Question

A 2.634g sample containing impure copper (II) chloride diydrate was heated. The sample mass after heating to drive off the H2O was 2.125g. What was the mass % of CuCl2 x 2 H2O in the original sample?

I think the following is correct. Can someone check my work?

2.634g-2.125g=0.509g H2O

H2O--> 0.509/2.634 x 100% = 19.32%
CuCl2--> 2.125/2.634 x 100 % = 80.68 %

Is 80.68 the correct answer, or would i need to provide both percentages for a correct outcome for the question asked? Thanks!

Answer 1

mass of H2O = 2.634 g - 2.125 g= 0.509 g

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 0.509 g

use

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(0.509 g)/(18.016 g/mol)

= 2.825*10^-2 mol

since 1 mol of CuCl2.2H2O has 2 moles of water

moles of CuCl2.2H2O = number of mol of H2O / 2

= 2.825*10^-2 / 2

= 1.4125*10^-2 mol

Molar mass of CuCl2.2H2O,

MM = 1*MM(Cu) + 2*MM(Cl) + 4*MM(H) + 2*MM(O)

= 1*63.55 + 2*35.45 + 4*1.008 + 2*16.0

= 170.482 g/mol

use

mass of CuCl2.2H2O,

m = number of mol * molar mass

= 1.413*10^-2 mol * 170.482 g/mol

= 2.408 g

So, in original sample, we have just 2.408 g of CuCl2.2H2O,

mass % = mass of CuCl2.2H2O * 100 / mass of sample

= 2.408*100/2.634

= 91.4 %

Answer: 91.4 %

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