Hello TDCIans!
Welcome to Inorganic and Organic Chemistry.
This course is intended to equip students of allied medical science and
science majors like you to deal with the structure, composition, properties,
and reactions of inorganic and organic materials.
At the end of the course you will be expected to:
1. explain and define the different terms encountered in the subject,
such as the different reactions, principles, and concepts;
2. apply the concepts in a real setting;
3. perform laboratory experiments related to theoretical aspect;
4. practice professionalism and proper attitude during laboratory
experiments and discussions.
LET’S BEGIN!
Chemistry is the study of matter and the changes it undergoes. Matter
surrounds us, thus, so as chemistry. Chemistry is said to be the central
science that connects the other branches of scientific fields such as, biology,
geology, ecology, medicine, forensics, and many other branches (Flowers et.
al, 2019). It is an ancient science that began even before the institution of
civilizations and founded intellectual and technological advances we now
study and enjoy.
Greeks, through Aristotle, used to believe that every matter is made up
of only four elements: water, earth, fire, and air. With this, they also
concluded that matter can only exhibit four properties: hot, cold, dry, and
wet.
Have you also read The Alchemist by Paulo Coelho? It is a novel about a
boy who left his life as a shepherd to find a treasure buried near the
Pyramids in the Egyptian desert. Along his way, he met different kinds of
people, and one of them is an alchemist. In the 17th century, the alchemists
believed that metals can be converted to gold, which they called the
Philosopher’s Stone. They also believed in the Elixir of Life, which they said
can cure any disease and can lengthen a person’s life. However, do you
believe that alchemy is one of the facets that led to where chemistry is now?
Although the practice of alchemy is now considered a pseudoscience, many
alchemists were able to discover several elements and the methods of
preparing mineral acids.
1
In 1661, chemistry was truly recognized as a science through the
quantitative experiment of Robert Boyle, which studied the relationship of
pressure and volume of air. Along with his contribution to gas laws, Boyle
was the one to establish the definition of an element, which he said is a
substance that cannot be broken down into simpler substances (A Brief
History of Chemistry, 2019). This breakthrough led to the discovery of more
elements.
Currently, chemists are working together with the other members of the
health and medical field in fighting against COVID-19. Therefore, as the
future and hope of your chosen field, learn as much as you can, keeping in
mind that this knowledge will equip you in serving the people better.
UNIT 1: Inorganic Chemistry
Intended Learning Outcomes
At the end of the unit, you are expected to:
1. discuss the definitions of inorganic chemistry and organic
chemistry, and differentiate them from each other; and
2. describe the contribution of inorganic chemistry to society.
Introduction
Inorganic Chemistry focuses on all elements except carbon. It is usually
concerned with chemicals such as paints, coating, fertilizer, surfactants,
disinfectants, and even food, beverage, and feeds. In this unit, you will be
introduced to Inorganic Chemistry and its contribution to modern society.
Unlocking of Difficulties
Inorganic Chemistry is just one of the five main branches of Chemistry.
However, Chemistry is an extremely broad field, covering various scientific
fields. To fuel your interest in Chemistry, and to understand its main
concepts,
read
the
article,
“What
is
Chemistry?”
(https://www.livescience.com/45986-what-is-chemistry.html).
2
Lecture Notes
CHEMISTRY DEFINED
Chemistry is the scientific study of:
- nature, composition, structure, and properties of matter
- physical and chemical changes which matter undergoes
- utilization of the different substances
Figure 1: Branches of Chemistry
(https://chemistrygod.com/chemistry-and-its-branches)
In this course, the focus will be on the two branches of chemistry,
Inorganic Chemistry, and Organic Chemistry.
Organic Chemistry
Organic Chemistry focuses only on carbon-containing compounds and
was once branded as the chemistry of life. However, this does not mean that
Inorganic Chemistry is the study of non-living things. It simply implies that
Inorganic Chemistry is the specific discipline in science that deals with the
non-carbon substances. Thus, this field is extremely broad and can be
considered as a wide network that connects other scientific fields.
Inorganic Chemistry
Inorganic Chemistry is concerned with the synthesis, structure,
reaction, and properties of all the elements, except carbon, and their
compounds. However, it goes beyond and touches organometallic chemistry
which focuses on metals that bond to carbon-containing substances.
3
Chemistry in the Modern Society
Currently, it is the foundational concept which governs many scientific
breakthroughs and technological advancement we enjoy:
1. Energy and the Environment
- Energy is a result of the chemical process. Due to the
increasing demand for energy and electricity due to
industrialization and other societal functions, chemists
together with the other agents of science continually seek
other alternatives for an energy source. Inorganic chemistry
is the core of energy conversion and storage and electronics.
2. Materials and Technology
- Polymers such as cellophane wraps and bags are products of
chemistry, together with ceramics, coatings, and now, the
superconductors. Chemists are also in action in the fields of
information, communication, automotive, aviation and space
industries.
But the real question, for an allied health student like you, is how does
chemistry contribute to the field of health and medicine? Inorganic
Chemistry is the frontier that leads to the invention of different diagnostic
equipment and analytical instruments, such as x-ray machines. Various
pharmaceutical companies and chemists are also exploring the medicinal
benefits of inorganic compounds.
Focus Questions
Instructions: Please submit your answers through Schoology. The document
should be in letter size. Use Calibri, 12 pt. 1.15 spacing. On the upper-left
corner of the first page, write your name, grade level, section, and the date
of submission. For uniformity purposes, name your document as
IOChem_Last Name_Unit1_FQ. Properly cite your references using APA
format. The basis for grading will be provided in Schoology.
1. Aside from those mentioned in the unit, what are the other
contributions of Inorganic and Organic Chemistry to the society? Cite
three applications of Inorganic and Organic Chemistry which are very
observable in our day-to-day life.
2. What is the coverage of Inorganic Chemistry as a scientific discipline?
3. As a student under the field of health and medicine, why do you think
is it important to learn Chemistry, especially Organic and Inorganic
chemistry? How will these subjects equip you in your future
profession?
4
Related Readings
● Visit the webpage below to further read about the contributions of
Inorganic Chemistry to the society
acs.org/content/acs/en/careers/college-to-career/areas-ofchemistry/inorganic-chemistry.html
● Follow the link below and watch the video attached to have a full
grasp of the coverage of inorganic chemistry
https://www.youtube.com/watch?v=MCYRhCA7j1s
Learning Assessment
Instructions: Please submit your answers through Schoology. The document
should be in letter size. Use Calibri, 12 pt. 1.15 spacing. On the upper-left
corner of the first page, write your name, grade level, section, and the date
of submission. For uniformity purposes, name your document as
IOChem_Last Name_Unit1_Assessment. Properly cite your references using
APA format. The basis for grading will be provided in Schoology.
Find online 2 journal articles about innovations related to Inorganic
Chemistry with an application to health and/or medicine.
1. Give the title of the journal article, the authors, the year it was
published (not more than 5 years ago), and the name of the journal it
was published under.
2. Give a brief description of what the article is about. Focus on how its
results and conclusion may able to help the health and medicine field.
For reliable references, you may go to Google Scholar, ScienceDirect, ACS
Publication, or PubMed.
5
UNIT 2: Atomic Theory and Symbolism
Intended Learning Outcomes
At the end of the unit, you are expected to:
1. outline the development of atomic theory;
2. describe the subatomic parts;
3. discuss and derive the atomic symbol; and,
4. calculate for the atomic mass of an element.
Introduction
Since atoms are the building blocks of matter. It is a core concept in
different scientific disciplines, especially Chemistry and Physics. In this unit,
you will learn about the historical development of atom that led to the
establishment of the modern atomic model to equip you in understanding
the succeeding units of this course. Moreover, you are introduced to the
modern atomic model. This accepted concept of atom and the subatomic
parts are the results of the past theories, models, and experiments done by
the early scientists.
Unlocking of Difficulties
Before going through the lecture notes, it is expected of you to read
and review the following concepts in this section so that you can fully
comprehend the topics that will be introduced to you in this unit.
First, review the following concepts below regarding matter.
Matter is anything that has mass and volume. It is everywhere. It
includes all things that can be perceived by our senses (water, earth, fire),
and even those that cannot be (air). Matter has weight, thus, we can tell that
it has mass. Matter also takes up space, so it follows that it has volume.
Classification of Matter
1. Pure Substances
- homogeneous matter (uniform throughout)
- definite or fixed composition
- have the same amount of matter
- have distinct properties
- cannot be separated by physical means
- example: pure water (composed only of H2O molecules); sucrose
or table sugar (composed of carbon and compound water: 42.1%
carbon, 6.5% hydrogen, and 51.4% oxygen)
- can be divided further into two groups: elements and compounds
6
1.1 Elements
- contains a single type of atom
- cannot be broken down or separated into simpler substances,
even by chemical means
1.2 Compounds
- contain more than one type of element
- chemically combined in fixed ratio and are represented by
formulas
- can be broken down chemically but not physically
- may be broken down into either elements or other compounds
- can either be a binary compound or a tertiary compound
o binary compounds - compounds that are composed of 2
elements (e.g. table salt: NaCl - sodium chloride)
o tertiary compounds - compounds that are composed of 3
elements (e.g. glucose: CHO – carbon, hydrogen, and
oxygen)
2. Mixtures
- contains two or more types of pure substances which preserve
their distinct properties
- can be broken down or separated into simpler substances through
a physical process, without altering the properties of its
components
- has variable composition
- can either be homogeneous or heterogeneous
2.1 Homogeneous
- uniformly mixed throughout the mixture, therefore, the
composition of the mixture is the same throughout
- also called as solutions
- can either be:
o dissolved particles in solution (e.g. salt added to water)
o aqueous solution dissolved in a liquid medium (e.g.
juice concentrate added to water)
2.2 Heterogeneous
- non-uniform composition
- have visibly distinguishable parts
Physical Properties and Chemical Properties
1. Physical Property
- a quality of a substance that can be observed or measured without
changing the substance’s composition
- examples: color, solubility, odor, hardness, density, melting point,
luster
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2. Chemical Property
- the ability of a substance to undergo a chemical reaction & to form
a new substance
- a substance must undergo a chemical change to observe a
chemical property
- examples: rusting, burning, decomposition, fermentation
Physical Changes and Chemical Changes
1. Physical Change
- changes in form, but does not result in something new
- no changes in the composition
- examples: dissolving, melting, boiling, freezing, cutting
2. Chemical Change
- changes that result in a new substance
- involves energy
- examples: burning, rusting, electrolysis
- pieces of evidence of a chemical change:
o change in color - change in the structure of components
o effervescence or bubbling - indicates the presence of gas
o changes in temperature - absorption (endothermic) and
generation (exothermic) of heat
o formation of a precipitate - indicates the formation of a new
substance
The development of the modern atomic theory is heavily founded on the
concept of electron configurations, which was fully understood through
Max Planck’s Quantum Theory.
Quantum Theory
- dual nature of matter at small-scales
o particle-like and wave-like
Quantum Theory and Wave Mechanical Model of Atom
- (1) states that energy is quantized: meaning it can only be
transferred in individual units called quanta
o quanta (singular: quantum) - smallest possible units by
which energy can be transferred
- exhibits the ability of electrons to absorb energy
o highly observable when atoms of metals are placed in a
flame
▪ electrons tend to absorb energy, thus, jumping to an
excited state (quantum leap), and then, return to its
ground state by emitting the absorbed energy
through light
8
-
(2) the theoretical basis of the electron configuration
o prediction of the organization and location of electrons in
the atom, thus, the establishment of quantum numbers
▪ principal quantum number (n): main energy levels or
shells of an orbital
▪ azimuthal quantum number (l): energy sublevels or
subshells
▪ magnetic quantum number (mL): spatial orientation
of the orbital in space
▪ spin quantum number (mS): electron spin – clockwise
or counterclockwise
Lecture Notes
DEVELOPMENT OF THE ATOMIC THEORY
1. Theory of Democritus
- The matter could not be divided into smaller and smaller pieces
forever, eventually, the smallest piece would be obtained.
- This smallest piece would be indivisible.
- smallest piece = atomos = not to be cut / unable to be cut
2. John Dalton’s Atomic Theory (1803)
- atom: a solid, indestructible, structureless sphere
- Dalton’s Atomic Theory
o has no scientific evidence, and based only on intuition
i. All matter is made up of tiny, indestructible particles
called atoms.
ii. All atoms of a given element have identical physical &
chemical properties.
iii. Atoms are neither created nor destroyed during a
chemical change, but only are rearranged.
▪ Law of conservation of mass
● Matter can neither be created nor destroyed.
iv. Atoms of different elements form compounds in wholenumber ratios.
▪ Law of Definite Proportions
● Joseph Proust
● All samples of a pure compound contain the
same elements in the same proportion by mass.
▪ Law of Multiple Proportions
● When two elements react to form more than
one compound, a fixed mass of one element will
react with masses of the other elements in a
ratio of a small, whole number.
9
o Some of these postulates now have exceptions:
- Atoms can be broken apart in nuclear reactions
- Atoms of a given element can have different physical and
chemical properties (isotopes)
3. J.J Thompson’s Plum-Pudding Model (1897)
- atomic model: electrons are viewed as plums in a matter pudding
- atom as a big ball of positive charge that contains small particles of
negative charge embedded in it.
- discovered the charge of an electron by observing cathode rays in
a cathode ray tube
- concluded that cathode rays are streams of negatively charged
particles with mass.
- Robert Millikan
o subsequently, determined the mass of an electron based
upon Thomson’s work
4. Ernest Rutherford’s Model (1909)
- gold foil experiment: discovery of the nucleus
- disproved Thomson’s model.
- alpha-scattering experiment: bombarded a thin piece of gold foil
with positively charged alpha (positive charge) particles (much
smaller than the atom), proving the nucleus to be positive.
- Observations:
i. almost all the alpha particles passed through foil without
deflection
ii. small percent slightly deflected
iii. some were largely deflected
iv. a few even reflected in the direction from where they
had come
- Thus, it was concluded in 1911 that:
o Atom is mostly space and all of the positive charges in an
atom is concentrated in a small, dense, or core (nucleus).
o This area is positive since positively charged particles were
deflected from it (repelled).
- Therefore:
o Atomic mass = Sum of protons + neutrons
o Atomic Number = Number of protons
o Number of protons = Number of electrons
o Last Level = Valence electrons
10
5. Neils Bohr’s Planetary Model (1913)
- the model displayed electrons traveling in orbits around the
nucleus.
- electrons are only found in orbitals (principal energy levels) not in
between
- as you move away from the nucleus, the energy in each PEL
increases.
- like climbing stairs, further, you go = more energy
o Ground State - when electrons are in the lowest energy level
o Quantum Leap - when electrons jump between energy levels
o electrons can only jump to levels that aren’t filled with
electrons
- electrons can only absorb a fixed amount of energy (quanta)to
move to a higher level
o Heat, light, and electricity are all stimuli that can excite an
electron
o Excited State - Electrons are in higher energy levels.
Acquired when an electron absorbs energy and becomes
unstable. Electrons quickly return to a ground state,
emitting the same amount of energy absorbed, usually in
some form of light.
- Every element gives off a unique pattern of colors (line spectrum)
which can be used to identify the element.
6. Erwin Schrödinger’s Quantum Mechanical Model
- wave model: based on the dual nature of light
- concluded that electron has a wave-like character or nature
(delocalized and always moving)
- a higher density of the electron cloud near the nucleus (contains
the positive protons and neutral neutrons) due to the electrostatic
attraction of the nucleus and the electron
THE ATOM
- the basic unit of an element that can engage in a chemical
combination
With all of the theories and models, it was then accepted that:
i. the atom consists of the electrons (-) and a nucleus
ii. the nucleus houses the protons (+) and the neutrons (no charge)
iii. the electron is responsible for the volume of the atom
iv. the nucleus is responsible for the mass of the atom
11
Figure 2: Subatomic Particles (Adapted from Flowers et al., 2019)
ATOMIC SYMBOL
ZX A
where: X = element
Z = atomic number
A = mass number
Atomic number
- number of protons
- the identifying number of an element
- in a neutral atom, the number of protons is also equal to the
number of electrons (Z = proton = electron)
Mass number
- rounded to the nearest whole number
- the total number of neutrons and protons in an atom (A = protons
+ neutrons)
- in a neutral atom, Z = proton, therefore, A = Z + neutrons
- therefore, the mass of neutrons is the mass number of an atom
subtracted by the mass of protons (neutron = A - protons)
Ions
Y
ZXA
-
Y = charge of an atom
charged atoms are formed by gaining or losing electrons
cation: positively charged; losing an electron
anion: negatively charged; gaining an electron
Isotopes
- atoms of elements with the same atomic number but the different
mass number
- results as atoms of the same element differ in the number of
neutrons
- determining the average atomic mass of an element (with multiple
isotopes)
o the sum of the masses of its isotopes, each multiplied by its
natural abundance
12
o Average atomic mass = f1M1 + f2M2 +… + fnMn
▪ f = natural abundance (in percent or fraction)
▪ M = mass of the isotope
Focus Questions
Instructions: Please submit your answers through Schoology. The document
should be in letter size. Use Calibri, 12 pt. 1.15 spacing. On the upper-left
corner of the first page, write your name, grade level, section, and the date
of submission. For uniformity purposes, name your document as
IOChem_Last Name_Unit2_FQ. Properly cite your references using APA
format. The basis for grading will be provided in Schoology.
1. Describe the contribution of each of the scientists mentioned in the
unit to the establishment of the modern atomic theory. Discuss the
basis of their experiments, and how they were able to come up with
their conclusions.
2. Give at least 3 examples of elements which has multiple isotopes.
Discuss the difference of each isotope in terms of their chemical
properties, origin, and use.
3. Explain how each of the atomic symbols is derived.
Related Readings
● Review the following articles to understand the quantum theory and
its relationship to the modern atomic theory:
- https://www.sciencedaily.com/terms/electron_configuration.
htm
- https://pressbooks.bccampus.ca/chem1114langaracollege/ch
apter/development-of-quantum-theory/
● To further know about the Historical Development of Atomic Theory,
search in your search engine Chemistry Libretexts 2.1: Historical
Development of Atomic Theory (August 10, 2020)
● Watch a video from Youtube (Introduction to the atom ǀ Chemistry
of life ǀ Biology ǀ Khan Academy) to supply your understanding about
atoms
● For further reading about Isotopes, search for Chemistry Libretexts
2.3 Isotopes –When the Number of Neutron Varies (June 18, 2020)
● In your search engine, type Average Atomic Mass – Biology
LibreTexts for further reading about determining the atomic mass of
an element given multiple isotopes.
13
Learning Assessment
General Instruction: Please submit the document containing your answer
for the following exercises through Schoology. Name your file
IOChemistry_Last Name_Unit2_Assessment.
I.
Supply the missing information in the table below.
Atomic
Mass
Element Symbol
Protons Neutrons Electrons
Number Number
Sodium
11
11
11
Iron
56
26
𝐹𝑒
50
𝑆𝑛
60
𝑃𝑑
Nickel
II.
Supply the missing information in the table below.
Ion
Proton Neutron
Electron
+3
1327𝐴𝑙
16𝑂-2
8
26
56𝐹𝑒
35
45𝐵𝑟
𝐺𝑎+3
31
39
III.
Calculate the atomic weight of silicon given the following
Isotope
Mass
28𝑆𝑖
27.97693
29𝑆𝑖
28.97649
30𝑆𝑖
29.97377
14
UNIT 3: The Periodic Table of Elements
Intended Learning Outcomes
At the end of the unit, you are expected to:
1. outline the development of the periodic table;
2. explain the organization of elements in the periodic table; and,
3. demonstrate the general properties of elements based on their
location on the periodic table.
Introduction
The development of chemistry began even before civilizations. It
progressed through the practices and experiments of many people. These
foundations are the basis for the scientific explorations of many intellectuals
that led to the theories and models that we study today.
The Periodic Table, one of the most essential tools of a chemist, or any
person practicing chemistry, is instituted through the theories and models
that presented by many, and the discovery and isolation of the different
elements. In this unit, you will learn about the history of the atom, the
building block of matter, the periodic table, and the basis for the
arrangement of elements in it.
Unlocking of Difficulties
The electron configuration of atoms goes beyond predicting the
location of electrons within the atom. It also influenced the arrangement of
elements in the periodic table. In the lecture notes below, periodic
variations will be discussed. There are similarities and trends in theV
properties of the elements because of the repeating pattern of the electron
configuration (s, p, d, f orbitals).
15
Lecture Notes
DEVELOPMENT OF THE PERIODIC TABLE
Figure 3: Trend of Electron Configuration in the Periodic Table
https://guidancecorner.com/electron-configuration/
1. Johann W. Dobereiner
- proposed elements can be grouped in triads based on the
similarities of their properties
2. John Newlands
- arranged the elements in octaves
- proposition based on his observation that properties of elements
tend to repeat every seventh element
3. Dimitri Mendeleev and Julius Lothar Meyer
- arranged the elements according to atomic mass, physical and
chemical properties
- elements are in order with reference to their atomic weight
4. Henry Moseley
- used x-rays to identify atomic mass.
