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Anthracene, a hydrocarbon obtained from coal, has an empirical formula of \(\mathrm{C}_{7} \mathrm{H}_{5} .\) To find its molecular formula you dissolve \(0.500 \mathrm{g}\) in \(30.0 \mathrm{g}\) of benzene. The boiling point of the pure benzene is \(80.10^{\circ} \mathrm{C},\) whereas the solution has a boiling point of \(80.34^{\circ} \mathrm{C} .\) What is the molecular formula of anthracene?

Short Answer

Expert verified
The molecular formula of anthracene is \(\mathrm{C}_{14}\mathrm{H}_{10}\).

Step by step solution

01

Determine Boiling Point Elevation

The boiling point elevation is calculated as the difference between the boiling point of the solution and pure solvent. Thus, the boiling point elevation is given by \( \Delta T_b = 80.34^{\circ} \mathrm{C} - 80.10^{\circ} \mathrm{C} = 0.24^{\circ} \mathrm{C} \).
02

Use Boiling Point Elevation Formula

The boiling point elevation can be expressed in terms of molality (\(m\)) as \( \Delta T_b = K_b \cdot m \), where \(K_b\) is the ebullioscopic constant of benzene (2.53 \( \text{°C kg/mol}\)). Rearrange the formula to find molality: \( m = \frac{\Delta T_b}{K_b} = \frac{0.24}{2.53} \approx 0.0949 \thinspace \text{mol/kg} \).
03

Determine Moles of Solute

Molality (\(m\)) is defined as \(m = \frac{\text{moles of solute}}{\text{kg of solvent}}\). With a molality of 0.0949 mol/kg and 30.0 g (or 0.030 kg) of benzene, calculate the moles of anthracene: \( \text{moles of anthracene} = 0.0949 \cdot 0.030 \approx 0.00285 \text{ mol} \).
04

Calculate the Molar Mass

Molar mass is calculated using the mass of anthracene and the moles determined: \( \text{Molar mass} = \frac{0.500 \thinspace \text{g}}{0.00285 \thinspace \text{mol}} \approx 175.44 \text{ g/mol} \).
05

Find the Molecular Formula

The empirical formula mass of anthracene (\(\mathrm{C}_{7}\mathrm{H}_{5}\)) is calculated as \( (7 \cdot 12.01) + (5 \cdot 1.01) = 89.09 \text{ g/mol} \). Determine the number of empirical units in the molecular formula using the ratio: \( \frac{\text{molar mass}}{\text{empirical formula mass}} = \frac{175.44}{89.09} \approx 2 \). This shows the molecular formula is \((\mathrm{C}_{7}\mathrm{H}_{5})_2 = \mathrm{C}_{14}\mathrm{H}_{10}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boiling Point Elevation
Boiling point elevation is a fascinating colligative property that refers to the increase in the boiling point of a solvent when a non-volatile solute is dissolved in it. This phenomenon occurs because adding a solute disrupts the solvent's ability to enter the gas phase, requiring a higher temperature to reach boiling. Thus, the boiling point of a solution (like benzene with anthracene) will always be higher than that of the pure solvent.
  • The boiling point elevation formula is: \( \Delta T_b = T_{ ext{solution}} - T_{ ext{pure solvent}} \).
  • Anthracene caused this elevation because its molecules mix with benzene, increasing the temperature needed for benzene to boil by 0.24°C.
Understanding this increase helps chemists determine the concentration and even the molecular weight of the dissolved substances.
Ebullioscopic Constant
The ebullioscopic constant, \(K_b\), is a specific property of solvents that reflects how much the boiling point changes per molal concentration of a solute. Every solvent has its own unique \(K_b\). For benzene, \(K_b\) is 2.53 °C kg/mol.
  • It forms the basis for the boiling point elevation equation \( \Delta T_b = K_b \cdot m \), where \(m\) is molality.
  • In the exercise example, this constant was crucial to calculate the molality of anthracene in benzene.
A precise understanding of the ebullioscopic constant allows chemists to predict and control the boiling points of solutions effectively.
Empirical Formula
The empirical formula of a compound is the simplest positive integer ratio of atoms present in a molecule. While not depicting the exact number of atoms—which the molecular formula does—it provides the basic proportions. For anthracene, this ratio is \(\mathrm{C}_7\mathrm{H}_5\).
  • By extending this ratio based on molecular weight, chemists can subsequently determine the molecular formula.
  • Anthracene's empirical formula mass is calculated to help figure out how many empirical units constitute one molecule.
So, while the empirical formula gives a fundamental understanding, the molecular formula shows the complete structure of molecules like anthracene.
Molality
Molality is a concentration measure used especially in colligative properties, denoted as \(m\). Unlike molarity, which relates to volume, molality uses mass, making it independent of temperature fluctuations. It is defined as the moles of solute per kilogram of solvent.
  • The calculation in the solution utilized: \(m = \frac{\Delta T_b}{K_b}\).
  • Molality helps in evaluating how the presence of a solute affects the boiling point and freezing point of a solution.
In the exercise, by determining the molality of anthracene in benzene, you acquire critical information necessary for product and reaction analysis.

