Chapter 5
Molecules
and
Compounds
Molecules and Compounds
• Salt
 Sodium – shiny, reactive, poisonous
 Chlorine – pale yellow gas, reactive,
poisonous
 Sodium chloride – table salt
• Sugar
 Carbon – pencil or diamonds
 Hydrogen – flammable gas
 Oxygen – a gas in air
 Combine to form white crystalline
sugar
2
Law of Constant Composition
• all pure substances have constant composition
all samples of a pure substance contain the same
elements in the same percentages (ratios)
mixtures have variable composition
3
Compounds Display
Constant Composition
If we decompose water by electrolysis, we find 16.0
grams of oxygen to every 2.00 grams of hydrogen.
Water has a constant Mass Ratio of Oxygen to
Hydrogen of 8.0.
mass of oxygen
Mass Ratio 
mass of hydrogen
16.0 g

 8.0
2.0 g
4
Why do Compounds Show
Constant Composition
• smallest piece of a compound is called
a molecule
• every molecule of a compound has the
same number and type of atoms as
determined by the electronic structures
of the atoms (more on that later in the
year)
• since all the molecules of a compound
are identical…
 every sample will have the same ratio of
the elements
 every sample of the compound will have
the same properties
5
EXAMPLE 5.1 Constant Composition of Compounds
Two samples of carbon dioxide, obtained from different sources, were decomposed into their constituent
elements. One sample produced 4.8 g of oxygen and 1.8 g of carbon, and the other sample produced 17.1 g
of oxygen and 6.4 g of carbon. Show that these results are consistent with the law of constant composition.
To show this, compute the
mass ratio of one element to
the other by dividing the
larger mass by the smaller
one. For the first sample:
Solution:
For the second sample:
Since the ratios are the same for the two samples, these results are consistent with the law of constant
composition.
6
5.3A What is the carbon-hydrogen mass ratio for
methane (CH4)?
1.
2.
3.
4.
5.
1
0.33
3
4
0.25
7
5.1B
1.
2.
3.
4.
5.
If the mass ratio of lead(II) sulfide is 270.0 g lead
and 41.8 g sulfur, how much lead is required to
completely react with 85.6 g of sulfur?
13.3
185
228
312
553
8
Formulas Describe Compounds
water = H2O \ two atoms of
hydrogen and 1 atom of oxygen
table sugar = C12H22O11 \12 atoms
of C, 22 atoms of H and 11 atoms O
9
Order of Elements in a Formula
• metals written first
NaCl
• nonmetals written in order from
Table 5.1
CO2
are occasional exceptions for
historical or informational reasons
H2O, but NaOH
Table 5.1
Order of Listing Nonmetals
in Chemical Formulas
C P N H S
I Br Cl O F
10
Molecules with
Polyatomic Ions
symbol of the polyatomic
ion called nitrate
symbol of the polyatomic
ion called sulfate
Mg(NO3)2
CaSO4
compound called
magnesium nitrate
compound called
calcium sulfate
implied “1” subscript
on magnesium
parentheses to group two NO3’s
implied “1” subscript
on calcium
no parentheses for one SO4
11
Molecules with
Polyatomic Ions
subscript indicating
two NO3 groups
no subscript indicating
one SO4 group
Mg(NO3)2
CaSO4
compound called
magnesium nitrate
compound called
calcium sulfate
implied “1” subscript
on nitrogen, total 2 N
stated “3” subscript
on oxygen, total 6 O
implied “1” subscript
on sulfur, total 1 S
stated “4” subscript
on oxygen, total 4 O
12
Classifying Materials
• atomic elements = elements whose
particles are single atoms
• molecular elements = elements whose
particles are multi-atom molecules
• molecular compounds = compounds
whose particles are molecules made
of only nonmetals
