FE CHEMISTRY REVIEW
STEVE DANIEL sdaniel@mines.edu

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, S 2, Ca 2+
3. increasing first ionization potential: Mg, K, S
4. increasing electron affinity: S, Si, Ga

1. increasing radius: P, N, F [F < N < P]
2. increasing radius: Ar, S 2, Ca 2+ [Ca 2+ < Ar < S 2]
3. increasing first ionization potential: Mg, K, S [K < Mg < S]
4. increasing electron affinity: S, Si, Ga [Ga < Si < S]

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 COH
2 with formula weight 12.0 + 16.0 + 2.02 =
30.0 or half the molecular weight. So molecular formula is C
2
O
2
H
4
.

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

SYSTEMATIC INORGANIC NOMENCLATURE
IONIC COMPOUNDS
First word is name of metal cation, second is name of anion.
CATION NAME
Element name followed by Roman numeral for ox. # (omitted for IA,IIA,B,Al)
ANION NAME
Monatomic anions (ox. # = Group # -8) Element name root + ide
Oxyanions contain a central element with positive ox # covalently bonded to oxygens. Max ox. # for central atom is group number, each other positive ox. # decreasing by 2 is possible. For example S in group 6A can have +6, +4, +2; Cl can have +7, +5, +3, +1. With four possibilities the largest has per- prefix and the smallest has hypo- prefix; two largest have –ate suffix and two lowest have
-ite suffix. With only three possibilities, per- is eliminated; with only two possibilities, hypo- is eliminated.
Oxyanion name = prefix + central element name root + suffix

EXAMPLES
+1 -3 +3 -2
Na
3
N – sodium nitride Fe
2
Se
3
– iron (III) selenide
+3 +6 -2 +4 +7 -2
Cr
2
(SO
4
)
3
– chromium (III) sulfate Sn(IO
4
)
4
– tin (IV) periodate
+1 +1 -2 +2 +3 -2
KBrO – potassium hypobromite Co(NO
2
)
2
– cobalt (II) nitrite

COVALENT COMPOUNDS
BINARY
More metallic element name then less metallic element root +ide with numerical prefix for number of each. Mono is often omitted,
Cl
2
O
5
– dichlorine pentoxide BF
3
– boron trifluoride
BINARY ACIDS
Add hydro- prefix to anion name, change –ide suffix to –ic an add the “acid”
H
2
S –hydrosulfuric acid HCl – hydrochloric acid
HYDROXY ACIDS
Change oxy anion name suffix –ate to –ic or –ite to –ous and add “acid”
+1 +4 +5
HIO – hypoiodous acid H
2
SeO
3
– selenous acid HNO
3
– nitric acid

BALANCE IN ACIDIC SOLUTION
Cr
2
O
7
2-
1. Assign oxidation numbers
+ N
2
H
5
+ → NO
2(g)
+ Cr 3+
+6 -2 -2 +1 +4 -2 +3
Cr
NOTE
2
O
7
2+ N
2
H
5
+ → NO
2(g)
+1 -2 +1 -2 -2 +1
2 H
2
O ↔ H
3
O + + OH -
+ Cr 3+

2. Balance oxidation number changes
+6 -2 -2 +1 +4 -2 +3
2Cr
2
O
(-6)
7
2+ N
2
H
5
+ → 2 NO
(+12)
2(g)
+ 4Cr 3+
3. Balance charge
2Cr
15H
3
2
O net 3-
7
2-
O + + 2Cr net 12+
+ N
2
2
O
H
5
+
7
2-
→ 2 NO
2(g)
+ 4Cr 3+ net 12+
+ N
2
H
5
+ → 2 NO
2(g) net 12+
+ 4Cr 3+
4. Balance remaining atoms (O and H)
15H
3
O + + 2Cr
2
O
7
2total 50 H atoms
15H
3
O + + 2Cr
2
O
7
2-
+ N
+ N
2
2
H
H
5
+
5
+
→ 2 NO
2(g) total 0 H atoms
→ 2 NO
2(g)
+ 4Cr 3+
+ 4Cr 3+ + 25H
2
O
5. Check last atoms (O)
15H
3
O + + 2Cr
2
O total 29 O atoms
7
2+ N
2
H
5
+ → 2 NO
2(g)
+ 4Cr 3+ + 25H
2
O total 29 O atoms

