A compound on analysis was found to have the following compositions,Na-14.13%,S-9.97%,oxygen-69.50%,hydrogen-6.22%..Calculate the molecular formula of the compound assuming that whole of hydrogen in the compound is present as water of crystallization and molar mass of compound is 322g

 

                                         moles                          Simplest Molar Ratio        simple ratio

Na        14.13%     23            14.13/23   = 0.614            0.614/0.311 = 1.97             2

S           9.97%     32             9.97/32    = 0.311            0.311 / 0.311 = 1               1

 O          69.5%      16             69.5/16    = 4.343            4.34 / .311   = 13.97           14

H          6.22%       1                6.22 / 1  = 6.22             6.22 / .311   = 20               20

 

 

Therefore, the emperical formula is = Na₂S₁H₂₀O₁₄

 

Therefore, emperical formula mass = 2x23 + 1x32 + 20x1 + 16x14 = 322

 

         Now,

                                       molecular mass               

                            n  =       emperical formula mass 

 

                               =  322/ 322 = 1

 

therefore molecular mass  = Na₂S₁H₂₀O₁₄

 

Finally, since all Hydrogen is present as water, the molecular formula would be, Na₂SO₄.10H₂0