zinc. Look up the average atomic mass of
calcium, Ca. Give at least one important use of Ca in the body.
EVALUATION (10 minutes)
Check-up quiz
Answer the following questions. Place the answers in the space provided. Show calculations
where applicable. Observe the use of significant figures for calculations and indicate the
appropriate units. Learners can use the periodic table to answer the questions.
______1. From the periodic table, look up the average atomic mass of bromine, Br.
______2. How much heavier is an atom of Br relative to an atom of carbon?
______3. Which element in the periodic table has an average atomic mass that is about ten times
that of fluorine?
Element A consists of isotope A-6 with natural abundance of 7.5% and a mass of 6.0151 amu, and
isotope A-7 with natural abundance 92.5% and mass of 7.0160 amu.
______4. Calculate the average atomic mass of element A.
______5. Identify Element A.
Naphthalene has the molecular formula C8H10.
______6. How many elements make up one molecule of naphthalene? What are they?
______7. What is the molecular mass of naphthalene?
______8. What is the formula mass of potassium chloride, KCl?
59
Answer Key
1. 79.90 amu
2. 6.653 times heavier
3. Osmium, Os
4. 6.94 amu
5. Lithium, Li
6. Two elements; Carbon and Hydrogen
7. 106.16 amu
8. 74.55 amu
General Chemistry 1
120 MINS
Lesson 8: The Mole Concept and Molar
Mass (Lecture)
Content Standard
Lesson Outline
The learners demonstrate an understanding of the mole concept in relation to
Avogadro’s number and mass.
Introduction
Communicating Learning Objectives
Performance Standards
Motivation
Inquiry
Instruction
The Mole Concept and Molar Mass
75
Enrichment
Relating the Mole to Real Life Situations
15
Evaluation
Check Up Quiz
15
The learners shall be able to design, using multimedia, demonstrations, or
models, a representation or simulation of any of the following:
1. Atomic structure
2. Mass relationships in reactions
Learning Competencies
At the end of the lesson, the learners:
1. Define a mole (STEM_GC11S-Ie-26);
2. Illustrate Avogadro’s number with examples (STEM_GC11S-Ie-27);
3. Determine the molar mass of elements and compounds (STEM_GC11SIe-28);
4. Calculate the mass of a given number of moles of an element or
compound, or vice versa (STEM_GC11S-Ie-29); and
5. Calculate the mass of a given number of particles of an element or
compound, or vice versa (STEM_GC11S-Ie-30).
Specific Learning Outcomes
At the end of the lesson, the learners shall be able to:
Resources
(1) Burdge, J & Overby, J. (2012). Chemistry: Atoms first. New York:
McGraw-Hill.
(2) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York:
McGraw-Hill.
(3) Moore, J.W., Stanitski, C.L. & Jurs, P.C. (2012). Chemistry: The
molecular science (4th ed.). Belmont, CA: Brooks Cole/Cengage
Learning.
(4) Zumdahl, SS. & Zumdahl, S.A. (2012). Chemistry: An atoms first
approach. Belmont, CA: Brooks/Cole Cengage Learning.
3. Define molar mass;
4. Determine the molar mass of elements and compounds; and
5. Perform calculations determining mass of a given number of
particles of an element or compound, or vice versa.
1. State the value of Avogadro’s number;
2. Perform calculations converting moles to number of entities and
vice versa;
60
12
3
INTRODUCTION (12 minutes)!
1. Introduce the following learning objectives using any of the suggested protocol (Verbatim,
Own Words, or Read-aloud):
a. State the value of Avogadro’s number
b. Perform calculations converting moles to number of entities and vice versa
c. Define molar mass
d. Determine the molar mass of elements and compounds
e. Perform calculations determining mass of a given number of particles of an element or
compound, or vice versa
2. Present the keywords for the concepts to be learned:
a. Avogadro’s number
b. Mole
c. Molar mass
3. Discuss the quiz given during the previous lesson. Show the answers with the corresponding
calculations.
4. Discuss the enrichment assignment given during the last lecture.
MOTIVATION (3 minutes)
1. What do chemists observe every October 23, from 6:02 am to 6:02 pm?
Teacher Tip
Ask the learners to wait until the end of the lesson
to find the significance of the date and time.
INSTRUCTION (75 minutes)
The Mole
What is a mole? What is Avogadro’s number?
Mole Day is observed by chemists every October
23, from 6:02 am to 6:02 pm.
Atoms have very small masses. We expect that macroscopic samples will contain a very large
number of atoms. A special unit of measure, called the mole, is used to deal with extremely large
numbers. In the SI system, the mole is the amount of substance that contains as many entities as
there are in exactly 12 g of C-12.
61
The number of atoms in 12 g of C-12 is experimentally determined to be 6.022 x 1023. This is
called Avogadro’s number.
Illustrate the mole with different counting units:
Misconception
Avogadro’s number is not a defined value. It is an
experimentally determined value. Mass
spectroscopy techniques are used to determine
the value of Avogadro’s number to a high degree
of accuracy.
a. 1 dozen = 12 entities or units
b. 1 dozen eggs = 12 eggs
Teacher Tip
Ask the learners to write all the zeroes for
Avogadro’s number.
c. 1 dozen papayas = 12 papayas
d. 1 dozen cars = 12 cars
Emphasize that while a dozen always has 12
entities, one dozen eggs will not have the same
mass as one dozen books, or one dozen oranges,
or one dozen cars.
e. 1 dozen books = 12 books
f.
1 pair = 2 entities
g. 1 gross = 12 dozens = 144 entities
h. 1 ream = 500 entities
i.
1 mole = 6.022 x
1023
entities = 6.022 x
1023
Teacher Tip
These are sample exercises showing conversion of
moles to number of atoms or molecules and vice
versa.
of anything
Ask them to answer the following practice exercises:
Answer Key
1. 6.022 x 1023 eggs
2. 6.022 x 1023 mongo beans
3. 6.022 x 1023 Na atoms
1. How many eggs are there in one mole of eggs?
2. How many mongo beans are there in one mole of mongo beans?
3. How many sodium atoms are there in 1 mole of Na atoms?
4. Calculate the number of atoms of argon in 0.500 moles Ar?
5. How many moles of Co are there in 4.960 x 1025 atoms of Co?
62
6. How many molecules of H2O are there in 1 mole of water molecules?
7. How many molecules of carbon dioxide, CO2, are there in 2.648 moles CO2?
Answer Key
6. 6.022 x 1023 H2O molecules
Ask them to use the unit factor method (also called
dimensional analysis) in their calculations.
8. How many atoms of oxygen are there in 2.648 moles CO2?
="3.189"x"10"24"O"atoms"
9. Determine the number of moles of ammonia, NH3, in 8.254 x 1025 molecules of ammonia.
="137.1"moles"NH3"
Molar Mass
Recall that the mole is the amount of substance that contains Avogadro’s number of units or
entities. But how much will one mole of a substance weigh? The molar mass is the mass in grams
of one mole of a substance.
One mole of C-12 has a mass of exactly 12 g and contains 6.022 x
C-12 is called the molar mass.
1023
C-12 atoms. This mass of
Notes:
1. The appropriate unit for molar mass is g/mol
2. The molar mass in grams is numerically equal to the atomic mass in amu. The molar mass in
grams is numerically equal to the molecular mass or the formula mass in amu.
63
Teacher Tip
Emphasize the use of the appropriate units in
calculations. For molar mass, the unit used is g/
mol.
For brevity, atomic mass is often used instead of
average atomic mass. It is understood that the
value in the periodic table is the average atomic
mass.
Ask them to answer the following examples:
1. What is the average atomic mass of Ca? What is the molar mass of Ca?
2. The atomic mass of Br is 79.90 amu. What is its molar mass?
3. The molecular mass of water, H2O, is 18.02 amu. What is the molar mass of water, H2O?
4. The formula mass of NaCl is 58.44 amu. What is the molar mass of NaCl?
Illustrate the relationship of amu and grams:
One mole of C-12 has a mass of exactly 12 g and one mole of C-12 has Avogadro’s number of
atoms. Calculate the mass of one atom of C-12 in grams.
Teacher Tip
Emphasize the use of the appropriate units in
calculations. For molar mass, the unit used is g/
mol.
For brevity, atomic mass is often used instead of
average atomic mass. It is understood that the
value in the periodic table is the average atomic
mass.
Answer Key
1. 40.08 amu; 40.08 g/mol
2. 79.90 g/mol
3. 18.02 g/mol
4. 58.44 g/mol
Recall the previous lesson on how to get the
molecular mass and the formula mass.
Calculate the mass in grams of 1 amu.
Therefore, 1 amu = 1.661 x 10-24 g.
Illustrate Avogadro’s number and molar mass:
SAMPLE
NUMBER OF ATOMS
in SAMPLE
MASS of 1 mole
1 mole of aluminium
6.022 x 1023 atoms
26.98 g
1 mole of copper
6.022 x 1023 atoms
63.55 g
1 mole of silver
6.022 x 1023 atoms
107.9 g
1 mole of gold
6.022 x 1023 atoms
197.0 g
64
Teacher Tip
Emphasize that while 1 mole Al, 1 mole Cu, 1 mole
Ag, and 1 mole Au will each contain the same
number of atoms, they will not weigh the same.
Similarly, one dozen apples and one dozen cars
will each have 12 units but will not weigh the
same.
Illustrate how to get the molar mass of elements and compounds through the following examples:
1. Determine the molar mass of silicon, Si.
2. Get the molar mass of zinc, Zn.
3. Which will have a higher mass: 0.500 mole zinc, Zn, or 0.250 mole lead, Pb?
Answer Key
1. 28.09 g/mol
2. 65.39 g/mol
3. Therefore, 0.250 mole of Pb has a higher mass
than 0.500 mole of Zn.
4. 78.12 g/mol
5. 45.07 g/mol
6. 73.89 g/mol
Therefore, 0.250 mole of Pb has a higher mass than 0.500 mole of Zn.
4. What is the molar mass of benzene, C6H6 ?
5. Find the molar mass of ethanol which has the following structural formula:
Teacher Tip
Emphasize that they should give the answers with
the appropriate units.
6. What is the molar mass of lithium carbonate, Li2CO3?
Using the above illustrations, ask them to do calculations involving moles, molar masses, and
Avogadro’s number.
1. How many grams of silver, Ag, are there in 1.34 moles? (This example illustrates the
conversion of moles ! grams)
65
Teacher Tip
Always observe the proper use of significant
figures in calculations. An Annex is included at the
end of this module for enrichment and review of
significant figures and rounding off in calculations.
Show how the proper use of units will facilitate the
solution of the problem through dimensional
analysis. The units cancel out, leaving the correct
unit required.
2. How many moles of copper, Cu, are there in 875 g Cu? (This example illustrates the
conversion of grams ! moles)
3. A bottle of calcium supplements in tablet form contains 268 g Ca. How many atoms
are present in 268 g calcium, Ca? (This example illustrates the conversion of grams !
moles ! number of atoms)
4. What is the mass in grams of 2.06 x 1023 atoms of potassium, K? (This example
illustrates the conversion of number of atoms ! moles ! grams)
5. Which has more atoms? 3.68 g neon atoms or 1.10 g sodium atoms?
Therefore, 3.68 g Ne will have more atoms than 1.10 g Na.
66
Teacher Tip
Allow them to analyze the way to solve the
problem using relationships before doing the
calculation. In Problem 3, for example, the grams
need to be converted to moles, then the moles
converted to number of atoms.
ENRICHMENT (15 minutes)
Teacher Tip
Return to the motivation question and ask them
why October 23, from 6:02 AM to 6:02 PM, is the
chosen date for Mole Day.
1. What do chemists observe every October 23 from 6.02 am to 6.02 pm?
2. Relate the mole to real life situations:
A. How many pesos are there in one mole of pesos? Do you think Manny Pacquiao will have
one mole of pesos? Does Bill Gates have one mole of dollars?
B. Ask them to check the Philippine national budget for one fiscal year. Does the Philippine
national budget reach one mole of pesos?
C. Ask them to look for the approximate age of the earth. Does the age of the earth
approximate one mole of years?
EVALUATION (15 minutes)
Check-up quiz
Answer the following questions. Place the answers in the space provided. Show calculations
where applicable. Observe the use of significant figures for calculations and indicate the
appropriate units. Learners can use the periodic table to answer the questions
This enrichment could be done for the more
advanced learners.
Answer Key for Enrichment
B. The Philippine National Budget for 2016 is
PHP 3.002 trillion or 3,002,000,000,000 or
3.002 x 1012 pesos. The Philippine national
budget does not reach one mole of pesos.
C. Current data show the earth to be about 4.54
billion years old. It is 4,540,000,000 years old
or 4.54 x 109 years old. The age of the earth
does not approximate one mole of years.
Answer Key for Evaluation
1. 20 x 1024 molecules
2. 41 x 1024 C atoms
_____1. How many molecules of acetylene, C2H4, are there in 2.00 moles acetylene?
_____2. How many atoms of carbon are there in 2.00 moles acetylene?
Complete the following table:
SUBSTANCE
MOLES
NO
2.88 moles
GRAMS
CCl4
MOLECULES
121.4 g
SO2
8.50 x 1024 molecules
67
Answer Key
SUBSTANCE
MOLES
GRAMS
MOLECULES
NO
2.88 moles
86.4 g
1.73 x 1024 molecules
CCl4
0.7893 mole
121.4 g
4.753 x 1023 molecules
SO2
14.11 moles
904 g
8.50 x 1024 molecules
Guidelines for Using Significant Figures
(from Chang, R. & Goldsby, K. (2016). Chemistry. (12thed.). New York: McGraw-Hill, Chapter 1, pp. 20-21)
In scientific work, significant figures are always to be observed. Here are the rules on the use of significant figures:
1. Any digit that is not zero is significant. (Eg. 483 g has three significant figures; 2,578 m has four significant figures)
2. Zeros between nonzero digits are significant. (Eg. 6.06 kg has three significant figures; 60,804 cm has five significant figures)
3. Zeros to the left of the first nonzero digit are not significant. (Eg. 0.078 L has two significant figures; 0.004 kg has one significant figure)
4. A. If a number is greater than 1, the zeros after the decimal point are significant. (Eg. 4.0 mg has two significant figures; 20.04 g has four
significant figures)
B. If a number is less than 1, only the zeros after the first nonzero digit are significant. (Eg. 0.0750 m has three significant figures; 0.4006 g
has four significant figures.
5. For numbers without decimal points, the zeroes at the end of nonzero digits may or may not be significant (ambiguous). For example, 600
g may have one or three significant figures. To avoid the ambiguity, we use scientific notation. We can say 6.00 g and this will have three
significant figures. Or we can say 6 x 102 and this will have only one significant figure.
68
How do you handle significant figures in calculations?
1. In addition and subtraction, the answer cannot have more digits to the right of the decimal point than either of the original numbers.
45.112
!
three digits after the decimal point
- 6.02
!
two digits after the decimal point
39.092
!
round-off to 30.09 so the answer will have two digits after the decimal point
2. For multiplication and division, the number of significant figures in the final product or quotient is determined by the original number that
has the smallest number of significant figures.
6.9 x 12.34 = 85.146
Round of the answer to 85, which has only two significant figures.
26.98/3.05 = 23.93
Round of the answer to 23.9, which has three significant figures because the smallest number of significant
figures in the operation is 3.
3. Remember that exact numbers are considered to have infinite number of significant figures.
Rules for Rounding Off:
1. To round off a number at a certain point, drop the digits that follow if the first of them is less than 5.
8.143 rounded off to only two significant figures becomes 8.1.
2.
To round off a number at a certain point, add 1 to the preceding digit if the number that follows is 5 or greater than 5.
7.378 rounded off to three significant digits becomes 7.38.
8.465 rounded off to three significant digits becomes 8.47.
0.575 rounded off to two significant digits becomes 0.58.
69
General Chemistry 1
120 MINS
Lesson 9: The Mole Concept and Molar
Mass (Laboratory)
Lesson Outline
Content Standard
The learners demonstrate an understanding of the mole concept in relation to
Avogadro’s number and mass.
Performance Standards
The learners shall be able to design, using multimedia, demonstrations, or
models, a representation or simulation of any of the following:
1. Atomic structure
2. Mass relationships in reactions
Learning Competencies
At the end of the lesson, the learners:
1. Define a mole (STEM_GC11S-Ie-26);
2. Illustrate Avogadro’s number with examples (STEM_GC11S-Ie-27);
3. Determine the molar mass of elements and compounds (STEM_GC11SIe-28);
4. Calculate the mass of a given number of moles of an element or
compound, or vice versa (STEM_GC11S-Ie-29); and
5. Calculate the mass of a given number of particles of an element or
compound, or vice versa (STEM_GC11S-Ie-30).
Specific Learning Outcomes
At the end of the lesson, the learners shall be able to:
1. Count the number of small objects by weighing;
2. Determine the number of moles in a given sample; and
3. Determine the number of atoms in a given sample
70
Introduction
Can you count objects by weighing
them?
10
Instruction and
Practice
Laboratory Work
80
Enrichment
Post-laboratory Discussion
20
Evaluation
Checking of Accomplished Data Tables
Materials
(1) Balance (triple beam or electronic balance)
(2) Paper cups
(3) Samples (kidney beans, mongo beans, rice, dried sago)
(4) Plastic spoons
(5) Aluminium metal or foil
(6) Iron (nails or filings)
(7) Sodium chloride (table salt, NaCl)
(8) Sucrose (table sugar, C12H22O11)
Resources
(1) Allan, Andy. The mole [PowerPoint presentation]. Retrieved from
http://www.sciencegeek.net/APchemistry/FlashPPT/3_TheMole/
index.html
(2) Burdge, J. & Overby, J. (2012). Chemistry: Atoms first. New York:
McGraw-Hill.
(3) Chang, R. & Goldsby, K. (2016). Chemistry (12th ed.). New York:
McGraw-Hill.
(4) Moore, J. W. & Stanitski, C.L. (2015). Chemistry: The Molecular
Science (5th ed.). Belmont, CA: Brooks Cole/Cengage Learning.
(5) Zumdahl, SS. &Zumdahl, S.A. (2012). Chemistry: An atoms first
approach. Belmont, CA: Brooks/Cole Cengage Learning.
INTRODUCTION (10 minutes)
Can you count objects by weighing them?
Ask the learners if they can count objects by weighing them. Present to them the following
situations:
1. Ms. Lilia sells shelled peanuts in a store. But she meets customers asking for ten peanuts,
another for 750 peanuts, and another for 2,000 peanuts. Obviously, it will take Ms. Lilia a very
long time to count the peanuts. What would be another way to count them?
Teacher Tip
1. Prepare the classroom or laboratory, the
materials, the laboratory sheets to be used.
2. Distribute the laboratory sheets at the start of
the lesson.
3. After the introduction and motivation, explain
the procedures of the activity.
4. Explain the safety precautions.
2. Mr. Jose goes to a hardware store and asks for 400 pieces of nails. What is an easier way to
approximate 400 pieces of nails without counting them one by one?
3. A candy factory sells chocolate chips in a bag. Each bag should have the same number of
chips. How does the candy factory count the number of chocolate chips in each bag?
INSTRUCTION and PRACTICE (80 minutes)
Safety Precautions
Teacher Tip
The activity can be performed individually or in
groups. All materials are household materials.
Nevertheless, caution must be observed in
handling any material in the lab. Instruct them
how to behave in the laboratory.
1. Never taste anything during a science activity.
2. Dispose of the samples as directed by your teacher.
3. Wash your hands with soap and water after the science activity.
4. Follow all laboratory instructions as directed by the teacher.
Part I. Counting by weighing: Relating mass to number
This method of counting by weighing is useful for counting very small objects, such as small
candies, beans, etc.
Ask them to perform the activity on relating mass to number of entities. The activity asks
them to determine the number of entities in a given sample by weighing a given amount of
sample and knowing the average mass of one entity. The activity uses common materials. Kidney
beans or other beans such as peanuts, squash seeds, or broad beans or patani may be used for
Sample 1. Use smaller sized samples like mongo, peas, or dried sago for Sample 2. Have them
answer the data tables and the questions. See the attached laboratory sheet.
71
Take note and emphasize that not all the kidney
beans will have the same mass since they are not
all identical. Some will be heavier and some will
be lighter. What was done was to obtain the
average mass of the kidney bean. However, for
purposes of counting, what is needed is only the
average mass.
Part II. Relating mass to moles
Ask them to perform the activity on relating mass to moles. The samples are common
household materials: Aluminium, iron, sodium chloride, and sucrose. Have them answer the data
tables and the questions. See the attached laboratory sheet.
Sample of Teacher’s Reference Table
A1
11 g
B1
14 g
A2
22 g
B2
27 g
A3
16 g
B3
18 g
A4
25 g
B4
30 g
1. For the Aluminium sample, crumple a sheet of Aluminium foil into a loose ball and place in
a small paper cup.
2. Be careful in using iron nails.
3. At the end of the activity, instruct them where to place the samples. Put separate
containers for each sample.
4. The samples may be reused for another class.
ENRICHMENT (20 minutes)
Post-laboratory discussion
1. Relate counting by weighing to finding the number of atoms in a weighed sample of material.
2. Give more exercises on calculating moles and molar masses of elements and compounds.
EVALUATION
Check their accomplished data tables and worksheet for correct use of units and significant
figures and the logical solutions.
72
Teacher Tip
Place the samples in small paper cups prior to the
class. There are four samples in this activity, and
make sure to prepare enough samples for the
class. Each group must work on all four samples.
Label the cups (e.g. Sample A1, A2, A3, A4 for
Group A, etc.).
You must also pre-determine the approximate
amount of sample to put in each cup and put
these in your notes. This will serve as a reference
for the masses measured by the learners. However,
they must measure the masses up to .01 g.
The masses do not have to be identical. For
example, the mass of Al in one group may have a
different value than the mass of Al in the other
group.
LABORATORY ACTIVITY: THE MOLE CONCEPT AND MOLAR MASS
Introduction
Atoms have very small masses. Macroscopic samples contain a very large number of atoms. The mole is used to deal with these extremely
large numbers of atoms in macroscopic samples. The mole is defined as the amount of substance that contains as many entities as there are in
exactly 12 grams of C-12. This is experimentally determined to be 6.022 x 1023 and is referred to as Avogadro’s number. The molar mass is
the mass in grams of one mole of a substance. It is possible and practical to count very small objects by determining an average mass then
weighing a given sample. You will be asked to determine the number of entities in a given sample of material through this technique. You will
also determine the number of moles of different substances and the corresponding number of atoms present in the sample.
Objectives
1. To determine the number of entities present in a given sample by weighing it and identifying the average mass of a single entity of the
sample.
2. To determine the number of moles and the number of atoms present in given samples of materials.
Materials
1. Balance – triple beam or electronic balance
5. Aluminium metal or foil
2. Paper cups
6. Iron (e.g. iron nails or iron filings)
3. Samples – e.g. kidney beans, mongo beans, rice, dried sago
7. Sodium chloride (NaCl)
4. Plastic spoons
8. Sucrose (table sugar, C12H22O11)
Safety Precautions
1. Never taste anything during a science activity.
2. Dispose of the samples as directed by your teacher.
3. Wash your hands with soap and water after the activity.
4. Follow all laboratory instructions as directed by your teacher.
73
Part I. Counting by weighing: Relating mass to number
Procedure
Sample 1
Sample 2
1. Count 20 beans (kidney, peanuts, patani, or other samples as
given by your teacher) and place them in a paper cup.
1. Count 20 mongo beans (rice, dried sago, peas, or any smaller
bean samples given by your teacher) and place them in a paper
cup.
2. Determine the mass of the 20 pieces of beans. Remember to
subtract the mass of the container. If using an electronic
balance, tare or set the balance to zero.
2. Determine the mass of the 20 pieces of beans. Remember to
subtract the mass of the container. If using an electronic
balance, tare or set the balance to zero.
3. Determine the mass of one bean by dividing the mass of the
sample by 20.
3. Determine the mass of one bean by dividing the mass of the
sample by 20.
Data Table
SAMPLE 1
Sample
Mass of 20 pieces of sample plus container
Mass of container
Mass of 20 pieces of sample
Mass of one piece of sample (Show calculation here)
Answer the following questions:
1. How much will 750 pieces of kidney beans weigh?
2. Calculate the mass of 5,500 mongo beans.
3. 158 grams of mongo beans is approximately how many pieces?
74
SAMPLE 2
Part II. Relating mass to moles
Procedure:
Determine the masses of Samples 1 to 4. Record these in the data tables provided. Calculate the number of moles in each sample. Show all
calculations and observe the correct use of units and significant figures.
Sample 1
Sample 2
Sample 3
Sample 4
ALUMINIUM
IRON
SODIUM CHLORIDE
(Table salt, NaCl)
SUCROSE
(Table sugar, C12H22O11)
1. Mass of sample + container, g
2. Mass of container, g
3. Mass of sample, g
4. Molar mass of sample, g/mol
5. Number of moles in sample
6. Number of atoms in sample
7. No. of atoms in 1.0 gram of
sample
75
General Chemistry 1
60 MINS
Lesson 10: Percent Composition and
Chemical Formulas
Lesson Outline
Content Standard
The learners demonstrate an understanding of percent composition and
chemical formulas.
Performance Standards
The learners shall be able to design, using multimedia, demonstrations, or
models, a representation or simulation of any of the following:
1. Atomic structure
2. Mass relationships in reactions
Learning Competencies
At the end of the lesson, the learners:
1. Calculate the percent composition of a compound from its formula
(STEM_GC11PC-If-31);
2. Calculate the empirical formula from the percent composition of a
compound (STEM_GC11PC-If-32); and
3. Calculate molecular formula given molar mass (STEM_GC11PC-If-33).
Specific Learning Outcomes
At the end of the lesson, the learners shall be able to:
1. Interpret the information provided by the chemical formula;
Introduction
Communicating Learning Objectives
5
Motivation
College Projections
5
Instruction
Percent Composition and Chemical
Formula
Enrichment
Determine the Sodium Percent in Snack
Food
Evaluation
Short Quiz
35
15
Resources
(1) Burdge, J.& Overby, J. (2012). Chemistry: Atoms first. New York:
McGraw-Hill.
(2) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York:
McGraw-Hill.
(3) Moore, J.W., Stanitski, C.L. & Jurs, P.C. (2012). Chemistry: The
molecular science (4th ed.). Belmont, CA: Brooks Cole/Cengage
Learning.
(4) Zumdahl, S.S. &Zumdahl, S.A. (2012). Chemistry: An atoms first
approach. Belmont, CA: Brooks/Cole Cengage Learning.
3. Explain the procedure used to determine the empirical formula
of a compound given the percent composition; and
2. Explain the procedure used to determine the percent
composition of a compound;
4. Utilize molar mass data to obtain the molecular formula from the
empirical formula.
76
INTRODUCTION (5 minutes)
1. Introduce the learning objectives using any of the suggested protocol (Verbatim, Own Words,
or Read-aloud)
a. Interpret the information provided by the chemical formula
b. Explain the procedure used to determine the percent composition of a compound
c. Explain the procedure used to determine the empirical formula of a compound given the
percent composition
Teacher Tip
Give examples of empirical formulas and molecular
formulas such as
a. Benzene, with molecular formula C6H6 and
empirical formula CH
b. Acetylene, with molecular formula C2H2 and
empirical formula CH
c. Ammonia, with molecular formula NH3 and
empirical formula NH3.
d. Utilize molar mass data to obtain the molecular formula from the empirical formula
2. Present the keyword for the concepts to be learned:
a. Percent composition mass
3. Review empirical formulas and molecular formulas and give examples.
MOTIVATION (5 minutes)
Percentage of the class planning (as first choice) to go to the different strands in STEM
1. Get total number of learners in class.
2. Get the number of learners who would like to pursue engineering in college. Get the
percentage.
3. Get the number of learners who would like to pursue the sciences. Get the percentage.
4. Get the number of learners who would like to pursue mathematics. Get the percentage.
INSTRUCTION (35 minutes)
1. Information from the chemical formula
What information can be obtained from a chemical formula? For example, what information
can you get from the formula of carbon dioxide, CO2?
= CARBON
= OXYGEN
77
Teacher Tip
This will give an indication of the interest of the
learners and their planned careers. This will also
review the concept of percentage and its
application in the real world.
Then, mention that the lesson will be about how
the concept of percentage is used in chemistry,
especially in chemical compounds.
a. The compound is made up of two elements, namely carbon and oxygen.
b. One molecule of CO2 is made up of one atom of carbon and two atoms of oxygen.
c. One mole of CO2 molecules will have one mole of C atoms and two moles of O atoms.
d. The ratio of the moles of C to the moles of O in CO2 is 1:2.
e. CO2 is composed of 27.29% carbon and 72.71% oxygen. The chemical formula provides
the percent composition of CO2.
Teacher Tip
You should motivate the learners to provide the
answers instead of just stating them.
It is important for them to understand the chemical
formula and all information that can be obtained
from it.
2. Percent Composition by Mass
The percent composition by mass is the percent by mass of each element in a compound.
Mathematically,
with n = the number of atoms of the element
For CO2,!
The answer indicates that CO2 is composed of 27.29% C atom and 72.71% O atom.
Ask them to answer the following practice exercises:
I.
Calculate the percent composition of NaCl.
II. The chemical formula of glucose is C6H12O6. Determine its percent composition.
III. Which element comprising Mg(OH)2 has the highest percentage by mass?
78
Answer Key
1. 39.34% Na, 60.66% Cl
2. 39.99% C, 6.727% H, 53.28% O
3. O; the composition is 41.68% Mg, 54.89% O,
and 3.46% H
Teacher Tip
You may want to connect the lesson to some real
world examples. Magnesium hydroxide, Mg(OH)2,
is used as a medication to treat symptoms brought
about by too much stomach acid such as heartburn
or indigestion.
3. Empirical Formula from Percent Composition
The empirical formula of a compound can be calculated from the percent composition.
Because percentage is given, it is convenient to assume 100.00 grams of the compound.
Illustrate using the following examples:
A. A compound is found to consist of 7.81% C and 92.19% Cl. What is the empirical formula
of the compound?
Assume 100.00 grams of the compound. The sample will therefore contain 7.81 g C and
92.19 g Cl. The grams are converted to moles to get the ratios of the moles of the
elements in the compound:
The compound is C0.650Cl2.601. But chemical formulas are expressed in whole numbers.
Empirical formulas are expressed as the lowest whole number ratio between the atoms.
To convert to whole numbers, divide the number of moles by the smallest value (that is
0.650).
The empirical formula is C1Cl4 or CCl4.
79
B. A compound is found to consist of 43.64% P and 56.36% O. The molar mass for the
compound is 283.88 g/mol. What is the empirical formula and molecular formula of the
compound?
Assume 100.00 grams of the compound. What is the mass of each element in 100.00
grams of compound?
What are the moles of each element in 100.00 grams of compound?
Divide the mole values by the smallest value to get
The compound is PO2.5. But the subscripts are still not whole numbers. Multiply the
subscripts by a factor to get the smallest whole number. When multiplied by 2, the
empirical formula is P2O5.
What is the molecular formula?
Compare the mass of the empirical formula to the molar mass:
Mass of P2O5 = 141.94 g/mol
Molar mass = 283.88 g/mol
80
Answer Key
1. 39.34% Na, 60.66% Cl
2. 39.99% C, 6.727% H, 53.28% O
3. O; the composition is 41.68% Mg, 54.89% O,
and 3.46% H
Therefore, the molecular formula is (P2O5)2 or P4O10.
ENRICHMENT
Determine the % sodium in snack food
Look at the food labels of some snack food like potato chips, peanuts, popcorn, etc. Fill up the
table below.
Teacher Tip
You may want to connect the lesson to some real
world examples. Magnesium hydroxide, Mg(OH)2,
is used as a medication to treat symptoms brought
about by too much stomach acid such as heartburn
or indigestion.
1. Identify your chosen snack food and brand.
2. Get the amount in grams of one serving of the snack food.
3. Get the amount of sodium in mg in one serving of the snack food.
4. Obtain the % sodium by mass in one serving of snack food.
EVALUATION (15 minutes)
Answer the following questions. Place the answers in the space provided. Show calculations
where applicable. Observe the use of significant figures for calculations and indicate the
appropriate units. Learners can use the periodic table to answer the questions.
Aspirin has the molecular formula C9H8O4.
_____1. What is the % C in aspirin by mass in aspirin?
_____2. What is the % O in aspirin by mass in aspirin?
_____3. An oxide of chromium is made up of 5.20 g chromium and 5.60 g oxygen What is the
empirical formula of the oxide? (Note: An oxide of nitrogen contains 63.1% oxygen and has a
molar mass of 76.0 g/mol.)
_____4. What is the empirical formula for this compound?
_____5. What is the molecular formula of the compound?
81
Answer Key
1. 60.00%
2. 35.53%
3. Cr2O7
4. N2O3
5. N2O3
General Chemistry 1
120 MINS
Lesson 11: Chemical Reactions and
Chemical Equations (Lecture)
Content Standard
The learners demonstrate an understanding of the use of chemical formulas to
represent chemical reactions.
Lesson Outline
Introduction
Review and Communicating Learning
Objectives
Motivation
Evidences of Chemical Change
Instruction
Chemical Reactions and Chemical
Equations
Enrichment
Inquiry
Performance Standards
The learners shall be able to design, using multimedia, demonstrations, or
models, a representation or simulation of any of the following:
1. Atomic structure
2. Mass relationships in reactions
Learning Competencies
At the end of the lesson, the learners:
1. Write equations for chemical reactions and balance the equations
(STEM_GC11CR-If-g-34);
2. Interpret the meaning of a balanced chemical reaction in terms of the Law
of Conservations of Mass (STEM_GC11CR-If-g-35);
3. Describe evidences that a chemical reaction has occurred (STEM_GC11CRIf-g-36); and
4. Perform exercises on writing and balancing chemical equations
(STEM_GC11CR-If-g-37).
Specific Learning Outcomes
At the end of the lesson, the learners shall be able to:
1. Write and balance chemical equations;
2. Derive pertinent information from a balanced chemical equation;
3. Determine whether a chemical reaction has occurred or not; and
4. Classify chemical reactions.
82
30
3
85
2
Resources
(1) Burdge, J & Overby, J. (2012). Chemistry: Atoms first. New York:
McGraw-Hill.
(2) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York:
McGraw-Hill.
(3) Chemical equations [Online lecture]. Retrieved from CK-12 website:
https://www.ck12.org/physical-science/Chemical-Equations-inPhysical-Science/
(4) Moore, J.W., Stanitski, C.L.& Jurs, P.C. (2012). Chemistry: The
molecular science (4th ed.). Belmont, CA: Brooks Cole/Cengage
Learning.
(5) Recognizing chemical reactions [Online lecture]. Retrieved from CK-12
website: https://www.ck12.org/physical-science/RecognizingChemical-Reactions-in-Physical-Science/
(6) Zumdahl, SS. & Zumdahl, S.A. (2012). Chemistry: An atoms first
approach. Belmont, CA: Brooks/Cole Cengage Learning.
INTRODUCTION (30 minutes)
Review
1. Discuss the enrichment assignment of last lesson (% sodium in snack food).
2. Discuss the quiz on percent composition given in the last lesson.
Communicating Learning Objectives
1. Introduce the following learning objectives using any of the suggested protocol (Verbatim,
Own Words, or Read-aloud):
a. Write and balance chemical equations
b. Derive pertinent information from a balanced chemical equation
c. Determine whether a chemical reaction has occurred or not
Teacher Tip
Ask the learners to discuss what they found to be
the % sodium in their snack samples. Alternatively,
these can be written on the board.
Call on some of them to show their calculations on
the board. Have the class comment on the
calculations, including the proper use of significant
figures.
Teacher Tip
List these keywords on the board or through
PowerPoint slides. Alternatively, you can write
them on flip charts.
d. Classify chemical reactions
2. Present the keywords for the concepts to be discussed:
a. Chemical equation
b. Reactant
c. Product
d. Aqueous
e. Decomposition reaction
f.
Synthesis reaction
g. Single displacement reaction
h. Double displacement reaction
i.
Combustion reaction
j.
Hydrocarbon
Note
Ask them to recall the Law of Conservation of
Mass and express their understanding of it.
3. Review the Law of Conservation of Mass
83
MOTIVATION (3 minutes)
Teacher Tip
You may bring some actual samples of rusty iron
nails, bleached and unbleached hair, or other
materials to show to the class.
Ask them what the following have in common:
a. Rusty iron nail
b. Change in color of leaves
These are evidences of chemical change.
c. Bleached hair
INSTRUCTION (85 minutes)
Writing and Balancing a Chemical Equation
In a chemical reaction, a substance (or substances) is converted to one or more new
substances. Chemical reactions follow the law of conservation of mass. No atoms are created
or destroyed; they are just rearranged. Chemists have a way of communicating chemical
reactions. They represent chemical reactions through chemical equations.
Consider the reaction of hydrogen gas (H2) with chlorine gas (Cl2) to yield hydrogen chloride.
The reaction is illustrated by the figure below.
Teacher Tip
Emphasize the Law of Conservation of Mass.
There must be the same type of atoms on both
sides of the arrow.
We can represent this reaction through a chemical equation. The reactants (starting
substances) are placed on the left side. The products (substances produced) are placed on
the right. An arrow points towards the direction of the reaction. The equation has to be
balanced so that the same number and types of atoms appear on the left and right side of the
equation. To balance, coefficients (numbers preceding the chemical formula) are used. For
additional information, the physical states of the reactants and products (s, l, g, for solid,
liquid, or gas, respectively)are indicated.
84
Hence, the balanced chemical equation is:
Teacher Tip
Show learners where to put the coefficients.
H2(g) + Cl2(g) ! 2 HCl(g)
Check if the equation is balanced:
Reactants
Products
H (2)
H (2)
Cl (2)
Cl (2)
When a substance is placed in water, we indicate this with aq, meaning it is in an aqueous
environment. For example, when KBr reacts with AgNO3 in an aqueous environment, KNO3
and solid AgBr are produced. This reaction is represented as
KBr(aq) + AgNO3(aq) ! KNO3(aq) + AgBr(s)
Show them the procedure of balancing equations through this example:
Ethane (C2H6) reacts with oxygen gas (O2) to produce carbon dioxide and water. Write the
balanced chemical equation for the reaction.
1. Identify reactants and products and write their correct formulas. Put reactants on the left
side and products on the right.
C2H6 + O2 ! CO2 + H2O
85
2. Balance the equation by changing the coefficients of the reactants or products. Do not
change the subscripts or the chemical formula.
C2H6 + 7/2 O2 ! 2CO2 + 3 H2O
To use the smallest whole number coefficients, we multiply the equation by 2 to give:
2C2H6 + 7O2 ! 4CO2 + 6H2O
3. Check to make sure that the number of each type of atom is the same on each side of the
equation.
Reactants
Products
4
4
C
C
12 H
12 H
14 O
14 O
Ask them to answer these exercises:
State if the illustrated equation below is balanced or not. If not, explain why it is not
balanced. Illustrate by a drawing how you would balance the equation.
1.
2.
3.
86
Teacher Tip
Show learners where to put the coefficients.
Balance the following equations
Teacher Tip
Show them where to put the coefficients.
1. ____ C + ____ O2 ! ___ CO
Hint:
Start with elements that appear only once on each
side.
2. ____ Mg + ____ O2 ! ____ MgO
3. ____ H2O2 ! ____ H2O + ____ O2
4. ____ CH4 + O2 ! ____ CO2 + ____ H2O
5. ____ N2O5 ! _____ N2O4 + _____ O2
Interpretation of a Chemical Equation
How can a balanced chemical equation be interpreted? See the example:
H2
+
Cl2
!
2HCl
One molecule
+
One molecule
!
Two molecules
One mole
+
One mole
!
Two moles
2 (1.008 g)
= 2.016 g
+
2 (35.45g)
= 70.90 g
!
Ask them to answer 72.929
this exercise:
2 (1.008 g + 35.45 g)
= 72.92 g
72.92 g
Interpret the balanced equation:
2C2H6 + 7O2 ! 4CO2 + 6H2O
Show that the Law of Conservation of Mass is followed.
87
Answer Key
1. 2, 1, 2
2. 2, 1, 2
3. 2, 2, 1
4. 1, 2, 1, 2
5. 2, 2, 1
Types and Evidences that a Chemical Reaction has Occurred
Teacher Tip
Ask them to give examples of evidences of
chemical changes they have observed around
them. Some examples are bleach turning hair
yellow, milk going sour, or apple slices becoming
brown.
Here are some evidences that a chemical reaction has occurred:
a. Change in color
b. Formation of a solid (a precipitate)
c. Evolution of gas (bubble formation)
d. Change in temperature (heat is released or absorbed)
Most chemical reactions can be classified into five types:
1. Decomposition reaction – a reactant breaks down into two or more products
Chemical reactions can be classified in other ways
such as acid-base reactions and oxidationreduction reactions. However, these concepts will
be introduced in later chapters.
AB ! A + B
Li2CO3 ! Li2O + CO2
2. Synthesis reaction – two or more reactants form a single product
A + B ! AB!
2NO + O2 ! 2NO2!
3. Single displacement reaction – one element replaces another in a compound
A + BC ! AC + B
Cu(s) + 2AgNO3(aq) ! Cu(NO3)2(aq) + 2Ag(s)
4. Double displacement – two ionic compounds exchange ions
AB + CD ! AD + CB
2KI(aq) + Pb(NO3)2(aq) ! 2KNO3(aq) + PbI2(s)
5. Combustion reaction – a hydrocarbon (a compound containing carbon and hydrogen) reacts
with oxygen to form carbon dioxide and water.
Hydrocarbon + O2 ! CO2 + H2O
2C2H6 + 7O2 ! 4CO2 + 6H2O
Teacher Tip
H2O2 breaks down into H2O and O2 aided by light.
This is a decomposition reaction. Write and
balance the equation.
See Practice Exercise 3 on balancing equations
above.
ENRICHMENT (2 minutes)
1. Why do you need to store hydrogen peroxide away from light often in dark colored bottles?
2. Learners may watch the videos in the sites given in the Resources section above. These can
be given as assignments.
88
Note
The Evaluation will be through the exercises in the
laboratory session hour.
General Chemistry 1
120 MINS
Lesson 12: Chemical Reactions and
Chemical Equations (Laboratory)
Content Standard
The learners demonstrate an understanding of the use of chemical formulas to
represent chemical reactions.
Performance Standards
The learners shall be able to design, using multimedia, demonstrations, or
models, a representation or simulation of any of the following:
1. Atomic structure
2. Mass relationships in reactions
Learning Competencies
At the end of the lesson, the learners:
1. Write equations for chemical reactions and balance the equations
(STEM_GC11CR-If-g-34);
2. Interpret the meaning of a balanced chemical reaction in terms of the Law
of Conservations of Mass (STEM_GC11CR-If-g-35);
3. Describe evidences that a chemical reaction has occurred (STEM_GC11CRIf-g-36); and
4. Perform exercises on writing and balancing chemical equations
(STEM_GC11CR-If-g-37).
Specific Learning Outcomes
At the end of the lesson, the learners shall be able to:
1. Write and balance chemical equations;
2. Derive pertinent information from a balanced chemical equation;
3. Determine whether a chemical reaction has occurred or not; and
4. Classify chemical reactions.
89
Lesson Outline
Practice
Laboratory Exercises
120
Resources
(1) Burdge, J & Overby, J. (2012). Chemistry: Atoms first. New York:
McGraw-Hill.
(2) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York:
McGraw-Hill.
(3) Chemical equations [Online lecture]. Retrieved from CK-12 website:
https://www.ck12.org/physical-science/Chemical-Equations-inPhysical-Science/
(4) Moore, J.W., Stanitski, C.L.& Jurs, P.C. (2012). Chemistry: The
molecular science (4th ed.). Belmont, CA: Brooks Cole/Cengage
Learning.
(5) Recognizing chemical reactions [Online lecture]. Retrieved from CK-12
website: https://www.ck12.org/physical-science/RecognizingChemical-Reactions-in-Physical-Science/
(6) Zumdahl, SS. & Zumdahl, S.A. (2012). Chemistry: An atoms first
approach. Belmont, CA: Brooks/Cole Cengage Learning.
PRACTICE
Laboratory exercises (120 minutes)
Give the following exercises in a separate time slot like the laboratory session. This provides
practice for the learners. Allow them 60 minutes to answer the exercises. Then go over the
exercises together. Ask them to show their answers on the board and explain their answers.
