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  • Solve it, An organic compound contains 49.3% C, 6.84% H and its vapour density is 73. Molecular formula of compund is

An organic compound contains 49.3% C, 6.84% H and its vapour density is 73. Molecular formula of compund is 

  • Option 1)

    C3H8O2

  • Option 2)

    C6H9O

  • Option 3)

    C4H10O2

  • Option 4)

    C6H10O4

 

Answers (1)

best_answer

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

 

 Molecular mass = 2\times 73 =146

 

C = \frac{\%}{100}\times \frac{Molecular \;mass}{Atomic \;mass} = \frac{49.3}{100}\times \frac{146}{12} = 6

H = \frac{\%}{100}\times \frac{Molecular \;mass}{Atomic \;mass} = \frac{6.84}{100}\times \frac{146}{1} = 10

O = \frac{\%}{100}\times \frac{Molecular \;mass}{Atomic \;mass} = \frac{43.86}{100}\times \frac{146}{16} = 4

Molecular Formula = C6H10O4

Molecular mass = 12\times 6 + 10\times 1 + 16\times 4 = 146


Option 1)

C3H8O2

Option 2)

C6H9O

Option 3)

C4H10O2

Option 4)

C6H10O4

Posted by

prateek

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