Chapter 3 Matter and Energy€¦ · Tro's "Introductory Chemistry", Chapter 3 15 Classification of...

Chapter 3

Matter and Energy

Tro's "Introductory Chemistry",

Chapter 3

2

Matter

• Matter is anything that occupies space and has mass.

• Even though it appears to be smooth and continuous, matter is actually composed of a lot of tiny little pieces we call atoms and molecules.


Chapter 3

3

Atoms and Molecules

• Atoms are the tiny particles

that make up all matter.

• In most substances, the

atoms are joined together in

units called molecules.


Chapter 3

4

Classification of Matter

Matter can be classified in two ways:

1- according to its physical state

2- according to its composition

4


Chapter 3

5

Classification of Matter According to its

Physical State (STATES OF MATTER)

Matter can be classified as solid, liquid, or gas

SOLID e.g. stone, charcoal,

diamond

LIQUID e.g. water, wine,

vinegar

GAS e.g. oxygen,

carbon dioxide

MATTER


Chapter 3

6

Solids • The particles in a solid are packed

close together and are fixed in position. Due to this close parking solids are

incompressible.

• The inability of the particles to move around results in solids retaining their shape and volume when placed in a new container and prevents the particles from flowing.

• Solids may be further classified as: Crystalline Solids

Amorphous Solids


Chapter 3

7

Crystalline vs Amorphous • Some solids have their particles

arranged in an orderly geometric pattern—we call these crystalline solids.

Salt and diamonds.

• Other solids have particles that do not show a regular geometric pattern over a long range—we call these amorphous solids.

Plastic and glass.


Chapter 3

8

Liquids

• The particles in a liquid are closely packed, but they have some ability to move around.

• The close packing results in liquids being incompressible.

• The ability of the particles to move allows liquids to take the shape of their container and to flow. However, they don’t have enough freedom to escape and expand to fill the container.


Chapter 3

9

Gases

• In the gas state, the particles have complete

freedom from each other.

• The particles are constantly flying around,

bumping into each other and the container.

• In the gas state, there is a lot of empty space

between the particles.

On average.


Chapter 3

10

Gases, Continued

• Because there is a lot of empty

space, the particles can be

squeezed closer together.

Therefore, gases are

compressible.

• Because the particles are not

held in close contact and are

moving freely, gases expand to

fill and take the shape of their

container, and will flow.


Chapter 3

11

Summary: Properties of Solids,

Liquids and Gases

Volume Shape Compressible?

Solids fixed fixed NO

Liquids fixed not fixed NO

Gases not fixed

not fixed

YES


Chapter 3

12

Classification of Matter by Composition

Pure Substance

Constant Composition

Mixture

Variable Composition

Matter


Chapter 3

13

Classification of Matter as

a pure substance

A pure substance is made of a single type of

particle (i.e., atom or molecule).

The composition of a pure substance does not

change from one sample to another and because

of this, all samples have the same

characteristics (or properties)


Chapter 3

14

Classification of Matter as a Pure Substance

•Pure substances can further be sub-divided into

two groups:

Elements

Compounds

Elements: Pure substances that cannot be broken down

into simpler substances e.g., copper, helium, sodium

Compounds: Pure substances that are made of two or

more elements in definite proportions e.g., sodium

chloride NaCl, Carbon dioxide CO2


Chapter 3

15

Classification of Matter as a Mixture

• A mixture is a combination of two or more substances in

which each substance retain their distinct characteristics (or properties). e.g., a mixture of rice and sand

• Mixtures can also be further sub-divided into two groups:

Homogeneous mixture

Heterogeneous mixture

Homogeneous mixture: Has uniform composition through out the

sample e.g., solutions such as tea with sugar, salt solutions

Heterogeneous mixture: Does not have a uniform composition

through out the sample. e.g., sand and water

Note: HOMO means the same while HETERO means different


Chapter 3

16

Example—Classify the Following as

Homogeneous or Heterogeneous

Mixtures

• A cup of coffee

• A mixture of table sugar and black pepper

• A mixture of sugar dissolved in water

• Sand and water

- homogeneous

- homogeneous

- heterogeneous

- heterogeneous

17

Classifying Matter


Chapter 3

18

How Do We Distinguish Matter?

