Chem 106, Prof. J.T. Spencer 1 CHE 106: General Chemistry u CHAPTER ONE Copyright © James T....

Chem 106, Prof. J.T. Spencer 11

CHE 106: General Chemistry

CHAPTER ONE

Copyright © James T. Spencer 1995 - 1999

All Rights Reserved


What is ChemistryWhat is Chemistry– Study of the “Physical” Properties Matter (Form and Function)

– Study of How Matter Changes (Reactivity)

Benefits of ChemistryBenefits of Chemistry– Pharmaceuticals

– Enhanced food production (fertilizers, herbicides, etc...)

– Plastics and Polymers

Why Study ChemistryWhy Study Chemistry– Core requirement (?)

– Central Science

EmploymentEmployment– Many fields

MatterMatter

Chapt. 1.1

BIO

Physics Medicine

GEO Engr

CHEM

Law

also: environmental economics electronics agriculture politics etc...

S.U. B.S.


Chemistry; Common ChemicalsChemistry; Common Chemicals

acetic acid .........................................acetic acid .........................................vinegarvinegarcalcium hypochloride ......................calcium hypochloride ......................bleaching powderbleaching powdercalcium sulfate .................................calcium sulfate ................................. plaster of parisplaster of pariscarbon tetrachloride .......................carbon tetrachloride ....................... cleaning fluidcleaning fluidferric oxide .......................................ferric oxide ....................................... iron rustiron rustgraphite ............................................graphite ............................................ pencil leadpencil leadmagnesium sulfate ..........................magnesium sulfate ..........................Epsom saltsEpsom saltsnaphthalene......................................naphthalene...................................... mothballsmothballssilicon dioxide...................................silicon dioxide................................... sandsandsodium bicarbonate .........................sodium bicarbonate .........................baking sodabaking sodasodium borate...................................sodium borate................................... boraxboraxsodium hydroxide ............................sodium hydroxide ............................ lyelyesulfuric acid......................................sulfuric acid...................................... battery acidbattery acidsucrose...............................................sucrose............................................... cane sugarcane sugar


Chemistry; Chemical ProductionChemistry; Chemical Production

Sulfuric Nitrogen Oxygen Ethylene Lime AmmoniaPropylene NaOH PhosphoricChlorine0

20

40

60

80

100

Bil

lion

s of

lbs

HH22SOSO44

NN22

OO22 CC22HH44

CaOCaO NHNH33

CC33HH66 NaOHNaOHHH33POPO44

ClCl22

19951995Chemical and Engineering NewsChemical and Engineering News


Nanoscale Chemistry

Use simpler molecular units are molecular-architectural elements


Nanoscale Chemistry


Nanosystems


Nanomachines

red blood cell

Interstellar Space Travel - Significant concepts in this area include: launch vehicles, the space elevator, interplanetary transportation, the swarm concept, smart dust, extraterrestrial materials utilization, terraforming, suspended animation, space telescopes and virtual sample return.Human Therapeutics - Nanotechnology has caused scientists to re-examine the problems of the human body from the perspective of atomic- engineering. By assuming a nanotechnological

point of view, the resolution of therapeutic ailments becomes simple.Nano-Robots and Nano-Computers with advanced Artificial Intelligence - Nanotechnology will operate under the control of nano-sized computers which will manage the process of Molecular Manufacturing. In order to achieve this, it will be necessary to devise advanced Artificial Intelligence that will be able to automate and regulate Molecular Manufacturing systems.


Matter; A ReviewMatter; A Review

Chapt. 1.2

Definition of MatterDefinition of Matter– anything that occupies space and has mass

StatesStates– gas (vapor); no fixed volume or shape, compressable

– liquid; fixed volume no fixed shape, mostly incompressable

– solid; fixed volume and shape, incompressable

FormsForms– Substances (pure or single); has a fixed composition and distinct

properties. Most things encountered are mixtures of substances.

PropertiesProperties– Physical Properties; can be measured without changing the substance,

i.e., color, density, melting point, etc...

