1. FE CHEMISTRY REVIEW STEVE DANIEL sdaniel@mines.edu

3. 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

4. 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]

5. 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.

6. RULES IN PRIORITY ORDER FOR OXIDATION NUMBERS  (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

7. Balance in basic solution: BaO 2(s) + Cr 3+ → CrO 4 2- + Ba 2+ 1. Assign oxidation numbers +2 -1 +3 +6 -2 +2 BaO 2(s) + Cr 3+ → CrO 4 2- + Ba 2+ NOTE +1 -2 +1 -2 -2 +1 2 H 2 O ↔ H 3 O + + OH -

8. 1. Assign oxidation numbers +2 -1 +3 +6 -2 +2 +1 -2 +1 -2 -2 +1 BaO 2 + Cr 3+ → CrO 4 2- + Ba 2+ and 2H 2 O ↔ H 3 O + + OH - 2. Balance oxidation number changes +2 -1 +3 +6 -2 +2 3BaO 2 + 2Cr 3+ → 2CrO 4 2- + 3Ba 2+ (-2) (+3) 3. Balance charge 3BaO 2 + 2Cr 3+ → 2CrO 4 2- + 3Ba 2+ net 6+ net 2+ 4OH - + 3BaO 2 + 2Cr 3+ → 2CrO 4 2- + 3Ba 2+ net 2+net 2+ 4. Balance remaining atoms (O and H) 4OH - + 3BaO 2 + 2Cr 3+ → 2CrO 4 2- + 3Ba 2+ total 10 O atomstotal 8 O atoms 4OH - + 3BaO 2 + 2Cr 3+ → 2CrO 4 2- + 3Ba 2+ + 2H 2 O 5. Check last atoms (H) 4OH - + 3BaO 2 + 2Cr 3+ → 2CrO 4 2- + 3Ba 2+ + 2H 2 Ototal 4 H atoms

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

10. STOICHIOMETRY 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 x g sample 262.0g Na 2 Cr 2 O 7 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) at 30.0 o C 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 Na 2 Cr 2 O 7 x mole Na 2 Cr 2 O 7 x g sample 262.0g Na 2 Cr 2 O 7 x 2mole NO 2 x.08205Latm x 303.15K x 760.0torr =.0873 L 2mole Cr 2 O 7 moleNO 2 K 620.0torr atm

11. 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 2 O 7 2- + N 2 H 5 + → 2 NO 2(g) + 4Cr 3+ (-6) (+12) So Na 2 Cr 2 O 7 has 6 eq/mole and N 2 H 5 Cl has 12eq/mole here ACID-BASE # eq/mole = # H + gained or lost per formula NH 3 + H 3 PO 3 → NH 4 + + HPO 3 2- NH 3 has 1eq/mole and H 3 PO 3 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 + 0.250 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

12. 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 x g sample 262.0g Na 2 Cr 2 O 7 x mole N 2 H 5 Cl x 12 eq N 2 H 5 Cl x L N 2 H 5 Cl x 1000mL = 5.73 mL 2mole Cr 2 O 7 2- mole N 2 H 5 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 solution? What is its normality? Ba(OH) 2 + 2 N 2 H 5 Cl → BaCl 2 + 2 H 2 O + 2 N 2 H 4.02763L N 2 H 5 Cl x 3.00 mole N 2 H 5 Cl x mole Ba(OH) 2 = 1.18 mole Ba(OH) 2 = 1.18M.03500LBa(OH) 2 L N 2 H 5 Cl 2 mole N 2 H 5 Cl L 1.18 mole Ba(OH) 2 x 2eq Ba(OH) 2 = 2.36N L Ba(OH) 2 mole Ba(OH) 2

13. 6. When 250 g CaCO 3 and 300 mL 3.00M H 3 PO 4 are mixed, how many L CO 2(g) at 30.0 o C and 700torr result? 3CaCO 3 + 2H 3 PO 4 → 3CO 2(g) + 3H 2 O + Ca 3 (PO 4 ) 2 250g CaCO 3 x mole CaCO 3 x 3mole CO 2 x.08205Latm x 303.15K x 760torr = 100 g CaCO 3 3mole CaCO 3 moleCO 2 K 700torr atm = 67.4 LCO 2.300L H 3 PO 4 x 3.00mole H 3 PO 4 x 3mole CO 2 x RT = 36.5L L H 3 PO 4 2mole H 3 PO 4 P

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

15. ELECTROCHEMISTRY 1.Write the anode, cathode, and cell reactions for this voltaic cell. E o = +0.34 v. for Cu 2+ + 2e - →Cu and +1.54 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 - +8H 3 O + → Mn 2+ + 4H 2 O) cathode 5Cu +2MnO 4 - + 16H 3 O + →2Mn 2+ + 5Cu 2+ + 8H 2 O cell E o cell = E o anode + E o cathode = -(+0.34) + 1.54 = 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.

16. 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 O ↔ H 3 O + + F - K a = [H 3 O + ][F - ]/[HF] = 7.1 x 10 -4 0.50-x x x7.1 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

17. 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 O + + F -.50-x 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 3 O + ][OH - ] = [HF] x [H 3 O + ][OH - ] = 1.0 x 10 -14 = 1.4 x 10 -11 [F - ] [H 3 O + ][OH - ] [H 3 O + ][F - ] 7.1 x 10 -4 F - + H 2 O ↔ 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 - ] [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