Amazing Chemistry Powerpoint Presentation! Aligned to the New York State Standards and Core Curriculum for “ The Physical Setting- Chemistry ” Produced

Amazing Chemistry Powerpoint Presentation! Aligned to the New York State Standards and

Core Curriculum for “The Physical Setting-Chemistry”

Produced and Edited by : Mark Rosengarten and Susan Katzoff

Outline for Review

1) The Atom (Electron Config, Nuclear)

2) Matter (Phases, Types, Changes)

3) Bonding (Periodic Table, Ionic, Covalent)

4) Compounds (Formulas, Reactions, IMAF’s)

5) Math of Chemistry (Formula Mass, Gas Laws, Neutralization, etc.)

6) Kinetics and Thermodynamics (PE Diagrams, etc.)

7) Acids and Bases (pH, formulas, indicators, etc.)

8) Oxidation and Reduction (Half Reactions, Cells, etc.)

9) Organic Chemistry (Hydrocarbons, Families, Reactions)Mark Rosengarten and Susan Katzoff

The Atom

1) Nucleons

2) Isotopes

3) Natural Radioactivity

4) Half-Life

5) Nuclear Power

6) Electron Configuation

7) Development of the Atomic ModelMark Rosengarten and Susan Katzoff

Nucleons - particles in the nucleus

Protons: +1 charge each, determines identity of element, # of protons = atomic number and nuclear charge, has a mass of 1 amu.

Neutrons: no charge, determines identity of isotope of an element, 1 amu, determined using mass number - atomic number (amu = atomic mass unit)

3216S and 33

16S are both isotopes of S

S-32 has 16 protons and 16 neutrons S-33 has 16 protons and 17 neutrons All atoms of S have a nuclear charge of +16 due to the 16

protons.Mark Rosengarten and Susan Katzoff

Isotopes

Atoms of the same element MUST contain the same number of protons.

Atoms of the same element can vary in their numbers of neutrons, therefore different atomic masses can exist for any one element. These are called isotopes.

The atomic mass on the Periodic Table is the weighted-average atomic mass, taking into account the different isotope masses and their relative abundance.

Rounding off the atomic mass on the Periodic Table will tell you what the most common isotope of that element is.

Mark Rosengarten and Susan Katzoff

Weight-Average Atomic Mass


Most Common Isotope

The weight-average atomic mass of Zinc is 65.39 amu. What is the most common isotope of Zinc? Zn-65!

What are the most common isotopes of:

– Co Ag– S Pb– * Most common isotopes are the

rounded mass number. *


Natural Radioactivity

Alpha Decay Beta Decay Positron Decay Gamma Decay Charges of Decay Particles

Natural decay starts with a parent nuclide that ejects a decay particle to form a daughter nuclide which is more stable than the parent nuclide was.


Alpha Decay

The nucleus ejects two protons and two neutrons. The atomic mass decreases by 4, the atomic number decreases by 2.

23892U 4

2He + 23490Th


Beta Decay

A neutron decays into a proton and an electron. The electron is ejected from the nucleus as a beta particle. The atomic mass remains the same, but the atomic number increases by 1.

146C


0-1e + 14

7N

Positron Decay

A proton is converted into a neutron and a positron. The positron is ejected by the nucleus. The mass remains the same, but the atomic number decreases by 1.

5326Fe 0

+1 e + 5325Mn


Gamma Decay

The nucleus has energy levels just like electrons, but they involve a lot more energy. When the nucleus becomes more stable, a gamma ray may be released. This is a photon of high-energy light, and has no mass or charge. The atomic mass and number do not change with gamma. Gamma may occur by itself, or in conjunction with any other decay type.


Charges of Decay Particles in an electric field


Half-Life

Half life is the time it takes for half of the nuclei in a radioactive sample to undergo decay.

Problem Types:– Going forwards in time

– Going backwards in time

– Radioactive Dating


Going Forwards in Time

How many grams of a 10.0 gram sample of I-131 (half-life of 8 days) will remain in 24 days?

#HL = t/T = total time elapsed/ half life time #HL = t/T = 24/8 = 3 so three half lives have gone by.. Cut 10.0g in half 3 times: once 5.00, twice 2.50, three

times 1.25g


Going Backwards in Time

How many grams of a 10.0 gram sample of I-131 (half-life of 8 days) would there have been 24 days ago?

#HL = t/T = 24/8 = 3 half lives Going back in time we would have to double our current

amount to find out what we had 3 half lives ago.

Double 10.0g 3 times: once 20.0, twice 40.0, three times 80.0 g is what you would have started with.


Radioactive Dating

A sample of an ancient scroll contains 50% of the original steady-state concentration of C-14. How old is the scroll?

It contains 50% and percent is out of 100 thus, 1 half life has gone by since half of the C-14 concentration is gone.

50% = 1 HL 1 HL X 5730 y/HL = 5730y


Nuclear Power

Artificial Transmutation Particle Accelerators Nuclear Fission Nuclear Fusion


Artificial Transmutation 40

20Ca + _____ -----> 4019K + 1

1H

9642Mo + 2

1H -----> 10n + _____

Nuclide + Bullet --> New Element + Fragment(s)

The masses and atomic numbers must add up to be the same on both sides of the arrow. Conservation of mass.

Susan Katzoff and Mark Rosengarten

Particle Accelerators

Devices that use electromagnetic fields to accelerate particle “bullets” towards target nuclei to make artificial transmutation possible!

Most of the elements from 93 on up (the “transuranium” elements) were created using particle accelerators.

Particles with no charge cannot be accelerated by the charged fields.


Nuclear Fission

23592U + 1

0n 9236Kr + 141

56Ba + 3 10n + energy

The three neutrons given off can be reabsorbed by other U-235 nuclei to continue fission as a chain reaction (almost like dominos, one will generate the whole chain to fall)

A tiny bit of mass is lost (mass defect) and converted into a huge amount of energy.


Chain Reaction


Nuclear Fusion

21H + 2

1H 42He + energy

Two small, positively-charged nuclei smash together at high temperatures and pressures to form one larger nucleus.

A small bit of mass is destroyed and converted into a huge amount of energy, more than even fission.


Electron Configuration

Basic Configuration Valence Electrons Electron-Dot (Lewis Dot) Diagrams Excited vs. Ground State What is Light?


Basic Configuration

The number of electrons is determined from the atomic number.

Look up the basic configuration below the atomic number on the periodic table. (PEL: principal energy level = shell)

He: 2 (2 e- in the 1st PEL) Na: 2-8-1 (2 e- in the 1st PEL, 8 in the 2nd and 1 in the

3rd) Br: 2-8-18-7 (2 e- in the 1st PEL, 8 in the 2nd, 18 in the

3rd and 7 in the 4th)


Valence Electrons

The valence electrons are responsible for all chemical bonding.

The valence electrons are the electrons in the outermost PEL (shell).

He: 2 (2 valence electrons) Na: 2-8-1 (1 valence electron) Br: 2-8-18-7 (7 valence electrons)

The maximum number of valence electrons an atom can have is EIGHT, called a STABLE OCTET.


Electron-Dot Diagrams

The number of dots equals the number of valence electrons.

The number of unpaired valence electrons in a nonmetal tells you how many covalent bonds that atom can form with other nonmetals OR how many electrons it wants to gain from metals to form an ion.

The number of valence electrons in a metal tells you how many electrons the metal will lose to nonmetals to form an ion.

EXAMPLE DOT DIAGRAMSMark Rosengarten and Susan Katzoff

Example Dot Diagrams

Carbon can also have this dot diagram, which ithas when it forms organic compounds.Mark Rosengarten and Susan Katzoff

Excited vs. Ground State

Configurations on the Periodic Table are all ground state electron configurations.

If electrons are given energy, they rise to higher energy levels (the excited state).

If the total number of electrons matches in the configuration, but the configuration doesn’t match, the atom is in the excited state.

