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Chapter 12: Problem 62

A solution of \(2.50 \mathrm{~g}\) of a compound having the empirical formula \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{P}\) in \(25.0 \mathrm{~g}\) of benzene is observed to freeze at \(4.3^{\circ} \mathrm{C}\). Calculate the molar mass of the solute and its molecular formula.

Short Answer

Expert verified
The molar mass of the compound is 10.6667 g/mol, and its molecular formula is C6H5P.

Step by step solution

01

Finding moles of the solute

First, we need to find the number of moles of the solute based on the given mass and its empirical formula. The empirical formula \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{P}\) means the molar mass is \(6(12.01 g/mol) + 5(1.008 g/mol) + (30.97 g/mol) = 92.056 g/mol\). So, the moles of the compound is given by: \(moles = \frac{mass}{molar ~mass} = \frac{2.5g}{92.056g/mol} = 0.02715 mol\).
02

Calculating the molality of the solution

Next, we calculate the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent. So, \(molality = \frac{moles ~of ~solute}{mass ~of ~solvent ~in ~kg} = \frac{0.02715 mol}{25.0 g * 1 kg/1000 g} = 1.086 mol/kg = 1.086 m\).
03

Calculating the molar mass using freezing point depression

The depression in freezing point is given by \(\Delta T_f = K_f * b\), where \(\Delta T_f\) is the freezing point depression, \(K_f\) is the cryoscopic constant, and \(b\) is the molality. For benzene, the freezing point and \(K_f\) at 1 atm pressure are \(5.5^{\circ} C\) and \(5.12 K*kg/mol\), respectively. The depression in freezing point is \(5.5^{\circ} C - 4.3^{\circ} C = 1.2^{\circ} C\). Rearranging the formula gives \(b = \frac{\Delta T_f}{K_f} = \frac{1.2}{5.12} = 0.234375 mol/kg\). However, we found the molality to be \(1.086 m\), which means that the molar mass of the compound is given by \(\frac{mass ~of ~solute}{moles ~of ~solute} = \frac{2.5 g}{0.234375 mol} = 10.6667g/mol\).
04

Finding the molecular formula

The molecular formula is calculated by dividing the molar mass of a compound by the molar mass of the empirical formula. We already know that the molar mass of the empirical formula is \(92.056 g/mol\), so the molecular formula is given by \(n = \frac{molar ~mass ~of ~compound}{molar ~mass ~of ~empirical ~formula} = \frac{10.6667 g/mol}{92.056 g/mol} = 0.1158\). This value is approximately equal to 1, which means the molecular formula is the same as the empirical formula, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{P}\).

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Most popular questions from this chapter

Making mayonnaise involves beating oil into small droplets in water, in the presence of egg yolk. What is the purpose of the egg yolk? (Hint: Egg yolk contains lecithins, which are molecules with a polar head and a long nonpolar hydrocarbon tail.)

Discuss the factors that influence the solubility of a gas in a liquid.

A solution of \(1.00 \mathrm{~g}\) of anhydrous aluminum chloride, \(\mathrm{AlCl}_{3}\), in \(50.0 \mathrm{~g}\) of water freezes at \(-1.11^{\circ} \mathrm{C}\). Does the molar mass determined from this freezing point agree with that calculated from the formula? Why?

Acetic acid is a polar molecule and can form hydrogen bonds with water molecules. Therefore, it has a high solubility in water. Yet acetic acid is also soluble in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right),\) a nonpolar solvent that lacks the ability to form hydrogen bonds. A solution of \(3.8 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{COOH}\) in \(80 \mathrm{~g} \mathrm{C}_{6} \mathrm{H}_{6}\) has a freezing point of \(3.5^{\circ} \mathrm{C}\). Calculate the molar mass of the solute and suggest what its structure might be. (Hint:Acetic acid molecules can form hydrogen bonds between themselves.)

Arrange the following solutions in order of decreasing freezing point: \(0.10 \mathrm{~m} \mathrm{Na}_{3} \mathrm{PO}_{4}, 0.35 \mathrm{~m} \mathrm{NaCl}\) \(0.20 \mathrm{~m} \mathrm{MgCl}_{2}, 0.15 \mathrm{~m} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}, 0.15 \mathrm{~m} \mathrm{CH}_{3} \mathrm{COOH}\)
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