STEM3 – GENERAL CHEMISTRY 1

TOPIC OUTLINE:
1. Module 3 2. Module 4
 Lesson 1: Stoichiometry  Lesson 1: Organic
 Lesson 2: The Gases Compounds
 Lesson 3: Lewis  Lesson 2: Structural
Structure of Molecular Isomerism &
Compound Organic Reaction
 Lesson 4: Molecular
Geometry & Polarity

1 Stoichiometry
Mole Sample Problem #1:
 (mol) a unit of amount or quantity of a Determine the number of particles and mass of
substance 0.85 mol Cu.
 It is the amount of substance that contains
Solution:
exactly 6.02 x10²³ atoms, molecules, or ions.
 1 mol = 602 000 000 000 000 000 000 000 (a) number of particles
000 pieces/particles 6.02𝑥1023 𝑎𝑡𝑜𝑚𝑠 𝐶𝑢
= 0.85 𝑚𝑜𝑙 𝐶𝑢 𝑥
 A quantity used to indicate very small species 1 𝑚𝑜𝑙 𝐶𝑢
= 5.1 𝑥 1023 𝑎𝑡𝑜𝑚𝑠 𝐶𝑢
that make up matter like atoms, ions, or (b) mass
molecules. 63.55 𝑔 𝐶𝑢
 A quantifying unit 0.85 𝑚𝑜𝑙 𝐶𝑢 𝑥 = 54 𝑔 𝐶𝑢
1 𝑚𝑜𝑙 𝐶𝑢
Molar Mass 1.1 Percentage Composition
 (g/mol) mass of one mole of substance  It is a way to express the relative abundance
 numerically equal to the atomic mass of each element in a chemical substance.
 used to convert mole to mass, or mass to  (percent composition by mass) percent by
mole mass of each element in a compound

 Total must be equal or approximately equal

to 100%.

Sample Problem #2:

Which element comprising Mg(OH)2 has the
highest percentage by mass? Determine the
percent composition of each element.
Mole Ratio Step 1.
 coefficients in a balanced equation which Mg = 1, O= 2, H= 2
can act as conversion factors to relate the Step 2.
reactants to the products Mg = 1 x 12.01 g/mol = 24.31g
O = 2 x 16.00 g/mol = 32.00 g
 2H₂ + O₂ → 2HO
H = 2 x 1.01 g/mol = 2.02 g
➢ “Two molecules of hydrogen gas Molar Mass = 58.33 g
reacts with 1 molecule of oxygen gas Step 3.
to produce 2 molecules of water.” 24.31 𝑔.𝑚𝑜𝑙
Mg: × 100% =41.676 = 41.68%
➢ “Two moles of hydrogen gas reacts 58.33 𝑔/𝑚𝑜𝑙
𝟑𝟐.𝟎𝟎 𝒈/𝒎𝒐𝒍
with 1 mole of oxygen gas to produce O: × 𝟏𝟎𝟎%=54.860 = 54.86%
𝟓𝟖.𝟑𝟑 𝒈/𝒎𝒐𝒍
2 moles of water.” H:
2.02 𝑔/𝑚𝑜𝑙
× 100%= 3.463 = 3.46 %
 Used to relate one substance to another
58.33 𝑔/𝑚𝑜𝑙

1-1 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

1.1.1 Empirical Formula

Note: Empirical formula may or may not be the
same as a molecular formula.

 simplest whole-number ratio of atoms

present in the compound
 The empirical formula of the compound
can then be determined from the Sample Problem #4:
percentage composition of the elements in For P₂O₅, the molar mass for the compound is
the compound. 283.88 g/mol. What is the molecular formula for
the compound?
Step 1.
P = 2 x 30.97 g = 61.94 g/mol
O = 5 x 16.00 g = 80.00 g/mol
Empirical Mass = 141.94 g/mol
Step 2.
283.88 𝑔/𝑚𝑜𝑙
Ratio = 141.94 𝑔/𝑚𝑜𝑙 = 2

Sample Problem #3: Step 3.

During an experiment, a compound is found to
consist of 7.81% C and 92.19% Cl. What is the (P₂O₅)2 = P₄O₁₀
empirical formula of the compound?
Step 1.
1.2 Stoichiometry
7.81 % = 7.81 g; 92.19 % = 92.19 g
7.81 g + 92.19 g = 100 g compound  the study of quantitative relationships
Step 2. between the amounts of reactants used and
1 𝑚𝑜𝑙 𝐶 amounts of products formed by a chemical
7.81 𝑔 × = 0.650 𝑚𝑜𝑙 𝐶
12.01 𝑔
1 𝑚𝑜𝑙 𝐶𝑙
reaction.
92.19 𝑔 × = 2.601 𝑚𝑜𝑙 𝐶𝑙  Make sure your chemical equation is
35.45 𝑔
Step 3. balanced.
0.650
C: 0.650 = 1
2.601
Cl: =4
0.650
Step 4.
𝐶𝐶𝑙4

1.1.2 Molecular Formula

 actual number of atoms of each element in
the compound 1.2.1 Mole to Mole Stoichiometry
 The molecular formula of a compound can
also be determined using the percentage
composition but only if the molar mass of
the compound is given.

1-2 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

4NH₃+5O₂→ 6H₂O + 4NO

Sample Problem #5:
How many moles of H₂O are produced if 2.00
moles of O₂ are used?
Given: 2.00 moles of O₂
Required: moles of H₂O
Solution:
6 𝑚𝑜𝑙 𝐻2 𝑂
4NH₃+5O₂→ 6H₂O + 4NO
2.00 𝑚𝑜𝑙 𝑂2 × = 2.40 𝑚𝑜𝑙 𝐻2 𝑂
5 𝑚𝑜𝑙 𝑂2 Sample Problem #8:
How many moles of H₂O are produces if there are
4NH₃+5O₂→ 6H₂O + 4NO 8.00 grams of O₂?

Sample Problem #6: Given: 8.00 grams O₂

How many moles of NO are produced in the
Required: moles of H₂O
reaction if 15.0 mol of H₂O are also produced?
Solution:
Given: 15 moles H₂O
Required: moles of NO
Solution:
4 𝑚𝑜𝑙 𝑁𝑂
15.0 𝑚𝑜𝑙 𝐻2 𝑂 × = 10.0 𝑚𝑜𝑙 𝑁𝑂
6 𝑚𝑜𝑙 𝐻2 𝑂
1.2.2 Mole-Mass Stoichiometry

1.2.3 Mass to Mass Stoichiometry

4NH₃+5O₂→ 6H₂O + 4NO

Sample Problem #7:

How many grams of H₂O are produced if 2.2 mol
of NH₃ are combined with excess oxygen? 4NH₃+5O₂→ 6H₂O + 4NO

Given: 2.2 moles NH₃ Sample Problem #9:

Required: grams of H₂O How many grams of NO is produced if 12 grams of
Solution: O₂ is combined with excess ammonia?
6 𝑚𝑜𝑙 𝐻2 𝑂
2.2 𝑚𝑜𝑙 𝑁𝐻3 × = 3.3 𝑚𝑜𝑙 𝐻2 𝑂 Given: 12 grams of O₂
4 𝑚𝑜𝑙 𝑁𝐻3
Required: grams of NO
Solution:

