Again Ca 3 P 2 O 8 -f 3(NH 4 N 3C2O4=3CaC 2 O4+2(NH4)3 PO 4 . 

 And 3CaC 2 O 4 (on ignition)=3CaCO 3 + 3CO 

 Ca 3 P a O 8 : 3CaCO 3 1 : percentage of Ca 3 P 2 O 8 present : 



CaCO 3 due to the Tricalcic 



phosphate, 



/.*., 310 : 300: -.3-49 : percentage of CaCO 3 coming from 

 the Tricalcic phosphate, 



/. CaC0 3 due to the Ca 3 P 2 O 8 = 3>49 * Q 300 =3'38% 



/.Taking away 8-56 + 3-38, or 11-94 from 43-63% 

 which is the total CaCO 3 obtained, we get 31-69% as 

 being due to the CaSO 4 



Now CaSO 4 + (NH 4 ) 2 C 2 O 4 =CaC 2 O 4 + (NH 4 ) 2 SO 4 

 CaC 2 O 4 (on ignition)=CaCO 3 + CO 

 /. CaCO 3 : CaSO 4 : : 31-69 : the % of CaSO 4 present 



.'. CaSO, present^ 31 ' 69 x I36 = 43 'i % 



100 - /0 



( i o ) ' ' Organic matter. "- 



The total loss on ignition (i.e. 29-05 / ) is made up 

 partly of H 2 O going off at iooC, partly of the H 2 O 

 that CaH 4 P 2 O 8 loses on ignition and the rest is the water 

 of crystallization of CaSO 4 which cannot be calculated, 

 together with the real organic matter. 



Now, CaH 4 P 2 O 8 (on ignition) =CaP 2 O 6 + 2 H 2 O (=36) 



(=234) 



CaH 4 P 2 O 8 I 2H 2 O::CaH 4 P 2 O 8 present : %of H 2 O lost by the 

 (=234) (=3 6 ) (''* 20 '03%^ rnonocalcic phosphate present on 



ignition. 

 .'. H 2 O lost by the monocalcium phosphate on ignition 



But the loss of H 2 O at iooC= 16-46 % 



.'.Taking away 16-46 + 3*08, or 19-54% from the total loss on 

 ignition, 



i. <?., 29-05 % we 8 et 9'S 1 % as t he proportion of " organfc matter 

 &c." present in the Super. 



