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 \(0.256\) -mol sample of naphthalene has a mass of \(32.8 \mathrm{~g}\). 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 \((g)\) \(K=4.29 \times 10^{-6}(\) 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?

Since we know that naphthalene is 93.71% carbon by mass, the remaining 6.29% is hydrogen. We are given that a 0.256-mol sample of naphthalene has a mass of 32.8 g. Using this information, we can find the molar mass of naphthalene. Molar mass of naphthalene = \( \frac{32.8 g}{0.256 mol}\) Molar mass of naphthalene = 128 g/mol Now, we can calculate the number of moles of carbon and hydrogen in naphthalene: Moles of carbon = \( \frac{93.71 g}{12.01 g/mol}\) Moles of carbon = 7.8 moles (approximately) Moles of hydrogen = \( \frac{6.29 g}{1.008 g/mol}\) Moles of hydrogen = 6.2 moles (approximately) Since we have rounded off the values, the molecular formula of naphthalene is approximately \(C_{7.8}H_{6.2}\), but we require integers for the molecular formula. To achieve this, divide both the number of moles of carbon and hydrogen by the smallest value, which is 6.2 (approximately): \( \frac{7.8}{6.2} \approx \frac{6.2}{6.2} = 1.25 \approx 1\) Thus, the molecular formula of naphthalene is \(C_{10}H_{8}\).

Given that 3.00 g of solid naphthalene is placed in an enclosed space of 5.00 L at 25°C and that the equilibrium constant K is \(4.29 \times 10^{-6}\) at 298 K, we can calculate the percentage of naphthalene that sublimed at equilibrium. First, convert the mass of naphthalene into the number of moles: Moles of naphthalene = \( \frac{3.00 g}{128 g/mol} = 0.0234 mol\) Next, we can set the number of moles of gaseous naphthalene equal to x at equilibrium while the remaining amount of solid naphthalene would be (0.0234 - x) moles. The equilibrium expression for the process is: \(K = \frac{[naphthalene(g)]}{[naphthalene(s)]}\) Since the total volume in the enclosed space is 5.00 L: \[ 4.29 \times 10^{-6} = \frac{x}{5}\] Solve for x: x = \(4.29 \times 10^{-6} \times 5 = 2.14 \times 10^{-5}\) This x value represents the moles of gaseous naphthalene at equilibrium. Now, we can calculate the percentage of solid naphthalene that sublimed at equilibrium: Percentage of sublimed naphthalene = \( \frac{2.14 \times 10^{-5} mol \times 100}{0.0234 mol}\) Percentage of sublimed naphthalene ≈ 0.091% Therefore, once equilibrium has been established in the enclosed space, approximately 0.091% of naphthalene is in the gaseous state.