1-3 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

2.2 Kinetic Molecular Theory

o The kinetic molecular theory (KMT) of
gases describes the nature of gases and the
behavior of the particles that comprise them.
o The theory makes the following
assumptions:
1. Gas consists of minute particles
4NH₃+5O₂→ 6H₂O + 4NO (atoms and molecules)
2. The gas particles are constantly
Sample Problem #10:
moving at a rapid and random motion
How many grams of NH₃ is needed if 18.0 g of H₂
3. The gas particles are considered as
O was produced after an experiment?
hard spheres such that their collisions
are perfectly elastic.
4. The interaction among particles
Given: 18.0 grams H₂O (intermolecular forces of attraction),
Required: grams of NH₃ either attractive or repulsive, is
negligible.
Solution: 5. The average kinetic energy of a gas is
directly proportional to its absolute
1 𝑚𝑜𝑙 𝐻 2 𝑂
18.0 𝑔 𝐻2 𝑂 × = 0.999 𝑚𝑜𝑙 𝐻2 𝑂 temperature in Kelvin.
18.02 𝑔 𝐻 2 𝑂

2.3 The Gas Laws

2.3.1 Boyle’s Law
 It is named after the Irish scientist Robert
Boyle, who first stated the law in 1662.
“The pressure of a fixed amount of gas is
inversely proportional to its volume at constant
temperature.”

o P₁V₁=P₂V₂
2 Gases o PV= constant

2.1 Pressure
Sample Problem #1:
 force that the gas exerts on the container A 20.0 L gas kept in a movable piston has an initial
boundaries pressure of 1.5 atm at 25°C. if the piston is allowed
1 atm, 760 torr 760 mm Hg to expand isothermally (at constant temperature) to
101325 Pa 14.7 psi, 1.01 bar 45.0 L, what will be the resulting pressure of the gas?
▪ atm = atmosphere
▪ mm Hg = millimeter Mercury
▪ Pa = pascal Given: V₁=20.0 L; P₁=1.5 atm; V₂=45.0 L
▪ psi = pounds per square inches
Requirement: P₂= ?
Barometer
 measures atmospheric pressure Equation : P₁V₁=P₂V₂

Manometer Solution: (1.5𝑎𝑡𝑚 ⋅ 20.0𝐿) = 𝑃2 (45.0𝐿)

 measures pressure in a closed system ]Answer: P₂= 0.67 atm

2-4 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

will the pressure be at noontime when the

temperature reaches 32°C?
2.3.2 Charles’ Law
 This law is named after the French scientist
Jacques Charles, who first formulated it in Given: P₁ = 5.0 atm; T₁ = 25°C; T₂ = 32°C
the late 18th century. Requirement: P₂ = ?
“The volume of a fixed amount of gas is Equation: P₁/T₁ = P₂/T₂ [°C + 273.15 = K]
directly proportional to its absolute Solution:
temperature at a constant pressure.” T₁ = 25.0 ° C + 273.15 T₂ = 32.0°C + 273.15
= 298.15 K = 305.15 K
 V₁/T₁ = V₂/T₂
 V/T = constant
5.0 𝑎𝑡𝑚 𝑃2
298.15 𝐾
= 305.15 𝐾

5.0 𝑎𝑡𝑚 𝑃2
Sample Problem #2: =
298.15 𝐾 305.15 𝐾
Consider a 25.0 L gas in a container initially at
25.0°C and 1 atm. If this gas is heated to 80.0°C, Answer: P₂ = 29.6 L
what will happen to its volume under constant
pressure?

Given: V₁ = 25.0 L; T₁ = 25.0 °C; T₂ = 80.0°C

Requirement: V₂ = ?
Equation: V₁/T₁ = V₂/T₂, [°C + 273.15 = K]
Solution:
T₁ = 25.0 ° C + 273.15 T₂ = 80.0°C + 273.15 2.3.4 Combined Gas Law
= 298.15 K = 353.15 K “The ratio of the product of pressure and
25.0 𝐿 𝑉2 volume and the absolute temperature of a gas
=
298.15 𝐾 353.15 𝐾 is equal to a constant.”
25.0 𝐿 𝑉2
=  PV/T = constant
298.15 𝐾 353.15 𝐾  P₁V₁/T₁ = P₂V₂/T₂
Answer: V₂ = 29.6 L
Sample Problem #4:
2.3.3 Gay-Lussac’s Law A gas sample held in a 1.0 L piston at 305 K has a
 also known as the Law of Combining pressure of 0.75 atm. The piston was heated to 387
Volumes. K at the same time the volume was expanded to 1.5
 Gay-Lussac's Law is named after the L. Will the pressure decrease or increase?
French chemist Joseph Louis Gay-Lussac. Given:
He formulated the law based on
experimental observations and first INITIAL P ₁ = 0.75 V₁ = 1.0 L T₁ = 305 K
atm
presented it in 1802. FINAL P₂ = ? V₂ = 1.5 L T₂ = 387 K
“The pressure of a gas is directly proportional Required: P₂ = ?
to its absolute temperature at a constant 𝑃1 𝑉1 𝑃2 𝑉2
Equation: =
volume.” 𝑇1 𝑇2

 P/T = constant Solution:

 P₁/T₁ = P₂/T₂ (0.75 𝑎𝑡𝑚)(1.0 𝐿) 𝑃2 (1.5 𝐿)
=
305 𝐾 387 𝐾
Sample Problem #3:
A 50.0 L tank of helium gas placed outside a 0.75 𝑎𝑡𝑚 ∙ 𝐿 1.5 𝐿
= 𝑃2
305 𝐾 387 𝐾
laboratory has a pressure of 5.0 atm at 25°C. What

2-5 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

Answer: P₂ = 0.63 atm (P₁>P₂ =decreased) 2.3.6 Graham’s Law of Effusion

 formulated by Scottish chemist Thomas
Graham in the 19th century
2.3.5 IDEAL GAS LAW  lighter gas molecules effuse more rapidly
“The volume of a given amount of gas is than heavier ones
directly proportional to the number of moles
“The rate of effusion of a gas is inversely
of gas, directly proportional to the temperature
proportional to the square root of its molar
and inversely proportional to the pressure.”
mass”
 PV = nRT Effusion
Ideal Gas Constant
 the process by which gas molecules escape
through a small opening into a vacuum.
 denoted as R.
𝐿 ∙ 𝑎𝑡𝑚 𝐽 Graham's Law can be stated mathematically as
𝑅 = 0.0821 𝑜𝑟 8.314 follows:
𝑚𝑜𝑙 ∙ 𝐾 𝑚𝑜𝑙 ∙ 𝐾
Standard Temperature & Pressure 𝑅𝑎𝑡𝑒 𝑜𝑓 𝐸𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝐺𝑎𝑠 𝐴
=√
𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐺𝑎𝑠 𝐵
𝑅𝑎𝑡𝑒 𝑜𝑓 𝐸𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝐺𝑎𝑠 𝐵 𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐺𝑎𝑠 𝐴
 T = 273.15 K
 P = 1.00 atm Sample Problem #7:
Sample Problem #5:
Two gases, hydrogen (H₂) and oxygen (O₂), are
How many moles of a gas occupies a volume of 0.68 placed in separate containers, and both containers
L at 0.85 atm and 301 K? have small openings. Calculate the ratio of the rates
of effusion of hydrogen to oxygen.

