1
Basic Concepts
Chapter Outline
•
•
•
•
•
•
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
•
•
•
•
•
Use of Numbers
The Unit Factor Method (Dimensional
Analysis)
Density and Specific Gravity
Heat and Temperature
Heat Transfer and the Measurement of
Heat
3
Mixtures, Substances,
Compounds, and Elements
• Matter-Anything that occupies space and has mass.
• Pure Substances or Substances-Cannot be separated
by physical processes.
-Elements-A substance which cannot be broken down
into simpler substances. e.g. Na, He, C, (atoms) or N2,
Cl2 (molecules)
-Compounds-A pure substance made up of two or more
elements. e.g. NaCl, H2O.
• Mixtures-Can be separated by physical processes.
– composed of two or more substances
– homogeneous mixtures-A mixture that is uniform throughout-e.g.
white wine, grape juice. Clear. Solutions.
– heterogeneous mixtures-A mixture that is not uniform
throughout-e.g. oil and water, orange juice. Cloudy.
4
Mixtures, Substances,
Compounds, and Elements
5
Substances and Mixtures
Matter
Pure Substances
Elements
Physical process
Chemical
Reaction Compounds
Mixtures
Homogeneous
Mixture
Heterogeneous
Mixture
6
7
Fig. 1-7, p. 10
States of Matter
• Change States
– heating
– cooling
Vaporization:
Evaporation
Boiling
Freezing:
Solidification
Crystallization
Melting:
Fusion
8
States of Matter
Solid
Liquid
heat
cool
Gas
heat
cool
Less attractive force and more disordered.
9
States of Matter
• Illustration of changes in state
– requires energy
10
Mixtures, Substances,
Compounds, and Elements
11
Elements you Need to Know:
1
2
3
4
5
6
7
IA
H
Li
Na
K
Rb
Cs
IIA
Be
Mg
Ca
Sr
Ba
"B" Groups
Sc
Ti
V
Cr
U
Mn
Fe
Co
Ni
Pd
Pt
Cu
Ag
Au
Zn
Cd
Hg
IIIA IVA VA VIA VIIA VIIIA
He
B
C
N
O
F
Ne
Al
Si
P
S
Cl Ar
Ga Ge As Se Br Kr
In Sn Sb Te
I
Xe
Pb Bi
Rn
Pu
12
Types of Solutions:
•
Liquid solutions are the most common,
but there are also gas and solid
solutions.
• Solutions have two components:
a) Solute - Solution component(s) present
in lesser amounts.
b) Solvent - Solution component present in
the greatest amount.
13
Types of Solutions:
Solute
Gas
Liquid
Solid
Gas
Solid
Solvent Appearance
Example
Liquid
Liquid
Carbonated Water
Liquid
Liquid
90 Proof Rum
Liquid
Liquid
Seawater
Gas
Gas
Air
Solid
Solid
14-karat gold
14
Characteristics of Solutions
•
•
•
•
•
Uniform distribution
Components do not separate upon standing.
Components cannot be separated by filtration.
Within certain limits its composition can vary.
Almost always transparent. (i.e. one can see
through it).
• An alloy is a homogeneous mixture of metals.
i.e. brass, bronze, sterling silver.
15
Chemical and Physical
Properties
• Physical Properties – A property that can be observed
in the absence of any change in composition. e.g. color,
odor, taste, melting point, boiling point, freezing point,
density, length, specific heat, density, solubility.
• Physical Changes-Changes observed without a change
in composition. i.e. cutting wood, melting of solids and
boiling of liquids.
• water, ice
water, liquid
water, steam
– changes of state
– dissolving
– polishing
16
Chemical and Physical Properties
• Chemical Properties-A property that matter exhibits as it undergoes
changes in composition. e.g. coal and gasoline can burn in air to
form carbon dioxide and water; iron can react with oxygen in the air
to form rust; bleach can turn blond hair blonde.
• Chemical Changes-Changes observed only when a change in
composition is occurring. e.g. reaction of sodium with chlorine,
rusting of iron, dying of hair, burning of wood, cooking an egg, rotting
food.
• Extensive Properties - depends on the amount of material present.
e.g. volume and mass.
• Intensive Properties – does not depend on the amount of material
present. e.g. melting point, boiling point, freezing point, color,
density.
17
Natural Laws
• Law of Conservation of Mass-Mass is neither
created nor destroyed.
• Law of Conservation of Energy-Energy is neither
created nor destroyed, only converted from one
form to another.
• Law of Definite Proportions-Different samples of
any pure compound contain the same element
in the same proportion by mass. e.g. water
(H2O) contains 11.1 % H and 88.9% O by mass.
Thus, a 25.0 sample of water would contain 2.78
g of H and 22.2 g of O.
