Question

An organic compound \[\left( {{C}_{x}}{{H}_{2y}}{{O}_{y}} \right)~\] was burnt with twice the amount of oxygen needed for complete combustion to \[C{{O}_{2}}~\] and \[{{H}_{2}}O.\] The hot gases when cooled to \[{{0}^{0}}C\] and $1$ atm pressure measured \[2.24L.\] The water collected during cooling weighs \[0.9g.\]The vapour pressure of pure water at \[{{20}^{0}}C\] is \[17.5mm\] of \[Hg\] and is lowered by \[0.104mm\]when \[50g\] of the organic compound is dissolved in \[1000g\]of water. Give the molecular formula of the organic compound.
A) ${{C}_{5}}{{H}_{10}}{{O}_{5}}$
B) ${{C}_{10}}{{H}_{20}}{{O}_{10}}$
C) ${{C}_{4}}{{H}_{8}}{{O}_{4}}$
D) None of these.

Answer 1

Hint: Empirical formula is the method to determine the simplest whole number ratio of atoms in a compound. Use the atomic weights of individual elements to calculate empirical formulas.

Complete step-by-step answer:
Empirical formula gives the “smallest whole number ratio between elements in a compound”. Since, ratio for Oxygen is not mentioned, we can calculate it using the following;
$=100-\left( 38.71+9.67 \right)%=51.62\text{ }%\text{ }$
For calculating empirical formulas, we need to calculate a simple ratio. We do this by calculating mole ratio for each element first and then dividing mole ratio of every element by the least mole ratio. In this case, mole ratio of both carbon and oxygen is the least, i.e. \[3.22.\] Therefore, we divide each mole ratio by \[3.22.\]to calculate the simple ratio. The process is given below;
The molecular formula of the compound is \[{{C}_{x}}{{H}_{2y}}{{O}_{y}}\], \[{{C}_{x}}{{H}_{2y}}{{O}_{y}}+x{{O}_{2}}\to xC{{O}_{2}}+y{{H}_{2}}O.\]

The amount of oxygen is twice the required amount i.e $2x.$
The hot gases when cooled to \[{{0}^{0}}C\] and $1$ atm pressure measured \[2.24L.\]
This corresponds to \[0.1\] mole as \[2.24L.\]of any gas at \[NTP\] corresponds to $1$ mole.
Thus, $x+x+y=0.1$ i.e. $\Rightarrow 2x+y=0.1..........(1)$
The water collected during cooling weighs \[0.9g.\]This corresponds to $\dfrac{0.9}{18}=0.05moles$
Thus, $y=0.05..........(2)$
Substitute equation $(1)$ in equation $(2)$ we get;
$2x+0.05=0.1$
On further simplifying, $2x=0.05$we get;
$\Rightarrow x=0.025...........(3)$
Hence, the empirical formula of the organic compound is \[C{{H}_{4}}{{O}_{2}}.~\] The empirical formula mass is \[12+4+32=48g/mol.\]
\[\dfrac{{{p}^{0}}-p}{{{p}^{0}}}=\dfrac{{{W}_{2}}}{{{M}_{2}}}\times \dfrac{{{M}_{1}}}{{{W}_{1}}}\]
Substituting the values in above formula;
\[\dfrac{0.104}{17.5}=\dfrac{50}{{{M}_{2}}}\times \dfrac{0.018}{1000}\]
\[\Rightarrow {{M}_{2}}=\dfrac{50\times 17.5}{0.104}\times \dfrac{0.018}{1000}=0.151kg\]
\[\Rightarrow {{M}_{2}}=151g\]
Now just we have to get the value of $r$,
$r=\dfrac{Molecular\left( mass \right)}{empirical\left( mass \right)}=\dfrac{151}{48}=3.1\simeq 3$
Therefore, the molecular formula of the organic gas is $3\times
C{{H}_{4}}{{O}_{2}}={{C}_{3}}{{H}_{12}}{{O}_{6}}.$

Therefore, the correct answer is option D, none of the given options are correct.

Note: Molecular formula is different from empirical formula. The molecular formula for a compound gives the actual whole number ratio between elements in a compound.