Na (ground, on table): 2-8-1 Example of excited states: 2-7-2, 2-7-1-1, 2-6-3 (note the total

number of electrons is still 11 if you add them up)Mark Rosengarten and Susan Katzoff

What Is Light?

Light is formed when electrons drop from the excited state to the ground state.

What goes up must eventually come back down. When the electrons fall back they release energy in the form of light.

The lines on a bright-line spectrum come from specific energy level drops and are unique to each element. (like fingerprints)


EXAMPLE SPECTRUM

This is the bright-line spectrum of hydrogen. The topnumbers represent the PEL (shell) change that produces the light with that color and the bottom number is thewavelength of the light (in nanometers, or 10-9 m).

No other element has the same bright-line spectrum ashydrogen, so these spectra can be used to identifyelements or mixtures of elements.


Development of the Atomic Model Thompson Model Rutherford Gold Foil Experiment and Mode

l Bohr Model Quantum-Mechanical Model


JJ Thompson Model

The atom is a positively charged diffuse mass with negatively charged electrons stuck in it. (like a Chocolate-chip cookie)


Rutherford Model

The atom is made of a small, dense, positively charged nucleus with electrons at a distance, the vast majority of the volume of the atom is empty space.

Alpha particles shotat a thin sheet of goldfoil: most go through(empty space). Somedeflect or bounce off(small + chargednucleus).


Bohr Model

Electrons orbit around the nucleus in energy levels (shells). Atomic bright-line spectra was the clue. (fills up by 2n2)


Wave-Mechanical Model (Electron Cloud Model) Electron energy levels are wave functions. Electrons are found in orbitals, regions of space where an

electron is most likely to be found. You can’t know both where the electron is and where it is

going at the same time. Electrons buzz around the nucleus like gnats buzzing

around your head.


Matter

1) Properties of Phases

2) Types of Matter

3) Phase Changes


Properties of Phases

Solids: Crystal lattice (regular geometric pattern), vibration motion only

Liquids: particles flow past each other but are still attracted to each other.

Gases: particles are small and far apart, they travel in a

straight line until they hit something, they bounce off container walls and other gas particles without losing any energy, they are so far apart from each other that they have almost no attractive forces and almost no volume, their speed is directly proportional to the Kelvin temperature (Kinetic-Molecular Theory, Ideal Gas Theory)


Solids

The positive and negative ions alternate in the ionic crystal latticeof NaCl.

Definite ShapeDefinite VolumeLow Entropy


Liquids

When heated, the ions or particles move faster and Eventually separate from eachother to form a liquid.

Held together loosely but areMoving to fast to maintain a Crystal lattice structure.


GasesSince all gas molecules spread outthe same way, equal volumes of gas under equal conditions of temperature and pressure will contain equal numbers of molecules of gas. 22.4 L of any gas at STP (1.00 atm and 273K)will contain one mole (6.02 X 1023) of gas molecules.

Since there is space between gasmolecules, gases are affected bychanges in pressure.


Types of Matter Pure Substances (Homogeneous composition)

– Elements (cannot be decomposed by chemical change): Al, Ne, O, Br, H

– Compounds (can be decomposed by chemical change): NaCl, Cu(ClO3)2, KBr, H2O, C2H6

Mixtures– Homogeneous: Solutions (solvent + solute)

– The same consistency throughout (salt water)

– Heterogeneous: soil, Italian dressing, etc.

– Can identify the different components.


Elements A sample of lead atoms (Pb). All

atoms in the sample consist of lead, so the substance is homogeneous.

A sample of chlorine atoms (Cl). All atoms in the sample consist of chlorine, so the substance is homogeneous.


Compounds Lead has two charges listed, +2

and +4. This is a sample of lead (II) chloride (PbCl2). Two or more elements bonded in a definite whole-number ratio is a COMPOUND.

This compound is formed from the +4 version of lead. This is lead (IV) chloride (PbCl4). Notice how both samples of lead compounds have consistent composition throughout? Compounds are homogeneous!


Mixtures A mixture of lead atoms and

chlorine atoms. They exist in no particular ratio and are not chemically combined with each other. They can be separated by physical means.

A mixture of PbCl2 and PbCl4 formula units. Again, they are in no particular ratio to each other and can be separated without chemical change.


Phase Changes

Phase Change Types Phase Change Diagrams Heat of Phase Change Evaporation


Phase Change Types


Phase Change Diagrams

AB: Solid PhaseBC: Melting (S + L)CD: Liquid PhaseDE: Boiling (L + G)EF: Gas Phase

Notice how temperature remains constant during a phase change? That’s because the PE is changing, not the KE. Where temperature changes potential energy remains the same. Mark Rosengarten and Susan Katzoff

Heat of Phase Change

How many joules would it take to melt 100. g of H2O (s) at 0oC?

q=mHf = (100. g)(334 J/g) = 33400 J

How many joules would it take to boil 100. g of H2O (l) at 100oC?

q=mHv = (100.g)(2260 J/g) = 226000 J


Evaporation

When the surface molecules of a gas travel upwards at a great enough speed to escape.

The pressure a vapor exerts when sealed in a container at equilibrium is called vapor pressure, and can be found on Table H.

When the liquid is heated, its vapor pressure increases. When the liquid’s vapor pressure equals the pressure in the

atmosphere, the liquid will boil. If the pressure exerted on a liquid increases, the boiling

point of the liquid increases (pressure cooker). If the pressure decreases, the boiling point of the liquid decreases (special cooking directions for high elevations).


Reference Table H: Vapor Pressure of Four Liquids


Bonding

1) The Periodic Table

2) Ions

3) Ionic Bonding

4) Covalent Bonding

5) Metallic Bonding


The Periodic Table

Metals Nonmetals Metalloids Chemistry of Groups Electronegativity Ionization Energy


Metals

Have luster, are malleable and ductile, good conductors of heat and electricity

Lose electrons to nonmetal atoms to form positively charged ions in ionic bonds

Large atomic radii compared to nonmetal atoms Low electronegativity and ionization energy Left side of the periodic table (except Hydrogen)


Nonmetals

Are dull and brittle, poor conductors of heat and electricity

Gain electrons from metal atoms to form negatively charged ions in ionic bonds

Share unpaired valence electrons with other nonmetal atoms to form covalent bonds and molecules

Small atomic radii compared to metal atoms High electronegativity and ionization energy Right side of the periodic table (except Group 18)


Metalloids Found lying on the jagged line between metals and nonmetals flatly touching

the line (except Al and Po). Remember like the dog food brand Alpo. Share properties of metals and nonmetals (Si is shiny like a metal, brittle like

a nonmetal and is a semiconductor used in solar panels and computers).


Chemistry of Groups

Group 1: Alkali Metals Group 2: Alkaline Earth Metals Groups 3-11: Transition Elements Group 17: Halogens Group 18: Noble Gases

Diatomic MoleculesMark Rosengarten and Susan Katzoff

Group 1: Alkali Metals

The Most active metals, only found in compounds in nature

React violently with water to form hydrogen gas and a strong base:

2 Na (s) + H2O (l) 2 NaOH (aq) + H2 (g)

1 valence electron Form +1 ion by losing that valence electron Form oxides like Na2O, Li2O, K2O


Group 2: Alkaline Earth Metals

Very active metals, only found in compounds in nature

React strongly with water to form hydrogen gas and a base:

– Ca (s) + 2 H2O (l) Ca(OH)2 (aq) + H2 (g)

2 valence electrons Form +2 ion by losing those valence electrons Form oxides like CaO, MgO, BaO


Groups 3-11: Transition Metals

Multiple positive oxidation states If there is more than one ion listed, give the charge as a

Roman numeral after the name Cu+1 = copper (I) Cu+2 = copper (II) Compounds containing these metals can be colored.