18
Law of Definite Proportions
11.1 % H and 88.9% O by mass, 25.0 g sample of water:
11.1gH
25.0 gwater(
)  2.78 gH
100 gwater
88.9 gO
25.0 gwater(
)  22.2 gO
100 gwater
19
Use of Numbers
• Exact numbers
– 1 dozen = 12 things for example
20
Rounding off Numbers
Previous digit
1.29
1.
4
Next digit
If the next digit is less than 5 the previous
digit remains the same.
.9946
1.294
.99
1.29
2. If the next digit is greater than 5 or 5
followed by non zeros then the previous digit
is increased by one.
.999
1.2951
1.00
1.30
3. If the next digit is 5 or 5 followed by all zeros
then the previous digit remains the same if it
is even or increased by one if it is odd.
1.285
1.295
1.22500
1.28
1.30
1.22
21
Scientific Notation
• Used to handle very large and very small
numbers.
Any number that is from + or – 1 to 9
Exponent-Power of 10
N. X 10x
For example: 3.21 x 103
-9.9 x 10-4
1.0 x 100 (Note that 100 is 1)
22
Scientific Notation
• To convert numbers to scientific notation
use the following guidelines:
Exponent increases by
3 powers of 10
1750.0 = 1750.0 x 100= 1.7500 x 103
Number decreases by 3 powers of 10
A you move the decimal place to the left (i.e. make the number smaller), the power
of ten (i.e., exponent) must increase by the same amount.
23
Scientific Notation
The number gets larger by 2 powers of 10
0.050 = 0.050 x 100 = 5.0 x 10-2
The exponent gets smaller by
2 powers of 10.
As you move the decimal place to the right (i.e. make the number larger),
the power of ten (i.e. exponent) must decrease by the same amount.
24
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
25
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
• Use scientific notation to remove doubt
2.40 x 103 has ? significant figures
26
Use of Numbers
• 3,380 ? significant figures
3.38 x 103
• 3,380. has ? significant figures
3.380 x 103
• Imbedded zeroes are always significant
3.0604 has ? significant figures
27
Use of Numbers
• Piece of Paper Side B – enlarged
– How long is the paper to the best of your ability to measure it?
11
12
13
14
13.36 in.
The second decimal place is
estimated
28
5
6
Use of Numbers
• Piece of Paper Side A – enlarged
8
7
– How wide is the paper to the best of your ability to measure it?
9
8.3 in
The first decimal place is estimated
29
Manipulating Powers of 10
• a) When multiplying powers of ten, the
exponents are added. For example:
105 x 10-4 = 105+(-4)=101
b) When dividing powers of ten, the exponents
are subtracted. For example:
104 = 104-(-4) = 108
10-4
c) When raising powers of ten to an exponent,
the exponents are multiplied. For example:
(104)3 = 10(4 x 3) = 1012
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
Multiplying and Dividing Numbers
with Powers of Ten
1. When using scientific notation:
a.) Place the powers of ten together.
(1.76 x 10200) x (2.650 x 10200)=
(1.76 x 2.650) x (10200 + 200)=
b.) The final answer has the same number of
significant figures as the number with the least
number of significant figures.
4.66 x 10400
c.) You must round off correctly.
d.) Preferably report the answer in scientific
notation.
34
Multiplying and Dividing Numbers
with Powers of Ten
(1.760 x 102) /(2.65 x 10-2)=
(1.760 / 2.65) x (102 – (-2))=
0.664 x 104 = 6.64 x 103
35
Use of Numbers
• Addition & Subtraction rule
More subtle than the multiplication rule
Answer contains smallest decimal place of the addends
36
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
8.7937
 2.123
6.6707
round off to 6.671
37
Addition and Subtraction with
Powers of Ten
a.) All numbers must have the same power or ten before
addition or subtraction is performed.
b.) Once the powers of ten are the same, the coefficients
can then be added or subtracted while the power of ten
remains the same.
c.) After adding or subtracting the coefficients, the answer
must have the same number of decimal places as the
coefficient with the fewest decimal places at the time of
the operation.
d.) You must round off correctly.
e.) Preferably report the answer in scientific notation.
38
Addition and Subtraction with
Powers of Ten
• 4.76 x 10200 + 9.6 x 10201 = ?
0.4 76 x 10201
+ 9.6
x 10201
10.0 76 x 10201
1.01 x 10202
(written in scientific notation and rounded off
to the correct number of significant figures)
39
Addition and Subtraction with
Powers of Ten
• 2.95 x 10-15 – 1.00 x 10-14 = ?
