Download Section 3.5 Ionic Compounds: Formulas and Names

Chapter 3
Molecules, Compounds and Chemical Equations
Chapter 3
Molecules, Compounds and Chemical Equations
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
3.11
3.12
Hydrogen, Oxygen and Water
Chemical Bonds
Representing Compounds: Chemical Formulas and Molecular
Models
An Atomic Level View of Elements and Compounds
Ionic Compounds: Formulas and Names
Molecular Compounds: Formulas and Names
Summary of Inorganic Nomenclature
Formula Mass and the Mole Concept for Compounds
Composition of Compounds
Determining a Chemical Formula from Experimental Data
Writing and Balancing Chemical Equations
Organic Compounds
2
Section 3.1
Hydrogen, Oxygen and Water
Mixtures vs Compounds
• When two or more elements combine to form a compound
an entirely new substance results.
– Balloon contains a mixture of hydrogen (H2) and oxygen gas (O2)
– Glass contains a compound composed of hydrogen and oxygen
(H2O)
3
Section 3.1
Hydrogen, Oxygen and Water
Properties of Hydrogen, Oxygen and Water
4
Section 3.1
Hydrogen, Oxygen and Water
Sodium Chloride NaCl
When two or more elements combine to form a compound
an entirely new substance results.
Sodium (Na) = Soft reactive metal
Chlorine (Cl) = toxic gas
Sodium Chloride (NaCl) =
Tasty condiment/preservative
5
Section 3.1
Hydrogen, Oxygen and Water
Ionic vs Covalent Bonds
• Compounds form through Bond formation
• 2H2 + O2 → 2H2O
covalent bond
• 2Na + Cl2 → 2NaCl
ionic bond
6
Section 3.2
Chemical Bonds
Ionic Bonds
•
Ionic bonds form when one atom loses
electrons and another atom gains electrons.
– Metal atoms tend to lose electrons and
become positive cations
– Nonmetal atoms tend to gain electrons and
become negative anions
7
Section 3.2
Chemical Bonds
Ionic Bonds
•
The oppositely charged ions attract one
another by electrostatic forces and form an
ionic bond
•
The result is an ionic compound, which
(in the solid phase) is composed of a
lattice – a regular three dimensional array
of cations and anions
8
Section 3.2
Chemical Bonds
Covalent Bonds
• Covalent bonds form when two atoms share
electrons
– Form between non-metals – neither are willing to give
up electrons so in order to form a bond – they share.
9
Section 3.2
Chemical Bonds
Covalent Bonds
• Water (H2O), ammonia (NH3), and Methane
(CH4) all have covalent bonds. A covalent
bond is a shared pair of electrons (represented
here by a line). Notice covalent compounds
are individual units (molecules)
10
Section 3.2
Chemical Bonds
Why do bonds form?
• Bonds form to lower the potential energy of the
atoms
– In the process each atom ends up with an octet of
electrons (8 electrons in outer shell)
– This is the result
– Not the driving force
• A major driving force in nature is to increase
stability (lower potential energy).
11
Section 3.3
Representing Compounds: Chemical Formulas and Molecular Models
Types of Chemical Formulas
• Empirical formula
– Simplest – tells what types of atoms are present in a
compound
– Empirical formula for hydrogen peroxide is HO
• Molecular Formula
– Give the actual number of atoms in a compound
– Molecular formula for hydrogen peroxide is H2O2
• Structural Formula
12
Section 3.3
Representing Compounds: Chemical Formulas and Molecular Models
Types of Chemical Formulas
• Structural Formula
– Can be simple (L)
– Can indicate geometry of the compound (R)
13
Section 3.3
Representing Compounds: Chemical Formulas and Molecular Models
Molecular Models
• Ball and Stick Models
– Atoms are balls and bonds are sticks
• Space filling Models
– Atoms fill the spaces between each
other
– Probably the most “accurate” way to
imagine a molecule if it was blown up
to a size we could see
14
Section 3.4
An Atomic-Level View of Elements and Compounds
Elements vs Compounds
•
Pure substances can be elements or
compounds (Ch 1)
H2, Ca, Ag
H2O, CaCl2
15
Section 3.4
An Atomic-Level View of Elements and Compounds
Elements vs Compounds
•
Elements and Compounds can be further categorized
based on the types of units that compose them
16
Section 3.4
An Atomic-Level View of Elements and Compounds
Elements vs Compounds
•
Elements can be atomic (exist as individual
atoms) or molecular (oxygen is a diatomic
molecule). There are even polyatomic
elements (next slide).
17
Section 3.4
An Atomic-Level View of Elements and Compounds
Molecular Elements
• H2, N2, O2,
F2, Cl2, Br2
and I2
• P4
• S8, Se8
• You should
learn the 7
diatomic
elements
18
Section 3.4
An Atomic-Level View of Elements and Compounds
Elements vs Compounds
•
•
•
Compounds can be
molecular (H2O) or
ionic (NaCl)
The smallest unit of a
molecular compound
is the molecule
The smallest unit of
an ionic compound is
a formula unit
19
Section 3.4
An Atomic-Level View of Elements and Compounds
Molecules vs Formula Units
• A water molecule is a discrete (individual)
stable unit that consists of three atoms tightly
bound together.
