Chapter 3

Matter – anything that has mass and takes up space
 Everything around us
 Mass: measurement that reflects the amount of matter
(usually in grams)
 Volume: the amount of space something takes up

Chemistry – the study of matter and the changes it
undergoes

Solids
 particles vibrate but can’t move around
 fixed shape
 fixed volume
 incompressible

Liquids
 particles can move around but
are still close together
 variable shape
 fixed volume
 Virtually incompressible

Gases
 particles can separate and move




throughout container
variable shape
variable volume
Easily compressed
Vapor = gaseous state of a substance
that is a liquid or solid at room
temperature

Plasma
 atoms collide with enough energy to break into
charged particles (+/-)
 gas-like, variable
shape & volume
 stars, fluorescent
light bulbs, TV tubes
II. Properties & Changes in Matter (p.73-79)
Extensive vs. Intensive
Physical vs. Chemical

Physical Property
 can be observed & measured without changing
the identity of the substance

Physical properties can be described as one of
2 types:

Extensive Property
 depends on the amount of matter present
(example: length, mass, volume)

Intensive Property
 depends on the identity of substance, not the
amount (example: scent, density, melting point)

Chemical Property
 describes the ability of a substance to be observed
reacting with or changing into another substance

Examples:
 melting point
physical
 flammable
chemical
 density
physical
 magnetic
physical
 tarnishes in air
chemical

Physical Change
 changes the form of a substance without
changing its identity
 properties remain the same

Examples: cutting a sheet of paper, breaking
a crystal, all phase changes
Evaporation =
Liquid -> Gas
Condensation =
Gas -> Liquid






Melting =
Solid -> Liquid
Freezing =
Liquid -> Solid
Sublimation =
Solid -> Gas
Deposition =
Gas -> Solid

Process that involves one or more substances
changing into a new substance
 Commonly referred to as a chemical reaction
 New substances have different compositions and
properties from original substances
 Reaction involves reactants reacting to
produce products

Signs of a Chemical Change
 change in color or odor
 formation of a gas (bubbles)
 formation of a precipitate (solid)
 change in light or heat

Examples:
 rusting iron
chemical
 dissolving in water
physical
 burning a log
chemical
 melting ice
physical
 grinding spices
physical
Exothermic- heat energy EXITS the system
 surroundings usually feel warmer
 1 g H2O (g)  1 g H2O (l) + 2260 J
 ex. Combustion, evaporation of water
Endothermic- heat energy ENTERS the system
- heat absorbed from surroundings
- surroundings usually feel cooler
- 1 g H2O (s) + 333 J  1 g H2O (l)
- 1 g H2O (l) + 2260 J  1 g H2O (g)
- ex. Cold packs, melting ice






Although chemical changes occur, mass is
neither created nor destroyed in a chemical
reaction
Mass of reactants equals mass of products
massreactants = massproducts
A+BC
III. Classification of Matter (pp. 80-87)
Matter Flowchart
Pure Substances
Mixtures
MATTER
yes
Can it be physically
separated?
MIXTURE
yes
Is the composition
uniform?
Homogeneous
Mixture
(solution)
no
PURE SUBSTANCE
no
Heterogeneous
Mixture
yes
Can it be chemically
decomposed?
Compound
no
Element

Examples:
 graphite
element
 pepper
hetero. mixture
 sugar (sucrose)
compound
 paint
hetero. mixture
 soda
solution

Element
 composed of one type of identical atoms

Atom: Composed of protons, electrons, and neutrons.
Smallest particle of matter that can be identified as one
element
 EX: copper wire, aluminum foil

Compound
 composed of 2 or more elements in a
fixed ratio (bonded together)
 properties differ from those of
individual elements
 EX: table salt (NaCl)

Variable combination of 2 or more pure
substances, each retains its chemical identity &
properties.
Heterogeneous
Homogeneous

Homogeneous: are uniform throughout
 Solutions
 very small particles
 particles don’t settle
 EX: rubbing alcohol, gasoline, soda

Heterogeneous
 medium-sized to
large-sized particles
 particles may or may
not settle
 EX: milk, freshsqueezed
lemonade

Examples:

Answers:
 tea
 Solution
 muddy water
 Heterogeneous
 fog
 Heterogeneous
 saltwater
 Solution
 Italian salad dressing
 Heterogeneous
+
Separation Methods
Ways to separate mixtures – Chapter 3: Matter
& Its Properties
+
Separating Mixtures

Substances in a mixture are physically combined, so
processes bases on differences in physical properties are
used to separate component

