Unformatted text preview: Chemical Foundations
Chapter 1
General Chemistry 1 Scientific Method
Make an observation
Look for patterns in the observation
Formulate theory
Design experiments to test the theory 1.
2.
3.
4.
•
• Hypothesis – tentative explanation or idea about
Hypothesis
how things work
how
Theory – explanation of general principles of certain
Theory
phenomena with considerable facts to support it
phenomena 2 Hypothesis vs Theory vs Fact Hypothesis
•
• “The moon is made of cheese”
How do we test this? Theory
• Remains valid only if every new piece of
Remains
information supports it
information Fact
• An indisputable truth
3 Development of Simple Theory
Development Observation:
Observation: Every swan I have ever seen is white.
is 4 Development of Simple Theory
Development Observation:
Observation: Every swan I have ever seen is white.
is Hypothesis: All swans must be
Hypothesis:
white.
white. 5 Development of Simple Theory
Development Observation:
Observation: Every swan I have ever seen is white.
is Hypothesis: All swans must be
Hypothesis:
white. Test: Random sampling of swans 6 Development of Simple Theory
Development Observation Hypothesis Test Publication:
Publication: “My global research has
indicated that swans are always white,
wherever they are observed.”
wherever 7 Development of Simple Theory
Development Observation Hypothesis Test Publication:
Publication: “My global research has
indicated that swans are always white ,
wherever they are observed.”
wherever Verification: Scientists observe white
Verification:
swans
swans
8 Development of Simple Theory
Development Observation Hypothesis Test Publication Verification Theory: All swans are white. 9 Development of Simple Theory
Development Observation Hypothesis Test Publication Verification Theory: All swans are white. Prediction: The next swan I see will be
Prediction:
white.
white.
10 Units of Measurement
Mass kilogram kg meter m Time second s Temperature Kelvin K ampere A Amt of substance mole mol Volume liter L candella cd Length Electric Current Luminous Intensity 11
11 SI System Prefixes
SI
Prefix
Prefix Symbol Meaning Exponential
Exponential
notation
notation giga G 1,000,000,000 109 mega M 1,000,000 106 kilo k 1,000 103 hecto h 100 102 deka da 10 101   1 100
12
12 SI System Prefixes
Prefix Symbol Meaning Exponential
Exponential
notation
notation   1 100 deci d 0.1 101 centi c 0.01 102 milli m 0.001 103 micro µ 0.000001 106 nano n 0.000000001 109
13
13 Exponential Scale
Powers of 10 perspective http://micro.magnet.fsu.edu/primer/java/scienceo 14
14 Mass vs Weight Mass
•
• Amount of matter
Measured by force necessary to give an
Measured
object a given acceleration
object Weight
•
•
•
• Gravitational effect on mass
On earth weigh 200 lbs.
On moon weigh 33.2 lbs
On Jupiter weigh 472.8 lbs
15
15 Uncertainty in Measurement Accuracy
Accuracy : agreement of a value with the
true value
true Precision: agreement among a set of
Precision:
measurements
measurements http://www.mhhe.com/physsci/chemistry/chang7/esp/folder_structure/ch/m2/s2/index.htm
16
16 Errors
Random: equal probability of being high or
low Systematic: occurs in the same direction
each time 17
17 Significant Figures Example: Postage Scale 3g Twopan balance 2.73 g Analytical How balance 2.731 g many significant figures?
18
18 Significant Figures
Significant 19 Significant Figures
Significant Ruler A Ruler B 20 Significant Figures
Significant Ruler A Ruler B 4.40 cm 21 Significant Figures
Significant Ruler A
4.40 cm Ruler B
4.5 cm 22 Significant Figures
Significant 23 Sig. Figs. In Measurements
A C B D 24
24 Sig. Figs. In Measurements A B C D E 25
25 Counting Significant Figures All numbers significant except zero 18.693 Zeros 5 sig. figs. within a number are significant 70.4006 6 sig. figs. For
For numbers < 1 count digits starting with
first nonzero digit on left and continue right
first 0.004289 4 sig. figs. 26
26 Counting Significant Figures Leading
Leading zeros left of decimal are not
significant
significant 0.0000915 3 sig. figs. Trailing
Trailing zeros to the right of decimal point
are significant
are 0.002700 4 sig. figs. Trailing
Trailing zeros to the left of decimal point
are ambiguous
are 473,000 3 – 6 sig. figs. 27
27 Counting Significant Figures Trailing
Trailing zeros to the left of decimal point
are ambiguous
are 230,000 230,000 230,000 230,000 26 sig. figs.
