Question

When copper(II) sulfate (CuSO4) is prepared from copper(II) nitrate and sodium sulfate, the reaction initially produces a hydrated product, copper(II) sulfate pentahydrate (CuSO4.5H2O). If the CuSO4.5H2O is heated, the water is removed, leaving anhydrous CuSO4. What mass (in grams ) of anhydrous CuSO4 will be produced from 2.05 g of CuSO4.5H2O?

Answer 1

Molar mass of CuSO4.5H2O,

MM = 1*MM(Cu) + 1*MM(S) + 9*MM(O) + 10*MM(H)

= 1*63.55 + 1*32.07 + 9*16.0 + 10*1.008

= 249.7 g/mol

mass of CuSO4.5H2O = 2.05 g

mol of CuSO4.5H2O = (mass)/(molar mass)

= 2.05/2.497*10^2

= 8.21*10^-3 mol

mol of CuSO4 formed = moles of CuSO4(H2O)5

= 8.21*10^-3 mol

Molar mass of CuSO4,

MM = 1*MM(Cu) + 1*MM(S) + 4*MM(O)

= 1*63.55 + 1*32.07 + 4*16.0

= 159.62 g/mol

mass of CuSO4 = number of mol * molar mass

= 8.21*10^-3*1.596*10^2

= 1.31 g

Answer: 1.31 g