chapter 5 nomenclature. chemical bonding chemical bond a bond results from the attraction of nuclei...

Chapter 5

Nomenclature

ChemicalBONDING

Chemical Bond• A bond results from the attraction of nuclei

for electrons– All atoms trying to achieve a stable octet

• IN OTHER WORDS– the p+ in one nucleus are attracted to the e- of

another atom• Electronegativity

• Molecule: 2 or more atoms joined by a chemical bond

• Compound: a molecule composed of atoms of 2 or more different elements bonded together in a fixed ratio

Diatomic Molecule

• Diatomic Molecule: a molecule containing 2 atoms

• The Diatomic molecules are:

• Hydrogen (H2) Nitrogen (N2)Oxygen (O2) Fluorine (F2)Chlorine (Cl2) Iodine (I2)Bromine (Br2)

• Chemical formula: represents the relative numbers of atoms of each kind in a chemical compound by using atomic symbols and numeric subscripts

• Bond energy: the energy required to break a chemical bond and form neutral atoms

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Types of Chemical Bonds: (4)

1. Ionic bonds

2. Covalent bonds

3. Metallic bonds

4. Hydrogen bonds

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Bond Formation

• exothermic process

Energy released

ENERGY

Reactants

Products

Breaking Bonds

• Endothermic reaction– energy must be put into the bond in order

to break itENERGY Reactants

Products

Energy Absorbed

Bond Strength• Strong, STABLE bonds require lots of

energy to be formed or broken

• weak bonds require little E

Two Major Types of Bonding

• Ionic Bonding

– forms ionic compounds

– transfer of e-

• Covalent Bonding– forms molecules

– sharing e-

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One minor type of bonding• Metallic bonding

– Occurs between like atoms of a metal in the free state

– Valence e- are mobile (move freely among all metal atoms)

– Positive ions in a sea of electrons

• Metallic characteristics– High mp temps, ductile, malleable, shiny– Hard substances– Good conductors of heat and electricity as (s) and (l)

It’s the mobile electrons that enable me-tals to

conduct electricity!!!!!!

IONic Bonding

• electrons are transferred between valence shells of atoms

• ionic compounds are made of ions

• ionic compounds are called Salts or Crystals

NOT MOLECULES

IONic bonding

• Always formed between metals and non-metals

[METALS ]+ [NON-METALS ]-

Lost e-Gained e-

• hard solid @ 22oC

• high mp temperatures

• nonconductors of electricity in solid phase

• good conductors in liquid phase or dissolved in water (aq)

SALTSCrystals

Properties of Ionic Compounds

Covalent Bonding

• Pairs of e- are shared between non-metal atoms

• electronegativity difference < 2.0

• forms polyatomic ions

molecules

Properties of Molecular Substances

• Low m.p. temp and b.p. temps

• relatively soft solids as compared to ionic compounds

• nonconductors of electricity in any phase

Covalent bonding

Covalent, Ionic, metallic bonding?

• NO2

• sodiumhydride

• Hg• H2S• sulfate

• NH4+

• Aluminum phosphate

• KH• KCl • HF

• CO• Co

Can You Tell What type of bond is formed

Drawing ionic compounds using Lewis Dot Structures• Symbol represents the KERNEL of the

atom (nucleus and inner e-)

• dots represent valence e-

NaCl• This is the finished Lewis Dot

Structure

[Na]+ [ Cl ]-

How did we get here?

• Step 1 after checking that it is IONIC

– Determine which atom will be the +ion– Determine which atom will be the - ion

• Step 2– Write the symbol for the + ion first.

• NO DOTS

– Draw the e- dot diagram for the – ion• COMPLETE outer shell

• Step 3– Enclose both in brackets and show each charge

Draw the Lewis Diagrams• LiF

• MgO

• CaCl2

• K2S

Drawing molecules using Lewis Dot Structures

• Symbol represents the KERNEL of the atom (nucleus and inner e-)

• dots represent valence e-

Always remember atoms are trying to complete their outer shell!

The number of electrons the atoms needs is the total number of bonds they can make.

Ex. … H? O? F? N? Cl? C?

one two one three one four

Methane CH4

• This is the finished Lewis dot structure

How did we get here?

• Step 1– count total valence e- involved

• Step 2– connect the central atom (usually the first in

the formula) to the others with single bonds • Step 3

– complete valence shells of outer atoms• Step 4

– add any extra e- to central atom

IF the central atom has 8 valence e- surrounding it . . YOU’RE DONE!

Sometimes . . . • You only have two atoms, so there is

no central atom, but follow the same rules.

• Check & Share to make sure all the atoms are “happy”.

