Yaacov Schlusselberg

Chemistry notesUnit 1Aim: Why use the metric system?

1. The metric systema) Length/distance

1) English – inch, foot, mile2) Metric – meter

b) Mass/weight1) Mass vs. weight – W (weight) = M (mass) X G(gravitational pull)

Mass doesn’t change, weight will change with the difference of gravity.2) English – ounces, pounds, ton3) Metric – gram

c) Volume1) Definition – amount of space that an object takes up.2) English – ounce, cup, pint, quart (2 pints), gallon (4 qs)3) Metric – liter (liquid), meter cubed (solid)

Metric system is much simpler.

In the metric system, one unit is used, regardless of the size of the object. A prefix is used to indicate the magnitude of the measurement: centi, milli, killo.

Aim: How do I convert 1 unit to another within the metric system?

1) Conversion factors:

King(kilo-1000 meters)Henry Drank Many(meter-100 centimeters-1000millileters) Delicious Chocolate (centi) Milk (milli) Malts.

2) Dimensional analysis/factor label method – the use of conversion factors to solve problems.a) Write down the known including units.b) Write down the unknownc) Multiply the known by the appropriate conversion factor so that the known units cancel out leaving your final answer.

3) Sample problems:1) How many inches are in 40 feet?

40ft X ( 12 in/1ft) = 480 in.

2) You are driving at a speed of 85 km/hr. How long will it take to travel 255 km?

256 km X (1hr/85km) = 3 hrs

Aim: Scientific notation: What is it and why do we need it?

Scientific notation – in scientific notation a number is expressed as a product of 2 factors. The first factor is a number between 1 and 10 and the second is a power of 10.

It helps us deal with very large and very small numbers.1) Rewrite the number as a number between 1 and 10.2) Count the number of places, the decimal was moved, that is the power of 10.3) Numbers larger than 1 have positive exponents. Numbers smaller than one have negative

exponents.

Number Scientific notation

205 2.05 X 102

.00039 3.9 X 10 -4

.000291 2.91 X 10 -4

Calculating with numbers written in scientific notation:1) Multiplication – multiply the number and add the exponents.2) Division – divide the numbers and subtract the exponents.3) Addition and subtraction – the exponents must be the same.

(8.41 X 103) + (9.71 X 104) = 10.55 X 104 = 1.055 X 105

Significant figures:A) Accuracy VS. Precision

Significant figures tell us the precision in which the measurement was made.

Accurate means that you are correct. Precise means that there are the most significant figures.

Rules for determining significant figures:1) All non-zero numbers are significant2) Zeros between 2 non-zero numbers are significant (101 – 3 sig fig)3) Zeros following both a decimal point and a number are significant.4) In numbers larger than 1, zeros at the end of the measurement each are significant, if a

decimal point is present.Rounding: Round the number of to the given number of significant figures:

384,275.297 significant figures 384,275.3 6 significant figures 384,2755 significant figures 384,280 4 significant figures 384,3003 significant figures 384,000 2 significant figures 380,0001 significant figures 400,000

Calculations using significant figures:1) Multiplication or division – the answer should have as many significant figures as the least

precise measurement.

2) Addition and subtraction – the answer should contain as many places after the decimal as the least precise measurement. We can only know the answer in a column if the column is complete.

160.1 12.23 1.046

173.376 173.4

Percent error:

A) Definition – when doing a lab experiment, there is usually a difference between calculated (observed) value and the actual/true value. Calculating a % error helps determine the accuracy of the experiment.

B) Formula % error = observed value – true value X 100%

True value

It’s impossible to record a negative percent error.The true value for the boiling point of methyl alcohol is 65 degrees. A student determines the boiling point to be 66 degrees, what is his percent error?

66-65 65 X100% = 1.54%

History behind the atom

A) Democritos – the first person to make up a name, he said that an atom is particles that matter can be broken into.

B) Dalton – all elements are composed of indivisible particles called atoms.- all atoms of the same element are identical in atoms. - Atoms of different element are different- Compounds are formed by 2 or more elements joined together in a

definite whole number ratio.C) Thompson – he did an experiment called cathode de ray tube experiment – he

took an empty tube, which had a positive and negative electrode on each side. The electric current was always deflected towards the positive electrode. So it must have contained negatively charged particles.

Plum pudding model – it states that negatively charged electrons are found amidst a sea of positive charged protons.

D) Rutherford:1) Gold foil experiment – sheet of gold foil was bombarded with positive

charged alpha particles. Most particles went straight through the sheet, while a few hit the sheet and were deflected.

2) Rutherford model – the model is made up of mostly empty spaces with a small, dense, positive charged nucleus in the center and negative charged particles orbiting at great distances.

D) Modern atomic theory – agrees with Dalton’s theory with the following exceptions:

1) Atoms are not indivisible; they’re made up of protons, neutrons, and electrons.2) Atoms of the same elements are not all exactly alike.3) Atoms can be changed from 1 element to another, but by nuclear reactions,

not ordinary chemical reactions.

A) Subatomic micro particles1) Nucleons – particles found in nucleus.

a) Protons – positively charged with a mass of 1 amu. (1.67 X 10 –24 g = 1 amu)

b) Neutron – neutral with a mass of 1 amu.2) Electrons – found in shells orbiting the nucleu, negatively charged, with a mass of 1/1836

amu.3) Atomic number – number of protons in the nucleus of an atom. It is the number that

determines the identity of an element.4) Atomic mass (atomic weight/ mass number) = number of protons + neutrons.

Atomic mass – atomic number = number of neutrons.

Atomic # Atomic mass # of protons # of neutrons # of electrons name of element

15 31 15 16 15 P

35 80 35 45 35 Br

53 127 53 74 53 I

Isotopes:1) Definitions – 2 or more forms of an element with the same number of protons but a

different number of neutrons. Example: C12, C13, C14 (C14 is radioactive and used for C-dating)

H1, H2, H3 (protium, deuterium, tritium)

2) Weighted average isotopic mass – the mass on the periodic table represents a weighted average of all isotopes found in nature. To calculate a weighted average, multiply the percentages of each isotope found in nature by the mass and add those values together for each isotope.

Chlorine has 2 naturally occurring isotopes Cl 35 and Cl37. If 25% of Cl is Cl37, and 75% is Cl35, calculate the average isotopic mass.

AMU = ½ mass of C12 atom.

StructureA) Rutherford – electrons orbit the nucleus at great distance.B) Bohr (planetary model) - electrons orbit nucleus in concentric circles (just like planets around the sun). Each electron in an atom must possess just the right amount of energy to keep it in place around the nucleus. Each orbit is another energy level. The first energy level holds 2 electrons, the second holds eight, the 3rd holds

lithium

Outer orbit

Inner orbit nuceleus Electron Wave-mechanical model – we can’t speak about specific orbits that electrons follow; instead we have to talk about general area where an electron is likely to be found. This model of the atom pictures the atom having a dense, positively charged nucleus. Instead of moving in definite, fixed orbits around the nucleus as suggested in the Bohr model, the wave-mechanical model portrays electrons with distinct amounts of energy moving in areas called orbitals.

An orbital is a region in which an electron of a particular amount of energy is most likely to be located.

Every orbital can hold a maximum of 2 electrons.

Sublevel # of orbitals Maximum # of electrons

S 1 2

P 3 6

d 5 10

f 7 14

Electron configuration describes the number of electrons in each sublevel of each Principle energy level.

Jr. High Method – only describes the number of electrons in each PEL.

2 4

3p+4n0

C

High school method – describes the electron in each sublevel of each PEL.

1S2 2 electrons

PEL # S sublevel

Principle energy level also refered to as a shell – refers to the distance of an electron from the nucleus.1. 2n2 – can be used to calculated the max number of electrons in any principle level.

Shell maximum number of electrons (2n2)1 2

2 8

3 18

4 32

1) Principle energy levels (PEL) – shells2) Sublevels – subshell – each PEL can be divided into sublevels (s,p,d,f)

PEL # of sublevels names of sublevels 1 1 s

2 2 s,p

3 3 s,p,d

4 4 s,p,d,df

5 5 s,p,d,f,g

Example: Electron configuration for Boron (atomic # 5)

Junior High school method - 5 2 3

High school method – 1S2 2S2 2P1

8 2 6

1S2 2S2 2P4

15 2 8 5

1S2 2S2 2P6 3S2 3P3

B

O

P

202 8 8 2

1S2 2S2 2P6 3S2 3P6 4S2Orbital notation – another method of electron configuration, which shows the number of electrons that occupy each orbital.

Example: Draw Junior high school, high school method, and orbital notations for the following elements.

6 2 41S2 2S2 2P2

orbital notation (each arrow represents an electron) 1S 2S 2P

1S 2S 2P

Lewis Structures – Electron dot diagram1) Valence electron – electron in shell furthest from the nucleus (outermost electron) P 2-8-5

valence electron2) Kernel electron – inner electrons (everything but the valence electron)

P 2-8-5

KernelLewis structures are a simple way of representing the valence electrons in an atom.

N #7 2 5

Electron dot diagram - NOrbital shapes:

1) S orbital – sphere shaped2) P orbital – double petal ( )

Quantum numbers – a set of four numbers that name each electron in an atom. No 2 electrons can have the same four quantum numbers.

i. N 1st quantum # - principle quantum # = shell that an e- is found in ii. L 2nd quantum # - sublevel s 0 ; p 1 ; d2 ; f3 [an e- in 3d

sublevel will have first 2 quantum #s (3,2)] iii. M 3rd quantum # - orbital

1. S – 1 orbital 0 – 1S orbital .2. D9

Ca

C

iv. 4th quantum # - spin # 1. if e- in an orbital spins clockwise (^) it’s 4th quantum # is + ½ 2. if e- spins counterclockwise ( ) , it’s 4th quantum is -½

Example: D10D!2 Atomic #27

- How many ½ filled orbitals? 3- How many occupied shells? 4- How many e- are in the 3d sublevel? 7- What are all 4 quantum #’s for the 25th e- in this atom? (3, 2, -2, +1/2)

ReviewP15 1S2 2S2 2P6 3S2 3P3

3P

13th e- (3, 1, +1, + ½)15th e- (3, 1, -1, + ½)12th e- (3, 0, 0, + + ½)

b. Ground vs. Excited State – The Bohr Atom

i.ii. In every atom in its normal (ground) state e- are in the lowest energy levels

available & the atom is stable. When an atom is heated e- absorb a specific amount of energy (quantum of energy) & jump to a higher energy level. This excited state is temporary & when the e- falls back to ground state the quantum of energy is released as light. Each element will give off light of diff. wavelengths & therefore diff. colors. When light is passed through a spectroscope it will break down into a series of colored lines, known as emission spectrum or bright line spectrum of that element.

II. RadioactivityDefinition – A natural, spontaneous process in which the nucleus of an atom breaks down emitting particles or rays.

a. Types of Radiation Name Symbol Atomic # Mass Charge DataAlpha Particle 2 4 +2 -Helium nucleus

-Weak penetrating power Beta particle -1 0 -1 -High speed e-

-Stronger penetrating powerGamma ray ------------ ------ --------- Penetrate all except for lead (Pb)Positron +1 0 +1 Pos. charged e- Neutron 0 1 0 1 of 3 subatomic particles

Balancing Nuclear EquationsLaw of conservation of matter & energy – Matter & Energy can’t be created or destroyed, it can only be changed from 1 form to another.

Alpha decay – when a radioisotope decays releasing alpha particles.220 4 216 Fr He + At87 2 85

235 4 231 U He + Th92 2 90

Beta decay – beta particle is released

16 0 16 N B + O 7 -1 8

Synthetic Transmutation – man made conversion of one atom to another.

27 4 30 1 Al + He P + n13 2 15 0

32 1 30 3 S + n P + H16 0 15 1

Fission – nuclear reaction that takes place in the atomic bomb.

