Midterm Examination Form IA

James Richard Fromm


Student Name (PLEASE PRINT, FAMILY NAME FIRST)...................

Student Name (SIGNATURE). . . . . . . . . . . . . . . . . . . . . . .

INSTRUCTIONS

Please enter your name, last or family name first, in the upper right-hand corner of this page, using block printing. You have 90 minutes to complete this test. You may use any silent calculator and your own annotated data sheet if you wish.

Answer all questions, placing your answers on the FRONT SIDE of each page, and SHOW YOUR WORK. Use the back sides of pages for scratch paper. Nothing on the back side of a page will be graded.


1. (10 pts) Name, using correct modern chemical nomenclature, each of these compounds: a. Mn(OH)2; b. CuO; c. HBr; d. AgSCN; e. BaCO3; f. CuSO4; g. Fe2(NO2)3; h. LiHCO3; i. CaF2; j. PbS.

2. (10 pts) Give the correct molecular formula for each of the following compounds. If no molecular formula exists, give the correct empirical formula: a. periodic acid; b. vanadium(V) oxide; c. manganese(IV) oxide; d. sodium sulfate decahydrate (Glauber's salt); e. uranium(VI) fluoride; f. divanadium pentoxide; g. bromous acid; h. hydrogen perchlorate; i. sodium nitride; j. zinc(II) iodate dihydrate.

3. (10 pts) Calculate the percentage composition by mass of disodium monosulfide nonahydrate,

Na2S.9H2O.
Percentage composition by mass is __________% Na, __________% S, __________% H, and __________ %O.

4. (15 pts) Combustion analysis of 5.7530 g of an organic compound known to contain hydrogen, carbon, and oxygen yielded the following results: increase in mass of CO2 absorber, 8.4312 g; increase in mass of water absorber, 3.4526 g.
a. Calculate the empirical formula of the compound.
The empirical formula is . . . . . . . . . . . . . . . . . . . . . ..
b. Are you certain that this is also the molecular formula? (Yes/No; circle one.)

5. (10 pts) A chemist started with 214.639 g of tetramminecopper(II) sulfate monohydrate,

Cu(NH3)4SO4.H2O and by a series of reactions prepared 53.21 grams of pure copper(II) oxide for use in semiconductor research.
a. Calculate the overall percentage yield for these reactions. The overall percentage yield was . . . . . . . . . . . . . . . . . %.
b. The copper oxide was precipitated from solution quantitatively as copper(II) hydroxide by addition of 0.030 molar aqueous NaOH solution. Calculate the minimum volume of NaOH solution needed to precipitate all of the copper (the total copper content) in the above sample. The minimum volume of aqueous NaOH solution is . . . . . . . . . . g.

6. (10 pts) In a university chemical laboratory, a cylinder of chlorine of 1.0 m3 capacity filled at a pressure of 150 atm fell to the floor, breaking the exit piping so that all of the Cl2 gas escaped. Reacting promptly, authorities surrounded the campus (1 km by 1 km) with a high impermeable wall to prevent the escape of dangerous materials. Assume a temperature of 25oC, an atmosphere which is only 1 km deep, and an atmospheric pressure of 1.0 atm. On that basis calculate the partial pressure of chlorine in the university atmosphere. (For your information, a partial pressure of

1 x 10-3 atm is likely to be fatal after a few deep breaths and a partial pressure of 1 x 10-6 atm is the maximum allowable concentration for prolonged exposure.) The partial pressure of chlorine would be . . . . . . . . . . . . . atm.

7. A chemist proposes to convert methane gas, CH4, to liquid methanol, CH3OH, by partial oxidation with atmospheric oxygen.
a. (3 pts) Write the complete balanced equation for this reaction, giving the phases of each reactant and product.
b. (7 pts) Calculate the enthalpy change for this reaction at 25oC and 1 atm: The enthalpy change is ____________________ kJ/mole reaction. Its sign is (positive/negative; circle one) because the reaction is (endothermic/exothermic; circle one).

