FTCE Chemistry SAEPreparation Course

Session 2

Lisa BaigInstructor

Session Norms

• Respect– No side bars– Work on assigned materials only– Keep phones on vibrate– If a call must be taken, please leave the

room to do so

Course OutlineSession 1

Review Pre TestCompetencies 6, 7 and 8

Competencies 1 & 2 Competency 5

Session 2Competency 3Competency 4

Post Test

Required Materials

• Scientific Calculator• 5 Steps to a 5: AP Chemistry

– Langley, Richard, & Moore, John. (2010). 5 steps to a 5: AP chemistry, 2010-2011 edition. New York, NY: McGraw Hill Professional.

• Paper for notes• State Study Guide

Chemistry Competencies

1. Knowledge of the nature of matter (11%)2. Knowledge of energy and its interaction with

matter (14%)3. Knowledge of bonding and molecular structure

(20%)4. Knowledge of chemical reactions and

stoichiometry (24%)5. Knowledge of atomic theory and structure (9%)6. Knowledge of the nature of science (13%)7. Knowledge of measurement (5%)8. Knowledge of appropriate laboratory use and

procedure (4%)

Electronegativity

• Fluorine is the most electronegative element.

• Pattern of increasing electronegativity moves from bottom to top, and from left to right across the periodic table.

Chemical BondMutual electrical attraction between the

nuclei and valence electrons of different atoms that bind the atoms together

Atoms would like to have 8 Valence electrons. These bonds help the atoms to achieve

their full valence shells

Three TypesIonic

CovalentMetallic

Ionic Bond• Force of attraction between oppositely

charged ions• Occurs between Metal and Non-Metal

elements• The Non-metal “steals” the valence

electron(s) from the Metal• Forms a crystalline structure of these

positive and negative charges• Typically solids at room temperature

Ionic Character

• Ionic Bonds are bonds with > 50% ionic character

• Difference in Electronegativity of involved atoms is >1.7

Covalent Bond• Sharing of valence electron pairs by 2 atoms• Occurs between 2 Non-metal elements

– Or the SAME non-metal element

• Can share one, two or three pairs of electrons– Single Bond = 1 pair (1 sigma)– Double Bond = 2 pairs (1 sigma, 1 pi)– Triple Bond = 3 pairs (1 sigma, 2 pi)

• Sharing can also be “unequal”– Called a POLAR covalent bond

• Typically liquids or gases at room temperature

Character• Ionic Character:

– Polar Covalent Bonds have between 5% and 50% ionic Character

– Non-Polar Covalent Bonds have less than 5% ionic character

• Difference in Electronegativities– Polar Covalent Bonds have between 0.3

and 1.7 as a difference in electronegativities

– Non-Polar Covalent bonds have less than 0.3 difference in electronegativities

Break Time

Take a 10 minute

break

Ionic Compounds

• Ion names are used in combination• Cation- same as the element

– Transitional Metals use Roman Numerals to represent Charge

• Anion- replace the ending syllable of the element name with –ide

• Polyatomic Ions- use the name of that ion- do not try to rename.

Use “criss-cross” to determine charges

CuCl2

Copper (II) ChlorideCuO

Copper (II) OxideNaCl

Sodium ChlorideKI

Potassium IodideMg3N2

Magnesium NitridePbO2

Lead (IV) Oxide

Lewis Structures

• A way to show the octet rule in molecules

Practice

• Draw the lewis structures for– Ammonia (NH3)

– Water (H2O)

– Phosphorus Trifluoride (PF3)– Hydrogen Cyanide (HCN)– Ozone (O3)

– Formaldehyde (CH2O)

VSEPR

•AB5

•Trigonal bipyramidal

•AB6

•Octahedral

VSEPR• AB4

– Tetrahedral– 109.50 Bond Angles

• AB3– Trigonal Planar– 1200 Bond Angles

• AB2– Linear– 1800 Bond Angles

VSEPR• AB2E

– Bent or Angular

• AB2E2

– Bent or Angular

• AB3E– Trigonal Pyramidal

Polarity?

• The potential for opposite charges at different areas of a molecule

Shape and Polarity?

