The mass percent of water in a hydrate of mncl2 is 36.41%, what is the formula of the hydrate?

The formula of the hydrate is MnCl2.4H2O

calculation

find the moles of MnCl2 and H2O

moles = % composition / molar mass

% composition of MnCl2 = 100 - 36.41=63.59 %

moles of MnCl2 = 63.59 / 125.84 = 0.505 moles

moles of H2O = 36.41/18 = 2.022 moles

find the mole ratio by dividing each moles with smallest moles (0.505)

that is MnCl2 = 0.505/0.505=1

H2O = 2.022/0505 = 4

therefore the formula = MnCl2.4H2O