Asked by Noelle
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?
Answers
Answered by
MathMate
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
=?
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
=?
Answered by
Shruti.
19.2