• Looking at both flasks, it is had to tell the difference between the two substances

• How can we tell if one flask contain water and the other contains an acid?

• We can only tell by studying their properties

water acid


Chapter 3

19

Properties of Matter

• Each sample of matter is distinguished by its characteristics.

• The characteristics of a substance are called its properties.


Chapter 3

20

Types of Properties of Matter

• Physical Properties

Properties of matter that can be observed

without changing its composition.

e.g., coke is dark brown

• Chemical Properties

Properties of matter that can be observed only

when matter changes its composition.

e.g., gasoline is a very flammable liquid


Chapter 3

21

Some Physical Properties Mass Volume Density

Solid Liquid Gas

Melting point Boiling point Volatility

Taste Odor Color

Texture Shape Solubility

Electrical

conductance

Thermal

conductance Magnetism

Malleability Ductility Specific heat

capacity


Chapter 3

22

Example: Some Physical Properties of Iron

• Iron melts at 1538 °C

• Iron boils at 4428 °C.

• Iron’s density is 7.87 g/cm3.

• Iron conducts electricity, but not as well as most other common metals.


Chapter 3

23

Some Chemical Properties

Acidity Basicity

Causticity Corrosiveness

Reactivity Stability

Inertness Explosiveness

Flammability Combustibility

Oxidizing ability Reducing ability


Chapter 3

24

Example: Some Chemical Properties of Iron

• Iron is easily oxidized in

moist air to form rust.

• When iron is added to

hydrochloric acid, it produces

a solution of ferric chloride

and hydrogen gas.

• Iron is more reactive than

silver, but less reactive than

magnesium.


Chapter 3

25

Example—Decide Whether Each of the Observations

About Table Salt Is a Physical or Chemical Property

• Salt is a white, granular solid.

• Salt melts at 801 °C.

• A liquid burns with a blue flame.

• 36 g of salt will dissolve in 100 g of water.

• Salt solutions conduct electricity.

• When a clear, colorless solution of silver nitrate is added to a salt solution, a white solid forms.

• When electricity is passed through molten salt, a gray metal forms at one terminal and a yellow-green gas at the other.

physical

physical

chemical

chemical

physical physical

chemical


Chapter 3

26

Changes in Matter

• Changes that alter the state or appearance of

matter without altering its composition are called

physical changes.

• Changes that alter the composition of matter are

called chemical changes.


Chapter 3

27

Physical Changes in Matter

• Physical Changes—Changes that do not affect composition of matter.

Heating water.

Raises its temperature, but it is still water.

Evaporating butane from a lighter.

Dissolving sugar in water.

Even though the sugar seems to disappear, it can easily be separated back into sugar and water by evaporation.


Chapter 3

28

Changes in Matter, Continued

• Chemical Changes— involve a

change in composition.

A chemical reaction.

Silver combines with sulfur in the air to make

tarnish.

Rusting is iron combining with oxygen to

make iron(III) oxide.

Burning results in butane from a lighter to

be changed into carbon dioxide and water.


Chapter 3

29

Is it a Physical or Chemical Change?

• A physical change results in a different form of the same substance.

The kinds of molecules don’t change. i.e., composition stays the same .

• A chemical change results in one or more completely new substances.

The new substances have different molecules than the original substances. i.e., composition changes.

Appearance may or may not change.

You will observe different physical properties because the new substances have their own physical properties.


Chapter 3

30

Phase Changes Are

Physical Changes

• Vaporizing = liquid to gas.

• Melting = solid to liquid.

• Subliming = solid to gas.

• Freezing = liquid to solid.

• Condensing = gas to liquid.

• Deposition = gas to solid.

• Changes in the state of matter require heating or cooling the substance.


Chapter 3

31

Example—Classify Each Change as Physical

or Chemical, Continued

• Evaporation of rubbing alcohol.

• Sugar turning black when heated.

• An egg splitting open and spilling out.

Physical change

Chemical change

Physical change


Chapter 3

32

Practice—Classify Each Change as Physical

or Chemical

• Sugar fermenting.