• ionic compounds = compounds whose
particles are cations and anions
13
Molecular Elements
• Certain elements occur as 2 atom molecules
• Rule of 7’s
 there are 7 common diatomic elements
 find the element with atomic number 7, N
 make a figure 7 by going over to Group 7A, then down
 don’t forget to include H2
VIIA
H2
N2
7
P4
O2
F2
S8
Cl2
Br2
I2
14
Molecular Compounds
• two or more
nonmetals
• smallest unit is a
molecule
15
Ionic Compounds
• metals + nonmetals
• no individual
molecule units, instead
have a 3-dimensional
array of cations and
anions made of
formula units
16
Molecular View of
Elements and Compounds
17
Classify each of the following as either an
atomic element, molecular element, molecular
compound or ionic compound
• aluminum, Al
• aluminum chloride, AlCl3
• chlorine, Cl2
• acetone, C3H6O
• carbon monoxide, CO
• cobalt, Co
18
Classify each of the following as either an
atomic element, molecular element, molecular
compound or ionic compound
• aluminum, Al = atomic element
• aluminum chloride, AlCl3 = ionic compound
• chlorine, Cl2 = molecular element
• acetone, C3H6O = molecular compound
• carbon monoxide, CO = molecular compound
• cobalt, Co = atomic element
19
Formula-to-Name
Step 1
Is the compound one of the
exceptions to the rules?
•H2O = water, steam, ice
•NH3 = ammonia
Formula-to-Name
Step 2
What major class of compound is it?
Ionic or Molecular
Major Classes
• Ionic
 metal + nonmetal
 metal first in formula
 Binary Ionic
 compounds with polyatomic ions
• Molecular
 2 nonmetals
 Binary Molecular (or Binary Covalent)
 Acids – formula starts with H
 though acids are molecular, they behave as ionic when
dissolved in water
 may be binary or oxyacid
22
Formula-to-Name
Step 3
What major subclass of compound is it?
Binary Ionic, Ionic with Polyatomic Ions,
Binary Molecular,
Binary Acid, Oxyacid
Classifying Compounds
• Compounds containing a metal and a nonmetal =
binary ionic
 Type I and II
• Compounds containing a polyatomic ion =
ionic with polyatomic ion
• Compounds containing two nonmetals =
binary molecular compounds
• Compounds containing H and a nonmetal =
binary acids
• Compounds containing H and a polyatomic ion =
oxyacids
24
Formula-to-Name
Step 4
Apply Rules for the Class and Subclass
Formula-to-Name
Rules for Ionic
• Made of cation and anion
• Name by simply naming the ions
If cation is:
Type I metal = metal name
Type II metal = metal name(charge)
Polyatomic ion = name of polyatomic ion
If anion is:
Nonmetal = stem of nonmetal name + ide
Polyatomic ion = name of polyatomic ion
26
Monatomic Nonmetal Anion
• determine the charge from position on the
Periodic Table
• to name anion, change ending on the element
name to –ide
4A = -4
5A = -3
6A = -2
7A = -1
C = carbide
N = nitride
O = oxide
F = fluoride
S = sulfide
Cl = chloride
Si = silicide P = phosphide
27
Metal Cations
• Type I
 metals whose ions can only have one
possible charge
 IA, IIA, (Al, Ga, In)
 determine charge by position on the
Periodic Table
 IA = +1, IIA = +2, (Al, Ga, In = +3)
• Type II
 metals whose ions can have more than How do you know a
one possible charge
metal cation is Type II?
 determine charge by charge on anion
its not Type I !!!
28
Determine if the following metals are Type I
or Type II. If Type I, determine the charge on
the cation it forms.
•
•
•
•
•
lithium, Li
copper, Cu
gallium, Ga
tin, Sn
strontium, Sr
29
Determine if the following metals are Type I
or Type II. If Type I, determine the charge on
the cation it forms.
