unit one. notes one unit one two classes of elements what are stable elements? stabilizing sodium...
TRANSCRIPT
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Unit One
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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
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Two Classes of ElementsWhat are the Two Main Classes of Elements?Metals and Nonmetals Noble Elements
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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 -1
Oxidation
Reduction
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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?
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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
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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
two Na
one O
High Electronegativity
Low Electronegativity
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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
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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
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Example OneFinding Oxidation Numbers
2 (+3)+ 3(S) = Zero
sum of the oxidation #’s =
Find Ox #’s for Al2S3?
zero
2 (Al)+ 3(S) = Zero
S = -2
+3 -2
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Example TwoFinding Oxidation Numbers
sum of the oxidation #’s = zero
+2 -2Find Ox #’s for Ca3(PO4)2?
3(Ca)+ 8(O) = Zero
+5
2(P)+
3(+2)+ 8(-2) = Zero2(P)+
= +5P
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Finding Oxidation #’s for Compounds
+1 -2
+1+5 -2HH33POPO44
H2O
HNO3
+1+5 -2
H2SO4
+1 -2+6
Hg2SO4
+6+1 -2
Na2Cr2O7
+1 +6 -2
H2CO3
+1 -2+4
(NH4)2CO3
-3 +1+4 -2
Ca3(AsO4)2
+2 +5 -2
Fe2(SO4)3
+6+3 -2
Ba(ClO4)2
+2 +7 -2
Al2(CO3)3
+3 +4 -2
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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.
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Lab A Setup
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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?
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Formula of water Lab A resultsQUESTIONS:• 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?
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Notes Two Unit One
• Naming Inorganic Salts
• Example One
• Example One Thinking
• Example Two
• Computer Assignment One
Pages 176-183
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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
• NO3-1
• SO4-2
• N-3
PositiveNegative
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Example One• Name the formula Fe2(CrO4)3
• Step #1 Find The + Ion(s).
Iron(II) Fe+2
Iron(III) Fe+3
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Example One• Step #2 Find The - Ion(s)
Chromate CrO4-2
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Fe2(CrO4)3 Fe+2 Fe+3 CrO4-2Iron(II) ChromateIron(III)
Fe+2 CrO4-2
Iron(II) Chromate
(+2) (-2)Y+ = 0
X=1 Y=1
FeCrO4
X
(+2) (-2)1+ = 01
Fe+3 CrO4-2
Iron(III) Chromate
(+3) (-2)Y+ = 0
X=2 Y=3
X
(+3) (-2)3+ = 02
Fe2(CrO4)3
Example One
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Al2(CO3)3 Al+3 CO3-2 CarbonateAluminum
Al+3 CO3-2
Aluminum Carbonate
(+3) (-2)Y+ = 0
X=2 Y=3
X
(+3) (-2)3+ = 02
Al2(CO3)3
Example Two
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Computer Assignment One/Two
• NAMING IONIC COMPOUNDS LEVELs ONE AND TWO
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Writing a Formula From a Name
HH33POPO44
LiNOLiNO33Lithium NitrateLithium Nitrate ( )_( )_LiLi+1+1 NONO33
-1-11111
Hydrogen PhosphateHydrogen Phosphate ( )_( )_HH+1+1 POPO44-3-3
1133
CaCa33(AsO(AsO44))22
(NH(NH44))22COCO33Ammonium carbonateAmmonium carbonate ( )_( )_NHNH44
+1+1 COCO33-2-2
1122
Calcium ArsenateCalcium Arsenate ( )_( )_CaCa+2+2 AsOAsO44-3-3
2233
HgHg22SOSO44
Fe(IOFe(IO44))33Iron(III) periodateIron(III) periodate ( )_( )_FeFe+3+3 IOIO44
-1-13311
Mercury(I) SulfateMercury(I) Sulfate ( )_( )_HgHg22+2+2 SOSO44
-2-22222
NaNa22CrCr22OO77
Ba(ClOBa(ClO44))22Barium PerchlorateBarium Perchlorate ( )_( )_BaBa+2+2 ClOClO44
-1-12211
Sodium DichromateSodium Dichromate ( )_( )_NaNa+1+1 CrCr22OO77-2-2
1122
Pb(SOPb(SO44))22Lead(IV) SulfateLead(IV) Sulfate ( )_( )_PbPb+4+4 SOSO44
-2-24422
(Cation+?)X(Anion-?)Y(+?) (-?)Y+ = 0X
Lowest Whole Number Ratio
If X or Y is 2 or greater...
and the ion is polyatomic.
