chem1

Throughout the test the following symbols have the definitions specified unless otherwise noted.

L, mL = liter(s), milliliter(s) mm Hg = millimeters of mercury

g = gram(s) J, kJ = joule(s), kilojoule(s)

nm = nanometer(s) V = volt(s)

atm = atmosphere(s) mol = mole(s)

ATOMIC STRUCTURE

E = hν

c = λν

E = energy

ν = frequency

λ = wavelength

Planck’s constant, h = 6.626 × 10−34 J s

Speed of light, c = 2.998 × 108 m s−1

Avogadro’s number = 6.022 × 1023 mol−1

Electron charge, e = −1.602 × 10−19 coulomb

EQUILIBRIUM

Kc =

[C] [D]

[A] [B]

c d

a b, where a A + b B c C + d D

Kp = C

A B

( ) ( )

( ) ( )

c dD

a b

P P

P P

Ka = [H ][A ]

[HA]

+ -

Kb = [OH ][HB ]

[B]

- +

Kw = [H+][OH−] = 1.0 × 10−14 at 25°C

= Ka × Kb

pH = − log[H+] , pOH = − log[OH−]

14 = pH + pOH

pH = pKa + log[A ]

[HA]

-

pKa = − logKa , pKb = − logKb

Equilibrium Constants

Kc (molar concentrations)

Kp (gas pressures)

Ka (weak acid)

Kb (weak base)

Kw (water)

KINETICS

ln[A] t − ln[A]0 = − kt

[ ] [ ]0A A

1 1

t

- = kt

t½ =

0.693

k

k = rate constant

t = time

t½ = half-life

, pOH = − log[OH−]

AP Chemistry Equations & Constants

GASES, LIQUIDS, AND SOLUTIONS

PV = nRT

PA = Ptotal × XA, where XA = moles A

total moles

Ptotal = PA + PB + PC + . . .

n = m

M

K = °C + 273

D =

m

V

KE per molecule = 1

2mv2

Molarity, M = moles of solute per liter of solution

A = abc

1 1

pressure

volume

temperature

number of moles

mass

molar mass

density

kinetic energy

velocity

absorbance

molarabsorptivity

path length

concentration

Gas constant, 8.314 J mol K

0.08206

P

V

T

n

m

D

KE

A

a

b

c

R

Ã

- -

=============

=

=

M

1 1

1 1

L atm mol K

62.36 L torr mol K 1 atm 760 mm Hg

760 torr

STP 0.00 C and 1.000 atm

- -

- -===

=

THERMOCHEMISTRY/ ELECTROCHEMISTRY

products reactants

products reactants

products reactants

ln

ff

ff

q mc T

S S S

H H H

G G G

G H T S

RT K

n F E

qI

t

D

D

D D D

D D D

D D D

=

= -Â Â

= -Â Â

= -Â Â

= -

= -

= -

�

heat

mass

specific heat capacity

temperature

standard entropy

standard enthalpy

standard free energy

number of moles

standard reduction potential

current (amperes)

charge (coulombs)

t

q

m

c

T

S

H

G

n

E

I

q

t

====

=

=

==

====

ime (seconds)

Faraday’s constant , 96,485 coulombs per mole

of electrons

1 joule1volt

1coulomb

F =

=

AP Chemistry Notes

Stephen Akiki

Colchester High School

Download at http://akiscode.com/apchem

♥♠♣♦

Special Thanks to Stephen Bosley (Boser)

Contents

1 FOREWORD/DISCLAIMER 4

2 Solubility Rules 52.1 Soluble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Insoluble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 Naming Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3 Periodic Table of Elements 5

4 Poly Atomic Naming 6

5 Common Units, Constants and Charges 65.1 Fundamental Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65.2 Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75.3 Radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

6 Atomic Theory 76.1 J.J. Thompson . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76.2 Robert Millikan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76.3 Ernest Rutherford . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76.4 Chadwick . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76.5 John Dalton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

7 Naming 87.1 Binary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

7.1.1 Greek Prefixes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87.2 Ionic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87.3 Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

7.3.1 Polyatomic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87.3.2 Binary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

8 Cations 9

9 Reaction Type 99.1 Combination (Synthesis) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99.2 Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

9.2.1 Special Binary Salt Splits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109.3 Combustion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

10 Blackbody Radiation 10

11 Bohr Model 1111.1 Energy Level Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

11.1.1 Energy Change during Level Jumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1

12 Wavelength 1112.1 De Broglie Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

13 Quantum Values 1213.1 Quantum Value Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1213.2 Special cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

14 Periodicity 1314.1 Electron Configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1314.2 Isoelectricity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

15 Nuclear Chemistry 1315.1 Isotopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1315.2 Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

15.2.1 Alpha Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1415.2.2 Beta Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1415.2.3 Gamma Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1415.2.4 Positron Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1515.2.5 Electron Capture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

15.3 Nuclear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1515.3.1 Radiation Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

15.4 Nuclear Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1615.4.1 Forces Invloved . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1615.4.2 Belt of Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1615.4.3 Magic Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1615.4.4 Half-Life . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

16 Ionization and Affinity 1716.1 Ionization Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1716.2 Electron Afinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

17 Reactions of Metals 17

18 Chemical Bonds 1718.1 Intramolecular . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

18.1.1 Ionic Bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1818.1.2 Covalent Bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1818.1.3 Metallic Bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

18.2 Intermolecular . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1818.2.1 Ion-Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1818.2.2 Dipole-Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1818.2.3 Hydrogen Bond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1818.2.4 London Dispersion/Van der Waals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1818.2.5 Intermolecular Flowchart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

18.3 Rule of Octet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

19 Lewis Structures 1919.1 Structures for Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1919.2 Structures for Ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1919.3 Structure for Ions of Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1919.4 Lewis Structures for Molecular Structures (Covalent) . . . . . . . . . . . . . . . . . . . . . . . 2019.5 Resonance Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

19.5.1 Formal Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

20 Lattice Energies of Ionic Solids 21

21 Bond Lengths of Covalent Bonds 22

22 Electronegativity 2222.1 Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

22.1.1 Dipole Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2

23 Bond Enthalpy 23

24 VSEPR 2324.1 Bond Shape Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

25 Organic Chemistry 2325.1 Polarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2325.2 Alkanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2425.3 Alkane Branch Structure Naming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

25.3.1 Branch Structure Naming Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2525.4 Alkenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

25.4.1 Alkene Naming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2625.5 Alkynes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

25.5.1 Alkyne Naming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

26 Functional Groups 2726.1 Alcohol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2726.2 Aldehyde . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2726.3 Carboxylic Acid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2826.4 Ester . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2826.5 Ketone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2826.6 Ether . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2826.7 Amine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2926.8 Amide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2926.9 Haloalkane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

27 Complex Ions 2927.1 Cations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2927.2 Anions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3027.3 Coordination Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3027.4 Naming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

27.4.1 Cations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3027.4.2 Anions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

28 Acidic and Basic Redox 3028.1 Acidic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3028.2 Basic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

29 Thermodynamics 3129.1 Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

29.1.1 Stoichiometry Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3129.1.2 Calorimetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3229.1.3 Hess Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3229.1.4 Standard Heat of Formation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

29.2 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3329.2.1 State of Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3329.2.2 Number of Moles of Gasses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3329.2.3 Pressure of Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

29.3 Gibbs Law of Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3329.3.1 ∆H, ∆S, ∆G, Relationship Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

30 Chemical Kinetics and Rate Laws 3430.1 Physical State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3430.2 Concentration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3430.3 Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3430.4 Pressure of Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3430.5 Catalysts and Inhibitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3430.6 Rate Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

30.6.1 Order Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3

31 Reaction Mechanisms 3631.1 Elementary Steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

32 Equilibrium 3732.1 Types of Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3732.2 Equilibrium Constant Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

32.2.1 Converting Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

33 Gas Laws 3733.1 Gas Units and Conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3733.2 Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3733.3 Real Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3833.4 Combined Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3833.5 Daltons Law of Partial Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3833.6 Gas Collection over a Water Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

34 ICE ICE (Baby) 39

35 Acids and Bases 3935.1 Definitions of Acids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3935.2 pH and pOH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

35.2.1 Changing Concentrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4035.3 Strong Acids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

35.3.1 Strong Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4035.3.2 Strong Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

35.4 Weak Acids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4035.4.1 Ka Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4135.4.2 Kb Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

35.5 Common Ion Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4135.6 Buffer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

36 Equilibrium of Saturated, Soluable Salts 42

37 Kinetic Molecular Theory 4237.1 Postulates: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4237.2 Root Mean Square Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4337.3 Effusion and Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

37.3.1 Effusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4337.3.2 Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4337.3.3 Finding the rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

38 Electro Chemistry 4338.1 Identifying Oxidation Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4338.2 Galvanic/Voltaic Cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4338.3 Calculating Cell Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

38.3.1 Nernst Equation to Find E◦cell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

39 Balancing Redox Reactions 4439.1 Acidic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4439.2 Basic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

1 FOREWORD/DISCLAIMER

First and formost, I am going to say what everone has on their minds. No you really should not just forgetabout taking notes anymore in AP Chemistry class because of this packet. This packet is meant to be areview and should be used as such. However that does not mean you can use this packet as your mainnotes and write notes in the margins to supplement your learning. Please take into account that this entirething was written over the course of 4 days. As such it is inevitable that I made mistakes in spelling and/orformulas.If you have any questions/comments/fixes to the text you can email me at [email protected] Luck

4

2 Solubility Rules

2.1 Soluble

• Nitrates NO−13 - All nitrates are soluble

• Chlorates ClO−13 - All chlorates are soluble

• Alkali metal Cations and Ammonium cation compounds NH+14 are all soluble

• Chlorides, Bromides, and Iodides are all soluble EXCEPT Ag+1, Pb+2, and Hg+2

• Acetates - All are soluble except Ag+

• Sulfates - All are soluble except Ba+2, Pb+2, Hg+2, Ca+2, Ag+1, and Sr+2

2.2 Insoluble

• Carbonates CO−23 - all carbonates are insoluble except alkali metals and ammonium compounds

• Chromates CrO−24 - all chromates are insoluble except alkali metals, ammonium, Ca+2, and Sr+2

• Hydroxides OH−1 - all hydroxides are insoluble except alkali metals, ammonium, Ba+2, Sr+2, andCa+2 although the last two (Sr+2 and Ca+2) are only slightly soluble so a precipitate can form.

• Phosphates PO−34 all are insoluble except alkali metals and ammonium

• Sulfites SO−23 all are insoluble except alkali metals and ammonium

• Sulfides S−2 all are insoluble except Alkali metals, alkali earth metals and ammonium

2.3 Naming Rules

• All strong acids and bases are soluble and should be written as the ions when completing net ionicreactions

. Sulfuric acid (H2SO4) should be written as H+ +HSO−14

• The strong acids are: HCL, HBR, HI, HNO3, HClO4, and H2SO4

• Strong bases are any alkali metal hydroxides (LiOH, NaOH, etc) and Ca(OH)2, Sr(OH)2, Ba(OH)2

• All acids and bases should be left in their molecular form:. Acetic acid → HC2H3O2

3 Periodic Table of Elements

5

4 Poly Atomic Naming

• Zinc Zn+2

• Silver Ag+1

• Ammonium NH+14

• Hydroxide OH−1

• Cyanide CN−1

• Nitrate NO−13

• Acetate C2H3O−12

• Chlorate ClO−13

• Bromate BrO−13

• Iodate IO−13

• Manganate MnO−13

• Sulfate SO−24

• Bisulfate (Hydrogen Sulfate) HSO−14

• Carbonate CO−23

• Bicarbonate (Hydrogen Carbonate) HCO−13

• Selenate SeO−24

• Biselenate (Hydrogen Selenate)HSeO−14

• Oxalate C2O−24

• Phosphate PO−34

• Hydrogen Phosphate HPO−24

• Dihydrogen Phosphate H2PO−14

• Chromate CrO−24

Per Ate Ate Ite Hypo ItePer Ic Ic Ous Hypo Ous

+1 Oxygen Most Common Ion -1 Oxygen -2 Oxygen

5 Common Units, Constants and Charges

5.1 Fundamental Constants

• Avogadros Number (N)

. 6.02214199 ∗ 1023mol−1

• Plancks Constant (h)

. 6.62606876 ∗ 10−34J ∗ s

• Speed of Light (c)

. 2.99792458 ∗ 108m/s

6

5.2 Charge

• e− charge = −1.602 ∗ 10−19 coulombs

• p+ charge = 1.602 ∗ 10−19 coulombs

• Atomic Mass Unit (amu) = 1.66054 ∗ 10−24

. p+ = 1.0073 amu

. n◦ = 1.0087 amu

. e− = 5.486 ∗ 10−4 amu

5.3 Radius

Angstroms (◦A) = 10−10 meters

6 Atomic Theory

6.1 J.J. Thompson

• Discovered e− and chargemass ratio

. Charge to Mass ratio: 1.76 ∗ 108 Coulombs/Gram (Charge of e−/mass)

• Plum Pudding Model of atom

6.2 Robert Millikan

• Found charge and mass of e−

• Millikan Oil Drop:

. Charge oil drops in a field and adjust field until drops levitate

6.3 Ernest Rutherford

• Discovered 3 types of radiation (Decay Particles)

. Alpha particles: He2+ size, very damaging, stoppable - α

. Beta particles - e− size, damaging, hard to stop - β

. Gamma particles - tiny, not so damaging, unstoppable - γ

• Also discovered proton and new dense nucleus model

. Rutherford worked with α particles most and discredited Thompsons model of the nucleus

6.4 Chadwick

• Discovers neutron by shooting radiation at light elements and it watching it kick out a neutral particle

6.5 John Dalton

• Four Postulates

. Everything made of atoms

. Atoms of one element differ from those of a different element

. Atoms will combine in whole number ratios

. Atoms can not be created or destroyed

• Law of Constant Composition

. In a compound, atom ratios are constant

7

7 Naming

7.1 Binary

• Smallest atomic number comes first

• Second element ends with -ide

7.1.1 Greek Prefixes

• 1-Mono

• 2-Di

• 3-Tri

• 4-Tetra

• 5-Penta

• 6-Hexa

• 7-Hepta

• 8-Octa

• 9-Nona

• 10-Deca

ExampleCl2O

Dichlorine Monoxide

7.2 Ionic

• Finding Charge:

Na?3Cl

+12

NadcCl

ab

(a∗b)c = d

7.3 Acids

7.3.1 Polyatomic

• Per...ate → Per...ic acid

. HNO4 → pernitric acid

• -ate → ic acid

. H +NO3 → HNO3 (Nitric Acid)

• -ite → ous acid

. HNO2 → nitrous acid

• Hypo...ite → hypo...ous acid

. HNO → hyponitrous acid

7.3.2 Binary

• Hydro + (stem)ic

. H +Br → Hydrobromic acid

. H +N → Hydronitric acid

. Hydrocarbonic acid → HC

. Carbonic Acid → HCO3

8

8 Cations

• Which cation forms a white precipate with HCL?

. Ag+ (reversed proves Cl−).

• What color is a typical Manganese solution?

. Pink/light purple. The precipate is dark black.

• How would you test for Al+ and what would it look like?

. Add Aluminom, it will make a precipate red and leave the solution clear.

• Which cation forms a gel like precipate?

. Aluminum.

• Which cation turns deep red with KSCN?

. Iron.

• How do you confirm the presence of zinc and what color is it?

. Add acid, then ammonia, which results in a white/bluish precipate.

• What cation turns a deep blue with ammonia?

. Copper.

9 Reaction Type

9.1 Combination (Synthesis)

When two or more chemicals react to form one product

Example2Mg +O2 → 2MgO2Na+ S → Na2S

• Metal + Non-Metal → Metal Nonmetal (Binary Salt)

• Metal Oxide + Water → Metal Hydroxide

. CaO +H2O → Ca(OH)2

. K2O +H2O → 2KOH

• Metal Oxide + CO2 → Metal Carbonate

. Na2O + CO2 → Na2CO3

• Nonmetallic Oxides + water → Acids (nonmetal oxides retains its oxide number)

. Na2O + SO3 → Na2SO4

9.2 Decomposition

When one chemical decomposes into 2 or more

Example

2Ag2O∆→ 4Ag +O2

∆ = Heat

• Metal Carbonate ∆→ Metal Oxide + CO2

. CaCO3∆→ CaO + CO2

• Metal Hydroxide ∆→ Metal Oxide + H2O

. Mg(OH)2∆→MgO +H2O

9

• Metal Nonmetal ∆→ Metal + Nonmetal (diatomic in nature)

. 2NaCl ∆→ 2Na+ Cl2

• Metal Chlorates ∆→ Metal Chlorides + O2

. Fe(ClO3)2∆→ FeCl3 +O2

9.2.1 Special Binary Salt Splits

These binary salts split into different elements

(NH4)2CO3 → NH3 +H2O + CO2

H2SO3 → H2O + SO2

H2CO3 → H2O + CO2

NH4OH → NH3 +H2O

H2O2 → H2O +O2

9.3 Combustion

Hydrocarbon+O2 → CO2 +H2O....⇓....CxHy → double x (multiply by 2) then add 2

• C1: meth

• C2: eth

• C3: pro

• C4: bu

• C5: pent

• C6: hex

• C7: hept

• C8: oct

• C9: non

• C10: dec

10 Blackbody Radiation

When an object is heated it will emmit radiant energy

E = hν

• E = Energy

• h = Max Plancks constant (6.626 ∗ 10−34J ∗ s)

• ν = frequency

Photoelectric effect: Metal will give off e−s if light shines on it. Light shining on a clean sheet ofmetals will release e−s if ν is strong enough.

10

11 Bohr Model

Neils Bohr:

1. Only orbits of certain radii, corresponding to certain definate energies are permitted for the electron ina hydrogen atom.

2. An electron in a permitted orbit has a specific energy and is in an allowed energy state. An electron inan allowed state will not radiate energy and therefore will not spiral into the nucleus.

3. Energy is emmitted or absorbed by the e− only as the e− changes from one allowed energy state toanother.

4. Flawed theory because it only works for hydrogen

11.1 Energy Level Formula

En = (−2.18 ∗ 10−18J)( 1n2 )

• E1: −2.18 ∗ 10−18J

• E2: −5.45 ∗ 10−19J

• E3: −2.42 ∗ 10−19J

• E4: −1.36 ∗ 10−19J

• E5: −8.72 ∗ 10−20J

• E6: −6.056 ∗ 10−20J

• E∞: 0

11.1.1 Energy Change during Level Jumps

∆E = EF − E0

• n = 3→ 2 | −3.03 ∗ 10−19J

• n = 4→ 2 | −4.09 ∗ 10−19J

• n = 5→ 2 | −4.578 ∗ 10−19J

• n = 6→ 2 | −4.844 ∗ 10−19J

12 Wavelength

12.1 De Broglie Formulas

λ = hmv

or

λ = hp

• λ = Wavelength

• h = Plancks Constant (6.626 ∗ 10−34J ∗ s)

• m = Mass of particle in Kg

• v = Velocity of particle (meterssecond )

• p = Momentum

Examplem = 9.11 ∗ 10−28gv = 5.97 ∗ 106m/s

λ = 6.626∗10−34J∗s(9.11∗10−31Kg)(5.97∗106m/s) = 1.22 ∗ 10−10m

11

13 Quantum Values

1. Principle Quantum number - (n)

n = 1 (lowest)

n = ∞ (at 8 or 9)

Follows Bohrs En = (−2.18 ∗ 10−18J)( 1n2 )

2. Azimuthal Quantum number - (l)

l = n - 1

if...

