Some Basic Principles Of Organic Chemistry Question 54
Question: An organic compound on analysis gave C = 48 gm, H = 8 gm and N = 56 gm. Volume of 1.0 g of the compound was found to be 200 ml at NTP. Molecular formula of the compound is [MP PET 1986]
Options:
A) $ C_4H_8N_4 $
B) $ C_2H_4N_2 $
C) $ C_{12}H_{24}N_{12} $
D) $ C_{16}H_{32}N_{16} $
Show Answer
Answer:
Correct Answer: A
Solution:
| Element | % | No.of Moles | Simple Ratio |
|---|---|---|---|
| C | 48 | 48/12 = | 4 :1 |
| H | 8 | 8/1 = | 8: 2 |
| N | 56 | 56/14 = | 4: 1 |
Empirical formula = $ CH_2N $
Empirical formula mass = 28 Now, 200 ml of compound = 1 gm 22400 ml of compound $ \frac{1}{200}\times 22400=112 $
$ n=\frac{12}{2}=6 $ Therefore, Molecular formula $ ={{(CH_2N)}_4}=C_4H_8N_4 $ .
