Question

A hydrate of copper (II) chloride has the following formula: CuCl2 - x H2O. The water in a 3.41-g sample of the hydrate was driven off by heating. The remaining sample had a mass of 2.69 g . Find the number of waters of hydration (x) in the hydrate.

Answer 1

Answer:

x=2 ($CuCl_{2} .2H_{2}O$

Explanation:

Mass of $CuCl_{2}$ is the mass of remaining sample, because it is a product of loss of drying from initial sample. This means that the mass of water is the mass has been lost.

after fing the masses of $CuCl_{2}$ and water you must to find the amount of moles in both cases.

$Mass of water=3.41g-2.69g=0.72gH_{2}O$

$molH_{2}O= 0.72g H_{2}O.\frac{1molH_{2}O}{18gH_{2}O} =0.04molH_{2}O$

$molCuCl_{2} =\frac{1molCuCl_{2}}{133.546gCuCl_{2}} = 0.02molCuCl_{2}$

Now you must to find the ratio between both molecules:

$x=\frac{molH_{2}O }{molCuCl_{2} }=\frac{0.04mol}{0.02mol} =2$

So the water is two times copper(II) chloride. ($CuCl_{2} .2H_{2}O$