Question

In an organic compound that is C=40.6%, H=6.6%, O=52.8%.If the vapor density is 30, find the molecular formula. ​

Answer 1

Given :

% composition of C = 40.6%

% composition of H = 6.6%

% composition of O = 52.8%

Vapour density = 30

To Find :

Molecular formula of the compound.

Solution :

$\boxed{\begin{array}{|c|c|c|c|c|}\cline{1-5}\bf Element&\bf\%&\bf Moles&\bf Mole\:ratio&\bf Whole\:no.\:ratio\\\cline{1-5}\sf C&\sf 40.6&\sf 40.6/12=3.38&\sf3.38/3.3=1.02&\bf1\\\cline{1-5}\sf H&\sf 6.6&\sf 6.6/1=6.6&\sf6.6/3.3=2&\bf2\\\cline{1-5}\sf O&\sf52.8&\sf 52.8/16=3.3&\sf 3.3/3.3=1&\bf1\\\cline{1-5}\end{array}}$

$\dag\:\underline{\boxed{\bf{\orange{Empirical\:Formula=CH_2O}}}}$

♦ Formula mass = C + 2(H) + O

» 12 + 2(1) + 16

» 28 + 2

» 30

♦ Molar mass = 2 × VD

» 2 × 30

» 60

$\sf:\implies\:Molecular\:Formula=n\times Empirical\:Formula$

$\sf:\implies\:Molecular\:Formula=\dfrac{60}{30}\times (CH_2O)$

$\sf:\implies\:Molecular\:Formula=2(CH_2O)$

$\sf:\implies\:\underline{\boxed{\bf{\gray{Molecular\:Formula=C_2H_4O_2}}}}$