Group 17: Halogens

Most reactive nonmetals React violently with metal atoms to form halide

compounds: 2 Na + Cl2 2 NaCl

Only found in compounds in nature Have 7 valence electrons Gain 1 valence electron from a metal to form -1

ions Share 1 valence electron with another nonmetal

atom to form one covalent bond.


Group 18: Noble Gases

Are completely nonreactive since they have eight valence electrons, making a stable octet.

Kr and Xe can be forced, in the laboratory, to give up some valence electrons to react with fluorine.

Since noble gases do not naturally bond to any other elements, one atom of noble gas is considered to be a molecule of noble gas. This is called a monatomic molecule. Ne represents an atom of Ne and a molecule of Ne.


Diatomic Molecules

Br, I, N, Cl, H, O and F are so reactive that they exist in a more chemically stable state when they covalently bond with another atom of their own element to make two-atom, or diatomic molecules.

Br2, I2, N2, Cl2, H2, O2 and F2

The decomposition of water: 2 H2O 2 H2 + O2


Electronegativity (E.N.) An atom’s attraction to another atoms electrons in a chemical

bond. F has the highest, at 4.0 Fr has the lowest, at 0.7 If two atoms have a difference in E.N. from each other by 1.7 or

more collide and bond (like a metal atom and a nonmetal atom), the one with the higher electronegativity will pull the valence electrons away from the atom with the lower electronegativity to form a (-) ion. The atom that was stripped of its valence electrons forms a (+) ion.

If the two atoms have an E.N.D of less than 1.7, they will share their unpaired valence electrons…covalent bond!


Ionization Energy

The energy required to remove the most loosely held valence electron from an atom in the gas phase.

High electronegativity means high ionization energy because if an atom is more attracted to electrons, it will take more energy to remove those electrons.

Metals have low ionization energy. They lose electrons easily to form (+) charged ions.

Nonmetals have high ionization energy but high electronegativity. They gain electrons easily to form (-) charged ions when reacted with metals, or share unpaired valence electrons with other nonmetal atoms.


Ions

Ions are charged particles formed by the gain or loss of electrons.– Metals lose electrons (oxidation) to form (+) charged cations.

– Nonmetals gain electrons (reduction) to form (-) charged anions.

Atoms will gain or lose electrons in such a way that they end up with 8 valence electrons (stable octet).– The exceptions to this are H, Li, Be and B, which are not

large enough to support 8 valence electrons. They must be satisfied with 2.


Metal Ions (Cations) Na: 2-8-1

Na+1: 2-8

Ca: 2-8-8-2

Ca+2: 2-8-8

Al: 2-8-3

Al+3: 2-8

Note that when the atom loses its valence electron, the next lower PEL becomes the valence PEL.

Notice how the dot diagrams for metal ions lack dots! Place brackets around the element symbol and put the charge on the upper right outside!


Nonmetal Ions (Anions)

F: 2-7 F-1: 2-8

O: 2-6 O-2: 2-8

N: 2-5 N-3: 2-8

Note how the ions all have 8 valence electrons. Also note the gained electrons as red dots. Nonmetal ion dot diagrams show 8 dots, with brackets around the dot diagram and the charge of the ion written to the upper right side outside the brackets.


Ionic Bonding

If two atoms that are different in EN (END) from each other by 1.7 or more collide and bond (like a metal atom and a nonmetal atom), the one with the higher electronegativity will pull the valence electrons away from the atom with the lower electronegativity to form a (-) ion. The atom that was stripped of its valence electrons forms a (+) ion.

The oppositely charged ions attract to form the bond. It is a surface bond that can be broken by melting or dissolving in water.

Ionic bonding forms ionic crystal lattices, not molecules.Mark Rosengarten and Susan Katzoff

Example of Ionic Bonding


Covalent Bonding

If two nonmetal atoms have an END of 1.7 or less, they will share their unpaired valence electrons to form a covalent bond.

A particle made of covalently bonded nonmetal atoms is called a molecule.

If the END is between 0 and 0.4, the sharing of electrons is equal, so there are no charged ends. This is NONPOLAR covalent bonding.

If the END is between 0.5 and 1.7, the sharing of electrons is unequal. The atom with the higher EN will be - and the one with the lower EN will be + charged. This is a POLAR covalent bonding. (means “partial”)Mark Rosengarten and Susan Katzoff

Examples of Covalent Bonding


Metallic Bonding

Metal atoms of the same element bond with each other by sharing valence electrons that they lose to each other.

This is a lot like an atomic game of “hot potato”, where metal kernals (the atom inside the valence electrons) sit in a crystal lattice, passing valence electrons back and forth between each other).

Since electrons can be forced to travel in a certain direction within the metal, metals are very good at conducting electricity in all phases.


Compounds

1) Types of Compounds

2) Formula Writing

3) Formula Naming

4) Empirical Formulas

5) Molecular Formulas

6) Types of Chemical Reactions

7) Balancing Chemical Reactions

8) Attractive ForcesMark Rosengarten and Susan Katzoff

Types of Compounds

Ionic: made of metal and nonmetal ions. Form an ionic crystal lattice when in the solid phase. Ions separate when melted or dissolved in water, allowing electrical conduction. Examples: NaCl, K2O, CaBr2

Molecular: made of nonmetal atoms bonded to form a distinct particle called a molecule. Bonds do not break upon melting or dissolving, so molecular substances do not conduct electricity. EXCEPTION: Acids [H+A- (aq)] ionize in water to form H3O+ and A-, so they do conduct.


Ionic CompoundsIonic Crystal Structure, then adding heat (or dissolving in water) to breakup the crystal into a liquid composed of free-moving ions.


Molecular Compounds


Formula Writing

The charge of the (+) ion and the charge of the (-) ion must cancel out to make the formula. Use subscripts to indicate how many atoms of each element there are in the compound, no subscript if there is only one atom of that element.

Na+1 and Cl-1 = NaCl Ca+2 and Br-1 = CaBr2

Al+3 and O-2 = Al2O3

Zn+2 and PO4-3 = Zn3(PO4)2

Try these problems!Mark Rosengarten and Susan Katzoff

Formulas to Write

Ba+2 and N-3 ->

NH4+1 and SO4

-2

Li+1 and S-2

Cu+2 and NO3-1

Al+3 and CO3-2

Fe+3 and Cl-1

Pb+4 and O-2

Pb+2 and O-2


Formula Naming

Compounds are named from the elements or polyatomic ions that form them.

KCl = potassium chloride Na2SO4 = sodium sulfate

(NH4)2S = ammonium sulfide

AgNO3 = silver nitrate

Notice all the metals listed here only have one charge listed? So what do you do if a metal has more than one charge listed? Take a peek!


The Stock System

CrCl2 = chromium (II) chloride Try

CrCl3 = chromium (III) chloride Co(NO3)2 and

CrCl6 = chromium (VI) chloride Co(NO3)3

FeO = iron (II) oxide MnS = manganese (II) sulfide Fe2O3 = iron (III) oxide MnS2 = manganese (IV) sulfide

The Roman numeral is the charge of the metal ion!Mark Rosengarten and Susan Katzoff

Empirical Formulas

Ionic formulas: represent the simplest whole number mole ratio of elements in a compound.

Ca3N2 means a 3:2 ratio of Ca ions to N ions in the compound.

Many molecular formulas can be simplified to empirical formulas

– Ethane (C2H6) can be simplified to CH3. This is the empirical formula…the ratio of C to H in the molecule.

All ionic compounds have empirical formulas.


Molecular Formulas

The count of the actual number of atoms of each element in a molecule.

H2O: a molecule made of two H atoms and one O atom covalently bonded together.

C2H6O: A molecule made of two C atoms, six H atoms and one O atom covalently bonded together.

