CHAPTER ONE
The Foundations of Chemistry
Chapter Outline
1.
2.
3.
4.
5.
6.
7.
Matter and Energy
States of Matter
Chemical and Physical Properties
Chemical and Physical Changes
Mixtures, Substances, Compounds, and
Elements
Measurements in Chemistry
Units of Measurement
2
Chapter Outline
8.
9.
10.
11.
12.
13.
Use of Numbers
The Unit Factor Method (Dimensional
Analysis)
Percentage
Density and Specific Gravity
Heat and Temperature
Heat Transfer and the Measurement of
Heat
3
Matter and Energy - Vocabulary

Chemistry


Matter


Anything that has mass and occupies space.
Energy


Science that describes matter – its properties, the
changes it undergoes, and the energy changes that
accompany those processes
The capacity to do work or transfer heat.
Scientific (natural) law

A general statement based the observed behavior of
matter to which no exceptions are known.
4
Natural Laws



Law of Conservation of Mass
Law of Conservation of Energy
Law of Conservation of Mass-Energy


Einstein’s Relativity
E=mc2
5
Scientific Method






Observation
Hypothesis
Observation or experiment
Theory
Observation or experiment
Law
6
States of Matter

Solids
7
States of Matter


Solids
Liquids
8
States of Matter



Solids
Liquids
Gases
9
States of Matter

Change States


heating
cooling
10
States of Matter

Illustration of changes in state

requires energy
11
Chemical and Physical Properties

Chemical Properties - chemical changes



Physical Properties - physical changes




rusting or oxidation
chemical reactions
changes of state
density, color, solubility
Extensive Properties - depend on quantity
Intensive Properties - do not depend on
quantity
12
Mixtures, Substances,
Compounds, and Elements

Substance


Elements


matter in which all samples have identical
composition and properties
substances that cannot be decomposed into
simpler substances via chemical reactions
Elemental symbols

found on periodic chart
13
Mixtures, Substances,
Compounds, and Elements
14
Mixtures, Substances,
Compounds, and Elements

Compounds


substances composed of two or more
elements in a definite ratio by mass
can be decomposed into the constituent
elements

Water is a compound that can be decomposed
into simpler substances – hydrogen and oxygen
15
Mixtures, Substances,
Compounds, and Elements
16
Mixtures, Substances,
Compounds, and Elements

Mixtures



composed of two or more substances
homogeneous mixtures
heterogeneous mixtures
17
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
18
Measurements in Chemistry
Metric Prefixes
Name
 mega
 kilo
 deka
 deci
 centi
Symbol
M
k
da
d
c
Multiplier
106
103
10
10-1
10-2
19
Measurements in Chemistry
Metric Prefixes
Name
 milli
 micro
 nano
 pico
 femto
Symbol
m

n
p
f
Multiplier
10-3
10-6
10-9
10-12
10-15
20
Units of Measurement
Definitions
 Mass


measure of the quantity of matter in a
body
Weight

measure of the gravitational attraction
for a body
21
Units of Measurement
Common Conversion Factors
 Length



Volume



1 m = 39.37 inches
2.54 cm = 1 inch
1 liter = 1.06 qt
1 qt = 0.946 liter
See Table 1-7 for more conversion factors
22
Use of Numbers

Exact numbers


Accuracy


1 dozen = 12 things for example
how closely measured values agree with
the correct value
Precision

how closely individual measurements agree
with each other
23
Use of Numbers

Significant figures



digits believed to be correct by the person
making the measurement
Measure a mile with a 6 inch ruler vs.
surveying equipment
Exact numbers have an infinite number
of significant figures
12.000000000000000 = 1 dozen
because it is an exact number
24
Use of Numbers

Significant Figures - Rules
Leading zeroes are never significant
0.000357 has three significant figures

Trailing zeroes may be significant
must specify significance by how the number
is written
1300 nails - counted or weighed?

Use scientific notation to remove doubt
2.40 x 103 has ? significant figures
25
Use of Numbers

Scientific notation for logarithms
take the log of 2.40 x 103
log(2.40 x 103) = 3.380
How many significant figures?