– Covalent bonds are very strong
• A sodium chloride formula unit is the smallest
unit of a sample of NaCl.
– Does not exist on its own as a stable discrete
unit.
– Single ionic bond is very weak
20
Section 3.5 and 3.6
Ionic and Molecular Compounds
Types of Compounds
•
Binary Ionic Compounds
–
Metal—nonmetal
NaCl, CaBr2
• Ionic Compounds with Polyatomic Ions
− Ions with more than one atom
• Hydrated Ionic Compounds
•
NaNO3, NH3Cl
MgSO47 H2O
Binary Covalent Compounds
–
Nonmetal—nonmetal
CO2, H2O, NH3
21
Section 3.5
Ionic Compounds: Formulas and Names
Writing Formulas for Ionic Compounds
• Binary Ionic Compounds form between a metal
and a nonmetal
– Metal forms a cation
– Non metal forms an anion
• All compounds are charge neutral
• For these two reasons it is fairly simple to
reason out the formula of most ionic compounds
just by knowing the identity of the component
elements.
22
Section 3.5
Ionic Compounds: Formulas and Names
Writing Formulas for Ionic Compounds
• If you have sodium (Na) and chlorine (Cl)
– Na forms a Na+ cation
– Cl forms a Cl– anion
– They combine 1:1 to form NaCl
• If you have calcium (Ca) and chlorine (Cl)
– Ca forms a Ca2+ cation
– Cl forms a Cl– anion
– They combine 1:2 to form CaCl2
23
Section 3.5
Ionic Compounds: Formulas and Names
Learning Check
• Write formulas for ionic compounds formed from
the following elements
1. Sodium and Oxygen
2. Calcium and bromine
3. Aluminum and Sulfur
24
Section 3.5
Ionic Compounds: Formulas and Names
Solution
• Write formulas for ionic compounds formed from
the following element
1. Sodium and Oxygen
•
Na+ and O2– combine to form Na2O
2. Calcium and bromine
•
Ca2+ and Br– combine to form CaBr2
3. Aluminum and sulfur
•
Al3+ and S2– combine to form Al2S3
25
Section 3.5
Ionic Compounds: Formulas and Names
Naming Ionic Compounds
• Ionic compounds have common names and
systematic names
• NaCl is sodium chloride or table salt (common
name)
• NaHCO3 is sodium bicarbonate or baking soda
(common name)
• We will learn the systematic naming system
26
Section 3.5
Ionic Compounds: Formulas and Names
Naming Ionic Compounds
• Binary Ionic compounds (metals and non metals)
• Metals are of two types
– Metals that only make one type of cation
• Groups I -3
– Metals that can from more than one type of cation
• Transition Group Metals and Main Groups 4 and above
(basically everything else)
27
Section 3.5
Ionic Compounds: Formulas and Names
Naming Binary Ionic Compounds Containing
a Metal that Forms Only One Type of Cation
1. The cation is always named first and the anion
second.
2. Cation takes its name from the name of the
parent element.
3. Anion is named by taking the root of the
element name and adding –ide.
28
Section 3.5
Ionic Compounds: Formulas and Names
Naming Binary Ionic Compounds Containing
a Metal that Forms Only One Type of Cation
KCl
Potassium chloride
MgBr2
Magnesium bromide
CaO
Calcium oxide
29
Section 3.5
Ionic Compounds: Formulas and Names
Concept Check
Write the names of the following compounds.
A. MgO
B. Al2S3
C. MgF2
30
Section 3.5
Ionic Compounds: Formulas and Names
Solution
Write the names of the following compounds.
A. MgO
magnesium oxide
B. Al2S3
aluminum sulfide
C. MgF2
magnesium fluoride
31
Section 3.5
Ionic Compounds: Formulas and Names
Naming Binary Ionic Compounds Containing
a Metal that Form More than One Type of
Cation
•
Metals in these compounds form more than
one type of positive ion
Charge on the metal ion must be specified.
Roman numeral indicates the charge of the
metal cation.
Transition metals and large metals (period IV
and higher usually require a Roman numeral
•
•
•
•
Exceptions: Ag+, Cd2+, Zn2+ (form only a single ion)
32
Section 3.5
Ionic Compounds: Formulas and Names
Naming Binary Ionic Compounds Containing
a Metal that Form More than One Type of
Cation
1. The cation is always named first and the anion
second.
2. A monatomic cation takes its name from the
name of the parent element.
3. Charge of cation in roman numerals (in
parenthesis)
3. A monatomic anion is named by taking the root
of the element name and adding –ide.
33
Section 3.5
Ionic Compounds: Formulas and Names
Naming Binary Ionic Compounds Containing
a Metal that Form More than One Type of
Cation
CuBr
Copper (I) bromide
FeS
Iron (II) sulfide
PbO2
Lead (IV) oxide
Figure out the charge on the cation based on the
charge of the anion.
