Chemical Nomenclature,
Formulas, and Equations
Formulas and Models
2
A molecular formula shows the exact number of
atoms of each element in the smallest unit of a
substance.
An empirical formula shows the simplest
whole-number ratio of the atoms in a substance.
molecular
empirical
H2O
H2O
C6H12O6
CH2O
O3
O
N2H4
NH2
3
2.2
Write the molecular formula of methylamine, a colorless gas
used in the production of pharmaceuticals and pesticides, from
its ball-and-stick model, shown below.
2.2
Solution
Refer to the labels (also see back end papers).
There are five H atoms, one C atom, and one N atom.
Therefore, the molecular formula is CH5N.
However, the standard way of writing the molecular formula for
methylamine is CH3NH2 because it shows how the atoms are
joined in the molecule.
2.3
Write the empirical formulas for the following molecules:
(a) biborane (B2H6), which is used in rocket propellants
(b) glucose (C6H12O6), a substance known as blood sugar
(c) nitrous oxide (N2O), a gas that is used as an anesthetic gas
(“laughing gas”) and as an aerosol propellant for whipped
creams.
2.3
Strategy
Recall that to write the empirical formula, the subscripts in the
molecular formula must be converted to the smallest possible
whole numbers.
2.3
Solution
(a) There are two boron atoms and six hydrogen atoms in diborane.
Dividing the subscripts by 2, we obtain the empirical formula BH3.
(b) In glucose there are 6 carbon atoms, 12 hydrogen atoms, and 6
oxygen atoms. Dividing the subscripts by 6, we obtain the
empirical formula CH2O. Note that if we had divided the
subscripts by 3, we would have obtained the formula C2H4O2.
Although the ratio of carbon to hydrogen to oxygen atoms in
C2H4O2 is the same as that in C6H12O6 (1:2:1), C2H4O2 is not the
simplest formula because its subscripts are not in the smallest
whole-number ratio.
2.3
(c) Because the subscripts in N2O are already the smallest
possible whole numbers, the empirical formula for nitrous
oxide is the same as its molecular formula.
Ionic compounds consist of a combination of cations
and anions.
• The formula is usually the same as the empirical formula.
• The sum of the charges on the cation(s) and anion(s) in
each formula unit must equal zero.
The ionic compound NaCl
10
The most reactive metals (green) and the most reactive
nonmetals (blue) combine to form ionic compounds.
11
Formulas of Ionic Compounds
2 x +3 = +6
3 x -2 = -6
Al2O3
Al3+
1 x +2 = +2
Ca2+
2 x +1 = +2
Na+
O22 x -1 = -2
CaBr2
Br1 x -2 = -2
Na2CO3
CO3212
2.4
Write the formula of
magnesium nitride, containing
the Mg2+ and N3− ions.
When magnesium burns in air,
it forms both magnesium oxide
and magnesium nitride.
2.4
Strategy Our guide for writing formulas for ionic compounds is
electrical neutrality; that is, the total charge on the cation(s)
must be equal to the total charge on the anion(s).
Because the charges on the Mg2+ and N3− ions are not equal,
we know the formula cannot be MgN.
Instead, we write the formula as MgxNy, where x and y are
subscripts to be determined.
2.4
Solution To satisfy electrical neutrality, the following
relationship must hold:
(+2)x + (−3)y = 0
Solving, we obtain x/y = 3/2. Setting x = 3 and y = 2, we write
Check The subscripts are reduced to the smallest wholenumber ratio of the atoms because the chemical formula of an
ionic compound is usually its empirical formula.
Chemical Nomenclature
• Ionic Compounds
– Often a metal + nonmetal
– Anion (nonmetal), add “-ide” to element name
BaCl2
barium chloride
K2O
potassium oxide
Mg(OH)2
magnesium hydroxide
KNO3
potassium nitrate
16
• Transition metal ionic compounds
– indicate charge on metal with Roman numerals
FeCl2
2 Cl- -2 so Fe is +2
iron(II) chloride
FeCl3
3 Cl- -3 so Fe is +3
iron(III) chloride
Cr2S3
3 S-2 -6 so Cr is +3 (6/2) chromium(III) sulfide
17
18
(a) Fe(NO3)2
(b) Na2HPO4
(c) (NH4)2SO3
19
2.5
Name the following compounds:
(a) Fe(NO3)2
(b) Na2HPO4
(c) (NH4)2SO3
2.5
Strategy Our reference for the names of cations and anions is
Table 2.3.
Keep in mind that if a metal can form cations of different
charges (see Figure 2.10), need to use the Stock system.
2.5
Solution
(a) The nitrate ion (NO3−) bears one negative charge, so the
iron ion must have two positive charges. Because iron
forms both Fe+ and Fe2+ ions, we need to use the Stock
system and call the compound iron(II) nitrate.
(b) The cation is Na+ and the anion is HPO42− (hydrogen
phosphate). Because sodium only forms one type of ion
(Na+), there is no need to use sodium(I) in the name. The
compound is sodium hydrogen phosphate.
(c) The cation is NH4+ (ammonium ion) and the anion is SO32−
(sulfite ion). The compound is ammonium sulfite.
2.6
Write chemical formulas
for the following
compounds:
(a) mercury(I) nitrite
(b) cesium sulfide
(c) calcium phosphate
2.6
Strategy
We refer to Table 2.3 for the formulas of cations and anions.
Recall that the Roman numerals in the Stock system provide
useful information about the charges of the cation.
2.6
Solution
(a) The Roman numeral shows that the mercury ion bears a +1
charge. According to Table 2.3, however, the mercury(I) ion
is diatomic (that is,