235 1 U + n 2 new elements and 2 or 3 neutrons and energy92 0

Fusion - 2 3 4 1H + H He + n and energy

1 1 2 0

This reaction occurs in the stars and the sun. Fusion releases more energy than fission. Also requires a lot of heat and energy to start this reaction.

Half – life - the amount of energy of the time for half of original mass of a radioactive element to decay.

Sample problems: Iodine has a half-life of 8.07 days. A 10.0 sample was allowed to decay for 32 days. What fraction will remain?

mass Time 1 0 10.0

8 1/2 5.0

16 1/4 2.5

24 1/8 1.25

32 1/16 .625

100g of a radioisotope decayed to 12.5 after 90.7 years. What was its half-life?

Time mass 0 100g

1 HL 50g2HL 25g 30.3

3HL 12.5g

A radioisotope has a ½ life of 10 days. 1g remains after 40 days. What was the initial amount?Time Mass

0 16g10 8g

20 4g16g

30 2g

40 1g

80g of a radioisotope decays to 5.0g over 100 years. What’s its half-life?

Time Mass 0 80g

1 HL 40g

2HL 20g 25 years

3HL 10g

4HL 5g

Benefits of radioisotopes

1) Tracers – used to follow the course of a chemical or biological reaction (C14)2) Medical – Tc used to pinpoint brain tumors.

Iodine – diagnosis and treatment for thyroid disorders. Treatment of cancers

3) Food storage – radiation keeps foods fresher – kills off bacteria4) Radioactive datinga) Geological dating (238 U 206 Pb)b) Dating living material I(14C 12C)

5) Nuclear power – a good source of energy

Risks of radioisotopes

1) Biological damage – exposure to radiation can damage cells of an organism.2) Long term storage – there could be an accident, and escape into the environment.3) Pollution – traces of radioactive material present in air, water, food, soil.

Chemical bonding1) Octet/duet rule – all atoms are stable if their valence shell is filled. Small atoms obey the duet

rule, while larger atoms obey the octet rule.

H 1 gain 1 electron H 2

He 2 neither gains or loses an electron, it is an inert or noble gas.

L 2 1 loses one electron L 2

Be 2 2 loses 2 electrons Be 2

B 2 3 loses 3 electrons B 2

P 2 8 5 atom (non-metal) gain 3 electrons P 2 8 8 -3 ion

Ca 2 8 8 2 atom (metal) loses 2 electrons Ca 2 8 8 +2 ion

2) Chemical bonda. Attraction between protons of one atom and electron of a second atom that holds them together.

A chemical bond is the result of the sharing or transferring of electrons.Energy is released when a bond is formed and energy must be supplied to break a bond.

b. Electronegativity - an atoms attraction for electrons in a chemical bond. The relative attraction of an atom for an electron.

Metals (on left of periodic table) have few valence electrons and therefore they have low Electronegativitys.Non-metals (on right of periodic table) have many valence electrons and therefore higher Electronegativitys.Noble gasses have a filled octet therefore they have no Electronegativity.

Metals and non-metalsMetals have 1,2, or 3 valence electrons and they will give them up forming positive ions.Non-metals have 4 or more valence electrons and will gain electrons forming negative ions.

Intramolecular Bonds – “strong bonds” (bonds within a molecule, bonds between 2 atoms)

Ionic bond – when a metal transfers its valence electrons to a non-metal forming a positive and negative ion, which attract each other.

Transfer of electrons – if the electronegativity difference between the 2 elements is greater than 1.7, the bond is ionic.

Exceptions to this rule: NaH (2.1 - .9 = 1.2) – ionic MgI2 = 1.4 – ionicHF =1.8 – not ionic (2 non-metal)

Difference in Electronegativity – this rule is used to determine which substance is the most or least ionic.

Example: Which is most ionic? A) KBr (22) B) KCl (24) C) Na (21) D) NaCl (23)

Properties of ionic compounds1) High melting point2) Conduct electricity when part of a solution but not in the solid form.3) Dissolve in polar substances such as water.4) No such thing as molecules – made of ions – not molecules.

Lewis structure for ionic compound.

X X -Na + X X

X

X X

-1

Ca +2 xCl

Mg to Mg +2

OCovalent bonds – 2 or more non-metals sharing a pair of electrons.

a) Non-polar covalent bond – equal sharing of electrons due to the same electronegativity value. – diatomic molecule.(Br2, I2, N2, Cl2, H2, O2, F2) X X X X Shared electrons X

X X

X XX

XX Triple covalent bond

XX

X X Double covalent bond

XX

b) Polar covalent bonds – unequal sharing of electrons due to different eletronegativity values.The more eletronegativity element will pull the electron stronger; therefore there will be a partial negative charge.

+ -

H FOne pair of shared elctrons

Cl

H F

H F Cl Cl

Polar covalent non-polar covalent(unequal sharing of electrons) (equal sharing of electrons)

Most common polar covalent bonding:HF>H2O>HCl>NH3>HBr

c) Coordinate covalent bond – both fo the shared electrons come from the same atom.

HXX coordinate covalent bond

H N H 3 Polar covalent bonds

HAmmonium

H Coordinate covalent bondX X

H X X H2 polar covalent bonds

X X

NH3, NH4+, H2O, H30+1. Which contain a coordinate covalent bond?

NH4+, H3O+2. Which can form a coordinate covalent bond?

NH3, H2O

Substances that have covalent bonds fall into the category of molecular substance. Molecular substances – CH2O, NH3, CO2, CH4, HCl

- soft- poor conductors- low melting point

Network solids – huge number of covalent molecule or non-metal atoms held together in a rigid, 2 –dimensional network (diamond – pure carbon atoms, quartz – SiO2, silicon carbide – SiC

Properties of network solids- Hard- High melting point- Poor conductors

Metallic bonds – positive nuclei immersed in a sea of mobile electrons.Metallic bonding is responsible for holding together a large number of atoms.Metals are good conductors of heat and electricity due to the mobile electrons.

Intermolecular bonds – (bonds between 2 electrons) “weak bonds”

1. Van Der Waals Forces – London dispersion forces – between 2 non-polar molecules or 2 noble gases, it is the weakest of the weak.VDWF are caused by the instantaneous polarization of the cloud.

Non-polar molecule:a) Diatomic moleculeb) Symmetrical molecule – the bonds are polar covalent but due to symmetry

the overall molecule is non-polar.

O<=C=>O – non-polar symmetrical – no positive or negative end, . forces are counter acting each other.Symmetrical molecule – (AB4 – 1 atom of element A in the center surrounded by 4 atoms of B) (Cl4, CH4, SiBr4, SnF4, Gd4)

The more electrons present, the bigger the molecule, the stronger VDWF, the higher the boiling point.

Molecule number of electrons Boiling Point

H2 2 20K

N2 14 77K

O2 16 90K

Cl2 34 239K

Molecule number of electrons Phase @ room temp.

F2 18 Gas

Cl2 34 Gas

Br2 70 Liquid

I2 106 Solid

Dipole Dipole Attraction – between the positive end of one polar molecule and the negative end of another molecule. (HCl)

Hydrogen Bonding – strong type of dipole attraction. A Hydrogen bond is between a hydrogen of one polar molecule and F,O,N of a second molecule with a hydrogen. Hydrogen bonding causes unusual high boiling points.

Molecule Number of electrons Boiling Point Type of Bond H2 2 20K VDWF

N2 14 77k VDWF

O2 16 90K VDWF

Cl2 34 239K VDWF

NH3 10 240K Hydrogen BondH2O 10 373K Hydrogen Bond

Molecule – ion attraction – a bond between a polar covalent molecule (dipole) and an ion from an ionic compound. (NaCl-aquis, KBr – aq, CuSo4 - aq)

H+ Na+

Cl ONa+

H+ ion surrounded by a cage of H2O . molecules it’s “hydrated”

Naming Compounds an writing formulas1. Writing Formulas – the sum of charges in any ionic compound. Most equal zero. If charges are equal in magnitude, but opposite in charge (+1,-1), the crisscross rule is used.

Ba +2 Cl-1 Ba+2 O-2-BaCl2 -Ba2O2 BaO

The metal (positive ion) is written first and the non-metal (negative ion) is written second.If there are 2 non-metals, the non-metal with lower En is written first.

Polyatomic ions – a group of atoms with an electric charge that tends to stick together. 3- 2- 2- -PO - Phosphate SO - Sulfate CO - Carbonate NO - Nitrate 4 4 3 3

- - + ClO - Chlorate OH - hydroxide NH - ammonium 3 4Mg +2 + PO -3

Mg3 (PO4)2

Write chemical formulas for the following compounds: -2

a) Sodium Carbonate Na+ CO Na2CO3 3 -2b) Aluminum Sulfate Al+3 SO Al2 (SO4)3

4 -

c) Calcium Nitrate Ca+2 NO Ca( NO3)2 3

Sodium Carbonate2-

Na+ CO 3

Na2CO3 = 2Na atoms 1 C atom 3 O atoms

6 atoms 3 elements 2-

3 ions (2Na+, 1 CO )3

+3 2- Aluminum Sulfate: AL SO = Al2 (SO4)3

4

2 Al 3 S 12 O

17 atoms3 elements

5 ions

Naming compoundsa) Binary Ionic compounds – ionic compound with only 2 elements.Name metal, name non-metal. The non-metal must end in “ide”.

Metal + non-metal + IDE

MgBr2 = Magnesium BromideNaI – Sodium Iodide All binary compounds end in “ide”

Ba3N2 – Barium Nitride

Not binaryMg (OH)2 – magnesium hydroxideKCN – Potassium CyanideNH4Cl – Ammonium Chloride

b) Naming compounds with polyatomic ions – use table “E” to get the name of the polyatomic ion.- “ATE” VS. “ITE” – “ITE” means 1 less oxygen then “ATE”.

3- 3-PO = Phosphate PO = Phosphite

4 3

Hyper Vs. Hypo – Hyper – 1 more oxygen Hypo – 1 less oxygen

ClO4 – PerchlorateClO3 – ChlorateClO2 – chlorite the same is true for Fluorine, Bromine, IodineClO - hypochlorite

BrO3 – BromateFO4 – perflorateIO – hypoiodite

Name the following compounds:a) Cu2SO4 – copper sulfateb) LiBrO – Lithium hypobromitec) MgF2 – magnesium Fluoride

Write formulas for the following compounds:a) Beryllium Oxide - BeOb) Ammonium Fluoride – NH4Fc) Barium Perchlorate – Ba (ClO4)2d) Calcium Acetate – Ca (C2H3O2)2

Naming ionic compounds in which metals have more than 1 possible charge. When naming compounds containing metals with more than 1 possible oxidation number, the stock system is used. A roman numeral in () after the metal indicates the charge of the metal.

FeO – Iron (II) Oxide – Oxygen has a –2 charge so Fe is +2.

HgCl – Mercury (I) chlorideHgCl2 – Mercury (II) chloride

Naming covalent compounds in which non-metals have more than 1 possible “charge”1) Stock system: N2O – Nitrogen (I) Oxide

NO – Nitrogen (II) oxideNO2 – Nitrogen (IV) oxide

Empirical formulasMolecular formula indicates the exact number of each type of atom in a compound. For example: C6H12O6

Empirical formula – represents the simplest ratio in which atoms combine. For example: CH2O.Empirical formulas are always used to represent ionic compounds. Either n empirical or a molecular formula can be used to represent a covalent compound.

Given molecular formula C4H10, what is the empirical formula? C2H5Which is both empirical and molecular? A)C12H2O11 B)C2H6

C) C2H4 D) C6H12O6

Types of equation1) Synthesis – (to make) 2 or more reactants combine to form 1 product.

2H2 + O2 2H2O2Ca +CO3 CaCO3

Decomposition – “to break” 1 compound broken down into 2 or more smaller elements.Ex: 2H2O 2H2 + O2

2KclO3 KCl + 3O2

3 single replacement element and compound new element and new compound

Mg + AlCl3 Al + MgCl2 – metals replace metalsF2 + NiCl2 NiF2 + Cl2 – non-metals replace non-metalsF2 + NaCl 2NaF + Cl2

Double replacement – 2 compounds combine to make 2 new compounds.