34
Section 3.5
Ionic Compounds: Formulas and Names
Learning Check
Name the following ionic compounds containing
metals that form two kinds of positive ions:
A. Fe2O3
B. SnCl2
B. PbI4
35
Section 3.5
Ionic Compounds: Formulas and Names
Solution
Name the following ionic compounds containing
metals that form two kinds of positive ions:
A. Fe2O3 Iron (III) oxide
• Oxygen is –2 x 3 = –6 so we need Fe3+ x 2 = +6
B. SnCl2
Tin (II) chloride
• Cl is –1 x 2 = –2 so we need Sn to be +2
B. PbI4
Lead (IV) iodide
• I is –1 x 4 = –4 so we need Pb to be +4
36
Section 3.5
Ionic Compounds: Formulas and Names
Naming Ionic Compounds Containing
Polyatomic Ions
• Naming these compounds is the same only now
we use the name of the polyatomic ion wherever
it occurs
37
Section 3.5
Ionic Compounds: Formulas and Names
Naming Ionic Compounds Containing
Polyatomic Ions
•
•
Must be memorized (polyatomic ion handout).
Examples of compounds containing polyatomic
ions:
NaOH
Sodium hydroxide
Pb(NO3)2
lead (II) nitrate
(NH4)2SO4
Ammonium sulfate
38
Section 3.5
Ionic Compounds: Formulas and Names
Learning Check
Write the names of the following compounds.
A. NaC2H3O2
B. Ca3(PO4)2
C. CuNO2
D. K2Cr2O7
E. Mg(ClO2)2
F. NH4CN
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
39
Section 3.5
Ionic Compounds: Formulas and Names
Solution
Write the names of the following compounds.
A. NaC2H3O2
sodium acetate
B. Ca3(PO4)2
calcium phosphate
C. CuNO2
copper (I) nitrite
D. K2Cr2O7
potassium dichromate
E. Mg(ClO2)2
magnesium chlorite
F. NH4CN
ammonium cyanide
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
40
Section 3.5
Ionic Compounds: Formulas and Names
Hydrated Ionic Compounds
• Some ionic compounds are hydrates
• This means they have a specific number of
water molecules associated with each formula
unit in the lattice structure
• An example is magnesium sulfate heptahydrate
• MgSO47H2O
41
Section 3.5
Ionic Compounds: Formulas and Names
Hydrated Ionic Compounds
• Cobalt chloride hexahydrate before and after
heating to drive off the water
42
Section 3.5
Ionic Compounds: Formulas and Names
Common hydrate prefixes
•
•
•
•
•
•
•
•
•
Hemi = ½
Mono = 1
Di = 2
Tri = 3
Tetra = 4
Penta = 5
Hexa = 6
Hepta = 7
Octa = 8
These prefixes are also used in
naming molecular compounds
For example: carbon dioxide
43
Section 3.6
Molecular Compounds: Formulas and Names
Molecular compounds vs Ionic Compounds
• Ionic compounds form between metal and non metal
– The ions in ionic compounds can only combine one way
because compounds are charge neutral
– Na+ and Cl– can only combine one way
• Molecular compounds form between two nonmetals
– The atoms in molecular compounds don’t form ions – they
share electrons. For this reason the same combination of
elements can form a number of different molecular
compounds.
44
Section 3.6
Molecular Compounds: Formulas and Names
Naming Molecular Compounds
1. The first element in the formula is named first,
using the full element name.
2. The second element is named as if 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.
45
Section 3.6
Molecular Compounds: Formulas and Names
Naming Molecular Compounds
Prefixes Used to
Indicate Number in
Chemical Names
46
Section 3.6
Molecular Compounds: Formulas and Names
Naming Molecular Compounds
CO2
Carbon dioxide
SF6
Sulfur hexafluoride
N2O4
Dinitrogen tetroxide
47
Section 3.6
Molecular Compounds: Formulas and Names
Learning Check
Write the name of each covalent compound.
A. SeO
B. NO2
C. PF3
D. CBr4
E. P2O5
48
Section 3.6
Molecular Compounds: Formulas and Names
Solution
Write the name of each covalent compound.
A. SeO
Selenium monoxide
B. NO2
Nitrogen dioxide
C. PF3
Phosphorus trifluoride
D. CBr4
Carbon tetrabromide
E. P2O5
Diphosphorus pentoxide
49
Section 3.6
Molecular Compounds: Formulas and Names
Conceptual Connection
• The compound NCl3 is nitrogen trichloride , but
AlCl3 is just aluminum chloride. Why?
50
Section 3.6
Molecular Compounds: Formulas and Names
Solution
• The compound NCl3 is nitrogen trichloride , but
AlCl3 is just aluminum chloride. Why?
• NCl3 is a covalent (molecular compound). Since
nitrogen and chlorine can combine more than
one way it is necessary to indicate the number of
chlorines.
• AlCl3 is an ionic compound. There is only one
combination of these two ions.
