A student performed the Separation by Fractional Crystallization experiment desc

Question

A student performed the Separation by Fractional Crystallization experiment described with the following results:
Initial mass of sample mixture: 5.692g, mass of salicylic acid recovered, 1.09g; mass of CuSO45H2O recovered, 3.78g.
The recovered CuSO45H2O was then heated for 20 minutes at 120C. The mass of the new hydrate was 2.55g.
(The molar mass of H2O = 18.015 g/mol, and the molar mass of CuSO4= 159.61g/mol)

Calculate the moles of water in the new hydrate (CuSO4XH2O).

PLEASE SHOW STEP BY STEP INSTRUCTION

Explanation / Answer

Mass of CuSO4.5H2O recovered, 3.78g

Molar mass of CuSO4.5H2O = 249.685 g

Molar mass of anhydrous CuSO4 = 159.609 g

Amount of water in CuSO4.5H2O = 249.685 - 159.609 = 90.076 g

% of water in CuSO4.5H2O = 90.076/249.685 x 100 = 36.07%

Amount of water in recovered CuSO4.5H2O = 36.07 x 3.78 / 100 = 1.36 g

Amount of CuSO4 in original recovered CuSO4.5H2O = 2.42 g

Mass of the new hydrate = 2.55 g

Therefore

Amount of water in the new hydrate (CuSO4.xH2O) = Mass of the new hydrate - Amount of CuSO4 in recovered CuSO4.5H2O

Amount of water in the new hydrate = 2.55 – 2.42 = 0.13 g

Number moles of water in the new hydrate (CuSO4.xH2O) = 0.007 moles

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Another way of looking at the problem:

Amount of water in recovered CuSO4.5H2O = 36.07 x 3.78 / 100 = 1.36 g

Water lost from CuSO4.5H2O = Mass of CuSO4.5H2O recovered - Mass of the new hydrate

Water lost from CuSO4.5H2O = 3.78 - 2.55 = 1.23 g

Amount of water in new hydrate, CuSO4.xH2O = Amount of water in recovered CuSO4.5H2O - Water lost from CuSO4.5H2O

Amount of water in new hydrate, CuSO4.xH2O = 1.36 – 1.23 = 0.13 g

Number moles of water in the new hydrate (CuSO4.xH2O) = 0.13/18 = 0.007 moles

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