Question

In the detection test of carbon and hydrogen with cupric oxide, carbon is oxidized to:
(A) Carbon monoxide
(B) Carbon dioxide
(C) Carbonic Acid
(D) Copper Sulphate

Answer 1

Hint: The Liebig’s Method is used for the estimation of elements in an organic compound. The organic compound is heated in presence of dry cupric oxide. The heating results in the oxidation of elements into its oxides. These obtained oxidized products are then absorbed by the caustic soda $\text{Ca(OH}{{\text{)}}_{\text{2}}}$and anhydrous $\text{CuS}{{\text{O}}_{\text{4}}}$ respectively. The mass adsorbed can be used to calculate the percentage of an element in the given known mass.

Complete step by step solution:
Carbon and hydrogen in the organic compound are estimated by Liebig's Method.
When an organic compound of known mass is strongly heated in presence of dry cupric oxide $\text{ (CuO)}$ the carbon $\text{ C}$ and hydrogen in the organic compound quantitatively oxidize to the carbon dioxide $\text{ C}{{\text{O}}_{\text{2}}}\text{ }$ and $\text{ }{{\text{H}}_{\text{2}}}\text{O}$ .
Let consider an organic compound having the molecular formula $\text{ }{{\text{C}}_{\text{x}}}{{\text{H}}_{\text{y}}}$ .it undergoes the reaction as follows:
${{\text{C}}_{\text{x}}}{{\text{H}}_{\text{y}}}\text{ + }\left( \text{x+}\dfrac{\text{y}}{\text{4}} \right){{\text{O}}_{\text{2}}}\to \text{ xC}{{\text{O}}_{\text{2}}}\text{ + }\dfrac{\text{y}}{\text{2}}{{\text{H}}_{\text{2}}}\text{O}$
We know that the given compound is mixed with dry copper oxide, and heated in a hard glass test tube. The products of the reaction are passed over (white) anhydrous copper sulfate and then bubbled through lime water.
If the copper sulfate turns blue due to the formation of $\text{CuS}{{\text{O}}_{\text{4}}}\text{.5}{{\text{H}}_{\text{2}}}\text{O}$ (water vapor) then the compound contains hydrogen.
If the lime water is turned milky by carbon dioxide, then the compound contains carbon. So, the carbon is oxidized to carbon dioxide.
The reaction associated with The Liebig’s Method for the oxidation of carbon and hydrogen are as follows:
$\text{C + 2CuO }\xrightarrow{\text{heat}}\text{ C}{{\text{O}}_{\text{2}}}\text{ + 2Cu}$
In this case, C is the organic compound.
$\text{C}{{\text{O}}_{\text{2}}}\text{ + Ca(OH}{{\text{)}}_{\text{2}}}\xrightarrow{{}}\text{ COC}{{\text{O}}_{\text{3}}}\text{ + }{{\text{H}}_{\text{2}}}\text{O}$
The $\text{Ca(OH}{{\text{)}}_{\text{2}}}$is also known as the lime water and $\text{CaC}{{\text{O}}_{\text{3}}}$is milky in color.
The hydrogen is estimated as the water molecule. The hydrogen oxidized to water and it changes the colour of anhydrous copper sulphate from colourless to blue.
$\text{2H + CuO }\xrightarrow{\text{heat}}\text{ }{{\text{H}}_{\text{2}}}\text{O + Cu}$
In this case, the H is the organic compound.
$\text{CuS}{{\text{O}}_{\text{4}}}\text{ + 5}{{\text{H}}_{\text{2}}}\text{O }\xrightarrow{{}}\text{ CuS}{{\text{O}}_{\text{4}}}\text{.5}{{\text{H}}_{\text{2}}}\text{O}$
Therefore, in Liebig’s Method or the estimation of carbon, the copper oxide oxidizes the carbon to the carbon dioxide.

Hence, (B) is the correct option.

Note: Copper sulfate is a salt that exists as a series of compounds that differ in their degree of hydration. Copper Sulphate is made up of copper and sulphuric acid, but simply adding copper to a diluted acidic solution will not promote the oxidation reaction. The Liebig’s Method also facilitates the oxidation of nitrogen to nitrogen dioxide.