Cl2 Br2 H2 O2 N2 HCl

• DOUBLE bond – atoms that share two e- pairs (4 e-)

O O• TRIPLE bond

– atoms that share three e- pairs (6 e-)

N N

Draw Lewis Dot Structures

You may represent valence electrons from different atoms with the following symbols x, ,

CO2

NH3

Draw the Lewis Dot Diagram for polyatomic ions

• Count all valence e- needed for covalent bonding

• Add or subtract other electrons based on the charge

REMEMBER! A positive charge means it LOST

electrons!!!!!

Draw Polyatomics

• Ammonium

• Sulfate

Types of CovalentCovalent BondsBonds• NON-Polar bonds

– Electrons shared evenly in the bond

– E-neg difference is zero

Between identical atomsDiatomic molecules

Types of Covalent BondsPolar bond

– Electrons unevenly shared

non-polar MOLECULES

• Sometimes the bonds within a molecule are polar and yet the molecule is non-polar because its shape is symmetrical. H

H

HH CDraw Lewis dot first andsee if equal on all sides

Polar molecules (a.k.a. Dipoles)

• Not equal on all sides– Polar bond between 2 atoms makes a

polar molecule– asymmetrical shape of molecule

H Cl -+

HHO

-

+

Water is asymmetrical+

Water is a bent molecule

O

H H H H

W - A - T - E - Ras bent as it can be!

Water’s polar MOLECULE!Water’s polar MOLECULE!

The H is positive The O is not - not - not - not

Making sense of the polar non-polar thing

BONDS

Non-polar Polar

Identical Different

MOLECULES

Non-polar PolarSymmetrical Asymmetrical

IONIC bonds ….

Ionic bonds are so polar that the electrons are not shared but transferred between atoms forming ions!!!!!!

4 Shapes of molecules

Linear (straight line)

Ball and stick model

Space filling model

Bent


Space filling model

Trigonal pyramidBall and stick model

Space filling model

Tetrahedral


Space filling model

• Attractions between molecules– van der Waals forces

• Weak attractive forces between non-polar molecules

– Hydrogen “bonding”• Strong attraction

between special polar molecules

Intermolecular attractions

van der Waals• Non-polar molecules can exist in liquid

and solid phases because van der Waals forces keep the

molecules attracted to each other

• Exist between CO2, CH4, CCl4, CF4,

diatomics and monoatomics

van der Waals periodicity• increase with molecular mass.

– Greater van der Waals force? • F2 Cl2 Br2 I2

• increase with closer distance between molecules– Decreases when particles are farther away

Hydrogen “Bonding”• Strong polar

attraction– Like magnets

• Occurs ONLY between H of one molecule and N, O, F of another

H “bond”

Why does H “bonding” occur?

• Nitrogen, Oxygen and Fluorine – small atoms with strong nuclear charges

• powerful atoms

– very high electronegativities

Intermolecular forces dictate chemical properties

• Strong intermolecular forces cause high b.p., m.p. and slow evaporation (low vapor pressure) of a substance.

Which substance has the highest boiling point?• HF

• NH3

• H2O

• WHY?

Fluorine has the highest e-neg, SO HF will experience the

strongest H bonding and

needs the most energy to weaken the i.m.f. and boil

Density????

H2O(s) is less dense than H2O(l)

• The hydrogen bonding in water(l) molecules is random. The molecules are closely packed.

• The hydrogen bonding in water(s) molecules has a specific open lattice pattern. The molecules are farther apart.

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Chemical Names and formulas• With all of the compounds and all of the

elements to be identified, a systematic method for writing formulas and naming compounds is necessary

• A correctly written chemical formula must represent the known facts about the composition of a compound

• Care must be taken so that subscripts are correct


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Using Chemical formulas

• Chemical formulas indicate the elements present in a compound and the relative numbers of atoms of each element in the compound

• In chemical formulas, the elements are given by their symbols and the relative number of atoms of each element by numerical subscript

• Ex H2SO4 the H, S & O are symbols, the 2 & 4 are subscripts


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• Ion: A charged particle due to loss or gain of electrons

• Cation: positive charge ion represented by a (+) after the chemical symbol (metal) Ex Na+

• Anion: negative charge ion represented by a (-) after the chemical symbol (metal) Ex Cl-

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Monatomic Ions• Positive ions are named by the element

name followed by the word “ion”

• Examples :

• K+ potassium ion

Mg+2 magnesium ion

• Al+3 aluminum ion


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• Negative ions are named by dropping the ending of the element name and adding the ending “ide” to it followed by the word “ion”

• Examples:

• F- fluoride ion

• S-2 sulfide ion

• I- Iodide ion


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74

Give the names of the following ions:

Ba2+ Al3+ K+

_________ __________ _________

N3 O2 F

_________ __________ _________

P3 S2 Cl

_________ __________ _________

Learning Check

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75

Ba2+ Al3+ K+

barium aluminum potassium

N3 O2 F

nitride oxide fluoride

P3 S2 Cl

phosphide sulfide chloride

Solution

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• Binary Compounds Composed of two elements

• Binary Ionic Compounds Metal—nonmetal

• Binary Covalent Compounds Nonmetal—nonmetal

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• Binary ionic compounds contain positive cations and negative anions. Type I compounds

• Metal present forms only one cation.

