Transcript Naming Compounds - Kowenscience.com

Chapter 5
Nomenclature
Chemical
BONDING
Chemical Bond
• A bond results from the attraction of nuclei
for electrons
– All atoms trying to achieve a stable octet
• IN OTHER WORDS
– the p+ in one nucleus are attracted to the e- of
another atom
• Electronegativity
• Molecule: 2 or more atoms joined by
a chemical bond
• Compound: a molecule composed of
atoms of 2 or more different elements
bonded together in a fixed ratio
Diatomic Molecule
• Diatomic Molecule: a molecule
containing 2 atoms
• The Diatomic molecules are:
• Hydrogen (H2) Nitrogen (N2)
Oxygen (O2)
Fluorine (F2)
Chlorine (Cl2)
Iodine (I2)
Bromine (Br2)
• Chemical formula: represents the
relative numbers of atoms of each kind
in a chemical compound by using
atomic symbols and numeric subscripts
• Bond energy: the energy required to
break a chemical bond and form
neutral atoms
Naming Compounds
Types of Chemical Bonds: (4)
1.
2.
3.
4.
Ionic bonds
Covalent bonds
Metallic bonds
Hydrogen bonds
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7
Bond Formation
• exothermic process
E
N
E
R
G
Y
Reactants
Energy
released
Products
Breaking Bonds
• Endothermic reaction
– energy must be put into the bond in order
to break it
E
N
E
R
G
Y Reactants
Products
Energy
Absorbed
Bond Strength
• Strong, STABLE bonds require lots of
energy to be formed or broken
• weak bonds require little E
Two Major Types of
Bonding
• Ionic Bonding
– forms ionic compounds
– transfer of e-
• Covalent Bonding
– forms molecules
– sharing e-
Naming Compounds
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One minor type of bonding
• Metallic bonding
– Occurs between like atoms of a metal in the
free state
– Valence e- are mobile (move freely among all
metal atoms)
– Positive ions in a sea of electrons
• Metallic characteristics
– High mp temps, ductile, malleable, shiny
– Hard substances
– Good conductors of heat and electricity as (s) and (l)
It’s the mobile electrons
that enable me tals to
conduct electricity!!!!!!
IONic Bonding
• electrons are transferred between
valence shells of atoms
• ionic compounds are
NOT MOLECULES
made of ions
• ionic compounds are called Salts or
Crystals
IONic bonding
• Always formed between metals and
non-metals
+
[METALS ] [NON-METALS ]
Lost e-
Gained e-
Properties of Ionic
Compounds
SALTS
Crystals
• hard solid @ 22oC
• high mp temperatures
• nonconductors of electricity in solid
phase
• good conductors in liquid phase or
dissolved in water (aq)
Covalent Bonding
molecules
• Pairs of e- are shared
between non-metal atoms
• electronegativity difference < 2.0
• forms polyatomic ions
Properties of Molecular
Substances
Covalent
bonding
• Low m.p. temp and b.p. temps
• relatively soft solids as compared
to ionic compounds
• nonconductors of electricity in
any phase
Covalent, Ionic, metallic
bonding?
• NO2
• sodium
hydride
• Hg
• H2S
• sulfate
• NH4+
• CO
• Aluminum • Co
phosphate
• KH
Can You Tell
• KCl
What type of
• HF
bond is
formed
Drawing ionic compounds
using Lewis Dot Structures
• Symbol represents the KERNEL of the
atom (nucleus and inner e-)
• dots represent valence e-
NaCl
• This is the finished Lewis Dot
Structure
How did we get here?
+
[Na]
-
[ Cl ]
• Step 1 after checking that it is IONIC
– Determine which atom will be the +ion
– Determine which atom will be the - ion
• Step 2
– Write the symbol for the + ion first.
• NO DOTS
– Draw the e- dot diagram for the – ion
• COMPLETE outer shell
• Step 3
– Enclose both in brackets and show each charge
Draw the Lewis Diagrams
• LiF
• MgO
• CaCl2
• K2S
Drawing molecules using
Lewis Dot Structures
• Symbol represents the KERNEL of the
atom (nucleus and inner e-)
• dots represent valence e-
Always remember atoms are
trying to complete their
outer shell!
The number of electrons the atoms
needs is the total number of bonds
they can make.
Ex. … H? O? F? N? Cl? C?
one two one three one four
Methane CH4
• This is the finished Lewis dot structure
How did we get here?
• Step 1
– count total valence e- involved
• Step 2
– connect the central atom (usually the first in
the formula) to the others with single bonds
• Step 3
– complete valence shells of outer atoms
• Step 4
– add any extra e- to central atom
IF the central atom has 8 valence e- surrounding
it . . YOU’RE DONE!
Sometimes . . .
• You only have two atoms, so there is
no central atom, but follow the same
rules.
• Check & Share to make sure all the
atoms are “happy”.
Cl2
Br2
H2
O2
N2
HCl
• DOUBLE bond
– atoms that share two e- pairs (4 e-)
O O
• TRIPLE bond
– atoms that share three e- pairs (6 e-)
N N
Draw Lewis Dot Structures
You may represent valence electrons
from different atoms with the
following symbols x, ,
CO2
NH3
Draw the Lewis Dot Diagram for
polyatomic ions
• Count all valence e- needed for
covalent bonding
• Add or subtract other electrons based
on the charge
REMEMBER!
