11

IonsIonsIonsIons: A charged particle formed when a neutral : A charged particle formed when a neutral

atom or group of atoms gain or lose one or atom or group of atoms gain or lose one or more electrons.more electrons.

ExampleExampleNa Na Na Na++ + e + e--

F + eF + e-- F F--

22

CationsCationsMg Mg Mg Mg2+2+ + 2e + 2e--

Al Al Al Al3+3+ + 3e + 3e--

Cation:Cation: A positively-charged ion. One or more A positively-charged ion. One or more electrons are lost from a neutral atomelectrons are lost from a neutral atom

oxidationoxidation

33

AnionsAnionsI + eI + e-- I I- - O + 2eO + 2e-- O O2- 2- S + 2eS + 2e-- S S2- 2-

AnionAnion: A negatively-charged ion. Electrons are : A negatively-charged ion. Electrons are gained by a neutral atomgained by a neutral atom

reductionreduction

44

IonsIons Ion charges can be predicted from the Periodic TableIon charges can be predicted from the Periodic Table

Main Group metal (IA-IVA) ion-charges correspond to group numberMain Group metal (IA-IVA) ion-charges correspond to group number Sodium (Na) in IA Sodium (Na) in IA +1 +1 ExceptionsExceptions

Tl, Sn, Pb, Sb, and BiTl, Sn, Pb, Sb, and Bi We’ll talk about non-main group metal ion-charges laterWe’ll talk about non-main group metal ion-charges later

Main Group non-metal (IIIA-VIIIA) ion-charges correspond to Main Group non-metal (IIIA-VIIIA) ion-charges correspond to (group# - 8)(group# - 8) Fluorine (F) in VIIA Fluorine (F) in VIIA (7-8) = -1 (7-8) = -1 Boron is exception Boron is exception -3 not -5 -3 not -5

All this has to do with electron configurationAll this has to do with electron configuration More anonMore anon

See website as well: See website as well: http://web.clark.edu/aaliabadi/CHEM131ions%20to%20memorize.htmhttp://web.clark.edu/aaliabadi/CHEM131ions%20to%20memorize.htm

55

Compounds That Contain Compounds That Contain IonsIons

Require metal and non-metalRequire metal and non-metal Form ionic bondsForm ionic bonds Called anCalled an ionic compound ionic compound

Characteristic PropertiesCharacteristic Properties1. Very high melting points1. Very high melting points2. Conduct an electric current when melted or 2. Conduct an electric current when melted or

when dissolved in waterwhen dissolved in water

66

Ionic compoundIonic compound The number of cations and anions must have a net charge of The number of cations and anions must have a net charge of

zero.zero.

77

Compounds that Contain Compounds that Contain IonsIons

Writing Formulas for Ionic CompoundsWriting Formulas for Ionic CompoundsGive the formulas for the compounds that contain the following Give the formulas for the compounds that contain the following

pairs of ions:pairs of ions:

(a) K and I(a) K and I

(b) Mg and N(b) Mg and N

(c) Al and O(c) Al and O

88

Types of Ionic CompoundsTypes of Ionic Compounds Type I CompoundsType I Compounds: The metal present forms : The metal present forms

only one type of cation.only one type of cation. Examples: the main group metalsExamples: the main group metals

Type II CompoundsType II Compounds: The metal present can : The metal present can form two or more cations that have different form two or more cations that have different charges or charges or oxidation statesoxidation states Oxidation state = ionic charge, if any, on speciesOxidation state = ionic charge, if any, on species

Examples include CrExamples include Cr2+2+, Cr, Cr3+3+, Cu, Cu++, Cu, Cu2+2+, etc., etc.

99

Naming Ionic CompoundsNaming Ionic Compounds1. The cation is always named first and the 1. The cation is always named first and the

anion second.anion second.2. The cation takes its name from the name of 2. The cation takes its name from the name of

the element.the element.3. The anion is named by taking the first part of 3. The anion is named by taking the first part of

the element name and adding –ide.the element name and adding –ide.