• l = 0 → S shape

• l = 1 → P shape

• l = 2 → D shape

• l = 3 → F shape

Examplen = 3l = 2⇓3d

3. Magnetic Quantum number (orbital) - (ml)

-l and l including zero

m0 = 0

m1 = −1, 0, 1

m2 = −2,−1, 0, 1, 2

4. Spin magnetic quantum number - (ms)

+ 12 or - 1

2

13.1 Quantum Value Table

n Possible l values Subshell ml values # of orbitals in subshell total # of orbitals in shell e− in shell1 0 1s 0 1 1 22 0 2s 0 1 4 8

1 2p -1,0,1 3 - -3 0 3s 0 1 9 18

1 3p -1,0,1 3 - -2 3d -2,-1,0,1,2 5 - -

4 0 4s 0 1 16 321 4p -1,0,1 3 - -2 4d -2,-1,0,1,2 5 - -3 4f -3,-2,-1,0,1,2,3 7 - -

13.2 Special cases

• Chromium has 6 half-filled orbitals

• Copper has one half-filled orbital and 5 filled orbitals

12

14 Periodicity

14.1 Electron Configuration

14.2 Isoelectricity

Two atoms are considered isoelectric when they gain or lose electrons to become ions and have the sameelectron configuration as each other.

ExampleNa+1: 1S2, 2S2, 2P6

Ne: 1S2, 2S2, 2P6

15 Nuclear Chemistry

Nuclear Chemistry involves changes in the nucleus of an atom.

Normal NuclearReactions involve electron transfer Reactions involve decay of nucleus i.e. transforming one element into anotherReaction affected by factors such Affected by the type of decay and the halflife of what is decaying

as pH, temp, pressure, [], etc.Reactions involve relatively small energy: Reactions deal with huge amounts of energy

400 kJ-1500kJ

15.1 Isotopes

Isotopes: Atoms of the same element that have a different number of neutrons

X −AAZX

AX

• X = Element Symbol

• A = Atomic Mass

• Z = Atomic Number

13

15.2 Radiation

15.2.1 Alpha Radiation

When a big nucleus ejects a He+2 size chunk of itself.

15.2.2 Beta Radiation

When a neutrally charged particle (equal amount of p+s and e−s) ejects its e−s leaving only the p+s.

15.2.3 Gamma Radiation

When a particle experiences some type of radiation (called * here) that causes the remaining nucleus tocollapse. This causes gamma (γ) rays to be emitted. Gamma radiation is also caused when a positron andan electron smash into each other.

14

15.2.4 Positron Radiation

When a positively charged nucleus emits its p+ leaving only the n◦.

15.2.5 Electron Capture

When an electron in orbit falls into the nucleus (positively charged) and makes it neutrally charged.

15.3 Nuclear Equations

15.3.1 Radiation Table

Neutron: 10n

Proton: 11p

+

Electron: 0−1e

−

Positron: 01e−

Alpha Particle: 42He or 4

2αBeta Particle: 0

−1e− or 0

−1β

ExampleAlpha

23892 U →234

90 Th+42 He

Beta13153 I →131

54 Xe+ 0−1e

−

10n→1

1 p+0−1 e

−

Positron116 C →11

5 B+01e−

11p→1

0 n+01e−

Electron Capture8137Rb+

0−1e

− →8136 Kr

11p+

0−1e

− →10 n

Positron-Electron Collision (Gamma)01e+

0−1e

− →00 γ

15

15.4 Nuclear Stability

Understanding why are some nuclides are radioactive while others are not.

15.4.1 Forces Invloved

• Electrostatic. Try to rip apart the nucleus because of like charges

• Strong Nuclear. Try to pull together the nucleus because subatomic particles naturally stick together

• The Glue. Neutrons act as the glue and more of it is required when the electrostatic force gets really strong

15.4.2 Belt of Stability

• Area A. More neutrons than protons - Beta decay → creates protons

• Area B. Less neutrons than protons - Positron emission (Smaller B) or Electron Capture (Larger B)

• Area C. Every element above 83 p+ is radioactive and no glue can hold it together - Alpha decay

15.4.3 Magic Numbers

The Magic Numbers tend to be stable if you have either a proton or neutron in those numbers. If you haveboth, they are very stable.

(p+) 2 8 20 28 50 82 -(n◦) 2 8 20 28 50 82 126

• If (p+) and (n◦) even → likely stable

• If either is odd → could go either way

• If (p+) and (n◦) odd → likely unstable

16

15.4.4 Half-Life

The time it takes 12 the amount of a substance to decay.

Example5g of nuclide

12 life of 15 years

How much of the original nuclide remains after 45 years?5⇓ (15 years)

2.5⇓ (30 years)

1.25⇓ (45 years)

0.625g

16 Ionization and Affinity

16.1 Ionization Energy

The energy needed to remove an e− (how easy it is to lose an e−). Needs energy (+).

16.2 Electron Afinity

How much a gaseous atom will be attracted to a free e− (how easy it is to gain an e−). Releases energy (-).

17 Reactions of Metals

Metal Oxides = Basic

• Metal + Water → Metal Hydroxide + H2

• Metal + O2(Li or any non-Alkali metal) → Metal Oxide

• K + O2(Any other Alkali metal) → Metal Peroxide (O−12 )

. K + O2 → KO2

• Metal Oxide + H2O → Metal Hydroxide

. Na2O +H2O → NaOH

• Metal Oxide + Acid → Salt + H2O

. Na2O +HCL→ NaCl +H2O

Nonmetal Oxides = Acidic

• Nonmetal Oxide + H2O → Acid

. CO2 +H2O → H2CO3

. SO2 +H2O → H2SO3

. P4O10 +H2O → H3PO4

• Nonmetal Oxide + Base → Salt + H2O

. CO2 +NaOH → Na2CO3 +H2O

18 Chemical Bonds

When 2 or more atoms are strongly attached (attracted) to each other.

18.1 Intramolecular

These forces act inside an atom or molecule:

17

18.1.1 Ionic Bonding

Gain/lose e−s (strong metal + strong nonmetal)

18.1.2 Covalent Bonding

Share e−s (weak metal or nonmetal + nonmetal)

18.1.3 Metallic Bonding

There are two models that explain metallic bonding:

• Electron Sea Model 1

. Metal atoms are floating in a sea of e−s. No one e− belongs to any particular atom.

• Orbital Bonding Model

. The valence e−s are overlapped and shared so much you have bonds of delocalized e−s that arefree to move but are still holding the atoms together.

Properties that result from metallic bonding include:

• Conductivity of electricity and heat

• Malibility and ductility

• Ability to form alloys

18.2 Intermolecular

These forces act between molecules:

18.2.1 Ion-Dipole

Ions bonding to molecules with a dipole (polver solvent). The strongest intermolecular force.

18.2.2 Dipole-Dipole

Polar near another polar. Weaker than Ion-Dipole but still strong, based on how strong thepolarity is.

18.2.3 Hydrogen Bond

Either (H −N), (H −O), or (H − F ). No shielding e−s on Hydrogen so central atoms e− pair gets full pullof Hydrogen nucleus.

18.2.4 London Dispersion/Van der Waals

An induced dipole between 2 polar molecules. An increase in pressure or decrease in temperature will causeone side to have a more positive force as the majority of e−s move to other side.

1Of the two theories, this is generally the more accepted one

18

18.2.5 Intermolecular Flowchart

18.3 Rule of Octet

Atoms tend to bond in such a way as to gain, lose, or share e−s in order to gain a complete valence (outer sand p).

19 Lewis Structures

19.1 Structures for Atoms

19.2 Structures for Ions

19.3 Structure for Ions of Molecules

19

19.4 Lewis Structures for Molecular Structures (Covalent)

1. Add valence e−s from all the atoms.

2. Write the symbols for the atoms. If there are more than 2 atoms, identify the central atom. Connectthem with a single line which represents 2 shared e−s. Subtract the number of e−s from total found instep 1.

. Central atom will be closest to Si, P or Metaloid staircase.

3. Complete octets around the atoms bonded to the central atom (Hydrogen does not get more than 2).

4. Place the remaining pairs around the central atom even if doing so gives more than an octet to thecentral atom.

5. If there are not enough pairs to complete an octet in the central atom, then you ned to try using doubleor triple bonds.

CH4

CH2Cl2

HNO3

CO2

HCN

20

19.5 Resonance Structures

Benzene

19.5.1 Formal Charge

Valence e−s of an atom - (total unbonded e−s + 12 total bonded e−s)

Molecular structures that tend to be the common one have a formal charge is closest to zero and any negativecharge is on the most electronegative element.

20 Lattice Energies of Ionic Solids

Coulombs Law

E =KQ1Q2

d

• Q1/Q2 = ion charges

• d = Distance between ions of the final crystalized lattice form.

a The greater the charge, the higher the energy.

a The closer the ions, the higher the energy.

ExampleWhich has a greater lattice energy?

+1

Na−1

Cl vs+2

Mg−1

Cl2+2

Mg−1

Cl2 has greater charges thus a higher lattice energy.+1

Li−1

Cl vs+1

Na−1

Cl+1

Li is smaller than+1

Na so+1

Li wil be closer

to−1

Cl than+1

Na will so+1

Li−1

Cl will have a higherlattice energy.

21

21 Bond Lengths of Covalent Bonds

• Single - Longest

• Double - Medium

• Triple - Shortest

Length

Single CO−44 1.42

◦A

Double CO2 1.24◦A

Triple CO 1.13◦A

22 Electronegativity

Difference in electronegativity determines the character of the bond.

• Large difference → Ionic Bond

. Biggest difference is 3.3

• Medium difference → Polar Covalent

. HF - 1.8

• Small/No difference → Non-Polar Covalent

. H2 - 0

22.1 Dipole

→H − F

Arrow points towards more electronegative atom.

22.1.1 Dipole Moment

Numeric value that represents how strong the dipole is

ExampleWhich has the greater dipole moment?

ORWhich has greater electronegative difference?

HI or HFAnswer: HF

22

23 Bond Enthalpy

∆H: Energy given off or taken in during a reaction.

• ∆H = -. Exothermic

• ∆H = +. Endothermic

a Breaking bonds requires energy

a Forming bonds releases energy

ExampleCH4 + 2O2 → CO2 + 2H2OO

Breaking Forming4 ∗ (C −H) = (4 ∗ 413) 2 ∗ (C=O) = (2 ∗ 799)

2 ∗ (O2) =(2*495) 4 ∗ (H −O) =(4*463)2642 3450

∆H = Broken - Formed∆H = 2642 - 3450 = -808 KJ

24 VSEPR

VSEPR stands for Valence Shell Electron Pair Repulsion. Make sure when counting bonds to treat doubleand triple bonds like a single bond. Also keep in mind that bonded pairs and lone pairs repel.

24.1 Bond Shape Table

Shape Example Total e− Bonded Pairs Lone e− Pairs Hybrid OrbitalLinear BeH2 2 2 0 sp

Trigonal Planar BCl3 3 3 0 sp2

Bent NO−12 3 2 1 sp2

Tetrahedral CH4 4 4 0 sp3

Trigonal Pyramidal NH3 4 3 1 sp3

Bent H2O 4 2 2 sp3

Trigonal Bipyramidal PCl5 5 5 0 sp3dSee-Saw SF4 5 4 1 sp3dT-Shape BrF3 5 3 2 sp3dLinear ICl2 5 2 3 sp3d

Octahedral SF6 6 6 0 sp3d2

Square Pyramidal BrF5 6 5 1 sp3d2

Square Planer ICl−4 6 4 2 sp3d2

25 Organic Chemistry

25.1 Polarity

Polarity in regards to organic chemistry relies on an element disrupting the symmetry of a molecule. Forexample the double bonded oxygen in Acetone allows it to be more polar than Propane.

Acetone Propane

23

25.2 Alkanes

a Spotted by seeing a single bond

• CH4 → Methane

• C2H6 → Ethane

• C3H8 → Propane

• C4H10 → Butane

• C5H12 → Pentane

• C6H14 → Hexane

• C7H16 → Heptane

• C8H18 → Octane

• C9H20 → Nonane

• C10H22 → Decane

PentaneCH3(CH2)3CH3

25.3 Alkane Branch Structure Naming

To name all single-bonded Carbon chains, see subsection on Alkanes above.

To name a branch structure first look for the longest unbroken Carbon chain, this is the root name. Thentake the root prefix of the alkyl (the branch of the root chain) and add -yl (for instance Methane becomesmethyl). Number the Carbon chain giving the side with an alkyl the lowest number. The end result shouldbe something such as 2 Methyl Butane.

2 Methyl Butane

24

25.3.1 Branch Structure Naming Table

25

25.4 Alkenes

a Spotted by seeing a double bond

• CH2 → Methene

• C2H4 → Ethene

• C3H6 → Propene

• C4H8 → Butene

• C5H10 → Pentene

• C6H12 → Hexene

• C7H14 → Heptene

• C8H16 → Octene

• C9H18 → Nonene

• C10H20 → Decene

25.4.1 Alkene Naming

Naming Alkenes is similar to naming Alkanes save for the naming of the root chain. To name the root chainyou must give side where the double bond is the lowest number and name all branches after using this numberscheme. You should end up with something like 2 Pentene

2 Pentene

25.5 Alkynes

a Spotted by seeing a triple bond

• CH → Methyne

• C2H2 → Ethyne

• C3H4 → Propyne

• C4H6 → Butyne

• C5H8 → Pentyne

• C6H10 → Hexyne

• C7H12 → Heptyne

• C8H14 → Octyne

• C9H16 → Nonyne

• C10H18 → Decyne

26

25.5.1 Alkyne Naming

Naming Alkynes is similar to naming Alkenes. Identify the root chain as you would using Alkenes exceptnow you identify the triple bond instead of the double bond.

2 Hexyne

26 Functional Groups

� When discussing functional groups, the letter R is used to signify any hydrocarbon or hydrocarbonchain.

26.1 Alcohol

• Root Name: -ol

• Identification: R-OH

Ethanol

26.2 Aldehyde

• Root Name: -al

• Identification: R-CHO

Ethanal

27

26.3 Carboxylic Acid

• Root Name: -oic

• Identification: R-COOH

Propanoic Acid

26.4 Ester

• Root Name: A-yl B-oate

• Identification: R-COO-R

Ethyl PropanoateB︷︸︸︷

CH3CH2CO

A︷︸︸︷OCH2CH3

26.5 Ketone

• Root Name: -one

• Identification: R-CO-R

Acetone

26.6 Ether

• Root Name: A-yl B-yl Ether

• Identification: R-O-R

Propyl Methyl EtherA︷︸︸︷

CH3CH2CH2O

B︷︸︸︷CH3

28

26.7 Amine

• Root name: -amine

• Identification: R-NH2

Methylamine

26.8 Amide

• Root Name: -amide

• Identification: R-CONH2

Ethanamide

26.9 Haloalkane

• Root Name: None, use standard naming of root chain

• Identification: Some Hydrogens in a a hydrocarbon are replaced with a halogen (F, Cl, Br, I)

2 Chloro Butane

27 Complex Ions

Complex Ions are usually metal ions with attached ligands (Lewis Bases).

27.1 Cations

[Cr(H2O)6]+3

a The charge of a cation is the charge of the transition metal (Cr in this case).

29

27.2 Anions

[Al(OH)4]−1

a The charge of a anion is determined by the individual charges of the elements.

. Al+3 + 4(OH)−1

. 3 - 4

. -1

27.3 Coordination Number

Generally (especially with cations) the coordination number is twice the charge of the transition metal.

Example[Cr(H2O)6]+3

Cr+3 → 3 ∗ 2 = 6

27.4 Naming

27.4.1 Cations

• Give the prefix associated with the coordination number

• Give appropriate name for ligand

• Name the transition metal

• Give roman numeral of transition metal

Example

[

Chromium︷︸︸︷Cr (H2O)︸︷︷︸

Aqua

Hexa︷︸︸︷6 ]+3

Hexa Aqua Chromium (III)

27.4.2 Anions

• Give prefix associated with the coordination number

• Give appropriate Ligand name

• Name transition metal with -ate ending

• Give roman numeral

Example[Al(OH)4]−1

Tetra Hydroxo AluminateNo roman numeral because Al is always +3

28 Acidic and Basic Redox

28.1 Acidic

• Find oxidation number

• Write 12 reaction with e−s

• Add H2O, then H+ and balance accordingly

• Balance for e−s and everything else

• Add together both balanced 12 reactions and cancel out where possibly to simplify

30

28.2 Basic

• Find oxidation number

• Write 12 reaction with e−s

• Add H2O, then H+ and balance accordingly

• Add OH amounts to both sides equal to the number of H+

• Cancel out the H+ with the OH to form H2O

• Move all H2O to one side

• Balance for e−s and everything else

• Add together both balanced 12 reactions and cancel out where possibly to simplify

Example+7

Mn O−4 ++3

C2

−2

O4

−2

→+1

Mn O−22 +

+4

C O−23

(4OH + C2O4 → 2CO3 + 2e− + 2H2O) ∗ 3

(2H2O +MnO4 + 3e− →MnO2 + 4OH) ∗ 212OH + 3C2O4 → 6CO3 + 6e− + 6H2O

4H2O + 2MnO4 + 6e− → 2MnO2 + 8OH4OH + 3C2O4 + 2MnO4 → 2MnO2 + 6CO−2

3

29 Thermodynamics

The study of energy and its transformationsUnits of Energy:

• Joules and Calories

. 1 cal = 4.184 J

The two main driving forces of thermodynamics is Enthalpy and Entropy:

29.1 Enthalpy

Enthalpy stands for the Heat of the reaction and is denoted by ∆H.

If:

• ∆H<0

. Reaction is exothermic

• ∆H>0

. Reaction is endothermic

There are 4 ways to find ∆H.

29.1.1 Stoichiometry Problems

ExampleHow much heat is released when 3.2 grams of Hydrogen is reacted with excess Oxygen?

2H2 +O2 → 2H2O∆H◦ = −572 KJ

3.2 g H21 ∗ 1 mole H2

2.02 g H2∗ −572 KJ

2 mole H2= −453.069 KJ

Ratio = Energy ReleasedCoefficient of Hydrogen in formula

31

29.1.2 Calorimetry

Find the ∆H by running a reaction and heating or cooling a substance.

q = m ∗ c ∗∆T

• q = Heat released or absorbed

• m = Mass of what is being heated (grams)

• c = Specific heat. Unique to every substance ( Jg∗C )

. Specific heat of water is 4.184

• ∆T = Change in temperature

ExampleBurn 0.1 grams of CH4 and it heats 100 grams H2O from 20◦ C to 33.29◦ C.

q = 100 ∗ 4.184 ∗ 13.29 = 5560 J = 5.560 KJ0.1 grams CH4

1 ∗ 1 mole CH416 g CH4

= 0.00625 moles CH45.560

0.00625 = 889.6 KJMole

29.1.3 Hess Law

Multiple reactions can be added together then ∆Hs can be added together.

ExampleSi+ 2H2 → SiH4 ∆H = +34 KJ

Mole

Si+O2 → SiO2 ∆H = −911 KJMole

H2 + 12O2 → H2O ∆H = −242 KJ

Mole

Find ∆H for:SiH4 + 2O2 → SiO2 + 2H2O

SiH4 →Si+2H2 ∆H = −34 KJMole

Si+O2 → SiO2 ∆H = −911 KJMole

2H2 + 2O2 → H2O ∆H = −484 KJMole

SiH4 + 2O2 → SiO2 + 2H2O ∆H = −1429 KJMole

29.1.4 Standard Heat of Formation

Standard heat (enthalpy) of formation (∆H◦f )2 is the energy involved in forming one mole of a chemical fromits elements under standard conditions.

a Elemental substances (O2, H2, etc.) always have a ∆H of zero.

ExampleFind the ∆H for:

2H2O2 → 2H2O +O2

∆HfH2O2 = −187∆HfH2O = −285

2∗(−187)

2H2O2︸︷︷︸−374

→2∗(−285)

2H2O +O2︸︷︷︸−570

∆H =∑

product−∑

reactant

∆H = −570− (−374) = −196 KJMole

2This symbol may be shortened to ∆H or ∆Hf in this subsection.

32

29.2 Entropy

Entropy stands for the Disorder of the reaction and is denoted by ∆S.

