edited by: JOHN J. ALEXANDER University of Cincinnati Cincinnati. Ohio 45221
exam que~tlon exchange
(Nm0)- [Nm = nonmetal]: one oxygen would give an atomic weight for Nm of 62 - 16 = 46 glmol. There are no nonmetals with this atomic weight. (NmO$: two oxygens would give an atomic weight of 62 32 = 30 glmol for the nonmetal. This is close to the atomic weieht of nhosohorus. but (PO?)- is not a common oxvanion. . . (NmO$: threeuxygens would mean that Um has an atomic weight of 14. Thks, rhe a n r m is nirrolr ton. !lr can be shown that i f n is 2 or :3, i.e.. if theanion i3dinegativeor trinegati~e, no other nonmetals fit the data).
A Multi-Topic Problem tor General Chemistry James H. Burness The Pennsylvania State University, York Campus York, PA 17403 One of t h e goals of a n introductory chemistry course is t o teach students how to solve unfamiliar problems using t h e concepts they have learned, a s opposed t o solving routine probl&ns w G r h test the knowledge of a single concept. A good way t o develop (and measure1 this ability is to present t h e student with a "arathon" problem which requires specific knowledge i n several areas, h u t a t t h e same time requires t h e student t o recognize how these areas are related. Our students should see t h e forest a s well as t h e trees. T h i s exam ouestion reouires t h e student t o perform calculations based on solution stoichiometry (using molarity or normality), atomic a n d formula weights, crystal structure (unit cell) concepts, electrolytic dissociation, a n d colligative properties for a (osmotic pressure). It might - be especiallv- appropriate .. . final exam. Question Usingonly a Periodic Chart, acalculator, a n d t h e information below, identify t h e strong electrolyte whose general formula is, M,(A),.zH,O
2. The information in point 2 can he used to identify the metal which makes the cation, M3+. (a) Since the metal crystallizes in a hody-centered cubic structure, where the atoms touch along the body diagonal of the cube, the radius of the atom is related to the edge length of the unit cell by the equation, r = (6/4)e,
or r = 0.433e
'Thus, the unit cell has an edge length given by
(b) Now the atomic weight for the metal can be determined from the equation
Ignore t h e effect of interionic attractions i n t h e solution. 1. An- is a common oxyanion. If 30.0 mg of the anhydrous sodium salt of this oxyanion ( N a a where n = 1,2, or 3) is reduced in a reaction that involves the gain of two moks of elerfrons per mole of myanion, it requrres 15.26 ml. of 0.04625 N solution 111the reducing agent to react rum~letelv . . with N a A . ( i i dmred, this section ;anbe reworded to emphasize molarity rather than normality.)' 2. The cation, W+, is derived from a silvery white metal that is relatively expensive. The metal itself crystallizes in a body-centered cubic unit cell, with an atomic radius of 198.4 pm. The metal has a density of 5.243 gIem3. The oxidation number of M in the strong rlectrulyre is t 3 . 3. When 33 4 mg uf the rompaund is dissolved in 10 mL of aqu~ous solut~mat25 T,the solutron hauan osmotic pressure of558 turr.
Acceptable Solution 1. The information given under point 1can be used to calculate the molecular weight of Na.A: (a) (15.26 mL) (0.04625 N) = 0.706 meq titrant, so 0.706 meq af Na.A must have been present in 30.0 mg. This gives a mass per equivalent of 30.0 mg10.706 meq = 42.5 gleq. (b) Since two moles of electrons are gained per mole of Na,A, there are two equivalents of Na,A per mole. In other words, the mass per equivalent is one-half the formula mass. This means that the formula mass is 85.0 glmol. (c) If n = 1 (one Na per formula unit), the formula mass of the oxvanion would be 85.0 - 23.0 = 62 elmol. Since there must heat least oneuxygen, we can see u,h~chnonmetal rould exist m themyanlon as t h numher ~ ofoqgenr wries:
where f is a factor that represents the number of atoms per unit cell, in this case 2 (far body-centered cubic). Substituting values gives
and therefore the atomic weight of the metal is 151.8. This is reasonably close to the atomic weight of europium, atomic number 63. (c) Since the cation and anion (and their charges) are known, the formula must be Eu(NO&.zH%O. 3. The informarion in pornt 3 can be used to determine the formula weight ofthccompmmd, uhich will thenallow thedctrrm~natim of the number of waters of hydration. The otmvtic pressure can be used to determine the molarity of the solution: T
= -=
RT
(558 tord760 torr atm-'1 (0.08205 L-atmmol-'K-')(298
33.4 mg17.50 X
' One possibility is the following: In e reaction where 2 rnol of me Oxyanion react with 3 moi of reducing agent, it requires 11.44 mL of a 0.04625 M solution of the reducing agent to react completely with 30.0 mg of NaA.
K)
Thus, the molarity is 0.0300 M with respect to solute particles. Since the compound is a strong electrolyte and the formula is known, it is clear that it dissociates into four ions per formula unit. Thus, the molarity with respect to the compound is 0.03001 4 = 0.00750 M. Since the compound was dissolved in 10 mL of solution, the number of moles of the compound in solution is 7.50 X The formula mass is then calculated to be
lo-'
mmol = 446 glmol
Subtracting the molarmasses of e europium ion and threenitrate ions from this value gives 108 g, which must be the mass of water (equivalent to 108/18 = 6 ma1 of water) in a mole of the compound. Thus, the formula of the compound is Eu(NOd36N~0. Volume 65
Number 2
February 1988
145
