Ch. 4: Molecules and

Compounds

Dr. Namphol Sinkaset

Chem 200: General Chemistry I

I. Chapter Outline

I. Introduction

II. Representing Compounds

III. Lewis Dot Symbols

IV. Ionic Bonds and Nomenclature

V. Covalent Bonds and Nomenclature

VI. Problems Involving Chemical Formulas

I. Introduction

• When elements form compounds, the

original properties of the elements are

lost.

I. To Lower Potential Energy

• A chemical bond is the force that holds

atoms together in a compound.

• But why would atoms want to join with

other atoms?

• It all comes back to positive-negative

attractions between particles in the

atom which lead to lower PE!

I. Two Main Ways to Lower PE

I. Metal + Nonmetal = Ionic

I. Nonmetal + Nonmetal =

Covalent

• Instead of transferring e-’s, covalent

bonding occurs via sharing of e-’s.

• Attraction to two nuclei lowers PE.

II. Chemical Formulas

• There are three types of

formulas.

molecular: gives the actual

number of atoms of each

element in a molecule of a

compound (e.g. H2O2)

empirical: gives the relative

number of atoms of each

element in a compound (e.g.

HO)

structural: uses lines to

represent covalent bonds and

shows interconnectivity

II. Chemical Models

• Formulas lead to models which give an idea

of the 3-D shape of a molecule.

II. From Names to Models

III. Lewis Dot Symbols

• Valence e-’s are the most important e-’s

in bonding.

• Lewis dot symbols are a way to depict

the valence e-’s of atoms.

• Lewis dot symbols have two parts:

1) element symbol: represents nucleus and

core e-

2) dots around symbol: represent valence e-’s

III. Lewis Dot Symbols

• The number of valence e- is given by the

element’s group number!!

III. The Octet Rule

• Noble gases are known for their lack of

reactivity – what do their e- configs have in

common?

• Lewis generalized bonding behavior by

observing that when atoms bond, they lose,

gain, or share e- to obtain 8 valence e-.

• Known as the octet rule (duet for H and He).

IV. Ionic Bonding

• In ionic bonding, metal transfers e- to

the nonmetal.

• Transferring e- achieves octet.

• Resulting ions attracted by +/- charge.

IV. Depicting Ionic Bonding

1. Draw Lewis dot structures for atoms involved.

2. Use harpoons to indicate e- transfer.

3. Fill octet of nonmetal, drawing additional Lewis dot structures as needed.

4. Use bracket notation on ions formed. Charges should cancel.

5. Write formula of ionic compound.

IV. Practice Problem 4.1

• Show the formation of the bonding that

occurs between magnesium and

chlorine using Lewis dot symbols.

IV. Energetics of Ionic Bonds

• Although transfer of e- achieves an

octet, ionic compounds are really stable

because of +/- attractions.

• The e- transfer process needs energy.

1st IE of Na = 496 kJ/mole

1st EA of Cl = -349 kJ/mole

• However, 411 kJ/mole of heat evolves

upon NaCl formation.

IV. Lattice Energy

IV. Electrical Conductivity

IV. Ionic Compounds Melt at

High Temperatures

• Why are such high

temperatures needed?

IV. Ionic Nomenclature

• Ionic compounds are named

systematically, broken into two groups.

IV. Type I Compounds

• Type I compounds are ionics that have a

metal from Groups 1 or 2 and a nonmetal

from Groups 14-17.

• Examples:

NaCl = sodium chloride

MgBr2 = magnesium bromide

K2S = potassium sulfide

IV. Type I Compounds

• To get a formula from a name,

remember that a compound must be

neutral.

• Ion charges can be found by locating

the element on the periodic table.

• “The charge on one becomes the

subscript of the other.”

IV. Sample Problem 4.2

• What are the formulas for sodium

nitride, calcium chloride, potassium

sulfide, and magnesium oxide?

IV. Transition Metals

• Transition metals are found in the

“Valley,” Groups 3-12, of the periodic

table.

• Transition metal cations often can carry

different charges, e.g. Fe2+ and Fe3+.

• Thus, a name like “iron chloride” is

ambiguous.

IV. Type II Compounds

• Type II compounds are ionics that have a

transition metal (Groups 3-12) and a

nonmetal (Groups 14-17).

• Examples:

FeCl2 = iron(II) chloride

FeCl3 = iron(III) chloride

IV. Sample Problem 4.3

• Give the correct name or formula for

the compounds below.

a) MnO2

b) copper(II) chloride

c) AuCl3

d) molybdenum(VI) fluoride

e) Hg2Cl2

IV. Type II Compounds

• An archaic naming system uses

common names for transition metal

cations of different charge.

Higher charge given –ic suffix

Lower charge given –ous suffix

• FeCl3 = ferric chloride

• FeCl2 = ferrous chloride

IV. Additional Complications

• To make naming ionic compounds

harder, sometimes polyatomic ions are

involved.

• polyatomic ion: an ion composed of two

or more atoms

IV. Common Polyatomic Ions

IV. Oxyanion Families

• Oxyanions are anions that contain

oxygen and another element.

• There are families of oxyanions, and

they have a systematic naming system.

• Have either two- or four-member

families.

e.g. NO2- and NO3

-

e.g. ClO-, ClO2-, ClO3

-, and ClO4-

IV. Two-Member Families

• For a two-member family, oxoanion with

fewer O atoms is given the “–ite” suffix

while the one with more O atoms is

given the “–ate” suffix.

e.g. NO2- = nitrite

and NO3- = nitrate

IV. Four-Member Families

• For the four-member families, the

prefixes “hypo-” and “per-” are used to

indicate fewer or more oxygen atoms.