Molecular formulas are whole-number multiples of empirical formulas:

– H2O = 1 X (H2O)

– C8H16 = 8 X (CH2)

Calculating Molecular FormulasMark Rosengarten and Susan Katzoff

Types of Chemical Reactions

Redox Reactions: driven by the loss (oxidation) and gain (reduction) of electrons. Any species that does not change charge is called the spectator ion.

– Synthesis

– Decomposition

– Single Replacement

Ion Exchange Reaction: driven by the formation of an insoluble precipitate. The ions that remain dissolved throughout are the spectator ions.

– Double ReplacementMark Rosengarten and Susan Katzoff

Synthesis

Two elements combine to form a compound 2 Na + O2 Na2O

Same reaction, with charges added in:

– 2 Na0 + O20 Na2

+1O-2

Na0 is oxidized (loses electrons), is the reducing agent O2

0 is reduced (gains electrons), is the oxidizing agent

Electrons are transferred from the Na0 to the O20.

No spectator ions, there are only two elements here.Mark Rosengarten and Susan Katzoff

Decomposition

A compound breaks down into its original elements. Na2O 2 Na + O2


– Na2+1O-2 2 Na0 + O2

0

O-2 is oxidized (loses electrons), is the reducing agent Na+1 is reduced (gains electrons), is the oxidizing agent

Electrons are transferred from the O-2 to the Na+1.

No spectator ions, there are only two elements here.Mark Rosengarten and Susan Katzoff

Single Replacement

An element replaces the same type of element in a compound. Ca + 2 KCl CaCl2 + 2 K


– Ca0 + 2 K+1Cl-1 Ca+2Cl2-1 + 2 K0

Ca0 is oxidized (loses electrons), is the reducing agent K+1 is reduced (gains electrons), is the oxidizing agent

Electrons are transferred from the Ca0 to the K+1.

Cl-1 is the spectator ion, since it’s charge doesn’t change.


Double Replacement

The (+) ion of one compound bonds to the (-) ion of another compound to make an insoluble precipitate. The compounds must both be dissolved in water to break the ionic bonds first.

NaCl (aq) + AgNO3 (aq) NaNO3 (aq) + AgCl (s)

The Cl-1 and Ag+1 come together to make the insoluble precipitate, which looks like snow in the test tube.

No species change charge, so this is not a redox reaction. Since the Na+1 and NO3

-1 ions remain dissolved throughout the reaction, they are the spectator ions.

How do identify the precipitate?


Identifying the Precipitate

The precipitate is the compound that is insoluble. AgCl is a precipitate because Cl- is a halide. Halides are soluble, except when combined with Ag+ and others.Mark Rosengarten and Susan Katzoff

Balancing Chemical Reactions

Balance one element or ion at a time Use a pencil Use coefficients only, never change formulas Revise if necessary

The coefficient multiplies everything in the formula by that amount

– 2 Ca(NO3)2 means that you have 2 Ca, 4 N and 12 O.

Examples for you to try!


Reactions to Balance

___NaCl ___Na + ___Cl2

___Al + ___O2 ___Al2O3

___SO3 ___SO2 + ___O2

___Ca + ___HNO3 ___Ca(NO3)2 + ___H2

__FeCl3 + __Pb(NO3)2 __Fe(NO3)3 + __PbCl2



__2_NaCl __2_Na + _1__Cl2

__4_Al + _3__O2 _2__Al2O3

_2__SO3 _2__SO2 + __1_O2

_1__Ca + _2__HNO3 _1__Ca(NO3)2 + _1__H2

2__FeCl3 + _3_Pb(NO3)2 _2_Fe(NO3)3 + _3_PbCl2

Reactions to Balance Answers

Attractive Forces

Molecules have partially charged ends. The + end of one molecule attracts to the - end of another molecule.

Ions are charged (+) or (-). Opposite charged ions attract to form ionic bonds, a type of attractive force.

Since partially charged ends result in weaker attractions than fully charged ends, ionic compounds generally have much higher melting points than molecular compounds.

Determining Polarity of Molecules Hydrogen Bond Attractions


Determining Polarity ofMolecules

-----------------------------------------------------------------------------


Hydrogen BondAttractions

A hydrogen bond attraction is a very strong attractive force between the H end of one polar molecule and the F, N, or O end of another polar molecule. This attraction is so strong that water is a liquid at a temperature where most compounds that are much heavier than water (like propane, C3H8) are gases.


Math of Chemistry

1) Formula Mass

2) Percent Composition

3) Mole Problems

4) Gas Laws

5) Neutralization

6) Concentration

7) Significant Figures and Rounding

8) Metric Conversions

9) CalorimetryMark Rosengarten and Susan Katzoff

Formula Mass

Gram Formula Mass = sum of atomic masses of all elements in the compound

Round given atomic masses to the nearest tenth H2O: (2 X 1.0) + (1 X 16.0) = 18.0 grams/mole

Na2SO4: (2 X 23.0)+(1 X 32.1)+(4 X 16.0) = 142.1 g/mole

Now you try:

– BaBr2

– CaSO4

– Al2(CO3)3


Percent Composition

What is the % composition, by mass,of each element in SiO2?

%Si = (28.1/60.1) X 100 = 46.8%%O = (2 X 16.0 = 32.0), (32.0/60.1) X 100 = 53.2%

The mass of part is the number of atoms of that element in the compound. The mass of whole is the formula mass of the compound. Don’t forget to take atomic mass to the nearest tenth! This is a problem for you to try.Mark Rosengarten and Susan Katzoff

Practice PercentComposition Problem What is the percent by mass of each element in Li2SO4?


Mole Problems

Grams <=> Moles Molecular Formula Stoichiometry


Grams <=> Moles

How many grams will 3.00 moles of NaOH (40.0 g/mol) weigh?

3.00 moles X 40.0 g/mol = 120. g

How many moles of NaOH (40.0 g/mol) are represented by 10.0 grams?

(10.0 g) / (40.0 g/mol) = 0.250 mol


Molecular Formula

Molecular Formula = (Molecular Mass/Empirical Mass) X Empirical Formula

What is the molecular formula of a compound with an empirical formula of CH2 and a molecular mass of 70.0 grams/mole?

1) Find the Empirical Formula Mass: CH2 = 14.0

2) Divide the MM/EM: 70.0/14.0 = 5 3) Multiply the molecular formula by the result:

5 (CH2) = C5H10


Stoichiometry Moles of Target = Moles of Given X (Coefficent of Target/Coefficient of given)

Given the balanced equation N2 + 3 H2 2 NH3, How many moles of H2 need to be completely reacted with N2 to yield 20.0 moles of NH3?

20.0 moles NH3 = 2 moles NH3

Xmoles H2 3moles H2

= 30.0 moles H2


Gas Laws

Make a data table to put the numbers so you can eliminate the words.

Make sure that any Celsius temperatures are converted to Kelvin (add 273).

Rearrange the equation before substituting in numbers. If you are trying to solve for T2, get it out of the denominator first by cross-multiplying.

If one of the variables is constant, then eliminate it. Try these problems!


Gas Law Problem 1

A 2.00 L sample of N2 gas at STP is compressed to 4.00 atm at constant temp-erature. What is the new volume of the gas?

V2 = P1V1 / P2

= (1.00 atm)(2.00 L) /(4.00 atm) = 0.500 L Mark Rosengarten and Susan Katzoff

Gas Law Problem 2

To what temperature must a 3.000 L sample of O2 gas at 300.0 K be heated to raise the volume to 10.00 L?

T2 = V2T1/V1

= (10.00 L)(300.0 K) / (3.000 L) = 1000. K


Gas Law Problem 3

A 3.00 L sample of NH3 gas at 100.0 kPa is cooled from 500.0 K to 300.0 K and its pressure is reduced to 80.0 kPa.

What is the new volume of the gas? V2 = P1V1T2 / P2T1

= (100.0 kPa)(3.00 L)(300. K) / (80.0 kPa)(500. K) = 2.25 L


Neutralization

10.0 mL of 0.20 M HCl is neutralized by 40.0 mL of NaOH. What is the concentration of the NaOH?

