Concept Explanation: Vapor Pressure Lowering and Molecular Formula Determination

This problem involves determining the molecular formula of a non-electrolyte, non-volatile solute based on its vapor pressure lowering effect. According to Raoult's Law, the relative lowering of vapor pressure is equal to the mole fraction of the solute.

Step 1: Understand the Given Data

Mass of solute (C_xH_2xO_x) = 20 g

Mass of solvent (water) = 100 g

Temperature = 100°C (vapor pressure of pure water at 100°C is important, but not directly needed as we use relative lowering)

Relative lowering of vapor pressure = $\frac{1}{100}$ of pure water's vapor pressure.

Let P° be the vapor pressure of pure water. Then, lowering of vapor pressure = $\frac{P^{\circ }}{100}$

So, relative lowering = $\frac{\mathrm{\Delta P}}{P^{\circ }}=\frac{1}{100}$

Step 2: Apply Raoult's Law for Relative Lowering of Vapor Pressure

For a non-volatile solute, Raoult's Law states:

$\frac{\mathrm{\Delta P}}{P^{\circ }}=x_{\text{solute}}=\frac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{solvent}}}$

Since the solute is non-electrolyte and non-volatile, and the solution is dilute, we can approximate:

$\frac{\mathrm{\Delta P}}{P^{\circ }}\approx \frac{n_{\text{solute}}}{n_{\text{solvent}}}$

Thus, $\frac{1}{100}=\frac{n_{\text{solute}}}{n_{\text{solvent}}}$

Step 3: Calculate Moles of Solvent (Water)

Molar mass of water (H₂O) = 18 g/mol

Moles of water, $n_{\text{solvent}}=\frac{\text{mass}}{\text{molar mass}}=\frac{100}{18}=\frac{50}{9}$ mol

Step 4: Express Moles of Solute and Substitute into Equation

Let M be the molar mass of solute C_xH_2xO_x.

Moles of solute, $n_{\text{solute}}=\frac{20}{M}$

From Step 2: $\frac{20/M}{100/18}=\frac{1}{100}$

Simplify: $\frac{20}{M}\times \frac{18}{100}=\frac{1}{100}$

$\frac{360}{100M}=\frac{1}{100}$

Cross-multiply: $360\times 100=100M$

$36000=100M$

Thus, $M=\frac{36000}{100}=360$ g/mol

Step 5: Determine Molecular Formula from Molar Mass

General formula of solute: C_xH_2xO_x

Molar mass M = (12 × x) + (1 × 2x) + (16 × x) = 12x + 2x + 16x = 30x

We found M = 360 g/mol, so:

$30x=360$

$x=\frac{360}{30}=12$

Therefore, molecular formula is C₁₂H₂₄O₁₂

Final Answer

The molecular formula of the compound is C₁₂H₂₄O₁₂.

Related Topics

• Colligative Properties
• Raoult's Law
• Vapor Pressure Lowering
• Molecular Mass Determination
• Ideal Solutions

Important Formulae and Theory

Raoult's Law for Vapor Pressure Lowering:

For a non-volatile solute, the relative lowering of vapor pressure is equal to the mole fraction of the solute:

$\frac{P^{\circ }-P}{P^{\circ }}=x_{\text{solute}}=\frac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{solvent}}}$

For dilute solutions, $\frac{P^{\circ }-P}{P^{\circ }}\approx \frac{n_{\text{solute}}}{n_{\text{solvent}}}$

Mole Calculation:

Number of moles, $n=\frac{\text{mass}}{\text{molar mass}}$