A 7.85-g sample of a compound with the empirical formula C5H4 is dissolved in 301 g of benzene. The freezing point of the solution is 1.05°C below that of pure benzene. What are the molar mass and molecular formula of this compound? Strategy Solving this problem requires three steps. First, we

Question

A 7.85-g sample of a compound with the empirical formula [latex]C_{5}H_{4}[/latex] is dissolved in 301 g of benzene. The freezing point of the solution is 1.05°C below that of pure benzene. What are the molar mass and molecular formula of this compound?

Strategy Solving this problem requires three steps. First, we calculate the molality of the solution from the depression in freezing point. Next, from the molality we determine the number of moles in 7.85 g of the compound and hence its molar mass. Finally, comparing the experimental molar mass with the empirical molar mass enables us to write the molecular formula.

Accepted Answer

The sequence of conversions for calculating the molar mass of the compound is

Freezing -point depression ⟶ molality ⟶ number of moles ⟶ molar mass

Our first step is to calculate the molality of the solution. From Equation (12.7) and Table 12.2 we write

[latex]ΔT_{f}=K_{f}m[/latex]

Solvent	Normal Freezing POINT©*	[latex]K_{f}[/latex] (C/m)	Normal Boiling Point ©*	[latex]K_{B}[/latex] (C/m)
Water	0	1.86	100	0.52
Benzene	5.5	5.12	80.1	2.53
Ethanol	-117.3	1.99	78.4	1.22
Acetic acid	16.6	3.9	117.9	2.93
Cycolohexane	6.6	20	80.7	2.79

[latex]molality=\frac{\Delta T_{f} }{K_{f} } =\frac{1.05°C}{5.12°C/m}=0.205 m [/latex]`

Because there is 0.205 mole of the solute in 1 kg of solvent, the number of moles of solute in 301 g, or 0.301 kg, of solvent is

[latex]0.301 kg imes \frac{0.205 mol}{1 kg}=0.0617 mol [/latex]

Thus, the molar mass of the solute is

[latex]molar mass=\frac{grams of compound}{moles of compound} [/latex]

[latex]=\frac{7.85 g}{0.0617 mol}=127 g/mol [/latex]

Now we can determine the ratio

[latex]\frac{molar mass}{empirical molar mass}=\frac{127 g/mol}{164 g/mol} \approx 2 [/latex]

Therefore, the molecular formula is [latex](C_{5}H_{4} )_{2}[/latex] or [latex]C_{10}H_{8}[/latex] (naphthalene).

Question 12.10: A 7.85-g sample of a compound with the empirical formula C5H...

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