Numerous techniques have been developed to separate
mixtures to study components

Visually

Magnetism

Filtration

Distillation

Crystallization

Chromatography
+
Filtration

Used to separate heterogeneous mixtures
composed of solids and liquids

Uses a porous barrier to separate the solid
from the liquid

Liquid passes through leaving the solid in the
filter paper
+
Distillation

Used to separate
homogeneous mixtures

Based on differences in
boiling points of substances
involved
+
Crystallization

Separation technique that results in
the formation of pure solid
particles from a solution containing
the dissolved substance

As one substance evaporates, the
dissolved substance comes out of
solution and collects as crystals

Produces highly pure solids

Rocky candy is an example of this
+
Chromatography

Separates components of a mixture based
on ability of each component to be drawn
across the surface of another material

Mixture is usually liquid and is usually
drawn across chromatography paper

Separation occurs because various
components travel at different rates

Components with strongest attraction for
paper travel the slowest
Thermochemistry
Chapter 17:1
Pages 505 – 510
A. Vocabulary
Thermochemistry:
The study of energy changes that occur during
chemical reactions and changes in state
Energy:
The capacity to do work or produce heat
A. Vocabulary
 TEMPERATURE
is a measure of the amount of
kinetic energy an object/substance has
 HEAT is energy that transfers from one
object/substance to another (thermal energy)
• Represented by the symbol q
• Transfers because of difference in temperature
• Always flows from a warmer to a cooler object
A. Vocabulary
Which
has more thermal
energy?
80ºC
A
80ºC
B
200 mL
400 mL
B. Heat Transfer
 Why does A feel hot and B feel cold?
• Heat flows from A to your hand = hot
• Heat flows from your hand to B = cold
80ºC
A
10ºC
B
C. Types of Energy
 Potential:
due to position or composition
– can be converted to work
 Kinetic: due to motion of the object
KE = ½ mv2
(m = mass, v = velocity)
 Law of Conservation of Energy –
energy can be neither created nor
destroyed
C. Types of Energy
Energy
that is stored in the
chemical bonds of a substance
is called CHEMICAL
POTENTIAL ENERGY
Types of atoms and their
arrangement determine amount
of energy stored in substance
D. Exothermic and Endothermic
 System
= the reaction (our focus)
 Surroundings = everything around the
reaction (rxn container, room, etc)
Surroundings
System
Universe = System + Surroundings
D. Exothermic and Endothermic
of (q) is a ‘signal’ to indicate
the direction of the heat transfer
Exothermic
Endothermic
• - q heat
• +q heat
transferred from
transferred into
a substance
a substance
Sign
E. Measuring Heat Flow
Two
Common Units
• Joule
• calorie
4.184 J = 1 cal
1J = 0.2390 cal
1Calorie = 1 kilocal = 1000 cal
F. Heat Capacity
“the
amount of heat needed to
increase the temperature of an
object by 1oC”
Heat Capacity depends on:
• The mass of the object
• The chemical composition of the
object
G. Specific Heat Capacity
heat capacity – amount of
heat needed to raise the temperature of
1g of a substance by 1oC
Specific
Heat of H2O (l) = 4.184 J/goC
 Specific Heat of H2O (s) = 2.02 J/goC
 Specific
heat capacity – amount of heat
needed to raise the temperature of
1 mole of a substance by 1oC
Molar
G. Specific Heat Capacity
q = m  C  T
q:
m:
C:
T:
heat (J)
mass (g)
specific heat (J/g·K) or (J/goC)
change in temperature (K or °C)
T = Tf - Ti
– q = heat loss
+ q = heat gain
G. Heat Transfer Problem
 A 32.0-g
silver spoon cools from 60.0°C
to 20.0°C. How much heat is lost by the
spoon?
GIVEN:
m = 32.0 g
Ti = 60.0°C
Tf = 20.0°C
q=?
C = .235 J/g·C
WORK:
q = m·C·T
T = 20°C - 60°C = – 40°C
q =(32.0g)(-40°C)(.235J/g·C)
q = – 301 J Exo
G. Heat Transfer Problem

The temperature of a 95.4g piece of copper
increases from 25.0 oC to 48.0 oC when it absorbs
849 J of heat. What is the specific heat of copper?
GIVEN:
m = 95.4 g
Ti = 25.0°C
Tf = 48.0°C
q = 849 J Endo
C=?
WORK:
q = m·C·T
C = q/m · T
T = 48°C – 25°C = 23°C
C = 849J/(95.4g)(23°C)
C = 0.387 J/g°C
Thermochemistry
Chapter 17:2
Pages 511 – 517
A. Calorimetry
 Measures
the heat flow into or out of a
system
 Heat released by the system is equal to
heat absorbed by the surroundings
 Calorimeter =
Insulated device used
to measure absorption
or release of heat
Coffee cup Calorimeter
A. Calorimetry
Enthalpy: (H) the heat absorbed or
released of a system at constant pressure
Δ H is the heat of a reaction
Heat or enthalpy change are used
interchangeably here
q=ΔH