-1.00 x 10-14
0.29 5 x 10-14
-0.70 5 x 10-14
-7.0 x 10-15
(written in scientific notation and rounded off
to the correct number of significant figures)
40
Mixing Addition/Subtraction with
Multiplication/Division
7.54 x 10-5 (99. x 10200 + 1.25 x 10201)
(1.75 x 10-3)3
7.54 x 10-5 (9.9 x 10201 + 1.25 x 10201) =
1.75 x 10-3 x 1.75 x 10-3 x 1.75 x 10-3
7.54 x 10-5 [(9.9 + 1.25) x 10201) =
1.75 x 1.75 x 1.75 x 10-3 x 10-3 x 10-3
7.54 x 10-5 (11.2 x 10201) =
5.36 x 10-9
7.54 x 11.2 x 10-5 x 10201 = 1.58 x 10206
5.36
10-9
41
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
42
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
43
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
44
Metric Conversions
•
•
•
•
1 km = 103 m
1 dL = 10-1 L
1 msec = 10-3 sec
1 m = 10-6 m
45
46
Fig. 1-20, p. 24
Metric
English Conversions
Common Conversion Factors
• Length
– 2.54 cm = 1 inch (exact conversion)
• Volume
– 1 qt = 0.946 liter (Rounded off)
• Mass
_ 1 lb = 454 g (Rounded off)
47
Use of Conversion Factors in
Calculations
• Commonly known
relationship (i.e. equality):
• 1 ft = 12 in
• Respective conversion
factors to above equality:
1 ft or 12 in
12 in
1 ft
Use the conversion factor
that allows for the
cancellation of units.
Convert 24 in to ft:
? ft = 24 in x ( 1 ft )  2 ft
12in
48
Conversion Factors
• Example 1-1: Express 9.32 yards in
millimeters.
49
Conversion Factors
9.32 yd  ? mm
3ft
9.32 yd ( )
1yd
50
Conversion Factors
9.32 yd  ? mm
3ft 12in
9.32 yd ( ) (
)
1yd 1ft
51
Conversion Factors
9.32 yd  ? mm
3ft 12in 2.54cm
9.32 yd ( ) (
)(
)
1yd 1ft
1in
52
Conversion Factors
9.32 yd  ? mm
3ft 12in 2.54cm
1m 1000mm
9.32 yd ( ) (
)(
)(
)(
)  8.52 103 mm
1yd 1ft
1in
100cm
1m
53
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.
54
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
55
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
? ft  2.6110 cm (
)
2.54 cm
2
4
2
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.
1 in 2 1 ft 2
? ft  2.61  10 cm (
) (
)
2.54 cm 12 in
2
4
2
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.
1in
2 1ft
2
? ft  2.6110 cm (
) (
)
2.54cm 12in
2
4
2
 28.09380619 ft  28.1 ft
2
2
58
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
59
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
60
Conversions of Double Units
mi
km
?
hr
min
km
mi 1hr
5280 ft 12in 2.54cm 10 2 m 1km
km
?
 55. (
)(
)(
)(
)(
)( 3 )  1.5
min
hr 60 min
1mi
1 ft
1in
1cm 10 m
min
55.
m3
qt
3.7
?
hr
min
3
3
qt
1hr
1cm
1mL 10 L
1qt
1 qt
m
?
 3.7
(
)(  2 ) (
)(
)(
)  6.5 x10
3
min
hr 60 min 10 m 1cm 1mL 0.946 L
min
3
61
Density and Specific Gravity
• density = mass/volume
• What is density?
• Why does ice float in liquid water?
62
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
3
3
density  m
V
63
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
742
g
density 
97.3 mL
density  7.63 g/mL
3
3
64
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
1mL
V  125 g x
 94.7 mL
1.32 g
65
Density and Specific Gravity
density (substance )
Specific Gravity 
density ( water )
• Water’s density is essentially 1.00 g/mL at room T.
• Thus the specific gravity of a substance is very nearly
equal to its density.
• Specific gravity has no units.
66
Density and Specific Gravity
• The density of lead is 11.4 g/cm3. What volume,
in ft3, would be occupied by 10.0 g of lead?
3
3 1 ft
3
1
1in
cm
3
3
5
? ft  10.0 g (
)(
(

3
.
10
x
ft
10
)
)
11.4 g 2.54cm 12in
67
Density and Specific Gravity
• What is the density (in g/mL) of a rectangular bar
of lead that weighs 173 g and has the following
dimensions:
length = 2.00 cm, w = 3.00 cm, h = 1.00 in?
2.54cm
)  15.2 cm3
V ( rec.solid)  lxwxh  (2.00cm)(3.00cm)(1.00inx
1in
m
173g
g
g
d 
 11.4 3  11.4
3
V 15.2 cm
mL
cm
68
Density and Specific Gravity
•
An irregularly shaped piece of metal with a mass of 0.251 lb was placed
into a graduated cylinder containing 50.00 mL of water; this raised the
water level to 67.50 mL. What is the density of the metal? What is the
density (in g/cm3) of the metal? Will the metal float or sink in water?