Type II compounds • Metal present can

form 2 or more cations with different charges.

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79

Metals (Groups I, II, and III) and Non-Metals

Metal _________ + Non-Metal _________ideSodium Chlorine

Sodium Chloride NaCl

Type I Compounds

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Common Simple Cations and Anions

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1. The cation is always named first and the anion second.

2. A simple cation takes its name from the name of the element.

3. A simple anion is named by taking the first part of the element name (the root) and adding –ide.

Rules for Naming Type I Ionic Compounds

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• Examples:

KCl Potassium chloride

MgBr2 Magnesium bromide

CaO Calcium oxide

Binary Ionic Compounds (Type I)

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Exercise

What is the name of the compound SrBr2?

a) strontium bromine

b) sulfur bromide

c) strontium dibromide

d) strontium bromide

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• Strontium bromide. Sr is the symbol for strontium.

• Br is the symbol for bromine,

• take the first part of the element name (the root) and add –ide to get the name bromide.


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Binary Ionic Compounds (Type II)

• Metals in these compounds can form more than one type of positive charge.

• Charge on the metal ion must be specified.

• Roman numeral indicates the charge of the metal cation.

• Transition metal cations usually require a Roman numeral.

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86

Metals (Transition Metals) and Non-Metals

Metal ______ +Roman Numeral (__) + Non-Metal ________ide Iron III Bromine

Iron (III) Bromide FeBr3

Compare with Iron (II) Bromide FeBr2Metals (Transition Metals) and Non-Metals

Older System

Metal (Latin) _______ + ous or ic + Non-Metal ________ide Ferrous Bromine

Ferrous Bromide FeBr2

Compare with Ferric Bromide FeBr3

Type II Compounds

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Different names are needed for positive ions of 2 different charges formed by the same metal

• Old system: “ous” ending for lower charge

• “ic” ending for higher charge

• New system: gives actual charge on the ion as a roman numeral


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Common Type II Cations

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1. The cation is always named first and the anion second.

2. Because the cation can assume more than one charge, the charge is specified by a Roman numeral in parentheses.

Rules for Naming Type II Ionic Compounds

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• Examples:

CuBr Copper(I) bromide

FeS Iron(II) sulfide

PbO2 Lead(IV) oxide

Binary Ionic Compounds (Type II)

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Exercise

What is the name of the compound CrO2?

a) chromium oxide

b) chromium(II) oxide

c) chromium(IV) oxide

d) chromium dioxide

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• Chromium(IV) oxide. Cr is the symbol for chromium. O is the symbol for oxygen, but

• take the first part of the element name (the root) and add –ide to get the name oxide.

• Since chromium can have more than one charge, a Roman numeral must be used to identify that charge.

• There are two oxygen ions each with a 2– charge, giving an overall charge of –4.

• Therefore, the charge on chromium must be +4.


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Exercise

What is the correct name of the compound that results from the most stable ion for sulfur and the metal ion that contains 24 electrons?

a) iron(III) sulfide

b) chromium(II) sulfide

c) nickel(III) sulfate

d) iron(II) sulfide

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• Iron(II) sulfide.

• For sulfur, take the first part of the element name (the root) and add –ide to get the name sulfide.

• Iron with a +2 charge (as the Roman numeral indicates) contains 24 electrons (26p – 24e = +2 charge).


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Naming Compounds

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• Formed between two nonmetals.

1. The first element in the formula is named first, and the full element name is used.

2. The second element is named as though it were an anion.

3. Prefixes are used to denote the numbers of atoms present.

4. The prefix mono- is never used for naming the first element.

Rules for Naming Type III Binary Compounds

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Type III Compounds

Non-Metals and Non-Metals

Use Prefixes such as mono, di, tri, tetra, penta, hexa, hepta, etc.

