Question

A 92.8 gram sample of CuSO4•5H2O.(copper sulfate pentahydrate) is heated until water is released. How many grams of water were released?

Answer 1

Mass of water released =

$92.8 g CuSO_{4}.5H_{2}O$×$\frac{5 * 18 g H_{2}O}{249.68 g CuSO_{4}.5H_{2}O}$

= 33.45 g $H_{2}O$

Answer 2

Answer: The mass of water released from the given amount of copper sulfate is 33.46 grams.

Explanation:

We are given a compound having chemical formula $CuSO_4.5H_2O$

The molar mass of this compound = $[63.55+32+(4\times 16)+5(16+(2\times 1))]=249.55g/mol$

Mass of water molecule = $[16+(2\times 1)]=18g/mol$

In 249.55 grams of the compound, $(5\times 18)g$ of water molecule is present.

So, in 92.8 grams of the compound, $\frac{5\times 18}{249.55}\times 92.8=33.46g$ of water molecule.

Hence, the mass of water released from the given amount of copper sulfate is 33.46 grams.