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A 1.547 g sample of blue copper(II) sulfate pentahydrate, ‍ , is heated carefully to drive off the water. The white crystals of that are left behind have a mass of g. How many moles of ‍ were in the original sample? Show that the relative molar amounts of and ‍ agree with the formula of the hydrate.

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Answer:

[tex]=6.2x10^{-3}mol CuSO_4[/tex]

Explanation:

Hello,

By developing the following stoichiometric relationship, the required amount could be found as follows:

- Moles of [tex]CuSO_4[/tex]:

[tex]1.547gCuSO_4.5H_2O *\frac{1molCuSO_4.5H_2O}{249.5gCuSO_4.5H_2O} *\frac{1mol CuSO_4}{1mol CuSO_4.5H_2O} =6.2x10^{-3}mol CuSO_4[/tex]

- Grams of [tex]CuSO_4[/tex]

[tex]1.547gCuSO_4.5H_2O *\frac{1molCuSO_4.5H_2O}{249.5gCuSO_4.5H_2O} *\frac{1mol CuSO_4}{1mol CuSO_4.5H_2O}*\frac{159.5g CuSO_4}{1mol CuSO_4} =0.989 g CuSO_4[/tex]

- Moles of water:

[tex]1.547gCuSO_4.5H_2O *\frac{1molCuSO_4.5H_2O}{249.5gCuSO_4.5H_2O} *\frac{5mol H_2O}{1mol CuSO_4.5H_2O}=0.031mol H_2O[/tex]

Finally, one could see that the mass of the anhydrous compound is less than the pentahydrated compound since it is waterless.

Best regards.