#H MaVa = #OH MbVb, so Mb = #H MaVa / #OH Vb

= (1)(0.20 M)(10.0 mL) / (1) (40.0 mL) = 0.050 M

How many mL of 2.00 M H2SO4 are needed to completely neutralize 30.0 mL of 0.500 M KOH?


Concentration

Molarity Parts per Million Percent by Mass Percent by Volume


Molarity

What is the molarity of a 500.0 mL solution of NaOH (FM = 40.0) with 60.0 g of NaOH (aq)?– Convert g to moles and mL to L first!

– M = moles / L = 1.50 moles / 0.5000 L = 3.00 M

How many grams of NaOH does it take to make 2.0 L of a 0.100 M solution of NaOH (aq)?– Moles = M X L = 0.100 M X 2.0 L = 0.200 moles

– Convert moles to grams: 0.200 moles X 40.0 g/mol = 8.00 g


Parts Per Million

100.0 grams of water is evaporated and analyzed for lead. 0.00010 grams of lead ions are found. What is the concentration of the lead, in parts per million?

ppm = (0.00010 g) / (100.0 g) X 1 000 000 = 1.0 ppm If the legal limit for lead in the water is 3.0 ppm, then the water

sample is within the legal limits (it’s OK!)


Percent by Mass

A 50.0 gram sample of a solution is evaporated and found to contain 0.100 grams of sodium chloride. What is the percent by mass of sodium chloride in the solution?

% Comp = (0.100 g) / (50.0 g) X 100 = 0.200%


Percent By Volume

Substitute “volume” for “mass” in the above equation.

What is the percent by volume of hexane if 20.0 mL of hexane are dissolved in benzene to a total volume of 80.0 mL?

% Comp = (20.0 mL) / (80.0 mL) X100 = 25.0%


Sig Figs and Rounding

How many Significant Figures does a number have?

What is the precision of my measurement?

How do I round off answers to addition and subtraction problems?

How do I round off answers to multiplication and division problems?


How many Sig Figs?

Start counting sig figs at the first non-zero. All digits except place-holding zeroes are sig figs.

Measurement # of Sig Figs

234 cm 3

67000 cm 2

_ 45000 cm

4

560. cm 3

560.00 cm 5

Measurement # of Sig Figs

0.115 cm 3

0.00034 cm 2

0.00304 cm 3

0.0560 cm 3

0.00070700 cm 5


What Precision?

A number’s precision is determined by the furthest (smallest) place the number is recorded to.

6000 mL : thousands place 6000. mL : ones place 6000.0 mL : tenths place 5.30 mL : hundredths place 8.7 mL : tenths place 23.740 mL : thousandths place


Rounding with addition and subtraction Answers are rounded to the least precise place. Find the data with the least number of decimal places, round to that

number of decimal places.

1) 4.732 cm 2) 17.440 mL 3) 32.0 MW 16.8 cm 3.895 mL + 0.0059 MW + 0.781 cm + 16.77 mL --------------- ---------- -------------- 22.313 cm 38.105 mL 32.0059 MW 22.3 cm 38.11 mL 32.0 MW


Rounding with multiplicationand division Answers are rounded to the fewest number of significant

figures from the data.

1) 37.66 KW 2) 14.922 cm 3) 98.11 kg x 2.2 h x 2.0 cm x 200 m ---------- ----------- ---------- 82.852 KWh 29.844 cm2 19 622 kgm 83 KWh 30. cm2 20 000 kgm


Calorimetry

This equation can be used to determine any of the variables here. You will not have to solve for C, since we will always assume that the energy transfer is being absorbed by or released by a measured quantity of water, whose specific heat is given above.

Solving for q Solving for m Solving for DT Mark Rosengarten and Susan Katzoff

Solving for q

How many joules are absorbed by 100.0 grams of water in a calorimeter if the temperature of the water increases from 20.0oC to 50.0oC?

q = mCDT = (100.0 g)(4.18 J/goC)(30.0oC) = 12500 J


Solving for m

A sample of water in a calorimeter cup increases from 25oC to 50.oC by the addition of 500.0 joules of energy. What is the mass of water in the calorimeter cup?

q = mCDT, so m = q / CDT = (500.0 J) / (4.18 J/goC)(25oC) = 4.8 g


Solving for DT

If a 50.0 gram sample of water in a calorimeter cup absorbs 1000.0 joules of energy, how much will the temperature rise by?

q = mCDT, so DT = q / mC = (1000.0 J)/(50.0 g)(4.18 J/goC) = 4.8oC

If the water started at 20.0oC, what will the final temperature be?– Since the water ABSORBS the energy, its temperature will

INCREASE by the DT: 20.0oC + 4.8oC = 24.8oC


Kinetics and Thermodynamics

1) Reaction Rate

2) Heat of Reaction

3) Potential Energy Diagrams

4) Equilibrium

5) Le Châtelier’s Principle

6) Solubility Curves


Reaction Rate

Reactions happen when reacting particles collide with sufficient energy (activation energy) and at the proper angle.

Anything that makes more collisions in a given time will make the reaction rate increase.

– Increasing temperature

– Increasing concentration (pressure for gases)

– Increasing surface area (solids)

Adding a catalyst makes a reaction go faster by removing steps from the mechanism and lowering the activation energy without getting used up in the process.


Heat of Reaction

Reactions either absorb PE (endothermic, +DH) or release PE (exothermic, -DH)

Exothermic, PEKE, Temp

Endothermic, KEPE, Temp

Rewriting the equation with heat included:

4 Al(s) + 3 O2(g) 2 Al2O3(s) + 3351 kJ

N2(g) + O2(g) +182.6 kJ 2 NO(g)Mark Rosengarten and Susan Katzoff

Potential Energy Diagrams (P.E.)

Steps of a reaction:

– Reactants have a certain amount of PE stored in their bonds (Potential Energy of Reactants)

– The reactants are given enough energy to collide and react (Activation Energy)

– The resulting intermediate peak has the highest energy that the reaction can make (P.E. of Activated Complex)

– The activated complex breaks down and forms the products, which have a certain amount of PE stored in their bonds (Potential Energy of Products)

DH = P.E.P – P.E.R. EXAMPLES Mark Rosengarten and Susan Katzoff

Making a PE Diagram

X axis: Reaction Coordinate (time, no units) Y axis: PE (kJ) Three lines representing energy (PEreactants, PEactivated complex,

PEproducts)

Two arrows representing energy changes:

– From PEreactants to PEactivated complex: Activation Energy

– From PEreactants to PEproducts : DH

ENDOTHERMIC PE DIAGRAM EXOTHERMIC PE DIAGRAM


Endothermic PE Diagram

If a catalyst is added?Mark Rosengarten and Susan Katzoff

Endothermic with Catalyst

The red line represents the catalyzed reaction.Mark Rosengarten and Susan Katzoff

Exothermic PE Diagram

What does it look like with a catalyst?Mark Rosengarten and Susan Katzoff

Exothermic with a Catalyst

The red line represents the catalyzed reaction. Lower A.E. and faster reaction time!Mark Rosengarten and Susan Katzoff

Equilibrium

When the rate of the forward reaction equals the rate of the reverse reaction.


Examples of Equilibrium

Solution Equilibrium: when a solution is saturated, the rate of dissolving equals the rate of precipitating.– NaCl (s) Na+1 (aq) + Cl-1 (aq)

Vapor-Liquid Equilibrium: when a liquid is trapped with air in a container, the liquid evaporates until the rate of evaporation equals the rate of condensation.– H2O (l) H2O (g)

Phase equilibrium: At the melting point, the rate of solid turning to liquid equals the rate of liquid turning back to solid.– H2O (s) H2O (l)


Le Châtelier’s Principle

If a system at equilibrium is stressed, the equilibrium will shift in a direction that relieves that stress.