If:

• ∆S<0

. Order is increasing

• ∆S>0

. Disorder is increasing

29.2.1 State of Matter

If:

• Solid → Liquid

. ∆S = +

• Gas → Solid

. ∆S = −

Solid Liquid GasLowest ∆S – Highest ∆S

29.2.2 Number of Moles of Gasses

a Solids and liquids do not apply

ExampleN2(g) + 3H2(g) → 2NH3(g)

4 moles gas → 2 moles gas | ∆S = −2 moles gas → 4 moles gas | ∆S = +

29.2.3 Pressure of Gas

• When pressure increases, disorder decreases.

• When pressure decreases, disorder increases.

ExampleWhat has more disorder?

N2 at 1 atmN2 at 0.001 atm

Answer: N2 at 0.001 atm

29.3 Gibbs Law of Free Energy

Gibbs Law determines ∆G which signifies whether a reaction is spontaneous or not.

∆G = ∆H − (T ∗∆S)

• ∆G = Free Energy in a system

• ∆H = Enthalpy (KJ)

• ∆S = Entropy (KJK )

. ∆S MUST be converted from JK to KJ

K .

• T = Temperature in Kelvin

. To convert C◦ → K add 273

33

If:

• ∆G<0

. Spontaneous

• ∆G>0

. Not spontaneous

29.3.1 ∆H, ∆S, ∆G, Relationship Table

∆H = - ∆S = + ∆G = - Always spontaneous∆H = - ∆S = - ∆G = - Spontaneous at low temperatures∆H = + ∆S = + ∆G = + Spontaneous at high temperatures∆H = + ∆S = - ∆G = + Never spontaneous

ExamplePOCL3 → 2PCl3 +O2

∆H = 542 KJ∆S = 179 J

K

What temperature is it spontaneous at?0 = 542 KJ − (T ∗ 0.179KJ

K )−542 = −0.179TT = 3027.93 K

30 Chemical Kinetics and Rate Laws

Factors that affect reaction rates

30.1 Physical State

• Solid

. An increase in surface area means in an increase in the rate.

• Gas - Gas

• Liquid - Gas

• Liquid - Liquid

30.2 Concentration3Molarity = moles

liter[HCL] = 3M

An increase in concentration is generally an increase in rate.

30.3 Temperature

An increase in temperature is an increase in rate.

30.4 Pressure of Gas

An increase in pressure is an bincrease in rate.

30.5 Catalysts and Inhibitors

A catalyst lowers the activation energy while an inhibitor increases the activation energy.3Molarity is signified by []s

34

30.6 Rate Laws

A+B → C +D

rate = k[A]m[B]n

• k = Constant

• m = Order of A

• n = Order of B

a Order of 0 → No effect

a Order of 1 → Linear - Double the concentration and you double the rate

a Order of 2 → Squared - Double the concentration and you quadruple the rate

Example:Trial [A] [B] Rate

1 0.1 M 0.1 M 0.04 M/s2 0.2 M 0.1 M 0.08 M/s3 0.1 M 0.2 M 0.04 M/s

Solve for m:

trial 2trial 1

= ([][]

)m =rate

rate= (

0.20.1

)m =0.080.04

2m = 2

m = 1

Solve for n:

(0.20.1

)n =0.040.04

1n = 1

n = 0

rate = k[A]1[B]0

Solve for k:

0.04 = k[0.1]1[0.1]0

k = 0.4

30.6.1 Order Table

Comments Zero Order First Order Second OrderRate Law rate = k rate = k[A]1 rate = k[A]2

Integrated Rate law [A]− [A]0 = −kt ln[A]− ln[A]0 = −kt 1[A] −

1[A]0

= kt

[A] = −kt+ [A]0 ln[A] = −kt+ ln[A]0 1[A] = kt+ 1

[A]0

Graph [A] vs Time ln[A] vs time 1[A] vs time

K = Slope Slope = −k Slope = −k Slope = k

Half-Life (t 12) t 1

2= [A]0

2k t 12

= 0.693k t 1

2= 1

k[A]0

35

Example:

2N2O5 → 4NO2 +O2

[N2O5] Time (s)0.1 0

0.0707 500.05 1000.025 2000.0125 3000.00625 400

1. What is the order of the reaction?

[A] 6= straight1

[A] 6= straight

ln[A] = straight

Order of 1

2. What is the k constant value?ln(0.0707)−ln(0.1)

50−0 = −0.34750 = 0.00693

k = 0.00693

3. What is the concentration of N2O5 at t = 150?ln[A] = −(0.00693)(150) + ln(0.1)ln[A] = −3.34[A] = 0.0354 M

4. What is the rate at 150 seconds?rate = k[A]rate = 0.00693 ∗ [0.0354]rate = 2.45 ∗ 10−4 M/s

5. What is the half life?t 1

2= 0.693

k

t 12

= 0.6930.00693

t 12

= 100 s

31 Reaction Mechanisms

Many/most reactions do not take place in one step. If a reaction were to react in one step, then you coulduse the balanced reaction to determine the rate law. For example, assume the following occured in one step.

MgCl2 + 2Hbr → 2HCl +MgBr2

rate = k[MgCl2]1[HBr]2

In reality though, things are not always as easy.

Through experimentation we figure out that the rate law for:

NO2 + CO → NO + CO2

is

rate = k[NO2]2

Because the rate law does not link up with the equation, it is not a single step reaction.

36

31.1 Elementary Steps

• Unimolecular - 1 reactant

• Bimolecular - 2 reactants

• Terrmolecular - 3 reactants

32 Equilibrium

The state where the concentration or partial pressures (if it is a gas) of all the reactants and products remainconstant with time. For equilibrium to occur, the forward reaction rate must equal the reverse rate. In otherwords, the amounts do not have to be equal, but the rates must be.

32.1 Types of Equilibrium

• Static → No movement

• Dynamic → Movement such as a sealed container of water

32.2 Equilibrium Constant Expressions

aA+ bB ⇀↽ cC + dD

Kc =[C]c[D]d

[A]a[B]b

Kp =(PCc)(PDd)(PAa)(PBb)

• Kc = Concentration constant

• Kp = Partial Pressure constant

32.2.1 Converting Constants

To convert between the two constants Kc and Kp use the formula:

Kp = Kc(RT )∆n

• ∆n =∑

Product Coefficients−∑

Reactant Coefficients

33 Gas Laws

33.1 Gas Units and Conversions

1 Atm = 760 Torr (mmHg) = 101.3 kPa = 14.7 PSI

33.2 Ideal Gas Law

Pv = nRT

• P = Pressure (Atm)

• v = Volume (L)

• n = Number of moles

• R = 0.0821 (constant)

• T = Temperature (Kelvin)

Example3 grams of HCl at 26◦ C in a 3 Liter container. What is the pressure?

P (3)3 =

( 3 grams36.5 g/mole

)(0.081)(26+273)

3P = 0.0664 Atm

37

33.3 Real Gas Law

For use when the ideal gas law fails. The ideal gas law fails when these two postulates fail:

• Molecules do have volume

• Molecules are attracted

This law is also used when there are conditions with high pressure and low temperature.

Pv = nRT

⇓

(P +n2a

v2) ∗ (v − nb) = nRT

• a = constant that fixes the intermolecular force issue

• b = constant that fixes the volume issue

a a and b are unique to each type of gas

a All other variables are the same as the ideal gas law

33.4 Combined Gas Law

P1V1

T1=P2V2

T2

33.5 Daltons Law of Partial Pressures

For a mixture of gases in a container, the total pressure (Ptot) is equal to the sum of the pressures each gasexerts as if it were alone.

Ptot = P1 + P2 + P3 + · · ·+ Pn

Example:A mixture of 1g H and 1g He in a 1 L container is at 27◦C. Calculate the mole fraction of each gas, partial

pressures of each and total pressure.

H2 1 g ∗ 12

= 0.5 moles

⇓

x =(H mole)

(H +He moles)=

0.50.75

= 0.667

He 1 g ∗ 14

= 0.25 moles

⇓

x =0.250.75

= 0.333

H2 HePv = nRT Pv = nRT

P (1) = (0.5)(0.0821)(300) P (1) = (0.25)(0.0821)(300)P = 12.3 Atm P = 6.15 Atm

Ptot = 12.3 + 6.15 = 18.45 Atm

38

33.6 Gas Collection over a Water Solution

Example:A 0.986 g sample has Zinc and some impurities. Excess HCl is added and reacts with thte Zinc but not theimpurities. Find the percent Zinc in the sample if 240 mL of H2 are collected over H2O at 30◦C and 1.032

Atm (HINT: This is the Ptot).

Ptot = P1 + PH2O

1.032 = P1 + 0.042P1 = 0.99 Atm H2

⇓Pv = nRT

(0.99)(0.240) = n(0.0821)(303)n = 0.0096 mole H2

⇓0.0096 mole H2 = 0.0096 mole Zn

0.0096 mole Zn1

∗ 65.4 g ZnMole

= 0.628 g Zn

0.628 g Zn0.986 g total

∗ 100 = 63.7%Zn

34 ICE ICE (Baby)

Given initial values for a system at equilibrium and one of the equilibrium values, you should find:

• a - The other equilibrium values

• b - The equilibrium constant

Example:A closed system initially containing 1 ∗ 10−3 M H2 and 2 ∗ 10−3 M I2 at 448◦ C is allowed to reach

equilibrium. Analysis of the equilibrium mixture shows the [HI] = 1.7 ∗ 10−3 M . Find the equilibriumconcentration for H2 and I2 as well as the Kc value.

H2 + I2 ⇀↽ 2HI

Initial 1 ∗ 10−3 M 2 ∗ 10−3 M 0 MChange −0.935 ∗ 10−3 0.935 ∗ 10−3 1.87 ∗ 10−3

Equilibrium 0.065 ∗ 10−3 1.065 ∗ 10−3 1.87 ∗ 10−3

[H2] = 0.065 M

[I2] = 1.065 ∗ 10−3 M

Kc =[1.87 ∗ 10−3]2

[1.065 ∗ 10−3][0.065 ∗ 10−3]

35 Acids and Bases

35.1 Definitions of Acids and Bases

1. Arrhenius. An acid dissociates in water to form H+ ions and a base dissociates to form OH− ions.

2. Bronsted-Lowry. Acids are proton donors (H+) and a base is a proton acceptor.. Conjugate acid base pair

Acid 1

HNO3 +Base 2

H2O→Acid 2

H3O+ +

Base 1

NO−3

3. Lewis Acid. Acid is an e− pair acceptor while a base is an e− pair donor.

39

35.2 pH and pOH

pH and pOH are measures of the amount of ions in a solution that either cause the solution to be acidic orbasic.

pH ScaleBasic ⇒ 0↔ 14⇐ AcidicImportant Formulas

pH = −log[H+]pOH = −log[OH−]pH + pOH = 14[H+] = 1 ∗ 10−pH

[OH−] = 1 ∗ 10−pOH

ExampleWhat is the concentration of HCl with a pH of 3?

[HCl] = 0.001 M

35.2.1 Changing Concentrations

M1V1 = M2V2

(0.25 M)(5 mL) = M2(50 mL)

M2 = 0.025 M

35.3 Strong Acids and Bases

Strong acids and bases completely dissociate in water.

35.3.1 Strong Acids

• HCl

• H2SO4

• HBr

• HI

• HNO3

• HClO4

35.3.2 Strong Bases

• Group 1 - Hydroxides

. NaOH

. KOH

• Group 2 - Heavier Hydroxides

. Ca(OH)2

. Sr(OH)2

. Ra(OH)2

35.4 Weak Acids and Bases

Weak acids and bases do not completely dissociate in water.

40

35.4.1 Ka Constant

HA ⇀↽ H+ +A−

HA+H2O ⇀↽ H3O+ +A−

Ka =[H+][A−]

[HA]

ExampleBenzoic acid dissociates as follows:

HC7H5O2 ⇀↽x

H+ +x

C7H6O−2︸︷︷︸

x2

[HC7H5O2] = 0.4 MKa = 6.3 ∗ 10−5

What is the pH?

Ka = [H+][C7H5O−2 ]

[HC7H5O2]

6.3 ∗ 10−5 = x2

0.4

35.4.2 Kb Constant

The Kb constant is used when bases are involved in a reaction (as opposed to Ka which is used in reactionswith acids). To convert between Kb and Ka use the following formula:

Ka ∗Kb = Kw

• Kw = 1 ∗ 10−14

ExampleF− +H2O ⇀↽ HF +OH−

Ka = 7.2 ∗ 10−4

What is the Kb constant?Kb = 1∗10−14

7.2∗10−4 = 1.39 ∗ 10−11

Find the pH and pOH.Kb = [HF ][OH−]

[F−]

1.39 ∗ 10−11 = x2

( 0.00220+13.3 )

x = 9.13 ∗ 10−7

pOH = 6.04pH = 7.96

35.5 Common Ion Effect

The effect of ionization of a weak electrolyte (acid/base) is decreased by adding a strong electrolyte thathas an ion in common with the weak electrolyte.

35.6 Buffer

Made of 2 components:

1. Weak acid

2. The salt of that acid

41

36 Equilibrium of Saturated, Soluable Salts

Solubility is how well a solute dissolves in a solvent4.

Example:CaCO3 (s) ⇀↽ Ca+2

(aq) + CO−23 (aq)

Ksp = [Ca+2][CO−23 ]

• Ksp is the solubility product

. A large Ksp means the solution is very soluable (meaning lots of products)

. A small Ksp means the solution is not very soluable.

1. Given Ksp, find the ion concentration.

Ksp = [Ca+2][CO−23 ] = 4.5 ∗ 10−9

[Ca+2] = [CO−23 ] =

√4.5 ∗ 10−9 = 6.7 ∗ 10−5 M

2. Given Ksp, find the solubility (g/L).

6.7 ∗ 10−5 M = 6.7∗10−5

1 ∗ 100.11mole = 6.37 ∗ 10−3g/L

3. Given solubility, find ion concentration.

Solubility of Silver Chloride at 25◦C is 1.3 ∗ 10−7 g100 mL

1.3 ∗ 10−7 g100 mL →

gL ∗

1010 = 1.3 ∗ 10−6 g

L

1.3∗10−6

L ∗ 1 mole143.35 g = 9.11 ∗ 10−9 m

L

4. Given solubility, find Ksp

Ksp = [Ag+][Cl−] = (9.11 ∗ 10−9)2 = 8.3 ∗ 10−17

37 Kinetic Molecular Theory

37.1 Postulates:

• The volume of the individual particales of a gas can be assumed to be negligible.

. So volume is determined by the space between molecules

• The gas particles are in constant motion. The pressure exerted by a gas is due to collisions of the gaswith the walls of the container.

• Gas particles are not attracted to one another.

• The average kinetic energy of a gas is directly proportional to the Kelvin temperature.

Kenergy = 32 (0.0821)T

ORKenergy = 1

2 (Molar Mass)(V elocity)2

A) CO at 760 torr and 0◦C

B) N2 at 760 torr and 0◦C

C) H2 at 760 torr and 0◦C

Q. Which will have the highest kinetic energy?

A. All will have the same kinetic energy

Q. Which will have a higher velocity?

A. H2 will because if all kinetic energies are constant according to the formula k =12mv2 the

smallest mass will yield the highest velocity to keep k constant.4Virtually every salt is soluable to some degree.

42

37.2 Root Mean Square Velocity

Urms =

√3RTM

• R = 8.314 JK∗Mole

• M = molar mass (Kg/mole)

. NOT g/mole

37.3 Effusion and Diffusion

37.3.1 Effusion

When you pass a gas through a small opening into an evacuated chamber.

37.3.2 Diffusion

When you mix gases

37.3.3 Finding the rate

The formula for finding the rate is as follows:

Rate of Effusion of Gas 1Rate of Effusion of Gas 2

=

√(Molar Mass 2)√(Molar Mass1)

• WHich effuses faster, He or NO2?

. He - it moves faster because it is smaller

• For the reaction: H2 +N2 at 20◦C and having a rate of effusion for H2 being 10 mL/min what is therate for N2?

. 10x =

√28√2

x = 10√

2√28

= 2.67 ml/min

• The rate for the gas is 24 mL/min, at the same temperature methane has a rate of 47.8 mL/min. Whatis the molar mass of the unknown gas?

. 2447.8 =

√16√x

x = 63.7 g/mole

38 Electro Chemistry

38.1 Identifying Oxidation Numbers

H2O H2SO4 Cl2H +1 H +1 Cl 0O -2 S +6

O -2

The chemical that has been oxidized is the reducing agent. The chemical that has been reduced isthe oxidizing agent.

38.2 Galvanic/Voltaic Cells

There are two beakers with salt and e−s in each solution. A salt bridge between the two solutions allowspassage of ions. One side is identified as the cathode an the other the anode. The cations go to the cathodeand the anions go to the anode. The e−s go to the cathode.

43

38.3 Calculating Cell Potential

E◦cell = E◦reduction+ E◦Oxidation

ExampleCu+2 + Zn→ Zn+2 + Cu

OxidationZn→ Zn+2 + 2e−

e◦ = 0.76Reduction

Cu+2 + 2e− → Cue◦ = 0.34

E◦cell = 0.34 + 0.76 = 1.1 volts

38.3.1 Nernst Equation to Find E◦cell

The Nernst equation to be used under standard conditions is:

E◦cell = E◦ − 0.0592n

log(Q)

• E◦ = Normal standard potential

• n = Number of moles of e−s changing

• Q = Reaction Quotient = [Product][Reactant]

ExampleCu(s) + Cu+2︸︷︷︸

1 M

→ Cu+2︸︷︷︸0.1 M

+Cu(s)

Q = 0.11 = 0.1

n = 2 (2 e− being transfered)E◦ = 0(1M − 1M = NovoltageE◦cell = 0− 0.0592

2 log(0.1)

This Nernst equation is to be used when the temperature is not standard and the concentrationsare not equal.

E◦cell = E◦ − RT

nFln(Q)

• R = 9.31 volt coulombmole Kelvin

• F = 96,500 per mole e−

• T = Temperature in Kelvin

ExampleZn+ Cu+2︸︷︷︸

2 M

→ Cu+ Zn+2︸︷︷︸0.5 M

E◦cell = 1.1− 0.05922 log( 0.5

2 )E◦cell = 1.1178 volts

39 Balancing Redox Reactions

39.1 Acidic

+2

Cr2

−2

O−27 +

−1

Cl−→+3

Cr+3 +0

Cl2

Half-Reactions

[2Cl− → Cl2 + 2e−] ∗ 3

14H+ + 6e−++6

Cr2O7→ 2Cr+3 + 7H2O

⇓14H+ + 6Cl− + Cr2O7 → 2Cr+3 + 7H2O + Cl2

44

39.2 Basic

CN− +MnO−4 → CNO− +MnO2

Half-Reactions

2OH +H2O + CN− → CNO− + 2e− + 2H+2OH−︸︷︷︸H2O

4OH− + 4H+︸︷︷︸2H2O

+3e− +MnO−4 →MnO2 + 2H2O + 4OH

⇓

(2OH +H2O + CN− → CNO− + 2e− + 2H+2OH−︸︷︷︸H2O

) ∗ 3

(4OH− + 4H+︸︷︷︸2H2O

+3e− +MnO−4 →MnO2 + 2H2O + 4OH) ∗ 2

⇓

6OH− + 3CN− → 3CNO− + 6e− + 3H2O

H2O + 6e− + 2MnO−4 → 2MnO2 + 2OH

⇓

H2O + 3CN− + 2MnO−4 → 3CNO− + 2MnO2 + 2OH

45

AP* Chemistry Chemical Foundations Chemistry: An Overview

• Matter – takes up space, has mass, exhibits inertia - composed of atoms only 100 or so different types - Water made up of one oxygen and two hydrogen atoms - Pass an electric current through it to separate the two types of atoms and they rearrange to become two different types of molecules

- reactions are reversible

Chemistry – is defined as the study of matter and energy and more importantly, the changes between them • Why study chemistry?

