Question

A sample of an Unknown sample has a mass of .4100 g, after titrated, boiled and ppt formed 1.9652 g CuS was filtered out. CuCl2X2H2O MW is 170.48256 g/mol while CuS MW is 159.609 g/mol using the balanced chemical equation and the given information find the percentage of CuCl2 x H2O in the sample

Answer 1

m = 0.41 g

m = 1.9652 g of CuS

CuCl2*2H2O

first. find Copper content:

from CuS:

Mw of Cu = 63.546 g/mol

NOTE: CuS mass is NOT 159.609 but 95.6110 !!!

then, Copper content fraction = Mw of Cu / MW o fCuS = 63.546 / 95.6110 = 0.664630

then..

from 0.41 g --> 0.664630*1.9652 = 1.30613 g of Copper (in 1.9652 g sample)

change to mol

mol of Cu = mass/MW = 1.30613 /63.546 = 0.020554 mol of Cu

then...

we can also assume that

0.020554 mol of Cu were in the CuCl2*2H2O sample..

so

0.020554 mol of Cu --> 0.020554 mol of CuCl2*2H2O

mass of CuCl2*2jh2O = mol*MW = 0.020554*170.48256 = 3.504 g/mol

therefore:

3.504 / 4.1 g --> 0.854 --> 85.4%