51
Section 3.6
Molecular Compounds: Formulas and Names
Acids
• Acids can be defined as molecular compounds
than release hydrogen ions (H+) when dissolved
in water.
• Acids are composed of hydrogen, usually written
first in the formula, followed by one or more non
metals.
• Two categories of acids for naming
– Binary acids
– Oxy Acids
52
Section 3.6
Molecular Compounds: Formulas and Names
Naming Binary Acids
•
•
Hydrogen plus a non metal, the acid is named
with the prefix hydro– followed by the root of the
anion and the suffix –ic.
Examples:
HCl
Hydro chlor ic acid
HBr
Hydro brom ic acid
H2S
Hydro sulfur ic acid
Notice sulfur is kind of a weirdo sulfur instead of sulf
53
Section 3.6
Molecular Compounds: Formulas and Names
Naming OxyAcids
•
Hydrogen plus an oxyanion (anion containing a
nonmetal and oxygen)


The suffix –ic is added to the root name if anion name ends in –ate.:
HNO3
Nitrate
Nitric acid
H2SO4
Sulfate
Sulfuric acid
HC2H3O2
Acetate
Acetic acid
The suffix –ous is added to the root name if anion name ends in –ite.
HNO2
Nitrite
Nitrous acid
H2SO3
Sulfite
Sulfurous acid
HClO2
Chlorite
Chlorous acid
Notice sulfate is kind of a weirdo sulfur instead of sulf
54
Section 3.6
Molecular Compounds: Formulas and Names
Flow Chart for Naming Acids
55
Section 3.6
Molecular Compounds: Formulas and Names
Learning Check
Name the following acids
a) HF (aq)
b) H2CO3 (aq)
c) HI (aq)
d) HClO2 (aq)
e) H2CrO4 (aq)
56
Section 3.6
Molecular Compounds: Formulas and Names
Solution
Name the following acids
a) HF (aq)
b) H2CO3 (aq)
c) HI (aq)
d) HClO2 (aq)
binary
oxy
binary
oxy
hydrofluoric acid
carbonic acid
hydroiodic acid
chlorous acid
e) H2CrO4 (aq)
oxy
chromic acid
57
Section 3.6
Molecular Compounds: Formulas and Names
58
Section 3.6
Molecular Compounds: Formulas and Names
Learning Check
Name the following compounds
a) KNO3
b) TiO2
c) Sn(OH)4
d) PBr5
e) H2SO3 (aq)
59
Section 3.6
Molecular Compounds: Formulas and Names
Solution
a) KNO3
b) TiO2
c) Sn(OH)4
d) PBr5
e) H2SO3 (aq)
potassium nitrate
titanium (IV) oxide
tin(IV) hydroxide
phosphorus pentabromide
sulfurous acid
60
Section 3.8
Formula Mass and the Mole Concept for Compounds
Formula Mass
• In Chapter 2 we defined the average mass of an
atom as its atomic mass
– Measured in atomic mass units (amu)
• The atomic mass of carbon is 12.01 amu
• Now that we have been introduced to ionic and
molecular compounds we can build on this
definition to something called a Formula Mass
61
Section 3.8
Formula Mass and the Mole Concept for Compounds
Formula Mass
• Formula mass is the average mass of a
molecule (or formula unit)
– Molecular mass and molecular weight are also
common (and interchangeable terms)
• What does it mean though?
• Formula mass is the sum of all the atomic
masses in the chemical formula
62
Section 3.8
Formula Mass and the Mole Concept for Compounds
Formula Mass
Formula Mass of H2O
(2 × 1.008 amu) + 16.00 amu = 18.02 amu
Formula Mass of Ba(NO3)2
137.33 g + (2 × 14.01 g) + (6 × 16.00 g) = 261.35
63
Section 3.8
Formula Mass and the Mole Concept for Compounds
Molar Mass of a Compound
• Also in chapter 2 we saw that the atomic mass in
amu/atom was equal to the molar mass in g/mole.
– This is because the definition for amu and mole are
related to each other.
• The same is true for a compound.
• The formula mass in amu/formula unit = molar
mass in g/mole
64
Section 3.8
Formula Mass and the Mole Concept for Compounds
Molar Mass of a Compound
• So if we know a formula mass for water = 18.02
amu/molecule
– 18.02 amu in one molecule
• We also know that the molar mass for water =
18.02 g/mole.
– 18.02 grams in one mole
65
Section 3.8
Formula Mass and the Mole Concept for Compounds
Using Molar Mass to Count by Weighing
• The fact that amu/molecule is related to g/mole
allows us to count molecules by weighing
them.
• This is the same thing we did when we counted
atoms by weighing them.
• Molar mass – g/mole – is a conversion factor
between mass and numbers of molecules.
66
Section 3.8
Formula Mass and the Mole Concept for Compounds
Example
• Glucose is one of the end products of photosynthesis,
the process that converts CO2 and H2O to complex
carbohydrates. The formula for glucose is C6H12O6.
Determine the molar mass of glucose.
Determine the number of moles in 50.0 g of glucose.
Determine the number of molecules in 50.0 g of glucose.
