Question

A mixture of CuSO4 and CuSO4.5 H2O has a mass of 1.270 g. After heating to drive off all the water, the mass is only 0.859 g. What is the mass percent of CuSO4.5 H2O in the mixture? Mass percent =

Answer 1

Mass of H2O = 1.270 g - 0.859 g

= 0.411 g

Molar mass of H2O,

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

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass of H2O = 0.411 g

mol of H2O = (mass)/(molar mass)

= 0.411/18.02

= 2.281*10^-2 mol

According to balanced equation

mol of CuSO4.5H2O present = (1/5)* moles of H2O

= (1/5)*2.281*10^-2

= 4.563*10^-3 mol

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 = number of mol * molar mass

= 4.563*10^-3*2.497*10^2

= 1.139 g

Use:

Mass % of CuSO4.5H2O = mass of CuSO4.5H2O * 100 / mass of mixture

= 1.139 * 100 / 1.270

= 89.69 %

Answer: 89.69 %