- concluded that the physical & chemical properties of elements
were listed by increasing atomic number.
- devised the modern classification system
THE PERIODIC TABLE
- shows all the names of all the known elements with all its
necessary information (abbreviations, atomic number, etc.)
- elements in the periodic table are arranged in vertical columns
according to similarities in their chemical properties
16
Figure 4: Periodic Table of Elements
https://www.brynmawr.edu/news/professor-chemistry-michelle-francl-talks-periodic-table-elements-bbc
The Periodic Law
- Henry Moseley: The chemical and physical properties of the
elements are periodic functions of their atomic numbers.
Groups of the Periodic Table
1. Group 1 A: ALKALI METALS
- active elements that readily form cations (1+) with nonmetals
- Lithium, Sodium, Potassium, Rubidium, Cesium, Francium
2. Group 2 A: ALKALINE EARTH METALS
- form cation (2+) with nonmetals
- Beryllium, Magnesium, Calcium, Strontium, Barium, Radium
3. Group 3-12: TRANSITION METALS
4. Group 13: BORON FAMILY
- Boron, Aluminum, Gallium, Indium, Thallium
5. Group 14: CARBON FAMILY
- Carbon, Silicon, Germanium, Tin, Lead
6. Group 15: NITROGEN FAMILY
- Nitrogen, Phosphorus, Arsenic, Antimony, Bismuth
7. Group 16: Oxygen Family
- Oxygen, Sulfur, Selenium, Tellurium, Polonium
8. Group 17: Fluorine Family/Halogens
- react with metals to form salts containing ions (1-)
- Fluorine, Chlorine, Bromine, Iodine, Astatine
9. Group 18: Noble Gases/Inert Gas Group
- inert and non-reactive
- Helium, Neon, Argon, Krypton, Xenon, Radon
17
CLASSIFICATION OF ELEMENTS
Metals
- most elements are metals
- most are conductors of electricity, malleable, ductile, has high
melting point and density and lustrous
- tend to lose electron easily forming a cation
Nonmetals
- found in the upper-right corner of the periodic table
- tend to gain electrons forming an anion
- bond to each other through covalent bonds
- poor conductor of heat and electricity, not ductile nor malleable,
brittle, dull, and most are gases
Metalloids
- have the properties of both metals & non-metals
- shiny, but brittle (silicon)
- better conductors than nonmetals, but not as good as metals
- ductile and malleable
PERIODIC VARIATIONS
- being familiar with the trends in the periodic table will enable you
to handle chemical compounds safely and will further equip you
with using chemistry in medical applications
- observable patterns in the periodic table with respect to the
element’s atomic number is because of the repeating pattern of
electron configurations (describes how electrons are distributed in
its atomic orbitals)
Atomic Number
- number of protons contained in the nucleus of an atom
- increasing: left to right; increasing: top to bottom
Atomic Radius
- the basis for atomic size, which can predict an element’s physical
properties (density, melting point, boiling point)
- half the distance of two nuclei of two adjacent atoms
- decreasing: left to right; increasing: top to bottom
Ionization Energy
- the attraction of an atom for its own electrons
- the energy required to remove an electron from an atom
- alkali metals and alkaline earth metals: low ionization energies
- noble gases: high ionization energies
18
- the difference in ionization energies
o metals readily form cation (lose electron)
o non-metals readily form anion (gain electron)
- ionization number decreases with increasing atomic number
- increasing: left to right; decreasing: top to bottom
Electron Affinity
- the attraction of an atom for an additional electron from other
atoms
- change in energy as an electron is accepted in an atom
- high electron affinity means that an atom has a greater tendency
to accept an electron
- metals: generally lower electron affinity than those of the nonmetals
- increasing: left to right; less variation in electron affinities from top
to bottom
Electronegativity
- the attraction of an atom for its electrons when in a chemical bond
- higher electronegativity = greater attraction for bonding electrons
- electrons: low ionization energy = low electronegativity; high
ionization energy = high electronegativity
- electronegativity decreases as the atomic number increases
Metallic Character
- decrease: left to right; increases from top to bottom
Atomic radius, ionization energy, and electron affinity, as said earlier,
is related to the electron configuration of the atom, and its capacity for
electron-nucleus attraction and electron-electron repulsion.
1. Higher electron-nucleus attraction = ↓ atomic size, ↑ ionization
energy, ↑ electron affinity
2. Higher electron-electron repulsion = ↑ atomic size, ↓ ionization
energy, ↓ electron affinity
19
Focus Questions
Instructions: Please submit your answers through Schoology. The document
should be in letter size. Use Calibri, 12 pt. 1.15 spacing. On the upper-left
corner of the first page, write your name, grade level, section, and the date
of submission. For uniformity purposes, name your document as
IOChem_Last Name_Unit3_FQ. Properly cite your references using APA
format. The basis for grading will be provided in Schoology.
1. How did each of the scientists mentioned in the unit contribute to the
establishment of the periodic table of elements?
2. Why is it more efficient to arrange the elements according to their
atomic numbers?
3. Describe the specific characteristics of each group in the periodic
table. Given their characteristics, how are they usually utilized?
4. What are the trends or patterns that can be observed in the periodic
table of elements? Why is it essential to know these trends?
Related Readings
● To recall your prior lesson regarding quantum numbers and electron
configuration, refer to the video and other supplementary articles
below:
o Khan Academy: Introduction to electron configurations
o Electron Configurations https://www.chem.fsu.edu/chemlab/chm1045/e_config.html
o Introduction: What are Electron Configurations? –
https://byjus.com/chemistry/electron-configuration/
● For further reading about The Periodic Table and its Development,
search for Chemistry Libretexts 2.1.1 The Periodic Table in your search
engine.
● To supplement your understanding about the organization of the
elements in the periodic table, visit the website below and read its
articles
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modu
les_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry
● For better copies of Periodic Table of Elements, please refer to
https://pubchem.ncbi.nlm.nih.gov/periodic-table/
20
Learning Assessment
General Instruction: Please submit the document containing your answer
for the following exercises through Schoology. Name your file
IOChemistry_Last Name_Unit3_Assessment.
I.
Group the following elements as metals, nonmetals, or
metalloids.
As, Xe, Fe, Li, B, Cl, Ba, P, I, Si, Tm, Dy, Ac, Zn, Uut, Rn, Rf, Rb, Pd,
Tl, Uup, Bk, Ds, Nb, Fr, W, Ru, Cn, Bi, Cm, Cf, Pu, U, Np, Os, Ta
II.
Make a graph or a table that highlights the similarities and
differences of metals, nonmetals, and metalloids n terms of their
physical and chemical properties.
III.
Arrange the following atoms in order of decreasing atomic radius:
Na, Al, P, Cl, Mg
IV. List the following in order of increasing ionic radius: N3-, Na+, F-,
Mg2+, O2V.
Arrange the elements on each of the following groups in
increasing order of the most positive electron affinity
a. Li, Na, K
b. F, Cl, Br, I
c. O, Si, P, Ca, Ba
VI.
Arrange the following in order of increasing first ionization
energy: Na, Cl, Al, S ad Cs.
UNIT 4: Nomenclature of Ionic, Covalent and Acid
Compounds
Intended Learning Outcomes
At the end of the unit, you are expected to:
1. explain ionic and covalent compounds;
2. determine formulas of ionic compounds; and,
3. name ionic compounds.
Introduction
Compounds are pure substances that contain two or more elements.
Thus, it follows that it also contains two or more types of atom. Bonds are
formed between these different types of atoms through transferring or
sharing of electrons. For this unit, you will learn about the basic concept
behind the formation of ionic, covalent, and acid compounds, and more
importantly, writing formulas and naming these compounds.
21
Unlocking of Difficulties
To efficiently learn how to write and name ionic, covalent, and acid
compounds, familiarize the symbols of the elements, and the name and
formula of the common cations and anions.
Figure 5: Elements and their Symbols
https://periodic.lanl.gov/images/periodictable-3-13-17.pdf
Table 1: Greek Prefixes Indicating Number of Atoms
Number
Greek Prefix
Number
1
Mono6
2
Di7
3
Tri8
4
Tetra9
5
Penta10
Greek Prefix
HexaHeptaOctaNonaDeca-
22
Figure 6: Common Cations and Anions
http://web.mst.edu/~tbone/subjects/tbone/chem1/Chem%201%20Compound%20Sheet%20FS08.pdf
Lecture Notes
IONIC COMPOUNDS
Ions are formed as an atom loses (cation) or gains (anion) an
electron or more. In the last unit, it was discussed that element classified
as metals tend to easily lose an electron, while nonmetals readily gain
electrons. Thus, an ionic compound is formed when an atom of a metal (M)
gives its electron/s to a nonmetal atom (N).
𝑀 → 𝑀+ + 𝑒 −
𝑁− + 𝑒− → 𝑁
Since the two atoms are charged oppositely, there will an
electrostatic attraction between them causing them to form an ionic bond.
Properties of Ionic Compounds
1. High melting point – due to strong ionic bond
2. Hard – also due to strong ionic bond
3. Brittle – caused by the repulsion of cations
4. Conductor in liquid and molten state – due to the dissociation of
cations and anions
23
Naming Ionic Compounds
1. Name of the cation followed by the anion.
2. Does not use prefixes to denote the number of ions, as it is already
implied by the respective charges of the cation and anion.
3. Monoatomic cations (has single-atom) are named after its parent
element.
4. Monoatomic anions: root name + ide
- KBr – potassium bromide
5. Polyatomic cations (cations with variable charges, e.g. Sn2+, Sn +4): two
systems are being followed – Stock system and Old system.
- Old system: tin (II), tin (IV)
- Stock system
o cation with higher charge: ends in –ic (stannic)
o cation with lower charge: ends in –ous (stannous)
6. Hydrates (ionic compounds containing water): name of ionic
compound + prefix indicating number of water molecules + -hydrate
o CuSO4٠5H2O – copper (II) sulfate pentahydrate / cupric
sulfate pentahydrate
Writing Formulas of Ionic Compounds
1. Write the symbol of the cation, followed by the symbol of the anion.
2. Take note of the charge of each ion for the subscript.
- cross-over rule
o cation: the absolute value of the charge of the anion
o anion: the absolute value of the charge of the cation
- then, simplify
- magnesium oxide - magnesium (Mg2+), oxide (O2-)
- Mg2O2 - MgO
3. Hydrates: similar steps + number of molecules, followed by H2O
COVALENT COMPOUNDS
Not all compounds are formed through electron transfer, in covalent
compounds; bonds are formed through sharing electrons, which occurs
between two non-metal atoms. There are two classifications of covalent
compounds: molecular covalent and network covalent.
Covalent Molecular Compounds
- bonded by two forces: covalent bond and intermolecular forces of
attraction (IMFA)
- are molecules: H2O, CO2, NH3, sucrose
24
Properties
1. Low melting point – to break the IMFA
2. Low density – due to the loosely-packed structure
3. Poor thermal and electrical conductivity – molecules are uncharged
4. Soft and brittle – because of the IMFA
Covalent Network Compounds
- forms crystal structure, thus, referred also as covalent crystal
structures
- bonded only by covalent bond
- quarts (SiO2), silicon carbide (SiC), sand
Properties
1. Very high melting point – due to the covalent bonds makes the
compound like a large molecule
2. Poor electrical conductor – no free electrons available (all electrons
used for the covalent bonding)
3. Hard and brittle – due to the covalent bonds
Naming Covalent Compounds
1. The more metallic element, followed by less metallic element + -ide
2. Prefixes are used to denote the number of atoms in the compound
- CO2
- carbon (group 14) - oxygen (group 16)
- monocarbon dioxide
(for most well-known compounds: mono- is omitted, e.g. carbon
dioxide, which automatically means that the ion is monoatomic)
Writing Covalent Compounds
1. Write the symbol of the more metallic element, followed by the
symbol of the less metallic one.
2. The subscript will follow the number denoted by the prefix of the ion.
- SO2 - monosulfur dioxide / sulfur dioxide
ACIDS
-
a molecule which contains H+ bonded to an anion
dissociates into a free H+ and an anion when in water
the formula can be easily recognized: compound has H in front (HX)
classified into two groups: binary acid and oxyacids
Binary Acids
- does not contain oxygen
- exclusively contain hydrogen and non-metal ions
25
Naming Binary Acids
1. Write hydro- as a prefix to the root name of the element, then –ic as
a suffix. Followed by the word acid.
- HF – hydrofluoric acid
Oxyacids
- contains hydrogen ion (H+) and an oxyanion (oxygen-containing
anion, e.g. ClO3- - chlorate; ClO2- - chlorite)
Naming Oxyacids
1. Oxyacid ending in –ate: replace –ate with –ic, and add acid
- HClO3 – chloric acid
2. Oxyacids ending in –ite: replace –ite with ous, and add acid
- HClO2 – chlorous acid
Focus Questions
Instructions: Please submit your answers through Schoology. The document
should be in letter size. Use Calibri, 12 pt. 1.15 spacing. On the upper-left
corner of the first page, write your name, grade level, section, and the date
of submission. For uniformity purposes, name your document as
IOChem_Last Name_Unit4_FQ. Properly cite your references using APA
format. The basis for grading will be provided in Schoology.
1. How are ionic, covalent, and acid compounds formed?
2. Using your own words, write a step-by-step process of writing and
naming ionic, covalent, and acid compounds. (Present answers in
bullet form.)
Related Readings
In your search engine, search for Chemistry LibreTexts – 2.7:
Nomenclature of Ionic, Covalent, and Acid Compounds to further read about
naming and writing compounds.
Learning Assessment
General Instruction: Please submit the document containing your answer
for the following exercises through Schoology. Name your file
IOChemistry_Last Name_Unit4_Assessment.
I.
1.
2.
3.
4.
5.
Name the following compounds.
FeCO3
6. Fe2O3
N2O5
7. NaOH
BeCl2 ٠ 4H2O
8. ClF3
(NH4)3BO3
9. ZnSO4 ٠ 7H2O
HNO2
10. H2O2
11. HCH3COO
12. As2(Cr2O7)5
13. Co(NO3)2
14. Ca3(PO4)2
15.H3PO3
26
16.AgI
17.Ca3As2
18.Mg(HSO4)2
19.Hg2Se
20.K2Te
21.H2SO4
22.Sb(C2H3O2)3
23.HF
24.P2Br4
25.H2C2O4
II.
Write the chemical formula of the following compound.
1. Disulfur difluoride
2. Magnesium chlorite
3. Copper (II) sulfate
4. Iron (III) chloride hexahydrate
5. Acetic acid
6. Ammonium phosphide
7. Sodium sulfate decahydrate
8. Arsenic (III) chloride
9. Hydrochloric acid
10.Copper (II) hypobromite
11.Dichlorine monoxide
12.Chlorous acid
13.Hydrocyanic acid
14.Manganese (II) sulfite
15.Chromium (III) sulfate
16.Phosphoric acid
17.Nitric acid
18.Beryllium chlorate
19.Iron (III) chloride hexahydrate
20.Hypochlorous acid
21.Carbon dioxide
22.Calcium oxalate
23.Ammonia
24.Hydrazine
25.Lead (IV) sulfide
27
UNIT 5: Lewis Structure
Intended Learning Outcomes
At the end of the unit, you are expected to:
1. know what a Lewis structure is and the information it conveys;
2. draw the Lewis structure of a given element, molecule or ion;
3. draw resonance structures of some molecules; and
4. assign formal charge to an atom in a Lewis structure.
Introduction
In the last unit, you have learned about the basic concept behind
chemical bonding, which involves transferring or sharing of electrons. For
this unit, you will be introduced to Lewis Structure to depict the bonding
scheme of the compounds.
Unlocking of Difficulties
Valence electrons, as previously defined, are the outermost electrons
which participate in a chemical reaction. G. N. Lewis in 1916 suggested a
visual representation of this concept through using dots. He theorized that
a single bond involves two electrons, and atoms tend to hold eight electrons
in its outermost subshell. Thus, the Octet rule, which states that for an atom
to be stable, it needs to have eight electrons in its valence shell. It is said that
if an atom does not have a complete eight electrons in its valence shell, it is
more reactive. Now, bear in mind that if we discuss about valence shell, we
are only considering the s and p orbitals.
Lewis structure goes beyond representing valence electrons of atoms, it
also used in explaining chemical linkages or chemical bonding.
Lecture Notes
Chemical Bonding
1. Ionic Bonding
- involves electron transfer
- between a metal and a nonmetal: large difference in ability to lose
(ionization energy) or gain (electron affinity) electrons
2. Covalent Bonding
- involves electron sharing
- between two nonmetals: both have high electron affinity, thus,
both tend to attract other electrons
- shared electron pair: localized between two the two atoms,
therefore linking two atoms in a covalent bond
28
Lewis Structure for Atoms
5. Element Symbol – the nucleus and the inner electrons
6. Surrounding dots – valence electrons
Lewis Structure for the Main Groups
1. Groups 1, 2, 13 – 18 (A –group number)
- dots equal to the number of valence electrons = electrons in the s
and p subshells
2. Place one dot at a time around the element symbol
- top, right, left, bottom
25.Add the dots until all four sides have 1 pair of electrons
Figure 1: Lewis Structure of Some Elements
https://www.chemistrylearner.com/germanium.html
The Lewis structures of the different elements give information about the
bonding behavior of its atoms
- for metals
o total number of dots = maximum number of electrons the
atom can give up
- for nonmetals
o total number of unpaired dots = number of electrons it can
accept (in ionic bonding) = number of electrons it has to
share with another atom (covalent bonding)
Lewis Structure for Ions
1. Element Symbol – nucleus and inner electrons
2. Dots – number of valence electrons present in the ion
3. Element symbol and dots enclosed in a square bracket
4. The charge of the ion is indicated as superscript
Figure 2: Lewis Structures for Common Ions
(Department of Food Science and Chemistry, 2016)
29
Atoms will either gain or lose electrons to be stable electron configuration.
Metals such as sodium (Na) and calcium (Ca)tend to lose electrons to form
cations. In Figure 3 below, sodium, an alkali metal, readily loses its only
valence electron to form a cation with a charge of 1+. While calcium, an
alkaline earth metal, easily loses its two valence electrons to form a cation
with a charge of 2+
Figure 3: Sodium and Calcium Losing Electrons
(Flowers, 2019)
Nonmetals such as chlorine and sulfur tend to gain electrons to be stable.
Chlorine (halogen) readily accepts 1 electron to complete its valence
electrons, thus gaining a charge of 1-. While, sulfur (chalcogen) readily
accepts 2 electrons to full fill its outermost shell, thus having a charge of 2-.
Figure 4: Chlorine and Sulfur Gaining Electrons
(Flowers, 2019)
Lewis Structure for Ionic Compounds
1. Number of electrons lost by the metal atom = Number of electrons
gained by the nonmetal atom
- therefore: overall charge of the compound should be zero
2. The Lewis structure of each ion will be used to make the Lewis
structure for the whole compound.
Examples:
1. Sodium Chloride (NaCl)
Sodium atom loses one electron to form the cation Na+
Chlorine atom gains one electron to form the anion Cl2. Magnesium oxide (MnO)
Magnesium atom loses two electrons to form the cation Mg2+
Oxygen atom gains two electrons to form the anion O23. Calcium Flouride
Calcium atom loses two electrons to form the cation Ca2+
Each of the Fluorine atom gains an electron to form 2 anions of F-
30
Figure 5: Lewis Structure of Ionic Compounds
(Flowers, 2019)
Lewis Structure for Covalent Compounds
- electrons are shared between two atoms to form the covalent
bond, thus, each atom should have a share in the number of
electrons to satisfy the octet rule
- atoms of the elements carbon (C), nitrogen (N), oxygen (O) and
fluorine (F)are strict followers of the octet rule
- lone pair: a pair of unbonded electrons
- single bond: a pair of electrons shared by two atoms
- double bond: two pairs of electrons shared by two atoms
- triple bond: three pairs of electrons shared by two atoms
1. Electrons of the two atoms are paired up to represent the bonding
2. Shared electrons are often enclosed in a circle, or it is either
represented as a dash (–)
Examples:
1. Hydrogen fluoride (HF)
Hydrogen atom has 1 valence electron
Fluorine has 7 valence electrons
- hydrogen will share its 1 electron with fluorine to form a covalent
bond
- therefore, hydrogen will now have 2 valence electrons and fluorine
8 valence electrons
Figure 6: Lewis Structure for Hydrogen Fluoride
2. Ammonia (NH3)
Nitrogen atom has 5 valence electrons
Each of the three Hydrogen atoms has 1 valence electron
- the three hydrogen atoms will share each of its electron with
nitrogen
- therefore, the three hydrogen will have 2 valence electrons, while
nitrogen will have 8 valence electrons
Figure 7: Lewis Structure for Ammonia
31
3. Oxygen molecule (O2)
Oxygen atom has 6 valence electrons
- each oxygen atom will share 2 of its valence electrons
- therefore, a double bond will be formed to satisfy the octet rule
Figure 8: Lewis Structure for Oxygen Molecule
4. Carbon monoxide (CO)
Carbon atom has 4 valence electrons
Oxygen atom has 6 valence electrons
- carbon will share one pair of its valence electrons with oxygen
- oxygen will share two pairs of its valence electrons with carbon
Figure 9: Lewis Structure for Carbon Monoxide
Writing Lewis Structure with Octet Rule
# of
atoms
1
3
Element
N
Valence
electrons
5
F
21
26es-
TOTAL
1. Determine the total number of valence electrons available.
- add electrons for anions
- subtract electrons for cations
# of
atoms
1
3
Element
N
Valence
electrons
8
F
24
TOTAL
32es-
2. Determine the number of electrons needed to satisfy the octet rule
per atom or ion.