67
Section 3.8
Formula Mass and the Mole Concept for Compounds
Example
• Glucose is one of the end products of photosynthesis,
the process that converts CO2 and H2O to complex
carbohydrates. The formula for glucose is C6H12O6.
Determine the molar mass of glucose.
(6 x 12.01) + (12 x 1.008) + (6 x 16.00) = 180.16 g/mol
Determine the number of moles in 50.0 g of glucose.
50.0 g x
1mol
180.16 g
 0.278 moles of glucose
Determine the number of molecules in 50.0 g of glucose.
6.022 x 1023 molecules
23
50.0 g x
x
 1.67x 10 molecules of glucose
mol
180.16 g
1mol
68
Section 3.8
Formula Mass and the Mole Concept for Compounds
Learning Check
Ethanol is produced from sugars by yeast in the process
called fermentation. The formula for ethanol is C2H5OH.
Determine the molar mass of ethanol.
Determine the number of moles in 525 g of ethanol.
Determine the number of molecules in 525 g of ethanol.
69
Section 3.8
Formula Mass and the Mole Concept for Compounds
Solution
Ethanol is produced from sugars by yeast in the process
called fermentation. The formula for ethanol is C2H5OH.
Determine the molar mass of ethanol.
(2 x 12.01) + (6 x 1.008) + (1 x 16.00) = 46.07 g/mol
Determine the number of moles in 525 g of ethanol.
525 g x
1mol
46.07 g
 11.4 moles of ethanol
Determine the number of molecules in 525 g of ethanol.
6.022 x 1023 molecules
24
525 g x
x
 6.86x 10 molecules of ethanol
mol
46.07 g
1mol
70
Section 3.9
Composition of Compounds
Mass Percent
• We use something called Mass Percent to
determine the relative amounts of elements in a
compound.
• Mass percent takes the different masses of the
different elements into consideration in
calculating relative amounts.
71
Section 3.9
Composition of Compounds
Mass Percent
• Mass Percent of hydrogen in H2O
Mass of H in 1 mole of H 2O
Mass Percent of H =
x 100%
Mass of 1 mole H 2O
Mass Percent of H =
2.016 g/mol
x 100%  11.19%
18.02 g/mol
72
Section 3.9
Composition of Compounds
Mass Percent (Mass Percent Composition)
• General Equation
Mass of Element X in 1 mole of Compound
Mass Percent of element X =
x100%
Mass of 1 mole of Compound
73
Section 3.9
Composition of Compounds
How is Mass Percent Composition Useful
• A lot of times we need to know how much of one
specific element is present in a compound
• A toxin
• An active ingredient
• Mass Percent (mass percent composition)
allows us to calculate this.
74
Section 3.9
Composition of Compounds
Learning Check
• Determine the mass percent of each element in
calcium carbonate (CaCO3)
75
Section 3.9
Composition of Compounds
Solution
• Determine the mass percent of each element in
calcium carbonate (CaCO3)
40.08 g/mol
Mass Percent of Ca =
x100%  40.04%
100.09 g/mol
12.01 g/mol
Mass Percent of C =
x100%  12.00%
100.09 g/mol
48.00 g/mol
Mass Percent of O =
x100%  47.96%
100.09 g/mol
76
Section 3.9
Composition of Compounds
Mass Percent Composition as a Conversion
Factor
• Mass percent can also be used as a conversion
factor
• From the last problem
• Mass Percent of Oxygen = 47.96%
47.96 g Oxygen
100 g CaCO 3
or
100 g CaCO 3
47.96 g Oxygen
77
Section 3.9
Composition of Compounds
Mass Percent Composition as a Conversion
Factor
• How do we use mass percent as a conversion
factor?
• To calculate the mass of an element in a given
mass of a compound.
78
Section 3.9
Composition of Compounds
Example
• Sodium chloride (table salt) is 39% sodium by
mass. How much sodium chloride is allowed per
the RDA (recommended daily allowance) if a
person is allowed to consume 2.4 g of sodium
per day.
79
Section 3.9
Composition of Compounds
Example
• Sodium chloride (table salt) is 39% sodium by
mass. How much sodium chloride is allowed per
the RDA (recommended daily allowance) if a
person is allowed to consume 2.4 g of sodium
per day.
100 g NaCl
2.4 g Na x
39 g Na
= 6.2 g NaCl
80
Section 3.9
Composition of Compounds
Learning Check
• Calcium carbonate (a common calcium
supplement) is 40.04% calcium by mass. How
many grams of calcium are present in one tablet
of calcium carbonate that has a mass of 500.
mg?
81
Section 3.9
Composition of Compounds
Solution
• Calcium carbonate (a common calcium
supplement) is 40.04% calcium by mass. How
much calcium is present in one tablet of calcium
carbonate that has a mass of 500. mg?
500. mg CaCO3 x
1g
40.04 g Ca
x
= 0.200 g Ca
1000 mg 100 g CaCO3
82
Section 3.9
Composition of Compounds
Conversion Factors from Chemical Formulas
• We just determined the amount of calcium in a
given mass of calcium carbonate using mass
percent as a conversion factor.
