Chapter 31 Molecules, Ions, and Their Compounds Chapter 3

Chapter 3 1

Molecules, Ions, and Their Molecules, Ions, and Their CompoundsCompounds

Chapter 3Chapter 3

Chapter 3 2

Molecular and Empirical FormulasMolecular and Empirical Formulas

Molecules and Molecular CompoundsMolecules and Molecular Compounds

Molecular formula – A formula which gives the actual number and type of atoms in a molecule.

Examples: H2O, CO2, CO, CH4, H2O2, O2, O3, and C2H4.

Empirical formula – A formula which gives the lowest whole number ratio of atoms in a molecule.

Examples:Substance Mol. formula Empirical

FormulaEthane C2H6 CH3

Water H2O H2O

Chapter 3 3

Molecular and Empirical FormulasMolecular and Empirical Formulas

Condensed formula – A formula which indicates how atoms are grouped together in a molecule.

Name Molecular Formula Condensed formulaEthane C2H6 CH3CH3

Diethyl ether C4H10O CH3CH2OCH2CH3

Molecules and Molecular CompoundsMolecules and Molecular Compounds

Chapter 3 4

Picturing MoleculesPicturing Molecules

Structural Formula – A formula which shows how the atoms of a molecule are joined.

• Structural formulas do not necessarily show the three dimensional shape of the molecule.

Molecules and Molecular CompoundsMolecules and Molecular Compounds

Chapter 3 5

Molecular ModelsMolecular ModelsThese are three-dimensional representations of molecules.

Molecules and Molecular CompoundsMolecules and Molecular Compounds

Chapter 3 6

• If an electron is removed or added to a neutral atom a charged particle or ion is formed.• A positively charged ion is called a cation.

Ions and Ionic CompoundsIons and Ionic Compounds

Chapter 3 7

• If an electron is removed or added to a neutral atom a charged particle or ion is formed.• A positively charged ion is called a cation.• A negatively charged ion is called an anion.

Ions and Ionic CompoundsIons and Ionic Compounds

Chapter 3 8

Predicting Ionic ChargePredicting Ionic Charge• Metals tend to form cations• Non-metals tend to form anions.

Ions and Ionic CompoundsIons and Ionic Compounds

Chapter 3 9

Ions and Ionic CompoundsIons and Ionic Compounds

Molecules can also gain or lose electrons and form ions, They are called polyatomic ions.

Chapter 3 10

Ions and Ionic CompoundsIons and Ionic Compounds

Ion Name Formula Ion Name FormulaPeroxide O2

2- Sulfate SO42-

Triiodide I3- Sulfite SO3

2-

Ammonium NH4+ Phosphate PO4

3-

Nitrate NO3- Acetate CH3CO2

-

Nitrite NO2- Perchlorate ClO4

-

Hydroxide OH- Permanganate MnO4-

Carbonate CO32- Dichromate Cr2O7

2-

Chapter 3 11

Ionic CompoundsIonic Compounds Ionic Compound – A compound that contains positively charged ions and negatively charged ions.

Ions and Ionic CompoundsIons and Ionic Compounds

Chapter 3 12

Predicting FormulasPredicting FormulasLet’s consider a compound containing Mg and N.

• The common charge on Mg is +2 (or Mg2+).• The common charge on N is –3 (or N3-).• Since we want to make a neutral (uncharged)

compound, the total charges from the cations and anions must cancel-out (or sum to zero).

• Therefore, Mg needs to lose 6 electrons (3 2+) and N gain those 6 electrons (2 3-).

• The resulting formula is: Mg3N2.

Ions and Ionic CompoundsIons and Ionic Compounds

Chapter 3 13

Names and Formulas of Ionic CompoundsNames and Formulas of Ionic Compounds Naming of compounds (nomenclature) is divided into:

• organic compounds (those containing C)• inorganic compounds (the rest of the periodic

table).

We will consider the naming rules of the Inorganic compounds.

Naming Inorganic CompoundsNaming Inorganic Compounds

Chapter 3 14

Naming Ionic CompoundsNaming Ionic Compounds

1. Cationsa) Cations from metal atoms have the same name as

the metal.

b) If the cation can form more than one ion, the positive charge is indicated by a roman numeral in parenthesis.

c) Cations of nonmetals end in –ium.

