ANALYSIS OF SORGHUM AXD MAIZE METHOD. 473 



All the wash waters were then passed in order through a filter, taking 

 care to bring as little as ix)ssible of the suboxide uix)u the filter. 



The suboxide on the filter and in the beakers \\as dissolved in an 

 acid solution of ferric sulphate, free fr»?m nitne acid and ferrous salt, 

 or more conveniently in an acid solution of ammonia ferric alum 

 (which is more easily obtained free from impurities), and poured ujwn 

 the suboxide in the original dish. All the copper suboxide being dis- 

 solved, it is brought into a liter flask, diluted with water to about 500'^'"--, 

 and acidified strongly with sulphuric acid. It is then ready to be 

 titrated in the usual manner for the amount of reduced iron, the num- 

 ber of *^"'"- of permanganate used giving easily the weight of glucose 

 represented by the suboxide of copper, as shown in report for 1879, p. 66. 



This methtxl for determining glucose dejiends upon the following facts : 



1. That two molecules (360 parts by weight) of glucose (CgH^jOe) 

 will reduce from Fehling's solution' five molecules of cuprous oxide 

 (5CuoO). 



2. That the five molecules of cuprous oxide thus precipitated will 

 reduce in acid sol. five molecules of ferric sulphate (Fco (804)3) *o 

 form ten molecules (1,520 parts by weight) of ferrous sulphate 

 (FeS04), as is explained by the following equation: 



f 5Cu, \^ rSFe, (804)3! _^ r5H,S0. 1 ^f lOCuS O^Xj^ 



\Ti5 parts/ ' \ 2.000 parts J ' \ 490 parts j" \ 1,593 pars J ' 



f 10FeSO4\^/ 5H, \ 



\ 1,520 parts/ ' \ 90 parts j 



The ten molecules of ferrous sulphate thus formed, will decolorize one mole- 

 cule (316.2 parts by weight) of potassium permanganate (K3 Mn, Oj), thus : 



f 10 Fe S O4 ) ^ r K, Mn, Og \ ^ f 8 H, S . \ , f 5 Fe, (S 64)3 \ , 

 \ 1,520 parts \ ' \ ^h> 2 parts / ' \ 7H-t parts j "^ \ 2,000 parts ( "^ 



(. 302 parts / ~ \ 174 2 parts / ' \ 144 parts / 



f 2 Mn S 0.. ■( , f K. S O4 1 , f S H, \ 



By following this explanation, it appears that two molecules of glucose are 

 exactly represented by one molecule of potassium permanjranate, as will ap- 

 pear from the following, by omitting the second and third members of the 

 series. Thus : 



f2 Ce Hi, OeX^f 5 Cu, O\^|10FeS 0.\^^ K, Mn, 0^ I 

 \ 360 parts / 1715 parts j 1 1.520 parts j \316 2 parts / 



In other words, 316.2 parts by weight of potassium permanganate are equiva- 

 lent to 3G0 parts of glucose, or one part of permanganate corresponds to 1.13S5 

 parts of glucose. If. then, the amount of permanganate decolorized be multi- 

 plied by 1.13S5, it will correctly represent the amount of glucose present. So 

 much for the theoretical explanation. In practice, it is found that each chem- 

 ist must determine for himself his titration error by estimations made upon 

 sugar of known purity. 



This individual error is due to the difficulty in determining the exact end re- 