1010

Naming Ionic CompoundsNaming Ionic CompoundsName the following Type I Name the following Type I

compoundscompounds

NaClNaCl RaBrRaBr22

RbRb22OO AlIAlI33

KK33NN CsCs44SiSi

Give the chemical formula for the Give the chemical formula for the following Type I compoundsfollowing Type I compounds

Strontium phosphideStrontium phosphide Calcium fluorideCalcium fluoride Beryllium carbideBeryllium carbide Lithium hydrideLithium hydride Barium sulfideBarium sulfide Magnesium tellurideMagnesium telluride

1111

Naming Ionic CompoundsNaming Ionic Compounds Type II compounds need to be identified by a Roman numeral Type II compounds need to be identified by a Roman numeral

(I), (IV), etc. (I), (IV), etc. Represents oxidation state of cationRepresents oxidation state of cation Not Not how many cations are present in compound!how many cations are present in compound!

Example: NaCl Example: NaCl sodium (I) chloride is INCORRECT sodium (I) chloride is INCORRECT Example: SnClExample: SnCl44 tin (IV) chloride is correct tin (IV) chloride is correct

1212

Naming Ionic CompoundsNaming Ionic CompoundsType II Ionic CompoundsType II Ionic Compounds

FeClFeCl2 2 and FeCland FeCl33

PbO and PbOPbO and PbO22

MnS and MnMnS and Mn22SS77

1313

A Mixed BagA Mixed Bag PbBrPbBr22 and PbBr and PbBr44 Aluminum arsenideAluminum arsenide FeS and FeFeS and Fe22SS33 Thallium (III) borideThallium (III) boride Mercury (II) carbideMercury (II) carbide NaNa22SS CoClCoCl33 Cerium (IV) phosphideCerium (IV) phosphide ScFScF33 Gold (I) selenideGold (I) selenide Vanadium (V) tellurideVanadium (V) telluride

1414

Naming Compounds that Naming Compounds that Contain Polyatomic IonsContain Polyatomic Ions

Polyatomic IonPolyatomic Ion: An ion that contains more than one atom. They : An ion that contains more than one atom. They are charged entities composed of several atoms bound are charged entities composed of several atoms bound together.together.

Consult my website for the list that must be Consult my website for the list that must be memorized:memorized:

http://web.clark.edu/aaliabadi/CHEM131ionshttp://web.clark.edu/aaliabadi/CHEM131ions%20to%20memorize.htm%20to%20memorize.htm

1515

TrendsTrends Sulfide, sulfite, sulfateSulfide, sulfite, sulfate Nitride, nitrite, nitrateNitride, nitrite, nitrate Phosphide, phosphite, phosphatePhosphide, phosphite, phosphate Chloride, hypochlorite, chlorite, chlorate, Chloride, hypochlorite, chlorite, chlorate,

perchlorateperchlorate Parenthesis required if more than one Parenthesis required if more than one

polyatomic ion presentpolyatomic ion present Ca(IOCa(IO33))2 2 is correctis correct Ca(I)Ca(I)22 is INCORRECT is INCORRECT

1616

Naming Compounds that Naming Compounds that Contain Polyatomic IonsContain Polyatomic Ions

Name or provide the chemical formula for each of the following Name or provide the chemical formula for each of the following compounds:compounds:

(a) Ca(OH)(a) Ca(OH)22 (e) Co(ClO (e) Co(ClO44))22

(b) Sodium phosphate (f) platinum (IV) bicarbonate(b) Sodium phosphate (f) platinum (IV) bicarbonate

(c) KMnO(c) KMnO44 (g) Cu(NO (g) Cu(NO22))22

(d) Ammonium carbonate(d) Ammonium carbonate (h) nickel (III) sulfite (h) nickel (III) sulfite

1717

Naming Compounds that Naming Compounds that Contain Polyatomic IonsContain Polyatomic Ions

Name or provide each of the following compounds:Name or provide each of the following compounds:

(a) calcium carbonate (e) MoO(a) calcium carbonate (e) MoO

(b) BaSO(b) BaSO44 (f) Iridium (VII) acetate (f) Iridium (VII) acetate

(c) CsClO(c) CsClO44 (g) ZnO (g) ZnO2 2

(d) Zirconium (IV) bisulfite(d) Zirconium (IV) bisulfite (h) lithium (h) lithium cyanidecyanide

1818

Naming AcidsNaming AcidsAcidsAcids: A substance that yields hydrogen ions : A substance that yields hydrogen ions

(protons, H(protons, H++) when dissolved in water.) when dissolved in water. HClHCl(aq) (aq) H H++

(aq)(aq) + Cl + Cl--(aq)(aq)

HH33POPO4(aq) 4(aq) 3H 3H++(aq)(aq) + PO + PO44

3-3-(aq)(aq)

1919

Rules for naming acidsRules for naming acids If the formula does not contain oxygen the prefix of the acid If the formula does not contain oxygen the prefix of the acid

is is hydrohydro and the suffix – and the suffix –icic is attached to the root name for is attached to the root name for the element.the element.

