Transcript Chapter 2

Chapter 2
Atoms, Molecules
& Ions
Dr. S. M. Condren
Quantum Corral
http://www.almaden.ibm.com/vis/stm/corral.html
Dr. S. M. Condren
Scanning Tunneling Microscope
Dr. S. M. Condren
Scanning Tunneling Microscope
Dr. S. M. Condren
Scanning Tunneling Microscope
Dr. S. M. Condren
http://www.cbu.edu/~mcondren/SeeAtoms.htm
Dr. S. M. Condren
Developed in collaboration with the
Institute for Chemical Education and the
Magnetic Microscopy Center
University of Minnesota
http://www.physics.umn.edu/groups/mmc/
Dr. S. M. Condren
http://mrsec.wisc.edu/
http://mrsec.wisc.edu/
Sample
http://www.nsf.gov/mps/dmr/mrsec.htm
http://www.nsf.gov/mps/dmr/mrsec.htm
Probe
Pull Probe Strip
Dr. S. M. Condren
Pull Probe Strip
Which best represents the poles?
(a)
(b)
(c)
North
South
Dr. S. M. Condren
Atoms & Molecules
Atoms
• can exist alone or enter into chemical
combination
• the smallest indivisible particle of an
element
Molecules
• a combination of atoms that has its own
characteristic set of properties
Dr. S. M. Condren
Law of Constant Composition
A chemical compound always contains the
same elements in the same proportions by
mass.
Dr. S. M. Condren
Law of Multiple Proportions
• the same elements can be combined to form
different compounds by combining the
elements in different proportions
Dr. S. M. Condren
Dalton’s Atomic Theory
Postulates
• proposed in 1803
• know at least 2 for first exam
Dr. S. M. Condren
Dalton’s Atomic Theory
Postulate 1
• An element is composed of tiny particles
called atoms.
• All atoms of a given element show the same
chemical properties.
Dr. S. M. Condren
Dalton’s Atomic Theory
Postulate 2
• Atoms of different elements have different
properties.
Dr. S. M. Condren
Dalton’s Atomic Theory
Postulate 3
• Compounds are formed when atoms of two
or more elements combine.
• In a given compound, the relative number
of atoms of each kind are definite and
constant.
Dr. S. M. Condren
Dalton’s Atomic Theory
Postulate 4
• In an ordinary chemical reaction, no atom
of any element disappears or is changed into
an atom of another element.
• Chemical reactions involve changing the
way in which the atoms are joined together.
Dr. S. M. Condren
Radioactivity
Dr. S. M. Condren
Radioactivity
• Alpha – helium-4 nucleus
• Beta – high energy electron
• Gamma – energy resulting from transitions
from one nuclear energy level to another
Dr. S. M. Condren
Alpha Radiation
•
•
•
•
•
•
composed of 2 protons and 2 neutrons
thus, helium-4 nucleus
+2 charge
mass of 4 amu
creates element with atomic number 2 lower
Ra226  Rn222 + He4(a)
Dr. S. M. Condren
Beta Radiation
• composed of a high energy electron which
was ejected from the nucleus
• “neutron” converted to “proton”
• very little mass
• -1 charge
• creates element with atomic number 1
higher
• U239  Np239 + b-1
Dr. S. M. Condren
Gamma Radiation
• nucleus has energy levels
• energy released from nucleus as the nucleus
changes from higher to lower energy levels
• no mass
• no charge
• Ni60*  Ni60 + g
Dr. S. M. Condren
Cathode Ray Tube
Dr. S. M. Condren
Thompson’s Charge/Mass Ratio
Dr. S. M. Condren
Millikin’s Oil Drop
Dr. S. M. Condren
Rutherford’s Gold Foil
Dr. S. M. Condren
Rutherford’s Model of the Atom
Dr. S. M. Condren
Rutherford’s Model of the Atom
• atom is composed mainly of vacant space
• all the positive charge and most of the mass
is in a small area called the nucleus
• electrons are in the electron cloud
surrounding the nucleus
Dr. S. M. Condren
Structure of the Atom
Composed of:
• protons
• neutrons
• electrons
Dr. S. M. Condren
Structure of the Atom
Composed of:
• protons
• neutrons
• electrons
• protons
– found in nucleus
– relative charge of +1
– relative mass of 1.0073 amu
Dr. S. M. Condren
Structure of the Atom
Composed of:
• protons
• neutrons
• electrons
• neutrons
– found in nucleus
– neutral charge
– relative mass of 1.0087 amu
Dr. S. M. Condren
Structure of the Atom
Composed of:
• protons
• neutrons
• electrons
• electrons
– found in electron cloud
– relative charge of -1
– relative mass of 0.00055 amu
Dr. S. M. Condren
Size of Nucleus
If the nucleus were
1” in diameter,
the atom would be
1.5 miles in diameter.
