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0.12g of an organic compound containing phosphorous gave 0.22g of Mg2P2O7 by usual analysis. calculate the percentage of phosphorous in the compound.

Asked by 23rd February 2013, 8:23 PM
Answered by Expert
Answer:
The mass of the compd = 0.12 g
Mass of Mg2P2O7     = 0.22 g
Since 1 mole of Mg2P2O7 has 2 g atoms of P, or
222 g of Mg2P2O7 = 62 g of P
 
So, 0.22 g of Mg2P2O7 contains P =  62/222  x 0.22  g 
So % P present in the compound = 62/222 x  0.22 x 100/ 0.12 

Answered by Expert 24th February 2013, 12:11 PM
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