During the conversion of one mole of gypsum to one mole of plaster of paris the percentage loss in weight is

${\mathrm{CaSO}}_4·2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CaSO}}_4·\frac{1}{2}{\mathrm{H}}_2\mathrm{O}+\frac{3}{2}{\mathrm{H}}_2\mathrm{O}$

Molecular weight of Gypsum $=40+32+64+36=172\mathrm{ }\mathrm{g}/\mathrm{mol}$

The molecular weight of Plaster of Paris $=40+32+64+9=145\mathrm{ }\mathrm{g}/\mathrm{mol}$

Loss in weight $=(172-145)$

$$\begin{gathered} =27 \\ \Rightarrow \%\text{ Loss }=\frac{27}{172}\times 100=15.7\% \end{gathered}$$

Hence, during the conversion of one mole of gypsum to one mole of plaster of paris the percentage loss in weight is $15.7\%$.