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

Asked by Braham | 20th Jun, 2013, 11:45: PM

Expert Answer:

 
                                         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
 
 

Answered by  | 25th Jun, 2013, 01:01: AM

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