Given: V = 0.68 L; P = 0.85 atm; T = 301 K

Requirement: n = ? Given: =𝐻2 , 𝑂2
Equation: PV = nRT Required: Rate ratio
Solution:
Equation:
𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑂𝑥𝑦𝑔𝑒𝑛
𝑅𝑎𝑡𝑒 𝑟𝑎𝑡𝑖𝑜 = √
𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐻𝑦𝑑𝑟𝑜𝑔𝑒𝑛

Solution:
Molar Mass
H = 2 x 1.01 = 2.02 g/mol O = 2 x 16.00 = 32.00
Sample Problem #6: g/mol
Calculate the volume that will be occupied by 20.0
g carbon dioxide at 25.0 °C and 1.25 atm 𝑅𝑎𝑡𝑒 𝑟𝑎𝑡𝑖𝑜 = √
32.00 𝑔/𝑚𝑜𝑙
2.02 𝑔/𝑚𝑜𝑙

𝑅𝑎𝑡𝑒 𝑟𝑎𝑡𝑖𝑜 = 3.98014 ≈ 4

Given: m = 20.0 g; P = 1.25 atm; T = 25.0 °C
Requirement: V = ?
Answer: The ratio of the rates of effusion of
hydrogen to oxygen is 4:1.
Equation: PV = nRT
Solution: This result indicates that hydrogen will effuse four
times faster than oxygen under the same conditions
of temperature and pressure.

2-6 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

2.4 Gas Stoichiometry 3 Lewis Structure of Molecular

 Gas stoichiometry involves applying the Compounds
principles of stoichiometry to reactions Everything around us is made up of matter, from the
involving gases. very simple air that we breathe to the most complex
 When dealing with gases, stoichiometry of things that we deal with every day. You may
considers the amounts of reactants and wonder how we got all these things that make our
products in terms of their volumes, life easy. This lesson will help you with that. This
pressures, and temperatures, in addition to lesson discusses how atoms combine and form
the usual considerations of moles and simple to complex substances.
masses.
Valence Electrons
 The number of valence electrons of an
element can be determined by looking at its
position on the periodic table.
 Valence electrons are the electrons in the
outermost energy level (shell) of an atom.
 The group number (vertical column) of an
element corresponds to the number of
valence electrons it
▪ Group 1 elements have 1 valence
Sample Problem #8: electron.
If a tank of LPG solely contains 11 kg C₄H₁₀, how ▪ Group 2 elements have 2 valence
much CO₂ (in L) will be produced at 1.0 atm and electrons.
32°C if all the butane has been burned? ▪ Group 13 elements have 3 valence
2C₄H₁₀+13O₂ →8CO₂+10H₂O electrons.
▪ Group 14 elements have 4 valence
Given: 11kg C₄H₁₀, 1.0 atm, 32°C electrons.
▪ Group 15 elements have 5 valence
Required: V of CO₂
electrons.
Solution: ▪ Group 16 elements have 6 valence
electrons.
C: (12.01 x 4) = 48.04 g ▪ Group 17 elements have 7 valence
H: (1.01 x 10) = 10.1g electrons.
▪ Group 18 elements (noble gases) have a
Molar Mass = 58.14 g/mol full outer shell and are considered to
11kg =11000 g have 8 valence electrons, except helium,
which has 2 valence electrons.
1 𝑚𝑜𝑙
11000 𝑔 C₄H₁₀ × = 189.3287 𝑚𝑜𝑙 C₄H₁₀  However, transition metals may have
58.1 𝑔 C₄H₁₀
varying numbers of valence electrons, and
8 𝑚𝑜𝑙 CO₂ some elements in the p-block may have
189.3287 𝑚𝑜𝑙 C₄H₁₀ × = 757.3148 𝑚𝑜𝑙 CO₂
2 𝑚𝑜𝑙C₄H₁₀
expanded valence shells.
𝑃𝑉 = 𝑛𝑅𝑇 32°𝐶 + 273.15 = 305.15 𝐾
3.1 Lewis Electron-Dot Structure
𝐿 ∙ 𝑎𝑡𝑚  Dots are often used to represent the valence
1 𝑎𝑡𝑚 (𝑉) = 757.3148 (0.0821 ) 305.15 𝐾
𝑚𝑜𝑙 ∙ 𝐾 (or outermost) electrons in atoms and
molecules.
Answer: 18 972.87 L or 19000 L CO₂
 These structures are referred to as Lewis
structures/symbols, electron-dot
structures, or Lewis electron-dot structures.

3-7 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

▪ the nucleus of the Cations (Positive Ions):

element is represented  Cations are formed when an atom loses one
by its symbol. or more electrons.
▪ valence (outermost)  To represent a cation in a Lewis dot
electrons are diagram, you remove electrons from the
represented by dots outermost shell (starting from the
(electron configuration) outermost and moving inward).
 The number of electrons removed is equal
to the positive charge of the cation.
Steps on creating a Lewis Dot Structure
1. Find your element on the periodic table.
2. Determine the number of valence electrons.
1 valence 2 valence electrons For example, consider the formation of a sodium
electron
cation (Na⁺):

 Neutral sodium (Na) has the electron

configuration: 1s² 2s² 2p⁶ 3s¹
 Sodium cation (Na⁺) has the electron
Lewis Dot Diagram of Main Group configuration: 1s² 2s² 2p⁶
Elementss
Anions (Negative Ions):
 Anions are formed when an atom gains one
or more electrons.
 To represent an anion in a Lewis dot
diagram, you add electrons to the
outermost shell (starting from the
innermost and moving outward).
 The number of electrons added is equal to
the negative charge of the anion.

3.1.1 Lewis Dot Diagram of Ions

After practicing how to write the Lewis electron-dot
symbol of atoms/elements, we will try writing those
of ions. Take note that a special stability is
associated with the electronic configurations of the Let's consider the formation of a chloride anion
noble gases. This means that atoms have the (Cl⁻):
tendency to gain or lose electrons in order to achieve Neutral Chlorine (Cl) has the
such electronic configurations, thus becoming ions. electron configuration: 1s² 2s² 2p⁶
Ion 3s² 3p⁵.
 An ion is formed whenever an uncharged The Lewis dot diagram for neutral
atom gains or loses an electron. chlorine would be written as:
 A positive ion is known as a cation and a
Chloride Anion (Cl⁻): To form the chloride anion,
negative ion is known as an anion. chlorine gains one electron.
Steps? The Lewis dot diagram for Cl⁻
1. Find your element on the periodic table. would be written as Cl⁻:
2. Determine the number of valence electrons
3. Determine the possible ion