) and the nitrite ion is
.
Therefore, the formula is Hg2(NO2)2.
(b) Each sulfide ion bears two negative charges, and each
cesium ion bears one positive charge (cesium is in Group
1A, as is sodium). Therefore, the formula is Cs2S.
2.6
(c) Each calcium ion (Ca2+) bears two positive charges, and
each phosphate ion (
) bears three negative charges.
To make the sum of the charges equal zero, we must adjust
the numbers of cations and anions:
3(+2) + 2(−3) = 0
Thus, the formula is Ca3(PO4)2.
• Molecular compounds
− Nonmetals or nonmetals + metalloids
− Common names
− H2O, NH3, CH4
− Element furthest to the left in a period
and closest to the bottom of a group on
periodic table is placed first in formula
− If more than one compound can be
formed from the same elements, use
prefixes to indicate number of each kind
of atom
− Last element name ends in -ide
27
Molecular Compounds
HI
hydrogen iodide
NF3
nitrogen trifluoride
SO2
sulfur dioxide
N2Cl4
dinitrogen tetrachloride
NO2
nitrogen dioxide
N2O
dinitrogen monoxide
28
2.7
Name the following molecular compounds:
(a) SiCl4
(b) P4O10
2.7
Strategy
We refer to Table 2.4 for prefixes.
In (a) there is only one Si atom so we do not use the prefix
“mono.”
Solution
(a) Because there are four chlorine atoms present, the
compound is silicon tetrachloride.
(b) There are four phosphorus atoms and ten oxygen atoms
present, so the compound is tetraphosphorus decoxide.
Note that the “a” is omitted in “deca.”
2.8
Write chemical formulas for the following molecular
compounds:
(a) carbon disulfide
(b) disilicon hexabromide
2.8
Strategy
Here we need to convert prefixes to numbers of atoms (see
Table 2.4).
Because there is no prefix for carbon in (a), it means that there
is only one carbon atom present.
Solution
(a) Because there are two sulfur atoms and one carbon atom
present, the formula is CS2.
(b) There are two silicon atoms and six bromine atoms present,
so the formula is Si2Br6.
33
An acid can be defined as a substance that yields
hydrogen ions (H+) when dissolved in water.
For example: HCl gas and HCl in water
•Pure substance, hydrogen chloride
•Dissolved in water (H3O+ and Cl−),
hydrochloric acid
34
35
An oxoacid is an acid that contains hydrogen,
oxygen, and another element.
HNO3
nitric acid
H2CO3
carbonic acid
H3PO4
phosphoric acid
36
Naming Oxoacids and Oxoanions
37
The rules for naming oxoanions, anions of
oxoacids, are as follows:
1. When all the H ions are removed from the
“-ic” acid, the anion’s name ends with “-ate.”
2. When all the H ions are removed from the
“-ous” acid, the anion’s name ends with “-ite.”
3. The names of anions in which one or more
but not all the hydrogen ions have been
removed must indicate the number of H ions
present.
For example:
– H2PO4- dihydrogen phosphate
– HPO4 2- hydrogen phosphate
– PO43- phosphate
38
39
2.9
Name the following oxoacid and oxoanion:
(a) H2SO3, a very unstable acid formed when SO2(g) reacts with
water (sulfurous acid  sulfite +2H+)
(b) H2AsO4−, once used to control ticks and lice on livestock
(dihydrogen arsenate)
(c) SeO32−, used to manufacture colorless glass. (selenite)
H3AsO4 is arsenic acid, and H2SeO4 is selenic acid.
H2SeO3 is selenous acid
2.9
Strategy We refer to Figure 2.14 and Table 2.6 for the
conventions used in naming oxoacids and oxoanions.
Solution
(a) We start with our reference acid, sulfuric acid (H2SO4).
Because H2SO3 has one fewer O atom, it is called sulfurous
acid.
(b) Because H3AsO4 is arsenic acid, the AsO43− is named
arsenate. The H2AsO4− anion is formed by adding two H+
ions to AsO43−, so H2AsO4− is called dihydrogen arsenate.
(c) The parent acid is H2SeO3. Because the acid has one fewer
O atom than selenic acid (H2SeO4), it is called selenous
acid. Therefore, the SeO32− anion derived from H2SeO3 is
called selenite.
A base can be defined as a substance that yields
hydroxide ions (OH-) when dissolved in water.
NaOH
sodium hydroxide
KOH
potassium hydroxide
Ba(OH)2
barium hydroxide
42
Hydrates are compounds that have a specific
number of water molecules attached to them.
BaCl2•2H2O
barium chloride dihydrate
LiCl•H2O
lithium chloride monohydrate
MgSO4•7H2O
magnesium sulfate heptahydrate
Sr(NO3)2 •4H2O
strontium nitrate tetrahydrate
CuSO4•5H2O
CuSO4
43
Organic chemistry is the branch of chemistry that
deals with carbon compounds.
Functional Groups:
H
H O
H
H
C
OH
H
methanol
H
C
NH2
H
methylamine
H
C
C
OH
H
acetic acid
44
45
Chemical Equations
• Symbolic representation of a chemical
reaction that shows:
1. reactants on left side of reaction
2. products on right side of equation
3. relative amounts of each using stoichiometric
coefficients
46
Chemical Equations
• Attempt to show on paper what is happening at the
laboratory and molecular levels.
47
Chemical Equations
• Law of Conservation of Matter
– There is no detectable change in quantity of matter in an
ordinary chemical reaction.
– Balanced chemical equations must always include the
same number of each kind of atom on both sides of the
equation.
– This law was determined by Antoine Lavoisier.
48
Chemical Equations
• Look at the information an equation provides:

