Answer
2.12872 grams of water
Explanation
Given:
Mass of copper (II) sulfate pentahydrate sample = 5.9 grams
What to find:
The grams of water that would be released if all of the water is removed from the 5.9 grams sample of copper (II) sulfate pentahydrate.
Step-by-step solution:
The chemical formula of copper (II) sulfate pentahydrate is: CuSO₄.5H₂O
Molar mass of CuSO₄.5H₂O = 249.68 g/mol
Mass of water in 1 mole of CuSO₄.5H₂O = 5 x molar mass of water = 5 x 18.01528 g/mol = 90.0764 g/mol
the percentage by mass of water in 1 mole of CuSO₄.5H₂O is
[tex]\begin{gathered} \%\text{ }by\text{ }mass\text{ }of\text{ }water=\frac{Mass\text{ }of\text{ }water}{Molar\text{ }mass\text{ }of\text{ }CuSO₄.5H₂O}\times100\% \\ \\ \%\text{ }by\text{ }mass\text{ }of\text{ }water=\frac{90.0764}{249.68}\times100\% \\ \\ \%\text{ }by\text{ }mass\text{ }of\text{ }water=36.08\% \end{gathered}[/tex]Hence, you can now use the % by mass of water in 1 mole of copper (II) sulfate pentahydrate to determine the mass of water that would be released in 5.9 grams of copper (II) sulfate pentahydrate as shown below.
[tex]\begin{gathered} Mass\text{ }of\text{ }water=\frac{36.08}{100}\times5.9\text{ }grams \\ \\ Mass\text{ }of\text{ }water=0.3608\times5.9grams \\ \\ Mass\text{ }of\text{ }water=2.12872\text{ }grams \end{gathered}[/tex]Therefore, the grams of water that would be released if all of the water is removed from the 5.9 grams of copper (II) sulfate pentahydrate is 2.12872 grams
