Matter and Energy

A. Introduction:

1. Chemistry The study of matter, its compositions,

structures, properties, changes it undergoes, and

energy accompanying these changes

2. Matter - Anything that has mass and takes up space

– Can be pure substances or mixtures

B. Types of Matter:

1. Pure Substance

Every sample has a definite and fixed composition

Has a unique set of properties

Each sample is the same (homogeneous)

Elements or compounds

SG pg.1 #1 & 2

2. Elements

Pure substance composed of identical atoms with the same atomic number

Cannot be decomposed into simpler substances by physical or chemical methods

SG pg.1 #3 & 4

3. Compounds

Pure substances that are composed of two or more different elements chemically combined

Can be decomposed into simpler substances by chemical methods

Properties of a compound are different from those of the elements that make up the compound

Law of definite composition

Elements in a compound are combined in a definite ratio by mass

SG pg.2 #5 & 7

4. Mixtures

Composed of two or more substances that are physically combined

Composition may vary from one sample to another

Can be separated by physical methods

Retains properties of the individual components

SG pg.2 #8 & 9

(a) Homogeneous Mixtures

Uniformly and evenly mixed throughout

Samples have definite and fixed composition

Aqueous solutions are homogeneous mixtures that

are made with water

Example: salt water NaCl(aq)

Solution - its components are all in the same phase (dissolve)

Suspension- its components are in different phases

(b) Heterogeneous Mixtures

Not uniformly nor evenly mixed throughout

Samples have different and varying composition

Matter

Pure Substance

Mixture

Element Compound

Homogeneous

Heterogeneous

solution suspension

SG pg.5 #10 – 19MC pg.119-120 Sets 1 & 2

B. Separation of Mixtures:

1. Boiling, distillation, evaporation

Every separates homogeneous mixtures (solutions) in which the components have different boiling points

Example: salt and water

2. Filtration

Used when the mixture has different particle sizes

3. Centrifugation

Spinning objects to separate heavier substances from a mixture

Example: blood

4. Dialysis

(Diffusion) – when a mixture moves from an area where it has a high concentration to an area where its concentration is lower

 Element / compound /

mixtureChemical symbol

How many sample of each

Element (diatomic)

Element

Compound

Compound

Mixture

Mixture

X2 5 X2

Y

XY

8 Y

5 XY

XY2 6 XY2

X2 and Y 5 X2 and 4 Y

XY2 and Y 4 XY2 and 4 Y

Atom X

Atom Y

SG pg.6 #20 – 22MC pg.121 Set 3 W/S 1: Types of Matter

A. Phases of Matter:

1. Solid Definite volume and definite shape Particles arranged orderly in a “regular

geometric pattern” Particles with strong attractive force to one

another Particles vibrating around fixed points Particles that cannot be easily compressed

(incompressible)

2. Liquid

definite volume, but no definite shape (it takes the shape of its container)

particles flow over each other

particles that cannot be easily compressed (incompressible)

3. Gas

No definite shape and no definite volume (it takes volume and shape of its container)

Particles that are most random

Particles move fast and freely

Particles have very weak attractive force to each other

Particles that can be easily compressed (compressible)

SG pg.7 #23 – 26MC pg.122 Sets 4 & 5

B. Phases Changes:

1. Phase change

A physical change

A substances changes form (state) without changing its chemical composition

Depends of temperature and pressure

Water can be found in nature in all three phases

2. Change in Phase

Melting H2O(s) ---> H2O(l)

Freezing H2O(l) ---> H2O(s)

Evaporations

C2H5OH(l) ---> C2H5OH(g)

Condensation

C2H5OH(g) ---> C2H5OH(l)

Sublimation CO2(s) ---> CO2(g)

Deposition CO2(g) ---> CO2(s)

Solid to liquid

Liquid to solid

Liquid to gas

Gas to liquid

Solid to gas

Gas to solid

MC pg.123 Set 6

3. Phase changes and energy

A phase change occurs when it has absorbed or released enough heat energy to rearrange its particles (atoms or molecules) from one form to another

Latent Heat – amount of heat needed to be absorbed or released during a change in phase

Units for heat energy: calories or joules

(a) Endothermic

describes a process that absorbs heat energy

includes fusion, evaporation, sublimation

(b) Exothermic

describes a process that releases heat energy

includes freezing, condensation and deposition

SG pg.9 #27 – 31MC pg.124 Sets 7 & 8

4. Phase changes and temperature

(a) Temperature

A measure of the average kinetic energy of particles in matter

(b) Kinetic energy

Due to movements of particles in matter

The higher the temperature the greater its kinetic energy

As temperature increases, the average kinetic energy increases.