- 8 electrons for most atoms
- 2 electrons for hydrogen
- 0 electrons for monoatomic cations
(32es– required) −
(26es– available)
6es – to be shared (bonding
electrons)
3. Find the difference between the result of step 2 and step 1. The
resulting value is equal to the number of bonding electrons.
32
4. Draw the skeleton structure of the
molecule by arranging the atoms
symmetrically.
- central atom: atom with the lowest
electronegativity
- hydrogen atom and halogens are usually outside atoms
- do not put more than four atoms around the central atom unless
the central atom belongs to third period or lower
5. Place the number of electrons to be
shared between atoms.
- place one pair of electrons between
each pair of atoms
- use the remaining pairs to make double/triple bonds
6. Distribute the remaining electrons as
lone pairs so that the octet rule will be
satisfied.
Resonance Structures
- a molecule which is represented by several structures with
different electron distribution among the bonds
- only the distribution of the electrons or the location of bonding
and lone pairs vary, but not the relative placement of the atoms
- combination of structures represents the real structure
- more stable structure contributes more than the less stable ones
Figure 10: Resonance Structure for Ozone (O3)
https://www.coursehero.com/sg/general-chemistry/formal-charges-and-resonance/
Exceptions to the Octet Rule
1. Electron – Deficient Molecules
- central atoms with less than an octet of electrons
- Be (4 valence es-), B (3 valence es-), Al (3 valence es-)
Figure 11: Lewis Structures for Beryllium chloride and Boron trichloride
https://www.thoughtco.com/exceptions-to-the-octet-rule-603993
33
Figure SEQ Figure \* ARABIC 12: Lewis Structures for Aluminum hydride
and Aluminum trichloride
https://www.chegg.com/homework-help/predict-geometries-following-species-alcl3-b-zncl2-c-chapter-10-problem-8p-solution9780072930276-exc
- compounds of boron and aluminum are never observed to share
5 electrons with other atoms, unlike what is expected according
to the octet rule
- octet rule may not be applicable for Group 13
Figure 13: Lewis Structures for Boron trifluoride
https://courses.lumenlearning.com/trident-boundless-chemistry/chapter/exceptions-to-the-octet-rule/
- higher electronegativity of halogens (such as fluorine and
chlorine) than Group 13 elements
- duet rule: hydrogen and lithium
- to attain an electron configuration the same as helium
Figure 14: Lewis Structures for Lithium chloride and Hydrochloric acid
https://study.com/academy/lesson/hydrogen-chloride-vs-hydrochloric-acid.html
- hydrogen: has one subshell (1s), thus, only needs one electrons to
be stable
- lithium: has one electron left in its outermost subshell (2s), thus,
needs to give up one electron to be stable
2. Expanded Valence Shells
- central atoms with more than an octet of electrons
- elements in Period 3 and higher: large central atoms with
available d orbitals
- sulfur, phosphorus, silicon, chlorine
Figure 15: Lewis Structures for Sulfur hexafluoride and Phosphorus
pentachloride
https://www.quora.com/What-is-the-Lewis-structure-of-PCl5
- SF6 : 12 bonded es-
- PCl5 : 10 bonded es-
o 2 es- to fill last subshell of S (3p4)
o 10 es- to fill the empty 3d orbital
o 3 es- to fill last subshell of S (3p3)
o 7es- to fill the empty 3d orbital
34
3. Free – radicals
- odd – electron molecules: odd electrons in the outermost shell
- odd – numbered groups: Group 15 (N) and Group 17 (Cl)
Figure 16: Lewis Structure for Nitric oxide
https://commons.wikimedia.org/wiki/File:Nitric_oxide_Lewis.png
Formal Charge
- since many Lewis structure of a specific molecule is possible, the
formal charge determines its stability
o Lower formal charge means more stable structure
o Adjacent atoms should have opposite formal charges
o Atoms with higher electronegativity should have a more
negative formal charge
Formal Charge = Number of Valence Electrons – Number of Unshared
Electrons – (Number of Bonding Electrons / 2)
Figure 17: Formal Charge for every Atom in Ozone
http://www.uwosh.edu/faculty_staff/gutow/Lewis_Tutorial/O3_7.html
Focus Questions
Instructions: Please submit your answers through LMS. The document
should be in letter size. Use Calibri, 12 pt. 1.15 spacing. On the upper-left
corner of the first page, write your name, grade level, section, and the date
of submission. For uniformity purposes, name your document as
IOChemistry_Last Name_Unit5. Properly cite your references using APA
format.
1. How are single, double, and triple bonds similar? How do they
differ?
2. How do you think will knowing how to write Lewis structure help
you understand more about chemical bonding of ions or
molecules?
Related Readings
To read more about Lewis Structure
- Read Chapter 33 of Silberberg’s Principles of General Chemistry
(2007)
35
Learning Assessment
General Instruction: Please submit the document containing your answer
for the following exercises through Schoology. Name your file
IOChemistry_Last Name_Unit5_Assessment.
1. Many monoatomic ions are found in the body, including the ions formed
from the following list of elements. Write the Lewis symbols for the
following elements.
a. Cl
b. Na
c. Mg
d. Ca
e. K
2.Identify the atoms that corresponds to the following electron
configuration. Then, write the Lewis structure for each atom.
a. 1s22s22p5
d. 1s22s22p63s23p64s23d104p4
b. 1s22s22p63s2
e. 1s22s22p63s23p64s23d104p1
c. 1s22s22p63s23p64s23d10
3. Write the Lewis symbols for the following compounds.
a. MgS
c. GaCl3
e. Li3N
g. HBr
i. CO
b. Al2O3
d. K2O
f. ClF3
h. N2
j. PCl5
4. Write all the possible resonance forms of the following compounds.
Determine the formal charge on each atom in each resonance structures
a. selenium dioxide
c. nitric acid
b. sulfur dioxide
5. Determine the formal charge of each element in the following
compounds.
a. HCl
b. CF4
c. PCl3
d. PF5
e. NH3
36
UNIT 6: Chemical Reactions
Intended Learning Outcomes
At the end of the unit, you are expected to:
1. write and balance chemical equations; and,
2. define precipitation, acid – base and oxidation – reduction
reactions.
Introduction
When matter undergoes as specific reaction leading to a production of a
new substance, chemical change occurs. In this unit, basic techniques in
balancing and writing equations will be taught.
Unlocking of Difficulties
Recall
1. Law of Conservation of Mass
- Matter can neither be created nor destroyed. Thus, arrangement
of atoms will only be rearranged after a chemical reaction.
2. Molecule
- a pure substance composed of two or more atoms
3. Atom
- basic / smallest unit of matter
4. Solute
- minor component of a solution = lesser quantity
5. Solvent
- substance that dissolves the solute
- component with the highest quantity
6. Oxidation State
- possible charge of an atom if it is in an ionic bonding
7. Mole
- SI Unit for amount of substance
- 1 mole = 6.022 x 1023 units of something (atoms, molecules)
- amount of matter that contains as many objects (atoms,
molecules, etc.) as the number of atoms in isotopically pure 12C
8. Reactants
- the original substances / starting material
- substances that undergoes change
9. Products
- new substances produced
10.Coefficient
- number of substances present
37
Lecture Notes
Chemical Equation
- represents the changes in the matter in a chemical reaction
- shows the reactants (left side) and products (right side)
- indicates the relative amount (coefficient) of the substances
involved in the reaction
What does a balanced equation mean?
- number of atoms per element in the reactant = number of atoms
per element in the product
Example: Two hydrogen molecules (H2) reacts to an oxygen molecule (O2),
forming 2 molecules of water (2 H2O)
Figure 18: Reaction of Hydrogen and Oxygen to Produce Water
https://exercicespdf.com/Telecharger_PDF_Cours_Exercices_Gratuit_5.php?Cours_Exercices_PDF=1982&PDF=2h2_o2=2h2o
Reactant
Product
Element
# of Atoms
Element
# of Atoms
H
2 molecules * 2 atoms = 4
H
2 molecules * 2 atoms = 4
O
2 atoms
O
2 molecules
Example: Cellular respiration – production of ATP, water and carbon
dioxide from glucose and oxygen
𝐶6 𝐻12 𝑂6 + 6𝑂2 → 6𝐶𝑂2 + 6𝐻2 𝑂 + 𝐴𝑇𝑃
Figure 19: Cellular Respiration
Element
C
H
O
Reactant
# of Atoms
6 atoms
12 atoms
6 atoms + (6 molecules *
2 atoms) = 18 atoms
Element
C
H
O
Product
# of Atoms
6 molecules
6 molecules * 2 atoms =
12
(6 molecules * 2 atoms) +
6 molecules = 18 atoms
38
Physical States of Substances in a Chemical Reaction
Symbol
Meaning
S
solid
L
liquid
G
gas
Aq
aqueous solution
↓
precipitate is produced
↑
gas is produced
∆
application of heat
Example: Hydrogen gas reacts with oxygen gas to produce water in
gaseous state or liquid state
2𝐻2 (𝑔) + 𝑂2 (𝑔) → 2𝐻2 𝑂(𝑔)
Figure 20: Physical State of the Substances in the Production of Water
Aqueous Solution
- solution where the solvent is water
- either: solid in water or liquid in water
- recall: ionic compounds dissociate in water
Example: Aqueous solutions of calcium chloride and silver nitrate subjected
to a chemical reaction to produce calcium nitrate and silver chloride
Molecular Equation
o simply referred as a balanced equation
𝐶𝑎𝐶𝑙2 (𝑎𝑞) + 2𝐴𝑔𝑁𝑂3 (𝑎𝑞) → 𝐶𝑎(𝑁𝑂3 )2(𝑎𝑞) + 2𝐴𝑔𝐶𝑙(𝑠)
Complete Ionic Equation
o balanced equation showing the dissociated ions
2+
−
𝐶𝑎𝐶𝑙2 (𝑎𝑞) → 𝐶𝑎(𝑎𝑞)
+ 2𝐶𝑙(𝑎𝑞)
+
2𝐴𝑔𝑁𝑂3 (𝑎𝑞) → 2𝐴𝑔(𝑎𝑞)
+ 2𝑁𝑂3−(𝑎𝑞)
2+
𝐶𝑎(𝑁𝑂3 )2 (𝑎𝑞) → 𝐶𝑎(𝑎𝑞)
+ 2𝑁𝑂3−(𝑎𝑞)
2+
+
2+
−
𝐶𝑎(𝑎𝑞)
+ 2𝐶𝑙(𝑎𝑞)
+ 2𝐴𝑔(𝑎𝑞)
+ 2𝑁𝑂3−(𝑎𝑞) → 𝐶𝑎(𝑎𝑞)
+ 2𝑁𝑂3−(𝑎𝑞) + 2𝐴𝑔𝐶𝑙(𝑠)
Net Ionic Equation
o balanced equation showing the substances involved in the
reaction in their most simplified coefficient
2+
+
2+
−
𝐶𝑎(𝑎𝑞)
+ 2𝐶𝑙(𝑎𝑞)
+ 2𝐴𝑔(𝑎𝑞)
+ 2𝑁𝑂3−(𝑎𝑞) → 𝐶𝑎(𝑎𝑞)
+ 2𝑁𝑂3−(𝑎𝑞) + 2𝐴𝑔𝐶𝑙(𝑠)
o remove the spectator ions found both in left and right side
of the equation
o spectator ions: ions that are not chemically nor physically
involved in the chemical reaction
39
+
−
2𝐶𝑙(𝑎𝑞)
+ 2𝐴𝑔(𝑎𝑞)
→ 2𝐴𝑔𝐶𝑙(𝑠)
+
−
𝐶𝑙(𝑎𝑞)
+ 𝐴𝑔(𝑎𝑞)
→ 𝐴𝑔𝐶𝑙(𝑠)
Types of Chemical Reactions
1. Precipitation Reaction
- one of the products should be solid
- reaction should produce a precipitate (solid substance)
Solubility Rules
● Soluble
- all alkali metals (Group 1)
- all ammonium compounds (NH4+)
- all compounds of nitrate (NO3-), chlorate (ClO3-) and perchlorate
(ClO4-), acetate (C2H3O2-), bicarbonate (HCO3-)
- compounds of chlorides (Cl-), bromides (Br-) and iodides (I-)
o 3 consecutive elements in Group 17 after fluoride
o except compounds containing: Ag+, Hg22+ and Pb2+
- compounds containing sulfates (SO42-), except:
o slightly soluble: CaSO4 and Ag2SO4
o insoluble: BaSO4, HgSO4 and PbSO4
● Insoluble
- all carbonates (CO32-), chromate (CrO42-), phosphates (PO43-), and
sulphides (S2-)
o except compounds of alkali metals and ammonium ions
- hydroxides (OH-), except
o compounds containing alkali metals and barium
o slightly soluble: Ca(OH)2
Figure SEQ Figure \* ARABIC 21: Calcium Stones in the Kidney
Practical Example: Calcium stones – in the form of calcium oxalate or
calcium phosphate
o ionized or free calcium: the physiologically active form in the
body (bones and feet, and cell and nerve function)
o oxalate can be acquired from diet (fresh vegetables and
fruits)
o phosphate: ion form of mineral phosphorus (bones and
teeth, and nerve and muscle function)
▪ phosphate build-up: phosphate ions binds with
calcium ions – Ca3(PO4)2
40
o Rewatkar et al., (2018): model to simulate the reactive
crystallization of calcium oxalate (major component of
kidney stones)
𝐶𝑎𝐶𝑙2 + 𝑁𝑎2 𝐶2 𝑂4 → 𝐶𝑎𝐶2 𝑂4 ↓ +2𝑁𝑎𝐶𝑙
Example: Reaction between aqueous NaOH and aqueous MgCl2
2𝑁𝑎𝑂𝐻(𝑎𝑞) + 𝑀𝑔𝐶𝑙2 (𝑎𝑞) → 2𝑁𝑎𝐶𝑙(𝑎𝑞) + 𝑀𝑔(𝑂𝐻)2 (𝑠)
+
2+
+
−
−
2𝑁𝑎(𝑎𝑞)
+ 2𝑂𝐻(𝑎𝑞)
+ 𝑀𝑔(𝑎𝑞)
+ 2𝐶𝑙 −(𝑎𝑞) → 𝑀𝑔(𝑂𝐻)2 (𝑠) + 2𝑁𝑎(𝑎𝑞)
+ 2𝐶𝑙(𝑎𝑞)
+
2+
+
−
−
2𝑁𝑎(𝑎𝑞)
+ 2𝑂𝐻(𝑎𝑞)
+ 𝑀𝑔(𝑎𝑞)
+ 2𝐶𝑙 −(𝑎𝑞) → 𝑀𝑔(𝑂𝐻)2 (𝑠) + 2𝑁𝑎(𝑎𝑞)
+ 2𝐶𝑙(𝑎𝑞)
2+
−
𝑀𝑔(𝑎𝑞)
+ 2𝑂𝐻(𝑎𝑞)
→ 𝑀𝑔(𝑂𝐻)2 (𝑠)
Example: Reaction between aqueous copper (II) chloride / cupric chloride
and aqueous sodium sulphate
𝐶𝑜𝐶𝑙2 (𝑎𝑞) + 𝑁𝑎2 𝑆𝑂4(𝑎𝑞) → 𝐶𝑜𝑆𝑂4(𝑎𝑞) + 2𝑁𝑎𝐶𝑙(𝑎𝑞)
+
+
2−
−
−
𝐶𝑜(𝑎𝑞)
+ 2𝐶𝑙(𝑎𝑞)
+ 2𝑁𝑎(𝑎𝑞)
+ 𝑆𝑂2−
4 (𝑎𝑞) → 𝑀𝑔(𝑂𝐻)2 (𝑠) + 2𝑁𝑎(𝑎𝑞) + 2𝐶𝑙(𝑎𝑞)
2. Acid – Base Reaction
- transfer of proton (hydrogen ion – H+)
- electrolytes: charged chemical species
o in the body: these species are charged minerals that balance
the body’s pH level
Acids: proton donor
- produces H3O+(when H2O is part of the equation) or H+ (H2O
excluded from the equation)
+
−
𝐻𝐶𝑙(𝑎𝑞) + 𝐻2 𝑂(𝑙) → 𝐻3 𝑂(𝑎𝑞)
+ 𝐶𝑙(𝑎𝑞)
+
−
𝐻𝐶𝑙(𝑎𝑞) → 𝐻(𝑎𝑞)
+ 𝐶𝑙(𝑎𝑞)
- sour taste; can conduct electricity when in an aqueous solution
o strong acids: acids that completely react with water
▪ complete dissociation (single arrow)
+
𝐻𝑁𝑂3 (𝑎𝑞) → 𝐻(𝑎𝑞)
+ 𝑁𝑂3 −
(𝑎𝑞)
▪ Hydrochloric acid (HCl), Hydrobromic acid (HBr),
Hydroiodic acid (HI), Nitric acid (HNO3), Sulfuric acid
(H2SO4), Perchloric acid (HClO4)
o weak acids: acids that partially react with water
▪ incomplete dissociation (double arrow)
+
−
𝐶𝐻3 𝐶𝑂𝑂𝐻(𝑎𝑞) ↔ 𝐻(𝑎𝑞)
+ 𝐶𝐻3 𝐶𝑂𝑂(𝑎𝑞)
▪ major quantity: original form
▪ lesser quantity: dissociated form (H+)
▪ Hydrofluoric acid (HF), Nitrous acid (HNO2),
Phosphoric acid (H3PO4), Acetic acid (CH3COOH)
41
Practical Example: The gastric juices synthesized by the stomach is naturally
acidic because of the presence of hydrochloric acid, which facilitates the
ease of digestion and aid in the elimination of foreign microorganisms taken
in such as bacteria (How does the stomach work?, 2016).
+
−
𝐻𝐶𝑙(𝑎𝑞) → 𝐻(𝑎𝑞)
+ 𝐶𝑙(𝑎𝑞)
Bases: proton acceptor
-
yields hydroxide ions (OH-) in the presence of water
+
−
𝐻(𝑎𝑞)
+ 𝑂𝐻(𝑎𝑞)
→ 𝐻2 𝑂(𝑙)
-
does not react with water chemically, instead it dissociates into
ions (including the OH-)
+
−
𝑁𝑎𝑂𝐻(𝑠) 𝐻2 𝑂 → 𝑁𝑎(𝑎𝑞)
+ 𝑂𝐻(𝑎𝑞)
- bitter taste; slippery; can conduct electricity when in an aqueous
solution
o strong bases: bases that completely dissociates in water
▪ mostly ionic compounds: alkali and alkaline earth metals
bonded with hydroxide
▪ Group 1: sodium hydroxide (NaOH), potassium hydroxide
(KOH)
▪ Group 2: calcium hydroxide (Ca(OH)2), barium hydroxide
(Ba(OH)2)
2+
−
𝐵𝑎(𝑂𝐻)2 (𝑠) 𝐻2 𝑂 → 𝐵𝑎(𝑎𝑞)
+ 2𝑂𝐻(𝑎𝑞)
o weak bases: bases that partially dissociates when introduced to
water
−
𝑁𝐻3 (𝑎𝑞) + 𝐻2 𝑂(𝑙) ↔ 𝑁𝐻4+(𝑎𝑞) + 𝑂𝐻(𝑎𝑞)
Practical Example: Patients suffering from metabolic acidosis, mostly due to
kidney failure, are administered with bicarbonate, a weak acid, to keep the
body fluids from being too acidic.
𝐻 + + 𝐻𝐶𝑂3− ↔ 𝐶𝑂2 + 𝐻2 𝑂
Neutralization Reaction
- reaction between an acid and a base
- acid – base reactions in aqueous solutions tend to produce water and
salt (salt: ionic compound with a cation other than H+ and anion other
than OH- or O2-)
𝑎𝑐𝑖𝑑 + 𝑏𝑎𝑠𝑒 → 𝑠𝑎𝑙𝑡 + 𝑤𝑎𝑡𝑒𝑟
Practical Example: Magnesium hydroxide, commercially available with
aluminum hydroxide and simeticone, is used as a medication for
hyperacidity (excess production of acid by the stomach).
