1. Unit One

2. Notes One Unit One Two Classes of Elements What Are Stable Elements? Stabilizing Sodium Stabilizing Oxygen Sodium Loses electrons to Oxygen Oxidation Numbers Key Elements Examples Pages 158-165

3. Two Classes of Elements What are the Two Main Classes of Elements? Metals and Nonmetals Noble Elements

4. What Makes Elements Stable? (Lose e -1 ) (Gain e -1 ) Losing or Gaining e -1. Do metals Lose or Gain e -1 ? Do nonmetals Lose or Gain e -1 ? 0 +1+2+3+4 +5+6+7+8+9 +10-10-9-8-7 -6-5-4-3-2 Oxidation Reduction

5. Stabilizing Sodium How many e -1 for Na? 11e -1 What is the noble element closest to Na? Ne How many e -1 for Ne? 10e -1 Sodium loses/gains how many electrons? 1e -1 Na  Na +1 + e -1 (protons) + (electrons)=charge (+11)(-10)+1 Oxidation or reduction?

6. Stabilizing Oxygen How many e -1 for O? 8e -1 What is the noble element closest to O? Ne How many e -1 for Ne? 10e -1 Oxygen loses/gains how many electrons? 2e -1 O + 2e -1  O -2 Oxidation or reduction? (protons) + (electrons)=charge (+8)(-10)-2

7. Sodium Loses electrons to Oxygen Na  Na +1 + e -1 (Stable Like Neon) Ox or Red? O + 2e -1  O -2 (Stable Like Neon) Ox or Red? How many sodium atoms are needed to satisfy oxygen’s electron hunger? 2e -1 means How many oxygen atoms are needed to satisfy sodium’s electron loss? 2e -1 means Na 2 O two Na one O High Electronegativity Low Electronegativity

8. Oxidation Numbers All elements Lose or Gain e -1. Some have multiple loss or gain possibilities. Fe +2 Fe +3 S -2 S +4 S +6

9. Key Elements (99%) H +1 H -1 (99%) O -2 O -1 (Always) Li +1, Na +1, K +1, Rb +1, Cs +1, Fr +1 (Always) Be +2, Mg +2, Ca +2, Ba +2, Sr +2, Ra +2 (Always) Al +3 (with only a metal) F -1, Cl -1, Br -1, I -1 (NO 3 -1 ) ion is always +5 (SO 4 -2 ) ion is always +6

10. Example One Finding Oxidation Numbers 2(+3)+3(S)=Zero sum of the oxidation #’s = Find Ox #’s for Al 2 S 3 ? zero 2(Al)+3(S)=Zero S = -2 +3-2

11. Example Two Finding Oxidation Numbers sum of the oxidation #’s = zero +2 -2 Find Ox #’s for Ca 3 (PO 4 ) 2 ? 3(Ca)+8(O)=Zero +5 2(P)+ 3(+2)+8(-2)=Zero2(P)+ =+5P

12. Finding Oxidation #’s for Compounds +1-2 +1+5-2 H 3 PO 4 H2OH2O HNO 3 +1+5-2 H 2 SO 4 +1-2 +6 Hg 2 SO 4 +6+1-2 Na 2 Cr 2 O 7 +1 +6 -2 H 2 CO 3 +1 -2 +4 (NH 4 ) 2 CO 3 -3+1+4-2 Ca 3 (AsO 4 ) 2 +2+5 -2 Fe 2 (SO 4 ) 3 +6 +3 -2 Ba(ClO 4 ) 2 +2 +7 -2 Al 2 (CO 3 ) 3 +3+4 -2

13. Formula of Water Lab A Water will be converted in to elements by passing and electric current through the water. Acid( contains + and -ions) is needed in order to pass an electric current through water.

14. Lab A Setup

15. Formula of water Lab A results 7. What gasses are being produced in the tubes? 8. When the electrode-tube with the most gas is 2/3 full, switch the alligator clips and finish filling the tubes. Observations 9. What is the identity of the gas with greater volume?

16. Formula of water Lab A results QUESTIONS: 1. What is the ratio of volumes for the gases collected 2. What is the formula for water? 3. Does the result of your work confirm the ratio of elements in water's formula? 4. Explain how it does so. 5. What mixture of hydrogen to oxygen should give off the best reaction using the spark device? 6. Why was the sulfuric acid solution used in this demonstration?

