A = 22; B = 16; C =11; D = 5. (Ignores +/-.)

mean = 21.9; median = 26.5; top score: 27.

Brief answers:

1. Se is a main group element, whereas Cu is a transition metal. Main group elements tend to form only one common ion, whereas transition metals often form multiple ions. Thus Se is predictable (at -2); the compound is copper(I) selenide. The predictability of Se2- comes from being in group VIA (16), in which atoms are two electrons short of having a noble gas electron configuration.

2. CaWO4. From the given formula, we learn that tungstate is WO42-, since Na ions are (predictably) 1+. Ca ions are 2+, so it takes only one of them to balance the tungstate ion. 

Some people finessed the question without ever explicitly showing the formula of the tungstate ion. They simply noted that two Na+ could be replaced by one Ca2+. That’s fine.

3. The formula is FeCl3·6H2O. The molar mass, reading from left to right, is 55.85 + 3*35.45 + 12 * 1.008 + 6 * 16.00 = 270.296 amu (per formula unit) or g/mol (more useful).

You can group the atoms any way you want, so long as you end up with the proper total.

Applying the rules for SF strictly would lead to rounding this to 270.30, since no data is available for the 3rd decimal place for some of the elements. I would encourage you not to round here in a real problem, but rather to keep all info until the final result; round only at the end. (I did not grade SF on this question, but if you did something that was not good SF practice, I noted it.)

The purpose of the question was to deal with hydrates. If you had a hydrate, even if incorrect or not properly written, I graded the formula mass based on your formula. If you did not have a hydrate at all, the formula mass calculation had no significance, and you got zero.

4. a. Li2SO4		b. Li2S

5. The compound is K2CO3, with molar mass = 138.21 g/mol.

218 g       mol
       x   =  1.58 mol (3 SF)
          138.21 g  

6. B2O3 (s) + 6 HF (l)  2 BF3 (g) + 3 H2O (l)

7. a.
850 g       mol     2.89x103 kJ
       x  x   =  4.23x104 kJ (3 SF)
          58.12 g       mol

What about the minus sign from the given H? The sign tells you whether heat is used or produced. The question asks how much is produced. It is unnecessary to use the sign during the calculation, and is actually confusing if you do.

b. C4H10 (l) + 13/2 O2 (g)  4 CO2 (g) + 5 H2O (l)

Some people included the heat term. That was not required, and I did not grade on it. But if you do that, be sure it is correct. For the equation I have above, it would be +2.89x103 kJ on the right. If your equation had double the coefficients, to get rid of the fraction, then the heat term is twice that.

c. (In the following setup, b = butane.)
1.00 kg b   1000 g b    1 mol b     4 mol CO2    44.01 g CO2
          x  x  x  x 
            1 kg b      58.12 g b    1 mol b      1 mol CO2

= 3.03x103 g CO2 (or 3.03 kg)

For water, you would get 1.55 kg.

8. a. Cl. It has 17 e (sum of the superscripts). OR: It is the 5th element of the p block in the 3rd period.

b. -1. It needs one electron to fill the p block.

c. 3s23p5. (The outer shell s and p electrons.)

d. F, Br, I, At. That is, any other member of the same group -- except H.

9. a. (NG = nitroglycerin)
8.7 mol NG   10 mol H2O
           x   =  22 mol H2O (2 SF)
             4 mol NG

b. decomposition. That is the simplest answer. Some people also identified this as a combustion reaction. Ok, but note that it is unusual for a combustion reaction, in that it does not use O2. In fact, it produces O2. It uses the internal O of the nitroglycerin as the oxidizing agent. The availability of internal O is one of the things that makes this an explosive. It can burn “inside” (everywhere at once) -- not just on the surface where O2 is available. The reaction is also an oxidation. However, the book presents only a simple view of what that means in Ch 8, so justifying that this is an oxidation reaction (better: redox = oxidation and reduction) is best left until Ch 19.

10. a. CaCl2 contains twice as many chloride ions as calcium ions. Thus a mole of CaCl2 contains 2 moles of chloride ions per one mole of calcium ions. The big issue is to focus on counting particles, as in balanced equations.

b. Ca2+ Cl- Cl-. That is, you should show one Ca2+ ion and two Cl- ions. 