Imbedded zeroes are always significant
3.0604 has five significant figures
26
Use of Numbers

Piece of Black Paper – with rulers beside the edges
2
3
4
5
6
7
8
9
10
11
12
13
14
9
8
7
6
5
4
3
2
1
1
27
Use of Numbers

Piece of Paper Side B – enlarged

11
How long is the paper to the best of your ability to measure it?
12
13
14
28
5
6
Use of Numbers
Piece of Paper Side A – enlarged
7
How wide is the paper to the best of your ability to measure it?
8

9

29
Use of Numbers


Determine the area of the piece of black
paper using your measured values.
Compare your answer with your classmates.


Where do your answers differ in the numbers?
Significant figures rules for multiplication
and division must help us determine where
answers would differ.
30
Use of Numbers

Multiplication & Division rule
Easier of the two rules
Product has the smallest number of
significant figures of multipliers
31
Use of Numbers

Multiplication & Division rule
Easier of the two rules
Product has the smallest number of
significant figures of multipliers
4.242
x 1.23
5.21766
round off to 5.22
32
Use of Numbers

Multiplication & Division rule
Easier of the two rules
Product has the smallest number of
significant figures of multipliers
4.242
x 1.23
2.7832
x 1.4
5.21766
round off to 5.22
3.89648
round off to 3.9
33
Use of Numbers


Determine the perimeter of the piece of
black paper using your measured values.
Compare your answer with your classmates.


Where do your answers differ in the numbers?
Significant figures rules for addition and
subtraction must help us determine where
answers would differ.
34
Use of Numbers

Addition & Subtraction rule
More subtle than the multiplication rule
Answer contains smallest decimal place of the
addends.
35
Use of Numbers

Addition & Subtraction rule
More subtle than the multiplication rule
Answer contains smallest decimal place of the
addends.
3.6923
 1.234
 2.02
6.9463
round off to 6.95
36
Use of Numbers

Addition & Subtraction rule
More subtle than the multiplication rule
Answer contains smallest decimal place of the
addends.
3.6923
 2.02
8.7937
 2.123
6.9463
round off to 6.95
6.6707
round off to 6.671
 1.234
37
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.
38
The Unit Factor Method

Change from a
following rules.
to a
by obeying the
39
The Unit Factor Method

1.
Change from a to a
by obeying the
following rules.
Must use colored fractions.
40
The Unit Factor Method

1.
2.
Change from a to a
by obeying the
following rules.
Must use colored fractions.
The box on top of the fraction must be
the same color as the next fraction’s
bottom box.
41
The Unit Factor Method
R

Fractions to choose from
R
O
B
O
B
B
O
R
O
B
B
B
42
The Unit Factor Method
O
R
R

Fractions to choose from
R
O
B
O
B
B
O
R
O
B
B
B
43
The Unit Factor Method
O
B
R
O
R

Fractions to choose from
R
O
B
O
B
B
O
R
O
B
B
B
44
The Unit Factor Method
O
B
B
R
B
R

O
B
Fractions to choose from
R
O
B
O
B
B
O
R
O
B
B
B
45
The Unit Factor Method
O
B
B
R
B
R

O
B
Fractions to choose from
R
O
B
O
B
B
O
R
O
B
B
B
46
The Unit Factor Method
O
B
B
R
B
R

O
B
Fractions to choose from
R
O
B
O
B
B
O
R
O
B
B
B
47
The Unit Factor Method
O
B
B
R
B
R

O
B
Fractions to choose from
R
O
B
O
B
B
O
R
O
B
B
B
48
The Unit Factor Method

colored fractions represent unit factors
1 ft = 12 in becomes

1 ft
12 in
or
12 in
1 ft
Example 1-1: Express 9.32 yards in
millimeters.
49
The Unit Factor Method
9.32 yd  ? mm
3ft
9.32 yd (
)
1yd
50
The Unit Factor Method
9.32 yd  ? mm
3ft 12in
9.32 yd ( ) (
)
1yd 1ft
51
The Unit Factor Method
9.32 yd  ? mm
3ft 12in 2.54cm
9.32 yd (
)(
)(
)
1yd 1ft
1in
52
The Unit Factor Method
9.32 yd  ? mm
3ft 12in 2.54cm 10mm
9.32 yd (
)(
)(
)(
)  8.52 103 mm
1yd 1ft
1in
1cm
53
The Unit Factor Method
9.32 yd  ? mm
3ft 12in 2.54cm 10mm
9.32 yd (
)(
)(
)(
)  8.52 103 mm
1yd 1ft
1in
1cm
O
B
B
T
R
T
R
O
B
B
54
The Unit Factor Method