- become a better problem solver in all areas of your life - safety – had the Roman’s understood lead poisoning, their civilization would not have fallen - to better understand all areas of science

The Scientific Method

• A plan of attack!

The fundamental steps of the scientific method

*AP is a registered trademark of the College Board, which was not involved in the production of this product. © 2013 by René McCormick. All rights reserved.

• Good experimental design coupled with repetition is key! Theory – hypotheses are assembled in an attempt at explaining “why” the “what” happened.

• Model – we use many models to explain natural phenomenon – when new evidence is found, the model changes!

• Robert Boyle

o loved to experiment with air o created the first vacuum pump o coin and feather fell at the same rate due to gravity o in a vacuum there is no air resistance to impede the fall of either

object! o Boyle defined elements as anything that cannot be broken into

simpler substances. Boyle’s Gas Law: P1V1 = P2V2

• Scientific Laws – a summary of observed (measurable) behavior [a theory is an explanation of behavior] A law summarizes what happens; a theory (model) is an attempt to explain WHY it happens. - Law of Conservation of Mass – mass reactants = mass products - Law of Conservation of Energy – (a.k.a. first law of thermodynamics) Energy CANNOT be created NOR destroyed; can only change forms.

- Scientists are human and subjected to • Data misinterpretations • Emotional attachments to theories • Loss of objectivity • Politics • Ego • Profit motives • Fads • Wars • Religious beliefs

• Galileo – forced to recant his astronomical observations in the face of strong religious resistance

• Lavoisier – “father of modern chemistry”; beheaded due to political affiliations.

• The need for better explosives; (rapid change of solid or liquid to gas where molecules become ≈2,000

diameters farther apart and exert massive forces as a result) for wars have led to

-fertilizers that utilizes nitrogen - Nuclear devices

Chemical Foundations 2

Units of Measure A quantitative observation, or measurement, ALWAYS consists of two parts: a number and a unit. Two major measurements systems exist: English (US and some of Africa) and Metric (the rest of the globe!) • SI system – 1960 an international agreement was reached to set up a system of units so scientists everywhere

could better communicate measurements. Le Système International in French; all based upon or derived from the metric system

KNOW THE UNITS AND PREFIXES shown in BLUE !!! • Volume – derived from length; consider a cube 1m on each edge ∴1.0 m3 - A decimeter is 1/10 of a meter so (1m)3 = (10 dm)3 = 103 dm3 = 1,000 dm3 1dm3 = 1 liter (L) and is slightly larger than a quart also 1dm3 = 1 L = (10 cm)3 = 103cm3 = 1,000 cm3 = 1,000 mL AND 1 cm3 = 1 mL = 1 gram of H2O (at 4ºC if you want to be picky)

Mass vs. Weight – chemists are quite guilty of using these terms interchangeably.

o mass (g or kg) – a measure of the resistance of an object to a change in its state of motion (i.e. exhibits inertia); the quantity of matter present

o weight (a force∴ has units of Newtons) – the response of mass to gravity; since all of our measurements will be made here on Earth, we consider the acceleration due to gravity a constant so we’ll use the terms interchangeably as well although it is technically incorrect! We “weigh” chemical quantities on a balance NOT a scale!!


Physics connection: Fw = ma Fw = mg

2

9.8 m

sw

mF

=

∴its units are

( ) 2

mkg

sN

=

Exercise 1 Precision and Accuracy To check the accuracy of a graduated cylinder, a student filled the cylinder to the 25-mL mark using water delivered from a buret and then read the volume delivered. Following are the results of five trials: Trial Volume Shown by Volume Shown Graduated Cylinder by the Buret 1 25 mL 26.54 mL 2 25 mL 26.51 mL 3 25 mL 26.60 mL 4 25 mL 26.49 mL 5 25 mL 26.57 mL Average 25 mL 26.54 mL Is the graduated cylinder accurate? Note that the average value measured using the buret is significantly different from 25 mL. Thus, this graduated cylinder is not very accurate. It produces a systematic error (in this case, the indicated result is low for each measurement).

Gravity – varies with altitude here on planet Earth • The closer you are to the center of the Earth, the stronger the gravitational field

SINCE it originates from the center of the Earth. • Every object has a gravitational field – as long as you’re on Earth, they are masked

since the Earth’s field is so HUGE compared to the object’s. • The strength of the gravitational field ∝ mass • Ever seen astronauts in space that are “weightless” since they are very far removed

from the center of Earth? Notice how they are constantly “drawn” to the sides of the ship and must push away?

• The ships’ mass is greater than the astronaut’s mass ∴ “g” is greater for the ship and the astronaut is attracted to the ship just as you are attracted to Earth! The moon has

61 the mass of the Earth ∴ you would experience6

1 the gravitational field you

experience on Earth and ∴ you’d WEIGH 61 of what you weigh on Earth.

Precision and Accuracy - Accuracy – correctness; agreement of a measurement with the true value - Precision – reproducibility; degree of agreement among several measurements. - Random or indeterminate error – equal probability of a measurement being high or low - Systematic or determinate error – occurs in the same direction each time

The results of several dart throws show the difference between precise and accurate. (a) Neither nor precise (large random errors). (b) Precise but not accurate (small random errors, large systematic error). (c) Bull’s-eye! Both precise and accurate (small random errors, no systematic error).


Significant Figures and Calculations

Determining the Number of Significant Figures (or Digits) in a Measurement • Nonzero digits are significant. (Easy enough to identify!) • A zero is significant IF and ONLY IF it meets one of the conditions below:

- The zero in question is “terminating AND right” of the decimal [must be both] - The zero in question is “sandwiched” between two significant figures

• Exact or counting numbers have an ∞ amount of significant figures as do fundamental constants (never to be confused with derived constants)

Reporting the Result of a Calculation to the Proper Number of Significant Figures • When × and ÷ , the term with the least number of significant figures (∴least accurate measurement)

determines the number of maximum number of significant figures in the answer. (It’s helpful to underline the digits in the least significant number as a reminder.)

4.56 × 1.4 = 6.38 →corrected 6.4

• When + and (−), the term with the least number of decimal places (∴least accurate measurement) determines the number of significant figures in the final answer.

12.11 18.0 ← limiting term (only 1 decimal place) 1.013 31.123 corrected→ 31.1 (limits the overall answer to only one decimal place)

• pH – the number of significant figures in least accurate measurement determines number decimal places on the reported pH (usually explained in the appendix of your text)

Rounding Guidelines for the AP Exam and This Course:

• Round ONLY at the end of all calculations (keep the numbers in your calculator) • Examine the significant figure one place beyond your desired number of significant figures.

IF > 5 round up; < 5 drop the remaining digits. • Don’t “double round”! Example: The number 7.348 rounded to 2 SF is reported as 7.3

In other words, DO NOT look beyond the 4 after the decimal and think that the 8 rounds the 4 up to a five which in turn makes the final answer 7.4. [Even though you may have conned a teacher into rounding your final average this way before!]

Exercise 2 Significant Figures (SF) Give the number of significant figures for each of the following experimental results. a. A student’s extraction procedure on a sample of tea yields 0.0105 g of caffeine. b. A chemist records a mass of 0.050080 g in an analysis. c. In an experiment, a span of time is determined to be 8.050 × 10‒3 s .

a. three; b. five; c. four


Dimensional Analysis Example: Consider a straight pin measuring 2.85 cm in length. Calculate its length in inches. Start with a conversion factor such as 2.54 cm = 1 inch ∴ you can write TWO

Conversion factors: 1 in

2.54 cm or

2.54 cm

1 in. Why is this legal? Both quantities

represent the exact same “thing” so the conversion factor is actually equal to “1”. To convert the length of the pin from cm to inches, simply multiply your given quantity by a conversion factor you engineer so that it “cancels” the undesirable unit and places the desired unit where you want it. For our example, we want inches in the numerator so our numerical answer is not reported in reciprocal inches! Thus,

2.85 cm1 in

2.54 cm× = 1.12 in

Let’s practice!

Exercise 3 A pencil is 7.00 in. long. Calculate the length in centimeters?

17.8 cm

Exercise 4 You want to order a bicycle with a 25.5-in. frame, but the sizes in the catalog are given only in centimeters. What size should you order? 64.8 in

Exercise 5 A student has entered a 10.0-km run. How long is the run in miles? We have kilometers, which we want to change to miles. We can do this by the following route:

kilometers → meters → yards → miles To proceed in this way, we need the following equivalence statements (conversion factors):

1 km = 1000 m 1 m = 1.094 yd

1760 yd = 1 mi

6.22 mi


Temperature I suspect you are aware there are three temperature scales commonly in use today. A comparison follows:

Notice a degree of temperature change on the Celsius scale represents the same quantity of change on the Kelvin scale.

Exercise 6 The speed limit on many highways in the United States is 55 mi/h. What number would be posted if expressed in kilometers per hour?

88 km/h

Exercise 7 A Japanese car is advertised as having a fuel economy of 15 km/L. Convert this rating to miles per gallon.

35 mi /gal


Density Classification of Matter States of Matter (mostly a vocabulary lesson) � Be very, very clear that changes of state involve altering IMFs not altering actual chemical bonds!! � solid – rigid; definite shape and volume; molecules close together vibrating about fixed points

∴ virtually incompressible � liquid – definite volume but takes on the shape of the container; molecules still vibrate but also have

rotational and translational motion and can slide past one another BUT are still close together ∴ slightly compressible

� gas – no definite volume and takes on the shape of the container; molecules vibrate, rotate and translate and are independent of each other ∴ VERY far apart ∴ highly compressible

- vapor – the gas phase of a substance that is normally a solid or liquid at room temperature - fluid – that which can flow; gases and liquids

• Mixtures – can be physically separated - homogeneous – have visibly indistinguishable parts, solutions including air - heterogeneous – have visibly distinguishable parts - means of physical separation include: filtering, fractional crystallization, distillation, chromatography

Density = volume

mass

Exercise 8 Determining Density A chemist, trying to identify the main component of a compact disc cleaning fluid, determines that 25.00 cm3 of the substance has a mass of 19.625 g at 20°C. Use the information in the table below to identify which substance may serve as the main component of the cleaning fluid. Justify your answer with a calculation.

Compound Density (g/cm3) at 20°C Chloroform 1.492 Diethyl ether 0.714 Ethanol 0.789 Isopropyl alcohol 0.785 Toluene 0.867

Density = 0.7850 g / cm3∴ isopropyl alcohol


Paper Chromatography: Distillation:

Paper chromatograph of ink. (a) A line of the mixture to be separate is placed at one end of a sheet of porous paper. (b) The paper acts as a wick to draw up the liquid. (c) The component with the weakest attraction for the paper travels faster than those that cling to the paper.

• Pure substances – compounds like water, carbon dioxide etc. and elements. Compounds can be separated into elements by chemical means - electrolysis is a common chemical method for separating compounds into elements - elements can be broken down into atoms which can be further broken down into

- nuclei and electrons - p+, n0 and e- - quarks

Electrolysis is an example of a chemical change. In this apparatus, water is decomposed to hydrogen gas (filling the red balloon) and Oxygen gas (filling the blue balloon).


AP* Chemistry ATOMS, MOLECULES & IONS

This is the highest honor given by the American Chemical Society. Priestly discovered oxygen. Ben Franklin got him interested in electricity and he observed graphite conducts an electric current. Politics forced him out of England and he died in the US in 1804. The back side, pictured below was given to Linus Pauling in 1984. Pauling was the only person to win Nobel Prizes in TWO Different fields: Chemistry and Peace.

THE EARLY HISTORY OF CHEMISTRY

• 1,000 B.C.—processing of ores to produce metals for weapons and ornaments; use of embalming fluids

• 400 B.C.—Greeks—proposed all matter was make up of 4 “elements”: fire, earth, water and air

• Democritus—first to use the term atomos to describe the ultimate, smallest particles of matter

• Next 2,000 years—alchemy—a pseudoscience where people sought to turn metals into gold. Much was learned from the plethora of mistakes alchemists made.

• 16th century—Georg Bauer, German , refined the process of extracting metals from ores & Paracelsus, Swiss, used minerals for medicinal applications

• Robert Boyle, English—first “chemist” to perform quantitative experiments of pressure versus volume. Developed a working definition for “elements”.

• 17th & 18th Centuries—Georg Stahl, German—suggested “phlogiston” flowed OUT of burning material. An object stopped burning in a closed container since the air was “saturated with phlogiston”

• Joseph Priestley, English—discovered oxygen which was originally called “dephlogisticated air”

FUNDAMENTAL CHEMICAL LAWS

• late 18th Century—Combustion studied extensively • CO2, N2, H2 and O2 discovered • list of elements continued to grow • Antione Lavoisier, French—explained the true

nature of combustion—published the first modern chemistry textbook AND stated the Law of Conservation of Mass. The French Revolution broke out the same year his text was published. He once collected taxes for the government and was executed with a guillotine as an enemy of the people in 1794. He was the

first to insist on quantitative experimentation. THE LAW OF CONSERVATION OF MASS:

Mass is neither created nor destroyed.


• 1808--John Dalton stated the Law of Definite proportions. He later went on to develop the Atomic Theory of Matter.

THE LAW OF DEFINITE PROPORTIONS: A given compound always contains exactly the same proportions of elements by mass.

THE LAW OF MULTIPLE PROPORTIONS: When two elements combine to form a series of compounds, the ratios of the masses of the second element that combine with 1 gram of the first element can always be reduced to small whole numbers. Dalton considered compounds of carbon and oxygen and determined:

Mass of Oxygen that combines with 1 gram of C

Compound I 1.33 g Compound II 2.66 g

Therefore, Compound I may be CO while Compound II may be CO2.

Exercise 1 Illustrating the Law of Multiple Proport ions The following data were collected for several compounds of nitrogen and oxygen: Mass of Nitrogen That Combines With 1 g of Oxygen Compound A 1.7500 g Compound B 0.8750 g Compound C 0.4375 g Show how these data illustrate the law of multiple proportions.

A = 1.7500 = 2 B 0.8750 1

B = 0.8750 = 2

C 0.4375 1

A = 1.750 = 4 C 0.4375 1


DALTON’S ATOMIC THEORY

Postulates of Dalton’s ATOMIC THEORY OF MATTER: (based on knowledge at that time) 1. All matter is made of atoms. These indivisible and indestructible objects are the ultimate chemical

particles. 2. All the atoms of a given element are identical, in both weight and chemical properties. However, atoms of

different elements have different weights and different chemical properties. 3. Compounds are formed by the combination of different atoms in the ratio of small whole numbers. 4. A chemical reaction involves only the combination, separation, or rearrangement of atoms; atoms are

neither created nor destroyed in the course of ordinary chemical reactions. **TWO MODIFICATIONS HAVE BEEN MADE TO DALTON’S THEORY: 1. Subatomic particles were discovered. Bet you can name them! 2. Isotopes were discovered. Bet you can define “isotope” as well!

• 1809 Joseph Gay-Lussac, French—performed experiments [at constant temperature and pressure] and measured volumes of gases that reacted with each other.

• 1811 Avogadro, Italian—proposed his hypothesis regarding Gay-Lussac’s work [and you thought he

was just famous for 6.02 × 1023] He was basically ignored, so 50 years of confusion followed. AVOGADRO’S HYPOTHESIS: At the same temperature and pressure, equal volumes of different gases contain the same number of particles.

Atoms, Molecules and Ions 3

AP* Chemistry

Stoichiometry

*AP is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

© 2013 by René McCormick. All rights reserved.

ATOMIC MASSES

12C—Carbon 12—In 1961 it was agreed that this isotope of carbon would serve as the standard used to

determine all other atomic masses and would be defined to have a mass of EXACTLY 12 atomic mass

units (amu). All other atomic masses are measured relative to this.

mass spectrometer—a device for measuring the mass of atoms or molecules

o atoms or molecules are passed into a beam of high-speed electrons

o this knocks electrons OFF the atoms or molecules transforming them into cations

o apply an electric field

o this accelerates the cations since they are repelled from the (+) pole and attracted toward the (−)

pole

o send the accelerated cations into a magnetic field

o an accelerated cation creates it’s OWN magnetic field which perturbs the original magnetic field

o this perturbation changes the path of the cation

o the amount of deflection is proportional to the mass; heavy cations deflect little

o ions hit a detector plate where measurements can be obtained.

o 13

13

12

Mass C1.0836129 Mass C (1.0836129)(12 amu) 13.003355 amu

Mass C

Exact by definition

average atomic masses—atoms have masses of whole numbers, HOWEVER samples of quadrillions of

atoms have a few that are heavier or lighter [isotopes] due to different numbers of neutrons present

percent abundance--percentage of atoms in a natural sample of the pure element represented by a

particular isotope

percent abundance = number of atoms of a given isotope × 100%

Total number of atoms of all isotopes of that element

counting by mass—when particles are small this is a matter of convenience. Just as you buy 5 lbs of

sugar rather than a number of sugar crystals, or a pound of peanuts rather than counting the individual

peanuts….this concept works very well if your know an average mass.

Stoichiometry Screencasts are available at www.apchemistrynmsi.wikispaces.com

2

mass spectrometer to determine isotopic composition—load in a pure sample of natural neon or other

substance. The areas of the “peaks” or heights of the bars indicate the relative abundances of Ne20

10 ,

Ne21

10, and Ne22

10

Exercise 1 The Average Mass of an Element

When a sample of natural copper is vaporized and injected into a mass spectrometer, the results shown in the

figure are obtained. Use these data to calculate the average mass of natural copper. (The mass values for 63

Cu

and 65Cu are 62.93 amu and 64.93 amu, respectively.)

63.55 amu/atom

THE MOLE

mole—the number of C atoms in exactly 12.0 grams of 12

C; also a number, 6.02 × 1023

just as the word

“dozen” means 12 and “couple” means 2.

Avogadro’s number—6.02 × 1023, the number of particles in a mole of anything

DIMENSIONAL ANALYSIS DISCLAIMER: I will show you some alternatives to dimensional analysis.

WHY? First, some of these techniques are faster and well-suited to the multi-step problems you will face on

the AP Exam. Secondly, these techniques better prepare you to work the complex equilibrium problems you

will face later in this course. Lastly, I used to teach both methods. Generations of successful students have

encouraged me to share these techniques with as many students as possible. They themselves did, once they got

to college, and made lots of new friends once word got out they had this “easy way” to solve stoichiometry

problems—not to mention their good grades! Give this a try. It doesn’t matter which method you use, I

encourage you to use the method that works best for you and lets you solve problems accurately and quickly!


3

ALTERNATE TECHNIQUE #1—USING THE MOLE MAP:

Simply reproduce this map on your scratch paper until you no longer need to since the image will be burned

into your brain!

MULTIPLY�[by�the�conversion�factor�on�the�ar row]�when�“traveling”�IN�THE�DIRECTION�OF�THE�ARROW�and�obviously,�divide�when�“traveling”�against�an�ar row.�


4

Exercise 2 Determining the Mass of a Sample of Atoms

Americium is an element that does not occur naturally. It can be made in very small amounts in a device known

as a particle accelerator. Calculate the mass in grams of a sample of americium containing six atoms.

2.42 × 1021

g

Exercise 3 Determining Moles of Atoms

Aluminum is a metal with a high strength-to-mass ratio and a high resistance to corrosion; thus it is often used

for structural purposes. Calculate both the number of moles of atoms and the number of atoms in a 10.0-g

sample of aluminum.

0.371 mol Al

2.23 × 1023

atoms

Exercise 4 Calculating the Number of Moles and Mass

Cobalt (Co) is a metal that is added to steel to improve its resistance to corrosion. Calculate both the number of

moles in a sample of cobalt containing 5.00 × 1020 atoms and the mass of the sample.

8.31 × 104

mol Co

4.89 × 102

g Co

MOLAR MASS AND FORMULA WEIGHT

molar mass, MM --the sum of all of the atomic masses in a given chemical formula in units of g/mol.

