Transcript Chapter One

CHAPTER 1
The Foundations of
Chemistry
Chapter Goals
1.
2.
3.
4.
5.
6.
7.
Matter and Energy
Chemistry – A Molecular
View of Matter
States of Matter
Chemical and Physical
Properties
Chemical and Physical
Changes
Mixtures, Substances,
Compounds, and
Elements
Measurements in
Chemistry
8. Units of Measurement
9. Use of Numbers
10. The Unit Factor Method
(Dimensional Analysis)
11. Percentage
12. Density and Specific
Gravity
13. Heat and Temperature
14. Heat Transfer and the
Measurement of
Heat
3
What is Chemistry?
• Chemistry
– Science that describes
matter – its properties, the
changes it undergoes, and
the energy changes that
accompany those processes
• Energy
– The capacity to do work or
transfer heat.
4
Branches of Chemistry
• Analytical Chemistry -studies composition of
substances.
• Organic Chemistry -compounds containing
carbon
• Inorganic Chemistry –mainly substances without
carbon
• Biochemistry- Chemistry of living things
• Physical Chemistry studies behavior of
substances
– rates and mechanisms of reactions
– energy transfers
Chemistry is……
• A natural science
• A language with its
own vocabulary
• A way of thinking
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.
9
Properties of Matter
• Chemical Properties - can be observed by
changing the type of substance (chemical
changes/chemical rxn)
– rusting or oxidation
– chemical reactions
• Physical Properties - a quality or condition of a
substance that can be observed or measured
without changing the substance’s composition -changes of state
- density, color, solubility
10
Types of Properties
• Intensive Properties…
– Are independent of the amount of the
substance that is present.
• Density, boiling point, color, etc.
• Extensive Properties…
– Depend upon the amount of the substance
present.
• Mass, volume, energy, etc.
© 2009, PrenticeHall, Inc.
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
13
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
14
simpler substances – hydrogen and oxygen
Mixtures, Substances,
Compounds, & Elements
15
States of Matter
© 2009, PrenticeHall, Inc.
States of Matter
• Changes in state
require changes in
energy.
– heating
– cooling
17
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.
19
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
20
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
22
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
24
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
25
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
26
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
27
Significant Figures Rules
• Counting Sig Figs (Appendix A)
– Imbedded zeroes are always significant
3.0604 has five significant figures
– 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
• 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
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
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.
35
Dimensional Analysis
• The “Factor-Label” Method
– Units, or “labels” are canceled, or “factored” out
g
cm 

g
3
cm
3
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
B. Dimensional Analysis
• You have 1.5 pounds of gold. Find its
volume in cm3 if the density of gold is 19.3
g/cm3.
cm3
lb
1.5 lb 1 kg 1000 g 1 cm3
2.2 lb
1 kg
19.3 g
= 35 cm3
B. Dimensional Analysis
• How many liters of water would fill a
container that measures 75.0 in3?
in3
L
75.0 in3 (2.54 cm)3
(1 in)3
1L
1000 cm3
= 1.23 L
The Unit Factor Method
Example 1-2: Express 627 milliliters in gallons.
You do it!
41
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
42
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.
43
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
44
The Unit Factor Method
Example 1-4: Express 2.61 ft3 in cm3.
Volume is three dimensional, thus units must
be in cubic terms.
You do it!
45
The Unit Factor Method
Example 1-4: Express 2.61 ft3 in cm3.
Volume is three dimensional, thus units must
be in cubic terms.
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
46
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!
47
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
49
• 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.
50
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
51
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!
52
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
53
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.
54
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
55
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
56
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!
57
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
58
= 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
59
Heat and Temperature
• Fahrenheit
• Celsius
• Kelvin
MP water
32 oF
0.0 oC
273 K
BP water
212 oF
100 oC
373 K
60
Relationships of the Three
Temperature Scales
Kelvin and Centigrade Relationships
K = C + 273
o
or
o
C = K - 273
61
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
63
Relationships of the Three
Temperature Scales
64
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.
65
Heat and Temperature
Example 1-10: Convert 211oF to degrees Celsius.
66
Heat and Temperature
Example 1-10: Convert 211oF to degrees Celsius.
F - 32
C=
1.8
211 - 32
o
C=
1.8
o
o
67
Heat and Temperature
Example 1-11: Express 548 K in Celsius degrees.
68
Heat and Temperature
Example 1-11: Express 548 K in Celsius degrees.
o
C = K - 273
o
C = 548 - 273
o
C = 275
69
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
70
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
71
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
72
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
4.184J
o
o
? J  200 g H 2O 

(55.0
C

10.0
C)
o
1 g H 2O C
 3.76104 J or 37.6 kJ
73
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!
74
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
75
Heat Transfer and the
Measurement of Heat
• Notice that it requires 30.3 times more heat
for water than for mercury.
• The specific heat of water (4.184 J/g oC) is
30.3 times greater than that of mercury
(0.138 J/g oC).
76
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)
77
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)
78
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.
79
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
80
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
81
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 mile3
-9
-6
ç
÷
=
(1.35
´
10
mile)
=
7.13
´
10
ft
2
ç
÷
58,910 mile
è 1 mile ø
82
Group Activity
• On a typical day, a hurricane expends the
energy equivalent to the explosion of two
thermonuclear weapons. A thermonuclear
weapon has the explosive power of 1.0 Mton of
nitroglycerin. Nitroglycerin generates 7.3 kJ of
explosive power per gram of nitroglycerin. The
hurricane’s energy comes from the evaporation
of water that requires 2.3 kJ per gram of water
evaporated. How many gallons of water does a
hurricane evaporate per day?
83