A stress is a factor that affects reaction rate. Since catalysts affect both reaction rates equally, catalysts have no effect on a system already at equilibrium.

Equilibrium will shift AWAY from what is added Equilibrium will shift TOWARDS what is removed. This is because the shift will even out the change in

reaction rate and bring the system back to equilibrium

» NEXTMark Rosengarten and Susan Katzoff

Steps to Relieving Stress

1) Equilibrium is subjected to a STRESS. 2) System SHIFTS towards what is removed from the system or

away from what is added. The shift results in a CHANGE OF CONCENTRATION for both the

products and the reactants.

– If the shift is towards the products, the concentration of the products will increase and the concentration of the reactants will decrease.

– If the shift is towards the reactants, the concentration of the reactants will increase and the concentration of the products will decrease.

» NEXT


Examples

For the reaction N2(g) + 3H2(g) 2 NH3(g) + heat

– Adding N2 will cause the equilibrium to shift RIGHT, resulting in an increase in the concentration of NH3 and a decrease in the concentration of N2 and H2.

– Removing H2 will cause a shift to the LEFT, resulting in a decrease in the concentration of NH3 and an increase in the concentration of N2 and H2.

– Increasing the temperature will cause a shift to the LEFT, same results as the one above.

– Decreasing the pressure will cause a shift to the LEFT, because there is more gas on the left side, and making more gas will bring the pressure back up to its equilibrium amount.

– Adding a catalyst will have no effect, so no shift will happen.Mark Rosengarten and Susan Katzoff

Solubility Curves

Solubility: the maximum quantity of solute that can be dissolved in a given quantity of solvent at a given temperature to make a saturated solution.

Saturated: a solution containing the maximum quantity of solute that the solvent can hold. The limit of solubility.

Supersaturated: the solution is holding more than it can theoretically hold OR there is excess solute which precipitates out. True supersaturation is rare.

Unsaturated: There are still solvent molecules available to dissolve more solute, so more can dissolve.

How ionic solutes dissolve in water: polar water molecules attach to the ions and tear them off the crystal.


Solubility

Solubility: go to the temperature and up to the desired line, then across to the Y-axis. This is how many g of solute are needed to make a saturated solution of that solute in 100g of H2O at that particular temperature.

At 40oC, the solubility of KNO3 in 100g of water is 64 g. In 200g of water, double that amount. In 50g of water, cut it in half.


Supersaturated

If 120 g of NaNO3 are added to 100g of water at 30oC:

1) The solution would be SUPERSATURATED, because there is more solute dissolved than the solubility allows

2) The extra 25g would precipitate out

3) If you heated the solution up by 24oC (to 54oC), the excess solute would dissolve.Mark Rosengarten and Susan Katzoff

UnsaturatedIf 80 g of KNO3 are added to 100g of water at 60oC:

1) The solution would be UNSATURATED, because there is less solute dissolved than the solubility allows

2) 26g more can be added to make a saturated solution

3) If you cooled the solution down by 12oC (to 48oC), the solution would become saturatedMark Rosengarten and Susan Katzoff

How Ionic Solutes Dissolve in Water

Water solvent molecules attach to the ions (H end to the Cl-, O end to the Na+)

Water solvent holds the ions apart and keeps the ions from coming back together


Acids and Bases

1) Properties of Acids

2) Properties of Bases

3) Neutralization

4) pH

5) Indicators

6) Alternate Theories


Formulas, Naming and Properties of Acids Arrhenius Definition of Acids: molecules that dissolve in

water to produce H3O+ (hydronium) as the only positively charged ion in solution.

HCl (g) + H2O (l) H3O+ (aq) + Cl-

Properties of Acids


Properties of Acids

Acids react with metals above H2 on Table J to form H2(g) and a salt.

Acids have a pH of less than 7. Dilute solutions of acids taste sour. Acids turn phenolphthalein CLEAR, litmus RED

and bromthymol blue YELLOW. Acids neutralize bases.


Naming of Acids


Properties of Bases Arrhenius Definition of Bases: ionic compounds that

dissolve in water to produce OH- (hydroxide) as the only negatively charged ion in solution.

NaOH (s) Na+1 (aq) + OH-1 (aq)

Properties of Bases


Properties of Bases

Bases react with animal fats to form soap and glycerol. The process of making soap is saponification.

Bases have a pH of more than 7. Dilute solutions of bases taste bitter. Bases turn phenolphthalein PINK, litmus BLUE and

bromthymol blue BLUE. Bases neutralize acids. Bases are formed when alkali metals or alkaline earth

metals react with water. The words “alkali” and “alkaline” mean “basic”, as opposed to “acidic”.


Naming of Bases


Formula Writing of Bases

Formula writing of bases is the same as for any ionic formula writing. The charges of the ions have to cancel out.

Calcium hydroxide = Ca+2 and OH-1 = Ca(OH)2 (aq)

Potassium hydroxide = K+1 and OH-1 = KOH (aq) Lead (II) hydroxide = Pb+2 and OH-1 = Pb(OH)2 (aq)

Lead (IV) hydroxide = Pb+4 and OH-1 = Pb(OH)4 (aq)


Neutralization

H+1 + OH-1 HOH which can be written as H2O water)

Acid + Base Water + Salt (double replacement)

HCl (aq) + NaOH (aq) HOH (l) + NaCl (aq)

H2SO4 (aq) + KOH (aq) 2 HOH (l) + K2SO4 (aq)

HBr (aq) + LiOH (aq) H2CrO4 (aq) + NaOH (aq) HNO3 (aq) + Ca(OH)2 (aq) H3PO4 (aq) + Mg(OH)2 (aq)


pH

A change of 1 in pH is a tenfold increase in acid or base strength.

A pH of 4 is 10 times more acidic than a pH of 5. A pH of 12 is 100 times more basic than a pH of 10.Mark Rosengarten and Susan Katzoff

Indicators

At a pH of 2:

Methyl Orange = red

Bromthymol Blue = yellow

Phenolphthalein = colorless

Litmus = red

Bromcresol Green = yellow

Thymol Blue = yellow

Methyl orange is red at a pH of 3.2 and below and yellow at a pH of 4.4 and higher. In between the two numbers, it is an intermediate color that is not listed on this table. Mark Rosengarten and Susan Katzoff

Alternate Theories

Arrhenius Theory: acids and bases must be in aqueous solution.

Alternate Theory: Not necessarily so!

– Acid: proton (H+1) donor…gives up H+1 in a reaction.

– Base: proton (H+1) acceptor…gains H+1 in a reaction.

HNO3 + H2O H3O+1 + NO3-1

– Since HNO3 lost an H+1 during the reaction, it is an acid.

– Since H2O gained the H+1 that HNO3 lost, it is a base.


Oxidation and Reduction

1) Oxidation Numbers

2) Identifying OX, RD and SI Species

3) Agents

4) Writing Half-Reactions

5) Balancing Half-Reactions

6) Activity Series

7) Voltaic Cells

8) Electrolytic Cells

9) ElectroplatingMark Rosengarten and Susan Katzoff

Oxidation Numbers

Elements have no charge until they bond to other elements.

– Na0, Li0, H20. S0, N2

0, C600

The formula of a compound is such that the charges of the elements making up the compound all add up to zero.

The symbol and charge of an element or polyatomic ion is called a SPECIES.

Determine the charge of each species in the following compounds:

NaCl KNO3 CuSO4 Fe2(CO3)3


Identifying OX, RD, SI Species

Ca0 + 2 H+1Cl-1 Ca+2Cl-12 + H2

0

Oxidation = loss of electrons. The species becomes more

positive in charge. For example, Ca0 Ca+2, so Ca0 is the species that is oxidized.

Reduction = gain of electrons. The species becomes more

negative in charge. For example, H+1 H0, so the H+1 is the species that is reduced.