•
•
•
•
•
lithium, Li
copper, Cu
gallium, Ga
tin, Sn
strontium, Sr
Type I
Type II
Type I
Type II
Type I
+1
+3
+2
30
Type I Binary Ionic Compounds
• Contain Metal Cation + Nonmetal Anion
• Metal listed first in formula & name
1. name metal cation first, name nonmetal anion
second
2. cation name is the metal name
3. nonmetal anion named by changing the ending
on the nonmetal name to -ide
31
Type II Binary Ionic Compounds
•
•
Contain Metal Cation + Nonmetal Anion
Metal listed first in formula & name
1. name metal cation first, name nonmetal anion second
2. metal cation name is the metal name followed by a
Roman Numeral in parentheses to indicate its charge
 determine charge from anion charge
 Common Type II cations in Table 5.5
3. nonmetal anion named by changing the ending on the
nonmetal name to -ide
32
Examples
•
•
•
•
•
•
•
LiCl = lithium chloride
AlCl3 = aluminum chloride
PbO = lead(II) oxide
PbO2 = lead(IV) oxide
Mn2O3 = manganese(III) oxide
ZnCl2 = zinc(II) chloride or zinc chloride
AgCl = silver(I) chloride or silver chloride
33
Compounds Containing
Polyatomic Ions
• Polyatomic ions are single ions that contain
more than one atom
• Name any ionic compound by naming cation
first and then anion
Non-polyatomic cations named like Type I and II
Non-polyatomic anions named with -ide
34
Fixed Charge Metals and Nonmetals
IA
IIA
IIIA
Li+1 Be+2
Na+1 Mg+2
K+1 Ca+2
Rb+1 Sr+2
Al+3
Zn+2 Ga+3
Ag+1 Cd+2 In+3
VA VIA VIIA
N-3 O-2
F-1
P-3 S-2
Cl-1
As-3 Se-2 Br-1
Te-2 I-1
Cs+1 Ba+2
35
Some Common Polyatomic Ions
Name
Formula
Name
Formula
acetate
carbonate
hydrogen carbonate
(aka bicarbonate)
hydroxide
nitrate
nitrite
chromate
dichromate
ammonium
C2H3O2–
CO32–
hypochlorite
chlorite
chlorate
perchlorate
sulfate
sulfite
hydrogensulfate
(aka bisulfate)
hydrogensulfite
(aka bisulfite)
ClO–
ClO2–
ClO3–
ClO4–
SO42–
SO32–
HCO3–
OH–
NO3–
NO2–
CrO42–
Cr2O72–
NH4+
HSO4–
HSO3–
36
Patterns for Polyatomic Ions
1. elements in the same column form similar
polyatomic ions
 same number of O’s and same charge
ClO3- = chlorate \ BrO3- = bromate
2. if the polyatomic ion starts with H, the name
adds hydrogen- prefix before name and add
1 to the charge
CO32- = carbonate \ HCO3-1 = hydrogencarbonate
37
Periodic Pattern of Polyatomic Ions
-ate groups
IIIA
-3
BO3
IVA
VA
VIA
VIIA
-2
CO3
-1
NO3
-2
SiO3
-3
PO4
-2
SO4
-1
ClO3
-3
AsO4
-2
SeO4
-1
BrO3
-2
TeO 4
-1
IO3
38
Binary Molecular Compounds
of 2 Nonmetals
1. Name first element in formula first
 use the full name of the element
2. Name the second element in the formula with an -ide
 as if it were an anion, however, remember these compounds
do not contain ions!
