Matter and Energy
Chapter 3
1
Properties
•
Characteristics of the substance under observation
•
Properties can be either
directly observable or
the manner something interacts with other substances
in the universe
2
Universe Classified
• Matter is the part of the universe that has
mass and volume
• Energy is the part of the universe that has
the ability to do work
• Chemistry is the study of matter
– The properties of different types of matter
– The way matter behaves when influenced by
other matter and/or energy
3
Properties of Matter
• Physical Properties are the characteristics of
matter that can be changed without changing its
composition
– Characteristics that are directly observable
• Chemical Properties are the characteristics that
determine how the composition of matter changes
as a result of contact with other matter or the
influence of energy
• Characteristics that describe the behavior of matter
4
Classify Each of the following as
Physical or Chemical Properties
The boiling point of ethyl alcohol is 78°C.
– Physical property – describes inherent characteristic of
alcohol – boiling point
Diamond is very hard.
– Physical property – describes inherent characteristic of
diamond – hardness
Sugar ferments to form ethyl alcohol.
– Chemical property – describes behavior of sugar –
forming a new substance (ethyl alcohol)
5
States of Matter
• solid, liquid, gas
State
Solid
Liquid
Gas
Shape
Keeps
Shape
Takes
Shape of
Container
Takes
Shape of
Container
Volume
Compress
Flow
Keeps
Volume
Keeps
Volume
No
No
No
Yes
Takes
Volume of
Container
Yes
Yes
6
Figure 3.2: The three states of water
7
Changes in Matter
• Physical Changes are changes to matter that do
not result in a change of the fundamental
components that make that substance
– State Changes – boiling, melting, condensing
• Chemical Changes involve a change in the
fundamental components of the substance
– Produce a new substance
– Chemical reaction
– Reactants  Products
8
Figure 3.3:
Electrolysis
9
Classify Each of the following as
Physical or Chemical Changes
Iron is melted.
– Physical change – describes a state change, but the
material is still iron
Iron combines with oxygen to form rust.
– Chemical change – describes how iron and oxygen
react to make a new substance, rust
Sugar ferments to form ethyl alcohol.
– Chemical change – describes how sugar forms a new
substance (ethyl alcohol)
10
Elements and Compounds
• Substances which can not be broken down into
simpler substances by chemical reactions are
called elements
• Most substances are chemical combinations of
elements. These are called compounds.
– Compounds are made of elements
– Compounds can be broken down into elements
– Properties of the compound not related to the properties
of the elements that compose it
– Same chemical composition at all times
11
Classification of Matter
Matter
Pure Substance
Constant Composition
Homogeneous
Mixture
Variable Composition
Heterogeneous
• Homogeneous = uniform throughout, appears to be one
thing
– pure substances
– solutions (homogeneous mixtures)
• Heterogeneous = non-uniform, contains regions with
different properties than other regions
12
Pure Substances vs. Mixtures
• Pure Substances
– All samples have the same physical and chemical properties
– Constant Composition  all samples have the same
composition
– Homogeneous
– Separate into components based on chemical properties
• Mixtures
– Different samples may show different properties
– Variable composition
– Homogeneous or Heterogeneous
– Separate into components based on physical properties
• All mixtures are made of pure substances
13
Figure 3.4: Table salt is stirred into water (left), forming a
homogeneous mixture called a solution (right)
14
Figure 3.5: Sand and water do not mix to form a
uniform mixture
15
Identity Each of the following as a Pure
Substance, Homogeneous Mixture or
Heterogeneous Mixture
Gasoline
– a homogenous mixture
A stream with gravel on the bottom
– a heterogeneous mixture
Copper metal
– A pure substance (all elements are pure substances)
16
Separation of Mixtures
• Separate mixtures based on different
physical properties of the components
– Physical change
Different Physical Property
Technique
Boiling Point
Distillation
State of Matter
(solid/liquid/gas)
Adherence to a Surface
Filtration
Chromatography
Volatility
Evaporation
17
Figure 3.6: Distillation of a solution consisting of salt
dissolved in water
18
Figure 3.7: No
chemical
change occurs
when salt
water is
distilled
19
Figure 3.8: Filtration separates a liquid from a solid
20
Figure 3.9: Separation of a sand-saltwater mixture
21
Figure 3.10: The organization of matter
22
Energy and Energy Changes
• Capacity to do work
– chemical, mechanical, thermal, electrical,
radiant, sound, nuclear
• Energy may affect matter
– e.g. raise its temperature, eventually causing a
state change
– All physical changes and chemical changes
involve energy changes
23
Units of Energy
• One calorie is the amount of energy needed to raise the
temperature of one gram of water by 1°C
– kcal = energy needed to raise the temperature of 1000 g of
water 1°C
• joule
– 4.184 J = 1 cal
• In nutrition, calories are capitalized
– 1 Cal = 1 kcal
24
Example - Converting Calories to Joules
Convert 60.1 cal to joules
1 cal  4.184 joules
4.184 J
60.1cal 
 251J
1 cal
25
Energy and the Temperature of Matter
• The amount the temperature of an object
increases depends on the amount of heat
added (Q).
– If you double the added heat energy the
temperature will increase twice as much.
• The amount the temperature of an object
increases depends on its mass
– If you double the mass it will take twice as
much heat energy to raise the temperature the
same amount.
26
Specific Heat Capacity
• Specific Heat (s) is the amount of energy
required to raise the temperature of one
gram of a substance by one Celsius degree
J
By definition , the specific heat of water is 4.184
g C
Amount of Heat = Specific Heat x Mass x Temperature Change
Q = s x m x T
27
Example – Calculate the amount of heat
energy (in joules) needed to raise the
temperature of 7.40 g of water from
29.0°C to 46.0°C
JJ
Specific Heat of Water = 4.184
gg-CC
Mass = 7.40 g
Temperature Change = 46.0°C – 29.0°C = 17.0°C
Q = s x m x T
J
Heat  4.184
 7.40g  17.0C  526 J
g C
28
Example – A 1.6 g sample of metal that
appears to be gold requires 5.8 J to raise the
temperature from 23°C to 41°C.
Is the metal pure gold?
Q  s  m  T
Q
s
m  T
T  41C - 23C  18C
5.8 J
J
s
 0.20
1.6 g x 18C
g C
Table 3.2 lists the specific heat of gold as 0.13
Therefore the metal cannot be pure gold.
J
g C
29