CHAPTER 1
The Foundations of Chemistry
What is Chemistry?
 Physical science that studies the
composition, structure, and properties of
matter and the changes it undergoes
 Includes many different branches of study
(focuses on a particular area, they do overlap)
 Organic
 Inorganic
 Physical
 Analytical
 Biochemistry
 Theoretical
Chemistry is……
 A natural science
 A language with its
own vocabulary
 A way of thinking
Chapter 1
Chemistry
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Visual Concept
Scientific Method
Observations
Theory
(Model)
Hypothesis
Modify
Experiment
Prediction
Law
Experiment
What is Matter?
 Matter is anything that takes up space and
has mass
 Mass is the amount of matter in an object
 Mass is resistance to change in motion
along a smooth and level surface
Matter
 Atoms are the building
blocks of all matter
 An atom is the smallest
particle of an element that
maintains its chemical
identity through all
chemical and physical
changes.
7
Properties of Matter
 Physical propertiescharacteristic that can
be observed without
changing the identity
 Observed with the
senses
 Changes of state
 Density, color solubility
 Chemical propertiesindicates how a
substance reacts with
something else
 When observed original
substance is changed
 Rusting or oxidation
 Chemical Rxn
Properties of Matter
Chapter
1
Comparing
Physical and
Chemical Properties
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Visual Concept
Characteristic Properties
 Quality of a substance that
never changes, used to
identify substance
 Extensive- depends on
amount of matter present
 Mass
 volume
 Intensive- does not depend
on amount of matter present
 Melting point
 Boiling point
Chapter
1
Comparing
Extensive and
Intensive Properties
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Visual Concept
Changes in Matter
 Physical changealter form of a
substance but NOT
identity
 Original substance
continues to exist
 Chemical changesubstances
combine/break apart
to form new
substance
 Original substance no
longer exists
Chemical Changes, continued
• A change in which one or more substances are
converted into different substances is called a chemical
change or chemical reaction.
• The reactants are the substances that react in a
chemical change.
• The products are the substances that are formed by the
chemical change.
carbon
+ oxygen
reactants
carbon dioxide
product
Carbon plus oxygen yields (or forms) carbon dioxide.
Chapter 1
Evidence of a Chemical Change
Chapter 1
Chemical Reaction
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Visual Concept
Chapter 1
Comparing Chemical and
Physical Changes
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Visual Concept
Physical vs. Chemical
 Examples:
 rusting iron
chemical
 dissolving in water
physical
 burning a log
chemical
 melting ice
physical
 grinding spices
physical
Mixtures, Substances,
Compounds, & Elements
 Substance
 matter in which all samples have identical
composition and properties
 Elements
 substances that cannot be decomposed into
simpler substances via chemical reactions
 Elemental symbols
 found on periodic table
20
Chapter 1
Element
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Visual Concept
Mixtures, Substances,
Compounds, & Elements
 Mixtures
 composed of two or more substances can be
separated by physical means
 homogeneous mixtures
 heterogeneous mixtures
 Compounds
 substances composed of two or more elements in a
definite ratio by mass
 can be decomposed into the constituent elements by
chemical means
 Water is a compound that can be decomposed into
22
simpler substances – hydrogen and oxygen
Chapter 1
Compounds
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Visual Concept
Chapter
1
Classification
Scheme for
Matter
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Visual Concept
States of Matter
 Solids
 Particles packed
tightly together
 Have definite volume
and definite shape
 Particles vibrate
 Gases
 Change volume very
easily
 Particles spread apart,
filling all space available
to them
 No definite shape nor
volume
 Liquids
 No shape of its own,
takes shape of its
container
 Has definite volume
States of Matter
© 2009, Prentice-Hall, Inc.
Chapter 1
Section 2 Matter and Its Properties
Liquid
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Visual Concept
States of Matter
 Changes in state
require changes in
energy.
 heating
 cooling
28
States of Matter
Definite Definite Temp.
Volume? Shape? increase
Solid
Liquid
Gas
YES
YES
NO
Compressible?
YES
Small
Expans.
NO
NO
Small
Expans.
NO
NO
Large
Expans.
YES
A Molecular View
Dalton’s Atomic Theory
Dalton’s atomic theory summarized the nature of
matter as known in 1808
1)
2)
3)
4)
5)
An element is composed of extremely small indivisible
particles called atoms.
All atoms of a given element have identical properties,
which differ from those of other elements.
Atoms cannot be created, destroyed, or transformed into
atoms of another element.
Compounds are formed when atoms of different
elements combine with each other in small wholenumber ratios.
The relative numbers and kinds of atoms are constant in
a given compound.