3-8 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

 All elements in Group IA (AlkalineMetals)  For example, metals may lose electrons
have the same electron-dot structure and to form positively charged cations, and
therefore each can lose one electron to form nonmetals may gain electrons to form
a +1 ion with the electron configuration of negatively charged anions.
the preceding noble gas.  The resulting oppositely charged ions
attract each other, forming an ionic
bond.
ii. Covalent Bonding:
3.1.2 Lewis Structure of Molecular  Atoms can share electrons to achieve a
Compounds
full outer shell. In covalent bonds,
electrons are shared between atoms,
leading to the formation of molecules.
Gilber N. Lewis 3. Stability Achieved:
 1875 - 1946  Through these bonding processes, atoms
 Atoms combine in order to achieve a more become part of molecules or compounds
stable electron configuration. (non- with a more stable overall electron
verbatim) configuration.
 This maximum stability is attained when an  This stability is achieved by reaching a state
atom is isoelectronic with a noble gas. similar to that of noble gases.
o This means that an atom achieves
Octet Rule
stability if and only if it has equal
number of electrons to the noble “Elements gain or lose electrons to attain an
gas closest to it; thus, having the electron configuration of the nearest noble gas”
same electron configuration.  Remember, in combining or bonding with
o Except for He, the noble gases other atoms, only the outer electrons, the
have eight outer electrons or an valence electrons, are involved.
octet of electrons.
Breakdown of the idea:
1. Stable Electron Configuration:
 Noble gases (e.g., helium, neon, argon)
have a stable electron configuration.
▪ They have a full outer electron
shell (valence shell), which makes
them relatively inert and stable. Covalent Bond
 Other atoms, especially those in the  is a type of chemical bond that is formed
main groups of the periodic table, aim by sharing of electrons.
to achieve a similar electron  This occurs between non-metals
configuration to become more stable.
2. Chemical Bonding:
 Atoms can achieve a stable electron
configuration by either gaining, losing,
or sharing electrons.
▪ There are two primary types of
chemical bonds: ionic and
covalent. The representation of the covalent compound above
i. Ionic Bonding: is called the Lewis structure.
 Atoms can transfer electrons to  In the Lewis structure, the shared electrons
achieve a full outer shell. that form a bond are represented by a line
or a pair of dots.

3-9 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

b. How many initial bonds are possible?

i. Place two electrons in each bond.
LONE PAIR 3. How much electrons are yet to be
➢ unshared pair of electrons distributed?
a. Subtract the electrons used in Step 2
from the total valence electrons and
BOND PAIR distribute.
➢ shared pair of electrons b. Make sure an octet of electrons is
formed.
4. Are there multiple bonds?
a. If multiple bonds are possible, add
electrons to construct the double or
 In a covalent bond, multiple bonds can be triple bond.
formed. b. If there are not enough electrons left
o To be specific, a covalent bond can to fill the structure, this indicates a
form any of the three types: single, multiple bond.
double, or triple bond. c. Take note:
Single covalent bond i. a shortage of two electrons
indicates a double bond;
 A single covalent bond is formed when a ii. a shortage of four electrons
pair of electrons between two atoms is indicates a triple bond or two
shared. Each atom provides one electron. It double bonds.
is illustrated as a short line (—).
Central Atom
Double covalent bond  an element where all other elements in a
compound is attached
 A double covalent bond is formed when
 usually the atom with the lowest subscript
two pairs of electrons between two atoms
in the molecular formula and the atom that
is shared. It is illustrated as a double short
can form the most bonds.
line (═).
 If all the atoms form the same number of
Triple covalent bond bonds, the least electronegative atom is
usually the central atom.
 A triple covalent bond is formed when
three pairs of electrons between two atoms a. Methane, CH₄
is shared. It is illustrated as a triple short
line.

TO DRAW THE LEWIS STRUCTURES OF

COMPOUNDS, TAKE NOTE OF THE
FOLLOWING:
1. What is the Lewis structure of each atom
involved in the compound?
a. How many total valence electrons
does the compound have?
2. Which atom are bonded to each other?
a. Arrange the atoms in a
preliminary diagram so that the
atoms that are bonded to each
other are next to each other.

3-10 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

b. Carbon tetrafluoride, CF₄ e. Nitric acid, HNO₃

c. Ammonia, NH₃

Exception to the OCTET Rule

1. Ions or molecules with:
a. odd number of electrons
b. less than an octet
c. more than 8 valence electrons

d. Carbon dioxide, CO₂ LESS THAN AN OCTET

➢ rare
➢ usually with B and Be
➢ Molecules have only 6 electrons around
central atom

MORE THAN 8 VALENCE ELECTRONS

➢ an expanded octet
➢ only for elements in 3rd row or below
➢ d orbitals in these atoms participate in
bonding

3-11 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

4 Molecular Geometry and  The 3D shapes of molecules can be

Polarity predicted from their Lewis structures, with
The way a substance behaves, such as that of the model known as the valence-shell
water, is actually affected by its molecule shape. electron pair repulsion (VSEPR) model.

4.1 Molecular Geometry An electron pair is attracted to the central atom

but is repelled by other electron pairs.
 Molecular geometry refers to the three-
dimensional arrangement of atoms in a
molecule.
 It describes the spatial arrangement of 4.1.1 Electron-Domain Geometry
atoms and the bond angles between them.  arrangement of electron domains
 The molecular geometry is influenced by surrounding the central atom.
the arrangement of electron pairs around  counts both bond pairs and lone pairs in
the central atom. the central atom.

VSEPR Theory
 The chemical formula of a compound has
4.1.2 Molecular Geometry
no direct relationship with its molecule.
 arrangement of the atoms surrounding the
 The three-dimensional shapes of molecules central atom
can be predicted from their Lewis
 counts only bond pairs in the central atom.
structures, with the model known as the
valence-shell electron pair repulsion
(VSEPR) model.
 According to 4.1.3 Geometry of Simple
VSEPR model, the Compounds
electron form pairs in  To determine the electron-domain
the valence shell of geometry and the molecular geometry of an
an atom. atom, the electron pairs on the central atom
 These electron pairs are counted.
can either be the  Tables containing the electron-domain and
bond pairs or the molecular geometries of simple molecular
lone pairs. compounds is shown below and on the
 Each of these electron pairs occupies its next page.
own domain but keeps other domains as far
away as possible (electron pairs repel each
other).
➢ This repulsion creates a three-
dimensional structure of the
compound; thus, a molecular
geometry.

To better understand the geometry of the

structure, several keywords will be used:
(1) lone pair,
(2) bond pair,
(3) central atom,
(4) electron-domain geometry or electron
geometry and
(5) molecular geometry.

4-12 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

➢ Two atoms bonded to the central atom,

with two lone pairs. The shape is bent or
V-shaped, and bond angles are less than
109.5 degrees.
Trigonal Bipyramidal (AX₅):
➢ Example: PCl₅
➢ Five atoms bonded to the central atom,
creating a trigonal bipyramidal
arrangement with bond angles of 90 and
120 degrees.

Octahedral (AX₆):
➢ Example: SF₆
➢ Six atoms bonded to the central atom,
forming an octahedral shape with bond
angles of 90 degrees.
Guide Questions during Practice
1. What element composes the central atom?
2. How many electron pairs are present in the
central atom?
a. This value provides the electron-
domain geometry.
From the tables, you would notice that both 3. Of the electron pairs, how many are the
the bond pairs and lone pairs, when added, bond pairs? The lone pairs?
provides the total number of electron pairs a. These values provide the
(electron domain). molecular geometry.
Linear (AX₂): Sample Problems
➢ Example: BeCl₂ Determine the electron-domain geometry and the
➢ Two atoms bonded to the central atom, molecular geometry of the following compounds:
with a bond angle of 180 degrees.

Trigonal Planar (AX₃):

➢ Example: BF₃ (1) PCl₅
➢ Three atoms bonded to the central atom,
forming a flat, triangular arrangement with
bond angles of 120 degrees.

Tetrahedral (AX₄):
➢ Example: CH₄
➢ Four atoms bonded to the central atom,
creating a three-dimensional tetrahedral (2) O₃
shape with bond angles of 109.5 degrees.

