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Chapter 1
The Atomic Nature of Matter
1-1 Chemistry: Science of Change
1-2 The Composition of Matter
1-3 The Atomic Theory of Matter
1-4 Chemical Formulas and Relative
Atomic Masses
1-5 The Building Blocks of the
Atom
1-6 Finding Atomic Masses the
Modern Way
1-7 The Mole Concept: Counting
and Weighing Atoms and Molecules
1-8 Finding Empirical and
Molecular Formulas the Modern
Way
1-9 Volume and Density
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Atomic Theory of Matter
•
•
Law of conservation of mass:
Mass is neither created nor
destroyed in a chemical reaction
Dalton’s Atomic Theory of
Matter (1808):
1. All matter consists of solid and
indivisible atoms
2. All atoms of a given chemical
element are identical in mass and in
all other properties
3. Different elements have different
kinds of atoms; these atoms differ in
mass from element to element
4. Atoms are indestructible and retain
their identity in all chemical reactions
5. The formation of a compound from
its elements occurs through the
combination of atoms of unlike
elements in small whole-number
ratio.
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Chemical Formulas and
Relative Atomic Masses
• Chemical Formulas display
symbols for the elements and
the relative number of atoms
– E.g., NH3, CO2, CH3CO2H or
C2H4O2
• Molecules are groupings of two
or more atoms bound closely
together by strong forces that
maintain them in a persistent
combination
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Building Blocks of the Atom
• Electrons, Protons and Neutrons
– Electrons discovered in 1897 by
Thomson
– Rutherford proposed that the
atomic nucleus was composed of
neutral particles called Neutrons
and positively charged particles
called protons
– Neutron number = N
– Atomic number = Z = number of
Protons
– Atomic mass number = A
A=Z+N
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http://www.chemsoc.org/viselements/pages/alc
hemist/alchemy.html
http://www.chemsoc.org
http://www.chemsoc.org/viselements/
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pages/pertable_j.htm
6
12.011
C
Carbon
7
14.007
N
Nitrogen
14
28.086
Si
Silicon
15
30.794
P
Phosphorus
Semi
Metal
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Non
Metal
OFB Chapter 1
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6
Atomic
Mass
12.011
C
Atomic
Number
Carbon
A=Z+N
Atomic Mass = # Protons + # Neutrons
For Carbon, 12 = 6 + Neutrons
Neutrons = 6
Every Carbon atom has 6 electrons, 6
protons and 6 neutrons
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• Mass Spectrometer
Mass
accelerates ions (or molecular
Spectrometry
ions) in an electric field and
and Isotopes
then separates those ions by
relative mass in a magnetic
field
• Theory and Experiment
M + e– (70 eV)
M+ + 2e–
M+
lower mass ions
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Mass Spectrometry and
Isotopes
• Mass Spectrometer
accelerates ions (or molecular
ions) in an electric field and
then separates those ions by
relative mass in a magnetic
field
17
35.453
Cl
Chlorine
Relative Amount
Mass Spectrometer Separation of Chlorine
100
80
60
40
20
0
35
37
Relative Mass
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Isotopes of Cl:
031
150 msec
EC/ECp,1
1.980
S-31/P30
30.9924
032
298 msec
EC/ECa/EC
p,12.685
S-32/Si28/P-31
31.985688
033
2.511
sec
EC,5.583
MeV
S-33
32.97745
034
1.5264
sec
EC,5.492
MeV
S-34
33.97376
035
75.77%
038
37.24 min
B-,4.917
MeV
Ar-38
37.96801
0
039
55.6 min
B-,3.442
MeV
Ar-39
38.968008
040
1.35 min
B-,7.480
MeV
Ar-40
39.97041
3
041
38.4 sec
B-,5.730
MeV
Ar-41
40.97064
9
042
6.8 sec
B-,9.430
MeV
Ar-42
41.9731
72
045
400 msec
B-/Bn,10.800
Ar-45/Ar44
44.97971
0
046
0.22 sec
B-/B-n,6.900
Ar-46/Ar-45
45.984111
047
200 nsec
B-/Bn,14.700
Ar-47/Ar46
46.98797
6
MASS
abund.
Halflife
Particle, Energy
Decay Product(s)
Isotopic Mass
Stable
34.9688
036
3.01E+5
yr
B/EC,10.4
13
Ar-36/S36
35.9683
037
24.23%
043
3.3 sec
B-,7.950
MeV
Ar-43
42.9742
02
044
0.43
sec
B-/Bn,3.920
Ar44/Ar43
43.9785
39
Stable
36.9659
35Cl
contains
protons and
neutrons
37Cl
contains
protons and
neutrons
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Atoms
• Avogadro’s Number is the number of
12C atoms in exactly 12 grams of
carbon
N0 = 6.0221420 X 1023
• The mass, in grams, of Avogadro's
number of atoms of an element is
numerically equal to the relative
atomic mass of that element
mass of atom C =
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Molecules
• Relative Molecular Mass of a
molecule equals the sum of the
relative atomic masses of all of
the atoms making up the
molecule
relative molecular mass of CO 2 =
mass
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CO
2
molecule
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Moles
• A mole measures the chemical amount of a
substance
• Mole is an abbreviation of gram molecular
weight
• One mole of a substance equals the
amount that contains Avogadro's number
of atoms, molecules.