A positive charge means it LOST
electrons!!!!!
Draw Polyatomics
• Ammonium
• Sulfate
Types of Covalent Bonds
• NON-Polar bonds
– Electrons shared evenly in the bond
– E-neg difference is zero
Between identical atoms
Diatomic molecules
Types of Covalent Bonds
Polar bond
– Electrons unevenly shared
non-polar MOLECULES
• Sometimes the bonds within a
molecule are polar and yet the
molecule is non-polar because its
shape is symmetrical. H
H C H
Draw Lewis dot first and
see if equal on all sides
H
Polar molecules (a.k.a.
Dipoles)
• Not equal on all sides
– Polar bond between 2 atoms makes a
polar molecule
– asymmetrical shape of molecule
+
H Cl

Water is asymmetrical
+
O

-
H
+
H
Water is a bent molecule
H
H
O
H
H
W-A-T-E-R
as bent as it can be!
Water’s polar MOLECULE!
Water’s polar MOLECULE!
The H is positive
The O is not - not - not - not
Making sense of the polar
non-polar thing
BONDS
Non-polar
Polar
Identical Different
MOLECULES
Non-polar
Symmetrical
Polar
Asymmetrical
IONIC bonds ….
Ionic bonds are
so polar that the electrons are not
shared but transferred between
atoms forming ions!!!!!!
4 Shapes of molecules
Linear (straight line)
Ball and stick
model
Space filling
model
Bent
Ball and stick
model
Space filling
model
Trigonal pyramid
Ball and stick
model
Space filling
model
Tetrahedral
Ball and stick
model
Space filling
model
Intermolecular attractions
• Attractions between
molecules
– van der Waals forces
• Weak attractive
forces between
non-polar
molecules
– Hydrogen “bonding”
• Strong attraction
between special
polar molecules
van der Waals
• Non-polar molecules can exist in liquid
and solid phases
because van der Waals forces keep the
molecules attracted to each other
• Exist between CO2, CH4, CCl4, CF4,
diatomics and monoatomics
van der Waals periodicity
• increase with molecular mass.
– Greater van der Waals force?
• F2 Cl2 Br2 I2
• increase with closer distance between
molecules
– Decreases when particles are farther away
Hydrogen “Bonding”
• Strong polar
attraction
– Like magnets
• Occurs ONLY
between H of one
molecule and N, O,
F of another
H “bond”
Why does H “bonding”
occur?
• Nitrogen, Oxygen and Fluorine
– small atoms with strong nuclear charges
• powerful atoms
– very high electronegativities
Intermolecular forces
dictate chemical properties
• Strong intermolecular forces cause
high b.p., m.p. and slow evaporation
(low vapor pressure) of a substance.
Which substance has the
highest boiling point?
• HF
• NH3
• H2O
Fluorine has the highest e-neg,
SO
HF will experience the
• WHY?
needs the most energy to
weaken the i.m.f. and boil
strongest H bonding and 
Density????
H2O(s) is less dense than
H2O(l)
• The hydrogen bonding in water(l) molecules
is random. The molecules are closely
packed.
• The hydrogen bonding in water(s) molecules
has a specific open lattice pattern. The
molecules are farther apart.
Naming Compounds
Chemical Names and formulas
• With all of the compounds and all of the
elements to be identified, a systematic method
for writing formulas and naming compounds is
necessary
• A correctly written chemical formula must
represent the known facts about the composition
of a compound
• Care must be taken so that subscripts are
correct
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Naming Compounds
Using Chemical formulas
• Chemical formulas indicate the elements present in a
compound and the relative numbers of atoms of each
element in the compound
• In chemical formulas, the elements are given by their
symbols and the relative number of atoms of each
element by numerical subscript
• Ex H2SO4 the H, S & O are symbols, the 2 & 4 are
subscripts
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Naming Compounds
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• Ion: A charged particle due to loss or
gain of electrons
• Cation: positive charge ion
represented by a (+) after the chemical
symbol (metal) Ex Na+
• Anion: negative charge ion
represented by a (-) after the chemical
symbol (metal) Ex Cl-
Naming Compounds
Monatomic Ions
• Positive ions are named by the element
name followed by the word “ion”
• Examples :
• K+
potassium ion
magnesium ion
Mg+2
aluminum ion
• Al+3
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Naming Compounds
• Negative ions are named by dropping the ending
of the element name and adding the ending “ide”
to it followed by the word “ion”
•
•
•
•
Examples:
FS-2
I-
fluoride ion
sulfide ion
Iodide ion
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Naming Compounds
Learning Check
Give the names of the following ions:
Ba2+
_________
Al3+
__________
K+
_________
N3
_________
O2
__________
F
_________
P3
_________
S2
__________
Cl
_________
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Naming Compounds
Solution
Ba2+
barium
Al3+
aluminum
K+
potassium
N3
nitride
O2
oxide
F
fluoride
P3
phosphide
S2
sulfide
Cl
chloride
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Naming Compounds
•
Binary Compounds

•
Binary Ionic Compounds

•
Composed of two elements
Metal—nonmetal
Binary Covalent Compounds

Nonmetal—nonmetal
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Naming Compounds
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Naming Compounds
•
Binary ionic compounds
contain positive cations
and negative anions.