Alternatively, part of the exercises can be taken as a quiz or as an assignment for evaluation.
Answer Key
1. B.
2. B.
3. D.
4. B.
5. D.
6. E.
7. C.
8. A.
9. A.
10. E.
11. A.
12. D.
13. A.
14. B.
15. A.
16. D.
17. C.
18. B.
19. E.
20. B.
LABORATORY ACTIVITY: CHEMICAL REACTIONS AND CHEMICAL EQUATIONS
Directions: Choose the best answer. Encircle the letter corresponding to your answer.
1. Balanced chemical equations imply which of the following?
A. Numbers of molecules are conserved in chemical change.
B. Numbers of atoms are conserved in chemical change.
C. Volume is conserved in chemical change.
D. A and B
E. B and C
3. The catalytic conversion of ammonia to nitric oxide is the first
step in a three-step process, which ultimately results in nitric
acid. Balance the equation for the reaction.
a NH3(g) + b O2(g) ! c NO(g) + d H2O(g)
A.
B.
C.
D.
E.
2. In balancing an equation, we change the __________ to make
the number of atoms on each side of the equation balance.
A. formulas of compounds in the reactants
B. coefficients of reactants and products
C. formulas of compounds in the products
D. subscripts of compounds
E. the reactants
90
a = 2, b = 1, c = 2, d = 1
a = 3, b = 2, c = 3, d = 3
a = 4, b = 3, c = 2, d = 6
a = 4, b = 5, c = 4, d = 6
a = 6, b = 15, c = 6, d = 9
4. In the reaction: a BaCl2 + b AgNO3 ! c Ba(NO3)2 + d AgCl
7. Which of the following equations is not balanced?
What is the coefficient, d, of silver chloride in the balanced
equation?
A. 1
B. 2
C. 3
D. 4
E. 5
A. 4Al + 3O2 ! 2Al2O3
B. C2H6 + O2 ! 2CO2 + 3H2O
C. 2KClO3 ! 2KCl + O2
D. 4P4 + 5S8 ! 4P4S10
E. P4 + 5O2 ! P4O10
8. The first step in the Ostwald process for making nitric acid is the
formation of NO as follows:
5. Balance the following equation with the smallest set of whole
numbers.
C4H10 + O2 ! CO2 + H2O
4NH3 + 5O2 ! 4NO + 6H2O
What is the coefficient for CO2 in the balanced equation?
A. 1
B. 4
C. 6
D. 8
E. 12
According to the equation, 5 moles NH3 will react with ________
moles O2 to form _______ moles of NO.
A. 5, 4
B. 4, 5
C. 25, 20
D. 5/4, 4/5
6. Balance the following equations:
E. 25/4, 5
_____ P4O10 + _______ H2O ! _______ H3PO4
9. Potassium metal and chlorine gas (Cl2) react in a combination
reaction to produce potassium chloride. What is the correct
balanced equation for this reaction?
What is the coefficient of H2O in the balanced equation?
A.
B.
C.
D.
E.
1
2
4
5
6
A. 2 K(s) + Cl2(g) !! 2 KCl(s)
B. K(s) + Cl2(g) ! KCl(s)
C. K(s) + Cl(g) !! KCl(s)
D. K2(s) + Cl2(g) ! 2 KCl(s)
E. K(s) + Cl2(g) !! KCl2(s)
91
10. In the reaction given below, how many grams of water are
consumed if 4.0 g hydrogen gas and 32.0 g oxygen gas are
produced?
13. Balance the following equation:
a NaNO3 ! b NaNO2 + c O2
What are the coefficients of the balanced equation for
2 H2O ! 2 H2 + O2
A.
B.
C.
D.
E.
a, b, and c?
2.0 g
4.0 g
18.0 g
20.0 g
36.0 g
A.
B.
C.
D.
E.
11. In the reaction given below, for every two molecules of
hydrogen peroxide (H2O2) consumed, how many molecules of
oxygen are produced?
14. Balance the following chemical reaction:
a CO + b NO ! c CO2 + d N2
The coefficients a, b, c, and d for the balanced chemical
equation are:
2H2O2 ! 2H2O + O2
A.
B.
C.
D.
E.
2, 2, 1
1, 1, 2
1, 2, 1
2, 3, 1
3, 1, 1
1
2
3
6
9
A. 2, 2, 2, 3
B. 2, 2, 2, 1
C. 1, 1, 1, 2
D. 2, 1, 2, 1
E. 1, 2, 2, 1
12. Balance the following reaction:
15. Classify the following reaction:
a Al2O3 ! b Al + c O2
2Na + Cl2 ! 2 NaCl
What is the sum of the coefficients of the reactant and products
(a + b + c) in the balanced equation using the smallest set of
whole numbers as coefficients?
A.
B.
C.
D.
E.
A. Synthesis
B. Decomposition
3
5
6
9
10
C. Combustion
D. Single Displacement
E. Double Displacement
92
16. Classify the following reaction:
19. Classify the following reaction:
Zn + 2HCl ! ZnCl2 + H2
NaCl(aq) + AgF(aq) ! NaF(aq) + AgCl(s)
A. Synthesis
A. Synthesis
B. Decomposition
B. Decomposition
C. Combustion
C. Combustion
D. Single Displacement
D. Single Displacement
E. Double Displacement
E. Double Displacement
17. Classify the following reaction:
H2SO3 ! H2O + SO2
20. Classify the following reaction:
CaCO3(s) ! CaO(s) + CO2(g)
A. Synthesis
B. Decomposition
A. Synthesis
C. Combustion
B. Decomposition
D. Single Displacement
C. Combustion
E. Double Displacement
D. Single Displacement
E. Double Displacement
18. Classify the following reaction:
CH4 + 2O2 ! CO2 + 2H2O
A. Synthesis
B. Decomposition
C. Combustion
D. Single Displacement
E. Double Displacement
93
General Chemistry 1
Lesson 13: Mass Relationships in
Chemical Reactions
Content Standard
The learners demonstrate an understanding of quantitative relationship of
reactants and products in a chemical reaction.
180 MINS
Lesson Outline
Introduction
Communicating Learning Objectives
5
Motivation
Let us Make Sandwiches
5
Performance Standard
The learners shall be able to design, using multimedia, demonstrations, or
models, a representation or simulation of any of the following:
1. Atomic structure
2. Mass relationships in reactions
Instruction,
Delivery and
Practice
I. Reactants and Products
II. Limiting Reagents
III. Theoretical Yield, Actual Yield, and
Percent Yield
140
Enrichment
Animation Videos of Limiting Reagents
15
Learning Competencies
At the end of the lesson, the learners:
1. Construct mole or mass ratios for a reaction in order to calculate the
amount of reactant needed or amount of product formed in terms of moles
or mass (STEM_GC11MR-Ig-h-38);
2. Calculate percent yield and theoretical yield of the reaction
(STEM_GC11MR-Ig-h-39);
3. Explain the concept of limiting reagent in a chemical reaction; identify the
excess reagent (STEM_GC11MR-Ig-h-40); and
4. Calculate reaction yield when a limiting reagent is present
(STEM_GC11MR-Ig-h-41).
Evaluation
Check up Quiz
15
Specific Learning Outcomes
At the end of the lesson, the learners shall be able to:
1. Identify mole ratios of reactants and products from balanced chemical
equations;
2. Perform stoichiometric calculations related to chemical equations;
3. Define theoretical, actual, and percent yield of reactions;
4. Calculate theoretical and percent yield of a reaction;
5. Identify the limiting and excess reagent(s) of a reaction; and
6. Calculate reaction yield in the presence of a limiting reagent.
94
Materials
Periodic table, calculator
Resources
(1) Allan, Andy. Stoichiometry [PowerPoint presentation]. Retrieved from
http://www.sciencegeek.net/APchemistry/FlashPPT/3_Stoichiometry/
index.html
(2) Burdge, J. & Overby, J. (2012). Chemistry: Atoms first. New York:
McGraw-Hill
(3) Chang, R. &Goldsby, K. (2016). Chemistry. (12th ed.). New York:
McGraw-Hill
(4) Limiting reagent [Vector animation]. Retrieved from McGraw Hill
Education web site: http://www.mhhe.com/physsci/chemistry/
essentialchemistry/flash/limitr15.swf
(5) Limiting reactant [Vector animation]. Retrieved from North Carolina
School of Medicine and Mathematics web site: http://
www.dlt.ncssm.edu/core/Chapter6-Stoichiometry/Chapter6Animations/LimitingReactant.html
(6) Moore, J.W., Stanitski, C.L. & Jurs, P.C. (2012). Chemistry: The
molecular science (4th ed.). Belmont, CA: Brooks Cole/Cengage
Learning.
(7) Reactants, products and leftovers [Simulation]. Retrieved from PhEt
Interactive Simulations web site: http://phet.colorado.edu/sims/html/
reactants-products-and-leftovers/latest/reactants-products-andleftovers_en.html
(8) Zumdahl, SS. &Zumdahl, S.A. (2012). Chemistry: An atoms first
approach. Belmont, CA: Brooks/Cole Cengage Learning.
INTRODUCTION (5 minutes)
1. Introduce the learning objectives using any of the suggested protocol (Verbatim, Own Words,
or Read-aloud):
a. Identify mole ratios of reactants and products from balanced chemical equations
b. Perform stoichiometric calculations related to chemical equations
c. Define theoretical, actual, and percent yield of reactions
d. Calculate theoretical and percent yield of a reaction
e. Identify the limiting and excess reagent(s) of a reaction
f. Calculate reaction yield in the presence of a limiting reagent
Teacher Tip
List these keywords on the board or through
PowerPoint slides. Alternatively, you can write
them on flip charts.
2. Present the keywords for the concepts to be learned:
a. Stoichiometry
b. Limiting reagent
c. Theoretical yield
d. Actual yield
e. Percent yield
3. Review the Law of Conservation of Mass
MOTIVATION (5 minutes)
Teacher Tip
Ask the learners to show the equation:
Let us make sandwiches!
Some learners are going on a road trip and they are to bring some food to eat along the way.
Karen was asked to bring hamburger sandwiches for which she will use two slices of bread and
one hamburger patty to make one sandwich. Show the equation.
INSTRUCTION/DELIVERY/PRACTICE(40 minutes)
Amounts of Reactants and Products
Stoichiometry is the study of the quantities of materials consumed and produced in chemical
reactions. From the balanced chemical equation, we will be able to:
a. Determine how much products will be produced from a specific amount of reactants
b. Determine the amount of reactants needed to produce a specific amount of products
95
two slices of bread + one hamburger patty "
one hamburger sandwich
This analogy will be used for mass relationships in
chemical equations.
1. Illustrate stoichiometry using the following examples:
a. Let us make hamburger sandwiches again. The equation is
Teacher Tip
Illustrate stoichiometry with real life applications.
two$slices$of$bread$+$one$hamburger$patty$$!$$one$hamburger$sandwich$
Suppose Karen has 14 hamburger patties, how many slices of bread will she need to consume
all the patties? The ratio of slices of bread to hamburger patty is 2: 1.
2. Suppose that instead of plain burgers, Karen is to make double cheeseburgers. Show the
equation so Karen can shop for enough ingredients.
two$slices$of$bread$$+$one$hamburger$patty$$+$$two$slices$of$cheese$$$!$$$one$double$cheeseburger$
How many slices of cheese, hamburger patties, and slices of bread will Karen need to make 25
double cheeseburgers?
Karen will therefore have to buy 50 slices of cheese, 25 hamburger patties, and 50 slices of
bread.
96
3. Ammonia, NH3, is a leading industrial chemical used in the production of agricultural fertilizers
and synthetic fibers. It is produced by the reaction of nitrogen and hydrogen gases:
3 H2(g) + N2(g) ! 2 NH3(g)
The balanced equation says that 3 moles H2 are stoichiometrically equivalent to 1 mole N2
and to 2 moles NH3. The ratio of moles H2 to moles NH3 is 3:2; the ratio of moles N2 to moles
NH3 is 1:2.
a. How many moles of NH3 will be produced if 10.4 moles H2 react completely with N2?
(moles H2 ! moles NH3)
b. How many moles of N2 are needed to produce 42.4 moles NH3? (moles NH3 ! moles N2)
c. How many grams of NH3 will be produced from 25.7 moles N2
(moles N2 ! moles NH3 ! g NH3)
d. How many grams of NH3 will be produced if 122 g N2 reacts completely with H2?
(g N2 ! moles N2 ! moles NH3 ! g NH3)
97
Teacher Tip
Before doing any calculations involving chemical
reactions, make sure that the chemical equation is
balanced.
Recall the interpretation of a balanced chemical
equation.
Teacher Tip
Give these examples but ask them to supply the
appropriate ratios or factors.
Before calculating, ask them the steps they will be
taking to get the answer.
4. Solid lithium hydroxide is used to remove carbon dioxide and is called a CO2 scrubber. This
technique has been used for space vehicles. The reaction is:
2 LiOH(s) + CO2(g) ! Li2CO3 (s) + H2O (l)
How many grams of CO2 can be absorbed by 785.0 g LiOH?
What are the steps of the solution?
a. Convert grams LiOH to moles LiOH.
b. Get the moles of CO2 stoichiometrically equivalent to moles LiOH
c. Convert moles CO2 to grams CO2.
(grams LiOH ! moles LiOH ! moles CO2 ! grams CO2)
Ask them to answer the following practice exercises:
1. The combustion of carbon monoxide gas in oxygen gas is represented by the following
balanced equation: 2 CO(g) + O2(g) ! 2CO2(g)
How many moles of carbon dioxide gas will be produced from the complete combustion of
4.60 moles CO(g)?
2. Consider the reaction: 2 KClO3 ! 2 KCl + 3 O2
a. How many moles of KClO3 are required to produce 22.8 moles oxygen gas, O2?
b. How many moles of KCl will be produced from the total decomposition of 18.8 moles
KClO3?
98
Teacher Tip
Assign the exercises to different groups. Ask them
to show the calculations on the board. If there is
no longer enough time, this could be given as an
assignment.
Alternatively, this can also be used as a quiz to
check on their understanding of the concept.
Answer Key
1. 4.60 moles CO2
2.a. 15.2 moles
2.b. 18.8 moles
3.a. 388 g Fe2O3
3.b. 6.43 x 103 g
4.a. 156.0 g MgO
4.b. 135.0 g MG
3. Given the reaction 4 Fe + 3 O2 ! 2 Fe2O3
a. How many grams of Fe2O3 will be formed from 4.86 moles Fe reacting with sufficient
oxygen gas?
b. How many grams of Fe are needed to react with sufficient oxygen to produce 28.8 moles
Fe2O3?
4. Consider the reaction 2Mg + O2 ! 2MgO
a. How many grams of MgO are produced from the complete reaction of 94.2 g Mg?
b. How many grams of Mg are needed to produce 224 g of MgO in the complete reaction of
Mg with oxygen gas?
Limiting Reagents
The reactant used up first in the chemical reaction is called the limiting reagent. Excess reagents
are reactants present in quantities greater than what is needed by the reaction.
Illustrate using the following examples:
1. Recall the example of the double cheeseburger. The equation is:
two$slices$of$bread$$+$$one$hamburger$patty$$+$$two$slices$of$cheese$$!$$$one$double$cheeseburger$
When Karen went shopping, she was able to buy 50 slices of cheese, 20 hamburger patties,
and 50 slices of bread. How many double cheeseburgers can she make? What is the limiting
material or reagent? What are the excess reagents?
To find the limiting reagent, determine which reagent will give the smallest amount of
product.
99
Therefore, the limiting reagent is the hamburger patty.
40$slices$of$bread$
10$slices$in$excess
+
20$patties +
40$slices$of$cheese$
10$slices$in$excess
!
20$double$cheeseburger
Karen can only make 20 double cheeseburgers. The limiting reagent is the hamburger patty.
There are ten slices of bread and ten cheese slices in excess. Karen cannot make more than
20 sandwiches because all the hamburger patties have been used up.
2. Consider again the reaction:
3H2(g) + N2(g) ! 2NH3(g)
a. If 6.60 moles H2 are made to react with 4.42 moles N2, what is the limiting reagent? How
many moles NH3 will be produced? What reagent is in excess and by how much?
Determine which reagent will produce the smallest amount of product:
Therefore, the limiting reagent is H2.
100
The amount of limiting reagent present at the start of the reaction determines the
theoretical yield. To determine the amount of NH3 produced, use the limiting reagent.
The excess reagent is N2. If you have 6.60 moles H2 then you will need
But you have 4.42 moles N2.
Therefore, the excess amount of N2 is 4.42 moles – 2.20 moles = 2.22 moles N2.
b. If 25.5 g H2 are made to react with 64.2 g N2, what is the limiting reagent? What is the
theoretical yield in g of NH3 that will be produced? How do you determine the limiting
reagent?
i. Get the number of moles of each reactant.
ii. Calculate the number of moles of product using each reagent.
iii. The one that yields the smallest number of moles of product is the limiting reagent.
From 12.6 moles of H2, how many moles of NH3 do we expect to get?
101
Teacher Tip
This example shows that even though the mass of
N2 was greater than the mass of H2, the limiting
reagent was still N2. This illustrates that the
limiting reagent is not determined by which
reactant is present in greater amount. It is only by
considering the mole ratios and relationships in
the balanced chemical reaction that the limiting
reagent can be determined.
From 2.29 moles of N2, how many moles of NH3 do we expect to get?
The limiting reagent is N2.
What amount of NH3 will be formed in this example?
The amount of product that can be produced is determined by the limiting reagent. Once
the limiting reagent is consumed, there is no further reaction. Hence, to calculate the
amount of NH3 produced, we use 2.29 moles N2, the limiting reagent.
Theoretical Yield, Actual Yield, and Percent Yield
The theoretical yield is the maximum amount of product that would result if the limiting reagent
is completely consumed. It is the amount of product predicted by stoichiometry (as shown in the
above example).
The actual yield is the quantity of the desired product actually formed.
If in the example given above, only 54.0 g NH3 were produced, then the actual yield is 54.0 g; the
theoretical yield is 78.0 g and the % yield is:
102
Teacher Tip
Explain why the theoretical yield is not obtained in
actual work. Ask them for possible reasons.
Ask them to answer this practice exercise:
1. Silver metal reacts with sulfur to form silver sulfide according to the following reaction:
2Ag (s) + S(s) ! Ag2S (s)
Answer Key
1. Ag
2. 57.5 g
3. 2.57 g
4. 78.3 %
a. Identify the limiting reagent if 50.0 g Ag reacts with 10.0 g S.
b. What is the theoretical yield in g of Ag2S produced from the reaction?
c. What is the amount in g of the excess reactant expected to remain after the reaction?
d. When the reaction occurred, the amount of Ag2S obtained was 45.0 g. What is the
percent yield of the reaction?
ENRICHMENT (15 minutes)
Watch the animation videos of limiting reagent from the following sources:
•
Limiting reagent [Vector animation]. Retrieved from McGraw Hill Education web site:
Teacher Tip
If this cannot be shown in the classroom, learners
can be asked to view the animation at home or in
the library.
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/limitr15.swf
•
Limiting reactant [Vector animation]. Retrieved from North Carolina School of Medicine and
Mathematics web site:
http://www.dlt.ncssm.edu/core/Chapter6-Stoichiometry/Chapter6-Animations/
LimitingReactant.html
•
Reactants, products and leftovers [Simulation]. Retrieved from PhEt Interactive Simulations
web site: http://phet.colorado.edu/sims/html/reactants-products-and-leftovers/latest/
reactants-products-and-leftovers_en.html
EVALUATION (15 minutes)
Encircle the letter of the best answer.
2. Given the reaction CH4 + 2O2 ! CO2 + 2H2O, what amount of
O2 is needed to completely react with 14.0 moles CH4?
A. 2.0 moles
B. 28.0 moles
C. 12.0 moles
D. 6.0 moles
E. 1.0 mole
1. Stoichiometry deals with
A. Combustion reactions
B. Rates of chemical reactions
C. Heat evolved or absorbed during chemical reactions
D. The study of amounts of materials consumed and products
formed in chemical reactions
E. Activation energy of chemical reactions
103
3. How much of SnF2 (stannous fluoride, active ingredient in
toothpaste) in g can be prepared from the reaction of 10.0 g
SnO with excess HF according to the following reaction?
SnO + 2HF ! SnF2 + H2O
A. 11.6 g
B. 10.0 g
C. 9.62 g
D. 26.0 g
E. 104.0 g
6. The reaction N2(g) + 2O2(g) ! N2O4(g) occurs in a closed
container. If 8.0 moles N2(g) are made to react with 12.0 moles
O2, the limiting reagent and the theoretical yield of N2O4 are:
A. The limiting reagent is N2; the theoretical yield of N2O4 is 8.0
moles
B. The limiting reagent is N2; the theoretical yield of N2O4 is
16.0 moles
C. The limiting reagent is O2; the theoretical yield of N2O4 is
12.0 moles
D. The limiting reagent is O2; the theoretical yield of N2O4 is
6.0 moles
E. The limiting reagent is O2; the theoretical yield of N2O4 is
8.0 moles
4. What is a limiting reagent?
A. The reactant that is used up last and prevents more product
from being made
B. The reactant that is never used up
C. The reactant that is used up first and prevents more products
from being made
D. The reactant that is in excess and does not get used up in
the reaction
E. The reactant that is always in greater quantity
7. The reaction of 5.0 g hydrogen with 5.0 g carbon monoxide
produced 4.5 g methanol. What is the percent yield for the
reaction 2H2 + CO ! CH3OH?
A. 11%
B. 79%
C. 96%
D. 24%
E. 63%
5. A mixture of 2.0 moles I2 and 4.0 moles Zn are reacted to
completion in a closed container according to the following
chemical equation: I2 + Zn ! ZnI2. What are the contents of
the container after the reaction?
A. Zn and ZnI2
B. I2 and ZnI2
C. Zn and I2
D. I2, Zn, and ZnI2
E. ZnI2
8. The reaction of 5.0 g fluorine with excess chlorine produced 5.6
g ClF3 in the reaction Cl2 + 3F2 ! 2ClF3. What was the percent
yield of the reaction?
A. 58%
B. 69%
C. 76%
D. 86%
E. 92%
104
General Chemistry 1
120 MINS
Lesson 14: Mass Relationships in
Chemical Reactions (Laboratory)
Content Standard
The learners demonstrate an understanding of the quantitative relationship of
reactants and products in a chemical reaction.
Performance Standards
The learners shall be able to design, using multimedia, demonstrations, or
models, a representation or simulation of any of the following:
1. Atomic structure
2. Mass relationships in reactions
Learning Competencies
The learners determine the mass relationship in a chemical reaction
(STEM_GC11MR-Ig-h-42).
Specific Learning Outcomes
At the end of the lesson the learner will be able to:
1. Prepare NaCl from the reaction of sodium bicarbonate and hydrochloric
acid;
2. Determine the actual yield of the reaction;
3. Illustrate the mass relationship in a chemical reaction by calculating the
theoretical yield of the reaction; and
4. Determine the percentage yield of the reaction.
105
Lesson Outline
Motivation
Why is Baking Soda Added to Cakes and
Cookies?
Introduction
Introduction to Laboratory Activity
17
Instruction,
Delivery and
Practice
Laboratory Activity
80
Enrichment
Post-laboratory Session
20
Evaluation
Data Sheet and Activity Sheet
3
Materials
Evaporating dish, watch glass, balance (triple beam or electronic balance),
Sodium bicarbonate, spatula or small plastic knife, dilute hydrochloric acid
(3 moles), beaker or glass container for the acid, long dropper, Bunsen
burner, wire gauze or mesh, iron stand, iron ring, wash bottle, and distilled
water
Resources
(1) Burdge, J. & Overby, J. (2012). Chemistry: Atoms first. New York:
McGraw-Hill.
(2) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York:
McGraw-Hill.
(3) Moore, J.W., Stanitski, C.L. & Jurs, P.C. (2012). Chemistry: The
molecular science (4th ed.). Belmont, CA: Brooks/Cole Cengage
Learning.
(4) Zumdahl, SS. &Zumdahl, S.A. (2012). Chemistry: An atoms first
approach. Belmont, CA: Brooks/Cole Cengage Learning.
MOTIVATION (3 minutes)
Why do we add baking soda, NaHCO3, in baking cookies and cakes?
INTRODUCTION (17 minutes)
Introduce the laboratory experiment as indicated in the Laboratory Sheet.
INSTRUCTION: Laboratory activity (80 minutes)
Safety Precautions:
1. Never taste anything during a science activity.
2. Wear appropriate laboratory attire; goggles and apron must be worn throughout the
experiment.
3. Dispose of the materials as directed by your teacher.
4. Wash your hands with soap and water after the science activity.
Teacher Tip
Baking soda is used to make cakes and cookies
‘rise’. When a weak acid such as lemon juice,
vinegar, or buttermilk is added to baking soda,
bubbles of carbon dioxide are produced. The
release of gas is what causes the cake to ‘rise’.
Teacher Tip
1. Prepare the classroom or laboratory, the
materials, the laboratory sheets to be used.
2. Distribute the laboratory sheets at the start of
the lesson.
3. After the introduction and motivation, explain
the procedures of the activity.
4. Explain the safety precautions.
5. This activity can be performed individually or
in groups.
Teacher Tip
It is important to discuss the safety precautions
thoroughly before starting the experiment.
5. Follow all laboratory instructions as directed by your teacher.
Procedure:
1. Discuss the procedure of the experiment. Demonstrate the setup to be used.
2. During the activity, they should record the data in the activity sheets.
3. After the activity, make sure that the learners clean the materials and equipment and properly
dispose of the product.
106
Note the precautions in handling the acid and in
lighting the burner.
The product, NaCl, can be disposed in the sink
during washing.
ENRICHMENT (20 minutes)
POST-LABORATORY SESSION
Give them enough time to accomplish the activity sheet of the experiment.
EVALUATION
EXCEEDS EXPECTATIONS
The learner performed the
experiment using proper
laboratory techniques while
observing safety precautions; and
was able to answer at least 75% of
the calculations and discussions in
the activity sheet.
MEETS EXPECTATIONS
NEEDS IMPROVEMENT
NOT VISIBLE
The learner needed to improve his/
her use of laboratory techniques;
but was able to observe safety
precautions; and was able to
answer at least 60% of the
calculations and discussions in the
activity sheet.
The learner needed to improve his/
her use of laboratory techniques
and observance of safety
measures; and was able to answer
at least 50% of the calculations and
discussions in the activity sheet.
The learner did not observe
proper safety procedures for
the experiment; did not use
proper laboratory techniques;
and was able to answer less
than 25% of the items in the
activity sheet.
LABORATORY EXPERIMENT
MASS RELATIONSHIPS IN CHEMICAL REACTIONS
Introduction
A reaction is said to have been completed if one of the reactants is completely consumed by the reaction. In this experiment, sodium
bicarbonate (baking soda) is made to react with hydrochloric acid to produce sodium chloride according to the reaction:
NaHCO3 +HCl → NaCl + H2O + CO2(g)
You will use an accurately measured amount of NaHCO3 and add enough HCl until the bicarbonate is completely used up. You will isolate the
product, NaCl, from the other products and determine its mass. This is the actual yield of the reaction. The theoretical yield can be calculated
by using the mass relationships in the balanced chemical equation above. The percentage yield can be determined from the ratio of the actual
yield to the theoretical yield.
Objectives
1. To perform a chemical reaction and measure the actual yield of sodium chloride from the chemical reaction.
2. To determine the percent yield of the reaction.
107
Materials
a. Evaporating dish
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
n.
Long dropper
Watch glass
Bunsen burner
Balance (triple beam or electronic balance)
Wire gauze or mesh
Sodium bicarbonate
Iron stand
Spatula or small plastic knife
Iron ring
Dilute hydrochloric acid (3 moles)
Wash bottle
Beaker or glass container for the acid
Distilled water (use commercially available distilled water)
Safety Precautions
1. Never taste anything during a science activity.
2.
3.
4.
5.
Wear appropriate laboratory attire; goggles and apron must be worn throughout the experiment.
Dispose of the materials as directed by your teacher.
Wash your hands with soap and water after the science activity.
Follow all laboratory instructions as directed by your teacher.
Procedure:
1.
2.
3.
4.
Clean and dry an evaporating dish and a watch glass. The watch glass will be used as the cover of the evaporating dish.
Weigh the combination of the evaporating dish and the watch glass to the nearest 0.01 g. Record the mass in the data table.
Put about 2.00 grams of pure NaHCO3 into the dish. Weigh the dish, the contents, and the cover watch glass to the nearest 0.01 g.
Set up the Bunsen burner, ring, and wire mesh, and set the evaporating dish on the wire mesh. To cover the dish, place the curved side
down and the glass slightly off center so that the lip of the dish is uncovered. Do not light the burner yet.
5. Add dilute hydrochloric acid drop wise down the lip of the dish to the bicarbonate sample in the dish. Continue adding the acid dropwise
until no more reaction takes place when a drop of acid is added. Gently swirl the contents of the dish so that all of the solid gets in contact
with the liquid. Do not add excess acid.
Caution: HCl acid is caustic and corrosive. Avoid contact with skin and eyes. Avoid breathing the vapors. Wear safety goggles and apron.
Wipe away all spills. If any acid spills on you, immediately flush the area with water and notify your teacher.
108
6. Carefully rinse the bottom of the watch glass with distilled water, a few drops at a time, and collect all the washings in the evaporating dish.
7. Gently heat the evaporating dish, contents, and cover with a low flame until the salt is completely dry. Move the burner back and forth to
avoid spattering. If the contents of the dish spatter, reduce the flame.
8. Turn off the flame. Allow the dish to cool to room temperature. Weigh the dish, contents, and cover watch glass to the nearest 0.01 g.
Caution: Before you light the burner, make sure that long hair and loose clothing have been confined. Remember to allow all apparatus to
cool before you handle it again.
9. Repeat Steps 7 and 8 to be sure that constant weight has been obtained. Two consecutive mass readings should agree within 0.02 g.
Calculations:
1. Calculate the theoretical yield of NaCl that should have been obtained from the reaction. Show your calculations.
2. Determine the percentage yield.
Discussion:
1. What is the cause of the effervescence that you observed during the reaction?
2. How can you conclude that the reaction has gone to completion?
3. How do you know if the product was completely dry?
4. List possible sources of error which you think affected the yield of your reaction. Did your error cause your result to be higher or lower than
the theoretical yield?
Additional Exercises - Practice:
1. Suppose you started with 6.4 g NaHCO3 and added sufficient HCl for a complete reaction, how much NaCl in g would you expect to
produce? Show your calculations.
2. If you wanted to produce 3.78 g NaCl, how much NaHCO3 in g would you start with, assuming no loss of product occurs?
REPORT SHEET: MASS RELATIONSHIPS - LABORATORY DATA TABLE
Mass of evaporating dish, watch glass, and NaHCO3
Mass of evaporating dish and watch glass
Mass of NaHCO3
Mass of evaporating dish, cover, and NaCl
Trial 1 ________________
Trial 2 ________________
Trial 3 _________________(if needed)
Mass of NaCl obtained (experimental)
109
General Chemistry 1
240 MINS
Lesson 15: Gases (Lecture)
Content Standard
The learners demonstrate an understanding of:
Lesson Outline
1. The mathematical relationship between the pressure, volume, and
temperature of a gas;
Introduction
Presentation of Learning Objectives and
Important Keywords
5
2. The partial pressures of a gas;
3. The quantitative relationships of the reactants and products in a gaseous
reaction; and
4. Behavior and properties of gases at the molecular level.
Motivation
Relate the Presence of Gases in Daily Life
5
Instruction,
Delivery and
Practice
I.
II.
III.
IV.
Enrichment
Problem Solving
Evaluation
Check-up Quiz
Performance Standards
The learners shall be able to:
1. Determine the volume or pressure of gas at different conditions;
2. Determine the pressure of the components or of the whole gas mixture;
3. Calculate the amount of products or reactants involved in a gaseous
reaction; and
4. Explain the properties and behavior of a gas in terms of its molecular
composition.
Learning Competency
At the end of the lesson, the learners:
1. Define pressure and give the common units of pressure
(STEM_GC11G-Ih-i-43);
2. Express the gas laws in equation form (STEM_GC11G-Ih-i-44);
3. Use the gas laws to determine pressure, volume, or temperature
of a gas under certain conditions of change (STEM_GC11G-Ihi-45);
4. Use the Ideal Gas Equation to calculate pressure, volume,
temperature, or number of moles of a gas (STEM_GC11G-Ihi-46);
Gas Laws
Gas Mixtures
Reaction Stoichiometry Involving Gases
Kinetic Molecular Theory of Gases
200
30
Materials
Projector, computer, flip charts
Resources
(1) Brown, T. L., Bursten, B. E., LeMay Jr., H. E., Murphy, C., & Woodward,
P. (2014). Chemistry: The central science. (13th ed.). Upper Saddle
River, NJ: Prentice Hall.
(2) Burdge, J. (2013).Chemistry. (3rd ed). New York: McGraw-Hill.
(3) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York:
McGraw-Hill.
6. Apply the principles of stoichiometry to determine the amounts
(volume, number of moles, or mass) of gaseous reactants and
products (STEM_GC11GS-Ii-j-48);
7. Explain the gas laws in terms of the kinetic molecular theory of
gases (STEM_GC11KMT-Ij-49); and
8. Relate the rate of gas effusion with molar mass
(STEM_GC11KMT-Ij-50).
5. Use Dalton’s Law of Partial Pressures to relate mole fraction and
partial pressure of gases in a mixture (STEM_GC11DL-Ii-47);
110
INTRODUCTION (5 minutes)
1. Introduce the learning objectives using the suggested protocol (Read-aloud):
a. I will be able to calculate the pressure or volume of a gas at different conditions
Teacher Tip
Display the objectives prominently on the board,
so that the learners can track the progress of their
learning. List these keywords on the board.
b. I will be able to determine the pressure of a gas mixture or of its components
c. I will be able to determine the amount of products or reactants involved in a gas phase
reaction
d. I will be able to discuss the properties and behavior of gases in terms of its molecular
composition
2. Present the keywords for the concepts to be learned:
a. Boyle’s Law
b. Charles’s Law
c. Avogadro’s Law
d. Ideal Gas Equation
e. Partial pressure
f.
Dalton’s Law
g. Gas reaction stoichiometry
h. Kinetic Molecular Theory
MOTIVATION (5 minutes)
Point out the abundance of gases in their surroundings, such as in the environment, at home, and
in other places. Ask them where gases are encountered or used in everyday life. Some expected
responses are:
a. In the air, which supplies us with the gases we breathe
b. In the kitchen, wherein a gas (liquid petroleum gas) is used for heating or cooking
c. In the hospital, wherein gases are used to aid the breathing of patients
d. In the automobile, wherein gases are burned in order to make the cars move
e. In carbonated drinks, wherein a gas (carbon dioxide) makes the drinks refreshing
111
Teacher Tip
The lesson is essentially a review of the basic
concepts presented and used in junior high school.
INSTRUCTION / DELIVERY / PRACTICE (200 minutes)
Note: The delivery will be done in four 50-minute sessions. It is best to summarize the concepts
learned at the end of each session.
PART ONE
Ask the learners to recall the definition of pressure – the amount of force exerted per unit area.
Let them give the unit for pressure.
Show them a balloon and ask them to point out the role of the pressure of the gas inside the
balloon.
Ask them to imagine the tire of a vehicle and the need to pump air into the tire up to a given
pressure.
a. What will happen if the pressure is much lower than what it should be?
b. What will happen if the pressure is much greater than what it should be?
Ask them if they know how the air pressure of the tire is measured and expressed. Point out the
various units used for pressure:
a. The old air pump in the gasoline stations used the unit pounds per square inch (psi), which is
widely used especially in the United States, but usage of which is supposedly discouraged.
b. Later on, the new air pumps used kilopascal (KPa) (or newton per square meter, N/m2),
which is the SI unit for pressure.
c. In chemistry, a widely used unit for pressure is the atmosphere (atm), but the International
Union of Pure and Applied Chemistry discourages its usage. However, it takes some time for
usage of this unit to be discontinued.
d. Another old and popular unit for pressure is the Torr (or mmHg), yet the International Union
of Pure and Applied Chemistry is also discouraging the usage of this unit.
112
Teacher Tip
Ask them how the balloon would look like if there
is no gas inside the balloon.
It would be helpful if a picture of an automobile
with tires is shown to them.An alternative example
is an air mattress.
Showing a picture of a gasoline station air pump
may also be beneficial to the learners.
Write on the board the relationship between the different units:
Teacher Tip
Ask them the origin of the unit Torr and the
experiment of Torricelli. If they are not familiar
with this, assign them to read about this from the
internet and to write a report on what they read.
1 atm = 760 Torr (mm Hg)
1 atm = 101.3 kPa
Point out to that aside from pressure, the other parameters (or variables) used to describe gases
are volume and temperature.
The common unit for volume is the liter (L), but the SI unit for volume is m3. The equivalence of
the liter in SI units is simple:
1 L = 1000 m3
1 L = 1 dm3
1 mL = 1 cm3
The common unit for temperature is degree Celsius (oC), but the SI unit is Kelvin (K). The
relationship between the units is
K = oC + 273
Point out that the relationship between these three parameter are expressed by the Gas Laws:
1. Boyle’s Law
2. Charles’s Law
3. Avogadro’s Law
Ask them to state Boyle’s Law and emphasize on expressing the law correctly: ‘The volume of a
given amount of gas is inversely proportional to its pressure at constant temperature.’
•
Highlight that Boyle’s Law is valid only if the amount of the gas and the temperature is
constant.
•
Write the mathematical expression for Boyle’s Law:
In terms of a proportion: V α 1/ P (at constant amount and temperature)
In terms of an equation:
PV = k
V = k/P
(at constant amount and temperature)
or
P1 V1 = P2 V2
113
Teacher Tip
They would be familiar with Boyle’s Law and
Charles’s Law from their Chemistry course in junior
high school. A review and an enrichment could be
done for these topics.
It would be helpful to demonstrate Boyle’s Law
through a 60-mL plastic syringe sealed at its inlet.
Show what happens to the volume of the trapped
gas once pressure of the gas is increased by
pushing the plunger inwards.
Draw the graph relating pressure and volume. Point out that the plot is called an isotherm, since
the relationship is exhibited only at constant temperature.
Teacher Tip
They should be familiar with the expressions, and
they could be asked to write them on the board.
They might be familiar with the plot from junior
high school, and they could be asked to sketch it
on the board.
They might have learned how to solve this type of
problem from the Chemistry course in junior high
school. Let them recall how to solve the problem.
Answer key:
1. 3.75 L
2. 3.0 atm
A graph showing the relationship between
volume and pressure, as stated by Boyle’s Law
Ask them to solve the following problems:
1. A gas sample occupies a volume of 2.5 L at a pressure of 1.5 atm. What would be the volume
of the gas if its pressure is reduced to 1 atm at the same temperature?
2. The gas inside a balloon has a volume of 15.0 L at a pressure of 2.0 atm. Calculate the
pressure of the gas if its volume is compressed to 10.0 L at the same temperature.
Ask them to state Charles’s Law and emphasize on expressing the law correctly: ‘The volume of
a given amount of gas is directly proportional to its absolute temperature at constant pressure.’
1. Highlight that Charles’s Law is valid only if the amount of the gas and the pressure is constant.
Also, point out that the temperature should be expressed in the unit Kelvin (K).
114
Teacher Tip
They should be familiar with the expressions, and
they could be asked to write them on the board.
Teacher Tip
They might be familiar with the plot from junior
high school, and they could be asked to sketch it
on the board.
Write the mathematical expression for Charles’s Law:
In terms of a proportion: V α T (at constant amount and pressure)
In terms of an equation:
V/T=k
V = k T (at constant amount and pressure)
or
V1 / T1 = V2 / T2
Draw the graph relating volume and temperature. Point out that the plot is called an isobar,
since the relationship is exhibited only at constant pressure.
A graph showing the relationship between volume and pressure, as stated by Charles’ Law
Ask them to solve the following problems:
1. At 30oC, the volume of a sample of air was 5.8 L. What would be the volume of the air sample
if it is heated to 60oC at the same pressure?
2. A given amount of oxygen gas has a volume of 25.0 L at a temperature of
and a
pressure of 1.0 atm. At what temperature would this gas occupy a volume of 22.0 L at a
pressure of 1.0 atm?
37oC
115
They might have learned how to solve this type of
problem from the Chemistry course in junior high
school. Let them recall how to solve the problem.
Answer key:
1. 6.37 L
2. 273 K
State Avogadro’s Law: ‘The volume of a gas at a given temperature pressure is directly
proportional to the number of moles contained in the volume.
•
Mention that this law is based on Avogadro’s hypothesis that ‘the same volume of two gases
at the same temperature and pressure contain the same number of molecules’.
•
Let them recall that the SI unit mole is related to the number of molecules in a substance.
•
Point out that experiments have shown that the volume of 1.0 mole of a gas at 0oC and 1 atm
is 22.4 L.
•
Write the mathematical expression for Avogadro’s Law:
In terms of a proportion: V α n
In terms of an equation:
V/n = k
V = kn
(at constant temperature and pressure)
(at constant temperature and pressure)
or
V1 / n1 = V2 / n2
Answer Key
1. 168 L
2. 0.446 mol
Ask them to solve the following problems:
1. 1.0 mole of a gas occupies a volume of 22.4 L gas at 0oC and 1 atm. What would be the
volume of 7.5 mol of the gas at the same temperature and pressure?
2. The volume of a gas sample at 0oC and 1.0 atm is 10.0 L. How many moles of gas are
contained in the sample?
The three gas laws can be combined into a single equation known as the Ideal Gas Equation:
PV = nRT
116
This equation can be rearranged into an equation known as the combined gas law, which holds
true for a given amount of gas:
PV
= nR = k
T
P 1 V1
P2V2
=
T1
T2
The combined gas law reduces to Boyle’s Law, if temperature is kept constant (i.e. T1#=#T2):
P1 V1 = P2V2
It also reduces to Charles’s Law, if pressure is kept constant (i.e. P1 = P2):
V1
=
T1
V2
T2
It will also show that pressure of a gas is directly proportional to its absolute temperature, if the
volume is kept constant (i.e. V1 = V2):
P1
T1
=
P2
T2
The value of R can be calculated from the molar volume at 0oC and 1 atm (V#=#22.4#L).
R
=
PV
nT
=
(1 atm) (22.4 L)
(1 mol) (273 K)
=
0.0821
atm L
mol K
Point out the importance of R, which is known as the gas constant, and that it is one of the few
universal constants, i.e. its value is the same anywhere and anytime.
117
Ask them to solve the following problems:
1. A gas sample occupies a volume of 12.0 L at 50oC and 700 Torr. How many moles of gas are
contained in the sample?
2. Calculate the volume that will be occupied by 20.0 g carbon dioxide at 25oC and 1.25 atm.
3. What would be the pressure of 6.40 g oxygen gas in a vessel with a volume of 4.5 L at 20oC?
The ideal gas equation can be transformed into an expression involving density. The number of
moles n can be expressed in terms of mass and molar mass (or weight and molecular weight,
respectively):
n = w/M
Introducing this into the ideal gas equation gives:
P V = (w / M) R T
which can be rearranged into
P M = (w / V) R T
The term w / V is recognized as equal to density, d, so that the equation
becomes:
PM = dRT
Note that if the value of R as 0.0821 (atm L) / (mol K) is used, the unit for density
in the equation should be g#/#L.
118
Answer Key
1. 0.0347 mol
2. 8.90 L
3. 1.07 atm
For a given gas (i.e. M = constant ) at a given pressure, the equation can be reduced into
dT = K
or
d1 T1 = d2 T2
This equation shows that the density of a gas is inversely proportional to its temperature. This
means that hot air has a lower density than cold air.
The relationship between density and temperature can explain:
a. The principle of the hot air balloon
b. The principle behind passive cooling in building design
PART TWO
Highlight that many of the gases encountered in the surrounding are mixtures.
Point out that the Ideal Gas Equation can also be applied to not only to pure gases, but also to
mixtures of gases.