– Chemical Properties; the way a substance changes (reacts), i.e., combustion


Matter; Elements and CompoundsMatter; Elements and Compounds

Chapt. 1.2

SubstancesSubstances– Elements - substances which cannot be decomposed into simpler

substances (see periodic table)

– Compounds- substances which can be separated into two or more elements

ElementsElements– 110 Known (periodic table to be revisited)

– make up all matter and composed of “subatomic particles”

– symbols used for abbreviations (from older or common names)

CompoundsCompounds– Elements combined in a definite proportion by mass (law of definite

proportion)

– properties different than consititutent elements

Water; example of mixtures, compound and elements?Water; example of mixtures, compound and elements?


Matter; Elements and Periodic TableMatter; Elements and Periodic Table

See Website: http://the-tech.mit.edu/Chemicool

Periodic TablePeriodic Table



Chapt. 1.2

Mixtures; combinations of substancesMixtures; combinations of substances

– MixtureMixture- combination of two or more substances in which each retains its own chemical identity (and properties). Vary widely by composition (infinite possibilities of combining ratios), can be separated using the different physical properties of the component substances.

– Homogeneous Homogeneous - appears the same throughout (solutions), liquid, gas and solid solutions are possible.

– Heterogeneous Heterogeneous - mixtures which do not have the same (uniform) appearance throughout.


Salt and Sand Mixture Ink from Cabbage Juicesolubility and filtrationsolubility and filtration chromatography chromatography

Water from Salt Water Iron and Gold Mixturedistillationdistillation magnetic propertiesmagnetic properties

melting point melting point differencesdifferences

chem. reactivity chem. reactivity (acids)(acids)

Iodine from Copper Chloridesolubility and filtrationsolubility and filtration


Chapt. 1.2

Separating Mixtures using Physical PropertiesSeparating Mixtures using Physical Properties– How would you separate;



Chapt. 1.2


Filtration Sand from SaltSand from Salt

Filter

Everyday ExamplesEveryday Examples;Auto Oil FilterAuto Air FilterAquarium Water FilterSpaghetti StrainerWindow ScreensRegistrar

Flow



Chapt. 1.2


Distillation Water from Salt WaterWater from Salt Water

NaCl(aq)

NaCl(s) + H2O(l)



Chapt. 1.2


Chromatograpgy Dyes from M&M’sDyes from M&M’s

Before After

Dyes



Chapt. 1.3

ChangesChanges

– Physical - Changes in appearance but not identity, i.e., evaporation, melting (all changes of state)

– Chemical - transformation into a different substance

Chemical ChangesChemical Changes Physical ChangesPhysical Changes

burning burning meltingmelting CC66HH1212OO66 + 6O + 6O22 6CO 6CO22 + 6H + 6H22OO H H22O(s)O(s) HH22O(l)O(l)

chemical reactionschemical reactions sublimationsublimation NaOH + HCl HNaOH + HCl H22O + NaCl O + NaCl H H22O(s)O(s) HH22O(g)O(g)

corrosioncorrosion dissolutiondissolution 4Fe + 3O4Fe + 3O22 2 Fe 2 Fe22OO3 3 H H22O(l ) + NaCl(s)O(l ) + NaCl(s) NaCl(aq)NaCl(aq)


MatterMatter

MatterMatter

Uniform ?Uniform ?HeterogeneousHeterogeneousMixtureMixture

HomogeneousHomogeneous

Can be separatedCan be separatedby physical methodsby physical methodsPure SubstancePure Substance

HomogeneousHomogeneousMixture (solution)Mixture (solution)

Decomposed ?Decomposed ?

CompoundCompoundElementElement

YesYesNoNo

NoNo

NoNo

YesYes

YesYes


ObservationsObservationsand Experimentsand Experiments

Patterns and Patterns and TrendsTrends

Form and testForm and testhypothesishypothesis

TheoryTheory

Scientific MethodScientific Method


Observations to TheoryObservations to Theory

ObservationsObservations Theory Theory








Matter; MeasurementMatter; Measurement

AA BB

Which is True?Which is True?A = BA = BA > BA > BA < BA < B



AA BB




AA BB




Chapt. 1.4

SystemsSystems

– Metric - base 10

– SI- international scientific system

– mass Kilogram

– length Meter

– time Second

– electric current Ampere

– temperature Kelvin

– light Candela

– Amount Mole Factor label method for conversionsFactor label method for conversions



Chapt. 1.4

PrefixesPrefixes

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



Chapt. 1.4

Sample exercise: What fraction of a second is a picosecond, ps?