Dr. S. M. Condren
Ions
• charged single atom
• charged cluster of atoms
Dr. S. M. Condren
Ions
• cations
– positive ions
• anions
– negative ions
• ionic compounds
– combination of cations and anions
– zero net charge
Dr. S. M. Condren
Atomic number, Z
• the number of protons in the nucleus
• the number of electrons in a neutral atom
• the integer on the periodic table for each
element
Dr. S. M. Condren
Isotopes
• atoms of the same element which differ in
the number of neutrons in the nucleus
• designated by mass number
Dr. S. M. Condren
Mass Number, A
• integer representing the approximate mass
of an atom
• equal to the sum of the number of protons
and neutrons in the nucleus
Dr. S. M. Condren
Masses of Atoms
Carbon-12 Scale
Dr. S. M. Condren
Isotopes of Hydrogen
H-1, 1H, protium
• 1 proton and no neutrons in nucleus
• only isotope of any element containing no
neutrons in the nucleus
• most common isotope of hydrogen
Dr. S. M. Condren
Isotopes of Hydrogen
H-2 or D, 2H, deuterium
• 1 proton and 1 neutron in nucleus
Dr. S. M. Condren
Isotopes of Hydrogen
H-3 or T, 3H, tritium
• 1 proton and 2 neutrons in nucleus
Dr. S. M. Condren
Isotopes of Oxygen
O-16
• 8 protons, 8 neutrons, & 8 electrons
O-17
• 8 protons, 9 neutrons, & 8 electrons
O-18
• 8 protons, 10 neutrons, & 8 electrons
Dr. S. M. Condren
The radioactive isotope 14C has how
many neutrons?
6, 8, other
Dr. S. M. Condren
The identity of an element is determined by the
number of which particle?
protons, neutrons, electrons
Dr. S. M. Condren
Mass Spectrometer
Dr. S. M. Condren
Mass Spectra of Neon
Dr. S. M. Condren
Measurement of Atomic Masses
Mass Spectrometer
a simulation is available at
http://www.colby.edu/chemistry/
OChem/DEMOS/MassSpec.html
Dr. S. M. Condren
Atomic Masses and
Isotopic Abundances
natural atomic masses =
sum[(atomic mass of isotope)
*(fractional isotopic abundance)]
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35 y = fraction Cl-37
x+y=1
y=1-x
(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453
Thus:
34.96885*x + 36.96590*y = 35.453
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35
y = fraction Cl-37
x + y = 1 <=> y = 1 - x
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35
y = fraction Cl-37
x + y = 1 <=> y = 1 - x
(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35
y = fraction Cl-37
y=1-x
(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453
x + y = 1 <=>
34.96885*x + 36.96590*y = 35.453
34.96885*x + 36.96590*(1-x) = 35.453
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35
y = fraction Cl-37
x + y = 1 <=> y = 1 - x
(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) =
35.453
34.96885*x + 36.96590*y = 35.453
34.96885*x + 36.96590*(1-x) = 35.453
(34.96885 - 36.96590)x + 36.96590 =
35.453
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35
y = fraction Cl-37
x + y = 1 <=> y = 1 - x
(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453
34.96885*x + 36.96590*y = 35.453
34.96885*x + 36.96590*(1-x) = 35.453
(34.96885 - 36.96590)x + 36.96590 = 35.453
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35
y = fraction Cl-37
x + y = 1 <=> y = 1 - x
(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453
34.96885*x + 36.96590*y = 35.453
34.96885*x + 36.96590*(1-x) = 35.453
(34.96885 - 36.96590)x + 36.96590 = 35.453
(34.96885 - 36.96590)x = (35.453 - 36.96590)
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35
y = fraction Cl-37
x + y = 1 <=> y = 1 - x
(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453
34.96885*x + 36.96590*y = 35.453
34.96885*x + 36.96590*(1-x) = 35.453
