The per cent, of available phosphoric acid is guaranteed ^^^d for 

 to be from 8 to lo per cent.: 8 X 4i cents = 36 cents. ^^^°*^ 

 insoluble phosphoric acid: 2X2 cents 1= 4 cents. ^^^ 



The guaranteed per cent, of potash is 3 to 5 per cent. 

 But the statement of analysis does not tell the form in which 

 the potash is present. AH we know is that there is from 

 3 to 5 per cent, of actual potash contained in the fertilizer, 

 so we will eive ourselves the benefit of the doubt and assume 

 the potash to be in the form of muriate (chloride) : 3 X 4^ 

 cents r= 13^ cents. 



We now have the value in cents of the Nitrogen, available 

 and insoluble phosphoric acid, and potash. Add these 

 together and the sum is the value in cents of the total fertiliz- 

 ing constituents in 100 pounds of fertilizer. This sum 

 multiplied by 20 gives the value in cents of one ton. 



Example: — 



Nitrogen i .65 X 15^ = 25.5 cenrs. 



Available phosphoric acid. 8 X 7i = 36.0 



Insoluble phosphoric acid 2X 2 = 4.0 



Potash 3X 4i=_i3J " 



Total value of 100 pounds 79° cents. 



79 X 20 =: 1580 cents, or ^15.80 value per ton. 



If the potash is given in the form of sulphate we find the 

 equivalent of actual potash by multiplying the per cent, of 

 sulphate by .54 and the result by the trade value, 5^ cents. 

 If the potash is given in the form of muriate (chloride), 

 multiply the per cent, of muriate (chloride) by .63 and the 

 result by the trade value, 4 j cents. 



Example I. — A manufacturer's guarantee-analysis is 

 8 to 10 per cent, of potash as sulphate: 8 X -54 ^= 4-32 per 

 cent, of actual potash; 4.32 X 5i cents = 23.7 cents, the 

 trade value of actual potash as sulphate in 100 pounds of 

 fertilizer. 



Example 2. — ^A manufacturer's guarantee-analysis is 6 to 

 8 per cent, of potash as muriate (chloriSe): 6 X -63 =^ 3.78 

 per cent, of actual potash; 3.78 X 4i ^= i7-0 cents, trade 

 value of actual potash as muriate in 100 pounds of fertilizer. 



Summary of the methods heretofore used in converting one 

 chemical compound into an equivalent of another chemical 

 compound. 



{a) To change Nitrogen into an equivalent amount of 

 ammonia, multiply the given amount of Nitrogen by 1.214. 