AgNO3 + NaCl AgCl + NaNO3 2 3Ca PO - Ca(NO3)2 + K3PO4 KNO3 + Ca3(PO4)2

4Combustion/burning/oxidation – burning an organic compound in the presence of O2, the products will always be CO3 + H2O + energy.

CH4 + O2 CO2 + H2O + energyC8H18 + O2 CO2 + H2O + energyC6H12O6 CO2 + H2O + energy

Balancing Equations

1) Law of conservation of matter and energy – matter or energy can’t be created or destroyed. It dictates that every chemical equation must be properly balanced.

Reactants = products When balancing an equation, a subscript can never be changed, only a coefficient can be changed.

2 KclO3 2 KCl + 3 O2

K 1 2 1 2

Cl 1 2 1 2

O 3 6 2 6

2 KOH + Mg (NO3)2 Mg(OH)2 + 2 KNO3Write a balanced equation for the combustion of Octane? (C8H18)

2 C8H18 + 25 O2 16 CO2 + 18 H2O

Butane (C4H10) – 2 C4H10 + 13 O2 8 CO2 + 10 H2O

Net ionic equationsMolecular equation – AgNO3 (aq) + BaCl2 (aq) Ba(NO3)2 (aq) + AgCl (s) + - +2 - +2 -Ionic equation – 2Ag + 2NO3 + Ba + 2Cl Ba + NO3 + 2AgClCross out because they are called spectator ions – it is equal on both sides

Net ionic equation – 2 Ag + 2Cl 2 AgCl

History Behind the periodic table1) Mendeleov (1869) – similar chemical and physical properties occur at regular intervals when elements are listed in order of increasing atomic mass.

a) K – 39.0, Ar – 39.9b) I – 126.9, Te – 127.6c) Left space under Si – called the element ekasilicon (beyond silicon) and described melting

point, boiling point, charge, density, atomic mass of the element. Later the element was discovered, geranium (ge)

2) Moseley - similarity in chemical properties is a function of atomic number, not atomic mass.

Arrangement of the periodic table1) Periodic law – properties of the elements are a periodic function of their atomic number,

therefore periodic table is an arrangement of elements from left right in order of increasing atomic number.

2) Period – horizontal rows – the number of each period indicates the P.E.L in which the valence electron is found.

3) Group/Family – vertical columns – elements within the same group have the same number of valence electrons and therefore have similar properties.

Metals, non-metals, and metalloidsA) Properties of Metals:

1) 2/3 of known elements2) Left of staircase3) Reactivity increases as you go down a group.4) Within the same period, elements in group 1 are more reactive than in group II,

because its easier to get rid of 1 valence electron, than it is to get rid of 2.5) Most metallic metals are found on bottom left of the table.

Atomic Radius – the original radius Ionic radius – radius of an ion is smaller than its atomc radius.

6) Metals loose electrons to form positive ions with radii smaller than the original atom.7) Malleable – hammer a sheet, Capable of being extended or shaped by beating with a

hammer, or by the pressure of rollers, applied to metals.8) Ductile – capable of being drawn into thin wires.9) Good conductors of heat and electricity because of their mobile electrons.10) They have metallic luster – they’re shiny.

B. Properties of non-metals:1) Right side of the staircase2) Reactivity decreases as you go down a group. Oxygen is therefore more reactive than

Sulfur.3) Non-metallic properties will be most pronounced in the upper right corner of the

periodic table.4) High electronegativity5) High ionization energy6) Gain electrons to form negative ions. The ionic radius will be larger.7) Usually gases or solids at room temperature except Bromine which is a liquid.8) Brittle – break easily, crack9) Poor conductors10) Lack of metallic lustor

C. Properties of metalloids – found between metals and non-metals and have some of each properties – (B, Si, As, Te, Ge, Sb)

D. Allotropes – forms of the same element with different chemical formulas (O2, O3) or a different arrangement of atoms. (different forms of Carbon) Therefore their chemical and physical properties differ.

Oxygen – O2 – oxygen we inhale O3 – Ozone

Carbon has many allotropes1) Diamond – network solid in which each carbon atom is bonded to 4 other.2) Graphite – Carbon atoms are arranged in sheets or layers3) Coal – no definite bonding pattern between Carbon atoms.4) Buckminster fullerene – rings of 5 or 6 Carbon atoms that are connected in a

ball shape (like a soccer ball)E. Properties from location in the periodic table

1) What can we tell about Na by its location on the periodic table?- 3 P.E.L- 1 valence electron, +1 charge- metals- very reactive (more reactive than Li, Mg)

As a scientist, where would you place the following elements with these properties?a) Poor conductor, but very reactive = near upper right cornerb) Most reactive element with 1 valence electron – bottom of group 1c) Complete outer shell, does not react – near top of group 8

Periodic propertiesA) Ionization energy – the amount of energy required to remove the most loosely bound electron

from an atom in the gas phase.

Metal Non – metal (ionization energy increases) – across a period, ionization energy increasesAs you go down a group, ionization energy decreases.Ionization energy values are listed on table S.

B) Electronegativity – The relative attraction of an atom for an electron.

Metals Non – metals – electronegativity will increaseDown a group

En will decrease

C) Atomic radius – ½ the distance between 2 adjacent nuclei.

Metal Non-metals – atomic radius decreases

Atomic radius increasesD) Ionic radius – as an atom becomes an ion it either gains or loses electrons. Therefore they differ in size from the original atom. In metals, ionic radius is smaller. In non-metals the ionic radius is bigger.

E) Atomic radii – measure in PM (pyrometers), found on table S.

Group ChemistryA) Group 1 – alkali metals – Li, Na, K, Rb, Cs, Fr - Highly reactive (low IE) – need to be stored under oil to prevent their violent reaction with water.

Li + H2O LIOH + H2 - Soft - Low melting point - 1 valence electron therefore they for +1 ion.B) Group 2 – Alkaline Earth metals – Mg, Ca, Ba, Be, Sr - Very reactive but not as reactive as group 1 metals - 2 valence electrons therefore form +2 ionsC) Group 15 - Nitrogen makes up 79% of the air we breathe - Phosphorous – Red P found in safety matches

White P is flammable and toxic Arsenic is a poison Nitrogen and phosphorous – non-metals

As and Sb – metalloidsBi –metal

- Normally reactivity of non-metals decrease as you go down a group.

N N – very stable and therefore not reactiveP4 is a less stable molecule therefore more reactive.D) Group 16 – O, S, Se, Te - Progression from Non – metals metalloid metal - O2 exists as only gas at room temperature – all other elements in group 16 are solids. - Allotropes of sulfur – S8 – yellow sulfur

Poly sulfur – is black and rubber like

E) Group 17 – halogens – f, Cl, Br, I, At - typical non-metals - At room temperature F and Cl - gases Br – liquid due to increasing VDWF

I – solid - Highly reactive and therefore occur only as diatomic molecules or in compounds.F) Group 18 – noble gasses – He, Ne, Ar, Kr, Xe, Rn - Monatomic gas molecule - a monatomic gas is one in which atoms are not bound to each other. - Most stable electron configuration - Xe, Kr can react with highly En elementsG) Transition metals - Groups 3-12 and sometimes called d block because their valence electrons are being added to the d

sublevel. - Typically hard with high melting point except for Hg (mercury), which is a liquid * Multiple positive oxidation states * Colored when dissolved in water (aq) or in hydrates (CuSO4 * 5H2O)CuSO4 (aq) – blue Cu – blueCuSO4 (s) White (because no water, anhydrous salt) Ni - green

Cr – yellow/orange

Matter and energy

Matter – anything that has a definite mass and occupies a definite volume.

Phases of matter:1) Gas (g) 1. Particles are very far apart

2. No definite shape3. No definite colume

2) Liquid (l) 1. Particles are closer together 2. No definite shape 3. Definite volume

3) Solid (s) –1.Particles are very close together 2. Definite shape and volume 3. Particles are locked in place but will vibrate

Temperature affects which state an element is in.

Matter Vocabulary1) Substance – any form of matter that will always have the same properties and compositions no matter where the sample was obtained from.A sample of any substance is homogenus (same properties)

Elements – cannot be composed into a simpler substance by ordinary means.Compounds – 2 or more elements chemically combined; definite ratio; whole number

ratio; molecule = smallest compound; you can only separate a compound by a chemical reaction.Compounds can only be decomposed by chemical reactions. The smallest particle of a compound is a molecule.2) Mixtures – like compounds, mixtures are composed of 2 different substances but there are some differences.

- Mixtures are physically combined and therefore it can be separated physically.- Ratios in a mixture are not fixed.- Components of a mixture can be either elements or compounds.- Properties of a mixture are intermediate between those of its components.

Homogenus mixture (solution) – each component is uniformly dispersed, so every sample will be identical. 3) Mixture a) Homogenus mixtureAlloy – solid solution (14 kt gold – mixture of Ni, Ay, Ag, Cu – melted together in an alloy – you cannot identify each individual component.

- Stainless - steel- Brass

(l) salt water, tea, apple juice, soda = liquid solutions(g) air – is a gaseous solution of N, O, etc.

Heterogeneous mixture – not uniformly dispersed therefore different samples will not be identical.(Orange Juice, Snapple, chocolate milk, soil)

Separation of mixtures – mixtures can be separated by differences in properties. (Density, boiling point, freezing point, molecular polarity)

1. Filtration – separated based on size, large particles will remain on top of the filter while smaller particles will pass through.

a) Solid suspended in a liquid – suspension – when left undisturbed, the solid will settle to the bottom and the liquid can be poured off.

b) 2 immiscible liquids (2 liquids that are not soluble in each other)Example: Water and oil can be separated by a separatory funnel. ( see pg. 57 fig. 4-9)

c) Mixture of solids and gases. (Filters in air conditioners and cars) 2. Distillation – separation based on boiling point.

a. Solution of a solid mixed with a liquid (salt water)b. 2 miscible liquids – 2 liquids that mix together

3. Chromatography – separate components of the mixture based on their attraction to the chromatography paper.4. Ultracentrifuge – separate based on density.5. Separation by magnet – separate mixture of metal and nonmetal.

Energy – ability to do workTypes of energy:

a. Potential energy- stored energy, energy do to position (ball on hill)b. Kinetic energy – energy of motion (ball rolling down hill)c. Other types of energy – heat, solar, atomic, electrical, nuclear, etc.

Energy and chemical reaction – all chemical reactions use or give off energy.1. Exothermic reaction – releases heat

A + B C + Heat

RHeat

P

Reaction

2) Endothermic reaction – absorbs / takes in heat

P Heat

R

Reaction

3) Activation energy – energy needed to start a reaction.

Temperature – A measure of the average kinetic energy, of the particles in a substance.When 2 objects at different temperatures are in contact with one another, heat will flow from object at higher temperature to object at lower temperature and will continue until the 2 objects are at the same temperature.

Temperature scales – temperature is measured in Celsius degrees or Kelvin. There are 2 fixed points on every thermometer – the freezing and boiling point of water.