 Type I compounds
•

Metal present forms
only one cation.
Type II compounds
•
Metal present can
form 2 or more
cations with
different charges.
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Naming Compounds
Type I Compounds
Metals (Groups I, II, and III) and Non-Metals
Metal _________
Sodium
+ Non-Metal _________ide
Chlorine
Sodium Chloride NaCl
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79
Naming Compounds
Common Simple Cations and Anions
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Naming Compounds
Rules for Naming Type I Ionic Compounds
1. The cation is always named first and the anion
second.
2. A simple cation takes its name from the name
of the element.
3. A simple anion is named by taking the first part
of the element name (the root) and adding –ide.
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Naming Compounds
Binary Ionic Compounds (Type I)
•
Examples:
KCl
Potassium chloride
MgBr2
Magnesium bromide
CaO
Calcium oxide
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Naming Compounds
Exercise
What is the name of the compound SrBr2?
a)
b)
c)
d)
strontium bromine
sulfur bromide
strontium dibromide
strontium bromide
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Naming Compounds
• Strontium bromide. Sr is the symbol for
strontium.
• Br is the symbol for bromine,
• take the first part of the element name (the
root) and add –ide to get the name bromide.
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Naming Compounds
Binary Ionic Compounds (Type II)
•
•
•
•
Metals in these
compounds can form
more than one type of
positive charge.
Charge on the metal ion
must be specified.
Roman numeral
indicates the charge of
the metal cation.
Transition metal cations
usually require a Roman
numeral.
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85
Naming Compounds
Type II Compounds
Metals (Transition Metals) and Non-Metals
Metal ______
Iron +Roman Numeral (__)
III + Non-Metal ________ide
Bromine
Iron (III) Bromide FeBr3
Compare with Iron (II) Bromide FeBr2
Metals (Transition Metals) and Non-Metals
Older System
Metal (Latin) _______
Ferrous + ous or ic + Non-Metal ________ide
Bromine
Ferrous Bromide FeBr2
Compare with Ferric Bromide FeBr3
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86
Naming Compounds
Different names are needed for positive
ions of 2 different charges formed by the
same metal
• Old system: “ous” ending for lower charge
•
“ic” ending for higher charge
• New system: gives actual charge on the
ion as a roman numeral
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Naming Compounds
Common Type II Cations
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Naming Compounds
Rules for Naming Type II Ionic Compounds
1. The cation is always named first and the anion
second.
2. Because the cation can assume more than one
charge, the charge is specified by a Roman
numeral in parentheses.
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Naming Compounds
Binary Ionic Compounds (Type II)
•
Examples:
CuBr
Copper(I) bromide
FeS
Iron(II) sulfide
PbO2
Lead(IV) oxide
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Naming Compounds
Exercise
What is the name of the compound CrO2?
a)
b)
c)
d)
chromium oxide
chromium(II) oxide
chromium(IV) oxide
chromium dioxide
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Naming Compounds
• Chromium(IV) oxide. Cr is the symbol for chromium. O is
the symbol for oxygen, but
• take the first part of the element name (the root) and add –ide
to get the name oxide.
• Since chromium can have more than one charge, a Roman
numeral must be used to identify that charge.
• There are two oxygen ions each with a 2– charge, giving an
overall charge of –4.
• Therefore, the charge on chromium must be +4.
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Naming Compounds
Exercise
What is the correct name of the compound that
results from the most stable ion for sulfur and
the metal ion that contains 24 electrons?
a)
b)
c)
d)
iron(III) sulfide
chromium(II) sulfide
nickel(III) sulfate
iron(II) sulfide
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Naming Compounds
• Iron(II) sulfide.
• For sulfur, take the first part of the element name
(the root) and add –ide to get the name sulfide.
• Iron with a +2 charge (as the Roman numeral
indicates) contains 24 electrons (26p – 24e = +2
charge).
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Naming Compounds
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Naming Compounds
Rules for Naming Type III Binary Compounds
• Formed between two nonmetals.
1. The first element in the formula is named
first, and the full element name is used.
2. The second element is named as though
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.
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Naming Compounds
Type III Compounds
Non-Metals and Non-Metals
Use Prefixes such as mono, di, tri, tetra, penta, hexa, hepta, etc.
CO2 Carbon dioxide
CO Carbon monoxide
PCl3 Phosphorus trichloride CCl4 Carbon tetrachloride
N2O5 Dinitrogen pentoxide
CS2 Carbon disulfide
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Naming Compounds
Prefixes Used to Indicate
Numbers in Chemical Names
Additional Prefixes
9
nona-
10
deca-
11
undeca-
12
dodeca-
13
trideca-
14
tetradeca-
15
pentadeca-
16
hexadeca-
17
heptadeca-
18
octadeca-
19
nonadeca-
20
icosa
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Naming Compounds
Binary Covalent Compounds (Type III)
•
Examples:
CO2
Carbon dioxide
SF6
Sulfur hexafluoride
N2O4
Dinitrogen tetroxide
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Naming Compounds
Exercise
What is the name of the compound SeO2?
a)
b)
c)
d)
selenium oxide
selenium dioxide
selenium(II) oxide
selenium(IV) dioxide
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Naming Compounds
•
•
•
•
Selenium dioxide.