• One mole = Molar mass (M) of that
element or molecule
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Exercise 1-6
•
Molecules of isoamyl acetate
have the formula C7H14O2.
Calculate (a) how many moles
and (b) how many molecules
are present in 0.250 grams of
isoamyl acetate.
Strategy:
•
1. Calculate molar mass of
C7H14O2
2. Calculate the number of moles
in 0.250 grams
3. Using Avogadro’s number to
calculate the number of
molecules in the number of
moles of C7H14O2
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Exercise 1-6
•
•
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Molecules of isoamyl acetate have
the formula C7H14O2. Calculate (a)
how many moles and (b) how
many molecules are present in
0.250g of isoamyl acetate.
Solution:
1.
Calculate molar mass of C7H14O2
2.
Calculate the number of moles in
0.250 grams
3.
Using Avogadro’s number calculate
the number of molecules in “n” moles
of C7H14O2
OFB Chapter 1
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Percentage Composition from
Empirical or Molecular Formula
Exercise 1-8
•
Tetrodotoxin, a potent poison
found in the ovaries and liver of
the globefish, has the empirical
formula C11H17N3O8. Calculate the
mass percentages of the four
element in this compound.
Strategy:
1. Calculate molar mass of C11H17N3O,
by finding the mass contributed by
each element
2. Divide the mass for each element by
the total mass of the compound.
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Exercise 1-8
Tetrodotoxin has the empirical formula
C11H17N3O8. Calculate the mass percentages of
the four element in this compound.
Solution:
1.
Calculate molar mass of C11H17N3O8, by
finding the mass contributed by each element
M=
M = 319.16
2.
Divide the mass for each element by the total
mass of the compound.
132.01
%C =
319.16
17.134
%H =
319.16
42.021
%N =
319.16
127.99
%O =
319.16
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Millimoles and Milligrams
1 mmol = 1 millimole
=1x10-3 mol
1mg = 1 milligram
=1x10-3 g
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Exercise 1-10
Moderate Heating of 97.44 mg of a
compound containing nickel,
carbon and oxygen and no other
elements drives off all of the
carbon and oxygen in the form of
carbon monoxide (CO) and leaves
33.50 mg of metallic nickel
behind. Determine the empirical
formula of the compound.
Strategy:
1 mmol = 1x10-3 mol
1mg = 1x10-3 g
1. Write the reaction
2. Use the Law of conservation of
mass to find the amount of CO
3. Find the number of moles (or
mmol) of CO and Nickel
4. Find the ratios of the moles for
each substance by dividing each
by the smallest
one, i.e.,
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normalize to the smallest.
Exercise 1-10
•
Moderate Heating of 97.44 mg of a compound
containing nickel, carbon and oxygen and no
other elements drives off all of the carbon and
oxygen in the form of carbon monoxide (CO)
and leaves 33.50 mg of metallic nickel behind.
Determine the empirical formula of the
compound.
Solution:
1.
Write the reaction
2.
Use the law of conservation of mass
to find the amount of CO
3.
Find the number of moles of CO and
Nickel
4. Find the ratios of the moles for each
substance by dividing each by the
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smallest one,
i.e., normalize
to the 20
smallest.
Exercise 1-10
•
Moderate Heating of 97.44 mg of a compound containing
nickel, carbon and oxygen and no other elements drives off
all of the carbon and oxygen in the form of carbon
monoxide (CO) and leaves 33.50 mg of metallic nickel
behind. Determine the empirical formula of the compound.
Solution:
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1.
Write the reaction
2.
Use the law of conservation of mass to find the
amount of CO
3.
Find the number of moles of CO and Nickel
4.
Find the ratios of the moles for each substance by
dividing each by the smallest one, i.e., normalize
to the smallest.
OFB Chapter 1
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Volume and Density
Exercise 1-13
The density of liquid mercury at 20
deg C is 13.594 g cm-3. A chemical
reaction requires 0.560 mol of
mercury. What volume (in cubic
centimeters) of mercury should be
measured out at 20°C?
Strategy:
1. Use density and mass to find
volume. Rearrange
m
d=
V
m
V =
d
2. Density is given, can find mass
from the number of moles of
mercury which is given
3. Solve for volume.
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Exercise 1-13
•
The density of liquid mercury at 20 deg C is
13.594 g cm-3. A chemical reaction requires 0.560
mol of mercury. What volume (in cubic
centimeters) of mercury should be measured out at
20°C?
Solution:
1.
Use density and mass to find volume.
m
V =
d
2.
Density is given, can find mass from the number
of moles of mercury which is given
 1mole Hg 
0.56 moles Hg = m grams Hg X 

 200.54 g Hg 
m = 112.3 grams Hg
3.
Solve for volume.
m
112.3 grams
3
V= =
= 8.26 cm
3
d 13.594 grams per cm
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Chapter 1
The Atomic Nature of Matter
Examples / Exercises (for your
practice only, not to be turned
in)
– All (1-1 thru 1-13)
HW Problems (to be turned in a
recitation, see due date)
Textbook errors
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