A VARIANT METHOD FOR SOLVING CHEMICAL PROBLEMS In the solution of chemical problems involving weights only, it may be of interest to compare the usual method with another in which the number of mols of the reacting substances is made the basis of the computation. Illustrative problems may, perhaps, make the matter clearer. Problem. How may grams of nitric acid are required to react with 12 grams of copper in the preparation of nitric oxide? (Cu = 64.) Usuzl method. (1) 3Cu (2) 3 X 64
+ 8HN03 8 X 63
(3) 12 192 504 (4) - = 12 z
---t
+
~ C U ( N O ~ )4~H 2 0 3 X 188 4 X 18
+
2N0 2 X 30
Z
(5) z = 504 l 2 or 31.5 grams. 192
Proposed Method It is clear that for every 3 mols of copper 8 mols of acid are used; the number of mols of acid is therefore 8/3 the number of mols of copper. We have 12 grams of copper but 64 is the weight of one rnol of copper; therefore we have 12/64 of a rnol of copper. The number of mols of acid is therefore 12/64 X 8/3 or 1/2 mol. One rnol of nitric acid is 63 grams so the weight of acid is 63 X 1/2 or 31.5 grams. Problem. How many grams of sulfuric acid are required to react with 12 grams of copper in the preparation of sulfur dioxide? By the usual method 64 196 of computation we have for step four, as shown above, - = -whence x 12 x equals 36.7 grams. By the proposed method the reasoning is: it requires 2 mols of acid for each rnol of copper and we have 12/64 mols of copper. Hence, the number of mols of acid is 12/64 X 2 or 3/8 rnol of acid. One rnol of acid is 98 grams, hence we have 3/8 X 98 grams or 36.7 grams. The advantages of this method are offset by the fact that it is not given in our high-school texts. The simplicity of the method is obvious and:ease of solution is evident since generally we have t o deal only with relatively small numbers.
