Question

A mixture of ${\text{CuS}}{{\text{O}}_{\text{4}}}$ and ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$ has a mass of ${\text{1}}{\text{.245g}}$ . After heating, it drives off all the water, the mass is only ${\text{0}}{\text{.832g}}$ . What is the mass (in mg) of ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$ in the mixture?

Answer 1

Hint: When copper sulphate pentahydrate is heated, the water of hydrations are removed from the crystal to form water vapour and the dehydrated form of copper sulphate is produced.
The balanced equation for this reaction is:
${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{s}} \right)\xrightarrow{{{\text{heat}}}}{\text{CuS}}{{\text{O}}_{\text{4}}}\left( {\text{s}} \right) + {\text{5}}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{g}} \right)$
Thus, one mole of copper sulphate pentahydrate gives one mole of dehydrated copper sulphate.

Complete step by step answer:
Given that the mass of a mixture of copper sulphate and copper sulphate pentahydrate is ${\text{1}}{\text{.245g}}$ .
Also given, after removal of all the water by heating, the mass of the remaining portion of the mixture is only ${\text{0}}{\text{.832g}}$ .
We need to find out the mass of copper sulphate pentahydrate in the mixture in milligrams.
Let us consider the mass of copper sulphate, ${\text{CuS}}{{\text{O}}_{\text{4}}}$ be x grams.
Since the total mass of the mixture is ${\text{1}}{\text{.245g}}$ , so the mass of the copper sulphate pentahydrate, ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$ will be equal to $\left( {{\text{1}}{\text{.245 - x}}} \right)$ grams.
Now, the molar mass of copper sulphate is the sum of the atomic masses of all the elements present which is equal to $63.5 + 32 + 4 \times 16 = 159.5$ gram per mole.
And the molar mass of copper sulphate pentahydrate is equal to $63.5 + 32 + 4 \times 16 + 5 \times \left( {2 + 16} \right) = 249.5$ gram per mole.
Therefore, from the balanced equation, it is seen that the mass of copper sulphate in ${\text{249}}{\text{.5g}}$ of copper sulphate pentahydrate is ${\text{159}}{\text{.5g}}$ .
Thus, the mass of copper sulphate in$\left( {{\text{1}}{\text{.245 - x}}} \right)$ grams copper sulphate pentahydrate is:
$ = \dfrac{{{\text{159}}{\text{.5}}}}{{{\text{249}}{\text{.5}}}} \times \left( {{\text{1}}{\text{.245 - x}}} \right){\text{g}}$
Therefore,
$
  \dfrac{{{\text{159}}{\text{.5}}}}{{{\text{249}}{\text{.5}}}} \times \left( {{\text{1}}{\text{.245 - x}}} \right){\text{ + x = 0}}{\text{.832}} \\
   \Rightarrow 0.796 - 0.6932{\text{x + x}} = {\text{0}}{\text{.832}} \\
   \Rightarrow 0.3608{\text{x}} = 0.036 \\
   \Rightarrow {\text{x}} = \dfrac{{0.036}}{{0.3608}} \\
   \Rightarrow {\text{x}} = 0.1{\text{g}} \\
 $
So, the mass of the copper sulphate pentahydrate, ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$ will be equal to $\left( {{\text{1}}{\text{.245 - 0}}{\text{.1}}} \right) = 1.145$ grams.
In milligrams, the mass of the copper sulphate pentahydrate, ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$ will be equal to $1.145 \times 1000 = 1145$ milligrams.

Note:
-Copper sulphate is used in Benedict’s solution and Fehling’s solution to test for reducing sugars, which reduce the soluble blue copper (II) sulphate into insoluble red copper (I) oxide.
-It is also used as a dehydrating agent in organic synthesis for forming acetal groups.