Question

The mass percent of water in a hydrate of MnCl2 is 36.41%. What is the empirical formula of the hydrate?

Answer 1

For calculations, we take a sample of the starting substance weighing 100 g:

$m(H20)=\frac{\omega(H20)\times100}{100}=36.41 g$

m( MnCl2)=100-36.41=63.59

find the amount of substance:

$n(H20)=\frac{m}{M}=\frac{36.41}{18}=2.023$

$n(Mncl2)=\frac{m}{M}=\frac{63.59}{125.8440}=0.505$

x:y=0.505:2.023=1:4

MnCl₂*4H₂0