1.

In an inorganic compound, the mass % of Na = 14.3, S = 9.97, H = 6.22 and O = 69.5. The hydrogen and oxygen are present as water. The vapour density of the compound is 161, what is its molecular formula?

Answer»
ANSWER : 
             The Molecular Formula Of Compound is Na2SH20O14



                 
                    
ANSWER :                                                       The Empirical Formula Mass Of the Compound = 2*23+1*32+20*1+14*16
Now Calculate the Value Of N =
MOLECULAR MASS/EMPIRICAL MASS =
322/322 = 
1

∴ Vapour density of compound is  = 161

∴ Molecular weight = 2 x vapour density

 = 2 x 161 = 322 g

∴ Number oxygen atoms = \(\cfrac{322\times\frac{69.5}{100}}{16}\) = 13.986 

 = 14 Oxygen atom

Number of hydrogen atoms = \(\cfrac{322\times\frac{6.22}{100}}{1}\) = 20.02

 = 20 Hydrogen atoms

∴ Maximum number of water molecule can present = 10 H2O

∴ remaining number of oxygen atoms = 4

Number of S atom = \(\frac{322\times9.97}{32\times100}\) = 1.003

 = 1 sulphur atom

Number of sodium atoms = \(\cfrac{322\times\frac{14.3}{100}}{23}\) 

 = 2.003 = 2 sodium atoms

Therefore,  the molecular formula will be 

Na2SO4.10H2O



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