• What is the shape and polarity of the following molecules?– Water– Ammonia– Carbon Tetrachloride– Carbon Dioxide– Hydrogen Chloride

Hybrids

• Atoms “don’t like” to have empty orbitals• Hybridization

– Mixing of 2 or more atomic orbitals of similar energies to produce new hybrid orbitals of equal energies

• It works like this– Methane: CH4 Normally: 1s22s22p2

– Through hybridization- it forms an “sp” orbital, with 4 electrons total

– New arrangement: 1s22(sp3) 4

Hybrid OrbitalsAtomic Orbitals

Type of Hybrid

Number of

Orbitals

MolecularGeomet

rys p sp 2 1800

Linear

s p p sp2 3 1200

Trigonal Planar

s p p p sp3 4 109.50

Tetrahedral

What type of hybrid?

• Beryllium fluoride– BeF2

– sp• Ammonia

– NH3

– sp2

• Methane– CH4

– sp3

Break Time

Take a 10 minute

break

Spectroscopy

• Devices that measure the interaction between matter and energy

• Absorption– Measures the wavelengths of

electromagnetic waves absorbed by a substance

• X-Ray spectroscopy – Used to determine elemental

composition and types of bonding

Spectroscopy

• UV– Used to quantify DNA and Protein

concentration• Infrared

– Used to determine bond type• Bonds resonate when exposed to the

radiation

• Nuclear Magnetic Imaging (NMR)– Used to determine bond structure

Simple Organics

• Alkanes (end in –ane)– Containing only single bonds– CnH2n+2

• Alkenes (end in –ene)– Containing at least one double bond– CnH2n

• Alkynes (end in –yne)– Containing at lease one triple bond– CnH2n-2

Simple OrganicsName Number of Carbons

Meth- 1

Eth- 2

Prop- 3

But- 4

Pent- 5

Hex- 6

Hept- 7

Oct- 8

Non- 9

Dec- 10

Functional GroupsCompound

Type Image Suffix or Prefix

Alcohol - -OH -ol

Haloalkane -X Halo-

Amine -CN- -amine

Aldehyde -COH -al

Ketone -COC- -one

Carboxylic Acid -COOH -oic acid

Ester -COOC- -oate

Amide -CON- -amide

Naming and Formulas

• Numbers are used in the name to designate locations of the following– Types of bonds– Branches– Attached functional groups

• For Example– 2,2,4- trimethylpentane– 1-pentyne– 2,3,4- trimethylnonane– 2-methyl 3-hexene– 2- propanol

Macromolecules

• Carbohydrates– Chains of carbon, hydrogen and oxygen.– Isomers

• Lipids– Fatty acids- Chains of Carbon and Hydrogen

• Proteins– Chains of Amino acids– Differ in their R group

• Nucleic Acids– Chains of Nucleic Acids

Organic Compound Naming• Numbers are used in the name to

designate locations of the following– Types of bonds– Branches– Attached functional groups

• For Example– 2,2,4- trimethylpentane– 1-pentyne– 2,3,4- trimethylnonane– 2-methyl 3-hexene– 2- propanol

Lunch Time

We startAgain

InONE HOUR

Determining Empirical Formulas

• Say you have 65.0g of compound containing Na and Cl.

• Determine the Empirical Formula if the compound is 39.3% Na and 60.7%Cl

Higher Level Practice

• 1st Step: Convert your percentages to mass of each element present

• Na: (.393)(65.0g)= 25.545g Na

• Cl: (.607)(65.0g) = 39.455g Cl

Higher Level Practice

• 2nd Step: Determine number of moles of each element in the sample

25.545g Na 1 mole = 1.11 mol Na 22.989 g/mol

39.455g Cl 1 mole = 1.11 mol Cl 35.453 g/mol

Higher Level Practice

• 3rd Step: Use these moles to determine the smallest whole number ratio of elements to each other. That is your empirical formula!