P+3 phosphorium

Names and Formulas of Ionic CompoundsNames and Formulas of Ionic Compounds

Chapter 3 15

Names and Formulas of Ionic CompoundsNames and Formulas of Ionic Compounds

Naming Inorganic CompoundsNaming Inorganic Compounds

Chapter 3 16

Naming Inorganic CompoundsNaming Inorganic Compounds

2. Anionsa) Monoatomic anions have names formed by

dropping the ending of the name of the element and adding –ide.

b) Polyatomic anions containing oxygen have names ending in –ate or –ite.

c) Anions derived by adding H+ to an oxyanion are named by adding as a prefix the word hydrogen- or dihydrogen-.

HSO4- Hydrogensulfate

H2PO4- Dihydrogenphsophate

Names and Formulas of Ionic CompoundsNames and Formulas of Ionic Compounds

Chapter 3 17

Naming Inorganic CompoundsNaming Inorganic Compounds

2. Anionsd) Oxyanions

There are rules for these, but they are confusing.

Ion Name

ClO4- perchlorate ion

ClO3- chlorate ion

ClO2- chlorite ion

ClO- hypochlorite ion

Names and Formulas of Ionic CompoundsNames and Formulas of Ionic Compounds

Chapter 3 18

Naming Inorganic CompoundsNaming Inorganic Compounds

3. Ionic Compounds

Name the compound by naming the cation followed by the anion.

Names and Formulas of Ionic CompoundsNames and Formulas of Ionic Compounds

Chapter 3 19

Naming Binary Molecular CompoundsNaming Binary Molecular Compounds

Binary molecular compounds have two elements.1. The name of the left-most element is written first.2. If the elements are in the same group the lower

element is written first.3. The name of the second element ends in –ide.4. Prefixes are used to indicate the number of atoms of

each element.

Naming Inorganic CompoundsNaming Inorganic Compounds

Chapter 3 20

Naming Binary Molecular CompoundsNaming Binary Molecular Compounds

Naming Inorganic CompoundsNaming Inorganic Compounds

Chapter 3 21

Naming Binary Molecular CompoundsNaming Binary Molecular Compounds

Binary molecular compounds have two elements.1. The name of the left-most element is written first.2. If the elements are in the same group the lower

element is written first.3. The name of the second element ends in –ide.4. Prefixes are used to indicate the number of atoms of

each element. • Mono is never used in the first element.

Naming Inorganic CompoundsNaming Inorganic Compounds

Chapter 3 22

Naming Inorganic CompoundsNaming Inorganic Compounds

N2O• This is a molecular compound.

• The first element (N), just takes its name, Nitrogen.

• The second compound takes its name, ending in -ide, Oxide.

• Now we must consider how to show that there are two nitrogen atoms, use di- as a prefix.

Dinitrogen Oxide

Chapter 3 23

Naming Inorganic CompoundsNaming Inorganic Compounds

N2O5

• This is a molecular compound.

• The first element (N), just takes its name, Nitrogen.

• The second compound takes its name, ending in -ide, Oxide.

• Now we must consider how to show that there are two nitrogen atoms, use di- as a prefix.

• Finally, we must consider how to show that there are five oxygen atoms, use penta- as a prefix.

Dinitrogen Pentoxide

Chapter 3 24

Formula and Molecular WeightsFormula and Molecular Weights

Molecular weight The sum of the atomic weights of each atom in the molecular formula.

• Formula weight is the general term, molecule weight refers specifically to molecular compounds.

Formulas, Compounds, and the MoleFormulas, Compounds, and the Mole

Chapter 3 25

Formula and Molecular WeightsFormula and Molecular Weights

Formula weight (FW)The sum of the atomic weights of each atom in the chemical formula.

Example: CO2

Formula Weight = 1(AW, carbon) + 2(AW, oxygen)

Formula Weight = 1(12.011amu) + 2(16.0amu)

Formula Weight = 44.0 amu

Formulas, Compounds, and the MoleFormulas, Compounds, and the Mole

Chapter 3 26

Formula and Molecular WeightsFormula and Molecular Weights

Chemistry “trick”• The masses of the atoms are on a “gram equivalent

scale”.1 atom (average) 1 mole

C 12.01 amu 12.01 gH 1.008 amu 1.008 g

• So, the mass of a single atom or a mole is numerically equvalent.