CO2 Carbon dioxide CO Carbon monoxide

PCl3 Phosphorus trichloride CCl4 Carbon tetrachloride

N2O5 Dinitrogen pentoxide CS2 Carbon disulfide

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Prefixes Used to Indicate Numbers in Chemical Names 9 nona-

10 deca-

11 undeca-

12 dodeca-

13 trideca-

14 tetradeca-

15 pentadeca-

16 hexadeca-

17 heptadeca-

18 octadeca-

19 nonadeca-

20 icosa

Additional Prefixes

http://home.comcast.net/~igpl/NWD.html












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• Examples:

CO2 Carbon dioxide

SF6 Sulfur hexafluoride

N2O4 Dinitrogen tetroxide

Binary Covalent Compounds (Type III)

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Exercise

What is the name of the compound SeO2?

a) selenium oxide

b) selenium dioxide

c) selenium(II) oxide

d) selenium(IV) dioxide

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• Selenium dioxide. • Se is the symbol for selenium. • O is the symbol for oxygen, • take the first part of the element name (the root) and

add –ide to get the name oxide. • Since they are both nonmetals, prefixes are used

to identify the elements (except mono- is not used for the first element).

• Two oxygen atoms require the use of the prefix di-, making the name dioxide.


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Flow Chart for Naming Binary Compounds

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103

Let’s Practice!

Name the following.

CaF2

K2S

CoI2

SnF2

SnF4

OF2

CuI2

CuI

SO2

SrS

LiBr

Strontium SulfideLithium Bromide

Copper (I) Iodide or Cuprous Iodide

Sulfur dioxide

Copper (II) Iodide or Cupric Iodide

Oxygen diflourideTin (IV) Fluoride or Stannic Fluoride

Tin (II) Fluoride or Stannous Fluoride

Cobalt (II) Iodide or Cobaltous IodidePotassium Sulfide

Calcium Flouride

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Naming Compounds

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• Polyatomic ions are charged entities composed of several atoms bound together.

• They have special names and must be memorized.

• We will be using our Fat Daddy Chart to help us with naming the polyatomic compounds

• Those used often enough will be memorized just out of sheer practice

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Names of Common Polyatomic Ions (page 130)

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• Naming ionic compounds containing polyatomic ions follows rules similar to those for binary compounds. Ammonium acetate

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NaOH Sodium hydroxide

Mg(NO3)2 Magnesium nitrate

(NH4)2SO4 Ammonium sulfate

FePO4 Iron(III) phosphate

Examples

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109

Learning CheckSelect the correct name for each.

A. Fe2S3

1) iron sulfide

2) iron(II) sulfide

3) iron(III) sulfide

B. CuO

1) copper oxide

2) copper(I) oxide

3) copper(II) oxide

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110

SolutionSelect the correct name for each.

A. Fe2S3

3) iron(III) sulfide Fe3+ S2–

B. CuO

3) copper(II) oxide Cu2+ O2–

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Overall Strategy for Naming Chemical Compounds

Naming Compounds

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Exercise

What is the name of the compound KClO3?

a) potassium chlorite

b) potassium chlorate

c) potassium perchlorate

d) potassium carbonate

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Exercise

Examine the following table of formulas and names. Which of the compounds are named correctly?

a) I, II

b) I, III, IV

c) I, IV

d) I only

Formula Name

I P2O5 Diphosphorus pentoxide

II ClO2 Chlorine oxide

III PbI4 Lead iodide

IV CuSO4 Copper(I) sulfate

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• Only Formula I is named correctly.

• Formula II is chlorine dioxide.

• Formula III is lead(IV) iodide.

• Formula IV is copper(II) sulfate.


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• Acids can be recognized by the hydrogen that appears first in the formula—HCl.

• Molecule with one or more H+ ions attached to an anion.

• Most lab acids are either:• binary acids ( composed of Hydrogen

and another element)• or • oxyacids (composed of Hydrogen, oxygen

and a third element

Acids

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Naming Compounds

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• If the anion does not contain oxygen, the acid is named with the prefix hydro– and the suffix –ic attached to the root name for the element.

• Examples:

HCl Hydrochloric acid

HCN Hydrocyanic acid

H2S Hydrosulfuric acid

Rules for Naming Acids

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Acids That Do Not Contain Oxygen

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• If the anion contains oxygen: The suffix –ic is added to the root name if

the anion name ends in –ate.• Examples:

HNO3Nitric acid

H2SO4 Sulfuric acid

HC2H3O2 Acetic acid


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• If the anion contains oxygen: The suffix –ous is added to the root name

if the anion name ends in –ite.• Examples:

HNO2Nitrous acid

H2SO3 Sulfurous acid

HClO2 Chlorous acid


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Some Oxygen-Containing Acids

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Flowchart for Naming Acids

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Exercise

Which of the following compounds is named incorrectly?

a) KNO3 potassium nitrate

b) TiO2 titanium(II) oxide

c) Sn(OH)4 tin(IV) hydroxide

d) PBr5 phosphorus pentabromide

e) H2SO3 sulfurous acid

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• The correct answer is “b”.

• The charge on oxygen is 2–.

• Since there are two oxygen atoms, the overall charge is 4–.