Ex: HCl = hydrochloric acid, HEx: HCl = hydrochloric acid, H22S = hydrosulfuric acidS = hydrosulfuric acid

When the anion contains oxygen, the acid name is formed When the anion contains oxygen, the acid name is formed from the anion name. The suffix –ic or –ous is added. from the anion name. The suffix –ic or –ous is added.

When the anion ends in –When the anion ends in –ateate, the suffix –, the suffix –icic is used. is used. HH22COCO33 = carbonic acid = carbonic acid

When the anion ends in –When the anion ends in –iteite, the suffix –, the suffix –ousous is used. is used. HH22SOSO33 = sulfurous acid = sulfurous acid

2020

MoreMoreAnionAnion Acid Acid ClOClO44

-- ______________ ______________ HClOHClO44 _______________ _______________

ClOClO33-- ______________ ______________ HClOHClO33 _______________ _______________

ClOClO22-- ______________ ______________ HClOHClO22 _______________ _______________

ClOClO-- _______________ _______________ HClO ________________ HClO ________________

2121

Naming Compounds that Contain Naming Compounds that Contain Only Nonmetals: Type IIIOnly Nonmetals: Type III

Type III compounds contain only nonmetals.Type III compounds contain only nonmetals. Form covalent bondsForm covalent bonds share electronsshare electrons

Rules for Naming Type III Binary CompoundsRules for Naming Type III Binary Compounds1. The first element in the formula is named first, and the full 1. The first element in the formula is named first, and the full

element name is used.element name is used.2. The second element is named as though it were an anion.2. The second element is named as though it were an anion.3. Prefixes are used to denote the numbers of atoms present.3. Prefixes are used to denote the numbers of atoms present.4. The prefix 4. The prefix monomono- is never used for naming the first element.- is never used for naming the first element.5. Drop the “a” when it is followed by an “o”5. Drop the “a” when it is followed by an “o” Tetraoxide should be tetroxideTetraoxide should be tetroxide

2222

Naming Compounds that Contain Naming Compounds that Contain Only Nonmetals: Type IIIOnly Nonmetals: Type III

Prefixes Used to Indicate Numbers in Chemical NamesPrefixes Used to Indicate Numbers in Chemical Names PrefixPrefix Number IndicatedNumber Indicated mono- 1mono- 1 di- 2di- 2 tri- 3tri- 3 tetra- 4tetra- 4 penta- 5penta- 5 hexa- 6hexa- 6 hepta- 7hepta- 7 octa- 8octa- 8 nona- 9nona- 9 deca- 10deca- 10

2323

PracticePractice CClCCl44

Silicon dioxideSilicon dioxide NONO22

Sulfur trioxideSulfur trioxide PP22OO55

Iodine pentafluorideIodine pentafluoride Dinitrogen tetroxideDinitrogen tetroxide SeISeI22

Xenon hexafluorideXenon hexafluoride

2424

The Name GameThe Name Game Extra credit opportunity Extra credit opportunity Mix-and-matchMix-and-match

2525

Remember the mole?Remember the mole?

Mole = amt that contains as many “things” as Mole = amt that contains as many “things” as there are atoms of 12 g of C-12there are atoms of 12 g of C-12

1 mole = 6.022 x 101 mole = 6.022 x 102323 particles particles Molar mass (MM) = mass in grams per 1 mole Molar mass (MM) = mass in grams per 1 mole

of particle (g/mol)of particle (g/mol)

2626

Molecular MassMolecular Mass Summation of molar masses from Periodic Summation of molar masses from Periodic

Table based on molecular formulaTable based on molecular formula NaClNaCl II22

VV22OO55

2727

Solutions Solutions

2

2 5

g g g: 22.98977 35.4527 58.4425mol mol molg g: 2 126.9045 253.8090mol mol

g g g: 2 50.9415 5 15.9994 181.8800mol mol mol

NaCl

I

V O

2828

Molar MassMolar MassExampleExampleCalculate the mass of 30.0 moles of polyvinyl chloride (PVC), Calculate the mass of 30.0 moles of polyvinyl chloride (PVC),

CC22HH33Cl.Cl.