• Bubbles escaping from soda.

• Bubbles that form when hydrogen peroxide is

mixed with blood.


Chapter 3

33

Separation of Mixtures • Separating mixtures based on different

physical properties of their components.

Decanting Density

Evaporation Volatility

Chromatography Adherence to a surface

Filtration State of matter (solid/liquid/gas)

Distillation Boiling point

Technique Different Physical Property

Decanting Density

Evaporation Volatility

Chromatography Adherence to a surface

Filtration State of matter (solid/liquid/gas)

Distillation Boiling point

Technique Different Physical Property


Chapter 3

34

Distillation


Chapter 3

35

Filtration


Chapter 3

36

Law of Conservation of Mass

• The law of conservation of mass states: “Matter is neither created nor destroyed in a chemical reaction.”

• The total amount of matter present before a chemical reaction is always the same as the total amount after.

• Example: 58 grams of butane burns in 208 grams of

oxygen to form 176 grams of carbon dioxide and 90 grams of water.

butane + oxygen carbon dioxide + water

58 grams + 208 grams 176 grams + 90 grams

266 grams = 266 grams


Chapter 3

37

Practice—A Student Places Table Sugar and

Sulfuric Acid into a Beaker and Gets a Total

Mass of 144.0 g. Shortly, a Reaction Starts that

Produces a “Snake” of Carbon Extending from

the Beaker and Steam Is Seen Escaping. If the

Carbon Snake and Beaker at the End Have a

Total Mass of 129.6 g, How Much Steam Was

Produced?


Chapter 3

38

Energy

• Energy is anything that has the capacity to do

work.

• Unlike matter, energy does not have mass and

does not occupy any space.

• Although chemistry is the study of matter, matter

is effected by energy.

It can cause physical and/or chemical changes in

matter.


Chapter 3

39

Law of Conservation of Energy

• “Energy can neither be created nor destroyed.”

• The total amount of energy in the universe is constant. There is no process that can increase or decrease that amount.

• However, we can transfer energy from one place in the universe to another, and we can also change energy from one form to another.


Chapter 3

40

Kinds of Energy

Kinetic and Potential

• Potential energy is energy that is stored. Water flows because gravity pulls it

downstream. However, the dam won’t allow it to

move, so it has to store that energy.

• Kinetic energy is energy of motion, or energy that is being transferred from one object to another. When the water flows over the dam,

some of its potential energy is converted to kinetic energy of motion.


Chapter 3

41

Some Forms of Energy • Electrical Energy

Kinetic energy associated with the flow of electrical charge.

• Heat or Thermal Energy

Kinetic energy associated with molecular motion.

• Light or Radiant Energy

Kinetic energy associated with energy transitions in an atom.

• Nuclear Energy

Potential energy stored in the nucleus of atoms.

• Chemical Energy

Potential energy in compounds.


Chapter 3

42

Converting Forms of Energy • When water flows over the dam, some of its

potential energy is converted to kinetic energy. Some of the energy is stored in the water because it is

at a higher elevation than the surroundings.

• The movement of the water is kinetic energy.

• Along the way, some of that energy can be used to push a turbine to generate electricity. Electricity is one form of kinetic energy.

• The electricity can then be used in your home. For example, you can use it to heat cake batter you mixed, causing it to change chemically and storing some of the energy in the new molecules that are made.


Chapter 3

43

How do We Measure Energy?

Units of Energy

• The SI unit for energy is the joule (J)

Other units such as calorie, Calorie, KWh are widely used

• calorie (cal) is the amount of energy needed to raise

one gram of water by 1 °C.

1 kcal = energy needed to raise 1000 g of water 1 °C.

1 food calories (Cal) = 1 kcals.

