Answer:
x=2 ([tex]CuCl_{2} .2H_{2}O[/tex]
Explanation:
Mass of [tex]CuCl_{2}[/tex] 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 [tex]CuCl_{2}[/tex] and water you must to find the amount of moles in both cases.
[tex]Mass of water=3.41g-2.69g=0.72gH_{2}O[/tex]
[tex]molH_{2}O= 0.72g H_{2}O.\frac{1molH_{2}O}{18gH_{2}O} =0.04molH_{2}O[/tex]
[tex]molCuCl_{2} =\frac{1molCuCl_{2}}{133.546gCuCl_{2}} = 0.02molCuCl_{2}[/tex]
Now you must to find the ratio between both molecules:
[tex]x=\frac{molH_{2}O }{molCuCl_{2} }=\frac{0.04mol}{0.02mol} =2[/tex]
So the water is two times copper(II) chloride. ([tex]CuCl_{2} .2H_{2}O[/tex]
