Question

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

Answer 1

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

This is your friend.

Answer 2

Answer:

molecular formula of compound is C6H10O4