516 



Mr. G. S. Johnson. 



0*1346 gram of the anhydrous base (C 4 H 7 N 2 0) in 60 c.c. were required 

 for the complete reduction of 40 c.c. of Pavy's ammoniacal cupric 

 solution. 



Now 40 c.c. of Pavy's solution = 0*02 gram of glucose. 

 And 11 2 c.c. of the kreatinin solution contain 0*0251 gram of 

 C 4 H 7 N 3 0. 



Therefore 12*5 grams of the effloresced kreatinin are equivalent to 

 10 grams of glucose in reducing action upon cupric oxide. 



Again 0*5 gram of the tabular hreatinin a of urine was dissolved in 

 250 c.c. of water. 



6 c.c. of this solution was required to decolourise 20 c.c. of Pavy's 

 ammoniacal cupric solution (= 0*01 gram of glucose). And 11*9 c.c. 

 was required for 40 c.c. of the cupric solution. 



Hence 12 grams of the tabular hreatinin a of urine are equivalent to 

 10 grams of glucose in reducing action upon cupric oxide. 



Now 2CH 7 N 3 : C fi H 12 6 : : 12 : 9*5. 



= 226 . = 180 



Prom which it appears that 2 mols. of the reducing kreatinin of 

 urine are about equal to 1 mol. of glucose in reducing action upon 

 cupric oxide. 



Reduction of Cupric Oxide by Kreatinin from Kreatin of Flesh prepared 

 by Liebig's Method. 



62 lbs. of fresh beef, treated by Liebig's process for the extraction 

 of the kreatin, gave some fine crystals of that substance. 



3*685 grams of the kreatin obtained in this way was converted into 

 kreatinin hydrochloride by the action of dry hydrogen chloride in 

 Liebig's drying tube. 



The resulting hydrochloride was dissolved in 24 parts of water as 

 recommended by Liebig, boiled, and treated with pure lead hydrate 

 added a little at a time. In short, Liebig's directions for preparing 

 kreatinin from the kreatin of flesh were followed a3 exactly as possible. 

 The kreatinin thus prepared was re-crystallised six times from the 

 smallest possible quantity of cold water by evaporation in vacuo over 

 sulphuric acid. 



Well-formed anhydrous tabular crystals resulted. In order to com- 

 pare the cupric oxide reduction of this substance with that of the 

 natural hreatinin of urine, a solution was made containing 0*1 gram of 

 the substance in 100 c.c. 



27*5 c.c. of this solution decolourised 30 c.c of Pavy's ammoniacal 



cupric solution (== 0*015 gram glucose). 

 Hence 180 parts by weight of glucose are equivalent to 329*94 parts 



by weight of kreatinin. 



