A = 19; B = 14; C = 9. (Ignores +/-.) 

Brief answers:

1. I2O5

2. One example: 2 Na(s) + Cl2(g)  2 NaCl(s)

Since the ion charges are consistent within each of these two groups, and all the Group 7A elements are diatomic, the equation would come out in the same general form for any other choice of elements. The only thing that would change would be the phase for the halogen: F2 and Cl2 are (g), Br2 is (l), I2 is (s).

3. The chemical is H2. It’s formula mass is 2.016 g/mol.

25 millimoles    1 mol       2.016 g
              x  x  = 0.050 g (2 SF)
                1000 mmol     1 mol

4. a. 
0.0270 mol Pt   2 mol NH4Cl    53.492 g NH4Cl
              x  x  = 0.963 g NH4Cl
                 3 mol Pt       mol NH4Cl

3 SF, like the data.

b. PtCl62-. Ammonium chloroplatinate is given as (NH4)2PtCl6. Ammonium is a familiar ion, NH4+. So the rest of the molecule must be chloroplatinate, and it must have charge 2- to charge-balance two ammonium ions.

5. a. Yes. There are 4.8 Al on the left side [4.8 Al(OH)3] and 3.27 + 1.53 = 4.80 Al on the right side [(Al3.27Fe2+0.90Mg0.39Ti0.03) and (Si6.47Al1.53)].

b. 0.9 moles Fe(OH)3 / 2.035 moles CO2

P Michalopoulos & R C Aller, Rapid clay mineral formation in Amazon delta sediments: Reverse weathering and oceanic elemental cycles. Science 270:614, 10/27/95. 

6. a. Cr(NO3)3; Cr2O3

b. The nitrate. All nitrates are soluble -- a very good rule of thumb.

7. K3PO4·8H2O. 

3K = 3 * 39.10; P = 30.97; 12 O (4 from the phosphate and 8 from the water) = 12 * 16.00; 16 H = 16 * 1.008. Adding them up: 356.398 g/mole.

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

Units amu is also ok, but g/mol is more useful.

Applying the rules for SF strictly would lead to rounding this to 356.40, 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.

8. a.
112 g CH4    mol CH4       802 kJ
          x  x  = 5.60x103 kJ (3 SF)
            16.042 g CH4   mol CH4

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. The H is negative, which indicates an exothermic reaction. This is basically a sign convention, but in this case familiarity with the reaction should help you realize what the convention is. (Note that the question asked you to answer this based on the H.)

9. Na2Se (alc) + CaI2 (alc)  2 NaI (alc) + CaSe (s)

The precipitate is CdSe; the big clue is that it is the intended product -- and the question says that it precipitates. You might also use the rule of thumb that sodium salts are soluble -- but remember that was intended for aqueous solutions, not alcoholic solutions.

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

B A Ridley et al, All-inorganic field effect transistors fabricated by printing. Science 286:746, 10/22/99. For more, on the use, see E Klarreich, Biologists join the dots. Nature 413:450, 10/4/01. This is a News Feature on quantum dots.

10. Estimate population of people on earth: 6 billion, 6x109.

6x109 persons   1014 bacteria   1 mol bacteria
              x  x  = 1 mole bacteria
                 person        6x1023 bacteria

R Gallagher, Immunology: Nota Bene: Monie a Mickle Maks a Muckle. Science 281:186, 7/10/98.
 
11. a. Si. Total number of electrons shown is 14, which is silicon. 
OR: It is the 2nd element of the p block in period 3.

b. 4

c. 3s23p2. (outer s and p electrons)

d. C (or any other element in group 4A).

12. CO2 (aq) + CaSiO3 (s)  SiO2 (s) + CaCO3 (s)

The phase of the CO2 was stated in the question. For the SiO2, the phase is clear if you recognize what quartz is. Otherwise, remember that phases of non-metal compounds really are hard to predict.

P F Hoffman, Oldest terrestrial landscape. Nature 375:537, 6/15/95.

B1. 7 and 15. The O atom has 8 electrons, so O+, which has lost one to give a positive charge, has 7. O2 has 16 electrons, so O2+, which has lost one, has 15.

J H Waite et al, Oxygen ions observed near Saturn’s A ring. Science 307:1260, 2/25/05.

B2. Ammonium nitrate is NH4NO3, formula mass = 80.05 g/mol. Consider one mole. Of that, 2 * 14.01 = 28.02 g is N.

1.00x102 g NH4NO3    28.02 g N
                 x  = 35.0 g N (3 SF)
                   80.05 g NH4NO3

N or N2? It doesn’t really matter in this case, so long as you are consistent. It’s simpler to use N. The question is about mass of nitrogen, in any form -- not specifically about the chemical nitrogen gas.

B3. a.
0.76 g Zn     mol Zn
          x  = 0.0116 mol Zn.
            65.39 g Zn

b. 2.83 g. Product is 3.59 g. Of that, Zn is 0.76 g. So the rest, 2.83 g, is I. You need to make “reasonable assumptions”, given the purpose of the experiment, to do this step; also see part e.

c.
2.83 g I     mol I
          x  = 0.0223 mol I.
            126.9 g I

d. ZnI2. The data show that the ratio of I atoms to Zn atoms is 0.0223 to 0.0116. This is 1.92:1. Since formulas (usually) involve small integers, the likely ratio is 2:1.

Since the common zinc ion is 2+, the result is reasonable. But remember, the question here is about the experimental measurement of the formula, not predicting it.

Why use I rather than I2 in part c? It doesn’t really matter, so long as you are clear and consistent about which one you are using. But many people find it simpler to count atoms here.

e. That it was in excess. If there weren’t enough I2 to react with all the zinc, the calculations above wouldn’t mean anything. The calculations use the given amount of Zn as the relevant amount.

B4. General idea Burn gasoline, get CO2 as a product. If we write a balanced equation for the burning, we can calculate the amount of CO2 made. To do that we need to know how much gasoline we used; to get that, we need the distance driven and the gas mileage. I’ll assume 20 miles for the trip, and 25 mpg (miles per gallon) for gas mileage. The nice thing about clear dimensional analysis set-ups is that they show the logic. You can easily substitute your numbers for mine.

Balanced equation: C8H18 (l) + 25/2 O2 (g)  9 H2O (l) + 8 CO2 (g)

Calculation logic: miles  gallons  moles octane  moles CO2  volume CO2. It’s a little worse, because of the non-metric units in my assumptions.

      ....................gasoline..................
20 mi   gal     3.785 L    1000 mL   0.7 g   1 mol   8 mol CO2    22.4 L
      x  x  x  x  x  x  x 
        25 mi    gal         L       mL      114 g    1 mol gas   mol CO2

= 3000 L (1 SF)

B5. They are all palindromes (they read the same forward and backwards) -- as is “noon”.

•	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 (Spring 2005-2)	
4/11/05