sulphur & phosphorous : Estimation
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sulpur & phosphorous : Estimation
Estimation of Sulphur By Carius Method-
In this method known amount of organic compd is heated with f HNO3 in a carius tube then sulphur is oxidised into H2SO4. This is precipitated as BaSO4 by adding BaCl2 Solution. Ppts. of BaSO4 are filtered, washed, dried & weighed, By knowing the wt. of organic compd & BaSO4, we can calculate % of S.
H2SO4 +BaCl2 —–> BaSO4 + 2HCl
white ppt
Calculation
Wt. of org. compd = w gm.
Wt. of BaSO4 = y gm
Mol. wt. of BaSO4 = 137+32+16×4=233
233 gm BaSO4 contains S = 32 gm
x gm BaSO4 contains S =(32 y) /233
w gm of o.c. gives S = (32 x y) /233
100 gm of o.c. gives S = (32 x y x 100 ) / ( 233 w)
% of S= (32 x wt.of BaSO4 x 100 ) / ( 233 x wt. of o.c.)
Q —Quantitative analysis of 0.1254 gm. of an org. compd gave 0.1292 gm BaSO4 in carius method . calculate % of S ?.
Solution : weight of BaSO4 =0.1292 gm.
wt. of o.c. =0.1254 gm
Percentage of S= (32 x wt.of BaSO4 x 100 ) / ( 233 x wt. of o.c.)
% of S= (32 x 0.1292 x 100 ) / ( 233 x 0.1254)
% of S= 14.15 %
Q — 0.32 gm of an org. compd when heated with conc. HNO3 & BaCl2 gave 0.932 gm. of BaSO4. Calculate % of S ?
Solution: wt. of o.c. 0.32 gm.
percentage of S= (32 x wt.of BaSO4 x 100 ) / ( 233 x wt. of o.c.)
= (32 x 0.932 x 100 ) / ( 233 x 0.32 )
% of S = 40%
% of Phosphorous
Mg2P2O7 = Magnesium pyrro Phosphate
% of P = 0.c + f HNO3 —-> Phosphorous is oxidised to H3PO4 ——> in presence of ammonical MgCl2 , ppts of magnesium ammonium phosphate (MgNH4PO4) ____> MgNH4PO4 on red hot gives Mg2P2 O7
% of P = (62 x wt. of Mg2P2O7 x 100 ) /( 222 x wt. of o.c.)
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