VOL 6, NO 5
MORE ABOUT PROBLEMS AND GRAPHS
943
MORE ABOUT PROBLEMS AND GRAPHS EUGENE R. DAVIS,WILLIAMI. DICKINSON HIGHSCHOOL, JERSEY CITY, NEW JERSEY Mr. Stone, in a recent issue of the JOURNAL OP CHEMICAL EDUCATION, described an interesting method of teaching the solution of problems by the use of graphs. The graph is helpful in reviewing problems of this type-How many grams of barium chloride, crystallized, will exactly precipitate 20 grams of ammonium sulfate? Say to the class: "A chemist dissolves exactly 20 grams of ammonium sulfate in water and adds a solution of barium chloride crystals, a little a t a time, waiting each time for the precipitate to settle, until a drop of the barium chloride makes no more precipitate. He had 40 grams of the barium chloride crystals dissolved in 250 cc. of water. By measuring the volume left, he calculates that he used 36.96 grams. His record reads: 20 grams of ammonium sulfate required 36.96 grams of barium chlorcqllmicln uacord ide crystals.'' After the class has the picture of this laboratory work sharply .a ' 60 defined, have these E weights p l o t t e d on 4 cross-section paper and 20 - - -L h Rerwd on the blackboard as 0 shown in the figure, and 36 96 244 mark point L, the point ?-Barium Chloride Crystals of Laboratory Record, because, knowing that point L means 20 grams of ammonium sulfate were matched chemically by 36.96grams of barium chloride crystals, we can solve problems of the following type: How many grams of barium chloride crystals would exactly use up 60 grams of ammonium sulfate. Ans.: 36.96 X 3, or 110.88 grams. Draw the line OL, extend it, and mark point P, representing 60 grams of ammonium sulfate and 110.88 grams of barium chloride crystals. Erase point L and the figures 20 and 36.96 from the graph on the blackboard, and propose this problem: If 110.88 grams of barium chloride crystals precipitate 60 grams of ammonium sulfate, how many grams of the barium chloride would be needed for 70 grams of ammonium sulfate? Let the class locate the point for 70 on line OLP and then trace down for the corresponding value. The classes can now see that the point L, that of the original laboratory record, can be dropped, but that problems can still be solved if we have some one point on the line OLP.
2 5
2
13 -------------'
Explain, then, that if crystals of barium chloride are made up using atoms or "standard packages," a weight of 244 must be taken, and that this weight is represented by BaClz.2Hz0. How many grams of ammonium sulfate will match 244 grams of the barium chloride crystals? Ans. 132 grams. This corresponds to the formula (NHI)L301. These figures (244 and 132) give us point E on the graph. Call this the point of Equation Record. Make certain that every one understands that this point is the only record pupils can find to use in solving these problems. Write: (NH& SO4 132
+ BaCL. 2H3O
4 Bas04
+?
244
and have the class practice problems such as: If 244 grams of barium choride crystals are equivalent to 132 grams of ammonium sulfate, how many grams of ammonium sulfate are equivalent to 110.88 grams of barium chloride crystals? Again, 300 grams of barium chloride crystals are equivalent to what weight of ammonium sulfate? The emphasis of this suggested drill is on the necessity for learning how to write equations if problems of the types given are to be solved.