Present a system composed of three gases contained in a vessel of volume V and kept at a
temperature T. The number of moles of each gas is n1, n2 and n3, for gases 1, 2 and 3,
respectively, so that the total number of moles of gases is
n total = n1 + n2 + n3
The pressure of the mixture is given by the Ideal Gas Equation:
P mixture V = n total R T
119
Teacher Tip
Assign them to read on ‘passive cooling’ from
internet resources. This reading will make them
realize the relevance of the gas laws in building
design.
Expressing ntotal in terms of the number of moles of each gas and solving for Pmixture will result in
P mixture
=
( n 1 + n 2 + n 3 ) RT
Teacher Tip
The learners could be asked to write the resulting
expression for each step of the derivation.
V
If the right-hand side of the equation is expanded, the expression becomes
P mixture
n 1 RT
=
V
+
n 2 RT
V
+
n 3 RT
V
Let them realize that the term ni R T / V is equal to P and see that the previous equation can be
written as
P mixture = P1 + P2 +P3
The pressures P1, P2, and P3, called partial pressure of each gas, corresponds to the pressure that
the gas will exert in a volume equal to that of the mixture.
According to this expression, the total pressure of a gas mixture is equal to the sum of the partial
pressure of each gas. This is known as Dalton’s Law of Partial Pressure.
Ask them to solve the following problems:
1. In a gas mixture composed of N2, Ne, and He, the partial pressure of N2 is 0.50 atm, that of
Ne is 1.1 atm, and that of He is 0.80 atm. What is the total pressure of the mixture?
2. A sample of oxygen gas, which is saturated with water vapor, is kept in a 10-L vessel at 30oC
and has a pressure of 758 Torr. If the pressure of the water vapor at this temperature is 31.8
Torr, what would be the pressure of the dry oxygen?
120
Answer Key
1. 2.4 atm
2. 726.2 Torr
The application of Dalton’s Law can also yield information about the composition of the mixture,
in terms of the mole fraction of each component. Let them write on the board the expression for
the pressure of gas 1 and that of the mixture:
P1
=
n 1 RT
P mixture
V
=
Teacher Tip
They could be asked to write the resulting
expression for each step of the derivation.
n total RT
V
Dividing P1 by Pmixture gives the following expression:
P1
P mixture
=
n1
n total
=
x1
The term at the right-hand of the equation is actually a fraction, i.e. part divided the whole, and is
known as mole fraction X1. Rearranging the expression leads to an important relationship:
P1 = Pmixture X1
Ask them to solve the following problems:
1. In a gas mixture composed of N2, Ne, and He, the partial pressure of N2 is 0.50 atm, that of
Ne is 1.1 atm, and that of He is 0.80 atm. Calculate the mole fraction of each gas.
2. A gas mixture contains 2.5 mol N2 and 9.7 mol CO2, and has a pressure of 2.3 atm. What is
the partial pressure of each gas?
PART THREE
As an introduction to this section, make the learners realize that reactions involving gases are
common, such as the burning of fuel and the digestion of sugars:
2 C4H10 (g) + 13 O2 (g) g 8 CO2 (g) + 10 H2O (l)
C6H12O6 (aq) + 6 O2 (g) g 3 CO2 (g) + 6 H2O (l)
121
Teacher Tip
1. N2: X = 0.21; Ne: X = 0.46;
He: X = 0.33
2.
N2: P = 0.47 atm;
CO2: P = 1.83 atm
Let them recall the basic principle of reaction stoichiometry, which is expressed by the balanced
chemical equation.
Note
Reaction stoichiometry – the relationship between
the moles of reactants and products in a reaction
Highlight the following principles learned in the previous discussion:
Answer Key
1. They can be guided in solving the problem by
asking them to solve first the number of moles
of NaN3:
mol NaN3 = 2
1. The Ideal Gas Equation enables the calculation of the number of moles of a gas from its
pressure, volume, and temperature:
n
=
PV
Then, ask them to solve the number of moles
of N2 produced, using the balanced equation:
mol N2 = 3
RT
2. The volume of a gas at standard temperature (0oC) and standard pressure (1 atm) can provide
information about the number of moles of the gas, through the known molar volume under
the standard condition:
V STP
n
Finally, ask them to recall the molar volume at
STP (22.4 L) and use it to solve for the final
answer:
VN2 = 67.2 L
2.
=
22.4
Point out that these principles are useful in calculating the amount of gases involved in a
reaction.
Ask them to solve the following problems:
1. The airbag is a safety device used in cars to cushion the passenger during a crash. It involves
the following chemical reaction which is triggered by an impact:
2 NaN3 (s) g 2 Na (s) + 3 N2 (g)
Calculate the volume of N2 gas (measured at STP) that can be produced from 130.0 g of NaN3
(molar mass = 65).
122
The mole of C2H2 is first calculated through
the molar volume (22.4 L) at STP:
mol C2H2 = 0.446
From this, the mole of CaC2 is calculated:
mol CaC2 = 0.446
The weight of CaC2 is calculated from the
number of moles:
g CaC2 = 28.6 g
2. Acetylene is formed by the reaction of water with calcium carbide, according to the following
equation:
CaC2 (s) + 2 H2O(l) g Ca(OH)2 (aq) + C2H2 (g)
Answer Key
3. The mole of C2H2 is first calculated through
the molar volume (22.4 L) at STP:
mol of C2H2 = 2.23
From this, the mole of CO2 is calculated:
mol CO2 = 4.46
How many grams of CaC2 would be needed to produce 10.0 L (measured at STP) acetylene?
3. The reaction involved in the explosive combustion of acetylene is:
The volume (measured at STP) of CO2 is
calculated from the number of moles:
V CO2 = 100.0 L
This answer can also be obtained through the
ratio of the mol CO2 to mol C2H2.
2 C2H2 (g) + 5 O2(g) g 4 CO2 (g) + 2 H2 (g)
How many L of CO2 gas (measured at STP) will be formed during the combustion of 50.0 L
C2H2 gas (measured at STP)?
Teacher Tip
They can be given a set of problems involving
reaction stoichiometry as homework.
PART FOUR
Emphasize to the learners that the gas laws summarize the general behavior of gases. Through
these laws, the behavior of gases can be predicted. However, no explanation is given for this
behavior.
Point out that the explanation is provided by the Kinetic Molecular Theory. The theory assumes
a model which can be used to explain why gases behave the way they do. The model is
described through a set of postulates:
1. Gases are made up of very small molecules,which are separated by a very great distance
between them. The dimension of the molecules is very much smaller than the distance
between them.
2. Because of the very great distance between them, the force of attraction between the
molecules is negligible. The molecules are independent of each other.
3. The molecules are in constant motion, moving in randomly in all directions.
4. Due to the great number of molecules and their random motion, it is unavoidable that the
molecules will collide with each other and with the walls of the container.
123
Teacher Tip
To highlight each postulate, write the keyword for
each statement on the board.
5. During these collisions, there is no change in the momentum of the molecules.
6. The average kinetic energy of the molecules is determined only by the absolute temperature
of the gas.
Ask them to draw a representation of the model of the Kinetic Molecular Theory. The model
should be:
Teacher Tip
The model would be familiar to them from the
Science course in junior high school. What might
not have been emphasized then is the motion of
the molecules.
Ask them to apply this model to explain some properties of gases:
a. Why can gases be compressed?
Make them see that because of the great distance between them, gases can be forced to be
close to each other by compressing it.
b. Why does the volume of a gas decrease as the pressure is increased at constant
temperature?
This question asks for an explanation for the behavior described by Boyle’s Law. The answer
would be similar to the previous question on the compressibility of gases. As the molecules
become closer to each other, the volume of the gas becomes smaller.
c. Why do gases exert pressure?
Remind them that pressure is actually a force acting on a unit area. Help them realize that the
collision of the molecules with the walls of the container produces a force acting on the wall.
124
d. Why does the volume of a gas increase as it is heated at constant pressure?
This question asks for an explanation for the behavior described by Charles’s Law. Help them
recognize that Postulate 5 expresses the effect of temperature on gases. According to this
postulate, if the temperature is increased, the kinetic energy of the molecule increases. The
increased kinetic energy makes the molecules to move faster and farther apart from each
other, leading to a greater volume.
Point out that a mathematical treatment of the Kinetic Molecular Theory would lead to an
equation for the root-mean-square velocity of the molecule:
This equation clearly shows that as the temperature increases, the velocity of the molecule
increases.The gas molecules move faster at a higher temperature. It also shows that as the
molar mass M of the molecule increases, the velocity of the molecule decreases.
If the velocity of two molecules of molar mass M1 and M2 are compared, the result is
2
1
The velocity of the molecules determines the rate of diffusion of the gases. The relationship
between diffusion rate and molar mass has been verified by experiments, and is known as
Graham’s Law of Diffusion.
Ask them to imagine that two bottles are placed at opposite ends of the room at equal distance
from them. One bottle contains ammonia gas, NH3 (M = 17) which has a pungent odor, and the
other contains hydrogen sulfide, H2S (M = 34) which has an odor like that of a rotten egg. Which
odor will they sense first?
125
Teacher Tip
The root-mean-square velocity is the squareroot of the mean of the square of the
velocities of the molecules:
ENRICHMENT
1. As mentioned in the Delivery, at the end of each session, summarize the concepts discussed in the session.
2. Conduct a session on problem solving to provide more exercises on the application of the gas equations.
3. Conduct a laboratory activity on Graham’s Law of Diffusion.
EVALUATION (30 minutes)
Check-up quiz
1. Under which of the following volumes will 1.00 mol of an ideal gas exhibit the greatest pressure at 300 K?
A. 0.01 L
C. 1.00 L
B. 0.10 L
D. 10.0 L
2. How will the volume of 0.50 mol of a gas behave if the temperature is raised from 30oC to 60oC at constant pressure?
A. The volume will increase.
C. The volume will be doubled.
B. The volume will decrease.
D. The volume will be halved.
3. Which among the following systems will have the greatest volume at STP?
A. 1.00 g N2 gas (M = 28 g/mol)
B. 1.00 g NH3 gas (M = 17 g/mol)
C. 1.00 g CO2 gas (M = 44 g/mol)
D. 1.00 g He gas (M = 4 g/mol)
4. How will the density of a gas vary if its temperature is increased from 25oC to 50oC at constant pressure?
A. The density of the gas will not change! !
B. The density of the gas will increase
C. The density of the gas will decrease
D. The density of the gas will double
126
5. Which of the following volumes of oxygen will contain the greatest number of molecules at 300K and 1 atm pressure?
A. 0.01 L
C. 1.00 L
B. 0.10 L
D. 10.0 L!
6.!In which of the following gas mixtures of N2(g) and He(g) is the partial pressure of He(g) the greatest?
A. 2 moles N2(g) and 3 mole He(g)
B. 3 moles N2(g) and 1 mole He(g)
C. 4 moles N2(g) and 2 mole He(g)
D. 5 moles N2(g) and 5 mole He(g)
7. Hydrogen, H2(g),reacts with oxygen, O2(g), to form water H2O(l): !
2 H2(g) + O2(g) ##!###2 H2O(g)#
How many liters of oxygen gas, measured at STP, will be needed to react completely with 10.0 L hydrogen gas, also measured at STP?
A. 5.00 L
C. 20.0 L
B. 10.0 L
D. 100.0 L
8. Which of the following postulates of the Kinetic Molecular Theory for gases can explain why gases exhibit pressure?
A. The molecules are in constant random motion!
B. The molecules collide with the walls of the vessel!
C. The distance between the molecules is great!
D. The molecular kinetic energy depends on temperature!
9. How will the velocity of a gas molecule vary if its molecular weight is increased from 32 g mol-1 to 64 g mol-1?
A. The velocity will increase
!
B. The velocity will decrease
C. The velocity will double
D. The velocity will remain the same
127
General Chemistry 1
120 MINS
Lesson 16: Gases (Laboratory)
Lesson Outline
Content Standard
The learners demonstrate the ability to explain experimental
observations using the laws and theories learned in the lecture course.
Introduction
Communicating Learning Objectives
5
Motivation
Inquiry
5
Instruction
Pre-activity
60
Enrichment
Problem solving
25
Learning Competencies
Evaluation
Post-laboratory
25
At the end of the lesson, the learners:
1. Observe and measure the difference in the diffusion rate of two
gases. (STEM_GC11KMT-Ij-51);
Materials
Laboratory glassware
Performance Standards
The learners shall be able to compare the rates of diffusion of two gases
and explain the observed behavior.
Resources
(1) Laboratory experiments found in the internet, such as: Fasano, Janet.
Graham’s Law lab [PDF document]. Retrieved from Needham Public Schools:
http://fcw.needham.k12.ma.us/~Janet/FOV1-00108AC5/Graham's%20Law
%20Lab.pdf
128
INTRODUCTION (5 minutes)
1. State the objective of the experiment that the learner will be performing.
2. Ask them to recall the diffusion property of gases and explain in terms of the kinetic molecular
theory.
3. Point out safety measures to be observed.
Teacher Tip
A laboratory experiment sheet has to be prepared
and distributed to the learners.
The experiment found in the internet could be
revised or simplified to suit the available facilities
in the laboratory.
Point out why we can smell the odor of a fruit (such as durian) or a flower (such as sampaguita)
from a distance.
Teacher Tip
An alternative experiential approach to the
motivation can be done using an open bottle of
perfume in front of the class.
INSTRUCTION (60 minutes)
Remind them to observe safety precautions during
the experiment.
MOTIVATION (5 minutes)
1. Provide each group with the necessary materials.
2. Ask them to follow the procedure in the experiment sheet.
Sample Problems
a. Gas X has a molar mass of 72 g/mol and Gas Y
has a molar mass of 4 g/mol. How much faster
or slower does Gas Y effuse from a small
opening than Gas X if they are at the same
temperature?
b. If the density of hydrogen is 0.090 g/L and its
rate of diffusion is 5.93 times that of chlorine,
what is the density of chlorine?
ENRICHMENT (25 minutes)
Assign them to solve some problems involving Graham’s Law.
Teacher Tip
Provide them with the worksheet that they have to
fill up. It could include some more questions.
EVALUATION (25 minutes)
Ask them to submit a report on the experiment.
129
DIFFUSION OF GASES
Introduction
One of the properties of gases is its ability to diffuse easily. This property can be explained by the motion of the gas molecules and the
absence of intermolecular forces of attraction. As a result of this property, a gas spreads easily in the air and fills up all available space.
In this experiment, the diffusion of two gases will be investigated and their relative rates of diffusion will be measured. The gases will be
confined in a glass tube and will be introduced at the opposite ends of the tube.The mixing of the two gases will be indicated by the formation
of a white solid in the tube.
Materials
a. Concentrated hydrochloric acid, HCl
b. Ammonia solution, NH3
c. Glass tube
d. Cotton buds (Q-tips)
Procedure
1. Set the glass tube against a black background and place markings on both ends to indicate where the cotton tips will be introduced.
2. Place two drops of concentrated HCl in one cotton bud, and two drops of NH3 solution in the second cotton bud.
Caution: These solutions can irritate your skin. Use gloves, if possible.
3. Simultaneously insert the cotton buds in the opposite ends of the glass tube.
4. Note which part of the tube a white ring will form. Mark this part and measure its distance from the HCl end and from the NH3 end.
5. Repeat Steps 1 to 4 to provide a duplicate measurement. This will be used to check the repeatability of the results.
6. Dispose the cotton buds in the designated waste container.
130
Treatment of results
1. Record the distance of the white ring formed in the tube from the ends where the two gases were introduced. Calculate the ratio of these
two distances. This ratio is equal to the ratio of the rates of diffusion of the two gases.
2. Obtain the molar mass of HCl and NH3, and calculate the ratio of the diffusion rates of the two gases using Graham’s Law of Diffusion.
3. Compare the observed and predicted ratio of the diffusion rates.
TRIAL 1
TRIAL 2
TRIAL 1
Distance from the HCl end (dHCl)
Molar mass of HCl (MHCl)
Distance from the NH3 end (dNH3)
Molar mass of NH3 (MNH3)
Ratio of distances (dHCl) / dNH3)
Calculated ratio of diffusion rates (dHCl) / dNH3)
TRIAL 2
EVALUATION
EXCEEDS EXPECTATIONS
MEETS EXPECTATIONS
NEEDS IMPROVEMENT
NOT VISIBLE
The learner:
The learner:
The learner:
The learner:
i.
i.
i.
i.
performed the experiment
correctly;
performed the experiment
correctly;
performed the experiment
correctly;
ii. described the results
correctly; and
ii. described the results correctly;
and
ii. described the results correctly;
but
iii. discussed the results of the
experiment very well.
iii. discussed the results of the
experiment well.
iii. did not discuss the results of
the experiment.
131
did not do the assigned
task.
General Chemistry 1
Lesson 17: Electromagnetic Waves, Planck’s
Quantum Theory, and Photoelectric Effect
120 MINS
Content Standard
The learners demonstrate an understanding of the quantum mechanical
description of the atom and its electronic structure.
Performance Standards
The learners can describe the dual nature of an electron.
Learning Competencies
At the end of the lesson, the learners:
1. Describe the quantum mechanical model of the atom
(STEM_GC11CB-IIa-b-52)
a. Identify the inadequacies of the classical physics in explaining
experimental results that brought about the quantum theory
b. Discuss Planck’s quantum theory
c. Describe the particle-wave duality of light in relation to the
photoelectric effect
d. Recognize the contribution of scientists to the development of the
Quantum Mechanical Model of the Atom
Specific Learning Competencies
At the end of the lesson, the learners will be able to:
1. Describe the characteristics of a wave;
2. Relate the order of the regions of the electromagnetic spectrum in
terms of their wavelength and frequency;
3. State Planck’s equation;
4. Solve problems related to electromagnetic radiation, its energy,
wavelength, and frequency;
5. Describe the particle-wave duality of light; and
6. Recognize technological applications of the photoelectric effect
132
Lesson Outline
Introduction
Presentation of Learning Objectives and
Keywords
10
Motivation
Thermal Imaging Infrared Photography
10
Instruction,
Delivery and
Practice
I.
II.
III.
IV.
Enrichment
Revisit Introductory Questions
Evaluation
Take Home Activity
The Characteristics of a Wave
Planck’s Quantum Theory
The Photoelectric Effect
The Particle-Wave Duality of Light
90
10
Materials
Calculator
Resources
(1) Chang, Raymond and Goldsby, Kenneth A. (2016). Chemistry (12th
ed). New York: McGraw-Hill.
(2) Petrucci, Herring, Madura, and Bissonnette (2011). General Chemistry
and Modern Applications, 10th Ed. Pearson Canada, Inc.
(3) Zumdahl, S.S. and Zumdahl, S.A (2013).Chemistry, 8th ed. Cengage
Learning
(4) Infrared Image Gallery: http://coolcosmos.ipac.caltech.edu/
image_galleries/
(5) PAGASA weather satellite maps: http://meteopilipinas.gov.ph/
INTRODUCTION (10 minutes)
1. Introduce the following learning objectives using any of the suggested protocol
At the end of Part I, I will be able to:
a. Describe the characteristics of a wave
b. Relate the order of the regions of the electromagnetic spectrum in terms of their
wavelength and frequency
c. State Planck’s equation.
d. Solve problems related to electromagnetic radiation, its energy, wavelength, and
frequency.
e. Describe the particle-wave duality of light
f.
Recognize technological applications of the photoelectric effect
2. Present the keywords for the concepts to be learned:
a. wave
h. Hertz
b. frequency
i.
speed of light
c. wavelength
j.
blackbody radiation
d. amplitude
k. quantum theory
e. crest
l.
f.
trough
photoelectric effect
m. particle-wave duality of light
g. period
3. Certain experimental results observed at the beginning of the 20th century could not be
explained by classical physics. These included the blackbody radiation, photoelectric effect,
and the emission spectrum of hydrogen. The new age of physics began when the German
physicist, Max Planck proposed his quantum theory of energy. The lesson will discuss the
emergence of this new theory and the contributions of Max Planck and Albert Einstein to the
development of the quantum theory.
133
4. Post on the board the following essential questions that will be answered after the discussion
a. What is a wave?
b. What is a particle?
c. Is the electron a wave or a particle?
MOTIVATION (5 minutes)
Engage the students in a discussion and ask them the following questions:
1. What is a thermal imaging infrared photography?
Thermal imaging infrared photography detects infrared light and converts this to an
electronic signal that is processed to produce a thermal image.
2. Give some uses of thermal imaging photography.
Examples of uses:
Military operations
Construction – check efficiency of insulation and detect where there are heat leaks; check
electrical wirings in houses to see where there are overheating joints
Fire fighters use this to locate hotspots in a building or locate people who are trapped.
PAGASA weather maps showing warm and cool areas of the ocean
INSTRUCTION (90 minutes)
I.
THE CHARACTERISTICS OF A WAVE
It was in the early 1900’s that a new way of looking at energy and matter began. It stemmed from
Max Planck’s idea about blackbody radiation and culminated in Schrodinger’s wave equation
known also as the wave function, ψ (psi), which described the hydrogen atom.
1. First let us define a particle and a wave.
a. A particle is an object which has distinct chemical or physical properties such as volume or
mass.
b. A wave is a disturbance that travels from one location to another location. The highest
peak of the wave is called the crest and the lowest point is named as the trough.
c. The wave has distinct characteristics that include amplitude, wavelength and frequency.
134
Teacher Tip
Check the Infrared Image Gallery site at http://
coolcosmos.ipac.caltech.edu/image_galleries/. If
possible, print a few infrared pictures beforehand
and pass these to the students during class.
Alternatively, ask the students to visit the website
from their homes, their mobile phones, or from the
school library.
Ask the students to go to the PAGASA website at
http://meteopilipinas.gov.ph/ to see the IR
weather satellite map.
The amplitude is defined as one-half the distance from crest to trough. The wavelength
(symbolized by the Greek letter lambda, λ) is the distance from crest to crest or from
trough to trough.
d. Frequency (symbolized by the Greek letter nu, ν) is defined as the number of waves
passing a fixed point in a specified period of time. Frequency has units of waves per
second or cycles per second. Another unit for frequency is the Hertz (abbreviated Hz)
where 1 Hz is equivalent to 1 cycle per second.
e. The period of a wave is the time for a particle on a medium to make one complete
vibrational cycle.
135
Teacher Tip
Remind the learners that
i. A wave is characterized by its wavelength,
frequency, and amplitude.
ii. The wavelength, λ , has units of length (i.e. m,
mm, nm, etc.)
iii. Frequency, v", has the units of Hz (1 Hz = 1cycle
per second). Sometimes it is also in terms of
(1/time) for example sec-1.
The figures above show two waves travelling between two points at a constant speed. Note
that the wave with longer wavelength has the lower frequency, and the wave with the shorter
wavelength has higher frequency. Thus, wavelength, λ, and frequency, v, are indirectly related
to one another. The wavelength of the wave multiplied by the frequency of the wave
corresponds to the speed, µ, of the wave. In an equation form,
λν = µ
Waves can be classified as mechanical or electromagnetic waves. A mechanical wave requires
a medium for it to travel, i.e. the sound wave, water wave, etc. An electromagnetic wave is a
wave that is capable of transmitting its energy through an empty space or vacuum. Light is
considered to be a electromagnetic wave. In electromagnetic radiation, the frequency of the
wave when multiplied by its wavelength corresponds to the speed of light, c, as shown in the
equation
λν = c
where c = 3.00 x 108 m/s.
The figure on the right shows the various types of
electromagnetic radiation, which differ from one another in
terms of wavelength and frequency. The shortest waves
which have the highest frequency, like the gamma rays,
result from the changes within the nucleus of the atom. The
visible light waves, with wavelength that range from about
400-700 nm, are produced by the motions of electrons
within the atoms and molecules. The longest waves are
those emitted by the antennas of broadcasting stations.
ELECTROMAGNETIC SPECTRUM. Retrieved (https://upload.wikimedia.org/wikipedia/commons/
thumb/f/f1/EM_spectrum.svg/2000px-EM_spectrum.svg.png) (07/02/2016 12:56 PM)
136
Give the learners some practice exercises:
1. If the wavelength is decreased to half its original length, what happens to the frequency?
Answer: The frequency is doubled.
2. A yellow light emitted by a sodium vapor lamp has a wavelength of 589 nm. What is the frequency of the yellow light?
v## =
c#
λ
=
3.00#x#10#8#m/s#
589#nm
x
10#9#nm#
1#m
!
5.09"x"10"14"s"+1""or##5.09"x"10"14"Hz
3. A radio station broadcasts at a frequency of 590 KHz. What is the wavelength of the radio waves?
λ## =
c#
v
3.00#x#10#8#m/s#
=
590#kHz
x
1#kHz#
1,000#Hz
1#Hz#
x
=##508#m##="""5.1"x"10"2"m
1#/#s
4. A particular electromagnetic radiation was found to have a frequency of 8.11 x 1014Hz. What is
the wavelength of this radiation in nm? To what region of the electromagnetic spectrum would
you assign it?
II. PLANCK’S QUANTUM THEORY
When an object is heated, the electrons on the surface are thermally agitated and begin to emit
radiation. Physics around the 1900s was concerned with the spectrum of the light emitted by
heated bodies, particularly by black bodies. A blackbody is a material that absorbs all radiation
that falls on it and is therefore a perfect absorber. When such a blackbody is heated, it was
expected to emit at every wavelength of light that it is able to absorb. Classical physics predicted
that the maximum wavelength emitted by the blackbody would be infinite. However, results
proved otherwise and classical physics could not explain the resulting spectrum of blackbody
radiation. Experimental results showed that while blackbodies emitted radiation at various
wavelengths, they showed a maximum wavelength (not infinite) that shifted toward lower
wavelengths as the temperature increased.
Planck made a radical proposal to explain the experimental results of the blackbody radiation. He
proposed that the atoms on the surface of the heated solid could absorb energy only in discrete
quantities or quanta, and not continuously as assumed by classical physics. The energy absorbed
137
Note
Give the practice exercises as a seatwork. Ask
some students to show their solutions on the
board and explain their answer.
or released by any oscillator are in integer multiples, or quanta, of hν. This became known as
Planck’s equation.
E"="hν"
Energy, E, is equal to frequency, ν, multiplied by Planck’s constant, h, with a value of
6.626 x 10-34 J s. According to quantum theory proposed by Planck, the amount of energy
emitted or absorbed by a body can have values of hν, 2 hν, 3 hν, 10 hν, but never 4.8 hν
or 0.25 hν.
Teacher Tip
Make sure the learner understands the meaning of
integral multiples and quanta. There is no need for
the learner to memorize Planck’s constant, h. The
value should be given to the student during exams
and quizzes.
And because v=c/λ, the equation can also be expressed as
E"="hν"="h""
c#
λ
Ask the learners the following questions:
a. Which is a quantized way of getting from the 1st to the 2nd floor of a building – using the stairs
or using a ramp?
b. Give some examples from daily life that shows quantization.
III. EINSTEIN’S EXPLANATION OF THE PHOTOELECTRIC EFFECT
Another stumbling block for classical physics was the photoelectric effect experiment. According
to classical physics, when light hits a metal surface, the electrons in the metal should slowly
absorb energy from the light until they have enough energy to be emitted to produce a current.
It also predicted that as the intensity of the incident light increases, the kinetic energy of the
emitted electrons should increase. However, the experiment did not support these predictions
but provided the following observations:
Answer Key
a. Using the stairs is quantized. One can take 1
step at a time or 2 steps at a time in going up
to the second floor but never 1.6 steps at a
time.!
b. Some answers could be: The smallest
denomination given by ATM machines; chairs
around the table, etc.
a. When light is made to hit a metal surface, there is a threshold frequency below which no
electrons can be ejected regardless of the intensity of the incident light.
b. Above the threshold frequency, the number of ejected electrons was proportional to the
intensity (or brightness) of the incident light but their energies were not.
c. Above the threshold frequency, the kinetic energy of the emitted electrons increased
linearly with the frequency of the incident light.
138
Source: Retrieved (https://upload.wikimedia.org/
wikipedia/commons/f/f5/Photoelectric_effect.svg)
11/02/16, 08:12 am
These results could not be explained by the wave theory of light. Waves can have any amount of
energy - big waves have a lot of energy, small waves have very little. And if light is a wave, then
the brightness of the light affects the amount of energy - the brighter the light, the bigger the
wave, the more energy it has.
IV. THE PARTICLE-WAVE DUALITY OF LIGHT
Einstein proposed that the only way to explain the photoelectric effect was to say that instead of
being a wave, as was generally accepted, light was actually made up of lots of small packets of
energy called photons that behaved like particles. Each photon has energy given by the equation
E#=#hν#
Where is the frequency of the light and h is Planck’s constant: h = 6.626 x 10-34 J s.
In explaining the results of the photoelectric effect experiment, the energy, hν, of the incident
light is used to remove the electron from the surface of the metal. If the electron is tightly held by
the metal and the energy of the incident light (corresponding to the threshold frequency) is not
sufficient, no electron will be ejected. If the energy of the incident light is sufficient, it will use the
energy to eject the electron; the rest will be given off as the kinetic energy of the electron. In
equation form, this is given by
hν#=#W#+#K.E.#
where W is the work function (the energy needed to eject the electron) and K.E. is the kinetic
energy.
K.E.="hν"+"W"
This explains the observation that the kinetic energy of the emitted electrons varied linearly with
the frequency of the incident light.
139
Light is a wave as shown by different experiments like the diffraction of light by a prism to yield
the visible spectrum. However, the photoelectric effect experiment showed that light also
behaves like a particle. Thus light has both wavelike and particlelike properties. This concept is
called the particle-wave duality of light.
Note
The apparent mass of a photon of light with
wavelength, λ, can be expressed from the
relationship of Einstein’s famous energy equation
from the theory of relativity:
E"="mc"2"
And the energy equation by Planck:
Ephoton"="hv"=""
hc"
λ
And is given by
E"
λc/"λ"
h"
m"=""""""""""""""=""""""""""""""""""""""="
2
2
c
c
λc
Note that the apparent mass of a photon depends
on its wavelength. However, a photon does not
have a mass in a classical sense.
Give the learners practice exercises:
1. The work function or the energy needed to eject an electron in cesium metal is 3.42 x 10-19 J.
If an incident light of frequency 1.00 x 1015 s-1 is used to irradiate the metal, will electrons be
ejected? Show your calculations.
The energy of the incident light can be calculated by
E""=""hv""=""(6.626"x"10"+34"Js)"(1.00"x"10"15"s"+1")""=""6.626"x"10+19"J"
This energy is greater than the work function of cesium metal. Therefore, electrons will be
ejected from the metal.
2. What will be the kinetic energy of the ejected electron?
KE""=""hv"+"W"=""(6.626"x"10"+19"J"+"3.42"x"10"+19"J""=""3.21"x"10+19"J"
140
Teacher Tip
Discuss some problems in class. The rest may be
given as a seatwork. The problems may be
assigned individually or to groups. Then ask the
learners to show their work on the board for
discussion with the entire class. This may also be
used for evaluation.
3. The blue color in fireworks is due to copper (I) chloride , (CuCl), is heated at a temperature of 1200
oC. What is the energy emitted at 4.50 x 102 nm by CuCl?
Solution:))The)quantum)of)energy)can)be)calculated)from)Planck’s)equation)!E"="hv."
a. The frequency can be calculated from the equation, c = λv ; rearranging the terms, we have:
v"" =
c"
λ
=
3.00"x"10"8"m"/"s"
4.50"x"10"+7"m
=""6.67"x"10"14"s"+1
b. Then, solve for the energy using Planck’s equation.
!"E"="hv"="(6.626"x"10"+34"Js")"("6.67"x"10"14"s"+1")"="4.41"x"10"+19"J""
This means that, CuCl emitting a blue light at 450 nm can lose energy only in the increments of
4.41 x 10-19 J, the size of the quantum in this case.
4. There are three types of UV radiation classified by wavelength: UVA (320 – 400 nm), UVB (290 – 320
nm), and UVC (180 – 280 nm). Which type of UV has the lowest energy?
5. A photon of ultraviolet (UV) light possesses enough energy to mutate a strand of human DNA. What
is the energy of a single UV photon having a wavelength of 5.00 nm?
6. Compare the energy (in joules) of (a) photon with a wavelength of 5.00 x 104 nm and (b) photon with
a wavelength of 5.00 x 10-2 nm. At what regions in the spectrum do the samples come from? Relate
the relationship of the wavelength of a radiation to the energy.
7. Chlorophyll absorbs light energies of 3.06 x 10-19J/photon and 4.41 x 10-19J/photon. To what color
and frequency do these absorptions correspond?
141
8. The protective action of ozone in the atmosphere comes through ozone’s absorption of UV radiation
in the 230 to 290 nm wavelength range. What is the energy, in kJ/mol, associated with radiation in
this wavelength range?
9. The work function of potassium metal is 3.68 x 10-19 J. Which of the following will cause electrons to
be ejected from the surface of potassium metal?
a. Red light ( λ = 7.00 x 10 -7 m)
b. Green light ( λ = 5.51 x 10 -7 m)
c. Violet light ( λ = 4.00 x 10 -7 m)
ENRICHMENT (10 minutes)
1. Go back to the essential questions presented during the introduction and ask the students to give
their answers to check their understanding of the lesson.
a. What is a wave?
b. What is a particle?
c. Is the electron a wave or a particle?
2. Return also to the motivation questions on thermal imaging infrared photography. Relate these to
the lesson.
a. Which has longer wavelength, IR or visible radiation?
b. Which will have more energy, IR or visible radiation?
3. Discuss the modern uses of the photoelectric effect.
After a century, Einstein's work on the photoelectric effect gave way to new and very useful
technologies
a. Photocell found in automatic door openers;
b. Ruby lasers, red light emitters used to read bar codes and night vision devices
c. Medical and dental devices
d. Other image processing technologies.
142
Answer Key
4. UVA
6.a. E = 3.98 x 10 -21 J
6.b. E = 3.98 x 10 -15 J from x-ray region
6.c. Wavelength is inversely proportional
to energy.
TAKE HOME ACTIVITY: Scientists on Parade
Explain to the students the take home activity. This may be assigned to individual students or to groups.
Ask the students to prepare a poster which illustrates or describes the role of the scientists listed below
in the development of the quantum mechanical description of the atom. If an LCD projector and laptop
are available, the students may present a 7-slide power point presentation.
Ask the students also to reflect on the lives of these scientists and relate what they liked in the life story
of the scientists and what attributes did they possess that are worthy to emulate.
Make a profile of each scientist and
explain their contributions to the
behavior of the electron.
Samples
1. Max Planck
2. Albert Einstein
3. Niels Bohr
Give the students about 1 to 2 weeks to work on the project. They are to present their work in class at
the end of the lesson on the quantum mechanical description of the atom.
4. Louis de Broglie
5. Werner Heisenberg
6. Erwin Schrödinger
EVALUATION
CRITERIA
EXCEEDS EXPECTATIONS
MEETS EXPECTATIONS
NEEDS IMPROVEMENT
NOT VISIBLE
Information
3-4 unique details or examples
provided; content is complete;
all information clearly relates to
topic
2-3 details are provided
information clearly relates to
topic; diagrams (if present) relate
to topic and add to clarity
1-2 details are provided; some
information provided is not
closely related to topic
Incomplete information;
irrelevant ideas or examples
included
Organization
Clear organizational method
chosen suits work; content
flows in a clear pattern; reader
is able to concentrate on the
information
Information generally organized;
the content flows nicely; the
reader has no difficulty following
the information despite a minor
error or two
2-3 gaps or out of sequence
information cause viewer or
reader to re-read numerous times
for clarity
Information is presented out of
sequence
Presentation
An occasional grammar or
spelling error may result from
risk-taking; materials are
organized and glued down;
presenting a polished, pleasing
result
3 or fewer minor grammar or
spelling errors; mostly
aesthetically pleasing, some
messy parts; evident care of
project
Many minor grammar or spelling
errors; messy; inconsistent care in
attaching materials or drawing or
writing care of the project is
inconsistent
Major and minor errors in
grammar and spelling;
information attached without
attention to pleasing effect;
care of project not evident
(edges rolled, etc.)
Creativity
Images or layout show original
ideas; reader is surprised,
interested and pleased
Images or layouts use a common
pattern which get intended
audience interested
Images or layout reproduce
common patterns, and give
viewer/reader what he/she is
accustomed to
Images or layout copied from
others or standard or so
sketchy that intent can’t be
determined
143
General Chemistry 1
Lesson 18: Emission Spectrum of Hydrogen,
and Dual Nature of Matter
Content Standard
The learners demonstrate an understanding of the quantum mechanical
description of the atom and its electronic structure.
Lesson Outline
Introduction
Communicating Learning Objectives
Motivation
Recall
Learning Competencies
At the end of the lesson, the learners:
1. Describe the quantum mechanical model of the atom (STEM_GC11CBIIa-b-52)
a. Identify the inadequacies of the classical physics in explaining the
emission spectrum of the hydrogen atom
b. Discuss the use of quantum theory in explaining the emission spectrum
of the hydrogen atom
c. Describe the Bohr model of the atom and the inadequacies of the Bohr
model
d. Explain the wave-particle duality of matter.
Instruction
I.
Enrichment
Revisit Essential Question
Evaluation
Quiz
Specific Learning Competencies
At the end of the lesson, the learners will be able to:
1. Explain the emission spectrum of hydrogen using the Bohr model of the
hydrogen atom;
2. Calculate the energy, wavelength, and frequencies involved in the electron
transitions in the hydrogen atom;
3. Relate the emission spectra to common occurrences like fireworks and
neon lights;
4. Describe the Bohr model of the atom and the inadequacies of the Bohr
model;
5. Explain the wave-particle duality of matter and
6. Perform calculations to determine wavelengths associated with moving
bodies.
Material
Calculator
144
120 MINS
The Emission Spectrum and the Bohr
Theory
II. Limitations of the Bohr Model
III. The Dual Nature of the Electron; De
Broglie Equation
IV. Calculating the De Broglie Wavelength
V. Experimental Evidence of De Broglie
Wavelength
12
3
85
5
15
Resources
(a) Chang, Raymond and Goldsby, Kenneth A. (2016). Chemistry (12th
ed). New York: McGraw-Hill.
(b) Petrucci, Herring, Madura, and Bissonnette (2011). General Chemistry
and Modern Applications, 10th Ed. Pearson Canada, Inc.
(c) Zumdahl, S.S. and Zumdahl, S.A (2013).Chemistry, 8th ed. Cengage
Learning
(d) Roque, et al. laboratory Manual in General Chemistry (2008). Philippine
Normal University.
INTRODUCTION (12 minutes)
1. Introduce the following learning objectives using any of the suggested protocol (Verbatim, Own
Words, or Read-aloud):
At the end of Part II, I will be able to:
a. Explain the emission spectrum of hydrogen using the Bohr model of the hydrogen atom
b. Calculate the energy, wavelength, and frequencies involved in the electron transitions in the
hydrogen atom.
c. Relate the emission spectra to common occurrences like fireworks and neon lights.
d. Describe the Bohr model of the atom and the inadequacies of the Bohr model
e. Explain the wave-particle duality of matter
f.
Perform calculations to determine wavelengths associated with moving bodies
2. Present the keywords for the concepts to be learned:
a. Emission spectrum
b. Rydberg’s constant
c. Ground state
d. Ground energy level
e. Excited state
f.
Excited energy level
g. Travelling wave
h. Standing wave
i.
De Broglie Equation
3. Post on the board the following essential questions that will be answered after the discussion
Why do elements emit different colors when heated?
What is the wave-particle duality of matter?
4. Review Rutherford’s nuclear model of the atom
145
MOTIVATION (3 minutes)
1. What causes the colors in fireworks displays? Ask the students to recall fireworks displays and ask
them what they think give the colors in the fireworks?
2. What causes the colors in neon lights?
INSTRUCTION (85 minutes)
I. THE EMISSION SPECTRUM AND THE BOHR THEORY OF THE HYDROGEN ATOM
When elements are energized by heat or other means, they give off a characteristic or distinctive
spectrum, called an emission spectrum, which can be used to differentiate one element from another.
While scientists recognized the usefulness of emission spectra in identifying elements, the origins of
these spectra were unknown.
From Rutherford’s theory, the atom was described to be mostly empty space having a very tiny but
dense nucleus that contained the protons. The electrons whirled around the nucleus in circular orbits at
high velocities. Classical mechanics and electromagnetic theory explained that any charged particle
moving on a curved path would emit electromagnetic radiation. This implies that electrons would lose
energy and spiral into the nucleus. Why this is not observed had to be explained.
Teacher Tip
Here is another occurrence that classical
mechanics is unable to explain.
In 1913, Niels Bohr proposed his model of the hydrogen atom to explain how electrons could stay in
stable orbits around the nucleus. This model is no longer considered to be correct in all its details.
However, it could explain the phenomenon of emission spectra. For his model of the hydrogen atom,
Bohr made the following postulates:
Note
Remember that the Bohr model is no
longer considered correct. However,
some of its features are still useful. One of
this is the explanation of the emission
spectrum. The limitations of the Bohr
model will be pointed out in a later
section.
a. Electrons go around the nucleus in circular orbits. However, not all circular orbits are allowed.
The electron is allowed to occupy only specific orbits with specific energies. Therefore, the
energies of the electron are quantized.
b. If the electron stays in the allowed orbit, its energy is stable. It will not emit radiation and it will
not spiral into the nucleus.
c. If an electron jumps from one orbit to another, it will absorb or emit energy in quanta equal to
#E"="hv
146
According to Bohr, the energy of the electron in the H atom is given by:
n%=%3
n%=%2
n%=%1
The negative sign is an arbitrary convention. A free electron is arbitrarily
considered to have an energy of zero. A negative energy means that the
energy of the electron is lower than the energy of a free electron.
RH is the Rydberg constant for hydrogen equal to 2.18 x 10-18J. The number
n is an integer equal to n = 1, 2, 3,…
Exercises
Teacher Tip
It is important for the learner to understand the negative values
for the energy. As the value gets more negative, the energy
gets lower. As the value of the energy gets less negative, the
energy gets higher. Comparing the energies for the first 3
energy levels, we see that E3 has the highest energy. It is less
negative. Therefore, as n increases, energy increases.
E1 = 2.18 x 10-18 J
E2 = 0.545 x 10-18 J
E3 = 0.242 x 10-18 J
1. What is the energy of the electron when it is in the first orbit, n=1?
E1 refers to the energy when the electron is in n=1.
2. What is the energy of the electron in orbit n = 2?
Do not make the learner memorise the value of RH. The value
should be given to the learner.
147
3. What is the energy of the electron in orbit n = 3?
4. Plot the energies of the electron in n=1, n=2, n=3.
5. In which orbit will the electron have the highest energy, n=1, n=2, or n=3? Answer: n=3
6. As the value of n increases, what happens to the energy value of the electron? Answer: As n
increases, energy increases.
148
E1 is the lowest energy and, therefore, the most stable state. It is called the ground state or the
ground level. E2, E3, E4, etc. have higher energies and are less stable than E1. They are called
excited states or excited levels. Note also that as the electron gets closer to the nucleus, it
becomes more stable.
When energy is absorbed by the atom, the electron gets excited and jumps from a lower orbit to
a higher orbit. When electrons go from a higher energy level to a lower energy level, it emits
radiation. According to Bohr, if an electron jumps from one orbit to another, it will absorb or emit
energy in quanta equal to:
∆Ε = hν = h
Teacher Tip
Note that this is the 3rd time that quantization of
energy is used to explain an experimental result.
The first was the blackbody radiation; the second
was the photoelectric effect; and now the third is
the emission spectrum of the hydrogen atom.
c
ν
The Bohr model can explain the experimental emission spectrum of hydrogen which includes a
wide range of wavelengths from the infrared to the UV region. These are summarized in the table
below:
SERIES
n final
n initial
Spectrum Region
Lyman
1
2, 3, 4
Ultraviolet
Balmer
2
3, 4, 5
Visible and ultraviolet
Paschen
3
4, 5, 6
Infrared
Brackett
4
5, 6, 7
Infrared
149
Teacher Tip
Do not ask the learners to memorize the H atom
emission spectrum series. This is only for
illustration purposes only. We want them to
understand the concept and not memorize.
Exercises
1. The electron in the hydrogen atom undergoes a transition from n=3 to n=2.
a. Is energy absorbed or emitted? Answer: E is emitted because the electron goes from a
higher energy level to a lower energy level.
b. What is the energy involved in the transition?
c. What is the wavelength (in nm) corresponding to this transition?
d. What region of the electromagnetic spectrum will this be?