• e.g. the chlorine oxoanions

ClO- = hypochlorite

ClO2- = chlorite

ClO3- = chlorate

ClO4- = perchlorate

IV. Hydrated Ionic Compounds

• Ionics with trapped

waters are are called

hydrates.

• Greek prefixes are

used to indicate #’s of

trapped waters.

• e.g. cobalt(II) chloride

hexahydrate.

IV. Practice Problem 4.4

• Give names or formulas for the

following compounds.

a) Na2CO3

b) magnesium hydroxide

c) CuSO4·5H2O

d) CoPO4

e) nickel(II) sulfate

f) NaClO2

V. Covalent Bonding

• In covalent bonding, nonmetals share

some (or all) of their valence electrons

to achieve an octet.

• Nonmetals can share two, four, or six

electrons in what are known as single,

double, or triple bonds.

V. Single Bonds

• In H2O, two unpaired e-’s on each H

combine with the two unpaired e- sites

on O.

V. Bonding vs. Lone Pairs

• Bonding pairs are shared; lone pairs are

not.

• Bonding pairs are often represented

with a line between the two atoms.

V. Lewis Model & Diatomics

V. Double Bonds

• When more e-’s need to be shared to

reach an octet, a double bond is

possible.

V. Triple Bonds

• When even more e-’s need to be

shared, a triple bond is possible.

e.g. molecular nitrogen, :N≡N:

• As the bond order increases, the bond

gets stronger and shorter.

V. Covalent Nomenclature

• For covalent compounds, many different

compounds can exist from the same

two elements.

e.g. NO, NO2, N2O, N2O3, N2O4, N2O5!

• Therefore, we need a systematic

naming method.

V. Type III Compounds

• Type III compounds are covalent (nonmetal bonded to nonmetal).

• Naming rules:1) Element w/ lower group # is named 1st using the

normal element name EXCEPT when halogens are bonded to oxygen.

2) If elements are in the same group, lower element named first.

3) Second element is named using its root and the “-ide” suffix.

4) #’s of atoms indicated with Greek prefixes EXCEPT when there is only one atom of the first element.

V. Greek Prefixes

V. Type III Compounds

• Some examples:

ClO2 = chlorine dioxide

N2O5 = dinitrogen pentoxide

S2Cl2 = disulfur dichloride

SeF6 = selenium hexafluoride

V. Practice Problem 4.5

• e.g. Give the correct formula or name of the compounds below.

a) CoCl3b) dichlorine heptaoxide

c) SrO

d) magnesium hydroxide

e) carbon tetrachloride

f) MgSO4·7H2O

g) sodium hydride

h) V2O5

i) Ru(ClO4)3

j) NI3k) titanium(IV) oxide

l) N2F2

VI. Composition of Compounds

• The ratio of elements in a compound is

given by its formula.

• We can calculate composition of

specific elements in different ways.

Mass percent

Inherent conversion factors

• We can also do the opposite: given

composition, determine formulas.

VI. Masses of Compounds

• Atomic masses are readily accessible via

the periodic table, e.g. H = 1.008 amu.

• Molecular masses or molecular weights

are calculated by adding up the masses

of each atom in the compound.

• Thus, molecular mass = sum of atomic

masses.

VI. Molecular Mass of Water

• The formula for water is H2O, so it is

comprised of 2 H atoms and 1 O atom.

VI. Mole Calculations

• Of course, we can use molar masses

and Avogadro’s number to calculate the

number of particles in a sample.

• e.g. How many water molecules are in a

sample of water that weighs 2100 g?

VI. Mass Percent

• As we know, elements account for a set

amounts by mass in a compound.

VI. Sample Problem 4.6

• Calculate the mass percent of nitrogen

in ammonium nitrate.

VI. Inherent Conversion Factors

• Formulas have conversion factors within

them to allow calculation of their

composition.

VI. Sample Problem 4.7

• How many grams of carbon are present

in 7.25 mL of butane (C4H10) if the

density of butane is 0.601 g/mL?

VI. Finding Formulas from

Mass Data

• Given elemental mass data of a

compound, it’s possible to find the

formula of the compound.

• Elemental analysis is a common test

performed on newly synthesized

compounds.

VI. Subscripts are Mole Ratios

• When finding formulas from mass data,

always go to moles of each element.

• Write a temporary formula using these

mole numbers.

• Divide by the smallest mole number to

get to empirical formula.

VI. Adjusting Subscripts to

Whole Numbers

VI. Empirical to Molecular

• The molecular molar mass is always a

whole number multiple (n) of the

empirical molar mass.

• Use this n to convert empirical to

molecular formula.

massmolar empirical

massmolar molecular n

VI. Sample Problem 4.8

• e.g. The carcinogen benzo[a]pyrene

(MW = 252.30 g/mole) is found to be

95.21% C and 4.79% H by mass. What

are its empirical and molecular

formulas?

VI. Combustion Analysis

• Empirical formulas of compounds containing

C, H, and one other element can be found via

combustion analysis.

• When sample is burned in O2, all the C

becomes CO2 and all the H becomes H2O.

VI. Sample Problem 4.9

• A 12.01 g sample of tartaric acid

(comprised of only C, H, and O) was

analyzed via combustion. If 14.08 g

CO2 and 4.32 g H2O are produced, find

the empirical formula of tartaric acid.