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101 chem

Instructor: Dr.

Course# and Name: Chem -101, General Chemistry

Semester Credit Hours: 2+1

Email Address:

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GENERAL CHEMISTRY (CHEM 101)

Recommended Books

1- Chemistry, The central Science 11E or 12E

BROWN, LEMAY, BURSTEN, MURPHY,

WOODWARD

Pearson International Edition

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2- Chemistry: The Molecular Nature of

Matter, 6E

Jespersen/Brady/Hyslop

Wiley Edition

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Chp. 1

Atomic Structure and

Periodic Table

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Complete Symbols of Element

Contain the symbol of the element, the

mass number and the atomic number.

4

X Mass

number

Atomic

number Subscript →

Superscript →

Symbols

a) number of protons

b)number of neutrons

c) number of electrons

d)Atomic number

e) Mass Number

Br 80

35

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Periodic Table

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The horizontal rows of the periodic table are

called PERIODS.

For a certain element, the number of occupied

shells indicated the number of period

The Periodic Table of Element

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The vertical columns of the periodic table are called

GROUPS, or FAMILIES. For a certain element, the number of outer most

shell’s electrons indicated the number of group

The elements in any group of the periodic table have

similar physical and chemical properties!

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Periodic Table

Group 3 – 7 A

Group 1 - 2A

For a certain element, the number of occupied shells

indicated the number of period

For a certain element, the number of outer most shell’s

electrons indicated the number of group

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Na

2, 8, 1 or 1s2, 2s2, 2p6, 3s1

Na in Group 1 and period 3

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Cl

2, 8, 7 or 1s2, 2s2, 2p6, 3s2, 3p5

Cl in Group 7 and period 3

31

Ga

2, 8, 18, 3 or 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p1

Ga in Group 3 and period 4

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General Properties of Periodic Table

The periodic trends of the following properties will

be studied here.

Atomic radius

Metallic and Non-metallic character

Ionization (energy) potential

Electron affinity

Electronegativity

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Atomic Radius

It is the distance from the centre of the

nucleus to the outermost shell of an atom.

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Trends in atomic radius

down a group

It is fairly obvious that the

atoms get bigger as you go

down groups. Due to adding

extra layers of electrons.

Trends in atomic radius

across periods

atomic radius decreases with

increase in atomic number

(in a period left to right. Due

to Electrons are pulled close

to the nucleus by the

increased Zeff

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Atomic Radius

Metallic character is the tendency of an element to

lose electrons and form positive ions (cations). For

e.g., alkali metals are the most electropositive

elements.

It is also known as electropositivity.

Metallic and Non-metallic Character

The tendency of an element to accept electrons to

form an anion is called its non-metallic or

electronegative character. For e.g., chlorine,

oxygen and phosphorous show greater

electronegative or non-metallic character."

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Trends across the period

Metallic character of elements decreases as we move

to the right. So, the elements at the left have metallic

character while those to the right have a non-metallic

character.

Trends across the group

As we move down the group the number of shells

increases. This causes the atomic size increases.

The electrons of the outermost shell experience less

nuclear attraction and so can lose electrons easily thus

showing increased metallic character.

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The amount of energy required to remove the most loosely

bound electron from an isolated gaseous atom is called

ionization energy (IE).

It is measured in the units of electron volts (eV) per atom

or kilo joules per mole of atoms (kJ mol-1).

Ionization Energy (IE)

Variation along a period

The ionization energy increases with increasing atomic

number in a period.

Variation down a group

The ionization energy gradually decreases in moving from

top to bottom in a group.

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Electron Affinity (EA) Electron affinity is the amount of energy released when

an electron is added to an isolated gaseous atom.

Electron affinity is the ability of an atom to hold an additional electron.

If the atom has more tendency to accept an electron then the energy

released will be large and consequently the electron affinity will be

high. Just like a strong magnet

Variation along a period The size of an atom decreases thus, Hence the electron affinity

increases in a period from left to right

Variation down a group As we move down a group the atomic size and nuclear size increases.

Consequently the electron affinity decreases.

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Electronegativity This the relative tendency of an atom in a molecule to

attract a shared pair of electrons towards itself.

Variation along a period

As the nuclear charge increases from going left to right in a

period because the electrons enter the same shell, the

shielding is less effective. Thus the increased nuclear charge

attract the shared pair of electrons more strongly resulting in

higher electronegativity from going left to right in a period.

Variation down the group

Electronegativity decreases down the group because the

atomic size increases. The larger the size of the atom the

lesser the tendency to attract the shared pair of electrons.

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Periodic Properties of the elements

Electron affinity

Ionization energy

Atomic radius

Atomic properties and the periodic table—a summary

Atomic Properties

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Chp. 2

Chemical Bonding

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Chemical Bonds

Attractive forces that hold atoms together in complex

substances

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•The tendency of an atom to take part in chemical combination is

determined by the number of valence electrons (electrons in the

outermost shell of an atom).

•The atoms acquire the stable noble gas configuration of having eight

electrons in the outermost shell (called octet rule) by mutual sharing or

by transfer of one or more electrons.

•The valency (number of electrons an atom loses, gains or mutually

shares to attain noble gas configuration) of an element is either equal to

the number of valence electrons or equal to 8 minus the number of

valence electrons.

The type of chemical bond developed between the two combining

atoms depends upon the way these atoms acquire a stable noble gas

configuration

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Octet Rule = atoms tend to gain, lose or share electrons so as to

have 8 electrons

C would like to

N would like to

O would like to

Gain 4 electrons

Gain 3 electrons

Gain 2 electrons 24

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Lewis electron dot symbols

The valence electrons are then written as

dots or (small cross marks) around the

symbol of the atom. They are spread in a

pair on four sides of the symbol.

In case of ions the charge is shown

with the symbol.

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Lewis Symbols • Electron bookkeeping method

• Way to keep track of e–’s

• Write chemical symbol surrounded by dots for each e–

H

Li

Na

Be

Mg

B

Al

C

Si

He

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Lewis Symbols

N O F Ne

P S Cl Ar

For the representative elements

Group # = # valence e–’s

He

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Give the dot structure for the following atoms:

1) Na 2) K 3) Al 4) B 5) N 6) P

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Learning Check

A. X would be the electron dot formula for

1) Na 2) K 3) Al

B. X would be the electron dot formula for

1) B 2) N 3) P

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Why are electrons important?

Elements have different electron configurations

– different electron configurations mean different levels

of bonding

– Chemical bonds: an attempt to fill electron shells

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1. Ionic Bonds When the chemical bond occurs by the complete transfer of

electron(s) from the atom(s) of metal to atoms of nonmetal is

called Ionic (electrovalent) bond.

2. Covalent Bonds When the shared electrons are contributed by the two

combining atoms (nonmetal with nonmetal), the bond formed

is called Covalent bond.

3. Metallic Bonds Metal atoms bonded to several other metal atoms

Types of Chemical Bonds

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Ionic Bond

• Between atoms of metals and nonmetals with very

different electronegativity

• Bond formed by transfer of electrons

• Produce charged ions all states.

• Conductors in a liquid state only and have high

melting point.

• Examples; NaCl, CaCl2, K2O

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Formation of Ions from Metals

Ionic compounds result when metals react with

nonmetals

Metals lose electrons to match the number of valence

electrons of their nearest noble gas

Positive ions form when the number of electrons are

less than the number of protons

Group 1 metals ion 1+

Group 2 metals ion 2+

• Group 13 metals ion 3+

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Formation of Sodium Ion

Sodium atom Sodium ion

Na – e Na +

2-8-1 2-8 ( = Ne)

11 p+ 11 p+

11 e- 10 e-

0 1+

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Formation of Magnesium Ion

Magnesium atom Magnesium ion

Mg – 2e Mg2+

2-8-2 2-8 (=Ne)

12 p+ 12 p+

12 e- 10 e-

0 2+

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Some Typical Ions with Positive

Charges (Cations)

Group 1 Group 2 Group 13

H+ Mg2+ Al3+

Li+ Ca2+

Na+ Sr2+

K+ Ba2+

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Learning Check

A. Number of valence electrons in aluminum

1) 1 e- 2) 2 e- 3) 3 e-

B. Change in electrons for octet

1) lose 3e- 2) gain 3 e- 3) gain 5 e-

C. Ionic charge of aluminum

1) 3- 2) 5- 3) 3+

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Solution

A. Number of valence electrons in aluminum

3) 3 e-

B. Change in electrons for octet

1) lose 3e-

C. Ionic charge of aluminum

3) 3+

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Learning Check

Give the ionic charge for each of the following:

A. 12 p+ and 10 e-

1) 0 2) 2+ 3) 2-

B. 50p+ and 46 e-

1) 2+ 2) 4+ 3) 4-

C. 15 p+ and 18e-

2) 3+ 2) 3- 3) 5-

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Ions from Non-metal Ions

• In ionic compounds, nonmetals in groups

5, 6, and 7 gain electrons from metals

• Nonmetal add electrons to achieve the

octet arrangement

• Nonmetal ionic charge: 3-, 2-, or 1-

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Fluoride Ion

unpaired electron octet

1 -

: F + e : F :

2-7 2-8 (= Ne)

9 p+ 9 p+

9 e- 10 e-

0 1 -

ionic charge

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Ions of Representative Elements Can use periodic table to predict ion charges

When we use North American numbering of groups:

Cation positive charge = group #

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Ionic Bonding

Na Cl The metal (sodium) tends to lose its one electron

from the outer level.

The nonmetal (chlorine) needs to gain one more

to fill its outer level, and will accept the one

electron that sodium is going to lose.

Formation of Sodium Chloride

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Ionic Bonding

Na+ Cl -

Note: Remember that NO DOTS are

now shown for the cation!

Formation of Sodium Chloride

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1). Ionic bond – electron from Na is transferred to

Cl, this causes a charge imbalance in each atom.

The Na becomes (Na+) and the Cl becomes (Cl-),

charged particles or ions. 46

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Formation of Magnesium fluoride

The Ionic Bond

[Ne] 3s2 1s22s22p5 1s22s22p6

[Ne] [Ne]

Mg2+ F - + F Mg F -

1s22s22p6 [Ne] 3s0

[Ne]

Mg2+ + 2e- Mg

2e- + F F - 2 2

MgF2 + 2F Mg

2

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Determining Ionic Formulas

“Cross-cross” rule

– Make magnitude of charge on one ion into subscript

for other

– When doing this, make sure that subscripts are

reduced to lowest whole number.

Ex. What is the formula of ionic compound

formed between aluminum and oxygen ions?

Al2O3 Al3+ O2–

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Determining Ionic Formulas

Ex. Formula of ionic compound formed when

magnesium reacts with oxygen

– Mg is group 2A

• Forms +2 ion or Mg2+

– O is group 6A

• Forms –2 ion or O2–

– To get electrically neutral particle need

• 1:1 ratio of Mg2+ and O2–

– Formula: MgO

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Your Turn! Which of the following is the correct formula for

the formula unit composed of potassium and

oxygen ions?

A.KO

B.KO2

C.K2O

D.P2O3

E.K2O2

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Your Turn! Which of the following is the correct formula for

the formula unit composed of Fe3+ and sulfide ions?

A.FeS

B.Fe3S2

C.FeS3

D.Fe2S3

E.Fe4S6

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Covalent Bond • Between nonmetallic elements of similar

electronegativity.