Fe2 O3 + 3 CO  2 Fe + 3 CO 2
Reactants
Yields
Products
49
Chemical Equations
• Look at the information an equation provides:

Fe2 O3 + 3 CO  2 Fe + 3 CO 2
Reactants
1 formula
3
unit
molecules
Yields
Products
2 atoms
3
molecules
50
Chemical Equations
• Look at the information an equation provides:

Fe2 O3 + 3 CO  2 Fe + 3 CO 2
Reactants
1 formula
3
unit
molecules
1 mole
3 moles
Yields
Products
2 atoms
2 moles
3
molecules
3 moles
51
How to “Read” Chemical Equations
2 Mg + O2
2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
52
Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on
the left side and the correct formula(s) for the
product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of
each element the same on both sides of the
equation. Do not change the subscripts.
2C2H6
NOT
C4H12
53
Balancing Chemical Equations
3. Start by balancing those elements that appear in
only one reactant and one product.
C2H6 + O2
2 carbon
on left
C2H6 + O2
6 hydrogen
on left
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
2CO2 + 3H2O
multiply H2O by 3
54
Balancing Chemical Equations
4. Balance those elements that appear in two or
more reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
multiply O2 by 7
2
4 oxygen + 3 oxygen = 7 oxygen
(3x1)
on right
(2x2)
C2H6 + 7 O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
remove fraction
multiply both sides by 2
55
Balancing Chemical Equations
5. Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
2C2H6 + 7O2
4CO2 + 6H2O
4 C (2 x 2)
4C
12 H (2 x 6)
12 H (6 x 2)
14 O (7 x 2)
14 O (4 x 2 + 6)
Reactants
4C
12 H
14 O
Products
4C
12 H
14 O
56
3.12
When aluminum metal is exposed
to air, a protective layer of
aluminum oxide (Al2O3) forms on its
surface. This layer prevents further
reaction between aluminum and
oxygen, and it is the reason that
aluminum beverage cans do not
corrode. [In the case of iron, the
rust, or iron(III) oxide, that forms is
too porous to protect the iron metal
underneath, so rusting continues.]
Write a balanced equation for the
formation of Al2O3.
An atomic scale image
of aluminum oxide.
3.12
Strategy Remember that the formula of an element or
compound cannot be changed when balancing a chemical
equation. The equation is balanced by placing the appropriate
coefficients in front of the formulas. Follow the procedure
described on p. 92.
Solution The unbalanced equation is
In a balanced equation, the number and types of atoms on
each side of the equation must be the same. We see that there
is one Al atom on the reactants side and there are two Al atoms
on the product side.
3.12
We can balance the Al atoms by placing a coefficient of 2 in
front of Al on the reactants side.
There are two O atoms on the reactants side, and three O
atoms on the product side of the equation. We can balance the
O atoms by placing a coefficient of
reactants side.
in front of O2 on the
This is a balanced equation. However, equations are normally
balanced with the smallest set of whole-number coefficients.
3.12
Multiplying both sides of the equation by 2 gives whole-number
coefficients.
or
Check For an equation to be balanced, the number and types
of atoms on each side of the equation must be the same. The
final tally is
The equation is balanced. Also, the coefficients are reduced to
the simplest set of whole numbers.
Combination Reactions
•
•
Combination reactions occur when two or more