Se is the symbol for selenium.
O is the symbol for oxygen,
take the first part of the element name (the root) and
add –ide to get the name oxide.
• Since they are both nonmetals, prefixes are used
to identify the elements (except mono- is not used for
the first element).
• Two oxygen atoms require the use of the prefix di-,
making the name dioxide.
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Naming Compounds
Flow Chart for Naming Binary Compounds
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Let’s Practice!
Naming Compounds
Name the following.
CaF2
K2S
CoI2
SnF2
SnF4
OF2
CuI2
CuI
SO2
SrS
LiBr
Calcium Flouride
Potassium Sulfide
Cobalt (II) Iodide or Cobaltous Iodide
Tin (II) Fluoride or Stannous Fluoride
Tin (IV) Fluoride or Stannic Fluoride
Oxygen diflouride
Copper (II) Iodide or Cupric Iodide
Copper (I) Iodide or Cuprous Iodide
Sulfur dioxide
Strontium Sulfide
Lithium Bromide
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Naming Compounds
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Naming Compounds
•
•
•
•
Polyatomic ions are charged entities composed
of several atoms bound together.
They have special names and must be
memorized.
We will be using our Fat Daddy Chart to
help us with naming the polyatomic
compounds
Those used often enough will be memorized
just out of sheer practice
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Naming Compounds
Names of Common Polyatomic Ions (page 130)
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Naming Compounds
•
Naming ionic compounds containing
polyatomic ions follows rules similar to those
for binary compounds.
 Ammonium acetate
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Naming Compounds
Examples
NaOH
Sodium hydroxide
Mg(NO3)2
Magnesium nitrate
(NH4)2SO4
Ammonium sulfate
FePO4
Iron(III) phosphate
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Naming Compounds
Learning Check
Select the correct name for each.
A. Fe2S3
1) iron sulfide
2) iron(II) sulfide
3) iron(III) sulfide
B. CuO
1) copper oxide
2) copper(I) oxide
3) copper(II) oxide
109
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Naming Compounds
Solution
Select the correct name for each.
A. Fe2S3
3) iron(III) sulfide
Fe3+ S2–
B. CuO
3) copper(II) oxide
Cu2+ O2–
110
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Naming Compounds
Overall Strategy for Naming Chemical Compounds
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Naming Compounds
Exercise
What is the name of the compound KClO3?
a)
b)
c)
d)
potassium chlorite
potassium chlorate
potassium perchlorate
potassium carbonate
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Naming Compounds
Exercise
Examine the following table of formulas and names.
Which of the compounds are named correctly?
a)
b)
c)
d)
I, II
I, III, IV
I, IV
I only
Formula
Name
I
P2 O 5
Diphosphorus pentoxide
II
ClO2
Chlorine oxide
III
PbI4
Lead iodide
IV
CuSO4
Copper(I) sulfate
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Naming Compounds
• Only Formula I is named
correctly.
• Formula II is chlorine dioxide.
• Formula III is lead(IV) iodide.
• Formula IV is copper(II) sulfate.
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Naming Compounds
Acids
•
•
•
•
•
•
Acids can be recognized by the hydrogen
that appears first in the formula—HCl.
Molecule with one or more H+ ions attached
to an anion.
Most lab acids are either:
binary acids ( composed of Hydrogen
and another element)
or
oxyacids (composed of Hydrogen, oxygen
and a third element
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Naming Compounds
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Naming Compounds
Rules for Naming Acids
•
•
If the anion does not contain oxygen, the
acid is named with the prefix hydro– and the
suffix –ic attached to the root name for the
element.
Examples:
HCl
Hydrochloric acid
HCN
Hydrocyanic acid
H2S
Hydrosulfuric acid
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Naming Compounds
Acids That Do Not Contain Oxygen
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Naming Compounds
Rules for Naming Acids
•
If the anion contains oxygen:
 The suffix –ic is added to the root name if
the anion name ends in –ate.
• Examples:
HNO3
Nitric acid
H2SO4
Sulfuric acid
HC2H3O2 Acetic acid
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Naming Compounds
Rules for Naming Acids
•
If the anion contains oxygen:
 The suffix –ous is added to the root name
if the anion name ends in –ite.
• Examples:
HNO2
Nitrous acid
H2SO3
Sulfurous acid
HClO2
Chlorous acid
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Naming Compounds
Some Oxygen-Containing Acids
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Naming Compounds
Flowchart for Naming Acids
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Naming Compounds
Exercise
Which of the following compounds is named
incorrectly?
a) KNO3
b) TiO2
c) Sn(OH)4
d) PBr5
e) H2SO3
potassium nitrate
titanium(II) oxide
tin(IV) hydroxide
phosphorus pentabromide
sulfurous acid
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Naming Compounds
• The correct answer is “b”.