• Formed by sharing electron pairs

• Stable non-ionizing particles, they are not

conductors at any state

• Examples; O2, CO2, C2H6, H2O, SiC

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A covalent bond is a chemical bond in which two

or more electrons are shared by two atoms.

Why should two atoms share electrons?

F F +

7e- 7e-

F F

8e- 8e-

F F

F F

Lewis structure of F2

lone pairs lone pairs

lone pairs lone pairs

single covalent bond

single covalent bond

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8e-

H H O + + O H H O H H or

2e- 2e-

Lewis structure of water

Double bond – two atoms share two pairs of electrons

single covalent bonds

O C O or O C O

8e- 8e- 8e- double bonds

double bonds

Triple bond – two atoms share three pairs of electrons

N N

8e- 8e-

N N

triple bond triple bond

or

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Electronegativity • The ability of atoms in a molecule to attract

electrons to itself.

• On the periodic chart, electronegativity increases

as you go…

– …from left to right across a row.

– …from the bottom to the top of a column.

.

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Types of covalent bonds

When two atoms bond their DIFFERENCE in

electronegativity determines the covalent bond

type.

1- Polar Covalent Bonds

2- Non Polar Covalent Bonds

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Polar Covalent Bonds 1- Occurs between metals and non metals

2-The greater the difference in electronegativity,

3- Electrons are unequallly shared

4- The more polar is the bond.

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Occurs between non metals

When electrons are shared equally

Has almost no electronegativity difference (0.0 to 0.4).

Examples:

Electronegativity

Atoms Difference Type of Bond

N-N 3.0 - 3.0 = 0.0 Nonpolar covalent

Cl-Br 3.0 - 2.8 = 0.2 Nonpolar covalent

H-Si 2.1 - 1.8 = 0.3 Nonpolar covalent

Non-polar Covalent Bonds

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Covalent bonds; Two atoms share one or more pairs of outer-

shell electrons. Oxygen Atom Oxygen Atom

Oxygen Molecule (O2)

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Your Turn! Which species is most likely covalently bonded?

A. CsCl

B. NaF

C. CaF2

D. CO

E. MgBr2

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Lewis Structures

Diatomic Gases:

• H and Halogens

H2

• H· + ·H H:H or HH

• Each H has 2 e–’s through sharing

• Can write shared pair of e–’s as line ()

• : =

• covalent bond

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Lewis Structures Diatomic Gases:

F2

• Each F has complete octet

• Only need to form one bond to complete octet

• Pairs of e–’s not included in covalent bond are

called Lone Pairs

• Same for rest of Halogens: Cl2, Br2, I2

F F+ F F F F

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Lewis Structures

Diatomic Gases:

HF

• Same for HCl, HBr, HI.

• Molecules are diatomics as need only 1e– to

complete octet

• Separate molecules

– Gas in most cases because very weak intermolecular

forces

H F+ H F H F

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Your Turn! How many electrons are required to complete the

octet around nitrogen, when it forms N2 ?

A. 2

B. 3

C. 1

D. 4

E. 6

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Lewis Structures • Many nonmetals form more than 1 covalent bond

C

H

CH H

H H

CH H

H

NH H

H

NH H

H

O H

H

O H

H

ON

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Double Bonds

• 2 pairs of e–’s shared between 2 atoms

Ex. CO2

Triple bond

• 3 pairs of e–’s shared between 2 atoms

Ex. N2

O OC C OO C OO

N N N N N N

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Your Turn! Which species is most likely to have multiple bonds

?

A. CO

B. H2O

C. PH3

D. BF3

E. CH4

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Metallic Bond

• Formed between atoms of metallic elements

• Electron cloud around atoms

• Good conductors at all states, lustrous, different

melting points

• Examples; Na, Fe, Al, Au, Co

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Intermolecular

Attractions and the

Properties of Liquids

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Intermolecular Forces (IMF)

• Important differences between gases,

solids, and liquids:

– Gases

• Expand to fill their container

– Liquids

• Retain volume, but not shape

– Solids

• Retain volume and shape

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Inter vs. Intra-Molecular Forces • Intramolecular forces

– Covalent bonds within molecule

– Strong

• Intermolecular forces

– Attraction forces between molecules

– Weak

Cl H Cl H

Covalent Bond (strong) Intermolecular attraction (weak)

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Intermolecular Forces

The attractions between molecules are not

nearly as strong as the intramolecular

attractions that hold compounds together.

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Intermolecular Forces

Intermolecular forces are attractive forces between molecules.

Intramolecular forces hold atoms together in a molecule.

Intermolecular vs Intramolecular

Generally, intermolecular forces are

much weaker than intramolecular forces.

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Types of intermolecular Forces

• Dipole-dipole forces

• Hydrogen bonding

• London dispersion forces

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Dipole-Dipole Forces

1- Attractive forces between polar molecules

Orientation of Polar Molecules in a Solid

2- Electrostatic interaction between the oppositely charged regions

of polar molecule and negative charged regions in another

molecules (dipoles).

Exampes: HCl, CO

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Ex: Which of the following molecules have dipole

interactions?

A) F2

B) CH4

C) H2O

D) CH3Cl

E) NH3

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Hydrogen bonds

What are they?

A special case of permanent dipole-dipole

interactions

They are stronger than dipole –dipole interaction.

Molecules with hydrogen bonds have higher boiling

points than molecules that don’t.

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Hydrogen bonds

What do you need?

A hydrogen atom covalently bonded to an

electronegative atom … N, O or F.

If only one of these conditions is met, you don’t

get hydrogen bonding.

A lone pair of electrons on the electronegative

atom.

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Hydrogen bonds

Give me an example!

This does not have any hydrogen bonds. Carbon is

not very electronegative, and it has no lone pairs of

electrons in methane.

methane, CH4 …

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Hydrogen bonds

Give me a real example!

This does have hydrogen bonds.

Nitrogen is very electronegative, and it has one

lone pair of electrons in ammonia.

ammonia, NH3 …

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Hydrogen bonds

Give me another example!

This has not one, but two hydrogen bonds.

Oxygen is very electronegative, and it has two

lone pairs of electrons in water.

water, H2O …

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Examples Hydrogen bonds

Remember, you need:

A hydrogen atom covalently bonded to an

electronegative atom … N, O or F.

If only one of these conditions is met,

you don’t get hydrogen bonding.

A lone pair of electrons on the electronegative

atom.

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Ex: Which of the following molecules can have

hydrogen bonding?

A) F2

B) CH4

C) H2O

D) CH3Cl

E) NH3

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London Dispersion Forces 1-They are the weakest kind of intermolecular attraction

2- occur between non polar molecules molecules.

3-They are thought to be caused by the motion of electrons.

This is because they are a temporary attractive force that results when the electrons in two adjacent atoms occupy a position that make the atoms form temporary dipoles.

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Dispersion Forces

• Dispersion Forces are also known as

• London forces and van

der Waals forces.

• They were named the London forces in honor of the German physicist; Fritz London who studied these forces!!

Johannes van Der Waals (1837-

1923)

Fritz London (1900-1954)

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Factors Affecting dispersion or

London Forces

The strength of dispersion forces tends to increase

with increased

1-molecular weight.

2- Larger atoms have larger electron clouds, which

are easier to polarize.

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S

Ex. What type(s) of intermolecular forces exist between

each of the following molecules?

HBr HBr is a polar molecule: dipole-dipole forces. There are also

dispersion forces between HBr molecules.

CH4

CH4 is nonpolar: dispersion forces.

SO2

SO2 is a polar molecule: dipole-dipole forces. There are also

dispersion forces between SO2 molecules.

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Intermolecular Forces and the Properties

of Liquids

• In summary, intermolecular forces play a large

role in many of the physical properties of liquids

and gases. These include:

• Vapor pressure

• Boiling point

• Surface tension

• Viscosity

• Solubility

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Table 12.1 Boiling Points of Halogens and Noble Gases

Larger molecules have stronger London forces and thus

higher boiling points.

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Vapor Pressure

Vapor pressures of liquids decreases with

decreases the intermolecular forces

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Propane, C3H8

BP –42.1oC

Hexane, C6H14

BP 68.7oC

More sites (marked with *) along its chain where attraction

to other molecules can occur

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Boiling Point

Your Turn!

Which species has a higher boiling point, Cl2 or

HCl; F2 or HF ?

A. HCl; F2

B. Cl2; F2

C. HCl; HF

D. Cl2 ; HF

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Surface Tension • Liquids containing molecules

with strong intermolecular

forces have high surface

tension

– Allows us to fill glass above rim

• Gives surface rounded appearance

• Surface acts as “skin” that lets

water pile up

• Surface resists expansion and

pushes back

Surface Tension

as IMF

Surface Tension

as IMF

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Viscosity • Resistance of a liquid

to flow is called

viscosity.

• It is related to the

ease with which

molecules can move

past each other.

• Viscosity increases

with stronger

intermolecular forces

and decreases with

higher temperature.

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Effect of Intermolecular Forces on Viscosity

Acetone

• Polar molecule

– Dipole-dipole and

– London forces

Ethylene glycol

• Polar molecule

– Hydrogen-bonding

– Dipole-dipole and

– London forces

Which is more viscous??

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Your Turn! For each pair given, which is more viscose ?

• CH3CH2CH2CH2OH, CH3CH2CH2CHO;

• C6H14, C12H26; NH3(l ), PH3(l )

A. CH3CH2CH2CH2OH; C6H14; NH3(l )

B. CH3CH2CH2CH2OH; C12H26; NH3(l )

C. CH3CH2CH2CHO; C6H14; PH3(l )

D. CH3CH2CH2CHO; C12H26; NH3(l )

E. CH3CH2CH2CH2OH; C12H26; PH3(l )

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Solubility • “Like dissolves like”

–To dissolve polar substance, use polar

solvent

–To dissolve nonpolar substance, use

nonpolar solvent

• Compare relative polarity of two substances

–Similar polarity means greater ability to

interact with each other

–Differing polarity means that they don’t

interact; move past each other 100

Factors Affecting Solubility

• Chemists use the axiom “like dissolves like."

– Polar substances tend to dissolve in polar

solvents.

– Nonpolar substances tend to dissolve in

nonpolar solvents.

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Factors Affecting Solubility

The more similar the intermolecular

attractions, the more likely one substance

is to be soluble in another.

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Factors Affecting Solubility

Glucose (which has hydrogen bonding) is very soluble in water, while cyclohexane

(which only has dispersion forces) is not.

103

Factors Affecting Solubility

• Vitamin A is soluble in nonpolar compounds

(like fats).

• Vitamin C is soluble in water.

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Temperature

Generally, the

solubility of solid

solutes in liquid

solvents

increases with

increasing

temperature.

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Temperature

• The opposite is true

of gases.

– Carbonated soft

drinks are more

“bubbly” if stored

in the refrigerator.

– Warm lakes have

less O2 dissolved

in them than cool

lakes.

106

Your Turn!

Ex: Which of the following are not expected

to be soluble in water?

A.HF

B.CH4

C.CH3OH

D.All are soluble

107

Chp. 3

The Mole

and Stoichiometry

Chemistry: The Molecular Nature of Matter, 6E

Jespersen/Brady/Hyslop

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The Mole • Number of atoms in exactly 12 grams of 12C

atoms

How many atoms in 1 mole of 12C ?