Answer: This will be in the visible region.
2. Which transition of the electron in the hydrogen atom will involve the highest frequency?
a. n = 5 to n = 3
b. n = 4 to n = 3
c. n = 5 to n = 2
150
Similarly, when substances like metal ions are subjected to heat, they absorb energy. The
electrons jump from their ground state to an excited state. Once the electrons return from the
excited state to the ground state, light is given off. The light emitted corresponds to the
energy released.
Example: Potassium emitted a pinkish purple color (approx. 400 nm) while lithium emitted a
dark red hue (approx. 700 nm) when heated.
a. What caused the color change during heating?
b. Which element required the greater energy absorption for the electrons to be in the
excited state and produce the observed emissions? Why?
c. What is the relationship between wavelength and energy?
II. THE LIMITATIONS OF THE BOHR MODEL OF THE ATOM
Although the Bohr model could explain the emission spectrum of hydrogen and was an important
step in the development of atomic theory, it has several limitations:
a. It cannot explain the spectrum of atoms with more than one electron.
b. It cannot explain the relative intensities of spectral lines (why are some lines more intense than
others)
c. It cannot explain why some lines are slit into several components in the presence of a
magnetic field (called the Zeeman effect)
Teacher Tip
In contrast to standing waves, travelling waves: are
waves that travel in one-dimensional direction.
Concrete examples of travelling waves can be
seen from
• Skipping rope held on one end, moved up and
down, forming waves, from one end to the
other
• Ocean waves: the wind produces waves on the
surface of water producing crests and troughs
that travel great distances.
d. According to the Bohr model, when electrons go around the nucleus in certain orbits, its
energy remains constant. But moving electrons would lose energy by emitting
electromagnetic waves and the electron is expected to spiral into the nucleus.
e. It violates the Heisenberg’s Uncertainty Principle. The Bohr model considers electrons to have
a known radius and orbit which is impossible according to Heisenberg. This will be explained
later in the next lesson.
For an allowed orbit, the circumference of the
orbit must be equal to an integral number of
wavelengths (a); otherwise the wave will cancel
itself (b).
151
III. THE DUAL NATURE OF THE ELECTRON; DE BROGLIE’S EQUATION
In 1924, Louis de Broglie made a bold proposition based on Planck’s and Einstein’s concepts. De
Broglie reasoned that if light could have particle-like properties, then particles like electrons could
also have wavelike properties. Why are only certain orbits allowed in the Bohr model? Following
De Broglie’s idea, if the electron going around the nucleus in a circular orbit behaves as a wave,
then it should behave as a standing wave as shown in Figure 1. In a standing wave, there are
fixed points, or nodes, where the amplitude is zero. The length of the wave must fit the
circumference of orbit (see Figure 2). Otherwise the wave would cancel itself.
The derivation of the De Broglie equation is only
for clarification for the teacher and need not be
included in the lesson. It is important that the
student understand the concept more than
knowing the derivation.
E%=%mc2%
E%=%hv%
mc2%=%hv%
How did de Broglie arrive at his hypothesis? He
combined the energy relationship of Einstein’s
relativistic equation and Planck’s energy of a
photon.
p%=%mc%
The momentum, p, of a photon is the product of
the relativistic mass of the photon, m, and the
speed of light, c, or
p%=%
(b)
(a)
Standing Waves
(c)
For an allowed orbit, the circumference of the orbit must equal to integral
multiple of wavelengths (b). Otherwise, the wave will cancel itself (c).
Mathematically, this means that the circumference of the allowed orbit (2r) must be equal to an
integral multiple of the wavelength.
where n = 1, 2, 3…. Because n is an integer, the radius, r, can only have certain values
corresponding to n. Therefore, only certain orbits with allowed r values are permissible.
152
hv%
c
Substituting the relationship c = λv, the
momentum expression becomes
p"="
h%
λ
For a material particle, such as the electron, de
Broglie substituted for the momentum its
equivalent, the product of the mass of the particle,
m, and its velocity, u. Therefore,
λ%de%Broglie%=
h%
mu
How are the particle and wave properties related according to De Broglie?
This is given by the De Broglie equation:
Where h is Planck’s constant, m is the mass of the particle, and u is the velocity. Therefore, a
particle in motion can be treated as a wave and a wave can exhibit properties of a particle. An
electron, for instance, has both particle and wavelike properties. This is referred to as the dual
nature of matter.
IV. CALCULATING THE DE BROGLIE WAVELENGTH
1. What is the relationship between the De Broglie wavelength and the mass of the moving
particle? What happens to the wavelength as the mass increases?
2. Without doing any calculations, compare the wavelength associated with a moving airplane
and an electron moving at the same speed. Which will have the smaller De Broglie
wavelength?
3. How will the wavelength vary if the velocity of the particles increases?
4. Calculate the wavelength of the following “particles”:
a. A 6.00 x 10-2 kg tennis ball travelling at 68 m/s.
λ
λ
b. An electron moving at the same speed (mass of electron is 9.1094 x 10-31 kg)
λ
λ
153
Answer Key
1. λ and mass are inversely proportional. As the
mass of the particle increases, λ becomes
smaller.
2. The moving airplane will have the smaller
wavelength.
The wavelength of the tennis ball is exceedingly small considering that the size of the
atom is in on the order of 1 x 10-10 m. This makes it difficult for a tennis ball to be detected
by any existing measuring device. Meanwhile, the wavelength of the electron is in the
infrared region. This shows that only small particles like the electrons and other
submicroscopic particles have measurable wavelengths.
5. What must be the velocity, in m/s, of a beam of electrons if they are to display a de Broglie
wavelength of 1µm?
6. 2. What is the de Broglie wavelength, in nm, of a 2.4 g bird flying at 1.20 x 102 mph? (1 mile
= 1.61 km)
7. What is the wavelength, in nm, associated with 1000 kg automobile travelling at a speed of 25
m/s. Comment on the experimental measurement of the wavelength associated with the
moving automobile.
V. EXPERIMENTAL EVIDENCE OF DE BROGLIE WAVELENGTH
Waves associated with material particles were called by de Broglie as “matter waves”. If matter
waves exist for small particles, then beams of particles, such as electrons, should exhibit the
properties of waves, like diffraction.
Diffraction refers to various phenomena which occur when a wave encounters an obstacle or a
slit. In classical physics, the diffraction phenomenon is described as the interference of waves. If
the distance between objects that the waves scatter from is about the same as the wavelength of
the radiation, diffraction occurs and an interference pattern occurs.
Although De Broglie was credited for his hypothesis, he had no actual experimental evidence for
his conjecture. In 1927, Clinton J. Davisson and Lester H. Germer, from the United States, shot
electron particles onto a crystal of nickel. What they saw was the diffraction of the electron similar
to waves diffraction against crystals (x-rays). In the same year, an English physicist, George P.
Thomson, from Scotland, fired electrons towards thin metal foil providing him with the same
results as Davisson and Germer. As a historical note, the father and son demonstrated the waveparticle duality of electrons. George P. Thomson is the son of J.J. Thomson, who won the Nobel
Prize in 1906 for discovering the electron. The father, J.J. Thomson, showed that the electron is a
particle and George P. Thomson, the son, showed that the electron is a wave.
154
Teacher Tip
A video on diffraction and applications can be
seen at https://www.youtube.com/watch?
v=F6dZjuw1KUo (4 minutes)
A 4-minute video on the Davisson and Germer
experiment can be seen at https://
www.youtube.com/watch?v=Ho7K27B_Uu8.
ENRICHMENT (5 minutes)
Return to the question posted on the board during the introduction.
Ask the learners to answer the question based on the preceding lesson.
Why do elements emit different colors when heated?
Pyrotechnic materials such as flares and fireworks also follow the atomic spectra concepts.
Inside a mortar are different chemicals.. These chemicals are ignited through a time fuse,
causing the electrons in the chemicals to be excited during the reaction in the atmosphere. As
the electrons go down a lower energy level, different colors are emitted from these different
chemicals. The red glow is light with the least energy and the violet glow has the most
energy.
What is the wave-particle duality of matter?
Ask the learners to answer this in their own words according to their understanding.
EVALUATION
Put a circle around the letter corresponding to the best answer.
1. Waves are characterized by frequency and wavelength. Frequency
A. is the distance between two consecutive peaks or troughs in a wave.
B. is the number of cycles or complete oscillations that pass a given point per second.
C. the vertical distance from the midline of a wave to the peak or trough.
D. has units of J-s.
E. has units of cm/s.
2. What is the relationship between energy and wavelength of a photon?
A. direct relation
D. inverse relation
B. logarithmic relation
E. quadratic relation
C. cubic relation
155
Answer Key
1. B
2. D
3. E
4. C
5. D
6. C
7. B
8. C
9. D
10. D
3. Which of the following types of electromagnetic radiation will have the least energy?
A. gamma rays
D. x-rays
B. visible light
E. radio waves
C. microwaves
4. What is the energy in joules of one photon of microwave radiation with a wavelength 0.122
m?
A. 2.70 x 10-43 J
D. 4.07 x 10-10 J
B. 5.43 x 10-33 J
E. 2.46 x 109 J
C. 1.63 x 10-24 J
5. In the Bohr model of the H atom
A. The atom is a mass of positive charge with electrons embedded in it.
B. The electron energy increases as it gets closer to the nucleus.
C. The electron goes around the nucleus in certain allowed circular orbits.
D. Energy is absorbed when an electron goes from an orbit of high energy to an orbit of low
energy
E. C and D
6. Complete this sentence: Atoms emit visible and ultraviolet light __________.
A. As electrons jump from lower energy levels to higher levels.
B. As the atoms condense from a gas to a liquid.
C. As electrons jump from higher energy levels to lower levels.
D. As they are heated and the solid melts to form a liquid.
E. As the electrons move about the atom within an orbit.
156
7. The line spectrum of hydrogen gives proof of the
A.
B.
C.
D.
E.
Shape of the orbits of the electron
Quantized nature of the H energy levels
Uncertainty of the momentum of the electron
Continuous emission of energy
B and D
8. Calculate the energy, in joules, required to excite a hydrogen atom by causing an electronic
transition from the n = 1 to the n = 4 principal energy level. Recall that the energy levels of the
H atom are given by En = -2.18 10-18 J(1/n2)
A.
B.
C.
D.
E.
2.07 10-29 J
2.19 105 J
2.04 10-18 J
3.27 10-17 J
2.25 10-18 J
9. Suppose that a tennis ball, a neutron, an electron, and a pingpong ball are all moving at the
same speed. The wavelengths associated with them will be of the order:
A.
B.
C.
D.
E.
10.
tennis ball > pingpong ball> electron> neutron
pingpong ball> tennis ball > electron > neutron
neutron > electron > pingpong ball > tennis ball
electron > neutron > pingpong ball > tennis ball
tennis ball > pingpong ball > neitron > electron
Calculate the wavelength of a neutron that has a velocity of 200. cm/s. (The mass of a
neutron = 1.675 10-27 kg.)
A.
B.
C.
D.
E.
1.98 10-9 m
216 nm
1.8 1050 m
198 nm
5.05 mm
157
General Chemistry 1
120 MINS
Lesson 19: Flame Test (Laboratory)
Content Standard
Lesson Outline
The learners demonstrate an understanding of the quantum mechanical
description of the atom and its electronic structure
Introduction
Explore Colorful Elements
Laboratory Activity; Flame Test
35
Performance Standards
Instruction
and Practice
The learners can illustrate the dual nature of an electron.
Enrichment
Post-Lab Activity
20
5
Materials
Learning Competencies
At the end of the lesson, the learners:
1. Describe the quantum mechanical model of the atom (STEM_GC11CB-IIab-52)
a. Discuss quantum theory
b. Discuss the use of quantum theory in explaining the emission spectrum
of the hydrogen atom
Specific Learning Competencies
At the end of the lesson, the learners will be able to:
1. Demonstrate the flame tests for various metal ions.
2. Calculate the energy, wavelength, and frequencies involved in the electron
transitions in the hydrogen atom.
3. Relate the emission spectra to common occurrences like fireworks and
neon lights.
158
Cream of tartar (potassium hydrogen tartrate), table salt
(sodium chloride), moisture absorber (calcium chloride),
barium chloride, lithium chloride, copper sulphate or copper
chloride, boric acid (sodium tetraborate), distilled water, 50mL beaker or a clean glass container, popsicle sticks, large
receptacle for used popsicle sticks, alcohol lamp, lighter/
splinter
Resources
(1) Chang, Raymond and Goldsby, Kenneth A. (2016). Chemistry (12th
ed). New York: McGraw-Hill.
(2) Petrucci, Herring, Madura, and Bissonnette (2011). General Chemistry
and Modern Applications, 10th Ed. Pearson Canada, Inc.
(3) Zumdahl, S.S. and Zumdahl, S.A (2013).Chemistry, 8th ed. Cengage
Learning
(4) Roque, et al. laboratory Manual in General Chemistry (2008). Philippine
Normal University.
INTRODUCTION (5 minutes)
Teacher Tip
When substances such as metal salts are heated to high temperatures, the electrons of the metal
ions are excited to higher energy levels. When these electrons return to their ground states,
energy is emitted in the form of light. Since each element emits a unique set of wavelengths, the
emission spectrum can be used as a tool to identify the elements.
2. Distribute the lab sheets at the start of
the lesson.
Explore Colorful Elements
One method of demonstrating the emission spectrum of substances is through a qualitative
analysis called the flame test. In this technique, a small amount of substance is heated. The heat
of the flame excites the electrons of the metals ions, causing them to emit visible light the color of
which is unique to the metal ion.
1. Prepare the classroom or laboratory,
the materials, the lab sheets to be
used.
3. After the introduction and motivation,
explain the procedures of the activity.
4. Explain the safety precautions.
Objectives
a. To be able to conduct a flame test for metal ions
b. To observe the flame colors emitted by selected metal ions.
c. To explain the origin of the flame colors.
Teacher Tip
INSTRUCTION and PRACTICE (35 minutes)
1. Safety Precautions
a. Do this activity with teacher supervision. Follow all laboratory instructions as directed by
the instructor.
b. Wear laboratory gown, goggles and mask.
c. Consider all metal salts as harmful materials. Do not taste the chemicals. Avoid skin
contact with the chemicals.
d. Do not eat or drink while doing the activity.
e. Dispose of all materials according to the instructions of your teacher.
159
The activity can be performed individually
or in groups. Nevertheless, caution must
be observed in handling any material in
the lab. Instruct the learners how to
behave in the laboratory.
Dispose the materials properly.
PROCEDURE
Teacher Tip
The teacher should prepare the samples ahead of
time. Place the salt samples in different watch
glass or paper/plastic plates. Label the samples.
1. Dip the popsicle stick in water.
2. Dip the wet popsicle stick into the solid sample.
3. Heat over the flame. Observe the color change in the flame.
Guide to the flame colors
4. Repeat procedures 1-3 with the other samples.
Note: It is advisable to repeat the test to ensure that the right color of the flame is observed.
5. Dispose of used popsicle sticks in a receptacle.
6. Note your observation in the data table given.
DATA TABLE
Sample Material
Metal Ion
Sodium chloride
sodium
Flame Color
There are many causes for the indicated color not to come out such as contaminants in the
material, contaminants in the water or in the popsicle stick. Or the flame may not be hot enough.
The teacher should NOT mark as wrong any observation. Encourage the students to be honest
with stating the result rather than getting the supposedly “correct” answer. The purpose of the
experiment is to record observations and try to explain the observations as well as possible
sources of error.
ENRICHMENT (20 minutes)
Post-Lab Activity
1. Have the students answer the following questions in their activity sheets. Then discuss the class results
for the post-lab activity. Compare results of the different groups.
a. Why do you think the chemicals have to be heated in the flame first before the colored light is
emitted?
160
Metal Ion
Flame Color
Lithium
Red
Sodium
Yellow
Potassium
Lilac
Calcium
Orange / Yellow-red
Strontium
Red
Barium
Pale green
Copper
Blue green
Note
Everyone sees and describes colors differently so
students may describe their colors as purple
instead of lilac or crimson instead of red.
b. Arrange the group of metals which produced the most easily identifiable colors. Start with those
that emitted the most intense color and end with those metals with colors that are least intense.
c.
Colorful light emissions are observed in everyday life. Where else have you observed light
emissions? Are these light emissions an evidence of excited electrons?
d. Cite at least 2 reasons why the flame test is sometimes inaccurate.
2. Write your conclusion and recommendations for the lab activity.
EVALUATION
Use the following rubric to rate the learner’s performance in the lab activity.
CONTENT
Observations (10)
PERFORMANCE
PRESENTATION
Attendance and Lab Attire (10)
Lab Results are presented well (10)
Lab Technique and Observance of
Safety Procedures (20)
Answers to Questions on
the Report (20)
Conclusions (5)
Housekeeping (5)
Participation in Oral Discussion / Oral
Presentation (10)
Lab Requirements / Materials (5)
Recommendations (5)
161
TOTAL
General Chemistry 1
120 MINS
Lesson 20: Electronic Structure
of the Atom
Content Standard
Lesson Outline
The learners demonstrate an understanding of the quantum mechanical
description of the atom and its electronic structure
Introduction
Review of the Quiz
Motivation
3D Models of the Orbitals
Performance Standard
Instruction
and Practice
I. Heisenberg’s Uncertainty Principle
II. The Schrondinger Equation
III. The Quantum Mechanical Description of
the Hydrogen Atom
IV. The Quantum Numbers
V. The Atomic Orbitals
80
Evaluation
Laboratory Activity
25
The learners can illustrate the distribution of the electrons in an atom.
Learning Competencies
At the end of the lesson, the learners:
1. Describe the electronic structure of atoms in terms of main energy levels,
sublevels and orbitals and relate this to energy (STEM_GC11CB-IIa-b-53)
12
3
2. Use quantum numbers to describe an electron in an atom
(STEM_GC11CB-IIa-b-54)
Material
Calculator
3. (LAB) Perform exercises on quantum numbers (STEM_GC11CB-IIa-b-55)
Resources
(1) Chang, Raymond and Goldsby, Kenneth A. (2016). Chemistry (12th
ed). New York: McGraw-Hill.
(2) Petrucci, Herring, Madura, and Bissonnette (2011). General Chemistry
and Modern Applications, 10th Ed. Pearson Canada, Inc.
(3) Zumdahl, S.S. and Zumdahl, S.A (2013).Chemistry, 8th ed. Cengage
Learning
(4) http://csi.chemie.tu-darmstadt.de/ak/immel/script/redirect.cgi?
filename=http://csi.chemie.tu-darmstadt.de/ak/immel/tutorials/
orbitals/hydrogenic.html
(5) http://winter.group.shef.ac.uk/orbitron/AOs/6g/
Specific Learning Competencies
At the end of the lesson, the learners will be able to:
1. Explain Heisenberg’s Uncertainty Principle
2. Describe how atomic orbitals arise from the Schrodinger equation
3. Relate orbital shapes to electron density distribution
4. Qualitatively sketch the orbital shapes
5. Interpret the information obtained from a set of four quantum numbers
6. Assign the correct set of quantum numbers for an electron
162
INTRODUCTION (12 minutes)
1. Review the quiz given in the last meeting
2. Introduce the following learning objectives using any of the suggested protocol (Verbatim,
Own Words, or Read-aloud):
3. At the end of the lesson, I will be able to:
a. Explain Heisenberg’s Uncertainty Principle
b. Describe how atomic orbitals arise from the Schrodinger equation
c. Relate orbital shapes to electron density distribution
d. Qualitatively sketch the orbital shapes
e. Interpret the information obtained from a set of four quantum numbers
f.
Assign the correct set of quantum numbers for an electron
4. Present the keywords for the concepts to be learned:
a. Heisenberg’s Uncertainty Principle
b. Schrodinger Equation
c. Wave function
d. Electron probability density
e. Atomic orbital
f.
Principal quantum number
g. Angular momentum quantum number
h. Magnetic quantum number
i.
Spin quantum number
j.
Shell
k. Subshell
MOTIVATION (3 minutes)
If available, show 3-dimensional models of the orbitals (s, p, and d) to the students to gain their
attention and curiosity. If 3-D models are not available, post large illustrations on the board.
163
INSTRUCTION and PRACTICE (80 minutes)
I.
HEISENBERG’S UNCERTAINTY PRINCIPLE
With the discovery that particles like electrons are wavelike (shown by De Broglie, Davisson and
Germer, and Thomson), how can the ‘position’ of a wave be specified? How can the precise
location of a wave be defined when a wave extends in space?
Werner Heisenberg, a German physicist, formulated what is now known as Heisenberg’s
Uncertainty Principle which states that “the position of a particle and its momentum cannot be
simultaneously measured with arbitrarily high precision.” In other words, it is not possible to
measure the exact position and the exact momentum of a particle at the same time.
Mathematically, this is stated as
where "x is the uncertainty in position, "p is the uncertainty in momentum, and h is Planck’s
constant.
To explain this equation, let us answer the following questions
1. What is the meaning of the ≥ (greater than or equal to) sign with respect to the uncertainties?
When conducting experiments, especially if conditions are crude, the uncertainties in position
and momentum can be large. The product of "x"p can be greater than h/4π. However,
even when you want to make very precise measurements, h/4π. The product "x"p can
never be smaller h/4π. Hence, there will always be uncertainties even under good conditions.
2. How are "x and "p related?
They are inversely related. Remember that the right side of the equation, h/4π, is a constant.
If we want to make very precise measurement of the position (meaning "x is very small), then
"p becomes large. Conversely, if we want a very small uncertainty in momentum, "p
becomes small, but the uncertainty in position ("x) becomes large.
164
3. According to the Bohr model, the electron goes around the nucleus in well-defined orbits, the
radius of which can be determined. How can you relate the Bohr model to Heisenberg’s
Uncertainty Principle?
The Bohr model violates Heisenberg’s Uncertainty Principle. Electrons do not go around
the nucleus in well-defined orbits. Otherwise, we will be able to determine the exact position
and momentum of the electron in the atom at the same time. A better model is needed to
fully describe the atom.
4. An electron is travelling at a speed of 2.05 x 106 m/s. Assuming that the precision
(uncertainty) of this value is 1.5%, with what precision can the position of the electron be
measured?
Uncertainty%in%velocity%=%u%=%(0.015)(%2.05%x%106%m/s)%=%3.1%x%104%m/s%%
To compute for the uncertainty in momentum, "p, multiply "u by the mass of the electron
"p%=%m(ru)%=%(9.109%x%10*31%kg)%(3.1%x%104%m/s)%=%2.8%x%10*26%kg*m/s%%
The uncertainty in position, "x, will be
"x%=
This value shows that the electron’s position is about 10 atomic diameters. Given the
uncertainty of the speed, there is no way to pin down the electron’s position with any greater
accuracy.
165
Teacher Tip
5. Why is the uncertainty principle not significant when applied to large objects such as a
transportation vehicle?
II. THE SCHRODINGER EQUATION
While the Bohr model of the atom could explain the emission spectrum of hydrogen, it could not
account for many observations and could not provide a complete description of the electronic
behavior in atoms.
In 1926, Erwin Schrodinger, an Austrian physicist, formulated a mathematical equation that
describes the behavior and energies of submicroscopic particles. The Schrodinger equation
incorporates particle behavior and wave behavior, treating the electron as a standing wave. The
solution to the Schrodinger equation is a wave function called ψ (psi). The wave functions are
also called atomic orbitals (as distinguished from the Bohr orbits). Aside from the wave
functions, energies are also obtained from solving the equation.
The wave function itself has no physical meaning. However, the probability of finding the electron
in a particular volume element in space is proportional to ψ2. In wave theory, the intensity of light
is proportional to the square of the amplitude of the wave or ψ2. Similarly, the most likely place
to find the particle is where the value of ψ2 is greatest.
The Schrodinger equation began a new field in physics and chemistry referred to as quantum
mechanics or wave mechanics. The Schrodinger equation can be solved exactly for the hydrogen
atom but not for atoms with more than one electron. For many-electron atoms, approximation
methods are used to solve the Schrodinger equation.
III. THE QUANTUM MECHANICAL DESCRIPTION OF THE HYDROGEN ATOM
It is not possible to pinpoint the exact location of the electron in an atom but ψ2 gives the region
where it can most probably be found. The electron density gives the probability that the
electron will be found in a particular region of an atom. Figure (a) is a representation of the
electron density distribution around the nucleus in the hydrogen atom. The darker the shade, the
166
The solution of the Schrodinger equation
involves advance calculus and differential
equations. The lesson will only deal with
the interpretation of the solution.
The Schrodinger equation for the
hydrogen atom looks like this:
higher the probability of finding the electron in that region. In this case, the probability
distribution is spherical. The probability can also be plotted versus the distance from the nucleus
as shown in Figure (b). It can be seen that there is a probability of finding the electron even very
far from the nucleus, although this probability is small. The closer to the nucleus, the higher the
probability.
Sources
(a) and (b) Probability of Finding the Electron in
the Ground State of the Hydrogen Atom at
Different Points in Space, “Atomic Orbitals and
Their Energies”, section 6.5 from the book
Principles of General Chemistry (v. 1.0), Retrieved
from http://2012books.lardbucket.org/books/
principles-of-general-chemistry-v1.0/s10-05atomic-orbitals-and-their-ener.html (2 Nov. 2016),
Creative Commons by-nc-sa 3.0 license.
(a)
(b)
As mentioned earlier, ψ is the solution to the Schrodinger equation. It is also referred to as an
atomic orbital. When we say that the electron is in an atomic orbital, we mean that it is described
by a wave function, ψ, and that the probability of locating the electron is given by the square of
the wave function associated with that orbital. Therefore, the atomic orbital has a characteristic
energy as well as a characteristic electron density distribution. This electron density distribution in
three-dimensions gives the shape of the atomic orbital.
167
IV. THE QUANTUM NUMBERS
In the mathematical solution of the Schrodinger equation, three quantum numbers are obtained.
These are the principal quantum number (n), the angular quantum number, (ℓ) ,and the
magnetic quantum number (ml). They describe the atomic orbitals. A fourth quantum number,
the spin quantum number (ms) completes the description of the electrons in the atoms.
The Principal Quantum Number (n)
a. Determines the energy of an orbital
b. Determines the orbital size
c. Is related to the average distance of the electron from the nucleus in a particular orbital; the
larger the n value, the farther the average distance of the electron from the nucleus
d. Can have the values: n = 1, 2, 3, …
e. Orbitals with the same n are said to be in the same shell.
The Angular Momentum Quantum Number (ℓ)
a. Describes the “shape” of the orbitals
b. Can have the following values: ℓ = 0, 1, 2, up to n-1.
Examples
n value
ℓ value
1
0
2
0, 1
3
0, 1, 2
c. Orbitals with the same n and values belong to the same subshell.
d. It is usually designated by letters s, p, d, f, … which have a historical origin from spectral lines.
The designations are as follows
Teacher Tip
The s, p, d, f designations of the orbitals refer to
sharp, principal, diffuse, and fundamental lines in
emission spectra.
!!!s!!!!!!!!!!!p!!!!!!!!!!d!!!!!!!!!!f!!!!!!!!!!!!g!!!!!!!!!!!h
168
The Magnetic Quantum Number (ml)
a. Describes the orientation of the orbital in space
b. Can have the values:
- ℓ, (-ℓ + 1), … 0, … (+ ℓ -1), + ℓ
The Electron Spin Quantum Number (ms)
a. The first three quantum numbers describe the energy, shape and orientation of orbitals. The
4th quantum number refers to two different spin orientations of electrons in a specified orbital.
b. When lines of the hydrogen spectrum are examined at very high resolution, they are found to
be closely spaced doublets and called as the Zeeman effect. This splitting is called fine
structure, and was one of the first experimental evidences for electron spin. The direct
observation of the electron's intrinsic angular momentum was achieved in the Stern–Gerlach
experiment.
c. Uhlenbeck, Goudsmit, and Kronig (1925) introduced the idea of the self-rotation of the
electron. The spin orientations are called "spin-up" or "spin-down" and is assigned the
number ms = ½ ms = -½, respectively.
d. The spin property of an electron would give rise to magnetic moment, which was a requisite
for the fourth quantum number. The electrons are paired such that one spins upward and one
downward, neutralizing the effect of their spin on the action of the atom as a whole. But in the
valence shell of atoms where there is a single electron whose spin remains unbalanced, the
unbalanced spin creates spin magnetic moment, making the electron act like a very small
magnet. As the atoms pass through the in-homogeneous magnetic field, the force moment in
the magnetic field influences the electron's dipole until its position matches the direction of
the stronger field.
The four quantum numbers compose the numbers that describe the electron in an atom. The
quantum numbers shall be in the order: energy level (n), sub-level or orbital type (ℓ), the
orientation of the orbital specified in ℓ (mℓ), and the orientation of the spin of the electron (ms). It
is written in the order (n, ℓ, mℓ, ms ).
169
For example
1. An electron is found in the first energy level. What is the allowed set of quantum numbers for
this electron?
a. The energy level, n = 1.
b. The orbital type is only s, its designation is 0, thus, ℓ = 0
c. From ℓ, the orbital type is s. There is only one orientation of an s orbital, designated as 0,
thus, mℓ = 0.m
d. An electron in the 1s orbital can have an up-spin or a down-spin. Therefore, ms could be
+1/2 or -1/2.
So the allowed set of quantum numbers for 1s electron are:
(1,0,0,1/2) and (1,0,0,-1/2)
How does (1,0,0,1/2) differ from (1,0,0,-1/2)? The first set corresponds to the electron with spin
up and the second set refers to the electron with spin down.
V. THE QUANTUM NUMBERS AND THE CORRESPONDING ATOMIC ORBITALS
The quantum numbers and corresponding atomic orbitals are given in the following table.
n
ℓ
mℓ
Number of Orbitals
Atomic Orbital
Designation
1
0
0
1
1s
2
0
0
1
2s
2
1
-1, 0, 1
3
2px, 2py 2pz
3
0
0
1
3s
3
1
-1, 0, 1
3
3px, 3py 3pz
3
2
-2, -1, 0, 1, 2
5
3dxy, 3dyz, 3dxz, 3dx2-y2, 3dz2
170
Exercises
1. What is the total number of orbitals associated with the principal quantum number n=1?
Answer: 1
What is the total number of orbitals associated with the principal quantum number n=2?
Answer: 4
What is the total number of orbitals associated with the principal quantum number n=3?
Answer: 9
We can therefore say that the total number of orbitals associated with a given principal
quantum number n is n2.
2. List the values of n, ℓ , mℓ for an orbital in the 4d subshell.
Answer: n=4; ℓ =2; ml can have the values of -2, -1, 0, 1, 2
The Representations of the Shapes of Atomic Orbitals
What are the shapes of the atomic orbitals? Strictly speaking, an orbital does not have a definite
shape because the wave function extends to infinity. However, while the electron can be found
anywhere, there are regions where the probability of finding it is much higher. Figure (a) shows
the electron density distribution of a 1s electron around the nucleus. Note that it does not have a
well-defined boundary; the more dots, the darker the shade, the higher the probability of finding
the electron in that region. Also note that the probability distribution is spherical. We can draw a
boundary surface that will enclose 90% of the total electron density in the orbital as shown in
Figure (c). This will result in a boundary surface diagram of the 1s orbital as shown in Figure (d).
Sources
(c) Circular boundary enclosing 90 percent of
electron density in a hydrogen atom 1s orbital.
From Electron Waves in the Hydrogen Atom,
Chemistry LibreTexts, National Science
Foundation. Retrieved from http://
chem.libretexts.org/Textbook_Maps/
General_Chemistry_Textbook_Maps/Map
%3A_ChemPRIME_(Moore_et_al.)/
05The_Electronic_Structure_of_Atoms/
5.06%3A_Electron_Waves_in_the_Hydrogen_Atom
(3 November 2016), Creative Commons
Attribution-Noncommercial-Share Alike 3.0 United
States License.
(d) The 1s, 2s, and 3s orbitals. From High School
Chemistry/Shapes of Atomic Orbitals. Retrieved
from https://en.wikibooks.org/wiki/
High_School_Chemistry/
Shapes_of_Atomic_Orbitals (3 November 2016),
Creative Commons Attribution-ShareAlike 3.0
License.
(c)
(d)
171
Figure (d) shows that all the s orbitals are spherical in shape but differ in size, which increases
as the value of n increases.
The p orbitals starts when n =2 for which ℓ has a value of 1 and mℓ has values -1, 0, +1.
Therefore, there are three 2p orbitals: 2px, 2py, 2pz indicating the axes along which they are
oriented. For the p orbitals, the electron probability density is not spherically symmetric but
has a double teardrop shape, or in some books, a dumbbell shape. The greatest probability of
finding the electron is within the two lobes of the dumbbell region; it has zero probability
along the nodal planes found in the axes. All three 2p orbitals are identical in shape and
energy but differ in orientation as shown in Figure (e). The p orbitals of higher principal
quantum numbers have similar shapes.
(e)
Figure (f) shows the d orbitals occur for the first time when n = 3. The angular function in these
cases possesses two angular (or planar) nodes. Four of the orbitals have the same basic shapes
except for the orientation with respect to the axes. The wave functions exhibit positive and
negative lobes along the axes and shows zero probability of finding the electron at the origin.
The fifth wave function, dx2 , has a similar shape with that of the p-orbital with a donut-shape
region along the x-axis.
Sources
(e) The boundary surface diagrams of the 2p orbitals.
From Atomic Orbitals and Their Energies. Retrieved
from http://2012books.lardbucket.org/books/
principles-of-general-chemistry-v1.0/s10-05-atomicorbitals-and-their-ener.html (3 November 2016), ),
Creative Commons by-nc-sa 3.0 license.
(f) The five 3d orbitals of the hydrogen atom. From
Atomic Orbitals and Their Energies. Retrieved from
http://2012books.lardbucket.org/books/principles-ofgeneral-chemistry-v1.0/s10-05-atomic-orbitals-andtheir-ener.html (3 November 2016), ), Creative
Commons by-nc-sa 3.0 license.
(f)
ASSESSMENT/LAB ACTIVITY (25 minutes)
QUANTUM NUMBERS Worksheet
Rearrange the letters of the correct term that is described by the corresponding statement.
1. Write your answer on the space provided.
___________
a
LAPNICRIP – It is the quantum number that represents the
energy level the electron is in.
___________
b
LATOBRI – It is a representation of the wave function of a
hydrogen-like atom.
___________
c
ALGANUR MUTMENMO – It is a quantum number that
represents the shape of orbitals.
___________
d
NOTRECLE – It is the particle that can be described by
four quantum numbers
___________
e
MEGATINC – It represents the quantum number that
describes the orientation of an orbital.
___________
f
NEREGY EVELL – It is being represented by n.
___________
g
RHEPES – It is the shape of the s orbital.
___________
h
ROGUND EATTS – It is the most stable state of the
electron in the hydrogen atom.
2. Give the n and ℓ values for the following orbitals
_______________________________
a. 1s
_______________________________
b. 3p
_______________________________
c. 5f
_______________________________
d. 4d
3. What is the mℓ values for the following types of orbitals?
_______________________________
a. s
_______________________________
b. p
_______________________________
c. d
_______________________________
d. f
173
Answer Key
1. Rearrange the letters
a. Principal
b. Orbital
c. Angular momentum
d. Electron
e. Magnetic
f. Energy level
g. Sphere
h. Ground state
2. Give the n and ℓ values
a. n=1, l = 0
b. n=3, l =1
c. n= 5, l= 3
d. n= 4, l=2
3. mℓ values
a. ml= 0
b. ml= -1, 0,1
c. ml= -2, -1, 0, 1, 2
d. ml= -3, -2, -1, 0, 1, 2, 3
4. Possible Orbitals
a. 32 electrons
b. 50 electrons
4. How many possible orbitals and how many electrons can inhabit the energy level n
_______________________________
a. 4
_______________________________
b. 5
5. State the number of possible electrons described by the following quantum numbers
a. n = 3, l = 0 _______________________________
b. n = 3, l = 1 _______________________________
c. n = 3, l = 2, ml = -1 ________________________
d. n = 5, l = 0, mℓ =-2, ms =-1/2 _________________
6. Which of the following is not a valid set of quantum numbers? Explain your answer.
a. n = 2, l = 2, ml = 0, and ms = -1/2
b. n = 2, l = 1, ml = -1, and ms = -1/2
c. n = 3, l = 0, ml = 0, and ms = 1
d. n = 3, l = 2, ml = 3, and ms = ½
7. What is the maximum electron pairs that can occupy an:
_______________________________
a. s orbital
_______________________________
b. the subshell of p orbitals
_______________________________
c. the subshell of d orbitals
_______________________________
d. the subshell of f orbitals
_______________________________
e. the subshell of g orbitals
8. Do as directed.
a. Sketch the shape of the orbital with the quantum numbers n=3, l=0 and mℓ = 0
b. The sketch of the shape of the subshell with the quantum numbers n=4, l=2 is
c. The highest orbital possible in n = 4
d. Sketch the orientation of the allowed values of l= 1 for the shell n=2.
e. Write the set of quantum numbers for the following
i. It is an up-spin 4d electron with an orbital orientation of 0.
ii. The electron is in the 3rd energy level, px-orbital, and down spin.
iii. When n=2, l is 1, mℓ = 1, ms = ½
f. What is the value of l for a 4f electron?
g. What is the orbital designation for an electron in the 3rd shell and p sublevel?
h. How many electrons have the following quantum numbers: n =4, l = 2, mℓ = -2?
174
5.
Number of possible electrons
a. 2
b. 6
c. 2
d. not possible
6. Valid Set of Quantum Numbers
a. l =2 is not allowed, maximum is 1
b. possible
c. ms should only be ½ or -1/2
d. mℓ should only be within the values of 2l+1;
mℓ should only be within the values of 2l+1
7. Maximum electron pairs
a. 1
b. 3
c. 5
d. 7
e. 9
8. “Do as directed”
a. Sphere
b. Any of the d orientations
c. f orbital
d. p orbitals
e. Answers:
i. (4, 2,0, ½)
ii. (3,1,-1,-1/2)
iii. (2,1,1,1/2)
f. 3
g. 3p
h. 2
General Chemistry 1
120 MINS
Lesson 21: Electron Configuration
Content Standard
The learners demonstrate an understanding of the electronic distribution in an
atom.
Performance Standards
The learners can illustrate the distribution of the electrons in an atom.
Learning Competencies
At the end of the lesson, the learners:
Lesson Outline
Introduction
Communicating Learning Objectives
7
Motivation
Addresses and Zip Codes
3
Instruction
and Practice
I.
II.
III.
IV.
V.
Evaluation
Exercises and Activity
1. Write the electronic configuration of atoms (STEM_GC11CB-IIa-b-56)
2. Determine the magnetic property of the atom based on its electronic
structure (STEM_GC11CB-IIa-b-57)
3. Draw an orbital diagram to represent the electronic configuration of atoms;
(STEM_GC11CB-IIa-b-58)
4. Perform exercises on writing electronic configuration (STEM_GC11CB-IIab-59)
Specific Learning Competencies
Energies of the Orbitals
Electron Configuration
Hand and Rule
Aufbau Principle
The Quantum Numbers and the
Arrangement of Elements in the Periodic
Table
80
30
Materials
Periodic Table
Resources
(a) Chang, Raymond and Goldsby, Kenneth A. (2016).
Chemistry (12th ed). New York: McGraw-Hill.
At the end of the lesson, the learners will be able to:
1. Explain the unique electron distribution of the atom;
(b) Petrucci, Herring, Madura, and Bissonnette (2011).
General Chemistry and Modern Applications, 10th Ed.
Pearson Canada, Inc.
2. Compare and contrast the orbital energies in a hydrogen atom
with that of the many-electron atom;
3. Write the electron configuration of an atom using the
conventional method as well as the core noble gas
configurations;
(c) Zumdahl, S.S. and Zumdahl, S.A (2013).Chemistry, 8th ed.
Cengage Learning
4. Illustrate the electron distribution using orbital diagrams;
6. Determine valence configuration and valence electrons.
5. Determine magnetic properties of an atom based on its
electronic configuration; and
7. Relate valence configuration of elements with position of
element in the periodic table.
175
INTRODUCTION/ REVIEW (7 minutes)
1. Review orbitals and their shapes.
2. Introduce the following learning objectives using any of the suggested protocol (Verbatim,
Own Words, or Read-aloud):
At the end of the lesson, I will be able to:
a. Explain the unique electron distribution of the atom;
b. Compare and contrast the orbital energies in a hydrogen atom with that of the manyelectron atom;
c. Write the electron configuration of an atom using the conventional method as well as
the core noble gas configurations;
d. Illustrate the electron distribution using orbital diagrams;
e. Determine magnetic properties of an atom based on its electronic configuration; and
f. Determine valence configuration and valence electrons.
g. Relate valence configuration of elements with position of element in the periodic table.
3. Present the keywords for the concepts to be learned:
a. Ground state
b. Excited state
c. Degenerate
d. Electron configuration
e. Orbital Diagram
f. Pauli Exclusion Principle
g. Paramagnetic
h. Diamagnetic
i. Hund’s Rule
j. Building-Up Principle (Aufbau Principle)
k. Noble gas
l. Transition metals
m. Valence configuration
n. Valence electrons
176
MOTIVATION (3 minutes)
1. Ask a few learners to give their home addresses.
2. What are zip codes? What is the zip code of the school?
3. What is the use of zip codes? Look for the zip code of a school outside your city or province
and compare with yours.
INSTRUCTION/ DELIVERY/ PRACTICE (80 minutes)
I. ENERGIES OF THE ORBITALS
After understanding the shapes and sizes of atomic orbitals, it is imperative to understand the
relative energies of the orbitals and how it affects the actual arrangement of electrons in atoms.
1. Orbital energy levels in a hydrogen atom
The energy of an electron in a hydrogen atom depends solely on its principal quantum number, n.
The energy of the electron in the hydrogen atom is given by:
where RH is equal to 2.18 x 10-18J. Therefore, the energies of the hydrogen atom increase
according to the following (see Figure 1):
1s"<"2s"="2p"<"3s"="3p"="3d"<"4s"="4p"="4d"="4f"<"…"
Orbitals with the same principal quantum number, n, have the same energy. It means that in a
hydrogen atom, the lowest energy is 1s. It is the most stable condition, or termed as the ground
state. An electron in the ground state is most strongly held by the nucleus.
Orbital energy levels in a hydrogen atom
177
The 2s, and the three 2p-orbitals have the same energy. We refer to orbitals with the same energy
as degenerate. When an external energy hits a hydrogen atom, the electron in the 1s orbital,
can jump to the 2s, 2p or higher orbitals and this electron is said to be in the excited state.
Similarly, the 3s, the three 3p-orbitals and the five 3d-orbitals are degenerate and have higher
energy than the orbitals in the 2nd energy level.
2. Orbital energy levels for many-electron atoms
For atoms containing more than one electron (many-electron atoms), the energy depends on
other factors. These include the potential energy of repulsion among the electrons, the attraction
between the nucleus and the other electrons, and the kinetic energies of the many electrons.
Thus the orbital energies of many-electron atoms depend not only on n but also on ℓ. Note that
the 3s, 3p, and 3d orbitals are no longer degenerate to each other. The 3d orbital energies are
even lower than those of the 4s orbitals.
Source
Generalized energy-level diagram for atomic
orbitals in an atom with two or more electrons (not
to scale). From Electronic Structure of Atoms
(Electron Configurations), Rice University.
Retrieved from https://opentextbc.ca/chemistry/
chapter/6-4-electronic-structure-of-atoms-electronconfigurations/ (3 November 2016), Creative
Commons Attribution 4.0 International License.
Orbital energy levels for many-electron atoms
III. ELECTRON CONFIGURATION
The four quantum numbers n, ℓ, ml, and ms are very useful in labelling an electron in any orbital in
an atom much like giving the address of an electron in an atom.
In the case of hydrogen, there is only one electron. In the ground state, the one electron of
hydrogen will occupy the 1s orbital, the one with the lowest energy. This electron is represented
178
Teacher Tip
There is an equal probability for the electron to
have a spin up (ms = ½) or spin down (ms = -½}.
by the set of quantum numbers: n = 1, ℓ =0, ml = 0, and ms = ½ or -½. By convention, the set of
quantum numbers is written as (1, 0, 0, ½) or (1, 0, 0, -½). The ms value does not affect the
energy, orientation, or size of the orbital but is important in describing the arrangement of
electrons in the atom.