3 sig. figs.
4 sig. figs.
6 sig. figs. Trailing zeros are not significant unless otherwise
specified by the line above the zero.
28
28 Class Problem #1
Number Significant Figures .0012
6500
408
408.02
408.020
0.0107
6500
29
29 Class Problem #1
Number Significant Figures .0012 2 6500
408
408.02
408.020
0.0107
6500
30
30 Class Problem #1
Number Significant Figures .0012
6500 24? 408
408.02
408.020
0.0107
6500
31
31 Class Problem #1
Number Significant Figures .0012
6500
408 3 408.02
408.020
0.0107
6500
32
32 Class Problem #1
Number Significant Figures .0012
6500
408
408.02 5 408.020
0.0107
6500
33
33 Class Problem #1
Number Significant Figures .0012
6500
408
408.02
408.020 6 0.0107
6500
34
34 Class Problem #1
Number Significant Figures .0012
6500
408
408.02
408.020
0.0107 3 6500
35
35 Class Problem #1
Number Significant Figures .0012
6500
408
408.02
408.020
0.0107
6500 4
36
36 Exact Numbers Unlimited number of significant figures Sources: Counting objects Defined quantities Numbers that are part of an equation Exact Numbers Determined by counting Never limit the number of significant
Never
figures in a calculation
figures Examples:
7 beakers 3 feet/yard 15 batches 38
38 Exact Numbers Examples Within English system: 16 oz. = 1 lb. 12 in. = 1 ft. Within 10 Metric system: mm = 1 cm Between English and Metric system: 1 in = 2.54 cm 454 g = 1 lb or 454 g/1 lb Significant Figures in Calculations Addition and Subtraction 14.105
100.9__
100.9__
115.005
115.005 Multiplication 115.0 and Division 36 * 1075 = 2580
15
15 2.6 x 103 40
40 Significant Figures in Calculations
6.404 x 2.91
18.7 – 17.1
18.7 41
41 Significant Figures in Calculations
6.404 x 2.91
18.7 – 17.1
18.7
18.6356
1.6
1.6 42
42 Significant Figures in Calculations
6.404 x 2.91
18.7 – 17.1
18.7
18.6356
1.6
1.6
11.64725
43
43 Significant Figures in Calculations
6.404 x 2.91
18.7 – 17.1
18.7
18.6356
1.6
1.6
11.64725 12
44
44 Rounding Off When
When working problems, carry the extra
digits through to the final result and then
round.
round. Less than 5 round down Equal to or more than 5 round up 45
45 Rounding Examples Round to the nearest tens position 1356 1353 1355 1345 46
46 Rounding Examples Round 1356 to the nearest tens position
1360 1353 1355 1345 47
47 Rounding Examples Round 1356 1353 to the nearest tens position
1360
1350 1355 1345 48
48 Rounding Examples Round 1356 1353 1355 to the nearest tens position
1360
1350
1360 1345 49
49 Rounding Examples Round 1356 1353 1355 1345 to the nearest tens position
1360
1350
1360
1350 50
50 Exponential Notation Simplifies numbers by getting rid of zeros Nonscientific Scientific Notation 10000 1 x 104 7354 7.354 x 103 0.01 1 x 102 0.00044 4.4 x 104
51
51 Class Problem #2
0.053
7025
(4.07*102)(6.39*104)
4.36*104  2.796*102
52
52 Class Problem #2
0.053 5.3 * 102 7025
(4.07*102)(6.39*104)
4.36*104  2.796*102
53
53 Class Problem #2
0.053
7025 7.025 * 103 (4.07*102)(6.39*104)
4.36*104  2.796*102
54
54 Class Problem #2
0.053
7025
(4.07*102)(6.39*104) 2.60 * 101 4.36*104  2.796*102
55
55 Class Problem #2
0.053
7025
(4.07*102)(6.39*104)
4.36*104  2.796*102 4.33 * 104
56
56 Dimensional Analysis Use
Use the equivalence
statement that relates the
two units
two Cancel the unwanted units Multiply the quantity to be
Multiply
converted by the unit
factor to give quantity with
desired units
desired
Industry Week, 1981 November 30 57
57 Dimensional Analysis Example #1
You are planning a party Friday night and
You
expect 30 people. How much pizza and
beer should you get?