3. Use a prefix in front of each name to indicate the
number of atoms
a) Never use the prefix mono- on the first element
39
Subscript - Prefixes
•
•
•
•
•
•
•
•
•
•
1 = mono 2 = di3 = tri4 = tetra5 = penta6 = hexa7 = hepta8 = octa9 = nona10 = deca-
not used on first nonmetal
drop last “a” if name
begins with vowel
40
Acids
• Contain H+1 cation and anion
in aqueous solution
• Binary acids have H+1 cation
and nonmetal anion
• Oxyacids have H+1 cation
and polyatomic anion
41
Formula-to-Name
Acids
• acids are molecular compounds that often behave
like they are made of ions
• All names have acid at end
• Binary Acids = hydro prefix + stem of the name of
the nonmetal + ic suffix
• Oxyacids
 if polyatomic ion ends in –ate = name of polyatomic
ion with –ic suffix
 if polyatomic ion ends in –ite = name of polyatomic ion
with –ous suffix
42
Example – Naming Binary Acids
HCl
1. Is it one of the common exceptions?
H2O, NH3, CH4, NaCl, C12H22O11 = No!
2. Identify Major Class
first element listed is H, \ Acid
3. Identify the Subclass
2 elements, \ Binary Acid
43
Sample - Naming Binary
Acids – HCl
4. Identify the anion
Cl = Cl-, chloride because Group 7A
5. Name the anion with an –ic suffix
Cl- = chloride  chloric
6. Add a hydro- prefix to the anion name
hydrochloric
7. Add the word acid to the end
hydrochloric acid
44
Example – Naming Oxyacids
H2SO4
1. Is it one of the common exceptions?
H2O, NH3, CH4, NaCl, C12H22O11 = No!
2. Identify Major Class
first element listed is H, \ Acid
3. Identify the Subclass
3 elements in the formula, \ Oxyacid
45
Example – Naming Oxyacids
H2SO4
4. Identify the anion
SO4 = SO42- = sulfate
5. If the anion has –ate suffix, change it to –ic. If the
anion has –ite suffix, change it to -ous
SO42- = sulfate  sulfuric
6. Write the name of the anion followed by the word acid
sulfuric acid
(kind of an exception, to make it sound nicer!)
46
Example – Naming Oxyacids
H2SO3
1. Is it one of the common exceptions?
H2O, NH3, CH4, NaCl, C12H22O11 = No!
2. Identify Major Class
first element listed is H, \ Acid
3. Identify the Subclass
3 elements in the formula, \ Oxyacid
47
Example – Naming Oxyacids
H2SO3
4. Identify the anion
SO3 = SO32- = sulfite
5. If the anion has –ate suffix, change it to –ic. If
the anion has –ite suffix, change it to -ous
SO32- = sulfite  sulfurous
6. Write the name of the anion followed by the
word acid
sulfurous acid
48
Formula-to-Name Flow Chart
49
Name – to – Formula
Writing the Formulas from the Names
•
•
For binary molecular compounds, use the
prefixes to determine the subscripts
For Type I, Type II, Ternary Compounds
and Acids
1. Determine the ions present
2. Determine the charges on the cation and anion
3. Balance the charges to get the subscripts
51
Example – Binary Molecular
dinitrogen pentoxide
• Identify the symbols of the elements
nitrogen = N
oxide = oxygen = O
• Write the formula using prefix number for
subscript
di = 2, penta = 5
N2O5
52
Compounds that Contain Ions
• compounds of metals with nonmetals are made
of ions
metal atoms form cations, nonmetal atoms for anions
• compound must have no total charge, therefore
we must balance the numbers of cations and
anions in a compound to get 0 charge
• if Na+ is combined with S2-, you will need 2 Na+
ions for every S2- ion to balance the charges,
therefore the formula must be Na2S
53
Writing Formulas for
Ionic Compounds
1.
2.
3.
4.
5.
Write the symbol for the metal cation and its charge
Write the symbol for the nonmetal anion and its charge
Charge (without sign) becomes subscript for other ion
Reduce subscripts to smallest whole number ratio
Check that the sum of the charges of the cation cancels
the sum of the anions
54
Write the formula of a compound made from
aluminum ions and oxide ions
1.
2.
3.
4.
5.