Energy Conversion Factors

1 calorie (cal) = 4.184 joules (J)

1 Calorie (Cal) = 1000 calories (cal)

1 kilowatt-hour (kWh) = 3.60 x 106 joules (J)


Chapter 3

44

Some Common Uses of Energy Use

Unit

Energy Required

to Raise

Temperature of 1 g

of Water by 1°C

Energy

Required to

Light 100-W

Bulb for 1

Hour

Energy

Used by

Average

U.S. Citizen

in 1 Day

joule (J) 4.18 3.6 x 105 9.0 x 108

calorie (cal) 1.00 8.60 x 104 2.2 x 108

Calorie (Cal) 1.00 x 10-3 86.0 2.2 x 105

kWh 1.1 x 10-6 0.100 2.50 x 102

Example 3.5—Convert 225 Cal to Joules

225 Cal = 9.41 x 105 J Round: 6. Significant figures and

round.

Solution: 5. Follow the solution map to

Solve the problem.

Solution

Map:

4. Write a Solution Map.

1 Cal = 1000 cal

1 cal = 4.184 J

Conversion

Factors:

3. Write down the appropriate

Conversion Factors.

? J Find: 2. Write down the quantity

you want to Find and unit.

225 Cal Given: 1. Write down the Given

quantity and its unit.

Cal

cal 1

J .1844

J 1041.9cal 1

J .1844

Cal 1

cal 1000Cal 252 5

cal J

Cal 1

cal 1000

3 sig figs

3 significant figures

46

Practice 1: The complete combustion of a wooden

match produces 512 cal of heat. How many

kilojoules are produced?

Practice 2: An energy bill indicates that the

customer used 955 KWh. How many calories did

the customer use?

Answer: 8.22 x exp(8) cal.

Answer: 2.14 kJ


Chapter 3

47

Practice 1: The complete combustion of a wooden match

produces 512 cal of heat. How many kilojoules are

produced?

Answer: 2.14 kJ

48

Practice 2: An energy bill indicates that the

customer used 955 KWh. How many

calories did the customer use?

Answer: 8.22 x exp(8) cal.


Chapter 3

49

Energy Changes

• Processes that involve energy

changes can be:

Exothermic

Endothermic


Chapter 3

50

Exothermic Processes

• When a change results in the release of energy it is

called an exothermic process.

• An exothermic chemical reaction occurs when the

reactants have more chemical potential energy

than the products.

• The excess energy is released into the surrounding

materials, adding energy to them.

Often the surrounding materials get hotter from the

energy released by the reaction.


Chapter 3

51

An Exothermic Reaction

Pote

nti

al e

ner

gy

Reactants

Products

Surroundings

reaction

Amount

of energy

released


Chapter 3

52

Endothermic Processes

• When a change requires the absorption of energy

it is called an endothermic process.

• An endothermic chemical reaction occurs when

the products have more chemical potential energy

than the reactants.

• The required energy is absorbed from the

surrounding materials, taking energy from them.

Often the surrounding materials get colder due to the

energy being removed by the reaction.


Chapter 3

53

An Endothermic Reaction

Pote

nti

al e

ner

gy

Products

Reactants

Surroundings

reaction

Amount

of energy

absorbed


Chapter 3

54

Temperature Scales

• Fahrenheit scale, °F.

Used in the U.S.

• Celsius scale, °C.

Used in all other countries.

A Celsius degree is 1.8

times larger than a

Fahrenheit degree.

• Kelvin scale, K.

Absolute scale.

Temperature Scales

Celsius Kelvin Fahrenheit -273°C -269°C

-183°C

-38.9°C

0°C

100°C

0 K 4 K

90 K

234.1 K

273 K

373 K

-459 °F -452°F

-297°F

-38°F

32°F

212°F

Absolute

zero

BP helium

Boiling

point

oxygen

Boiling

point

mercury

Melting

point ice

Boiling

point water

0 R 7 R

162 R

421 R

459 R

671 R

Rankine

Room temp 25°C 298 K 75°F 534 R


Chapter 3

56

The Fahrenheit Temperature Scale

• The Fahrenheit temperature scale was setup

by assigning 0 °F to the freezing point of

concentrated saltwater and 96 °F for normal

body temperature.

• Room temperature is about 72 °F.


Chapter 3

57

The Celsius Temperature Scale

• Was setup by assigning 0 °C to the freezing

point of distilled water and 100 °C to the

boiling point of distilled water.

Most commonly used in science.

More reproducible standards.

• Room temperature is about 22 °C.