beer
Estimate that one person eats a third of a
Estimate
pizza and drinks 4 beers
pizza 58
58 Dimensional Analysis Example #1 0.333 pizza 30humans * ÷= 10 pizza 1human 4beers 30humans * ÷= 120beers 1human 59
59 Dimensional Analysis Example #1 0.333 pizza 30humans * ÷= 10 pizza 1human 4beers 30humans * ÷= 120beers 1human 60
60 Dimensional Analysis Example #1
Should you buy the beer in sixpacks or in cases? 1sixpack 120beers * ÷ = 20 sixpacks 6beers 1case 120beers * ÷ = 5cases 24beers 61
61 Dimensional Analysis Example #1 How do we know the equation was set up
How
correctly?
correctly? 6beers 2
120beers * ÷ = 720beers / sixpack 1sixpack 62
62 Class Problem #3
The bike distance in an Olympic length
The
triathlon is 40 km. How many miles is this?
triathlon 63
63 Class Problem #3
The bike distance in an Olympic length
The
triathlon is 40 km. How many miles is this?
triathlon
Need to go from kilometers to miles
kilometers
meters
yards miles 64
64 Class Problem #3
The bike distance in an Olympic length
The
triathlon is 40 km. How many miles is this?
triathlon
Need the following equivalence statements:
1 km
1000 m
1m
1.094 yd
1.094
1760 yd
1 mii
1760
m
65
65 Class Problem #3
The bike distance in an Olympic length
The
triathlon is 40 km. How many miles is this?
triathlon
40 km * 1000 m =
40
1000
1 km 66
66 Class Problem #3
The bike distance in an Olympic length
The
triathlon is 40 km. How many miles is this?
triathlon
40 km * 1000 m =
40
1000
1 km 40000 m 67
67 Class Problem #3
The bike distance in an Olympic length
The
triathlon is 40 km. How many miles is this?
triathlon
40 km * 1000 m = 40000 m
40
1000
1 km
40000 m * 1.094 yd =
1.094
1m
68
68 Class Problem #3
The bike distance in an Olympic length
The
triathlon is 40 km. How many miles is this?
triathlon
40000 m * 1.094 yd = 43760 yd
40000
1.094
1m
43760 yd * 1 mi
=
1760 yd
1760
69
69 Class Problem #3
The bike distance in an Olympic length
The
triathlon is 40 km. How many miles is this?
triathlon
40000 m * 1.094 yd = 43760 yd
40000
1.094
1m
43760 yd * 1 mi
= 24.864 mi
1760 yd
1760
70
70 Class Problem #3
The bike distance in an Olympic length
The
triathlon is 40 km. How many miles is this?
triathlon
43760 yd * 1 mi
1760 yd
1760 = 24.864 mi 24.864 mi rounds to 25 mi
24.864
71
71 Dimensional Analysis Example #3
You have a 2 in x 2 in x 2 in cube. What is
You
the volume of the cube?
the 72
72 Dimensional Analysis Example #3
You have a 2 in x 2 in x 2 in cube. What is
You
the volume of the cube?
the
Need the following equivalence statements:
1 inch
2.54 cm
100 cm
1m 73
73 Dimensional Analysis Example #3
You have a 2 in x 2 in x 2 in cube. What is
You
the volume of the cube?
the
2 in x 2 in x 2 in = 8 in3
in 74
74 Dimensional Analysis Example #3
You have a 2 in x 2 in x 2 in cube. What is
You
the volume of the cube?
