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, S2-, Ca2+
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, S2-, Ca2+ [Ca2+< Ar < S2-]
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 COH2 with formula weight 12.0 + 16.0 + 2.02 =
30.0 or half the molecular weight. So molecular formula is C2O2H4.
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
INORGANIC NOMENCLATURE
IONIC COMPOUNDS
Metal element name (Ox. Number unless IA,IIA, or B,Al)
Monatomic anions - element name root + ide
Oxyanions – a prefix + central element name root + a suffix
Suffix and prefix determined by oxidation number. Max ox. # for an element is
group number, then other positive ox. # decrease by 2. 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.
Na3N – sodium nitride
Fe2Se3 – iron (III) selenide
Cr2(SO4)3 – chromium (III) sulfate Sn(IO4)4 – tin (IV) periodate
Ca(IO2)2 – calcium hypoiodite
COVALENT COMPOUNDS
BINARY
More metallic element name and less metallic element root +ide with
numerical prefix for each.
N2O3 – dinitrogen trioxide
BF3 – boron trifluoride
HYDROXY ACIDS
Change oxy anion name suffix –ate to –ic or –ite to –ous and add “acid”
H2S –hydrosulfuric acid
HI – hydroiodic acid
BINARY ACIDS
Add hydro- prefix to anion name, change –ide suffix to –ic an add the “acid”
HIO2 – hypoiodous acid
H2SeO3 – selenous acid
HNO3 – nitric acid
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+
NOTE
+1 -2
+1 -2
-2 +1
2 H2O ↔ H3O+ + OH-
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 3net 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
STOICHIOMETRY
1. 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
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+ + HPO32NH3 has 1eq/mole and H3PO3 has 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 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 Cr2O72mole 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
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
HESS’ LAW
1. Standard heats of formation for C2H5OH(l) is -277.7, for CO2(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 = Σ nprod∆Hof,prod - Σnreact∆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
ELECTROCHEMISTRY
1. Write the anode, cathode, and cell reactions for this
voltaic cell. Eo = +0.34 v. for Cu2+ + 2e-→Cu and +1.51 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) + 1.51 = 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 = Eo –(.0592/n)log Q here n = 10, Q = [Mn2+]2[Cu2+]5/[MnO4-]2[H3O+]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 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 MnO4Therefore 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 = [H3O+][F-]/[HF] = 7.1 x 10-4
0.50-x
7.1 x 10-4 = (x)(x)/(.50-x)
x= 1.85 x 10-2
x
x
pH = -log[H3O+] = -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 + H2O ↔ H3O+ + F.50-x
x
(.20/.300)+x
7.1 x 10-4= [H3O+][ F-]/[HF] = x(.667+x)/(.50-x)
x = [H3O+] = 5.3 X 10-4
pH = 3.28
3. Calculate the pH of 0.50 M NaF solution.
F- + H2O ↔ HF + OH2 H2O ↔ H3O+ + OH- Kw = 1.0 x 10-14 = [H3O+][OH-]
Kh = [HF][OH-] x [H3O+][OH-] = [HF]
x [H3O+][OH-] = 1.0 x 10-14 = 1.4 x 10-11
[F-]
[H3O+][OH-] [H3O+][F-]
7.1 x 10-4
F- + H2O ↔ HF + OH- 1.4 x 10-11 = [HF][OH-] = (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