788 



SCIENCE. 



[N. S. Vol. XXIII. No. 594. 



percentage. Nearly all rules and tables that 

 have come to my notice limit the starting 

 point either to 100 c.e. of the alcohol on hand 

 or to some other quantity which may not be 

 a mathematical factor of the volume that one 

 desires to change. Just one possible excep- 

 tion to this statement has come to my notice. 

 Professor John H. Schaffner in his book 

 (' Laboratory Outlines for General Botany ') 

 gives the general pharmaceutical rule which 

 works out very well : ' Take of the grade at 

 hand as many volumes as the number of the 

 per cent, you wish to make, then add to this 

 enough volumes of pure water to make the 

 total number of volumes agree with the num- 

 ber of the per cent, at hand.' This is quite 

 simple and is really a special ease of what I 

 have to offer. While recently wrestling with 

 this problem I determined to work it out 

 algebraically, and I believe with success, evolv- 

 ing a formula that is simple and which gives 

 results in an abstract number, or miultiplier, 

 with which one can find the amount of water 

 to be added to any given volume of the alcohol 

 at hand, to obtain the per cent, desired. This 

 simple formula is, v^ (P — P') -=- P' and is 

 translated into words at the end of this 

 article. 



To get at the starting point of my formula 

 I took a special case : Make 25 per cent, alcohol 

 from 95 per cent, alcohol. Take 100 c.c. of 

 95 per cent, alcohol. This contains 95 per 

 cent, of pure alcohol and 5 per cent, of water, 

 or there are 19 parts of pure alcohol and one 

 of water. To make 25 per cent, alcohol from 

 one part of pure alcohol requires 3 parts of 

 water. In order then to make 25 per cent, 

 alcohol from the 19 parts of pure alcohol (in 

 the 100 c.c. of 95 per cent, alcohol) we must 

 multiply each part of pure alcohol by 3, ex- 

 cepting the nineteenth part, which must be 

 multiplied by 2, since there is already one 

 part of water, namely the twentieth part pres- 

 ent. In figures this gives 18 X 3 ■+ 1 X 

 (3 — 1) ^ 56 parts of water to be added. But 

 we began with 19 parts (or 95 per cent.) of 

 pure alcohol and 1 part (5 per cent.) of water, 

 so that our total number of parts will be 

 56 + 19 + 1 = 76 parts of 25 per cent, alcohol. 

 Proof : 19 -^ Y6 = 25 per cent as required. 



Or, using per cent., we have 90 per cent. X 

 3 + 5 per cent. X (3 — 1) + 5 per cent. + 95 

 per cent. = 380 per cent. 95 -f- 380 = 25 per 

 cent, as required. Now it is quite evident 

 that a similar course of reasoning can with 

 more or less difficulty be applied to any case 

 imaginable. Therefore, let P represent the 

 per cent of the alcohol on hand, P' the per 

 cent, required, v the multiplier with which to 

 multiply any volume of the alcohol on hand 

 to obtain the volumes of water to be added, 

 y the number of volumes of water to be added 

 to a volume of pure alcohol to obtain the per 

 cent, required, z the per cent, of water in the 

 alcohol on hand, and 100 per cent, the volume 

 taken of the alcohol on hand. Then, following 

 our original course of reasoning we have: 

 (P — z)y + z(y — l)=v 100 per cent (by 

 definition), or Py — z = v 100 per cent. But 

 z^lOO per cent — Pj y = (100 per cent. — 

 P') -f- P'j substituting and simplifying we get 

 P'v = P — P'. This formula is clearly the 

 pharmaceutical rule above quoted. Simplify- 

 ing this we have, i; ^ (P — P') -^ P', a simple 

 formula, independent of any volume of alcohol 

 that we choose to take, and easy to keep in 

 mind, in which v represents the multiplier 

 with which to multiply any volume of the 

 alcohol P that we choose to take, to obtain 

 the volumes of water necessary for making 

 alcohol P'. Or we may regard v as represent- 

 ing the number of volumes of water to be 

 added to one volume of P in order to make 

 P'. Thus, if we desire to make 40 per cent, 

 alcohol from 95 per cent, alcohol, (95 • — 40) -^ 

 40 ^ ■!; = If volumes of water to be added to 

 one volume of 95 per cent, alcohol. 



Rule: To find the number of volumes {v) 

 of water to he added to one volume of alcohol 

 of the grade per cent. (P) on hand, divide the 

 difference between the number (P) denoting 

 the grade per cent, on hand and the number 

 (P') denoting the grade per cent, required by 

 the latter number (P'). Or, which is simpler, 

 v=(P — P')h-P'. 



E. W. Berger. 



Biological Hall, 

 Ohio State Univebsitt, 

 February 22, 1906. 