the
2 in x 2 in x 2 in = 8 in3
in
8 in3 x 2.54 cm 3 =
in
2.54
1 in 75
75 Dimensional Analysis Example #3
You have a 2 in x 2 in x 2 in cube. What is
You
the volume of the cube?
the
2 in x 2 in x 2 in = 8 in3
in
8 in3 x 2.54 cm 3 = 1.310965 * 102 cm3
in
2.54
1 in 76
76 Dimensional Analysis Example #3
You have a 2 in x 2 in x 2 in cube. What is
You
the volume of the cube?
the
8 in3 x 2.54 cm 3 = 1.310965 * 102 cm3
in
2.54
1 in
1.310965 * 102 cm3 x 1 m 3 =
1.310965
100 cm
100
77
77 Dimensional Analysis Example #3
You have a 2 in x 2 in x 2 in cube. What is the
You
volume of the cube?
volume
8 in3 x 2.54 cm 3 = 1.310965 * 102 cm3
in
2.54
1 in
1.310965 * 102 cm3 x 1 m 3 =
1.310965
100 cm
78
78 Dimensional Analysis Example #3
You have a 2 in x 2 in x 2 in cube. What is the
You
volume of the cube?
volume
1.3110 * 102 cm3 x 1.3110 * 104 m3 1m
100 cm
100 3 = 1.3110*104 m3
1.3110*10 1 * 104 m3
10
79
79 Dimensional Analysis Summary Used in numerical calculations Used in converting units Can help identify if an equation is set up
Can
correctly
correctly If a variable in an equation is squared,
If
then the associated dimensions are
squared
squared
80
80 Temperature http://www.chem.ufl.edu/~itl/2045/matter/FG01_018.GIF 81
81 Temperature Conversions
TK = TC + 273.15
TC = TK  273.15
(TF – 32°F)(5/9) = TC
TF = TC x (9/5) + 32°F
Where do these equations come from?
82
82 Temperature Conversions
212 – 32 = 180°F
100 – 0 = 100°C
180°F = 9°F
9°F
100°C
5°C 83
83 Temperature Conversions
Different zero points:
32°F = 0°C
Apply unit factor of 5/9
(TF – 32°F)(5°C/9°F) = TC 84
84 Temperature Conversions
Different zero points:
32°F = 0°C
Apply unit factor of 5/9
(TF – 32°F)(5°C/9°F) = TC 85
85 Temperature Problem
432°C = ? K
432°C 86
86 Temperature Problem
432°C = ? K
432°C
TK = TC + 273.15 87
87 Temperature Problem
432°C = ? K
432°C
TK = TC + 273.15
432 + 273.15 = 705.15 K 705 K 88
88 Temperature Problem
2835 K = ? °C
2835 89
89 Temperature Problem
2835 K = ? °C
2835
TC = TK  273.15 90
90 Temperature Problem
2835 K = ? °C
2835
TC = TK  273.15
2835 – 273.15 = 2561.85°C 2562°C 91
91 Temperature Problem
92°F = ? °C 92
92 Temperature Problem
92°F = ? °C
(TF – 32°F)(5°C/9°F) = TC 93
93 Temperature Problem
92°F = ? °C
(TF – 32°F)(5°C/9°F) = TC
(92 – 32)(5/9) =
(92 94
94 Temperature Problem
92°F = ? °C
(TF – 32°F)(5°C/9°F) = TC
(92 – 32)(5/9) = 33.3°C 33°C 95
95 Temperature Problem
3°C = ? °F 96
96 Temperature Problem
3°C = ? °F
TF = TC x (9/5) + 32°F 97
97 Temperature Problem
3°C = ? °F
TF = TC x (9/5) + 32°F
TF = 3 x (9/5) + 32°F 98
98 Temperature Problem
3°C = ? °F
TF = TC x (9/5) + 32°F
TF = 3 x (9/5) + 32°F
TF = 37.4°F rounds to 37°F 99
99 Volume Represent
Represent amounts in three dimensional space
space 1 mL = 1 cm3 1 L = 1000 cm3 100
100 Density Mass per unit volume Used as an “identification
Used
tag”
tag”
_mass_
Density = volume http://www.chem.com.au/products/lstheavyliquid/ancillary/hydrometer.jpeg 101
101 Density Problem #1 Stirring
Stirring bar with a mass of 7.68 g is put in
a graduated cylinder containing 6.6 mL of
water. The level of water was raised such
that the new volume was 8.5 mL.