It is also equal mass in grams of Avogadro’s number of molecules; i.e. the mass of a mole

empirical formula--the ratio in the network for an ionic substance

formula weight--same as molecular weight, just a language problem � “molecular” implies covalent

bonding while “formula” implies ionic bonding {just consider this to be a giant conspiracy designed to

keep the uneducated from ever understanding chemistry—kind of like the scoring scheme in tennis}.

Just use “molar mass” for all formula masses.

A WORD ABOUT SIG. FIG.’s—It is correct to “pull” from the periodic table the least number of sig.

figs for your MM’s as are in your problem—just stick with 2 decimal places for all MM’s —much

simpler!


5

Exercise 5 Calculating Molar Mass I

Juglone, a dye known for centuries, is produced from the husks of black walnuts. It is also a natural herbicide

(weed killer) that kills off competitive plants around the black walnut tree but does not affect grass and other

noncompetitive plants [a concept called allelopathy]. The formula for juglone is C10H6O3.

(a) Calculate the molar mass of juglone.

(b) A sample of 1.56 × 102 g of pure juglone was extracted from black walnut husks. Calculate the number of

moles of juglone present in this sample.

a. 174.1 g

b. 8.96 × 105

mol juglone

Exercise 6 Calculating Molar Mass II

Calcium carbonate (CaCO3), also called calcite, is the principal mineral found in limestone, marble, chalk,

pearls, and the shells of marine animals such as clams.

(a) Calculate the molar mass of calcium carbonate.

(b) A certain sample of calcium carbonate contains 4.86 moles. Calculate the mass in grams of this sample.

Calculate the mass of the CO32

ions present.

a. 100 g/mol

b. 486 g; 292g CO32


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Exercise 7 Molar Mass and Numbers of Molecules

Isopentyl acetate (C7H14O2), the compound responsible for the scent of bananas, can be produced commercially.

Interestingly, bees release about 1µg (1 × 10-6 g) of this compound when they sting. The resulting scent attracts

other bees to join the attack.

(a) Calculate the number of molecules of isopentyl acetate released in a typical bee sting.

(b) Calculate the number of carbon atoms present.

5 × 10

15 molecules

4 × 1016

carbon atoms

ELEMENTS THAT EXIST AS MOLECULES

Pure hydrogen, nitrogen, oxygen and the halogens exist as DIATOMIC molecules under normal conditions. MEMORIZE!!! Be sure you compute their molar masses as diatomics. We lovingly refer to them as the “gens”,

“Hydrogen, oxygen, nitrogen & the halogens!”

Others to be aware of, but not memorize:

P4—tetratomic form of elemental phosphorous; an allotrope

S8—sulfur’s elemental form; also an allotrope

Carbon—diamond and graphite � covalent networks of atoms

PERCENT COMPOSITION OF COMPOUNDS

There are two common ways of describing the composition of a compound: 1) in terms of the number of its

constituent atoms and 2) in terms of the percentages (by mass) of its elements.

Percent Composition (by mass): The Law of Constant Composition states that any sample of a pure compound always consists of the same elements combined in the same proportions by mass. Remember, all

percent calculations are simply part

100%whole

% comp = mass of desired element × 100%

total mass of compound

Consider ethanol, C2H5OH

Mass of C = 2 mol × mol

g01.12 = 24.02 g

Mass of H = 6 mol × mol

g01.1 = 6.06 g

Mass of O = 1 mol × mol

g00.16 = 16.00g

Mass of 1 mol of C2H5OH = 46.08 g


7

NEXT, THE MASS PERCENT CAN BE CALCULATED:

Mass percent of C = 24.02 g C × 100% = 52.14%

46.08 g

Repeat for the H and O present.

Exercise 8 Calculating Mass Percent I

Carvone is a substance that occurs in two forms having different arrangements of the atoms but the same

molecular formula (C10H14O) and mass. One type of carvone gives caraway seeds their characteristic smell, and

the other type is responsible for the smell of spearmint oil. Calculate the mass percent of each element in

carvone.

C = 79.96%

H = 9.394%

O = 10.65%

Exercise 9 Calculating Mass Percent II

Penicillin, the first of a now large number of antibiotics (antibacterial agents), was discovered accidentally by

the Scottish bacteriologist Alexander Fleming in 1928, but he was never able to isolate it as a pure compound.

This and similar antibiotics have saved millions of lives that might have been lost to infections. Penicillin F has

the formula C14H20N2SO4. Calculate the mass percent of each element.

C = 53.82%

H = 6.47%

N = 8.97%

S = 10.26%

O = 20.49%


8

DETERMINING THE FORMULA OF A COMPOUND

When faced with a hydrocarbon compound of “unknown” formula, one of the most common techniques is to

combust it with oxygen to produce oxides of the nonmetals CO2 and H2O which are then collected and

weighed.

Calculating empirical and molecular formulas: empirical formulas represent the simplest or smallest ratio of elements within a compound while molecular formulas represent the actual numbers of elements within a compound. The empirical mass is the least common multiple of the molar mass.

Example: CH2O is the empirical for a carbohydrate—get it? “carbon waters”.

Anyway, glucose is a perfect example of a carbohydrate (a sugar to be exact) with an empirical molar

mass of 12 + 2(1) + 16 = 30 g/mol and since glucose is 6 units of CH2O which is equivalent to (CH2O)6

or C6H12O6; the empirical mass of 30 is also multiplied by 6. Thus the MM of glucose is 180 g/mol.

Make your problem solving life easy and assume a 100 gram sample if given %’s—that way you can

convert the percents given directly into grams and subsequently into moles in order to simplify your life!

Other twists and turns occurring when calculating molar masses involve:

hydrates—waters of hydration or “dot waters”. They count in the calculation of molar masses for

hydrates and used to “cement” crystal structures together

anhydrous—means without water—just to complete the story—just calculate the molar masses of

anhydrous substances as you would any other substance

Example:

A compound is composed of carbon, nitrogen and hydrogen. When 0.1156 g of this compound is reacted with

oxygen [a.k.a. “burned in air” or “combusted”], 0.1638 g of carbon dioxide and 0.1676 g of water are collected.

Determine the empirical formula of the compound.

So, Compound + O2 oxides of what is burned. In this case Compound + O2 CO2 + H2O + N2

(clearly not balanced)

You can see that all of the carbon ended up in CO2 so…when in doubt, calculate THE NUMBER OF MOLES!!

0.1638 g CO2 ÷ 44.01 g/mol CO2 = 0.003781 moles of CO2 = 0.003781 moles of C (why?)


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Next, you can see that all of the hydrogen ended up in H2O, so….calculate THE NUMBER OF MOLES!!

So, 0.1676 g H2O ÷ 18.02 g/mol H2O = 0 .009301 moles of H2O, BUT there are 2 moles of H for each mole of

water [ Think “organ bank” one heart per body, one C per molecule of carbon dioxide while there are 2 lungs

per body, 2 atoms H in water and so on…] thus, DOUBLE THE NUMBER OF MOLES of H2O GIVES THE

NUMBER OF MOLES OF HYDROGEN!! moles H = 2 × 0 .009301 moles of H2O = 0.01860 moles of H

Therefore, the remaining mass must be nitrogen, BUT we only have mass data for the sample so convert your

moles of C and H to grams:

grams C = 0.003781 moles C × 12.01 g

mol = 0.04540 grams C

grams H = 0.01860 moles H × 1.01 g

mol = 0.01879 grams H

Total grams : 0.06419 total grams accounted for thus far

What to do next? SUBTRACT!

0.1156 g sample – 0.06419 total grams accounted for thus far = grams N left = 0.05141 g N so….

0.05141 g N ÷ 14.01 g

mol = 0.003670 moles N

Next, realize that chemical formulas represent mole to mole ratios, so…divide the number of moles of each by

the smallest # of moles for any one of them to get a guaranteed ONE in your ratios…multiply by 2, then 3, etc

to get to a ratio of small whole numbers. Clear as mud? WATCH THE SCREENCAST!!

Element # moles ALL Divided by the

smallest (0.003670

moles)

C 0.003781 1

H 0.01860 5

N 0.003670 1

Therefore, the correct EMPIRICAL formula based on the data given is CH5N.

Finally (this is drumroll worthy), IF we are told that the MM of the original substance is 31.06 g/mol, then

simply use this relationship:

(Empirical mass) × n = MM

(12.01 + 5.05 + 14.01) × n = 31.07 g/mol n = 0.999678

This is mighty close to 1.0! Thus, the empirical formula and the molecular formula are one and the same.


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Exercise 10

Determine the empirical and molecular formulas for a compound that gives the following analysis in mass

percents:

71.65% C1 24.27% C 4.07% H

The molar mass is known to be 98.96 g/mol.

Empirical formula = CH2 C1

Molecular formula = C2H4C12

Exercise 11

A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. The

compound has a molar mass of 283.88 g/mol. What are the compound’s empirical and molecular formulas?

Empirical formula = P2O5

Molecular formula = (P2O5)2 or P4O10


11

Exercise 12

Caffeine, a stimulant found in coffee, tea, and chocolate, contains 49.48% carbon, 5.15% hydrogen,

28.87% nitrogen, and 16.49% oxygen by mass and has a molar mass of 194.2 g/mol. Determine the molecular

formula of caffeine.

Molecular formula = C8H10N4O2

BALANCING CHEMICAL EQUATIONS

Chemical reactions are the result of a chemical change where atoms are reorganized into one or more new

arrangements. Bonds are broken [requires energy] and new ones are formed [releases energy]. A chemical

reaction transforms elements and compounds into new substances. A balanced chemical equation shows the

relative amounts of reactants [on the left] and products [on the right] by molecule or by mole.

Subtle details:

s, l, g, aq—state symbols that correspond to solid, liquid, gas,

aqueous solution

NO ENERGY or TIME is alluded to

Antoine Lavoisier (1743-1794)—The Law of Conservation of Matter: matter can be neither created nordestroyed � this means you having to “balance equations” is entirely his fault!!

BALANCING CHEMICAL EQUATIONS

Begin with the most complicated-looking thing (often the scariest, too).

Save the elemental thing for last.

If you get stuck, double the most complicated-looking thing.

MEMORIZE THE FOLLOWING:

metals + halogens � MaXb

CH and/or O + O2 � CO2(g) + H2O(g)

H2CO3 [any time formed!] � CO2 + H2O; in other words, never write carbonic acid as a

product, it spontaneously decomposes [in an open container] to become carbon dioxide andwater

metal carbonates � metal OXIDES + CO2


12

Exercise 13

Chromium compounds exhibit a variety of bright colors. When solid ammonium dichromate, (NH4)2Cr2O7,

a vivid orange compound, is ignited, a spectacular reaction occurs. Although the reaction is actually somewhat

more complex, let’s assume here that the products are solid chromium(III) oxide, nitrogen gas (consisting of N2

molecules), and water vapor. Balance the equation for this reaction.

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g)

(4 × 2) H (4 ×2) H

http://www.youtube.com/watch?v=CW4hN0dYnkM

Exercise 14

At 1000ºC, ammonia gas, NH3(g), reacts with oxygen gas to form gaseous nitric oxide, NO(g), and water vapor.

This reaction is the first step in the commercial production of nitric acid by the Ostwald process. Balance the

equation for this reaction.

4 NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)


13

STOICHIOMETRIC CALCULATIONS: AMOUNTS OF REACTANTS AND PRODUCTS

Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions.

Stoichiometry is the most important thing you can learn as you embark upon AP Chemistry! Get good at

this and you will do well all year. This NEVER goes away!

It’s time to repeat my dimensional analysis disclaimer.

DIMENSIONAL ANALYSIS DISCLAIMER: I will show you some alternatives to dimensional analysis. WHY? First, some of

these techniques are faster and well-suited to the multi-step problems you will face on the AP Exam. Secondly, these techniques better

prepare you to work the complex equilibrium problems you will face later in this course. The first problem you must solve in the free

response section of the AP Exam will be an equilibrium problem and you will need to be able to work them quickly. Lastly, I used to

teach both methods. Generations of successful students have encouraged me to share these techniques with as many students as

possible. They did, once they got to college, and made lots of new friends once word got out they had this “cool way” to solve

stoichiometry problems—not to mention their good grades! Give this a try. It doesn’t matter which method you use, I encourage you

to use the method that works best for you and lets you solve problems accurately and quickly!

First you have to be proficient at the following no matter which method you choose!:

Writing CORRECT formulas—this requires knowledge of your polyatomic ions and being able to use

the periodic table to deduce what you have not had to memorize. Review section 2.8 in your Chapter 2

notes or your text.

Calculate CORRECT molar masses from a correctly written formula

Balance a chemical equation

Use the mole map to calculate the number of moles or anything else!

Remember the mole map? It will come in mighty handy as well!


14

Here’s the “template” for solving the problems…you’ll create a chart. Here’s a typical example:

Example: Calculate the mass of oxygen will react completely with 96.1 grams of propane?

[notice all words—you supply the chemical formulas!]

Molar Mass: (44.11) (32.00) (44.01) (18.02) Balanced Eq’n C3H8 + 5 O2 � 3 CO2 + 4 H2O

mole:mole 1 5 3 4 # moles

Amount

1. Write a chemical equation paying special attention to writing correct chemical formulas!

2. Calculate the molar masses and put in parentheses above the formulas—soon you’ll figure out you don’t

have to do this for every reactant and product, just those in which you are interested.

3. Balance the equation! Examine the coefficients on the balanced equation, they ARE the mole:mole

ratios! Isolating them helps you internalize the mol:mol until you get the hang of this.

4. Next, re-read the problem and put in an amount—in this example it’s 96.1 g of propane.

Molar Mass: (44.11) (32.00) (44.01) (18.02)

Balanced

Eq’n C3H8 + 5 O2 � 3 CO2 + 4 H2O

mole:mole 1 5 3 4 # moles 2.18 10.9 6.53 8.71 amount 96.1 grams

5. Calculate the number of moles of something, anything! Use the mole map. Start at 96.1 grams of C3H8,

divide the 96.1 g [against the arrow on the mole map] by molar mass to calculate the # moles of

propane.

6. USE the mole: mole to find moles of EVERYTHING! If 1 = 2.18 then oxygen is 5(2.18) etc…. [IF the

first mol amount you calculate is not a “1”, just divide appropriately to make it “1” before moving on to

calculate the moles of all the rest!] Leave everything in your calculator—I only rounded to save space!

7. Re-read the problem to determine which amount was asked for…here’s the payoff….AP problems ask

for several amounts! First, we’ll find the mass of oxygen required since that’s what the problem asked.

10.9 moles × 32.00 g/mol = 349 g of oxygen


15

Now, humor me…What if part (b) asked for liters of CO2 at STP [1 atm, 273K]?

Use the mole map. Start in the middle with 6.53 moles × [in direction of arrow] 22.4 L/mol = 146 L

What if part (c) asked you to calculate how many water molecules are produced?

Use the mole map , start in the middle with 8.71 mol water × 6.02 × 1023

molecules

mol = 5.24 × 10

24 molecules

of water.

Try these two exercises with whichever method you like best!

Exercise 15

Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living

environment by forming solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can be

absorbed by 1.00 kg of lithium hydroxide?

920. g

Molar Mass: (44.11) (32.00) (44.01) (18.02)

Balanced

Eq’n C3H8 + 5 O2 � 3 CO2 + 4 H2O

mole:mole 1 5 3 4 # moles 2.18 10.9 6.53 8.71

amount 96.1 grams 349 g 146 L


16

Exercise 16

Baking soda (NaHCO3) is often used as an antacid. It neutralizes excess hydrochloric acid secreted by the

stomach:

NaHCO3(s) + HC1(aq) → NaC1(aq) + H2O(l) + CO2(aq)

Milk of magnesia, which is an aqueous suspension of magnesium hydroxide, is also used as an antacid:

Mg(OH)2(s) + 2HC1(aq) → 2H2O(l) + MgC12(aq)

Which is the more effective antacid per gram, NaHCO3 or Mg(OH)2? Justify your answer.

Mg(OH)2

CALCULATIONS INVOLVING A LIMITING REACTANT

Ever notice how hot dogs are sold in packages of 10 while the buns come in packages of 8? What’s up with

that?! The bun is the limiting reactant and limits the hot dog production to 8 as well! The limiting reactant [or

reagent] is the one consumed most entirely in the chemical reaction.

Let’s use a famous process [meaning one the AP exam likes to ask questions about!], the Haber process.

This reaction is essentially making ammonia for fertilizer production from the nitrogen in the air reacted with

hydrogen gas. The hydrogen gas is obtained from the reaction of methane with water vapor. This process has

saved millions from starvation!! The reaction is shown below.

Exercise17

Examine the particle views and explain the differences between the two situations pictured below with regard to

what is or is not reacting and total yield of ammonia.

N2(g) + 3H2(g) 2NH3(g)

Situation 1 Situation 2


17

Plan of attack: First, you must realize that you even need a plan of attack! IF ever you are faced with TWO

starting amounts of matter reacting, you have entered “The Land of Limiting Reactant”.

When faced with this situation …calculate the number of moles of everything you are given. Set up your table

like before, only now you’ll have TWO amounts and thus TWO # ‘s of moles to get you started.

Cover one set of moles up (pretending you only had one amount to work from) and ask yourself, “What if all of

these moles reacted?” “How many moles of the other reactants would I need to use up all of these moles?”

Next, do the calculation of how many moles of the “other” amount(s) you would need. Do you have enough?

If so, the reactant you began with IS the limiting reactant. If not repeat this process with the “other” reactant

amount you were given.

It doesn’t matter where you start the “What if?” game….you get there either way.

Clear as mud? Read on…(and consider listening to the SCREENCAST!)

Let’s revisit the Haber process:

Molar Mass: (28.02) (2.02) (17.04)

Balanced Eq’n N2 + 3 H2 � 2 NH3

mole:mole 1 3 2 # moles

amount

Suppose 25.0 kg of nitrogen reacts with 5.00 kg of hydrogen to form ammonia. What mass of ammonia can be

produced? Which reactant is the limiting reactant? What is the mass of the reactant that is in excess?

**Insert the masses in the amount row and find the number of moles of BOTH! Molar Mass: (28.02) (2.02) (17.04)

Balanced

Eq’n N2 + 3 H2 � 2 NH3

mole:mole 1 3 2 # moles 892 moles 2,475 moles

amount 25,000 g 5,000 g

WHAT IF I used up all the moles of hydrogen? I’d need 1/3 × 2,475 moles = 825 moles of nitrogen.

Clearly I have EXCESS moles of nitrogen!! Therefore, hydrogen limits me.

OR WHAT IF I used up all the moles of nitrogen? I’d need 3 × 892 moles = 2,676 moles of hydrogen.

Clearly I don’t have enough hydrogen, so it limits me!! Therefore nitrogen is in excess.

Continued on next page.


18

Either way, I’ve established that hydrogen is the limiting reactant so I modify the table:

In English, that means I’ll use up all the hydrogen but not all the nitrogen!

Molar Mass: (28.02) (2.02) (17.04)

Balanced Eq’n N2 + 3 H2 � 2 NH3

mole:mole 1 3 2

# moles 825 mol used

892 moles

2,475 moles

1650 mol

produced

amount 825 mol (28.02) =

23,116 g used

25,000 g

5,000 g

1650 mol (17.04)

= 28,116 g

produced

1,884 g excess!!

Here’s the question again, let’s clean up any sig.fig issues:

Suppose 25.0 kg of nitrogen reacts with 5.00 kg of hydrogen to form ammonia. (3 sig. fig. limit)

What mass of ammonia can be produced? 28,100 g produced = 28.1 kg

(It is always polite to respond in the unit given).

Which reactant is the limiting reactant? Hydrogen—once that’s established, discard the nitrogen amounts and

let hydrogen be your guide!

What is the mass of the reactant that is in excess? 1,884 g = 1.88 kg excess nitrogen!!

Exercise 18

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures.

The other products of the reaction are solid copper and water vapor. If a sample containing 18.1 g of NH3 is

reacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N2 will be formed?

