Find the number of waters of hydration (x) in the hydrate.

Question

A hydrate of copper (II) chloride has the following formula:

CuCl_2 *xH_2 O.

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.

Please Help! Thanks!

Answer 1

Actually

CuCl2 x 2H2O.

Answer 2

Calculate the mass of water, the mass of anhydrous salt remaining, convert those to moles, and calculate the mole ratio of water to anhydrous salt.

Mass H2O = 3.41 - 2.69 = 0.72 g

Mass anhydrous CuCl2 = 2.69 g

moles H2O = 0.72 g H2O x (1 mole H2O / 18.0 g H2O) = 0.040 moles H2O

moles anhydrous CuCl2 = 2.69 g x (1 mole CuCl2 / 134.5 g CuCl2) = 0.020 moles CuCl2

moles H2O / moles CuCl2 = 0.040 / 0.020 = 2  so x = 2

The formual of the hydrate is CuCl2 x 2H2O.

Answer 3

at wt of Cu=63.5

Cl=35.5

2CL=70

dehydrate CuCl2weigh= 63.5+70=133.5g

2.69gm CuCL carried (3.41-2.69)gm OR 0.72gm water

1gm carried 0.72/2.69=0.2676 gm

133.5 gm carrying  133.5*0.2676 or35.72  or approx  36 Gm

therefore  2 molucles of H2O  or 36gm carried by CuCl2

x=2 ans