Trigonal Pyramidal (AX₃E):

➢ Example: NH₃
➢ Three atoms bonded to the central atom,
with one lone pair. The shape is pyramidal,
and bond angles are less than 109.5
degrees.
Bent or V-Shaped (AX₂E₂):
➢ Example: H₂O

4-13 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

(3) OF₂ Polar covalent bond (or simply polar

bond)
➢ is formed when electrons are unequally
shared between two atoms in a compound.
➢ Polar bond occurs because one atom has a
stronger affinity for electrons that the other
(yet not enough to pull the electrons away
completely and form ions)
➢ A good example is the bond between
Fluorine and Chlorine in FCl
4.2 Polarity of Simple Compounds
It is, however, important to note that polarity of
molecules differs from bond polarity. Let us
differentiate the two.

4.2.1 Polarity of Chemical Bonds Other than covalent bond, another type of chemical
Covalent Bond bond is the ionic bond.
 Covalent bond is a type of chemical bond Ionic bonds
that is formed by sharing of electrons in  form as a result of electrostatic attraction
atoms. between oppositely charged ions.
 It occurs because the atoms in a compound  This type of bond occurs between a metal
have similar tendency to gain electrons. and a nonmetal.
 With this, there are two subtypes of
covalent bonds – the nonpolar covalent The bond between two atoms can be a polar
bond and the polar covalent bond. covalent bond, a nonpolar covalent bond, or an
ionic bond.
Nonpolar covalent bond (or simply
nonpolar bond)  To determine the nature of a bond, one
➢ forms when two atoms equally share the specific property is used.
bonding electrons.
➢ Take H2 molecule as an example. Both
atoms have equal attraction (Affinity) for This property is electronegativity.
electrons; thus, forming a nonpolar bond.
4.2.2 Electronegativity
 is an atom’s ability to attract an electrons
towards itself.
 Stronger electronegativity means an atom
has more tendency to gain electrons.
 In chemical bonds, the electronegativity
difference between the two atoms help
determine bond polarity.
 Greater difference means one of the two
atoms has a stronger affinity for electrons.
 Generally, if the electronegativity
difference is:
▪ 0 – less than 0.4 = nonpolar covalent
bond
▪ 0.4 – 1.78 = polar covalent bond
▪ Greater than 1.78 = ionic bond

4-14 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

Example: Given the following pairs, ➢ The more electronegative atom attracts the
determine the polarity of the bond: shared electrons more strongly, creating a
 Note: Electronegativity difference is an partial negative charge (δ-) on that atom
absolute value and a partial positive charge (δ+) on the
less electronegative atom.
Example: Water (H₂O) is a polar molecule.
Oxygen is more electronegative than hydrogen,
leading to a partial negative charge on oxygen
Water and oil do not mix due to their and partial positive charges on the hydrogens.
polarities.
Nonpolar Molecules
Molecular Polarity ➢ Nonpolar molecules have an even
 results from the uneven partial charge distribution of electrons, resulting in no
distribution between various atoms in a significant charge separation.
compound. ➢ This occurs when there is little or no
 Molecular polarity differs from bond difference in electronegativity between the
polarity atoms in a covalent bond, or when the
 The polarity of a molecule affects its molecular geometry cancels out the dipole
behavior. moments.
Example: Diatomic molecules like oxygen (O₂) and
nitrogen (N ₂ ) are nonpolar because the
4.2.3 Polarity of Molecules electronegativity of the two atoms is similar, and the
 A molecule is considered polar if there is a linear molecular geometry cancels out the dipole
distortion of electron cloud around a moments.
molecule. Dipole Moment:
 This happens due to several reason: ➢ A dipole moment is a measure of the
 the presence of alone pair on the central polarity of a molecule. It is the product of
atom, the charge (Q) and the distance of
 a different element in one of its terminal separation (r) between the positive and
atoms (atom attached to the central atom), negative charges: μ = Q × r.
 or a combination of any of these. ➢ The dipole moment is a vector quantity,
 To determine the polarity of a molecule, and its direction points from the positive
two things are considered: to the negative charge.
▪ polarity of bond Determining Polarity:
▪ molecule shape. (1) To determine the polarity of a molecule,
you can look at its molecular geometry and
 The polarity of a molecule is determined by the electronegativity of the atoms involved.
the distribution of electrons within the (2) If a molecule has polar bonds but the
molecule and the resulting charge geometry results in a symmetrical
separation. distribution of those bonds, the dipole
 A molecule can be either polar or nonpolar, moments may cancel out, leading to a
and this classification depends on the nonpolar molecule.
electronegativity of the atoms and the (3) If the molecular geometry and
molecular geometry. electronegativity differences result in an
Polar Molecules unsymmetrical distribution of charge, the
➢ Polar molecules have an uneven molecule is likely polar.
distribution of electrons, leading to a
separation of charges.
➢ This charge separation arises when there is
a significant difference in electronegativity A molecule’s shape can be symmetrical or
between the atoms in a covalent bond. asymmetrical.

4-15 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

Bonds: contains at least one polar bonds

(bonds are different)
Symmetrical Shape Shape: symmetrical
➢ no lone pairs on the central atom HCN is a polar molecule
(3) CO2
Bonds: contains at least on
polar bonds (all bonds are the
same)
Shape: symmetrical
CO2 is a nonpolar molecule
(4) H2O
Asymmetrical Shape
Bonds: contains at least one
➢ the central atom has at least one lone pair
polar bonds
Shape: asymmetrical
H2O is a polar molecule

5 Organic Compounds

To determine whether a molecule is polar or

Carbon
nonpolar, simply use this table:
 a chemical element with the symbol C and
POLARITY Contains Contains at least atomic number 6.
only one polar bond  Classified as a nonmetal,
nonpolar  carbon is a solid at room temperature.
bonds
Symmetric Nonpolar Nonpolar – if all  Though it is just one out of the 118 elements
Shape Molecule bonds are the same in the modern periodic table, a whopping 10
Polar – if bonds million compounds (a large majority) is
are different formed from carbon.
Asymmetric Polar Polar Molecule
CARBON
Shape Molecule
Weak Atomic Mass 12.01
Note that only covalent compounds exist as Electron 1s²2s²2p²
Configuration
molecules. It’s a good practice to draw the Lewis
Oxidation State +4, +2, -4
structure of the compounds first before going
Group 14 (IVA)
through the table.
Period 2
Block p
Key Isotopes ¹²C¹³C¹⁴C
Sample Problems
Determine the polarity of the following molecules:
 The origin of the name ‘carbon’ comes
(1) CF4 from the Latin name carbo, meaning
Bonds: contains at least charcoal
one polar bonds (all  Ever heard of the term ‘carbon-based life
bonds are the same) forms’ from Sci-Fi films? Well, carbon
occurs in all living organisms.
Shape: symmetrical  Humans are carbon-based life forms.
CF4 is a nonpolar
molecule
Organic Chemistry
(2) CHN
 is the study of carbon-based compounds
known as organic compounds.