• Therefore, the charge on titanium must be 4+ (not 2+ as the Roman numeral indicates).


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• Sodium hydroxide NaOH

• Potassium carbonate K2CO3

• Sulfuric acid H2SO4

• Dinitrogen pentoxide N2O5

• Cobalt(III) nitrate Co(NO3)3

Examples

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Exercise

A compound has the formula XCl3 where X could represent a metal or nonmetal. What could the name of this compound be?

a) phosphorus trichloride

b) carbon monochloride

c) tin(IV) chloride

d) magnesium chloride

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• Phosphorus trichloride.

• Carbon monochloride has the formula CCl.

• Tin(IV) chloride has the formula SnCl4.

• Magnesium chloride has the formula MgCl2.

• Phosphorus trichloride has the formula PCl3 and is therefore the correct answer


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128

Lets Practice Some More!HFNa2CO3

H2CO3

KMnO4

HClO4

H2S

NaOH

CuSO4

PbCrO4

H2O

NH3

Hydrooxic acid (no……just water)

Nitrogen trihydride (no..just ammonia)

Copper (II) sulfate or Cupric sulfate

Lead (II) chromate or Plubous chromate

Sodium hydroxide

Hyrdogen sulfuric acidPerchloric acid

Potassium permanganate

Sodium carbonate

Hydroflouric acid

Carbonic acid

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Naming Compounds

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131

Identifying Ionic Charges

• Group A elements – use the periodic table to determine ionic charge

* elements in same group have same ionic charge

* Group 4A and Noble gases – almost never form ions• Group B elements – many have more than one ionic

charge

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132

Identifying Ionic Charges

http://wps.prenhall.com/wps/media/objects/476/488316/ch04.html

Charge on cations corresponds to group #.Charge on anions is found by subtracting 8 by group number the number 8 is used b/c it represents # of valence e- in Noble gases

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133

Naming Cations and Anions

• Potassium ion• Copper (II) ion• Chloride ion• Oxide ion• Ba2+

• S2-

• Au3+

• Nitrite ion• Hydroxide ion• Phosphate ion

• SO42-

• CrO42-

• ClO32-

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134

Binary Ionic Compounds

• Compounds composed of 2 different monatomic elements

• To write binary formulas – write cation first, then anion

*criss-cross charges to determine how many of each ion you need *use subscripts to denote number of ions ex: Ca2+ + Cl1- CaCl2

Na1+ + Cl1- NaCl

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135

Ternary Ionic Compounds

• Compounds containing at least one polyatomic ion; at least 3 different elements

• To write ternary formulas: write cation first, then anion *criss-cross charges to determine how many of each ion you need *use subscripts to denote number of ions

*must use parentheses around polyatomic if more than one is

needed!!!ex: Na1+ + SO3

2- Na2SO3

Mg2+ + OH1- Mg(OH)2 [not same as MgOH2]

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136

Ionic Compounds

• NaNO3

• CaSO4

• (NH4)2O

• CuSO3

• Fe(OH)3

• NaF

• Lithium sulfide• Iron (III) phosphide• Magnesium fluoride• Barium nitrate• Aluminum hydroxide• Potassium phosphate

Practice making ionic compounds!

http://antoine.frostburg.edu/chem/senese/101/kits/kit_ionic_compound.html

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137

Molecular Compounds

• P2O5

• N2O

• NO2

• CBr4

• CO2

• tetraiodine nonoxide• sulfur hexafluoride• nitrogen trioxide• carbon tetrahydride• phosphorus trifluoride

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138

Formula Ions Name Cation Anion

NaCl Na+ Cl– sodium chloride

K2S K+ S2– potassium sulfide

MgO Mg2+ O2– magnesium oxide

CaI2 Ca2+ I– calcium iodide

Al2O3 Al3+ S2– aluminum sulfide

Examples of Ionic Compounds with Two Elements

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139

Learning CheckWrite the formulas and names for compounds of the following ions:

Br– S2− N3−

Na+

Al3+

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140

Solution Br− S2− N3−

Na+

Al3+

NaBr

sodium bromide

Na2S

sodium sulfide

Na3N

sodium nitride

AlBr3

aluminum bromide

Al2S3

aluminum sulfide

AlN

aluminum nitride

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141

Transition Metals Form Positive IonsMost transition metals and Group 4(14) metals,

Form 2 or more positive ions Zn2+, Ag+, and Cd2+ form only one ion.

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142

Guide to Writing Formulas from the Name

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143

Writing FormulasWrite a formula for potassium sulfide.

STEP 1 Identify the cation and anion.

potassium = K+

sulfide = S2−

STEP 2 Balance the charges.

K+ S2−

K+

2(1+) + 1(2–) = 0

STEP 3 Write the cation first.