• The charge on oxygen is 2–.
• Since there are two oxygen atoms,
the overall charge is 4–.
• Therefore, the charge on titanium
must be 4+ (not 2+ as the Roman
numeral indicates).
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124
Naming Compounds
Examples
•
•
•
•
•
Sodium hydroxide
 NaOH
Potassium carbonate
 K2CO3
Sulfuric acid
 H2SO4
Dinitrogen pentoxide
 N2O5
Cobalt(III) nitrate
 Co(NO3)3
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Naming Compounds
Exercise
A compound has the formula XCl3 where X
could represent a metal or nonmetal. What
could the name of this compound be?
a)
b)
c)
d)
phosphorus trichloride
carbon monochloride
tin(IV) chloride
magnesium chloride
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Naming Compounds
•
•
•
•
•
Phosphorus trichloride.
Carbon monochloride has the formula CCl.
Tin(IV) chloride has the formula SnCl4.
Magnesium chloride has the formula MgCl2.
Phosphorus trichloride has the formula PCl3
and is therefore the correct answer
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Naming Compounds
Lets Practice Some More!
HF
Hydroflouric acid
Na2CO3 Sodium carbonate
H2CO3
KMnO4
HClO4
H2S
NaOH
CuSO4
PbCrO4
Carbonic acid
Potassium permanganate
Perchloric acid
Hyrdogen sulfuric acid
Sodium hydroxide
Copper (II) sulfate or Cupric sulfate
Lead (II) chromate or Plubous chromate
H2O
Hydrooxic acid (no……just water)
NH3
Nitrogen trihydride (no..just ammonia)
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Naming Compounds
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129
Naming Compounds
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130
Naming Compounds
Identifying Ionic Charges
• Group A elements – use the periodic table to determine
ionic charge
* elements in same group have
same ionic charge
* Group 4A and Noble gases –
almost never form ions
• Group B elements – many have more than one ionic
charge
131
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Naming Compounds
Identifying Ionic Charges
http://wps.prenhall.com/wps/media/objects/476/488316/ch04.html
Charge on cations corresponds to group #.
Charge on anions is found by subtracting 8 by group number
132
the number 8 is used b/c it represents # of valence e- in Noble gases
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Naming Compounds
Naming Cations and Anions
•
•
•
•
•
•
•
Potassium ion
Copper (II) ion
Chloride ion
Oxide ion
Ba2+
S2Au3+
•
•
•
•
•
•
Nitrite ion
Hydroxide ion
Phosphate ion
SO42CrO42ClO32133
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Naming Compounds
Binary Ionic Compounds
• Compounds composed of 2 different monatomic
elements
• To write binary formulas – write cation first, then
anion
*criss-cross charges to determine how
many of each ion you need
*use subscripts to denote number of ions
ex: Ca2+ + Cl1CaCl2
Na1+ + Cl1-
NaCl
134
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Naming Compounds
Ternary Ionic Compounds
• Compounds containing at least one polyatomic ion; at
least 3 different elements
• To write ternary formulas: write cation first, then anion
*criss-cross charges to determine how
many of each ion you need
*use subscripts to denote number of ions
*must use parentheses around polyatomic if more
than one is
needed!!!
ex: Na1+ + SO32Na2SO3
Mg2+ + OH1MgOH2]
Mg(OH)2 [not same as
135
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Naming Compounds
Ionic Compounds
•
•
•
•
•
•
NaNO3
CaSO4
(NH4)2O
CuSO3
Fe(OH)3
NaF
•
•
•
•
•
•
Lithium sulfide
Iron (III) phosphide
Magnesium fluoride
Barium nitrate
Aluminum hydroxide
Potassium phosphate
Practice making ionic compounds!
136
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Naming Compounds
Molecular Compounds
•
•
•
•
•
P2O5
N2O
NO2
CBr4
CO2
•
•
•
•
•
tetraiodine nonoxide
sulfur hexafluoride
nitrogen trioxide
carbon tetrahydride
phosphorus trifluoride
137
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Naming Compounds
Examples of Ionic Compounds
with Two Elements
Formula
Ions
Cation Anion
NaCl
Na+
Cl–
sodium chloride
K2S
K+
S2–
potassium sulfide
MgO
Mg2+ O2–
magnesium oxide
CaI2
Ca2+ I–
calcium iodide
Al2O3
Al3+
aluminum sulfide
S2–
Name
138
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Naming Compounds
Learning Check
Write the formulas and names for compounds of
the following ions:
Br–
S2−
N3−
Na+
Al3+
139
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Naming Compounds
Solution
Br−
S2−
N3−
Na+
NaBr
Na2S
sodium bromide sodium sulfide
Na3N
sodium nitride
Al3+
AlBr3
aluminum
bromide
AlN
aluminum
nitride
Al2S3
aluminum
sulfide
140
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Naming Compounds
Transition Metals Form Positive
Ions
Most transition metals and Group 4(14) metals,
 Form 2 or more positive ions
 Zn2+, Ag+, and Cd2+ form only one ion.