𝑀𝑔(𝑂𝐻)2 (𝑠) + 2𝐻𝐶𝑙(𝑎𝑞) → 𝑀𝑔𝐶𝑙2 (𝑎𝑞) + 2𝐻2 𝑂(𝑙)
2+
+
2+
−
−
𝑀𝑔(𝑎𝑞)
+ 2𝑂𝐻(𝑎𝑞)
+ 2𝐻(𝑎𝑞)
+ 2𝐶𝑙 −(𝑎𝑞) → 𝑀𝑔(𝑎𝑞)
+ 2𝐶𝑙(𝑎𝑞)
+ 2𝐻2 𝑂(𝑙)
2+
+
2+
−
−
𝑀𝑔(𝑎𝑞)
+ 2𝑂𝐻(𝑎𝑞)
+ 2𝐻(𝑎𝑞)
+ 2𝐶𝑙 −(𝑎𝑞) → 𝑀𝑔(𝑎𝑞)
+ 2𝐶𝑙(𝑎𝑞)
+ 2𝐻2 𝑂(𝑙)
+
−
𝐻(𝑎𝑞)
+ 𝑂𝐻(𝑎𝑞)
→ 𝐻2 𝑂(𝑙)
42
Practical Example: A patient experiencing heartburn (excess stomach acid –
HCl - leaking into the esophagus) will be recommended to intake an antacid.
One of the most common antacids commercially available is calcium
carbonate (with famotidine and magnesium hydroxide).
𝐶𝑎𝐶𝑂3 (𝑠) + 2𝐻𝐶𝑙(𝑎𝑞) → 𝐶𝑎𝐶𝑙2 (𝑎𝑞) + 𝐻2 𝑂(𝑙) + 𝐶𝑂2 (𝑔)
2+
+
2+
𝐶𝑎(𝑎𝑞)
+ 𝐶𝑂32−(𝑎𝑞) + 2𝐻(𝑎𝑞)
+ 2𝐶𝑙 −(𝑎𝑞) → 𝐶𝑎(𝑎𝑞)
+ 2𝐶𝑙 −(𝑎𝑞) + 𝐻2 𝑂(𝑙) + 𝐶𝑂2 (𝑔)
2+
+
2+
𝐶𝑎(𝑎𝑞)
+ 𝐶𝑂32−(𝑎𝑞) + 2𝐻(𝑎𝑞)
+ 2𝐶𝑙 −(𝑎𝑞) → 𝐶𝑎(𝑎𝑞)
+ 2𝐶𝑙 −(𝑎𝑞) + 𝐻2 𝑂(𝑙) + 𝐶𝑂2 (𝑔)
+
𝐶𝑂32−(𝑎𝑞) + 2𝐻(𝑎𝑞)
→ 𝐻2 𝑂(𝑙) + 𝐶𝑂2 (𝑔)
Example: Reaction between hydrochloric acid (strong acid) and sodium
hydroxide (strong base)
𝐻𝐶𝑙(𝑎𝑞) + 𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝑁𝑎𝐶𝑙(𝑎𝑞) + 𝐻2 𝑂(𝑙)
+
+
+
−
−
𝐻(𝑎𝑞)
+ 𝐶𝑙 −(𝑎𝑞) + 𝑁𝑎(𝑎𝑞)
+ 𝑂𝐻(𝑎𝑞)
→ 𝑁𝑎(𝑎𝑞)
+ 𝐶𝑙(𝑎𝑞)
+ 𝐻2 𝑂(𝑙)
+
+
+
−
−
𝐻(𝑎𝑞)
+ 𝐶𝑙 −(𝑎𝑞) + 𝑁𝑎(𝑎𝑞)
+ 𝑂𝐻(𝑎𝑞)
→ 𝑁𝑎(𝑎𝑞)
+ 𝐶𝑙(𝑎𝑞)
+ 𝐻2 𝑂(𝑙)
+
−
𝐻(𝑎𝑞)
+ 𝑂𝐻(𝑎𝑞)
→ 𝐻2 𝑂(𝑙)
Example: Reaction between hydrocyanic acid (weak acid) and sodium
hydroxide (strong base)
𝐻𝐶𝑁(𝑎𝑞) + 𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝑁𝑎𝐶𝑁(𝑎𝑞) + 𝐻2 𝑂(𝑙)
+
+
−
−
𝐻𝐶𝑁(𝑎𝑞) + 𝑁𝑎(𝑎𝑞)
+ 𝑂𝐻(𝑎𝑞)
→ 𝑁𝑎(𝑎𝑞)
+ 𝐶𝑁(𝑎𝑞)
+ 𝐻2 𝑂(𝑙)
+
+
−
−
𝐻𝐶𝑁(𝑎𝑞) + 𝑁𝑎(𝑎𝑞)
+ 𝑂𝐻(𝑎𝑞)
→ 𝑁𝑎(𝑎𝑞)
+ 𝐶𝑁(𝑎𝑞)
+ 𝐻2 𝑂(𝑙)
−
−
𝐻𝐶𝑁(𝑎𝑞) + 𝑂𝐻(𝑎𝑞)
→ 𝐶𝑁(𝑎𝑞)
+ 𝐻2 𝑂(𝑙)
3. Oxidation – Reduction Reaction
- redox reaction
- transfer of electrons between reactants to produce ionic
compounds
- loss of electrons (increase in oxidation number) =
oxidation/oxidized = reducing agent (LEORA)
- gain of electrons (decrease in oxidation number) =
reduction/reduced = oxidizing agent (GEROA)
Example: Reaction between two hydrogen molecules and one oxygen
molecule to produce two molecules of water.
2 𝐻2 + 𝑂2 → 2𝐻2 𝑂
0
0
+1 -2
Hydrogen = oxidized = increase of oxidation number from 0 to +1 (loss of 1
electron) = reducing agent
Oxygen = reduced = decrease of oxidation number from 0 to -2 (gain of -2
electrons) = oxidizing agent
43
Practical Example: Fenton Reaction - Hydrogen peroxide, commonly found
in beverages such as instant coffee, reacts with transition metal ions in the
body forming oxidized iron and a highly reactive hydroxyl radical. Over
production of hydroxyl radicals in the body can lead to cell death and
mutations. Thus, it is linked to several degenerative diseases such as
atherosclerosis, cancer, and neurological disorders.
3+
𝐹𝑒 2+ + 𝐻2 𝑂2 + 𝐻+ → 𝐹𝑒(𝑎𝑞)
+ 𝐻2 𝑂 + 𝑂𝐻 •
+2
+1 -1
+1
+3
(Beiras, 2018)
+1 -2 -2 +1
Iron = oxidized = increase of oxidation number from +2 to +3 (loss of 1
electron) = reducing agent
Hydrogen Peroxide (H2O2) = reduced = decrease of oxidation number of
oxygen ion from -1 to -2 (gain of 1 electron) = oxidizing agent
Oxidation Number of Atoms
1. Atoms in elemental state (neutral atom): zero (0)
examples: Ne, H2, O2, F2, Cl2, Br2, I2, S8
2. Monoatomic ions (one-atom ions): charge of the ion
example: Na+ = +1
3. Fluorine (in all of its compounds): -1
4. Group 1 elements: +1
example: Na2SO3 = each atom of Na has an oxidation number of +1
5. Group 2 elements: +2
example: CaCl2 = oxidation number of Ca is +2
6. Group 17 elements (in binary compounds with metals): -1
example: FeCl3 = each atom of Cl has an oxidation number of -1
7. Oxygen in most compounds: -2, except when
- in compounds with fluorine: +2 (OF2)
- peroxides: -1 (H2O2)
- superoxide ion (O2-): - ½
8. Hydrogen (in most compounds): +1 (H2O, NH3)
- in hydrides with less electronegative atoms: -1 (NaH, LiH, CaH2)
9. Accounting the oxidation number of atoms of other elements. First, it
should always be remembered that the sum of the oxidation numbers of all
atoms in any neutral molecule is equal to zero. Thus, the oxidation number
of other atoms not mentioned above can be easily solved algebraically.
example: oxidation number of Mn in KMnO4
K = 1 * (+1) = +1
O = 4 * (-2) = - 8
Mn = (-8) + (+1) = +7
thus, for the molecule to be neutral (net charge = 0), charge of Mn
should be +7.
44
Balancing Equations by Change in the Oxidation Number Method
𝐾2 𝐶𝑟2 𝑂7 + 𝑆 + 𝐻2 𝑂 → 𝑆𝑂2 + 𝐾𝑂𝐻 + 𝐶𝑟2 𝑂3
1. Determine the atoms that change in oxidation number and indicate
the oxidation numbers of these atoms on both sides of the equation.
𝐾2 𝐶𝑟2 𝑂7 + 𝑆 + 𝐻2 𝑂 → 𝑆 𝑂2 + 𝐾 𝑂 𝐻 + 𝐶𝑟2 𝑂3
+1 +6 -2
0 +1 -2
+4 -2 +1 -2 +1 +3 -2
gain of 3 e- per K atom or gain of 6 e- per K2Cr2O7
loss of 4 e- per S atom
2. Determine the number of electrons transferred per formula unit of
reducing and oxidizing agents.
Sulfur = oxidized = increase in the oxidation number from 0 to +4 (loss of 4
electrons) = reducing agent LEORA
Potassium permanganate (K2Cr2O7) = reduced = decrease in the oxidation
number of Cr from +6 to +3 (gain of 3 electrons) = oxidizing agent GEROA
3. Adjust the coefficient of the reducing and oxidizing agents in such
a way that the number of electrons lost will be equal to the
number of electrons gained.
Sulfur (S):
4 es- lost per formula unit
Potassium permanganate (K2Cr2O7): 6 es- gained per formula unit
Thus, find the least common multiple of both values. This value will be the
number of electrons lost and gained. Subsequently, it will be used to
determine the number of moles of the reducing and oxidizing agents needed
for the equation to be balanced. For the case of sulfur and potassium
permanganate, the least common multiple is 12.
For S (lost electrons):
12 es- ÷ 4 es- = 3 moles of S
For K2Cr2O7 (gained electrons): 12 es-÷ 6es- = 2 moles of K2Cr2O7
2 𝐾2 𝐶𝑟2 𝑂7 + 3𝑆 + 𝐻2 𝑂 → 𝑆𝑂2 + 𝐾𝑂𝐻 + 𝐶𝑟2 𝑂3
4. Balance the elements that changed in oxidation number by
adjusting the coefficients of the substances.
Consider that the number of sulfur (S) atoms in the reactants’ side is 3.
Therefore, there should also be 3 S atoms in the products’ side. So,
coefficient of SO2 should be 3.
2 𝐾2 𝐶𝑟2 𝑂7 + 3 𝑆 + 𝐻2 𝑂 → 3 𝑆𝑂2 + 𝐾𝑂𝐻 + 𝐶𝑟2 𝑂3
For the chromium (Cr), total number of atoms in the reactants’ side is 4.
Thus, there should also be 4 atoms of Cr in the products’ side. So, coefficient
of Cr2O3 must be 2.
45
2 𝐾2 𝐶𝑟2 𝑂7 + 3 𝑆 + 𝐻2 𝑂 → 3 𝑆𝑂2 + 𝐾𝑂𝐻 + 2 𝐶𝑟2 𝑂3
5. Balance the rest of the elements in the equation by inspection.
Balance all the other elements first before H and O. Thus, start with
balancing potassium (K)
Since there are a total of 4 atoms of potassium in the reactants’ side, balance
the equation by assigning 4 as a coefficient of KOH.
2 𝐾2 𝐶𝑟2 𝑂7 + 3 𝑆 + 𝐻2 𝑂 → 3 𝑆𝑂2 + 4 𝐾𝑂𝐻 + 2 𝐶𝑟2 𝑂3
Next, balance the hydrogen atoms. In the products’ side, there are 4 H
atoms, therefore, the coefficient of H2O must be 2.
2 𝐾2 𝐶𝑟2 𝑂7 + 3 𝑆 + 2 𝐻2 𝑂 → 3 𝑆𝑂2 + 4 𝐾𝑂𝐻 + 2 𝐶𝑟2 𝑂3
Lastly, inspect the number of oxygen (O) atoms in both sides.
Reactants
Products
K2Cr2O7
2 * 7 = 14
SO2
3*2= 6
H2O
2*1= 2
KOH
4*1= 4
Cr2O3
2*3= 6
TOTAL
16
TOTAL
16
6. Check if all atoms are balanced.
Example: Balancing Redox Reactions
Step 1:
𝐶𝑢 + 𝐻𝑁𝑂3 → 𝐶𝑢(𝑁𝑂3 )2 + 𝑁𝑂 + 𝐻2 𝑂
Step 2:
𝐶𝑢 + 𝐻𝑁𝑂3 → 𝐶𝑢(𝑁𝑂3 )2 + 𝑁𝑂 + 𝐻2 𝑂
0
+5*
+2
+2
loss of 2 es- per mole of Cu
gain of 3 es- per moles of HNO3
* HNO3 = H+ + NO3NO3- = N5+ + O3-2
Step 3:
Copper (Cu):
Nitric acid (HNO3):
LCM = 6
For Cu (lost electrons):
For HNO3 (gained electrons):
2 es- lost per formula unit
3 es- gained per formula unit
6 es- ÷ 2 es- = 3 moles of Cu
6 es- ÷ 3 es- = 2 moles of HNO3
3 𝐶𝑢 + 2 𝐻𝑁𝑂3 → 𝐶𝑢(𝑁𝑂3 )2 + 𝑁𝑂 + 𝐻2 𝑂
46
Step 4:
3 𝐶𝑢 + 2 𝐻𝑁𝑂3 → 3 𝐶𝑢(𝑁𝑂3 )2 + 2 𝑁𝑂 + 𝐻2 𝑂
Step 5:
However, observe that even after adjusting the number of atoms of nitrogen
that underwent change in oxidation number, the overall number of nitrogen
atoms is still not balanced. Why is this?
Note that the nitrate ion coming from the nitric acid is also responsible for
the nitrate component of the copper (II) nitrate. Thus, it should suffice both
the nitrogen atoms in the copper (II) nitrate and nitrogen monoxide.
N atoms in the product side
Cu(NO3)2 = 3 * 2 = 6
NO =
2*1=2
TOTAL =
8
Therefore, for the N atoms to be balanced, the coefficient of HNO3 should
be 8.
3 𝐶𝑢 + 8 𝐻𝑁𝑂3 → 3 𝐶𝑢(𝑁𝑂3 )2 + 2 𝑁𝑂 + 𝐻2 𝑂
Balance the hydrogen atoms.
8 H atoms in the reactants’ side, thus, the coefficient of H2O should be 4.
3 𝐶𝑢 + 8 𝐻𝑁𝑂3 → 3 𝐶𝑢(𝑁𝑂3 )2 + 2 𝑁𝑂 + 4 𝐻2 𝑂
Account for the last element, oxygen.
Reactants’ side = 8 * 3 = 24
Products’ side = [3 * (3 * 2)] + (2 * 1) + (4 * 1) = 24
Therefore, the oxygen atoms are already balanced. Subsequently, the whole
chemical equation is finally balanced.
Balancing Net Ionic Equations (Redox Reactions)
𝐶𝑢 + 𝐻3 𝑂+ + 𝑁𝑂3− → 𝐶𝑢2+ + 2 𝑁𝑂3− + 𝑁𝑂 + 𝐻2 𝑂
However, we must remove the spectator ions (ions that do not participate
in the chemical reaction) to account for the net ionic equation.
𝐶𝑢 + 𝐻3 𝑂+ + 𝑁𝑂3− → 𝐶𝑢2+ + 2 𝑁𝑂3− + 𝑁𝑂 + 𝐻2 𝑂
𝐶𝑢 + 𝐻3 𝑂+ + 𝑁𝑂3− → 𝐶𝑢2+ + 𝑁𝑂 + 𝐻2 𝑂
Step 1 – 4:
𝐶𝑢 + 𝐻3 𝑂+ + 𝑁𝑂3− → 𝐶𝑢2+ + 𝑁𝑂 + 𝐻2 𝑂
0
+5
+2
+2
loss of 2 es- per mole of Cu
gain of 3 es- per moles of HNO3
LCM = 6
For Cu (lost electrons):
6 es- ÷ 2 es- = 3 moles of Cu
For NO3- (gained electrons):
6 es- ÷ 3 es- = 2 moles of NO33 𝐶𝑢 + 𝐻3 𝑂+ + 2 𝑁𝑂3− → 3 𝐶𝑢2+ + 2 𝑁𝑂 + 𝐻2 𝑂
47
Step 5: Balance the charge by adjusting the coefficient of H3O+ or OH-,
whichever is present in the reaction.
Reactant
Charge
Cu
0
NO3
2 * (-1) = -2
TOTAL
-2
Product
Charge
Cu2+
3 * (+2) = +6
NO
0
TOTAL
+6
Therefore, to achieve a +6 net charge in the reactants’ side, the total charge
coming from the hydronium ion (H3O+) should be equal to +8 (8 – 2 = 6).
Thus, the coefficient of H3O+ should be 8.
3 𝐶𝑢 + 8 𝐻3 𝑂+ + 2 𝑁𝑂3− → 3 𝐶𝑢2+ + 2 𝑁𝑂 + 𝐻2 𝑂
Step 6: Balance the remaining atoms.
Start with the hydrogen atoms in the products’ side. There are a total of 24
atoms in the reactants’ side, thus, the coefficient of the H2O should be 12,
for the hydrogen atoms on both side to be equal.
3 𝐶𝑢 + 8 𝐻3 𝑂+ + 2 𝑁𝑂3− → 3 𝐶𝑢2+ + 2 𝑁𝑂 + 12 𝐻2 𝑂
Account for the oxygen atoms.
Reactants’ side : (8 * 1) + (2 * 3) = 14; Products’ side: (2 * 1) + (12 * 1) = 14
Focus Questions
Instructions: Please submit your answers through LMS. The document
should be in letter size. Use Calibri, 12 pt. 1.15 spacing. On the upper-left
corner of the first page, write your name, grade level, section, and the date
of submission. For uniformity purposes, name your document as
IOChemistry_Last Name_Unit6. Properly cite your references using APA
format.
1. Explain the difference between chemical and ionic equations.
2. What are the characteristics of an acid-base neutralization reaction?
3. Is it possible to have a reaction involving oxidation, but not reduction?
Related Readings
To read more about the mole concept, please refer to the webpage below
- 7.1 The Mole Concept
https://courses.lumenlearning.com/suny-mccintroductorychemistry/chapter/formula-mass-and-mole-conceptfrom-che151/
48
To read more about the oxidation states
- Oxidation States (Oxidation Numbers)
http://chemguide.co.uk/inorganic/redox/oxidnstates.html#:~:tex
t=Peroxides%20include%20hydrogen%20peroxide%2C%20H,of%2
0%2D1%20to%20balance%20it.
To further read about Precipitation Reaction
- Precipitation Reactions (Chemistry LibreText)
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/M
odules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemi
stry/Main_Group_Reactions/Reactions_in_Aqueous_Solutions/Pr
ecipitation_Reactions?fbclid=IwAR3sntUOWUfof-Ii8gb0AfdM6yGvSXjW2LWDwsTz55Svx5hH8eqLXcZMDk
Learning Assessment
General Instruction: Please input all your answers in the same document as
the Focus Questions.
A. Balance the following equations:
1. 𝐶𝑢(𝑠) + 𝐻𝑁𝑂3 (𝑎𝑞) → 𝐶𝑢(𝑁𝑂3 )3 (𝑎𝑞) + 𝐻2 𝑂(𝑙) + 𝑁𝑂(𝑔)
2. 𝑃𝐶𝑙5 (𝑠) + 𝐻2 𝑂(𝑙) → 𝑃𝑂𝐶𝑙3 (𝑙) + 𝐻𝐶𝑙(𝑎𝑞)
3. 𝐴𝑙(𝑠) + 𝐻2 𝑆𝑂4 (𝑎𝑞) → 𝐴𝑙2 (𝑆𝑂4 )3 (𝑠) + 𝐻2 (𝑔)
4. Aqueous solutions of magnesium chloride and sodium hydroxide
react to produce solid magnesium hydroxide and aqueous sodium
chloride
B. Given the balanced equations below, write the complete ionic and net
ionic equations for the following:
1. 𝐾2 𝐶2 𝑂4 (𝑎𝑞) + 𝐵𝑎(𝑂𝐻)2 (𝑎𝑞) → 2𝐾𝑂𝐻(𝑎𝑞) + 𝐵𝑎𝐶2 𝑂4 (𝑠)
2. 𝑃𝑏(𝑁𝑂3 )2 (𝑎𝑞) + 𝐻2 𝑆𝑂4 (𝑎𝑞) → 𝑃𝑏𝑆𝑂4 (𝑠) + 2𝐻𝑁𝑂3 (𝑎𝑞)
C. Balance the following equation by change in oxidation number method.
Also, account for the net ionic equations.