Spectator Ion = no change in charge. The species does not gain or lose any electrons. For example, Cl-1 Cl-1, so the Cl-1 is the spectator ion.Mark Rosengarten and Susan Katzoff

Agents

Ca0 + 2 H+1Cl-1 Ca+2Cl-12 + H2

0

Since Ca0 is being oxidized and H+1 is being reduced, the electrons must be going from the Ca0 to the H+1.

Since Ca0 would not lose electrons (be oxidized) if H+1 weren’t there to gain them, H+1 is the cause, or agent, of Ca0’s oxidation. H+1 is the oxidizing agent.

Since H+1 would not gain electrons (be reduced) if Ca0 weren’t there to lose them, Ca0 is the cause, or agent, of H+1’s reduction. Ca0 is the reducing agent.


Writing Half-Reactions

Ca0 + 2 H+1Cl-1 Ca+2Cl-12 + H2

0

Oxidation: Ca0 Ca+2 + 2e- Reduction: 2H+1 + 2e- H2

0

The two electrons lost by Ca0 are gained by the two H+1 (each H+1 picks up an electron).

PRACTICE SOME!Mark Rosengarten and Susan Katzoff

Practice Half-Reactions

Don’t forget to determine the charge of each species first!

4 Li + O2 2 Li2O

Oxidation Half-Reaction: Reduction Half-Reaction:

Zn + Na2SO4 ZnSO4 + 2 Na

Oxidation Half-Reaction: Reduction Half-Reaction:


Balancing Half-Reactions

Ca0 + Fe+3 Ca+2 + Fe0

– Ca’s charge changes by 2, so double the Fe.

– Fe’s charge changes by 3, so triple the Ca.

– 3 Ca0 + 2 Fe+3 3 Ca+2 + 2 Fe0

Try these: __Na0 + __H+1 __Na+1 + __H2

0

– (hint: balance the H and H2 first!)

__Al0 + __Cu+2 __Al+3 + __Cu0


Activity Series For metals, the higher up the chart the

element is, the more likely it is to be oxidized. This is because metals like to lose electrons, and the more active a metallic element is, the more easily it can lose them.

For nonmetals, the higher up the chart the element is, the more likely it is to be reduced. This is because nonmetals like to gain electrons, and the more active a nonmetallic element is, the more easily it can gain them.


Metal Activity

Metallic elements start out with a charge of ZERO, so they can only be oxidized to form (+) ions.

The higher of two metals MUST undergo oxidation in the reaction, or no reaction will happen.

The reaction 3 K + FeCl3 3 KCl + Fe WILL happen, because K is being oxidized, and that is what Table J says should happen.

The reaction Fe + 3 KCl FeCl3 + 3 K will NOT happen.

3 K0 + Fe+3Cl-13

REACTION

Fe0 + 3 K+1Cl-1

NO REACTION


Voltaic Cells

Produce electrical current using a spontaneous redox reaction Used to make batteries! Materials needed: two beakers, piece of the oxidized metal

(anode, - electrode), solution of the oxidized metal, piece of the reduced metal (cathode, + electrode), solution of the reduced metal, porous material (salt bridge), solution of a salt that does not contain either metal in the reaction, wire and a load to make use of the generated current!

Use Reference Table J to determine the metals to use– Higher = (-) anode Lower = (+) cathode


Making Voltaic Cells

Create

Your

Own

Cell!!!!

More

Info!!!


How It Works

The Zn0 anode loses 2 e-, which go up the wire and through the load. The Zn0 electrode gets smaller as the Zn0 becomes Zn+2 and dissolves into solution. The e- go into the Cu0, where they sit on the outside surface of the Cu0 cathode and wait for Cu+2 from the solution to come over so that the e- can jump on to the Cu+2 and reduce it to Cu0. The size of the Cu0 electrode increases. The negative ions in solution go over the salt bridge to the anode side to complete the circuit.

Since Zn is listed above Cu, Zn0 will be oxidized when it reacts with Cu+2. The reaction: Zn + CuSO4 ZnSO4 + Cu


You Start At The Anode


Electrolytic Cells Use electricity to force a nonspontaneous redox reaction to

take place. Uses for Electrolytic Cells:

– Decomposition of Alkali Metal Compounds

– Decomposition of Water into Hydrogen and Oxygen

– Electroplating Differences between Voltaic and Electrolytic Cells:

– ANODE: Voltaic (-) Electrolytic (+)

– CATHODE: Voltaic (+) Electrolytic (-)

– Voltaic: 2 half-cells, a salt bridge and a load

– Electrolytic: 1 cell, no salt bridge, IS the loadMark Rosengarten and Susan Katzoff

Decomposing AlkaliMetal Compounds

2 NaCl 2 Na + Cl2

The Na+1 is reduced at the (-) cathode, picking up an e- from the battery

The Cl-1 is oxidized at the (+) anode, the e- being pulled off by the battery (DC)


Decomposing Water

2 H2O 2 H2 + O2

The H+ is reduced at the (-) cathode, yielding H2 (g), which is trapped in the tube.

The O-2 is oxidized at the (+) anode, yielding O2 (g), which is trapped in the tube.


Electroplating

The Ag0 is oxidized to Ag+1 when the (+) end of the battery strips its electrons off.

The Ag+1 migrates through the solution towards the (-) charged cathode (ring), where it picks up an electron from the battery and forms Ag0, which coats on to the ring.Mark Rosengarten and Susan Katzoff

Organic Chemistry

1) Hydrocarbons

2) Substituted Hydrocarbons

3) Organic Families

4) Organic Reactions


Hydrocarbons

Molecules made of Hydrogen and Carbon Carbon forms four bonds, hydrogen forms one bond Hydrocarbons come in three different homologous series:

– Alkanes (single bond between C’s, saturated)

– Alkenes (1 double bond between 2 C’s, unsaturated)

– Alkynes (1 triple bond between 2 C’s, unsaturated) Count the number of carbons and add the appropriate

suffix!


Alkanes

CH4 = methane

C2H6 = ethane

C3H8 = propane

C4H10 = butane

C5H12 = pentane

To find the number of hydrogens, double the number of carbons and add 2.


Methane

Meth-: one carbon

-ane: alkane

The simplest organic molecule, also known as natural gas!


Ethane

Eth-: two carbons

-ane: alkane


Propane

Prop-: three carbons

-ane: alkane

Also known as “cylinder gas”, usually stored under pressure and used for gas grills and stoves. It’s also very handy as a fuel for Bunsen burners!


Butane

But-: four carbons

-ane: alkane

Liquefies with moderate pressure, useful for gas lighters. You have probably lit your gas grill with a grill lighter fueled with butane!


Pentane

Pent-: five carbons

-ane: alkane

Your Turn!!!

Draw Hexane:

Draw Heptane:Mark Rosengarten and Susan Katzoff

Alkenes

C2H4 = Ethene

C3H6 = Propene

C4H8 = Butene

C5H10 = Pentene

To find the number of hydrogens, double the number of carbons.


EtheneTwo carbons, double bonded. Notice how each carbon has four bonds? Two to the other carbon and two to hydrogen atoms.

Also called “ethylene”, is used for the production of polyethylene, which is an extensively used plastic. Look for the “PE”, “HDPE” (#2 recycling) or “LDPE” (#4 recycling) on your plastic bags and containers!


Propene

Three carbons, two of them double bonded. Notice how each carbon has four bonds?

If you flipped this molecule so that the double bond was on the right side of the molecule instead of the left, it would still be the same molecule. This is true of all alkenes.

Used to make polypropylene (PP, recycling #5), used for dishwasher safe containers and indoor/outdoor carpeting!


ButeneThis is 1-butene, because the double bond is between the 1st and 2nd carbon from the end. The number 1 represents the lowest numbered carbon the double bond is touching.

This is 2-butene. The double bond is between the 2nd and 3rd carbon from the end. Always count from the end the double bond is closest to.