5-16 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

 These compounds are basically made up of ➢ The general formula is CnH2n+3, where n
carbon atoms bonded mostly to hydrogen, is the number of carbon atoms.
oxygen, nitrogen, and sulfur. ➢ In naming an alkane, the suffix -ane is
 However, it is important to note that all added
organic compounds have carbons, but not
all carbon-containing compounds are
organic compounds.
C Atoms, Prefix
1 meth- Alkenes
2 eth- ➢ unsaturated hydrocarbon (olefins)
3 prop-
➢ General Formula: CnH2n
4 but-
➢ Naming Rule: Carbon chain prefix + -ene
5 penta-
6 hexa- ➢ are nonpolar organic compounds
7 hepta- containing a carbon to carbon double
8 octa- bond.
9 nona-
10 deca-
➢ Suffixes in the name indicate the functional
group present in the chain.
➢ To better understand, let us begin the
discussion with the simplest class of
organic compounds – the hydrocarbons.  To name a straight chain alkene with only
one double bond, number the carbon
atoms of the parent chain such that the
carbon with the double bond has the lowest
5.1 Hydrocarbons
possible number.
 Hydrocarbons are organic compounds
 The suffix -ene is then added in the name.
containing only two elements – carbon and
hydrogen.
 All other organic compounds are derived
from hydrocarbons in which one or more
hydrogen atoms have been replaced by
other atoms or groups of atoms.
 Hydrocarbons can be classified as aliphatic
hydrocarbon or aromatic hydrocarbon.
Alkynes
➢ unsaturated hydrocarbon
➢ General Formula: CnH2n-2
5.1.1 Aliphatic hydrocarbon
➢ Naming Rule: Carbon chain prefix + -yne
 Aliphatic hydrocarbon can be subdivided
➢ are nonpolar organic compounds
into alkanes, alkene, alkynes, and
containing a carbon to carbon triple bond
cycloalkanes.

Alkanes
➢ saturated hydrocarbon (paraffins)
➢ General Formula: CnH2(n+2)
➢ Naming Rule: Carbon chain prefix + -ane
➢ Alkanes (also called saturated
➢ the rules in naming alkynes are the same
hydrocarbon) are the most basic type of
with those if alkenes except for the suffix
organic compound.
for alkynes is -yne.
➢ These contain only single bonds.

5-17 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

a. CH2═CHCH2CH3

⇒ The location of the double bond and triple bond

along an alkene and alkyne chain is indicated by a
b. CH3CH2C(triple)CCH3
prefix number that designates the double-bond atom
that is nearest an end of the chain.
In propene and propyne the only possible
location for the multiple bond is between the
first and second carbons; thus, a prefix
indicating its location is unnecessary.
c. CH3CH═CHCH3
→ The same is true with ethene and ethyne.

In naming hydrocarbon (alkanes, alkenes, and

alkynes to be specific), take note of the following:

Guide Questions during Practice

Extra information:
1. How many carbon atoms are present in the 5.1.2 Aromatic Hydrocarbons
chain? (The number of carbon atoms will
 Aromatic hydrocarbons are a class of
determine the prefix necessary.)
organic compounds that contain a specific
2. Are there multiple bonds between two carbon
cyclic structure known as a benzene ring.
atoms?
 The benzene ring consists of six carbon
a. If so, is it a double bond or a triple
atoms arranged in a planar hexagonal ring,
bond?
with alternating single and double bonds
i. If all bond are single bonds, simply
between the carbon atoms.
add –ane to the prefix;
ii. If one bond is a double bond, add –  The molecular formula for benzene is
ene; C₆H₆.
iii. If one bond is a triple bond, add –  Aromatic hydrocarbons exhibit unique
yne.) properties due to the resonance and
3. If Step 2 is a Yes; which side is the multiple stability of the benzene ring.
bond closer to?
a. (Make sure that the carbon with the
multiple bonds has the lowest possible Here are some key characteristics of aromatic
number.) hydrocarbons:
Benzene Ring:
 The hallmark feature of aromatic
Sample Problems
hydrocarbons is the presence of a benzene
Write the name and write the expanded structural
ring.
formula of the following compounds:
 The structure of the benzene ring is often
represented by a hexagon with a circle
inside to indicate the delocalization of
electrons.

5-18 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

Stability and Resonance: Discussed below are the classes under hydrocarbons:
 Aromatic hydrocarbons are characterized
by a high degree of stability attributed to 5.2.1 Alkanes
resonance in the benzene ring.  also called paraffins
 The electrons in the double bonds of the  are nonpolar compounds
benzene ring are delocalized, creating a ▪ this means that these compounds are
more stable structure. hydrophobic.
 In liquid form, alkanes are commonly used as
Aromaticity: organic solvents in laboratories.
 Aromatic hydrocarbons exhibit a property  alkanes are utilized in general anesthetics and
known as aromaticity, which is related to as fuels.
the stability of the benzene ring.
 A compound is considered aromatic if it 5.2.2 Alkenes
meets certain criteria, including having a  are nonpolar organic compounds and show
planar, cyclic, and fully conjugated trends in properties similar to those of alkanes
structure.  example of alkenes are ethene and propene (or
propylene).
Examples:
 Examples of aromatic hydrocarbons Ethene
include benzene (C₆H₆), toluene (C₇H₈), ➢ commonly known as ethylene
and xylene (C₈H₁₀). ➢ is used in synthesizing many plasticcs and
commercially important alcohols
The primary distinction between aromatic and
aliphatic hydrocarbons lies in the presence of Propene
the benzene ring. ➢ used in manufacturing polypropene
polymer
Aromatic hydrocarbons contain benzene rings o a type of synthetic polymer
and exhibit unique stability and resonance
properties, while aliphatic hydrocarbons do 5.2.3 Alkynes
not contain benzene rings and can have a  are nonpolar organic compounds and
variety of open-chain or cyclic structures. exhibit the same trends in boiling point and
physical states as alkanes and alkenes.
 a good example is ethyne
5.2 Organic Compound Families Ethyne
 The major families of organic compounds are ➢ the simplest and smallest alkyne
characterized by their functional groups. ➢ commonly known as acetylene
 For this lesson, we will be discussing three ➢ a gaseous substance commonly burned with
organic compound families: oxygen gas in a welding torch.
▪ the hydrocarbons,
▪ the oxygen-containing compounds, Alkenes and alkynes are called unsaturated
▪ the nitrogen-containing compounds. hydrocarbon because they do not contain the
maximum number of hydrogen atoms due to
Summarized below are the classes of organic the presence on the carbon to carbon double
compounds in each family. and triple bond.
Hydrocarbon Nitrogen- Oxygen-
containing containing
Alkane Amine Alcohol Other than hydrocarbons containing just carbon and
Alkene Amide Ether hydrogen, organic compounds can exist as
Alkyne Aldehyde compounds containing oxygen and nitrogen. The
Ketone kind or the family in which an organic compound
Carboxylic Acid belong to depends on the functional group present
Ester in it.

5-19 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

C atoms in Alkyl group name

alkyl group
This discussion will use key terms: 1 methyl
Functional group 2 ethyl
➢ is an atom or group of atoms that imparts 3 propyl
characteristic chemical properties to an 4 butyl
5 pentyl
organic compound
6 hexyl
➢ it is the site of a chemical reaction
7 heptyl
Alkyl group 8 octyl
➢ is a group that is formed by removing a 9 nonyl
hydrogen atom from an alkane. 10 decyl
➢ this is the less reactive portions of the
molecules. The name of an organic compound containing
➢ Alkyl groups are often treated as a functional group depends on the family the
substituents in organic chemistry and are compound belongs to.
attached to a parent molecule, replacing
one or more hydrogen atoms.
➢ The presence of alkyl groups can affect the
5.2.4 Alcohol
physical and chemical properties of organic
compounds.  An alcohol is a compound that has a hydroxyl
group (–OH) bonded to a carbon atom. Its
➢ Alkyl groups are named by adding the
general formula is R–OH where –OH is the
suffix "-yl" to the name of the
functional group.
corresponding alkane.
▪ For example, the methyl group comes Naming alcohols
from methane, the ethyl group from  In naming alcohols, the suffix –ol may be
ethane, and so on. added to the carbon chain prefix.
To better understand, examples are given below.  Another way is by naming the alkyl group and
The ones in red are the functional group. then adding ‘alcohol’.

