1. FTCE Chemistry SAE Preparation Course Session 2 Lisa Baig Instructor

2. 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

3. Course Outline Session 1 Review Pre Test Competencies 6, 7 and 8 Competencies 1 & 2 Competency 5 Session 2 Competency 3 Competency 4 Post Test

4. 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

5. 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%)

6. 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.

7. Chemical Bond Mutual 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 Types Ionic Covalent Metallic

8. 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

9. Ionic Character Ionic Bonds are bonds with > 50% ionic character Difference in Electronegativity of involved atoms is >1.7

10. 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

11. 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

12. Break Time Take a 10 minute break

13. 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

14. CuCl 2 Copper (II) Chloride CuO Copper (II) Oxide NaCl Sodium Chloride KI Potassium Iodide Mg 3 N 2 Magnesium Nitride PbO 2 Lead (IV) Oxide

15. Lewis Structures A way to show the octet rule in molecules

16. Practice Draw the lewis structures for – Ammonia (NH 3 ) – Water (H 2 O) – Phosphorus Trifluoride (PF 3 ) – Hydrogen Cyanide (HCN) – Ozone (O 3 ) – Formaldehyde (CH 2 O)

17. VSEPR AB 5 Trigonal bipyramidal AB 6 Octahedral

18. VSEPR AB 4 – Tetrahedral – 109.5 0 Bond Angles AB 3 – Trigonal Planar – 120 0 Bond Angles AB 2 – Linear – 180 0 Bond Angles

19. VSEPR AB 2 E – Bent or Angular AB 2 E 2 – Bent or Angular AB 3 E – Trigonal Pyramidal

20. Polarity? The potential for opposite charges at different areas of a molecule

21. Shape and Polarity? What is the shape and polarity of the following molecules? – Water – Ammonia – Carbon Tetrachloride – Carbon Dioxide – Hydrogen Chloride

22. 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: CH 4 Normally: 1s 2 2s 2 2p 2 – Through hybridization- it forms an “sp” orbital, with 4 electrons total – New arrangement: 1s 2 2(sp 3 ) 4

23. Hybrid Orbitals Atomic Orbitals Type of Hybrid Number of Orbitals Molecular Geometry s p sp2 180 0 Linear s p p sp 2 3 120 0 Trigonal Planar s p p p sp 3 4 109.5 0 Tetrahedral

24. What type of hybrid? Beryllium fluoride – BeF 2 – sp Ammonia – NH 3 – sp 2 Methane – CH 4 – sp 3

25. Break Time Take a 10 minute break

26. 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

27. 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

28. Simple Organics Alkanes (end in –ane) – Containing only single bonds – C n H 2n+2 Alkenes (end in –ene) – Containing at least one double bond – C n H 2n Alkynes (end in –yne) – Containing at lease one triple bond – C n H 2n-2

29. Simple Organics NameNumber of Carbons Meth-1 Eth-2 Prop-3 But-4 Pent-5 Hex-6 Hept-7 Oct-8 Non-9 Dec-10

30. Functional Groups Compound Type Image Suffix or Prefix Alcohol - -OH-ol Haloalkane -XHalo- Amine -CN--amine Aldehyde -COH-al Ketone -COC--one Carboxylic Acid -COOH-oic acid Ester -COOC--oate Amide -CON--amide

31. 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

32. 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

33. 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

34. Lunch Time We start AgainIn ONE HOUR

35. 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

36. Higher Level Practice 1 st Step: Convert your percentages to mass of each element present Na: (.393)(65.0g)= 25.545g Na Cl: (.607)(65.0g) = 39.455g Cl

37. Higher Level Practice 2 nd 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

38. Higher Level Practice 3 rd 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 Cl 1 mol Na : 1 mol Cl Empirical Formula = NaCl

39. Balancing Equations __ C 3 H 8 + __ O 2  __ CO 2 + __ H 2 O __ Ca 2 Si + __ Cl 2  __ CaCl 2 + __ SiCl 4 __ C 7 H 5 N 3 O 6  __ N 2 + __ CO + __ H 2 O + __ C __ C 2 H 2 + __ O 2  __ CO 2 + __ H 2 O __ Fe(OH) 2 + __ H 2 O 2  __ Fe(OH) 3 __ FeS 2 + __ Cl 2  __ FeCl 3 + __ S 2 Cl 2 __ Al + __ Hg(CH 3 COO) 2  __ Al(CH 3 COO) 3 + __ Hg __ Fe 2 O 3 + __ H 2  __ Fe + __ H 2 O __ NH 3 + __ O 2  __ NO + __ H 2 O

40. 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 **(CO 2 + H 2 O) Single Displacement – ACTIVITY SERIES – AB + C  AC + B Double Displacement – AB + CD  AD + CB

41. Predict the Product CaO + H 2 O  H 2 SO 3 + O 2  CaCO 3  KClO 3  C 6 H 10 + O 2  C 6 H 12 O 6 + O 2  Al + CuCl 2  Ca + KCl  Na 2 SO 4 + CaCl 2  KCl + NaOH  Ca(OH) 2 H 2 SO 4 CaO + CO 2 KCl + O 2 CO 2 + H 2 O AlCl 3 + Cu No Reaction NaCl + CaSO 4 KOH + NaCl