1) Fahrenheit scale - not used in science2) Celcius scale – normal freezing point of water – 0 degrees celcius

- Boiling point of water – 100 degrees celcius3) Kelvin/absolute scale – based on celcius – sized degrees but starts at absolute zero, the

temperature at which there is no motion.Kelvin = Celcius + 273

Temp ScalesBoling point of water

212 100 373

Normal body temp 98.6 37 320 310

Freezing point of water 32 0 273

Fahrenheit Celsius Kelvin

Sample Problems1) What temperature on the Kelvin scale corresponds to 37 degrees Celsius?37 + 273 = 310

2) Convert the following to Celsius temperature:a) 25k - 248 degrees Celsius b) 300k 27 degrees Celsius

273 Absolute zero

T – The difference in temperature10 degrees Celsius = 283 K

T – 10C – but if something changes by 10C, it will also

T – 10K – change by 10K==

Calorimetry – study of heatUnit of measuring heat: 1) Calorie, Kcalorie (cal, kcal), Joule, kilojoule (J, KJ)

2) Calorie – the amount of heat needed to raise the temperature of 1 gram of water by 1 degree Celsius. 1 cal = 4.18 Joules

Calorimeter – device used to determine the amount of heat absorbed or released by a physical or chemical change.The heat is measured in-directly by measuring the temperature change in a known mass of H2O that is in contact with the reaction vessel.

O = M X C X T Amount change in temperature of heat (J) Mass specific

of H2O heat

Sample problems:1) If 420J of heat is added to water at 20 degrees Celsius and the temperature of the water rises

to 25 degrees Celsius, What was the mass of H2O?

420J = (x) (4.18J/g)(5) = 20.95/ g X = 420J = 20.092) If 1680J are added to 20g of H2O at 25 degrees Celsius, what will the final temperature be?

1680 = (20)(4.18)(x)1680/83.6 = x = 20.09 45 degrees Celsius

How much heat is released when 50.0 g of water is cooled from 75 degrees celsius to 30 degrees celsius?

X=(50)(4.18)(45) = 9405 9000J = 9KJ

Phase Changes – in order to change from one phase to another, the intermolecular forces must be overcome, therefore intermolecular forces will determine boiling point and freezing point of a substance. Stronger intermolecular force – higher boiling point.

1) Fusion – melting – H2O (s) H2O (l)2) Freezing – H2O (l) H2O (s)3) Crystallization – glucose (aq) glucose (s)4) Vaporization – boiling – H2O (l) H2O (g) Evaporation is the change from a liquid to a gas.

Evaporation takes place at any temperature, while vaporization only takes place at the boiling point.

5) Condensation – H2O (g) H2O (l)

6) Sublimation – A solid - directly to the gas phase without passing through the liquid phase.

CO2 (s) CO2 (g) dryiceI2 (s) I2 (g)

Heat/cooling curve -

GTemperature

VaporizationL

FusionS Endothermic – adding heat

Time

GCondensation exothermic

L Freezing

S

CoolingTemperature (kinetic energy) remains constant during a phase change. A phase change is accompanied by the absorption of heat ( potential energy) but without a change in temperature (kinetic energy)

Phases change as you move up a heating curve – endothermicAs you move down a cooling curve – exothermic

During a phase change, both phases exist at equilibrium.

Heat of fusion – amount of energy needed to convert a given quantity of a substance from S L at constant temperature.

Heat fusion of ice at 0 degrees celsius – 334J/gphase Q = (M)(Hf)change Heat of vaporization – 2259 j/g

Q = (M)(Hv)Used to calculate amount of heat absorbed or released during a phase change.

Temp change – Q=MC THow many joules are required to melt 150g of H2O at its melting point?Q=M(Hf) = (150)(334) = 50,100

Gas Laws A) Inroduction – discuss the relationship between 3 factors:

1) Temperature2) Pressure3) Volume of a gas

We use these to determine how changing one variable will affect another.We speak about gas in terms of volume, not mass.STP – Standard Temperature Pressure

Standard temperature – 0 degrees Celsius, 273 KelvinStandard pressure – 101.3 Kpa, 760 torr, 1 atm

B) Boyle’s Law – deals with the relationship between pressure and volume. Inverse relationship – when pressure increases volume decreases.

P1V1 = P2V2

Volume

Pressure

Sample Problems1. If 100 ml of H2 at 2 atm is compressed to 25 ml at constant temperature. What is the new pressure?

(2)(100) = 25X X = 8 atm2. If 10 ml of O2 gas at 25 degrees Celsius and 740 torr has its pressure changed to 760 torr, what will the resulting volume be?P1 – 720 V1 - 10 (740)(10) = 760X = 9.74 mlP2 – 760 V2 - ?

C) Charles’s Law – discusses the relationship with volume and temperature of a gas is directly proportional to the Kelvin temperature, if pressure remains constant.

V1 = V2V T1 T2

T

1. What will the new temperature be of 50 ml of gas that has its temperature raised from 27 degrees C to 127 degrees C at constant pressure?

V1 = 50 ml T1 = 27 50 = X = 235V2 = ? T2 = 127 27 127

27 + 273 = 300 50 = X X = 66.7 ml

127 + 273 = 400 300 400

D) Gay Lussac’s Law – the pressure exerted by a gas varies directly with the Kelvin temperature when volume is kept constant.

P1 = P2T1 T2

E) Combined gas law – Used when more than one variable is being changed.P1V1 = P2V2

T1 T2F) Dalton’s Law of partial pressure – in a mixture of gases, the total pressure is equal to the sum of the pressure that each gas would exert on its own in the same volume.

Pt = P1 + P2 + P3

TotalKinetic molecular theory – defines the behavior of an ideal gas.A) Axioms or postulates

1) The kinetic energy of the particles in a gas is directly proportional to the Kelvin temperature.2) Gas particles travel in a random, straight – line motion (Brownian motion) until they hit an

obstacle. (Container wall, another molecule)3) Collisions between gas particles are elastic (there is a transfer of energy during collisions)4) Gas particles have no significant volume5) There are no attractive forces between gas particles.

B) Properties of an ideal Gas – A gas that would follows the kinetic molecular theory exactly. Hydrogen and helium are closest to ideal gases.

C) Properties of a real gas – real gases differ from ideal has because:1. Gas particles have a small but significant volume.2. Gas particles do exert some attractive force on each other.

At high pressure and low temperature, these differences are most significant.

LiquidsA) Vapor Pressure – Although evaporation occurs at the surface of all liquids, at every temperate, the rate of evaporation increases as temperature increases. Volatile (easily evaporates) liquids evaporate easily due to weak intermolecular forces.

In a closed system, sealed container, the vapor produced during evaporation will exert a pressure known as vapor pressure. Vapor pressure increases as temperature increases (due to evaporation). Vapor pressure has a specific value for each substance at a given temperature. The quantity of the liquid is irrelevant with regard to vapor pressure, only temperature matters. (table H)

Propanone evaporates fastest (highest vapor pressure at any given temperature) because of its weak intermolecular forces.

Ethonoic acid has the strongest intermolecular forces because of its lowest vapor pressure or because it evaporates the slowest.

B) Boiling point – temperature at which vapor pressure is equal to atmospheric pressure. (pressure on surface of liquid)

Mathematics of Chemistry

What the mole?1) Avogadros number of particles – 6.02 X 1023

2) A mole of any gas at STP occupies a volume of 22.4 liters.1 mole gas – 22.4 L2 mole gas – 44.8 L1/10 mole gas – 2.24 L

3) A mole is equal to the Gram Molecular Weight (G.M.W) of a substance.1 molecule of water has a mass of 18 AMU.1 mole of water has a mass of 18 grams.

Mole wheel

LitersMolecules

/ X (6.02 X 1023)

X22.4 /

Mole (n)

X / GMW N = weight

Grams molecular weight

Compound moles (n) # of molecules liters at STP grams

H2 1 mole 6.02 X 1023 22.4 12 O2 2 mole 12.04 X 1023 44.8 64 N2 10 6.02 X 1024 224 280 NO .25 1.5 X 1023 5.6 7.5 CO2 5 3.0 X 1024 112 220

CH4 .5 3.01 X 1023 11.2 8

CO 1.5 33.6 4.21

GMW NO2 – 46 gMass of one molecule – 46gMass of 6.02 X 1023 molecule – 46gMass that will occupy 22.4 L of STP – 46 g

Practice problems with moles

1. What is the mass of 1 mole of ammonium phosphate? (NH4)3PO4149g

2. I have 3,86g of H2O, how many molecules do I have?

Grams GMW Mole 6.02 X 1023 Molecule

(3.86g)(1 mole/18g)(6.02 X 1023 mol./1 mole) - 1.29 X 1023

3. How many molecules are in 5.6 liters of gas A at STP?

Liters Mole Molecule

5.6L X (1 mole / 22.4 L) X (6.02 X 1023 molecules / 1 mole) 1.5

Counting atomsOne molecule of O2 gas has 2 oxygen atoms.

How many Oxygen atoms are in 5 molecules of Oxygen? 10# of atoms = (# of molecules)(# of atoms/molecules)

Sample QuestionsHow many Nitrogen atoms are in 80.0g of NH4NO3

GMW Grams Mole Molecule atoms

N = 28.0 H = 4.04 80.0g x (1 mole / 80.0g)(6.02 x 1023 molecules / 1 mole)(2 atoms / 1 molecule)O = 48.0 80.0

1.20 x 1024

Composition by MassA) % Composition by mass of a compound.

% = total weight of the elementX 100%

Molecular weight

1. What is the % composition by mass of each element in sodium sulfate? (Na2SO4)Na – 46 = 46/142 = 32%S – 32 = 32/142 = 23%O – 64 = 64/142 = 45%

2. What is the % by mass of Mg in MgO?Mg – 24 = 24/40 = 60%O – 16

B) % Composition of H2O in a hydrate?

Mass of H2OMass of hydrate

What is the % mass of H2O in Sodium Carbonate crystals? (Na2CO3 . 10H2O)Na = 46C = 12 180/286 = 63 %O = 48H2O = 180

286

C) Laboratory % composition Problems1. 8.80g of a hydrate was heated, after heating only 660g remained. What was the % if H2O in the hydrate?

Hydrate = 8.80 Anhydrous salt – 6.60 Mass of H2O = 2.20g 2.20/8.80 = 25 %

2. A student collected the following data during an experiment:Mass of crucible and lid = 2.50gMass of crucible and hydrate = 7.50gMass of crucible and anhydrous salt = 6.25 gCalculate the % by mass of H20 in the hydrate?

Weight of hydrate – 7.5 – 2.5 = 5Anhydrous salt – 6.25 – 2.5 = 3.75Water – 5 – 3.75 = 1.25Water/hydrate – 1.25/5 = 25%

Empirical FormulaA) Identifying empirical & molecular formula B) Finding Empirical Formulas from % composition

1) A compound is composed of 58% Ba, 13.75% S, 27.45% O What is its empirical formula?

Step 1: Assume a 100g sample Step 2: convert g mol

Ba = 58.8g / 137 = 4.29 S = 13.75 / 32 = .429 O = 27.45 / 16 = 1.72

Step 3: Divide all answers to step 2 by the smallest #Ba= 4.29/.429 = 1S = .429/.429 = 1 BaSO4O = 1.72/.429 = 4

C) Finding Molecular Formulas from empirical fomulas & molecular mass 1) The empirical formula of a compound is CH & the molecular mass is

26. Find the molecular formula.Step 1: Calculate mass of empirical formula C= 12 + H = 1 = 13Step 2: Divide molecular mass by empirical mass 26/13 C2H2

1. A compound is composed of 28.8% K, 25.6% Cl; 46.2% O, Calculate the empirical formula:Step 1: Assume 100g sample: K = 28.8 – 39.1 = .737 moles

Cl = 25.6 – 35.5 = .721 molesO = 46.2 – 16.0 = 2.89 moles

Step 2: Convert g moleStep 3: Divide by smallest number to get whole number of moles.

.737/.721 = 1 (round to closest whole number)

.721/.721 = 12.89/.721 = 4 KclO4

2. A compound is composed of 65.2% As, and 34.8% O. Calculate the empirical formula of this compound.

65.2g (1 mole/75g) = .87 moles / .87 = 134.8g (1mole/16g) = 2.175 / .87 = 2.5

AsO33. The molecular mass of a compound is 150g. If its empirical formula is CH2O, find the molecular formula.

