Answer to Question #112141 in General Chemistry for Lauren Pantelone

Question #112141
The mass percent of water in a hydrate of MnCl2 is 36.41%. What is the empirical formula of the hydrate?
1
Expert's answer
2020-04-27T01:45:54-0400

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



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