2929

Solution Solution

2 3g g g g: 2 12.011 3 1.00794 35.4527 62.499mol mol mol mol

62.49930.0 1874.97 1870

C H Cl

gmol gmol

3030

Molar MassMolar MassExampleExampleA sample of NaA sample of Na22SOSO4.4.with a mass of 300.0 grams with a mass of 300.0 grams

represents what number of moles of Narepresents what number of moles of Na22SOSO44??

ExampleExampleCalculate the number of grams of caffeine, Calculate the number of grams of caffeine,

CC88HH1010NN44OO22, in 8.13 x 10, in 8.13 x 102323 molecules. molecules.

3131

Solution Solution

2 4

8 10 4 2

1323

g g g g: 2 22.98977 32.066 4 15.9994 142.043mol mol mol mol

300.0 2.112g142.043 mol

g g g g g:8 12.011 10 1.00794 4 14.0067 2 15.9994 194.193mol mol mol mol mol18.13 10

6.022 10

Na SO

molg mol

C H N O

molmoleculesmo

8194.193 2.62 10g glecules mol

3232

Percent Composition Percent Composition of Compounds: NaClof Compounds: NaCl

22.99g 58.44gNa: [(1 ) ] 100% = 39.34%mol mol

35.45g 58.44gCl: [(1 ) ] 100% 60.66%mol mol

partmass percent for a given element = ( ) 100%whole

22.99 gNa = mol

35.45gCl = mol

22.99 g 35.45g 58.44gThus, NaCl = mol mol mol

3333

Percent CompositionPercent Compositionof Compoundsof Compounds

ExampleExampleCompute the mass percent of each element in Compute the mass percent of each element in

sodium sulfide, Nasodium sulfide, Na22S.S.

ExampleExampleCompute the mass percent of each element in Compute the mass percent of each element in

nitric acid, HNOnitric acid, HNO3(aq)3(aq)..

3434

Solution Solution

2

: 2 22.98977 45.97954

:32.066

: 45.97954 32.066 78.046

45.97954Thus, Na = 100% 58.913%

78.046

32.066And, S = 100% 41.086%

78.046

g gNa mol molgS mol

g g gNa S mol mol molgmol

gmol

gmolgmol

3535

SolutionSolution

3

:1.00794

:14.0067

: 3 15.9994 47.9982

:1.00794 14.0067 47.9982 63.0128

1.00794Thus, H = 100% 1.59958%

63.0128

14.0067And, N = 100% 22.2283%

63.0128

And,

gH molgN molg gO mol molg g g gHNO mol mol mol molgmolgmol

gmolgmol

47.9982 O = 100% 76.1721%

63.0128

gmolgmol

3636

Empirical FormulasEmpirical FormulasEmpirical FormulaEmpirical Formula: or the simplest formula; the smallest : or the simplest formula; the smallest whole-number ratio of the atoms present.whole-number ratio of the atoms present.

Molecular FormulaMolecular Formula: the actual formula of a compound. It gives : the actual formula of a compound. It gives the composition of the molecules that are present.the composition of the molecules that are present.

Empirical FormulaEmpirical Formula Molecular FormulaMolecular Formula CHCH CC66HH66

CHCH22O O CC66HH1212OO66

HH22OO HH22OO

3737

Calculation of Empirical Calculation of Empirical FormulasFormulas

Steps for Determining the Empirical Formula of a CompoundSteps for Determining the Empirical Formula of a Compound

1. Obtain the mass of each element present (in grams).1. Obtain the mass of each element present (in grams).2. Determine the number of moles of each type of atom present.2. Determine the number of moles of each type of atom present.3. Divide the number of moles of each element by the smallest 3. Divide the number of moles of each element by the smallest

number of moles to convert the smallest number to 1. If all of number of moles to convert the smallest number to 1. If all of the numbers are integers (whole numbers), these are the the numbers are integers (whole numbers), these are the subscripts in the empirical formula. If one or more of these subscripts in the empirical formula. If one or more of these numbers are not integers, go to step 4.numbers are not integers, go to step 4.