It is possible to represent this arrangement of the electron in hydrogen in terms of the electron
configuration or in terms of the orbital diagram. The electron configuration shows how the
electrons of an atom are distributed among the atomic orbitals. The orbital diagram shows the
spin of the electron. For the electron in the ground state of hydrogen, the electron configuration
is given as
In an orbital diagram, a 1s orbital can be represented as a box with 1 arrow up (up-spin) or arrow
down (down-spin)
In filling up the orbitals, the lower energy levels are filled up first before the higher energy levels.
For many-electron atoms, the Pauli Exclusion Principle is used. This states that in an atom or
molecule, no two electrons can have the same four electronic quantum numbers. Consequently,
an orbital can contain a maximum of only two electrons, the two electrons must have opposing
spins. This means if one is assigned an up-spin (+1/2), the other must be down-spin (-1/2).
179
Consider the case of He with 2 electrons.
Teacher Tip
1. What are the sets of quantum numbers that
describe the first and second electrons in Cases
A, B, and C?
2. Why do Case A and Case B violate the Pauli
Exclusion Principle?
3. Why is the arrangement in Case C acceptable?
Exercises
1. What are the possible sets of quantum numbers that can describe a 2p electron in an atom?
Answer: For a 2p electron, n = 2; = 1; ml can be -1, 0, +1; and ms can be ½ or -½.
(2, 1, -1, ½)
(2, 1, -1, -½)
(2, 1, 0, ½)
Answer Key
1. All six representations are possible.
2. 1s2 2s1
The outermost electron in Li can be described
by the quantum numbers (2, 0, 0, ½)
3.
1s2 2s2
4.
1s2 2s2 2p1
(2, 1, 0, -½)
(2, 1, 1, ½)
(2, 1, 1, -½)
2. Give the electron configuration of Li. Give the set of quantum numbers that describe the
outermost electron in lithium as shown in the orbital diagram below.
Note that the unpaired electron in boron can go to
2px, or 2py, or 2pz since they have equal energies.
3. Give the electron configuration of Be. Draw the orbital diagram.
4. Give the electron configuration of B. Draw the orbital diagram.
180
III. HUND’S RULE
For carbon, the electronic configuration is 1s2 2s2 2p2. But the orbital diagram shows three ways
in which the last electron can be placed in the orbitals which do not violate the Pauli’s exclusion
principle as shown in the following:
However, each arrangement provides a different energy value. The one with the lowest energy
has the greatest stability. Hund’s rule is the guide in determining the most stable distribution.
Hund’s rule: The most stable arrangement of electrons in the subshells is the one with the
most number of parallel spins.
Based on Hund’s rule, the third option is the most favorable arrangement for the electron to
attain the greatest stability. In the first option, the presence of two electrons with opposing
spins in one orbital results in a greater mutual repulsion than when they occupy separate
orbitals. Hund’s Rule is followed in d and f orbitals as well.
Pauli’s Exclusion Principle can be tested by simple observation. Measurements of magnetic
properties provide the most direct evidence for specific electronic configurations of elements.
Paramagnetic materials are those that contain unpaired electrons or spins and are attracted by a
magnet. Diamagnetic materials are those with paired spins and are repelled by a magnet.
Any atom with an odd number of electrons will contain one or more unpaired spins, and are
therefore attracted by a magnet, thus, can be classified as paramagnetic. For an even number
of electrons like helium, if the two electrons in the 1s orbitals had parallel spins, their net
181
magnetic fields should strengthen each other. But experimental results showed that the helium
atom in its ground state has no net magnetic field.
This observation supports the pairing of two electrons with opposite spins in the 1s orbital. Thus,
helium gas is diamagnetic. Lithium, on the other hand, has an unpaired electron and is
paramagnetic. The orbital diagram provides information on the diamagnetic or paramagnetic
characteristic of an element.
Exercises:
Answer Key
Fill in the following table:
ATOMIC
ELEMENT
NUMBER
H
1
He
2
Li
3
Be
4
B
5
C
6
N
7
O
8
F
9
Ne
10
NUMBER
ORBITAL
OF
DIAGRAM
ELECTRONS
ELECTRON
CONFIGURATION
PARAMAGNETIC
OR
DIAMAGNETIC
Which of the 10 elements has the highest magnetic properties (most paramagnetic)?
182
No. of
UNPAIRED
ELECTRONS
IV. AUFBAU PRINCIPLE
The Aufbau principle dictates that as protons are added one by one to the nucleus to
build up the elements, electrons are similarly added to the atomic orbitals. The order of
filling up the atomic orbitals is from lowest energy to highest energy. Within the same
principal quantum number, the order of energies of the atomic orbitals is
s"<"p"<"d"<"f"
For example, for n = 3, the order is E3s < E3p < E 3d.
For multi-electron atoms, the general order of filling up orbitals can be diagrammed as
follows:
The electron configuration of elements higher than hydrogen and helium can be
represented using the noble gas core. In the periodic table, the noble gases are found
in the last column named as Group 8A (or Group 18 in the IUPAC convention). These
are 2He, 10Ne, 18Ar, 36Kr, 54Xe, 86Rn.
183
The smallest noble element is helium, so the shortened electronic configuration can be
written as follows for the given elements:
ELEMENT
ATOMIC
NUMBER
NO. OF
ELECTRONS
ELECTRON
CONFIGURATION
NOBLE GAS
CONFIGURATION
He
2
2
1s2
[He]
Li
3
3
1s2 2s1
[He] 2s1
Be
4
4
1s2 2s2
[He] 2s2
B
5
5
1s2 2s2 2p1
[He]2s2 2p1
C
6
6
1s2 2s2 2p2
[He]2s2 2p2
N
7
7
1s2 2s2 2p3
[He]2s2 2p3
O
8
8
1s2 2s2 2p4
[He]2s2 2p4
F
9
9
1s2 2s2 2p5
[He]2s2 2p5
Ne
10
10
1s2 2s2 2p6
[Ne]
Na
11
11
1s2 2s2 2p63s1
[Ne]3s1
Mg
12
12
1s2 2s2 2p6 3s2
[Ne]3s2
K
19
19
1s2 2s2 2p6 3s2 3p6 4s1
[Ar] 4s1
The elements in the 4th period, starting from potassium will have argon as the noble
gas core
19K:
[Ar]4s1
20Ca:
[Ar] 4s2
184
Teacher Tip
Please keep in mind that the electron configuration is a tool
used by chemists to explain various properties and phenomena.
Learners should not be made to recite electron configuration of
very large atoms as there are bound to be many cases when the
general guide to filling up orbitals is not followed. It is better
to emphasize the concept rather than rote memorization.
The 4s orbital has lower energy than the 3d orbitals; it is first filled with electrons
before the 3d orbitals.
Elements scandium to copper are transition metals. These elements will have
incompletely filled d subshells or readily gives electrons and form cations that have
incomplete filled d subshells. There will be some irregularities in the electron
distribution of this series as seen in Cr and Cu.
21Sc:
22Ti:
24Cr:
[Ar]4s23d1
[Ar]4s23d2
[Ar]4s13d5
29Cu:
[Ar]4s13d10
30Zn:
[Ar] 4s23d10
The irregularities in Cr and Cu are due to experimental results that show that there is a
greater stability associated with the half-filled (3d5) and the completely filled (3d10)
subshells. Similar observations are also found in the higher d and f-orbitals.
Gallium is the next element after Zn, its electronic configuration is:
31Ga:
[Ar] 4s23d104p1
Important data that can be gathered from the shortened electronic configuration are
the following:
a. Valence configuration: The electronic configuration representing the outermost
subshells.
b. Valence electrons: the number of electrons in the outermost subshells.
Determining the valence electrons is important to understand the behavior of the
elements especially in their bonding patterns to be discussed in the next sessions.
185
ELEMENT
NOBLE GAS
CONFIGURATION
VALENCE
CONFIGURATION
VALENCE ELECTRONS
Li
[He] 2s1
2s1
1
Be
[He] 2s2
2s2
2
B
[He]2s2 2p1
2s2 2p1
3
N
[He]2s2 2p3
2s2 2p3
5
Ne
[Ne]
[Ne] or 2s2 2p6
8
Sc
[Ar]4s23d1
4s23d1
3
Cr
[Ar]4s13d5
4s13d5
6
Ga
[Ar]4s23d104p1
4s24p1
3
V. THE QUANTUM NUMBERS AND THE ARRANGEMENTS OF ELEMENTS IN
THE PERIODIC TABLE
It was mentioned earlier that the complete set of quantum numbers specifies the
address of an electron in an atom. This can be seen in the arrangement of elements in
the periodic table. The periodic table is designed such that elements with valence
configurations in the s orbitals are found in the first two columns on the left, the ones
with p-orbitals are found on the right. The transition metals have d-orbitals and are
found at the middle and the elements with f-orbitals as valence configurations are
found at the bottom.
Note
In the example on the left, there is no need to include the
completely filled 3d subshells in the valence configuration and
the inclusion of the 3d electrons in counting the valence
electrons.
Teacher Tip
Recall that there are no elements with the same set of quantum
numbers. Similarly, in the periodic table, each known element
has its own corresponding place.
Source
The periodic table showing the s, p, d, and f sublevel blocks.
From High School Chemistry/The Periodic Table and Electron
Configuration. Retrieved from https://en.wikibooks.org/wiki/
High_School_Chemistry/
The_Periodic_Table_and_Electron_Configurations (3November
2016), Creative Commons Attribution-ShareAlike 3.0 License.
186
ENRICHMENT/EVALUATION (30 minutes)
1. Which of the four quantum numbers (n, l, ml , ms) determine (a)
the energy of an electron in a hydrogen atom and in a manyelectron atom, (b) the size of an orbital, (c) the shape of an
orbital, (d) the orientation of an orbital in space?
8. The atomic number of an element is 73. Is this element
diamagnetic or paramagnetic?
9. Indicate the number of unpaired electrons present in each of the
following atoms: B, Ne, P, Sc, Mn, Se, Kr, Fe, Cd, I, Pb.
Ans: (a) n (b) ms (c) l (d) ml
10. Draw the orbital diagrams for atoms with the following electron
configurations:
2. Calculate the total number of electrons that can occupy (a) one s
orbital, (b) three p orbital, (c) five d orbitals, (d) seven f orbitals.
(a) 1s22s22p5
Ans: (a) 2, (b) 6, (c) 10, (d) 14
(b) 1s22s22p63s23p3
(c) 1s22s22p63s23p64s23d7
3. List the values of n, l, and ml for the orbital in the 4d subshell.
Ans: n = 4; l = 2; ml = -2, -1, 0, 1, 2
11. What is the maximum number of electrons in an atom that can
have the following quantum numbers? Specify the orbitals in
which the electrons would be found.
(a) n= 2, ms = +½; (b)
n= 4, ml = +1; (c) n= 3, l = 2; (d) n= 2, l = 0, ms = -½; (e) n= 4, l
= 3, ml = -2
4. Write the four quantum numbers for an electron in a 3p orbital.
Ans: (3, 1, -1, +½) (3, 1, 0, +½) (3, 1, 1, +½)
(3, 1,-1, -½) (3, 1, 0, -½) (3, 1, 1, -½)
5. What is the total number of orbitals associated with the principal
quantum number n = 3?
12. Shown below are portions of orbital of diagrams representing
the ground-state electron configurations of certain elements.
Which of them violate the Pauli Exclusion Principle? Hund’s rule?
Ans: 9
6. Indicate which of the following sets of quantum numbers in an
atom are acceptable and explain why:
(a) (1, 0, ½, ½), (b) (3, 0, 0, +½), (c) (2, 2, 1, +½), (d) (4, 3, -2,
+½), (e) (3, 2, 1, 1).
7. The ground-state electron configurations listed here are
incorrect. Explain what mistakes have been made in each and
write the correct electron configurations.
Al: 1s22s22p43s23p3
B: 1s22s22p5
F: 1s22s22p6
187
Electron Configuration Worksheet
1. Complete the table below with the appropriate information as asked.
ELEMENT
ATOMIC NUMBER
ELECTRONIC
CONFIGURATION
NO. OF ELECTRONS
NOBLE GAS
CONFIGURATION
Na
Al
Si
Cl
Ar
Ca
Ti
Mn
Fe
Cu
Ge
Os
Au
U
Cf
2. From A, List 5 elements that exhibit paramagnetism and 5 elements that exhibit diamagnetism in its ground state.
3. Write the four quantum numbers of each electron in a nitrogen atom.
188
ORBITAL DIAGRAM OF
VALENCE
CONFIGURATION
4. Determine the element whose outermost valence electron is
represented by the following quantum numbers.
5. Write the electron configurations for the elements in number 4.
Encircle the valence configuration.
a. n=1, l= 0, ml= 0, ms=-1/2
6. Plot the elements in number 4 in the blank periodic table
provided for.
b. n=2, l=1, ml= 0, ms= +1/2
7. Pick 3 elements from number 4. Give their sources and a
minimum of three uses of the elements you picked.
c. n=3, l=1, ml= 0, ms= +1/2
d. n=4, l=2, ml= 0, ms= +1/2
e. n= 6, l=0, ml= 0, ms= -1/2
f.
n=3, l=1, ml = -1, ms= +1/2
g. n=5, l= 3, ml = 0, ms= +1/2
h. n=4, l=1, ml = -1, ms= -1/2
i.
n=4, l=1, ml = 0, ms= -1/2
j.
n=5, l=1,ml = 1, ms = ½
8A
1A
2A
3A
189
4A
5A
6A
7A
General Chemistry 1
Lesson 22: Periodic Relationships
Among the Elements
Content Standard
The learners demonstrate an understanding of the arrangement of elements in
the periodic table and trends in the properties of the elements in terms of
electronic structure.
Performance Standard
The learners can arrange elements and explain their properties through the
knowledge of electron structure.
Learning Competencies
At the end of the lesson, the learners:
1. Explain the periodic recurrence of similar properties among elements to
their group number in terms of electronic structure (STEM_GC11CB-IIcd-60)
2. Relate the number of valence electrons of elements to their group number
in the periodic table (STEM_GC11CB-IIc-d-61)
3. Compare the properties of families of elements (STEM_GC11CB-IIc-d-62)
4. Predict the properties of individual elements based on their position in the
periodic table (STEM_GC11CB-IIc-d-63)
5. Describe and explain the trends in atomic properties in the periodic table
(STEM_GC11CB-IIc-d-64)
Specific Learning Competencies
At the end of the lesson, the learners will be able to:
1. Sketch the periodic table showing the groups and periods.
2. Identify the metals, metalloids and nonmetals in the periodic
table.
3. Identify the representative elements, the transition metals, the
lanthanides and actinides in the periodic table.
4. Give the electron configuration of cations and anions.
5. Determine the trends in the physical properties of elements in a
group
120 MINS
Lesson Outline
Introduction
Communicating Learning Objectives and
Development of the Periodic Table
Motivation
The Periodic Table
Instruction
and Practice
I. Periodic Classification of Elements
II. Electron Configuration of Cations and
Anions
III. Periodic Variation in Physical Properties
90
Evaluation
Exercises and Activity
15
3
Materials
Periodic Table
Resources
(1) Chang, Raymond and Goldsby, Kenneth A. (2016). Chemistry (12th ed).
New York: McGraw-Hill.
(2) Petrucci, Herring, Madura, and Bissonnette (2011). General Chemistry
and Modern Applications, 10th Ed. Pearson Canada, Inc.
(3) Zumdahl, S.S. and Zumdahl, S.A (2013).Chemistry, 8th ed. Cengage
Learning
(4) Cox, P.A. Inorganic Chemistry Second Edition (2004). Inorganic
Chemistry Laboratory, New College, Oxford, UK
(5) Roque, et al. Laboratory Manual in General Chemistry (2008).
Philippine Normal University.
6. Describe and explain the trends in atomic properties in the
periodic table
7. Compare the properties of families of elements
8. Predict the properties of individual elements based on their
position in the periodic table
9. Perform exercises and collaborative work with peers.
190
12
INTRODUCTION (12 minutes)
1. Introduce the following learning objectives using any of the
suggested protocol (Verbatim, Own Words, or Read-aloud):
At the end of the lesson, I will be able to:
a. Sketch the periodic table showing the groups and periods.
b. Identify the metals, metalloids and nonmetals in the periodic
table.
c. Identify the representative elements, the transition metals,
the lanthanides and actinides in the periodic table.
d. Give the electron configuration of cations and anions.
e. Determine the trends in the physical properties of elements
in a group
f. Describe and explain the trends in atomic properties in the
periodic table
g. Compare the properties of families of elements
h. Predict the properties of individual elements based on their
position in the periodic table
i. Perform exercises and collaborative work with peers.
2. Present the keywords for the concepts to be learned:
a. Periodic table
b. Metals
c. Non-metals
d. Metalloids
e. Alkali metals
f. Alkaline earth metals
g. Halogens
h. Noble gases
i. Representative elements/main group elements
j. Transition elements
k. Lanthanides
l. Actinides
m. Isoelectronic
n. Effective nuclear charge
o. Shielding or screening
p. Atomic radius
q. Ionic radius
r. Ionization Energy (First, Second, Third,…)
s. Electron affinity
191
3. Development of the Periodic Table
The arrangement of elements in the modern periodic table was made possible through the
efforts of several chemists, such as; Dobereiner, John Newlands, Dmitri Mendeleev, and Henry
Moseley.
It started with Dobereiner’s “Law of Triads”. He found a relationship among three elements
where the atomic weight of the middle element is nearly the same as average of the atomic
weights of other two elements. John Newlands arranged the elements in what is known as the
“law of octaves”. He noted that the eighth element has similar chemical properties with the first
element.
Mendeleev prepared a tabulation of elements based on equivalent weights (atomic mass) and
the regular recurrence of properties of the elements. In a few cases, the mass and the
properties did not go the same directions. But Mendeleev rationalized that the properties were
more accurate than the masses since technology used to determine the mass was still
improving. Henry Moseley discovered that each element in Mendeleev’s table was arranged in
an order such that their integral positive charge (atomic number) increased numerically from
left to right and top to bottom.
The present periodic table is arranged according to increasing atomic number which also equals
the number of electrons. The electron configuration helps to predict and explain the recurrence
of chemical and physical properties.
MOTIVATION (3 minutes)
1. Show the learners some periodic tables. Ask them what they know about the periodic table.
2. Ask them to identify a few of the elements in the table.
192
INSTRUCTION/DELIVERY/PRACTICE (90 minutes)
I.
Majority of the elements are metals (shown in light gray shade
in the figure). Metals are good conductors of electricity. The
non-metals are shown as boxes with no shade. The metalloids
(shown in boxes with dark grey shade) have properties that are
intermediate between metals and nonmetals.
PERIODIC CLASSIFICATION OF ELEMENTS
Some groups have been given collective names. Group 1A
elements are called alkali metals; Group 2A elements are
referred to as alkaline earth metals; Group 7A elements are
called halogens; Group 8A elements are known as noble
gases.
The Periodic Table: Metals, Nonmetals, and Metalloids
The periodic table is a chart in which elements having similar chemical and
physical properties are grouped together. The elements are arranged
according to increasing atomic number. The rows are called periods. The
vertical columns are called groups or families according to the similarities in
their properties. At present, it contains 118 elements; however, elements
113 to 118 have just recently been synthesized and naming is not yet fully
complete.
There are 18 groups or families. There are two conventions in designating
the groups. The Union of Pure and Applied Chemistry (IUPAC) refers to the
columns are Groups 1-18. Many books, however, continue to refer to the
columns as Groups 1A, 2A, 3B, and so on as shown in the figure above. This
teaching guide will use the Groups A and B convention.
193
The Group A elements are classified as representative
elements or main group elements. These elements have
unfilled or filled s and p orbitals in the highest principal
quantum number. The Group B elements are called the
transition elements where the d subshells are being filled up.
This, however, is not the universal convention. In Europe, the
convention is to use B for representative elements and A for
transition elements; the opposite of the American convention.
However, in this lesson, the American convention will be
followed.
The two separate rows at the bottom of the periodic table are
lanthanides and the actinides. Sometimes, they are referred to
as the f-block elements.
Exercises
1. Write the electron configuration (using noble gas notation)
of the elements in Group 1A.
2. Comment on the outermost electron configuration of
Group 1A elements.
3. How many valence electrons do Group 1A elements have?
4. Write the electron configuration (using noble gas notation)
of the halogens.
5. Comment on the outermost electron configuration of the
halogens.
6. How many valence electrons do the halogens have?
7. Comment on the arrangement of the representative
elements in the periodic table with respect to their electron
configuration.
The Periodic Table: Representative, Transition, and f-block Elements
Exercises
1. Give the electron configuration of Na and Na+.
Na: [Ne] 3s1
Na atom has 11 protons and 11 electrons.
Na+: [Ne]
Na+ ion has 11 protons and 10 electrons. Na+ is
isoelectronic with Ne.
II. ELECTRON CONFIGURATION OF CATIONS AND ANIONS
Ions derived from representative elements
In the formation of cations from representative elements, the
electrons are removed from the outermost shell to achieve a noble
gas configuration. In the formation of anions, electrons are added
to the highest partially filled n shell so that they become
isoelectronic (same number of electrons) with the noble gas.
2. Give the electron configuration of Ca and Ca2+
Ca: [Ar] 4s2
Ca2+: [Ar]
3. Give the electron configuration of F and F
F: 1s2 2s2 2p5
F : 1s2 2s2 2p6 or [Ne]
F is isoelectronic with Ne.
4. Give the electron configuration of O and O2
O: 1s2 2s2 2p4
O2 : 1s2 2s2 2p6 or [Ne]
O2 is isoelectronic with Ne.
194
Ions derived from transition elements
1. Give the electron configuration of Mn and Mn2+.
Mn: [Ar]4s2 3d5
Mn2+ [Ar] 3d5
Note that for transition elements, the ns electrons are removed first. In filling up the orbitals,
the ns orbitals are filled first before the (n-1)d orbitals because the ns orbitals are more stable
and lower energy. However, the electron-electron interactions are different in a neutral atom
from that in an ion. For transition metal ions the 3d orbital is more stable. Hence, the 4s
electrons are removed first before the 3d electrons.
III. PERIODIC VARIATION IN PHYSICAL PROPERTIES
1. The Effective Nuclear Charge
In many-electron atoms, the inner or core electrons shield the outer electrons from the nucleus
reducing the electrostatic attractions between the nucleus and the outer electron. The effective
nuclear charge, Zeff, is given by Zeff = Z - σ where Z is the nuclear charge and σ is the shielding
constant. Screening or shielding refers to how an outer electron is blocked from the nuclear charge
by the inner electrons. It means that the attraction of the outer electron to the nucleus is not felt
100% because of the effect of the inner electrons. Electrons in the inner shells are very effective in
shielding the nucleus. Electrons in the same shell as the electron of interest provide a relatively
smaller screening effect. To see how the effective nuclear charge varies across a period, answer the
following questions.
a. Approximate the Zeff for the outermost electron of Li.
Li (electron configuration [He] 2s1) has 2 inner core electrons to shield the valence electrons.
Qualitatively, the +3 charge of the Li nucleus will be neutralized by the 2 inner electrons; the
outer electron of Li will feel an effective nuclear charge of about +1.
b. Approximate the Zeff for boron (Z=5).
Boron (electron configuration [He] 2s2 2p1) also has 2 inner core electrons to shield the 3
valence electrons. The +5 charge of the boron nucleus is neutralized by the 2 inner
electrons leaving a net of +3. However, there are 3 valence electrons that also provide a
lesser amount of shielding to one another. One of these outer electrons will feel the
shielding effect of the 2 other valence electrons. Therefore, the Zeff for boron is expected to
be below 3.
195
Teacher Tip
The shielding constant, σ , can be calculated using
some specified rules. In 1930, J. C. Slater devised a
simple set of guidelines to determine shielding or
screening constants. However, this will not be done
in this lesson. Shielding will be treated qualitatively.
Suffice it to say that core electrons shield the outer
electrons and lessen the force of attraction felt by
the electron for the nucleus. Electrons in the same
shell as the electron of interest provide a relatively
smaller screening effect. Guide the learners with
the questions on Zeff for them to reason out the
trend.
c. Approximate the Zeff for F (Z=9).
F (electron configuration [He] 2s2 2p5) also has 2 inner core electrons to shield the 7 valence
electrons. The +9 charge of the F nucleus is neutralized by the 2 core electrons leaving a net
charge of +7. However, there are 7 valence electrons. A valence electron will also feel the
shielding effect of the other 6 valence electrons. Therefore, the Zeff is expected to be less
than 7 but larger than the Zeff of the outer electron in boron because there are more valence
electrons to provide shielding.
d. From the above analysis, how will Zeff vary across a period?
Zeff increases as you go from left to right across a period.
e. To validate the approximations made, values of Zeff for Li to Ne are given below:
3Li
Z
Zeff
4Be
5B
6C
7N
8O
9F
10Ne
3
4
5
6
7
8
9
10
1.28
1.91
2.42
3.14
3.83
4.45
5.10
5.76
2. Atomic Radius
Atomic size is difficult to define because there is no distinct outer boundary to an atom. The
probability of finding the electron decreases with increasing distance from the nucleus but the
probability does not fall to zero. Thus, it is safe to describe the effective atomic radius which is the
distance of the electron from the nucleus within which 95% of the electron charge density is found.
A more specific way to get atomic radius values is to get one-half the distance between two nuclei
in adjacent atoms (the internuclear distance) in a metal solid or in a diatomic molecule.
The covalent radius is one-half the distance between two identical
atoms joined together by a single bond.
196
Teacher Tip
There are various types of radii used in inorganic
chemistry. However, the lesson focuses only on
general trends in the sizes of atoms. There is no
need to further distinguish among these types of
radii.
The metallic radius: it is one-half the distance between the nuclei of
the two atoms in contact in the crystalline solid metal.
How does the atomic radius vary within a period for representative elements? Explain.
The atomic radius decreases from left to right through a period of elements for representative
elements. This corresponds to the increase in Zeff across a period. With the increase in Zeff, the
outer electrons are pulled in and attracted towards the nucleus resulting in a decrease in the size of
the atoms.
How does the atomic radius vary down a group for representative elements? Explain.
The more electronic shells (n) in an atom, the larger is the atom. Atomic radius increases from top
to bottom through a group of elements.
Exercises
Using the periodic table, arrange the following atoms in order of increasing atomic radius. Explain
your reasoning.
Teacher Tip
For additional information, atomic size does not
change very much for a transition series within a
period. Additional electrons in a transition series
go into an inner electron shell where they
participate in shielding the outer shell-electrons
from the nucleus. The number of electrons in the
outer shell remains constant; they experience a
comparable force of attraction to the nucleus
throughout the transition series.
For example, for Fe (Z=25), Co (Z=26) and Ni (Z =
27), Zeff for the 4s electrons of the first transition
series is approximately constant, thus the atomic
radii do not change very much for this series of
three elements, Fe (124 pm), Co (125 pm) and Ni
(125 pm).
a. C, Li, Be
b. As, I, S
c. P, Si, N
3. Ionic Radius
Ionic radii are very difficult to measure with certainty because they are affected
by their immediate environment. They can be measured by x-ray diffraction.
The sizes vary depending on the environment. However, we are going to
discuss the general trends and relative sizes.
197
Atomic radius versus ionic radius
Cations are smaller than the atoms from which they are formed. When a metal atom loses one or
more electrons to form a positive ion, the positive nuclear charge exceeds the negative charge of
the electrons in the resulting cation. For isoelectronic cations, the more positive the ionic charge,
the smaller the ionic charge.
Anions are larger than the atoms from which they are formed
When a non-metal gains one or more electrons, it forms a negative ion termed as anion. The
nuclear charge remains constant, but Zeff is reduced because of the additional electrons. The
additional electrons results in increase repulsions among the electrons in the outer shell. This results
to the tendency of the electrons to spread out more, thus increasing the size of the anion.
For isoelectronic anions, the more negative charge, the larger is the ionic radius.
a. Compare the size of a neutral atom of Na and a Na+ ion. Which is larger. Explain.
The Na atom has 11 protons attracting 11 electrons. Its electron configuration is [Ne] 3s1. This
outer electron is lost when it forms the Na+ ion.
The Na+ ion has 11 protons attracting only 10 electrons. Therefore the electrons are pulled
closer to the nucleus.
The Na atom is larger than the Na+ ion: Na > Na+
b. Compare the size of a Mg atom and a Mg2+ ion. Which is larger?
The Mg atom has 12 protons attracting 12 electrons.
The Mg2+ ion has 12 protons attracting 10 electrons. The electrons feel a larger attractive force
towards the nucleus.
The Mg atom is larger than the Mg2+ ion:
Mg > Mg2+
c. Compare the sizes of Na+, Mg2+, and Al3+. Arrange according to increasing size.
Na+, Mg2+, and Al3+ are isoelectronic; that is, they all have the same number of electrons. They
have 10 electrons outside the nucleus. But for Al3+, the 10 electrons are pulled by 13 protons;
for Mg2+, the 10 electrons are attracted by 12 protons; and for Na+, the 10 electrons are pulled
by only 11 protons.
198
Therefore, the sizes of the ions increase according to: Al3+ < Mg2+ < Na+.
d. Compare the size of a F atom and a F– ion. Which is larger?
F has 9 protons attracting 9 electrons. A fluoride ion, F– ion has 9 protons attracting 10
electrons. Zeff decreases for the fluoride ion.
The fluoride ion, F–, is larger than the F atom: F– > F
e. Which is larger, the O atom or the O2– ion? Which is larger?
The O atom has 8 protons attracting 8 electrons. The oxide ion, O2–, has 8 protons attracting
10 electrons resulting in a decrease of Zeff for the outer electrons. Therefore, the oxide ion, O2–,
is larger than the O atom: O2– > O
f.
Compare the sizes of F–, O2–, and N3–. Arrange according to increasing size.
F–, O2–, and N3– are isoelectronic. All have 10 electrons. However, only 7 protons are attracting
the 10 electrons in the nitride ion; 8 protons are pulling in the 10 electrons in the oxide ion;
while 9 protons are attracting the 10 electrons in the fluoride ion. Therefore, the ionic sizes
increase according to F– > O2– > N3–. .
Exercises
Arrange the following set of ions and atoms in increasing size and explain your answer
a. K+, Cl-, S2-, Ca2+
b. N, Cs, As, Mg2+, BrIonization Energy
Ionization energy (IE) is the minimum amount of energy (in kJ/mol) required to remove an electron
from a gaseous atom in its ground state.
Energy + X(g) —> X+ (g) + e–
The energy required to remove the first electron is called the first ionization energy. The first
ionization energy, IE1, has the lowest value. The second ionization energy, IE2, is the energy
199
required to strip the second electron from the atom; it has higher energy value, and so on.
IE1 < IE2 < IE3 < …
Examples of ionization energies:
Al(g) —> Al+(g) + e–
IE1 = 577.9 kJ/mol
Al+(g) —> Al2+(g) + e–
Al2+(g)
—>
Al3+(g)
+
IE2 = 1,820 kJ/mol
e–
IE3 = 2,750 kJ/mol
Al3+(g) —> Al4+(g) + e–
IE4 = 11,600 kJ/mol
Ask the learners to:
a. Explain why IE for Al increases from IE1 to IE2 to IE3 to IE4.
b. Why is there a drastic increase in energy from IE3 to IE4 for Al?
When electrons are removed from the same shell, the main effect is that with each successive
ionization there is one less electron left to repel the others. The electrons are more attracted to the
nucleus and are harder to remove. The fourth IE of Al is very much higher because now the outer
shell is exhausted and the tightly bound inner shell, 2p, is being ionized.
Ionization energies decrease as atomic radii increases. The farther an electron is from the nucleus,
the easier it is to be released. Down a group, as n increases and atomic size increases, electrons
are easily released. Thus, ionization energy decreases from top to bottom of a group. Across a
period, as Zeff increases and size decreases, ionization energy increases.
The following tables provide some first ionization energies for representative elements.
IE1
Li
Be
kJ/
mol
520
IE1
Li
kJ/mol
B
899
801
C
N
1,086
1,400
Na
520
K
495.9
O
1,314
F
Ne
1,680
2,080
Rb
418.7
Cs
403.0
375.7
200
Teacher Tip
There are various irregularities seen in the trends.
These can be further discussed if time permits.
They are contained in the enrichment section.
Electron Affinity
The electron affinity of an atom may be defined as the negative of the energy change that
occurs when a gaseous atom accepts an electron. For example, for the F atom,
F(g) + e–—> F– (g)
Energy involved = -328 kJ/mol
The electron affinity is
F– (g) —> F(g) + e–
Electron Affinity = EA = +328 kJ/mol
This is the reason why electron affinity is sometimes defined as the ionization energy of a
negative ion. The more positive the electron affinity, the greater the tendency to accept an
electron and form an ion.Generally, the electron affinity increases across a period from left to
tight. The electron affinity generally decreases going down a group.
ENRICHMENT
Some applications of metal ions
1. Knowledge of atomic and ionic radii is used to vary physical properties of materials. For
example:
a. Strengthening Glass. Normal glass windows that contain Na+ and Ca2+ ions are brittle and
shatters easily. Replacing the Na+ ions with bigger K + ions results in surfaces where
surface sites are being filled up leaving less opportunity for cracking.
b. Colors in gemstones. Pure Al2O3 is colorless. Substituting Al3+ with a little amount of Cr3+
in Al2O3 gives a red color in ruby.
2. Explanation for the irregularities in the ionization potential trends for beryllium and boron
(This is optional topic)
a. Boron has lower ionization energy than Be. The ionization energy of B is slightly less than
that of Be because boron removes an electron from a 2p orbital, which is less tightly
bound than the 2s involved in lithium and beryllium.
b. Hund’s rule play an important role in explaining the ionization energies of nitrogen and
oxygen. Remember, there are three 2p electrons that can be accommodated in different
orbitals with parallel spin so as to minimize their mutual repulsion. For O (2p)4 and
subsequent elements in the period some electrons are paired and repel more strongly,
leading to IE values less than would be predicted by extrapolation from the previous
three elements.
201
Teacher Tip
The teacher should be very careful in explaining
electron affinity because the sign conventions may
vary in different textbooks. The convention used
here is from Chang and Goldsby.
Note that there are many irregularities in the trends
for electron affinities. Teachers need only to focus
on the general trends at this level.
Electronegativity will be discussed in relation to
covalent bonding as is not included in this section.
EVALUATION
Directions: Fill up the blank periodic table with the Element as described by each statement below:
1. Element A is the biggest in Group 1A.
2. Element B forms the biggest anion in period 2.
3. Element C has complete d electrons in period 4.
4. Element D is the most electronegative in period 2.
5. Element E will be isoelectronic with the noble gas in period 3 when it loses two electrons.
6. Element F has the highest ionization energy in period 4.
7. Element G has the least electron affinity in group 6.
8. Element H has the 4f14 configuration
9. Element I is the first member of the actinide series
8A
1A
2A
3A
4A
5A
6A
B
E
C
D
J
F
G
A
H
I
202
7A
General Chemistry 1
60 MINS
Lesson 23: Periodic
Relationships of Main
Group Elements (Lab)
Lesson Outline
Content Standard
The learners demonstrate an understanding of the arrangement of elements in
the periodic table and trends in the properties of the elements in terms of
electronic structure.
Performance Standard
The learners can arrange elements and explain their properties through the
knowledge in electronic structure.
Learning Competencies
At the end of the lesson, the learners:
1. Compare the properties of families of elements (STEM_GC11CB-IIc-d-62)
2. (LAB) Investigate reactions of ions and apply these in qualitative analysis
(STEM_GC11CB-IIc-d-65)
3. (LAB) Determine periodic properties of the main group elements
(STEM_GC11CB-IIc-d-66)
Specific Learning Competencies
At the end of the lesson, the learners will be able to:
1. Explain the periodic properties of the main group elements;
2. Investigate qualitatively the trends in chemical reactivity of metallic
elements;
3. Analyze patterns in data and draw conclusions that are consistent with
evidence;
4. Compare and contrast the properties of main group elements; and
5. Perform exercises and collaborative work with peers.
203
Activity 1
Properties of Group 2A Elements: Reactions
of their Ions
30
Activity 2
Electron Configuration and Periodicity
30
Materials
Materials for Activity No. 1
A. Reagents: 0.5 M magnesium nitrate, 0.5 M calcium nitrate, 0.5 M
strontium nitrate, 0.2 M barium nitrate, 1.0 M sodium hydroxide, 0.5 M
sodium fluoride, 0.5 M sodium chloride, 0.2 M potassium bromide, 0.2 M
potassium iodide, 0.5 M sodium sulfate, 0.5 M sodium carbonate.
B. Equipment: spot plate (use plastic film like that used for overhead
transparencies or plastic wrap in the absence of a spot plate or wax paper),
respective droppers and small beakers for each solution
Materials for Activity No. 2
Giant Periodic table posted on the board, flash cards of elements with
atomic number, whiteboard markers, manila paper
Resources
(1) Chang, Raymond and Goldsby, Kenneth A. (2016). Chemistry (12th
ed). New York: McGraw-Hill.
(2) Petrucci, Herring, Madura, and Bissonnette (2011). General Chemistry
and Modern Applications, 10th Ed. Pearson Canada, Inc.
(3) Zumdahl, S.S. and Zumdahl, S.A (2013).Chemistry, 8th ed. Cengage
Learning
(4) Silberberg (2006). Chemistry. The Molecular Nature of Matter and
Change. McGraw-Hill:
(5) Roque, et al. Laboratory Manual in General Chemistry (2008).
Philippine Normal University.
(6) Periodic Trends Laboratory. classrooms.tacoma.k12.wa.us/Wilson/
krichardson/documents/ ptrendlab.pdf, July4, 2007
(7) http://www.rsc.org/learn-chemistry/resource/res00000513/theperiodic-table-properties-of-group-2-elements?cmpid=CMP00000583
(8) http://www.oresomeresources.com/resources_view/resource/
experiment_metal_reactivity
ACTIVITY NO. 1
PROPERTIES OF GROUP 2A ELEMENTS – REACTIONS OF THEIR IONS
Arrangements of elements in the periodic table have been painstakingly thought of by
scientists. We learned that elements are arranged in order of increasing atomic number.
Elements are grouped in s, p, d, f blocks with the same valence electrons in a group.
We found out that elements in the same group have similar physical properties. Is there
periodicity of chemical reactivity? It’s time for you to find out.
Pre-Activity
1. Recall
In the periodic table, elements that are found in the left are metals and in the far right
are the non-metals. The elements found in the boundary of the two are called
metalloids.
2. Materials
Reagents: 0.5 M magnesium nitrate, 0.5 M calcium nitrate, 0.5 M strontium nitrate, 0.2
M barium nitrate, 1.0 M sodium hydroxide, 0.5 M sodium fluoride, 0.5 M sodium
chloride, 0.2 M potassium bromide, 0.2 M potassium iodide, 0.5 M sodium sulfate, 0.5
M sodium carbonate.
Equipment: spot plate (or use plastic film like that used for overhead transparencies or
plastic wrap in the absence of a spot plate or wax paper), respective droppers and
small beakers for each solution
3. Precautions
a. Always wear your laboratory gown and safety goggles.
b. Do not contaminate the reagents. Use respective droppers for each solution.
c. Avoid skin contact with the reagents
d. Dispose all solutions in the containers provided by your teacher. Wash your hands
before leaving the laboratory.
204
Teacher Tip
Prior to the activity, the instructor needs to:
1. Prepare the room for the laboratory activity. Ensure that it
is equipped with safety equipment especially eyewash and
shower stations, fire extinguisher, sand buckets, first aid
kits, running water.
2. Place posters on safety measures around the room as
reminders.
3. Prepare the reagents, equipment, and materials.
4. Prepare the waste receptacles.
5. At the start of the activity, explain the activity. Go through
the entire procedure. Identify the reagents and equipment.
6. Clearly discuss the safety precautions to be strictly
followed by the students. All students need to wear safety
goggles and lab gowns. No open footwear. Long hair
should be tied back.
Instruction / Activity
Procedure
1. Place the spot plate or OHP film at the center of an 8 x 11 paper. Label the rows
and columns as indicated below:
F–
Cl–
Br –
I–
CO32-
Teacher Tip
Ask the learners:
1. What group of elements is being investigated?
2. What particular property of this group is investigated?
SO42-
Mg2+
Ca2+
Sr+
Ba2+
2. Following the grid on the first row of the spot plate put two drops of magnesium
nitrate solution on the six holes (or boxes drawn on the OHP film). Then add two
drops of fluoride, chloride, bromide, iodide, carbonate, and sulfate
solutions to each of the holes or boxes as shown on the grid.
3. Repeat with calcium nitrate on the second row, then strontium nitrate on
the third row, and barium nitrate on the fourth row.
4. Record your observations in the data table.
F–
Cl–
Br –
Mg2+
Ca2+
Sr+
Ba2+
Additional Questions
5. What are the indications that a chemical reaction took place?
6. List the group IIA elements in order of increasing chemical reactivity
205
I–
CO32-
SO42-
ACTIVITY NO. 2
Teacher Tip
ELECTRON CONFIGURATION AND PERIODICITY
Rules for Task 1
1. Each group earns 5 points for each correct answer.
2. An additional 5 point will be given to the group which finishes within 10 minutes.
3. A group which finishes the task will raise the group flag. Time started and time
done should be recorded.
Task 1
1. The teacher will distribute flash cards of different colors randomly in the class.
2. Ask the learners to form groups according to the colors of the paper of the
flashcards.
3. Let the students write the long configuration and noble gas configuration of the
given elements on a response sheet.
NAMES:
SCORE:
TIME STARTED:
TIME DONE:
Element
Atomic
number
Long
Configuration
Noble
Configuration
Valence
Configuration
206
The teacher will need to prepare:
1. Colored flash cards where element symbols and atomic
numbers are present (preferably laminated flashcards).
2. Use the following colors for the elements:
a. On yellow paper: Na, K, Cs, Cr, Mn, Fe
b. On blue paper: Mg, Ca, Ba, Mo, Tc, Ru
c. On green paper: B, Al, Ga, Cu, Ag, Au
d. On red paper: C, Si, Sn ,U, Am, Cf
e. On white paper: F, Cl, Br, Nd, Eu, Dy
f. On orange paper: Ne, Ar, Kr, Ni, Pd, Pt
3. Prepare group flags corresponding to the color of the
flashcards.
4. Response Sheets enough for all the groups
5. A big giant periodic table can be posted on the board
Task 2
1. Ask the students to write the valence configuration of elements in the flash cards. (if
the flash cards are not laminated, let students write the configuration on separate
strips of paper)
Teacher Tip
There are two types of valence configurations in the group of
elements:
1. (ns) (np) elements
2. (ns) (n-1) d elements or (ns) (n-2) f elements
2. Ask the students what they observe about the valence configuration of the
elements in their group. Study what are the similarities and differences?
3. Ask the students to arrange the elements in the periodic table on the board. What
is the relationship of the configuration with their arrangement in the periodic table?
•
Elements found in the same family have the same number of valence electrons
but different period/energy level.
•
Elements with ns np configurations are called representative elements and the
(ns) (n-1)d elements are called transition elements, and the ones with (n-2)f
configurations are in the lanthanide or actinide series.
Exercises
A. Determine the valence configuration, the valence electrons, the block, the family,
and the period to which the following elements belong.
1. [Ar]4s23d3
2. [Xe]4f145d106s26p1
3. [Ar]4s23d104p3
4. 1s22s22p63s23p64s2
5.
[Kr]4d105s1
6. [Rn]7s25f3
B. Identify the following elements and indicate their position in a blank periodic table.
To the teacher, provide a blank periodic table.
1. [Ar]4s23d3
2. [Xe]4f145d106s26p1
3. [Ar]4s23d104p3
4. 1s22s22p63s23p64s2
5. [Kr]4d105s1
207
Answer Key
A.
1. 4s23d3 , 5, d-block, 5B, 4
2. 6s26p1 , 3, p-block, 3A, 6
3. 4s24p3, 5, p block, 5A, 4
4. 4s2 , 2, s-block, 2A, 4
5. 4d10 5s1, 11, d-block, 1B, 4
6. 7s25f3, 5, f-block, Actinide series, 7
B.
1.
2.
3.
4.
5.