(34.96885 - 36.96590)x + 36.96590 = 35.453
(34.96885 - 36.96590)x = (35.453 - 36.96590)
- 1.99705x = - 1.5129
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35
y = fraction Cl-37
x + y = 1 <=> y = 1 - x
(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453
34.96885*x + 36.96590*y = 35.453
34.96885*x + 36.96590*(1-x) = 35.453
(34.96885 - 36.96590)x + 36.96590 = 35.453
(34.96885 - 36.96590)x = (35.453 - 36.96590)
- 1.99705x = - 1.5129
1.99705x = 1.5129
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35
y = fraction Cl-37
x + y = 1 <=> y = 1 - x
(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453
34.96885*x + 36.96590*y = 35.453
34.96885*x + 36.96590*(1-x) = 35.453
(34.96885 - 36.96590)x + 36.96590 = 35.453
(34.96885 - 36.96590)x = (35.453 - 36.96590)
- 1.99705x = - 1.5129
1.99705x = 1.5129
x = 0.7553 <=> 75.53% Cl-35
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35
y = fraction Cl-37
x + y = 1 <=> y = 1 - x
(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453
34.96885*x + 36.96590*y = 35.453
34.96885*x + 36.96590*(1-x) = 35.453
(34.96885 - 36.96590)x + 36.96590 = 35.453
(34.96885 - 36.96590)x = (35.453 - 36.96590)
- 1.99705x = - 1.5129
1.99705x = 1.5129
x = 0.7553 <=> 75.53% Cl-35
y=1-x
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35
y = fraction Cl-37
x + y = 1 <=> y = 1 - x
(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453
34.96885*x + 36.96590*y = 35.453
34.96885*x + 36.96590*(1-x) = 35.453
(34.96885 - 36.96590)x + 36.96590 = 35.453
(34.96885 - 36.96590)x = (35.453 - 36.96590)
- 1.99705x = - 1.5129
1.99705x = 1.5129
x = 0.7553 <=> 75.53% Cl-35
y = 1 - x = 1.0000 - 0.7553
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35
y = fraction Cl-37
x + y = 1 <=> y = 1 - x
(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453
34.96885*x + 36.96590*y = 35.453
34.96885*x + 36.96590*(1-x) = 35.453
(34.96885 - 36.96590)x + 36.96590 = 35.453
(34.96885 - 36.96590)x = (35.453 - 36.96590)
- 1.99705x = - 1.5129
1.99705x = 1.5129
x = 0.7553 <=> 75.53% Cl-35
y = 1 - x = 1.0000 - 0.7553 = 0.2447
Dr. S. M. Condren
Example: Chlorine has two isotopes, Cl-35 and Cl-37,
which have masses of 34.96885 and 36.96590 amu,
respectively. The natural atomic mass of chlorine is
35.453 amu. What are the percent abundances of the
two isotopes?
let x = fraction Cl-35
y = fraction Cl-37
x + y = 1 <=> y = 1 - x
(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453
34.96885*x + 36.96590*y = 35.453
34.96885*x + 36.96590*(1-x) = 35.453
(34.96885 - 36.96590)x + 36.96590 = 35.453
(34.96885 - 36.96590)x = (35.453 - 36.96590)
- 1.99705x = - 1.5129
1.99705x = 1.5129
x = 0.7553 <=> 75.53% Cl-35
y = 1 - x = 1.0000 - 0.7553 = 0.2447
24.47% Cl-37
Dr. S. M. Condren
Development of Periodic Table
Newlands - English
1864 - Law of Octaves - every 8th element
has similar properties
Dr. S. M. Condren
Development of Periodic Table
Dmitri Mendeleev - Russian
1869 - Periodic Law - allowed him to
predict properties of
unknown elements
Dr. S. M. Condren
Mendeleev’s Periodic Table
the elements are arranged according to
increasing atomic weights
Dr. S. M. Condren
Mendeleev’s Periodic Table
Missing elements: 44, 68, 72, & 100 amu
Dr. S. M. Condren
Properties of Ekasilicon
Dr. S. M. Condren
Modern Periodic Table
Moseley, Henry Gwyn Jeffreys
1887–1915, English physicist.
Studied the relations among bright-line spectra of different
elements.
Derived the ATOMIC NUMBERS from the frequencies of
vibration of X-rays emitted by each element.
Moseley concluded that the atomic number is equal to the
charge on the nucleus.