Write the symbol for the metal
cation and its charge
Write the symbol for the
nonmetal anion and its charge
Charge (without sign) becomes
subscript for other ion
Reduce subscripts to smallest
whole number ratio
Check that the total charge of
the cations cancels the total
charge of the anions
Al+3 column IIIA
O2- column VIA
Al+3 O2Al2 O3
Al = (2)∙(+3) = +6
O = (3)∙(-2) = -6
55
Practice - What are the formulas for
compounds made from the following ions?
• potassium ion with a nitride ion
• calcium ion with a bromide ion
• aluminum ion with a sulfide ion
56
Practice - What are the formulas for
compounds made from the following ions?
• K+ with N3-
K3N
• Ca+2 with Br-
CaBr2
• Al+3 with S2-
Al2S3
57
Example – Ionic Compounds
manganese(IV) sulfide
1.
2.
3.
4.
5.
Write the symbol for the cation
and its charge
Write the symbol for the anion
and its charge
Charge (without sign) becomes
subscript for other ion
Reduce subscripts to smallest
whole number ratio
Check that the total charge of
the cations cancels the total
charge of the anions
Mn+4
S2Mn+4 S2-
Mn2S4
MnS2
Mn = (1)∙(+4) = +4
S = (2)∙(-2) = -4
58
Example – Ionic Compounds
Iron(III) phosphate
1.
2.
3.
4.
5.
Write the symbol for the cation
and its charge
Write the symbol for the anion
and its charge
Charge (without sign) becomes
subscript for other ion
Reduce subscripts to smallest
whole number ratio
Check that the total charge of
the cations cancels the total
charge of the anions
Fe+3
PO43Fe+3 PO43- Fe3(PO4)3
FePO4
Fe = (1)∙(+3) = +3
PO4 = (1)∙(-3) = -3
59
Example – Ionic Compounds
ammonium carbonate
1.
2.
3.
4.
5.
Write the symbol for the cation
and its charge
Write the symbol for the anion
and its charge
Charge (without sign) becomes
subscript for other ion
Reduce subscripts to smallest
whole number ratio
Check that the total charge of
the cations cancels the total
charge of the anions
NH4+
CO32NH4+ CO32- (NH4)2CO3
(NH4)2CO3
NH4 = (2)∙(+1) = +2
CO3 = (1)∙(-2) = -2
60
Practice - What are the formulas for
compounds made from the following ions?
• copper(II) ion with a nitride ion
• iron(III) ion with a bromide ion
• aluminum ion with a sulfate ion
61
Practice - What are the formulas for
compounds made from the following ions?
• Cu2+ with N3-
Cu3N2
• Fe+3 with Br-
FeBr3
• Al+3 with SO42-
Al2(SO4)3
62
Example – Binary Acids
hydrosulfuric acid
1.
2.
3.
4.
5.
Write the symbol for the cation
and its charge
Write the symbol for the anion
and its charge
Charge (without sign) becomes
subscript for other ion
Reduce subscripts to smallest
whole number ratio
Check that the total charge of
the cations cancels the total
charge of the anions
H+
S2H+ S2-
in all acids the
cation is H+
hydro means
binary
H2S
H 2S
H = (2)∙(+1) = +2
S = (1)∙(-2) = -2
63
Example – Oxyacids
carbonic acid
1.
2.
3.
4.
5.
Write the symbol for the cation
and its charge
Write the symbol for the anion
and its charge
Charge (without sign) becomes
subscript for other ion
Reduce subscripts to smallest
whole number ratio
Check that the total charge of
the cations cancels the total
charge of the anions
H+
CO32-
in all acids the
cation is H+
no hydro means
polyatomic ion
-ic means -ate ion
H+ CO32-
H2CO3
H2CO3
H = (2)∙(+1) = +2
CO3 = (1)∙(-2) = -2
64
Practice - What are the formulas for the
following acids?
• chlorous acid
• phosphoric acid
• hydrobromic acid
65
Practice - What are the formulas for the
following acids?
