Question

A $17.6\,gram$ of unknown solute is dissolved in $100\,gram$ of a solvent $({{K}_{b}}=2K\,kg\,molality{{y}^{-1}})$ to prepare a solution. Boiling point of pure solvent is $225{}^\circ C$ whereas the boiling point of the solution is $229{}^\circ C$. Predict molecular formula of the solute if it contains $54.54%\,C$ and $9.09%\,H$ (by weight)

Answer 1

Hint: Elevation in boiling point is a colligative property which is directly proportional to molality and its constant ${{K}_{b}}$ is called molal elevation constant or molal ebullioscopic constant.

Formula used: $\Delta Tb={{K}_{b}}m$ or\[\Delta {{T}_{b}}=\dfrac{{{K}_{b}}\times {{W}_{B}}\times 1000}{{{M}_{B}}\times {{W}_{A}}}\]

Complete step by step solution:
When we add a non volatile solute to a volatile solvent, then an increase in the boiling point is experienced which is called elevation in boiling point.
As this elevation in boiling point is directly proportional to molality, so through this, the molar mass of the solute can be taken out.
Given in the question:
Amount of solute \[=17.6\,g\]
Amount of solvent $=100\,g$
${{K}_{b}}=2K\,\,kg\,molality{{y}^{-1}}$
Boiling point of solution ${{T}_{b}}=229{}^\circ C$
Boiling point of solvent ${{T}_{b}}{}^\circ =225{}^\circ C$
\[%\,C\,\] in the solute (by mass)$=54.54%$
\[%\,H\] in the solute (by mass)$=9.09%$
So, putting the quantities in the formula:\[~~\]
$\Delta {{T}_{b}}={{K}_{b}}m$ \[=\dfrac{{{K}_{b}}\times {{W}_{B}}\times 1000}{{{M}_{B}}\times {{W}_{A}}}\]
\[{{T}_{b}}-{{T}_{b}}{}^\circ ={{K}_{b}}m\] \[=\dfrac{{{K}_{b}}\times {{W}_{B}}\times 1000}{{{M}_{B}}\times {{W}_{A}}}\]
Where the quantities are:
\[\Delta {{T}_{b}}=\] elevation in boiling point (${{T}_{b}}=229{}^\circ C$and ${{T}_{b}}{}^\circ =225{}^\circ C$)
${{K}_{b}}$is molal elevation constant $=2K\,\,kg\,molality{{y}^{-1}}$
${{W}_{B}}=$mass of solute\[=17.6\,g\]
${{W}_{A}}=$mass of solvent$=100\,g$
${{M}_{B}}=$molar mass of solute
\[229-225=2\times \dfrac{17.6\times 1000}{100\times {{M}_{B}}}\]
\[4=2\times \dfrac{176}{{{M}_{B}}}\]
\[{{M}_{B}}=\dfrac{2\times 176}{4}\]
\[{{M}_{B}}=88\,g\,mo{{l}^{-1}}\]
Now, to take out the amount of carbon and hydrogen, the given percentage of carbon and hydrogen is multiplied by this molar mass of solute and divided by hundred.
Amount of carbon $=\dfrac{88\times 54.54}{100}$
Amount of carbon $=48\,g\,mo{{l}^{-1}}$
As we know that molecular mass of carbon is 12, so this amount of carbon is divided by 12 to get the number of carbon atoms as:
Number of carbon atoms$=\dfrac{48}{12}$
Number of carbon atoms= 4
Similarly, we take out the amount and number of hydrogen.
Amount of hydrogen $=\dfrac{88\times 9.09}{100}$
Amount of hydrogen$=8\,g\,mo{{l}^{-1}}$
So, the number of hydrogen atoms will be the amount of hydrogen upon the molar mass of hydrogen.
Number of hydrogen atoms$=\dfrac{8}{1}$
Number of hydrogen atoms= 8
Amount of third compound has to be taken out, as these together do not make up the molar mass of solute. So,
Amount of third compound $=88-(8+48)$
Amount of third compound $=32\,g\,mo{{l}^{-1}}$
Since, this number 32 is a multiple of oxygen atom and is the mass of di atoms of oxygen, so the third compound will be oxygen.
Number of atoms of oxygen $=\dfrac{32}{16}$
Number of atoms of oxygen=2
So, the molecular formula of the solute is derived as ${{C}_{4}}{{H}_{8}}{{O}_{2}}$

Note: Check the mass that is contributed by the given percentage of oxygen and hydrogen, if it does not add up to the mass of solute then always take out the third compound present in the solute.