30
Natural Laws
 Scientific (natural) law
 A general statement based the
observed behavior of matter to
which no exceptions are known.
 Law of Conservation of Mass
 Law of Conservation of Energy
 Law of Conservation of Mass
and Energy
 Einstein’s Theory of
Relativity
 E=mc2
31
Number vs. Quantity
 Quantity = number + unit
UNITS MATTER!!
Measurements in Chemistry
Quantity
 length
 mass
 time
 current
 temperature
 amt. substance
Unit
meter
kilogram
second
ampere
Kelvin
mole
Symbol
m
kg
s
A
K
mol
33
Measurements in Chemistry
Metric Prefixes
Prefix
mega-
Symbol
M
Factor
106
kilo-
k
103
BASE UNIT
---
100
deci-
d
10-1
centi-
c
10-2
milli-
m
10-3
micro-

10-6
nano-
n
10-9
pico-
p
10-12
Units of Measurement
Definitions
 Mass
 measure of the quantity of matter in a
body
 Weight
 measure of the gravitational attraction
for a body
35
Units of Measurement
Common Conversion Factors
 Length
 1 m = 39.37 inches
 2.54 cm = 1 inch
 Volume
 1 liter = 1.06 qt
 1 qt = 0.946 liter
 See Table 1-8 for more conversion factors
36
Use of Numbers
 Exact numbers
 1 dozen = 12 things
 Accuracy
 how closely measured
values agree with the
correct value
 Precision
 how closely individual
measurements agree
with each other
37
Accuracy and Precision
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Visual Concept
Significant Figures
 Indicate precision of a measurement.
 Consists of all the digits known with
certainty plus one final digit, which is
somewhat uncertain or estimated
2.35 cm
Significant Figures Rules
 Counting Sig Figs Count all numbers
EXCEPT:
Leading zeros -- 0.0025
Trailing zeros without
a decimal point -- 2,500
Calculating with Significant Figures
 Exact Numbers do not limit the # of
sig figs in the answer.
Counting numbers: 12 students
Exact conversions: 1 m = 100 cm
“1” in any conversion: 1 in = 2.54
cm
Significant Figures, continued
Rounding
Counting Sig Fig Examples
1. 23.50
4 sig figs
2. 402
3 sig figs
3. 5,280
3 sig figs
4. 0.080
2 sig figs
Calculating with Significant
Figures
 Multiply/Divide - The # with the fewest
sig figs determines the # of sig figs in
the answer.
(13.91g/cm3)(23.3cm3) = 324.103g
4 SF
3 SF
3 SF
324 g
Calculating with Significant
Figures
 Add/Subtract - The # with the lowest
decimal value determines the place of
the last sig fig in the answer.
3.75 mL
+ 4.1 mL
7.85 mL  7.9 mL
224 g
+ 130 g
354 g  350 g
Practice Problems
(15.30 g) ÷ (6.4 mL)
4 SF
2 SF
= 2.390625 g/mL  2.4 g/mL
2 SF
18.9 g
- 0.84 g
18.06 g  18.1 g
Introduction to the Periodic Table
• The vertical columns of the periodic table are called
groups, or families.
• Each group contains elements with similar
chemical properties.
• The horizontal rows of elements in the periodic
table are called periods.
• Physical and chemical properties change
somewhat regularly across a period.
The Unit Factor Method
 Simple but important method to get correct
answers in word problems.
 Method to change from one set of units to
another.
 Visual illustration of the idea.
48
B. Dimensional Analysis
 Steps:
1. Identify starting & ending units.
2. Line up conversion factors so units cancel.
3. Multiply all top numbers & divide by each
bottom number.
4. Check units & answer.
B. Dimensional Analysis
 How many milliliters are in 1.00 quart of
milk?
qt
mL
1.00 qt

1L
1000 mL
1.057 qt
1L
= 946 mL
The Unit Factor Method
Example 1-2: Express 627 milliliters in gallons.
You do it!
51
The Unit Factor Method
Example 1-2: Express 627 milliliters in gallons.
? gal =627 mL
1L
1.06qt 1gal
? gal = 627 mL (
)(
)(
)
1000mL
1L
4qt
? gal = 0.166155 gal » 0.166 gal
52
The Unit Factor Method
Example 1-3: Express 2.61 x 104 cm2 in ft2.
Area is two dimensional, thus units must be in
squared terms.
53
The Unit Factor Method
Example 1-3: Express 2.61 x 104 cm2 in ft2.
Area is two dimensional, thus units must be in
squared terms.
1in 2 1ft 2
? ft = 2.61 ´10 cm (
) (
)
2.54cm 12in
2
4
2
= 28.0938061 9 ft » 28.1 ft
2
2
54
Derived Units
 Combination of units.
