Question

A mixture of NiSO4 and its hydrate NiSO4•6H2O has a mass of 1.9319 g. After heating to drive off all the water, the mass is only 1.6438 g. What is the weight percentage of NiSO4•6H2O in the original mixture?  

Molar masses: NiSO4 154.76 g/mol; NiSO4•6H2O 262.85 g/mol.

Answer 1

after heating all water will evaporate only NiSO4 remains

mass of H2O = 1.9319 - 1.6438
= 0.2881 g
number of mol of H2O = (mass of H2O)/(molar mass of H2O)
= 0.2881/18
= 0.016 mol

6 mol of H2O is contained by 1 mol of NiSO4.6H2O
1 mol of H2O is contained by 1/6 mol of NiSO4.6H2O
0.016 mol of H2O is contained by (1/6)*0.016 mol of NiSO4.6H2O
0.016 mol of H2O is contained by (2.67*10^-3) mol of NiSO4.6H2O
so,
number of mol of NiSO4.6H2O = (2.67*10^-3) mol

mass of NiSO4.6H2O = (number of mol of NiSO4.6H2O)*(mlar mass of NiSO4.6H2O)
= (2.67*10^-3)*262.85
= 0.7018 g

mass % of NiSO4.6H2O = {(mass of NiSO4.6H2O)/(mass of mixture)}*100
= (0.7018/1.9319)*100
= 36.3 %

Answer: 36.3 %