VOL.
9, No. 5
CORRESPONDENCE
941
PREPARATION OF RHOMBIC SULFUR CRYSTALS DEAR EDITOR: It is well known that the size of crystals formed from a solution depends upon the rate of evaporation of the solvent: slow evaporation producing large crystals. In order to obtain large crystals i t is necessary only to prolong the period of evaporation. In the case of sulfur, proceed as follows: dissolve in 20 cc. carbon disulfide as much powdered roll sulfur as possible. Filter through a dry filter paper into a small flat-bottomed glass dish. Cover the dish with two or more layers of hlotting paper, the sheets being large enough to extend completely over the top of the dish. Lay a book or other heavy flat body on the top of the blotting paper. Set the dish away in a cool place for a week. Under the above conditions, evaporation must proceed very slowly and crystals half an inch long may he obtained. The general shape may readily be observed and the angles and beveled sides easily seen through a lens, but i t is not a simple matter to get perfect large crystals. For exhibition purposes, the smaller but more perfect sulfur crystals may be shown behind a good reading glass of high magnification so arranged from the viewpoint of the observer as to focus upon the crystals. The dish in which the crystals formed should he supported a t the eyelevel and so that the bottom of the dishjs in a vertical position. The crystals adhere to the bottom of the dish and will not fall off. A suitable P label should be appropriately attached. The above might serve as a project-"A Study of Crystallization as Affected by Timew-the two samples, large and small, being shown side by side. CHARLESH. STONE MEMOKIAL Hrcn SCHOOL BOSTON,Mass.
A SIMPLE METHOD FOR CALCULATING THE PROPORTIONS -- . - - OR .VARIOUS SUBSTANCES OF KNOWN COMPOSITION TO GIVE A MIXTURE OF DEFINITE COMPOSITION DEAREDITOR: Pharmacists, chemists, many factory workmen where chemical processes are used, teachers and workers in the field of science in general (either for the purposes of instruction or for their own use in laboratory work), as well as all others who wish to mix various quantities of materials of known composition of any one ingredient to obtain a material of certain definite composition will find the following simple method a great help in cal-
942
JOURNAL OF CHEMICAL EDUCATION
MAY. 1932
culating the proportions or quantities of materials to be used. Four examples are worked out herewith to illustrate the method. A few applications may be indicated here. These, with the examples, will suggest the application of the method to a wide range of operations. The method will apply to the formation of alloys, to mixing ores, mixing fuels, calculating quantities of batches of materials to be mixed, diluting and mixing all kinds of solutions such as milk, acids, alkalies, salts, etc., to the determination of the proportions of various isotopes to give the "average" atomic weight as given in our atomic weight tables, and, in fact, to all determinations of the proportions by weight of given materials to be used in making a product of a certain definite composition. The mathematics of the method will be found discussed in "Van Nostrand's Chemical Annual" (1913), Olsen, and a.rectangle method for making determinations where only two quantities are used for obtaining a third is given there, also. However, the method given herewith is just as simple as the rectangle method for calculations and may be used for any number of quantities. It is based upon the principle that the mixing of materials is strictly an additive property which holds true as far as weight relations go and in many cases there is no practical difference for volume relations in dilute solutions (e. g., adjusting normalities). However, volume relations may easily be calculated by the method if densities are taken into consideration. The method in "set-up" form may be,easily compared to and recalled from the fact that it closely resembles a season schedule-sheet for a league of baseball or football teams. I t can best be explained by taking concrete problems. 1. What weight of milk of 3.5% butter fat should be mixed with 100 pounds of milk of 4.2y0 butter fat to give a milk of 4.0% butter fat? Ilcrirod
Given
Per cent butter fat
3.5
3.5 4.2 Parts by wt., whole nos. Given and wanted
... 0.2 2 X lb.
4.2 0.5
4.0
...
5 100 ib.
7
Solution 4.0 - 3.5 4.2 - 4.0
= =
0.5 0.2
2 + 5 = 7
5X = 200 X = 40 1b.
The data are arranged in a table of horizontal rows and vertical columns as shown. It is evident that 4.0% milk cannot be made from either 3.5 or 4.2% milk alone so dashes are placed in these spaces. The quantity of 3.5% milk can be calculated readily from the proportion, 2 : 5 = X :100, or by "cross-multiplying" and solving for X which represents the unknown amount. Of course, the total amount of 4.0y0 milk represented by the
VOL.9, No. 5
CORRESPONDENCE
943
+
7 parts is 40 100 = 140 pounds. Though the parts are given by weight no appreciable error would be introduced by substituting gallons for pounds for the relative quantities. 2. In what proportions should 4, 7, 10, 15, and 18% hydrochloric acid solutions be mixed to give a 12y0 acid solution (all by weight)? Desired
Per cent by weight
18 4 7 10 15 18
12
8 5 2
12- 4 12- 7 12 - 10 15 - 12 18 - 12
..