This work explained discrepancies in Mendeleev’s Periodic
Law.
Dr. S. M. Condren
Modern Periodic Table
the elements are arranged according to
increasing atomic numbers
Dr. S. M. Condren
Periodic
Table
of the
Periodic
Table
of
the
Elements
Elements
IA
1
1
3
4
5
6
7
III B
IV B
VB
VI B
VII B
VIII B
IB
II B
III A
IV A
VA
VI A
VII A
1
VIII A
2
H
H
He
1.008
1.008
4.0026
10
3
2
II A
4
5
6
7
8
9
Li
Be
B
C
N
O
F
Ne
6.939
9.0122
10.811
12.011
14.007
15.999
18.998
20.183
11
12
13
14
15
16
17
18
Na
Mg
Al
Si
P
S
Cl
Ar
22.99
24.312
26.982
28.086
30.974
32.064
35.453
39.948
19
20
31
32
33
34
35
36
21
22
23
24
25
26
27
28
29
30
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
39.102
40.08
44.956
47.89
50.942
51.996
54.938
55.847
58.932
58.71
63.54
65.37
69.72
72.59
74.922
78.96
Br
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
85.468
87.62
88.906
91.224
92.906
95.94
* 98
101.07
102.91
106.42
107.9
112.41
114.82
118.71
121.75
127.61
126.9
131.29
55
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
79.909
Kr
83.8
Cs
Ba
**La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
132.91
137.33
138.91
178.49
180.95
183.85
186.21
190.2
192.22
195.08
196.97
200.29
204.38
207.2
208.98
* 209
* 210
* 222
87
88
89
104
105
106
107
108
109
110
111
112
113
114
115
116
Rf
Ha
Sg
Ns
Hs
Mt
* 261
* 262
* 263
* 262
* 265
* 268
Fr
* 223
Ra ***Ac
226.03 227.03
58
* Designates that **Lanthanum
all isotopes are
Series
radioactive
*** Actinium
Series
59
60
61
62
Uun Uuu Uub
* 269
* 272
63
64
* 277
65
Uut
118
Uuq Uup Uuh
Uuo
*284
*285
*288
*292
Based on symbols used by ACS
66
67
68
69
S.M.Condren 2007
*294
70
71
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
140.12
140.91
144.24
* 145
150.36
151.96
157.25
158.93
162.51
164.93
167.26
168.93
173.04
174.97
90
91
92
93
94
95
96
97
98
99
100
101
102
103
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
232.04
231.04
238.03
237.05
* 244
* 243
* 247
* 247
* 251
* 252
* 257
* 258
* 259
* 260
Dr. S. M. Condren
Organization of Periodic Table
• period - horizontal row
• group - vertical column
Dr. S. M. Condren
Family Names
Group IA
alkali metals
Group IIA
alkaline earth metals
Group VIIA
halogens
Group VIIIA
noble gases
transition metals
inner transition metals
• lanthanum series rare earths
• actinium series
trans-uranium series
Dr. S. M. Condren
Types of Elements
metals
nonmetals
metalloids - semimetals
Dr. S. M. Condren
Elements, Compounds, and
Formulas
Elements
• can exist as single atoms or molecules
Compounds
• combination of two or more elements
• molecular formulas for molecular
compounds
• empirical formulas for ionic compounds
Dr. S. M. Condren
Organic Compounds
Organic Chemistry
• branch of chemistry in which carbon
compounds and their reactions are
studied.
• the chemistry of carbon-hydrogen
compounds
Dr. S. M. Condren
Inorganic Compounds
Inorganic Chemistry
• field of chemistry in which are studied
the chemical reactions and properties of
all the chemical elements and their
compounds, with the exception of the
hydrocarbons (compounds composed of
carbon and hydrogen) and their
derivatives.
Dr. S. M. Condren
Molecular and Structural Formulas
Dr. S. M. Condren
Bulk Substances
• mainly ionic compounds
– empirical formulas
– structural formulas
Dr. S. M. Condren
Models of Sodium Chloride
NaCl “table salt”
Dr. S. M. Condren
How many atoms are in the formula Al2(SO4)3?
3, 5, 17
Dr. S. M. Condren
Naming Binary
Molecular Compounds
• For compounds composed of two nonmetallic elements, the more metallic
element is listed first.