17. Notes Two Unit One Naming Inorganic Salts Example One Example One Thinking Example Two Computer Assignment One Pages 176-183

18. Naming Inorganic Salts TWO parts to the name 1) Cation 2) Anion Cation Examples Ca +2 Al +3 Fe +2 Na +1 Anion Examples Cl -1 NO 3 -1 SO 4 -2 N -3 Positive Negative

19. Example One Name the formula Fe 2 (CrO 4 ) 3 Step #1 Find The + Ion(s). Iron(II) Fe +2 Iron(III) Fe +3

20. Example One Step #2 Find The - Ion(s) Chromate CrO 4 -2

21. Fe 2 (CrO 4 ) 3 Fe +2 Fe +3 CrO 4 -2 Iron(II) Chromate Iron(III) Fe +2 CrO 4 -2 Iron(II) Chromate (+2) (-2) Y+=0 X=1Y=1 Fe CrO 4 X (+2) (-2) 1+=01 Fe +3 CrO 4 -2 Iron(III) Chromate (+3) (-2) Y+=0 X=2Y=3 X (+3) (-2) 3+=02 Fe 2 (CrO 4 ) 3 Example One

22. Al 2 (CO 3 ) 3 Al +3 CO 3 -2 Carbonate Aluminum Al +3 CO 3 -2 Aluminum Carbonate (+3) (-2) Y+=0 X=2Y=3 X (+3) (-2) 3+=02 Al 2 (CO 3 ) 3 Example Two

23. Computer Assignment One/Two NAMING IONIC COMPOUNDS LEVELs ONE AND TWO

24. Writing a Formula From a Name H 3 PO 4 LiNO 3 Lithium Nitrate ( )_( )_ Li +1 NO 3 -1 11 Hydrogen Phosphate ( )_( )_ H +1 PO 4 -3 13 Ca 3 (AsO 4 ) 2 (NH 4 ) 2 CO 3 Ammonium carbonate ( )_( )_ NH 4 +1 CO 3 -2 1 2 Calcium Arsenate ( )_( )_ Ca +2 AsO 4 -3 2 3 Hg 2 SO 4 Fe(IO 4 ) 3 Iron(III) periodate ( )_( )_ Fe +3 IO 4 -1 31 Mercury(I) Sulfate ( )_( )_ Hg 2 +2 SO 4 -2 22 Na 2 Cr 2 O 7 Ba(ClO 4 ) 2 Barium Perchlorate ( )_( )_ Ba +2 ClO 4 -1 2 1 Sodium Dichromate ( )_( )_ Na +1 Cr 2 O 7 -2 1 2 Pb(SO 4 ) 2 Lead(IV) Sulfate ( )_( )_ Pb +4 SO 4 -2 4 2 (Cation +? ) X (Anion -? ) Y (+?) (-?) Y+=0X Lowest Whole Number Ratio If X or Y is 2 or greater... and the ion is polyatomic. Ba +2 Cr 2 O 7 -2 Hg 2 +2 Pb +4

25. Notes Three Unit One Standard Amounts One Gopher One Mole Formula mass Percent Composition Empirical Formula Pages 286-297

26. Standard Amounts How many dollars is… A) 120 pennies? 1.2 dollars B) 2 quarters? 0.5 dollars C) 15 nickels? 0.75 dollars How many dozens is… D) 48 eggs? 4 dozen E)18 apple fritters 1.5 dozen

27. One Gopher One Gopher equals 12 items What is the mass of one gopher of… A) white beads? 2.81g/G B) blue beads? 0.50g/G C) orange Beads? 1.67g/G

28. Eight Rows Seven Rows

29. One Gopher(12 items) In Six groups (1) How many gophers of beads are in… (2) How many beads are in… A) ___g B) ___g C) ___g D) ___g E) ___g F) ___g 2.00 3.49 2.51 3.75 1.75 5.82 4.00G 1.24G 1.50G 2.25G 3.50G 2.07G 48 beads 15 beads 18 beads 27 beads 42 beads 25 beads