Chapt. 1.4

Sample exercise: What fraction of a second is a picosecond, ps?

10-12 second



Chapt. 1.4

Common Units:Length and Mass

Length - unit of distance measured in meters

Mass - measures the amount of matter in an object in grams

TemperatureKelvinCelsius °C = 5/9 (°F -32)

K = °C + 273.15



Chapt. 1.4

Sample exercise: Ethylene glycol, the major ingredient in antifreeze, freezes at -11.5°C. What is the freezing point in

a) K

b) °F



Chapt. 1.4


a) K

b) °FK = °C + 273.15

= -11.5 + 273.15



Chapt. 1.4


a) K

b) °FK = °C + 273.15

= -11.5 + 273.15

= 261.65 K



Chapt. 1.4


a) K

b) °FK = °C + 273.15

= -11.5 + 273.15

= 261.65 K

= 261.7 K



Chapt. 1.4


a) K

b) °F °C = 5/9 (°F -32)



Chapt. 1.4


a) K

b) °F °C = 5/9 (°F - 32)

-11.5 = 5/9(x - 32)



Chapt. 1.4


a) K

b) °F °C = 5/9 (°F - 32)

-11.5 = 5/9(x - 32)

9(-11.5) + 32 = x

5



Chapt. 1.4


a) K

b) °F °C = 5/9 (°F - 32)

-11.5 = 5/9(x - 32)

9(-11.5) + 32 = x

5

11.3°F = x



Chapt. 1.4

Derived Units:Volume

Length x length x lengthmeasured in cm3, which is equal to mL



Chapt. 1.4

Derived Units:Density

amount of mass per unit volumemeasured in g/cm3, or g/mL



Chapt. 1.4

Sample exercise: A student needs 15.0 g of ethanol (ethyl alcohol) for an experiment. If the density of the alcohol is 0.789 g/mL, how many milliliters of alcohol are needed?



Chapt. 1.4


D = m/V so V = m/D



Chapt. 1.4


D = m/V so V = m/D

= 15.0 g 0.789 g/mL



Chapt. 1.4


D = m/V so V = m/D

= 15.0 g 0.789 g/mL

= 19.0 mL


Matter; Uncertainty in MeasurementMatter; Uncertainty in Measurement

Chapt. 1.5

Precision and AccuracyPrecision and Accuracy– Precision - how closely individual measurements

agree– Accuracy- how closely the measurements agree with

the true value

Significant FiguresSignificant Figures– All measurements are inaccurate intrinsically

– measured quantities are reported such that the last figure is uncertain



Good PrecisionPoor Accuracy

Good PrecisionGood Accuracy

Poor PrecisionPoor Accuracy


Determining Significant FiguresDetermining Significant Figures

–all non zero digits are significantall non zero digits are significant

– zeros between nonzero digits are significantzeros between nonzero digits are significant

– zeros to the left of first nonzero digit are not zeros to the left of first nonzero digit are not significantsignificant

– zeros at the end of a number and to the right zeros at the end of a number and to the right of a decimal point are significantof a decimal point are significant

–when a number ends in a zero but with no when a number ends in a zero but with no decimal point, the zero may or may not be decimal point, the zero may or may not be signigicant (use scientific notation)signigicant (use scientific notation)


Chapt. 1.5


Determining Significant FiguresDetermining Significant Figures

3.573 has 4 significant figures3.573 has 4 significant figures




- - multiplication and divisionmultiplication and division; result can have no ; result can have no more than the figure with the fewest significant more than the figure with the fewest significant figuresfigures

- - addition and subtractionaddition and subtraction; result can have the ; result can have the same number of decimal places as the term with same number of decimal places as the term with the least number of decimal placesthe least number of decimal places


Chapt. 1.5



Chapt. 1.5

Sample exercise: A balance has a precision of 0.001 g. A sample that weighs about 25 g is weighed on this balance. How many significant figures should be reported for this measurement?



