A QUESTION OF PROCEDURE CHARLES H. STONE, ENGLISH HIGHSCHOOL, BOSTON, MASSACHUSETTS
In the solution of a series of related mathematical problems it would appear that the best method is one which is simple, logical, and capable of application to any variations of such problems. If one is teaching a class in percentage, one might assign such a problem as this: "Mr. B. withdrew from his savings account the sum of $36 which was 9% of his total deposit. How much did he have in the bank before the withdrawal?" If Johnny divides 9 by 36 his procedure and reasoning are wrong. One would expect him to reason that if $36 is 9%, 1% can be found by dividing 36 by 9 and loo%, the total or base of the problem, is to be found by multiplying the result thus obtained by 100, which gives 400. Therefore, Mr. B. had $400 in the bank before his withdrawal. Take another illustration. "The atomic weight of sodium is 23. A certain compound contains 43.4% of sodium. What are some possible molecular weights of the compound?' Reasoning as before, we divide 23 by 43.4 and multiply the result by 100 which gives approximately 53. If the compound contains a single sodium atom per molecule, this is the molecular weight: if the compound contains two sodium atoms per molecule, the molecular weight is 53 X 2 or 106, etc. The principle involved is that the base is found by dividing the percentage by the rate. But now consider this problem: "Find the formula of a compound containing sodium 43.4%, carbon 11.3%, and oxygen 45.3%." Our textbook tells us that we are to divide each percentage by its respective atomic weight to get the ratios in which the atoms of the different atoms are present. But this procedure is directly opposite to that which Johnny has used in every preceding problem involving rate and percentage. Is it necessary thus to proceed contrary to the reasoning which he has been accustomed to follow? No, i t isn't. Proceeding just as shown in paragraph three of this article we find the molecular weight to be 53 or some multiple of 53. From this we get: 53 X 0.434 = 23 = 1Na 53 X 0.113 = 6 = '/zC 53 X 0.453 = 24 = 8 / z 0 Since we cannot have halves of atoms it is evident that we must multiply our result by two which gives 2Na, lC, and 3 0 or NaC03. If we start with carbon instead of sodium the result is the same. (12 + 11.3) X 100 = 106, the molecular weight if one carbon atom is present. From this we get: 106 X 0.434 = 46 = 2Na 106 X 0.113 = 12 = 1C or NanCOs. 106 X 0.453 = 48 = 3 0 4
If we begin with oxygen we get a possible molecular weight of 35.3. Proceeding from this as illustrated above we get 2/3Na, 1/8C, and 10. Since thirds of atoms are impossible we must multiply our result by three which gives the same formula obtained in the other two cases. It appears, therefore, that without abandoning the reasoning and procedure to which pupils have been accustomed in previous work in percentage we may derive the formula of the compound by the use of any one of the percentages given in the problem. The writer knows of no text which presents this method. All texts are unanimous in giving the method of dividing the percentages by the related atomic weights to get the ratios of the numbers of atoms; these ratios often do not come out as whole numbers but must be divided by the smallest one of the ratios to reduce the values to whole numbers. But to divide percentage by percentage is contrary to the rules of arithmetic as the pupil has been accustomed to think of it. Why, then, puzzle Johnny by giving him a new method for doing problems when methods already familiar will do the same work satisfactorily? To thinking teachers the question is submitted whether in the writing of chemistry texts there is anything gained by abandoning the line of reasoning and procedure which has been used in all other problems involving percentage for a method which is no simpler, gives no different answer, and is neither necessary nor more logical.
