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