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Unit One: Classes of Elements, Periodic Table, and Stable Elements

Learn about the two main classes of elements, their properties, and the factors that make elements stable. Understand the role of oxidation numbers and explore key elements and examples.

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Unit One: Classes of Elements, Periodic Table, and Stable Elements

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  1. Unit One

  2. Notes One Unit One • Two Classes of Elements • Periodic Table Info? • What Are Stable Elements? • Stabilizing Sodium • Stabilizing Oxygen • Sodium Loses electrons to Oxygen • Oxidation Numbers • Key Elements and Examples

  3. Two Classes of Elements • What are the Two Main Classes of Elements? • Metals and Nonmetals (Stair-Step)

  4. What information doe the Periodic Table give us? Atomic Number # Protons = # of Electrons Electron structure Mass Number = # P + # N

  5. What Makes Elements Stable? Which orbitals? • Eight Electrons…… • Highest S and P…… • Noble Gases (electronically stable) • He, Ne, Ar, Kr, Xe, and Rn • Lose or Gain to get 8e-1 (to become stable) • Do metalsLose or Gain electrons? • Do nonmetalsLose or Gain electrons? Which Family? (Lose) (Gain)

  6. Stabilizing Sodium • How many electrons does sodium have? • 11Na • What is Sodium’s Short-Hand Element? • 10Ne • How many Valence Electrons for Sodium? • 1e-1 • What is the shorthand electron configuration? • 11Na • Sodium loses/gain how many electrons? • 1e-1 • Na Na+1 + e-1 Oxidation or reduction? • (protons) + (electrons)=charge 1 [10Ne] 3s (+11) (-10) +1

  7. Stabilizing Oxygen • How many electrons does oxygen have? • 8O • What is oxygen’s Short-Hand Element? • 2He • How many Valence Electrons for oxygen? • 6e-1 • What is the shorthand electron configuration? • 8O • Oxygen loses/gains how many electrons? • 2e-1 • O + 2e-1 O-2 Oxidation or reduction? • (protons) + (electrons)=charge 2p [2He] 2s 2 4 (+8) (-10) -2

  8. 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 • Na2O High Electronegativity two Na Low Electronegativity One O

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

  10. 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 • (NO3-1) ion is always +5 • (SO4-2) ion is always +6

  11. Example One • Find the oxidation numbers. • Al2S3 • Algebra is useful! • Al+3key element • 2(+3) + 3(S)=0 • S= -2 2(Al) + 0 3(S)=

  12. Example Two • Find the oxidation numbers. • Ca(NO3)2 • Algebra is useful ! • Ca+2 and O-2key elements • (+2)+2(N)+6(-2)=0 • (+2)+2(N)+(-12)=0 • 2(N)+(-10)=0 • 2(N)=10 • N= +5 (Ca)+ 6(O)= 2(N)+ 0

  13. Finding Oxidation #’s for Compounds +1 -2 +1 +4 -2 H2O H2CO3 +1 +5 -2 -3 +1 +4 -2 H3PO4 (NH4)2CO3 -2 +2 +5 +1 +5 -2 Ca3(AsO4)2 HNO3 +1 -2 +3 +6 +6 -2 H2SO4 Fe2(SO4)3 +1 +6 -2 +2 -2 +7 Hg2SO4 Ba(ClO4)2 +3 +4 -2 +6 +1 -2 Na2Cr2O7 Al2(CO3)3

  14. Now it is time for class work !!! • A101: paper practice (Work Together) • CA101 : computer assignment • Both will be due at beginning of class next time • Ready Set Break !!!

  15. Notes Two Unit One • Naming Inorganic Salts • Example One • Example One Thinking • Example Two • Computer Assignment One

  16. Naming Inorganic Salts Positive Negative • TWO parts to the name • 1) Cation • 2) Anion • Cation Examples • Ca+2 • Al+3 • Fe+2 • Na+1 • Anion Examples • Cl-1 • NO3-1 • SO4-2 • N-3

  17. Example One • Name the formula Fe2(CrO4)3 • Step #1 Find The + Ion(s). Iron(II) Fe+2 Iron(III) Fe+3

  18. Example One • Step #2 Find The - Ion(s) Chromate CrO4-2

  19. Example One Iron(II) Fe+3 Iron(III) CrO4-2 Fe2(CrO4)3 Fe+2 Chromate Chromate Chromate Iron(II) Iron(III) Fe+2 Fe+3 CrO4-2 CrO4-2 X (+2) + Y = 0 X (+3) + Y = 0 (-2) (-2) X=1 Y=1 X=2 Y=3 1 (+2) + 1 = 0 2 (+3) + 3 = 0 (-2) (-2) Fe Fe2(CrO4)3 CrO4

  20. Example Two Al+3 Aluminum CO3-2 Al2(CO3)3 Carbonate Carbonate Aluminum Al+3 CO3-2 X (+3) + Y = 0 (-2) X=2 Y=3 2 (+3) + 3 = 0 (-2) Al2(CO3)3