2K+ and 1S2− = K2S1 = K2S

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144

Writing FormulasWrite a formula for iron(III) chloride.STEP 1 Identify the cation and anion. iron (III) = Fe3+ (III = charge of 3+) chloride = Cl−

STEP 2 Balance the charges. Fe3+ Cl−

Cl− Cl−

1(3+) + 3(1–) = 0STEP 3 Write the cation first.

1Fe3+ and 3Cl− = FeCl3

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145

Learning CheckThe correct formula for each of the following is:

A. copper(I) nitride

1) CuN 2) CuN3 3) Cu3N

B. lead(IV) oxide

1) PbO2 2) PbO 3) Pb2O4

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146

SolutionThe correct formula for each of the following is:

A. copper(I) nitride

3) Cu3N 3Cu+ + N3– = 3(1+) + (3–) = 0

B. lead(IV) oxide

1) PbO2 Pb4+ + 2O2– = (4+) + 2(2–) = 0

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Naming Compounds

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Percent Composition, Empirical Formulas, Molecular Formulas

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Formula Masses and Molar masses:

• Molecular mass or molecular weight are used instead of the term formula mass.

• The formula mass of any compound is the sum of the average atomic masses of all of the atoms present in the formula

149

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Example of formula mass

• H2O

• 2 H atom weigh 1.0079 each• 1 O atom weighs 15.9994 each

• 2 x 1.oo79• +1x 15.9994• 18.0153 formula mass for water

150

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Molar mass as a conversion factor

• Moles x grams/mole = mass in grams

• Mass in grams x 1 mol/grams = moles

• Thus 2 conversions relate mass in grams to numbers of moles of a substance


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Example

• What is the molar mass of Barium nitrate Ba(NO3)2

• Solution• 1 mol Ba x 137.33 g/1 mol Ba = 137.33 g Ba• 2 moles N x 14.0067 g/1mole N = 28.0134g N• 6 moles O x 15.999g/1mol O = 95.9964g

• Molar mass Ba(NO3)2 = 261.34


Naming Compounds

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Percent Composition

• Percent Composition – the percentage

by mass of each element in a compound

Percent =_______Part

Wholex 100%

Percent composition

of a compound or =

molecule

Mass of element in 1 mol____________________

Mass of 1 mol

x 100%

Percent Composition

Example: What is the percent composition of Potassium Permanganate

(KMnO4

)?

Molar Mass of KMnO4

K = 1(39.1) = 39.1

Mn = 1(54.9) = 54.9

O = 4(16.0) = 64.0

MM = 158 g

Percent Composition

Example: What is the percent composition of Potassium Permanganate

(KMnO4

)?

= 158 g

% K

Molar Mass of KMnO4

39.1 g K

158 gx 100 = 24.7 %

% Mn54.9 g Mn

158 gx 100 = 34.8 %

% O64.0 g O

158 gx 100 = 40.5 %

K = 1(39.10) = 39.1

Mn = 1(54.94) = 54.9

O = 4(16.00) = 64.0

MM = 158

Percent Composition

Determine the percentage composition of sodium carbonate (Na2

CO3

)?

Molar Mass Percent Composition

% Na =46.0 g

106 gx 100% = 43.4 %

% C =12.0 g

106 gx 100% = 11.3 %

% O =48.0 g

106 gx 100% = 45.3 %

Na = 2(23.00) = 46.0

C = 1(12.01) = 12.0

O = 3(16.00) = 48.0

MM= 106 g

Percent Composition

Determine the percentage composition of ethanol (C2

H5

OH)?

% C = 52.13%, % H = 13.15%, % O = 34.72%

_______________________________________________

Determine the percentage composition of sodium oxalate

(Na2

C2

O4

)?

% Na = 34.31%, % C = 17.93%, % O = 47.76%

Percent Composition

Calculate the mass of bromine in 50.0 g of Potassium bromide.

1. Molar Mass of KBr

K = 1(39.10) = 39.10

Br =1(79.90) =79.90

MM = 119.0

79.90 g ___________

119.0 g

= 0.6714

3. 0.6714 x 50.0g = 33.6 g Br

2.

Percent Composition

Calculate the mass of nitrogen in 85.0 mg of the amino acid lysine, C6

H14

N2

O2

.

1. Molar Mass of C6

H14

N2

O2

C = 6(12.01) = 72.06

H =14(1.01) = 14.14

MM = 146.2

28.02 g ___________

146.2 g

= 0.192

3. 0.192 x 85.0 mg = 16.3 mg N

2.

