Question:
An organic compound is heated with \(Na_2O_2\) then boiled with \(HNO_3\). The solution is then treated with ammonium molybdate. The yellow precipitate obtained is due to the presence of the element
An organic compound is heated with \(Na_2O_2\) then boiled with \(HNO_3\). The solution is then treated with ammonium molybdate. The yellow precipitate obtained is due to the presence of the element
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Yellow precipitate with ammonium molybdate is the characteristic test for Phosphate ions (\(PO_4^{3-}\)).
Updated On: Apr 29, 2026
- nitrogen
- sulphur
- phosphorus
- carbon
- molybdenum
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
The detection of non-metals in organic compounds involves specific chemical tests after mineralizing the compound.
Step 3: Detailed Explanation:
1. When a compound containing phosphorus is heated with an oxidizing agent like sodium peroxide (\(Na_2O_2\)), phosphorus is converted to sodium phosphate (\(Na_3PO_4\)).
2. This is extracted with water and boiled with nitric acid (\(HNO_3\)) to give phosphoric acid.
3. When ammonium molybdate is added, it reacts to form a canary-yellow precipitate of ammonium phosphomolybdate.
Chemical formula: \((NH_4)_3PO_4 \cdot 12MoO_3\).
Step 4: Final Answer:
The element detected is phosphorus.
The detection of non-metals in organic compounds involves specific chemical tests after mineralizing the compound.
Step 3: Detailed Explanation:
1. When a compound containing phosphorus is heated with an oxidizing agent like sodium peroxide (\(Na_2O_2\)), phosphorus is converted to sodium phosphate (\(Na_3PO_4\)).
2. This is extracted with water and boiled with nitric acid (\(HNO_3\)) to give phosphoric acid.
3. When ammonium molybdate is added, it reacts to form a canary-yellow precipitate of ammonium phosphomolybdate.
Chemical formula: \((NH_4)_3PO_4 \cdot 12MoO_3\).
Step 4: Final Answer:
The element detected is phosphorus.
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