substances combine to form a compound.
There are three basic types of combination
reactions.
1. Two elements react to form a new compound
2. An element and a compound react to form one new
compound
3. Two compounds react to form one compound
61
Combination Reactions
1. Element + Element  Compound
A. Metal + Nonmetal  Binary Ionic Compound
2 Na s   Cl2g   2 NaCls 
62
Combination Reactions
1. Element + Element  Compound
A. Metal + Nonmetal  Binary Ionic Compound
2 Mg s   O 2g   2 MgO s 
63
Combination Reactions
1. Element + Element  Compound
A. Metal + Nonmetal  Binary Ionic Compound
2 Al s   3 Br2    2 AlBr 3s 
64
Combination Reactions
1. Element + Element  Compound
B. Nonmetal + Nonmetal  Covalent Binary
Compound
P4 s   5 O2 g   P4O10s 
65
Combination Reactions
1. Element + Element  Compound
B. Nonmetal + Nonmetal  Covalent Binary
Compound
P4 s   6 Cl2 g   4 PCl3 
66
Combination Reactions
1. Element + Element  Compound
B. Nonmetal + Nonmetal  Covalent Binary
Compound
• Can control which product is made with the reaction
conditions.
2 As s   3 Cl2g   2 AsCl 3s 
in limited chlorine
2 As s   5 Cl2g   2 AsCl 5s 
in excess chlorine
67
Combination Reactions
1. Element + Element  Compound
B. Nonmetal + Nonmetal  Covalent Binary
Compound
• Can control which product is made with the reaction
conditions.
Se s   2 F2g   SeF4s 
in limited fluorine
Se s   3 F2g   SeF6g 
in excess fluorine
68
Combination Reactions
2. Compound + Element  Compound
69
Combination Reactions
The reaction of oxygen with oxides of nonmetals
is an example of this type of combination
reaction.
2 COg   O 2g   2 CO2g 
catalyst & 
2 SO2g   O2g   2 SO3g 
P4O6  2 O2  P4O10
70
Combination Reactions
3. Compound + Compound  Compound
–
gaseous ammonia and hydrogen chloride
NH3g   HCl g   NH4Cls 
– lithium oxide and sulfur dioxide
Li 2O  SO3  Li 2SO4
71
Decomposition Reactions
•
Decomposition reactions occur when one
compound decomposes to form:
1. Two elements
2. One or more elements and one or more
compounds
3. Two or more compounds
72
Decomposition Reactions
1. Compound  Element + Element
–
decomposition of dinitrogen oxide

2 N 2Og  
 2 N 2g   O2g 
• decomposition of calcium chloride
CaCl2  
 Ca    Cl2g 
electricity
 decomposition of silver halides
h
2 AgBr s   2 Ag s   Br2 
73
Decomposition Reactions
2. Compound  One Element +
Compound(s)
–
decomposition of hydrogen peroxide
hν or Fe 3 or Mn
2 H 2O2 aq     2 H 2O   O2 g 
74
Decomposition Reactions
3. Compound  Compound + Compound
–
decomposition of ammonium hydrogen carbonate

NH4 HCO3s  
 NH3g   H 2Og   CO2g 
75
Displacement Reactions
• Displacement reactions occur when one
element displaces another element from a
compound.
– These are redox reactions in which the more
active metal displaces the less active metal of
hydrogen from a compound in aqueous solution.
– Activity series is given in Table 4-14.
76
Displacement Reactions
[More Active Metal + Salt of Less Active Metal]  [Less Active
Metal + Salt of More Active Metal]
1.
–
molecular equation
AgNO 3aq  + Cu (s)  CuNO 3aq   Ag (s)
77
Displacement Reactions
[Active Metal + Nonoxidizing Acid]  [Hydrogen +
Salt of Acid]
2.
–
–
•
Common method for preparing hydrogen in the laboratory.
HNO3 is an oxidizing acid.
Molecular equation
2 Al (s) + 3H 2SO 4aq   Al 2 (SO 4 )3aq  + 3 H 2g 
79
Displacement Reactions
• Total ionic equation
You do it!
2 Al (s) + 6 H

