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

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 –
$\dfrac{{{w}_{1}}}{{{m}_{1}}{{V}_{1}}}=\dfrac{{{w}_{2}}}{{{m}_{2}}{{V}_{2}}}$

& \Rightarrow \dfrac{9}{180}=\dfrac{6}{x} \\\ & \Rightarrow x=\dfrac{6\text{x}180}{9} \\\ \end{aligned}$$ So, x = 120 g To calculate a molecular formula with the help of empirical formulae we have to find the value of n. $$\begin{aligned} & \dfrac{\text{Molecular mass}}{\text{Empirical mass}}=n \\\ & \Rightarrow n=\dfrac{120}{30}=4 \\\ \end{aligned}$$ Molecular formula = $${{\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.