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Most popular questions from this chapter

The melting point of pure biphenyl \(\left(\mathrm{C}_{12} \mathrm{H}_{10}\right)\) is found to be \(70.03^{\circ} \mathrm{C} .\) If \(0.100 \mathrm{g}\) of naphthalene is added to \(10.0 \mathrm{g}\) of biphenyl, the freezing point of the mixture is \(69.40^{\circ} \mathrm{C} .\) If \(K_{\mathrm{fp}}\) for biphenyl is \(-8.00^{\circ} \mathrm{C} / \mathrm{m},\) what is the molar mass of naphthalene?

Fill in the blanks in the table. Aqueous solutions are assumed. $$\begin{array}{llll} \hline & & \text { Weight } & \text { Mole } \\ \text { Compound } & \text { Molality } & \text { Percent } & \text { Fraction } \\\ \hline \mathrm{KN} 0_{3} & & 10.0 & \\ \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} & 0.0183 & & \\ \mathrm{H} 0 \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & & 18.0 & \\ \hline \end{array}$$

An unopened soda can has an aqueous \(\mathrm{CO}_{2}\) concentration of \(0.0506 \mathrm{M}\) at \(25^{\circ} \mathrm{C} .\) What is the pressure of \(\mathrm{CO}_{2}\) gas in the can?

Some lithium chloride, LiCl, is dissolved in 100 mL of water in one beaker and some \(\mathrm{Li}_{2} \mathrm{SO}_{4}\) is dissolved in 100 mL of water in another beaker. Both are at \(10^{\circ} \mathrm{C}\) and both are saturated solutions; some solid remains undissolved in each beaker. Describe what you would observe as the temperature is raised. The following data are available to you from a handbook of chemistry: $$\begin{array}{lll} \hline & \text { Solubility }(\mathrm{g} / 100 \mathrm{mL}) \\ \hline \text { Compound } & 10^{\circ} \mathrm{C} & 40^{\circ} \mathrm{C} \\ \hline \mathrm{Li}_{2} \mathrm{SO}_{4} & 35.5 & 33.7 \\\ \mathrm{LiCl} & 74.5 & 89.8 \\ \hline \end{array}$$

Cigars are best stored in a "humidor" at \(18^{\circ} \mathrm{C}\) and \(55 \%\) relative humidity. This means the pressure of water vapor should be \(55 \%\) of the vapor pressure of pure water at the same temperature. The proper humidity can be maintained by placing a solution of glycerol \(\left[\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\right]\) and water in the humidor. Calculate the percent by mass of glycerol that will lower the vapor pressure of water to the desired value. (The vapor pressure of glycerol is zero.)

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