CuO is limiting; 10.6 g N2


19

Theoretical Yield: The amount of product formed when a limiting reactant is completely consumed. This

assumes perfect conditions and gives a maximum amount!! Not likely!

Actual yield: That which is realistic.

Percent yield: The ratio of actual to theoretical yield.

Actual Yield × 100% = Percent yield

Theoretical Yield

Exercise 19

Methanol (CH3OH), also called methyl alcohol, is the simplest alcohol. It is used as a fuel in race cars and is a

potential replacement for gasoline. Methanol can be manufactured by combination of gaseous carbon monoxide

and hydrogen. Suppose 68.5 kg CO(g) is reacted with 8.60 kg H2(g). Calculate the theoretical yield of

methanol. If 3.57 × 104 g CH3OH is actually produced, what is the percent yield of methanol?

Theoretical yield is 6.82 × 104 g

Percent yield is 52.3%

(1) AP® is a registered trademark of the College Board. The College Board was not involved in the production of and does not endorse this product. (2) Test Questions are Copyright © 1962-2008 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Web or Mass distribution prohibited.

WRITING AP* NET IONIC EQUATIONS

AP equation sets are found in the free-response section of the AP test. You are given eight equations and you must choose to answer five of these.** The equations are of mixed types. The section is worth 15 points and is 15 % of the free response grade. Free response is 55% of the total AP test grade.

All AP equations "work". In each case, a reaction will occur. These equations

need to be written in net ionic form. All spectator ions must be left out and all ions must be written in ionic form. All molecular substances and nonsoluble compounds must be written together (not ionized!). Weak electrolytes, such as acetic acid, are not ionized. Solids and pure liquids are written together, also. A saturated solution is written in ionic form while a suspension is written together. AP equations do not need to be balanced. Don't waste your time on balancing!

Each equation is worth a total of 3 points. One point is given for the correct

reactants and two points for all correct products. If a reaction has three products, one point is given for two correct products and two points for all correct products. Leaving in the spectator ions will result in a one point deduction on the equation set (not 1 point per problem).

The best way to prepare for the equation section of the AP test is to practice lots

of equations. The equation sets are similar and some equations show up year after year. When you are reading an equation, first try to classify it by type. If it says anything about acidic or basic solution, it is redox. If you are totally stuck, look up the compounds in the index of your book or other reference books and try to find information that will help you with the equation. All reactions do not fit neatly into the five types of reactions that you learned in Chemistry I. Save the reactions that you write and practice them again before the AP test in May.

Kristen Henry Jones 01/28/99 ** There was a format change in 2007. Now students are given 3 equations and only 3 equations, each worth 5 points. They must write the net ionic equation (3 points total: 1 point for set of reactants and 2 points for products), balance the net ionic equation ( the 4th point), and answer a descriptive question about the reaction (the 5th and final point). The methods presented in this document still hold true, but be sure and require students to balance the equations by mass and charge even though they are not balanced on the key!

Kristen Henry Jones 01/28/99 Used with permission.

Double Replacement (metathesis)

Two compounds react to form two new compounds. No changes in oxidation numbers occur. All double replacement reactions must have a "driving force" that removes a pair of ions from solution.

Formation of a precipitate: A precipitate is an insoluble substance formed by the reaction of two aqueous substances. Two ions bond together so strongly that water can not pull them apart. You must know your solubility rules to write these net ionic equations (see next page)

Ex. Solutions of silver nitrate and lithium bromide are mixed.

Ag + Br AgBr

Formation of a gas: Gases may form directly in a double replacement reaction or can form from the decomposition of a product such as H2CO3 or H2SO3. Common gasses: CO2, SO2, SO3, H2S, NO2, NH3, O2, H2

Ex. Excess hydrochloric acid solution is added to a solution of potassium sulfite.

H+ + SO32- H20 + SO2

Ex. A solution of sodium hydroxide is added to a solution of ammonium chloride.

OH- + NH4+ NH3 + H2O

Formation of a molecular substance: When a molecular substance such as water or acetic acid is formed, ions are removed from solution and the reaction "works".

Ex. Dilute solutions of lithium hydroxide and hydrobromic acid are mixed.

OH- + H+ H2O (HBr, HCI, and HI are strong acids)

Ex. Gaseous hydrofluoric acid reacts with solid silicon dioxide.

HF + SiO2 SiF4 + H2O

Writing Net Ionic Equations 3


Solubility Rules These are strong electrolytes (100% ionized ) and written as ions 1. Strong Acids: HCl, HBr, HI, H2SO4, HNO3, HClO4, HClO3 2. Strong Bases: Hydroxides of group IA and IIA(Ba, Sr, Ca are marginal Be and Mg are WEAK) 3. Soluble Salts (see table): (ionic compounds: metal/nonmetal)

ALWAYS SOUBLE IF IN A COMPOUND

EXCEPT WITH

NO3-, Group IA, NH4

+, C2H3O2-, ClO4

-, ClO3- No Exceptions

Cl- , Br- , I- Pb, Ag, Hg22+

SO42- Pb, Ag, Hg2

2+

Ca, Sr, Ba If it does not fit one of the three rules above, assume it is INSOLUBLE or a WEAK ELECTROLYTE (and written together). This won’t always be correct, but will cover most of the situations. Also, GASES, PURE LIQUIDS, and SOLIDS are non-electrolytes. Remember H2CO3 decomposes into H2O(l) and CO2(g) Remember NH4OH decomposes into H2O(l) and NH3(g)

Remember H2SO3 decomposes into H2O and SO2



Single Replacement

Reaction where one element displaces another in a compound. One element is oxidized and another is reduced.

A + BC B + AC Active metals replace less active metals or hydrogen from their compounds in aqueous solution. Use an activity series or a reduction potential table to determine activity. The more easily oxidized metal replaces the less easily oxidized metal. The metal with the most negative reduction potential will be the most active. Ex. Magnesium turnings are added to a solution of iron(III) chloride.

Mg + Fe 3+ Fe + Mg2+

Ex. Sodium is added to water.

Na + H2O Na+ + OH- + H2

Active nonmetals replace less active nonmetals from their compounds in aqueous solution. Each halogen will displace less electronegative (heavier) halogens from their binary salts. Ex. Chlorine gas is bubbled into a solution of potassium iodide.

C12 + I- I2 + Cl- Tricky redox reactions that appear to be ordinary single replacement reactions: • Hydrogen reacts with a hot metallic oxide to produce the elemental metal and water.

Ex. Hydrogen gas is passed over hot copper(II) oxide.

H2 + CuO Cu2+ + H2O • A metal sulfide reacts with oxygen to produce the metallic oxide and sulfur dioxide. • Chlorine gas reacts with dilute sodium hydroxide to produce sodium hypochlorite,

sodium chloride and water.



• Copper reacts with concentrated sulfuric acid to produce copper(II) sulfate, sulfur

dioxide, and water. • Copper reacts with dilute nitric acid to produce copper(II) nitrate, nitrogen monoxide

and water. • Copper reacts with concentrated nitric acid to produce copper(II) nitrate, nitrogen

dioxide and water


Anhydrides • Anhydride means "without water" • Water is a reactant in each of these equations.

Nonmetallic oxides (acidic anhydrides) plus water yield acids.

Ex. Carbon dioxide is bubbled into water.

CO2 + H2O H2CO3 Metallic oxides (basic anhydrides) plus water yield bases. Ex. Solid sodium oxide is added to water.

Na2O + H2O Na+ + OH-

Metallic hydrides (ionic hydrides) plus water yield metallic hydroxides and hydrogen gas. Ex. Solid sodium hydride is added to water.

NaH + H2O Na+ + OH- + H2 Phosphorus halides react with water to produce an acid of phosphorus (phosphorous acid or phosphoric acid) and a hydrohalic acid. The oxidation number of the phosphorus remains the same in both compounds. Phosphorus oxytrichloride reacts with water to make the same products. Ex. Phosphorus tribromide is added to water.

PBr3 + H2O H3PO3 + H+ + Br- Group I&II nitrides react with water to produce the metallic hydroxide and ammonia.


OXIDATION-REDUCTION REACTIONS

Redox reactions involve the transfer of electrons. The oxidation numbers of at least two elements must change. Single replacement, some combination and some decomposition reactions are redox reactions.

To predict the products of a redox reaction, look at the reagents given to see if there is both an oxidizing agent and a reducing agent. When a problem mentions an acidic or basic solution, it is probably redox.

Common oxidizing agents Products formed MnO4

- in acidic solution Mn2+ MnO2 in acidic solution Mn2+ MnO4

- in neutral or basic solution MnO2 (s) Cr2O7

- in acidic solution Cr3+ HNO3, concentrated NO2 HNO3, dilute NO H2SO4, hot, concentrated SO2 metal-ic ions metal-ous ions free halogens halide ions Na2O2 NaOH HC1O4 Cl- H2O2 H2O Common reducing agents Products formed halide ions free halogen free metals metal ions sulfite ions or SO2 sulfate ions nitrite ions nitrate ions free halogens, dilute basic solution hypohalite ions free halogens, conc. basic solution halate ions metal-ous ions metal-ic ions H2O2 O2 C2O4

2- CO2

Ex. A solution of tin(II) chloride is added to an acidified solution of potassium permanganate. Sn 2+ + H+ + MnO4

- Sn 4+ + Mn2+ + H2O

Ex. A solution of potassium iodide is added to an acidified solution of potassium dichromate. I- + H+ + Cr2O7

- Cr3+ + I2 + H2O Ex. Hydrogen peroxide solution is added to a solution of iron(II) sulfate.

H2O2 + Fe2+ Fe3+ + H2O

Ex. A piece of iron is added to a solution of iron(III) sulfate. Fe + Fe3+ Fe2+


COMPLEX ION REACTIOINS

Complex ion- the combination of a central metal ion and its ligands

Ligand- group bonded to a metal ion

Coordination compound- a neutral compound containing complex ions [Co(NH3)6]Cl3

(NH3 is the ligand, [Co(NH3)6]3+ is the complex ion)

Common complex ions formed in AP equations: Complex ion Name Formed from: [AI(OH)4]

- tetrahydroxoaluminate ion (Al or Al(OH)3 or Al 3+ + OH-)

[Al(H 2O)6]3+ hexaquaaluminum ion (Al3+ in H2O)

[Ag(NH3)2]+ dimminesilver(I) ion (Ag+ + NH3)

[Zn(OH)4]2- tetrahydroxozincate ion (Zn(OH)2 + OH-)

[Zn(NH3)4]

2+ tetramminezinc ion (Zn2+ + NH3)

[Cu(NH3)4]2+

tetramminecopper(II) ion (Cu+2 + NH3)

[Cd(NH3)4]2+

tetramminecadmium(II) ion (Cd2+ + NH3)

[FeSCN]+2 thiocyanoiron(III) ion (Fe +3 + SCN-)

[Ag(CN)2]- dicyanoargentate (I) ion (Ag+ + CN-)

Adding an acid to a complex ion will break it up. If HCl is added to a silver complex, AgCl(s) is formed. If an acid is added to an ammonia complex, NH4

+ is formed.



Kristen Henry Jones 01/28/99

DECOMPOSITION REACTIONS

Reactions where a compound breaks down into two or more elements or compounds. Heat, electrolysis, or a catalyst is usually necessary.

A compound may break down to produce two elements.

Ex. Molten sodium chloride is electrolyzed.

NaCl Na+ + C12 A compound may break down to produce an element and a compound. Ex. A solution of hydrogen peroxide is decomposed catalytically. H2O2 H2O + O2

A compound may break down to produce two compounds. Ex. Solid magnesium carbonate is heated. MgCO3 Mg + CO2 Metallic carbonates break down to yield metallic oxides and carbon dioxide, Metallic chlorates break down to yield metallic chlorides and oxygen. Hydrogen peroxide decomposes into water and oxygen.

Ammonium carbonate decomposes into ammonia, water and carbon dioxide.

Sulfurous acid decomposes into water and sulfur dioxide.

Carbonic acid decomposes into water and carbon dioxide.



ADDITION REACTIONS

Two or more elements or compounds combine to form a single product. A group IA or IIA metal may combine with a nonmetal to make a salt. Ex. A piece of lithium metal is dropped into a container of nitrogen gas. Li + N2 Li3N Two nonmetals may combine to form a molecular compound. The oxidation number of the less electronegative element is often variable depending upon conditions. Generally, a higher oxidation state of one nonmetal is obtained when reacting with an excess of the other nonmetal. Ex. P4 + 6 Cl2 4 PCl3 limited Cl P4 + 10 Cl2 4 PCl5 excess Cl When an element combines with a compound, you can usually sum up all of the elements on the product side. Ex. PCl3 + C12 PCl5 Two compounds combine to form a single product. Ex. Sulfur dioxide gas is passed over solid calcium oxide. SO2 + CaO CaSO3 Ex. The gases boron trifluoride and ammonia are mixed. BF3 + NH3 H3NBF3 A metallic oxide plus carbon dioxide yields a metallic carbonate. (Carbon keeps the same oxidation state) A metallic oxide plus sulfur dioxide yields a metallic sulfite. (Sulfur keeps the same oxidation state) A metallic oxide plus water yields a metallic hydroxide. A nonmetallic oxide plus water yields an acid.



ACID-BASE NEUTRALIZATION REACTIONS

Acids react with bases to produce salts and water. One mole of hydrogen ions react with one mole of hydroxide ions to produce one mole of water. Watch out for information about quantities of each reactant! Remember which acids are strong (ionize completely) and which are weak (write as molecule). Sulfuric acid (strong acid) can be written as H+ and SO4

2- or as H+ and HSO4-.

Ex. A solution of sulfuric acid is added to a solution of barium hydroxide until the same number of moles of each compound as been added.

H+ + SO42- + Ba2+ + OH- BaSO4 + H2O

Ex. Hydrogen sulfide gas is bubbled through excess potassium hydroxide solution. H2S + OH- H2O + S2- Watch out for substances that react with water before reacting with an acid or a base. These are two-step reactions. Ex. Sulfur dioxide gas is bubbled into an excess of a saturated solution of calcium hydroxide. SO2 + Ca2+ + OH- CaSO3 + H2O



COMBUSTION REACTIONS

Elements or compounds combine with oxygen.

Hydrocarbons or alcohols combine with oxygen to form carbon dioxide and water.

Ammonia combines with limited oxygen to produce NO and water and with excess oxygen to produce N02 and water.

Nonmetallic hydrides combine with oxygen to form oxides and water.

Nonmetallic sulfides combine with oxygen to form oxides and sulfur dioxide.

Ex. Carbon disulfide vapor is burned in excess oxygen.

CS2 + O2 CO2 + SO2 Ex. Ethanol is burned completely 'in air.

C2H5OH + O2 CO2 + H2O

Adapted from an original document by Kristen Henry Jones

AP CHEMISTRY EQUATIONS BY TYPE

Double Replacement 1. Hydrogen sulfide is bubbled through a solution of silver nitrate. 2. An excess of sodium hydroxide solution is added to a solution of magnesium nitrate. 3. Solutions of sodium iodide and lead nitrate are mixed. 4. A solution of ammonia is added to a solution of ferric chloride. 5. Solutions of silver nitrate and sodium chromate are mixed. 6. Excess silver acetate is added to a solution of trisodium phosphate. 7. Manganese(II) nitrate solution is mixed with sodium hydroxide solution. 8. A saturated solution of calcium hydroxide is added to a solution of magnesium chloride. 9. Hydrogen sulfide gas is added to a solution of cadmium nitrate. 10. Dilute sulfuric acid is added to a solution of barium acetate. 11. A precipitate is formed when solutions of trisodium phosphate and calcium chloride are mixed. 12. A solution of copper(II) sulfate is added to a solution of barium hydroxide. 13. Equal volumes of dilute equimolar solutions of sodium carbonate and hydrochloric acid are

mixed. 14. Solid barium peroxide is added to cold dilute sulfuric acid. 15. Excess hydrochloric acid solution is added to a solution of potassium sulfite. 16. Dilute sulfuric acid is added to a solution of barium chloride. 17. A solution of sodium hydroxide is added to a solution of ammonium chloride. 18. Dilute hydrochloric acid is added to a solution of potassium carbonate. 19. Gaseous hydrogen sulfide is bubbled through a solution of nickel(II) nitrate. 20. A solution of sodium sulfide is added to a solution of zinc nitrate. 21. Concentrated hydrochloric acid is added to solid manganese(II) sulfide. 22. Solutions of tri-potassium phosphate and zinc nitrate are mixed. 23. Dilute acetic acid solution is added to solid magnesium carbonate. 24. Gaseous hydrofluoric acid reacts with solid silicon dioxide. 25. Equimolar amounts of trisodium phosphate and hydrogen chloride, both in solution, are mixed. 26. Ammonium chloride crystals are added to a solution of sodium hydroxide. 27. Hydrogen sulfide gas is bubbled through a solution of lead(II) nitrate. 28. Solutions of silver nitrate and sodium chromate are mixed. 29. Solutions of sodium fluoride and dilute hydrochloric acid are mixed. 30. A saturated solution of barium hydroxide is mixed with a solution of iron(III) sulfate. 31. A solution of ammonium sulfate is added to a potassium hydroxide solution. 32. A solution of ammonium sulfate is added to a saturated solution of barium hydroxide. 33. Dilute sulfuric acid is added to solid calcium fluoride. 34. Dilute hydrochloric acid is added to a dilute solution of mercury(I) nitrate. 35. Dilute sulfuric acid is added to a solution of lithium hydrogen carbonate. 36. Dilute hydrochloric acid is added to a solution of potassium sulfite. 37. Carbon dioxide gas is bubbled through water containing a suspension of calcium carbonate. 38. Excess concentrated sulfuric acid is added to solid calcium phosphate. 39. Hydrogen sulfide gas is bubbled into a solution of mercury(II) chloride. 40. Solutions of zinc sulfate and sodium phosphate are mixed. 41. Solutions of silver nitrate and lithium bromide are mixed.


42. Solutions of manganese(II) sulfate and ammonium sulfide are mixed. 43. Excess hydrochloric acid solution is added to a solution of potassium sulfite. Single Replacement 1. A piece of aluminum metal is added to a solution of silver nitrate. 2. Aluminum metal is added to a solution of copper(II) chloride. 3. Hydrogen gas is passed over hot copper(II) oxide. 4. Small chunks of solid sodium are added to water. 5. Calcium metal is added to a dilute solution of hydrochloric acid. 6. Magnesium turnings are added to a solution of iron(III) chloride. 7. Chlorine gas is bubbled into a solution of sodium bromide. 8. A strip of magnesium is added to a solution of silver nitrate. 9. Solid calcium is added to warm water. 10. Liquid bromine is added to a solution of potassium iodide. 11. Chlorine gas is bubbled into a solution of potassium iodide. 12. Lead foil is immersed in silver nitrate solution. 13. Solid zinc strips are added to a solution of copper(II) sulfate. 14. Sodium metal is added to water. 15. A bar of zinc metal is immersed in a solution of copper(II) sulfate. 16. A small piece of sodium metal is added to distilled water. Anhydrides 1. Excess water is added to solid calcium hydride. 2. Solid lithium hydride is added to water. 3. Liquid phosphorus trichloride is poured into a large excess of water. 4. Solid sodium carbide is added to an excess of water. 5. Solid magnesium nitride is added to excess deuterium oxide. 6. Water is added to a sample of pure phosphorus tribromide. 7. Water is added to a sample of pure sodium hydride. 8. Dinitrogen trioxide gas is bubbled into water. 9. Solid phosphorus pentachloride is added to excess water. 10. Solid dinitrogen pentoxide is added to water. 11. Sulfur trioxide gas is added to excess water. 12. Solid sodium oxide is added to water. 13. Phosphorus(V) oxytrichloride is added to water. 14. Water is added to a sample of solid magnesium nitride. 15. Solid potassium oxide is added to water. 16. Solid sodium cyanide is added to water. 17. Trisodium phosphate crystals are added to water. 18. Solid lithium oxide is added to excess water. 19. Solid barium oxide is added to distilled water. 20. Solid calcium hydride is added to distilled water. 21. Phosphorous(V) oxide powder is sprinkled over distilled water. Combustion 1. Lithium metal is burned in air. 2. The hydrocarbon hexane is burned in excess oxygen.


