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}(\( at \)298 \mathrm{~K})\( If \)3.00 \mathrm{~g}$ solid naphthalene is placed into an en with a volume of \(5.00 \mathrm{~L}\) at $25^{\circ} \mathrm{C},$ what percentage thalene will have sublimed once equilibriur estahlished?

Step by step solution

01

Calculate the empirical formula of naphthalene

To find the empirical formula, we will first convert the mass percentage of carbon (93.71%) to grams. Assume a 100g sample, then we have 93.71g of Carbon. The remaining mass would be hydrogen. Now, convert the mass of carbon and hydrogen to moles, and then find the mole ratio to determine the empirical formula. Carbon mass in 100g sample: 93.71g Hydrogen mass in 100g sample: 100g - 93.71g = 6.29g Moles of Carbon: \( \frac{93.71}{12.01} \approx 7.80 \) Moles of Hydrogen: \( \frac{6.29}{1.008} \approx 6.24 \) Mole ratio: C: \( \frac{7.80}{6.24} \approx 1.25 \) H: \( \frac{6.24}{6.24} \approx 1 \) The empirical formula is C\( _{1.25} \)H. Since the empirical formula should contain whole numbers, we will multiply by 4 to obtain the whole number ratio: Empirical formula: C5H4

02

Calculate the molecular formula of naphthalene

Now, we can determine the molecular formula by comparing the molar mass of the empirical formula with the given mass for a 0.256-mole sample of naphthalene. Empirical formula mass: 5(12.01) + 4(1.008) = 64.08 g/mol Mass of 0.256-mole sample: 32.8g Molar mass of naphthalene: \( \frac{32.8}{0.256} \approx 128.12 \) Now divide the molar mass of naphthalene with the empirical formula mass: \( \frac{128.12}{64.08} \approx 2 \) The molecular formula is 2 times the empirical formula: Molecular formula: C10H8

03

Set up the equilibrium expression

In this step, we will set up the equilibrium expression for the sublimation of naphthalene. The equilibrium constant expression for the reaction is: \( K = \frac{[naphthalene(g)]}{[naphthalene(s)]} \) At 298 K, the equilibrium constant (K) is given as 4.29 x 10^(-6).

04

Calculate the equilibrium concentrations

Since the initial mass of solid naphthalene is given as 3g, we can convert this to moles using the molecular formula and then determine the equilibrium concentrations. Moles of solid naphthalene: \( \frac{3}{128.12} \approx 0.0234 \) Let x be the moles of naphthalene that has sublimed. Then, at equilibrium: Moles of solid naphthalene: 0.0234 - x Moles of gaseous naphthalene: x Now, convert the moles of gaseous naphthalene to concentration by dividing by the volume (5L) of the container. Equilibrium concentration of gaseous naphthalene: \( \frac{x}{5L} \)

05

Solve for x

Now, plug the values into the equilibrium expression, and solve for x. \( 4.29 \times 10^{-6} = \frac{\frac{x}{5L}}{0.0234 - x} \) Solve for x = 1.15 x 10^(-4) moles

06

Calculate the percentage of naphthalene that has sublimed

To find the percentage of naphthalene that has sublimed, we will use the moles of sublimed naphthalene (x) and the initial moles of solid naphthalene (0.0234) as follows: Percentage of sublimed naphthalene: \( \frac{1.15 \times 10^{-4}}{0.0234} \times 100 \approx 0.49 \% \) Thus, approximately 0.49% of naphthalene will have sublimed once equilibrium is established.

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