For example,

We can think of organic molecules as

being composed of functional groups bonded
to one or more alkyl groups.
 Alcohols widely occurs in nature and have
many applications.
 In describing general features of organic  Ethanol or ethyl alcohol (CH3CH2OH),
compounds, chemists often use the designation ▪ is a fuel additive, an industrial solvent,
R to represent any alkyl group: methyl, ethyl, and an ingredient in alcohol beverages.
propyl, and so on. ▪ It is also the only type of alcohol that
▪ Alkanes, for example, which contain no humans can safely drink.
functional group, are represented as R– ▪ Other alcohols are considered
H. poisonous for human consumption.
▪ Naming alkyl groups are simply adding a
suffix –yl to the carbon chain prefixes.

5-20 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

5.2.5 Ether For example,

 is an organic compound in which two
hydrocarbon groups are bonded to the same
atom of oxygen.
 its general formula is R-O-R’ where R and R’
are both alkyl groups that may differ from one
another. To further explain the nomenclature rules for both
aldehyde and ketone, here’s a summary.
 ethers, like alkanes, are not very reactive
compounds.
 they are commonly used as solvents
Aldehyde
 ethers are named by naming the alkyl groups R
 Aldehydes take their name from their parent
and R’, then adding the word ‘ether’. If both
alkane chains.
alkyl groups are identical, a prefix di- is added
otherwise you have to arrange the alkyl groups  The -e is removed from the end and is replaced
alphabetically. with -al.
 Aldehydes are more reactive than ketones.
 Aldehydes are also considered the most
For example, important functional group.
 An example of aldehyde is formaldehyde
(methanal).

Formaldehyde
➢ Formaldehyde is soluble in water, and a 40
percent concentration called formalin has
been used as an embalming agent and to
preserve biological specimens.
➢ Formaldehyde is also a raw material used to
Diethyl ether make plastics.
➢ commonly known as ether
➢ is the best-known ether
➢ it is extremely flammable and can Ketone
sometimes exhibit explosive properties.  Ketones take their name from their parent
alkane chains. The ending -e is removed and
replaced with -one.
5.2.6 Aldehyde and Ketone  The common name for ketones are simply the
Both aldehydes and ketones contain the carbonyl substituent groups listed alphabetically +
group C═O as their functional group. ketone.
 Ketones are less reactive than aldehydes.
 The difference between the two is the
 An example of ketones are acetone
substituents on the carbonyl group.
(propanone), which is used as a solvent in paint
 In aldehydes, the substituents are R and H; in
removers and nail polish removers.
ketones, the substituents are both R groups.
 The general formula for aldehydes and ketones
are R–COH and R–CO–R’ respectively.
Many common odors and flavors come from
 Aldehydes are named by replacing the terminal – aldehyde and ketones. Examples for aldehydes
e of the corresponding alkane name with –al. In are vanillin from vanilla bean and citral (or
ketones, the suffix –one is used. citrus?) from lemongrass

5-21 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

5.2.7 Carboxylic Acid 5.2.8 Esters

 also known as organic acids  Esters are organic compounds that have
 are organic compounds that contain the carboxylic acid groups in which the hydrogen
carboxyl functional group (shown below) in OH is replaced by an alkyl group; thus, the
 The general formula for carboxylic acids is general formula is R—COO—R’.
R—COOH.  Esters are named by giving the name of the
 Carboxylic acids are a class of organic alkyl group attached to the oxygen first, then
compounds that contain the carboxyl group, identifying the carboxylic acid.
which consists of a carbonyl group (C=O) and  In doing so, replace the terminal –ic acid of the
a hydroxyl group (-OH) attached to the same corresponding carboxylic acid name with –ate.
carbon atom.
For example,
 These compounds have higher boiling points
than alcohols of comparable molecular mass
and have weak acidic properties compared to
binary acids.
 Carboxylic acids are named by replacing the
terminal –e of the corresponding alkane name
with –oic acid.
 Esters are responsible for many pleasant
smells and tastes of fruits and flowers.
Examples of carboxylic acids are formic acid and  Esters are also used in perfumes and
acetic acid. artificial flavorings.
 The best example of ester is ethyl ethanoate
commonly known as ethyl acetate.
 Other flavor-producing examples of esters
include amyl acetate (pentyl ethanoate) in
banana, octyl acetate in oranges, and ethyl
formate in rum.

Ethyl ethanoate/acetate
➢ It is used as an artificial fruit essence and
Formic acid aroma enhancer, as an artificial flavor for
➢ Formic acid or methanoic acid are secreted ice cream, cakes and confectionary, as a
by ants when they bite producing a ‘sting’ solvent and many more.
and causing the pain.
5.2.9 Amine
Acetic acid or ethanoic acid  Amines are organic compounds where an alkyl
➢ Acetic acid or ethanoic acid is the acid in group is bonded to a derivative of ammonia,
vinegar. NH3.
Stearic acid  Hydrogen atoms in ammonia is replaced by
➢ Many carboxylic acids, such as stearic acid, alkyl group.
are used in soap-making.  This means that the number of alkyl
substituents can be up to three.
Acetylsalicylic acid  The formula for amines can be R—NH2,
➢ Organic acids are also used in preparing R2—NH, or R3—N depending on the
several drug like aspirin also known as number of alkyl substituents.
acetylsalicylic acid, a medication used to
reduce pain, fever, or inflammation.

5-22 | P a g e December 14, 2023

 STEM3 – GENERAL CHEMISTRY 1

 Amines are named the same way as ethers.

▪ When there are identical alkyl groups,
Alkenes CnH_2n Ethene Carbon chain
prefixes di- and tri- are used otherwise alkyl prefix + -ene*
groups have to be arranged alphabetically. Alkynes CnH_2n-2 Ethyne Carbon chain
 The suffix –amine is then added. Note that the prefix + -yne*
names of amines are written as a single word. ORGANIC COMPOUNDS CONTAINING OXYGEN
(O)
 Simple amines are very soluble in water and Alcohol R-OH Methanol or Corresponding
have low boiling points. Methyl alkane name – e
Alcohol + -ol
or
Alkyl group
name + alcohol
Ether R-O-R’ Dimethyl Alkyl group
Ether names + ether
If identical, add
5.2.10 Amide prefix di-; If
 Examples of amines include caffeine and not,
nicotine. arrange
alphabetically
 Amides are derived from carboxylic acids Aldehyde R-COH Methanal Corresponding
where the —OH part is replaced by —NH2 alkane name – e
group. + -al
 The general formula is R—CONH2. Keton R-CO-R’ Propanone Corresponding
alkane name – e
 Simple amides are named after the + -one
corresponding carboxylic acids by changing the Carboxylic R-COOH Methanoic Corresponding
terminal –oic acid with the suffix –amide. Acid Acid alkane name – e
+ -oic acid
Ester R-COO-R’ Methyl Alkyl group
Ethanoate name +
corresponding
carboxylic acid
name – -ic acid
+ ate
ORGANIC COMPOUNDS CONTAINING
NITROGEN (N)
Amine R-NH2, Methylamine Alkyl group
R2-NH, names + -amine
R3-N If identical, add
prefix di-; If
not,
 Amides have a very high melting points and are arrange
alphabetically
used in plastic and rubber industry, paper
Amide R-CONH2 Ethanamide Corresponding
industry, water and sewage treatment, and in carboxylic acid
colors and inks. name +
-amide
(* location of the multiple bond is sometimes
The table below summarizes all the organic necessary)
compound families and their corresponding classes. In identifying and describing organic compounds
The table presents the general formula, examples, containing functional groups, take note of the
and a brief idea on the systematic naming method. following:
Class General Example Systematic Naming
Formula Method
HYDROCARBONS
Alkanes CnH_2n+2 Methane Carbon chain
prefix + -ane

