198 QUANTITATIVE ANALYSIS. 



No. 1.) The calculation of the amount of glucose is made as 

 follows : 



e7TK = Number of grams of sugar in 100 cc. of urine, 



in which a = observed rotation using sodium light, 

 I = length of tube in decimeters. 



(b) Determination by Means of Reduction. 



Under certain definite conditions one molecule of glucose 

 reduces very nearly five molecules or ten equivalents of 

 copper oxide to cuprous oxide, hence 180 parts of anhydrous 

 glucose will reduce the oxide from 1248.8 parts of crystallized 

 copper sulphate, CuS0 4 -f 5H 2 0, to cuprous oxide. 



1. FEHLING'S ITERATION METHOD. 



Preparation of the Solutions, (a) 34.639 g. of pure copper 

 sulphate in crystals, which have not effloresced, are accurately 

 weighed off on a large watch-glass, dissolved by warming 

 with water in a dish, the solution placed in a 500-cc. measur- 

 ing-flask and, after it is perfectly cold, filled up to the mark. 



(6) About 173 g. of potassium sodium tartrate (Rochelle 

 salts) are dissolved by warming with a little water, the 

 solution placed in a 500-cc. measuring-flask, 100 cc. of sodium 

 hydroxide solution of 1.34 specific gravity added, and, after 

 the mixture is cold, filled up to the volume of 500 cc. 



Mix equal volumes of the two fluids about 25 cc. meas- 

 ured with a pipette in a dry beaker: deep-blue fluid, Feh- 

 ling's solution, 10 cc. of which is equivalent to 0.05 g. of 

 glucose. Test the solution by diluting a portion of it in a 

 test-tube with about four times its volume of water and 

 heating to boiling: no cuprous oxide should precipitate. 