 Volume amount of space occupied by an object
 length  length  length
1 cm3 = 1 mL
 (m3 or cm3)
1 dm3 = 1 L
Density (kg/m3 or g/cm3)
mass per volume
M
D=
V
Density and Specific Gravity
 density = mass/volume
D=M/V
 How heavy something is for its size.
 The ratio of mass to volume for a
substance.
 Independent of how much of it you have
 gold - high density
56
 air low density.
Density and Specific Gravity
Example 1-6: Calculate the density of a
substance if 742 grams of it occupies 97.3 cm3.
57
Density and Specific Gravity
Example 1-6: Calculate the density of a
substance if 742 grams of it occupies 97.3 cm3.
1 cm3 = 1 mL \ 97.3 cm3 = 97.3 mL
density = m
V
742
g
density =
97.3 mL
density = 7.63 g/mL
58
Density and Specific Gravity
Example 1-7: Suppose you need 125 g of a
corrosive liquid for a reaction. What volume do
you need? (liquid’s density = 1.32 g/mL)
You do it!
59
Density and Specific Gravity
Example 1-7 Suppose you need 125 g of a
corrosive liquid for a reaction. What volume do
you need? (liquid’s density = 1.32 g/mL)
m
m
density = \V =
V
density
125 g
V=
=
94.7
mL
1.32 g mL
60
Density and Specific Gravity
density (substance )
Specific Gravity =
density ( water )
 Water’s density is essentially 1.00 at room T.
 Thus the specific gravity of a substance is
very nearly equal to its density.
 Specific gravity has no units.
61
Density and Specific Gravity
Example 1-8: A 31.0 gram piece of chromium is
dropped into a graduated cylinder that contains
5.00 mL of water. The water level rises to 9.32 mL.
What is the specific gravity of chromium?
You do it
62
Density and Specific Gravity
Example 1-8: A 31.0 gram piece of chromium is
dropped into a graduated cylinder that contains
5.00 mL of water. The water level rises to 9.32 mL.
What is the specific gravity of chromium?
31.0 g
density of Cr 
4.32 mL
 7.17593 g
7.18
Specific Gravity of Cr 
1.00
g
mL
g
mL
 7.18 g
mL
 7.18
mL
63
Density and Specific Gravity
Example 1-9: A concentrated hydrochloric acid
solution is 36.31% HCl and 63.69% water by mass.
The specific gravity of the solution is 1.185. What
mass of pure HCl is contained in 175 mL of this
solution?
You do it!
64
Density and Specific Gravity
Example 1-9: A concentrated hydrochloric acid
solution is 36.31% HCl and 63.69% water by mass.
The specific gravity of the solution is 1.185. What
mass of pure HCl is contained in 175 mL of this
solution?
Specific Gravity = 1.185
g
g
\ density = 1.185
= 1185
mL
L
1.185 g sol' n
36.31 g HCl
? g HCl = 175 mL sol' n ´
´
1 mL
100.00 g solution
65
= 75.3 g HCl
Heat and Temperature
 Heat and Temperature
are not the same thing
 Temperature- measure
of the average kinetic
energy
 Temperature is which
way heat will flow. (from
hot to cold)
 3 common temperature
scales - all use water as
a reference
66
Heat and Temperature
 Fahrenheit
 Celsius
 Kelvin
MP water
32 oF
0.0 oC
273 K
BP water
212 oF
100 oC
373 K
67
Relationships of the Three
Temperature Scales
Kelvin and Centigrade Relationships
K = C + 273
or
o
o
C = K - 273
68
How much it
changes
100ºC = 212ºF
0ºC = 32ºF
100ºC = 180ºF
0ºC 100ºC
212ºF
32ºF
Relationships of the Three
Temperature Scales
Fahrenheit and Centigrade Relationships
180 18 9
= = = 1.8
100 10 5
70
Relationships of the Three
Temperature Scales
71
Relationships of the Three
Temperature Scales
Easy method to remember how to convert
from Centigrade to Fahrenheit.
1. Double the Centigrade temperature.
2. Subtract 10% of the doubled number.
3. Add 32.
72
Heat and Temperature
Example 1-10: Convert 211oF to degrees Celsius.
73
Heat and Temperature
Example 1-10: Convert 211oF to degrees Celsius.
F - 32
C=
1.8
211 - 32
o
C=
1.8
o
o
74
Heat and Temperature
Example 1-11: Express 548 K in Celsius degrees.