B. Thermochemical Equations
 In
a thermochemical equation, the enthalpy of
change for the reaction can be written as either
a reactant or a product
(positive ΔH)
2NaHCO3 + 129kJ
Na2CO3 + H2O + CO2
ΔH = 129 kJ
 Exothermic (negative ΔH)
CaO + H2O
Ca(OH)2 + 65.2kJ
ΔH = - 65.2 kJ
 Endothermic
A. Calorimetry
•∆H
= Hfinal - Hinitial
•∆Hreaction (rxn) = Hproducts – Hreactants
OR
•For endothermic reactions Hfinal>Hinitial &
∆H is positive (+∆H)
•For exothermic reaction Hfinal<Hinitial and
∆H is negative (-∆H)
A. Calorimetry
When solving for the heat transfer of a system
(between 2 objects), assume that:
q initial soln = - qfinal rxn
heat goes in = heat goes out
+∆H
=
-∆H
or
CH20 x m H20 x ∆T H20 = -1(Cmetal x mmetal x ∆Tmetal)
When 25.0mL of water containing HCl at 25.0 oC is added
to 25.0mL of water containing NaOH at 25.0 oC in a
calorimeter a rxn occurs. Calculate the enthalpy change
(in kJ) during the rxn if the highest temperature
observed was 32.0 oC. Assume all densities =1.00g/mL
KNOWN:
 ΔH= -mCΔT=
Cwater = 4.184 J/g oC
 -(50.0g)(4.184J/goC)
V = 25.0mL + 25.0mL
oC)
(7.0
ΔT = 32.0 – 25.0 = 7.0 oC
 ΔH= -1463 J = -1460J
Density= 1.00g/mL
M = (50mL) x (1.00g/mL) = 50g
Exo
ΔH = ?
Write the thermochemical equation for the
oxidation of Iron (III) if its ΔH= -1652 kJ Exo
4Fe(s) + 3 O2(g)→ 2 Fe2O3(s) + 1652 kJ
How much heat is evolved when 10.00g of Iron is
reacted with excess oxygen?
10.00g Fe
1 mol
1652 kJ
55.85g Fe 4 mol Fe
=73.97 kJ of heat
Thermochemistry
Chapter 17:3
Pages 520 – 526
A. Heat of Fusion

Heat of Fusion (Hfus)
• Heat absorbed by one mole of a solid
substance when it melts to a liquid at a
constant temperature

• Hfus of ice = 6.009 kJ/mol
Heat of Solidification (Hsolid)
• Heat lost by one mole of a liquid substance
when it solidifies at a constant temperature
• Hfus = - Hsolid
• Hsolid of water = -6.009 kJ/mol
B. Heating Curves
Gas - KE 
Boiling - PE 
Liquid - KE 
Melting - PE 
Solid - KE 
B. Heating Curves
 Temperature
Change
• change in KE (molecular motion)
• depends on heat capacity
 Phase
Change
• change in PE (molecular arrangement)
• temp remains constant
C. Heat of Vaporization
 Heat
of Vaporization (Hvap)
• energy required to boil 1 gram of a
substance at its b.p.
• Hvap for water = 40.79 kJ/mol
• usually larger than Hfus…why?
 EX:
sweating,
steam burns
D. Practice Problems
• How much heat energy is required
to melt 25 grams of ice
ice at 0oC to
liquid water at a temperature of
0oC?
25 g H2O 1 mol H2O
6.009 kJ
18.02 g H2O 1 mol H2O
= 8.3 kJ
D. Practice Problems
• How much heat energy is required
to change 500.0 grams of liquid
steam at 100oC?
water at 100oC to steam
500.0 g H2O 1 mol H2O
40.79 kJ
18.02 g H2O 1 mol H2O
= 1132 kJ
D. Practice Problems
• How many kJ are absorbed when
0.46g of C2H5Cl vaporizes at its
normal boiling point? The molar
Hvap is 26.4 kJ/mol.
0.46 g C2H5Cl 1 mol C2H5Cl
26.4 kJ
64.52 g C2H5Cl 1 mol C2H5Cl
= 0.19 kJ
E. Heat of Solution
 During
the formation of a solution,
heat is either released or absorbed
 Enthalpy change caused by
dissolution of 1 mol of a substance
is the molar heat of solution Hsoln
 Examples: hot packs, cold packs
E. Heat of Solution
→ Na+(aq) + OH-(aq)
• Hsoln = -445.1 kJ/mol
 NaOH(s)
→ NH4+(aq) + NO3-(aq)
• Hsoln = 25.7 kJ/mol
 NH4NO3