Ch. 5: 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 Bonding1. 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. Some Practice

• e.g. 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. Type I Compounds

• e.g. 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. Type II Compounds

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

a) MnO2

b) copper(II) chloride

c) AuCl3d) 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

- = nitriteand 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. Naming Practice

• e.g. 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. Naming Practice• e.g. Give the correct formula or name of the

compounds below.a) CoCl3b) dichlorine heptaoxidec) SrOd) magnesium hydroxidee) carbon tetrachloride

f) MgSO4·7H2Og) sodium hydride

h) V2O5

i) Ru(ClO4)3

j) NI3

k) 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.

amu 18.02 OH

amu 16.00 amu 16.00 1 :O

amu 2.016 amu 1.0082 :H

2

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.

%100

compound massmolar

X massmolar formulain X moles X % mass

VI. Sample Problem

• e.g. 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

• 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

• 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

• 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.