150°C
X*5H₂O   →   X + 5H₂O↑
mâ‚€=100 g
m(X)=8 g
M(Hâ‚‚O)=18.0 g/mol
n(Hâ‚‚O)=5 mol
the mass of the waterÂ
m(Hâ‚‚O)=M(Hâ‚‚O)n(Hâ‚‚O)
m(Hâ‚‚O)=18.0*5=90 g
the mass of the residue after removal of water
m'(X)=100-90=10 g
the percentage of impurity in X*5Hâ‚‚O
w(imp)=[m'(X)-m(X)]/mâ‚€
w(imp)=[10-8]/100=0.02 (2%)