
For the detection of phosphorus, the organic compound after fusion with $N{a_2}{O_2}$ is extracted with water, boiled with $HN{O_3}$ and then ammonium molybdate added to it. A yellow ppt. is obtained which is due to formation of:
A) Ammonium sulphate
B) Ammonium phosphomolybdate
C) Ferric phosphate
D) Disodium ammonium phosphate
Answer
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Hint: The question itself the reactants are given from the start. One can write the possible reaction scheme and decide which will form the product at the last step which will also be the precipitate which is yellow coloured.
Complete step by step answer:
1) First of all we will analyze the data which is given in the question and try to write the general reaction for the detection of phosphorus and try to discover the product which is as follows,
$N{a_3}P{O_{3(s)}} + 3HN{O_{3(l)}} \to {H_3}P{O_{4(l)}} + 3NaN{O_{3(s)}}$
${H_3}P{O_4} + 12{(N{H_4})_2}Mo{O_4} + 21HN{O_3} \to {(N{H_4})_3}P{O_4} \cdot 12Mo{O_3} + 21N{H_4}N{O_3} + 12{H_2}O$
The product formed as ${(N{H_4})_3}P{O_4} \cdot 12Mo{O_3}$ which is a yellow coloured precipitate.
2) In the above reaction when the organic compound is fused with the $N{a_2}{O_2}$ that is sodium peroxide the aqueous solution is obtained as a product.
3) Now the aqueous solution which was obtained in the above step is then boiled with ammonium molybdate solution and the concentrated nitric acid which forms the yellow precipitate containing solution as the final product.
4) The yellow solution containing the precipitate shows the presence of phosphorus in the initiative which was taken as an organic compound. The name of the yellow precipitate formed is ammonium phosphomolybdate.
5) Therefore, the yellow ppt. is obtained as the reaction final product which is due to the formation of ammonium phosphomolybdate
Therefore, the correct option is B
Note:
The ammonium phosphomolybdate is an inorganic salt that contains the phosphomolybdate ion complex in it. The chemical method used above is the efficient method which is widely used for the detection of phosphorus in any unknown organic compound.
Complete step by step answer:
1) First of all we will analyze the data which is given in the question and try to write the general reaction for the detection of phosphorus and try to discover the product which is as follows,
$N{a_3}P{O_{3(s)}} + 3HN{O_{3(l)}} \to {H_3}P{O_{4(l)}} + 3NaN{O_{3(s)}}$
${H_3}P{O_4} + 12{(N{H_4})_2}Mo{O_4} + 21HN{O_3} \to {(N{H_4})_3}P{O_4} \cdot 12Mo{O_3} + 21N{H_4}N{O_3} + 12{H_2}O$
The product formed as ${(N{H_4})_3}P{O_4} \cdot 12Mo{O_3}$ which is a yellow coloured precipitate.
2) In the above reaction when the organic compound is fused with the $N{a_2}{O_2}$ that is sodium peroxide the aqueous solution is obtained as a product.
3) Now the aqueous solution which was obtained in the above step is then boiled with ammonium molybdate solution and the concentrated nitric acid which forms the yellow precipitate containing solution as the final product.
4) The yellow solution containing the precipitate shows the presence of phosphorus in the initiative which was taken as an organic compound. The name of the yellow precipitate formed is ammonium phosphomolybdate.
5) Therefore, the yellow ppt. is obtained as the reaction final product which is due to the formation of ammonium phosphomolybdate
Therefore, the correct option is B
Note:
The ammonium phosphomolybdate is an inorganic salt that contains the phosphomolybdate ion complex in it. The chemical method used above is the efficient method which is widely used for the detection of phosphorus in any unknown organic compound.
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