141
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Naming Compounds
Guide to Writing Formulas from
the Name
142
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Naming Compounds
Writing Formulas
Write a formula for potassium sulfide.
STEP 1 Identify the cation and anion.
potassium = K+
sulfide
= S2−
STEP 2 Balance the charges.
K+
S2−
K+
2(1+) + 1(2–) = 0
STEP 3 Write the cation first.
2K+ and 1S2− = K2S1 = K2S
143
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Naming Compounds
Writing Formulas
Write a formula for iron(III) chloride.
STEP 1 Identify the cation and anion.
iron (III) = Fe3+ (III = charge of 3+)
chloride = Cl−
STEP 2 Balance the charges.
Fe3+
Cl−
Cl−
Cl−
1(3+) + 3(1–) = 0
STEP 3 Write the cation first.
1Fe3+ and 3Cl− = FeCl3
144
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Naming Compounds
Learning Check
The correct formula for each of the following is:
A. copper(I) nitride
1) CuN
2) CuN3
3) Cu3N
B. lead(IV) oxide
1) PbO2
2) PbO
3) Pb2O4
145
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Naming Compounds
Solution
The correct formula for each of the following is:
A. copper(I) nitride
3) Cu3N
3Cu+ + N3– = 3(1+) + (3–) = 0
B. lead(IV) oxide
1) PbO2
Pb4+ + 2O2– = (4+) + 2(2–) = 0
146
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Naming Compounds
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147
Naming Compounds
Percent Composition, Empirical Formulas,
Molecular Formulas
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Naming Compounds
Formula Masses and Molar masses:
• Molecular mass or molecular weight are used instead of
the term formula mass.
• The formula mass of any compound is the sum of the
average atomic masses of all of the atoms present in the
formula
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149
Naming Compounds
Example of formula mass
• H2O
• 2 H atom weigh 1.0079 each
• 1 O atom weighs 15.9994 each
• 2 x 1.oo79
• +1x 15.9994
• 18.0153 formula mass for water
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150
Naming Compounds
Molar mass as a conversion factor
• Moles x grams/mole = mass in grams
• Mass in grams x 1 mol/grams = moles
• Thus 2 conversions relate mass in grams to
numbers of moles of a substance
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151
Naming Compounds
Example
•
What is the molar mass of Barium nitrate Ba(NO3)2
•
•
•
•
Solution
1 mol Ba x 137.33 g/1 mol Ba = 137.33 g Ba
2 moles N x 14.0067 g/1mole N = 28.0134g N
6 moles O x 15.999g/1mol O = 95.9964g
• Molar mass Ba(NO3)2 = 261.34
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152
Naming Compounds
Example
• What is the mass in grams of 2.5
moles of oxygen gas (O2)
• Solution
• 80.0g
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Naming Compounds
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Percent Composition
• Percent
Composition – the
percentage
by mass of
each element in a
compound
Percent =
Percent composition
of a compound or =
molecule
Part
_______
Whole
x 100%
Mass of element in 1 mol
____________________
Mass of 1 mol
x 100%
Percent Composition
Example: What is the percent composition of Potassium
Permanganate (KMnO )?
4
Molar Mass of KMnO
4
K=
1(39.1) = 39.1
Mn = 1(54.9) = 54.9
O = 4(16.0) = 64.0
MM = 158 g
Percent Composition
Example: What is the percent composition of Potassium
Permanganate (KMnO )?
4
Molar Mass of KMnO
% K
% Mn
K=
1(39.10) = 39.1
Mn = 1(54.94) = 54.9
O = 4(16.00) = 64.0
MM = 158
% O
39.1 g K
158 g
54.9 g Mn
158 g
64.0 g O
158 g
= 158 g
4
x 100 =
x 100 =
x 100 =
24.7 %
34.8 %
40.5 %
Percent Composition
Determine the percentage composition of sodium carbonate (Na CO )?
2 3
Molar Mass
Na = 2(23.00) = 46.0
C = 1(12.01) = 12.0
O = 3(16.00) = 48.0
MM= 106 g
Percent Composition
% Na =
% C =
% O =
46.0 g
106 g
12.0 g
106 g
48.0 g
106 g
x 100% =
43.4 %
x 100% =
11.3 %
x 100% =
45.3 %
Percent Composition
Determine the percentage composition of ethanol (C H OH)?
2 5
% C = 52.13%, % H = 13.15%, % O = 34.72%
_______________________________________________
Determine the percentage composition of sodium oxalate
(Na C O )?
2 2 4
% Na = 34.31%, % C = 17.93%, % O = 47.76%
Percent Composition
Calculate the mass of bromine in 50.0 g of Potassium bromide.
1. Molar Mass of KBr
K = 1(39.10) = 39.10
Br =1(79.90) =79.90
MM = 119.0
2.
79.90 g
___________
= 0.6714
119.0 g
3.
0.6714 x 50.0g = 33.6 g Br
Percent Composition
Calculate the mass of nitrogen in 85.0 mg of the amino acid lysine, C H N O .
6 14 2 2
1. Molar Mass of C H N O
6 14 2 2
C = 6(12.01) = 72.06
H =14(1.01) = 14.14
N = 2(14.01) = 28.02
O = 2(16.00) = 32.00
MM = 146.2
2.