8. (10 pts) A chemical engineer designs a plant to produce the white paint pigment TiO2(s) using the reaction TiCl4(g) + 2H2O(g) --> TiO2(s) + 4HCl(g).
a. (2 pts) For each metric tonne (1 tonne = 1000 kg = 1 Mg) of TiO2 produced, what mass of

TiCl4(g) reactant is required? The mass of TiCl4 required is ____________________ tonnes.
b. For each metric tonne of TiO2 produced, what is the amount of heat produced at 1 atm and

25oC? Use DHof (TiO2, s) = -945 kJ/mol, DHof (TiCl4, g) = -763 kJ/mol. The amount of heat produced would be _______________ kJ/tonne TiO2.
c. For each metric tonne of TiO2 produced, what would be the volume of the byproduct HCl(g) at 300oC and 1 atm pressure? The volume of HCl gas would be . . . . . . . . . . litres/tonne TiO2

9. (15 pts) The equilibrium constant for the formation of the toxic gas phosgene, COCl2(g), from carbon monoxide and chlorine gas has the value 6.60 x 10+11 atm at 25oC. Compute the equilibrium partial pressures of carbon monoxide, chlorine, and phosgene in a tank originally filled with phosgene if the equilibrium total pressure is 50 atm and the temperature is 25oC.
The partial pressure of CO(g) is ____________________ atm.
The partial pressure of Cl2(g) is ____________________ atm.
The partial pressure of COCl2 is ____________________ atm.


Answer Key

1. a, manganese(II) hydroxide or manganese dihydroxide; b, copper(II) oxide or copper monoxide; c, hydrogen bromide or hydrobromic acid; d, silver(I) thiocyanate [the I oxidation state for silver is sometimes omitted]; e, barium carbonate; f, copper(II) sulfate; g, iron(III) nitrite; h, lithium hydrogen carbonate; i, calcium fluoride or calcium difluoride; j, lead(II) sulfide or lead monosulfide.

2. a, HIO4; b, V2O5; c, MnO2; d, Na2SO4.10H2O; e, UF6; f, V2O5; g, HBrO2; h, HClO4; i, Na3N; j, Zn(IO3)2.H2O.

3. mass Na = 2 x 23.99 = 45.98 g
mass S = 1 x 32.06 = 32.06 g
mass H = 18 x 1.008 = 18.14 g
mass O = 9 x 15.999 = 143.999 g
total mass = 240.17 g
45.98/240.17 x 100 = 19.14% Na
32.06/240.17 x 100 = 13.35% S
18.14/240.17 x 100 = 7.55% H
143.99/240.17 x 100 = 59.95% O

4. a. mass CO2 = 8.4312 g; 8.4312/44.00 = 0.1916 mol CO2
mass H2O = 0.1655 g; 3.4526/18.02 = 0.1916 mol H2O
mol C = 1 x 0.1916; mass C = 2.3013 g
mol H = 2 x 0.1916; mass H = 0.3863 g
There are therefore 2.6876 g C + H
5.7530 - 2.6876 g = 3.0654 g = mass O; mol O = 3.0654/15.9994 = 0.1916
The ratio is C: 0.1916; H: 0.3832, O: 0.1916, so the empirical formula is CH2O.
b. No. The empirical formula is a multiple (submultiple) of the actual molecular formula. The actual compound here is methanol, CH3COOH, whose empirical formula is CH2O or C2H4O2.

5. a. molar mass of hydrate = 245.693 g
molar mass of CuO = 79.54 g
mol Cu in hydrate = 214.639/245.693 = 0.8736 mol
mol Cu in oxide = 53.21/79.54 = 0.6690 mol
% yield = 100 x (mol Cu in oxide)/(mol Cu in hydrate) = 76.6%
The overall percentage yield was 76.6%.
b. Each mole of Cu requires 2 mol OH-, so Cu2+ + 2OH- --> Cu(OH)2
mol OH- required = 2 x 0.8736 = 1.7472 mol
0.030 mol/L = 1.7472 mol/x L
The minimum volume is 58.24 L = 58,240 mL (which is 58,240 g, since 1 mL weighs about 1 g for any dilute aqueous solution). Now you know why you did this with only 5 drops of solution! The minimum volume of aqueous NaOH solution is 58,240 g or mL.