aq 
+ 3 SO
24aq 
 2 Al
3
aq 
+ 3 SO
24aq 
+ 3 H 2 g 
• Net ionic equation
You do it!
2 Al (s) + 6 H

aq 
 2 Al
3
aq 
+ 3 H 2 g 
80
Displacement Reactions
• The following metals are active enough to
displace hydrogen
– K, Ca, Na, Mg, Al, Zn, Fe, Sn, & Pb
• Notice how the reaction changes with an
oxidizing acid.
– Reaction of Cu with HNO3.
• H2 is no longer produced.
81
Displacement Reactions
3.
[Active Nonmetal + Salt of Less Active Nonmetal]  [Less Active
Nonmetal + Salt of More Active
Nonmetal]
•
Molecular equation
Cl2g  + 2 NaIaq   I 2s   2 NaCl(aq)
Total
ionic equation
You do it!
Cl2g  + 2 Na

aq 
+2I
aq 
 I 2s   2 Na

aq 
+ 2 Cl
82
aq 
Metathesis Reactions
•
Metathesis reactions occur when two ionic
aqueous solutions are mixed and the ions switch
partners.
AX + BY  AY + BX
•
Metathesis reactions remove ions from solution in
two ways:
1. form predominantly unionized molecules like H2O
2. form an insoluble solid
•
Ion removal is the driving force of metathesis
reactions.
84
Metathesis Reactions
1. Acid-Base (neutralization) Reactions
– Formation of the nonelectrolyte H2O
– acid + base  salt + water
85
Metathesis Reactions
• Molecular equation
HBr(aq) + KOH(aq)  KBr(aq) + H 2O(  )
Total
ionic equation
You do it!
H

aq 
aq 
+ Br
+K
Net ionic

aq 
aq 
+ OH
K

aq 
aq 
+ Br
+ H 2 O(  )
equation
You do it!
H

aq 
aq 
+ OH
 H 2 O(  )
86
Metathesis Reactions
• Molecular equation
Ca(OH) 2 (aq) + 2 HNO3(aq)  Ca(NO3 ) 2 (aq) + 2 H 2O()
Total ionic equation
You do it!
Ca 2aq  + 2 OH-aq  + 2 Haq  + 2 NO3- aq   Ca 2aq  + 2 NO3- aq  + 2 H2O()
Net
ionic equation
You do it!
2 OH-aq  + 2 H aq   2 H 2 O (  )
or better
OH-aq  + H aq   H 2 O (  )
87
Metathesis Reactions
2. Precipitation reactions are metathesis
reactions in which an insoluble compound is
formed.
– The solid precipitates out of the solution much
like rain or snow precipitates out of the air.
88
Metathesis Reactions
• Precipitation Reactions
• Molecular equation
Ca(NO3 ) 2 (aq) + K 2CO3(aq)  2 KNO3(aq ) + CaCO3(s)
Total ionic reaction
You do it!
Ca
2
aq 
 2 NO
3aq 
2K
2K

aq 

aq 
 CO
23aq 
 2 NO

3 aq 
 CaCO3s 
89
Metathesis Reactions
• Molecular equation
3 CaCl2(aq) + 2 Na3PO4(aq)  6 NaCl(aq ) + Ca 3 PO4 2(s)
Total
ionic reaction
You do it!
3 Ca
2
aq 
 6 Cl
1aq 
1
aq 
 2 PO
1
aq 
 6 Cl
+ 6 Na
6 Na
34 aq 
1aq 

+ Ca3 PO4 2 s 
91
Metathesis Reactions
• Molecular equation
2 HCl(aq) + Na2SO3( aq)  2 NaCl( aq ) + H2O   SO2g 
Total
ionic reaction
You do it!
1
aq 
2H
 2 Cl
1aq 
1
aq 
+ 2 Na
 SO
23aq 


12 Na1aq

2
Cl

aq  + H 2 O    SO 2g 
93
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•
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•
•
•
LiBF4 + H2O  H3BO3 + HF + LiF
Al(OH)3 + HCl  AlCl3 + H2O
C4H9SO + O2  CO2 + SO2 + H2O
Cd + NiO2 + H2O  Cd(OH)2 + Ni(OH)2
SiO2 + C + Cl2  CO + SiCl4
Br2 + H2O  HBr + HBrO3
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