VOL. 6, NO. 11 SOLVING AND F O ~ T I N G PROBLEMS 1979

SIMPLIFIED METHODS FOR SOLVING AND FORMULATING PROBLEMS IN INDIRECT ANALYSIS

EARLE R. CALEY, PRINCETON UNIVERSITY, PRINCETON, NEW JERSEY

The task of computing or formulating problems in indirect analysis is one that consumes considerable time in calculation if the usual algebraic method of solution is followed. The time taken in such computations is especially felt when checking or formulating a series of such problems for student use. The writer has found that a short arithmetical method may be substituted for the usual algebraic solution in handling this type of analytical problem. In order to show the greater simplicity of the method as compared to the algebraic solution, as well as to show the principle of the method, the two types of solution are given below as applied to specific problems under the two general cases that occur in indirect analysis.

Case I

General Statement: Two diierent elements united to a common element or radical are weighed together and then the amount of the common ele- ment or radical is estimated in a suitable manner. From the weight of the mixture and the weight of the common element found, the percentages or weights of the two components of the mixture are computed.

Spec& Problem: A residue of potassium chloride and sodium chloride weighed 0.2230 gram. The total chlorine in the residue was determined gravimetrically and was found to amount to 0.1285 gram. Find the per- centage of each chloride present in the mixture.

USUAL METHOD OP SOLUTION

Let r equal the weight of NaCl present And 0.2230-x the weight of KC1 present

35.5 35.5 Then - x + (0.2230-x) - = 0.1285 58.5 74.6

0.606% + (0.2230-x)(0.4759) = 0.1285

~ - -

0.1711 ---- x loo = 7 6 . n % ~ ~ c l 0.2230

STEPS

I

100.00-76.72 = 23.28%KC1 8

SHORTENED XETHOD OP SOLUTION STEPS

I . If the residue consisted of pure NaCl the amount of C1 found would be: 35 5

0.2230 X - = 0.1353 gram 1 58.5

2. If the residue consisted of pure KC1 the amount of C1 found would be: 35.5

0.2230 X - = 0.1061grvn 2 74.6

3. Then the difference between the amount of C1 found and the amount that could theoretically be present if the residue consisted of only the first component, divided by the total possible range of C1, mul- tiplied by 100 gives the percentage of the second component:

4. And 100.00-23.28 = 76.72% NaCl 4

Case I1

General Statement: Two different elements united to a common element or radical are weighed together. The original mixture is then converted into a second mixture which contains the two elements united to a different common element or radical. From the weights of the two mixtures the percentages or weights of the components of the original mixture are com- puted.

Specific Problem: A mixture of calcium oxide and strontium oxide weighed 0.3200 gram. When evaporated with sulfuric acid and weighed again as the sulfates the weight was found to be 0.6511 gram. Find the percentage of each oxide in the original mixture.

USUAL METHOD OF SOLVTION

Let x equal the weight of CaO present And 0.3200-x the weirht of SrO present

136.15 18.3.65 Then -- x+(0.3200-x)- = 0.6511

56.1 103.6 2 . 4 2 6 9 ~ + (0.3200-x)(1.7727) = 0.6511

r = 0.1281 0.1281 ---- x I00 = 40.03% CaO 0.3200

SHORTENED METHOD O F SOLUTION

136.15 0 3 2 W X - = 0.7766

56.1

0.7766-0.6511 X I00 = 59.97% SrO

0.7766-0.5673 100.00-59.97 = 40.03% CaO

STEPS

1

STEPS

In the above shortened solution the same principles are followed as in Case I; i . e., the weights of the sulfates that would be obtained if the original mixture consisted of either pure oxide are first found, and then the difference between the actual given weight of mixed sulfates and the theo-

VOL. 6, NO. 11 SOLVING AND FORMULATING PROBLEMS 1981

retical weight of the sulfate if only the first oxide was present, divided by the total possible range in the weight of the sulfates multiplied by one hundred, gives the percentage of the second oxide in the given mixture.

It is apparent, then, that this type of solution is general for problems in indirect analysis since the amount of the common element found or the weight of a second mixture containing a common element lies somewhere between certain possible limits and its position within these limits is a linear function of the percentage composition of the given mixture. The general method involved in this shortened solution is to calculate first the possible limits in a problem and then from the position of the given quantity within these limits compute the composition of the given mixture by simple pro- portion.

Further Applications

This method can also be extended to those problems in which the weight of the common element is given in terms of the weight of precipitate ob- tained in its determination or in the volume of standard solution consumed in its estimation, as the following example shows.

Example 1: A mixture of NaCl and KC1 weighed 0.2550 gram. The total chloride in the sample was determined by titration with tenth-normal silver nitrate, 40.1 cc. being required. Find the percentage of each chloride in the mixture.

Solution:

100.0-37.2 = 62.8% NaCl

There are also certain types of problems in indirect analysis which, at first glance, do not appear to belong to either of the two general classes mentioned above. These almost invariably can be grouped under one of the above general cases upon closer inspection, and in any event they can be solved by the proportional method. The following example is illustrative.

Example 2: Given a mixture of potassium chloride and potassium sul- fate weighing 0.2500 gram. The mixture was converted entirely to potas- sium sulfate by evaporation with sulfuric acid and the weight of pure potas- sium sulfate obtained was 0.2615 gram. Find the percentage of potas- sium chloride present in the original mixture.

Solution: On inspection this is seen to he an example coming under Case 1 since two different groups are united to a common element, which is

estimated in a form which happens to be identical with one of the con- stituents of the original mixture. The proportional solution is simple.

Thus the proportional method of solution is seen to be applicable to all types of problems in indirect analysis regardless of their form and is to be preferred to the algebraic method of solution on the bases of brevity and simplicity.

The Graphic Solution of Problems in Indirect Analysis

A graphic solution of problems of this nature, based upon the same prin- ciples used in the above-described short numerical method, is even of

.+% q w . .. w 0 a *I yl 20 iC 1 .+a o sa s u, 00 m l o o % ~ c ~ o r ~ s o +

ILLUSTRATI~ GRAPE SHOWING COMPOSITION OP ALL POSSIBLE MIXTURES OP EITHER T a p CEWRIDES OR THE SULPATES OP SODIUM AND POTASSIUM POR USE IN CHECKING SOLUTIONS OP PROBLEMS I N INDIRECT ANALYSIS. BASED UPON A ONE-GW SAMPLE

greater convenience for rapidly checking a series of such problems for cor- rectness of solution or formulation. The accompanying figure shows a graph adapted to mixtures of the chlorides or sulfates of sodium and potas- sium. The method of construction and use is evident from the figure. When the weight of the mixture differs from one gram it is then necessary to multiply the graphic values by the sample weight or to convert the given values to the one-gram basis. It is apparent that separate graphs must be drawn for the different mixtures that are to be considered. To be of any value such graphs must he drawn accurately and be of ample size.

VOL. 6. No. 11 So~vmo AND E I O ~ W L A T I N G PROBLEMS 1983

They are not to be recommended for the purpose of solving problems in indirect analysis, but simply as a means of rapidly checking solutions or formulations of such problems.

Conclusion

The graphic method described above is rather to be regarded as a teacher's aid than as a method for student use, but the short numerical method given in this paper is to be recommended for the use of both the teacher and the student of quantitative analysis. From the teacher's viewpoint, in addition to time saved in calculation, this proportional method simplifies the matter of formulating new problems in indirect analysis or forming variations on the same problem, since the effect of altering the quantity of the common element or the second mixture is readily calculated from the given proportion. For the student it is simpler and shorter than the usual algebraic method of solving problems in indirect analysis.