N = 2(14.01) = 28.02

O = 2(16.00) = 32.00

Hydrates

Hydrated salt – salt that has water molecules trapped within the crystal lattice

Examples: CuSO4

•5H2

O , CuCl2

•2H2

O

Anhydrous salt – salt without water molecules

Examples: CuCl2

Can calculate the percentage of water in a hydrated salt.

Percent Composition

Calculate the percentage of water in sodium carbonate decahydrate, Na2

CO3

•10H2

O.

1. Molar Mass of Na2

CO3

•10H2

O

Na = 2(22.99) = 45.98

C = 1(12.01) = 12.01

MM = 286.2

H = 20(1.01) = 20.2

O = 13(16.00)= 208.00

H = 20(1.01) = 20.2

Water

O = 10(16.00)= 160.00

MM = 180.2

2.

3.

180.2 g _______

286.2 g

67.97 % x 100%=

orH = 2(1.01) = 2.02

O = 1(16.00) = 16.00

MM H2O = 18.02

So…

10 H2

O = 10(18.02) = 180.2

Percent Composition

Calculate the percentage of water in Aluminum bromide hexahydrate, AlBr3

•6H2

O.

1. Molar Mass of AlBr3

•6H2

O

Al = 1(26.98) = 26.98

Br = 3(79.90) = 239.7

MM = 374.8

H = 12(1.01) = 12.12

O = 6(16.00) = 96.00

H = 12(1.01) = 12.1

Water

O = 6(16.00)= 96.00

MM = 108.1

2.

3.

108.1 g _______

374.8 g

28.85 % x 100%=

or

MM = 18.02

For 6 H2O = 6(18.02) = 108.2

Percent Composition

If 125 grams of magnesium sulfate heptahydrate is completely dehydrated, how many grams

of anhydrous magnesium sulfate will remain?

MgSO4

. 7 H2

O

1. Molar Mass

Mg = 1 x 24.31 = 24.31 g

S = 1 x 32.06 = 32.06 g

O = 4 x 16.00 = 64.00 g

MM = 120.37 g

H = 2 x 1.01 = 2.02 g

O = 1 x 16.00 = 16.00 g

MM = 18.02 g

MM H2O =

7 x 18.02 g = 126.1 g

Total MM =

120.4 g + 126.1 g = 246.5 g

2. % MgSO4

120.4 g

246.5 g

X 100 = 48.84 %

3. Grams anhydrous MgSO4

0.4884 x 125 = 61.1 g

Percent Composition

If 145 grams of copper (II) sulfate pentahydrate is completely dehydrated, how many grams of

anhydrous copper sulfate will remain?

CuSO4

. 5 H2

O

1. Molar Mass

Cu = 1 x 63.55 = 63.55 g

S = 1 x 32.06 = 32.06 g

O = 4 x 16.00 = 64.00 g

MM = 159.61 g

H = 2 x 1.01 = 2.02 g

O = 1 x 16.00 = 16.00 g

MM = 18.02 g

MM H2O =

5 x 18.02 g = 90.1 g

Total MM =

159.6 g + 90.1 g = 249.7 g

2. % CuSO4

159.6 g

249.7 g

X 100 = 63.92 %

3. Grams anhydrous CuSO4

0.6392 x 145 = 92.7 g

Percent Composition

A 5.0 gram sample of a hydrate of BaCl2

was heated, and only 4.3 grams of the anhydrous

salt remained. What percentage of water was in the hydrate?

1. Amount water lost

5.0 g hydrate

- 4.3 g anhydrous salt

0.7 g water

2. Percent of water

0.7 g water

5.0 g hydrate

x 100 = 14 %

Percent Composition

A 7.5 gram sample of a hydrate of CuCl2

was heated, and only 5.3 grams of the anhydrous

salt remained. What percentage of water was in the hydrate?


7.5 g hydrate


2.2 g water

2. Percent of water

2.2 g water

7.5 g hydrate

x 100 = 29 %

Percent Composition

A 5.0 gram sample of Cu(NO3

)2

•nH2

O is heated, and 3.9 g of the anhydrous salt remains.

What is the value of n?


5.0 g hydrate


1.1 g water

2. Percent of water

1.1 g water

5.0 g hydrate

x 100 = 22 %

3. Amount of water

0.22 x 18.02 = 4.0

Percent Composition

A 7.5 gram sample of CuSO4

•nH2

O is heated, and 5.4 g of the anhydrous salt remains. What

is the value of n?


7.5 g hydrate


2.1 g water

2. Percent of water

2.1 g water

7.5 g hydrate

x 100 = 28 %

3. Amount of water

0.28 x 18.02 = 5.0

Formulas

Empirical Formula – formula of a compound that expresses lowest whole number

ratio of atoms.

Molecular Formula – actual formula of a compound showing the number of atoms

present

Percent composition allow you to calculate the simplest ratio among the atoms found in

compound.