V
Tl
As
Ca
Ag
General Chemistry 1
120 MINS
Lesson 24: Ionic Bonds
Lesson Outline
Content Standard
Introduction
The learners demonstrate an understanding of ionic bond formation
in terms of atomic properties.
Communicating Learning Objectives
5
Motivation
Inquiry
5
Instruction
I.
II.
III.
IV.
V.
Enrichment
Ions in the Human Body
Evaluation
Worksheet
Performance Standards
The learners shall be able to identify ionic compounds used in daily
life.
Learning Competencies
At the end of the lesson, the learners:
1. Relate the stability of noble gases to their electron configuration
(STEM_GC11CB-IId-g-67);
Octet Rule and Lewis Dot Symbol
Ions formed from Representative Elements
Ionic Bond Formation
Lattice Energy of Ionic Compounds
Properties of Ionic Compounds
85
5
20
Resources
(1) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York: McGraw-Hill.
(2) Joesten, M., Castellion, M. & Hogg, J. (2007). The World of Chemistry:
Essentials. Belmont, CA: Brooks/Cole Cengage Learning
(3) Petrucci, Herring, Madura, and Bissonnette (2011). General Chemistry and
Modern Applications, 10th Ed. Pearson Canada, Inc.
(4) Zumdahl, S.S. and Zumdahl, S.A (2013).Chemistry, 8th ed. Cengage Learning
2. State the Octet Rule (STEM_GC11CB-IId-g-68);
3. Determine the charge of ions formed by the representative
elements and relate this to their ionization energy or electron
affinity, valence electron configuration, and position in the
periodic table. (STEM_GC11CB-IId-g-69);
4. Draw the Lewis structure of ions (STEM_GC11CB-IId-g-70);
Specific Learning Outcomes
5. Predict the formula of the ionic compound formed by a metal and
non-metal among the representative elements. (STEM_GC11CBIId-g-71);
At the end of the lesson, the learners shall be able to:
1. State the Octet Rule.
2. Draw the Lewis dot symbols of atoms and ions.
6. Draw the Lewis structure of ionic compounds. (STEM_GC11CBIId-g-72);
3. Identify elements that easily form cations and anions based on
ionization energies and electron affinities
7. List the properties of ionic compounds and explain these
properties in terms of their structure (STEM_GC11CB-IId-g-73);
and
4. Explain the formation of ionic bonds using electron configuration
and the Lewis dot symbols.
8. Perform exercises on writing Lewis structures of ions/ionic
compounds and molecules. (STEM_GC11CB-IId-g-74).
5. Relate lattice energy with physical properties of ionic compounds
such as melting point.
208
INTRODUCTION (5 minutes)
1. Introduce the learning objectives by using the suggested protocol (Read-aloud):
a. I will be able to state the Octet Rule.
b. I will be able to draw the Lewis dot symbols of atoms and ions.
c. I will be able to identify elements that easily form cations and anions based on ionization
energies and electron affinities
d. I will be able to explain the formation of ionic bonds using electron configuration and the Lewis
dot symbols.
e. I will be able to relate lattice energy with physical properties of ionic compounds such as melting
point.
2. Present the keywords for the concepts to be learned:
a. Octet Rule
b. Lewis dot symbol
c. Ionic bond
d. Lattice energy
3. Connecting and reviewing essential knowledge
a. Briefly review the electron configuration of the elements found in Group 1A to Group 8A
(representative elements).
b. Ask learners to identify the valence electron of these elements.
c. Ask the learners what are valence electrons and why is their importance.
MOTIVATION (5 minutes)
1. Let them deliver the following statement using the read-aloud protocol: “A chemist named
Gilbert Lewis noticed something interesting about elements that were very stable.”
209
Teacher Tip
The group number of the element is equal to
the number of valence electrons using the old
system of group numbering. Using the IUPAC
system of group numbering, the number of
valence electrons is equal to (Group number
-10). For example, chlorine is in Group 17, so
its valence electrons are (17-10 =7). Chlorine
has seven valence electrons.
2. Show them the following table and ask them to answer the following questions:
Element
Atomic
Number
He
2
1s2
Ne
10
1s2 2s2 2p6
Ar
18
1s2
Kr
36
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6
Xe
54
1s2
Ra
86
Electron Configuration
2s2
2s2
2p6
2p6
3s2
3s2
3p6
3p6
3d10
4s2 4p6
4d10 4f145s2 5p6
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6
4d10 4f145s2 5p6 5d10 6s2 6p6
Group
Number (Old)
Group Number Period
(IUPAC)
Number
8A
18
1
8A
18
2
8A
18
3
8A
18
4
8A
18
5
8A
18
6
a. To what group in the periodic table do these elements belong?
b. These elements are called the noble gases. Why? What is the common characteristic of
these gases?
c. Light bulbs are filled with Ar gas rather than oxygen gas. Why?
d. From the electron configuration, how many valence electrons do noble gases have?
e. Can we relate the number of valence electrons with the stability of the element?
Answer Key
a. Group 8A
b. Historically, they are called noble because
they generally do not react with other
elements and are stable.
c. In light bulbs, current is passed through a
wire to heat it up until it emits light. Ar is
used for its inert atmosphere. If oxygen
gas were to be used, it would react and
explode.
d. 8
Teacher Tips
In the past, noble gases were thought to be
inert and unreactive. But recently, many
compounds of noble gases have been
synthesized. Stress that the word stable is
related to why elements combine with
another element. Mention that this will be
discussed later on.
This is what the lesson will be about.
INSTRUCTION/PRACTICE/DELIVERY (85 minutes)
I.
OCTET RULE AND LEWIS DOT STRUCTURE
According to Lewis, atoms combine in order to achieve a more stable electron configuration. And
this maximum stability is attained when an atom is isoelectronic with a noble gas. Except for He,
the noble gases have eight outer electrons or an octet of electrons. The Octet Rule states that
when atoms of elements combine to form compounds, they tend to lose, gain, or share electrons
to achieve the same electron arrangement as the noble gas nearest them in the periodic table.
210
Teacher Tip
Ask the learners to recall the meaning of
isoelectronic (means having the same number
of electrons).
In combining with other atoms, only outer electrons, the valence electrons, are involved. To keep
track of these valence electrons, the Lewis dot symbol is used. The Lewis dot symbol consists of
the symbol of an element and one dot for each valence electron in an atom of the element. See
examples below.
Element
Electron Configuration
Number of
valence electrons
Lewis Dot Symbol
H
1s1
1
H
Li
1s22s1
1
Li
1s2 2s2 2p6 3s2
2
Mg
B
1s2 2s2 2p1
3
C
1s2 2s2 2p2
4
C
N
1s2 2s2 2p3
5
N
O
1s2 2s2 2p4
6
O
F
1s2 2s2 2p5
7
F
Mg
211
B
Teacher Tip
Ask the learners to give the Lewis dot symbols
of some of the elements in the table shown at
the left..
The Lewis dot symbols of the representative elements are shown in the table below.
2. Ask the learners what common features exist for the dot symbols of the elements.
All elements belonging to the same group have the same number of valence electrons. They
have the same number of dots around the element symbol. For example, all Group 1A elements
have only one valence electron represented by one dot. All Group 7A elements have 7 valence
electrons represented by 7 dots.
II. IONS FORMED FROM REPRESENTATIVE ELEMENTS
Ionization energy and electron affinity have been discussed in earlier lessons. Review the learners
on these concepts.
212
Teacher Tip
The teacher can prepare a power point slide
or a poster to show the Lewis dot symbols of
the representative elements as shown in the
figure on the left.
Ionization energy (IE) is the minimum amount of energy (in kJ/mol) required to remove an electron
from a gaseous atom in its ground state.
Teacher Tip
Briefly discuss ionization energy and electron
affinity for the learners to recall the concepts.
Energy + X(g) —> X+ (g) + e–
Ionization energy increases from left to right across a period. It increases from bottom to top in a
group as shown in the figure below. Ask the learners the reason for the trend.
Increasing ionization energy
Increasing ionization energy
Examples
Li
1s22s1
[He] 2s1
Ionization energy trend based on the periodic table
Elements with small ionization energies tend to easily give up electrons to form positive ions or
cations. From the positions in the periodic table, these elements would be the metals particularly
those in Groups 1A and 2A. The larger the metal atom, the easier it is to lose valence electrons
and the more reactive the metal.
The electron affinity of an atom may be defined as the negative of the energy change that
occurs when a gaseous atom accepts an electron, or the ionization energy of a negative ion. The
more positive the electron affinity, the greater the tendency to accept an electron and form an ion.
Generally, the electron affinity increases across a period from left to right. The electron affinity
generally decreases going down a group.
213
Li+ +
1s2
[He]
e–
Na
→
1s22s22p63s1
[Ne] 3s1
Na+ + e–
1s22s22p6
[Ne]
Ca
→
[Ar] 4s2
Ca2+ + 2 e–
[Ar]
Increasing tendency to accept electrons
Examples
:F
+
e–
→
Increasing tendency to accept
electrons
1s22s22p5
[He] 2s22p5
:O
+
1s22s22p4
[He] 2s22p4
Electron affinity trends
Therefore, the elements on the right hand side of the periodic table, the non-metals, have a high
tendency to accept electrons and form negative ions, or anions. The smaller the nonmetal atom,
the greater the tendency to attract electrons, and the higher the reactivity of the nonmetal. Using
dot symbols and their electron configuration, the formation of the anions are shown in the
equations below.
Exercises
1. Give the charge and draw the Lewis dot symbol
a. of the anion formed when a sulphur atom accepts electrons.
b. of the cation formed when a Rb atom loses an electron
c. of the anion formed when a nitrogen atom accepts electrons
d. of the anion formed when an iodine atom accepts an electron
214
:F:
1s22s22p6
[Ne]
2e
→
: F :2
1s22s22p6
[Ne]
III. IONIC BOND FORMATION
From the previous section we saw that atoms with low ionization energies tend to form cations
while atoms with high electron affinities tend to form anions. These cations and anions combine to
form ionic compounds.
An ionic bond is “the electrostatic force that holds ions together in an ionic bond.” The
formation of the ionic compound, LiF, can be represented using the Lewis dot symbols.
Li
+
:F
→
Li+
:F
Ionic bonds are formed by the strong
interaction of ions.
:-
The two ions, Li+ and F-, now attract each other to form Li+F- or LiF.
Exercises
1. Using the Lewis dot symbol, show the ionic bond formation for Ca2+O2- or CaO.
Ca
+
[Ar]2s2
:O
→
1s22s22p4
Ca2+
: O :2-
[Ar]
[Ne]
2. Using the Lewis dot symbol, show the ionic bond formation for Na2O.
2 Na
[Ne]3s1
+
:O
→
1s22s22p4
2 Na+
[Ne]
Teacher Tip
Remember that the Bohr model is incorrect
and should no longer be used to show the
formation of ionic bonds. Do not use
electrons orbiting around the nucleus where
one is transferred to another atom also with
electrons orbiting the nucleus. This model is
not the true picture of the formation of the
ionic bond.
: O : 2[Ne]
IV. LATTICE ENERGY OF IONIC COMPOUNDS
The stability of the ionic compound depends on the strength of the interaction among all the ions
in an ionic solid. This stability is measured by the lattice energy of the compound. The lattice
energy is defined as the amount of energy required to completely separate one mole of a
solid ionic compound into gaseous ions.
215
Therefore, the stronger the interaction among the ions in the ionic compound, the harder to
separate them, the larger the lattice energy, the stronger the ionic bond.
The lattice energy is proportional to the product of the charges of the ions and inversely
proportional to r, the distance of separation between the ions (Coulomb’s law). In the case of LiF,
for example, the lattice energy is proportional to
Teacher Tip
The lattice energy can be calculated using the
Born-Haber cycle. However, this is beyond
the scope of the lesson. Nevertheless, it is
important to emphasize the meaning of lattice
energy and its relationship to the strength of
the ionic bond.
Relate lattice energy to the melting points of
ionic compounds.
where QLi+ and QF%!are charges of Li+ and F, k is the proportionality constant. Therefore, the higher
the ion charges, the stronger the bond; the shorter the distance between ions, the stronger the
bond.
The lattice energy is correlated to the physical properties of ionic compounds such as the melting
points. The larger the lattice energy, the harder to separate the ions, the higher the melting point.
Exercises
1. Which is expected to have a higher melting point? LiF or NaF?
Both ions have +1 and -1 charges but the distances between ions are different. Na+ is larger than
Li+. Therefore the internuclear distance in LiF is shorter; hence, LiF will have the higher melting
point.
2. Arrange the melting points of the following ionic compounds in decreasing order: LiF, LiBr, LiI,
and LiCl.
Because all ions have +1 and -1 charges, the internuclear distance will affect the melting point.
The order of melting points will be: LiF > LiCl > LiBr > LiI.
3. Which will have the higher melting point, NaCl or MgO? Explain.
Answer: MgO
216
V. PROPERTIES OF IONIC COMPOUNDS
Ionic compounds have the following general properties:
Ionic substances form crystalline solids. In the solid state, the ions are in rigid formation in
relatively fixed positions in a crystal lattice. This makes them immobile and poor conductors of
electricity and heat. However, when they melt or are dissolved in solution, they become good
electrical conductors. Note, however, that not all ionic solids are soluble in water.
Because of the strong electrostatic force of attraction among ions in the solid, they have high
melting points and high boiling points. (See discussion on lattice energy).
Ionic solids are hard and brittle. The electrostatic forces have to be overcome to move the ions
and shift them away from one another.
ENRICHMENT (5 minutes)
Ions in the human body
Ions play an important role in the body. Calcium, potassium, sodium, chloride, and copper ions are
some key ions involved in the electrical events inside the body. Potassium is the major positive ion
inside the cell, while sodium is the major positive ion found in the fluid outside the cell. Ionic chlorine
is the most abundant negative ion.
What will happen to our body if there is an imbalance of any of these ions or certain trace ions
in the body?
Imbalances of any of these ions, certain traces of ions in the body, or inhibition of sodium ion
transport across the cell membranes can lead to dysfunction in the conduction of electrical messages.
This dysfunction quickly leads to a general body disturbance and loss of ability to maintain somewhat
stable internal conditions.
EVALUATION/ACTIVITY (20 Minutes)
Evaluation will be based on the activity provided. See attached sheet.
217
ACTIVITY
Drawing Lewis Dot Symbols
Name: ___________________________________
I.
Section: _______________________________
Date: ____________________________________
Using a simple periodic table, accomplish the following table. Write the atomic number, electron configuration (long form), number of
valence electrons, and Lewis dot symbol for each of the elements listed below.
Element
Atomic Number
Electron Configuration Number of Valence Electrons
1. Fluorine
2. Phosphorus
3. Magnesium
4. Iodine
5. Carbon
6. Barium
7. Selenium
8. Silicon
9. Bromine
10. Aluminum
218
Lewis Dot Symbol
II. Complete the following table.
Sample
Element
Number of
electrons lost
Number of
electrons gained
Number of
electrons in the ion
Ion
Name of Ion
1
2
3
4
5
6
7
III. Draw the Lewis dot symbol for the ions in Part II.
Sample 1
Sample 2
Sample 3
Sample 4
219
Sample 5
Sample 6
Sample 7
General Chemistry 1
180 MINS
Lesson 25: Covalent Bonds and Lewis Structures
Content Standard
Lesson Outline
The learners demonstrate an understanding of covalent bond formation in
terms of atomic properties.
Performance Standard
The learners can identify covalent compounds and describe their general
properties.
Learning Competencies
At the end of the lesson, the learners:
1. Describe covalent bonding in terms of electron sharing (STEM_GC11CBIId-g-75)
2. Apply the octet rule in the formation of molecular covalent compounds
(STEM_GC11CB-IId-g-76)
3. Write the formula of molecular compounds formed by the nonmetallic
elements of the representative block (STEM_GC11CB-IId-g-77)
4. Draw Lewis structure of molecular covalent compounds (STEM_GC11CBIId-g-78)
5. Determine the polarity of a bond based on the electronegativities of the
atoms forming the bond (STEM_GC11CB-IId-g-80)
Specific Learning Competencies
At the end of the lesson, the learners will be able to:
1. Illustrate the formation of covalent bonds in terms of electron
sharing;
2. Apply the octet rule in forming covalent compounds;
3.
4.
5.
6.
Introduction
Communicating Learning Objectives
Motivation
Samples of Covalent Bonds
Instruction
and Practice
I.
II.
III.
IV.
V.
VI.
Enrichment
Assignment
Evaluation
Multiple Choice
10
5
Formation of the Covalent Bond
130
Electronegativity
Writing Lewis Structures
Lewis Structure and Resonance
Exceptions to the Octet Rule
Naming Covalent Compounds: A Review
5
30
Resources
(1) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York:
McGraw-Hill.
(2) Petrucci, Herring, Madura, and Bissonnette (2011). General Chemistry
and Modern Applications, 10th Ed. Pearson Canada, Inc.
(3) Zumdahl, S.S. and Zumdahl, S.A (2013). Chemistry, 8th ed. Cengage
Learning
(4) Chemical Education Digital Library http://www.chemeddl.org/
resources/models360/models.php?pubchem
7. Draw the resonance structures of covalent compounds;
8. Determine the polarity of a bond based on the
electronegativities of the bonding atoms;
Define electronegativity;
Describe the electronegativity trends in the periodic table;
9. Determine whether a bond is ionic, polar covalent, or covalent
based on the differences in electronegativities of the bonding
atoms; and
Draw Lewis structure of covalent compounds;
Identify lone pairs and bond pairs;
220
INTRODUCTION (10 minutes)
1. Introduce the following learning objectives using any of the suggested protocol (Verbatim, Own
Words, or Read-aloud):
At the end of the lesson, I will be able to:
a. Illustrate the formation of covalent bonds in terms of electron sharing.
b. Apply the octet rule in forming covalent compounds.
c. Define electronegativity.
d.
e.
f.
g.
h.
i.
Describe the electronegativity trends in the periodic table.
Draw Lewis structure of covalent compounds.
Identify lone pairs and bond pairs
Draw the resonance structures of covalent compounds
Determine the polarity of a bond based on the electronegativities of the bonding atoms
Determine whether a bond is ionic, polar covalent, or covalent based on the differences in
electronegativities of the bonding atoms.
2. Present the keywords for the concepts to be learned:
a. Lewis structure
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
n.
Covalent bond
Lone pair
Bond pair
Single bond
Double bond
Triple bond
Nonpolar covalent bond
Polar covalent bond
Electronegativity
Percent ionic character
Resonance
Incomplete octet
Expanded octet
221
3. Post on the board the following essential questions that will be answered after the discussion
a. How are covalent bonds formed?
b. Why are the electrons often unequally shared by the atoms in a covalent bond?
c. How do we represent covalent compounds?
d. How do we name covalent compounds?
MOTIVATION (5 minutes)
1. Bring to the class samples of covalent compounds. Tell the learners that they will be finding out
about how the atoms in these compounds are linked. This will be related to the assignment
activity at the end of the class.
Samples could be: water, naphthalene balls, sugar, acetone, ethyl alcohol, ammonia.
INSTRUCTION (130 minutes)
I.
FORMATION OF THE COVALENT BOND
It was Gilbert Lewis who suggested that the chemical bond is formed by sharing of electrons in
atoms. For the hydrogen molecule, this is depicted by
The two electrons are shared equally between the two atoms forming a covalent bond. The
bond is typically depicted by a single line, H - H. The electrons are attracted to the nuclei of
both atoms keeping the atoms together to form a molecule. Show the formation of the covalent
bond for the F2 molecule
222
Teacher Tip
Clearly illustrate for the learners the formation of
the covalent bond through electron sharing.
The representation of the covalent compound above is called the Lewis structure. In the Lewis
structure, shared electrons that form a bond is represented by a line or a pair of dots; lone pairs
are represented by dots above the atom. Only valence electrons are included in Lewis
structures.
a. From the Lewis structure of F2, how many electrons are around each fluorine atom in F2?
Answer: There are eight electrons fulfilling the octet rule. By sharing the electrons, each
fluorine atom fulfils the octet rule.
Note: The octet rule works mainly for elements in the second period (2s and 2p subshells
can hold 8 electrons). For hydrogen, only two electrons are needed to fulfill the noble gas
configuration.
b. How many bond pairs are there in the F2 molecule? Answer: One
c. How many lone pairs are there in the F2 molecule? Answer: Six lone pairs
d. Further illustrate the formation of the covalent bond in Cl2. How many bond pairs are there?
How many lone pairs?
e. Illustrate the formation of the covalent bond in HCl.
223
Exercises
1. Draw the Lewis structure for H2O, CH4 (methane), and for NH3.
2. Which of the three molecules has the largest number of bond pairs (covalent bonds)?
Answer: CH4 has four bond pairs, NH3 has three, and H2O has two.
3. Draw the Lewis structure for carbon dioxide, CO2.
The unpaired electrons of O
and C will pair up.
Note that there are 8
electrons around each of the
atoms of C and O fulfilling the
octet rule.
The unpaired electrons of N
will pair up.
The examples of CO2 and N2 show that there are different types of covalent bonds that are formed. Single bonds are formed when two atoms
are held together by one pair of electrons. Multiple bonds can be formed. A double bond is from the sharing of two pairs of electrons such as
in the case of O and C in CO2. A triple bond exists in N2 where the two N atoms are held by three pairs of electrons.
II. ELECTRONEGATIVITY
Consider the covalent bond in the molecule
Experimental evidence has shown that electrons are not equally shared between H and F; the
electrons spend more time near F rather than H. Therefore the electron density is shifted more
towards F rather than H. This leaves the F end of the molecule partially negative, δ - , and the
H end of the molecule partially positive, δ+. Such a bond is referred to as a polar covalent
bond. The polar covalent bond is somewhere between a purely covalent (nonpolar) bond and an
ionic bond (where there is almost complete transfer of electrons).
224
Teacher Tip
Ask the learners what they understand with the
word “polar”. What is the meaning of “polarized”?
Ask them to give examples of daily situations that
illustrate polarity. It is important that the learners
understand the meaning of the word rather than
just memorizing it.
Note that the F end of the H—F bond is partially
negative. Why not totally negative? What would it
mean if the F end is totally negative? Likewise, the
H end of the H—F bond is partially positive. There
are other electronegativity scales used in inorganic
chemistry such as the Muliken scale and the AllredRochow scale. What is important to note is that
A property that distinguishes the polarity of bonds is electronegativity, the tendency of an atom
in a chemical bond to attract electrons toward itself. Electronegativity is a theoretical concept
and devised as a relative scale. That is, it can be estimated relative to, or in comparison to,
other elements in chemical bonds. Linus Pauling developed a relative scale of
electronegativities which is widely used in General Chemistry textbooks. In contrast, ionization
energies and electron affinities are physically measurable properties of elements.
In general, electronegativity increases from left to right across a period. It increases as atomic
radius decreases. The most electronegative elements are those in the upper right hand side of
the periodic table with fluorine as the most electronegative. Metals especially the ones with
large atomic radii are the least electronegative.
The difference in the electronegativity values (!EN) of two bonded atoms determines the
percent ionic character of the bond. If the bond is between two identical elements, for
example F—F, then the bond is purely covalent with 0 percent ionic character. The difference in
electronegativity is 0.
Teacher Tip
Note that the F end of the H—F bond is partially
negative. Why not totally negative? What would it
mean if the F end is totally negative? Likewise, the
H end of the H—F bond is partially positive. There
are other electronegativity scales used in inorganic
chemistry such as the Muliken scale and the AllredRochow scale. What is important to note is that
unlike ionization energy or electron affinity,
electronegativity is a theoretical concept where a
“ranking” of elements in chemical bonds is
established in terms of their ability to attract
electrons. This is why different scales have been
used. The most commonly used scale in General
Chemistry is that devised by Linus Pauling.
Increasing electronegativity
8A
Increasing electronegativity
1A
2.1
2A
3A
4A
5A
6A
7A
1.0
1.5
2.0
2.5
3.0
3.5
4.0
0.9
1.2
1.5
1.8
2.1
2.5
3.0
0.8
1.0
1.3
1.5
1.6
1.6
1.5
1.8
1.9
1.9
1.9
1.6
1.6
1.8
2.0
2.4
2.8
3.0
0.8
1.0
1.2
1.4
1.6
1.8
1.9
2.2
2.2
2.2
1.9
1.7
1.7
1.8
1.9
2.1
2.5
2.6
0.7
0.9
2.0
2.2
0.7
0.9
Electronegativities of selected elements and electronegativity trend
225
For the molecule H—Cl, the difference in electronegativity is 0.9 showing that the bond is a polar
covalent bond. A 50% ionic character corresponds to EN=1.7. While there is no bond that is 100%
ionic, an electronegativity difference of 2.0 or greater is usually classified to be predominantly ionic.
When !EN ≥ 2.0, the bond is predominantly ionic.
Even without electronegativity values, it is possible to predict the polarity of a bond by examining
the position of the bonded elements in the periodic table.
Exercises
1. Classify the following bonds as ionic, polar covalent, or covalent. Explain your answers.
A. The C-C bond in H3CCH3
B. The K-I bond in KI
C. The C-F bond in CF4
D. The N-H bond in NH3
2. Arrange the flowing bonds according to increasing bond polarity: Cs to F, Cl to Cl, Br to Cl, Si to
C.
III. WRITING LEWIS STRUCTURES
1. The following guidelines are used in writing the Lewis structure of covalent molecules:
a. Draw a skeletal structure of the molecule putting bonded atoms next to each other. In
general, the least electronegative atom occupies the central position. H and F usually
occupy terminal (end) positions.
b. Count the total number of valence electrons from all the atoms in the structure. Add
electrons corresponding to the charge for negative ions; subtract electrons corresponding to
the charge for positive ions.
c. Distribute the valence electrons to the non-central atoms such that these atoms fulfill the
octet rule. Remaining electrons are assigned to the central atom. Remember that bonds are
equivalent to 2 electrons.
d. If the valence electrons are not enough, multiple bonds may be formed.
226
Teacher Tip
Different textbooks use different cutoffs for
classifying a bond to be ionic. Chemistry by
Chang and Goldsby uses the cutoff at 2.0 while
other textbooks use a value of EN greater than
1.7 to classify a bond to be ionic. This lesson
has chosen the cutoff value of 2.0.
It is important to emphasize that there is no
bond that is 100% ionic.
Exercises:
1. Write the Lewis structure for NCl3.
a. Skeleton structure is
!Cl!
Cl!!!!!!!!!!!!!N!!!!!!!!!!!Cl!
b. Count valence electrons:
N=5
3 Cl = 3 (7) = 21
Total = 26
c. Distribute the 26 electrons to the atoms such that they fulfill the octet rule. Bonds are
equivalent to 2 electrons. Check if all atoms have 8 electrons around them.
2. Write the Lewis structure of OCS. C is the central atom.
a. Skeleton structure is: O!!!!!!!!!!!C!!!!!!!!!!!S!
b. Valence electrons: 6 for oxygen, 4 for carbon, and 6 for sulfur = 16 electron
c. Distribute the 16 electrons such that all atoms obey octet rule.
d. The Lewis structure is
227
3. Write the Lewis structure of CN–.
a. Skeleton structure is C!!!!!!!!!!N
b. Valence electrons: 4 for carbon, 5 for nitrogen, and 1 for the negative charge = 10
c. Distribute the 10 electrons to the skeleton structure. The Lewis structure is:
4. Write the Lewis structure of the following molecules:
a. Ethylene, C2H4
b. Acetylene, C2H2
c. Carbon tetrachloride, CCl4
d. COBr2 (for the skeletal structure, C is bonded to O and Br atoms)
Teacher Tip
The octet rule works best for second period
elements. Be cautious in giving exercises. Limit
them to simple examples since there are too
many exceptions. Avoid elements beyond the
second period. Keep in mind that the Lewis
structure is a tool to use for understanding
bonding and properties.
IV. LEWIS STRUCTURE AND RESONANCE
1. Write the Lewis structure for the ozone molecule, O3.
a. The skeletal structure is
O!!!!!!!!!!!!O!!!!!!!!!!O
b. Valence electrons: 3 (6) = 18
c. Lewis structure is:
d. The Lewis structure can also be written as:
Note that the Lewis structure above has one double bond between oxygen atoms and one single
bond also with oxygen atoms. The double bonds are expected to be shorter than a single bond.
When measuring the bond lengths, we expect two values: one for the shorter double bond and one
for the longer single bond.
228
For a more accurate Lewis structure, we need to
determine the formal charges of the atoms in
the structure. Formal charges are not actual
charges. They are assigned following rules. The
best Lewis structure is chosen from structures
that comply with formal charge assignments.
Note that the discussion on formal charges is
not included in the learning objectives and not
included in the lesson. For this reason, keep the
examples simple.
However, experimental results show that there is only one bond length obtained for ozone. The
bond length is between that of a single bond and a double bond. This means the above Lewis
structure is not an accurate representation of ozone. In fact, we are unable to write the accurate
representation using either the first or second Lewis structure shown above.
To resolve this discrepancy, we represent the ozone molecule using the two structures presented as
follows:
Each of the above structures is called a resonance structure. The double sided arrow shows that
the structures are resonance structures. A resonance structure is one of two or more Lewis
structures for a molecule that cannot be represented accurately by only one Lewis structure.
What is the correct representation for ozone? The two resonance structures shown with the double
arrow.
2. Draw the resonance structures for the carbonate ion, CO32–. Practice with this ion. The correct
resonance structures are:
How many resonance structures will the NO3– have? Draw them.
229
Misconception
The molecule does not oscillate between the
two structures. We use the resonance structures
as a way to resolve the inability to represent
accurately the structure of a molecule. In the
case of ozone, the accurate picture is the
combination of the two structures.
V. EXCEPTIONS TO THE OCTET RULE
1. The octet rule works best for second-period elements. Hence there are many exceptions. They
fall into three categories:
a.
Incomplete octet
b.
Odd number of electrons
c.
Expanded Octet
Incomplete octet
An example of a molecule with incomplete octet is BeH2, beryllium hydride. Its structure is
There are only 4 electrons around Be and not 8. Boron and aluminum also form molecules with
incomplete octets.
Exercise
Draw the Lewis structure of aluminum triiodide, AlI3, showing the incomplete octet.
Molecules with Odd Number of Electrons
Examples are nitric oxide, NO, and dinitrogen dioxide, N2O. Their Lewis structures are shown
below. Because they have an odd number of electrons, the octet rule cannot be followed.
The odd numbered molecules are sometimes referred to as radicals. They are generally highly
reactive.
Expanded Octets
Atoms belonging to the second period cannot have more than eight valence electrons around the
central atom because they only have the 2s and 2p subshells. This is different for atoms of
elements in the 3rd period and beyond.
230
Misconception
The molecule does not oscillate between the
two structures. We use the resonance structures
as a way to resolve the inability to represent
accurately the structure of a molecule. In the
case of ozone, the accurate picture is the
combination of the two structures.
These elements have 3d orbitals that can participate in the bonding. Hence they can have more
than eight valence electrons around the central atom. An example is SF6, sulfur hexafluoride, with
the Lewis structure shown below. SF6 has 12 electrons around the central atom.
Another example is phosphorus pentafluoride, PF5, where the central atom has 10 electrons
around it.
VI. NAMING COVALENT COMPOUNDS: A REVIEW
1. Here are some guidelines for naming covalent compounds and for writing their formulas. For
binary compounds, state the name of the first element. The name of the second element ends
in –ide.
HF
Hydrogen fluoride
HI
Hydrogen iodide
SiC
Silicon carbide
231
Teacher Tip
For Lewis structures and 3-D representations of
molecules, please see Chemical Education
Digital Library http://www.chemeddl.org/
resources/models360/models.php?pubchem
Sources
From Lewis Symbols and Structures, Rice
University. Retrieved from https://
opentextbc.ca/chemistry/chapter/7-3-lewissymbols-and-structures/ (3 November 2016),
Creative Commons Attribution 4.0 International
License
2. Prefixes (as shown in the table below) are used to denote the number of atoms in the formula.
The prefix “mono” usually omitted for the first element in the formula.
No.
Prefix
No.
Prefix
1
mono -
6 hexa -
2
di -
7 hepta -
3
tri -
8 octa -
4
tetra -
9 nona -
5
penta -
10 deca -
Samples
CO
carbon monoxide
CO2
carbon dioxide
NO2
nitrogen dioxide
N2O4
dinitrogen tetroxide
CCl4
carbon tetrachloride
SF6
sulfur hexafluoride
ENRICHMENT (5 minutes)
1. This is an assignment to be submitted the following meeting.
2. Learners will look for at least 2 examples of covalent compounds that can be found in nature or
used in everyday life. They must include the following information:
a. Brief information about the use of the covalent compound
b. Chemical formula and chemical name of the covalent compound
c. Structure of the compound
232
Teacher Tip
This can be given as an assignment at the end of
the first meeting on covalent compounds. This
can be discussed in class the following meeting
so the learners can all participate.
EVALUATION (30 minutes)
4. The electron pair in a C - F bond could be considered…
Multiple Choice: Encircle the letter corresponding to the best
answer. You will be provided a simple periodic table.
a. Closer to C because Carbon has a larger radius and thus
exerts greater control over the shared electron pair
b. Closer to F because Fluorine has a higher electronegativity
than Carbon
1. Which element will have 5 electrons in its Lewis dot symbol?
a. Argon
c. Closer to C because Carbon has a lower electronegativity
than Fluorine
b. Boron
c. Carbon
d. An inadequate model since the bond is ionic
d. Phosphorus
e. Centrally located directly between the C and F
e. Sulfur
5. Considering the position of the elements in the periodic table
and their relative electronegativities and bond polarities, which
bond is longest?
2. Which of the following elements can only form one bond in a
Lewis structure?
a. O
a. carbon - Oxygen triple bond
b. C
b. carbon - Oxygen single bond
c. N
c. carbon - Carbon single bond
d. Al
d. carbon - Carbon double bond
e. H
e. carbon - Nitrogen triple bond
3. Write the correct Lewis dot structure for O2. Which statement
correctly describes the structure of the whole molecule?
6. Which bond is the strongest?
a. carbon - Nitrogen triple bond
a. There is a double bond and four lone pairs.
b. carbon - Nitrogen double bond
b. There is a double bond and six lone pairs.
c. carbon - Hydrogen bond
c. There is a single bond and four lone pairs.
d. carbon - Carbon triple bond
d. There is a single bond and six lone pairs.
e. carbon - Carbon single bond
e. There is a single bond, a double bond, and six lone pairs.
233
7. Predict qualitatively the relative bond lengths of the four single
bonds given below and arrange them from shortest to longest:
C-N
N-O
N-Si
10. Which element is the least electronegative?
a. Calcium
b. Cesium
O-O
c. Iron
d. Barium
a. O-O < N-SI < C-N < N-O
e. Potassium
b. O-O < N-O < C-N < N-Si
c. O-O < C-N < N-O < N-Si
11. Which of the following statements about resonance is true?
Resonance hybrids occur because a compound changes back
and forth between two or more resonance structures.
d. N-O < O-O < C-N < N-Si
e. N-Si < C-N < N-O < O-O
I.
8. Which of the following represents a non-polar covalent bond?
Resonance structures differ in the arrangement of electrons
but not in the arrangement of atoms.
II. Resonance hybrids contain delocalized electrons.
a. H-O
III. Resonance structures for a given compound always
contribute equally to the resonance hybrid.
b. C-N
c. C-C
IV. Resonance structures occur when there are two or more valid
Lewis structures for a given compound.
d. Li-F
e. S-O
V. Resonance hybrids are a composite of resonance structures.
9. Based on electronegativities, which of the following would you
expect to be most ionic?
a. I, II, V, VI
b. I, II, V, VI
a. N2
c. II, III, IV, VI
b. CaF2
d. II, III, V, VI
c. CO2
e. II, IV, V, VI
d. CH4
e. CF4
234
12. How many resonance forms will nitrate ion (NO3-) have?
Answer Key
1. D
2. E
3. A
4. B
5. C
6. A
7. B
a. -1
b. 0
c. 1
d. 2
e. 3
13. A list of non-metals is given below. Which elements cannot
exceed the octet rule? B Si N P O S F Cl
a. Si, P, S, Cl
b. B, N, O, F
c. O, S, F, Cl
d. B, Si, N, P
e. All eight elements can exceed the octet rule.
14. Write the singly bonded Lewis dot structure for BF3. Which of
the following statements best describes this structure?
a. It obeys the octet rule on all atoms.
b. It has less than an octet on at least one atom.
c. It has a lone pair of electrons on the boron atom.
d. It has less than an octet of electrons on all atoms.
e. It exceeds the octet rule.
235
8. C
9. B
10. B
11. D
12. E
13. B
14. B
General Chemistry 1
120 MINS
Lesson 26: Geometry of Molecules and
Polarity of Compounds
Lesson Outline
Content Standard
Introduction
Communicating Learning Objectives
5
The learners demonstrate an understanding of the properties of molecular
covalent compounds in relation to their structure.
Motivation
Models of Solid Geometrics
5
Instruction
and Practice
I. Molecular Geometry
II. Molecular Geometry of Sample
Molecules
III. Summary of Molecular Geometrics
90
Learning Competencies
Evaluation
Quiz
20
At the end of the lesson, the learners:
Materials
Periodic Table of Elements
Performance Standard
The learners can demonstrate the properties of molecular compounds resulting
from their structure.
1. Explain the properties of covalent molecular compounds in terms of their
structure (STEM_GC11CB-IId-g-79)
2. Describe the geometry of simple compounds (STEM_GC11CB-IId-g-81)
3. Determine the polarity of simple molecules (STEM_GC11CB-IId-g-82)
Specific Learning Competencies
At the end of the lesson, the learners will be able to:
1. Apply the Valence Shell Electron Pair Repulsion Theory to predict the
geometry of simple molecules.
2. Define dipole moment.
3. Predict the polarity of molecules.
236
Resources
(1) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York:
McGraw-Hill.
(2) Petrucci, Herring, Madura, and Bissonnette (2011). General Chemistry
and Modern Applications, 10th Ed. Pearson Canada, Inc.
(3) Zumdahl, S.S. and Zumdahl, S.A (2013). Chemistry, 8th ed. Cengage
Learning
(4) Chemical Education Digital Library at http://www.chemeddl.org/
resources/models360/models.php?pubchem!
INTRODUCTION (5 minutes)
1. Introduce the following learning objectives using any of the suggested protocol (Verbatim, Own
Words, or Read-aloud):
At the end of the lesson, I will be able to:
a. Apply the Valence Shell Electron Pair Repulsion Theory to predict the geometry of simple
molecules.
b. Define dipole moment.
c. Predict the polarity of molecules
2. Present the keywords for the concepts to be learned:
a. Molecular geometry
b. Valence Shell Electron Pair Repulsion Theory
c. Bond angle
d. linear
e. trigonal planar
f.
tetrahedral
g. trigonal bipyramidal
h. octahedral
i.
dipole moment
j.
polar bond
k. polar molecule
MOTIVATION (5 minutes)
1. Bring to the class models of different solid geometries such as a tetrahedron, a trigonal
bipyramid, and an octahedron. If models are unavailable, show drawings of these geometries
to the learners. Ask the learners if they know the names of these geometries.
2. There are some fruit juices being sold in tetrahedral packs in the supermarket. The learners can
be shown some of these.
237
INSTRUCTION and PRACTICE (90 minutes)
I.
MOLECULAR GEOMETRY
What is molecular geometry? Why do we need to know about the geometry of molecules?
Molecular geometry pertains to the three-dimensional arrangement of atoms in a molecule.
Geometry affects the physical and chemical properties of molecules and their reactivity towards
other molecules.
How can we know the geometry of a molecule?
Molecular geometry can be determined by experiment such as x-ray diffraction. However, the
geometry of simple molecules can be predicted even without experimentation. While the results of
the prediction is only qualitative and not as accurate as experiment, they still help in explaining the
properties of chemical substances.
What is the basis of the prediction?
The prediction rests on the assumption that all electron pairs in the valence shell around a
central atom repel one another. They want to be as far apart from one another as possible. These
valence shell electron pairs are the ones involved in bonding. They assume a geometry or
orientation that will minimize the repulsions. This is the stable orientation and the one with lowest
energy. This approach in predicting molecular geometry is called the Valence Shell Electron Pair
Repulsion Theory (VSEPR).
How do we apply the VSEPR theory to predict molecular geometry?
The key ideas of the VSEPR theory are:
1. Electron pairs stay as far apart from each other as possible to minimize repulsions.
2. Molecular shape is determined by the number of bond pairs and lone pairs around the central
atom.
3. Treat multiple bonds as if they were single bonds (in making the prediction).
4. Lone pairs occupy more volume than bond pairs. Lone pair-lone pair repulsions are greater than
lone pair-bond pair repulsions which in turn are greater than bond pair-bond pair repulsions
5. Molecular geometry is a very important concept. Ask the learners to explain the VSEPR theory
in their own words or even in Filipino. Make sure they are not memorizing the theory but can
explain it.
238
What are the common orientations of electrons pairs (bond pairs and lone pairs) that minimize
repulsions?
Number of
Electron Pairs
Orientation of Electron Pairs
2
Linear
3
Trigonal Planar
4
Tetrahedal
5
Trigonal bipyramidal
6
Octahedral
Is the orientation of the electron pair the same as molecular geometry?
They are not necessarily the same. The molecular geometry is determined by the position of the
nuclei of the atoms. We do not “see” lone pairs.
II. MOLECULAR GEOMETRY OF SAMPLE MOLECULES
For this lesson, the following notation is adopted: A refers to the central atom and X refers to
another atom bonded to it. If there are lone pairs attached to the central atom, this is indicated by
the letter E. Hence, AX2E2 means that A has two atoms of X bonded to it and A also has two lone
pairs of electrons.
1. Predict the molecular geometry of the molecule BeCl2. This is of the type AX2.
a. The first thing to do before we can predict the molecular geometry is to draw the Lewis
structure of the molecule. This is shown below:
b. How many bond pairs surround the central atom of Be? Two bond pairs surround Be.
c. How will two electron pairs orient themselves such that they will be as far apart from one
another as possible? Remember VSEPR Theory says they repel one another. To minimize
repulsion, the two electron pairs will be arranged in a linear arrangement as shown above.
239
Teacher Tip
To be able to visualize the molecular geometries, it
is highly recommended that models be used.
There are several commercial models that can be
purchased. However, in their absence, other
modelling materials can be used. These include
sticks for the bonds and modelling clay that can be
shaped into balls to represent atoms. Small
Styrofoam balls can also be used.
An activity using balloons as electron pairs can
easily illustrate molecular geometry.
Also refer to websites showing three-dimensional
models of molecules such as the Chemical
Education Digital Library http://www.chemeddl.org/
resources/models360/models.php?pubchem
d. What is the molecular geometry?
The molecular geometry is determined by the arrangement of the nuclei of the atoms in the
molecule. The molecular geometry of BeCl2 is linear.
e. What is the Cl-Be-Cl bond angle? It will be 180o.
2. Predict the molecular geometry of CO2. This is also of the type AX2 but with double bonds.
a. In determining molecular geometry, always start with the Lewis structure.
b. How many electron pairs are around the central atom of carbon? We have indicated earlier
that in applying the VSEPR theory, we will treat multiple bonds to be like single bonds.
Therefore, there will be two pairs around carbon.
c. What will be the orientation of the electron pairs: Answer: Linear
d. What will be the molecular geometry of CO2? Answer: Linear
e. What will be the O – C – O bond angle? Answer: 180o.
3. Predict the molecular geometry of the molecule BCl3. This is of the type AX3.
a. Again, the first step is to get the Lewis structure.
trigonal planar
240
b. How many bond pairs surround the central atom of boron? Three bond pairs surround B.
c. How will three electron pairs orient themselves such that they will be as far apart from one
another as possible?
To minimize repulsion, the two electron pairs will be arranged in a trigonal planar
arrangement as shown above.
d. What is the molecular geometry?
The molecular geometry of BCl3 is trigonal planar. This is a flat molecule as shown in the
figure on the right above.
e. What is the Cl – B – Cl bond angle? The bond angle is 120o.
4. Predict the molecular geometry of ozone, O3. This molecule is of the type AX2E.
a. Lewis structure
For predicting geometry, we may use only one of the resonance structures.
b. Number of electron pairs around central oxygen atom (treat multiple bonds as single
bonds): three electron pairs
c. Orientation of three electron pairs: trigonal planar
d. Molecular geometry: bent
We only use the positions of the nuclei of the atoms. We are unable to “see” the lone pair.