1.𝐾𝑀𝑛𝑂4 + 𝐻𝐶𝑙 + 𝐾𝐼 → 𝑀𝑛𝐶𝑙2 + 𝐼2 + 𝐾𝐶𝑙 + 𝐻2 𝑂
2. 𝑁𝑎𝐶𝑙 + 𝐻2 𝑆𝑂4 + 𝑀𝑛𝑂2 → 𝑁𝑎2 𝑆𝑂4 + 𝑀𝑛𝐶𝑙2 + 𝐻2 𝑂 + 𝐶𝑙2
49
UNIT 7: Acid – Base Chemistry
Intended Learning Outcomes
At the end of the unit, you are expected to:
1. identify acids and bases according to Arrhenius concept, Lewis
concept, and Bronsted – Lowry concept;
2. assess the relative strengths of acids and bases;
3. perform calculations relating to pH and pOH; and
4. calculate the pH of a buffer solution.
Introduction
In the previous chapter, it was discussed to balance acid–base
equations. However, since the scope of acid–base is very wide, thus, a
separate unit is set to discuss the concepts it includes.
Unlocking Difficulties
Definition of Terms:
1. Chemical equilibrium
- a state where there is no net change in the concentrations of the
reactants and products
2. Molarity
- indicated by symbol M
- number of moles (m) per liter (L) solution
3. Amphoteric
- a substance that can both act as an acid and a base
4. Dissociation
- the process of breaking up of compounds into its components
5. Hydrolysis
- chemical breakdown of substances through water
6. Electrolyte
- result of dissociation
- charge species
7. Auto – ionization of H2O
- a process where water produces hydroxide ion (OH-) and hydrogen
ion (H+)
50
Lecture Notes
Concepts of Acid and Base
1. Arrhenius Concept
- Acid: produces H+ when in solution with H2O
- Base: produces OH- when in solution with H2O
2. Lewis Concept
- Acid: electron pair acceptor (should contain available orbital to
accept an electron pair)
- Base: electron pair donor (should have an available electron pair)
3. Bronsted-Lowry Concept
- Acid: proton donor (produces conjugate base)
- Base: proton acceptor (produces conjugate acid)
Conjugate Acid – Base Pairs
- based on Bronsted – Lowry Concept of acid – base reactions
- two substances (one acid and one base) in acid – based equilibrium
Example: Reaction between ammonia and water
𝑁𝐻3 + 𝐻2 𝑂 ↔ 𝑁𝐻4+ + 𝑂𝐻−
NH3 (base) – NH4+ (conjugate acid); H2O (acid) – OH- (conjugate base)
Strength of Acid
1. Strong acids: undergoes 100% dissociation
- Hydrohalic acids = Hydrogen Halides
o hydrogen + halide ion (negatively charged halogen atom)
o HCl, HBr, HI (HF only hydrohalic acid not considered strong)
- Mineral acids = acids that do not contain atoms of carbon
o HNO3, H2SO4, HClO4
+
−
𝐻𝐶𝑙 𝐻2 𝑂 → 𝐻(𝑎𝑞)
+ 𝐶𝑙(𝑎𝑞)
0.1M
0.1 M 0.1M
*only the dissociated ions of HCl will be present in the solution
2. Weak acids: undergoes less than 100% dissociation
- CH3COOH, HCN, H3PO4
𝐶𝐻3 𝐶𝑂𝑂𝐻 𝐻2 𝑂 ↔ 𝐶𝐻3 𝐶𝑂𝑂− + 𝐻 +
0.1 M
> 0.1 m
> 0.1 M
–
*molecules of CH3COOH and the ions, CH3COO and H+, will be present in
A strong acid produces a weak conjugate base. On the other hand, a weak
acid produces a strong conjugate base.
Given the hypothetical acids whose acid is the following: HX > HY > H2O > HZ
produces the following bases: X– < Y – < OH –< Z –
51
Strength of Base
1. Strong base: undergoes 100% dissociation
- Alkali Metal Hydroxides (Group 1)
o NaOH, LiOH, KOH, RbOH, CsOH
- Alkaline Earth Metal Hydroxides (Group 2)
o Ca(OH)2, Mg(OH)2, Sr(OH)2, Ba(OH)2
+
−
𝑁𝑎𝑂𝐻 + 𝐻2 𝑂 → 𝑁𝑎(𝑎𝑞)
+ 𝑂𝐻(𝑎𝑞)
0.1 M
0.1 M
0.1 M
*only the dissociated ions of NaOH will be present in the solution
2. Weak base: undergoes less than 100% dissociation
o NH3
𝑁𝐻3 + 𝐻2 𝑂 ↔ 𝑁𝐻4+ + 𝑂𝐻−
0.1 M
> 0.1 M > 0.1 M
*molecules of NH3 and the ions, NH4+and OH– will be present in the solution
Comparing Acid Strength
- comparing the tendency to dissociate or remove a proton
1. Binary Acid
- (H + X): X is any non-metal
a. For the same group: ↑ size of X = ↑ acidity
o NH3< PH3< AsH3
b. For the same period: ↑ electronegativity of X = ↑ acidity
o PH3< H2S <HCl
o H2O < H2S <HCl
2. Oxyacids
- Tertiary acids (H + X + O)
a. If X belongs to the same group and have the same oxidation
number: ↑ electronegativity of X = ↑ high acidity
o HIO, HBrO, HClO
H I O
H Br O
H Cl O
+1 +1 -2
+1 +1 -2
+1 +1 -2
Therefore: HIO < HBrO < HClO
b. The same X, but different number of oxygen attached: ↑
oxidation numberof X = ↑acidity
o HClO, HClO2, HClO3, HClO4
H Cl O
H Cl O2 H Cl O3 H Cl O4
+1 +1 -2
+1 +3 -2
+1 +5 -2
+1 +7 -2
Therefore: HClO < HClO2 < HClO3 < HClO4
o another way: ↑ number of attached oxygen = ↑acidity
52
The pH and pOH of a Solution
Since the concentrations of H+ and OH- involve minute values, it is usually
expressed in terms of pH (power of Hydrogen).
- Acidic = Low pH / more [H+] = High pOH / less [OH-]
- Basic = High pH / less [H+] = Low pOH / more [OH-]
Auto-ionization of H2O
+
−
𝐻2 𝑂(𝑙) ↔ 𝐻(𝑎𝑞)
+ 𝑂𝐻(𝑎𝑞)
Figure 22: The pH Scale
(https://www.news-medical.net/health/pH-in-the-Human-Body.aspx)
Solving for pH, given the concentration of H+ or H3O+ : 𝑝𝐻 = − 𝑙𝑜𝑔 𝑙𝑜𝑔 [𝐻+ ]
+
Solving for concentration of H+ or H3O+, given the pH: [𝐻 ] = 𝑎𝑛𝑡𝑖𝑙𝑜𝑔(−𝑝𝐻)
Solving for the pOH, given the concentration of OH-: 𝑝𝑂𝐻 = − 𝑙𝑜𝑔 𝑙𝑜𝑔 𝑂𝐻 −
−
Solving for concentration of OH-, given the pOH: [𝑂𝐻 ] = 𝑎𝑛𝑡𝑖𝑙𝑜𝑔 (−𝑝𝑂𝐻)
Ionization constant of water
𝐾𝑤 = [𝐻 + ][𝑂𝐻− ]
𝑎𝑡 25℃, 𝐾𝑤 = 1 × 10−14
𝐾𝑤 = 𝐾𝑎 𝐾𝑏
Solving for pH, given pOH and vice versa: 𝑝𝐻 + 𝑝𝑂𝐻 = 14
Examples:
1. Solve for the pH of 0.015 M HNO3
𝐻𝑁𝑂3 → 𝐻 + + 𝑁𝑂3−
0.015 M 0.015 M 0.015 M
𝑝𝐻 = − 𝑙𝑜𝑔 𝑙𝑜𝑔 [𝐻 + ] = − 𝑙𝑜𝑔 𝑙𝑜𝑔 0.015 = 1.82
2. Solve for the pOH, pH and concentration of H+ of 0.01 M NaOH
𝑁𝑎𝑂𝐻 → 𝑁𝑎+ + 𝑂𝐻−
0.01 M
0.01 M 0.01 M
𝑝𝑂𝐻 = − 𝑙𝑜𝑔 𝑙𝑜𝑔 [𝑂𝐻− ] = − 𝑙𝑜𝑔 𝑙𝑜𝑔 0.01 = 2
𝑝𝐻 + 𝑝𝑂𝐻 = 14
𝑝𝐻 = 14 − 𝑝𝑂𝐻 = 14 − 2 = 12
𝐾𝑤 = [𝐻 + ][𝑂𝐻− ]
−14
𝐾𝑤
1 × 10
[𝐻 ] =
− =
0.01
[𝑂𝐻 ]
+
= 1 × 10
−12
53
Dissociation of Weak Acids
+
𝐻𝐴(𝑎𝑞) ↔ 𝐻(𝑎𝑞)
+ 𝐴−
(𝑎𝑞)
acid hydrogen ion conjugate base
Acid Ionization Constant
𝐾𝑎 =
+
[𝐻(𝑎𝑞)
][𝐴−
(𝑎𝑞) ]
[𝐻𝐴(𝑎𝑞) ]
𝑝𝐾𝑎 = − 𝑙𝑜𝑔 𝑙𝑜𝑔 𝐾𝑎
𝐾𝑎 = 𝑎𝑛𝑡𝑖𝑙𝑜𝑔 (−𝑝𝐾𝑎 )
- high Ka = stronger acid = lower pKa
Example: Solve for the pH of the weak acid, 2 M CH3COOH (Ka = 1.8 x 10-5).
↔
CH3COOH
CH3COOH+
Initial
2
Change
–x
+x
+x
Equilibrium Conc.
2 –x
x
x
−
+
[𝐶𝐻3 𝐶𝑂𝑂 ][𝐻 ]
𝐾𝑎 =
[𝐶𝐻3 𝐶𝑂𝑂𝐻]
(𝑥)(𝑥)
1.8 × 10−5 =
2−𝑥
assume change in the concentration of CH3COOH is negligible
(𝑥)(𝑥)
1.8 × 10−5 =
2
−5
1.8 × 10 (2) = (𝑥)(𝑥)
3.6 × 10−5 = 𝑥 2
√3.6 × 10−5 = √𝑥 2
6 × 10−3 = 𝑥
to check if assumption is valid
𝑥
6 × 10−3
× 100 =
× 100 = 0.3%
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑐𝑜𝑛𝑐
2
since 0.3 < 5%, therefore, the assumption is valid
𝑝𝐻 = − 𝑙𝑜𝑔 𝑙𝑜𝑔 [𝐻+ ] = − 𝑙𝑜𝑔 𝑙𝑜𝑔 6 × 10−3 = 2.22
Dissociation of Weak Base
−
−
𝐵(𝑎𝑞)
↔ 𝐻𝐵(𝑎𝑞) + 𝑂𝐻(𝑎𝑞)
base conjugate acid hydroxide ion
Base Ionization Constant
𝐾𝑏 =
−
[𝐻𝐵(𝑎𝑞) ][𝑂𝐻(𝑎𝑞)
]
−
[𝐵(𝑎𝑞)
]
𝑝𝐾𝑏 = − 𝑙𝑜𝑔 𝑙𝑜𝑔 𝐾𝑏
𝐾𝑏 = 𝑎𝑛𝑡𝑖𝑙𝑜𝑔 (−𝑝𝐾𝑏 )
- high Kb = stronger base = lower pKb
54
Example: Solve for pOH and pH of the weak base, 1.0 M CH3NH2
(methylamine, Kb = 4.38 x 10-4)
𝐻2𝑂 ↔
CH3NH2
CH3NH3+
OHInitial
1
Change
–x
+x
+x
Equilibrium Conc.
1 –x
x
x
[𝐶𝐻3 𝑁𝐻3+ ][𝑂𝐻− ]
𝐾𝑏 =
[𝐶𝐻3 𝑁𝐻2 ]
(𝑥)(𝑥)
4.38 × 10−4 =
1−𝑥
assume change in the concentration of CH3NH2 is negligible
(𝑥)(𝑥)
4.38 × 10−4 =
1
−4
4.38 × 10 (1) = (𝑥)(𝑥)
4.38 × 10−4 = 𝑥 2
√4.38 × 10−4 = √𝑥 2
0.02 = 𝑥
to check if assumption is valid
𝑥
0.02
× 100 =
× 100 = 2%
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑐𝑜𝑛𝑐
1
since 2% < 5%, therefore, the assumption is valid
𝑝𝐻 = − 𝑙𝑜𝑔 𝑙𝑜𝑔 [𝐻+ ] = − 𝑙𝑜𝑔 𝑙𝑜𝑔 0.02 = 1.7
𝑝𝐻 + 𝑝𝑂𝐻 = 14
𝑝𝑂𝐻 = 14 − 𝑝𝐻 = 14 − 1.7 = 12.3
Percent Ionization
- accounts for the extent of dissociation in weak acids and bases
[𝑋]
% 𝐷𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛 =
× 100%
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
- Acids: [X] = [H+]
- Bases: [X] = [OH-]
Example: Solve the percent dissociation of the weak acid, 0.1 M HCOOH (Ka
= 1.77 x 10-4)
Initial
Change
Equilibrium Conc.
HCOOH
0.1
–x
0.1 –x
↔
HCOO+x
x
H+
+x
x
[𝐻𝐶𝑂𝑂− ][𝐻+ ]
𝐾𝑎 =
[𝐻𝐶𝑂𝑂𝐻]
(𝑥)(𝑥)
1.77 × 10−4 =
0.1 − 𝑥
assume change in the concentration of CH3NH2 is negligible
55
(𝑥)(𝑥)
0.1
−4
1.77 × 10 (0.1) = (𝑥)(𝑥)
1.77 × 10−5 = 𝑥 2
1.77 × 10−4 =
√1.77 × 10−5 = √𝑥 2
4.21 × 10−3 = 𝑥
to check if assumption is valid
𝑥
4.21 × 10−3
× 100 =
× 100 = 4.21%
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑐𝑜𝑛𝑐
0.1
since 4.21% < 5%, therefore, the assumption is valid
𝑥
4.21 × 10−3
% 𝐷𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛 =
× 100 =
× 100 = 4.21%
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑐𝑜𝑛𝑐
0.1
pH of Salts
- some ions undergo hydrolysis resulting in a acidic or basic solution
- only ions for weak electrolyte (weak acid and base) can undergo
hydrolysis
Hydrolysis of Cation: 𝑀+ + 𝐻2 𝑂 ↔ 𝑀𝑂𝐻 + 𝐻+
- because of the presence of free H+, resulting solution is acidic
Hydrolysis of Anion: 𝐵− + 𝐻2 𝑂 ↔ 𝐻𝐵 + 𝑂𝐻−
- because of the presence of free OH-, resulting solution is basic
1. Case I: Only the cation will undergo hydrolysis
- example: ammonium chloride in water: 𝑁𝐻4 𝐶𝑙 𝐻2 𝑂 → 𝑁𝐻4+ + 𝐶𝑙 −
- NH4+ is an electrolyte of the weak base, NH3, while Cl- is an electrolyte
of the strong acid, HCl. Thus, only NH4+ will undergo hydrolysis.
𝑁𝐻4+ + 𝐻2 𝑂 ↔ 𝑁𝐻4 𝑂𝐻 + 𝐻+
- Therefore, aqueous ammonium chloride solution is acidic.
2. Case II: Only the anion will undergo hydrolysis
- example: sodium fluoride in water: 𝑁𝑎𝐹 𝐻2 𝑂 → 𝑁𝑎+ + 𝐹 −
- Na+ is an electrolyte of the strong base, NaOH, while F- is an electrolyte
of the weak acid, HF. Thus, only F- will undergo hydrolysis
𝐹 − + 𝐻2 𝑂 ↔ 𝐻𝐹 + 𝑂𝐻−
- Therefore, aqueous sodium fluoride solution is basic.
3. Case III: Both the cation and anion will not undergo hydrolysis
−
+
- example: sodium chloride in water: 𝑁𝑎𝐶𝑙 𝐻2𝑂 → 𝑁𝑎 + 𝐶𝑙
- Na+ is an electrolyte of the strong base, NaOH, while Cl- is an
electrolyte of the strong acid, HCl. Therefore, both electrolytes will
not undergo hydrolysis.
- Thus, the pH of the solution will be neutral. It is determined by the
auto – ionization of H2O
56
4. Case IV: Both the cation and anion will undergo hydrolysis
- example: ammonium fluoride in water: 𝑁𝐻4 𝐹 𝐻2 𝑂 → 𝑁𝐻4+ + 𝐹 −
- NH4+ is an electrolyte of the weak base, NH3, while F- is an electrolyte
of the weak acid, HF. Thus, both electrolytes will undergo hydrolysis.
𝑁𝐻4+ + 𝐻2 𝑂 ↔ 𝑁𝐻4 𝑂𝐻 + 𝐻+
- to account for the actual strength of NH4+, Ka should be calculated
𝐹 − + 𝐻2 𝑂 ↔ 𝐻𝐹 + 𝑂𝐻−
- similarly, to account for the actual strength of HF, Kb
o if Ka = Kb, the solution is neutral
o if Ka> Kb, the solution is acidic
o if Ka< Kb, the solution is basic
Examples: Predict the acidity of the following solutions.
1. Na2CO3 = presence of weak electrolyte CO3-2
− Case II = Basic
2. FeCl3 = presence of weak electrolyte Fe +3
− Case I = Acidic
3. NH4CH3COO = presence of two weak electrolytes, NH4+ and CH3COO− Case IV = pH is dependent on the Ka and Kb of the corresponding weak
acid (CH3COOH) and base (NH3)
4. Ba(NO3)4
− both strong electrolytes, coming from a strong acid (HNO3) and a
strong base (Ba(OH)2)
− Case III = pH is neutral
Example: Solve for the pH of 0.30 M NaF aqueous solution (Ka = 7.2 x 10-4)
NaF will undergo dissociation
𝑁𝑎𝐹 𝐻2 𝑂 → 𝑁𝑎+ + 𝐹 −
0.30 M
0.30 M 0.30 M
-
F is a weak electrolyte of HF, thus, it will undergo hydrolysis
↔
FH2O
HF
Initial
0.30
Change
–x
+x
Equilibrium Conc. 0.30 –x
x
OH+x
x
𝐾𝑤 = 𝐾𝑎 𝐾𝑏
𝐾𝑤
1 × 10−14
𝐾𝑏 =
=
= 1.39 × 10−11
−4
𝐾𝑎 7.2 × 10
[𝐻𝐹][𝑂𝐻− ]
𝐾𝑏 =
[𝐹 − ]
(𝑥)(𝑥)
1.39 × 10−11 =
0.30 − 𝑥
assume change in the concentration of CH3NH2 is negligible
(𝑥)(𝑥)
1.39 × 10−11 =
1
57
1.39 × 10−11 (0.30) = (𝑥)(𝑥)
4.17 × 10−12 = 𝑥 2
√4.17 × 10−12 = √𝑥 2
2.04 × 10−6 = 𝑥
to check if assumption is valid
𝑥
2.04 × 10−6
× 100 =
× 100 = 6.8 × 10−6 %
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑐𝑜𝑛𝑐
0.30
−6
since 6.8 × 10 % < 5%, therefore, the assumption is valid
𝑝𝑂𝐻 = − 𝑙𝑜𝑔 𝑙𝑜𝑔 [𝑂𝐻− ] = − 𝑙𝑜𝑔 𝑙𝑜𝑔 2.04 × 10−6 = 5.7
𝑝𝐻 + 𝑝𝑂𝐻 = 14
𝑝𝐻 = 14 − 𝑝𝐻 = 14 − 5.7 = 8.3
Buffer
- solutions that resist drastic change in pH even with the addition of
small amounts of H+ and OH- consists of either a weak acid and its conjugate base, or a weak base
and its conjugate acid
- amount of H+ or OH- ions a buffer can absorb without significant
change in pH
Henderson – Hasselbalch Equation
[𝐻𝐴]
[𝐴− ]
[𝐴− ]
𝑝𝐻 = 𝑝𝐾𝑎 +𝑙𝑜𝑔 𝑙𝑜𝑔
[𝐻𝐴]
[𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑐𝑎𝑡𝑖𝑜𝑛]
𝑝𝑂𝐻 = 𝑝𝐾𝑏 +𝑙𝑜𝑔 𝑙𝑜𝑔
[𝑏𝑎𝑠𝑒]
[𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑐𝑎𝑡𝑖𝑜𝑛]
𝑝𝑂𝐻 = 𝑝𝐾𝑏 −𝑙𝑜𝑔 𝑙𝑜𝑔
[𝑏𝑎𝑠𝑒]
𝑝𝐻 = 𝑝𝐾𝑎 −𝑙𝑜𝑔 𝑙𝑜𝑔
Example: A buffered solution contains 0.03 M ammonia (Kb = 1.8 x 10-5) and
0.50 M ammonium chloride. Solve for the pH of the solution.
solution contains a weak base (NH3) and its conjugate acid (NH4Cl →
NH4+ + Cl-)
𝑁𝐻3 − 𝑁𝐻4+
0.03 M 0.50 M
𝐾𝑤 = 𝐾𝑎 𝐾𝑏
𝐾𝑤
1 × 10−14
𝐾𝑎 =
=
= 5.56 × 10−10
−5
𝐾𝑏 1.8 × 10
𝑝𝐾𝑎 = − 𝑙𝑜𝑔 𝑙𝑜𝑔 𝐾𝑎 = − 𝑙𝑜𝑔 𝑙𝑜𝑔 5.56 × 10−10 = 9.26
[𝐴− ]
0.03
𝑝𝐻 = 𝑝𝐾𝑎 +𝑙𝑜𝑔 𝑙𝑜𝑔
= 9.26 +𝑙𝑜𝑔 𝑙𝑜𝑔
= 8.04
[𝐻𝐴]
0.50
therefore, the solution is basic
58
Focus Questions
Instructions: Please submit your answers through LMS. The document
should be in letter size. Use Calibri, 12 pt. 1.15 spacing. On the upper-left
corner of the first page, write your name, grade level, section, and the date
of submission. For uniformity purposes, name your document as
IOChemistry_Last Name_Unit6. Properly cite your references using APA
format.