75
Heat and Temperature
Example 1-11: Express 548 K in Celsius degrees.
o
C = K - 273
o
C = 548 - 273
o
C = 275
76
Heat Transfer and the
Measurement of Heat
 Heat is energy, ability to do work.
 SI unit J (Joule)
 calorie
Amount of heat required to heat 1 g of water 1 oC
1 calorie = 4.184 J
 Calorie
Large calorie, kilocalorie, dietetic calories
Amount of heat required to heat 1 kg of water 1 oC
 English unit = BTU
 Specific Heat
amount of heat required to raise the T of 1g of a substance by
1o C
unit = J/goC
77
Heat Transfer and the
Measurement of Heat
 Heat capacity
amount of heat required to raise the T of 1 mole
of a substance by 1oC
 unit = J/mol oC
 Heat transfer equation
necessary to calculate amounts of heat
amount of heat = amount of substance x
specific heat x DT
q = m ´ C ´ DT
78
Heat Transfer and the
Measurement of Heat
Example 1-12: Calculate the amt. of heat to
raise T of 200.0 g of water from 10.0oC to
55.0oC
79
Heat Transfer and the
Measurement of Heat
Example 1-12: Calculate the amt. of heat to
raise T of 200.0 g of water from 10.0oC to
55.0oC
q  m  C  DT
4.184 J
o
o
? J  200 g H 2 O 

(55.0
C

10.0
C)
o
1 g H 2 O C
 3.76 10 4 J or 37.6 kJ
80
Heat Transfer and the
Measurement of Heat
Example 1-13: Calculate the amount of heat to
raise the temperature of 200.0 grams of mercury
from 10.0oC to 55.0oC. Specific heat for Hg is
0.138 J/g oC.
You do it!
81
Heat Transfer and the
Measurement of Heat
Example 1-13: Calculate the amount of heat to
raise the temperature of 200.0 grams of mercury
from 10.0oC to 55.0oC. Specific heat for Hg is
0.138 J/g oC.
q = m ´ C ´ DT
0.138 J
o
o
? J = 200 g Hg ´
´
(55.0
C
10.0
C)
o
(1 g Hg) C
= 1.24 kJ
82
Heating Curve for 3 Substances
Heating Curve
Which
substance has
the largest
specific heat?
140
120
Temperature (celsius degree)
100
80
Substance 1
Substance 2
Substance 3
60
40
20
0
0
50
100
150
200
250
300
Which
substance’s T
will decrease
the most after
the heat has
been removed?
Tim e (s)
83
Heating Curve for 3 Substances
Temperature (deg C)
Heating Curve
140
120
100
80
60
40
20
0
Substance 1
Substance 2
Substance 3
0
200
400
600
Time (s)
84
Synthesis Question
 It has been estimated that 1.0 g of seawater
contains 4.0 pg of Au. The total mass of
seawater in the oceans is 1.6x1012 Tg, If all of
the gold in the oceans were extracted and
spread evenly across the state of Georgia,
which has a land area of 58,910 mile2, how tall,
in feet, would the pile of Au be?
Density of Au is 19.3 g/cm3. 1.0 Tg = 1012g.
85
Synthesis Question
12
10 g
24
(1.6 ´ 10 Tg) (
) = 1.6 ´ 10 g of H 2 O
Tg
12
-12
4.0 ´ 10 g Au
12
(1.6 ´ 10 g of H 2 O)(
) = 6.4 ´ 10 g Au
g of H 2 O
24
86
Synthesis Question
12
10
g
12
(1.6 ´ 10 Tg) (
) = 1.6 ´ 10 24 g of H 2 O
Tg
-12
4.0
´
10
g Au
24
(1.6 ´ 10 g of H 2 O)(
) = 6.4 ´ 1012 g Au
g of H 2 O
3
æ
ö
1cm
12
÷÷ = 3.3 ´ 1011 cm 3 Au
6.4 ´ 10 g Au çç
è 19.3 g Au ø
æ 5280 ft öæ 12 in öæ 2.54 cm ö
÷÷çç
÷÷çç
÷÷ = 160,934 cm
(1 mile)çç
è 1 mile øè 1 ft øè 1 in ø
(
)
(160,934 cm )3 = (1 mile)3\ 4.16 ´ 1015 cm 3 = 1 mile3
87
Synthesis Question
3
æ
ö
1
mile
11
3
-5
3
÷
(3.3 ´ 10 cm Au)çç
=
7.96
´
10
mile
15
3 ÷
4.16
´
10
cm
è
ø
æ 5280 ft ö
7.96 ´ 10 -5 mile 3
-9
-6
ç
÷
=
(1.35
´
10
mile)
=
7.13
´
10
ft
2
ç
÷
58,910 mile
è 1 mile ø
88