1. C - 12 2. Molecular mass/empirical mass – 150/30 = 5 H – 2 O – 16 3. Multiply subscripts of empirical by answer to 2. (5) C5H10O5

30DensityA. Introduction – Density = Mass/volume – since one mole of any gas at STP has a volume of 22.4, the density of any gas at STP can be calculated by D=GMW/22.4.

Find the density of CO2 at STP?C=12O=32 44/22.4 = 1.96 44

StoichiometryA. Rules for stoichiometry problems:

1. Write an equation for the reaction.2. Balance the equation.3. Write the given information above the equation.4. Below the equation, write: a) Coefficient X GMW if given g

b) Coefficient X 22.4 if given Lc) Rewrite coefficient if given moles.

5. Top/Bottom = Top/Bottom (top and bottom must be in same unit, but you can have different units on either side of the equation)6. Solve for unknown

B. Mole – mole problems1. How many moles of O2 are used to produce 1000 moles of H2O?

X 10002H2 + O2 = 2H2O X/1 = 1000/2 X=500

1 2

2)How many moles of H2O are produced by the combustion of 3 mol of ethane (C2H6)?C2H6 + O2 H2O + CO2 2C2H6 + 7O2 6H2O + 4CO2

3 mol x mol2C2H6 + 7O2 6H2O + 4CO2

2 mol 6 mol 3 mol = x mol 2 mol 6 mol

C. Mass - Mass1) How many grams of H2 are used in the production of 8g of H2O?

2H2 + O2 2H2OStep 1: convert 8g mole 8g X 1 mole/18g = .44 mole Step 2: Set up a proportion .44/2 =X/2 X = .44 mole H2 Step 3: convert .44 mol H2 grams .44 mole X 2g/mole = .88g H2

Another method: Step 3: xg 8g 2H2 + O2 2H2OStep 4: 2 mole X 2g/mole = 4g Step 5: x/4 = 8/36 x= 88g

2) How many grams of N are needed to produce 34g of NH3? N2 + 3H2 2NH3 x 34gN2 + 3H2 2NH 2 X 17g/1mole = 34g

X/28 = 34/34 28g

4) Given the equation N2 + O2 2NO How many grams of NO can be made from 8.0g O2? Step 1: Convert grams O2 mole O2 8.0g * 1 mole/32g = .25 molStep 2: set up a proportion .25/1 = x/2 x= .50 moles NOStep 3: convert mole g .50 mol X 30g/1mole = 15g

D) Mole - Mass 1) How many mol of H2O can be made from 64.0g of CH4 in the combustion of methane (CH4)?

64.0g X moleCH4 + 2O2 2H2O + CO2

2 mole C-12H - 4 1 mole X 16g/1mole = 16g

16 64g/16g = x mole/2mole

8 moles of H2O2) How many g of Al will combine with 1.5 moles of O2 in the synthesis of Al2O3? Xg 1.5

4Al + 3O2 2Al2O3 4(29)=128 3

1.5/3 = x/108 x = 54g

E) Mass – Volume 1) How many L of CO2 (g) at STP are produced by the combustion of 342.0g of

octane (C6H18)?

342.0g xL C = 12 X 8 = 96 2C8H18 + 25O2 18H2O + 16CO2

H = 1 X 18 = 18 2 X 114 = 228g 16 X 22.4 = 358.4 114

342.0/228 = x/358.4 x = 537.6L of CO2

F) Volume – Volume 1) How many L of C2H6 can be completely oxidized by the reaction with 63L of O2? xL 63L2C2H6 + 7O2 6H2O + 4CO2

2 7 X/2 = 63/7 7x/z = 126/7 x= 18L 6) Mole – Volume

1) What volume of ammonia (NH3) is produced by 1 mole of H2? N2 + 3H2 2NH3

3Solutions1. 1) Solution – homogenous mixture

2) Solute - substance being dissolved (sugar, salt, etc.)3) Solvent – substance doing the dissolving, when the solvent is water the solution is referred to the aqueous solution.4) Dilute Solution – little solute (watery)5) Concentrated Solution – more solute 6) Saturated Solution – contains the maximum amount of possible solute

- any point on the solubility curve7) Unsaturated Solution – contains less than maximum amount of possible solute

- any point below the solubility curve8) Supersaturated Solution – contains more then the maximum amount of possible solute. A supersaturated solution is formed when a saturated solution is cooled but not disturbed. If it is disturbed the excess solvent will settle forming a precipitant.

B) Solubility Curves / Table G- as temperature increases, solubility of a solid (KNO3, NaNO3, KCl). increase as well (coffee dissolves better in hot water than cold water)- as temperature increases, solubility of a gas (SO2, NH3, HCl) will decrease (soda will lose its fizz at room temperature faster than in the fridge).- pressure does not effect solubility of a solid or the solubility of a liquid As pressure increases solubility of a gas increases as well.

Sample Questions 1) According to table G, which compounds solubility decreases most rapidly from 50oC to 70oC?

(a) NH3 (b) HCl (c)SO2 (d)KNO3

a2) a. How many grams of ammonium chloride (NH4Cl) will dissolve in 100g of water at 50oC to form a saturated solution? 52g b. What is 200g of H2O were used? 104g3) If a saturated solution of KNO3 at 60oC is cooled to 40oC, how much precipitate will form?

at 60oC 108g at 40oC 65g

43g4) If 27g of KCl is added to 100g of H2O at 10oC, what kind of Solution is formed?

Concentration1) Molarity = moles of solute / Liters of solution (M = n/L)

Sample Problems1) What is the molarity of a solution that contains 4 moles of solute in 800 ml of solution?

Convert 800 ml L M = 4/.8 = 5M2) What is the molarity of a solution that contains 170g of sodium nitrate in 2.0L of solution?Na = 23N = 14 170g (1 mole/85g) = 2.0 moles / 2.0 L = 1.0 MO = 48

85 3) How many grams of HCl are in 5.0L of a 6.0 molarity solution?

H – 1Cl – 35_

364) What volume will 140g of 6 Molarity of NaOH solution occupy?Na = 23 140g (1 mole/40g) = 3.5 moleO = 16H = 1_ 6 = 3.5/X X = .6

405) If 2.0 liters of a 6.0 molarity of HCl is diluted to 12 L, what is the new molarity? (M1V1 = M2V2)

(6.0)(2.0) = (X)(12.0) X = 1 molarity

2) Percent by mass = weight of solute / 100g of solution5 % glucose solution = 5g glucose + 95g H2O1 % hydrocortisone = 1g hydrocortisone + 99g “cream”

If 8g of NaCl is dissolved in 100g of solution, what is the % by mass? 8/100 = 8%

3) % by volumeBeer = 3% alcohol by volume – 3ml alcohol + 91ml H2OWine = 9-15%Whiskey = 40-45%

4) Parts per million (PPM) = g solute / g solution X 1,000,000A CuSO4 solution has .50g of salt in 100g of solution, express the concentration in % by mass = .5%PPM = .5/100 X 1,000,000 = 5,000 PPM 5) Molality (m) = mass of solute / 1 Kg solventSample problem – If 45g of C6H12O6 is dissolved in 500g of H2O, what is the molality?C = 12 X 6 = 72H = 1 X 12 = 12 45g (1 mole/180) = .25 molesO = 16 X 6 = 96_ 500g (1kg / 1000g) = .5 kg 180 .25 mole / .5 kg = .5 molality

D) Colligative Properties – refer to certain properties of a liquid that are affected by the presence of solute.

Colligative properties depend on the number of dissolved particles, not the types of particles.1) Boiling Point Elevation – the presence of solute will raise the B.P. of any solvent.

B.P. Elevation = molality X Kb

Kb = molal B.P.E Constant = .52 degrees Celsius Sample problems – 1) Find the B.P. of a 2 molal glucose solution. 2(.52) = 1.04

100 + 1.04 = 101.04 degrees Celsius

2) Find the B.P. of 10 molal antifreeze. 10 (.52) = 5.2 = 105.2 2) Freezing point depression – the presence of solute lowers the F.P. of the solvent.

F.P.D = molality X KfKf = molal F.P.D constant = 1.86 degrees Celsius

Sample problems – 1) Find the F.P. of a 2 molal glucose solution? 2(1.86) = 3.72 0 – 3.72 = -3.72 degrees Celsius

2) Find the F.P. of a 10 molal solution of antifreeze. 10(1.86) = 18.6 = -18.6

3) Electrolyte vs. Non – electrolyte – (ionic compound) – ionizes and breaks down when put in a solution. It will cause a larger change in the properties of a solvent than a non-electrolyte.

1 mole Na +1 mole NaCl 2 moles of particles will be added to the solution.

1 mole Cl -

1 mole C6H12O6 1 mole particles added to solution.Ionic compounds cause a larger change because they break down into individual ions adding more particles to the solution than the same quantity of a covalent compound.

Chemical Molality Effective molality (me = m X # of ions) F.P B.P ionic/covalentC6H12O6 1 m 1 -1.86 100.52 Cov

C2H6O2 3 m 3 - 5.58 101.56 Cov

Ca(NO3)2 1 m 3 -5.58 101.56 ionic

Al2(SO4)3 2 m 10 -18.6 105.2 ionic

NaCl 2 m 4 - 7.44 102.08 ionic

CaCl2 1 m 3 - 5.58 101.56 ionic

KCl 1 m 2 - 3.72 101.04 ionic

Gases

A) Combined gas law revisited = P 1 V 1 = P 2 V 2 T1 T2

Pressure – torr, atm, kpaVolume – litersTemperature – always Kelvin

B) Dalton’s law of partial pressure – Pt = P1 + P2 + P3The pressure in a sealed container is 1000 torr. The container is holding N and O gas in a 3:2

ratio; calculate the partial pressure of each gas? N = 3X O = 2X 5X = 1000 N = 600 torr, O = 400 torr

C) Graham’s law of diffusion – lighter gases diffuse faster than heavier gases.D) Ideal gas law – PV = nRT

Temperature moles

.082

Sample problems – 1) Calculate the volume of 4 moles of a gas at a pressure of 1 atm and a temperature of 273K. (1.00)(x) = (4)(.082)(273) = X = 89.5

Unit 7 – Kinetics, equilibrium, and spontaneous reaction

Kinetics – deals with the rates of chemical reactions and the mechanisms by which they occur.Rate is measured in terms of moles of reactants consumed or moles of product formed in a given time.Mechanism describes the actual steps by which the reaction takes place.

Potential energy diagrams1) What are they? – diagrams that show the progression of a reaction from reactants to products.2) Exothermic reaction – a reaction that releases heat, therefore reactants have more potential energy

than products. A + B C + heat

RPotential H (heat of product – heat of reactants) (neg. value) energy

P

reaction

3) Endothermic – absorbs heat, therefore products have more P.E than reactants.A + B + Heat C H = positive value

P

R

4) Vocabulary HR – Heat of reactants, P.E of reactants HP – Heat of products, P.E of products H – HP – HR, heat absorbed or released during a chemical reaction. EA/AE = Activation energy – energy needed to start a chemical reaction. Catalys – speeds up rate of reaction, by lowering A.E needed HAC – activated complex = put in all of energy (ball on of hill ready to roll down)5) Labeling a potential energy diagram: AC

1 = HR

2 = HP R 5 7 8

3 = HAC 3

4 = H P.E5 = EA 4 6 Exothermic6 = EA of reverse 17 = EA with catalyst P

8 = EA of reverse with catalyst 2

reaction

AC

5

3 P

4 2R

1

C) Math problems using P.E and diagrams1) If the A.E for the reaction A + B C + 31 Kcal is 22 Kcal, find the A.E for the reverse reaction.

AC

2222 + 31 = 53

R

31P

2) A.E = 40 Kcal for the reaction A + B + 31 Kcal + HR = 16 Kcal Find: A) HP = 31 + 16 = 47 B) HAC = 40 + 16 = 56 C) A.E of reverse = 56 – 47 = 9Kcal

Ae

40 9

P

56 31R 47

16

3) H = 10 Kcal + E.A = 15 Kcal – find the AE of Reverse = 25

AE

15 25R

10P

D) Heat of formation ( HF) – amount of heat absorbed or released when one mole of product is formed from its elements.