4. Multiply the numbers derived in step 3 by the smallest integer 4. Multiply the numbers derived in step 3 by the smallest integer that will convert all of them to whole numbers. This set of that will convert all of them to whole numbers. This set of whole numbers represents the subscripts in the empirical whole numbers represents the subscripts in the empirical formula.formula.

3838

Calculation of Empirical Calculation of Empirical FormulasFormulas

ExperimentExperiment::Suppose we weigh out 6.50 grams of Cr. We decide to heat this Suppose we weigh out 6.50 grams of Cr. We decide to heat this

Cr in the air so that the Cr can react with O to form CrCr in the air so that the Cr can react with O to form CrxxOOyy . . After the sample cools, we weigh it again and find its mass to After the sample cools, we weigh it again and find its mass to be 9.50 grams. How do we find the mass of oxygen?be 9.50 grams. How do we find the mass of oxygen?

What is the empirical formula of this compound?What is the empirical formula of this compound?Let’s work on this together.Let’s work on this together.

3939

Calculation of Empirical Calculation of Empirical FormulasFormulas

ExampleExampleNylon-6 consists of 63.68% C, 12.38% N, 9.80% H, and Nylon-6 consists of 63.68% C, 12.38% N, 9.80% H, and

14.14%O. 14.14%O. Calculate the empirical formula for Nylon-6.Calculate the empirical formula for Nylon-6.

4040

Solution Solution

6 11

molC: 63.68g 5.302mol 0.8838mol=5.999 612.011g

molN: 12.38g 0.8839mol 0.8838mol=1.000 114.0067g

molH: 9.80g 9.72mol 0.8838mol=11.00 111.00794g

molO: 14.14g 0.8838mol 0.8838mol=1.000 115.9994g

C H NO

4141

Calculation of Calculation of Molecular FormulasMolecular Formulas

We need to know the empirical formula and molar mass of the We need to know the empirical formula and molar mass of the molecular compound.molecular compound.

Molecular Formula = (empirical formula)Molecular Formula = (empirical formula)nn

where n is a small whole number.where n is a small whole number.

Molecular Formula = n x Empirical FormulaMolecular Formula = n x Empirical FormulaMolar Mass = n x Formula WeightMolar Mass = n x Formula Weight

n = Molar Mass/Formula Weightn = Molar Mass/Formula Weight

4242

Calculation ofCalculation ofMolecular FormulasMolecular Formulas

ExampleExampleA compound shows the following percentage compostion:A compound shows the following percentage compostion:

71.65% Cl 24.27% C 4.07% H71.65% Cl 24.27% C 4.07% HThe molar mass is known to be 98.96 g/mol. Determine the The molar mass is known to be 98.96 g/mol. Determine the

empirical formula and the molecular formula for this empirical formula and the molecular formula for this compound.compound.

4343

Solution Solution 71.65% Cl 24.27% C 4.07% HThe molar mass is known to be 98.96 g/mol. Determine the empirical formula and the molecular formula

molCl: 71.65g 2.021 2.021 135.4527g

molC: 24.27g12.011g

mol mol

2

2.021 2.021 1

molH: 4.07g 4.038 2.021 1.998 21.00794g

empirical formula = CH Clg g g gempirical formula mass = 12.011 2 1.00794 35.4527 49.476mol mol mol mol

98.96molar massempirical formula mass

mol mol

mol mol

2 4 2

gmol 2g49.476 mol

molecular fomrula = C H Cl

4444

Calculation ofCalculation ofMolecular FormulasMolecular Formulas

ExampleExampleVitamin C consists of 40.92% C, 4.58% H, and 54.50% O on a Vitamin C consists of 40.92% C, 4.58% H, and 54.50% O on a

mass basis, and has a molar mass of 176.12 g/mol. Determine mass basis, and has a molar mass of 176.12 g/mol. Determine the molecular formula of the compound.the molecular formula of the compound.

4545

Solution Solution