Question

The empirical formula of a nonelectrolyte is $C{{H}_{2}}O$ . A solution containing 6 g of the compound exerts the same osmotic pressure as that of 0.05 M glucose solution at the same temperature. The molecular formula of the compound is:
a.) ${{C}_{2}}{{H}_{4}}{{O}_{2}}$
b.) ${{C}_{3}}{{H}_{6}}{{O}_{3}}$
c.) ${{C}_{5}}{{H}_{10}}{{O}_{5}}$
d.) ${{C}_{4}}{{H}_{8}}{{O}_{4}}$

Answer 1

Hint: Empirical formula is the method to determine the simplest whole-number ratio of atoms in a compound. We will use the atomic weights of individual elements to calculate molecular formulas, by comparing data for unknown compounds and glucose.

Complete step by step answer:
We know that the molecular formula for a compound gives the actual whole number ratio between elements in a compound. If we talk about isotonic solution:
\[\dfrac{{{w}_{1}}}{{{m}_{1}}{{V}_{1}}}=\dfrac{{{w}_{2}}}{{{m}_{2}}{{V}_{2}}}\]
- Now given in the question,
Molecular mass of glucose = 180 g
Weight of 1 M glucose = 180 g
Weight of 0.05 M glucose = 0.05 x 180 g = 9 g
Weight of unknown compound = 6 g (given)
Molecular mass of unknown compound = x (Assume)
Putting the values in the above equation, we get –
\[\dfrac{{{w}_{1}}}{{{m}_{1}}{{V}_{1}}}=\dfrac{{{w}_{2}}}{{{m}_{2}}{{V}_{2}}}\]
\[\begin{align}
 & \Rightarrow \dfrac{9}{180}=\dfrac{6}{x} \\
 & \Rightarrow x=\dfrac{6\text{x}180}{9} \\
\end{align}\]
So, x = 120 g
To calculate a molecular formula with the help of empirical formulae we have to find the value of n.
\[\begin{align}
 & \dfrac{\text{Molecular mass}}{\text{Empirical mass}}=n \\
 & \Rightarrow n=\dfrac{120}{30}=4 \\
\end{align}\]
Molecular formula = \[{{\left( C{{H}_{2}}O \right)}_{n}}\]
Molecular formula = \[{{C}_{4}}{{H}_{8}}{{O}_{4}}\]
So the correct answer is “D”:

Note: Molecular formula is different from empirical formula. Empirical formula gives the “smallest whole number ratio between elements in a compound”. 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.