434 The Neglect of Chemistry hy Practical Farmers. 



solved thus: — Quantity contained in ash of specimen X 100 

 and -f- amount of ash in specimen = amount of ingredient in 100 

 parts of ash. 



Example 2. — To show the necessity of Form 1, in order to be 

 ahle to institute a comparison between the composition of the 

 dry matter of any two substances, suppose the organic matter of 

 wheat grain and turnip-bulb to contain in each 2 per cent, of 

 nitrogen ; required the amount of nitrogen per cent, in the whole 

 grain and turnip-bulb, the wheat containing 87 per cent., and the 

 turnip-bulb 9 per cent, of organic matter, in its natural state. 



Here, as in the previous case, we multiply the nitrogen by the 

 amount of organic matter in the specimen, and divide by 100. 



87 (organic matter in wheat), X 2 (nitrogen in 100 parts of 

 organic matter) = 174, which, divided by 100, = 1.74, the 

 quantity of nitrogen contained m 100 parts of the whole 

 wheat grain. 



Again: 9 (organic matter in turnip) X 2 (nitrogen in 100 

 parts of organic matter) = 18, which, divided by 100, = 0.18, 

 the quantity of nitrogen contained in 1 00 parts of turnips. 



Here we see, that although, by Form 2 of these tables, it would 

 appear that turnip-balb contained scarcely 1-1 0th part the quan- 

 tity of nitrogen that the same weight of wheat grain did ; yet, by 

 referring to Form 1, we find that their respective organic matter 

 contains the same amount of nitrogen. 



Example 3. — To determine the actual relative amount of either 

 organic matter or ash, in any two specimens. 



Suppose, for example, the one specimen to contain 20.0 water, 

 79.0 organic matter, 1.0 ash, and the other to be composed of 10.0 

 water, 88.9 organic matter, 1.1 ash, and that it is desired to 

 ascertain the relation that exists between these two substances, 

 when the variable amount of water is left out of consideration. 



The method of ascertaining their relative proportions when dry, 

 is as follows : — 



Multiply the amount of organic matter, or ash, by 100, and 

 divide the amount by 100, minus the quantity of water contained 

 in the original specimen, and the converse problem, viz., " to find 

 their proportion when any given quantity of water is present, from 

 their proportion when dry," is solved l3y inverting the preceding 

 method, viz., multiplying by 100, minus the given quantity of 

 water, and divide by 100. 



In the case in question, where it is required to find their pro- 

 portion when dried, at 212*^ F., the operation would be as 

 follows : — 