1. Read the following articles:
7. Acid-Base Balance. (n.d.). Functions of Cells and Human Body.
Retrieved October 7, 2020 from http://fblt.cz/en/skripta/viivylucovaci-soustava-a-acidobazicka-rovnovaha/7acidobazickarovnovaha/
Acidosis and Alkalosis. (n.d). AACC Lab Tests Online. Retrieved
October
7,
2020
from
https://labtestsonline.org/conditions/acidosis-and-alkalosis
Castro, D., Keenaghan, M. (2020, March 21). Arterial Blood Gas.
Retrieved
October
7,
2020
from
https://www.ncbi.nlm.nih.gov/books/NBK536919/
Interpreting ABGs (Arterial Blood Gases) Made Easy. (n.d.). Ausmed.
Retrieved
October
7,
2020
from
https://www.ausmed.com/cpd/articles/interpreting-abgs
2. How are the four articles related to Acid-Base Chemistry?
3. Based on the articles, how will learning about Acid-Base Chemistry equip
you in becoming effective Medical Technologists?
4. Cite at least 4 points from the articles that you think is essential to
remember as a future Medical Technologist.
Related Readings
To read more about Molarity
- https://www.khanacademy.org/science/ap-chemistry/states-ofmatter-and-intermolecular-forces-ap/mixtures-and-solutionsap/a/molarity
- https://courses.lumenlearning.com/introchem/chapter/molarity/
To read about Ka, Kb, pKa, and pKb
- Find in your search engine, Chemistry LibreTexts 7.11: Relationship
between Ka, Kb, pKa, pKb(Stephen Lower, June 20, 2020)
59
To read more about ICE table/chart
- Making an ICE Chart
https://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/IC
Echart.htm
- 13.4 Equilibrium Calculations
https://openstax.org/books/chemistry-2e/pages/13-4-equilibriumcalculations
To read more about the 5% rule for calculations involving weak acids and
bases
- The 5% rule for weak acids and bases
https://www.chemteam.info/AcidBase/five-percent-rule.html
Key Equations involving Acid – Base Chemistry
- Flowers, P., Theopold, K. Langley, R., Neth, E. J., Robinson, W. R.
(2019).
Chemistry: Atoms First 2e. Houston, Texas: OpenStax.
p.787
Learning Assessment
General Instruction: Please input all your answers in the same document as
the Focus Questions.
1. Identify the Bronsted-Lowry acid and its conjugate base, and the
Bronsted-Lowry base and its conjugate acid in the following chemical
equations.
a. 𝐻𝐵𝑟 + 𝐻2 𝑂 → 𝐻3 𝑂+ + 𝐵𝑟 −
c. 𝐻2 𝑆 + 𝑁𝐻2− → 𝐻𝑆 − + 𝑁𝐻3
b. 𝐻𝑁𝑂3 + 𝐻2 𝑂 → 𝐻3 𝑂+ + 𝑁𝑂3− d. 𝐻2 𝑃𝑂4− + 𝑂𝐻 − → 𝐻𝑃𝑂4−2 + 𝐻2 𝑂
2. Solve the pH and pOH of the following:
a. 0.2 M HCl
b. 0.0031 M Ca(OH)2
3. Calculate the concentration of H+ and OH- of the following:
a. 3.0 M HNO3 b. 2.5 M KOH
4. Propionic acid, C2H5CO2H (Ka=1.34 x 10-5), is used in the manufacture of
calcium propionate, a food preservative. What is the pH of a 0.698 M
solution of C2H5CO2H?
5. Solve for the pOH and pH of 0.20 M ammonia solution (Kb = 1.8 x 10-5).
6. The first disinfectant used by Joseph Lister was the carbolic acid (C6H5OH,
pKa = 10.0). What is the pH of 0.10 M carbolic acid solution?
7. Calculate the percent ionization of a 0.20 M aspirin (HC9H7O4) solution,
given its Ka = 3.0 x 10-4.
8. Solve for the pH of 0.20 M sodium acetate (Ka of acetic acid = 1.8 x 10-5.
9. Solve for the pH of 0.10 M NH4Cl (Kb for NH3 = 1.8 x 10-5).
10. Calculate for the pH and pOH of a buffer solution that contains 0.20 M
and 0.40 M NaF. (Ka of HF= 6.6 x 10-4)
60
UNIT 8: Functional Groups and Organic Reactions
Intended Learning Outcomes
At the end of the unit, you are expected to:
1. describe the different players in organic chemistry;
2. identify the type of a given reaction;
3. know the major classifications of organic compounds; and
4. identify important functional groups.
Introduction
Organic molecules surround us. It is involved in the manufacture of
clothes, foods and feeds and medicine and drugs. It is also the molecules
responsible for our senses, emotions, and fight and flight responses (Klein,
2015).
Initially, the term organic chemistry referred to compounds obtained
from living organisms. However, soon, by 1800s, it was established that
organic compounds are carbon-containing molecules. McMurry (2010)
described these special compounds to be low-melting solids. They are also
difficult to isolate, purify and work-with.
Unlocking Difficulties
Definition of Terms:
1. Organic compounds
- carbon-containing compounds
2. Functional groups
- classifications of organic compounds
3. Bond Cleavage
- splitting of a chemical bond
Recall:
1. Electronegativity
- tendency or capacity to attract a pair of electrons
2. Double bond
- a type of chemical bond where two electron pairs are being
shared
3. Triple bond
- another chemical bond which involves three electron pairs
61
Lecture Notes
Functional Groups
Organic compounds which have the similar structural features and
share the same chemical behavior are groups into families called functional
groups (McMurry, 2010). Compounds belonging to the same functional
group are expected to react chemically in the same way. Thus, it is that: “the
chemistry of every organic molecule, regardless of the size and complexity,
is determined by the functional groups it contains”.
Organic
Compounds
With
Heteroatoms
Hydrocarbons
Aliphatic
Aromatic
Alkanes
With Oxygen
With Sulfur
Alcohols and
Phenols
Alkenes
Ethers
Alkynes
Aldehydes and
Ketones
Cyclic Analogs
Carboxylic acids,
Ethers, and
Anhydrides
With Halogen
With Nirigen
Thiols
Alkyl and Aryl
Halides
Amines
Sulfides
Acid chloride
Amides and
Amino acids
Figure 1: Families of Organic Compounds
Figure 2: Different Representations of the Molecular Structure of Organic Compounds
(https://openstax.org/books/chemistry-2e/pages/20-1-hydrocarbons)
62
Figure 3: Functional Groups
(https://www.compoundchem.com/2014/01/24/functional-groups-in-organiccompounds/)
Common Functional Groups
1. Hydrocarbons
- only contains C and H
a. Alkane
- saturated aliphatic hyrocarbon
- single-bonded carbons
- structure: C-C
- substituent: -yl; suffix: -ane
heptane
3-methylheptane
63
b. Alkene
- unsaturated aliphatic hydrocarbon 2-heptene
- double-bonded carbons
- structure: C=C
Ethenylbenzene (Styrene)
- substituent: -enyl; suffix: -ene
c. Alkyne
- also unsaturated aliphatic hydrocarbon
- triple-bonded carbons
- structure: C≡C
- always included in the main chain
- suffix: -yne
1-heptyne
2-methyl-3-octyne
d. Arene
- contains benzene ring
2. Halide
- contains halides (F, Cl, Br, I)
- structure: R-X
- substituents of the main
- fluoro-, chloro-, bromo-,
5-bromo-3-heptene
1-chloro-4-
chain iodobenzene
iodo-
3. Alcohols
- contains hydroxyl group (OH)
1-bromo-3-chloro-4-methyl-2-pentene
3-ethyl-1-pentanol
- structure: R-OH
cyclohexane-1,2-diol
- substituent: hydroxy4-hydroxybutanal
- suffix: -ol
4. Ether
- containes alkoxy group (-OR)
- alkyl group bonded to oxygen
- structure: R-O-R
- substituent of the main chain
- substituents: -oxy-
methoxyethane
5. Aldehydes
- contains carbonyl group (C=O)
- structure: R-CHO
- H: single-bond with C
- O: double-bond with the same C
- always terminal group
- always part of the main chain
- suffix: -al
ethoxybenzene
1-methoxyheptane
5-methyl-4-hexenal
heptanal
64
6. Ketones
- contains carbonyl group (C=O)
- structure: R-CO-R
2,5-heptadione
- C and O are double bonded
- substituent: -oxo
- suffix: -one
2-pentanone
2,2-dimethyl-3oxopentanal
7. Carboxylic acid
- structure: R-COOH
- O: double bond with C
heptanoic acid
- OH: single bond with the same C (enanthic acid)
- always terminal group
- always part of the main chain
benzoic acid
- suffix: -oic acid
octanedioic acid
8. Ester
- a derivative of carboxylic acid
- formed by carboxylic acid and alcohol
- H in the OH group is replaced with C
- structure: R-COOC
- terminal: part of the main chain
- suffix: -oate
ethyl ethanoate
ethyl benzoate
9. Amide
- also a derivative of carboxylic acid
- OH group is completely replaced
- structure: R-CON
- terminal: part of the main chain
- suffix: -amide
- substituents attached to nitrogen
atoms indicated by writing N
ethanamide
with N
N, N-diethylmethylpropanamide
N-methylpropanamide
10.Amines
- contains nitrogen
atom
- structure: R-N
- substituent: -aminop-aminobenzoic acid 2-methylcyclopentanamine
- indicated by N
- suffix: -amine
butanamine
65
11.Thiols
- contains SH
- structure: R-SH
- substituent: -mercapto- suffix: -thiol
methanethiol
2-mercaptoethanol
cyclohexanethiol
Other Carbonyl-containing Groups
- groups that act as substituents
1. Acyl group
- represented by R or C
- structure: -COR
- an aliphatic group
- alkyl or alkenyl group
2. Acetyl group
- structure: -COCH3
3. Formyl group
- structure: -CHO
4. Aroyl group
- structure: -COAr
- Ar = aromatic molecule
5. Benzoyl group
- structure: COC6H5
Acyl group
Acetyl group
Ar
Formyl group
Aroyl group
Benzoyl group
Examples
1. Aspirin
- Acetylsalicylic acid
- 2-Acetylbenzoic acid
- Analgesic, antipyretic (lower fever), anti-inflammatory
Carboxylic acid
Ester
Benzene ring
2. Advil or Motrin
- Ibuprofen
- 2-[4-(2-Methylpropyl)phenyl]propanoic acid
- Analgesic, antipyretic, anti-inflammatory
66
Alkane
substituents
Carboxylic acid
Benzene ring
3. Tylenol
- Acetaminophen
- N-(4-Hydroxyphenyl)ethanamide
- Analgesic, antipyretic
Amide
Alcohol
Benzene ring
4. Tamoxifen
- 2-[4-[(Z)-1,2-diphenylbut-1-enyl]phenoxy]-N,N-dimethylethanamine
- a selective estrogen receptor modulator (SERM)
- treatment or prevention of breast cancer
Benzene rings
Alkane
Amine
Ether
Alkene
Benzene ring
67
5. Valium or Diastat or Novazam
- Diazepam
- 7-chloro-1-methyl-5-phenyl-3H-1,4-benzodiazepin-2-one
- Anti-anxiety, sedative, hypnotic, and anticonvulsant
Amide
Amine
Benzene rings
Halide - chloro
6. Dopamine
- 4-(2-aminoethyl)benzene-1,2-diol
- the chemical hormone that mediates pleasure in the brain
- a neurotransmitter that sends messages from nerve cell to nerve cell
- precursor of norepinephrine and epinephrine
Amine
Alcohol
Benzene ring
7. Formaldehyde
- methanal
- a gas with unpleasant odor
- commercially available as formalin (37% formaldehyde by weight)
- for preservation of biological samples and for sterilization
Aldehyde
8. Acetone
- 2-propanone
- a colorless liquid commonly used as a solvent for various coatings,
and in pharmaceutical and chemical manufacture
68
Ketone
9. Isoamyl acetate
- 3-methylbutyl acetate
- responsible for the distinct flavor of banana
- especially high in concentration as banana becomes over-riped
Ester
10.
Monosodium glutamate
- sodium 2-aminopentanedioate
- sodium salt of glutamic acid
- enhances the meaty flavor in food
Carboxylic acid
Ketone
Amine
Nature of Organic Reactions
Types of Bond Cleavage
1. Homolytic bond cleavage
● covalent bond is broken equally: each atom gets exactly one
electron
● results in the formation of free-radicals (odd number electron)
● negligible difference in the electronegativity of both atoms
2. Heterolytic bond cleavage
● covalent bond is broken unequally: only one of the two atoms
sharing the electron pair will get the two electrons
● results in the formation of ions
● large difference in the electronegativity of both atoms
Figure 4: Bond Cleavage
(http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/radikale/radikale_einfuehrung.vlu.
html)
69
Examples:
https://wou.edu/chemistry/courses/online-chemistry-textbooks/ch450-and-ch451-biochemistry-defining-life-at-the-molecularlevel/chapter-7-catalytic-mechanisms-of-enzymes/
Homolytic Cleavage
Heterolytic Cleavage
https://www.chegg.com/homework-help/questions-and-answers/1-match-reactions-1-2-witht-type-bond-cleavage-homolyticheterolytic-cleavage---homolytic--q14589849
Species Involves in a Reaction
1. Substrate: primary reactant – an organic compound
2. Reagent: acts upon the substrate – usually inorganic compound(s)
● Nucleophillic reagent
o nucleus-loving: attack positively-charged positions
o negatively-charged chemical species
o electron-rich, therefore, can donate an electron pair
● Electrophillic reagents
o electron-loving: attack negatively-charged positions
o positively-charged chemical species
o electron-deficient, therefore, can accept an electron pair
● Radical-like reagents
o free-radicals: chemical species with odd number of
electrons
o react readily with negatively-charged positions
3. Catalyst: substance that hasten a chemical reaction
4. Transition state/Activated complex: unstable chemical species that
has partially broken or partially formed bonds
70
5. Intermediate: chemical species that are more stable than the
transition state/activated complex, but still highly reactive due to the
presence of unpaired electrons (charged)
● Carbocation
o heterolytic cleavage involving carbon
o carbon is bonded with a more electronegative atom
o results in carbon atom being positively-charged
● Carbanion
o heterolytic cleavage of carbon and a positively-charged
atom
o results in carbon atom being negatively-charged
● Free radicals
o homolytic cleavage involving a carbon atom
o induced by introducing the substance to light, high
temperature or peroxide
6. Product: the result of the chemical reaction between the substrate
and the reagent
7. By-product: obtained when the intermediate undergoes an alternate
route
Figure 5: Propene reaction with Water (Tolentino, 2017)
Overall Reaction and Reaction Mechanism
● Overall Reaction: shows the substrate, reagent and products
𝑠𝑢𝑏𝑠𝑡𝑟𝑎𝑡𝑒 + 𝑟𝑒𝑎𝑔𝑒𝑛𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠 → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡(𝑠)
● Reaction Mechanism: step by step description of how a reaction takes
place
𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 → → → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡(𝑠)
Organic Reactions
1. Addition
● reaction occurs with two carbon atoms bonded by double or
triple bonds
● atoms of the reagent attaches to the carbon atoms
● addition of atoms to carbon
71
Figure 6: Ethylene Reaction with Diatomic Chlorine
https://chemstuff.co.uk/unit-2/functional-groups/alkenes/
Figure 7: Ethylene Reaction with Water
https://chemstuff.co.uk/unit-2/functional-groups/alkenes/
2. Elimination
● Opposite of addition reaction
● Atoms attached in the carbon atoms are eliminated
● Formation of double or triple bonds between the two carbon
atoms
Figure 8: Dehydration of Ethanol
https://easychem.com.au/production-of-materials/renewable-ethanol/dehydration-of-ethanol/
Figure 8: Dehydrohalogenation of Iodoethane with Potassium hydroxide
https://intl.siyavula.com/read/science/grade-12/organic-molecules/04-organic-molecules-06
3. Substitution
● replacement or substitution of an atom (or group of atoms)
attached to carbon by another atom (or group of atoms)
Figure 9: Chloroethane Reaction with Water
https://intl.siyavula.com/read/science/grade-12/organic-molecules/04-organic-molecules-06
Figure 10: Bromoethane Reaction with Potassium Hydroxide
https://intl.siyavula.com/read/science/grade-12/organic-molecules/04-organic-molecules-06
72
4. Rearrangement
● movement of an atom (or group of atoms) within the organic
molecule or the substrate
Figure 11: 1-bromopropane to 2-bromopropane (Tolentino, 2017)
Figure 12: cis-but-2-ene to trans-but-2-ene by Treatment of an Acid Catalyst
(https://brainly.in/question/1706230)
Additional Examples:
https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/react2.html
5. Redox
● identified based on the increase or decrease in oxygen and/or
hydrogen atoms
● oxidation: addition of O or removal of H (or both)
● reduction: addition of H or removal or O (or both)
73
6. Acid-Base
● Bronsted-Lowry concept
o Acid: H+ donor
o Base: H+ acceptor
● Lewis concept
o Acid: electron pair acceptor
o Base: electron pair donor
Focus Questions
Instructions: Please submit your answers through LMS. The document
should be in letter size. Use Calibri, 12 pt. 1.15 spacing. On the upper-left
corner of the first page, write your name, grade level, section, and the date
of submission. For uniformity purposes, name your document as
IOChem_Last Name_Unit8. Properly cite your references using APA format.
Please refer to the LMS for the rubrics.
1. Cite at least 3 practical points on how organic chemistry will equip
you with your future profession.
2. Currently, spectroscopic techniques are used to determine the
structure of an unknown organic compound in just a day. These
technologies became available in the first half of the twentieth
century. Thus, how did the earlier scientists determine structures
of organic samples?
Related Readings
Functional Group Names, Properties, and Reactions
● https://courses.lumenlearning.com/boundlesschemistry/chapter/functional-group-names-properties-andreactions/?fbclid=IwAR0ytdz9lLghFs7_f3feXHuiAJBwcYoVTfo0zJ_
HiVM-hmNHYgP7Yi-sDk8
24.1 Functional Groups and Classes of Organic Compounds (August 13,
2020)
● https://chem.libretexts.org/Bookshelves/General_Chemistry/Boo
k%3A_Chemistry_(Averill_and_Eldredge)/24%3A_Organic_Comp
ounds/24.1%3A_Functional_Groups_and_Classes_of_Organic_Co
mpounds?fbclid=IwAR1DvjwDulWU51XuGEgGBgOXjzLlxaYOkAiE5SnWstCXJ4R_NpZ8ST5w4c
Examples of Organic Reactions
● https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/react
2.htm
Addition, elimination, and substitution reactions
● https://intl.siyavula.com/read/science/grade-12/organicmolecules/04-organic-molecules-06
74
Learning Assessment
General Instruction: Please submit the document containing your
answer for the following exercises through LMS.
1. Many compounds with desirable medicinal properties are isolated
from natural resources and are thus referred to as natural products.
However, a compound’s medicinal properties is known before the
structure of the compound has been determined. Below are
examples of compounds where the first proposed structure was
incorrect. In each case, the corresponding structure is also shown.