3. Gaseous diborane, B2H6, is burned in excess oxygen. 4. A piece of solid bismuth is heated strongly in oxygen. 5. Solid zinc sulfide is heated in an excess of oxygen. 6. Propanol is burned completely in air. 7. Excess oxygen gas is mixed with ammonia gas in the presence of platinum. 8. Gaseous silane, SiH4, is burned in oxygen. 9. Ethanol is completely burned in air. 10. Solid copper(II) sulfide is heated strongly in oxygen gas. 11. Ethanol is burned in oxygen. 12. Carbon disulfide vapor is burned in excess oxygen. Redox 1. Iron(III) ions are reduced by iodide ions. 2. Potassium permanganate solution is added to concentrated hydrochloric acid. 3. Magnesium metal is added to dilute nitric acid, giving as one of the products a compound in which the oxidation number for nitrogen is -3. 4. A solution of potassium iodide is electrolyzed. 5. Potassium dichromate solution is added to an acidified solution of sodium sulfite. 6. Solutions of potassium iodide, potassium iodate, and dilute sulfuric acid are mixed. 7. A solution of tin(II) sulfate is added to a solution of iron(III) sulfate. 8. Metallic copper is heated with concentrated sulfuric acid. 9. Manganese(IV) oxide is added to warm, concentrated hydrobromic acid. 10. Chlorine gas is bubbled into cold dilute sodium hydroxide. 11. Solid iron(III) oxide is heated in excess carbon monoxide. 12. Hydrogen peroxide solution is added to acidified potassium iodide solution. 13. Hydrogen peroxide is added to an acidified solution of potassium dichromate. 14. Sulfur dioxide gas is bubbled through an acidified solution of potassium permanganate. 15. A solution containing tin(II) ions is added to an acidified solution of potassium dichromate. 16. Solid silver sulfide is warmed with dilute nitric acid. 17. A dilute solution of sulfuric acid is electrolyzed between platinum electrodes. 18. Pellets of lead are dropped into hot sulfuric acid. 19. Potassium permanganate solution is added to a solution of oxalic acid, H2C2O4, acidified with a few drops of sulfuric acid. 20. Powdered iron is added to a solution of iron(III) sulfate. 21. A concentrated solution of hydrochloric acid is added to powdered manganese dioxide and gently heated. 22. A strip of copper metal is added to a concentrated solution of sulfuric acid. 23. Copper(II) sulfide is oxidized by dilute nitric acid. 24. A solution of copper(II) sulfate is electrolyzed using inert electrodes. 25. A solution of potassium iodide is added to an acidified solution of potassium dichromate. 26. Hydrogen peroxide solution is added to a solution of iron(II) sulfate. 27. Solid silver is added to a dilute nitric acid (6M) solution. 28. A solution of formic acid, HCOOH, is oxidized by an acidified solution of potassium dichromate. 29. A piece of iron is added to a solution of iron(III) sulfate. 30. An acidified solution of potassium permanganate is added to a solution of sodium sulfite.


31. A solution of tin(II) chloride is added to a solution of iron(III) sulfate. 32. Concentrated hydrochloric acid solution is added to solid manganese(IV) oxide and the reactants are heated. 33. A strip of copper is immersed in dilute nitric acid. 34. Potassium permanganate solution is added to an acidic solution of hydrogen peroxide. 35. Solid copper is added to a dilute nitric acid solution. 36. Chlorine gas is bubbled into a cold solution of dilute sodium hydroxide. 37. A solution of potassium permanganate is mixed with an alkaline solution of sodium sulfite. 38. Solid sodium dichromate is added to an acidified solution of sodium iodide. 39. Hydrogen gas is passed over hot iron(III) oxide. 40. Solutions of potassium iodide and potassium iodate are mixed in acid solution. 41. Hydrogen peroxide is added to an acidified solution of sodium bromide. 42. Chlorine gas is bubbled into a cold, dilute solution of potassium hydroxide. 44. A solution of iron(II) nitrate is exposed to air for an extended period of time. 45. A stream of chlorine gas is passed through a solution of cold, dilute sodium hydroxide. 46. A solution of tin(II) chloride is added to an acidified solution of potassium permanganate. 47. A concentrated solution of hydrochloric acid is added to solid potassium permanganate. 48. A solution of potassium dichromate is added to an acidified solution of iron(II) chloride. Acid-Base Neutralizations 1. Solutions of ammonia and hydrofluoric acid are mixed. 2. Hydrogen sulfide gas is bubbled through a solution of potassium hydroxide. 3. A solution of sulfuric acid is added to a solution of barium hydroxide until the same number of moles of each compound has been added. 4. A solution of sodium hydroxide is added to a solution of sodium dihydrogen phosphate until the same number of moles of each compound has been added. 5. Dilute nitric acid is added to crystals of pure calcium oxide. 6. Equal volumes of 0.1-molar sulfuric acid and 0.1-molar potassium hydroxide are mixed. 7. A solution of ammonia is added to a dilute solution of acetic acid. 8. Excess sulfur dioxide gas is bubbled through a dilute solution of potassium hydroxide. 9. Sulfur dioxide gas is bubbled into an excess of a saturated solution of calcium hydroxide. 10. A solution of sodium hydroxide is added to a solution of calcium hydrogen carbonate until the number of moles of sodium hydroxide added is twice the number of moles of the calcium salt. 11. Equal volumes of 0.1M hydrochloric acid and 0.1M sodium monohydrogen phosphate are mixed. 12. Hydrogen sulfide gas is bubbled through excess potassium hydroxide solution. 13. Ammonia gas and carbon dioxide gas are bubbled into water. 14. Carbon dioxide gas is bubbled through a concentrated solution of sodium hydroxide. 15. Acetic acid solution is added to a solution of sodium hydrogen carbonate. 16. Excess potassium hydroxide solution is added to a solution of potassium dihydrogen phosphate. 17. Carbon dioxide gas is bubbled through a concentrated solution of potassium hydroxide. Complex Ions 1. Concentrated (15M) ammonia solution is added in excess to a solution of copper(II) nitrate.


2. An excess of nitric acid solution is added to a solution of tetraaminecopper(II) sulfate. 3. Dilute hydrochloric acid is added to a solution of diamminesilver(I) nitrate. 4. Solid aluminum nitrate is dissolved in water. 5. A suspension of copper(II) hydroxide is treated with an excess of ammonia water. 6. A solution of diamminesilver(I) chloride is treated with dilute nitric acid. 7. An excess of concentrated ammonia solution is added to freshly precipitated copper(II) hydroxide. 8. Excess dilute nitric acid is added to a solution containing the tetraaminecadmium(II) ion. 9. An excess of ammonia is bubbled through a solution saturated with silver chloride. 10. Solid aluminum oxide is added to a solution of sodium hydroxide. 11. A concentrated solution of ammonia is added to a solution of zinc iodide. 12. An excess of sodium hydroxide solution is added to a solution of aluminum chloride. 13. A concentrated solution of ammonia is added to a solution of copper(II) chloride. 14. Excess concentrated sodium hydroxide solution is added to solid aluminum hydroxide. 15. Excess concentrated ammonia solution is added to a suspension of silver chloride, 16. Pellets of aluminum metal are added to a solution containing an excess of sodium hydroxide. 17. A suspension of zinc hydroxide is treated with concentrated sodium hydroxide solution. 18. Silver chloride is dissolved in excess ammonia solution. 19. Sodium hydroxide solution is added to a precipitate of aluminum hydroxide in water. 20. A drop of potassium thiocyanate is added to a solution of iron(III) chloride. 21. A concentrated solution of ammonia is added to a suspension of zinc hydroxide. 22. Excess concentrated potassium hydroxide solution is added to a precipitate of zinc

hydroxide. 23. A solution of ammonium thiocyanate is added to a solution of iron(III) chloride. 24. Excess sodium cyanide is added to a solution of silver nitrate. Addition 1. The gases boron trifluoride and ammonia are mixed. 2. A mixture of solid calcium oxide and solid tetraphosphorus decaoxide is heated. 3. Solid calcium oxide is exposed to a stream of carbon dioxide gas. 4. Solid calcium oxide is heated in the presence of sulfur trioxide gas. 5. Calcium metal is heated strongly in nitrogen gas. 6. Excess chlorine gas is passed over hot iron filings. 7. Powdered magnesium oxide is added to a container of carbon dioxide gas. 8. A piece of lithium metal is dropped into a container of nitrogen gas. 9. Magnesium metal is burned in nitrogen gas. 10. Sulfur dioxide gas is passed over solid calcium oxide. 11. Samples of boron trichloride gas and ammonia gas are mixed. Decomposition 1. A solution of hydrogen peroxide is heated. 2. Solid magnesium carbonate is heated. 3. A solution of hydrogen peroxide is catalytically decomposed. 4. Solid potassium chlorate is heated in the presence of manganese dioxide as a catalyst. 5. Sodium hydrogen carbonate is dissolved in water. 6. Solid ammonium carbonate is heated.

AP* Gas Law Free Response Questions

(1) AP® is a registered trademark of the College Board. The College Board was not involved in the production of and does not endorse this product. (2) Test Questions are Copyright © 1984-2008 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Web or Mass distribution prohibited.

1971 2 HCOONa + H2SO4 → 2 CO + 2 H2O + Na2SO4

A 0.964 gram sample of a mixture of sodium formate and sodium chloride is analyzed by adding sulfuric acid. The equation for the reaction for sodium formate with sulfuric acid is shown above. The carbon monoxide formed measures 242 milliliters when collected over water at 752 torr and 22.0°C. Calculate the percentage of sodium formate in the original mixture. 1971 At 20°C the vapor pressure of benzene is 75 torr, and the vapor pressure of toluene is 22 torr. Solutions in both parts of this question are to be considered ideal. (a) A solution is prepared from 1.0 mole of biphenyl, a nonvolatile solute, and 49.0 moles of benzene. Calculate

the vapor pressure of the solution at 20°C. (b) A second solution is prepared from 3.0 moles of toluene and 1.0 mole of benzene. Determine the vapor

pressure of this solution and the mole fraction of benzene in the vapor.

1972 A 5.00 gram sample of a dry mixture of potassium hydroxide, potassium carbonate, and potassium chloride is reacted with 0.100 liter of 2.0 molar HCl solution. (a) A 249 milliliter sample of dry CO2 gas, measured at 22°C and 740 torr, is obtained from the reaction. What is

the percentage of potassium carbonate in the mixture? (b) The excess HCl is found by titration to be chemically equivalent to 86.6 milliliters of 1.50 molar NaOH.

Calculate the percentages of potassium hydroxide and of potassium chloride in the original mixture. 1973 A 6.19 gram sample of PCl5 is placed in an evacuated 2.00 liter flask and is completely vaporized at 252°C. (a) Calculate the pressure in the flask if no chemical reaction were to occur. (b) Actually at 252°C the PCl5 is partially dissociated according to the following equation:

PCl5(g) → PCl3(g) + Cl2(g) The observed pressure is found to be 1.00 atmosphere. In view of this observation, calculate the partial

pressure of PCl5 and PCl3 in the flask at 252°C. 1976 When the molecular weight of a volatile liquid is calculated from the weight, volume, temperature, and pressure of a sample of that liquid when vaporized, the assumption is usually made that the gas behaves ideally. In fact at a temperature not far above the boiling point of the liquid, the gas is not ideal. Explain how this would affect the results of the molecular weight determination.

AP* Gas Law Free Response Questions page 2 1982 (a) From the standpoint of the kinetic-molecular theory, discuss briefly the properties of gas molecules that

cause deviations from ideal behavior. (b) At 25°C and 1 atmosphere pressure, which of the following gases shows the greatest deviation from ideal

behavior? Give two reasons for your choice. CH4 SO2 O2 H2

(c) Real gases approach ideality at low pressure, high temperature, or both. Explain these observations. 1984 The van der Waals equation of state for one mole of a real gas is as follows:

(P + a/V2)(V - b) = RT

For any given gas, the values of the constants a and b can be determined experimentally. Indicate which physical properties of a molecule determine the magnitudes of the constants a and b. Which of the two molecules, H2 or H2S, has the higher value for a and which has the higher value for b? Explain.

One of the van der Waals constants can be correlated with the boiling point of a substance. Specify which constant and how it is related to the boiling point. 1986 Three volatile compounds X, Y, and Z each contain element Q. The percent by weight of element Q in each compound was determined. Some of the data obtained are given below.

Compound Percent by Weight

Element Q of

Molecular

eight W X

64.8%

?

Y 73.0% 104. Z 59.3% 64.0

(a) The vapor density of compound X at 27 degrees Celsius and 750. mm Hg was determined to be 3.53 grams per liter. Calculate the molecular weight of compound X.

(b) Determine the mass of element Q contained in 1.00 mole of each of the three compounds. (c) Calculate the most probable value of the atomic weight of element Q. (d) Compound Z contains carbon, hydrogen, and element Q. When 1.00 gram of compound Z is oxidized and all of the

carbon and hydrogen are converted to oxides, 1.37 grams of CO2 and 0.281 gram of water are produced. Determine the most probable molecular formula.

AP* Gas Law Free Response Questions page 3 1990 A mixture of H2(g), O2(g), and 2 millilitres of H2O(l) is present in a 0.500 litre rigid container at 25°C. The number of moles of H2 and the number of moles of O2 are equal. The total pressure is 1,146 millimetres mercury. (The equilibrium vapor pressure of pure water at 25°C is 24 millimetres mercury.) The mixture is sparked, and H2 and O2 react until one reactant is completely consumed. (a) Identify the reactant remaining and calculate the number of moles of the reactant remaining. (b) Calculate the total pressure in the container at the conclusion of the reaction if the final temperature is 90°C.

(The equilibrium vapor pressure of water at 90°C is 526 millimetres mercury.) (c) Calculate the number of moles of water present as vapor in the container at 90°C.

1993 Observations about real gases can be explained at the molecular level according to the kinetic molecular theory of gases and ideas about intermolecular forces. Explain how each of the following observations can be interpreted according to these concepts, including how the observation supports the correctness of these theories. (a) When a gas-filled balloon is cooled, it shrinks in volume; this occurs no matter what gas is originally placed in

the balloon. (b) When the balloon described in (a) is cooled further, the volume does not become zero; rather, the gas

becomes a liquid or solid. (c) When NH3 gas is introduced at one end of a long tube while HCl gas is introduced simultaneously at the other

end, a ring of white ammonium chloride is observed to form in the tube after a few minutes. This ring is closer to the HCl end of the tube than the NH3 end.

(d) A flag waves in the wind.

1994

A student collected a sample of hydrogen gas by the displacement of water as shown by the diagram above. The relevant data are given in the following table.

GAS SAMPLE DATA Volume of sample

90.0 mL

Temperature 25°C

Atmospheric Pressure 745 mm Hg

Equilibrium Vapor Pressure of H2O (25°C)

23.8 mm Hg

AP* Gas Law Free Response Questions page 4

(a) Calculate the number of moles of hydrogen gas collected. (b) Calculate the number of molecules of water vapor in the sample of gas. (c) Calculate the ratio of the average speed of the hydrogen molecules to the average speed of the water

vapor molecules in the sample. (d) Which of the two gases, H2 or H2O, deviates more from ideal behavior? Explain your answer. 1995 Propane, C3H8, is a hydrocarbon that is commonly used as fuel for cooking. (a) Write a balanced equation for the complete combustion of propane gas, which yields CO2(g) and H2O(l). (b) Calculate the volume of air at 30°C and 1.00 atmosphere that is needed to burn completely 10.0 grams of

propane. Assume that air is 21.0 percent O2 by volume. (c) The heat of combustion of propane is −2,220.1 kJ/mol. Calculate the heat of formation, ΔHf

o, of propane given that ΔHf

o of H2O(l) = −285.3 kJ/mol and ΔHfo of CO2(g) = −393.5 kJ/mol.

(d) Assuming that all of the heat evolved in burning 30.0 grams of propane is transferred to 8.00 kilograms of water (specific heat = 4.18 J/g.K), calculate the increase in temperature of water.

1996

Represented above are five identical balloons, each filled to the same volume at 25°C and 1.0 atmosphere pressure with the pure gas indicated.

(a) Which balloon contains the greatest mass of gas? Explain.

(b) Compare the average kinetic energies of the gas molecules in the balloons. Explain.

(c) Which balloon contains the gas that would be expected to deviate most from the behavior of an ideal gas? Explain.

(d) Twelve hours after being filled, all the balloons have decreased in size. Predict which balloon will be the smallest. Explain your reasoning. 2002B A rigid 8.20 L flask contains a mixture of 2.50 moles of H2, 0.500 mole of O2, and sufficient Ar so that the partial pressure of Ar in the flask is 2.00 atm. The temperature is 127ºC. (a) Calculate the total pressure in the flask. (b) Calculate the mole fraction of H2 in the flask. (c) Calculate the density (in g L−1) of the mixture in the flask. The mixture in the flask is ignited by a spark, and the reaction represented below occurs until one of the reactants is entirely consumed.

2 H2(g) + O2(g) → 2H2O(g) (d) Give the mole fraction of all species present in the flask at the end of the reaction.

AP* Chemistry GASES The gaseous state of matter is the simplest and best-understood state of matter. You inhale approximately 8,500 L of air each day. This amounts to about 25 lbs of air. Breathing is a three-step process: inhaling, gas exchange with the circulatory system, and exhaling. Approximately 80% pressure differences created by your body allow you to breathe. Nearly the Earth’s entire atmosphere is made up of only five gases: nitrogen, oxygen, water vapor, argon, and carbon dioxide. PROPERTIES OF GASES Only FOUR quantities are needed to define the state of a gas:

1. the quantity of the gas, n (in moles) 2. the temperature of the gas, T (in KELVINS) 3. the volume of the gas, V (in liters) 4. the pressure of the gas, P (in atmospheres)

A gas uniformly fills any container, is easily compressed & mixes completely with any other gas.

GAS PRESSURE is a measure of the force that a gas exerts on its container. Force is the physical quantity that interferes with inertia. Gravity is the force responsible for weight. Force = mass × acceleration; (Newton’s 2nd Law) The SI units follow: N = kg × m/s2 Pressure-- Force/ unit area; N/m2; which is the definition of 1.0 Pascal

Barometer--invented by Evangelista Torricelli in 1643; uses the height of a column of mercury to measure gas pressure (especially atmospheric P) 1 mm of Hg = 1 torr 1.00 atm = 760.00 mm Hg = 760.00 torr = 29.92 in Hg = 14.7 psi = 101.325 kPa ≈ 105 Pa

At sea level all of the above define STANDARD PRESSURE. The SI unit of pressure is the Pascal (named after Blaise Pascal); 1 Pa = 1 N / m2 The manometer—a device for measuring the pressure of a gas in a container. The pressure of the gas is given by h [the difference in mercury levels] in units of torr (equivalent to mm Hg).

a) Gas pressure = atmospheric pressure – h b) Gas pressure = atmospheric pressure + h

*AP is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product. © 2014 by René McCormick. All rights reserved. Some art copyright Pearson Education Inc. 2014

Exercise 1 Pressure Conversions The pressure of a gas is measured as 49 torr. Represent this pressure in both atmospheres and Pascals. Exercise 2 Pressure Comparisons Rank the following pressures in decreasing order of magnitude (largest first, smallest last): 75 kPa, 300. torr, 0.60 atm and 350. mm Hg.