that Calculate the density of the stirring rod
Initial 10 5 After addition of
stir bar 10 5 102
102 Density Problem #1
Volume 8.5 – 6.6 = 1.9 mL
Density 7.68 g / 1.9 mL = 4.042 g/mL
round to 4.0 g/mL 103
103 Density Problem #2 Diamonds
Diamonds are measured in carats, and 1
carat = 0.200 g. The density of diamond is
3.51 g/cm3.
3.51 What is the volume of a 5.0 carat
What
diamond?
diamond? 104
104 Density Problem #2 Diamonds
Diamonds are measured in carats, and 1
carat = 0.200 g. The density of diamond is
3.51 g/cm3.
3.51 What is the volume of a 5.0 carat
What
diamond?
diamond?
First need mass of 5.0 carat diamond
(5.0 carat)(0.200 g/1 carat) = 1.000 g
105
105 Density Problem #2 Diamonds
Diamonds are measured in carats, and 1
carat = 0.200 g. The density of diamond is
3.51 g/cm3.
3.51 What is the volume of a 5.0 carat
What
diamond?
diamond?
Density
= mass / volume
3.51 g/cm3 = (1.000 g / V)
106
106 Density Problem #2 Diamonds
Diamonds are measured in carats, and 1
carat = 0.200 g. The density of diamond is
3.51 g/cm3.
3.51 What is the volume of a 5.0 carat
What
diamond?
diamond?
3.51 g/cm3 = (1.000 g / V)
V = 0.2849 cm3
0.2849
107
107 Density Problem #2 Diamonds
Diamonds are measured in carats, and 1
carat = 0.200 g. The density of diamond is
3.51 g/cm3.
3.51 What is the volume of a 5.0 carat
What
diamond?
diamond?
V
V = 0.2849 cm3
= 0.28 cm3 round to 108
108 State of Matter
Solid: Definite volume
Fixed shape Liquid: Definite volume
Flexible shape Gas: No definite volume or shape
Occupies entire container http://library.tedankara.k12.tr/chemistry/vol1/matter/trans7.jpg 109
109 Matter
Can be separated by a
physical process? No Mixture Pure Substance
Uniform material of definite
composition w/ characteristic
properties Can it be decomposed
by a chemical process? No
Element
Cannot be
broken down
into simpler
substances
hydrogen,
zinc, sulfur Yes Yes
Compound
Can be broken
down into two
or more
elements
Water, zinc
sulfide Variable composition whose parts
can be separated by physical
means Is it uniform
throughout? Yes Homogeneou
s Mixture
(Solution)
Uniform mixture
of different
substances
seawater, air No Heterogeneous
Mixture
Nonuniform
mixture of different
substances that
remain distinct
wood, concrete
110
110 Matter Examples Cookie Yogurt Lead Sinker 111
111 Physical Changes Change
Change in the form of a substance, not in
its chemical composition
its Water
Water Steam Liquid Ice 112
112 Physical Changes Piece of paper Cut Tear Folded Written on Painted 113
113 Chemical Change Substance
Substance becomes a new substance with
different properties and different
composition
composition Electrolysis of water 114
114 Chemical Change New matter formed Burning Rusting
Rusting Cooking Dehydration of sugar
115
115 Separation Methods Distillation 116
116 Separation Methods Filtration: 117
117 Chromatography: 118
118 Separation Methods Paper Chromatography: 119
119 Learning Objectives Scientific Method Units of Measurement Significant Figures Dimensional Analysis Temperature Density Matter Classification
Classification Separation 120
120 ...
View
Full
Document
This note was uploaded on 11/22/2010 for the course CHM 121 taught by Professor Scheru during the Spring '09 term at Oakton.
 Spring '09
 SCHERU
 Chemistry

Click to edit the document details