– Based on experimental evidence

1 mole of 12C = 6.022 × 1023 atoms = 12.011 g

Avogadro’s number = NA

– Number of atoms, molecules or particles in one

mole

• 1 mole of X = 6.022 × 1023 units of X

• 1 mole Xe = 6.022×1023 Xe atoms

• 1 mole NO2 = 6.022×1023 NO2 molecules 109

The Mole The amount of substance that contains as many atoms as there

are in exactly 12 grams of pure 12C. That is 6.022 x 1023 atoms

1 mole of particles contains 6.022 x1023 particles (Avogadro’s

number )

For example

One mole of hydrogen atoms = 6.023 x 1023 atoms of

hydrogen

One mole of hydrogen molecules = 6.023 x 1023 molecule of

hydrogen

One mole of electrons = 6.023 x 1023 electrons

One mole of sodium ions (Na+) = 6.023 x 1023 Na+ ions

110

Avogadro’s Number

• NA = 6.02 x 1023

1 mole of 12C has a mass of 12 g. 111

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Molecular & Formula Weight

Formula Mass is sometimes used in a more general sense to include Molecular Mass, but its formal definition refers to the sum of the atomic weights of the atoms in ionic bonded compounds

112

Molecular Mass (also referred to Molecular Weight (MW)) is the sum of the atomic weights of all atoms in a covalently bonded molecule – organic compounds, oxides, etc.

Molar Mass A substance’s molar mass (molecular weight) is the mass

in grams of one mole of the compound.

C=12 , O=16 amu

CO2 = 44.01 grams / mole

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 atom of 12C = 12.00 amu

1 mole 12C atoms = 12.00 g 12C

For any element atomic mass (amu) = molar mass (grams)

113

Ex: What is the molar mass of caffeine, C8H10N4O2?

C = 12.0107 g/mol H = 1.00794 g/mol

N = 14.0067 g/mol O = 15.9994 g/mol

12.0107 g/mol C x 8 mol C/mol C8H10N4O2 = 96.0856 g/mol C8H10N4O2

1.00794 g/mol H x 10 mol H/mol C8H10N4O2 = 10.0794 g/mol C8H10N4O2

14.0067 g/mol N x 4 mol N/mol C8H10N4O2 = 56.0268 g/mol C8H10N4O2

15.9994 g/mol O x 2 mol O/mol C8H10N4O2 = 31.9988 g/mol C8H10N4O2

Add total the elemental masses to get the molecular mass of caffeine

96.0856 + 10.0794 + 56.0268 + 31.9988 =

194.1906 g/mol C8H10N4O2

114

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Moles of Compounds Atoms

– Atomic Mass

• Mass of atom (from periodic table)

• 1 mole of atoms = gram atomic mass

= 6.022×1023 atoms

Molecules

– Molecular Mass

• Sum of atomic masses of all atoms in compound’s

formula

1 mole of molecule X = gram molecular mass of X

= 6.022 × 1023 molecules

115

Moles of Compounds Ionic compounds

– Formula Mass

• Sum of atomic masses of all atoms in ionic compound’s

formula

1 mole ionic compound X = gram formula mass of X

= 6.022 × 1023 formula units

General

Molar mass (MM)

• Mass of 1 mole of substance (element, molecule, or ionic

compound) under consideration

1 mol of X = gram molar mass of X

= 6.022 × 1023 formula units

116

Ex: How many H atoms are in 72.5 g of C3H8O?

1 mol C3H8O molecules = 8 mol H atoms

1 mol H = 6.023 x 1023 atoms H

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

x 8 mol H atoms x 6.023 x 1023 H atoms =

= 5.82 x 1024 atoms H

72.5 g C3H8O

60 g C3H8O

117

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Ex: How many moles of iron (Fe) are in 15.34 g

Fe?

• What do we know?

1 mol Fe = 55.85 g Fe

• What do we want to determine?

15.34 g Fe = ? Mol Fe

• Set up ratio so that what you want is on top & what you

start with is on the bottom

118

Fe g 55.85

Fe mol 1Fe g 15.34 = 0.2747 mole Fe

Start End

Ex: If we need 0.168 mole Ca3(PO4)2 for an

experiment, how many grams do we need to weigh

out?

• Calculate MM of Ca3(PO4)2

3 × mass Ca = 3 × 40.08 g = 120.24 g

2 × mass P = 2 × 30.97 g = 61.94 g

8 × mass O = 8 × 16.00 g = 128.00 g

1 mole Ca3(PO4)2 = 310.18 g Ca3(PO4)2

• What do we want to determine?

0.168 g Ca3(PO4)2 = ? Mol Ca3(PO4)2

119

Start End

• Set up ratio so that what you want is on the top &

what you start with is on the bottom

120

= 52.11 g Ca3(PO4)2

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Your Turn! Ex: How many moles of CO2 are there in 10.0 g?

A.1.00 mol

B.0.0227 mol

C.4.401 mol

D.44.01 mol

E.0.227 mol

121

= 0.227 mol CO2

2

22

CO g 44.01

CO mol 1 CO g 0.10

Molar mass of CO2

1 × 12.01 g = 12.01 g C

2 × 16.00 g = 32.00 g O

1 mol CO2 = 44.01 g CO2

Your Turn! Ex: How many grams of platinum (Pt) are in 0.475

mole Pt?

A.195 g

B.0.0108 g

C.0.000513 g

D.0.00243 g

E.92.7 g

Pt mol 1

Pt g 195.08Pt mol 475.0

= 92.7 g Pt

Molar mass of Pt = 195.08 g/mol

122

Percent Composition Percent composition by mass:

The mass of one element in a compound divided by

the mass of the entire compound

Steps to determine percentage composition:

1. Calculate the mass of each individual element

in the compound

2. Add up all the masses of each element to get

the total mass of compound

3. Divide the mass of each individual element

with the total mass of compound

123

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42

% element = x 100 (number of atoms)(atomic weight)

(FW of the compound)

Percent Composition

The Percent of C in C2H6

%C = (2)(12.0 amu)

(30.0 amu)

= 24.0 amu x 100

30.0 amu

= 80.0 %

124

Percent composition of an element in a compound

n x molar mass of element molar mass of compound

x 100%

n is the number of moles of the element in 1 mole of the

compound

C2H6O

%C = 2 x (12.01 g)

46.07 g x 100% = 52.14%

%H = 6 x (1.008 g)

46.07 g x 100% = 13.13%

%O = 1 x (16.00 g)

46.07 g x 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.00%

Percent composition

=

125

Types of Formulas •Empirical Formula: (E.F.)

•Simplest whole-number ratio of atoms in a

compound where “empirical” means derived from

experiment

•Molecular Formula: (M.F.)

• Chemical formula of a compound that expresses the

actual number of atoms present in one molecule.

-The molecular formula will either be exactly the

same or some multiple of the empirical formula!

H2O E.F same as M.F 126

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Empirical Formula Ex: The compound para-aminobenzoic acid is composed of

carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and

oxygen (23.33%). Find the empirical formula of PABA.

Assuming 100.00 g of para-aminobenzoic acid,

C: 61.31 g x = 5.105 mol C

H: 5.14 g x = 5.09 mol H

N: 10.21 g x = 0.7288 mol N

O: 23.33 g x = 1.456 mol O

1 mol

12.01 g

1 mol

14.01 g

1 mol

1.01 g

1 mol

16.00 g

127

Calculate the mole ratio by dividing by the smallest number of

moles:

C: = 7.005 7

H: = 6.984 7

N: = 1.000

O: = 2.001 2

5.105 mol

0.7288 mol

5.09 mol

0.7288 mol

0.7288 mol

0.7288 mol

1.458 mol

0.7288 mol

These are the subscripts for the empirical formula:

C7H7NO2

128

Molecular Formula

Molecular formula of a compound that

expresses the actual number of atoms present

in one molecule.

The molecular formula will either be exactly

the same or some multiple of the empirical

formula

129

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Determining Molecular Formula

1. Follow the same steps for determining

empirical formula.

2. For the Molecular Formula, you need to be

given the molar mass of the compound. Find

the molar mass of the empirical formula.

3. Divide the molar mass of the compound by

the molar mass of the empirical formula to get

the factor with which to multiply each

subscript in the empirical formula

130

Ex: A compound contains 75% carbon and 25% hydrogen.

Determine its empirical formula. The molecular mass of this

compound is 16 amu. Determine its molecular formula also.

The atomic masses are: C = 12 amu, H = 1 amu.

So, The empirical formulae of the compound = C1H4 or CH4

The empirical formula mass = (1 x 12) + (4 x 1) = 12 + 4 = 16 amu

Molecular mass (given) = 16 amu

Therefore, Molecular formula = 1 x Empirical formula = 1 x CH4 = CH4

Molecular mass

Empirical formula mass

16 amu

16 amu

131

= So, N =

Chemical Equation

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)

Reactants appear on the left side of the equation

Products appear on the right side of the equation 132

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Chemical Equation

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)

The states of the reactants and products are written in

parentheses to the right of each compound 133

Chemical Equation

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)

Coefficients are inserted to balance the equation

134

Subscripts and Coefficients

•Subscripts tell the number of atoms of each element in a

molecule

•Coefficients tell the number of molecules.

135

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Balancing Equation

Balancing by Inspection GUIDELINES

1. Count the number of elements on both sides

of the equation

2. Change the coefficients (NEVER the

subscripts) to get the same number of elements

on both sides of the equation

136

Balancing Equation

A representation of a chemical reaction:

C2H5OH + O2 → CO2 + H2O

reactants products

Unbalanced !

C2H5OH + 3O2 → 2CO2 + 3H2O

The equation is balanced.

1 mole of ethanol reacts with 3 moles of oxygen

to produce

2 moles of carbon dioxide and 3 moles of water 137

138

Stoichiometry

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Ex: Methanol burns in air according to the equation

2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what

mass of water is produced?

= 235 g H2O

2CH3OH + 3O2 2CO2 + 4H2O

2 mol 4 mol

2x(12+4+16) g 4x(2+16) g

209 g m

209 x 4(2+16)

m =

2(12+4+16)

139

Ex: Without using a calculator, arrange the

following samples in order of increasing

numbers of carbon atoms:

12 g 12C, 1 mol C2H2, 9 x 1023 molecules of

CO2

12 g 12C (6 x 1023 C atoms) < 9 x 1023 CO2

molecules (9 x 1023 C atoms) < 1 mol C2H2 (12

x 1023 C atoms).

140

Ex: What is the mass in grams of 1.000 mol of glucose,

C6H12O6?

C6H12O6 has a molar mass of 180.0 g/mol

Ex: Calculate the number of moles of glucose (C6H12O6) in

5.380 g of C6H12O6

141

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Ex: Calculate the mass, in grams, of 0.433 mol of calcium

nitrate

Ex: How many glucose molecules are in 5.23 g of C6H12O6?

(b) How many oxygen atoms are in this sample?

142

Ex: Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and

54.50% O by mass. What is the empirical formula of ascorbic

acid?

