Question

If a student determines that the amount of anhydrous copper(II) sulfate remaining in the crucible is 0.648, how many moles of copper(II) sulfate are present?

molar mass, g/mol

CuSO4•5H2O 249.72

H2O 18.02

CuSO4 ??

Answer 1

one mole of anhydrous copper(II) sulphate (CuSO4) comes from one mole of Copp(II) sulphate (CuSO4. 5H2O)

Therefore

no of moles of CuSO4. 5H2O = No of moles of CuSO4

no of moles = mass/molar mass

mass of CuSO4 remains in the crucible = 0.648g

Molar mass of CuSO4 = Molar mass of CuSO4. 5H2O - (5×Molar mass of H2O)

Molar mass of CuSO4 = 249.72g/mol - (5× 18.02g/mol)

Molar mass of CuSO4 = 159.62g/mol

no of moles of CuSO4 = 0.648g/159.62g/mol = 0.004060mol

Therefore

moles of Copper(II) sulphate = 0.004060 mol