FE CHEMISTRY REVIEW

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FE CHEMISTRY REVIEW   STEVE DANIEL   sdaniel_at_mines
.edu
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CHEMICAL PERIODICITY   Arrangement of elements in
the Periodic Table allows prediction of relative
properties of atoms based on the overly
simplistic ideas 1. Atoms in the same row have
outermost electrons in the same shell. 2.
Strength of attraction of the outermost electrons
depends on the effective nuclear charge and which
shell the outermost electrons occupy (nominal
distance from the nucleus). 3. Effective nuclear
charge (number of protons in the nucleus
partially decreased by electronic repulsions)
increases left to right in the Periodic
table.   Relative atomic radii, ionic radii,
ionization potentials, electron affinities can be
rationalized. For example, arrange in order
of   1. increasing radius P, N, F 2.
increasing radius Ar, S2-, Ca2 3. increasing
first ionization potential Mg, K, S 4.
increasing electron affinity S, Si, Ga    
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1. increasing radius P, N, F F lt N lt P 2.
increasing radius Ar, S2-, Ca2 Ca2lt Ar lt
S2- 3. increasing first ionization potential
Mg, K, S K lt Mg lt S 4. increasing electron
affinity S, Si, Ga Ga lt Si lt S
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EMPIRICAL AND MOLECULAR FORMULAE Atomic weight
mass of one mole of atoms molecular weight
mass of one mole of molecules. A compound is
40.0 carbon, 53.3 oxygen, and 6.70 hydrogen
and has a molecular weight of 60.0. What are its
empirical and molecular formulae? 40.0g C x
mole C x 16.0g O 1.00 mole C 53.3g O 12.0gC
mole O mole O 6.70g H x mole H x
16.0g O 1.99 mole H 53.3g O 1.01g H mole
O mole O Therefore empirical formula
is COH2 with formula weight 12.0 16.0 2.02
30.0 or half the molecular weight. So
molecular formula is C2O2H4.
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RULES IN PRIORITY ORDER FOR OXIDATION NUMBERS 1.
S(oxidation numbers) charge 2. Group IA
(Li,Na,K,etc) assign 1 3. Group 2A
(Be,Mg,Ca,etc) assign 2 4. B, Al assign 3 5.
Hydrogen assign 1 6. Oxygen assign -2
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Balance in basic solution BaO2(s) Cr3 ?
CrO42- Ba2 1. Assign oxidation numbers 2 -1
3 6 -2 2 BaO2(s) Cr3 ? CrO42-
Ba2 NOTE 1 -2 1 -2 -2
1 2 H2O ? H3O OH-
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1. Assign oxidation numbers 2 -1 3
6 -2 2 1 -2 1 -2
-2 1 BaO2 Cr3 ? CrO42- Ba2 and 2H2O
? H3O OH- 2. Balance oxidation number
changes 2 -1 3 6
-2 2 3BaO2 2Cr3 ? 2CrO42-
3Ba2 (-2) (3) 3. Balance
charge 3BaO2 2Cr3 ? 2CrO42- 3Ba2
net 6 net 2 4OH- 3BaO2 2Cr3 ?
2CrO42- 3Ba2 net 2 net 2 4.
Balance remaining atoms (O and H) 4OH- 3BaO2
2Cr3 ? 2CrO42- 3Ba2 total 10 O
atoms total 8 O atoms 4OH- 3BaO2 2Cr3 ?
2CrO42- 3Ba2 2H2O 5. Check last atoms (H)
4OH- 3BaO2 2Cr3 ? 2CrO42- 3Ba2
2H2O total 4 H atoms total 4 H atoms
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Balance in acidic solution Cr2O72- N2H5 ?
NO2(g) Cr3 1. Assign oxidation numbers 6
-2 -2 1 4 -2 3 Cr2O72-
N2H5 ? NO2(g) Cr3 2. Balance oxidation number
changes 6 -2 -2 1 4 -2
3 2Cr2O72- N2H5 ? 2 NO2(g)
4Cr3 (-6) (12)  3. Balance
charge 2Cr2O72- N2H5 ? 2 NO2(g) 4Cr3
net 3- net 12 15H3O 2Cr2O72- N2H5 ? 2
NO2(g) 4Cr3 net 12 net 12  4.