STOICHIOMETRY
1. A solid sample is 15.0% Na
2
Cr
2
O
7
. How many grams of N
2
H
5
Cl is needed to react with 5.00 g of this sample.
15H
3
O + + 2Cr
2
O
7
2+ N
2
H
5
+ → 2 NO
2(g)
+ 4Cr 3+ + 25H
2
O
5.00g sample x .150g Na
2
Cr
2
O
7 x mole Na
2
Cr
2
O
7 g sample 262.0g Na
2
Cr
2
O
7 x x mole N
2
H
5
Cl x 68.5g N
2
H
5
Cl = 0.0980 g N
2
H
5
Cl
2 mole Na
2
Cr
2
O
7 mole N
2
H
5
Cl
2. How many liters NO
2(g) sample is reacted?
at 30.0
o C and 620.0torr are formed when the 5.00 g of
IDEAL GAS LAW PV = nRT V = nRT/P
5.00g sample x .150g Na
2
Cr
2
O
7
2mole Cr
2
O
7 x mole Na
2
Cr
2
O
7 x g sample x 2mole NO
2
262.0g Na
2
Cr
2
O
7 x .08205Latm x 303.15K x 760.0torr = .0873 L moleNO
2
K 620.0torr atm

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
2Cr
(-6)
2
O
7
2+ N
2
H
5
+
(+12)
→ 2 NO
2(g)
+ 4Cr 3+
So Na
2
Cr
2
O
7
ACID-BASE has 6 eq/mole and N
2
H
5
Cl has 12eq/mole here
# eq/mole = # H + gained or lost per formula
NH
3
NH
3
+ H
3
PO
3
→ NH has 1eq/mole and H
3
PO
3
4
+ + HPO
3
2has 2eq/mole in this reaction
3. 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 + 0.0250 L
Total moles = (.0250L)(1.50mole/L) + V(12.0mole/L) = .0375 +12.0V
(.0375+12.0V)/(V+.0250) = 3.00 mole/L
V = 0.00417 L or 4.17 mls

4. How many mLs of 3.00N N
2
H
5
Cl solution would be required to react with 5.00g of the sample?
5.00g sample x .150g Na
2
Cr
2
O
7 x mole Na
2
Cr
2
O
7 g sample 262.0g Na
2
Cr
2
O
7 x x mole N
2
H
5
Cl x 12 eq N
2mole Cr
2
O
7
2mole N
2
2
H
5
H
5
Cl x L N
2
H
5
Cl x 1000mL = 5.73 mL
Cl 3.00eq N
2
H
5
Cl L
5. Titration of 35.00 mL of a Ba(OH)
2 solution requires 27.63 mL of 3.00 M N
2
H
5
Cl.
What is the molarity of Ba(OH)
2
Ba(OH)
2
+ 2 N
2
H
5 solution? What is its normality?
Cl → BaCl
2
+ 2 H
2
O + 2 N
2
H
4
.02763L N
2
H
5
Cl x 3.00 mole N
2
.03500LBa(OH)
2
L N
2
H
5
H
5
Cl x mole Ba(OH)
Cl 2 mole N
2
H
5
Cl
2
= 1.18 mole Ba(OH)
2
L
= 1.18M
1.18 mole Ba(OH)
2
L Ba(OH)
2 x 2eq Ba(OH)
2 mole Ba(OH)
2
= 2.36N

6. When 250 g CaCO
3 and 700torr result?
3CaCO
3
+ 2H and 300 mL 3.00M H
3
PO
4
→ 3CO
2(g)
+ 3H
2
3
PO
4 are mixed, how many L CO
O + Ca
3
(PO
4
)
2
2(g) at 30.0
o C
250g CaCO
3 x mole CaCO
3
100 g CaCO
3 x 3mole CO
2
3mole CaCO
3 x .08205Latm x 303.15K x 760torr = moleCO
2
K 700torr atm
=
67.4 LCO
2
.300L H
3
PO
4 x 3.00mole H
3
PO
4
L H
3
PO
4 x 3mole CO
2 x RT = 36.5L
2mole H
3
PO
4
P