CaCO3 + Heat CaO + CO2

C + O2 CO2 + heat This equation can be used to represent HF because It shows CO2 being formed from its elements.

2CO2 + O2 CO2 + heat

Sample problems

1. What is HF of NH3 in the following reaction?N2 + 3H2 2NH3 + 22Kcal -11 kcal/mole

2. 2A + 3B 2C + 80 Kcal -40Kcal/mole

3. Hf of Al2O3 = -400.5 Kcal/mole: write a balanced equation that supports this statement:4Al + 3O2 2Al2O3 + 801 Kcal

E. Collision theory / factors affecting reaction rate:1) direct factors

- Direct collisions – as collisions increase, rate of reaction increases.- Effectiveness of collisions- angle of collision

2) Indirect factors (factors that will influence direct factors thereby effecting the rate of reaction)- Temperature – the higher the temperature the more K.E the particles have, the faster they will move.More effectiveness and more collision.- Concentration (liquid) – if you increase concentration, you force particles more together, so there are more collisions.- Pressure (g) cause more collisions- Surface area- catalyst – lowers the activation energy by optimizing the angle of collision thereby increasing the reaction rate.- reaction mechanism – describes the route that the reaction takes. (2A + B C)

Mechanism #1 – A + A + B C

Mechanism #2 – A + A 2A 2A + B C

- Nature of reactants:Na + H2O NaOH + H2K + H2O KOH + H2 The more active the metal, the faster the reactionMg + H2O Mg(OH)2 + H2

Ionic compounds react faster than covalent compounds.Equilibrium – 2 processes occurring at the same rate

A) Types of equilibrium: 1) Phase equilibrium – water in a closed container will reach phase equilibrium when the rate of condensation = rate of vaporization.2) Solution equilibrium – rate of dissolving = rate of crystallization. 3) Chemical equilibrium – rate if a forward reaction is = to the rate of the reverse reaction.

N2(g) + 3H2(g) 2NH3(g) only in a sealed container. B) Reactions that go to completion – there are 3 common reasons why a reaction would go to

completion:

( Becomes equilibrium will never be reached)

1) A gas is formed – if a gas is allowed to escape, equilibrium can’t be reached.2) A precipitate is formed (reference table F)

AgNO3 + NaCl AgCl + NaNO3AgCl is not soluble, therefore it will form a precipitate which will fall to the bottom and equilibrium will not be reached.

KNO3 + NaCl KCl + NaNO3 Since both products are soluble, equilibrium will be reached.3) Water is formed – HCl + NaOH H2O + NaCl (completion)

D) Le CHATelier’s Principle – when stress is applied to a system at equilibrium, the system will respond by relieving the stress and equilibrium will be re-established.

Stress = change in pressure (gas)Change in temperatureChange in concentration of a product or reactant

To help solve the problems, the following 4 steps are used. Step 1: What did I do?Step 2: What does the system do? (the opposite of what you did)Step 3: Will equilibrium shift left or right?Step 4: Answer the question

Sample Problems ( [ ] = concentration )Given the equation – 2A (g) + B(g) 2C(g) + 2D(g) + heat1) What will happen to B if [ C ] is increased?

Step 1: I increase [C]Step 2: System will decrease [C]Step 3: equilibrium will shift leftStep 4: [B] will increase

2) If pressure is increased, what will happen to [D]?Step 1: I increase pressureStep 2: System will decrease pressureStep 3: Equilibrium will shift leftStep 4: [D} will decrease

3) If temperature is decreased, what will happen to [A]?Step 1: I decrease temperatureStep 2: System will increase temperatureStep 3: Equilibrium will shift rightStep 4: [A] will increase

Given the equation – N2(g) + 3H2(g) 2NH3(g) + 22KcalHow will an increase in temperature affect the [N2], [H2], [NH3]?

Step 1: I increase temperatureStep 2: System will decrease temperatureStep 3: equilibrium will shift leftStep 4: [N2 + H2] will increase

[NH3] will decreaseGiven the equation – 2SO2 + O2 2SO3 + 47 KcalWhat affect will a decrease in [O2] have on [SO3]? Step 1: I decrease [O2] Step 2: System will increase [O2] Step 3: equilibrium will shift left Step 4: [SO3] will decreaseEquilibrium Constant – when a reaction reaches equilibrium, there is a mathematical relationship between the concentration of the products and the reactants.Equilibrium constant (Keq) describes the ration of products to reactants of a reaction at equilibrium. [ products ] coefficient * only gas and aquius chemicals can enter an equilibrium expression

Keq = [ reactants ] coefficient

Sample Problems – write an equilibrium expression for the following reactions:1) 2A(g) + 3B (g) C (g) + 4D (g)

Keq = [C] [D] 4

[A] 2 [B] 32) What is the significance of equilibrium constant? If K is huge (108) then the reaction goes to completion If K is large (102) then the reaction is in equilibrium, but it favors the product. If K is small (10-16) the reaction is in equilibrium, but it favors the reactants. In order to change the value of K, there must be a change in temperature.Sample Problems – 1. Which reaction would contain the least amount of products?a) K = 5.6 X 10 –6 b) K = 6.5 X 10 –5 c)K = 6.5 X 10 -6

3) Types of equilibrium constants A. Ka = equilibrium constant for an acid/ acid ionization constant. Related to how many ions the acid puts in solution. More ions the higher the Ka. Strong acids ionize completely putting more ions into solution, yielding a greater Ka.5 strong acids – HCl, Hbl, HI, H2SO4, HNO3HCl H+ (aq) + Cl- (aq) B. Kb = equilibrium constant for a base/base ionization constant. Mg(OH)2 Mg +2 (aq) + 2OH- (aq) - base usually gives off OH- ions in solution Group 1 and 2 metals form strong bases. C. Kw – water ionization equilibrium constant

H2O H+ + OH-Kw = [ H+ ] [ OH-] = 1.0 X 10 –14

H2O + H2O H3O + + OH-

D. Ksp – solubility product constant – used to compare the solubility of compounds.AgCl (s) Ag+ (aq) + Cl- (aq)Ksp = [Ag+] [Cl-] = 1.8 X 10 -10

2-CaSO4 (s) Ca +2 + SO4

2-Ksp = [Ca +2][SO4 ] = 9.1 X 10 -6

2-PbCO3 (s) Pb +2 + CO3 2-Ksp = [Pb +2][CO3 ] = 7.4 X 10 -14

PbCO3 is least soluble of the 3 because it has the smallest Ksp.CaSO4 is most soluble.

F) Common Ion Effect (directly related to Lechatlier’s principle) AgCl Ag+ + Cl-By adding NaCl Na+ - has no effect

Cl - will cause the reaction to shift left (Le Chatlier)During the common ion effect, equilibrium will shift left decreasing the solubility of AgCl.

Spontaneous reactionsA) Definition – a reaction that once started, keeps going and doesn’t stop. The reaction

occurs without being forced.B) H = (Hp – HR) – change in enthalpy. It is logical to assume that exothermic reactions

are spontaneous, but enthalpy is only one factor. On the basis of enthalpy of one, we would expect endothermic reactions to be non-spontaneous and exothermic to be spontaneous.

C) H – (Sp - SR) – change in entropy. Entropy is a measure of randomness or disorder of the molecule in a substance.

Solid – low entropyLiquid – medium entropyGas – high entropy

On the basis of entropy, we would expect reactions that go toward increasing entropy to be spontaneous.Low high entropy - S = T,

D) T = Kelvin temperature, the deciding factor.H = + (endothermic) and S negative ( entropy) = Nonspontaneous

H = negative (exothermic) and S is positive ( entropy) = spontaneous 3-

H3PO4 + 3H2O PO4 + 3H3O +If Na3PO4 is added to the system, the concentration of which will decrease?

3-a) H3PO4 B) H2O C) PO4 D) H3O+

S H Spontaneous+ more entropy - exothermic Spontaneous

- less entropy + endothermic non- Spontaneous

__ ___Temperature decides

At low temperature, enthalpy decidesmelting ice + + Temperature decides At high temperature, entropy decides

E) G = Free energy change G determines whether a reaction is spontaneous or not. G = + = reaction is not spontaneous

G = - = reaction is spontaneousG = 0 = reaction is in equilbrium Spontaneous (?) G H T S

G = H – T S Spontaneous - - +

Acids, Bases, and salts1) Acids:

A) Arrehenous – any substance that gives off H+ ions as the only positive ion in solution. (HCl, HNO3, HBr)B) Bronsted Lowry (alternate acid – base theory) – any substance which acts as a proton donor in a reaction.

HF + H2O H3O + + F-

HF is the proton donor

C) Properties of acids:1) Sour2) Turn blue litmus paper red3) Phenolphthalein is colorless when tested with an acid. Changes color in a base.4) React with a base (OH-) to produce a salt and water – this reaction is known as

neutralization. 5) Electrolytes – dissolve in water to form a solution that conducts an electric current.6) PH less than 77) React with active metals to produce hydrogen gas.

Active metal + Acid H2Na + HCl NaCl + H2

2) Bases:A) Arrhenius – any chemical that gives off an OH- ion as the only negative ion when dosolved in H2O. (LiOH, KOH)B) Bronsted Lowry – any chemical that acts as a proton acceptor.

Brownsted Lowry acid

NH3 + H2O NH4+ + OH-

Brownsted Lowry base

HF + H2O H3O - + F-

Base

Not spontaneous

??

+??

++-

-+-

Bronsted Lowry conjugates – every bronsted Lowry acid has a conjugate baseNH3 and NH4+ - conjugate acid-base pair

The conjugate of a strong acid is a weak base and vise versa.

B) Properties of bases:1) Bitter2) Turn red litmus paper blue3) Phenothalien turns pink in the presence of a base4) Reacts with an acid in neutralization reaction5) Electrolytes6) PH greater than 77) Corrosive

Reference table L lists same common bases.3. Naming acids and bases A) Binary acids – contain H and one other element (HCl, HBr)- the names of binary acids begins with hydro followed by the name of the second element, ending in ic

Hydro + element 2 + ic acidHCl - Hydrochlorine ic acidHBr - Hydrobromine ic acid

2- B) Ternary acids – contain a polyatomic ion containing oxygen. (NO3 - , SO4 )

- the anion (negative ion) suffix ate ic- Acid is added as second name

Sample problems – Name the followingHNO3 – nitrate nitric acidH2SO4 – sulfate Sulfuric acidH2SO3 – sulfite Sulfurous acid

C) Bases – the first name of a base is the positive ion (cation) which is not changed at all. The second name if any base is hydroxide.

Ca(OH)2 – calcium hydroxideKOH – potassium hydroxideAl(OH)3 – aluminum hydroxide

Exceptions – NaH – metal hydride CH3OH – alcohol not baseCH3COOH – organic acid

Acid BaseHCl Cl- The bronsted Lowry acid will alwaysH2SO4 HSO4 – have one more H than its conjugate base.H2SO4- SO4 2-

H2O OH-H3O + H2ONH3 NH2NH4+ NH3 Base

2- Acid Acid BaseS + H2O HS - + OH-

Amphoterism – a substance that can act as both a bronsted lowry acid and base, is said to be amphoteric or amphiprotic. acid

H2O + H2O H3O+ + OH-

BaseA) WaterB) Ammonia (NH3)C) Any negative ion containing hydrogen. (HSO4-, HCO3-)

Neutralization/TitrationA) Neutralization

1) Arrhenius Definition - a reaction in which an acid and base combine to produce salt and water. A salt is a positive ion other than H+ combined with a negative ion other than OH-.Example of neutralization reactions: NaOH + HBr NaBr + H2O Ca(OH)2 + HCl CaCl2 + 2H2O

2) Bronsted Lowry definition – acid and base Acid + baseH2O + NH3 NH4+ + OH-

B) Titration 1) Definition – a lab procedure in which a base is precisely added to a known amount of acid or vice

versa, Until an indicator changes color signaling the endpoint of the reaction. Ionization of an acid – a reaction of an acid with H2O to produce H3O+, the hydronium ion.