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

  22. (Cation+?)X(Anion-?)Y X (+?) + Y = 0 (-?) Writing a Formula From a Name If X orY is 2 or greater... Lowest Whole Number Ratio and the ion is polyatomic. Ba+2 Cr2O7-2 Hg2+2 Pb+4 ( )_( )_ Lithium Nitrate Li+1 1 NO3-1 1 LiNO3 ( )_( )_ Hydrogen Phosphate H3PO4 H+1 3 PO4-3 1 ( )_( )_ Ammonium carbonate NH4+1 CO3-2 1 2 (NH4)2CO3 ( )_( )_ Calcium Arsenate Ca3(AsO4)2 2 Ca+2 3 AsO4-3 ( )_( )_ Iron(III) periodate Fe+3 1 IO4-1 3 Fe(IO4)3 ( )_( )_ Mercury(I) Sulfate Hg2SO4 Hg2+2 2 SO4-2 2 ( )_( )_ Barium Perchlorate Ba+2 ClO4-1 2 1 Ba(ClO4)2 ( )_( )_ Sodium Dichromate Na2Cr2O7 1 Na+1 2 Cr2O7-2 ( )_( )_ Lead(IV) Sulfate Pb(SO4)2 4 Pb+4 2 SO4-2

  23. Notes Three Unit One • Standard Amounts • One Gopher • One Mole • Formula mass • Percent Composition • Empirical Formula

  24. 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

  25. 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

  26. Seven Rows Eight Rows

  27. One Gopher(12 items) • In Six groups • (1) How many gophers of beads are in…(2) How many beads are in…A) ___gB) ___gC) ___gD) ___gE) ___gF) ___g 2.00 48 beads 4.00G 3.49 15 beads 1.24G 2.51 18 beads 1.50G 27 beads 2.25G 3.75 42 beads 1.75 3.50G 5.82 2.07G 25 beads

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

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

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

  31. Formula Mass Example One Calculate the formula mass for 1 mole of C6H12O6. # E Mass 12.011 C 12.0 = 72.0 6 x 1.0079 H 12 x 1.0 = 12.0 15.9994 O 6 x 16.0 = 96.0 180.0g/mol How many molecules of C6H12O6 is 180.0g/mol? 6.022x10+23 molecules

  32. Empirical Formula Example One What is the empirical (simplest) formula containing 36.8% N, 63.2% O? 1) Calculate moles of each element. 3) Write the formula Q Mass E Mass N 2.63 mol N 14.0067 36.8 g ÷ 14.0 = 15.9994 O 63.2 g ÷ 16.0 = 3.95 mol O 2) Calculate the lowest ratio. Moles E Lowest Ratio N 1.00 2.63 mol N ÷ 2.63 mol = O 3.95 mol O ÷ 2.63 mol = 1.50 X by 2 to get whole numbers N2O3

  33. Percent Composition Example One 1)Calculate the formula mass for 1 mole of H2O # E Mass Calculate the percentage composition of H2O. H 2.0 2 x 1.0 = 1.0079 O 1 x 16.0 = 16.0 15.9994 18.0g/mol 2) Divide each contribution by the total mass. ( x 100) = 11% H 0.11 2.0 ÷ 18.0 = Answer ( x 100) = 88.9% O 16.0 ÷ 18.0 = 0.889 3) Add the percentages to check work. 11% 88.9% 100.%

  34. Empirical Formula Example Two What is the empirical (simplest) formula containing 69.58% Ba, 6.090% C, 24.32% O? 1) Calculate moles of each element. 3) Write the formula Q Mass E Mass Ba 0.50666 mol Ba 69.58 g ÷ 137.33 = C 6.090 g ÷ 12.01 = 0.50708 mol C O 24.32 g ÷ 16.00 = 1.520 mol O 2) Calculate the lowest ratio. E Moles Lowest Ratio Ba 0.50666 mol ÷ 1.000 0.50666 mol = C 0.50708 mol ÷ 0.50666 mol = 1.001 O 1.520 mol ÷ 0.50666 mol = 3.00 X by 1 to get whole numbers BaCO3

  35. Percent Composition Example Two 1)Calculate the formula mass for 1 mole of Fe(ClO4)3. # E Mass Calculate the percentage composition of Fe(ClO4)3. 55.847 Fe 55.8 1 x 55.8 = 35.453 Cl 106.5 3 x 35.5 = 15.9994 12 x O 16.0 = 192.0 354.3g/mol 2) Divide the each contribution by the total mass. ( x 100) = 15.8% Fe 0.1575 55.8 ÷ 354.3 = Cl 0.3006 106.5 ÷ 354.3 = ( x 100) = 30.06% Answer O 192.0 ÷ 354.3 = ( x 100) = 54.19% 0.5419 3) Add the percentages to check work. 15.8 % 30.1 % 54.2 % 100.1%

  36. Formula Mass Example Three Calculate the formula mass for 1 mole of Al2O3 # E Mass Al 54.0 2 x 27.0 = 26.98154 O 3 x 16.0 = 48.0 15.9994 102.0g/mol How many molecules of Al2O3 is102.0g/mol? 6.022x10+23 molecules

  37. Formula Mass Example Two Calculate the formula mass for 1 mole of CaCO3. # E Mass 40.078 Ca 40.1 = 40.1 1 x 12.011 C 1 x 12.0 = 12.0 15.9994 O 3 x 16.0 = 48.0 100.1g/mol How many molecules of CaCO3 is100.1g/mol? 6.022x10+23 molecules

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