How many molecules of water are lost by Gypsum to form plaster of Paris?

The plaster of Paris (${\mathrm{CaSO}}_4.\frac{1}{2}{\mathrm{H}}_2\mathrm{O}$) is formed by heating Gypsum (${\mathrm{CaSO}}_4.2{\mathrm{H}}_2\mathrm{O}$).

${\mathrm{CaSO}}_4.2{\mathrm{H}}_2\mathrm{O}\stackrel{∆}{\to }{\mathrm{CaSO}}_4.\frac{1}{2}{\mathrm{H}}_2\mathrm{O} + 1\frac{1}{2}{\mathrm{H}}_2\mathrm{O}$

Thus, 1.5 molecules of water are lost.

Hence, 1.5 molecules of water are lost by Gypsum to form the plaster of Paris.