Chapter 3

58

The Kelvin Temperature Scale

• Both the Celsius and Fahrenheit scales have negative numbers.

• The Kelvin scale avoids negative numbers and therefore the lowest temperature is 0 K.

• 0 K is called absolute zero. It is too cold for matter to exist because all molecular motion would stop.

0 K = -273 ° C = -459 °F.

Absolute zero is a theoretical value obtained by following patterns mathematically.


Chapter 3

59

• Celsius to Kelvin

K = ° C + 273

• Fahrenheit to Celsius

1.8

32-F C

Converting from one temperature

scale to another

Example 3.7—Convert –25 °C to Kelvins

248 K Round: 6. Significant figures and

round.


Solve the problem.

Solution

Map:


Equation: 3. Write down the appropriate

Equations.

K ? Find: 2. Write down the quantity


-25 °C Given: 1. Write down the Given


° C K

273C K

K 248273C) 25(K

K = ° C + 273

Example 3.8—Convert 55° F to Celsius

Units and magnitude are correct.

Check: 7. Check.

12.778 °C = 13 °C Round: 6. Significant figures and

round.


Solve the problem.

Solution

Map:



Equations.

° C ? Find: 2. Write down the quantity


55 °F Given: 1. Write down the Given

quantity and its unit. units place

and 2 sig figs

units place and 2 sig figs

° F ° C

1.8

32-F C

1.8

32-F C

C 778.12

1.8

32-F 55 C

Example 3.9—Convert 310. K to Fahrenheit

Units and magnitude are correct.

Check: 7. Check.

98.6 °F = 99 °F Round: 6. Significant figures and

round.


Solve the problem.

Solution

Map:



Equations.

°F ? Find: 2. Write down the quantity


310 K Given: 1. Write down the Given

quantity and its unit. units place

and 3 sig figs

units place and 2 sig figs

1.8

32-F C

32C1.8 F

F 6.9832C 37.81 F

K = °C + 273

°F °C K

°C = K - 273

C 37273310 C


Chapter 3

63

Practice 1- A sick child has a body

temperature of 40.00 °C. What is the child’s

temperature in Kelvins (K) ?

Practice 2- During one summer day in

Gettysburg, a record temperature of 92.1 ° F

was reported. What is the temperature in K ?


Chapter 3

64

Practice 1- A sick child has a body

temperature of 40.00 °C. What is the child’s

temperature in Kelvins (K) ?


Chapter 3

65

Practice 2- During one summer day in

Gettysburg, a record temperature of 92.1 ° F

was reported. What is the temperature in K ?


Chapter 3

66

Energy and the Temperature of Matter

• The increase in temperature of an object

depends on the amount of heat energy

added (q) and the mass of the object.

If you double the added heat energy the

temperature will increase twice as much.

If you double the mass, it will take twice as

much heat energy to raise the temperature the

same amount.

67

Heat Capacity • Heat capacity is the amount of heat a substance

must absorb to raise its temperature by 1 °C.

cal/°C or J/°C.

Metals have low heat capacities; insulators have

high heat capacities.

• Specific heat = heat capacity of 1 gram of the

substance.

cal/g°C or J/g°C.

Water’s specific heat = 4.184 J/g°C for liquid.

Or 1.000 cal/g°C.


Chapter 3

68

Specific Heat Capacity • Specific heat is the amount of energy required to raise

the temperature of one gram of a substance by 1 °C.

• The larger a material’s specific heat is, the more energy it takes to raise its temperature a given amount.

• Like density, specific heat is a property of the type of matter.

It doesn’t matter how much material you have.

It can be used to identify the type of matter.

• Water’s high specific heat is the reason it is such a good cooling agent.

It absorbs a lot of heat for a relatively small mass.