V(disp)l= 67.50 mL – 50.00 mL = 17.50 mL
454 g
0.251lb (
)
m
g
g
1
lb
d 
 6.51
 6.51 3
V
17.50mL
mL
cm
The metal will sink in water because its density is greater than that of water. (1.00 g/mL)
69
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
31.0 g
density of Cr 
4.32 mL
70
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?
31.0 g
density of Cr 
4.32 mL
g
 7.17593
7.18
Specific Gravity of Cr 
1.00
g
mL
g
mL
mL
g
 7.18
mL
 7.18
71
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
72
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
73
Heat and Temperature
• Fahrenheit
• Celsius
• Kelvin
MP water
32 oF
0.0 oC
273 K
BP water
212 oF
100 oC
373 K
74
Relationships of the Three
Temperature Scales
Kelvin and Centigrade Relationsh ips
K  C  273
or
o
o
C  K  273
75
Relationships of the Three
Temperature Scales
Fahrenheit and Centigrade Relationsh ips
180 18 9
   1.8
100 10 5
76
Relationships of the Three
Temperature Scales
Fahrenheit and Centigrade Relationsh ips
180 18 9
   1.8
100 10 5
o
o
F  1.8  C  32
or
F  32
C
1.8
o
o
77
Heat and Temperature
• Example 1-10: Convert 211oF to
degrees Celsius.
F  32
C
1.8
211  32
o
C
1.8
o
o
=
99.4
78
Heat and Temperature
• Example 1-11: Express 548 K in
Celsius degrees.
o
C  K  273
o
C  548  273
o
C  275
79
Heat Transfer and the
Measurement of Heat
• Chemical reactions and physical changes occur with
either the simultaneous evolution of heat
(exothermic process), or the absorption of heat
(endothermic process).
• The amount of heat transferred is usually expressed
in calories (cal) or in the SI unit of joules (J).
1 cal = 4.184 J
• Specific heat is defined as the amount of heat
necessary to raise the temperature of 1 g of
substance by 1o C.
• Each substance has a specific heat, which is a
physical intensive property, like density and melting
point.
80
Heat Transfer and the
Measurement of Heat
• From a knowledge of a substance’s specific
heat, the heat (q) that is absorbed or released in
a given process can be calculated by use of the
following equation:
q = s x m x DT
q (heat energy)
cal, kcal, J or kJ
m (mass)
g
s (specific heat)
cal
g oC (kcal, J, or kJ can be used in
lieue of cal).
DT = T2 – T1 (change in temp-make
DT a positive #)
oC
81
Heat Transfer and the
Measurement of Heat
• Substances with large specific heats require
more heat to raise their temperature.
• Water has one of the highest specific heats,
1.00 cal/goC. The high specific heat of water
(which constitutes ~60% of our body weight)
makes our body’s task of maintaining a
constant body temperature of ~37oC much
easier. Thus, our body has the ability to
absorb or release considerable amounts of
energy with little change in temperature.
82
Heat Transfer and the
Measurement of Heat
Substance
water
specific heat (cal/goC)
1.00
wood
0.421
gold
graphite
0.0306
0.172
83
Heat Transfer and the
Measurement of Heat
Calculate the amount of heat to raise T of
200.0 g of water from 10.0oC to 55.0oC.
You need to know that the specific heat for water (swater) is 1.00 cal/goC
q  m  s  DT
4.184 J
? J  200 g H 2 O  o
 (55.0 o C  10.0o C)
g C
 3.76 10 4 J
1kJ
3.76 x 10 Jx 3  3.76 x 101 kJ  37.6kJ
10 J
4
84
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  s  DT
? J  200 g Hg 
0.138 J
o
o

(55.0
C

10.0
C)
o
g C
 1.24 x 103 J
1.24 x 103 Jx
1kJ
 1.24kJ
3
10 J
• Requires 30.3 times more heat for water
• 4.184 is 30.3 times greater than 0.138
85
Heat Transfer and the
Measurement of Heat
• If we add 450 cal of heat to 37 g of ethyl alcohol
(s=0.59 cal/goC) at 20oC, what would its final
temperature be?
q = m x s x DT
cal
450 cal = 37 g x 0.59 g oC x DT
DT
450cal
cal
(0.59 o ) x37 g
gC
DT=T2- T1

Since heat was added, the final
o
21 C temperature must be greater than
the initial temperature.
21oC = T2 – 20oC
T2 = 21oC + 20oC = 41oC
86