..
Parts by weight Parts by wt., smallest whole numbers
Solution
15
= 8 = 5 = 2 = 3 = 6
57
It is obvious that 12% acid could not be made by mixing 4, 7, and 10% acids in any proportions and this is also true of mixing 15 and 18yo acids, hence the dashes in the spaces as shown. The calculations at the right of the table give the figures in the other spaces which readily indicate the parts for mixing any two acids (e. g., 3 parts of 4y0 acid mixed with 8 parts of 15y0 acid give 3 8 = 11 parts of 12y0 acid) and likewise the numbers in the last row (3, 3, 3, 5, 5).give the relative quantities in smallest whole numbers of the acid as indicatsd by the numbers (%) a t the tops of the columns which should be used and the relative quantity of 12y0 acid obtained (19). For convenience in measuring these, parts by weight (grams, say) may be changed t o parts by volume (cc.) by dividing each by its corresponding specific gravity. These specific gravities may be obtained directly or by interpolation in tables such as are given in the "Handbook of Chemistry and Physics," Chemical Rubber Co., Cleveland, Ohio.
+
Dcsircd
Given
110 111 112 113 114 115 116 112.4 110 111 112 113 114 115 116
.........
2.4 2 . 4 2.4 2.4
. . . . . . . . . 1.4 1 . 4 1.4 1.4
......... 0.6 1.6 2.6 3.6 8.4
Partsbywt. Parts by wt., relative whole nos. 2
..
..
0 . 4 0 . 4 0 . 4 0.4 .. 0.6 0.6 . . . . . . . . . . . . . . 1.6 1.6 . . . . . . . . . . . . . . 2.6 2.6 . . . . . . . . . . . . . . 3.6 3.6 . . . . . . . . . . . . . . 8 . 4 8 . 4 4 . 2 4 . 2 4.2 4.2 42 2
2
1
1
1
1
10
112.4 - 110 112.4 - 111 112.4 - 112 113 - 112.4 114 -112.4 115 - 112.4 116 - 112.4
=2+2+2+1+1+1+1
JOURNAL OF CHEMICAL EDUCATION
944
.
MAY.1932
In an exactly similar way the relative numbers of atoms of the various isotopes which go to make up the "average" atomic weight may he calculated. 3. What are the proportions of the various isotopes of cadmium which give the atomic weight 112.4? The atomic weights of the isotopes are given as 110, 111, 112, 113, 114, 115, and 116. The dashes are placed where i t is obvious mixtures of two such isotopes could not give the atomic weight, 112.4. The numbers for the other spaces are obtained as indicated a t the right of the table. The last row shows that a mixture of 10 atoms of twoeach of 110,111, and 112 and one each of 113, 114, 115, and 116 atomic weight, respectively, give the desired ratio. Adjustments of normalities of solutions may be made quite reasonably accurate even by measuring by volumes instead of by weight. 4. How much 0.1012 N solution must he added to 1000 cc. of 0.5009 N solution to make N / 5 solution? Gioen
Normality 0.1012 0.5009 Parts by vol.
0.1012' A
.... 0.3009 3009
X cc.
Waded
0.5009 N 0.0988
0.2000 N
....
988 1000 cc.
Where water is used in adjusting solutions the concentration of active reagent in i t would be indicated as 0. For example, if water is used in place of one of the acids in problem 2 the strength of the acid is 0.0 per cent. In problem 4 if water is used in place of the 0.1012N solution O.ON should be substituted for 0.1012N. Otherwise the calculation for the desired quantity would be carried out as indicated. H. V. HOUSEMAN MENLOJUNIOR COLLEGE MENLOPARK.C A L I P O ~ A
BALANCING CHEMICAL EQUATIONS DEAREDITOR: I have been following the discussion of balancing chemical equations in the recent issues of the JOURNAL OF CHEMICAL EDUCATION with rather keen interest. May I offer a few comments for publication? A chemical equation is not only the shorthand writing of the chemist, but it should be a mental picture of an actual reaction. To the observant investigator, the equation should immediately remind him as to the
