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Question: An organic compound contains 49.30\% \( carbon, \) 6.84\% \( hydrogen and its vapour density is \) 7...

An organic compound contains 49.30% carbon,carbon, 6.84% hydrogenanditsvapourdensityishydrogen and its vapour density is 73 .Molecularformulaofthecompoundis:a).Molecular formula of the compound is: a) {C_3}{H_8}{O_2} b) b) {C_6}{H_{10}}{O_4} c) c) {C_6}{H_9}O d) d) {C_4}{H_{10}}{O_2} $

Explanation

Solution

From the gaseous state chapter we know that vapour density is the ratio of density of a gas to that of hydrogen at the same pressure and same temperature. Also it can be defined as the mass of compound divided by molar mass of hydrogen.

Formula used: V.D=M.M2V.D = \dfrac{{M.M}}{2} Moles = GivenmassMolarMass\dfrac{{Given mass}}{{Molar Mass}}

Complete step by step solution:
Empirical Formula may be defined as the formula which gives the simplest whole number ratio of atoms in a given compound.
The following steps can be copied to determine empirical formula-
a) Assume the given percentage as the amount of elements out of 100.
b) Then by using molar mass we have to calculate the number of moles.
c) Then the lowest value of moles is used to divide all other moles and calculate the simple whole number.
d) Respective numbers are to be placed beside the elements in the formula.
e) We have to calculate the ‘n’ value to find out the molecular formula as n is the ratio of molecular mass to that of empirical mass.
Given: %\% of C=49.30C = 49.30 , %\% of H=6.84H = 6.84 , %\% of O=100(49.3+6.84)=43.86O = 100 - (49.3 + 6.84) = 43.86
Hence the mass will be as follows:
Mass of C=49.30gC = 49.30g , mass of H=6.84gH = 6.84g , mass of O=43.86gO = 43.86g
Moles = GivenMassMolarMass\dfrac{{Given Mass}}{{Molar Mass}}
The values can be tabulated as-

S.noELEMENTGIVEN MASS(g)Molar mass(g)MolesSimplest numberWhole number
1C49.3124.11.52
2H6.8416.842.53
3O43.86162.711

Therefore the empirical formula can be written as C2H3O{C_2}{H_3}O .
The empirical mass will be 2×12+3×1+1×16=43g2 \times 12 + 3 \times 1 + 1 \times 16 = 43g
From the vapour density formula, molar mass = 2×73=146g2 \times 73 = 146g .
Hence the n=M.ME.Mn = \dfrac{{M.M}}{{E.M}} ,
n=12643 n=2.93 \begin{gathered} \Rightarrow n = \dfrac{{126}}{{43}} \\\ \Rightarrow n = 2.9 \approx 3 \\\ \end{gathered}
Therefore the molecular formula is n×E.F=3×C2H3O=C6H9On \times E.F = 3 \times {C_2}{H_3}O = {C_6}{H_9}O
The correct option is (c).

Note:
Density is defined as the ratio of mass and volume. The term density is used for mostly solids and liquids. But vapour density is defined as the weight of a gas relative to that of hydrogen and indicates mainly to gas or vapour.