GAS LAWS: THE EXPERIMENTAL BASIS BOYLE’S LAW: “Father of Chemistry”--the volume of a confined gas is inversely proportional to the pressure exerted on the gas. ALL GASES BEHAVE IN THIS MANNER! • Robert Boyle was an Irish chemist. He studied PV relationships

using a J-tube set up in the multi-story entryway of his home. (Thus his was MUCH larger than the one shown right.) o P µ 1/V o ∴ pressure and volume are inversely proportional o Volume ↑ Pressure ↓ at constant temperature, the

converse is also true o for a given quantity of a gas at constant temperature, the product of pressure

and volume is a constant PV = k

Therefore, P

kPkV 1

==

which is the equation for a straight line of the type y = mx + b, where m = slope, and b is the y-intercept

In this case, y = V, x = 1/P and b = 0. Check out the plot on the right (b).

The data Boyle originally collected is graphed on (a) above on the right.

o P1V1 = P2V2 is the easiest form of Boyle’s law to memorize

o Boyle’s Law has been tested for over three centuries. It holds true only at low pressures.

Gases 2

Exercise 3 Boyle’s Law I Sulfur dioxide (SO2), a gas that plays a central role in the formation of acid rain, is found in the exhaust of automobiles and power plants. Consider a 1.53- L sample of gaseous SO2 at a pressure of 5.6 × 103 Pa. If the pressure is changed to 1.5 × 104 Pa at a constant temperature, what will be the new volume of the gas?

Ideal Gases At “normal” conditions such as standard temperature and pressure, most real gases behave qualitatively like an ideal gas. Many gases such as nitrogen, oxygen, hydrogen, noble gases, and some heavier gases like carbon dioxide can be treated like ideal gases within reasonable tolerances. Generally, a gas behaves more like an ideal gas at higher temperature and lower pressure, as the work which is against intermolecular forces becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the empty space between them. An ideal gas is expected to have a constant value of PV, as shown by the dotted line on the graph pictured above. CO2 shows the largest change in PV, and this change is actually quite small: PV changes from about 22.39 L·atm at 0.25 atm to 22.26 L·atm at 1.00 atm. Thus Boyle’s Law is a good approximation at these relatively low pressures. So, why does CO2 deviate from ideal behavior the most? It has more electrons, thus is more polarizable, thus has higher dispersion forces (a type of intermolecular force a.k.a. London dispersion forces or LDFs), therefore the molecules are more attracted to each other, so carbon dioxide gas deviates from ideal behavior more than oxygen or carbon dioxide do. Gases 3

Exercise 4 Boyle’s Law II In a study to see how closely gaseous ammonia obeys Boyle’s law, several volume measurements were made at various pressures, using 1.0 mol NH3 gas at a temperature of 0ºC. Using the results listed below; calculate the Boyle’s law constant for NH3 at the various pressures. Calculate the deviation from ideal behavior in each case. Account for the trend apparent in the data.

Experiment Pressure (atm) Volume (L) 1 0.1300 172.1 2 0.2500 89.28 3 0.3000 74.35 4 0.5000 44.49 5 0.7500 29.55 6 1.000 22.08

Exercise 5 Boyle’s Law III Next, PLOT the values of PV vs. P for the six experiments in Exercise 4. Extrapolate to determine what PV equals at a hypothetical 0.00 atm pressure. Compare it to the PV vs. P graph on page 3 of these lecture notes. What is the value of the y-intercept?

Gases 4

CHARLES’S LAW: If a given quantity of gas is held at a constant pressure, then its volume is directly proportional to the absolute temperature. Must use KELVINS Why? • Jacques Charles was a French physicist and the first

person to fill a hot “air” balloon with hydrogen gas and made the first solo hot air balloon flight! o V µ T plot = straight line o V1T2 = V2T1 o Temperature ∝ Volume at constant pressure o This figure shows the plots of V vs. T (Celsius) for

several gases. The solid lines represent experimental measurements on gases. The dashed lines represent extrapolation of the data into regions where these gases would become liquids or solids. Note that the samples of the various gases contain different numbers of moles.

o What is the temperature when the volume extrapolates to zero? Sound familiar?

Exercise 6 Charles’s Law A sample of gas at 15ºC and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38ºC and 1 atm? Each of the balloons below contains the same number of particles in the gas phase. This is yet another example of “heat ‘em up and speed ‘em up!”. As the molecules warm, they gain kinetic energy, move faster, and thus collide with the walls of their container with more energy. (Umph! If you prefer) In this case, the “walls” of the container are made of rubber which can expand and contract.

Gases 5

GAY-LUSSAC’S LAW of combining volumes: volumes of gases always combine with one another in the ratio of small whole numbers, as long as volumes are measured at the same T and P. - P1T2 = P2T1

- Avogadro’s hypothesis: equal volumes of gases under the same conditions of temperature and pressure

contain equal numbers of molecules.

AVOGADRO’S LAW: The volume of a gas, at a given temperature and pressure, is directly proportional to the quantity of gas. - V ∝ n - n ∝ Volume at constant T & P HERE’S AN EASY WAY TO MEMORIZE ALL OF THIS! Start with the combined gas law:

P1V1T2 = P2V2T1 Memorize just this use a simple pattern to figure the rest out: Place the scientist names in alphabetical order. Boyle’s Law uses the first 2 variables, Charles’ Law the second 2 variables & Gay-Lussac’s Law the

remaining combination of variables. Whichever variable doesn’t appear in the formula is being held CONSTANT!

Exercise 7 Avogadro’s Law Suppose we have a 12.2-L sample containing 0.50 mol oxygen gas (O2) at a pressure of 1 atm and a temperature of 25ºC. If all this O2 were converted to ozone (O3) at the same temperature and pressure, what would be the volume of the ozone?

These balloons each hold 1.0 L of gas at 25°C and 1 atm. Each balloon contains 0.041 mol of gas, or 2.5 × 1022 gas molecules.

Gases 6

THE IDEAL GAS LAW Four quantities describe the state of a gas: pressure, volume, temperature, and # of moles (quantity). Combine all 3 laws: V µ nT P Replace the µ with a constant, R, and you get: PV = nRT The ideal gas law is an equation of state. R = 0.8206 L• atm/mol • K also expressed as 0.8206 L atm mol−1 K−1

Useful only at low Pressures and high temperatures! Guaranteed points on the AP Exam! These next exercises can all be solved with the ideal gas law, BUT, you can use another if you like! Exercise 8 Ideal Gas Law I A sample of hydrogen gas (H2) has a volume of 8.56 L at a temperature of 0ºC and a pressure of 1.5 atm. Calculate the moles of H2 molecules present in this gas sample.

Exercise 9 Ideal Gas Law II Suppose we have a sample of ammonia gas with a volume of 3.5 L at a pressure of 1.68 atm. The gas is compressed to a volume of 1.35 L at a constant temperature. Use the ideal gas law to calculate the final pressure.

Exercise 10 Ideal Gas Law III A sample of methane gas that has a volume of 3.8 L at 5ºC is heated to 86ºC at constant pressure. Calculate its new volume.

Gases 7

Exercise 11 Ideal Gas Law IV A sample of diborane gas (B2H6), a substance that bursts into flame when exposed to air, has a pressure of 345 torr at a temperature of -15ºC and a volume of 3.48 L. If conditions are changed so that the temperature is 36ºC and the pressure is 468 torr, what will be the volume of the sample?

Exercise 12 Ideal Gas Law V A sample containing 0.35 mol argon gas at a temperature of 13ºC and a pressure of 568 torr is heated to 56ºC and a pressure of 897 torr. Calculate the change in volume that occurs.

GAS STOICHIOMETRY Use PV = nRT to solve for the volume of one mole of gas at STP: Look familiar? This is the molar volume of a gas at STP. Work stoichiometry problems using your favorite method, dimensional analysis, mole map, the table way…just work FAST! Use the ideal gas law to convert quantities that are NOT at STP. Exercise 13 Gas Stoichiometry I A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N2 are present?

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Exercise 14 Gas Stoichiometry II Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO3). Calculate the volume of CO2 at STP produced from the decomposition of 152 g CaCO3 by the reaction

CaCO3(s) → CaO(s) + CO2(g)

Exercise 15 Gas Stoichiometry III A sample of methane gas having a volume of 2.80 L at 25ºC and 1.65 atm was mixed with a sample of oxygen gas having a volume of 35.0 L at 31ºC and 1.25 atm. The mixture was then ignited to form carbon dioxide and water. Calculate the volume of CO2 formed at a pressure of 2.50 atm and a temperature of 125ºC.

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THE DENSITY OF GASES

d = m = P(MM) {for ONE mole of gas}= MM AND Molar Mass = MM = dRT V RT 22.4 L P “Molecular Mass kitty cat”—all good cats put dirt [dRT] over their pee [P]. Corny? Yep! Crude and socially unacceptable? You bet! But … you’ll thank me later when you’re flying through such gas law problems. Just remember that densities of gases are reported in g/L NOT g/mL. What is the approximate molar mass of air expressed in g/L? List 3 gases that float in air: List 3 gases that sink in air: Exercise 16 Gas Density/Molar Mass The density of a gas was measured at 1.50 atm and 27ºC and found to be 1.95 g/L. Calculate the molar mass of the gas.

GAS MIXTURES AND PARTIAL PRESSURES The pressure of a mixture of gases is the sum of the pressures of the different components of the mixture: Ptotal = P1 + P2 + . . . Pn John Dalton’s Law of Partial Pressures also uses the concept of mole fraction, χ χ A = moles of A . moles A + moles B + moles C + . . . so now, PA = χ A Ptotal The partial pressure of each gas in a mixture of gases in a container depends on the number of moles of that gas. The total pressure is the SUM of the partial pressures and depends on the total moles of gas particles present, no matter what they are!

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Exercise 17 Dalton’s Law I Mixtures of helium and oxygen are used in scuba diving tanks to help prevent “the bends.” For a particular dive, 46 L He at 25ºC and 1.0 atm and 12 L O2 at 25ºC and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25ºC.

Exercise 18 Dalton’s Law II The partial pressure of oxygen was observed to be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of O2 present. Exercise 19 Dalton’s Law III The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N2 in air when the atmospheric pressure is 760. torr. WATER DISPLACEMENT It is common to collect a gas in the laboratory by water displacement. The confounding factor is that some of the pressure in the collection vessel is due to water vapor collected as the gas was passing through! You must correct for this in order to report the P of “dry” gas. How? You simply look up the partial pressure due to water vapor at a given temperature and subtract that value from the total pressure. The experiment pictured is a classic! You may have done it with Mg rather than zinc and used a eudiometer or inverted and sealed buret to measure the volume of the gas collected “over water”.

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Exercise 20 Gas Collection over Water A sample of solid potassium chlorate (KClO3) was heated in a test tube (see the figure above) and decomposed by the following reaction: 2 KClO3(s) → 2 KCl(s) + 3 O2(g) The oxygen produced was collected by displacement of water at 22ºC at a total pressure of 754 torr. The volume of the gas collected was 0.650 L, and the vapor pressure of water at 22ºC is 21 torr. Calculate the partial pressure of O2 in the gas collected and the mass of KClO3 in the sample that was decomposed. KINETIC MOLECULAR THEORY OF GASES Assumptions of the MODEL: 1. All particles are in constant, random, motion. 2. All collisions between particles are perfectly elastic. 3. The volume of the particles in a gas is negligible 4. The average kinetic energy of the molecules is its Kelvin temperature. This theory neglects any intermolecular forces as well. And it is important to note that gases expand to fill their container, solids/liquids do not. And that gases are compressible; solids/liquids are not appreciably compressible.

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This helps explain Boyle’s Law: If the volume is decreased that means that the gas particles will hit the wall more often, thus increasing pressure

V

nRTP 1)(=

And also helps explain Charles’ Law When a gas is heated, the speeds of its particles increase and thus hit the walls more often and with more force. The only way to keep the P constant is to increase the volume of the container.

TP

nRV

=

Yep, you guessed it! It also helps explain Gay-Lussac’s Law When the temperature of a gas increases, the speeds of its particles increase, the particles are hitting the wall with greater force and greater frequency. Since the volume remains the same this would result in increased gas pressure.

TVnRP

=

And it even helps explain Avogadro’s Law An increase in the number of particles at the same temperature would cause the pressure to increase if the volume were held constant. The only way to keep constant P is to vary the V.

nP

RTV

=

What about Dalton’s Law? The P exerted by a mixture of gases is the SUM of the partial pressures since gas particles are acting independent of each other and the volumes of the individual particles DO NOT matter.

Constant

Constant

Constant

Constant

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DISTRIBUTION OF MOLECULAR SPEEDS Plot # of gas molecules having various speeds vs. the speed and you get a curve. Changing the temperature affects the shape of the curve NOT the area beneath it. Change the # of molecules and all bets are off! Maxwell’s equation:

2 3rms

RTu uMM

= =

Use the “energy R” or 8.314510 J/K• mol for this equation since kinetic energy is involved.

Exercise 21 Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of helium gas at 25ºC.

If we could monitor the path of a single molecule it would be very erratic. Mean free path—the average distance a particle travels between collisions. It’s on the order of a tenth of a micrometer. WAAAAY SMALL! Examine the effect of temperature on the numbers of molecules with a given velocity as it relates to temperature. HEAT ‘EM UP, SPEED ‘EM UP!!

Drop a vertical line from the peak of each of the three bell shaped curves—that point on the x-axis represents the AVERAGE velocity of the sample at that temperature. Note how the bells are “squashed” as the temperature increases. You may see graphs like this on the AP exam where you have to identify the highest temperature based on the shape of the graph!

GRAHAM’S LAW OF DIFFUSION AND EFFUSION Effusion is closely related to diffusion. Diffusion is the term used to describe the mixing of gases. The rate of diffusion is the rate of the mixing. Effusion is the term used to describe the passage of a gas through a tiny orifice into an evacuated chamber as shown on the right. The rate of effusion measures the speed at which the gas is transferred into the chamber.

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The rates of effusion of two gases are inversely proportional to the square roots of their molar masses at the same temperature and pressure.

REMEMBER rate is a change in a quantity over time, NOT just the time! Exercise 22 Effusion Rates Calculate the ratio of the effusion rates of hydrogen gas (H2) and uranium hexafluoride (UF6), a gas used in the enrichment process to produce fuel for nuclear reactors.

Exercise 23 A pure sample of methane is found to effuse through a porous barrier in 1.50 minutes. Under the same conditions, an equal number of molecules of an unknown gas effuses through the barrier in 4.73 minutes. What is the molar mass of the unknown gas?

Diffusion This is a classic! https://www.youtube.com/watch?v=GRcZNCA9DxE (animation) Effusion of a Gas: http://www.youtube.com/watch?v=0uBK7VxT00E https://www.youtube.com/watch?v=o7C4lo5n0zU (actual, but not great)

Distance traveled by NH3 = urmsfor NH3 = Distance traveled by HCl urms for HCl

2

1

Rate of effusion of gas 1 = Rate of effusion of gas 2

MMMM

3

36.5 1.517

HCl

NH

MMMM

= =

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https://www.youtube.com/watch?v=GRcZNCA9DxE

http://www.youtube.com/watch?v=0uBK7VxT00E

https://www.youtube.com/watch?v=o7C4lo5n0zU

The observed ratio is LESS than a 1.5 distance ratio—why? This diffusion is slow despite considering the molecular velocities are 450 and 660 meters per second—which one is which? This tube contains air and all those collisions slow the process down in the real world. Speaking of real world…. REAL, thus NONIDEAL GASES Most gases behave ideally until you reach high pressure and low temperature. (By the way, either of these can cause a gas to liquefy, go figure!) van der Waals Equation--corrects for negligible volume of molecules and accounts for inelastic collisions leading to intermolecular forces (his real claim to fame).

The coefficients a and b are van der Waals constants; no need to work problems, it’s the concepts that are important! Notice pressure is increased (intermolecular forces lower real pressure, you’re correcting for this) and volume is decreased (corrects the container to a smaller “free” volume). These graphs are classics and make great multiple choice questions on the AP exam.

When PV/nRT = 1.0, the gas is ideal This graph is just for nitrogen gas. All of these are at 200K. Note that although nonideal behavior Note the P’s where the curves is evident at each temperature, the cross the dashed line [ideality]. deviations are smaller at the higher Ts. Don’t underestimate the power of understanding these graphs. We love to ask question comparing the behavior of ideal and real gases. It’s not likely you’ll be asked an entire free-response gas problem on the real exam in May. Gas Laws are tested extensively in the multiple choice since it’s easy to write questions involving them! You will most likely see PV = nRT as one part of a problem in the free-response, just not a whole problem! GO FORTH AND RACK UP THOSE MULTIPLE CHOICE POINTS!!

[ ][ ]2nP + a( V bn = nRT)V

−

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And, just for fun: Collapsing Can: http://www.graspr.com/videos/The-Collapsing-Can-1 Shall we try a bigger can? http://www.youtube.com/watch?v=Uy-SN5j1ogk&NR=1 How about the biggest can we can find? http://www.youtube.com/watch?v=E_hci9vrvfw Boiling water with ice: http://www.youtube.com/watch?v=zzVtbvVS2lQ Cooling gases with liquid nitrogen MIT: http://www.youtube.com/watch?v=ZvrJgGhnmJo Getting a boiled egg into a bottle: http://www.youtube.com/watch?v=xZdfcRiDs8I&NR=1&feature=fvwp Gravity has nothing to do with it! http://www.youtube.com/watch?v=BofIBaYk5e0&feature=related Getting egg OUT of bottle! http://www.youtube.com/watch?v=x--4l-SL77Y&feature=related Peeps in a vacuum: http://www.youtube.com/watch?v=lfNJJEdKgLU&NR=1 and another: http://www.youtube.com/watch?v=ciPr4Tg9k78&feature=related Ideal Gas Law: http://www.youtube.com/watch?v=Mytvt0wlZK8&feature=related Diffusion of ammonia and HCl (animation): http://www.youtube.com/watch?v=L41KhBPBymA&feature=related Diffusion of ammonia and HCl (real deal): http://www.youtube.com/watch?v=WAJAslkwolk&feature=related Putting it all together: http://www.mhhe.com/physsci/chemistry/animations/chang_7e_esp/gam2s2_6.swf Why R as in PV = nRT? This is the kind of stuff that drives me crazy as well! I went in search of the answer some time ago, so I'll share what I found: As usual, the answer relates to a history lesson. It was Clausius in the mid 1800’s that refined the conversion factor for converting degrees Celsius to Kelvins (from adding 267 to adding 273 to the Celsius temperature. He did so using the careful experimental data of another French scientist, Regnault. Clausius also noted that Regnault’s data indicated that the farther the temperature and pressure conditions were from the condensation point of the gas, the more correctly the Ideal Gas Law applies. So, there is speculation that the “constant” was assigned the letter “R” to honor Regnault’s work. In the spirit of giving credit where credit is due, my source was a Journal of Chem. Ed article written by William B. Jensen Department of Chemistry, University of Cincinnati, Cincinnati, OH 45221-0172

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http://www.graspr.com/videos/The-Collapsing-Can-1

http://www.youtube.com/watch?v=Uy-SN5j1ogk&NR=1

http://www.youtube.com/watch?v=E_hci9vrvfw

http://www.youtube.com/watch?v=zzVtbvVS2lQ

http://www.youtube.com/watch?v=ZvrJgGhnmJo

http://www.youtube.com/watch?v=xZdfcRiDs8I&NR=1&feature=fvwp

http://www.youtube.com/watch?v=BofIBaYk5e0&feature=related

http://www.youtube.com/watch?v=x--4l-SL77Y&feature=related

http://www.youtube.com/watch?v=lfNJJEdKgLU&NR=1

http://www.youtube.com/watch?v=ciPr4Tg9k78&feature=related

http://www.youtube.com/watch?v=Mytvt0wlZK8&feature=related

http://www.youtube.com/watch?v=L41KhBPBymA&feature=related

http://www.youtube.com/watch?v=WAJAslkwolk&feature=related

http://www.mhhe.com/physsci/chemistry/animations/chang_7e_esp/gam2s2_6.swf

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