• H+ with ClO2–
HClO2
• H+ with PO43–
H3PO4
• H+ with Br–
HBr
66
Formula Mass
• the mass of an individual molecule or
formula unit
• also known as molecular mass or molecular
weight
• sum of the masses of the atoms in a single
molecule or formula unit
whole = sum of the parts!
mass of 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu
67
EXAMPLE 5.1 Constant Composition of Compounds
Two samples of carbon dioxide, obtained from different sources, were decomposed into their constituent
elements. One sample produced 4.8 g of oxygen and 1.8 g of carbon, and the other sample produced 17.1 g
of oxygen and 6.4 g of carbon. Show that these results are consistent with the law of constant composition.
To show this, compute the
mass ratio of one element to
the other by dividing the
larger mass by the smaller
one. For the first sample:
Solution:
For the second sample:
Since the ratios are the same for the two samples, these results are consistent with the law of constant
composition.
68
EXAMPLE 5.2 Writing Chemical Formulas
Write a chemical formula for each of the following:
(a) the compound containing two aluminum atoms to every three oxygen atoms
(b) the compound containing three oxygen atoms to every sulfur atom
(c) the compound containing four chlorine atoms to every carbon atom
Since aluminum is the metal,
it is listed first.
Solution:
(a)
Al2O3
Since sulfur is below oxygen
on the periodic table and since
it occurs before oxygen in
Table 5.1, it is listed first.
(b)
SO3
Since carbon is to the left of
chlorine on the periodic table
and since it occurs before
chlorine in Table 5.1, it is
listed first.
(c)
CCl4
69
EXAMPLE 5.3 Total Number of Each Type of Atom in a
Chemical Formula
Determine the number of each type of atom in Mg3(PO4)2.
Solution:
Mg: There are three Mg atoms, as indicated by the subscript 3.
P:
There are two P atoms, as we see by multiplying the subscript outside the parentheses (2) by
the subscript for P inside the parentheses, which is 1 (implied).
O: There are eight O atoms, as we see by multiplying the subscript outside the parentheses (2) by
the subscript for O inside the parentheses (4).
70
EXAMPLE 5.4 Classifying Substances as Atomic Elements, Molecular
Elements, Molecular Compounds, or Ionic Compounds
Classify each of the following substances as an atomic element, molecular element, molecular
compound, or ionic compound.
(a) krypton
(b) CoCl2
(c) nitrogen
(d) SO2
(e) KNO3
Solution:
(a) Krypton is an element that is not listed as diatomic in Table 5.2; therefore, it is an atomic
element.
(b) CoCl2 is a compound composed of a metal (left side of periodic table) and nonmetal (right side
of the periodic table); therefore, it is an ionic compound.
(c) Nitrogen is an element that is listed as diatomic in Table 5.2; therefore, it is a molecular
element.
(d) SO2 is a compound composed of two nonmetals; therefore, it is a molecular compound.
(e) KNO3 is a compound composed of a metal and two nonmetals; therefore, it is an ionic
compound.
71
EXAMPLE 5.5 Writing Formulas for Ionic Compounds
Write a formula for the ionic compound that forms from aluminum and oxygen.
1. Write the symbol for the metal and
its charge followed by the symbol of
the nonmetal and its charge. For
many elements, these charges can
be determined from their group
number in the periodic table (refer to
Figure 4.14).
Solution:
Al3+
O2–
2. Make the magnitude of the charge
on each ion (without the sign)
become the subscript for the other
ion.
3. Reduce the subscripts to give a
ratio with the smallest whole
numbers.
4. Check that the sum of the charges
of the cations exactly cancels the
sum of the charges of the anions.
In this case, the numbers cannot be
reduced any further; the correct formula
is Al2O3.
Cations: 2(3+) = 6+
Anions: 3(2–) = 6–
The charges cancel.
72
EXAMPLE 5.6 Writing Formulas for Ionic Compounds
Write a formula for the ionic compound that forms from magnesium and oxygen.
