%of carbon=69.77%
%of hydrogen=11.63%
%of oxygen={100-(69.77+11.63)}%
=18.6%
Thus,the ratio of the number of carbon,hydrogen, and oxygen atoms in the organic compound can be given as:
\(C:H:O=\frac{69.77}{12}:\frac{11.63}{1}:\frac{18.6}{16}\)
\(=5.81:11.63:1.16\)
\(=5:10:1\)
Therefore,the empirical formula of the compound is \(C_5H_{10}O\).Now,the empirical formula mass of the compound can be given as:
5×12+10×1+1×16
=86
Molecular mass of the compound=86
Therefore,the molecular formula of the compound is given by \(C_5H_{10}O\).
Since the given compound does not reduce Tollen's reagent,it is not an aldehyde.Again,the compound forms sodium hydrogen sulphate addition products and gives a positive iodoform test.Since the compound is not an aldehyde,it must be a methyl ketone.
The given compound also gives a mixture of ethanoic acid and propanoic acid.
Hence,the given compound is Pentan-2-one.
The given reactions can be explained by the following equations:
Ethanal to But-2-enal
A certain reaction is 50 complete in 20 minutes at 300 K and the same reaction is 50 complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction. Given: \[ R = 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1}, \quad \log 4 = 0.602 \]
Read Also: Aldehydes, Ketones, and Carboxylic Acids
Read More: Chemistry Named Reactions