CHAPTER ONEThe Foundations of Chemistry

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Why is Chemistry Important?

Materials for our homes

Components for computers and other electronic devices

Cooking Fuel Body functions

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Some definitions / Vocabulary

Chemistry Science that describes matter – its properties,

the changes it undergoes, and the energy changes that accompany those processes

Matter Anything that has mass and occupies space.

(In other words: anything that has mass and volume)

Energy The capacity to do work or transfer heat.

Types of energy Kinetic and potential energy Heat energy, light energy,

chemical energy, mechanical energy

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Natural Laws

The Law of Conservation of Mass During a chemical or physical change the mass

of the system remains constant

The Law of Conservation of Energy Energy cannot be created or destroyed in a

chemical reaction or in a physical change. It can only be converted from one form to another.

The Law of Conservationof Matter and Energy Read at home

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States of Matter

Liquid Solid Gas

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States of Matter

Change States heating cooling

Ice

Steam

Water

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Substances

Substance matter all samples of which have

identical composition and properties

Examples water sulfuric acid

Properties physical properties – physical changes chemical properties – chemical

changes

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Physical Properties

Physical properties changes of state density, color, solubility always involve only one substance

A substance cannot be broken down or purified by physical means!

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Mixtures Mixture

a combination of two or more substances can be separated by physical means

Homogeneous mixtures have uniform properties throughout examples: salt water; air

Heterogeneous mixtures do not exhibit uniform properties throughout examples: iron+sulfur; water+sand

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Chemical Properties

Chemical properties chemical reactions always involve changes in composition always involve more than one

substance

Examples burning of methane rusting of iron oxidation of sugar

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Decomposition of Water

Element

Element

Compound

oxygen

hydrogen

water

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Compounds and Elements Compounds

If a substance can be decomposed into simpler substances through chemical changes, it is called a compound

Elements If a substance cannot be decomposed

into simpler substances by chemical means, it is called an element

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Important to remember both compounds and elements are substances a compound consists of 2 or more elements

Law of Definite Proportions different samples of any pure compound contain

the same elements in the same proportion by mass

Symbols of elements found on the periodic chart (learn Table 1-2) www.webelements.com

Compounds and Elements

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Scientific Notation

Use it when dealing with very large or very small numbers:

42,800,000. =

0.00000005117 =

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Measurements in Chemistry

QuantityQuantity UnitUnit SymbolSymbol length meter m mass kilogram kg time second s current ampere A temperature Kelvin K amt. substance mole mol

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Metric Prefixes

NameName SymbolSymbol MultiplierMultiplier mega- M 106

kilo- k 103

deci- d 10-1

centi- c 10-2

milli- m 10-3

micro- 10-6

nano- n 10-9

pico- p 10-12

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1000 m =

0.008 s =

30,000,000 g =

0.07 L =

Metric Prefixes: Examples

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Use of Numbers

Exact numbers obtained from counting or by definition 1 dozen = 12 things for example

Measured numbers numbers obtained from measurements

are not exact every measurement involves an

estimate

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Significant Figures Significant figures

digits believed to be correct by the person making the measurement

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Significant Figures

Side B:

13.6 mm

>13.5 mm but <13.7 mm in my judgement!

13.6 mm

certain figures estimated figure

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Significant Figures

13.6 mm

certain figures + estimated figure

significant figures

we always report only 1 estimated figure the estimated figure is always the last one

of the significant figures

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1) Exact numbers (defined quantities) have an unlimited number of significant figures. We do not apply the rules of significant figures to them.

2) Leading zeroes are never significant: 0.000357 has three significant figures

3) Zeros between nonzero digits are always significant:

20.034 1509 1.0000005

Significant Figures - Rules

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4) Zeroes at the end of a number that contains a decimal point are always significant: 35.7000 0.07200 40.0 41.0

5) Zeroes at the end of a number that does not contain a decimal point may or may not be significant (use scientific notation to remove doubt): 173,700 may have 4, 5, or 6 significant

figures

Significant Figures - RulesTrailing zeros

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6) The position of the first doubtful digit dictates the last digit retained in the sum or difference.

Significant Figures - Rules

Addition/Subtraction Rule

Multiplication/Division Rule

7) In multiplication or division, an answer contains no more significant figures than the least number of significant figures used in the operation.

Study examples 1-1 & 1-2 in the book

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The Unit Factor Method The basic idea of the method:

multiplication by unity (by 1) does not change the value of the expression

Principles: construct unit factors from any two

terms that describe identical quantity the reciprocal of a unit factor is also a

unit factor

Study examples 1-3 through 1-9 in the book

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1 ft = 12 inUnit factors:

Example: Express 77.5 inches in feet

77.5 in = 77.5 in x

in 12ft 1

ft 1in 12

in 12

ft 1= 6.46 ft

The Unit Factor Method

See Table 1-7 for various conversion factors

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More examples

9.32 yrd = ? mm

1. We use the following knowledge to build unit factors:

1 yrd = 3 ft 1 in = 2.54 cm1 ft = 12 in 1 cm = 10 mm

2. Multiply 9.32 yrd by unit factors to get the value expressed in mm:

3 ft1 yrd

12 in1 ft

x2.54 cm

1 inx9.32 yrd x10 mm1 cm

x = 8.52·103 mm

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Density

density =mass

volume

tells us how heavy a unit volume of matter is

usually expressed as “g/ml” for liquids and solids and as “g/L” for gases

Table 1-8 lists densities of some common substances

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Density: Example

Example: Calculate the density of a substance if 742 grams of it occupies 97.3 cm3.

mL 97.3 cm 97.3 mL 1 cm 1 33

g/mL 637mL 97.3g 742

.d

Learn examples 1-11 through 1-13 in the book

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Specific Gravity

Sp. Gr. =d (substance)

d (water)

tells us how much heavier or lighter a substance is compared to water:

Sp. Gr. < 1 – lighter than waterSp. Gr. > 1 – heavier than water

specific gravity has no units – it is a dimensionless quantity

See example 1-14 in the book

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Specific Gravity: Example

Example 1-15: Battery acid is 40% sulfuric acid, H2SO4, and 60% water by mass. Its specific gravity is 1.31. Calculate the mass of pure H2SO4 in 100.0 mL of battery acid.

