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  • Solve it, A compoundon analysis was found to have the following composition : (i) Sodium 14.31%, (ii) Sulphur 9.97%, (iii) Oxygen 69.5%, (iv) Hydrogen 6.22%. Calculate the molecular formula of the compund assuming that whole of Hydrogen in the compou

A compound on analysis was found to have the following composition : (i) Sodium 14.31%, (ii) Sulphur 9.97%, (iii) Oxygen 69.5%, (iv) Hydrogen 6.22%. Calculate the molecular formula of the compund assuming that whole of Hydrogen in the compound is present as water of Crystallization. Molecular mass of compund is 322.

  • Option 1)

    Na2SH20O10

  • Option 2)

    Na2SH20O14

  • Option 3)

    Na2SH20O7

  • Option 4)

    Na2SH30O16

 

Answers (1)

As we have learnt

 

Molecular Formula -

The molecular formula shows the exact number of different types of atoms present in a molecule of a compound.

- wherein

For glucose, empirical formula is CH2O .its molar mass is 180 gram.

n = molar mass/empirical formula mass = 180/30=6

Hence molecular formula= C6H12O6

 

 

Element %age Atomic mass Relative no of atoms Simplest Ratio
Na 14.31 23 0.622 \frac{0.622}{0.311} = 2
S 9.97 32 0.311 \frac{0.311}{0.311} = 1
H 6.22 1 6.22 \frac{6.22}{0.311} = 20
O 69.50 16 4.34 \frac{4.34}{0.311} = 14

Empirical formula = Na2SH20O14

Empirical formula mass = 2\times 23 + 32 + 20\times 1 + 14\times 16 = 322

Molecular mass = 322

Molecular Formula = Na2SH20O14

 


Option 1)

Na2SH20O10

Option 2)

Na2SH20O14

Option 3)

Na2SH20O7

Option 4)

Na2SH30O16

Posted by

subam

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