A) Strong vs. weak acids - Strong acids will ionize completely in solution yielding many ions; therefore considered to be strong electrolytes.

HCl 100% H+ + Cl-HCl + H2O H3O+ + Cl- Same reaction

HF 1% H+ + F-HF + H2O H3O+ + F- same reaction

5 strong acids – HCl, HBr, HF, H2SO4, HNO3Bases containing group 1 or 2 metals are strong

Ionization constant of water – KwA) Definition – Kw = [H+][OH-] = 1.0 X 10 -14

H2O H+ + OH-Since pure water is neutral, [H+] = [OH-] + [H+] = 1.0 X 10 -8

PH measures the acidity or basicity of a solution.A) Definition – the negative log of the H+ ion concentration. (Take the exponent of [H+] and change it to a positive #) Smallest value for [H+] = 1.0 X 10 -14 PH 14 Largest value for [H+] = 1.0 X 10 0 PH 0B) PH scale – PH values can range from 0-14 with values below 7 representing acids and above 7, bases, 7 = neutral.C) Sample problems – 1) Given a solution with a PH of 5:

a) What is [H+]? 1.0 X 10 -5b) What is [OH-]? 1.0 X 10 –9

c) Which is greater? H+d) Is this solution an acid or base? Acid

2) An unknown is found to have an [OH-] of 1.0 X 10 –8, what’s its PH? 63) What is the PH of a .0001m NaOH solution assuming 100% ionization?[OH-] = 1.0 X 10 –4 [H+] = 1.0 X 10 –10 PH – 104) As compared with a PH of 3, a solution with a PH of 5 contains:

a) 100 more H+ ions c) 20 more H+ ionsb) 100 less H+ ions d) 20 less H+ ions

D) PH indicators – Reference table MAcid – base indicators are used to help determine the PH range of a specific solution.

Hydrolysis – the reaction of some salts and water to produce an acidic or basic solution.Ba(NO2)2 + 2H2O Ba(OH)2 + HNO2

B) Types of salts1) Neutral salt – produced from a strong acid and a strong base

Nacl – NaOH – strong base HCl – strong acid

2) Acidic salts – produced from a strong acid and weak base 0

AgCl – Ag(OH) – weak baseHCl – strong acid

3) Basic salts – produced from a weak acid and strong base.NaF – NaOH – strong base

HF – weak acid

Salt Parent base Parent acid Type of Salt

Ba(NO2)2 Ba(OH)2 HNO2 Basic

CuSO4 Cu(OH)2 H2SO4 acidic

LiCl LiOH HCl neutral

FeCl3 Fe(OH)3 HCl acidic

KI KOH HI neutral

Redox and ElechtrochemistryVocabA) Oxidation – combination of an element with oxygen (burning, combustion) The definition of oxidation has been extended to include the increase in charge due to the loss of electrons.

Mg Mg+2 + 2 electrons - the substance losing electrons (Mg) is oxidized.B) Reduction – the decrease in charge caused by the gain of electrons.

Cl2 + 2 electrons 2Cl- The substance gaining electrons (Cl2) is being reduced.C) Redox – when one element loses an electron another takes it in, therefore neither oxidation nor reduction can occur alone. A redox reaction is the transfer of electrons from one atom to the next.

Oxidation numbersA) An oxidation number is the charge an atom or ion has or appears to have. If it is an ion, it’s ok and we know they’re charged particles. But NH3 isn’t charged (covalent) so we say “appears to have”. Imaginary numbers are used to balance redox equations.

B) Rules for assigning oxidation numbers 1) The oxidation number of any element by itself is 0. (Na, He, O2, etc) – uncombined with another

element.2) The sum of the oxidation numbers in any compound must be equal to 0.3) Group I metals have an oxidation number of +1.4) Group II metals have an oxidation number of +2.5) Hydrogen has an oxidation number of +1 in all of its compounds except for metal hydrides (LiH,

CaH2) where its oxidation number is –1.6) Oxygen has an oxidation number of –2 in all compounds with 2 exceptions; peroxide (H2O2) O

is –1 and in oxygen-fluorine compounds (OF2) O is +2. 7) a) Oxidation number of a monatomic ion is its charge. Cl- = -1 , Al+3 = +3

b) In a polyatomic ion, he total oxidation number must be equal to the charge of the ion. 3- -PO4 , NO3

8) Oxidation number of a halogen (group 17) in a binary compound is –1. KI (k = +1, I = -1) vs. KlO3 (K = +1, I = +5, O= -2)

Practice problems – assign oxidation numbers to each element. +1 -21) Na2 S

+4 -6 2-2) S O3

D) Charges in oxidation numbers – a redox reaction can be recognized by the charge in oxidation numbers of some elements from the beginning to the end of the reaction. 0 0 +3 -2 0 +3

Fe + O2 Fe2 O3 Fe is oxidized Fe Fe+6 -6

0 -2 O is reduced O2 O

0 0 +2 -1 0 +2Mg + Cl2 Mg Cl2 Mg is oxidized Mg Mg

-2 0 -2

Cl is reduced Cl2 Cl

+4 -2 +1 -1 +2 -1 +1 -2 0MnO2 + 4HCl MnCl2 + 2H2O + Cl2

-4 -2 +2

+4 +2Mn is reduced Mn Mn

- 0Cl is oxidized Cl Cl2

E) Oxidizing and reducing agents – the substance being oxidized is called the reducing agent. Substance being reduced is called the oxidizing agent.

Half reactions – both oxidation and reduction are ½ of the redox process and separate equations can be written for each.

0 0 +2 -1Mg + Cl2 MgCl2

0 +2Oxidation ( ): Mg Mg + 2e-

Both mass and charge are balanced in each ½ reaction 0

Reduction ( ): Cl2 + 2e- 2Cl-

Identifying ½ reactions1) Assign oxidation numbers to each element.2) Identify the element undergoing oxidation and reduction.3) Label the 2 ½ reaction (oxidation or reduction)4) Balance elements undergoing redox5) Balance charge by adding e- to more positive side.

LEO the lion goes GER

oxidation gain lose electron electron reduction

Write ½ reactions for the following equations: +1 +7 -2 +1 -1 +1 -1 +2 -1 +1 -2 01) KMnO4 + HCl KCl + MnCl2 + H2O + Cl2

-8 -2 +2

+7 +2Reduction - Mn + 5e- Mn

-1 0Oxidation - 2Cl Cl2 + 2e-

+1 -2 +1 +5 –2 +1 +5 –2 +2 –2 0 +1 -22)Ag2S + HNO3 Ag NO3 + NO + S + H2O +2 -6 -6 +2

+5 +2Reduction - N + 3e- N

-2 0Oxidation - S S + 2e- -3 +1 0 +2 –2 +1 -23) NH3 + O2 NO + H2O +3 +2

0 -2Reduction – O + 4e- O

-3 +2Oxidation – N N + 5e-

0 +1 +6 -2 +3 +8 -8 04) Al + H2SO4 Al2 (SO4)3 + H2 +2 -8 +6

0 +3Oxidation – Al Al + 3e-

+1 0Reduction – 2H + 2e- H2

Balancing Redox reactionsA) Rules – 1) Write out the ½ reactions

2) The number of e- lost during oxidation must equal the number of electrons gained during reduction. (LCM)3) Multiply the entire ½ reaction when balancing electrons put new coefficient into original

reaction.4) Balance remainder by inspection

a) Metals firstb) Balance non-metalsc) Balance H and OH last.

B) Sample Problems1) 4 _ NH3 + 5_ O2 4_ NO + 6_ H2O

-3 +2 -3 +2Oxidation – 4( N N + 5e- ) 4N 4N + 20 e-

0 -2 0 -2Reduction – 5(O2 + 4e- 2O ) 5O2 + 20 e- 10 O

+1 +7 -2 +1 -1 +1 -1 +2 -1 +1 -2 02) 2_ KMnO4 + 16_ HCl 2_ KCl + 2_ MnCl2 + 8_ H2O + Cl2

-8 -2 +2

-1 0 - 0Oxidation – 5(2Cl Cl2 + 2e- ) 10 Cl 5Cl2 + 10 e-

+7 +2 +7 +2Reduction – 2 ( Mn + 5 e- Mn ) 2 Mn + 10 e- 2 Mn

0 +1 +5 -2 +2 -2 +2 -2 +1 -23) 3_ Zn + 8_ HNO3 3_ Zn (NO3)2 + 2_ NO + 4_ H2O

-6 +2

0 +2 0 +2Oxidation – 3 (Zn Zn + 2e-) 3 Zn 3Zn + 6e-

+5 +2 +5 +2

Reduction – 2 (N + 3e- N ) 2N + 6e- 2N

+2 +5 - +3 +1 -2 +1 -24) 4_ Fe + 1_ NO3 + 4_ H+ 4_ Fe + 1_NO - + 2_ H2O

+2

+5 +1Reduction - N + 4e- N

+2 +3 +2 +3Oxidation - 4 (Fe Fe + 1e- ) 4Fe 4Fe + 4e-

VoltageA) Introduction – Electricity / electric current is the movement of electrons through a circuit. The number of electrons is measured in amperes, while the force that pushes electrons through a circuit is measured in volts using a voltmeter.Positive voltage – spontaneous reaction (battery)Negative voltage – non-spontaneous reaction

B) Old reference Table N (p.133)Li+/Li – reduction process – Li + + 1e- Li - 3.0 V (non-sponatneous)Oxidation – Li Li+ + 1e- 3.04 V (spontaneous)

Reduction – F2 +2e- 2F- 2.87 VOxidation – 2F- F2 + 2e- - 2.87 V

C) Calculating voltage of a battery – add together voltage of oxidation and reduction half reactions.+2 +2

1) Zn + Cu Zn Cu+2

Oxidation – Zn Zn + 2e- V=.76+2

Reduction – Cu + 2e- Cu V= .34 1.10 V

+4 +22) Find the voltage for the Al, Sn battery: given the following Sn + 2e- Sn .15

+3 Al + 3e- Al 1.66

+3

Al Al +3e- = 1.66

1.81 VBattery, Galvanic cell, voltaic cell, electrochemical cell

A) Definition – a device that converts chemical energy into electrical energy using a redox reaction. A system that contains 2 electrodes separated by an electrolyte.

e- V

salt bridgeCl- Na+

- +

Zn Cu

Zn Cuanode cathode

Zn Zn+2 + 2 e- Cu +2 + 2e- Cu

- As a general rule, oxidation/anode is always on the left- A salt bridge – 1) completes the circuit

2) allows ions to migrate between solutions 3) maintains electrical neutrality- An electrode is the site at which oxidation or reduction occurs. Electrons will flow from anode to cathode (left right) because anode is negatively charged and cathode is positively charged.- Oxidation occurs at anode red cat and an ox Reduction occurs at cathode

Given the reaction Zn + Cu+2 Zn+2 + Cu as time goes on, reactants turn into products therefore amount of reactant decreases while product increases. The anode (Zn) gets smaller, while the cathode (Zn) gets larger. The Zn+2 ions increases while the Cu+2 ions decrease since Cu is a transition metal, the solution will change from deep blue to lighter blue to colorless as the [Cu+2] decreases. The battery runs out when the anode is used up.C) Spontaneous Reactions 1) Chapter VII method – positive G – not spontaneous, negative G – spontaneous2) Sum of the voltage – (maximum voltage, net voltage, potential difference, maximum potential, net potential, E0 )

+ voltage – Spontaneous - voltage – non-spontaneous 3) Dental metals/Jewelry metals – Au, As, Cu, Hg – these metals don’t react and therefore will not undergo oxidation.4) Reference Table J – In a spontaneous reaction the metal higher up on Table J is oxidized and replaces the metal below it, which is reduced.