28.02 g
___________
= 0.192
146.2 g
3.
0.192 x 85.0 mg = 16.3 mg N
Hydrates
Hydrated salt – salt that has water molecules trapped
within the crystal lattice
Examples: CuSO •5H O , CuCl •2H O
4
2
2
2
Anhydrous salt – salt without water molecules
Examples: CuCl
2
Can calculate the percentage of water in a hydrated
salt.
Percent Composition
Calculate the percentage of water in sodium carbonate decahydrate, Na CO •10H O.
2 3
2
1. Molar Mass of Na CO •10H O
2 3
2
Na = 2(22.99) = 45.98
C
= 1(12.01) = 12.01
H = 20(1.01) = 20.2
3.
O = 13(16.00)= 208.00
180.2 g
_______
MM = 286.2
2.
286.2 g
Water
H = 20(1.01) = 20.2
O = 10(16.00)= 160.00
MM = 180.2
or
H = 2(1.01) = 2.02
O = 1(16.00) = 16.00
MM H2O = 18.02
So…
10 H O = 10(18.02) = 180.2
2
x 100%=
67.97 %
Percent Composition
Calculate the percentage of water in Aluminum bromide hexahydrate, AlBr •6H O.
3
2
1. Molar Mass of AlBr •6H O
3
2
Al = 1(26.98) = 26.98
Br = 3(79.90) = 239.7
H = 12(1.01) = 12.12
O = 6(16.00) = 96.00
MM = 374.8
2.
Water
H = 12(1.01) = 12.1
O = 6(16.00)= 96.00
MM = 108.1
or
MM = 18.02
For 6 H2O = 6(18.02) = 108.2
3.
108.1 g
_______
374.8 g
x 100%=
28.85 %
Percent Composition
If 125 grams of magnesium sulfate heptahydrate is completely dehydrated, how many
grams of anhydrous magnesium sulfate will remain?
MgSO
1. Molar Mass
Mg = 1 x 24.31 = 24.31 g
S = 1 x 32.06 = 32.06 g
O = 4 x 16.00 = 64.00 g
MM = 120.37 g
2. % MgSO
4
. 7 H O
2
4
120.4 g
246.5 g
X 100 =
3. Grams anhydrous MgSO
H = 2 x 1.01 = 2.02 g
O = 1 x 16.00 = 16.00 g
MM = 18.02 g
MM H O =
2 g = 126.1 g
7 x 18.02
Total MM =
120.4 g + 126.1 g = 246.5 g
0.4884 x 125 =
48.84 %
4
61.1 g
Percent Composition
If 145 grams of copper (II) sulfate pentahydrate is completely dehydrated, how many
grams of anhydrous copper sulfate will remain?
CuSO . 5 H O
4
2
1. Molar Mass
Cu = 1 x 63.55 = 63.55 g
S = 1 x 32.06 = 32.06 g
O = 4 x 16.00 = 64.00 g
MM = 159.61 g
2. % CuSO
4
159.6 g
249.7 g
X 100 =
3. Grams anhydrous CuSO
H = 2 x 1.01 = 2.02 g
O = 1 x 16.00 = 16.00 g
MM = 18.02 g
MM H O =
2 g = 90.1 g
5 x 18.02
Total MM =
159.6 g + 90.1 g = 249.7 g
0.6392 x 145 =
63.92 %
4
92.7 g
Percent Composition
A 5.0 gram sample of a hydrate of BaCl was heated, and only 4.3 grams of the
2
anhydrous salt remained. What percentage
of water was in the hydrate?
1. Amount water lost
5.0 g hydrate
4.3 g anhydrous salt
-0.7
g water
2. Percent of water
0.7 g water
5.0 g hydrate
x 100 =
14 %
Percent Composition
A 7.5 gram sample of a hydrate of CuCl was heated, and only 5.3 grams of the
2
anhydrous salt remained. What percentage
of water was in the hydrate?
1. Amount water lost
7.5 g hydrate
5.3 g anhydrous salt
-2.2
g water
2. Percent of water
2.2 g water
7.5 g hydrate
x 100 =
29 %
Percent Composition
A 5.0 gram sample of Cu(NO ) •nH O is heated, and 3.9 g of the anhydrous salt
32
remains. What is the value of
n? 2
1. Amount water lost
5.0 g hydrate
3.9 g anhydrous salt
-1.1
g water
3. Amount of water
0.22 x 18.02 =
2. Percent of water
1.1 g water
5.0 g hydrate
x 100 =
22 %
4.0
Percent Composition
A 7.5 gram sample of CuSO •nH O is heated, and 5.4 g of the anhydrous salt remains.
4
2
What is the value of n?
1. Amount water lost
7.5 g hydrate
5.4 g anhydrous salt
-2.1
g water
3. Amount of water
0.28 x 18.02 =
2. Percent of water
2.1 g water
7.5 g hydrate
x 100 =
28 %
5.0
Formulas
Percent composition allow you to calculate the simplest
in compound.
ratio among the atoms found
Empirical Formula – formula of a compound that expresses
ratio of atoms.
lowest whole number
Molecular Formula – actual formula of a compound showing
present
the number of atoms
Examples:
C H
4 10
C H
2 5
- molecular
C H O
6 12 6
- molecular
- empirical
CH O
2
- empirical
Formulas
Is H O an empirical or molecular formula?
