Question

An organic compound on analysis gave C = 42.8%, H = 7.2% and N = 50%. Volume of 1 g of the compound was found to be 50 mL at STP. Molecular formula of the compound is ______.

Options

• C₂H₄N₂

• C₁₂H₂₄N₁₂

• C₁₆H₃₂N₁₆

• C₄H₈N₄

Answer 1

An organic compound on analysis gave C = 42.8%, H = 7.2% and N = 50%. Volume of 1 g of the compound was found to be 50 mL at STP. Molecular formula of the compound is C₁₆H₃₂N₁₆.

Explanation:

Given, C = 42.8%, H = 7.2% and N = 50%

Element	%	Moles	Simplest ratio	Whole number
C	42.8	`42.8/12 = 3.56`	1	1
H	7.2	`7.2/1 = 7.2`	2	2
N	50	`50/14 = 3.57`	1	1

∴ Empirical formula = CH₂N

Empirical formula mass = 28

Now, the number of moles = `"volume given at STP"/(22400 " mL")`

`= 50/22400`

= `2.23 xx 10^-3`

Also, the number of moles = `"weight"/"molecular weight"`

`=> "Molecular weight" = 1/(2.23 xx 10^-3)` = 448

`therefore "n" = "molecular weight"/"empirical weight"`

`= 448/28`

= 16

Hence, molecular formula = (empirical formula)_n

= \[\ce{(CH2N)16 = C16H32N16}\]