Chapter 3

69

Specific Heat Capacities Substance Specific Heat

J/g°C

Aluminum 0.903

Carbon (dia) 0.508

Carbon (gra) 0.708

Copper 0.385

Gold 0.128

Iron 0.449

Lead 0.128

Silver 0.235

Ethanol 2.42

Water (l) 4.184

Water (s) 2.03

Water (g) 2.02


Chapter 3

70

Heat Gain or Loss by an Object

• The amount of heat energy gained or lost by an object

depends on 3 factors:

how much material there is

what the material is

how much the temperature changed

Amount of Heat = Mass x Specific Heat Capacity x Temperature Change

q = m x C x DT

J 557.4

C25.0-29.9 0.372g 2.5 Cg

J

q

q

Example 3.10—Calculate Amount of Heat Needed to Raise Temperature of 2.5 g Ga from 25.0 to 29.9 °C

Units and magnitude are

correct. Check: 7. Check.

4.557 J = 4.6 J Round: 6. Significant figures and

round.


Solve the problem.

Solution

Map:



Equations.

q, J Find: 2. Write down the quantity


m = 2.5 g, T1 = 25.0 °C,

T2= 29.9 °C, C = 0.372 J/g°C

Given: 1. Write down the Given


2 significant figures

m, C, DT q

TCmq D

TCmq D

Example—Calculate the Amount of Heat Released

When 7.40 g of Water Cools from 49° to 29 °C

q = m ∙ Cs ∙ DT

Cs = 4.18 J/gC (Table 3.4)

The unit and sign are correct.

T1 = 49 °C, T2 = 29 °C, m = 7.40 g

q, J

Check: • Check.

Solution: • Follow the

concept

plan to

solve the

problem.

Solution Map:

Relationships:

• Strategize

Given:

Find:

• Sort

Information

TC mq s Δ

J 106.2J 64.816

C 02- 4.18g 7.40

Δ

2

Cg

J

TCmq s

C 02-

C94 - C 29

12

D

D

T

TTT

Cs m, DT q


Chapter 3

73

Practice 1- If you hold gallium in your hand, it melts

from body heat. How much heat must 3.5 g of gallium

absorb from your hand to raise its temperature from 25.0

°C to 30.5 °C? The heat capacity of gallium is 0.372

J/g°C.

Practice 2- A bottle containing 24.1 g of an alcohol was

removed from a refrigerator and placed on top of a

counter in a kitchen. The amount of heat absorbed by the

alcohol was determined to be 1.17 kJ. Given that the

specific heat of the alcohol is 2.42 J/g°C

1-Calculate the change in temperature

2-If the initial temperature of the alcohol was 5 °C, find

the final temperature


Chapter 3

74

Practice 1- If you hold gallium in your hand, it melts

from body heat. How much heat must 3.5 g of gallium

absorb from your hand to raise its temperature from 25.0

°C to 30.5 °C? The heat capacity of gallium is 0.372

J/g°C.


Chapter 3

75

Practice 2- A bottle containing 24.1 g of an alcohol was

removed from a refrigerator and placed on top of a

counter in a kitchen. The amount of heat absorbed by the

alcohol was determined to be 1.17 kJ. Given that the

specific heat of the alcohol is 2.42 J/g°C

1-Calculate the change in temperature

2-If the initial temperature of the alcohol was 5 °C, find

the final temperature


Chapter 3

76

Practice 3-

25.3 g of iron was heated to 53.2 °C and let to cool

slowly. If the amount of heat lost by iron during the

cooling process is 353.5 J, find the final temperature of

the iron. The specific heat of iron is 0.499 J/g°C.


Chapter 3

77

Recommended Study Problems Chapter 3

NB: Study problems are used to check the student’s understanding

of the lecture material. Students are EXPECTED TO BE ABLE

TO SOLVE ALL THE SUGGESTED STUDY PROBLEMS.

If you encounter any problems, please talk to your professor or seek

help at the HACC-Gettysburg learning center.

Questions from text book Chapter 3, p 81

1, 3, 5, 15, 17, 19, 21, 23, 27, 28, 29, 31, 35-37, 39, 41, 44, 45, 47,

51, 52, 55, 56, 57, 60, 63-65, 67, 71, 73, 75, 77, 79, 81, 83, 87, 93,

97, 99, 107, 109, 110

ANSWERS

-The answers to the odd-numbered study problems are found at

the back of your textbook

Chapter 3 Matter and Energy€¦ · Tro's "Introductory Chemistry", Chapter 3 15 Classification of...

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