Question

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.

Answer 1

Answer:

$=6.2x10^{-3}mol CuSO_4$

Explanation:

Hello,

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

- Moles of $CuSO_4$:

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

- Grams of $CuSO_4$

$1.547gCuSO_4.5H_2O *\frac{1molCuSO_4.5H_2O}{249.5gCuSO_4.5H_2O} *\frac{1mol CuSO_4}{1mol CuSO_4.5H_2O}*\frac{159.5g CuSO_4}{1mol CuSO_4} =0.989 g CuSO_4$

- Moles of water:

$1.547gCuSO_4.5H_2O *\frac{1molCuSO_4.5H_2O}{249.5gCuSO_4.5H_2O} *\frac{5mol H_2O}{1mol CuSO_4.5H_2O}=0.031mol H_2O$

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

Best regards.