What do we know?

1. The mass percentage of H2SO4 and H2O in the sample of battery acid.

2. Specific gravity of battery acid.3. Density of water (1.00 g/mL).

To find the mass of H2SO4, we need to know the mass of 100.0 mL of battery acid.

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Specific Gravity: Example

311g/mL001

(bat.acid) O)(H

(bat.acid) .acid)Sp.Gr.(bat

2

..

dd

d

Vm

d Vd m

Therefore,

g/mL 1.31 g/mL 001311 (bat.acid) ..d

g 131 mL 100.0 g/mL 311 (bat.acid) .m

g 52.4 100%40%

g 131 )SO(H 42 m

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Heat and Temperature

Heat and Temperature are not the same thing: Heat is a form of energy T is a measure of the intensity of heat in a body

Heat always flows spontaneously from a hotter body to a colder body – never in the reverse direction

Body 1

T1

Body 2

T2

hotter T1 > T2 colder

Heat

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Temperature Scales

3 common temperature scales

0ºF – freezing (salt+H2O)

30ºF – freezing H2O

90ºF – human body

0ºC – freezing H2O

100ºC – boiling H2O

0 K – absolute zero273.15 K – freezing H2O

Fahrenheit Celcius Kelvin

http://home.comcast.net/~igpl/Temperature.html

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Temperature Scales & Water

Melting (MP) and boiling (MP) points of water on different temperature scales

MP BP Fahrenheit 32 oF 212 oF Celsius 0.0 oC 100 cC Kelvin 273 K 373 K

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Temperature Conversion

degrees Kelvin degrees Celcius

? K = ?ºC + 273 ?ºC = ? K - 273

degrees Fahrenheit degrees Celcius

?ºF = (?ºC)·1.8 + 32 ?ºC = (?ºF – 32)/1.8

Examples 1-16 & 1-17 in the book

http://www.lenntech.com/unit-conversion-calculator/temperature.htm

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Heat

Chemical and Physical changes: evolution of heat (exothermic processes) absorption of heat (endothermic processes)

Units of measurement: joule (J) – SI units calorie (cal) – conventional units 1 cal = 4.184 J

A “large calorie” (1 large cal = 1000 cal = 1 kcal) is used to express the energy content of foods

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Specific Heat

The specific heat (Cp) of a substance: the amount of heat (Q) required to raise the

temperature of 1 g of the substance 1ºC (or 1 K)

Units of measurement:

Tmp Δ Q

C

KgJ

or Cg

J

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Specific Heat: Example 1

Knowing specific heat, we can determine how much energy we need in order to raise the temperature of a substance by T = T2 – T1:

Calculate the amount of heat necessary to raise the temperature of 250 mL of water from 25 to 95ºC given the specific heat of water is 4.18 J·g-1 ·ºC-1.

What do we know? the temperature change the specific heat of water the volume of water the density of water

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Specific Heat: Example 1

Examples 1-18 through 1-20 in the book

Tm heat) (specific heatTm Δ

heat heat specific

C70 C25 - C95 T

g 250 mL 250g/mL 001 .Vdm

J 107.3 J 73150 C70g 250Cg

J 4.18 heat 4

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Specific Heat: Example 2

Given specific heats of two different substances, we can also calculate the heat transfer between them:

0.350 L of water at 74.0ºC is poured into an aluminum pot at room temperature (25.0ºC). The mass of the pot is 200 g. What will be the equilibrium temperature of water after it transfers part of its heat energy to the pot? The specific heats of aluminum and water are 0.900 and 4.18 J·g-1 ·ºC-1, respectively.

You might encounter this kind of problem at your first exam

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What do we know? the pot and water come to equilibrium, that is

eventually they have the same temperature the specific heat of aluminum and water the mass of aluminum the volume of water the density of water finally, the Law of conservation of energy

which tells us that the amount of heat lost by water is the same as the amount of heat gained by the aluminum pot

Specific Heat: Example 2

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Specific Heat: Example 2

Let’s denote the final temperature as Tf. Then the changes in temperature for water and aluminum are:

Tm heat) (specific heatTm Δ

heat heat specific

C025(Al) and - C047O)(H ff2 .TTT.T

g 350 L 1mL 1000

L 3500g/mL 001O)(H2 ..Vdm

Note that we used the unit factor method to convert L to mL

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Specific Heat: Example 2

C)25.0-(g 200 Cg

J0.900)-C(74.0g 350

CgJ

4.18 ff

TT

Solving this equation with respect to Tf, we obtain

(Al) gain heat O)(H loss heat 2

Tm heat) (specific heat

Tf = 68.6ºC

Try to solve the equation yourself and analyze why the answer is given with 3 significant figures

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Reading Assignment

Read Chapter 1 Learn Key Terms (pp. 40-41) Go through Chapter 2 notes

available on the class web site If you have time, read Chapter 2

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Homework Assignment

Textbook problems (optional, Chp. 1): 11, 13, 15, 18, 27, 29, 30, 32, 36, 41, 43,

47, 49, 57, 62, 68, 80

OWL: Chapter 1 Exercises and Tutors – Optional Introductory math problems and Chapter

1 Homework problems – Required (homework #1; due by 9/13)