5-23 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

Guide Questions during Practice There are several types of isomers and these are
1. Are there functional groups in the shown below.
structure?
2. If Step 1 is a Yes; what family does the
compound belong to?
a. Each family organic compounds has a
unique method of naming.

Examples of identifying and naming organic 6.1.1 Constitutional isomers

compounds are given below.  also known as structural isomers
 are those whose atoms are connected
differently; thus having different structures.
 There are three types of structural isomers,
namely skeletal, functional, and positional
isomers.

Skeletal isomers
➢ differ in carbon linkages found in the
“skeleton” chains.
➢ The skeleton chain or the parent chain is
the longest carbon chain.

Positional isomers
➢ only differ in the attachment of the
functional group to the parent chain.
6 Structural Isomerism and Functional isomers
Organic Reactions ➢ are two or more compounds having the
same molecular formula but different
functional groups.
6.1 Isomerism
Isomers
 are different compounds that have the same
Skeletal Isomers
chemical formula.
 also known as chains isomers, these isomers
 In other words, these are group of atoms
arise because of the possibility of branching
with the same molecular formula but
in carbon chains.
different arrangement of atoms.
 an example is butane, C4H10, which has
For example, have a C4H10 molecular formula, but two chain isomers.
have different structures. o In one of them, the carbon atoms
lie in a “straight chain” whereas in
the other the chain is branched

➢ A molecular formula can provide the actual

number of atoms in a compound, but never
the arrangement of atoms. Guide Questions during Practice
➢ This mean that two or more compounds 1. How many carbon atoms are present in the
may have exactly the same molecular longest chain?
formula but may not have the same a. How many carbon atoms are present
structure. in the branch?

6-24 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

 The number of carbon atoms Functional Isomers

will determine the prefix  In this variety of structural isomerism, the
necessary for both the longest isomers contain different functional groups
chain and the branch. - that is, they belong to different families
 Branch name will use the suffix of compounds (different homologous
‘–yl’ series).
2. Are there multiple bonds between two  In the previous lesson, the families of
carbon atoms or functional group? compounds were discussed by pair.
a. What family does the compound o Except for amine and amide, these
belong to? families are actually functional
 Each family of organic isomer of one another: alcohol
compounds has a unique and ether, aldehyde and ketone,
method of naming. and carboxylic acid and ester.
o Alkenes, alkynes, and
alcohols considers the
location of the functional
group.
3. What is the name of the compound?
a. The branch is usually written first.

Sample Problems

6.2 Organic Reactions

 Organic reactions are chemical reactions
involving organic compounds.
 Some of the basic organic chemistry
reaction types are addition reactions,
elimination reactions, and redox reactions.
 Organic reactions are used in the
construction of new organic molecules.

Positional Isomers 6.2.1 Combustion

 In positional isomerism, the basic carbon ➢ Combustion reaction happens when an
skeleton (longest carbon chain) remains element or a compound reacts with oxygen,
unchanged, but important groups are often producing energy in the form of heat
moved around on that skeleton. and light.
 For example, there are two structural ➢ The complete combustion of a
isomers with the molecular C3H7Br. In hydrocarbon produces CO2 and H2O.
one of them the bromine atom is on the end The general equation for combustion
of the chain, whereas in the other it’s reaction is
attached in the middle

➢ For example, in the reaction,

6-25 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

➢ propane is burned to produce CO2 and Halogenation

H2O. If the oxygen is not sufficient, ➢ halogenation is the addition of a diatomic
combustion will be incomplete and carbon halogen molecule to an unsaturated
and toxic carbon monoxide are produced. compound (containing a carbon-carbon
double bond).

6.2.2 Condensation
 A condensation reaction is characterized by the
joining of two molecules to form a larger
molecule and the elimination of a small
molecule such as water or methanol.
 The general formula of a condensation Hydrohalogenation
reaction is ➢ Hydrohalogenation involves the addition
 A + B → C + water/methanol* of a hydrogen atom and a halogen atom to
an unsaturated compound (containing a
carbon-carbon double bond)

6.2.3 Addition
 Addition reactions occur with unsaturated
compounds –compounds containing a carbon
to carbon double and triple bonds. Hydration
 Atoms or groups of atoms are simply added to ➢ a hydration reaction involves the addition
a multiple bond without the elimination of of water (H20) to an unsaturated
atoms or other molecules producing a single compound.
compound with no atoms (from the original ➢ this is one way of preparing an alcohol
compound) removed or displaced. from the corresponding alkene.
 The general equation is addition reaction is
▪ compound A + compound B →
compound C 6.2.4 Saponification
▪ Notice that C is the final product with no  This is process by which triglycerides (a
A or B remaining as a residue. type of fat and most often animal fats or
 Addition reactions can be any of the following: vegetable oils) are reacted with sodium or
hydrogenation, halogenation, potassium hydroxide(lye) to produce
hydrohalogenation, and hydration. glycerol and a fatty acid salt called "soap.”
 When sodium hydroxide is used, a hard
soap is produced. The general formula of a
Hydrogenation saponification process is
➢ Hydrogenation involves adding hydrogen
(H2) to an alkene.
➢ During hydrogenation the double bond is
broken and more hydrogen atoms are
added to the molecule

6-26 | P a g e December 14, 2023

STEM3 – GENERAL CHEMISTRY 1

THE SCIENCE BEHIND SOAP MAKING

➢ Soap making has remained unchanged over
the centuries.
➢ All soap is made from fats and oils, mixed
with alkaline (basic) solutions.
➢ There are many kinds of fats and oils, both
animal and vegetable.
➢ Fats are usually solid at room temperature,
but many oils are liquid at room
temperature.
o Liquid cooking oils originate
from corn, peanuts, olives,
soybeans, and many other plants.
➢ For making soap, all different types of fats
and oils can be used – anything from lard
to exotic tropical plant oils.

SAPONIFICATION REACTION:
➢ Fat+Lye → Soap+Glycerol
➢ The saponification reaction breaks the oil
or fat molecule into glycerin and a sodium
or potassium salt of the long hydrocarbon
chains that have a carboxylate end
functional group.

The modern technique to form solid soap

uses plant oils and/or animal fat and sodium
hydroxide to form solid bars or potassium
hydroxide to form liquid soap.

6-27 | P a g e December 14, 2023