ISOMERS: Molecules that share the same molecular formula, but have different structural formulas.


Pentene

This is 1-pentene. The double bond is on the first carbon from the end.

This is 2-pentene. The double bond is on the second carbon from the end.

This is not another isomer of pentene. This is also 2-pentene, just that the double bond is closer to the right end.Mark Rosengarten and Susan Katzoff

Alkynes

C2H2 = Ethyne

C3H4 = Propyne

C4H6 = Butyne

C5H8 = Pentyne

To find the number of hydrogens, double the number of carbons and subtract 2.


Ethyne

Now, try to draw propyne! Any isomers? Let’s see!

Also known as “acetylene”, used by miners by dripping water on CaC2 to light up mining helmets. The “carbide lamps” were attached to miner’s helmets by a clip and had a large reflective mirror that magnified the acetylene flame.

Used for welding and cutting applications, as ethyne burns at temperatures over 3000oC!Mark Rosengarten and Susan Katzoff

Propyne

This is propyne! Nope! No isomers.

OK, now draw butyne. If there are any isomers, draw them too.


Butyne

Well, here’s 1-butyne!

And here’s 2-butyne!

Is there a 3-butyne? Nope! That would be 1-butyne. With four carbons, the double bond can only be between the 1st and 2nd carbon, or between the 2nd and 3rd carbons.

Now, try pentyne!Mark Rosengarten and Susan Katzoff

Pentyne

1-pentyne

2-pentyne

Now, draw all of the possible isomers for hexyne!


Substituted Hydrocarbons

Hydrocarbon chains can have three kinds of “dingly charms” attached to the chain. If the dingly-charm is made of anything other than hydrogen and carbon, the molecule ceases to be a hydrocarbon and becomes another type of organic molecule.

– Alkyl groups

– Halide groups

– Other functional groups To name a hydrocarbon with an attached group, determine

which carbon (use lowest possible number value) the group is attached to. Use di- for 2 groups, tri- for three.


Alkyl Groups


Halide Groups


Organic Families

Each family has a functional group to identify it.

– Alcohol (R-OH, hydroxyl group)

– Organic Acid (R-COOH, primary carboxyl group)

– Aldehyde (R-CHO, primary carbonyl group)

– Ketone (R1-CO-R2, secondary carbonyl group)

– Ether (R1-O-R2)

– Ester (R1-COO-R2, carboxyl group in the middle)

– Amine (R-NH2, amine group)

– Amide (R-CONH2, amide group)

These molecules are alkanes with functional groups attached. The name is based on the alkane name.


Alcohol

On to DI and TRIHYDROXY ALCOHOLSMark Rosengarten and Susan Katzoff

Di and Tri-hydroxy Alcohols


Positioning of Functional Group

PRIMARY (1o): the functional group is bonded to a carbon that is on the end of the chain.

SECONDARY (2o): The functional group is bonded to a carbon in the middle of the chain.

TERTIARY (3o): The functional group is bonded to a carbon that is itself directly bonded to three other carbons.


Organic Acid

These are weak acids. The H on the right side is the one that ionized in water to form H3O+. The -COOH (carboxyl) functional group is always on a PRIMARY carbon.

Can be formed from the oxidation of primary alcohols using a KMnO4 catalyst.


Aldehyde

Aldehydes have the CO (carbonyl) groups ALWAYS on a PRIMARY carbon. This is the only structural difference between aldehydes and ketones.

Formed by the oxidation of primary alcohols with a catalyst. Propanal is formed from the oxidation of 1-propanol using pyridinium chlorochromate (PCC) catalyst.*Mark Rosengarten and Susan Katzoff

Ketone

Ketones have the CO (carbonyl) groups ALWAYS on a SECONDARY carbon. This is the only structural difference between ketones and aldehydes.

Can be formed from the dehydration of secondary alcohols with a catalyst. Propanone is formed from the oxidation of 2-propanol using KMnO4 or PCC catalyst.*


Ether

Ethers are made of two alkyl groups surrounding one oxygen atom. The ether is named for the alkyl groups on “ether” side of the oxygen. If a three-carbon alkyl group and a four-carbon alkyl group are on either side, the name would be propyl butyl ether. Made with an etherfication reaction.


Ester

Esters are named for the alcohol and organic acid that reacted by esterification to form the ester. If the alcohol was 1-propanol and the acid was hexanoic acid, the name of the ester would be propyl hexanoate. Esters contain a COO (carboxyl) group in the middle of the molecule, which differentiates them from organic acids.


Amine

- Component of amino acids, and therefore proteins, RNA and DNA…life itself!

- Essentially ammonia (NH3) with the hydrogens replaced by one or more hydrocarbon chains, hence the name “amine”!


Amide

Synthetic Polyamides: nylon, kevlar

Natural Polyamide: silk!

For more information on polymers, go here.


Organic Reactions

Combustion Fermentation Substitution Addition Dehydration Synthesis

– Etherification

– Esterification Saponification Polymerization


Combustion

Happens when an organic molecule reacts with oxygen gas to form carbon dioxide and water vapor. Also known as “burning”.


Fermentation

Process of making ethanol by having yeast digest simple sugars anaerobically. CO2 is a byproduct of this reaction.

The ethanol produced is toxic and it kills the yeast when the percent by volume of ethanol gets to 14%.


Substitution Alkane + Halogen Alkyl Halide + Hydrogen Halide The halogen atoms substitute for any of the hydrogen atoms in the alkane. This happens one atom at a time. The halide generally replaces an H on the end of the

molecule.

C2H6 + Cl2 C2H5Cl + HCl

The second Cl can then substitute for another H:

C2H5Cl + HCl C2H4Cl2 + H2


Addition

Alkene + Halogen Alkyl Halide The double bond is broken, and the halogen adds at either

side of where the double bond was. One isomer possible.


Etherification* Alcohol + Alcohol Ether + Water

A dehydrating agent (H2SO4) removes H from one alcohol’s OH and removes the OH from the other. The two molecules join where there H and OH were removed.

Note: dimethyl ether and diethyl ether are also produced from this reaction, but can be separated out.


Esterification

Organic Acid + Alcohol Ester + Water

A dehydrating agent (H2SO4) removes H from the organic acid and removes the OH from the alcohol. The two molecules join where there H and OH were removed.


Saponification

The process of making soap from glycerol esters (fats).

Glycerol ester + 3 NaOH soap + glycerol

Glyceryl stearate + 3 NaOH sodium stearate + glycerol

The sodium stearate is the soap! It emulsifies grease…surrounds globules with its nonpolar ends, creating micelles with - charge that water can then wash away. Hard water replaces Na+ with Ca+2 and/or other low solubility ions, which forms a precipitate called “soap scum”.

Water softeners remove these hardening ions from your tap water, allowing the soap to dissolve normally.


Polymerization

A polymer is a very long-chain molecule made up of many monomers (unit molecules) joined together.

The polymer is named for the monomer that made it.

– Polystyrene is made of styrene monomer

– Polybutadiene is made of butadiene monomer Addition Polymers Condensation Polymers Rubber


Addition PolymersJoining monomers together by breaking double bonds

Polyvinyl chloride (PVC): vinyl siding, PVC pipes, etc.

Vinyl chloride polyvinyl chloride

n C2H3Cl -(-C2H3Cl-)-n

Polytetrafluoroethene (PTFE, teflon):

TFE PTFE

n C2F4 -(-C2F4-)-nMark Rosengarten and Susan Katzoff

Condensation Polymers

Condensation polymerization is just dehydration synthesis, except instead of making one molecule of ether or ester, you make a monster molecule of polyether or polyester.


THE ENDGood Luck = ]


Documents

Amazing Chemistry Powerpoint Presentation! Aligned to the New York State Standards and Core Curriculum for “ The Physical Setting- Chemistry ” Produced