11. Mg(NO3)2 (aq) + 2 NaOH (aq)  Mg(OH)2 (s) + 2 NaNO3 (aq)

Phases. The two reactants are given as aqueous. For the products, NaNO3 is predictably aqueous by two rules of thumb: It is both a sodium salt and a nitrate. Thus the precipitate is likely magnesium hydroxide (milk of magnesia).

Do not show ion charges when you write formulas for compounds. For example, sodium iodide is best written as NaI, not Na+I-.

12. FeTiO3 (s) + CO (g)  Fe (s) + TiO2 (s) + CO2 (g)

Phases. The two C oxides are familiar gases. The three chemicals with metals in them are all likely solids. (There is no mention that any water is present; anyway, these are all insoluble.)

K M Blyth et al, Reduction of ilmenite with charcoal. J Chem Educ 82:456, 3/05. The reaction shown here is only one part of the process. C, in the form of charcoal, is used -- but probably is converted to CO before reacting with the mineral. TiO2 is a useful product.

B1. +7. Just add up the expected charges. There are no big ideas here, just a big ion.

G Furrer et al, The origin of aluminum flocs in polluted streams. Science 297:2245, 9/27/02.

B2. The compound is K2CO3, with molar mass = 138.21 g/mol. Of that, 48.00 g is O. Therefore, there are 48.00 g O per 138.21 g K2CO3.
218 g K2CO3    48.00 g O
            x   =  75.7 g oxygen (3 SF)
              138.21 g K2CO3 

Some of you used a longer pathway, explicitly showing: g K2CO3  mol K2CO3  mol O  g O. This is fully logical, and fine. The one CF I used above combines all three CF into one. If you see that, ok. If not, the full set-up is fine.

B3.
1 ZL  1021 L    1 kg ocean  1000 g ocean    10 g Au         oz     $450
    x   x  x  x  x  x 
       ZL        1 L        1 kg ocean     1012 g ocean    31 g     1 oz

= $1.5x1014 (though 1 SF is really fine, given the approximate data)

In the set-up above, L refers to the ocean, and all terms after “g Au” is introduced refer to Au.

Assumption: Ocean is given in volume, and concentration of gold is given on a mass basis, so need to convert volume of ocean to mass of ocean. This requires using the density of the ocean. I assumed that it is the density of water (1 g/mL = 1 kg/L). It is actually a bit higher (1.025 g/mL, says my CRC handbook), with the dissolved salt, but given the other approximate data here, this assumption does not make a big difference. It is important to know what assumptions/approximations you make, so that you (or someone else) can adjust them if desired/needed. With a clear set-up, it is easy to substitute better numbers.

So why don’t we harvest all this gold -- over 100 trillion dollars worth? Economics. The concentration (Ch 16 term) is so low that it would cost more to harvest than it is worth. This is an issue even with ordinary gold mines. One thing that happens as the price of Au increases is that more mines become economical to work. When the price of Au declines, some mines shut down, as they can no longer be worked at a profit.

B4. a. SO2 (g) + 2 H2O (l) + 2 CaS (s)  2 Ca(OH)2 (aq) + 3 S (s)

Phase of SO2 is “common knowledge”; recall our discussion of combustion of S. Ca(OH)2 is slightly soluble.

b. The key point is to find the limiting reactant (LR). There are various ways to do that. One “intuitive” way is to note that we need a 2:1 ratio of CaS to SO2, but have only 1.5:1. Thus the CaS is the LR, and is the chemical that will determine the amount of product. Then it is a simple stoichiometry problem:
9.00 mol CaS    3 mol S
             x   =  13.5 mol S (3 SF)
               2 mol CaS

c. As good an answer as any is that this reaction does not fall into any of the simple classes we have discussed. It is a complex reaction, which undoubtedly has several little steps along the way. It is a redox reaction (Ch 19), though not typical of those from Ch 8.

B5. Assume a 250 mL (0.25 L) glass of water (about 8 ounces).
0.25 L water   1.5x10-5 g Pb   1 mol    6.02x1023 atoms
             x   x  x 
               L water         207 g        mol

= 1.1x1016 atoms of lead (2 SF -- though 1 SF would be fine)

•	Formatting of chemical formulas (e.g., superscripts and subscripts) in these handouts is done with ChemFormula, a macro for Microsoft Word. See my “Downloads” web page for more information; you can download yourself a copy from there.

pq2a (Fall 2005-2)	
11/14/05