143

Ex: Calculate the percentage of carbon, hydrogen,

and oxygen (by mass) in C12H22O11

144

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49

Ex: Calculate the number of H atoms in 0.350 mol of C6H12O6

145

The percentage by mass of

phosphorus in Na3PO4 is:

a. 44.0%

b. 11.7%

c. 26.7%

d. 18.9%

146

Your Turn! Calculate the number of moles of calcium in 2.53 moles of

Ca3(PO4)2

A.2.53 mol Ca

B.0.432 mol Ca

C.3.00 mol Ca

D.7.59 mol Ca

E.0.843 mol Ca

2.53 moles of Ca3(PO4)2 = ? mol Ca

3 mol Ca 1 mol Ca3(PO4)2

243243

)PO(Ca mol 1

Ca mol 3 )PO(Ca mol 53.2

= 7.59 mol Ca 147

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Ex: How many g of iron are required to use up all of 25.6 g

of oxygen atoms (O) to form Fe2O3?

A.59.6 g

B.29.8 g

C.89.4 g

D.134 g

E.52.4 g

148

Fe mol 1

Fe g 55.845

O mol 3

Fe mol 2

O g 16.0

O mol 1O g 25.6

mass O mol O mol Fe mass Fe

25.6 g O ? g Fe

3 mol O 2 mol Fe

= 59.6 g Fe

Ex: Balance the equation

__C3H8(g) + __O2(g) __CO2(g) + __H2O(ℓ)

Assume 1 in front of C3H8

3C 1C 3

8H 2H 4

1C3H8(g) + __O2(g) 3CO2(g) + 4H2O(ℓ)

2O 5 =10 O = (3 2) + 4 = 10

8H H = 2 4 = 8

1C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(ℓ)

149

___KClO3(s) → ___KCl(s) +___ O2(g) 2 2 3

__ Ba(OH)2(aq) +__ Na2SO4(aq) → __ BaSO4(s) + __ NaOH(aq) 1 1 2 1

__H3PO4(aq) + __ Ba(OH)2(aq) → __Ba3(PO4)2(s) + __H2O(ℓ) 3 1 6 2

150

Ex: Balance each of the following equations.

What are the coefficients in front of each compound?

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151

Chp. 4

States of Matters

State of Matters

How many state of matter found in

nature? Matter exists on earth in three physical states :

-Gases -Liquid -Solids

-Example water

-In the solid state H2O is known as ice .

-In the liquid state is called water.

-In the gases state is known as steam or water

Vapour.

152

The various points of distinction

between the three states

153

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52

Solid Liquid Gas

Melt Evaporate

Condense Freeze

154

The Gaseous State

155

Characteristics of Gases

•Unlike liquids and solids, gases

–expand to fill their containers;

–are highly compressible;

–have extremely low densities.

• Gas volume changes greatly with pressure

• Gas volume changes greatly with temperature

• Gas volume is a function of the amount of gas

156

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Introduction to Pressure

Pressure = Force

Area

Barometer

Units of Pressure

1 pascal (Pa) = 1 N/m2 = 1 kg/m.s2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa(~105) = 101.325 kPa

Mercury Barometer At sea level,

h = 760 mm Hg 157

Example: a) Convert 0.357 atm to torr. 1 atm = 760 torr 0.357 atm = ? torr

760 x 0.357 P= ____________ = 271 torr 1

b) Convert 0.066 torr to atm c) Convert 147.2 kPa to torr

158

The Gas Laws

Four variables define the state of a gas

1)Temperature T

2) Pressure P

3) Volume V

4) Number of moles n

159

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Boyle’s Law

The volume of a fixed quantity of gas at

constant temperature is inversely

proportional to the pressure.

160

Decrease the volume Increase the pressure

P and V are inversely proportional

Constant temperature

Constant amount of gas 161

A plot of V versus P results in a curve

A plot of V versus 1/P will be a straight line

P a 1/V P = k (1/V )

P x V = constant

P1 x V1 = P2 x V2 This means a plot of P versus

1/V will be a straight line 162

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Ex: A sample of chlorine gas occupies a volume of 946 mL

at a pressure of 726 mmHg. What is the pressure of the gas

(in mmHg) if the volume is reduced at constant temperature

to 154 mL?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2 = ?

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL 154 mL

= = 4460 mmHg

163

Charles’ Law

The volume of a fixed

amount of gas at

constant pressure is

directly proportional

to its absolute

temperature i.e., V = k T

A plot of V versus T will be a straight line 164

The volume occupied by any sample of gas at constant

pressure is directly proportional to its absolute temperature

V a T

V = constant x T

V1/T1 = V2/T2

Note:

Temperature must be

in ____________

T (K) = t (°C) + 273.15 165

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Ex: A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of

1.54 L if the pressure remains constant?

V1 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398.15 K 3.20 L

= = 192 K

V1/T1 = V2/T2

T2 = 192 – 273 = -81 0C 166

Gay-Lussac’s Law

V= constant, n= Constant

The pressure of a fixed amount of gas at constant

volume is directly proportional to its absolute

temperature

P / T = K

P1T2 = P2 T1

Constant V Constant n

167

Combined Gas Law

In the event that all three

parameters, P, V, and T,

are changing, their

combined relationship is

defined as follows

assuming the mass of

the gas (number of

moles) is constant.

•For Fixed Amount of Gas

• n = constant

• P . V = k

• V / T = k

• P / T = k

•P.V / T = k

•P1V1 / T1 = P2V2 / T2

168

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Ex: The pressure of 4.0 L of Nitrogen in a flexible container is

decreased to one-half its original pressure, and its absolute

temperature is increased to double the original temperature. The

new volume is now

a. 2.0 L b. 4.0 L c. 8.0 L d. 16.0 L e. 32.0 L

169

Avogadro’s Law

The volume of a gas at constant temperature and

pressure is directly proportional to the number of

moles of the gas

Mathematically, this means V = k.n

170

– The volume of one mole of gas is called the: molar gas volume, Vm

– Volumes of gases are often compared at standard temperature and pressure (STP), chosen to be

0 °C (273.15 ° K) and 1 atm pressure – At STP, the molar volume, Vm, that is, the volume occupied by one mole of any gas, is 22.4 L/mol

Avogadro’s Law

171

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58

V a number of moles (n)

V = constant x n

V = k.n

V1/n1 = V2/n2

172

Ex: A sample of Fluorine gas has a volume of 5.80 L at

150.0 °C and 10.5 atm of pressure. How many moles of

Fluorine gas are present?

First, use the combined empirical gas law to determine

the volume at STP. Then, use Avagadro’s law to

determine the number of moles

173

Ideal Gas Equation

174

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59

Ex: Ammonia burns in oxygen to form nitric oxide (NO) and

water vapor. How many volumes of NO are obtained from

one volume of ammonia at the same temperature and

pressure?

175

176

The Ideal Gas Law

The numerical value of “R” can be derived using

Avogadro’s law, which states that one mole of any

gas at STP will occupy 22.4 liters

177

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The Ideal Gas Law

The ideal gas equation is usually expressed in the

following form

178

Ex: Argon is an inert gas used in light bulbs to retard the

vaporization of the filament. A certain light bulb containing

argon at 1.20 atm and 18 °C is heated to 85 °C at constant

volume. What is the final pressure of argon in the light bulb

(in atm)?

179

Ex: A steel tank has a volume of 438 L and is filled with

0.885 kg of O2. Calculate the pressure of oxygen in the tank

at 21oC

180

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Ex: A 50-L cylinder of nitrogen, N2, has a pressure of 17.1 atm

at 23 °C. What is the mass (g) of nitrogen in the cylinder?

181

Molecular Mass

Molecular Weight Determination

182

Density of Gas

183

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62

Ex: What is the density of Methane gas (natural gas),

CH4, at 125 °C and 3.50 atm?

184

Gas Stoichiometry Ex: Suppose you heat 0.0100 mol of Potassium Chlorate,

KClO3, in a test tube. How many liters of Oxygen can you

produce at 298 K and 1.02 atm?

185

Gas Stoichiometry

Ex: What is the volume of CO2 produced at 37°C and 1.00

atm when 5.60 g of glucose are used up in the reaction:

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

g C6H12O6 mol C6H12O6 mol CO2 V CO2

186

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Ex: Calcium carbonate, CaCO3(s), decomposes upon heating

to give CaO(s) and CO2(g). A sample of CaCO3 is

decomposed, and the carbon dioxide is collected in a 250 mL

flask. After the decomposition is complete, the gas has a

pressure of 1.3 atm at a temperature of 31 °C. How many

moles of CO2 gas were generated?

187

Ex: The gas pressure in an aerosol can is 1.5 atm at 25 °C.

Assuming that the gas inside obeys the ideal-gas equation,

what would the pressure be if the can were heated to 450 °C?

188

Ex: An inflated balloon has a volume of 6.0 L at sea level (1.0

atm) and is allowed to ascend in altitude until the pressure is

0.45 atm. During ascent the temperature of the gas falls from

22 °C to –21 °C. Calculate the volume of the balloon at its

final altitude.

189

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Ex: What is the density of carbon tetrachloride

(CCl4) vapor at 714 torr and 125 °C?

190

191

192

Chp. 5

Redox Reactions

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65

193

Oxidation–Reduction Reactions

Involves 2 processes:

Oxidation = Loss of Electrons (LEO)

Na Na+ + e Oxidation Half-Reaction

Reduction = Gain of electrons (GER)

Cl2 + 2e 2Cl Reduction Half-Reaction

Net reaction:

2Na + Cl2 2Na+ + 2Cl

– Oxidation & reduction always occur together

– Can't have one without the other

Oxidation Reduction Reaction Oxidizing Agent

• Substance that accepts e's

– Accepts e's from another substance

– Substance that is reduced

– Cl2 + 2e 2Cl–

Reducing Agent

• Substance that donates e's

– Releases e's to another substance

– Substance that is oxidized

– Na Na+ + e–

194

Your Turn! Which species functions as the oxidizing agent in the

following oxidation-reduction reaction?

Zn(s) + Pt2+(aq) Pt(s) + Zn2+(aq)

A.Pt(s)

B.Zn2+(aq)

C.Pt2+(aq)

D.Zn(s)

E.None of these, as this is not a redox reaction.

195

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Rules for Assigning Oxidation Numbers

1. Oxidation numbers must add up to charge on

molecule, formula unit or ion.

2. Atoms of free elements have oxidation numbers of

zero.

3. Metals in Groups 1A, 2A, and Al have +1, +2, and

+3 oxidation numbers, respectively.

4. H & F in compounds have +1 & –1 oxidation

numbers, respectively.

5. Oxygen has –2 oxidation number.

6. Group 7A elements have –1 oxidation number.

196

Rules for Assigning Oxidation Numbers

7. Group 6A elements have –2 oxidation number.

8. Group 5A elements have –3 oxidation number.

9. When there is a conflict between 2 of these rules or

ambiguity in assigning an oxidation number, apply

rule with lower oxidation number & ignore

conflicting rule.