Identify all functional groups in each pair of compounds and then
compare the similarities and differences between their molecular
structures.
2.
3.
The following compound is an intermediate in the synthesis of a
gelator, which is a compound capable of self-assembling to form a
gel in an organic liquid. Identify each functional group in the
molecule.
Nonsteroidal anti-inflammatory drugs (NSAIDs) are commonly used
to treat minor aches and pains. Four of the most common NSAIDs
are:
a.
What is the molecular formula of acetaminophen?
b.
Aspirin, ibuprofen, and naproxen are all believed to
target the same enzyme, cyclooxygenase, which produces
substances called prostaglandins, that mediate inflammation.
Discuss any similarities in the structure of these drugs.
* Questions lifted from Klein (2015) and McMurry (2010)
75
UNIT 9: Hydrocarbons
Intended Learning Outcomes
At the end of the unit, you are expected to:
1. determine the name and structure of alkane, alkenes and alkynes;
2. predict the physical properties of the hydrocarbon based on their
chemical structure; and
3. write chemical reactions involving hydrocarbons.
Introduction
This unit will be focused on discussing the structure and general features
of the hydrocarbons, as well as physical and chemical properties.
Unlocking Difficulties
Definition of Terms:
1. Hydrocarbons
- organic compounds composed only of carbon and hydrogen
atoms
2. Alphatic hydrocarbons
- a hydrocarbon that contains carbon and hydrogen atoms
arranged in either a straight-chain, branched-chain or ringlike
3. Saturated aliphatic hydrocarbons
- aliphatic hydrocarbons joined by single bonds
4. Unsaturated aliphatic hydrocarbons
- aliphatic hydrocarbons joined by double bond and/or triple
bond
5. IUPAC
- International Union of Pure and Applied Chemistry
Lecture Notes
Alkanes
- aliphatic hydrocarbons
- can be a straight chain or branched
- also referred to as saturated hydrocarbons
- general molecular formula: CnH2n+2
- IUPAC name: always end in -ane
76
Figure 13: Straight-Chain Alkanes
(https://link.springer.com/chapter/10.1007/978-3-319-07605-8_1)
Figure 14: Branched Alkane
(https://socratic.org/questions/what-is-the-skeletal-structure-for-c-6h-14)
Physical Properties
- nonpolar substances
- increase in the carbon chain length = increase in the intermolecular
forces of attraction (attraction between atoms or ions) = increase in
melting and boiling points
- carbon chain length = IMFA = melting and boiling points
- C1 - C4 - gaseous state
- C5 - C17 - liquid state
- C18 and above - solid state
- branching weakens the IMFA, subsequently, lowering the melting
and boiling points
Chemical Properties of Alkanes
- alkanes do not have enough electron affinity, thus, they have low
reactivity towards common reagents
- alkanes sa undergo pyrolytic cracking (heating without the presence
of oxygen), combustion and halogenation
Halogenation
- takes place in the presence of heat or light
- alkanes react with halogens to yield alkyl halides
- a substitution reaction
77
- reactivity of halogens: F2 > Cl2 > Br2 > I2
- however, in the laboratory, only Cl2 and Br2 are used
- F2 is highly reactive, while I2 is nonreactive
Figure 15: Halogenation General Reaction (Tolentino, 2017)
Figure 16: Chlorination of Methane
(https://www.cliffsnotes.com/study-guides/chemistry/organic-chemistry-i/reactions-of-alkanes/alkaneshalogenation)
Cycloalkanes
- still classified as saturated hydrocarbons
- however, carbons are in rings
- general molecular formula: CnH2n
- name: cyclo- + Greek prefix + -ane
- structure is represented by polygons, where the number of vertices
is equal to the number of carbons in the molecule
- cycloalkanes containing less than four carbons are not highly
available due to angle strain (occurs when bond angles are forced to
deviate from the ideal angle)
- cyclohexane: strain-free
- cycloalkanes of intermediate size = moderate strain
- rings containing more than 14 carbons = strain-free
Chemical Properties
- cycloalkanes can undergo the same chemical reactions as
alkanes, however, up to some extent
Halogenation
- mechanism: substitution
Figure 17: Light-initiated Reaction of Cyclohexane with Chlorine
(https://www.chegg.com/homework-help/write-mechanism-light-initiated-reaction-cyclohexane-chlorin-chapter-4problem-44sp-solution-9780321971128-exc)
78
Figure 18: Light-initiated Reaction Methylcyclohexane with Bromine
(https://www.chegg.com/homework-help/organic-chemistry-1st-edition-chapter-12-problem-6p-solution9781118137512)
- cyclopropane when undergoes addition or substitution
reactions yield open chain products
Figure 19: Cyclopropare Subjected to Addition and Substitution Reactions
(https://www.askiitians.com/iit-jee-chemistry/organic-chemistry/cycloalkanes.aspx)
Alkenes
- classified as unsaturated aliphatic hydrocarbons
- general molecular formula: CnH2n
- IUPAC name: always end in -ene
- in nomenclature: have higher priority than alkanes
https://www.quora.com/What-are-some-examples-of-alkenes
Physical Properties
- nonpolar substances
- have low melting and boling points
- however melting and boiling points increases as number of
carbon atoms attached increases
- less dense than water
79
Chemical Properties
1.
Reduction-Catalytic Hydrogenation
- syn addition: addition of two substituents to the same
side of a double bond or a triple bond
http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/alkene/hydrierung.vl
u/Page/vsc/en/ch/12/oc/alkene/hydrierungen/hydrierungen.vscml.html
Figure 20: General Reactions for Reduction-Catalytic Hydrogenation
(https://www.masterorganicchemistry.com/2013/04/02/epoxidation-hydroxylation-cyclopropanation-alkenemechanism/)
2.
Electrophilic Addition
Figure 21: Electrophillic Addition with Alkenes (Tolentino,
2017)
3.
Free Radical Addition of HBr (hydrobromic acid)
Figure 22: Free Radical Addition of HBr (Tolentino, 2017)
80
4.
Oxidation
Figure 23: Oxidation of Alkenes (Tolentino, 2017)
Alkynes
- also classified as unsaturated aliphatic hydrocarbons
- general molecular formula: CnH2n-n
- IUPAC name: always end in -yne
Physical Properties
- nonpolar substances
- have low melting and boling points
- melting and boiling points increases as number of carbon
atoms attached increases
- less dense than water
Chemical Properties
1.
Hydrogenation
Figure 24: General Reaction for Hydrogenation of Alkynes
(https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wa
de)/09%3A_Alkynes/10.06_Reduction_of_Alkynes)
2.
Electrophilic Addition
a.
Halogenation
Figure 25: Halogenation of Alkynes
(https://www.chemistrysteps.com/hydrohalogenation-of-alkynes/)
81
b.
Hydrohalogenation
Figure 26: Hydrohalogenation of Alkynes
(https://www.chemistrysteps.com/hydrohalogenation-of-alkynes/)
c.
Hydration
Figure 27: Hydration of Alkynes
(https://www.chemistrysteps.com/acid-catalyzed-hydration-alkynes/)
3.
Acidity of Terminal Alkynes
Figure 28: Obtaining Metal Acetylides from Terminal Alkynes
(Tolentino, 2017)
4.
Oxidation
Figure 30: Oxidation of Alkynes
(https://chem.libretexts.org/Courses/Sacramento_City_College/SCC%3A_Chem_420__Organic_Chemistry_I/Text/10%3A_Alkynes/10.07%3A_Oxidation_of_Alkynes)
Aromatic Compounds
- compounds that contain six-membered benzene-like rings with
three double bonds
- general formula: C6H6
82
Figure 31: Structure of Benzene Ring
(https://www.indiamart.com/proddetail/benzene-2246634433.html)
The Hückle Rule
- criteria for Aromaticity
1. The molecule is cyclic.
2. The molecule is planar.
3. The molecule is conjugated.
4. The molecule has (4n+2)=π. The value of n should be any positive
integer.
Figure 32: Most Common Benzene-Containing Compounds
(https://www.researchgate.net/figure/Structures-of-benzene-and-the-simplest-benzenoids-used-in-the-currentstudy_fig2_263238378)
Other Benzene-Containing Compounds
Figure 33: Taxol - Anti-cancer Drug
(http://www.chem.ucla.edu/~harding/IGOC/B/benzene_ring.html)
83
Figure 20: Benzene-containing Medical Drugs
(https://laney.edu/corlett/wp-content/uploads/sites/234/2019/11/ch17.pdf)
Figure 21: Aromatic Compounds Other than Benzenze
(https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/Chapter_17%3A
_Benzene_and_Aromatic_Compounds/17.08_Examples_of_Aromatic_Compounds)
Chemical Properties
1.
Electrophilic Substitution of Benzene
a.
Halogenation
https://slideplayer.com/slide/5693672/
84
b.
Nitration
https://www.masterorganicchemistry.com/2018/04/30/electrophilic-aromatic-substitutions-2-nitration-andsulfonation/
c.
Sulfonation
https://www.masterorganicchemistry.com/2018/04/30/electrophilic-aromatic-substitutions-2-nitration-andsulfonation/
d.
Friedel-Crafts Alkylation
https://www.chemistrylearner.com/friedel-crafts-alkylation.html
e.
Friedel-Crafts Acylation
https://www.chemistrylearner.com/friedel-crafts-acylation.html
2.
Electrophilic Substitution of Monosubstituted Benzene
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/Chapte
r_18%3A_Electrophilic_Aromatic_Substitution/18.7%3A_Electrophilic_Aromatic_Substitution_of_Substitut
ed_Benzenes
85
3.
Side Chain Reactions of Substituted Benzene
- occurs at the group/s attached to the chain
a.
Halogenation of alkyl side chain
- Alkyl portion - undergoes free-radical substitution
(same way as alkanes does)
- Aromatic portion - undergoes electrophilic aromatic
substitution
https://www.clutchprep.com/organic-chemistry/side-chain-halogenation
b.
Oxidation of alkyl side chain
- conversion of -R group into -COOH group by addition of
hot KMnO4 (oxidizing agent)
- resulting product will always be benzoic acid
https://www.clutchprep.com/organic-chemistry/side-chain-oxidation
Qualitative Analysis of Hydrocarbons
- detection of the presences of hydrocarbons and aromatic
compounds
1.
Detection of Unsaturated Hydrocarbons
- for alkenes and alkynes
- Baeyer’s Test
a. 5 drops of test compound
b. Add 1 ml of 0.05 M KMnO4
- positive result: loss of the purple color and formation of brown
precipitate
- can be used to detect alkenes and alkynes
- alkenes and alkynes will undergo oxidation with the addition
of the oxidixing agent (KMnO4)
- can also be used to confirm that a compound is not a saturated
hydrocarbon
- since, alkanes have very low reactivity towards common
reagents, including oxidizing agents such as KMnO4
86
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_Lab_Techniques_(Nichol
s)/06%3A_Miscellaneous_Techniques/6.04%3A_Chemical_Tests/6.4D%3A_Individual_Tests
2.
Detection of Terminal Alkynes
- reaction with Ammoniacal Silver Nitrate
a. 5 drops of 0.1 N AgNO3
b. 6 N ammonium hydroxide until formed precipitates redissolves
c. Pour the solution to the test compound
- positive result: formation of insoluble solids
- can be used to detect alkynes
- terminal proton (H+) of the alkyne is released, forming
acetylide ion, which subsequently reacts with Ag+ to result to
white to gray precipitate (silver acetylide)
3.
Detection of Aromatic Compounds (Benzene Ring)
- Friedel-Crafts Alkylation
- reaction with CHCl2 (or CHCl3/chloroform) in AlCl3
a. 2 mL of dichloromethane/chloroform
b. Add the test compound
c. Mix and incline the tube to the side
d. Add 0.5-1g fine crystals of
anhydrous aluminum chloride at the sides
- positive result: formation of colored products
(yellow-‐-orange, red, blue, green, or
purple)
- detection of the presence of the
relatively stable product formed by the
benzene ring and the carbocation
http://kimiagar2010.blogspot.com/2010/04/friedel-crafts-test-for-aromatic.html?m=1
87
Focus Questions
Instructions: Please submit your answers through LMS. The document
should be in letter size. Use Calibri, 12 pt. 1.15 spacing. On the upper-left
corner of the first page, write your name, grade level, section, and the date
of submission. For uniformity purposes, name your document as
IOChem_Last Name_Unit9. Properly cite your references using APA format.
Please refer to the LMS for the rubrics.
1. Explain why is it possible to use Baeyer’s test for qualitative analysis
of alkenes and alkynes.
2. Explain why is it possible to detect the presence of aromatic
compounds using chloroform in aluminum chloride.
Related Readings
Hydrocarbons
● https://www.youtube.com/watch?v=1UE3hZ7cOP0
Alkanes and Cycloalkanes
● https://www.angelo.edu/faculty/kboudrea/molecule_gallery/01_
alkanes/00_alkanes.htm
Alkenes
● https://www.angelo.edu/faculty/kboudrea/molecule_gallery/02_
alkenes/00_alkenes.htm
Alkynes
● https://www.angelo.edu/faculty/kboudrea/molecule_gallery/03_
alkynes/00_alkynes.htm
Short Summary of IUPAC Nomenclature of Organic Compounds
● https://www.angelo.edu/faculty/kboudrea/organic/IUPAC_Hand
out.pdf
Baeyer’s Test to Test Presence of Unsaturated Compounds
● https://www.youtube.com/watch?v=0szJH6LoZFw
Qualitative Analysis of Compounds
● https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Boo
k%3A_Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Mis
cellaneous_Techniques/6.04%3A_Chemical_Tests/6.4D%3A_Indiv
idual_Tests
88
Learning Assessment
General Instruction: Please submit the document containing your
answer for the following exercises through LMS.
1.
Make a table that will highlight the general characteristics and
properties of the hydrocarbons.
Hydrocarbon
General
Physical
Detection
Properties
Properties
Alkane
Alkene
Alkyne
Aromatic
General properties = general molecular structure, IUPAC name
Physical properties = polarity, melting and boling points, density
Detection = What test/s to do to detect its presence?
2.
Find five examples of each:
a.
alkanes
b.
alkenes
c.
alkynes
d.
aromatic
Write the name of each compound, and draw the structure of its
molecule.
UNIT 10: Alcohols and Phenols, and Ethers and Epoxides
Intended Learning Outcomes
At the end of the unit, you are expected to:
1. identify the name and structure of alcohols and ethers;
2. predict physical properties of alcohols and phenols, and ethers and
epoxides; and
3. know how to conduct quantitative analysis for the presence of
alcohols and phenols.
Introduction
Unit 10 will focus on describing the physical and chemical properties of
the oxygen-containing functional groups, alcohols and phenols, and ethers
and epoxides. In addition, qualitative analysis of the said functional groups
will be specified.
89
Unlocking Difficulties
Recall:
1. Oxidation
- gain in oxygen atoms and/or loss of hydrogen atoms
2. Dehydration
- reverse of hydration reaction
- a reaction that involves loss of water
3. Primary Alcohol
- when the hydroxyl group is attached to a single alkyl group
4. Secondary Alcohol
- when the hydroxyl group is attached to a carbon that is also
attached to two other carbons
5. Tertiary Alcohol
- when the hydroxyl group is attached to a carbon that is also
attached to three other carbons
Lecture Notes
Alcohols
- an organic derivative of water
- -OH: hydroxyl group attached to a carbon atom
- OH bond is highly polar (responsible for the physical and
chemical properties of alcohol)
- names are the same as of alkane, however:
- suffix (if highest priority): -ol
- prefix (as substituent): hydroxyPhysical Properties
- alcohols are polar substances, however, polarity decreases with
increasing number of carbons in the main chain
- small alcohols as miscible in water, however, solubility
decreases as the number of carbons increases
- increase in the number of carbons in the main chain and branching
decreases intermolecular forces of attraction
- subsequently, lowering the melting and boiling points
- have higher boiling points and are more soluble in water than
hydrocarbons
Chemical Properties
1. Oxidation
- addition of oxygen atoms or removal of hydrogen atoms
- reagent can be:
- warm/hot KMnO4 or acidic Na2Cr2O7 or K2Cr2O7
90
http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/oxi_alk.vlu.html
2. Conversion into Alkyl Halides
- via Nucleophilic Substitution
Figure 1: Reaction of Alcohols to Hydrogen Halides
(http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch04/ch4-5.html)
Figure 2: Conversion of Alcohol to Alkyl Halide Using Thionyl chloride and
Phosphorus tribromide
(https://www.masterorganicchemistry.com/2015/03/20/pbr3-and-socl2/)
3. Dehydration of Alcohols to Alkene
- Elimination of H2O from alcohol
- acid-catalyzed reaction
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)/14%
3A_Reactions_of_Alcohols/14.04%3A_Dehydration_Reactions_of_Alcohols
Phenols
- compounds that have hydroxyl group (-OH) bonded directly to
benzene ring
- -OH group allows hydrogen bonding to other phenol molecules and
to water
Figure 3: Hydrogen bonding between Two Phenols
(https://www.zigya.com/study/book?class=12&board=hbse&subject=Chemistry&book=Chemistry+II&chapter=Alcoh
ols,+Phenols+and+Ethers&q_type=&q_topic=Alcohols+and+Phenols&q_category=&question_id=CHEN12071651)
91
Figure 4: Hydrogen bonding of Phenol and Water
(https://chemistrypartner.blogspot.com/2011/01/hydrogen-bonding.html)
Physical Properties
- compared to compounds of same size and molecular weight, the
hydrogen bonding in phenols raises its melting point, boiling point
and solubility in water
- phenols are more acidic than water and alcohols
- they can give up H+ (from the -OH group) easier
Chemical Properties
1. Electrophilic Aromatic Substitution
Figure 5: Nitration of Phenols
(https://byjus.com/chemistry/phenol-electrophilic-substitution/)
2. Oxidation
- phenols undergo oxidation more readily than simple alcohols
- common oxidizing agents: Ag2O, Na2Cr2O7, KMnO4
http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch24/ch24-3-5.html
3. Reaction with FeCl3
- general test for phenols
https://www.toppr.com/ask/question/neutral-fecl3-gives-purple-colour-with
Ethers
- oxygen atom bonded to two hydrocarbon groups
- R can be any alkyl, and it can also be an aromatic group
- a substituent: - oxyFigure 6: General Structure of Ether
(https://en.wikipedia.org/wiki/Ether)
92
Physical Properties
- C-O-C group is slightly polar
- thus, boiling point does not significantly differ from alkane
- hydrogen bonding is possible between ether and water
- small ethers are soluble in water
- similar with alcohols, solubility in water decreases as the number of
carbons in the main chain increases
Preparation of Ethers
1. Williamson Ether Synthesis
- ethers can be synthesized through reacting metal alkoxides
with alkyl halides
https://www.chemistrysteps.com/williamson-ether-synthesis/
2. Acid-catalyzed Dehydration of Alcohols
- symmetrical ethers with small alkyl groups
- a nucleophilic substitution reaction
https://www.masterorganicchemistry.com/2014/11/14/ether-synthesis-via-alcohols-and-acid/
Chemical Properties
1. Cleavage of Ethers by Acids
- since ether linkage is more stable, thus, it is less reactive to
bases, oxidizing agents and reducing agents
- however, when introduced to high temperature and
concentrated acid, ether linkage can undergo cleavage to yield
an alcohol and an alkyl halide
- a substitution reaction
https://www.masterorganicchemistry.com/2014/11/19/ether-cleavage/
93
Epoxides
- a three-membered ring ether
- a cyclic compound with two carbons and one oxygen
Preparation of Epoxides
1. When alkenes react with peroxy acids, it yields epoxides
https://www.masterorganicchemistry.com/2015/01/26/epoxides-the-outlier-of-the-ether-family/
Chemical Properties
1. Epoxide Opening with Acid
- epoxides can undergo cleavage using acids (still substitution)
- however, process can be done in a milder condition since
epoxides tend to be more reactive due to bond strain
Figure 7: Ether reacting with aqueous acid to produce a diol
(https://www.masterorganicchemistry.com/2015/02/02/opening-of-epoxides-with-acid/)
Figure 8: Ethers react with Hydrohalic acids to a compound containing
alcohol and halide
(https://www.masterorganicchemistry.com/2015/02/02/opening-of-epoxides-with-acid/)
2. Epoxide Opening with Base
- ring opening of epoxide can also occur using base and other
nucleophiles
- undergoes a substitution reaction using a base, alkoxide
(oxygen-containing alkyl), amine or Grignard reagents
(RMgX)
Figure 9: Epoxides introduced to basic condition
(https://www.masterorganicchemistry.com/2015/02/10/opening-of-epoxide-with-base/)
Figure 10: Epoxide treated with alkoxide
(https://www.masterorganicchemistry.com/2015/02/10/opening-of-epoxide-with-base)
94
Qualitative Analysis
1. Detection of Tertiary and Secondary Alcohols
- Lucas Test
- reaction is via nucleophilic substitution, hydroxyl group is
substituted by chlorine atom, therefore, forming chloroalkane
- different alcohols react to Lucas reagent at different rates
- primary alcohols: low reactivity, thus, must be heated
first
- secondary alcohol: reaction can be observed within 3 to
5 minutes
- tertiary alcohol: reaction can be observed immediately
http://www.chem.uiuc.edu/organic/Alcohols/Chapter%206/sec6-6/66.htm#:~:text=The%20Lucas%20test%20is%20a,water%20insoluble%20t%2Dbutyl%20chloride.
