Question

A mixture of $CuS{O_4}.5{H_2}O$ and $MgS{O_4}.7{H_2}O$ was heated until all the water was driven off. If $5.0g$ of mixture gave $3g$ of anhydrous salt, what was the percentage by mass of $CuS{O_4}.5{H_2}O$ in the original mixture?

Answer 1

In this question, we can calculate the required mass of ($CuS{O_4}.5{H_2}O$) by finding the corresponding moles of water. We can find the mass of water from the difference of mass of mixture including water and mass of anhydrous salt.

Complete step by step answer:
In the question, $CuS{O_4}.5{H_2}O$ is the chemical formula for Copper sulphate pentahydrate and is the chemical formula for magnesium sulphate heptahydrate. When copper sulphate pentahydrate is heated then it is decomposed to anhydrous copper sulphate and water.
When sulphate heptahydrate is heated then it is decomposed to form anhydrous magnesium sulphate and water. These two reactions can be shown as:
$CuS{O_4}.5{H_2}O \to CuS{O_4} + 5{H_2}O \\\ MgS{O_4}.7{H_2}O \to MgS{O_4} + 7{H_2}O \\\$
As given in the question, when two given compounds are mixed with each other then the total mass (including mass of water) is $5g$ and the mass of produced anhydrous salt (excluding mass of water).
Therefore, Mass of water $5g - 3g = 2g$
Now,
Moles of water $= \dfrac{{{\text{Given mass}}}}{{{\text{Molecular mass}}}}$
As we know that the molecular mass of water is $18$. So, we can write
Moles of water (${H_2}O$) $= \dfrac{2}{{18}} = 0.111mole$(This is the total moles of water for $12$ mole of water when both compounds are mixed)
Now, As we know that $CuS{O_4}.5{H_2}O$ includes $5$ moles of water and there are $12$ moles of water in the mixture ($5$ from $CuS{O_4}.5{H_2}O$ and $7$ from $MgS{O_4}.7{H_2}O$) .
Therefore, Moles of $({H_2}O)$ from $CuS{O_4}.5{H_2}O$ $= \dfrac{5}{{12}} \times 0.111$
$= 0.047mole$
Now, As we know that $MgS{O_4}.7{H_2}O$ includes $7$ moles of water and there are $12$ moles of water in the mixture ($5$ from $CuS{O_4}.5{H_2}O$ and $7$ from $MgS{O_4}.7{H_2}O$) .
Therefore, Moles of $({H_2}O)$ from $MgS{O_4}.7{H_2}O$ $= \dfrac{7}{{12}} \times 0.111$
$= 0.066mole$
As we know that $1mole$ of $CuS{O_4}.5{H_2}O$ produces $5moles$ of $({H_2}O)$.
Thus, $1mole$ of $({H_2}O)$ $= \dfrac{1}{5}mole$ of $CuS{O_4}.5{H_2}O$
Hence, $0.047mole$ of water $= \dfrac{1}{5} \times 0.047mole$ of $CuS{O_4}.5{H_2}O$
Hence, we can say that $0.047mole$ of water will be from $0.0093mole$ of $CuS{O_4}.5{H_2}O$.
Now, the molecular mass of $CuS{O_4}.5{H_2}O$ is
$= 63.5 + 32 + 4(16) + 5(18) \\\ = 249.5gm/mol \\\$
Now, Given mass $=$ Number of moles $\times$Molecular mass
Given mass $= 0.0093 \times 249.49$
$= 2.332g$
Now, the percentage by mass of the $CuS{O_4}.5{H_2}O$ can be given as :
$= \dfrac{{{\text{Mass of }}CuS{O_4}.5{H_2}O}}{{{\text{Total mass of the mixture}}}}$
$= \dfrac{{2.332}}{5} \times 100 \\\ = 46.64\% \\\$
Hence, the required percentage by mass is $46.64\%$.

Note:
The most common form of copper sulphate is the copper sulphate pentahydrate which is having the chemical formula as $CuS{O_4}.5{H_2}O$. This form of copper sulphate is characterized by its bright blue colour.