HESS’ LAW
1. Standard heat of formation for C
2
H
5
OH
(l)
- Σn react
∆H o f,react is -277.7, for CO
2(g) is -393.5, and for H
2
O
(l) is -285.8 (all kJ/mole). Calculate the standard heat of combustion of C
2
2 C
3 H
C
C
2
2
H
H
(gr)
2(g)
5
5
+ 2 O
OH
OH
(l)
(l)
2(g)
+ 3/2 O
→ 2 C
+ 3 O
→ 2 CO
2(g)
(gr)
2(g)
→3 H
+ 3 H
2
2(g)
O
(l)
2(g)
→ 2 CO
+1/2 O
2(g)
+ 3 H
2(g)
2
O
(l)
∆H o = 2(-393.5)
∆H o = 3(-285.8)
∆H o = -1(-277.7)
H
5
OH
(l)
∆H o = -1366.7 kJ/mole
In general:
D
H rxn o = Σ n prod
∆H o f,prod
2. For the reaction: 2 C
2
H
5
OH
(l)
+ O
2(g)
→ 2 CH
3
CHO
(l)
+ 2 H
2
O
(l)
∆H o = -348.6 kJ. Calculate the standard heat of combustion of CH
3
CHO
(l)
.
2 C
2
H
5
OH
(l)
2 CH
3
CHO
(l)
+ 6 O
2(g)
→ 4 CO
+ 2 H
2
O
(l)
2(g)
+ 6 H
→ 2 C
2
H
5
OH
(l)
2
O
+ O
(l)
2(g)
∆H
∆H
So ∆H o comb
2 CH
3
CHO
(l)
+ 5 O
2(g)
→ 4 CO
= -2384.8kJ/(2 mole CH
3
2(g)
+ 4 H
2
O
(l)
CHO) = -1192.4 kJ/mole
∆H o o o
= 2(-1366.7)
= -(-348.6)
= -2384.8 kJ

ELECTROCHEMISTRY
1. Write the anode, cathode, and cell reactions for this voltaic cell. E o = +0.34 v. for Cu 2+ + 2e →Cu and +1.51 v. for MnO
4
+5e + 8H
3
O + → Mn 2+ + 4H
2
O .
Since voltaic cell potential must be positive:
5x(Cu → Cu 2+ + 2e ) anode
2x(MnO
4
+ 5e -
5Cu +2MnO
4
-
+8H
3
+ 16H
3
O + → Mn 2+ + 4H
2
O) cathode
O + →2Mn 2+ + 5Cu 2+ + 8H
2
O cell
E o cell
= E o anode
+ E o cathode
= -(+0.34) + 1.51 = 1.17 v.
NOTE: Potentials are not multiplied by the coefficients used in balancing.
2. What is the cell potential at the initial concentration given?
E = E o –(.0592/n)log Q here n = 10, Q = [Mn 2+ ] 2 [Cu 2+ ] 5 /[MnO
4
] 2 [H
3
O + ] 16
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.

3. How long could the cell operate at 2.00 amp if the Cu electrode initially weighs
5.00g and the volume of the MnO
4
solution is 500.0mL?
(5.00g Cu)(mole Cu/63.54g Cu)(2 eq Cu/mole Cu) = 0.157 eq. Cu
(.500L)(.10 mole MnO
4
/L)(5 eq MnO
4
/mole MnO
4
) = 0.250 eq MnO
4
-
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 10 3 sec
ACID-BASE EQUILIBRIA
1. Calculate the pH of 0.50 M HF solution. K a
= 7.1 x 10 -4
HF + H
2
0.50-x
O ↔ H
3
O + + F x x
K a
= [H
3
O + ][F ]/[HF] = 7.1 x 10 -4
7.1 x 10 -4 = (x)(x)/(.50-x) x= 1.85 x 10 -2 pH = -log[H
3
O + ] = -log(1.85 x 10 -2 ) = 1.73

2. If 0.20 mole NaF is dissolved in 300.0mL 0.50M HF, what is the solution pH?
HF + H
2
O ↔ H
3
.50-x
O + + F x (.20/.300)+x
7.1 x 10 -4 = [H
3
O + ][ F ]/[HF] = x(.667+x)/(.50-x) x = [H
3
O + ] = 5.3 X 10 -4 pH = 3.28
3. Calculate the pH of 0.50 M NaF solution.
F + H
2
O ↔ HF + OH 2 H
2
O ↔ H
3
O + + OH K w
= 1.0 x 10 -14 = [H
3
O + ][OH ]
K h
= [HF][OH ] x [H
[F ] [H
3
3
O + ][OH ] = [HF] x [H
3
O + ][OH ] [H
3
O + ][F ]
O + ][OH ] = 1.0 x 10 -14 = 1.4 x 10 -11
7.1 x 10 -4
F + H
2
O ↔ HF + OH -
.50-x x x
1.4 x 10 -11 = [HF][OH ] = (x)(x)
[F ] .50-x x= 2.6 x 10 -6 =[OH ] [H
3
O + ] = 1.0 x 10 -14 /2.6 x 10 -6 = 3.8 x 10 -9 pH = -log(3.8 x 10 -9 ) = 8.42