42. Identifying Redox Reactions 2 KNO 3 (s)  2 KNO 2 (s) + O 2 (g) +1 -1 +1 -1 0 H 2 (g) + CuO(s)  Cu(s) + H 2 O(l) 0 -2 +2 0 2(+1) -2 NaOH(aq) + HCl(aq)  NaCl(aq) + H 2 O(l) +1 -1+1 -1 +1 -1 2(+1) -2 H 2 (g) + Cl 2 (g)  2HCl(g) 0 0 +1 -1 SO 3 (g) + H 2 O(l)  H 2 SO 4 (aq) +6 3(-2) 2(+1) -2 2(+1) -2 Redox Not Redox Redox Not Redox

43. Balancing Redox Reactions The following unbalanced equation represents a redox reaction that takes place in a basic solution containing KOH. Balance the redox reaction. Br 2 (l) + KOH (aq)  KBr (aq) + KBrO 3 (aq)

44. Ionic Reaction: Br 2  Br - + BrO 3 - 0-1 +5 3(-2) - Reduction ½ Rxn: Br 2  Br - Br 2 + 2e -  2Br - 5(Br 2 + 2e -  2Br - ) Oxidation ½ Rxn: Br 2  BrO 3 - 12OH - + Br 2  2BrO 3 - + 6H 2 O + 10e - Combined Rxn: 5Br 2 + 12OH - + Br 2 + 10e -  10Br - + 2BrO 3 - + 6H 2 O + 10e - 6Br 2 + 12KOH  10KBr + 2KBrO 3 + 6H 2 O 3Br 2 + 6KOH  5KBr + KBrO 3 + 3H 2 O

45. 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. Fe 3+ (aq) +3e -  Fe (s) E 0 =-0.04V Ag + (aq) +e -  Ag (s) E 0 =+0.80V E 0 cell = E 0 cathode - E 0 anode E 0 cell = (+0.80 V)- (-0.04 V)= +0.84 V E 0 cell = positive = spontaneous

46. 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

47. Break Time Take a 10 minute break

48. Mass-Mass Stoichiometry 3 Cu + 8 HNO 3  3 Cu(NO 3 ) 2 + 4 H 2 O + 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?

49. Step 3: Compute 150g Cu(NO 3 ) 2 1 mole8 mol HNO 3 63.012 g = 187.554g3 mol Cu(NO 3 ) 2 1 mole 134 g HNO 3

50. Gas Stoichiometry Xenon gas reacts with fluorine gas according to the shown reaction. If a researcher needs 3.14L of XeF 6 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 F 2 (g)  XeF 6 (g)

51. Gas Stoichiometry Xenon 3.14L XeF 6 1mole 1Xe 22.4L = 22.4L 1XeF 6 1 mole 3.14L Xe Fluorine 3.14L XeF 6 1 mole 3 F 2 22.4L = 22.4L 1 XeF 6 1 mole 9.42L F 2

52. Solution Stoichiometry How many milliliters of 18.0M Sulfuric Acid are required to react with 250mL of 2.50M Aluminum Hydroxide? H 2 SO 4 + Al(OH) 3  H 2 O + Al 2 (SO 4 ) 3 3 H 2 SO 4 + 2 Al(OH) 3  6 H 2 O + Al 2 (SO 4 ) 3

53. 250mL Al(OH) 3 1L 2.5 mol 3 H2SO4 1L 1000mL 1000mL1 L 2 Al(OH)3 18.0 mol 1L 52.1 mL H2SO4

54. 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

55. 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?

56. Limiting Reactant 0.960g O 3 1 mole 1 NO 2 44.0g NO 2 48g O 3 1 O 3 1 mole 0.880g NO 2 0.900gNO 1 mole 1 NO 2 44.0g NO 2 30g O 3 1 O 3 1 mole 1.32g NO 2

57. Break Time Take a 10 minute break

58. 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

59. 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

60. Equilibrium Constant nA + mB ↔ xC + yD K= [C] x [D] y [A] n [B] m

61. Factors affecting Reaction Rates

62. Rate Laws A 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 of Formation of C 10.20 M 2.0 x 10 -4 M/min 20.20 M0.40 M8.0 x 10 -4 M/min 30.40 M 1.6 x 10 -3 M/min

63. Rate Law for the Reaction A + 2B  C R = k[A][B] 2 Order of the Reaction with respect to B B is of a 2 nd order reaction A is of a 1 st order reaction

64. Calculating pH and pOH pH + pOH = 14 pH = -log[H + ] pOH = -log[OH - ] What is the pH of a 2.5x10 -6 M HNO 3 solution? pH = -log [2.5x10 -6 ] pH = 5.6

65. Break Time Take a 10 minute Break When we return… POST TEST

66. 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.

67. Good Luck! When finished, turn in test to instructor, and you may leave.