CHAPTER ONEThe Foundations of Chemistry

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Why is Chemistry Important?

Materials for our homesComponents for computers

and other electronic devices

Cooking Fuel Body functions

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Some definitions / Vocabulary

ChemistryScience that describes matter – its properties, the changes it undergoes, and the energy changes that accompany those processes

MatterAnything that has mass and occupies space.(In other words: anything that has mass and volume)

EnergyThe capacity to do work or transfer heat.

Types of energyKinetic and potential energyHeat energy, light energy,chemical energy, mechanical energy

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Natural Laws

The Law of Conservation of MassDuring a chemical or physical change the mass of the system remains constant

The Law of Conservation of EnergyEnergy 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

LiquidSolid Gas

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States of Matter

Change Statesheatingcooling

Ice

Steam

Water

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Substances

Substancematter all samples of which have identical composition and properties

Exampleswatersulfuric acid

Propertiesphysical properties – physical changeschemical properties – chemical changes

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Physical Properties

Physical propertieschanges of statedensity, color, solubilityalways involve only one substance

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

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MixturesMixture

a combination of two or more substancescan be separated by physical means

Homogeneous mixtureshave uniform properties throughoutexamples: salt water; air

Heterogeneous mixturesdo not exhibit uniform properties throughoutexamples: iron+sulfur; water+sand

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Chemical Properties

Chemical propertieschemical reactionsalways involve changes in compositionalways involve more than one substance

Examplesburning of methanerusting of ironoxidation of sugar

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Decomposition of Water

Element Element

Compound

oxygen

hydrogen

water

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Compounds and Elements

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

ElementsIf 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 substancesa compound consists of 2 or more elements

Law of Definite Proportionsdifferent samples of any pure compound contain the same elements in the same proportion by mass

Symbols of elementsfound 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 SymbolSymbollength meter mmass kilogram kgtime second scurrent ampere Atemperature Kelvin Kamt. substance mole mol

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Metric Prefixes

NameName SymbolSymbol MultiplierMultipliermega- M 106

kilo- k 103

deci- d 10-1

centi- c 10-2

milli- m 10-3

micro- µ 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 numbersobtained from counting or by definition1 dozen = 12 things for example

Measured numbersnumbers obtained from measurements are not exactevery measurement involves an estimate

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Significant FiguresSignificant 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 Figures13.6 mm

certain figures + estimated figure

significant figures

we always report only 1 estimated figurethe 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 - RulesAddition/Subtraction Rule

Multiplication/Division Rule7) 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 MethodThe 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 quantitythe 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

in12ft1

ft 1in12

in 12ft 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 ftx

2.54 cm1 inx9.32 yrd x

10 mm1 cmx = 8.52·103 mm

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Density

density =mass

volume

tells us how heavy a unit volume of matter isusually 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.

mL97.3 cm97.3 mL1 cm 1 33 =⇒=

g/mL 637mL97.3g742 .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: ExampleExample 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 H2SO4in 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.(bat2

..

dd

d===

Vmd = Vdm ⋅=

Therefore,

g/mL1.31 g/mL001311(bat.acid) =⋅= ..d

g131mL100.0g/mL311(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 energyT 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 H2O90ºF – human body

0ºC – freezing H2O100ºC – 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 BPFahrenheit 32 oF 212 oFCelsius 0.0 oC 100 cCKelvin 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 bookhttp://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 unitscalorie (cal) – conventional units1 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 ∆Q C⋅

=

KgJ or

CgJ

⋅°⋅

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Specific Heat: Example 1Knowing 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 changethe specific heat of waterthe volume of waterthe density of water

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Specific Heat: Example 1

Examples 1-18 through 1-20 in the book

Tm ∆⋅⋅= heat)(specific heatTm ∆

heat heat specific⋅

=

C70C25- C95 °=°°=∆Tg250 mL250g/mL001 =⋅=⋅= .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 temperaturethe specific heat of aluminum and waterthe mass of aluminumthe volume of waterthe density of waterfinally, 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 ∆

heat heat specific⋅

=

C025(Al) and -C047O)(H ff2 °−=∆°=∆ .TTT.T

g 350 L1mL1000L 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 Cg

J 4.18 ff °⋅⋅°⋅

=°⋅⋅°⋅

TT

Solving this equation with respect to Tf, we obtain

(Al)gainheatO)(Hlossheat 2 =

Tm ∆⋅⋅= heat)(specificheat

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 1Learn Key Terms (pp. 40-41)Go through Chapter 2 notes available on the class web siteIf 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 – OptionalIntroductory math problems and Chapter 1 Homework problems – Required (homework #1; due by 9/13)