Formulas, Compounds, and the MoleFormulas, Compounds, and the Mole

Chapter 3 27

Converting Between Mass, Moles, Molecules and Converting Between Mass, Moles, Molecules and AtomsAtoms

Formulas, Compounds, and the MoleFormulas, Compounds, and the Mole

Chapter 3 28

Moles Moles Numbers of Particles Numbers of Particles

The MoleThe Mole

mol

particlesmolesparticlesofnumbers

1

1002.6 23

Chapter 3 29

Mass Mass Moles Moles

weightformula

substanceofgramsmoles

The MoleThe Mole

Chapter 3 30

Moles Moles Mass Mass

sampleofmolesweightformulamass

The MoleThe Mole

Chapter 3 31

A sample of hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain?

The MoleThe Mole

Chapter 3 32

A sample of hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain?

24H:18C

Hto Cofratio

The MoleThe Mole

Chapter 3 33

A sample of hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain?

H

C

24

18

or

24H:18C H,to Cofratio

The MoleThe Mole

Chapter 3 34

A sample hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain?

Hatoms

x

H

C20100.324

18

or

24H:18C H,to Cofratio

The MoleThe Mole

Chapter 3 35

A sample hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain?

Catomsx

Hatoms

x

H

C

20

20

103.2

100.324

18

or

24H:18C H,to Cofratio

The MoleThe Mole

Chapter 3 36

A sample hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain?

Hto moleculesofratio

The MoleThe Mole

Chapter 3 37

A sample hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain?

24H:molecule 1

Hto moleculesofratio

The MoleThe Mole

Chapter 3 38

A sample hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain?

H

molecule

24

1

or

24H:molecule 1

Hto moleculesofratio

The MoleThe Mole

Chapter 3 39

A sample hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain?

Hatoms

x

H

molecule20100.324

1

or

24H:molecule 1

Hto moleculesofratio

The MoleThe Mole

Chapter 3 40

A sample hormone, estradiol, C18H24O2, contains 3.0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain?

moleculesx

Hatoms

x

H

molecule

19

20

103.1

100.324

1

or

24H:molecule 1

Hto moleculesofratio

The MoleThe Mole

Chapter 3 41

Percentage Composition from FormulasPercentage Composition from Formulas

100%

moleculeofweightformula

elementofmasselement

Describing Compound FormulasDescribing Compound Formulas

Chapter 3 42

Percentage Composition from FormulasPercentage Composition from FormulasExample:

Calculate the percent oxygen in CH3CH2OH.

Formula weight of ethanol: 2(12.01amu) + 6(1.01amu) + 1(16.00amu) = 46.08amuMass of oxygen: 1(16.00amu) = 16.00amu% oxygen

100

08.46

00.16

amu

amu

Describing Compound FormulasDescribing Compound Formulas

Chapter 3 43

Percentage Composition from FormulasPercentage Composition from FormulasExample:

Calculate the percent oxygen in CH3CH2OH.

Formula weight of ethanol: 2(12.01amu) + 6(1.01amu) + 1(16.00amu) = 46.08amuMass of oxygen: 1(16.00amu) = 16.00amu% oxygen

%72.3410008.46

00.16

amu

amu

Describing Compound FormulasDescribing Compound Formulas

Chapter 3 44

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 45

AnalysisHg 73.9%Cl 26.1%

-assume 100g sampleHg 73.9 gCl 26.1g

-convert grams to molesHg 73.9g / 200.59g/mol = Cl 26.1g/ 35.45g/mol =

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 46

AnalysisHg 73.9%Cl 26.1%

-assume 100g sampleHg 73.9 gCl 26.1g

-convert grams to molesHg 73.9g / 200.59g/mol = 0.368 mol Cl 26.1g/ 35.45g/mol = 0.736 mol

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 47

AnalysisHg 73.9%Cl 26.1%

-assume 100g sampleHg 73.9 gCl 26.1g

-convert grams to molesHg 73.9g / 200.59g/mol = 0.368 mol Cl 26.1g/ 35.45g/mol = 0.736 mol

-determine the empirical formula by using the moles of theelements to get the smallest whole number ratio of the elements.