• There is actually another way to do this
calculation
• Using the Mole Relationships in the Chemical
Formula
• What does this mean and how do we use it?
83
Section 3.9
Composition of Compounds
Mole Relationships
• Chemical formulas tell us the number of atoms
in a compound
• For a single formula unit of calcium carbonate
(CaCO3)
1 atom Ca
1 formula unit CaCO3
1 atom C
1 formula unit CaCO3
3 atoms O
1 formula unit CaCO3
• But what if we have a mole of CaCO3?
• Mole Relationships for CaCO3
84
Section 3.9
Composition of Compounds
Mole Relationships
• Mole Relationships for CaCO3
• There is a 1:1 relationship between moles of
calcium or carbon atoms and moles of
compound
1 mol Ca
1 mol C
1 mol CaCO 3
1 mol CaCO 3
• There is a 3:1 relationship between moles of
oxygen atoms and moles of compound

3 mol O
1 mol CaCO 3
85
Section 3.9
Composition of Compounds
Conversion Factors from Chemical Formulas
• How do we use mole relationships to determine
composition of a compound?
• The mole relationship is a conversion factor.
• So lets say we want to know how many grams of
calcium are in a 500. mg sample of CaCO3
86
Section 3.9
Composition of Compounds
Conversion Factors from Chemical Formulas
• How many grams of calcium are in a 500. mg
sample of CaCO3
0.500 g CaCO3
g Ca
• We cannot convert directly from grams to grams.
• The relationship between atoms in a compound
is moles to moles
• How do we use the mole:mole conversion to do
this.
87
Section 3.9
Composition of Compounds
Conversion Factors from Chemical Formulas
• How many grams of calcium are in a 500. mg
sample of CaCO3
• Grams of CaCO3 to moles CaCO3
• Moles CaCO3 to moles of Ca
• Moles of Ca to grams of Ca
• GMMG
0.500 g CaCO 3 x
1 mol CaCO 3
1 mol Ca
40.08 g Ca
x
x
= 0.200 g Ca
100.09 g
1 mol CaCO 3
1 mol Ca
88
Section 3.9
Composition of Compounds
Learning Check
• The book uses the compound CCl2F2
(chlorofluorocarbons) for several examples
• Write out the mole relationships for this
compound
89
Section 3.9
Composition of Compounds
Solution
• The book uses the compound CCl2F2
(chlorofluorocarbons) for several examples
• Write out the mole relationships for this
compound
1 mol C
1 mol CCl2 F2
2 mol Cl
2 mol F
1 mol CCl2 F2 1 mol CCl2 F2
90
Section 3.9
Composition of Compounds
Learning Check
• Use the mole relationships from the last problem
to determine how many grams of fluorine (F) are
in 1500 g of the compound CCl2F2.
91
Section 3.9
Composition of Compounds
Solution
• Use the mole relationships from the last problem
to determine how many grams of fluorine (F) are
in 1500 g the compound CCl2F2.
1500 g CCl2 F2 x
1 mol CCl2 F2
2 mol F
19.00 g F
x
x
= 470 g F
120.91 g
1 mol CCl2 F2
1 mol F
92
Section 3.10
Determining a Chemical Formula from Experimental Data
Decomposition Analysis
• How do we determine the formula of a
compound that is newly isolated?
• This is commonly done in the lab when new
chemicals/drugs are synthesized
• Also common when new compounds are
isolated
93
Section 3.10
Determining a Chemical Formula from Experimental Data
Decomposition Analysis
• First step is to decompose the compound into its
constituent parts
• Weigh them
• But this gives us a ratio of masses (in grams)
• Does not really tell us anything about the
relationship of the number of atoms in the
compound
94
Section 3.10
Determining a Chemical Formula from Experimental Data
Decomposition Analysis
•
•
•
•
If we decompose 7.717 g of water (H2O) we get
6.868 g of oxygen
0.857 g of hydrogen
If we divide these it gives an 8:1 ratio of oxygen
to water (well we know that’s not right)
• To get the ratio of atoms in the compound we
have to correct for the different in their
atomic weight
• We have to convert to moles
95
Section 3.10
Determining a Chemical Formula from Experimental Data
Decomposition Analysis
• We have to convert to moles
1 mol H
0.857 g H x
= 0.850 mol H
1.008 g
1 mol O
6.86 g O x
= 0.429 mol O
16.00 g
• Now this looks more like it
• H0.850 O0.429
• Divide by the smaller number = H1.98O = H2O
96
Section 3.10
Determining a Chemical Formula from Experimental Data
Empirical vs Molecular Formula
• A decomposition analysis breaks a complex
compound down to its elements to determine the
relationship between them
• This relationship is called an empirical formula
• The simplest whole number ratio of atoms in the
compound
• The form of the compound before it was broken
down might actually be a multiple of this
empirical formula
• This is called the molecular formula
97
Section 3.10
Determining a Chemical Formula from Experimental Data
Empirical vs Molecular Formula
• To determine the molecular formula you need
the empirical formula and the molecular weight
of the original compound.