Balance remaining atoms (O and H) 15H3O
2Cr2O72- N2H5 ? 2 NO2(g) 4Cr3 total 50 H
atoms total 0 H atoms  15H3O 2Cr2O72- N2H5
? 2 NO2(g) 4Cr3 25H2O  5. Check last atoms
(O) 15H3O 2Cr2O72- N2H5 ? 2 NO2(g) 4Cr3
25H2O total 29 O atoms total 29 O atoms    
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STOICHIOMETRY A solid sample is 15.0 Na2Cr2O7.
How many grams of N2H5Cl is needed to react with
5.00 g of this sample. 15H3O 2Cr2O72- N2H5 ?
2 NO2(g) 4Cr3 25H2O   5.00g sample x .150g
Na2Cr2O7 x mole Na2Cr2O7 x g sample
262.0g Na2Cr2O7 x mole N2H5Cl x 68.5g N2H5Cl
0.0980 g N2H5Cl 2 mole Na2Cr2O7 mole
N2H5Cl   2. How many liters NO2(g) at 30.0oC and
620.0torr are formed when the 5.00 g of sample is
reacted?   IDEAL GAS LAW PV nRT V
nRT/P 5.00g sample x .150g Na2Cr2O7 x mole
Na2Cr2O7 x g sample 262.0g
Na2Cr2O7   x 2mole NO2 x .08205Latm x 303.15K x
760.0torr .0873 L 2mole Cr2O7 moleNO2K
620.0torr atm  
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• CONCENTRATION UNITS
• Molarity M moles solute/L solution
•  Molality m moles solute/kg solvent
•  Normality N equivalents solute/L solution
•  REDOX
• equivalents/mole ox. change per formula
•  2Cr2O72- N2H5 ? 2 NO2(g) 4Cr3
• (-6) (12)
• So Na2Cr2O7 has 6 eq/mole and N2H5Cl has
  12eq/mole here
•  ACID-BASE
• eq/mole H gained or lost per formula
•   NH3 H3PO3 ? NH4 HPO32-
•  NH3 has 1eq/mole and H3PO3 has 2eq/mole in this
  reaction
• How many mLs 12.0M HCl must be mixed (assuming
  additive volumes) with 25.0 mL 1.50M HCl to yield
  3.00M HCl?
• Total volume V 0.250 L
• Total moles (.0250L)(1.50mole/L)
  V(12.0mole/L) .0375 12.0V
• (.037512.0V)/(V.0250) 3.00 mole/L
• V 0.00417 L or 4.17 mls

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4. How many mLs of 3.00N N2H5Cl solution would
be required to react with 5.00g of the sample?
 5.00g sample x .150g Na2Cr2O7 x mole Na2Cr2O7
x g sample 262.0g Na2Cr2O7   x mole
N2H5Cl x 12 eq N2H5Cl x L N2H5Cl x
1000mL 5.73 mL 2mole Cr2O72-
mole N2H5Cl 3.00eq N2H5Cl L
5. Titration of 35.00 mL of a Ba(OH)2 solution
requires 27.63 mL of 3.00 M N2H5Cl. What is the
molarity of Ba(OH)2 solution? What is its
normality? Ba(OH)2 2 N2H5Cl ? BaCl2 2 H2O
2 N2H4 .02763L N2H5Cl x 3.00 mole N2H5Cl x
mole Ba(OH)2 1.18 mole Ba(OH)2 1.18M
.03500LBa(OH)2 L N2H5Cl
2 mole N2H5Cl L  1.18 mole Ba(OH)2 x 2eq
Ba(OH)2 2.36N L Ba(OH)2 mole Ba(OH)2
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6. When 250 g CaCO3 and 300 mL 3.00M H3PO4 are
mixed, how many L CO2(g) at 30.0oC and 700torr
result? 3CaCO3 2H3PO4 ? 3CO2(g) 3H2O
Ca3(PO4)2  250g CaCO3 x mole CaCO3 x 3mole CO2
x .08205Latm x 303.15K x 760torr
100 g CaCO3 3mole CaCO3 moleCO2 K 700torr
atm   67.4 LCO2 .300L H3PO4 x 3.00mole
H3PO4 x 3mole CO2 x RT 36.5L L
H3PO4 2mole H3PO4 P  
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HESS LAW 1. Standard heats of formation for
C2H5OH(l) is -277.7, for CO(g) is -393.5 and for
H2O(l) is -285.8 (all kJ/mole),
respectively. Calculate the standard heat of
combustion of C2H5OH(l) 2 C(gr) 2 O2(g) ? 2
CO2(g) ?Ho 2(-393.5) 3 H2(g) 3/2 O2(g) ?3
H2O(l) ?Ho 3(-285.8) C2H5OH(l) ? 2 C(gr) 3
H2(g) 1/2 O2(g) ?Ho -1(-277.7) C2H5OH(l) 3
O2(g) ? 2 CO2(g) 3 H2O(l) ?Ho -1366.7
kJ/mole In general DHrxno S nprod?Hof,prod -
Snreact?Hof,react 2.  For the reaction 2
C2H5OH(l) O2(g) ? 2 CH3CHO(l) 2 H2O(l) ?Ho
-348.6 kJ. Calculate the standard heat of
combustion of CH3CHO(l). 2 C2H5OH(l) 6 O2(g) ?