Temperature

(c) Thermometer

Instrument used to measure temperature

Degree Celsius (°C) and Kelvin (°K)

Phase change of water is used as reference points

0°C and 273°K = freezing point

100°C and 373°K = boiling point

Kelvin = °C + 273°

SG pg.10 #32 – 40MC pg.130-131 Sets 14 & 15

While the substance is a solid, liquid or gas, the Temperature –Kinetic energy – Potential energy -

Increases

Increases

Remains the same

solid / liquid

liquid / gas

Time (minutes)

Heating curve - Shows changes of a substance starting from the solid phase

Heat is being absorbed / endothermic

When two phases are present (solid/liquid or liquid/gas), the Temperature –Kinetic energy – Potential energy -

Remains the same

Remains the same

Increases

solid / liquid

liquid / gas

Time (minutes)

While the substance is a gas, liquid or solid, the Temperature –Kinetic energy – Potential energy -

Decreases

Decreases

Remains the same

solid / liquid

liquid / gas

Time (minutes)

Cooling curve - Shows changes of a substance starting from the gas phase

Heat is being released / exothermic

When two phases are present (solid/liquid or liquid/gas), the Temperature –Kinetic energy – Potential energy -

Remains the same

Remains the same

Decreases

solid / liquid

liquid / gas

Time (minutes)

SG pg.13 #41 – 50MC pg.125-129 Sets 9 – 13W/S 2: Phases of Matter

A. Introduction:

1. Heat A form of energy that can flow (or transfer) from

one object to another

Heat flows from an area of higher temperature to an area of lower temperature until equilibrium is reached

Energy is either absorbed or released during a chemical or physical changes

2. Exothermic A process that releases (emits or loses) heat

3. Endothermic A process that absorbs (or gains) heat

4. Joules & Calories unites for measuring heat

2. Calorimeter Device used for measuring heat during physical

and chemical changes

SG pg.14 #51 – 53MC pg.131 Set 16

B. Heat constants and heat equations:

1. Specific Heat capacity (C)

The amount of heat needed to change the temperature of one gram sample of the substance by adding one degree Celsius

Depends on the substances

Specific heat for water is 4.18 J/g ∙ °C

MC pg.132 Set 17

Calculating Heat gained or released (Table T)

Determining the amount of heat absorbed or released by a substance

Heat (q) = m × C × ΔT

Example: How much heat is released by a 7 gram sample of water to change its temperature from 15°C to 10°C?

q = m × C × ΔT

= 7g × 4.18 J/g ∙ °C × 5 °C

= 146.3 J

SG pg.17 #54 – 56MC pg.133 Set 18

2. Heat of fusion (Table B)

Amount of heat needed to melt or freeze one gram of the substance at a constant temperature

Heat of fusion for water is 334 J/g

334 J/g – absorbed to melt one gram of ice

334 J/g – released to freeze one gram of water

Calculating Heat of fusion (Table T)

Determining the amount of heat absorbed or released by freezing or melting

Heat (q) = m × Hf

Example: What is the number of joules needed to melt a 16g sample of ice to water at °C?

q = m × Hf

q = 16g 334 J/g

= 5344 J

SG pg.17 #57 – 59MC pg.133 Set 19

3. Heat of vaporization (Table B)

Amount of heat needed to vaporize (evaporate) or condense one gram of the substance at a constant temperature

Heat of fusion for water is 2260 J/g

2260 J/g – absorbed needed to vaporize one gram of ice

2260 J/g – released to condense one gram of water

Calculating Heat of vaporization (Table T)