Oxidation State

– Used interchangeably with oxidation number

– Indicates charge on monatomic ions

– Iron (III) means +3 oxidation state of Fe or Fe3+

197

Ex: Assigning Oxidation Number

1. Li2O

Li (2 atoms) × (+1) = +2 (Rule 3)

O (1 atom) × (–2) = –2 (Rule 5)

sum = 0 (Rule 1)

+2 –2 = 0 so the charges are balanced to zero

2. CO2

C (1 atom) × (x) = x

O (2 atoms) × (–2) = –4 (Rule 5)

sum = 0 (Rule 1)

x 4 = 0 or x = +4

C is in +4 oxidation state

198

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199

Learning Check

Assign oxidation numbers to all atoms:

Ex. ClO4

O (4 atoms) × (–2) = –8

Cl (1 atom) × (–1) = –1

(molecular ion) sum ≠ –1 (violates Rule 1)

Rule 5 for O comes before Rule 6 for halogens

O (4 atoms) × (–2) = –8

Cl (1 atom) × (x) = x

sum = –1 (Rule 1)

–8 + x = –1 or x = 8 –1

So x = +7; Cl is oxidation state +7

Learning Check Ex: Assign Oxidation States To All Atoms:

• MgCr2O7

Mg =+2; O = –2; and Cr = x (unknown)

+2 + 2x + {7 × (–2)} = 0

2x – 12 = 0 x = +3

Cr is oxidation # of +3

• KMnO4

K =+1; O = – 2; so Mn = x

+1 + x + {4 × (–2)} = 0

x – 7 = 0 x = +7

Mn is oxidation # of +7

200

Your Turn! Ex: What is the oxidation number of each atom in

H3PO4?

A. H = –1; P = +5; O = –2

B. H = 0; P = +3; O = –2

C. H = +1; P = +7; O = –2

D. H = +1; P = +1; O = –1

E. H = +1; P = +5; O = –2

201

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Redefine Oxidation-Reduction

in Terms of Oxidation Number • A redox reaction occurs when there is a change

in oxidation number.

Oxidation

– Increase in oxidation number

– e loss

Reduction

– Decrease in oxidation number

– e gain

202

Using Oxidation Numbers to

Recognize Redox Reactions • Sometimes literal electron transfer:

Cu: oxidation number decreases by 2

reduction

Zn: oxidation number increases by 2

oxidation

203

+ ++2 +20 0

increase oxidation

decrease reduction

Cu2+ Zn Zn2+ Cu

Balancing Redox Reactions Some Redox reactions are simple:

Ex. Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)

Break into half-reactions

Zn(s) Zn2+(aq) + 2e oxidation

LEO

Reducing agent

Cu2+(aq) + 2e Cu(s) reduction

GER

Oxidizing agent 204

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Example Zn(s) Zn2+(aq) + 2e oxidation

Cu2+(aq) + 2e Cu(s) reduction

• Each half-reaction is balanced for atoms

– Same # atoms of each type on each side

• Each half-reaction is balanced for charge

– Same sum of charges on each side

• Add both equations algebraically, canceling e’s

• NEVER have e's in net ionic equation

Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)

205

Redox in Aqueous Solution Ex. Mix solutions of K2Cr2O7 & FeSO4

– Dichromate ion, Cr2O72–, oxidizes Fe2+ to Fe3+

– Cr2O72– is reduced to form Cr3+

– Acidity of mixture decreases as H+ reacts with oxygen

to form water

Skeletal Eqn. Cr2O72– + Fe2+ Cr3+ + Fe3+

206

Ox. # Cr = +6 Fe = +2 Cr = +3 Fe = +3

Balance Ion-Electron Method –

Acidic Solution

1. Divide equation into 2 half-reactions

2. Balance atoms other than H & O

3. Balance O by adding H2O to side that needs O

4. Balance H by adding H+ to side that needs H

5. Balance net charge by adding e–

6. Make e– gain equal e– loss; then add half-reactions

7. Cancel anything that is the same on both sides

207

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Ion Electron Method

Ex. Balance in Acidic Solution

Cr2O72– + Fe2+ Cr3+ + Fe3+

1. Break into half-reactions

Cr2O72 Cr3+

Fe2+ Fe3+

2. Balance atoms other than H & O

Cr2O72 2Cr3+

– Put in 2 coefficient to balance Cr

Fe2+ Fe3+

– Fe already balanced 208

Ex. Ion-Electron Method in Acid

3. Balance O by adding H2O to the side that needs

O.

Cr2O72 2Cr3+

• Right side has 7 O atoms

• Left side has none

• Add 7 H2O to left side

Fe2+ Fe3+

• No O to balance

209

+ 7 H2O

Ex. Ion-Electron Method in Acid

4. Balance H by adding H+ to side that needs H

Cr2O72 2Cr3+ + 7H2O

• Left side has 14 H atoms

• Right side has none

• Add 14 H+ to right side

Fe2+ Fe3+

• No H to balance

210

14H+ +

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Ex. Ion-Electron Method in Acid

5. Balance net charge by adding electrons.

14H+ + Cr2O72 2Cr3+ + 7H2O

– 6 electrons must be added to reactant side

Fe2+ Fe3+

– 1 electron must be added to product side

• Now both half-reactions balanced for mass &

charge

211

6e +

+ e

Net Charge = 2(+3)+7(0) = 6

Net Charge = 14(+1) (–2) = 12

Ex. Ion-Electron Method in Acid 6. Make e– gain equal e– loss; then add half-

reactions

6e + 14H+ + Cr2O72– 2Cr3+ + 7H2O

Fe2+ Fe3+ + e

7. Cancel anything that's the same on both sides

6[ ]

212

6e + 6Fe2+ + 14H+ + Cr2O7

2 6Fe3+ + 2Cr3+

+ 7H2O + 6e

6Fe2+ + 14H+ + Cr2O7

2 6Fe3+ + 2Cr3+

+ 7H2O

Ex. Ion-Electron Method in Base Returning to our example of Cr2O7

2 & Fe2+

8. Add to both sides of equation the same number of

OH– as there are H+.

9. Combine H+ and OH– to form H2O.

10. Cancel any H2O that you can

213

6Fe2++ 14H+

+ Cr2O72

6Fe3+ + 2Cr3+

+ 7H2O + 14 OH– + 14 OH–

6Fe2+ + 14H2O + Cr2O7

2 6Fe3+ + 2Cr3+

+ 7H2O + 14OH

7

6Fe2+ + 7H2O + Cr2O7

2 6Fe3+ + 2Cr3+

+ 14OH

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Your Turn! Which of the following is a correctly balanced

reduction half-reaction?

A.Fe3+ + e– Fe°

B.2Fe + 6HNO3 2Fe(NO3)3 + 3H2

C.Mn2+ + 4H2O MnO4– + 8H+ + 5e–

D.2O2– O2 + 4e–

E.Mg2+ + 2e– Mg°

214

Ex: Ion-Electron Method Balance the following equation in acidic solution:

MnO4– + HSO3

– Mn2+ + SO42

1. Break it into half-reactions

MnO4– Mn2+

HSO3– SO4

2–

2. Balance atoms other than H & O

MnO4 Mn2+

– Balanced for Mn

HSO3 SO4

2

– Balanced for S

215

Ex: Ion-Electron Method

3. Add H2O to balance O

MnO4 Mn2+

HSO3 SO4

2

4. Add H+ to balance H

MnO4 Mn2+ + 4H2O

H2O + HSO3 SO4

2

216

+ 4H2O

H2O +

8H+ +

+ 3H+

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Ex: Ion-Electron Method

5. Balance net charge by adding e–.

8H+ + MnO4 Mn2+ + 4H2O

8(+1) + (–1) = +7 +2 + 0 = +2

Add 5 e– to reactant side

H2O + HSO3 SO4

2 + 3H+

0 + (–1) = –1 –2 + 3(+1) = +1

Add 2 e– to product side

5e– +

+ 2 e–

217

Ex: Ion-Electron Method

6. Make e– gain equal e– loss

5e– + 8H+ + MnO4 Mn2+ + 4H2O

H2O + HSO3 SO4

2 + 3H+ + 2e–

– Must multiply Mn half-reaction by 2

– Must multiply S half-reaction by 5

– Now have 10 e– on each side

218

2[ ]

5[ ]

10e– + 16H+ + 2MnO4

+ 5H2O + 5HSO3

2Mn2+ + 8H2O +

5SO42 + 15H+ + 10e

Ex: Ion-Electron Method

6. Then add the two half-reactions

10e– + 16H+ + 2MnO4 2Mn2+ + 8H2O

5H2O + 5HSO3 5SO4

2 + 15H+ + 10e–

7. Cancel anything that is the same on both sides.

Balanced in acid. 219

3 1

H+ + 2MnO4

+ 5HSO3

2Mn2+ + 3H2O + 5SO4

2

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220

Balance each equation in Acid & Base using the Ion

Electron Method.

MnO4– + C2O4

2– MnO2 + CO32–

Acid: 2MnO4– + 3C2O4

2– + 2H2O 2MnO2 + 4H+ + 6CO32–

ClO– + VO3– ClO3

– + V(OH)3

Acid: ClO– + 2H2O + 2VO3– + 2H+ ClO3

–+ 2V(OH)3

Your Turn!

Oxidation–Reduction in Biological

Systems In biological systems, oxidation may involve

the loss of H

the gain of O

In biological systems, reduction may involve

the gain of H

the loss of O

221

Oxidation–Reduction in Biological

Systems

• Plants represent one of the most basic

examples of biological oxidation and

reduction. The chemical conversion of carbon

dioxide and water into sugar (glucose) and

oxygen is a light-driven reduction process:

6CO2 + 6H2O → C6H12O6 + 6O2

• The light-driven reduction of CO2

222

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Aerobic respiration • Aerobic respiration may be represented by the general

equation

C6H12O6 + 6O2 → 6CO2 + 6H2O

• About 3000 kJ mol-1 of energy is released. Burning

glucose in air would release this amount of energy in one

go. However, it is not as simple as this in aerobic

respiration. Aerobic respiration is a series of enzyme-

controlled reactions that release the energy stored up in

carbohydrates and lipids during photosynthesis and make

it available to living organisms

223

Chp. 6

Chemical

Equilibrium

224

Reversible Reaction

Reversible reactions : Reaction in which entire amount of

the reactants is not converted into products is termed

as reversible reaction.

Characteristics of reversible reactions

(a) These reactions can be started from either side,

(b) These reactions are never complete,

(c) These reactions have a tendency to attain a state of

equilibrium

Example of reversible reactions

Salt hydrolysis, e.g., FeCl3 + 3H2O ⇄ Fe(OH)3 + 3HCl

225

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Irreversible Reaction

• Irreversible reactions : Reaction in which entire amount of the

reactants is converted into products is termed as irreversible reaction.

Characteristics of irreversible reactions

(a) These reactions proceed only in one direction (forward

direction),

(b) These reactions can proceed to completion,

(d) The arrow ( ) is placed between reactants and products,

Example of irreversible reactions

Neutralisation between strong acid and strong base e.g.,

NaOH + HCl → NaCl + H2O + 13.7 kcal

226

What is Equilibrium?

227

This is not Equilibrium?

228

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The Concept of Equilibrium

• As a system approaches

equilibrium, both the

forward and reverse

reactions are occurring.

• At equilibrium, the

forward and reverse

reactions are proceeding at

the same rate.

229

A System at Equilibrium

Once equilibrium is

achieved, the amount

of each reactant and

product remains

constant.

230

Expression of Equilibrium Constant

Forward reaction:

Backward reaction:

Reverse reaction:

Rate Law Rate law

231

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The Equilibrium Constant

At equilibrium

Rearranging gives:

232

The Equilibrium Constant

The ratio of the rate constants is a constant (as long as T is constant).