1.11 mol Na : 1.11 mol Cl1 mol Na : 1 mol Cl

Empirical Formula = NaCl

Balancing Equations

• __ C3H8 + __ O2 __ CO2 + __ H2O

• __ Ca2Si + __ Cl2 __ CaCl2 + __ SiCl4

• __ C7H5N3O6 __ N2 + __ CO + __ H2O + __ C

• __ C2H2 + __ O2 __ CO2 + __ H2O

• __ Fe(OH)2 + __ H2O2 __ Fe(OH)3

• __ FeS2 + __ Cl2 __ FeCl3 + __ S2Cl2

• __ Al + __ Hg(CH3COO)2 __ Al(CH3COO)3 + __ Hg

• __ Fe2O3 + __ H2 __ Fe + __ H2O

• __ NH3 + __ O2 __ NO + __ H2O

Types of Chemical Reactions• Synthesis

– A+B AB• Decomposition

– AB A + B• Combustion

– Burn in the presence of O2, to form dioxide gas, and other products **(CO2 + H2O)

• Single Displacement– ACTIVITY SERIES– AB + C AC + B

• Double Displacement– AB + CD AD + CB

Predict the Product

CaO + H2O

H2SO3 + O2

CaCO3

KClO3

C6H10 + O2

C6H12O6 + O2

Al + CuCl2

Ca + KCl Na2SO4 + CaCl2

KCl + NaOH

Ca(OH)2

H2SO4

CaO + CO2

KCl + O2

CO2 + H2O

CO2 + H2O

AlCl3 + Cu

No ReactionNaCl + CaSO4

KOH + NaCl

Identifying Redox Reactions2 KNO3(s) 2 KNO2(s) + O2(g)

+1 -1 +1 -1 0H2(g) + CuO(s) Cu(s) + H2O(l)

0 -2 +2 0 2(+1) -2NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

+1 -1 +1 -1 +1 -1 2(+1) -2H2(g) + Cl2(g) 2HCl(g)

0 0 +1 -1SO3(g) + H2O(l) H2SO4(aq)+6 3(-2) 2(+1) -2 2(+1) -2

Redox

Redox

Not Redox

Redox

Not Redox

Balancing Redox Reactions

• The following unbalanced equation represents a redox reaction that takes place in a basic solution containing KOH. Balance the redox reaction.

Br2(l) + KOH(aq) KBr(aq) + KBrO3(aq)

Br2(l) + KOH(aq) KBr(aq) + KBrO3(aq)Ionic Reaction: Br2 Br- + BrO3

-

0 -1 +5 3(-2)-

Reduction ½ Rxn:Br2 Br-

Br2 + 2e- 2Br-

5(Br2 + 2e- 2Br-)Oxidation ½ Rxn:

Br2 BrO3-

12OH- + Br2 2BrO3- + 6H2O + 10e-

Combined Rxn:5Br2 + 12OH- + Br2 + 10e- 10Br- + 2BrO3

- + 6H2O + 10e- 6Br2 + 12KOH 10KBr + 2KBrO3 + 6H2O 3Br2 + 6KOH 5KBr + KBrO3 + 3H2O

Standard Reduction Potentials in Voltaic Cells

Write the overall cell reaction and calculate the cell potential for a voltaic cell consisting of the following half-cells: an Iron electrode in an Iron (III) Nitrate solution, and a Silver electrode in a Silver(I) Nitrate solution.

• Fe3+(aq)+3e-Fe(s) E0=-0.04V• Ag+(aq)+e-Ag(s) E0=+0.80V• E0

cell= E0cathode- E0

anode

• E0cell= (+0.80 V)- (-0.04 V)= +0.84 V

• E0cell= positive = spontaneous

Acid/Base Properties

• Strong Acids and Bases– Will ionize completely in a solvent

• Weak Acids and Bases– Will ionize partially in a solvent

• Buffer Systems– Solution containing a weak acid, and a

salt of the weak acid• Acetic Acid and Sodium Acetate• Carbonic Acid and Bicarbonate

Break Time

Take a 10 minute

break

Mass-Mass Stoichiometry

3 Cu + 8 HNO3 3 Cu(NO3)2 + 4 H2O + 2NO

• Copper Nitrate is used in creation of some light sensitive papers

• Specialty photographic film

• Your company needs 150 grams of Copper nitrate to fill an order. How many grams of Nitric Acid are needed to undergo reaction?

• Step 3: Compute

150g Cu(NO3)2 1 mole 8 mol HNO3 63.012 g =

187.554g 3 mol Cu(NO3)2 1 mole

134 g HNO3

Gas Stoichiometry

Xenon gas reacts with fluorine gas according to the shown reaction. If a researcher needs 3.14L of XeF6 for an experiment, what volumes of Xenon and Fluorine should be reacted? Assume all volumes are measured under the same temperatures and pressures.