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 48

AnalysisHg 73.9%Cl 26.1%

-assume 100g sampleHg 73.9 gCl 26.1g

-convert grams to molesHg 73.9g / 200.59g/mol = 0.367 mol Cl 26.1g/ 35.45g/mol = 0.736 mol

-determine the empirical formula by using the moles of theelements to get the smallest whole number ratio of the elements.

736.0367.0 ClHg

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 49

AnalysisHg 73.9%Cl 26.1%

-assume 100g sampleHg 73.9 gCl 26.1g

-convert grams to molesHg 73.9g / 200.59g/mol = 0.368 mol Cl 26.1g/ 35.45g/mol = 0.736 mol

-determine the empirical formula by using the moles of theelements to get the smallest whole number ratio of the elements.

368.0

736.0

368.0

368.0 ClHg

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 50

AnalysisHg 73.9%Cl 26.1%

-assume 100g sampleHg 73.9 gCl 26.1g

-convert grams to molesHg 73.9g / 200.59g/mol = 0.368 mol Cl 26.1g/ 35.45g/mol = 0.736 mol

-determine the empirical formula by using the moles of theelements to get the smallest whole number ratio of the elements.

21ClHg

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 51

Determine the empirical formula of the compound with the followingcompositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 52

Determine the empirical formula of the compound with the followingcompositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.

Analysis:C 10.4%S 27.8%Cl 61.7%

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 53

Determine the empirical formula of the compound with the followingcompositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.

Assume 100g sampleC 10.4gS 27.8gCl 61.7g

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 54

Determine the empirical formula of the compound with the followingcompositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.

Moles of each elementC 10.4g/12.011g/mol = 0.866 molS 27.8g/32.066g/mol = 0.867 molCl 61.7g/35.453g/mol = 1.74 mol

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 55

Determine the empirical formula of the compound with the followingcompositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.

Moles of each elementC 10.4g/12.011g/mol = 0.866 molS 27.8g/32.066g/mol = 0.867 molCl 61.7g/35.453g/mol = 1.74 mol

74.1867.0866.0 ClSC

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 56

Determine the empirical formula of the compound with the followingcompositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.

Moles of each elementC 10.4g/12.011g/mol = 0.866 molS 27.8g/32.066g/mol = 0.867 molCl 61.7g/35.453g/mol = 1.74 mol

866.0

74.1

866.0

867.0

866.0

866.0 ClSC

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 57

Determine the empirical formula of the compound with the followingcompositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.

Moles of each elementC 10.4g/12.011g/mol = 0.866 molS 27.8g/32.066g/mol = 0.867 molCl 61.7g/35.453g/mol = 1.74 mol

211 ClSC

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 58

Molecular Formula from Empirical FormulaMolecular Formula from Empirical Formula

To determine the molecular formula from an empirical formula, you must have the molecular weight of the substance.

weightformulaempirical

weightmolecularunitsformulaofnumber

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

formulaempiricalunitsformulaofnumberformulamolecular

Chapter 3 59

Molecular Formula from Empirical FormulaMolecular Formula from Empirical Formula

Empirical formula: CH Empirical formula weight: 13.019 g/mol

Molecular weight: 78.114g/mol

molg

molgunitsformula

/019.13

/114.78

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 60

Molecular Formula from Empirical FormulaMolecular Formula from Empirical Formula

Empirical formula: CH Empirical formula weight: 13.019 g/mol

Molecular weight: 78.114g/mol

6/019.13

/114.78

molg

molgunitsformula

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 61

Molecular Formula from Empirical FormulaMolecular Formula from Empirical Formula

Empirical formula: CH Empirical formula weight: 13.019 g/mol

Molecular weight: 78.114g/mol

116

6/019.13

/114.78

HCformulamolecular

molg

molgunitsformula

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 62

Molecular Formula from Empirical FormulaMolecular Formula from Empirical Formula

Empirical formula: CH Empirical formula weight: 13.019 g/mol

Molecular weight: 78.114g/mol

66

116

6/019.13

/114.78

HC

HCformulamolecular

molg

molgunitsformula

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Chapter 3 63

End of Chapter ProblemsEnd of Chapter Problems

6, 8, 12, 14, 20, 22, 28, 32, 36, 42, 48, 54, 68, 86