Example 1-2: Express 627 milliliters in
gallons.
You do it!
55
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
56
The Unit Factor Method


Area is two dimensional thus units must
be in squared terms.
Example 1-3: Express 2.61 x 104 cm2 in
ft2.
57
The Unit Factor Method
 Area is two dimensional thus units must
be in squared terms.
 Example 1-3: Express 2.61 x 104 cm2 in
ft2.
1 in
? ft  2.61  10 cm (
)
2.54 cm
2

4
2
common mistake
58
The Unit Factor Method
 Area is two dimensional thus units must
be in squared terms.
 Example 1-3: Express 2.61 x 104 cm2 in
ft2.
1 in
? ft  2.61  10 cm (
)
2.54 cm
2
4
2
O
R
P
59
The Unit Factor Method
 Area is two dimensional thus units must
be in squared terms.
 Example 1-3: Express 2.61 x 104 cm2 in
ft2.
1 in
2
4
2
2
? ft  2.6110 cm (
)
2.54 cm
O
R
R
60
The Unit Factor Method
 Area is two dimensional thus units must
be in squared terms.
 Example 1-3: Express 2.61 x 104 cm2 in
ft2.
1 in
2
4
2
2 1 ft
2
? ft  2.61  10 cm (
) (
)
2.54 cm 12 in
61
The Unit Factor Method
 Area is two dimensional thus units must
be in squared terms.
 Example 1-3: Express 2.61 x 104 cm2 in
ft2.
1in
2 1ft
2
? ft  2.6110 cm (
) (
)
2.54cm 12in
2
4
2
 28.09380619 ft  28.1 ft
2
2
62
The Unit Factor Method


Volume is three dimensional thus units
must be in cubic terms.
Example 1-4: Express 2.61 ft3 in cm3.
You do it!
63
The Unit Factor Method


Volume is three dimensional thus units
must be in cubic terms.
Example 1-4: Express 2.61 ft3 in cm3.
12 in 3 2.54 cm 3
? cm  2.61 ft (
) (
)
1 ft
1 in
3
3
 73906.9696 cm  7.39 10 cm
3
4
3
64
Percentage


Percentage is the parts per hundred of
a sample.
Example 1-5: A 335 g sample of ore
yields 29.5 g of iron. What is the
percent of iron in the ore?
You do it!
65
Percentage


Percentage is the parts per hundred of a
sample.
Example 1-5: A 335 g sample of ore yields 29.5
g of iron. What is the percent of iron in the
ore?
grams of iron
? % iron 
x 100%
grams of ore
29.5 g Fe

x 100%
335 g ore
 8.81%
66
Density and Specific Gravity



density = mass/volume
What is density?
Why does ice float in liquid water?
67
Density and Specific Gravity



density = mass/volume
What is density?
Why does ice float in liquid water?
H
H
H C
H
H
H C
H
H
68
Density and Specific Gravity

Example 1-6: Calculate the density of a
substance if 742 grams of it occupies
97.3 cm3.
1 cm  1 mL  97.3 cm  97.3 mL
density  m
V
3
3
69
Density and Specific Gravity

Example 1-6: Calculate the density of a
substance if 742 grams of it occupies
97.3 cm3.
1 cm 3  1 mL  97.3 cm 3  97.3 mL
density  m
V
742
g
density 
97.3 mL
density  7.63 g/mL
70
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!
71
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
72
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
73
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.
74
Density and Specific Gravity

Example 1-8: A 31.0 gram piece of chromium is
dipped 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
75
Density and Specific Gravity

Example1-8: A 31.0 gram piece of chromium is
dipped 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?
Volume of Cr  9.32 mL - 5.00 mL
 4.32 mL
31.10 g
density of Cr 
4.32 mL
76
Density and Specific Gravity

