A mixture of CuSO4 and CuSO4·5 H2O has a mass of 1.235 g. After heating to drive off all the water, the mass is only 0.830 g. What is the mass percent of CuSO4·5 H2O in the mixture?

Answer 1

Mass of H2O expelled = 1.235-0.830=0.405 g

RMM of 5H2O = 5*(2*1+16)=90

RMM of crystalline CuSO4
=63.55+32+4*16 + 90
=249.55

Therefore, by proportion, mass of crystalline CuSO4.5H2O
=0.405*(249.55/90)
=1.123g
Mass percentage of CuSO4.5H2O
=mass of CuSO4.5H2O / total mass
=?

Answer 2

19.2