Electrolysis/Electrolytic cellA) Definition – when the sum of the half-cell potentials is negative the reaction will not be

spontaneous. Although this reaction can’t be used to produce an electric current, a current from an external source (battery) can be used to drive the reaction.

B) Electrolysis of a molten salt – solid salts don’t conduct an electric current. +1 -1 0 0NaCl Na + Cl2

+1 0Na + 1e- Na - -2.71 -12Cl- Cl2 + 2e- -1.36

V= - 4.07 – not spontaneous

In order for this to occur, a battery with at least a voltage of 4.07 or more must be used.

C) Electroplating Battery

+ -0 + + 0

AgAg + 1e- Ag + 1e- Aganode cathode

Electrochemical cell Electrolytic cell 0

D) Electrochemical vs. Electrolytic cell E + -Spont yes no

Require voltage produce current requires currentProduce voltage

Anode - +Organic Chemistry

A) Definition – the study of carbon compounds. All organic compounds contain carbon, but not all carbon compounds are organic. (CO2, CO not organic)

B) Properties of organic compounds:1) Non – polar covalent molecules2) Insoluble in water or other polar solvents (like dissolves like)3) Contain covalent bonds and therefore have a low boiling and melting point4) Most are non-electrolytes (carboxylic acid COOH is a weak electrolyte)5) Organic reactions are slow

C) Vocabulary: 1) Structural formula – shows the bonding pattern within a molecule. CH4 – molecular formula

H

H C H - Structural formula

H

2) Isomers – 2 compounds with the same molecular formula but different structural formulas; Different I.U.P.A.C name

H H H H H H H Butane - notConsidered

Ag Spoon

cathode+ -

H - C – C – C – C – H - Butane H - C - C - C - H an isomer

H H H H H H H - C - H C4H10

H H H H

H - C – C – C - H 2 Methyl Propane

H H - C – H H

H

3) Saturated Vs. unsaturated compoundsSaturated compounds have only single carbon to carbon bonds. Unsaturated compounds

have at least 1 double or triple carbon to carbon bond. C C, C – C

HydrocarbonsA) Defintion – organic compounds containing only carbon and hydrogen.B) Forms of hydrocarbons:

1) Cyclic – closed chain structures Benzene

CH3Tolvene

2) Open chain, aliphatic hydrocarbonsa) Straight chains – C – C – C – C - C

b) Branched chains - C – C - C

C

C

c) Homologous series of hydrocarbons – hydrocarbons are classified into different series based on the type of bonding found between Carbon atoms. Members of a series will differ from one another by the addition of CH2.

1) Alkanes – aliphatic saturated hydrocarbons with a general formula CnH2n + 2All members of this series have names ending in “ane”.

H

H- C – H Methane CnH2n+2 - CH4

H

H H

H - C – C – H Ethane - C2H6

H H

H H H H - C – C – C – H Propane – C3H8

H H H

H H H H

H - C – C – C – C - H Butane - C4H10

H H H H

H H H H H

H – C – C – C – C – C – H Pentane – C5H12

H H H H H

H H H H H H

H – C – C – C – C – C – C – H Hexane – C6H14

H H H H H H

2) Alkenes – Aliphatic, unsaturated hydrocarbons with on C C and the general formula CnH2n. Alkenes have names ending in “ene”. When more than one possibility exists for the placement of the C Cit’s location must be specified in the name.

H H

C C - ethene C2H4

H H

H H

H C C – C – H Propene – C3H6

H H H H H H

H C C – C – C – H 1 - butene

H H HIsomers – C4H8

H H H H

H – C - C C – C - H 2 - butene

H H

3) Alkynes – Aliphatic unsaturated hydrocarbons with one C -- C and the general formula CnH2n-2.Alkynes have names ending in “yne”. H

H – C -- C - C - H C2H2

H H H

C – C – C – H C2H4

H

Sample Problems1) a. What is the chemical formula for the alkyne with 7 atoms? C7H12 b. Draw the structural formula.

H H H H H H H

H - C - C - C -- C – C – C – C H

H H H H H

c. Name the structure that you drew: 3 - heptyne 2) Which series does C8H16 belong to? Alkenes

4. Aromatics/Benzene – cyclic unsaturated hydrocarbons. Benzene family has a general formula of CnH2n –6.

IUPAC naming systemA) Rules for Naming

1) Count the number of Carbon atoms in the longest connected C chain, to determine the name of the parent chain. (ref table P)

2) Circle all groups attached to the parent chain.3) Number the C atoms in the chain in either direction (left to right or right to left) so that the

circled groups are attached to the C atom with the lowest number.4) Follow the steps outlined below when naming circled groups:

a. Name all circled groups alphabeticallyb. The name of each circled group ends in “yl” (methyl, ethyl, propyl) except for

halogens, which end in O (chloro, bromo, floro).c. If more than one of any given circled group exists use the prefixes: di, tri, etc. to

indicate the number present.d. Indicate the number of C atom that the circled group is attached to.

B) Sample Problems1) Name the following compounds: H H H H H

4 ethyl, 2 hexeneH – C - C C – C – C – C - H (parent chain)

H H H H -C - H H

C

2) Br Cl H

Br - C - C - C - H TetraBromo Chloropropane

Br H Br H

H H- C -H F

Name the following: C – C – C – C - F 1 butyne, 4 bromo, 3 chloro, 4, 4 difluoro, 3 methyl (?)

Cl Br

Draw 4-methyl 2 pentyne: H

H- C - H

H H- C - H H

H -C – C ----- C – C – C - H

H H H

3 isomers of pentane exist, draw and name the 3 structures H H H H H H H

H H-C - H HH - C – C – C – C – C - H H H-C–H H H

H – C – C – C - H H H H H H H- C – C – C – C - H H- C - H

H H Pentane H H H H

2,2 dimethyl propane2 methyl butane

C5H12

Organic compounds containing functional groupsA) Alcohol – 1 or more H in a hydrocarbon is replaced with an “OH”. Alcohols look similar to bases but they have different properties.

1) Monohydroxy alcohol – containing only one OH.a) Primary alcohol – carbon attracted to OH is attached to only one other Carbon (OH is attached at end of chain)

H H H H H - C – C – C – C – OH Primary alcohol

H H H H

- R is often used to represent a hydrocarbon chain. Therefore a primary alcohol can be represented by R OH.- To name a primary alcohol, the name of the hydrocarbon chain is used with the ending changed to “ol”. The location of where the OH is attached must be included.

b) Secondary alcohol – carbon attached to OH is attached to 2 other carbons. (OH is attached in middle of chain)

H H H

H – C – C – C – OH 1 Propanol

H H H

H H H

H – C – C – C – H 2 Propanol

H OH H

c) Tertiary alcohol – carbon attached to OH is attached to 3 other C atoms (OH is attached to a C with an alkyl branch)

H C H

H – C – C – C – H Methyl 2, 2 propanol

H OH H H

H H H-C-H H

H – C – C – C – C – H 3 bromo, 2-methyl, 2 butanol

H Br OH H

2) Dihydroxy alcohols (glycols) – alcohols with 2 OH groups. To name a dihydroxy alcohol “ol” is replaced with a “diol”

C – C - C 1,2 propanodiolEthelyne glycolo

OH OH3) Trihydroxy alcohols – (glycerols) contain 3 OH groups. Ex: C – C – C = 1,2,3 propanetriol

O OH OH OH

B) Organic Acids - COOH R – C - OH

To name an organic acid drop “e” and add “oic acid”

O

H – C – OH Methanoic acid O

C – C – C – OH Propanoic acid CH3CH2COOH

C) Ether - formed when 2 primary alcohols are joined by dehydration synthesis.

H H H H

H – C – C – C – OH + OH – C – C - H

H H H H H

Propanol ethanol

C – C – C – O – C – C + H2OEthyl propyl ether

CH3CH2CH2OCH2CH3

D) Aldehyde – formed when a primary alcohol is oxidized. To name an aldehyde “al” is added to the base name.

H O

H – C – C ethanol CH3CHO

H HH H H O

H - C – C – C – C - H

H H H H

E) Ketone – formed when a secondary alcohol undergoes oxidation. Ketones are named by dropping the final “e” and adding “one”, the location of the double bonded O must be specified.

H O H

H - C – C – C – H CH3COCH3 Propanone

H H

H O H H H

H – C – C – C – C – C – H 2-Pentanone CH3COCH2CH2CH3

H H H H

F) Ester – formed when an organic acid reacts with an alcohol. The first name of an ester is the name of the alcohol, the 2nd name is the name of the organic acid but “ic” is changed to an “oate”. H O H H H O H H

H – C – OH + OH – C – C – C – H H – C – O – C – C – C – H + H2O

H H H H HMethanol + Propanoic acid

Methyl ProponoateCH3CH2COOCH3

Draw structure for: CH2COOCH2CH2CH2CH3H H O H H H H

H – C – C – C – O – C – C – C – C - H

H H H H H H

Identify: Butyl Propanoate

H H O H H H H H

H – C – C – O – C – C – C – C – C – C - H

H H H H H H H

CH3CH2CH2CH2CH2COOCH2CH3

G) Amines – formed when one or more H from ammonia (NH3) is represented with an alkyl group.To name an amine the “e” ending of the alkane is dropped and changed to amine.

H H

H – C – N – H Methyl amine

HH H H

H – C – N – C – H Dimethyl amine

H HH H

H – C – N – C – H Trimethyl amine

H H-C-H H

H H H H H H H

H – C – C – C – C – C – N – H Pentylamine (Pentanamine)

H H H H H

H H H H H

H – C – C – C – C – C – H 3 - pentanamine

H H N H H

H) Amide – when 2 amino acids come together an amide is formed. An amide is simply a peptide bond.H H O H

H – C – C – C – N - H

H H H H O H H O H H O H H O

H - N - C – C – OH + H – N – C – C - OH = H – N – C – C – N – C – C - OH H

acid H H HAmino group Peptide bond

Organic ReactionsA) Substitution – 1 or more of the H atoms from an alkane is substituted with another group or another atom most commonly a halogen. Solo reactions have 2 products and begin with an alkane. H H H H H H

H – C – C – C – H + Br2 H – C – C – C – H + HBr

H H H H Br H

B) Addition – adding one or more atoms of a halogen at a double or triple bonds of an alkene or an alkyne. Addition reactions start with an unsaturated compound and have only one product. C3H6 + Br2 C3H6Br2

H H H H H H

H – C C – C – H + Br2 H – C – C – C - H

H H Br Br H

C) Esterification – Organic acid + alcohol Ester + waterH O H H O H

CH3OH + CH3COOH H – C – OH + OH – C – C – H H – C – O – C – C - HH H H H

D) Saponification – the process of making soap.Fat (ester) + NaOH (inorganic base) soap + glycerol

E) Fermentation – The breakdown of glucose into ethanol and CO2. Yeast cells secrete the enzyme necessary for this reaction. Glucose ethanol + CO2

F) Oxidation (combustion) – When sufficient O2 is present, hydrocarbons will burn to produce water and CO2. CH4 + O2 CO2 + H2O H H H H H O CH18 + O2 CO2 + H2OOxidation of a primary alcohol aldehyde H – C – C – C – OH H – C – C – C - HOxidation of a secondary alcohol KetoneTertiary alcohols can’t be oxidized. H H H H H

G) Polymerization – polymers are large molecules made up of chains of many smaller repeating subunits. Natural polymers include proteins (amino acids), starch, cellulose, DNA, RNA.Synthetic Polymers – plastic, rubber, nylon, polyester

1) Addition polymerization – Joining unsaturated compounds by breaking double or triple bonds. N(not nitrogen)C2H2 (C2H2)n2) Condensation polymerization – simply dehydration synthesis

Amino acid + a.a + a.a. protein + water