Determining the amount of heat absorbed or released by vaporizing or condensing

Heat (q) = m × Hv

Example: Liquid ammonia has a heat of vaporization of 1.35 KJ/g. How many kilojoules of heat are needed to evaporate a 5 gram sample of ammonia at its boiling point?

q = m × Hv

q = 5g x 1.35 KJ/g

= 6.75 KJ

SG pg.18 #60 – 62MC pg.134-135 Sets 21-22

Review of Equations (Table B and Table T)

If two temperatures are given, change temperature from

q = mCΔT

To melt/freeze, changes from liquid to solid, at 0°C

q = mHf

To boil/condense/evaporate, at 100°C

q = mHv

SG pg.18 #63 – 64W/S 3: Heat & Heat Calculations

A. Introduction:

1. Behavior of gases

Influenced by three key factors

Volume (space)

Pressure

Temperature

2. Kinetic molecular theory Gas is composed of individual particles

Distances between particles are far apart

Gas particles are in continuous, random,

straight-line motion

When two particles collide energy is transferred

from one particle to another

Particles of gasses have no attraction to each

other

Individual gas particle has no volume (negligible

or insignificant)

3. An ideal gas A theoretical (or assumed) gas that has all

properties previously summarized

4. An real gas A gas that actually exists

Oxygen, carbon dioxide, hydrogen, helium, etc.

Has particles that attract each other

Does have volume

Real gases with small molecular mass behave

most like an ideal gas (H and He)

SG pg.20 #65 – 72MC pg.136-137 Sets 23-25

B. Gas Laws:

1. Avogadro’s Law

Under the same conditions of temperature and pressure equal volume of different gasses contains equal number of molecules (particles)

If the number of helium gas molecules are counted in Container A and the number of oxygen gas molecules are counted in Container B, you will find that the number of molecules of helium in A is the same as the number of moleculesof oxygen in B.

SG pg.21 #77 – 80MC pg.138 Set 26

2. Dalton’s Law of Partial Pressure The total pressure (Ptotal) of a gas mixture is the

sum of all the partial pressures.)

Partial Pressure (P) is a pressure exerted by individual gas in a gas mixture

Total Pressure from Partial Pressures:

A three gas mixture PgasΔ = .2 atm

PgasO = .4 atm

Pgas = .5 atm

Ptotal = PgasA + PgasB + Pgas C.2 atm + .4 atm + .5 atm = 1.1 atm

Total pressure when gas X is collected over water

Ptotal = Pgas X + VPH2O (at temp)

VPH2O is the vapor pressure of water at the given

water temperature. (Table H)

Example: Oxygen gas is collected over water at

45oC in a test tube. If the total pressure of the gas

mixture in the test tube is 26 kPa, what is the

partial pressure of the oxygen gas ?

26 kPa = Pgas O + VPH2O at 45°C

26 kPa = Pgas O + 10

16kPa = Pgas O

Partial Pressure of gas X from mole fraction:

P gas X = Moles of gas X (P total)

Total moles

Example: A gas mixture contains 0.8 moles of O2 and 1.2 moles of N2. If the total pressure of the mixture is 0 5 atm, what is the partial pressure of N2 in this mixture?

P gas N2 = 1.2 (0.5)

2.0

= 0.3 atm

3. Graham’s Law of Diffusion

a lighter gas will diffuse faster than a heavier gas

4. Boyle’s Law At constant temperature, volume of a gas is

inversely proportional to the pressure on the gas.

As pressure increases, volume (space) of the gas decreases by the same factor

MC pg.138 Set 27

Boyle’s Law (continued)

equation to calculate the new volume of a gas when pressure on the gas is changed atconstant temperature

P1V1 = P2V2

[ P= pressure, V = volume, T = Kelvin temperature, 1 = initial condition, 2 = new condition]

Example: At constant temperature, what is the new of volume of a 3 L sample of O gas if its pressure is changed from 0.5 atm to 0.25 atm?