The expression becomes

233

The Equilibrium Constant

To generalize, the reaction:

Has the equilibrium expression:

234

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Homogeneous Equilibrium The equilibrium constant always has the same value (provided you

don't change the temperature), irrespective of the amounts of A, B, C

and D you started with. It is also unaffected by a change in pressure

or whether or not you are using a catalyst

235

Equilibrium-Constant Expressions

Write the equilibrium expression Kc for the following:

236

Heterogeneous Equilibrium The equilibrium established if steam is in contact with red hot

carbon. Here we have gases in contact with a solid. Everything is

exactly the same as before in the equilibrium constant expression,

except that you leave out the solid carbon.

Writing an expression for Kc for a heterogeneous equilibrium.

The important difference this time is that you don't include any

term for a solid in the equilibrium expression 237

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Equilibrium Constant Using Partial Pressures

• Consider the following reaction in gaseous phase:

wA + xB ⇄ yC + zD

The equilibrium constant of gases are expressed in terms of

partial pressures, instead of its concentrations and denoted

by Kp

• Kp =

Example:

2 SO2(g) + O2(g) ⇄ 2 SO3(g)

y z

C D

w x

A B

(P ) (P )

(P ) (P )

O2

2

SO2

2

SO3

PP

PKP

238

Relationship Between Kc and Kp

• Consider a general reaction:

• wA + xB ⇄ yC + zD

• Kp =

• PA = [A]RT; PB = [B]RT; PC = [C]RT; PD = [D]RT

• Kp =

• Kp = Kc(RT)∆n; (where ∆n = (y+z)-(w+x))

y z

C D

w x

A B

(P ) (P )

(P ) (P )

y z y z(y+z)-(w+x)

w x w x

([C]RT) ([D]RT) [C] [D](RT)

([A]RT) ([B]RT) [A] [B]

Dn = number of moles of gaseous products – number of moles of

gaseous reactants in chemical equation

239

Relationship between Kc and Kp

For other reactions:

1.2NO2(g) ⇄ N2O4(g); Kp = Kc(RT)-1

2.H2(g) + I2(g) ⇄ 2 HI(g); Kp = Kc

3.N2(g) + 3H2(g) ⇄ 2 NH3(g); Kp = Kc(RT)-2

240

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241

Interpreting KC

• Large K (K>>1)

– Means product rich mixture

– Reaction goes far toward

completion

Ex.

2SO2(g) + O2(g) 2SO3(g)

Kc = 7.0 1025 at 25 ° C

1

100.7

][O][SO

][SO 25

22

2

23

cK

242

Interpreting KC

• Small K (K<<1)

– Means reactant rich mixture

– Only very small amounts of

product formed

Ex.

H2(g) + Br2(g) 2HBr(g)

Kc = 1.4 10–21 at 25 °C

1

104.1

]][Br[H

[HBr] 21

22

2 cK

243

Interpreting KC

• K 1

– Means product and

reactant concentrations

close to equal

– Reaction goes only ~

halfway

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244

• Size of K gives measure of how reaction

proceeds

• K >> 1 [products] >> [reactants]

• K = 1 [products] = [reactants]

• K << 1 [products] << [reactants]

245

Ex: Consider the reaction of

2NO2(g) N2O4(g)

If Kp = 0.480 at 25°C, does the reaction favor product or reactant?

K is small (K < 1)

Reaction favors reactant

Since K is close to 1, significant amounts of

both reactant and product are present

Le Châtelier's Principle

• Le Châtelier's principle states that,

when a system at equilibrium is subjected to a change, the system will response by shifting in the direction that tends to minimize the change while reaching a new equilibrium position.

246

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247

Factors Affect Equilibrium According

Le Châtelier’s Principle

1. Concentration

2. Pressure and volume

3. Temperature

4. Catalysts

5. Adding inert gas to system at constant volume

Effect of Change in Concentration

2SO2(g) + O2(g) → 2SO3(g)

Kc = 2.4 x 10-3 at 700 oC

• Which direction will the reaction move if 0.125 moles of O2 is added to an equilibrium mixture ?

the equilibrum shift Towards the products

Adding or removing concentration change equilibrium concentration and not affect on KC or KP values (remain constant)

248

Shift Due to Pressure Change

• For the reaction: N2(g) + 3H2(g) ⇄ 2NH3(g),

• Increasing p, the equilibrium shift to lower

moles (shift to right) and KC or KP remain

constant

• decreasing p, the equilibrium shift to higher

moles (shift to left) and KC or KP remain

constant

249

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Shift Due to Pressure Change

For the reaction: CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g),

– Compression will increase pressure; equilibrium

shifts left, reducing the number of molecules and

pressure.

– Expansion causes pressure to decrease; equilibrium

shifts right, increasing the number of molecules

pressure.

– This type of reactions favor low pressure for product

formation.

250

Shift Due to Pressure Change

• For the reaction: CO(g) + H2O(g) ⇄ CO2(g) + H2(g),

Increasing or decreasing P ont affect equilibrium

position or Kc or Kp due to the number of

gaseous mole of reactant = number of mole of

products).

251

Shift Due to Volume Change

• For the reaction: N2(g) + 3H2(g) ⇄ 2NH3(g),

• Increasing V of reaction container, the equilibrium

shift to higher moles (shift to left) and KC or KP

remain constant

• decreasing P, the equilibrium shift to lower moles

(shift to right) and KC or KP remain constant

252

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85

Shift due to Temperature Change

• Consider the following reactions:

(1) N2(g) + 3H2(g) ⇄ 2NH3(g); DHo = -92 kJ

(2) CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g); DHo = 206 kJ

– For exothermic reactions, like reaction (1), raising the

temperature causes a shift to the left;

– For endothermic reactions, like reaction (2),

increasing the temperature causes a shift to the right;

– Exothermic reactions favor low temperature, while

endothermic reactions favor high temperature for

products formation. 253

254

Shift due to Temperature Change

• T shifts reaction in direction that produces

endothermic (heat absorbing) change

• T shifts reaction in direction that produces

exothermic (heat releasing) change

• Changes in T change value of mass action

expression at equilibrium, so K changed

– K depends on T

– T of exothermic reaction makes K smaller

• More heat (product) forces equilibrium to reactants

– T of endothermic reaction makes K larger

• More heat (reactant) forces equilibrium to products

255

Effect of Catalysts

• Catalyst lowers Ea for both forward and reverse reaction

• Change in Ea affects rates kr and kf equally

• Catalysts have no effect on equilibrium

• Equilibrium position and

KC or KP remain constant

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256

Effect of Adding Inert Gas

Inert gas

– One that does not react with components of reaction

Ex. Argon, Helium, Neon, usually N2

• Adding inert gas to reaction at fixed V (n and T),

P of all reactants and products

• Since it doesn’t react with anything

– No change in concentrations of reactants or products

– No net effect on reaction

Chp. 7

Chemistry: The Molecular Nature of Matter, 6E

Jespersen/Brady/Hyslop

Acids and Bases

257

Arrhenius Acids and Bases Acid produces H3O

+ in water

Base gives OH–

Acid-base neutralization

– Acid and base combine to produce water and a salt.

Ex. HCl(aq) + NaOH(aq) H2O + NaCl(aq)

H3O+(aq) + Cl–(aq) + Na+(aq) + OH–(aq)

2H2O + Cl–(aq) + Na+(aq)

• Many reactions resemble this without forming

H3O+ or OH– in solution

258

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87

Brønsted-Lowry Definition • Acid = proton donor

• Base = proton acceptor

• Allows for gas phase acid-base reactions

Ex. HCl + H2O H3O+ + Cl–

– HCl = acid

• Donates H+

– Water = base

• Accepts H+ 259

Conjugate Acid-Base Pair • Species that differ by H+

Ex. HCl + H2O H3O+ + Cl–

• HCl = acid

• Water = base

• H3O+

– Conjugate acid of H2O

• Cl–

– Conjugate base of HCl

260

Formic Acid is Bronsted Acid

• Formic acid (HCHO2) is a weak acid

• Must consider equilibrium

– HCHO2(aq) + H2O CHO2–(aq) + H3O

+(aq)

• Focus on forward reaction

261

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Formate Ion is Bronsted Base • Now consider reverse reaction

• Hydronium ion transfers H+ to CHO2

262

H3O+ + CHO2

HCHO2 + H2O

conjugate pair

conjugate pair

acid base acid base

Learning Check

263

• Ex: Identify the Conjugate Partner for Each

Cl–

NH4+

C2H3O2–

HCN

F–

Learning Check

• Ex: Write a reaction that shows that HCO3– is a

Brønsted acid when reacted with OH–

• HCO3–(aq) + OH–(aq)

• Write a reaction that shows that HCO3– is a

Brønsted base when reacted with H3O+(aq)

• HCO3–(aq) + H3O

+(aq)

264

H2CO3(aq) + H2O(ℓ)

H2O(ℓ) + CO32–(aq)

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Your Turn! Ex: In the following reaction, identify the acid/base

conjugate pair.

(CH3)2NH + H2SO4 → (CH3)2NH2+ + HSO4

-

A. (CH3)2NH / H2SO4 ; (CH3)2NH2+ / HSO4

-

B. (CH3)2NH / (CH3)2NH2+ ; H2SO4 / HSO4

-

C. H2SO4 / HSO4- ; (CH3)2NH2

+ / (CH3)2NH

D. H2SO4 / (CH3)2NH ; (CH3)2NH2+ / HSO4

-

265

Amphoteric Substances • Can act as either acid or base

– Also called amphiprotic

– Can be either molecules or ions

Ex. hydrogen carbonate ion:

– Acid:

HCO3–(aq) + OH–(aq) CO3

2–(aq) + H2O(ℓ)

– Base:

HCO3–(aq) + H3O

+(aq) H2CO3(aq) + H2O(ℓ)

2H2O(ℓ) + CO2(g)

266

Your Turn! Ex: Which of the following can act as an

amphoteric substance?

A. CH3COOH

B. HCl

C. NO2-

D. HPO42-

267

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Strengths of Acids and Bases

Strength of Acid

– Measure of its ability to transfer H+

– Strong acids • React completely with water Ex. HCl and HNO3

– Weak acids • Less than completely ionized Ex. CH3COOH and CHOOH

Strength of Base classified in similar fashion: – Strong bases

• React completely with water Ex. Oxide ion (O2) and OH

– Weak bases • Undergo incomplete reactions

Ex. NH3 and NRH2 (NH2CH3, methylamine)

268

Strong vs. Weak Acid

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 508 269

Reactions of Strong Acids and Bases

In water

• Strongest acid = hydronium ion, H3O+

– If more powerful H+ donor added to H2O

– Reacts with H2O to produce H3O+

Similarly,

• Strongest base is hydroxide ion (OH)

– More powerful H+ acceptors

– React with H2O to produce OH

270

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91

In General • Stronger acids and bases tend to react with each

other to produce their weaker conjugates

– Stronger Brønsted acid has weaker conjugate

base

– Weaker Brønsted acid has stronger conjugate

base

• Can be applied to binary acids (acids made from

hydrogen and one other element)

271

Autoionization of Water • Self Ionization of Water

• Pure water undergoes auto-ionization to produce

Hydronium and Hydroxide ions

2 H2O(l) H3O+(aq) + OH-(aq)

• The extent of this process is described by an auto-

ionization (ion-product) constant, Kw

• The concentrations of Hydronium and Hydroxide

ions in any aqueous solution must obey the auto-

ionization equilibrium

+ - -14w 3K = [H O ][OH ] = 1.0 10

272

273

• In aqueous solution,

• Product of [H+] and [OH] equals Kw

• [H+] and [OH] may not actually equal each other

Solution Classification

Autoionization of water

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Learning Check

EX: In a sample of blood at 25 °C, [H+] = 4.6 108

M. Find [OH] and determine if the solution is acidic,

basic or neutral.