1. Write the symbol for the metal and
its charge followed by the symbol of
the nonmetal and its charge. For
many elements, these charges can
be determined from their group
number in the periodic table (refer to
Figure 4.14).
Solution:
Mg2+
O2–
2. Make the magnitude of the charge
on each ion (without the sign)
become the subscript for the other
ion.
3. Reduce the subscripts to give a
ratio with the smallest whole
numbers.
4. Check that the sum of the charges
of the cations exactly cancels the
sum of the charges of the anions.
To reduce the subscripts, divide both
subscripts by 2.
Mg2O2 ÷ 2 = MgO
Cations: 2+
Anions: 2–
The charges cancel.
73
EXAMPLE 5.7 Writing Formulas for Ionic Compounds
Write a formula for the compound that forms from potassium and oxygen.
Solution:
We first write the symbol for each ion along with its appropriate charge from its group number in the
periodic table.
K+
O2–
We then make the magnitude of each ion’s charge become the subscript for the other ion.
K+
O2– becomes K2O
No reducing of subscripts is necessary in this case. Finally, we check to see that the sum of the
charges of the cations [2(1+) = 2+] exactly cancels the sum of the charges of the anion (2–). The
correct formula is K2O.
74
EXAMPLE 5.8 Naming Type I Ionic Compounds
Give the name for the compound MgF2.
Solution:
The cation is magnesium. The anion is fluorine, which becomes fluoride. The correct name is
magnesium fluoride.
75
EXAMPLE 5.9 Naming Type II Ionic Compounds
Give the name for the compound PbCl4.
Solution:
The name for PbCl4 consists of the name of the cation, lead, followed by the charge of the cation in
parenthesis (IV), followed by the base name of the anion, chlor-, with the ending -ide. The full name
is lead(IV) chloride. We know the charge on Pb is 4+ because the charge on Cl is 1–. Since there are
4 Cl– anions, the Pb cation must be Pb4+.
PbCl4
lead (IV) chloride
76
EXAMPLE 5.10 Naming Ionic Compounds That Contain a Polyatomic Ion
Give the name for the compound K2CrO4.
The name for K2CrO4 consists of the name
of the cation, potassium, followed by the
name of the polyatomic ion, chromate.
Solution:
K2CrO4 potassium chromate
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EXAMPLE 5.11 Naming Molecular Compounds
Name each of the following: CCl4
BCl3
SF6.
Solution:
CCl4
The name of the compound is the name of the first element, carbon, followed by the base name of the
second element, chlor, prefixed by tetra- to indicate four, and the suffix -ide. The entire name is
carbon tetrachloride.
BCl3
The name of the compound is the name of the first element, boron, followed by the base name of the
second element, chlor, prefixed by tri- to indicate three, and the suffix -ide. The entire name is boron
trichloride.
SF4
The name of the compound is the name of the first element, sulfur, followed by the base name of the
second element, fluor, prefixed by hexa- to indicate six, and the suffix -ide. The entire name is sulfur
hexafluoride.
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EXAMPLE 5.12 Naming Binary Acids
Give the name of H2S.
The base name of S is sulfur, so the name is
hydrosulfuric acid.
Solution:
H2S hydrosulfuric acid
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EXAMPLE 5.13 Naming Oxyacids
Give the name of HC2H3O2.
The oxyanion is acetate, which ends in
-ate; therefore, the name of the acid is
acetic acid.
Solution:
HC2H3O2 acetic acid
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EXAMPLE 5.14 Nomenclature Using Figure 5.17
Name each of the following: CO, CaF2, HF, Fe(NO3)3, HClO4, H2SO3
Solution:
For each compound, the following table shows how to use Figure 5.17 to arrive at a name for
the compound.