P1V1 = P2V2

(0.5 atm)(3 L) = (0.25 atm)(V2) 1.5 = 0.25 V2

6.0 = V2SG pg.22 #81 – 82MC pg.139 Sets 28-29

5. Charle's Law At constant pressure, the volume of a gas is

directly proportional to the Kelvin temperature of the gas.

as temperature increases, volume (space) increases by the same factor

equation to calculate the new volume of a gas when temperature of the gas is changed at constant pressure

V1/T1 = V2 / T 2

Charle's Law (continued)

Example: The volume of a confined gas is 25 ml at 280 K. At what temperature would the gas volume be 75 ml if the pressure is held constant?

V1/T1 = V2/T2

(25mL) / (280K) = (75mL) / (T2)

840 K = T2

SG pg.23 #83 – 84MC pg.140 Sets 30-31

6. Gay-Lussac’s Law

At constant volume, pressure of a gas is directly proportional to the Kelvin temperature of the gas.

As temperature increases, pressure increases by the same factor

Equation to calculate the new pressure of a gas when temperature of the gas is changed at constant volume

P1/T1 = P2/T2

Gay-Lussac’s Law (continued)

Example: At constant volume, pressure on a gas changes from 45 kPa to 50 kPa when the temperature of the gas is changed to 340°K.What was the initial temperature of the gas?

P1/T1 = P2/T2

(45 kPa) / (T1) = (50 kPa) / (340 °K)

T1 = 306 °K

SG pg.24 #85 – 86MC pg.141 Sets 32-33

7. Combined Gas Law

describes a gas behavior when all three factors (volume, pressure, and temperature) of the gas are changing:

the only constant is the mass of the

equation to solve any problem related to the above three gas

P1 V1/T1 = P2 V2/T2

Combined Gas Law (continued)

Example: A 30 mL sample of H2 is gas at 1 atm and 200 K. What will be its new volume at 2.0 atm and 600 K

P1 V1/T1 = P2 V2/T2

(1) (30) / 200 = (V2) (2) / 600

45 mL = V2

SG pg.25 #87 – 91MC pg.142-143 Sets 34-36

C. Pressure, Volume, and Temperature:

1. Pressure

Pressure of gas is a measure of how much force is put on a confined gas

Units: Atmosphere (atm) or Kilopascal (kPa)

1 atm = 101.3 kPa

2. Volume

Volume of a gas is the space of the container the gas is placed.

Units: Milliliters (mL) or centimeters cube (cm3)

1 atm = 101.3 kPa

1 mL = 1 cm3

3. Temperature (gas)

a measure of the average kinetic energy of the gas particles

As temperature increases, gas particles move faster, and their average kinetic energy increases.

Units: degrees Celsius (°C) or Kelvin (K)

°K = °C + 273

D. Standard Temperature and Pressure: STP

Reference Table A

Standard Temperature 273 °K (or) 0 °C

Standard Pressure 1 atm (or) 101.3 kPa

NOTE: Always use Kelvin temperature in all gas law calculations.

Example: Hydrogen gas has a volume of 100 mL at STP. If temperature and pressure are changed to 546 K and 0.5 atm respectively, what will be the new volume of the gas?

V1= 100 mL

V2 = ?

T1 = 273 K

T2 = 546 K

P1 = 1 atm

P2 = 0.5 atm

P1V1

T1

=P2V2

T2

(1)(100)273

=(0.5)(V2)

546

400 mL = V2

W/S 4: Gasses

A. Introduction:

1. Properties

set of characteristics that can be used to identify and classify matter

Two types of properties of matter are physical and chemical properties.

2. Physical Property can be observed or measured without changing

chemical composition

(a) extensive properties

depend on sample size or amount

Example: mass, weight and volume

(b) intensive properties

2. does not depend on sample size or amount

3. Example: Melting, freezing and boiling points,

density, solubility, color, odor, conductivity,

luster, and hardness

3. Physical change

a change of a substance from one form to another without changing its chemical composition

Examples:

Phase change

Size change

Dissolving NaCl(s) Na + (aq) + Cl –(aq)

4. Chemical Property

a characteristic of a substance that is observed or measured through interaction with other substances.

(a) Examples:

it burns, it combusts, it decomposes, it reacts with, it combines with, or, it rusts

4. Chemical Change

a change in composition and properties of one substance to those of other substances.

Chemical reactions are ways by which chemical changes of substances occur

SG pg.26 #92 – 96MC pg.144 Set 37