Example1-8: A 31.0 gram piece of chromium is
dipped 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?
31.10 g
density of Cr 
4.32 mL
 7.19907 g
mL
 7.20 g
mL
7.20 g mL
Specific Gravity of Cr 
 7.20
g
1.00 mL
77
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!
78
Density and Specific Gravity
Some Possible Unit Factors from this Problem
36.31 g HCl
36.31 g HCl
63.69 g H 2 O
or
or
63.69 g H 2 O
100.00 g solution
100.00 g solution
79
Density and Specific Gravity
Specific Gravity  1.185 from problem
g
g
 density  1.185
 1185
mL
L
80
Density and Specific Gravity
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
 75.3 g HCl
81
Try This One…

Battery acid is 40.0% sulfuric acid and
60% water by mass. Its specific gravity
is 1.31.
82
Solution:

From the given specific value number
1.31
We may write Density = 1.31g/mL

Next…

83

The solution is 40%
sulfuric acid and
60% water by
mass…from this
information we get…
40.0 g
100 g
sulfuric acid
solution
Because 100g
of solution
contains 40.0
g of sulfuric
acid
84
We can now solve the problem:
__g H2SO4
=
1.31g soln
40 g H2SO4
100.00mL solution x
x
 52.4 g H2SO4
1 mL soln
100 g soln
85
Heat and Temperature

Heat and Temperature are not the same thing
T is a measure of the intensity of heat in a body

3 common temperature scales - all use water as
a reference
86
Heat and Temperature

Heat and Temperature
are not the same thing
T is a measure of the
intensity of heat in a
body

3 common temperature
scales - all use water as
a reference
87
Heat and Temperature



Fahrenheit
Celsius
Kelvin
MP water
32 oF
0.0 oC
273 K
BP water
212 oF
100 cC
373 K
88
Relationships of the Three
Temperature Scales
Kelvin and Centigrade Relationsh ips
K  C  273
or
o
o
C  K  273
89
Relationships of the Three
Temperature Scales
Fahrenheit and Centigrade Relationsh ips
180 18 9
   1.8
100 10 5
90
Relationships of the Three
Temperature Scales
Fahrenheit and Centigrade Relationsh ips
180 18 9
   1.8
100 10 5
o
F  1.8  o C  32
or
F  32
C
1.8
o
o
91
Relationships of the Three
Temperature Scales

1.
2.
3.
Easy method to remember how to
convert from Centigrade to Fahrenheit.
Double the Centigrade temperature.
Subtract 10% of the doubled number.
Add 32.
92
Heat and Temperature

Example 1-10: Convert 211oF to
degrees Celsius.
F  32
C
1.8
211  32
o
C
1.8
o
o
93
Heat and Temperature

Example 1-11: Express 548 K in Celsius
degrees.
o
C  K  273
o
C  548  273
o
C  275
94
Heat Transfer and The
Measurement of Heat


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 1oC
units = J/goC
95
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


units = J/mol oC
Example 1-12: Calculate the amt. of
heat to raise T of 200.0 g of water
from 10.0oC to 55.0oC
96
Heat Transfer and the
Measurement of Heat

Heat transfer equation
necessary to calculate amounts of heat
amount of heat = amount of substance x
specific heat x DT
q  m  C  DT
97
Heat Transfer and the
Measurement of Heat

Heat transfer equation
necessary to calculate amounts of heat
amount of heat = amount substance x
specific heat x DT
q  m  C  DT
4.184 J
o
o
? J  200 g H 2 O 
 (55.0 C  10.0 C)
1 g H 2O
98
Heat Transfer and the
Measurement of Heat

Heat transfer equation
necessary to calculate amounts of heat
amount of heat = amount substance x
specific heat x DT
q  m  C  DT
4.184 J
? J  200 g H 2 O 
 (55.0 o C  10.0o C)
1 g H 2O
 3.76 10 4 J or 37.6 kJ
99
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!
100
Heat Transfer and the
Measurement of Heat

Example 1-13: Calculate the amount of heat
to raise T of 200.0 g of Hg 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


Requires 30.3 times more heat for water
4.184 is 30.3 times greater than 0.138
101
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)
102
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)
103