Yellow precipitate in the detection of phosphorous when an organic compound is heated with and then boiled with conc. Followed by the addition of ammonium molybdate is due to the formation of :
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To check whether phosphorus is present in the organic compound or not:
The organic compound is treated with sodium peroxide(Na2O2NaX2OX2). Phosphorus gets converted to sodium phosphate. Then we carry out the same procedure that we do to identify phosphate anion.
The aqueous extract is heated with concentrated nitric acid and ammonium molybdate. A canary yellow precipitate of ammonium phosphomolybdate is obtained. This yellow precipitate helps us conclude that Phosphorus was present in the organic compound.
Reactions:
Na3PO4+3HNO3⟶H3PO4+3NaNO3NaX3POX4+3HNOX3⟶HX3POX4+3NaNOX3
H3PO4+12(NH4)2MoO4+21HNO3⟶(NH4)3PO4⋅12MoO3 (yellow ppt⋅)+21NH4NO3+12N2OHX3POX4+12(NHX4)X2MoOX4+21HNOX3⟶(NHX4)X3POX4⋅12MoOX3 (yellow ppt⋅)+21NHX4NOX3+12NX2O
The organic compound is treated with sodium peroxide(Na2O2NaX2OX2). Phosphorus gets converted to sodium phosphate. Then we carry out the same procedure that we do to identify phosphate anion.
The aqueous extract is heated with concentrated nitric acid and ammonium molybdate. A canary yellow precipitate of ammonium phosphomolybdate is obtained. This yellow precipitate helps us conclude that Phosphorus was present in the organic compound.
Reactions:
Na3PO4+3HNO3⟶H3PO4+3NaNO3NaX3POX4+3HNOX3⟶HX3POX4+3NaNOX3
H3PO4+12(NH4)2MoO4+21HNO3⟶(NH4)3PO4⋅12MoO3 (yellow ppt⋅)+21NH4NO3+12N2OHX3POX4+12(NHX4)X2MoOX4+21HNOX3⟶(NHX4)X3POX4⋅12MoOX3 (yellow ppt⋅)+21NHX4NOX3+12NX2O
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Answer:
Ammonium phosphomolybdate
Explanation:
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