BaBa+2 +2 CrCr22OO77-2 -2 HgHg22
+2 +2 PbPb+4+4
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Notes Three Unit One• Standard Amounts
• One Gopher
• One Mole
• Formula mass
• Percent Composition
• Empirical Formula
Pages 286-297
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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
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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
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Eight Rows
Seven Rows
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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
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
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One Mole• One mole is 6.022x10+23 items.• Each element on the period table has
a mass per mole.
NOC
14.0g16.0g12.0g
6.022x10+23atoms6.022x10+23atoms6.022x10+23atoms
N
O
C
7.0g
4.0g
18.0g
=0.50m
=0.25m
=1.50m
• How many moles are in each?
=3.01x10+23atoms
=1.51x10+23atoms
=9.03x10+23atoms
÷14.0g/m
÷16.0g/m
÷12.0g/m
• How many atoms are in each?
x6.022x10+23atoms/m
x6.022x10+23atoms/m
x6.022x10+23atoms/m
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Calculations Bases on Chemical Formulas
•Formula mass (Molecular Mass or Gram-Formula Mass)•Empirical Formula•Percent Composition
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Rounding Atomic Mass
CFeO
12.01155.84715.9994
BiKAu
208.98083739.0983196.96654
OsMgNa
190.2324.305022.98968
12.055.816.0209.039.1197.0190.224.323.0
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Formula Mass Example OneCalculate the formula mass for 1 mole of C6H12O6.
CHO
6 x12 x6 x
12.0 = 1.0 =16.0 =
72.012.096.0
180.0g/mol
E # Mass
How many molecules of C6H12O6 is 180.0g/mol?
6.022x10+23 molecules
12.0111.007915.9994
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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.
NO
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.
NO
2.63 mol N ÷3.95 mol O ÷
2.63 mol =2.63 mol =
1.00 1.50
E Moles Lowest Ratio
N2O3
14.006715.9994
Mass
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Percent Composition Example OneCalculate the percentage composition of H2O.
HO
2 x1 x
1.0 =16.0 =
2.0 16.018.0g/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 H2O
HO
2.0 ÷16.0 ÷
18.0 =18.0 =
0.11 0.889
11% 88.9%100.%
1.007915.9994
( x 100) = 11%( x 100) = 88.9% Answer
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End
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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.
BaC
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.
BaC
0.50666 mol ÷0.50708 mol ÷
0.50666 mol =0.50666 mol =
1.000 1.001
E Moles Lowest Ratio
BaCO3
O 24.32 g ÷ 16.00 = 1.520 mol O
O 1.520 mol ÷ 0.50666 mol = 3.00
Mass
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Percent Composition Example TwoCalculate the percentage composition of Fe(ClO4)3.
ClO
3 x12 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(ClO4)3.
ClO
106.5 ÷192.0 ÷
354.3 =354.3 =
0.30060.5419
30.1 % 54.2 %100.1%
Fe 1 x 55.8 = 55.8
Fe 55.8 ÷ 354.3 = 0.1575
15.8 %
55.84735.45315.9994
( x 100) = 15.8%( x 100) = 30.06%( x 100) = 54.19%
Answer
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Formula Mass Example ThreeCalculate the formula mass for 1 mole of Al2O3
AlO
2 x3 x
27.0 =16.0 =
54.0 48.0102.0g/mol
E # Mass
How many molecules of Al2O3 is102.0g/mol?
6.022x10+23 molecules
26.9815415.9994
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Formula Mass Example Two
Calculate the formula mass for 1 mole of CaCO3.
CaCO
1 x1 x3 x
40.1 =12.0 =16.0 =
40.112.0
48.0100.1g/mol
E # Mass
How many molecules of CaCO3 is100.1g/mol?
6.022x10+23 molecules
40.07812.01115.9994