The hydrocarbon naphthalene was frequently used in mothballs until recently, when it was discovered that human inhalation of naphthalene vapors can lead to hemolytic anemia. Naphthalene is \(93.71 \%\) carbon by mass, and a 0.256 -mole sample of naphthalene has a mass of 32.8 g. What is the molecular formula of naphthalene? This compound works as a pesticide in mothballs by sublimation of the solid so that it fumigates enclosed spaces with its vapors according to the equation Naphthalene( \(s) \rightleftharpoons\) naphthalene \((g)$$$K=4.29 \times 10^{-6}(\text {at } 298 \mathrm{K})$$.If \)3.00 \mathrm{g}\( solid naphthalene is placed into an enclosed space with a volume of \)5.00 \mathrm{L}\( at \)25^{\circ} \mathrm{C},$ what percentage of the naphthalene will have sublimed once equilibrium has been established?

Step by step solution

01

Find the empirical formula of naphthalene

Since naphthalene consists of 93.71% carbon by mass, the rest (100% - 93.71%) must be hydrogen. Convert the percentages to grams, assuming a 100g sample: Carbon: \(0.9371 \times 100 g = 93.71 g\) Hydrogen: \(0.0629 \times 100 g = 6.29 g\) Now, to find the moles of each element, divide the mass by the molar mass: Carbon: \(\frac{93.71 g}{12.01 g/mol} = 7.80 mol\) Hydrogen: \(\frac{6.29 g}{1.01 g/mol} = 6.23 mol\) Next, divide the moles of each element by the lowest number of moles: Carbon: \(\frac{7.80}{6.23} \approx 1.25\) Hydrogen: \(\frac{6.23}{6.23} = 1\) Round the two numbers to the nearest whole number to get the empirical formula: \(\text{C}_{10}\text{H}_{8}\).

02

Find the molar mass of the empirical formula

Calculate the molar mass of the empirical formula (C10H8): \(10 \times 12.01 + 8 \times 1.01 = 120.1 + 8.08 = 128.18 g/mol\)

03

Find the molecular formula of naphthalene

Using the molar mass of the empirical formula, calculate the ratio between it and the molar mass of the 0.256-mole sample of naphthalene: \(\frac{32.8 g}{0.256 mol} = 128.12 g/mol\) The molar mass of the empirical formula and the sample is roughly the same, indicating that the molecular formula and empirical formula are the same: \(\text{C}_{10}\text{H}_{8}\).

04

Calculate the initial pressure of naphthalene

To find the initial pressure of naphthalene, use the ideal gas law with partial pressure, \(P=\frac{nRT}{V}\): Initial pressure: \(P_i = \frac{3.00 g \times (1 mol/128.18 g)}{5.00 L \times 0.0821 L\cdot atm/mol \cdot K \times 298 K} = 0.00152 atm\)

05

Determine equilibrium pressure and calculate mass of naphthalene that has sublimed

Using the equilibrium constant, find the equilibrium pressure: \(K = \frac{P}{P_i - P}\) \(P = \sqrt{K \times P_i} = \sqrt{4.29 \times 10^{-6} \times 0.00152} = 3.707 \times 10^{-4} atm\) With the equilibrium pressure, find the mass of naphthalene that has sublimed: Mass sublimed = \(\frac{P \times V}{R \times T} \times M = \frac{3.707 \times 10^{-4} \times 5.00}{0.0821 \times 298} \times 128.18 = 0.962 g\)

06

Calculate the percentage of naphthalene that has sublimed

To find the percentage of naphthalene that has sublimed at equilibrium, divide the mass of naphthalene that has sublimed by the total mass and multiply by 100: Percentage sublimed = \(\frac{0.962}{3.00} \times 100 \% \approx 32.1 \% \) Once equilibrium has been established, approximately 32.1% of the naphthalene will have sublimed.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!