Chapt. 1.5


25.XXX



Chapt. 1.5


25.XXX

5 sig figs



Chapt. 1.5

Sample exercise: How many significant figures are in each of the following measurements?

A) 3.549 g

B) 2.3 x 104 cm

C) 0.00134 m3



Chapt. 1.5


A) 3.549 g 4 sig figs

B) 2.3 x 104 cm

C) 0.00134 m3



Chapt. 1.5



B) 2.3 x 104 cm 2 sig figs

C) 0.00134 m3



Chapt. 1.5



B) 2.3 x 104 cm 2 sig figs

C) 0.00134 m3 3 sig figs



Chapt. 1.5

Sample exercise: There are exactly 1609.344 m in a mile. How many meters are in a distance of 1.35 mi?



Chapt. 1.5


1.35 mi = 1 mi

x 1609.344 m



Chapt. 1.5


1.35 mi = 1 mi

x 1609.344 m

x = 2172.6144 m



Chapt. 1.5


1.35 mi = 1 mi

x 1609.344 m

1.35 has 3 sig figs x = 2172.6144 m 1609.344 has 7 sig figs 1 is infinitely significant



Chapt. 1.5


1.35 mi = 1 mi

x 1609.344 m

1.35 has 3 sig figs x = 2172.6144 m 1609.344 has 7 sig figs x = 2170 m 1 is infinitely significant


Use Units throughout the calculation (helps “guide” calculation.

Should always yield the proper units Uses conversion factors Example; How fast is 50 mph in in/sec.?

Dimensional AnalysisDimensional Analysis

50 mi.50 mi. 1 hour1 hour 5280 ft5280 ft 12 in.12 in. = = inin1 hour1 hour 3600 sec.3600 sec. 1 mi.1 mi. 1 ft1 ft sec.sec.



Chapt. 1.6

Sample exercise: By using a conversion factor from the back inside cover, determine the length in kilometers of a 500.0 mi automobile race.



Chapt. 1.6


500.0 mi



Chapt. 1.6


500.0 mi 1 km

0.62137 mi



Chapt. 1.6


500.0 mi 1 km = 804.674 km

0.62137 mi



Chapt. 1.6


500.0 mi 1 km = 804.674 km

0.62137 mi

* answer can only have 4 sig figs; 804.7 km



Chapt. 1.6

Sample exercise: The distance between carbon atoms in a diamond is 154 pm. Convert this distance to millimeters.



Chapt. 1.6


154 pm



Chapt. 1.6


154 pm 1 m 103 mm

1012 pm 1 m



Chapt. 1.6


154 pm 1 m 103 mm = 1.54 x 10-7 mm

1012 pm 1 m



Chapt. 1.6

Sample exercise: A car travels 28 mi to the gallon of gasoline. How many kilometers per liter will it go?



Chapt. 1.6


28 mi

gal



Chapt. 1.6


28 mi 1 km

gal 0.62137 mi



Chapt. 1.6


28 mi 1 km 1 gal

gal 0.62137 mi 3.7854 L



Chapt. 1.6


28 mi 1 km 1 gal = 11.9041 km

gal 0.62137 mi 3.7854 L L



Chapt. 1.6


28 mi 1 km 1 gal = 11.9041 km

gal 0.62137 mi 3.7854 L L

* 2 sig figs = 12 km L


Matter: Chemical and Physical ChangesMatter: Chemical and Physical Changes Elements and CompoundsElements and Compounds Units of MeasurementUnits of Measurement Uncertainty and Significant FiguresUncertainty and Significant Figures Precision and AccuracyPrecision and Accuracy ““Factor Label” Method (Dimensional Factor Label” Method (Dimensional

Analysis)Analysis)

Chapter One; ReviewChapter One; Review

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Chem 106, Prof. J.T. Spencer 1 CHE 106: General Chemistry u CHAPTER ONE Copyright © James T....

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