%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:

(i) Explain Aldol condensation with example.
(ii) How are the following conversions achieved:
(a) Benzene Benzaldehyde, (b) Ethanoic acid ethanol.
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).






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