Question

Na=14.31% ,S =9.97% , O=69.50% , H=6.22% . Calculate the molecular formula of compound assuming that the whole of hydrogen in the compound is present as water of crystallization . Molecular mass of compound is 322. (related to empirical formula) .............please tell ​

Answer 1

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Question:

Na=14.31% ,S =9.97% , O=69.50% , H=6.22% . Calculate the molecular formula of compound assuming that the whole of hydrogen in the compound is present as water of crystallization . Molecular mass of compound is 322. (related to empirical formula)

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Answer:

Take the approximate ratio between the percentage composition and the molar mass of the elements given

Element  mass % Atomic mass  mass percent/Atomic mass  ratio  

Na        14.31    23                          0.622                                    2

S        9.97            32                          0.311                                    1

H        6.22             1                                  6.22                                   20

O        69.5             16                          4.34                                   14

Empirical formula is therefore Na₂SH₂₀O₁₄​  

 

Empirical mass = (2×23) + (32 )+ (20×1) + (14×16) =322g

Given:

Molecular mass = 322 g

∴ molecular formula = empirical formula = Na₂SH₂₀O₁₄​  

 

Given that all the hydrogen in the compound is reset in combination with oxygen as water of crystallisation

If water of crystallization are nH₂O​  

Then, 2n = 20

or n=10

Crystallized water is 10H₂O

 

therefore, the molecule is Na₂SO₄​ . 10H₂O

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