2 2
Molecular, it can be reduced to HO
HO = empirical formula
Calculating Empirical Formula
An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen.
Calculate the empirical formula.
1. Determine the number of grams of each element in the compound.
4.151 g Al
and
3.692 g O
2. Convert masses to moles.
4.151 g Al
1 mol Al
=
0.1539 mol Al
=
0.2308 mol O
26.98 g Al
3.692 g O
1 mol O
16.00 g O
Calculating Empirical Formula
An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen.
Calculate the empirical formula.
3. Find ratio by dividing each element by smallest amount of moles.
0.1539 moles Al
= 1.000 mol Al
0.1539
0.2308 moles O
= 1.500 mol O
0.1539
4. Multiply by common factor to get whole number. (cannot have fractions of atoms in compounds)
O = 1.500 x 2 = 3
Al = 1.000 x 2 = 2
therefore,
Al O
2 3
Calculating Empirical Formula
A 4.550 g sample of cobalt reacts with 5.475 g chlorine to form a binary compound. Determine the
empirical formula for this compound.
4.550 g Co
1 mol Co
= 0.07721 mol Co
58.93 g Co
5.475 g Cl
1 mol Cl
= 0.1544 mol Cl
35.45 g Cl
0.07721 mol Co
=1
0.07721
0.1544 mol Cl
0.07721
CoCl
2
=2
Calculating Empirical Formula
When a 2.000 g sample of iron metal is heated in air, it reacts with oxygen to achieve a final mass of
2.573 g. Determine the empirical formula.
Fe = 2.000 g
2.000 g Fe
1 mol Fe
O = 2.573 g – 2.000 g = 0.5730 g
= 0.03581 mol Fe
55.85 g Fe
0.573 g O
1 mol O
= 0.03581 mol Fe
16.00 g
1:1
FeO
Calculating Empirical Formula
A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead,
0.00672g of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula
for lead arsenate.
1.3813 g Pb
1 mol Pb
= 0.006667 mol Pb
207.2 g Pb
0.00672 gH
1 mol H
= 0.00667 mol H
1.008 g H
0.4995 g As
1 mol As
= 0.006667 mol As
74.92 g As
0.4267g Fe
1 mol O
16.00 g O
= 0.02667 mol O
Calculating Empirical Formula
A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead,
0.00672g of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula
for lead arsenate.
0.006667 mol Pb
0.006667
0.00667 mol H
0.006667
0.006667 mol As
0.006667
0.02667 mol O
0.006667
= 1.000 mol Pb
= 1.00 mol H
= 1.000 mol As
= 4.000 mol O
PbHAsO
4
Calculating Empirical Formula
The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and
14.14% oxygen. Calculate the empirical formula for Nylon-6.
Step 1:
In 100.00g of Nylon-6 the masses of elements present are 63.38 g C, 12.38 g n, 9.80 g H, and 14.14 g O.
Step 2:
63.38 g C
1 mol C
= 5.302 mol C
9.80 g H
12.01 g C
12.38 g N
1 mol N
14.01 g N
1 mol H
= 9.72 mol H
1.01 g H
= 0.8837 mol N
14.14 g O
1 mol O
16.00 g O
= 0.8832 mol O
Calculating Empirical Formula
The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and
14.14% oxygen. Calculate the empirical formula for Nylon-6.
Step 3:
5.302 mol C
0.8837
0.8837 mol N
0.8837
9.72 mol H
0.8837
0.8837 mol O
0.8837
= 6.000 mol C
= 1.000 mol N
= 11.0 mol H
= 1.000 mol O
6:1:11:1
C NH O
6
11
Calculating molecular formula
• It is not possible to determine the correct
molecular formula unless the molecular mass
of the substance has been determined
• The relationship between the simplest
formula and the molecular mass is:
• (simple formula)x = molecular formula
• Where x is a whole number multiple of the
simple formula
Calculating Molecular Formula
A white powder is analyzed and found to have an empirical formula of P2O5. The
compound has a molar mass of 283.88g. What is the compound’s molecular formula?
Step 1: Molar Mass
P = 2 x 30.97 g = 61.94g
O = 5 x 16.00g = 80.00 g
141.94 g
Step 2: Divide MM by
Empirical Formula Mass
238.88 g
141.94g
=2
Step 3: Multiply
(P O ) =
2 52
P O
4 10
Calculating Molecular Formula
A compound has an experimental molar mass of 78 g/mol. Its empirical formula is
CH. What is its molecular formula?
(CH) =
6
C = 12.01 g
H = 1.01 g
13.01 g
C H
6 6
78 g/mol
13.01 g/mol
=6
Oxidation Numbers
• Are used to indicate general distributions of
electrons among bonded atoms.
• Refer to handout for rules of oxidation
numbers
Ex find oxidation # of following:
• UF6
• ClO3• Solution
•
•
U = +6
Cl =+5
F = -1
O =-2