• For water the empirical formula and molecular
formula are the same
• Not true for other compounds
• Molecular formula = empirical formula x n
molar mass
n
empirical formula mass
98
Section 3.10
Determining a Chemical Formula from Experimental Data
Example
• The empirical formula for a compound by
decomposition analysis is CH2O and the molar
mass is determined to be 180.2. What is the
molecular formula?
• Molecular formula = (CH2O) x n
molar mass
180.2
n

= 6
empirical formula mass 30.03
• Molecular formula = C6H12O6
99
Section 3.10
Determining a Chemical Formula from Experimental Data
Learning Check
• A compound with the following percent
composition has a molar mass of 60.10 g/mol.
Determine its molecular formula. Assume 100 g
of the compound.
• C, 39.97%
• H, 13.41%
• N, 46.62%
100
Section 3.10
Determining a Chemical Formula from Experimental Data
Solution
1 mol
= 3.33 mol C
12.01 g C
1 mol H
13.41 g H x
= 13.3 mol H
1.008 g H
1 mol N
46.62 g N x
= 3.33 mol N
14.01 g N
39.97 g C x
C 3.33H13.3N 3.33 CH 4 N = empirical formula
Molecular formula = empirical formula x n
molar mass
60.10
n
=
= 2
empirical formula mass
30.05
molecular formula = (CH 4 N) x 2 = C 2H 8N 2
101
Section 3.10
Determining a Chemical Formula from Experimental Data
Combustion Analysis
• One way to determine hydrogen and carbon
content of a compound is to react it with oxygen
then collect the H2O and CO2 produced
102
Section 3.10
Determining a Chemical Formula from Experimental Data
Combustion Analysis
•
•
•
How can this information be used to determine a formula?
If a substance has been isolated that contains only C, H and N.
If 0.1156 g of this sample is reacted with oxygen and 0.1638 g of CO2
and 0.1676 g of H2O are collected.
0.1638 g CO2 x
12.01 g C
44.01 g CO2
 0.04470 g C
0.1676 g H2O x
2.016 g H
18.02 g H2O
 0.01875 g H
•
•
The nitrogen is what is left over
0.1156 – (0.04470 + 0.01875)
 = 0.05215 g N
•
To determine formulas though we need numbers of atoms and grams
don’t really help us with that
103
Section 3.10
Determining a Chemical Formula from Experimental Data
Combustion Analysis
0.04470 g C x
0.01875 g H x



0.05215 g N x
•
•
1 mole C
12.01 g C
1 mole
1.008 g H
1 mole
14.01 g N
 0.00372 mole C
 0.0186 mole H
 0.00372 mole N
Ratio of moles is the same as ratio of atoms (Section 3.3 The
Mole)
Divide by the smallest value
104
Section 3.10
Determining a Chemical Formula from Experimental Data
Combustion Analysis
•
So we need to convert grams to moles
0.04470 g C x
0.01875 g H x
0.05215 g N x
•
1 mole C
12.01 g C
1 mole
1.008 g H
1 mole
14.01 g N
 0.00372 mole C/0.00372 = 1
 0.0186 mole H/0.00372 = 5
 0.00372 mole N/0.00372 = 1
Empirical Formula = C1H5N1
105
Section 3.11
Writing and Balancing Chemical Equations
How to Write Balanced Chemical Equations
• Chemical equations include the states of the
reactants and products
• CH4 (g) + O2 (g) → CO2 (g) + H2O (g)
• All of the reactants and products in this
reaction are gases
106
Section 3.11
Writing and Balancing Chemical Equations
States of reactants and products
• Notice the
difference
between (l)
liquid and (aq)
aqueous
• Aqueous is
something
dissolved in
water.
107
Section 3.11
Writing and Balancing Chemical Equations
How to Write Balanced Chemical Equations
• Balanced equation
• CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
108
Section 3.11
Writing and Balancing Chemical Equations
Tips for balancing
• Balance polyatomic ions as a group
• Always balance oxygen last
• Can used fractions to balance – then multiply
them out at the end
• Practice, practice, and practice.
109
Section 3.11
Writing and Balancing Chemical Equations
Example
Write a balanced equation for the reaction of
aqueous lead (II) nitrate with aqueous sodium
phosphate to produce solid lead (II) phosphate and
aqueous sodium nitrate.