4 CO2(g) 6 H2O(l) ?Ho 2(-1366.7) 2 CH3CHO(l)
2 H2O(l) ? 2 C2H5OH(l) O2(g) ?Ho
-(-348.6) 2 CH3CHO(l) 5 O2(g) ? 4 CO2(g) 4
H2O(l) ?Ho -2384.8 kJ So ?Hocomb -2384.8kJ/(2
mole CH3CHO) -1192.4 kJ/mole  
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• ELECTROCHEMISTRY
• Write the anode, cathode, and cell reactions for
  this
• voltaic cell. Eo 0.34 v. for Cu2 2e-?Cu
  and 1.54 v.
• For MnO4- 5e- 8H3O ? Mn2 4H2O .
• Since voltaic cell potential must be positive
• 5x(Cu ? Cu2 2e- ) anode
• 2x(MnO4- 5e- 8H3O ? Mn2 4H2O) cathode
• 5Cu 2MnO4- 16H3O?2Mn2 5Cu2 8H2O cell

• Eocell Eoanode Eocathode -(0.34) 1.54
  1.17 v.
• NOTE Potentials are not multiplied by the
  coefficients used in balancing.
• What is the cell potential at the initial
  concentration given?
• E Eo (.0592/n)log Q here n 10, Q
  Mn22Cu25/MnO4-2H3O16
• Cu metal does not have a variable concentration
  and that of water is assumed to be
• nearly constant, so these do not appear in Q.
• E 1.17 (.0592/10)log(.10)2(.10)5/(.10)2(.10)16
  1.24 v.

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3. How long could the cell operate at 2.00 amp
if the Cu electrode initially weighs 5.00g and
the volume of the MnO4- solution is
500.0mL? (5.00g Cu)(mole Cu/63.54g Cu)(2 eq
Cu/mole Cu) 0.157 eq. Cu (.500L)(.10 mole
MnO4-/L)(5 eq MnO4-/mole MnO4-) 0.250 eq
MnO4- Therefore Cu is the limiting reactant and
0.157 mole of electrons will flow (.157 eq)(96487
coul/eq)(sec/2.00coul) 7.57 x 103
sec ACID-BASE EQUILIBRIA 1. Calculate the pH of
0.50 M HF solution. Ka 7.1 x 10-4 HF H2O ?
H3O F- Ka H3OF-/HF 7.1 x
10-4 0.50-x x x 7.1 x 10-4
(x)(x)/(.50-x) x 1.85 x 10-2 pH -logH3O
-log(1.85 x 10-2) 1.73
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• 2. If 0.20 mole NaF is dissolved in 300.0mL 0.50M
  HF, what is the solution pH?
• HF H2O ? H3O F-
• .50-x x (.20/.300)x
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• 7.1 x 10-4 H3O F-/HF x(.667x)/(.50-x)
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• x H3O 5.3 X 10-4 pH 3.28
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• Calculate the pH of 0.50 M NaF solution.
• F- H2O ? HF OH- 2 H2O ? H3O
  OH- Kw 1.0 x 10-14 H3OOH-
• Kh HFOH- x H3OOH- HF x
  H3OOH- 1.0 x 10-14 1.4 x 10-11
• F- H3OOH-
  H3OF- 7.1 x 10-4
• F- H2O ? HF OH- 1.4 x 10-11
  HFOH- (x)(x)
• .50-x x x
  F- .50-x
• x 2.6 x 10-6 OH- H3O 1.0 x 10-14/2.6 x
  10-6 3.8 x 10-9
• pH -log(3.8 x 10-9) 8.42