Some Basic Principles Of Organic Chemistry Question 137
Question: An organic compound contains 49.3% carbon 6.84% hydrogen and its vapour density is 73. Molecular formula of the compound is [MP PET 2000; Kerala PMT 2004; Pb. CET 2004]
Options:
A) $ C_3H_5O_2 $
B) $ C_6H_{10}O_4 $
C) $ C_3H_{10}O_2 $
D) $ C_4H_{10}O_2 $
Show Answer
Answer:
Correct Answer: B
Solution:
Element. No.of moles Simple ratio
C 12 49.3/12 = 4.1 4.1/2.7 = 1.3 ´ 2 = 2.6 = 3
H 1 6.84/1= 6.84 6.84/2.7=2.5´2=5
O 16 43.86/16 = 2.7 2.7/2.7=1´2=2
Empirical formula = $ C_3H_5O_2 $ E.F.
wt.
= 12 ´ 3 + 1 ´ 5 + 16 ´2 = 73 Molecular wt = V.D.
´ 2 = 73 ´ 2 = 146 $ n=\frac{M.wt}{E.F.wt}=\frac{146}{73}=2 $ Molecular formula = (E.F)n $ ={{(C_3H_5O_2)} _2}=C_6H _{10}O_4 $ .