Therefore, the molecular geometry is bent.!!
The lone pair occupies more volume and pushes the bond pair closer. Therefore, the bond
angle is slightly less than 120o.
241
Teacher Tip
You may relate the significance of ozone. Briefly
ask the learners what they know about ozone. What
is its role in the environment? Do they know its
chemical formula? Do they know its structure?
5. Predict the molecular geometry of the molecule methane, CH4. This is of the type AX4.
a. Draw the Lewis structure of methane.
b. Methane has four bonding pairs of electrons around C.
c. The four bonding pairs will arrange themselves to be as far apart from one another as
possible. This is achieved through a tetrahedral arrangement where the four H atoms are at
the corners of a tetrahedron.
d. The molecular geometry is tetrahedral.
Teacher Tip
Ask the students if they are familiar with a
tetrahedral shape. The teacher can bring some
samples of tetrahedra to class. There are some fruit
juices being sold in tetrahedral packs in the
supermarket. The learners can be shown some of
these. Show the bond angles using models.
Source
The tetrahedral methane molecule. From
Introduction to Chemistry: General, Organic, and
Biological. Retrieved from http://
2012books.lardbucket.org/books/introduction-tochemistry-general-organic-and-biological/s15organic-chemistry-alkanes-and-.html (3 November
2016), Creative Commons by-nc-sa 3.0 license.
e. The H-C-H bond angle is 109.5o.
6. Predict the geometry and bond angles in ammonia, NH3. This molecule is of the type AX3E.
a. Draw the Lewis structure.
b. NH3 has three bond pairs and one lone pair around nitrogen.
c. The electron pairs are arranged in a tetrahedral orientation.
d. Since the lone pair is not considered, the molecular geometry is pyramid.
Teacher Tip
There are many other types of geometries.
However, at this level, it is important for the learner
to understand the geometries of the simple
molecules such as methane, water, ammonia and
relate their properties to structure. These will be
taken up in the next lessons.
Source
Valence Shell Electron Pair Repulsion, University of
Liverpool. Retrieved from http://
www.chemtube3d.com/VSEPRShapeNH3.html (3
November 2016, Creative Commons AttributionNoncommercial ShareAlike 2.0, UK.
e. Again, since the lone pair occupies more volume, it will push the bond pair in and the
resulting H-N-H bond angle is slightly less than 109.5. Experimental results show it is 107o.
242
7. Predict the molecular geometry of water, H2O. This is of the type AX2E2.
a. Draw the Lewis structure of water.
b. There are four electron pairs around the central atom: two bond pairs and two lone pairs.
c. The electron pairs are tetrahedrally oriented.
d. The molecular geometry is bent.
Source
Valence Shell Electron Pair Repulsion, Water,
University of Liverpool. Retrieved from http://
www.chemtube3d.com/VSEPRShapeH2O.html (7
November 2016, Creative Commons AttributionNoncommercial ShareAlike 2.0, UK.
e. Because there are two lone pairs occupying more volume and pushing in the bond pairs,
the H-O-H bond angle is less than 1200. Experiment shows this to be 104.5o. This is
smaller than the bond angle in NH3. Remember that lone pair-lone pair repulsions > lone
pair-bond pair repulsions > bond pair-bond pair repulsions.
8. Predict the molecular geometry of PF5. This is of the type AX5.
a. Draw the Lewis structure of PF5.
b. There are five electron pairs around phosphorus.
c. The orientation of the five electron pairs is trigonal bipyramidal.
d. The molecular geometry is trigonal bipyramidal.
e. The bond angles are 90o and 120o.
243
Source
Valence Shell Electron Pair Repulsion, Water,
phosphorus pentafluoride, University of Liverpool.
Retrieved from http://www.chemtube3d.com/
VSEPRShapePF5.html (7 November 2016, Creative
Commons Attribution-Noncommercial ShareAlike
2.0, UK.
9. Predict the molecular geometry of SF6. This is of the type AX6.
Source
Valence Shell Electron Pair Repulsion, Water, sulfur
hexaflouride, University of Liverpool. Retrieved
from http://www.chemtube3d.com/
VSEPRShapeSF6.html (7 November 2016, Creative
Commons Attribution-Noncommercial ShareAlike
2.0, UK.
a. Draw the Lewis structure of SF6.
b. There are six electron pairs around S.
c. The electrons pairs are oriented in an octahedral manner.
d. The molecular geometry is octahedral.
e. The bond angles are 90o and 180o.
III. SUMMARY OF MOLECULAR GEOMETRICS
SUMMARY TABLE OF MOLECULAR GEOMETRY
VSEPR Type
No. of Electron Pairs
Around Central Atom
No. of Bond Pairs
No, of Lone Pairs
AX2
AX3
AX2E
AX4
AX3E
AX2E2
AX5
AX4E
AX3E2
AX2E3
AX6
AX5E
AX4E2
2
3
3
4
4
4
5
5
5
5
6
6
6
2
3
2
4
3
2
5
4
3
2
6
5
4
0
0
1
0
1
2
0
1
2
3
0
1
2
244
Orientation of
Electron Pairs
linear
trigonal planar
trigonal planar
tetrahedral
tetrahedral
tetrahedral
trigonal bipyramidal
trigonal bipyramidal
trigonal bipyramidal
trigonal bipyramidal
octahedral
octahedral
octahedral
Molecular
Geometry
linear
trigonal planar
bent
tetrahedral
pyramidal
bent
trigonal bipyramidal
seesaw
T-shaped
linear
octahedral
square pyramidal
square planar
Examples
BeCl2
BF3
O3
CH4
NH3
H2O
PCl5
SF4
ClF3
I3
SF6
BrF5
XeF4
Exercises
1. Using the VSEPR theory, give the electron pair orientation and predict the geometry of the
following:
a. CH3I
b. SiH4
c. NF3
d. SCN– (C is the middle atom)
e. H2S
2. Give the bond angles for the molecules given in #1.
3. The molecule, acetone, has the following Lewis structure.
a. What is the geometry of the first carbon?
b. What are the bond angles around the first carbon?
c. What is the geometry of the middle carbon?
d. What are the bond angles around the middle carbon?
245
Dipole Moments and Polarity of Molecules
As earlier discussed in polar covalent bonds, the electrons are not equally shared by the bonding
atoms. Instead, there is a shift in electron density towards the move electronegative atom. Such is
the case with the bond in HF. This shift is symbolized by a crossed arrow (
) with the arrow
pointing toward the direction of the shift.
H#——#F
or
The polarity of the bond can be experimentally measured in terms of the dipole moment,
.
By definition, the dipole moment is the product of the charge, Q, and the distance between the
charges, r. To maintain neutrality, the charges on the ends of the molecule must be equal in
magnitude but opposite in sign.
Polar molecules exhibit dipole moments. In the presence of an electric field, the positive end of
the molecules orient themselves towards the negative plate. Nonpolar molecules have no dipole
moments. Remember to distinguish between polar bonds versus polar molecules. Some
molecules have polar bonds but are not polar. The molecular geometry determines whether the
molecule is polar or not.
The unit of the dipole moment is in terms of the Debye¸D¸"where 1 D = 3.336 x 10-30 C m.
Examples
1. Is carbon dioxide a polar molecule or not?
To answer this, first determine the Lewis structure followed by the molecular geometry. Then
determine the net dipole moment for the molecule.
246
2. Is ammonia a polar molecule? Is so, which is the partially negative end?
Again draw the Lewis structure and determine the molecular geometry. The molecular
geometry of ammonia is pyramidal. Because N is more electronegative than H, the N H
bond is polar with the N end as the more negative end. The three dipole moment vectors
point towards N. There is a resultant dipole moment. The nitrogen end is the - end.
Therefore, NH3 is a polar molecule; the N end is the - end.
3. Is NF3 polar? If so, which end is partially negative?
The molecular geometry os NF3 is pyramidal like NH3. However, in NF3 the F atoms are more
electronegative than N; therefore, the dipole moment vectors point towards the F atoms.
NF3 is a polar molecule; the F end is partially negative while the N end is partially positive.
4. Is water, H2O polar? Which is the partially negative end?
The molecular geometry of water is bent. The oxygen is more electronegative than
hydrogen; the dipole moment vectors point toward oxygen. Water is a polar molecule; the
oxygen end is the partially negative end. for H2O = 2.0967 D
5. Is carbon tetrachloride, CCl4, polar? What is the partially negative end?
CCl4 is nonpolar. The dipole moment vectors cancel each other out. for CCl4 = 0.
6. Determine the polarity of the following molecules. If polar, determine the partially negative
end.
a. CH4
b. H2CCl2
c. BF3
d. H2S
7. Arrange the following molecules according to increasing polarity: HF, HCl, HBr, HI.
247
Note
Dipole moments are vector quantities; they have
magnitude and direction. CO2 Has a linear
geometry. The C O bond is polar with the oxygen
end as the - end. But the dipole moment vectors
are equal in magnitude but point to opposite
directions. Hence, they cancel out each other. The
resultant dipole moment is zero. Therefore, CO2 is
a nonpolar molecule. This is an example where
you have a polar bond but the resulting molecule is
nonpolar.
Answer
The space-filling model for NH3 and its resultant
dipole moment equal to 1.9113 Debye pointing
towards the nitrogen end.
ENRICHMENT
There is a laboratory activity to illustrate bond polarity. This serves as enrichment.
EVALUATION (20 minutes)
1. Which statement correctly describes the Valence Shell Electron
Pair Repulsion (VSEPR) Theory?
3. If there are four (4) electron pairs around the central atom of a
molecule, these electron pairs are in a _________________
arrangement.
A. The valence shell electrons are given by the group number
in the periodic table.
A. linear
B. The valence shell electrons are the outermost electrons of
the atom that are involved in bonding.
B. trigonal planar
C. The valence shell electron pairs repel one another and tend
to stay as far apart as possible.
D. octahedral
C. trigonal pyramidal
E. tetrahedral
D. The valence shell electron pairs are the lone pairs of the
atom.
4. The geometry of the molecule NF3 is
E. A and B
A. linear
B. tetrahedral
2. It is important to know the geometry of a molecule because the
geometry _________________.
C. pyramidal
D. trigonal planar
A. will give the Lewis structure of the molecule
E. bent
B. affects the physical and chemical properties of the substance
C. will determine whether the molecule is ionic or covalent
5. The molecule boron trifluoride, BF3, assumes a trigonal planar
geometry with boron as the central atom. Which statement
correctly describes the polarity of the B-F bond and the polarity of
the molecule?
D. B and C
E. A, B, and C
A. The B-F bond is polar; the BF3 molecule is nonpolar.
B. The B-F bond is nonpolar; the BF3 molecule is nonpolar.
C. The B-F bond is nonpolar; the BF3 molecule is polar.
D. The B-F bond is polar; the BF3 molecule is polar.
E. The polarity cannot be determined.
248
6. The H - SH bond angle in H2S is approximately
9. If a compound has a polar bonds, then
A. 90o
B.
I.
180o
It is polar overall.
II. There is an electronegativity difference between the
bonded atoms.
C. 109.5o
III. It is ionic.
D. 120o
IV. It doesn't have resonance.
E. 360o
A. II only
7. Which of the following species will exhibit tetrahedral geometry?
B. II, IV
A. CCl4
C. I, II, IV
B. CO32 -
D. I, III
C. O3
E. All of the above statements are correct.
D. PCl5
E. SF6
10. Predict the geometry and polarity of the CS2 molecule.
A. linear, nonpolar
8. Which of the following types of molecules always has a dipole
moment?
B. linear, nonpolar
C. tetrahedral, polar
A. Linear molecules with two identical bonds.
D. bent, polar
B. Tetrahedral molecules (four identical bonds equally spaced).
E. bent, nonpolar
C. Trigonal planar molecules (three identical bonds equally
spaced).
D. A and B
E. None will have a dipole moment
249
General Chemistry 1
60 MINS
Lesson 27: Geometry of Molecules and Polarity of
Molecules (Laboratory)
Content Standard
The learners demonstrate an understanding of the properties of molecular
covalent compounds in relation to their structure.
Performance Standard
Lesson Outline
Introduction
Communicating Learning Objectives
Instruction
and Practice
Part I: Demonstration of Polarity of Water
Part II: Construction of Paper Models
The learners can demonstrate the properties of molecular compounds resulting
Evaluation
from their structure.
Learning Competency
Materials
Plastic cup with a hole at the bottom
PVC Pipe, Ball pen, Comb
Paper
At the end of the lesson, the learners:
1. Determine and/or observe evidence of molecular polarity (LAB)
(STEM_GC11CB-IId-g-83)
Resources
(1) https://www.youtube.com/watch?v=VhWQ-r1LYXY#t=39.
(2) https://mail.google.com/mail/u/0/#inbox/155df1bb449aae66?
compose=155dd64396ef0842&projector=1
(3) https://www.youtube.com/watch?v=phhVl-N9M4Y
Specific Learning Competencies
At the end of the lesson, the learners will be able to:
1. Demonstrate the polarity of water.
2. Construct paper molecular models
250
5
45
15
Teacher Tip
Prepare the laboratory materials needed before the
class.
INTRODUCTION (5 minutes)
1. Introduce the activity to the class citing the objectives.
Molecular geometry affects physical and chemical properties of molecules, one of which is the
polarity. For this reason, it is important to determine the geometry of molecules.
Always stress laboratory safety rules and proper conduct in the laboratory including proper
disposal of waste.
2. Safety Precautions
a. Never taste anything during a science activity.
b. Dispose of the samples and materials as directed by your instructor.
c. Wash your hands with soap and water after the activity.
d. Follow all laboratory instructions as directed by your instructor.
INSTRUCTION (45 minutes)
The activity has two parts:
PART I. DEMONSTRATION ON THE POLARITY OF WATER
Materials
1. Plastic cup with a small hole at the bottom or a plastic bottle of water with a hole at the bottom
2. PVC pipe
3. Comb
4. Ballpen or balloon
5. Receptacle to catch the water
Procedure
1. Assemble the plastic cup on a ring stand or holder about 12 inches above the table. Place a
receptacle beneath the cup to catch the flowing water.
2. Pour water into the plastic cup with a hole at the bottom.
251
If learners need to bring some materials, assign
these to them before the day of the activity.
Prepare the activity sheets for the learners.
3. Rub the PVC pipe or any of the given objects against your hair or any surface so that it becomes
charged.
4. Bring the charged pipe near the stream of water. Observe what happens to the stream of water.
5. Repeat the process by using other objects (ballpen, balloon and comb) and observe also what
happens to the stream of water.
Answer the following questions
1. What happens to the stream of water as the charged object comes near it?
2. Why is water attracted to a charged object?
After the learners perform the experiment and answer the questions, they may watch the video on
static electricity and water at https://www.youtube.com/watch?v=VhWQ-r1LYXY#t=39.
PART II. MODEL MAKING USING ORIGAMI
1. Watch the video on You Tube “How to Make Paper Tetrahedron –Origami Tutorial” at https://
mail.google.com/mail/u/0/#inbox/155df1bb449aae66?
compose=155dd64396ef0842&projector=1
2. Watch the video on You Tube “Seamless Octahedron” at https://www.youtube.com/watch?
v=phhVl-N9M4Y
3. Make at least 2 tetrahedra and one octahedron.
4. Make models of tetrahedrally shaped molecules like CH4, CHCl3, SiH4 by labeling the ends of
the tetrahedral with toothpick wrapped in colored paper or cellophane. Different colors
represent different atoms attached to the central atom. Learners can use their imagination in
labeling the atoms.
5. Make a model of an octahedrally shaped molecule like SF6 using the paper octahedron.
6. If time permits, make models of ethane, H3CCH3 where two tetrahedral are joined at one end.
252
Teacher Tip
Prepare data and observation tables for the activity.
EVALUATION (15 minutes)
1. Ask the students to accomplish the data and observation tables for the polarity of water.
2. Discuss the results in class. Ask the learners their observations and their conclusions.
3. Ask the students to submit all paper models. Show all the paper models to the class. Learners
may make models of other molecules.
4. Each learner should have constructed at least one paper model.
253
General Chemistry 1
180 MINS
Lesson 28: Carbon Compounds
Content Standard
Lesson Outline
The learners demonstrate an understanding of the properties of
organic compounds.
Introduction
Communicating Learning Objectives
5
Performance Standards
Motivation
Organic Compounds and Smells
5
The learners can illustrate the properties and uses of organic
compounds.
Instruction
I.
II.
III.
IV.
V.
Enrichment
Identifying the Functional Groups
Evaluation
Organic Compound Report
Learning Competencies
At the end of the lesson, the learners:
1. Describe the special nature of carbon (STEM_GC11OC-IIg-j-84)
2. List general characteristics of organic compounds
(STEM_GC11OC-IIg-j-85)
3. Describe the bonding of ethane, ethene, and ethyne and
explain their geometry in terms of hybridization and and
carbon-carbon bonds (STEM_GC11OC-IIg-j-86)
4. Describe the different functional groups (STEM_GC11OC-IIg-j-87)
The Valence Bond Theory
Hybridization of Atomic Orbitals
The Special Nature of Carbon
Organic Compounds: Hydrocarbons
Organic Compounds: Functional Groups
165
5
20
Resources
(1) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York: McGraw-Hill.
(2) Burdge, J. and Overby, J. (2012). Chemistry: Atoms First. New York: McGrawHill.
(3) Zumdahl, S.S. and Zumdahl, S.A (2013). Chemistry, 8th ed. Cengage Learning
(4) http://www.chemeddl.org/resources/models360/models.php?pubchem=222.
5. Cite uses of representative examples of compounds bearing the
different functional groups (STEM_GC11OC-IIg-j-88)
6. Describe structural isomerism; give examples (STEM_GC11OCIIg-j-89)
Specific Learning Outcomes
The Specific Learning Outcomes are detailed in the Introduction
Part of this lesson.
7. Describe some simple reactions of organic compounds;
combustion of organic fuels, addition, condensation,
saponification of fats (STEM_GC11OC-IIg-j-90)
254
INTRODUCTION (5 minutes)
1. Introduce the learning objectives by using the suggested protocol (Read-aloud):
At the end of the lesson, I will be able to:
a. Use the Valence Bond Theory to explain the hybridization of atomic orbitals and bonding in
covalent compounds.
b. Relate the molecular geometries and bond angles to the hybridization of atomic orbitals used
in bonding.
c. Describe the formation of sigma bonds and pi bonds.
d. Describe the bonding in ethane, ethene, and ethyne and other covalent compounds
containing single, double, and triple bonds.
e. Discuss the special nature of carbon and its ability to form compounds.
f.
Describe hydrocarbons, its properties and reactions.
g. Identify structural and geometric isomers.
h. Identify basic functional groups in organic compounds.
i.
Describe simple reactions of organic compounds.
Teacher Tip
There are many terms that the learner will
encounter in the lesson. To save time, place
the terms on the board, on a flip chart, on
posters, or on power point slides. Check the
terms as you encounter them in the lesson.
2. Present the keywords for the concepts to be learned:
a. Hybridization
b. sp3 hybrid orbitals
c. sp2 hybrid orbitals
d. sp hybrid orbitals
The learner will only be introduced to many of
the terms.
e. Sigma bonds
f.
Pi bonds
g. Organic chemistry
h. Hydrocarbons
i.
Alkanes
j.
Alkenes
k. Alkynes
l.
Cycloalkanes
255
m. Aliphatic hydrocarbons
n. Aromatic hydrocarbons
o. Saturated hydrocarbons
p. Unsaturated hydrocarbons
q. Straight chain hydrocarbons
r.
Branched hydrocarbons
s. Structural isomers
t.
Geometric isomers
u. Cis and trans
v. Functional groups
w. Alcohols
x. Aldehydes
y. Carboxylic acids
z. Esters
aa. Amines
bb. Amides
cc. Hydrogenation
dd. Condensation reaction
ee. Saponification reaction
MOTIVATION (5 minutes)
1. Introduce the learners to some organic compounds through their odor. Some organic
compounds have foul smell while others have a sweet or fresh smell. Ask the learners if they are
familiar with the scents of the compounds below. While these molecules are meant to motivate
the learners, their structures and bonding will be discussed in the lesson.
256
Organic Compounds and Smells
Putrescine (butane-1,4-diamine) and cadaverine (pentane-1,5-diamine) are foul-smelling organic
compounds found in decaying animals.
Putrescine
Cadaverine
NH2(CH2)4NH2
NH2(CH2)5NH2
Limonene is the source of the scent of lemons; vanillin gives the distinctive scent of vanilla;
cinnamaldehyde is the scent of cinnamon; and methyl salicylate is responsible for the smell of oil of
wintergreen
Limonene
Vanillin
C10H16
C8H8O3
Cinnamaldehyde
Methyl Salicylate
C9H8O
C8H8257
O3
INSTRUCTION/DELIVERY/PRACTICE (165 minutes)
I.
THE VALENCE BOND THEORY
Lewis structures and VSEPR theory provide simple descriptions of bonding in molecules. They treat
all bonding to be due to the pairing up of electrons which, if accurate, should provide similar
properties for bonds of the same type (e.g. all single bonds). However, experimental results show
that properties (such as bond energies, bond lengths, etc.) vary. These cannot be explained by the
simple Lewis structure.
A more accurate description of bonding comes from quantum mechanics. There are two quantum
mechanical theories of bonding: the valence bond (VB) theory and the molecular orbital theory. This
lesson will briefly touch on the valence bond theory and its application to bonding for the carbon
atom.
According to the VB theory, when two single atoms of hydrogen approach each other, there will be an
optimum distance between them where the attractive forces of the nuclei will be greatest and
repulsion will be least. In this state, the energy of the system is at a minimum (lowest). Therefore the
system is most stable in this state and we say that a bond has been formed, the H—H bond. VB
theory says that the bond is formed from the overlap of the s orbitals of the H atoms. Here, overlap
means that the electrons occupy a common region in space.
For the F2 molecule, the overlap is between the 2p orbitals of the F atoms. Remember that the
electron configuration of F is 1s2 2s2 2p5. A p orbital of F is partially filled. This will overlap with the
partially filled orbital of the second F atom.
258
For the HF molecule, the overlap is between the s orbital of hydrogen and the p orbital of fluorine.
Because different orbitals overlap, the differences in the properties of these bonds (e.g. bond length
and bond strength) can be explained by VB theory unlike the Lewis structures that treat all bonds
alike.
II. HYBRIDIZATION OF ATOMIC ORBITALS
sp3 hybridization
Consider the molecule CH4 where C is bonded to four H atoms in a tetrahedral geometry. The
valence electron configuration of C is
How can carbon form four bonds with hydrogen in CH4 when it only has two unpaired electrons?
Because the energy gap between the 2s and the 2p orbitals is small, one of the electrons in the 2s
orbital can be promoted to the 2p orbital as shown below.
Now the four unpaired electrons can form four bonds of different types: one bond will be the overlap
of the 1s orbital of hydrogen and the 2s orbital of carbon; the other three will be from the overlap of
the 1s orbital of H and the 2p orbitals of C. The 2p orbitals are 90o from one another. However,
experimental results show only one type of CH bond at angles of 109.5o and not 90o. To explain the
bonding in CH4, valence bond theory uses a theoretical concept of hybrid orbitals. Hybrid orbitals
are obtained from the combination or mixing of two or more nonequivalent orbitals of the same
atom. Hybridization produces hybrid orbitals which have the same energies.
259
Teacher Tip
Ask the learners to look up the meaning of
the term “hybrid”. Give examples of hybrids.
When one s orbital and three p orbitals are combined through hybridization, four equivalent sp3
hybrid orbitals result. These sp3 hybrid orbitals are tetrahedrally oriented. The shape of an sp3
orbital is not symmetrical; it has a larger probability on one side of the nucleus compared to the
other.
Source
sp3 hybridization from Electronic structure.
Retrieved from https://
jahschem.wikispaces.com/electronic+structure
(5 November 2016), Creative Commons
Attribution Share-Alike 3.0 License.
The formation of CH4 with the overlap of the 1s orbital of hydrogen with the four sp3 hybrid orbitals
of carbon are shown in the figure below.
Source
The CH4 molecule. From Hybridization.
Retrieved from http://
mrstinechemistry.wikispaces.com/covalent
+bonding (5 November 2016), Creative
Commons Attribution Share-Alike 3.0 License.
The CH4 molecule
The four sp3 orbitals are oriented towards the corners of a tetrahedron. The CH4 molecule is
tetrahedral with bond angles of 109.5o.
260
Note that other atoms also exhibit hybridization. NH3 is pyramidal and the N atom is sp3 hybridized.
The lone pair occupies an sp3 orbital. H2O is bent with bond angles close to 109.5. The O atom is
sp3 hybridized.
sp2 hybridization
Consider the bonding in BF3. What is the electron configuration of boron? Answer: 1s2 2s2 2p1.
How can boron form three bonds with fluorine in BF3 when it only has only one unpaired electrons?
Because the energy gap between the 2s and the 2p orbitals is small, one of the electrons in the 2s
orbital can be promoted to the 2p orbital as shown below.
The 2s and two 20 orbitals can be mixed to form three hybrid orbitals called the sp2 hybrid orbitals.
The sp2 hybrid orbitals have a trigonal planar orientation. Therefore, all are on a plane with
angles of 120o.
The three sp2 hybrid orbitals
261
Source
sp2 hybridization. from Electronic structure.
Retrieved from https://
jahschem.wikispaces.com/electronic+structure
(5 November 2016), Creative Commons
Attribution Share-Alike 3.0 License.
Ask the learners to draw the bonding in BF3 showing the overlap of the 2p orbitals of fluorine and the
sp2 orbitals in boron.
Describe the bonding in ethylene, C2H4.
From the Lewis structure we deduce that the geometry around each carbon atom in ethylene is
trigonal planar.
The valence electron configuration of carbon is
For trigonal planar planar, the hybridization used is sp2 as shown in the example for BF3. One electron
from the 2s orbital of carbon is promoted to the 2p. One 2s orbital and two 2p orbitals are mixed to
form the three sp2 orbitals leaving one unpaired electron in a 2p orbital.
Source
https://www.utdallas.edu/~scortes/ochem/
OChem1_Lecture/Class_Materials/
05_orbitals_hybrid_geom.pdf
Teacher Tip
Some textbooks refer to the end-to-end
overlap as head-to-head overlap or head-on
overlap.
There are hybridizations involving d orbitals.
However, they will no be discussed in this
lesson.
Side view of sp2 hybridized C atom showing the unhybridized p orbital
262
We now make a distinction between two types of covalent bonds in C2H4: the sigma (σ) bonds and
the pi (π) bonds.
Sigma bonds are formed by end-to-end overlap of the atomic orbitals with electron density
concentrated between the nuclei of the bonding atoms. Pi bonds, on the other hand, are formed by
the sideways overlap of orbitals with the electron density concentrated above and below the plane
of the nuclei of the bonding atoms. An end-to-end overlap is the most efficient way to bond
compared to a sideways overlap. Hence, sigma bonds are relatively stronger than pi bonds.
Source
Bonding orbitals in Ethene (ethylene) sp2.
Retrieved from http://www.chemtube3d.com/
orbitalsethene.htm (5 November 2016), by
Nick Greeves, Creative Commons AttributionNoncommercial-Share Alike 2.0 UK: England
& Wales License.
Formation of sigma and pi bonds in ethylene
How many sigma bonds are there in C2H4? Name them.
Answer: There are five sigma bonds in C2H4. These are the C-C bond and the four C-H bonds.
How many pi bonds are there in C2H4? There is only one pi bond. Note that a pi bond consists of
two lobes – one above the plane and another below the plane.
263
sp hybridization
Describe the bonding in ethyne (also called acetylene, C2H2).
From the Lewis structure we deduce that the geometry around each carbon atom in acetylene is
linear. The valence electron configuration about each carbon atom is
One electron from the 2s orbital of carbon is promoted to the 2p. One 2s orbital and one 2p orbital
are mixed to form the two sp orbitals leaving unpaired electrons in two 2p orbitals. The unhybridized
p orbitals are perpendicular to each other.
Source
http://cnx.org/contents/[email protected]/
Multiple-Bonds
The sp hybridized carbon showing the two unhybridized p orbitals
264
The hybridized sp orbitals of each carbon atom overlap end-to-end forming a σ bond. The
unhybridized p orbitals of each carbon atom overlap sideways forming two π bonds.
The bonding in C2H2 howing the formation of the σ and π bonds)
The table shows the geometrical arrangements of hybrid orbitals.
Atomic Orbitals
Hybrid Orbitals
Geometry
Bond Angle
s, p
sp
linear
180o
s, p, p
sp2
trigonal planar
120o
s, p, p, p
sp3
tetrahedral
109.5o
s, p
sp
linear
180o
Exercises
1. Determine the hybridization of each carbon atom (going left to right) in the following molecules:
a. H3C — CH3
Ask the learners to draw the Lewis structure.
Deduce the geometry around the carbon atoms.
From the geometry, determine the hybridization.
Answer: Both carbons have tetrahedral geometry. Therefore the hybridization of both carbon
atoms is sp3. All the bond angles are 109.5o.
265
Source
The bonding in C2H2 showing the formation of
sigma and pi bonds. From Multiple Bonds,
Advanced Theories of Covalent Bonding.
Retrieved from https://opentextbc.ca/
chemistry/chapter/8-3-multiple-bonds/ (5
November 2016), Creative Commons
Attribution 4.0 International License.
b. H3C — CHCH2
c.
H3C — C
Answer Key
1.
b. C1 is sp3 hybridized
C2 is sp2 hybridized.
C3 is sp2 hybridized.
C — CH3
d. H3C — CH = O
e. H2C = C = CH2
c.
C1 is sp3 hybridized.
C2 is sp hybridized.
C3 is sp hybridized.
C4 is sp3 hybridized.
d.
C1 is sp3 hybridized
C2 is sp2 hybridized.
e.
C1 is sp2 hybridized.
C2 is sp hybridized.
C3 is sp hybridized.
C4 is sp2 hybridized.
2. How many sigma bonds and pi bonds are in each of the molecules in #1?
3. What is the hybridization of N in NH3?
From the Lewis structure, we find that the geometry of NH3 is pyramidal with bond angles close to
109.5o. Therefore, the hybridization of N in NH3 is sp3.
4. What orbitals overlap in the formation of the O — H bond in H2O?
5. What orbitals overlap in the formation of the C — Cl bond in CH3Cl?
III. THE SPECIAL NATURE OF CARBON
About 200 year ago, organic chemistry was defined as the study of compounds produced by living
things like plants and animals. Organic compounds needed a ‘life force’ to be produced. Compounds
that were from nonliving things like rocks were referred to as inorganic. All these changed in 1828
with the experiment of Friedrich Wöhler. In his laboratory, Wöhler synthesized urea (an organic
compound) from ammonium cyanate (an inorganic compound). This marked a turning point in
organic chemistry. It dispelled the belief that organic compounds could only be formed by nature.
From the discussion on the bonding properties of carbon in the previous sections, it can be seen that
carbon has a unique nature
The electron configuration of carbon is 1s2 2s2 2p2. Carbon completes its octet by sharing electrons
and not by forming ions. It shares its electrons with other carbon atoms forming single, double, and
triple bonds. It also shares its electrons and readily forms bonds with atoms of other elements like O,
H, N, and the halogens.
266
2.
a.
b.
c.
d.
e.
7 sigma bonds; 0 pi bonds
8 sigma bonds; 1 pi bond
9 sigma bonds; 2 pi bonds
6 sigma bonds; 1 pi bond
6 sigma bonds; 2 pi bonds
3. sp3
4. 1s of H and sp3 of O
5. sp3 of C and 3p of Cl
The small radius of carbon allows it to approach another carbon atom closely, giving rise to short and
strong covalent bonds and stable compounds. Because the sp or sp2 hybridized orbitals can
approach each other closely, the unhybridized p orbitals containing unpaired electrons can form pi
bonds resulting in stronger bonds.
Carbon can form four covalent bonds. This allows it to form chains (straight, branched or cyclic) in
endless arrays.
Carbon can form millions of different compounds. To date, over 20 million organic compounds, both
synthetic and natural, are known compared with only about 100,000 inorganic compounds. Carbon
can form more compounds than any other element in the periodic table.
IV. ORGANIC COMPUNDS: HYDROCARBONS
A major group of organic compounds is the hydrocarbons. Hydrocarbons are made up of only
carbon and hydrogen atoms. Hydrocarbons are further classified into aliphatic hydrocarbons (those
that do not contain a benzene ring) and aromatic hydrocarbons (those that contain a benzene ring).
267
Teacher Tip
This section on organic compounds is meant
as an overview. Only the general concepts
are presented. Note that the time allotment
for this section in the learning competencies is
very short.
1. Alkanes
Alkanes have the general formula CnH2n+2 where n=1, 2, 3…. Alkanes only have single bonds.
They are also known as saturated hydrocarbons. They are referred to as saturated hydrocarbons
because they contain the maximum number of hydrogen atoms that can bond to the carbon
atoms present; that is, they are saturated with hydrogen atoms. In naming alkanes, the –ane suffix
(ending) is used. The name of the parent compound is determined by the number of carbon
atoms in the longest chain. Examples are ethane, propane, butane.
Only the first ten alkanes are named here to
show the use of prefixes. Ask the learners the
meaning of the prefixes and how they denote
the number of carbon atoms,
Tell the learners to note the ability of carbon
to from chains.
Prefixes Used in Naming Hydrocarbons
Number of Atoms
Prefix
Number of Atoms
Prefix
1
Meth-
6
Hex-
2
Eth-
7
Hept-
3
Prop-
8
Oct-
4
But-
9
Non-
5
Pent-
10
Dec-
The table below shows the first ten straight-chain hydrocarbons. Ask the learners the following:
a. Fill in the molecular formula and the structural formula of straight chain pentane up to decane.
b. How many bonds does each carbon atom have in the compounds?
c. What is the geometry of each carbon atom?
d. What is the bond angle around each carbon atom?
e. What is the hybridization of each carbon atom in hydrocarbons?
f.
Teacher Tip
It is not the objective of the lesson to learn
naming of organic compounds. Leave the
examples to the ten hydrocarbons
Describe the boiling points of the hydrocarbons as the number of carbon atoms increases and
the chain gets longer.
g. Which of the hydrocarbons will be gases at room temperature (Room Temperature = 25oC)?
Answer: methane, ethane, propane, butane
268
If you have a ball and stick model, show the
structures to the learners. Alternatively, the
learners can view the ball and stick models at
http://www.chemeddl.org/resources/
models360/models.php?pubchem=222.
Ask the learners if they are familiar with any of
the 10 hydrocarbons. In what way?
Number of Atoms
Prefix
Number of Atoms
Boiling Point
methane
CH4
-161.6oC
ethane
C2H6
-88.6oC
propane
C3H8
-42.1oC
butane
C4H10
-0.5oC
pentane
36.1oC
hexane
68.7oC
heptane
98.4oC
octane
125.7oC
nonane
150.8oC
decane
174.0oC
269
Structural Isomers
Isomers are different compounds that have the same chemical formula. There are two ways of writing
the structure of butane: n-butane (where n stands for normal) and isobutene. These are called
structural isomers. Structural isomers are molecules that have the same molecular formula but
different structures. Alkanes are described as having straight chains (such as n-butane) or branched
chains (such as isobutane).
n-butane
isobutane
straight chain
branched chain
Boiling Point =
-42.1oC
Teacher Tips
Ask the learners what is the meaning of “iso”.
In what lesson did they first meet this prefix?
(Iso comes from the Greek word meaning
“equal” or same.)
Some books refer to structural isomers as
constitutional isomers.
Boiling Point = -11.7oC
For alkanes, the number of isomers increases as the number of carbon atoms increases. While butane
has only 2 isomers, decane has 75 isomers and the alkane C30H62 has over 400 million possible
isomers. While many of these do not exist in nature, this illustrates how carbon forms more
compounds than any other element.
Exercise: Pentane has three structural isomers. Can you draw them?
270
Teacher Tip
Alkanes also exhibit stereoisomerism.
However, this lesson will not deal with this
topic given the time limit and scope.
Cycloalkanes
Alkanes whose carbon atoms are joined in rings are called cycloalkanes. They have the general
formula CnH2n. The simplest cycloalkane is cyclopropane.
cyclopropane
cyclobutane
cyclopentane
cyclohexane
Reactions of Alkanes
a. Under suitable conditions, alkanes undergo combustion reactions to produce carbon dioxide
and water.
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l)
b. Alkanes undergo halogenation reaction where one or more hydrogen atoms are replaced by
halogen atoms.
CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
methyl chloride
Under excess chlorine, the reaction proceeds further:
CH3Cl(g) + Cl2(g)
→
CH2Cl2(g)
+ HCl(g)
methylene chloride
271
Teacher Tip
Cycloalkanes exhibit different geometries
(called conformations) such as “chair” or
“boat”. However, these are outside the scope
of the lesson.
CH2Cl2(g) + Cl2(g) → CHCl3(g) + HCl(g)
chloroform
CHCl3(g) + Cl2(g)
→
CCl4(g)
+
HCl(g)
carbon tetrachloride
2. Alkenes
Alkenes are hydrocarbons that contain at least one carbon-carbon double bond. They are also
called olefins. Their formula is CnH2n where n = 2, 3…. Alkenes are classified as unsaturated
hydrocarbons as opposed to the alkanes which are saturated hydrocarbons. In naming alkenes, the –
ene suffix (ending) is used. The name of the parent compound is determined by the number of
carbon atoms in the longest chain. Examples are ethene, propene, butene.
Ethene (ethylene)
Molecular
Formula
C2H4.
-102.4oC
Propene (propylene)
C3H6
-48oC
1-butene
C4H8
-6.47oC
cis-2-butene
C4H8
4oC
Trans-2-butene
C4H8
1oC
Name
Boiling Point
272
Note that in butene, there are two molecules: one with the double bond in the first carbon, and the
other with the double bond in the second carbon. The numbers in the names refer to the lowest
numbered carbon atoms in the chain that is part of the CC double bond.
Geometric Isomers of Alkenes
Alkanes have single bonds. The CC single bonds are free to rotate with relatively small energies.
However, this is not true for CC double bonds which are restricted and cannot rotate without breaking
the bonds. Therefore, the placement of substituents is crucial to the properties of the molecules and
alkenes exhibit geometric isomers. In the cis isomer, two particular atoms or group of atoms are
adjacent to each other (same side of the double bond). In the trans isomer, the two groups are
across from each other. The cis and trans isomers exhibit distinctly different chemical and physical
properties.
Reactions of Alkenes
Alkenes are classified as unsaturated hydrocarbons which are compounds that have double or
triple bonds that enable them to add hydrogen atoms.
273
a. Addition Reactions: Unsaturated hydrocarbons commonly undergo addition reactions where
one molecule adds to another to form a single product.
Hydrogenation is an example of an addition reaction where hydrogen is added to
compounds containing double bonds usually in the presence of a catalyst.
Hydrogenation is very important in the food industry particularly for vegetable oils.
Alkenes also undergo addition reactions involving hydrogen halide, HX (where X is a halogen).
C2H4(g) + HX(g) → H3CCH2X(g)
C2H4(g) + X2 (g) → CH2XCH2X(g)
274
3. Alkynes
Alkynes contain at least one CC triple bond. They have the general formula CnH2n-2 where n = 2, 3,…
In naming alkenes, the –yne suffix (ending) is used. The name of the parent compound is determined
by the number of carbon atoms in the longest chain. Like the alkenes, the names of alkynes indicate
the position of the carbon-carbon triple bond. Examples are ethyne, propyne, butyne.
Name
Molecular
Formula
Ethyne (acetylene)
C2H2
propyne
C3H6
HC
C
1-butyne
C4H6
HC
CCH2CH3
2-butyne
C4H6
Boiling Point
H
H3C
C
C
C
H
-84.7oC
CH3
C
-23oC
8oC
CH3
27oC
Reactions of alkynes
a. Combustion
2C2H2(g) + 5 O2(g) → 4CO2(g) + 2H2O(l)
This reaction gives off a large amount of heat; thus its use in oxyacetylene torches for welding
metals.
b. Addition reaction
Hydrogenation: C2H2(g) + H2(g) → C2H4(g)
Reaction with halogens and hydrogen halides:
C2H2(g) + HX(g) → C2H2CHX(g)
C2H2(g) + H2 (g) → CHXCHX(g)
275
Teacher Tip
Ask the learners to give common uses of
acetylene.
4. Aromatic Hydrocarbons
Aromatic hydrocarbons are a class of hydrocarbons whose molecules contain a ring of six carbon
atoms (benzyl ring) attached.
Its simplest member is benzene, C6H6, with the following resonance structures:
The benzene structure is often written as:
! or!
The group containing benzene less one hydrogen atom (C6H5) is called a phenyl ring. Other
examples of aromatic hydrocarbons are shown below.
Toluene or
Methylbenzene
2-phenylpropane
naphthalene
276
Simple Reactions of aromatic hydrocarbons
a. Substitution reactions – an atom or group of atoms replaces an atom or group of atoms in
another molecule
V. ORGANIC COMPOUNDS: FUNCTIONAL GROUPS
Organic compounds may also be classified according to the functional groups they contain. A
functional group is a group of atoms that is largely responsible for the chemical behavior of the
parent molecule. Compounds containing the same functional groups undergo similar reactions.
277
Teacher Tip
These functional groups should not be
memorized by the learners. The purpose is
for them to be aware of the functional groups
in relation to common substances.
The teacher should provide the students a
copy of the table. In answering the exercises,
the students can be guided by the handout.
Common Functional Groups
Class
General
Formula
Alcohol
ROH
Carboxylic acid
RCOOH
Carboxyl group
RCOOR’
Ester group
RCHO
Carbonyl group
RCOR’
Carbonyl group
Ester
(R’=hydrocarbon)
Aldehyde
Ketone
(R’=hydrocarbon)
Amine
(R’, R” = H or hydrocarbon)
Amide
(R’, R” = H or hydrocarbon)
Functional Group
—O—H
Hydroxyl group
Amino group
RNR’R”
RCONR’R”
Amide group
278
Alcohols
Some common alcohols are shown below.
Methanol is the simplest alcohol. It is highly toxic and causes blindness. Ethyl alcohol is a common
solvent and starting material for various commercial uses. It is produced commercially by the addition
reaction of ethylene with water at high pressure and temperature. It is also produced from the
fermentation of sugar. An isomer, isopropyl alcohol, is commonly called rubbing alcohol. Ethylene
glycol is used as an antifreeze.
Ethyl alcohol can be oxidized by inorganic oxidizing agents to acetaldehyde and acetic acid.
279
Ethers
Ethers are usually prepared by a condensation reaction. A condensation reaction is characterized by
the joining of two molecules and the elimination of a small molecule, usually water.
Aldehydes and Ketones
The functional group in aldehydes and ketones is the carbonyl group. A common aldehyde is
formaldehyde. An aqueous solution of formaldehyde is used in the laboratory to preserve animal
specimens. A common ketone is acetone, which is mainly used as solvent for organic compounds
and as nail polish remover. Alcohols can be oxidized to produce aldehydes and ketones:
280
Carboxylic Acids
The functional group in carboxylic acids is the carboxyl group, -COOH. Among the common
carboxylic acids are formic acid, acetic acid, and butyric acid.
Carboxylic acids can be produced by the oxidation of alcohols and aldehydes.
Carboxylic acids also react with alcohols to produce esters.
CH3COOH + HOCH2CH2 — CH3COCH2CH3 + H2O
Acetic acid
ethanol
ethyl acetate
Esters
Esters are sued in flavoring and perfumery owing to their characteristic smells. The smell of many
fruits come from esters such as those given in the motivation section.
A common reaction of esters is saponification. In this reaction, an ester reacts with aqueous NaOH
solution to produce back the carboxylic acid and the alcohol. This reaction originates from
soapmaking. Soap (Latin “sapo”) was originally produced by the hydrolysis of fats.
281
ENRICHMENT (5 minutes)
Go back to the molecules in the motivation section.
1. Using the table provided, identify if any the functional groups present in:
a. Putrescine
b. Cadaverine
c. Cinnemaldehyde
d. Limonene
e. Vanillin
f.
Methyl salicylate
2. Which of the above molecules is a hydrocarbon?
EVALUATION
Organic Compound Report: This may be assigned by groups or individually.