- reagent: hydrochloric acid - zinc chloride
1. Add 1 ml of sample
2. Add 6 ml pf HCl-ZnCl2 reagent
3. Shake and allow to stand
- positive result: formation of insoluble layer or emulsion
Figure 11:Formation of Insoluble Layer
(https://upload.wikimedia.org/wikipedia/commons/9/97/Lukastest_etoh_tbutoh.JPG)
2. Detection of Primary and Secondary Alcohols
- Baeyer’s Test = oxidation
- only primary and secondary alcohols can undergo oxidation
- primary alcohol will be converted to aldehyde and
carboxylic acid
- secondary alcohol will be converted to ketone
- tertiary alcohol lacks C-H bonding, thus it cannot
undergo oxidation
95
- reagent: neutral potassium permanganate
1. Add 1 ml of sample
2. Add 1 ml of neutral KMnO4
- positive result: loss of purple color
Figure 12: Loss of Purple Color
(https://www.sciencephoto.com/media/4335/view)
*positive both in Lucas Test and Baeyer’s Test = Secondary Alcohol
** positive in Lucas Test but negative in Baeyer’s Test = Tertiary Alcohol
*** negative in Lucas Test but positive in Baeyer’s Test = Primary Alcohol
Test
Lucas
Baeyer’s
Primary Alcohol
+
Secondary Alcohol
+
+
Tertiary Alcohol
+
-
3. Detection of Phenols and Derivatives
- Ferric Chloride Test
- reagent: 1% FeCl2
1. Add 5 ml of H2O
2. Add 1-2 drops of sample
3. Add 1-2 drops of 1% FeCl2
4. Shake
- positive result: formation of intense red, blue, purple and
green color
Figure 13: Formation of Intense Color
(https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_Lab_Techniques_(Nicho
ls)/06%3A_Miscellaneous_Techniques/6.04%3A_Chemical_Tests/6.4D%3A_Individual_Tests)
96
Focus Questions
Instructions: Please submit your answers through LMS. The document
should be in letter size. Use Calibri, 12 pt. 1.15 spacing. On the upper-left
corner of the first page, write your name, grade level, section, and the date
of submission. For uniformity purposes, name y our document as
IOChem_Last Name_Unit10. Properly cite your references using
APA format. Please refer to the LMS for the rubrics.
1. Explain in your own words the mechanism for the detection of
primary, secondary, and tertiary alcohols, as well as the detection
of phenols and its derivatives.
2. Find a specific product (can be a food or skin
care product, medicine or cleaning agent)
that contains either an alcohol, phenol,
ether, or epoxide. Present a picture of the
ingredient’s list of the product. And cite what
is the specific mode of action of the
functional group in the product.
Example:
Celeteque
Hydration
Facial
Moisturizer
 contains Panthenol
 Panthenol is the alcohol form of Vitamin B5. It acts as a
humectant, thus, it holds the moisture in the skin so that the
skin may not turn dry nor oily.
Related Readings
Alcohols and Phenols (August 11, 2020)
 https://chem.libretexts.org/Bookshelves/Organic_Chemistry/
Map%3A_Organic_Chemistry_(McMurry)/17%3A_Alcohols_an
d_Phenols
Ethers and Epoxides; Thiols and Sulfides (August 11, 2020)
 https://chem.libretexts.org/Bookshelves/Organic_Chemistry/
Map%3A_Organic_Chemistry_(McMurry)/18%3A_Ethers_and
_Epoxides_Thiols_and_Sulfides
Chapter 16: Ethers, Epoxides and Sulfides
 http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch16/ch
16-0.html
97
Learning Assessment
General Instruction: Please submit the document containing your answers
for the following exercises through LMS.
1. Make a table that will highlight the general characteristics and
properties of the functional groups presented in the unit.
Functional
General
Physical
Detection
Group
Properties
Properties
Alcohol
Phenol
Ether
Epoxide
General properties = general molecular structure, IUPAC name etc
Physical properties = polarity, melting point and boiling points, density
Detection = test to detect its presence
2. Find five alcohol-containing compounds, five phenol-containing
compounds, five ether-containing compounds, five epoxide-containing
compounds. Give the name of the compound and draw it molecular
structure.
UNIT 11: Aldehydes and Ketones, Carboxylic acid and
derivatives, and Amines
Intended Learning Outcomes
At the end of the unit, you are expected to:
1. determine the name and structure of aldehydes and
carboxylic acid and derivatives, and amines;
2. predict the physical properties of the aldehydes and
carboxylic acid and derivatives, and amines based
chemical structure; and
3. write chemical reactions involving aldehydes and
carboxylic acid and derivatives, and amines.
ketones,
ketones,
on their
ketones,
Introduction
This unit will discuss the rest of the common functional groups. These
are the oxygen-containing functional groups, aldehydes, ketones and
carboxylic acid, and its derivatives, including the nitrogen-containing
functional group, amine.
98
Unlocking Difficulties
Recall:
1. Carbonyl group
- an organic group containing a carbon atom double-bonded to
oxygen
Figure 14: General structure of Carbonyl group
(https://www.thoughtco.com/definition-of-carbonyl-605835)
Lecture Notes
Aldehydes and Ketones
 both are carbonyl-containing
 Aldehyde IUPAC name:
 always a terminal group
 highest priority = suffix = -al
 substituent = prefix = formly Ketone IUPAC name:
https://www.daviddarling.info/encyc
 always internal
lopedia/K/ketone.html
 highest priority = suffix = -one
 substituent = prefix = oxoPhysical Properties
 the carbonyl group contained in aldehydes are ketones is a polar
group
 have higher boiling point compared to ethers and hydrocarbons
(given similar molecular weight)
 cannot undergo hydrogen bonding with itself but can undergo
hydrogen bonding with water
 thus, small aldehydes and ketones are soluble in water
Preparation of Aldehydes
1. Hydration of Alkynes
 addition reaction
 two atoms of hydrogen and one atom of oxygen will be added to
alkyne
 triple bond of the alkyne will be broken
 site of triple bond = where the double-bonded oxygen will attach
99
Figure 15: Hydration of alkyne in the presence of H2SO4 and Hg2SO4
(Tolentino, 2017)
2. Ozonolysis of Alkenes
 substitution reaction
 oxygen atom from ozone (reagent) will attach to the site of the
double carbons
 the original alkene molecule will be divided
 product can be an aldehyde or ketone depending on the site of the
double bond
Tolentino, 2017
3. Oxidation of Alcohols
 oxidation of primary alcohols (alcohol attached to only one alkyl
group) will result to an aldehyde
 oxidation of secondary alcohol (hydroxyl group is bonded to a
carbon which at the same time is bonded to another two carbons
atoms) will result to a ketone
Tolentino, 2017
100
Chemical Properties
1. Addition of HCN
https://www.cliffsnotes.com/study-guides/chemistry/organic-chemistry-ii/aldehydes-and-ketones/reactions-ofaldehydes-and-ketones
2. Addition of Grignard reagent
http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch17/ch17-3-2-2.html
3. Addition of alcohol
https://courses.lumenlearning.com/suny-potsdam-organicchemistry2/chapter/21-3-formation-of-hydrateshemiacetals-acetals/
4. Addition of derivatives of ammonia
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/17%3A_Aldehy
des_and_Ketones_-_The_Carbonyl_Group/17.09%3A_Nucleophilic_Addition_of_Ammonia_and__Its__Derivatives
5. Oxidation of Aldehydes
https://2012books.lardbucket.org/books/introduction-to-chemistry-general-organic-andbiological/s17-10-properties-of-aldehydes-and-ke.html
101
6. Reduction to Alcohol
http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch15/ch15-2-6.html
7. Aldol Condensation
https://www.chemistrysteps.com/crossed-aldol-and-directed-aldol-reactions/
Carboxylic Acid
 contains the carboxyl group
 combination of carbonyl group and hydroxyl group
 derivatives: acid halide, acid anhydride, ester, amide
Physical properties
 carboxylic acids have higher boiling points and melting points than
alcohols
 acid chlorides, acid anhydrides and ester have similar boiling points
with aldehydes and ketones, but have lower boiling points than
alcohols and carboxylic acids
 amides have high boiling points because of the presence of
hydrogen bonding
Preparation of Carboxylic acids
1. Oxidative methods
 alkenes to aldehydes to carboxylic acid
 alkynes to carboxylic acid
 alkyl benzene to benzoic acid
 primary alcohols to aldehydes to carboxylic acid
 aldehyde to carboxylic acid
2. Hydrolysis of Nitriles (-CN)
http://www.ilpi.com/msds/ref/carboxylicacid.html
102
Acid Halide
https://chem.libretexts.org/Courses/Sonoma_State_University/SSU_Chem_335B/Material_for_Exam_1/Chapter_22
%3A_Carboxylic_Acids_and_Their_Derivatives%E2%80%94_Nucleophilic_Acyl_Substitution/22.3_Nomenclature
Acid Anhydride
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Anhydrid
es/Nomenclature_of_Anhydrides
Ester
http://www.chemguide.co.uk/organicprops/alcohols/esterification.html
Amide
https://slideplayer.com/slide/5309463/
Amines
 organic derivatives of ammonia
 contains amino group
 can be in a chain (straight or branched), or in a ring
(heterocyclic amine
 if highest priority: suffix = -amine
 if as a substituent: prefix = amino-
Figure 16: Amino-containing molecules
(https://www.chemistrysteps.com/naming-amines-systematic-and-common-nomenclature/)
103
Physical Properties
 polar substances (similar with ammonia)
 can undergo hydrogen bonding with water and with itself (but not
all)
 thus, small amine molecules are soluble in water
 have higher boiling point than nonpolar compounds
 however, have lower boiling point than alcohols and carboxylic
acids
 soluble in less polar solvents such as ether, alcohol, benzene
 smells like ammonia
 aromatic amines: generally toxic
 similar with ammonia: has basic pH
Qualitative Analysis
1.
Detection of Aldehydes and Ketones
- 2, 4 DNP Test
1. Add 1 ml 95% ethanol
2. Add 1-2 drops of sample
3. Add 1 ml of 2, 4 DNP
4. Shake and let it stand for 15 minutes
- positive result: formation of yellow-orange-red precipitate
- aliphatic = yellow
- aromatic = red
Figure 17: Formation of yellow-to-red-colored precipitate
(https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Mi
scellaneous_Techniques/6.04%3A_Chemical_Tests/6.4D%3A_Individual_Tests)
2.
Detection of Aldehydes
- Tollen’s Test
1. Add 1.5 ml of Tollen’s reagent
2. Add 4 drops of sample
3. Shake
4. Heat in water bath
- positive result: formation of silver mirror at the bottom
Figure 18: Formation of silver mirror at the bottom or a colloidal
precipitate
(https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Mi
scellaneous_Techniques/6.04%3A_Chemical_Tests/6.4D%3A_Individual_Tests)
104
3. Detection of Aliphatic Aldehydes
- Benedict’s Test
- reagent: CuSO4 in sodium citrate
1. Add 1 ml of Benedict’s reagent
2. Add 5 drops of sample
3. Put the solution in water bath
- positive result: discoloration of light blue solution and
formation of red-brown precipitate
Figure 19: Formation of red-brown precipitate
(https://slideplayer.com/slide/4366040/)
4. Detection of Methyl Ketone
- Iodoform Test
- reagent: KI/I2
1. Add 1 ml of NaOH
2. Add 4 drops of sample
3. Add KI/I2 dropwise with shaking
4. Warm to 60°C
5. If solution is colorless, repeat step 3 and 4
- positive result: formation of yellow precipitate
Figure 20: Formation of yellow precipitate
(https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Mi
scellaneous_Techniques/6.04%3A_Chemical_Tests/6.4D%3A_Individual_Tests)
5. Detection of Carboxylic acids
- Esterification - ester formation
- reagent: ethanol
1. Add 2 drops of sample
2. Add 5 drops of ethanol
3. Add 1 drop of concentrated sulfuric acid
4. Warm the solution
- positive result: the solution has a pleasant smell
105
6. Detection of Esters
- Hydroxamic acid for Esters
- reagent: 0.5 M hydroxylamine in 95% ethanol
1. Add 2 drops of sample
2. Add 3 drops of alcohol NH2OH ‫ ـ‬HCl
3. Add 2 drops of NaOH
4. Heat to boil
5. Add drops of 2 M HCl until it becomes acidic
6. If solution becomes cloudy: add 1 ml of 95% ethanol
7. Add 2 drops of FeCl3 solution
- positive result: formation of magenta-colored complex with
FeCl3
Figure 21: Formation of magenta-colored complex
(https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Mi
scellaneous_Techniques/6.04%3A_Chemical_Tests/6.4D%3A_Individual_Tests)
Focus Questions
Instructions: Please submit your answers through LMS. The document
should be in letter size. Use Calibri, 12 pt. 1.15 spacing. On the upper-left
corner of the first page, write your name, grade level, section, and the date
of submission. For uniformity purposes, name y our document as
IOChem_Last Name_Unit11. Properly cite your references using
APA format. Please refer to the LMS for the rubrics.
1. Explain in your own words the mechanism for the detection of
aldehydes, ketones, carboxylic acids, and esters.
Related Readings
Nomenclature of Aldehydes and Ketones (September 13, 2020)
 https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supple
mental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/No
menclature_of_Aldehydes_and_Ketones#:~:text=Aldehydes%20deri
ve%20their%20name%20from,functional%20group%2C%20C%3DO.
Carboxylic Acid
 https://courses.lumenlearning.com/introchem/chapter/carboxylicacids/
Amines
 https://courses.lumenlearning.com/introchem/chapter/amines/
106
Learning Assessment
General Instruction: Please submit the document containing your
answer for the following exercises through LMS.
1. Make a table that will highlight the general characteristics and
properties of the functional groups presented in the unit.
Functional
General
Physical
Detection
Group
Properties
Properties
Aldehyde
Ketone
Carboxylic acid
Ester
Amide
Amine
General properties = general molecular structure, IUPAC name etc
Physical properties = polarity, melting point and boiling points, density
Detection = test to detect its presence
2. Find two aldehyde-containing compounds, two ketone-containing
compounds, two carboxylic acid-containing compounds, two estercontaining compounds, two amide-containing compounds, two aminecontaining compounds. Give its name and draw its molecular structure.
UNIT 12: Overview of Biomolecules
Intended Learning Outcomes
At the end of the unit, you are expected to:
1. determine the name and structure of carbohydrates and amino
acids;
2. identify carbohydrates and amino acids into classes; and
3. identify glycosidic linkage between monosaccharides, as well as
peptide bond formation in amino acids.
Introduction
Unit 12 is a short introduction to Biochemistry. It will discuss the
different properties and general characteristics of the biomolecules,
carbohydrates, and amino acids.
107
Unlocking Difficulties
Definition of Terms:
1. Biomolecules
- organic molecules that are synthesized by living organisms
- includes carbohydrates, amino acids and proteins, nucleic
acids, and lipids
Lecture Notes
Carbohydrates
 synthesized by autotrophs (generally by plants and some protists)
 also referred to as sugars = polyhydroxy aldehydes and ketones
 glucose was the first sugar to be extracted purely
 C6H12O6 = C6(H2O)6 = hydrate of carbon
Fischer Projection and Haworth Projection
https://www.sciencedirect.com/topics/biochemistry-genetics-and-molecular-biology/monosaccharides
Classification Based on Size
1. Monosaccharide
 simple sugars
 has three to seven carbons
 most common: glucose, galactose, fructose
 glyceraldehyde, dihydroxyacetone, ribose
https://www.chemistrylearner.com/fischer-projection.html
108
https://www.chemistrylearner.com/haworth-projection.html
2. Disaccharide
 has two monosaccharide units attached to each other by an
ester linkage called glycosidic bond
 sucrose: composed of glucose and fructose monomers
 most abundant disaccharide
https://www.toppr.com/ask/question/a-write-the-haworth-structure-of-sucroseb-sucrose-is-a/
 maltose: two glucose monomers
 present in malt
https://www.toppr.com/ask/question/write-the-haworth-structure-of-maltose/
 lactose: glucose and galactose monomer
 present in milk
https://simple.wikipedia.org/wiki/Lactose
3. Oligosaccharide
 has 3 to 10 monosaccharide units
 raffinose: one galactose unit, one glucose unit and one fructose
unit
https://www.toppr.com/content/story/amp/oligosaccharides-64074/
109
4. Polysaccharide
 has a very large number of monosaccharide units
 also referred to as glycans
 common examples: glycogen, cellulose, starch
 starch: can be separated into amylose and amylopectin
 made up of glucose units
 a form of stored energy found in plants
 glycogen: also a polymer of glucose
 the energy-reserve carbohydrate for humans and
animals
 cellulose: still composed of glucose units
 gives the structure of plants
https://medicinalherbals.net/polysaccharides/
Classification Based on the Type of Carbonyl Group
1. Aldose
 polyhydroxy aldehyde
2. Ketose
 polyhydroxy ketone
Classification Based on the Number of Carbon
1. Triose – contains 3 carbons
2. Tetraose – contains 4 carbons
3. Pentose – contains 5 carbons
4. Hexose – contains 6 carbons
Amino Acids
 building blocks of proteins or polypeptides
 contains an amino group (-NH2), carboxylic acid (-COOH) and an
alkyl side chain
110
Figure 22: General Structure of Amino Acid
(https://biochemden.com/what-is-amino-acids/)
Physical Properties
 amino acids are soluble in water, but not in hydrocarbons
 highly available in crystalline form
 has high melting point
 amphiprotic: can act both as an acid and as a base
Classification of Amino Acids
 the 20 amino acids can be classified into 5 groups depending on their
R side chain
1. Nonpolar, aliphatic R groups
 7 amino acids
 glycine, alanine, proline, valine, leucine, isoleucine, methionine
http://chemistry.creighton.edu/~jksoukup/lec5-aminoacids.pdf
2. Nonpolar, aromatic R groups
 3 amino acid
 phenylalanine, tyrosine, tryptophan
http://chemistry.creighton.edu/~jksoukup/lec5-aminoacids.pdf
3. Polar, uncharged R groups
 5 amino acids
 serine, threonine, cysteine, asparagine, glutamine
111
http://chemistry.creighton.edu/~jksoukup/lec5-aminoacids.pdf
4. Positively charged R groups
 3 amino acids
 lysine, arginine, histidine
http://chemistry.creighton.edu/~jksoukup/lec5-aminoacids.pdf
5. Negatively charged R groups
 2 amino acids
 aspartate, glutamate
http://chemistry.creighton.edu/~jksoukup/lec5-aminoacids.pdf
Peptide bonds
 bonds shared by amino acids in a polypeptide chain
 formed through dehydration or condensation reaction
https://skowalskibiology.weebly.com/carbohydrates-and-proteins.html
112
Focus Questions
Instructions: Please submit your answers through LMS. The document
should be in letter size. Use Calibri, 12 pt. 1.15 spacing. On the upper-left
corner of the first page, write your name, grade level, section, and the date
of submission. For uniformity purposes, name your document as
IOChem_Last Name_Unit9. Properly cite your references using
APA format. Please refer to the LMS for the rubrics.
1. How do humans acquire carbohydrates? How are carbohydrates utilized
by the human body?
2. Briefly explain each of the functions of the 20 amino acids in the human
body.
Related Readings
14: Carbohydrates (Bennett, 2019)
 https://chem.libretexts.org/Courses/Sacramento_City_College/SCC%
3A_Chem_309__General_Organic_and_Biochemistry_(Bennett)/Text/14%3A_Carbo
hydrates
Introducing Amino Acids
 https://www.chemguide.co.uk/organicprops/aminoacids/backgroun
d.html#:~:text=The%20amino%20acids%20are%20crystalline,molecu
les%2C%20this%20is%20very%20high.Fischer
and
Haworth
projections
Learning Assessment
General Instruction: Please submit the document containing your
answer for the following exercises through LMS.
1. Amino acids can also be grouped into essential and nonessential amino
acids. Briefly explain the difference between the two.
2. There are nine amino acids classified as essential amino acids. Give a food
product or drink where these amino acids can be acquired.
Example: Serine can be obtained from consuming soybeans and nuts.
113
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2.7: Nomenclature of Ionic, Covalent, And Acid Compounds. (2019, August
11).
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20.3 Aldehydes, Ketones, Carboxylic acids, and Esters. (n. d.). Retrieved
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