Xe (g) + 3 F2 (g) XeF6 (g)

Gas Stoichiometry

• Xenon3.14L XeF6 1mole 1Xe 22.4L =

22.4L 1XeF6 1 mole

3.14L Xe • Fluorine3.14L XeF6 1 mole 3 F2 22.4L =

22.4L 1 XeF6 1 mole

9.42L F2

Solution Stoichiometry

• How many milliliters of 18.0M Sulfuric Acid are required to react with 250mL of 2.50M Aluminum Hydroxide?

• H2SO4 + Al(OH)3 H2O + Al2(SO4)3

• 3 H2SO4 + 2 Al(OH)3 6 H2O + Al2(SO4)3

250mL Al(OH)3 1L 2.5 mol 3 H2SO4 1L 1000mL 1000mL 1 L 2 Al(OH)3 18.0 mol 1L

52.1 mL H2SO4

Titrations• In a titration, 27.4mL of 0.0154M Ba(OH)2 is added

to a 20.0mL sample of HCl solution with unknown concentration until the equivalence point is reached. What is the molarity of the acid solution?

0.0154M Ba(OH)2 x 27.4mL Ba(OH)2 x 2 mol HCl x 1 = 1 1 mol Ba(OH)2 20.0mL

4.22 x 10-2 M HCl

Limiting Reactant

• The reaction of Ozone with Nitrogen Monoxide to form Oxygen and Nitrogen Dioxide in the atmosphere is responsible for the Ozone hole over Antarctica.

• If 0.960g of Ozone reacts with 0.900g of Nitrogen Monoxide, how many grams of Nitrogen Dioxide are produced?

Limiting Reactant

0.960g O3 1 mole 1 NO2 44.0g NO2

48g O3 1 O3 1 mole

0.880g NO2

0.900gNO 1 mole 1 NO2 44.0g NO2

30g O3 1 O3 1 mole

1.32g NO2

Break Time

Take a 10 minute

break

Chemical Equilibrium

• Chemical Equilibrium– Point in a reversible chemical reaction

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

– The concentrations of its products and reactants remain unchanged

• Le Chatelier’s Principle– If a system at equilibrium is stressed, the

equilibrium is shifted in the direction that relieves the stress

How to Affect Equilibrium• Change in Pressure

– Only affects reactions with gases– Increased pressure increases concentration– Decreased pressure decreases concentration

• Change in Concentration– Of reactants or products.

• Increase one- it moves to the other• Decrease one- it moves towards the one you lowered

• Change in Temperature– Exothermic

• Increase temperature will direct in reverse• Decrease temperature will direct forward

– Endothermic• Increase temperature will direct forward• Decrease temperature will direct in reverse

Equilibrium Constant

nA + mB ↔ xC + yD

K= [C]x[D]y

[A]n[B]m

Factors affecting Reaction Rates

Rate LawsA chemical reaction is expressed by the

balanced chemical equation A + 2B C

Three reaction rate experiments yield the following data.

What is the Rate Law for the Reaction?What is the Order of the reaction with

respect to B?

Experiment Number

Initial[A]

Initial[B]

Initial Rate ofFormation of C

1 0.20 M 0.20 M 2.0 x 10-4 M/min

2 0.20 M 0.40 M 8.0 x 10-4 M/min

3 0.40 M 0.40 M 1.6 x 10-3 M/min

Rate Law for the ReactionA + 2B CR = k[A][B]2

Order of the Reaction with respect to BB is of a 2nd order reactionA is of a 1st order reaction

Calculating pH and pOHpH + pOH = 14 pH = -log[H+] pOH = -log[OH-]

• What is the pH of a 2.5x10-6M HNO3 solution?

• pH = -log [2.5x10-6]• pH = 5.6

Break Time

Take a 10 minute Break

When we return…

POST TEST

Post-Test

• You will have one and a half hours to complete the post-test

• This test will include examples from all the competencies.

• Scores will be posted on the Quia Website tomorrow as a class file.

• Also to be posted- a reference key of the correct answers AND which competency and skill were covered for each question.

Good Luck!

When finished, turn in test to instructor, and you may leave.