• To designate the multiplicity of an element,
Greek prefixes are used:
mono => 1; di => 2; tri => 3; tetra => 4;
penta => 5; hexa => 6; hepta => 7;
octa => 8
Dr. S. M. Condren
Common Compounds
H2O
water
NH3
ammonia
N2O
nitrous oxide
CO
carbon monoxide
CS2
SO3
sulfur trioxide
CCl4
carbon tetrachloride
PCl5
phosphorus pentachloride
SF6
sulfur hexafluoride
carbon disulfide
Dr. S. M. Condren
Alkanes - CnH2n+2
•
•
•
•
•
methane - CH4
ethane - C2H6
propane - C3H8
butanes - C4H10
pentanes - C5H12
•
•
•
•
•
hexanes - C6H14
heptanes - C7H16
octanes - C8H18
nonanes - C9H20
decanes - C10H22
Dr. S. M. Condren
Burning of
Propane Gas
Dr. S. M. Condren
Butanes
Dr. S. M. Condren
Ionic Bonding
Characteristics of compounds with ionic
bonding:
• non-volatile, thus high melting points
• solids do not conduct electricity, but melts
(liquid state) do
• many, but not all, are water soluble
Dr. S. M. Condren
Ion Formation
Dr. S. M. Condren
Valance
Charge on Ions
• compounds have electrical neutrality
• metals form positive monatomic ions
• non-metals form negative monatomic ions
Dr. S. M. Condren
Valence of Metal Ions
Monatomic Ions
Group IA
=> +1
Group IIA
=> +2
Maximum positive valence
equals
Group A #
Dr. S. M. Condren
Valence of Non-Metal Ions
Monatomic Ions
Group VIA
Group VIIA
=> -2
=> -1
Maximum negative valence
equals
(8 - Group A #)
Dr. S. M. Condren
Charges of Some Important Ions
Dr. S. M. Condren
Polyatomic Ions
• more than one atom joined together
• have negative charge except for NH4+ and
its relatives
• negative charges range from -1 to -4
Dr. S. M. Condren
Polyatomic Ions
ammonium
perchlorate
cyanide
hydroxide
nitrate
NH4+
ClO41CN1OH1NO31-
sulfate
carbonate
phosphate
Dr. S. M. Condren
SO42CO32PO43-
Names of Ionic Compounds
1. Name the metal first.
If the metal has more than one oxidation
state, the oxidation state is specified by
Roman numerals in parentheses.
2. Then name the non-metal,
changing the ending of the non-metal to
-ide.
Dr. S. M. Condren
Nomenclature
NaCl
sodium chloride
Fe2O3
iron(III) oxide
N2O4
dinitrogen tetroxide
KI
potassium iodide
Mg3N2
magnesium nitride
SO3
sulfur trioxide
Dr. S. M. Condren
Nomenclature
NH4NO3
ammonium nitrate
KClO4
potassium perchlorate
CaCO3
calcium carbonate
NaOH
sodium hydroxide
Dr. S. M. Condren
Nomenclature Drill
Available for PCs:
– on your disk to use at home or in the dorm
– in the Chemistry Resource Center
– off the web under Chapter 2, Links
http://www.cbu.edu/~mcondren/c115lkbk.html
Dr. S. M. Condren
How many moles of ions are there per
mole of Al2(SO4)3?
2, 3, 5
Dr. S. M. Condren
Chemical Equation
• reactants
• products
• coefficients
reactants -----> products
Dr. S. M. Condren
Writing and Balancing
Chemical Equations
• Write a word equation.
• Convert word equation into formula
equation.
• Balance the formula equation by the use of
prefixes (coefficients) to balance the
number of each type of atom on the reactant
and product sides of the equation.
Dr. S. M. Condren
Example
Hydrogen gas reacts with oxygen gas to
produce water.
Step 1.
hydrogen + oxygen -----> water
Step 2.
H2 + O2 -----> H2O
Step 3.
2 H2 + O2 -----> 2 H2O
Dr. S. M. Condren
Example
Iron(III) oxide reacts with carbon monoxide to
produce the iron oxide (Fe3O4) and carbon
dioxide.
iron(III) oxide + carbon monoxide -----> Fe3O4 + carbon dioxide
Fe2O3 + CO -----> Fe3O4 + CO2
3 Fe2O3 + CO -----> 2 Fe3O4 + CO2
Dr. S. M. Condren