1. Choose an organic compound and present it as a poster. You must indicate the following:
a. Chemical name and chemical formula of the organic compound
b. Structure of the organic compound (identify the functional group/s if any)
c. Properties of the organic compound (physical and chemical)
d. Use of the organic compound in everyday life
e. Effects to humans and other living things
f.
Precautions in using the compound (if any)
2. A rubric will be used to evaluate the learners’ work.
282
Teacher Tips
Materials that will students use can be colored
paper, bond paper and magazines.
Encourage students to be creative in
presenting their work.
Post the outputs on the walls around the room
or corridor for all students to read and learn.
CRITERIA
1 (NOT VISIBLE)
2 (NEEDS
IMPROVEMENT)
Few of the required
elements are clearly visible,
organized and well placed.
May be missing elements.
3 (MEETS EXPECTATIONS)
4 (EXCEEDS
EXPECTATIONS)
Most of the required
elements are clearly visible,
organized and well placed.
All of the required elements
are clearly visible, organized
and well placed.
The project needs
improvement in design,
layout or neatness.
The project has a nice
design and layout. It is neat
and easy to read.
The project has an excellent
design and layout. It is neat
and easy to understand the
content.
The student’s drawings are
not clear or relevant.
Few of the drawings and
graphics are clear and
relevant.
Most drawings and graphics
are clear and relevant.
Drawings and graphics are
clear and relevant.
Many grammatical, spelling
or punctuation errors.
A few grammatical, spelling
or punctuation errors.
Almost no grammatical,
spelling or punctuation
errors.
No grammatical, spelling or
punctuation errors.
REQUIRED
ELEMENTS
Missing most or all of the
required elements.
VISUAL CLARITY
AND APPEAL
The project needs
significant improvement in
design, layout and neatness.
DRAWINGS/
GRAPHICS
MECHANICS
283
General Chemistry 1
60 MINS
Lesson 29: Polymers
Lesson Outline
Content Standard
Introduction
Communicating Learning Objectives
5
The learners demonstrate an understanding of the properties of polymers in
terms of their structure.
Motivation
Polymers Used in Everyday Life
5
Instruction
I.
II.
III.
IV.
Learning Competencies
Evaluation
Recycling and Universal Recycling Codes
At the end of the lesson, the learners:
Resources
(1) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York:
McGraw-Hill.
(2) Zumdahl, S.S. and Zumdahl, S.A (2013). Chemistry, 8th ed. Cengage
Learning
(3) Joesten, Melvin & Hogg, John. The World of Chemistry. 2012.
Cengage Learning
(4) http://pslc.ws/macrog/maindir.htm
(5) https://www.nobelprize.org/educational/chemistry/plastics/
readmore.html
(6) http://www.cmu.edu/gelfand/k12-teachers/polymers/natural-syntheticpolymers/
Performance Standard
The learners can illustrate the reactions at the molecular level in enzyme action,
protein denaturation and separation of components in coconut milk.
1. Describe the formation and structure of polymers. (STEM_GC11OC-IIgj-91)
2. Give examples of polymers (STEM_GC11OC-IIg-j-92)
3. Explain the properties of some polymers in terms of their structure.
(STEM_GC11OC-IIg-j-93)
Specific Learning Competencies
At the end of the lesson, the learners will be able to:
1. Define polymers;
2. Give the basic classification of polymers;
3. Distinguish between addition polymerization and condensation
polymerization;
4. Describe the arrangement of polymers;
5. Relate the arrangement of polymers to their properties; and!
6. Illustrate the appropriate uses of polymers.
284
What are Polymers?
Making Polymers
Polymer Arrangements and Structures
Plastics and Polymers
40
10
INTRODUCTION (5 minutes)
1. Introduce the following learning objectives using any of the suggested protocol (Verbatim, Own
Words, or Read-aloud):
At the end of the lesson, I will be able to:
a. Define polymers.
b. Give the basic classification of polymers.
c. Distinguish between addition polymerization and condensation polymerization.
d. Describe the arrangement of polymers.
e. Relate the arrangement of polymers to their properties.
f.
Illustrate the appropriate uses of polymers.
2. Present the keywords for the concepts to be learned:
a. Polymers
b. Natural polymers
c. Synthetic polymers
d. Monomers
e. Macromolecules
f.
Addition polymerization
g. Condensation polymerization
h. Homopolymer
i.
Copolymer
j.
Isotactic
k. Syndiotactic
l.
Atactic
m. Plastic
n. Thermoplastic
o. Thermoset
285
MOTIVATION (5 minutes)
Connect to a real life situations. Ask the learners to give at least five (5) objects or materials that
they think are made of polymers. Ask the learners a brief description of polymers to check on prior
knowledge.
The following website called The Macrogalleria (http://pslc.ws/macrog/maindir.htm) is devoted
solely to polymers. This website is useful in studying polymers. The teacher may assign the
learners to visit the Macrogalleria website before the lesson. The website has several levels of
discussion. The learners can view the introduction, Level 1.
INSTRUCTION (40 minutes)
I.
WHAT ARE POLYMERS?
Polymers are large molecular compounds made up of many repeating units called monomers.
They can be natural or synthetic. They are sometimes called macromolecules because of their high
molar masses.
The word polymer comes from the Greek “poly” (meaning many) and
“mer” (meaning part or segment). Therefore a polymer means many parts. Polymers are generally
classified as synthetic or natural. Natural polymers occur in nature. Synthetic polymers are manmade and synthesized in the laboratory.
286
Notes
The site is developed and maintained by the
polymer department of the University of Southern
Mississippi. It uses the store of a modern shopping
mall to illustrate the variety of polymers and their
practical uses before allowing you to explore the
chemistry of the polymers, including structure,
properties, and synthesis, in as much detail as you
wish.
II. MAKING POLYMERS
How are polymers made? The chemical reaction by which the monomers are linked together to
form polymers is called polymerization. There are several types of polymerization reactions. The
basic types are the addition polymerization and the condensation polymerization reactions.
1. Addition polymerization
In addition polymerization, the entire monomer becomes part of the polymer. They involve
molecules with double bonds or triple bonds. Consider the formation of polyethylene, a stable
polymer used widely as packaging wrap. The polymerization reaction consists of three steps:
Step 1: Initiation
An initiator molecule is heated to produce free radicals. These free radicals have one unpaired
electron each and are therefore very reactive seeking other electrons from other molecules to pair
up with. The free radicals react with the ethylene molecule monomer breaking the double bond
and forming a new radical.
Step 2: Propagation
The radical will react with another ethylene molecule monomer. This is repeated many times over
and a long chain is formed.
Step 3: Termination
The process gets terminated when two long-chained radicals combine to form the polyethylene
polymer.
where (CH2 — CH2)n is a shorthand convention that represents n repeating units.
287
Polyethylene is an example of a homopolymer. This is a type of polymer where there is only one
type of monomer. Other examples of monomers used to produce polymers are shown in the table
below:
MONOMER
POLYMER
Tetrafluoroethylene
Polytetrafluoroethylene (Teflon)
Vinyl Chloride
Polyvinylchloride (PVC)
Polystyrene
Propene
Polypropene (or polypropylene)
In the examples given above, ethylene (CH2 = CH2) and tetrafluoroethylene (CF2 = CF2) are
symmetric monomers (the carbons have the same substituents) while vinyl chloride, styrene, and
propene are asymmetric monomers (the carbons in the monomer have different substituents).
The examples (polyethylene, polystyrene, polypropylene, and Teflon) are synthetic polymers.
288
Practice. Ask the learners the following:
Answer Key
1. Write the structure of the polymer, tetrafluoroethylene (up to 10 carbons in length).
2. Write the structure of polyvinylchloride (up to 10 carbons in length)
3. Write the structure of polystyrene (up to 10 carbons in length)
2. Condensation Polymerization
Condensation polymers are those formed through a condensation reaction; that is, where
monomers join together at the same time losing a small molecule like water as by-product.
Recall that when a carboxylic acid reacts with an alcohol, an ester is formed. Now when a
monomer which has two carboxylic acid groups, one at each end, is made to react with a second
monomer containing two –OH groups, one at each end, then many ester linkages are formed
resulting in a polyester.
The reaction of a dicarboxylic acid and a dialcohol to produce a polyester
For example, the polymer polyethylene terephthalate (PET or sometimes called PETE) is formed by
the reaction of terephthalic acid and ethylene glycol. PET is a polyester.
n
Polyethylene terephthalate (PET)
289
Teacher Tip
Emphasize that this type of chemical reaction
does not depend on the presence of carboncarbon double bond in the reacting molecules.
Instead, it requires the presence of two different
kinds of functional groups on two different
molecules.
If the dicarboxylic acid is made to react with a diammine (instead of a dialcohol), then a
polyamide is formed such as nylon.
The reaction of a dicarboxylic acid and a diamine to produce a polyamide.
nylon
III. Polymer Arrangements and Structures
Polymers can be arranged in a number of ways. The arrangement of the polymer
chains affects their properties such as whether they are stiff or rigid, crystalline or
amorphous.. For asymmetric monomers.
A linear polymer is a one where the arrangement of atoms is like that of a long chain.
This long chain is often referred to as the backbone. Atoms or small groups of atoms
attached to the long chain are called pendant atoms.
290
Teacher Tip
The polymer chains are very long. They can get
entangled with each other just like strands of
spaghetti pasta get entangled in a bowl of
pasta. These molecular level features are
manifested in the bulk properties of the
polymers.
The long chains are not stiff; they are flexible. They twist and bend and wrap around one another .
They become entangled like strands of spaghetti especially in the solid state. This makes the
polymer strong. When polymers are dissolved in solvents, the chains move very slowly compared to
small molecules. This is due to the entanglement of the long chain strands. Thus, polymers
dissolved in solvents can be more viscous than the pure solven.
The arrangement of the pendant atoms or pendant groups attached to the backbone gives different
properties to the polymer. Three distinct arrangements are observed: syndiotactic, isotactic, or
atactic.
The isotactic arrangement is where all the pendant groups or substituents (represented by R — ) are
on the same side of the polymer chain. They pack efficiently resulting in polymers with high melting
point, high crystallinity, and superior mechanical strength. A syndiotactic polymer chain is one
where the substituent group alternates from left to right of the asymmetric carbons. They pack less
efficiently than isotactic chains. In atactic polymers, the substituents occur randomly. Therefore,
they do not pack well. These polymers are rubbery, not crystalline, and relatively weak.
Industrial synthesis is capable of producing pure syndiotactic or only isotactic polymers through the
use of special catalysts.
Depending on synthesis conditions, the chains can also be branched. Branched chain polymers are
more flexible and less dense than straight chained polymers. For example, high density
polyethylene (HDPE) polymers are used for firm plastic bottles and containers while low density
polyethylene (LDPE) are used for plastic food bags and plastic wraps.
291
Syndiotactic
Atactic
Sometimes, the polymer chains are cross-linked as in the case of vulcanized rubber. Rubber is a
natural organic polymer formed by the addition of the monomer isoprene. In vulcanized rubber, the
polymer strands of isoprene are crossed linked or bridged by short sulfur chains. The cross-linked
polymers can be visualized by the following diagram:
292
The crosslinks tie or bind the polymer strands together. Therefore, when these crosslinked polymers
are heated, the strands cannot flow past each other. They do not melt or break apart.
Sometimes, there are two or more different monomers that are joined together to form a polymer.
Such is called a copolymer. Let us say that the two monomers are monomer A and monomer B.
These two monomers may be arranged in several ways in a polymer giving different physical
properties to the polymer:
-A-B-A-B-A-B-A-B-
Alternating copolymer
-A-A-B-A-B-B-A-B-A
Random copolymer
-A-A-A-A-A-B-B-B-B-B-
Block copolymer
Examples of copolymers are Saran wrap, styrene butadiene rubber (used for soles of shoes)
III. PLASTICS AND POLYMERS
What is the meaning of plastic? Are all polymers plastic? Are all plastic polymers?
The word ‘plastic’ comes from the Greek ‘plastikos’ meaning ‘to mold’. Generally, plastics refer to
synthetic polymers. Plastics are polymers but not all polymers are plastic.
Plastics are classified into two types: thermoplastics and thermosets. Thermoplastics are those
that keep their plastic properties: they melt when heated and harden when cooled. On the other
hand, thermosets are permanently “set” once they are formed. They cannot be melted or
293
reshaped; if enough heat is added, they will crack or become charred. Thermoplastic materials are
made of long linear polymer chains that are weakly bonded to each other. When heated, the bonds
are easily broken and the polymer chains easily glide past each other. Therefore, they are readily
remolded. On the other hand, thermosets are made up of linear chains that are cross-linked to one
another preventing the material from being melted and reformed.
ENRICHMENT
Show the universal recycling codes to the learners. Or hang a poster of the codes in the classroom
for the learners to examine. Ask the learners to bring a plastic product with any of the recycling
code listed below. Ask them to describe the material:
1. What is the sample polymer material used for?
2. What recycling code is printed on the sample?
Teacher Tip
This may be a class activity where the students
discuss the different polymer materials in class.
Alternatively, learners may submit a brief
written report describing the material. The
teacher may add a few other items to be
included in the report.
3. Describe the material in terms of hardness and/or flexibility.
4. Why are recycling codes used in the plastic materials?
Source
From Recycling codes. Retrieved from https://
is.muni.cz/el/1431/podzim2013/C7935/
Recycling_codes_WIKI_ENG.pdf (5 November
2016), Creative Commons Attribution-Share
Alike 3.0.
294
EVALUATION (10 minutes)
Multiple Choice: Choose the best answer. Encircle the letter corresponding to your chosen answer.
1. A large molecule that consists of many repeating units is called a
A. Monomer
B. Polymer
C. Polypeptide
D. Polyethylene
E. Polystyrene
5. In a ____________ reaction for the synthesis of a polymer, a small
molecule is removed from the reacting materials and produced
as a by product.
A. Elimination
B. Substitution
C. Addition
D. Condensation
E. Oxidation
2. Each unit that comes together to form a polymer is called a ___.
A. Monomer
B. Copolymer
C. Thermoset
D. Amino acid
E. Ethylene
6. An example of a natural polymer is
A. Nylon
B. Dacron
C. Polyethylene
D. Rubber
E. Lucite
3. The reactions in which polymers are produced are called _____
reactions.
A. Synthesis
B. Polymerization
C. Decomposition
D. Single displacement
E. Oxidation
7. When asymmetric alkenes polymerize (such as chloroethene to
produce PVC), the pendant group like chlorine can be arranged
in different ways. An arrangement in which the –Cl groups are
arranged on the same side of the polymer chain is called a(n)
___________ arrangement.
A. Atactic
B. Isotactic
C. Syndiotactic
D. Sporadic
E. Random
4. Which of the following is a requirement for monomers to be
able to undergo addition polymerization?
A. The presence of an oxygen atom
B. The presence of a lone pair
C. The presence of a carbon to carbon double bond
D. The presence of a nitrogen atom
E. The presence of an electronegative atom
295
8. Kevlar is a type of polymer which has a great resistance to tear.
Its application includes its use in making bulletproof vests and
radial tires. The two monomers used to make Kevlar are the
following:
10. This is one of the coding symbols adopted by the plastics
industry for packaging materials. What is the purpose of these
coding symbols?
A. To make recycling easier by making the identification of
the plastics easier
B. To create a code that shows how the polymer ranks in
tons produced per year
C. To indicate the hardness of the polymer.
D. To indicate the solvent to use for the degradation
process.
E. C and D
When these two monomers join
A. A molecule of CO2 will be eliminated and an ester
linkage will form
B. Nothing will be eliminated and the two monomers will
be linked through a hydrogen to hydrogen bond.
C. A water molecule will be eliminated and a peptide
linkage will be formed.
D. Nothing will be eliminated. The two monomers will be
linked by a carbon-carbon bond.
E. The two monomers cannot be joined.
11. A polyester can be formed from the reaction of
A. Dicarboxylic acids and dialcohols
B. Dicarboxylic acids and diammes
C. Dialcohols and diammines
D. Alkenes
E. Alkynes
9. Polymers that can be heated easily to form other shapes are
known as _______.
A. Thermoplastics
B. Malleables
C. Foams
D. Gels
E. Rubber
12. All units in a polymer need to be from the same monomer.
A. True
B. False
13. Thermoplastics can only be melted once.
A. True
B. False
14. All polymers are considered as plastics.
A. True
B. False
296
General Chemistry 1
120 MINS
Lesson 30: Biomolecules
Lesson Outline
Content Standard
The learners demonstrate an understanding of the properties of organic
compounds and polymers in terms of their structure.
Introduction
Communicating Learning Objectives
5
Motivation
The Food Pyramid
5
Performance Standard
Instruction
I.
II.
III.
IV.
Carbohydrates
Proteins
Nucleic Acids
Lipids
90
Learning Competencies
Enrichment
Additional Information
5
At the end of the lesson, the learners:
Evaluation
Quiz
The learners can illustrate the reactions at the molecular level in protein
denaturation.
1. Describe some biomolecules: proteins, nucleic acids, lipids, and
carbohydrates. (STEM_GC11OC-IIg-j-94)
2. Describe the structure of proteins, nucleic acids, lipids, and carbohydrates,
and relate them to their function (STEM_GC11OC-IIg-j-95)
Specific Learning Competencies
At the end of the lesson, the learners will be able to:
1. Distinguish the biomolecules of proteins, carbohydrates, nucleic
acids, and lipids;
15
Resources
(1) Chang, R. & Goldsby, K. (2016). Chemistry. (12th ed.). New York:
McGraw-Hill.
(2) Zumdahl, S.S. and Zumdahl, S.A (2013). Chemistry, 8th ed. Cengage
Learning
(3) Joesten, Melvin & Hogg, John. The World of Chemistry. 2012.
Cengage Learning
(4) http://pslc.ws/macrog/maindir.htm
(5) http://www.cnpp.usda.gov/sites/default/files/archived_projects/
FGPPamphlet.pdf
7. Describe the basic general features of the different levels of
protein structure;
2. Give the monomers involved in the formation of the
biopolymers of carbohydrates, proteins, and nucleic acids;
8. Explain the denaturation of proteins in terms of protein
structure;
3. Give examples of common monosaccharides, disaccharides, and
polysaccharides used in daily life;
4. Distinguish the properties of starch, glycogen and cellulose;
9. Describe the general features of DNA and RNA;
5. Define an amino acid and give the functional groups present in
the molecule;
11. Give the general features of a triglyceride.
10. Differentiate between saturated and unsaturated fatty acids; and
6. Describe the formation of the peptide linkage;
297
INTRODUCTION (5 minutes)
1. Introduce the following learning objectives using any of the
suggested protocol (Verbatim, Own Words, or Read-aloud):
2. Present the keywords for the concepts to be learned:
a. Carbohydrates
b. Monosaccharides
c. Disaccharides
d. Polysaccharides
e. Glucose
f. Fructose
g. Galactose
h. Hexoses
i. Sucrose
j. Starch
k. Glycogen
l. Cellulose
m. Amino acid
n. Glycine
o. Peptide bond
p. Dipeptide
q. Primary protein structure
r. Secondary protein structure
s. Tertiary protein structure
t. Quaternary protein structure
u. Denaturation
v. Nucleic acids
w. DNA
x. RNA
y. Nucleotides
z. Lipids
aa. Fatty acids
bb. Unsaturated fatty acid
cc. Saturated fatty acid
dd. Triglycerides
At the end of the lesson, I will be able to:
a. Distinguish the biomolecules of proteins, carbohydrates,
nucleic acids, and lipids.
b. Give the monomers involved in the formation of the
biopolymers of carbohydrates, proteins, and nucleic acids.
c. Give examples of common monosaccharides, disaccharides,
and polysaccharides used in daily life.
d. Distinguish the properties of starch, glycogen and cellulose.
e. Define an amino acid and give the functional groups present
in the molecule.
f.
Describe the formation of the peptide linkage.
g. Describe the basic general features of the different levels of
protein structure.
h. Explain the denaturation of proteins in terms of protein
structure.
i.
Describe the general features of DNA and RNA.
j.
Differentiate between saturated and unsaturated fatty acids.
k. Give the general features of a triglyceride.
298
MOTIVATION (5 minutes)
1. Ask the learners if they are familiar with the food pyramid.
2. Show the food pyramid to the class and label the food pyramid. Discuss the food pyramid.
3. Which group will predominantly be carbohydrates?
4. Which group will predominantly be proteins?
5. Which group will predominantly provide triglycerides?
Source
USDA Food Pyramid. Retrieved from https://commons.wikimedia.org/wiki/
File:USDA_Food_Pyramid.gif (5 November 2016), Creative Commons AttributionShareAlike License.
Mention to the learners that the lesson will be on biomolecules: carbohydrates, proteins, nucleic
acids, and lipids which are all found in the food pyramid.
299
INSTRUCTION (90 minutes)
I.
CARBOHYDRATES
What are carbohydrates?
Carbohydrates are compounds made up of carbon, hydrogen, and oxygen. They are also known as
saccharides. They have the general formula Cx (H2O)y. Carbohydrates function as the energy source of
the body.
A simple general classification of carbohydrates is according to the number of sugar units (saccharides)
present in the molecule: monosaccharides, disaccharides, and polysaccharides.
Classification of carbohydrates
300
Teacher Tip
The lesson will not dwell on the different
forms of glucose (α-glucose and
β-glucose) as well as on the different ways
of representing sugars (linear and ring
forms). These are beyond the scope of the
lesson. The purpose of the lesson is to
serve as an introduction to different types
of biomolecules.
The learners can be shown the structures
of the carbohydrates to see commonalities
and differences. However, please do not
require them to draw the structures.
Learners are also not expected to
memorize the formulas of the
carbohydrates. The learners are only
expected to understand that
carbohydrates are natural polymers
formed by linkages of monomers. They
should be familiar with the general
features of the carbohydrate molecule.
Monosaccharides: their formula, structure and sources
MONOSACCHARIDE
FORMULA
Glucose
C6H12O6
Fructose
C6H12O6
Galactose
C6H12O6
STRUCTURE
SOURCES
Fruits
Fruits
Honey
Not naturally
occurring
Ask the learners to look at the table above and answer the following questions:
1.
2.
3.
4.
What is the formula of glucose? What is the formula of fructose? What is the formula of galactose?
What do you call compounds with the same formula but different structures?
What is the difference between the structure of glucose and the structure of fructose?
How many carbon atoms do glucose, fructose, and galactose have?
Answer: They have 6 carbon atoms. Hence they are called hexoses.
5. What functional groups are present in glucose, fructose, and galactose? Ask the learners to point
them out.
Answer: Alcohol groups and ether group.
301
When two monosaccharides join together through a condensation reaction, a disaccharide is produced
along with a molecule of water. For instance, when two glucose units react via condensation reaction,
the disaccharide, maltose, is formed.
Glu$+$Glu$$!$$$Glu$—$O$—$Glu$$$$+$$$$H2O$
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$maltose$
When glucose and fructose are joined, sucrose (table sugar) is formed.
Glu$+$Fru$$$!$$$$Glu$—$O$—$Fru$$$$+$$$$H2O$
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$Sucrose$$
When glucose and galactose are joined, lactose (milk sugar) is formed.
Glu$+$Gal$$$!$$$$Glu$—$O$—$Gal$$$$+$$$$H2O$
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$lactose!
The reactions are condensation (dehydration) reactions shown below:
Source
Maltose formation and sucrose formation.
Retrieved rom https://
commons.wikimedia.org/wiki/
File:Maltose_formation_and_sucrose_form
Creative
Commons Attribution-ShareAlike
License.
ation.png (5 November 2016),
Disaccharides: their formula, structure and sources
DISACCHARIDE
FORMULA
STRUCTURE
SOURCES
Sucrose
C12H22O11
Glu - O - Fru
Lactose
C12H22O11
Glu - O - Gal
Milk
Maltose
C12H22O11
Glu - O - Glu
Germinating grain
Sugar cane
sugar beet
Polysaccharides are polymers of monosaccharides. Examples of polysaccharides are starch, glycogen,
and cellulose which are all polymers of glucose. Their general properties are given in the table below.
Ask the learners what they notice about the names of carbohydrates?
Answer: They end with –ose.
Starch is found in plants and used as source of energy. Starch consists of two types of glucose polymers:
amylose and amylopectin which differ from each other in structure. Amylose which consists of about
20% of all starch polymers has the simpler structure.
Source
From A Closer Look at Carbohydrates, An
Introduction to Nutrition. Retrieved from
http://2012books.lardbucket.org/books/
an-introduction-to-nutrition/s08-01-acloser-look-at-carbohydrates.html (5
November 2016),
Creative
Commons by-nc-sa 3.0 license.
303
The structure shown may be represented more simply by
Glu$$4$$(OGlu)n$$4$$OH$
where n is very large (about 1000 glucose units linked together by oxygen bridges). The long chains of
starch are tightly wound in a coil. When amylose reacts with water through hydrolysis, these oxygen
bridges break releasing glucose units.
Most of starch is made up of amylopectin, the molecules of which are even larger than amylose. Unlike
the more linear amylose, the amylopectin molecule consists of amylose molecules that are linked by
oxygen bridges from the end of one amylose unit to a site in another amylose chain.
Starch is not found in animals. Instead, animals use glycogen for energy storage. Glycogen has shorter
chains than starch and is more highly branched. Because of the shorter chains and branching, glycogen
is more readily hydrolyzed than starch
Like amylose, cellulose is a straight chain polymer consisting of glucose units. However. In cellulose. The
glycosidic links between the glucose units in cellulose alternate in direction. This produces a long,
straight and rigid molecule. There is no branching in cellulose. Cellulose the major structural material
which plants are made of. Wood is largely cellulose. Humans cannot digest cellulose but animals can.
304
II. PROTEINS
Proteins are natural polymers. They are very large molecules that are critical for the functions of the
human body. They are made from the linkage of monomers called amino acids. Amino acids have the
following structure:
Notice that amino acids have two functional groups. What are they?
a. Carboxylic group (—COOH)
b. Amine group (—NH2)
There are 20 kinds of amino acids depending on the —R group. The simplest amino acid is glycine
where R is a hydrogen atom. The body cannot make all the amino acids required by the body and is
dependent on protein taken through food.
Two amino acids can link together through a condensation reaction with the removal of a water
molecule. The linkage is called a peptide bond. Take the case of two amino acids reacting to form a
dipeptide.
Teacher Tip
There is no need at this point for the learners
to memorize the 20 amino acids. However, it
is important for them to realize that because
there are 20 amino acids, there will be
millions of different ways to connect them to
form peptides.
When many amino acids are linked together through peptide bonds, the resulting molecule is called a
polypeptide. A very large number of amino acids linked together results in a protein. Some proteins
are made up of only one polypeptide while most proteins involve assemblies of two or more
polypeptides. The term polypeptide is usually used for shorter, unstructured chains while proteins fold
into fixed structures.
It is therefore possible to think of a protein as a strong of beads strung together where each bead is an
amino acid.
Different Levels of Protein Structure
The sequence and the structure of proteins are important in determining their functions. There are
four levels of protein structures: the primary structure, secondary structure, tertiary structure, and
quaternary structure.
306
Teacher Tip
There is no need at this point for the learners
to memorize the 20 amino acids. However, it
is important for them to realize that because
there are 20 amino acids, there will be
millions of different ways to connect them to
form peptides.
The primary structure refers to the linear sequence of amino acids joined by peptide bonds such as
the sequence of amino acids below.
Gly
Ile
Val
Glu
Gln
Cys
Gly
Cys
Ala
Ser
Val
Cys
Ser
Leu
The single bonds in the polypeptide chain allow rotation. Therefore, the polypeptide chain can twist
and fold in a variety of ways. These folded structures are referred to as secondary protein structures.
The two essential secondary structures are the alpha helix and the beta pleated sheets. The
structures are stabilized by hydrogen bonds between amino acids. In the alpha-helix structure, the
chain twists like a corkscrew while the chain takes the form of a folded sheet in the beta pleated
structure. Several secondary structures come together forming tertiary structures. When several
tertiary structures come together, a quaternary protein structure is formed. For example, the protein
hemoglobin is a quaternary structure formed by four tertiary structures.
Teacher Tip
In the diagrams, the amino acids are
abbreviated as follows: Gly (glycine), Ile
(isoleucine), Val (valine), Glu (glutamic acid),
Gln (glutamine), Cys (cysteine), Ala (alanine),
Ser (serine), Leu (leucine). Again, the
learners should not memorize these amino
acids. The enumeration is given only to
clarify the diagram.
Protein Functions
Proteins are important molecules in cell. Each protein in the body has a specific function. Some of the
types of proteins and their functions are :
a. Antibodies - proteins involved in defending the body against antigens. They are the molecules of
the immune system.
b. Contractile proteins – responsible for body movement such as muscle contraction
c. Enzymes – proteins that catalyze (speed up) or facilitate biochemical reactions
d. Hormonal proteins – serve as messenger proteins to help coordinate some body functions. An
example is insulin (which controls blood sugar concentration).
e. Structural proteins – are fibrous and provide support. An example is collagen which provides
support to connective tissues.
f. Storage proteins – store amino acids like casein in milk.
g. Transport proteins – are carrier proteins which move molecules from one place to another in the
body. An example is hemoglobin which transports oxygen.
Protein Denaturation
Denaturation is a process in which a protein loses its secondary, tertiary, or quaternary structures. This
may be caused by physical or chemical agents like strong acid, base, inorganic salt, heat, or solvent
which disrupt the bonds that hold the protein structures together. Denaturation does not cause the
cleavage of the peptide bond (the primary structure). Note that a protein will lose its biological activity
if it loses its 3-dimensional shape.
Examples of Protein Denaturation
DENATURING AGENT
Heat
Acids and Bases
Heavy metal ions like Ag+,
Pb2+, Hg2+
Organic compounds
Mechanical agitation
EXAMPLES
Cooking food (boiling egg, frying an egg)
Acid denatures milk proteins in the preparation of cheese
Mercury and lead poisoning
Chemicals used in hairstyling or hair straightening or hair
curling
Preparation of whipped cream or meringue from egg whites
308
III. NUCLEIC ACIDS
Nucleic acids are natural polymers with very large molar masses. The two main types of nucleic acids
are deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). DNA and RNA are polymers made up
of monomers in the form of nucleotides. When these nucleotides combine, they form
polynucleotides. DNA carries the genetic information passed on from parents to children. RNA plays
an important role in protein synthesis.
Each nucleotide is made up of three parts:
1. A nitrogen base
a. Purines - adenine (A), and guanine (G)
b. Pyrimidines - cytosine (C) , thymine (T), uracil (U)
2. A five-carbon sugar
a. Deoxyribose
b. Ribose
3. Phosphate
Features of DNA and RNA
Function
Structure
Sugar used
Bases used
DNA
Repository of genetic information
Double helix
deoxyribose
C,T,A,G
RNA
Involved in protein synthesis
Usually single-strands
Ribose
C, U, A, G
IV. LIPIDS
Lipids are a class of naturally occurring organic compounds distinguished by their solubility in an
organic solvent (and not in water). Lipids are hydrophobic, nonpolar, and made up mostly of
hydrocarbon chains. Some of the more important lipids are: triglycerides (fats), saturated and
unsaturated fatty acids, waxes, phospholipids, and steroids. Some of these are further discussed
below.
309
1. Fatty Acids
a. Fatty acids consist of a long hydrocarbon chain (typically about 12 – 18 carbons) attached to a
carboxyl group.
i.
Saturated fatty acid – contains only single C-C bonds because the carbon atoms are saturated
or filled up with hydrogens. Because their structure is straight, they can pack well and are solid
at room temperature (e.g. fat in butter).
ii. Unsaturated fatty acids – contain carbon-carbon double bonds. When there is only one C-C
double bond, it is called monounsaturated; if there are several C-C double bonds, they are
called polyunsaturated. Remember that when there are double bonds, there will be
geometric isomers (cis and trans). Because of the double bonds, they do not pack as tightly as
saturated fatty acids. They are usually liquids at room temperature. An example of an
unsaturated fatty acid is olive oil.
b. Fats and Oils
Fat molecules have two parts: a glycerol backbone and three fatty acid chains. The resulting
molecule is called a triglyceride.
Glycerol
3 fatty acids
Triglyceride
310
ENRICHMENT
The process of denaturation is used as an antidote to lead or mercury poisoning. Egg whites can be
given to the person who has ingested a heavy metal. The egg whites are denatured by the heavy
metals and a precipitate is formed. Then, vomiting is induced to eliminate the metal-protein
precipitate.
EVALUATION (15 minutes)
1. What elements make up a carbohydrate?
A. hydrogen, calcium, oxygen
B. hydrogen, carbon, oxygen
4. __________ is a carbohydrate that cannot be digested by
humans.
A. Cellulose
B.
C.
D.
E.
C. carbon, potassium, oxygen
D. carbon, magnesium, hydrogen
E. nitrogen, carbon, oxygen
2. _________ is known as “animal starch”.
A. Glucose
B.
C.
D.
E.
Starch
Glucose
Fructose
Maltose
5. Long chains of sugars are called _________.
A. Polypeptides
Cellulose
Fructose
B.
C.
D.
E.
Glycogen
Lactose
3. ____________ is the monosaccharide found in starch.
A. Glucose
B. Fructose
Polysaccharides
Polynucleotides
Polyunsaturated
Dipeptide
6. One function of a carbohydrate is _______________.
A. To provide the body with immediate energy
B. To keep the heart functioning smoothly
C. Maltose
D. Lactose
E. Amylose
C. To store and transport genetic material
D. To control the rate of the biochemical reactions
E. To regulate the body’s metabolism
311
7. The carbohydrate that provides support in plants is called
__________.
A. Chitin
B.
C.
D.
E.
11. Biopolymers formed from the linkage of monomers in the form
of nucleotides are called
A. nucleic acids
B.
C.
D.
E.
Dextrose
Lipids
Cellulose
Amylose
8. Glucose, galactose and fructose are __________.
A. Disaccharides
B. Isotopes
C. proteins
D. nucleic Acids
E. polysaccharides
13. Fats and oils are composed of what two groups of molecules?
A. glucose and fructose
9. The small repeating units that make up proteins are called
__________.
A. Fatty acids
B.
C.
D.
E.
Amino acids
Monosaccharides
Ethylene
Styrene
starch and sugar
water and cellulose
glycerol and fatty acids
RNA and DNA
14. Which of the following is a polymer of glucose?
A. starch
B. glycogen
10. The sequence of amino acids in a polypeptide is called the
protein’s __________.
A. primary structure
B.
C.
D.
E.
lipids
proteins
12. Enzymes are _______
A. monosaccharides
B. lipids
C. Polymers
D. Isomers
E. Amines
B.
C.
D.
E.
carbohydrates
rubber
C. cellulose
D. A nd B
E. A, B, and C
secondary structure
tertiary structure
ouarternary structure
crystal structure
312
15. What kind of molecule is represented by the structure below?
19. The structure on the left is a(n) __________ and the structure on
the right is a(n) ________.
CH3CH2CH2CH2CHCHCH2CH2CH2CH2CH2CH2CH2COOH
A. A sugar
B. A disaccharide
C. A dipeptide
D. A saturated fatty acid
E. An unsaturated fatty acid
A.
B.
C.
D.
E.
16. Table sugar is a form of
A. protein
B. lipid
C. carbohydrate
D. nucleic acid
E. steroid
B.
C.
D.
E.
dehydrated
denatured
plasmolyzed
folded
Nucleotide, carbohydrate
Lipid
Fatty Acid
Nucleic Acid
Protein
Answer Key
1. B
2. D
3. A
4. A
5. B
6. A
7. D
8. D
9. B
10. A
18. The group of biologically important organic compounds
responsible for storage and transfer of information is
A. carbohydrates
B.
C.
D.
E.
Carbohydrate, amino acid
Nucleotide, amino acid
20. DNA is a ________
A. Carbohydrate
17. When a protein is boiled, it loses all levels of organization
except the primary level. When this happens, the protein is said
to be:
A. hydrolyzed
B.
C.
D.
E.
Lipid, polypeptide
Carbohydrate, lipid
phospholipids
polypeptides
nucleic acids
polysaccharides
313
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
A
C
D
E
E
C
C
D
A
D
Biographical Notes
WYONA C. PATALINGHUG, PH.D.
Team Leader
VIC MARIE I. CAMACHO
Writer
Dr. Wyona C. Patalinghug served as a Professor of Chemistry at
the De La Salle University for 26 years. She finished her doctoral
and master’s degrees in Chemistry at the University of Hawaii;
and her bachelor’s degree in Chemistry at the College of the
Holy School, Manila where she finished Cum Laude. She is a
member of the ASEAN University Network Quality Assurance
Council (2013-2016) ; Lead Assessor of the ASEAN University
Network Quality Assurance in Education (2007-2016); Director of
the International Quality Assurance Office DLSU (2013-2015);
Director of the Institutional Assessment and Accreditation, DLSU
(2009-2013); Vice Chancellor for Research DLSU (2007-2008); and
Chair of the Chemistry Department DLSU (2001-2005).
Prof. Vic Marie I. Camacho is currently finishing her doctoral
degree in Science Education at the Philippine Normal University,
and finished her master’s degree in Chemistry from the De La
Salle University under a DOST ESEP Scholarship Grant. Prof.
Camacho is currently an Associate Professor 1 at the Philippine
Normal University, where she served as the Head of the Centre of
Science for the Mathematics and Technology. She served as an
Author and Technical Contributor for Sci-Tech Magazine, LET
Reviewers, and of various Learning Resource Materials for
Advanced Physical Sciences. She has actively participated in
Institutional, National, and International conferences and
workshops.
Dr. Patalinghug held consultancy work at the ASEAN University
Network and Asian Development Bank (2014-2016); Department
of Education (2015-2016); and Quality Assurance and
Accreditation (2015-2016).
She received the Award for Continuing Excellence and Service
from the Metrobank Foundation (2009), the Juan Medrano
Service Award (2008). the National Research Council of the
Philippines Lifetime Achievement Award in Chemistry (1998),
Metrobank Foundation’s Outstanding Teacher (1994), British
Council Exchange Scientist from the University of Southampton
UK (1995), and a JSPS Exchange Scientist at Waseda University,
Sophia University, and Tokyo University. She has a wide
experience in materials production, teaching, assessment, quality
assurance, and a deep appreciation of Chemistry.
FORTUNATO B. SEVILLA III, PH.D.
Writer
Dr. Fortunato Sevilla III is a professor in the chemistry department
at the University of Santo Tomas. Dr. Sevilla’s research specializes
on instrumentation and analytical science including the
development of optical chemical sensors and biosensors,
piezoelectric chemical sensors based on molecularly imprinted
polymers and conducting polymers, and low-cost designs of
chemical equipment and instruments for chemical education. Dr.
Sevilla received his master’s and doctoral degree in
Instrumentation and Analytical Science from the University of
Manchester Institute of Technology. He received the G.F.
Kirkbright Award from the University of Manchester and the
Gregorio Zara for Applied Science Philippine Association for the
Advancement of Science Award in 1996.
MARIA CRISTINA D. SINGSON
Writer
Ms. Maria Cristina D. Singson has been teaching Chemistry at
Pasay City West High School since 2007. She graduated with a
degree in Bachelor in Secondary Education, Major in Chemistry
from the Philippine Normal University Manila, and a degree in
Master of Arts in Science Education, specializing in Chemistry
from the same institution. She received various awards in the
production of materials and in Science Teaching, including the
1st Place Winner for Strategic Intervention Material for 3rd Year
High School Science (Division Level). She has attended
workshops on developing metacognitive readers across levels,
investigatory project making, and robotics.
MARLENE B. FERIDO, PH.D.
Technical Editor
Dr. Marlene B. Ferido is a Science Education Specialist V and the
Chair of the Chemistry Group of the University of the Philippines
National Institute for Science and Mathematics Education
Development (UP NISMED). She served as Deputy Director for
Research and Extension of UP NISMED for six years and as
Deputy Director for Administration for three years.
She served as UP NISMED’s coordinator in the development of
the Curriculum Guide of Science for K to 10, in collaboration with
the Department of Education. She is one of the writers of
DepEd’s Grade 7 and Grade 8 Learner’s Modules and Teacher’s
Guides in Science.
She was one of the curriculum developers for Physical Science for
the Senior High School curriculum and for the Chemistry portion
of the Grades 11 & 12 Senior High School Science, Technology,
Engineering & Mathematics (STEM) strand of the K to 12
program. She was appointed as one of the Curriculum Program
Leaders for of the Assessment, Curriculum, and Technology
Research Centre, a a partnership between the University of
Melbourne and the University of the Philippines, supported by
the Australian government.
JANETH M. FUENTES, PH.D.
Technical Editor
Dr. Janeth Morata-Fuentes is a Special Science Teacher IV and
the Curriculum Development Coordinator of the Philippine
Science High School System - Department of Science and
Technology. Dr. Fuentes graduated as doctoral degree in
Chemistry Education from the University of the Philippines Open
University, and with a master’s degree in Education (Teaching and
Curriculum Studies) from the University of Sydney, New South
Wales, Australia.
She was recognised as an Outstanding Educator by the Ministry
of Education, Singapore, and a 2002 Australian Development
Scholarship Awardee of the Australian Agency for International
Development. She also received the Metrobank Outstanding
Teacher Award in 2010, and the 2011 Presidential Lingkod Bayan
Award from the Civil Service Commission Honor Awards
Program.
She has authored and edited books for Integrated Science for
Grade 7, and served as a resource person and speaker to various
local, national, and international teacher and student training
seminars and workshops.
PATRICIA MARIE W. BAUN
Copyreader
Ms. Patricia Marie W. Baun graduated with a degree in Bachelor
of Arts in Communication Arts from the De La Salle University
(Cum Laude), and received the Gawad Magaaral Service Merit
Award. She worked as a Project Manager of the Clark
International Motor Show 2015, Creative Director of Breakfast
Magazine, and Production Trainee of Star Cinema. She is highly
interested in photojournalism, media communication, whole
brain self mastery, and on Philippine media.
JUAN MIGUEL M. RAZON
Illustrator
Mr. Juan Miguel M. Razon graduated with a degree in Bachelor
of Science in Management and Bachelor of Science in
Information Technology Entrepreneurship, Minor in Literature
from the Ateneo de Manila University. He worked at IBM
Philippines and contributed in the ideation and implementation
of the intranet-based “knowledge hub” for the employees of
IBM. He also served as the Finance Commissioner of the Ateneo
Commission on Elections and the Vice President for Public
Relations for Ateneo Kaingin. He intends to pursue a long-term
career in business intelligence, corporate finance, and graphic
design.
RACHELLE ANN J. BANTAYAN
Illustrator
Ms. Rachelle Ann J. Bantayan is a full time graphic designer at
Kalibrr and in charge of designing company collaterals, online
and offline ads, infographics and graphics. She also served as the
Graphic Artist at Edlir Pharma where she did the layout for the
employee handbook, design and layout for advertisements.
She also served as a contributing illustrator for The Fat Kid
Inside, Wordplay, FringeMNL, Tripda, and Woman, Create. She
finished her Bachelor of Fine Arts, Major in Information Design;
and Bachelor of Science, Minor in Management at the Ateneo de
Manila University.
DANIELLE CHRISTINE QUING
Illustrator
Ms. Danielle Christine Quing graduated with a degree in
Bachelor of Arts in Multimedia Arts at the De La Salle College of
Saint Benilde. She served as the illustrator and developer for elearning of Green Jakobsen and Rakso CT. She also participated
in the Philippine Centre for Creative Imaging Workshop in 2016,
and the Icon Manila Workshop in 2015. She won the Successful
Coffee Table Book Design: Case Studies on Award-Winning Book
Design at the Fiera de Manila 18th Graphic Expo in 2013.
RENAN U. ORTIZ
Illustrator
Mr. Renan U. Ortiz is a teacher and visual artist who has
collaborated in local and international art exhibitions such as the
SENSORIUM at the Ayala Museum, Populus in Singapore,
Censorship_2013 Move On Asia in South Korea, and the Triumph
of Philippine Art in New Jersey, USA. Mr. Ortiz’s solo exhibitions
include versereverse at the Republikha Art Gallery. He first
completed his bachelor’s degree in Political Science at the
University of the Philippines Manila before finishing his bachelor’s
degree in Fine Arts major in Painting at the University of the
Philippines Diliman. Mr. Ortiz is an awardee of the Cultural
Center of the Philippines’ CCP Thirteen Artists Awards in 2012.
IMAGES
Download