•So 2.2 107 M > 4.6 108 M

•[OH] > [H3O+]

•Solution slightly basic 274

14w 101OHH ]][[K

7

8

14

10221064

1001

HOH

.

.

.

][][ wK

The pH Concept • With most weak acids and bases [x] small

• Must compare values and exponents

• Easier to compare if you take base 10 logarithm

of each side

• Define

• Get back to concentration by

• In general

– Can adapt to many values

275

]log[ HpHpH10H ][

XpX log

]log[ OHpOH

0014p ww .log KK

Redefine Acidic, Basic and Neutral

Solutions in terms of pH!

• As pH , [H+] ; pOH , and [OH]

• As pH , [H+] ; pOH , and [OH]

276

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93

Your Turn! Ex: Kw increases with increasing temperature. At

50 oC, Kw = 5.476 x 10-14. What is the pH of a

neutral solution at 50 oC ?

A. 7.00

B. 6.63

C. 7.37

D. 15.3

277

Learning Check Ex: What are [H+] and [OH] of pH = 3.00 solution?

• [H+] = 103.00 = 1.0 103 M

• [OH] = = 1.0 1011 M

Ex: What are [H+] and [OH] of pH = 4.00 solution?

• pH = 4.00 [H+] = 1.0 104 M

• [OH] = = 1.0 1010 M

• Or pH 4.00 solution has 10 times less H+ than pH 3.00

solution

4

14

1001

1001

.

.

278

3

14

1001

1001

.

.

Sample pH Calculations Ex: Calculate pH and pOH of blood in Ex. 1.

• We found [H+] = 4.6 108 M

[OH] = 2.2 x 107 M

pH = log(4.6 x 108) = 7.34

pOH = log(2.2 x 107) = 6.66

14.00 = pKw

Or

pOH = 14.00 pH = 14.00 – 7.34 = 6.66

279

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94

Ex: What is the pH of NaOH solution at 25 °C in

which the OH concentration is 0.0026 M?

[OH] = 0.0026 M

pOH = log(0.0026) = 2.59

pH = 14.00 – pOH

= 14.00 – 2.59

= 11.41

280

Your Turn! Ex: A sample of fresh pressed apple juice has a pH

of 3.76. Calculate [H+].

A.7.6 x 103 M

B.3.76 M

C.10.24 M

D.5.9 x 109 M

E.1.7 x 104 M

281

H10]H[ p

= 103.76 = 1.7 x 104 M

Learning Check EX: What is the [H3O

+] and pH of a solution that has [OH–] = 3.2 × 10–3 M?

• [H3O+][OH-] = 1 x 10-14

• [H3O+] = 1 x 10-14/3.2 x 10-3 =3.1 x 10-12 M

• pH = -log [H3O+] = -log(3.1 x 10-12)= 11.50

282

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95

Your Turn!

283

Ex: What is the [OH–] and pH of a solution that has

[H3O+] = 2.3 × 10–5 M?

Learning Check

284

Ex: What is the pOH and the [H3O+] of a solution

that has a pH of 2.33?

[H3O+]= 4.7×10–3

pOH = 11.67

Your Turn!

285

Ex: What is the pH and the [H3O+] of a solution that

has a pOH of 1.89?

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pH of Dilute Solutions of Strong

Bases

• 1 mole OH for every 1 mole B

• [OH] = [B] for strong bases

• 2 mole OH for every 1 mole B

• [OH] = 2*[B] for strong bases

286

Learning Check Ex: Calculate the pH of 0.011 M Ca(OH)2.

Ca(OH)2(s) + H2O Ca2+(aq) + 2 OH(aq)

• [OH] = 2*[Ca(OH)2] = 2*0.011M = 0.022M

• pOH = – log (0.022) = 1.66

• pH = 14.00 – pOH

• = 14.00 – 1.66 = 12.34

• What is this in the [H+] of the solution?

• [H+] = 1012.34 = 4.6 x 1013 M

287

Learning Check

Ex: What is the pH of 0.1M HCl?

• Assume 100% dissociation

HCl(aq) + H2O(ℓ) H+(aq) + OH(aq)

I 0.1 N/A 0 0

C -0.1 -0.1 0.1 0.1

End 0 N/A 0.1 0.1

288

pH = –log(0.1) = 1

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Learning Check

Ex: What is the pH of 0.5M Ca(OH)2?

• Assume 100% dissociation

Ca(OH)2 (aq) Ca2+ (aq) + 2 OH– (aq)

I 0.5 0 0

C -0.5 +0.5 +0.52

E 0 0.5 1.0

289

pOH = -log(1.0) = 0

pH = 14.00 – pOH = 14.00 – 0 = 14

Chp. 8

Organic Chemistry

Chemistry: The Molecular Nature of Matter, 6E

Jespersen/Brady/Hyslop

290

What Organic Chemistry

• Chemistry of the compounds present

in living organisms.

• They all contain carbon.

• Organic Chemistry is the Chemistry

of Carbon.

291

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98

Living

things

Carbohydrates /

Proteins / Fats /

Vitamins / Antibiotics

Natural Sources of Organic Compounds

A variety of organic products

obtained from living things

292

The Unique Nature of Carbon CATENATION is the binding of an element to itself through covalent bonds to

form chain or ring molecules.

Carbon forms chains and rings, with single, double and triple covalent bonds, because it is

able to FORM STRONG COVALENT BONDS WITH OTHER CARBON ATOMS

Carbon forms a vast number of carbon compounds because of the strength of

the C-C covalent bond. Other Group IV elements can do it but their chemistry

is limited due to the weaker bond strength.

BOND ATOMIC RADIUS BOND ENTHALPY

C-C 0.077 nm +348 kJmol-1

Si-Si 0.117 nm +176 kJmol-1

The larger the atoms, the weaker the bond. Shielding due to filled inner orbitals

and greater distance from the nucleus means that the shared electron pair is held

less strongly.

293

CHAINS AND RINGS

CARBON ATOMS CAN BE ARRANGED IN

STRAIGHT CHAINS

BRANCHED CHAINS

and RINGS

The Unique Nature of Carbon

You can also get a combination of rings and chains

MULTIPLE BONDING AND SUBSTITUENTS

CARBON-CARBON COVALENT BONDS CAN BE SINGLE, DOUBLE OR

TRIPLE

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DIFFERENT ATOMS / GROUPS OF ATOMS CAN BE PLACED ON THE

CARBONS

The basic atom is HYDROGEN but groups containing OXYGEN, NITROGEN, HALOGENS

and SULPHUR are very common.

CARBON SKELETON FUNCTIONAL CARBON SKELETON FUNCTIONAL

GROUP GROUP

The chemistry of an organic compound is determined by its FUNCTIONAL GROUP

The Unique Nature of Carbon

295

Classification of Hydrocarbons

Section 14.2

296

Aliphatic Hydrocarbons

• Aliphatic hydrocarbons are hydrocarbons having

no benzene rings.

• Aliphatic hydrocarbons can be divided into four

major divisions:

– Alkanes

– Cycloalkanes

– Alkenes

– Alkynes

Section 14.3

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100

Alkanes • Alkanes are hydrocarbons that contain only single

bonds.

• Alkanes are said to be saturated hydrocarbons

– Because their hydrogen content is at a maximum.

• Alkane general formula CnH2n + 2

• The names of alkanes all end in “-ane.”

• Methane butane are gases

• Pentane C17H36 are liquids

• C18H38 and higher are solids

Section 14.3

298

Straight Chained Alkanes

299

Alkyl Groups Alkane type groups added to parent chain are

known as alkyl groups. Consist of alkane, minus

one H atom. Name always ends in -yl

Example

CH4 : now remove one H which yields –CH3

• Naming of –CH3

• Start with parent name, which is methane

• Drop –ane and add –yl

• So methane becomes methyl group

300

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Hydrocarbon Nomenclature Rules for naming alkanes

Established by IUPAC

1. Name ends in “-ane”

2. Complete name uses that of parent

compound with constituent groups added.

3. Parent is longest continuous carbon chain.

4. Name of longest chain based on the number

of carbons.

5. Carbon atoms are numbered starting at the

end that gives the lowest number for the

first branch.

301

6. Aryl groups names are prefixed to parent

name.

7. Multiple aryl groups on a parent are

numbered and named alphabetically.

8. When there are multiple identical groups

add di, tri, tetra to the aryl name.

9. If multiple, identical aryl groups are

attached to the same carbon repeat the

carbon number.

302

Hydrocarbon Nomenclature

Example • The longest continuous chain of C atoms is

five

• Therefore this compound is a pentane

derivative with an attached methyl group

– Start numbering from end nearest the

substituent

– The methyl group is in the #2 position

• The compound’s name is 2-methylpentane.

303

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102

Example What is the name of the compound shown?

1. The longest carbon chain (parent) is

four. Parent name is butane.

2. Start numbering from the left to get the

smallest number for the attached group.

304

Example

3. The attached alkyl group is a methyl group.

– Thus, the correct name is:

• 2-methylbutane

– What is the name of the following

compound?

305

Example • The parent chain contains five carbons.

• Thus, the parent name is pentane.

• Number from the left to obtain the smallest

number for the first alkyl group.

• The alkyl groups are at the 2 and 3 positions.

• The 2 and 3 positions each contain a methyl

group. 306

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Example

• Thus, the correct name is:

• 2,3-dimethylpentane

• Let’s consider an alkane with two substituents

on the same carbon.

307

Example

• The parent chain is six carbons long.

• The lowest correct numbering of positions is

shown below.

• There are methyl and ethyl groups attached to

carbon 3.

308

Example

• The correct name is:

3-ethyl-3-methylhexane

309

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104

Your Turn!

What is the correct name for the molecule shown

below?

A. 3-butylpentane

B. 1,1-diethylpentane

C. 3-ethylheptane

D. 5-ethylheptane

310

Your Turn!

What is the name of the compound shown

below?

A. 3-methyl-3-methyloctane

B. 3,3-dimethyloctane

C. 2-ethyl-2-methylheptane

D. 6,6-dimethyloctane 311

Chemical Properties of Alkanes

• Alkanes are relatively unreactive

• Not reactive in conc. NaOH or H2SO4 at room

temperature.

• React with hot HNO3

• Will react with Cl2 and Br2 to form

halogenated hydrocarbons.

• Examples are CH3Cl, CH2Cl2 and CHCl3

• Can crack molecules like ethane under

controlled conditions to form CH2CH2

• Will react with O2 to form CO2, CO, and H2O 312

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105

Alkenes and Alkynes

• Alkenes contain one or more double bonds

– General form: CnH2n

• Alkynes contain one or more triple bonds

– General form: CnH2n-2

• Non-polar compounds are not water soluble

• Examples:

313

Alkenes and Alkynes

• Nomenclature

– The parent chain must contain the multiple

bond even if it is a smaller chain length than

one without a multiple bond

– Number from end that gives the lowest

number to the first carbon of the multiple

bond

– The number is given as -x- and placed just

before the –ene or –yne of the parent name.