Formula
Flow Chart Path
Name
CO
CaF2
HF
Fe(NO3)3
HClO4
H2SO3
Molecular
Ionic
Acid
Ionic
Acid
Acid
carbon monoxide
calcium fluoride
hydrofluoric acid
iron(III) nitrate
perchloric acid
sulfurous acid
Type I
Binary
Type II
Oxyacid
Oxyacid
-ate
-ite
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EXAMPLE 5.15 Calculating Formula Mass
Calculate the formula mass of carbon tetrachloride, CCl4.
Solution:
To find the formula mass, we sum the atomic masses of each atom in the chemical formula.
Formula mass = 1 x (Formula mass C) + 4 x (Formula mass Cl)
= 12.01 amu + 4(35.45 amu)
= 153.81 amu
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EXAMPLE 5.16 Constant Composition of Compounds
Two samples said to be carbon disulfide (CS2) are decomposed into their constituent elements. One
sample produced 8.08 g S and 1.51 g C, while the other produced 31.3 g S and 3.85 g C. Are these
results consistent with the law of constant composition?
Solution: Sample 1
These results are not consistent with the law
of constant composition and the data given
must therefore be in error.
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EXAMPLE 5.17 Writing Chemical Formulas
Write a chemical formula for the compound containing one nitrogen atom for every two oxygen atoms.
Solution:
NO2
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EXAMPLE 5.18 Total Number of Each Type of Atom in a
Chemical Formula
Determine the number of each type of atom in PB(ClO3)2.
Solution:
One Pb atom
Two Cl atoms
Six O atoms
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EXAMPLE 5.19 Classifying Elements as Atomic or Molecular
Classify each of the following elements as atomic or molecular: Na, I, N.
Solution:
Na: atomic
I: molecular (I2)
N: molecular (N2)
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EXAMPLE 5.20 Classifying Compounds as Ionic or Molecular
Classify each of the following compounds as ionic or molecular. If they are ionic, classify them as Type I
or Type II ionic compounds:
FeCl3, K2SO4, CCl4
Solution:
FeCl3: ionic, Type II
K2SO4: ionic, Type I
CCl4: molecular
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EXAMPLE 5.21 Writing Formulas for Ionic Compounds
Write a formula for the compound that forms from lithium and sulfate ions.
Solution:
Li+
SO
2–
4
Li2(SO4)
In this case, the subscripts cannot be further reduced.
Li2SO4
Cations:
2(1+) = 2+
Anions:
2–
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EXAMPLE 5.22 Naming Type I Binary Ionic Compounds
Give the name for the compound Al2O3
Solution:
aluminum oxide
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EXAMPLE 5.23 Naming Type II Binary Ionic Compounds
Give the name for the compound Fe2S3
Solution:
3 sulfide ions x (2–) = 6–
2 iron ions x (?) = 6+
? = 3+
Charge of each iron ion = 3+
iron(III) sulfide
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EXAMPLE 5.24 Naming Compounds Containing a Polyatomic Ion
Give the name for the compound Co(ClO4)2.
Solution:
3 perchlorate x (1–) = 2–
Charge of cobalt ion = 2+
cobalt(II) perchlorate
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EXAMPLE 5.25 Naming Molecular Compounds
Name the compound NO2.
Solution:
nitrogen dioxide
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EXAMPLE 5.26 Naming Binary Acids
Name the acid HI.
Solution:
hydroiodic acid
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EXAMPLE 5.27 Naming Oxyacids with an Oxyanion Ending in -ate
Name the acid H2SO4.
Solution:
The oxyanion is sulfate. The name of the acid is sulfuric acid.
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EXAMPLE 5.28 Naming Oxyacids with an Oxyanion Ending in -ite
Name the acid HClO2.
Solution:
The oxyanion is chlorite. The name of the acid is chlorous acid.
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EXAMPLE 5.29 Calculating Formula Mass
Calculate the formula mass of Mg(NO3)2
Solution:
Formula mass = 24.31 + 2(14.01) + 6(16.00)
= 148.33 amu
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