6. pV = nRT, p(1)V(1) = p(2)V(2)
150 atm x 1.0 m3 = p(2) x 1000 x 1000 x 1000 m3
p(2) = 150 x 10-9 = 0.15 x 10-6 atm
You could also do this a longer way by calculating the moles of chlorine present in the cylinder, but you don't need to. This partial pressure would be unpleasant but below the maximum allowable concentration. The partial pressure of chlorine would be 0.150 x 10-6 atm.

7. a. 2CH4(g) + O2(g) --> 2CH3OH(l)
b. CH4(g), DHof = -75 kJ/mol
O2(g), DHof = 0 kJ (element in standard state)
CH3OH(l), DHof = -238 kJ/mol
DHøR = 2(-238 kJ) - [2(-75 kJ) + 1 (0 kJ)]
DHøR = -476 kJ - [-150 kJ] = -476 kJ + 150 kJ = -326 kJ
The enthalpy change is -326 kJ/mol reaction. Its sign is negative because the reaction is exothermic.

8. a. molar mass TiCl4 = 47.88 + 4(35.453) = 189.692 g
molar mass TiO2 = 47.88 + 2(15.999) = 79.878 g
The amount of TiO2 = 1000 103/79.878 = 12.519 x 10+3 mol, and this is also the amount of TiCl4 since reaction is 1:1. The molar mass of TiCl4 = 2.375 x 10+6 g, so the mass of TiCl4 required is 2.375 tonnes.
b. DHof (H2O(g)) = -242 kJ; DHof (HCl) = -92 kJ
DHoR = S DHof (products) - S DHof (reactants)
DHøR = -945 kJ + 4(-92 kJ) - [-763 kJ + 2(-242 kJ)]
DHøR = -1363 kJ - [-1247 kJ] = -116 kJ/mole
This is per mole reaction or per mole TiO2 in a 1:1 reaction.
mol TiO2 = 12.519 x 10+3, so DHo = -116 kJ/mole x 12.519 x 10+3 mol
The amount of heat produced would be (-)1452.2 kJ/tonne TiO2.
c. The reaction would produce 4 x 12.519 x 10+3 mol HCl, so pV = nRT
T = 300 + 273.15 = 573.15 K
V = 50,076 mol x 0.08206 L atm/K mol x 573.15 K/1.0 atm = 218,406 L
The volume of HCl gas would be 218. x 10+3 litres/tonne TiO2.

9. CO(g) + Cl2(g) --> COCl2(g)
Kp = 6.60 x 10+11 = p(COCl2)/p(CO)p(Cl2)
Since only phosgene was originally added, p(CO) = p(Cl2) but not = p(COCl2).
50 atm = p(COCl2) + p(CO) + p(Cl2)
50 atm = p(COCl2) + 2p(CO)
p(COCl2) = 50 atm - 2p(CO)
6.60 x 10+11 p2(CO) = 50 atm - 2p(CO)
6.60 x 10+11 p2(CO) + 2p(CO) - 50 = 0
Assume all phosgene (little dissociation) or just use the quadratic formula here. I used the assumption p(COCl2) equals about 50 atm, then 6.60 x 10+11 p2(CO) = 50, p2(CO) = 76 x 10-12, p(CO) = 8.7 10-6 atm. Quadratic formula gives identical answers.
The partial pressure of CO(g) is 8.7 x 10-6 atm.
The partial pressure of Cl2(g) is 8.7 x 10-6 atm.
The partial pressure of COCl2 is 50 atm.


Copyright 1997 James R. Fromm