30. One Mole One mole is 6.022x10 +23 items. Each element on the period table has a mass per mole. N O C 14.0g 16.0g 12.0g 6.022x10 +23 atoms N O C 7.0g 4.0g 18.0g =0.50m =0.25m =1.50m How many moles are in each? =3.01x10 +23 atoms =1.51x10 +23 atoms =9.03x10 +23 atoms ÷14.0g/m ÷16.0g/m ÷12.0g/m How many atoms are in each? x6.022x10 +23 atoms/m

31. Calculations Bases on Chemical Formulas Formula mass (Molecular Mass or Gram-Formula Mass) Empirical Formula Percent Composition

32. Rounding Atomic Mass C Fe O 12.011 55.847 15.9994 Bi K Au 208.980837 39.0983 196.96654 Os Mg Na 190.23 24.3050 22.98968 12.0 55.8 16.0 209.0 39.1 197.0 190.2 24.3 23.0

33. Formula Mass Example One Calculate the formula mass for 1 mole of C 6 H 12 O 6. C H O 6 x 12 x 6 x 12.0 = 1.0 = 16.0 = 72.0 12.0 96.0 180.0g/mol E # Mass How many molecules of C 6 H 12 O 6 is 180.0g/mol? 6.022x10 +23 molecules 12.011 1.0079 15.9994

34. Empirical Formula Example One 3) Write the formula What is the empirical (simplest) formula containing 36.8% N, 63.2% O? X by 2 to get whole numbers 1) Calculate moles of each element. N O 36.8 g ÷ 63.2 g ÷ 14.0 = 16.0 = 2.63 mol N 3.95 mol O E Q Mass 2) Calculate the lowest ratio. N O 2.63 mol N ÷ 3.95 mol O ÷ 2.63 mol = 1.00 1.50 E Moles Lowest Ratio N 2 O 3 14.0067 15.9994 Mass

35. Percent Composition Example One Calculate the percentage composition of H 2 O. H O 2 x 1 x 1.0 = 16.0 = 2.0 16.0 18.0g/mol E # Mass 2) Divide each contribution by the total mass. 3) Add the percentages to check work. 1)Calculate the formula mass for 1 mole of H 2 O H O 2.0 ÷ 16.0 ÷ 18.0 = 0.11 0.889 11% 88.9% 100.% 1.0079 15.9994 ( x 100) = 11% ( x 100) = 88.9% Answer

36. End

37. Empirical Formula Example Two 3) Write the formula What is the empirical (simplest) formula containing 69.58% Ba, 6.090% C, 24.32% O? X by 1 to get whole numbers 1) Calculate moles of each element. Ba C 69.58 g ÷ 6.090 g ÷ 137.33 = 12.01 = 0.50666 mol Ba 0.50708 mol C E Q Mass 2) Calculate the lowest ratio. Ba C 0.50666 mol ÷ 0.50708 mol ÷ 0.50666 mol = 1.000 1.001 E MolesLowest Ratio BaCO 3 O24.32 g ÷16.00 = 1.520 mol O O 1.520 mol ÷0.50666 mol = 3.00 Mass

38. Percent Composition Example Two Calculate the percentage composition of Fe(ClO 4 ) 3. Cl O 3 x 12 x 35.5 = 16.0 = 106.5 192.0 354.3g/mol E # Mass 2) Divide the each contribution by the total mass. 3) Add the percentages to check work. 1)Calculate the formula mass for 1 mole of Fe(ClO 4 ) 3. Cl O 106.5 ÷ 192.0 ÷ 354.3 = 0.3006 0.5419 30.1 % 54.2 % 100.1% Fe 1 x55.8 = 55.8 Fe 55.8 ÷354.3 = 0.1575 15.8 % 55.847 35.453 15.9994 ( x 100) = 15.8% ( x 100) = 30.06% ( x 100) = 54.19% Answer

39. Formula Mass Example Three Calculate the formula mass for 1 mole of Al 2 O 3 Al O 2 x 3 x 27.0 = 16.0 = 54.0 48.0 102.0g/mol E # Mass How many molecules of Al 2 O 3 is102.0g/mol? 6.022x10 +23 molecules 26.98154 15.9994

40. Formula Mass Example Two Calculate the formula mass for 1 mole of CaCO 3. Ca C O 1 x 3 x 40.1 = 12.0 = 16.0 = 40.1 12.0 48.0 100.1g/mol E # Mass How many molecules of CaCO 3 is100.1g/mol? 6.022x10 +23 molecules 40.078 12.011 15.9994