Pb(NO3)2 (aq) + Na3PO4 (aq) → Pb3(PO4)2 (s) + NaNO3 (aq)
3 Pb(NO3)2 (aq) + 2 Na3PO4 (aq) → Pb3(PO4)2 (s) + 6 NaNO3
(aq)
Section 3.11
Writing and Balancing Chemical Equations
Learning Check
Write a balanced equation for the reaction of
ammonia (NH3) with the oxygen to produce
nitrogen monoxide and water
Section 3.11
Writing and Balancing Chemical Equations
Solution
Write a balanced equation for the reaction of
ammonia (NH3) with the oxygen to produce
nitrogen monoxide and water
NH3 + O2 → NO + H2O
4NH3 + 5O2 → 4NO + 6H2O
Section 3.11
Writing and Balancing Chemical Equations
Learning Check
Write a balanced equation for the reaction of
aqueous cobalt (I) nitrate with sulfuric acid to
produce solid cobalt (I) sulfate and nitric acid. Both
acids are aqueous solutions
Section 3.11
Writing and Balancing Chemical Equations
Solution
Write a balanced equation for the reaction of
aqueous cobalt (I) nitrate with sulfuric acid to
produce solid cobalt (I) sulfate and nitric acid. Both
acids are aqueous solutions
CoNO3 (aq) + H2SO4 (aq) → Co2SO4 (s) + HNO3 (aq)
2 CoNO3 (aq) + H2SO4 (aq) → Co2SO4 (s) + 2 HNO3 (aq)
Section 3.11
Writing and Balancing Chemical Equations
Conceptual Connection
• Which quantity or quantities must always be the
same on both sides of a chemical equation?
• (a) the number of atoms of each kind
• (b) the number of molecules of each kind
• (c) the number of moles of each kind of
molecule
• (d) the sum of the masses of all the substances
involved
115
Section 3.11
Writing and Balancing Chemical Equations
Solution
• Which quantity or quantities must always be the
same on both sides of a chemical equation?
• (a) the number of atoms of each kind
• (b) the number of molecules of each kind
• (c) the number of moles of each kind of
molecule
• (d) the sum of the masses of all the
substances involved
116
Section 3.12
Organic Compounds
Organic vs Inorganic Compounds
• Early chemists divided compounds into two
types
• Organic - Originated from living things
– sugars
• Inorganic - Originated from the earth
– salts
117
Section 3.12
Organic Compounds
Organic vs Inorganic Compounds
• Organic and inorganic compounds have different
chemical properties
• Organic – Easy to decompose
– Very hard to synthesize – very complex
• Inorganic –
– Hard to decompose
– Easier to synthesize
118
Section 3.12
Organic Compounds
What are organic compounds
• Major components of living organisms
– Proteins, fats, DNA
•
•
•
•
Smells, tastes, fuels
Drugs
Food
Pretty much everything we consume
119
Section 3.12
Organic Compounds
What are organic compounds made of?
• Organic compounds are composed primarily of
carbon and hydrogen with a few other elements
– Nitrogen, oxygen and sulfur
• Key element in an organic compound is carbon
120
Section 3.12
Organic Compounds
What are organic compounds made of?
• Carbon makes 4 bonds and often bonds to itself
to make chain, branched and ring structures.
121
Section 3.12
Organic Compounds
Hydrocarbons
• Organic compounds that contain only carbon
and hydrogen.
• These are all the common fuels that we use
• Gasoline, propane, natural gas
• Can have single, double or triple bonds
– Alkane – single bonds
– Alkenes – double bonds
– Alkynes – triple bonds
122
Section 3.12
Organic Compounds
Naming Hydrocarbons
• Base name determined by a prefix that indicates
the number of carbons
– Meth = 1, eth = 2, prop = 3, but = 4, pent = 5
– hex = 6, hept = 7, oct = 8, non = 9, dec = 10
• Suffix indicates the presence of a multiple bond
– Single bonds - ane
– Double bond - ene
– Triple bond - yne
123
Section 3.12
Organic Compounds
Naming Hydrocarbons
• Butane CH3CH2CH2CH3
• Butene CH3CH2CH=CH2
124
Section 3.12
Organic Compounds
Common Hydrocarbons
125
Section 3.12
Organic Compounds
Learning Check
• Write the name for the following hydrocarbons
• CH3CH2CH2CH2CH2CH3
126
Section 3.12
Organic Compounds
Solution
• Write the formula or the name for the following
hydrocarbons
• CH3CH2CH2CH2CH2CH3
6 carbons hexane
7 carbons with
double bond
heptene
127
Section 3.12
Organic Compounds
Functionalized Hydrocarbons
• A functionalized hydrocarbon is a carbon
hydrogen chain that has a functional group
added to it.
• What is a functional group?
• A functional group is a group of atoms that
change the characteristics of the hydrocarbon.
128
Section 3.12
Organic Compounds
Functionalized Hydrocarbons
• Example of a functionalized hydrocarbon.
• Ethane
• Ethanol
129
Section 3.12
Organic Compounds
Functional Groups
Ethanoic Acid *
*Acetic Acid
(Vinegar)
130
Section 3.12
Organic Compounds
Learning Check
• For the following compounds, give the family of
the functional group (try to name the
compound!)
131
Section 3.12
Organic Compounds
Solution
• For the following compounds, give the family of
the functional group
Alcohol - Propanol
Carboxylic
Acid - Butyric Acid
Amine
Ethyl amine
132
Section 3.12
Organic Compounds
Suggested Homework
• Chapter 3
• Review Questions 1 – 5, 8, 14 – 22
• Odd numbered problems 23 – 118 (skip 25, 63,
69, 73, 79, 87, 89, 97)
133