Question

Carbon and hydrogen in an organic compound are detected as:
(A)- $CaHC{{O}_{3}},CuS{{O}_{4}}.5{{H}_{2}}O$
(B)- $CaC{{O}_{3}},CuS{{O}_{4}}.5{{H}_{2}}O$
(C)- $CaC{{O}_{3}},Cu{{(OH)}_{2}}$
(D)- $CaHC{{O}_{3}},CaC{{O}_{3}}$

Answer 1

Hint: Leibeg’s test is used to detect the presence of carbon and hydrogen in an organic compound. Carbon dioxide turns lime water milky. Hydrated copper sulphate is blue in colour.

Complete step by step solution:
Let’s look at the answer.
For the detection of carbon and hydrogen, the sample is treated with cuprous oxide.
Carbon will react with cuprous oxide and release carbon dioxide.
Hydrogen will hydrate the white anhydrous copper sulphate into blue hydrous copper sulphate.
Carbon dioxide turns calcium hydroxide milky due to the formation of calcium carbonate. This reaction confirms the presence of carbon in the inorganic compound.
Anhydrous copper sulphate which is white in colour becomes blue in colour as the water molecules attach to it.
In option (A) calcium carbonate ($CaC{{O}_{3}}$) is not present so it is incorrect.
In option (C) calcium carbonate ($CaC{{O}_{3}}$) is not present so it is incorrect.
In option (D) only calcium carbonate ($CaC{{O}_{3}}$) is present not calcium carbonate ($CaC{{O}_{3}}$) so it is also in correct.
Only option (B) contains both $CaC{{O}_{3}} and CuS{{O}_{4}}.5{{H}_{2}}O$. So, it is the correct answer.
Hence, the answer for the given question is option (B).

Note: The water of crystallization for copper sulphate is 5. It means that 5 water molecules are linked to the copper sulphate molecule. When excess of carbon dioxide is passed through lime water then its white colour vanishes due to the formation of calcium hydrogen carbonate.