Question

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?

Short answer

The molecular formula of naphthalene is C10H8. When 3g of solid naphthalene is placed in a 5L container at 25°C, approximately 0.49% of naphthalene will have sublimed once equilibrium is established.

Step-by-step solution

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

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

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).

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: \( \frac{x}{5L} \)

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

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.

Key concepts

Empirical Formula

The empirical formula of a compound provides the simplest whole-number ratio of the elements within it. For naphthalene, which has been found to contain 93.71% carbon by mass, determining the empirical formula involves converting this percentage into a mass format, assuming a hypothetical 100 g sample.

• Convert the mass of each element into moles. For carbon, 93.71 g converts to approximately 7.80 moles using its molar mass, 12.01 g/mol.
• For hydrogen, after subtracting carbon's mass, 6.29 g of hydrogen converts to about 6.24 moles because hydrogen's molar mass is 1.008 g/mol.
• Establish the mole ratio by dividing each element's mole quantity by the smallest number of moles obtained, in this case, hydrogen.
• The result yields a ratio of C: 1.25 to H: 1. Multiply by 4 to get whole numbers, resulting in the empirical formula C5H4.

Equilibrium Expression

In chemical reactions, equilibrium expressions are special formulas that describe how concentrations of products and reactants relate to each other at equilibrium. This is crucial for understanding sublimation processes in compounds like naphthalene. The sublimation equilibrium for naphthalene is:
\[ \text{Naphthalene}(s) \rightleftharpoons \text{Naphthalene}(g) \]The corresponding equilibrium expression is expressed as:\[ K = \frac{[\text{Naphthalene}(g)]}{[\text{Naphthalene}(s)]} \]At 298 K, this equilibrium constant \( K \) is given as \( 4.29 \times 10^{-6} \). Notice that the expression mainly centers on the concentration of the gas phase, as solids' activity in a saturated system is considered constant and, thus, not factored into the expression.

Sublimation

Sublimation is the phase change where a solid directly transitions to a gas without passing through a liquid state. For naphthalene, this transition occurs when it sublimates from solid to vapor. This property is exploited when naphthalene is used in mothballs to ensure its vapor fills an enclosed space and acts as a pesticide.

• The sublimation equilibrium equation illustrates how naphthalene can exist both in solid and gaseous forms. It is represented by \( \text{Naphthalene}(s) \rightleftharpoons \text{Naphthalene}(g) \).
• At equilibrium, only a small fraction, about 0.49% in this context, of the initial 3 g of solid naphthalene sublimates into a gas. This low percentage reflects a small equilibrium constant value, signifying a negligible amount of gaseous naphthalene in comparison to its solid state at room temperature.

Molar Mass

Molar mass is a measure of the mass of one mole of a substance, and is crucial for calculations related to molecular formulas. For naphthalene, determining the molar mass is essential to derive its molecular formula from the empirical formula.
The empirical formula mass is obtained by summing the atomic masses of all atoms in the formula C5H4:\[ 5(12.01) + 4(1.008) = 64.08 \text{ g/mol} \]By utilizing the information that a 0.256-mole sample of naphthalene weighs 32.8 g, we calculate the actual molar mass:\[ \frac{32.8}{0.256} \approx 128.12 \text{ g/mol} \]By comparing the empirical formula mass (64.08 g/mol) and the actual molar mass (128.12 g/mol), we find that the actual compound contains twice the number of molecules as the empirical formula suggests. Thus, the molecular formula of naphthalene is C10H8.