For example, but-2-ene. The double bond

starts on carbon 2 of the chain. 314

Alkene Examples

• Start numbering from the left to get the lowest

number for the first carbon with the double

bond

• The parent is heptene and the correct naming

including the double bond location would be

hep-2-ene

315

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Example

• The parent chain is four carbons

• 2,3-dimethylbut-2-ene

• We would not name this 2-methyl-3-methylbut-2-

ene

316

Naming Polyenes

• How do we name compounds such as the

following?

• This compound contains two double bonds and

is known as a diene

• We want the lowest number for the first carbon

of each of the double bonds

• Start numbering from the right

317

Naming Polyenes

• The correct name would be hex-1,3-diene

• Three double bonds would be a triene

hex-1,3,5-triene

318

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Cyclic Alkenes

• Number ring to obtain lowest number for first

carbon of the double bond

319

Cyclic Alkenes

• Correct name is 1,6-dimethylcyclohex-1-ene

• Other ring examples

320

Your Turn! What is the correct name for the compound

shown

below?

A. 1,4-dimethylcyclopent-1-ene

B. 1,3-dimethylcyclopent-1-ene

C. 1-methyl-4-methylcyclopent-1-ene

D. 1,3-dimethylcyclo-1-pentene

321

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108

Your Turn! What is the correct structure for 3,3-dimethylpro-1-

ene?

A.

B.

C.

D.

322

Geometric Isomers

• Groups cannot freely rotate about a double

bond

• Therefore, it is possible to have geometric

isomers

• Examples:

323

Reactions of Alkene

• Alkenes readily add across the double bond

• Examples of an addition reaction:

– CH2CH2 + H2 CH3CH3

hydrogenation

– CH2CH2 + HCl → CH3CH2Cl

– CH2CH2 + H2O → CH3CH2OH

– CH CH + Cl → CH ClCH Cl

Pt

324

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Aromatic Hydrocarbons

• The most common aromatic compound is

benzene and its derivatives

• Representation of bonding

• Delocalized π bonds create unique stability,

called resonance stabilization. The circle in the

ring represents delocalization.

325

Aromatic Hydrocarbons

• Aromatic hydrocarbons contain one or

more benzene ring.

• Benzene (C6H6) is the most important

aromatic hydrocarbon.

• It is a clear, colorless liquid with a distinct

odor, and is a carcinogen (cancer-causing

agent.)

• Traditional Lewis Structure

Section 14.2

326

Other Aromatic Hydrocarbons

• Toluene is used in modeling glue. Naphthalene is use

in mothballs, and Phenanthrene are used in the

synthesis of dyes, explosives, and drugs.

Section 14.2

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When Other Atoms are Substituted for

the H’s in the Benzene Ring A vast array of other compounds can be produced

Section 14.2

328

Naming for Benzene Derivatives

• Draw the structural formula for 1,3-dibromobenzene.

• First, Draw a benzene ring.

Section 14.2

329

Drawing Structures for Benzene

Derivatives

• Second, attach a bromine atom (“bromo”) to

the carbon atom at the ring position you

choose to be number 1.

Section 14.2

330

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Drawing Structures for Benzene

Derivatives

• Third, attach a second (“di”) bromine atom to ring

position 3 (you may number either clockwise or

counterclockwise from carbon 1) and you have the

answer.

• 1,3-dibromobenzene

Section 14.2

331

Drawing Structures for Benzene

Derivatives

• Draw the structural formula for 1-chloro-2-fluorobenzene.

1. Draw a benzene ring.

3. Attach a fluorine atom to

ring position 2 and you have

the answer.

• 1-chloro-2-fluorobenzene

Section 14.2

Cl

F

2. Attach a chlorine atom (“chloro”) to the carbon atom at the ring position you choose to

be number 1.

332

Reactions of Aromatics

• Substitution reactions maintain benzene’s

resonance structure.

• Addition reactions, like those of alkenes,

destroy resonance structure

• Substitution reaction:

333

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Addition Reaction

• Notice that you have reduced the double

bonding in the ring and altered the resonance

stabilization of the ring

334

Learning Check What product would form if benzene reacted with

nitric acid using an appropriate catalyst?

• Sulfuric acid is the catalyst

• A substitution reaction occurs

335

Your Turn! Which product is most likely formed when

sulfuric acid reacts with benzene?

A.

B.

C.

D.

336

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Organic Compounds Containing Oxygen

Important functional groups:

337

Alcohols and Ethers • Common alcohols: names end in -ol

– CH3OH methanol

– CH3CH2OH ethanol

– CH3CH2CH2OH propan-1-ol

• If the –OH group was attached to the

central carbon then the alcohol would be

propan-2-ol

– Alcohols form hydrogen bonds, causing

their boiling points to be higher than

predicted. 338

Alcohols and Ethers

• Primary alcohols:

• Secondary alcohols:

• Tertiary alcohols: 339

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Alcohols and Ethers

• Common ethers:

– CH3OCH3 dimethyl ether

– CH3CH2OCH2CH3 diethyl ether

– CH3OCH2CH3 methyl ethyl ether

– No hydrogen bonding occurs, thus, boiling

points are lower than corresponding

alcohols

– Like alkanes, ethers are not very reactive 340

Reactions of Alcohols

• Alcohols can undergo oxidation to form a

variety of products. Oxidation removes an H

atom from the alcoholic carbon as well as the

H on the –OH group.

• Primary alcohols can be oxidized to aldehydes

and carboxylic acids

341

Reactions of Alcohols • Aldehydes are more readily oxidized than

alcohols

• Secondary alcohols can be oxidized to ketones

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Reactions of Alcohols

• Ketones are not further oxidized

• Tertiary alcohols have no H atom on the

alcoholic carbon and thus, do not undergo

oxidation

• Alcohols undergo elimination reactions in the

presence of concentrated H2SO4 forming water

and alkenes

• -OH group readily accepts a proton from

sulfuric acid

343

Elimination Reaction

• Dehydration of an alcohol

• During the reaction a very unstable

carbocation is formed. This ion eliminates a

proton to form the alkene.

344

Substitution Reactions of Alcohols

• Using heat and concentrated HBr, HI, or HCl,

a halogen will replace the –OH group

• A proton adds to the –OH forming –OH2+

• Water leaves and the halogen ion attaches to

the carbon site where the –OH was attached

345

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Aldehydes and Ketones

• Naming Aldehydes

– Parent name ends in –al, replacing –e in the

alkane name

– The aldehyde group is always at the end of a

chain and numbering starts with that end of the

chain 346

Naming Aldehydes

• Number from the Aldehyde end

• Do not use -1- for Aldehyde:

• 3-methylpropan-1-al, or 3-methyl-1-propanal

would be wrong 347

Learning Check What is the name of the following aldehyde?

4-ethylhexanal

348

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Naming Ketones

• Parent name ends in –one

• Parent chain must contain carbonyl group

• Numbering so carbonyl carbon has lowest

possible number

4-ethylheptan-3-one

NOT: 4-ethylheptan-5-one

349

Your Turn! What is the correct name for the aldehyde shown

below?

A. 2,4-dimethylpentanal

B. 2,4-dimethyl-1-pentanal

C. 2-methyl-4-methylpropanal

D. 2,4-dimethyl-5-pentanal 350

Your Turn! - Solution

• Aldehydes are numbered from the aldehyde

end of the molecule

• There are two identical groups, (methyl) so we

use –di in the naming

351

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Your Turn!

What is the correct name for the ketone shown

below?

A. 4-methyl-3-ethylhexan-2-one

B. 4-ethyl-3-methylhexan-5-one

C. 3-ethyl-4-methylhexan-2-one

D. 3,4-diethylpentan-2-one 352

Your Turn! - Solution • Number to give lowest number to keto group

so you start from the right

• Alkyl groups are ordered alphabetically so

ethyl comes before methyl 353

Reactions of Aldehydes and Ketones

• Aldehydes and ketones add hydrogen across

the C=O bond

• Process is hydrogenation or reduction

354

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Carboxylic Acids and Esters

355

Naming Carboxlic Acids

• Name ends in –oic, replacing –e in the parent

name

• Numbering begins with carboxyl group

• -COOH or -CO2H is the condensed form

• CH3COOH is ethanoic acid (acetic acid)

356

Naming Carboxylic Acids

• Benzoic acid

• Propanoic acid

357

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Naming Ethers

• Name begins with alkyl group attached to the –O

• Name of parent acid is separate from the alkyl

group name and –oic is replaced with –ate

• Ethyl propanate

358

Learning Check

• What is the name of the following ester?

• Alkyl group is propyl

• Number, starting with

the ester carbon

• Propyl 4-methylpentanate

359

Your Turn!

What is the correct name for the product when 3-

methylbutan-1-ol is completely oxidized?

A. 3-methylbutanoic acid

B. 2-methyl-1-butanoic acid

C. 2-methlybutan-1-oic acid

D. 3-methylbutan-1-oic acid

360

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Reactions of Carboxylic Acids

• The –COOH is weakly acidic and therefore

reacts with base

• RCOOH + OH- → RCOO- + H2O

361

Formation of Esters • Esters give fruits their characteristic odor

ethyl pentanoate

362

Saponification • Strong base reacts with an ester to form alcohol

and the ester’s anion forms pentanoate ion

363

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Your Turn!

Name the ester formed when methanol reacts

with

hexanoic acid.

A. 1-methyl hexanoate

B. methylhexanoate

C. methyl hexanoate

D. methyl hexan-1-oate

364

Organic Derivatives of Ammonia

• Amines are derived from ammonia with one or

more H atoms replaced with organic groups

• Like ammonia, amines are weakly basic

• Amines react with acids

365

Acid Property of Protonated Amines

• Ethylmethylammonium ion is the conjugate

acid of ethylmethylamine

pKa = 10.76 pKb= 3.24

366

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Amides

• General form

• Where (H)R indicates either an H atom or an R

group attached

• Naming

– The name of the parent acid is amended

dropping the –oic ending and replacing it

with -amide 367

Amides

• Propanamide

• 4-ethylhexamide

• These are examples

of simple amides

368

Synthesis of Simple Amides • An organic acid reacts with aqueous NH3 to

form an amide

2-methylpropanoic acid yields 2-methylpropanamide

369

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Amide Reactions • Amides can be hydrolyzed back to their acid

form producing ammonia in the process

370

Amide Reactions • Urea, an amide, ultimately hydrolyzes to NH3,

CO2 and water

• Carbonic acid is formed, which then

decomposes to carbon dioxide and water

• The overall reaction is:

371

Basicity of Amides • Amides are not basic like amines

• The lone pair on the N atom is delocalized and

thus not readily available for donation to a

proton

• Amides are neutral in an acid-base sense

372

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Your Turn!

What is the correct name for the molecule shown

below?

A. 4,5-dimethylhexanamide

B. 2,3-dimethyl-6-hexanamide

C. 4-methyl-5-methylhexanamide

D. 4-isopropyl-4-methylpropanamide

373