Examples:

C4

H10

- molecular

C2

H5

- empirical

C6

H12

O6

- molecular

CH2

O - empirical

Formulas

Is H2

O2

an empirical or molecular formula?

Molecular, it can be reduced to HO

HO = empirical formula

Calculating Empirical Formula

An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate

the empirical formula.

1. Determine the number of grams of each element in the compound.

4.151 g Al and 3.692 g O

2. Convert masses to moles.

4.151 g Al 1 mol Al

26.98 g Al

= 0.1539 mol Al

3.692 g O 1 mol O

16.00 g O

= 0.2308 mol O


An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate

the empirical formula.

3. Find ratio by dividing each element by smallest amount of moles.

0.1539 moles Al

0.1539

= 1.000 mol Al

0.2308 moles O

0.1539

= 1.500 mol O

4. Multiply by common factor to get whole number. (cannot have fractions of atoms in compounds)

O = 1.500 x 2 = 3

Al = 1.000 x 2 = 2

therefore, Al2

O3


A 4.550 g sample of cobalt reacts with 5.475 g chlorine to form a binary compound. Determine the empirical

formula for this compound.

4.550 g Co 1 mol Co

58.93 g Co

= 0.07721 mol Co

5.475 g Cl 1 mol Cl

35.45 g Cl

= 0.1544 mol Cl

0.07721 mol Co 0.1544 mol Cl

0.07721 0.07721

= 2= 1

CoCl2


When a 2.000 g sample of iron metal is heated in air, it reacts with oxygen to achieve a final mass of 2.573

g. Determine the empirical formula.

2.000 g Fe 1 mol Fe

55.85 g Fe

= 0.03581 mol Fe

0.573 g O 1 mol O

16.00 g

= 0.03581 mol Fe

Fe = 2.000 g O = 2.573 g – 2.000 g = 0.5730 g

1 : 1

FeO


A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead, 0.00672g

of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate.

1.3813 g Pb 1 mol Pb

207.2 g Pb

= 0.006667 mol Pb

0.00672 gH 1 mol H

1.008 g H

= 0.00667 mol H

0.4995 g As 1 mol As

74.92 g As

= 0.006667 mol As

0.4267g Fe 1 mol O

16.00 g O

= 0.02667 mol O


A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead, 0.00672g

of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate.

0.006667 mol Pb

0.00667 mol H

0.006667 mol As

0.02667 mol O

0.006667

0.006667

0.006667

0.006667

= 1.000 mol Pb

= 1.00 mol H

= 1.000 mol As

= 4.000 mol O

PbHAsO4


The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14%

oxygen. Calculate the empirical formula for Nylon-6.

Step 1:

In 100.00g of Nylon-6 the masses of elements present are 63.38 g C, 12.38 g n, 9.80 g H, and 14.14 g O.

Step 2:

63.38 g C 1 mol C

12.01 g C

= 5.302 mol C

12.38 g N 1 mol N

14.01 g N

= 0.8837 mol N

9.80 g H 1 mol H

1.01 g H

= 9.72 mol H

14.14 g O 1 mol O

16.00 g O

= 0.8832 mol O


The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14%

oxygen. Calculate the empirical formula for Nylon-6.

Step 3:

5.302 mol C

0.8837

= 6.000 mol C

0.8837 mol N

0.8837

= 1.000 mol N

9.72 mol H

0.8837

= 11.0 mol H

0.8837 mol O

0.8837

= 1.000 mol O

6:1:11:1

C6

NH11

O

Calculating molecular formula

• It is not possible to determine the correct molecular formula unless the molecular mass of the substance has been determined

• The relationship between the simplest formula and the molecular mass is:

• (simple formula)x = molecular formula• Where x is a whole number multiple of the

simple formula

Calculating Molecular Formula

A white powder is analyzed and found to have an empirical formula of P2O5. The

compound has a molar mass of 283.88g. What is the compound’s molecular formula?

Step 1: Molar Mass

P = 2 x 30.97 g = 61.94g

O = 5 x 16.00g = 80.00 g

141.94 g

Step 2: Divide MM by

Empirical Formula Mass

238.88 g

141.94g= 2

Step 3: Multiply

(P2O

5)2

=

P4O

10

Calculating Molecular Formula

A compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH.

What is its molecular formula?

C = 12.01 g

H = 1.01 g

13.01 g

78 g/mol

13.01 g/mol= 6

(CH)6

=

C6H

6

Oxidation Numbers

• Are used to indicate general distributions of electrons among bonded atoms.

• Refer to handout for rules of oxidation numbers

Ex find oxidation # of following:

• UF6

• ClO3-

• Solution

• U = +6 F = -1

• Cl =+5 O =-2

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