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)$$$K=4.29 \times 10^{-6}(\text {at } 298 \mathrm{K})$$.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?

Short answer

The molecular formula of naphthalene is \(\text{C}_{10}\text{H}_{8}\). When 3.00 g of solid naphthalene is placed into an enclosed space with a volume of 5.00 L at \(25^{\circ} \mathrm{C}\), approximately 32.1% of the naphthalene will have sublimed once equilibrium has been established.

Step-by-step solution

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: \(\text{C}_{10}\text{H}_{8}\).

Find the molar mass of the empirical formula

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

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: \(\text{C}_{10}\text{H}_{8}\).

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: \(P_i = \frac{3.00 g \times (1 mol/128.18 g)}{5.00 L \times 0.0821 L\cdot atm/mol \cdot K \times 298 K} = 0.00152 atm\)

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 = \(\frac{P \times V}{R \times T} \times M = \frac{3.707 \times 10^{-4} \times 5.00}{0.0821 \times 298} \times 128.18 = 0.962 g\)

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.

Key concepts

Empirical Formula

The empirical formula of a compound gives the simplest whole-number ratio of the elements present in the compound. To find the empirical formula, start by determining the mass of each element if you assume a 100g sample. For naphthalene, which is 93.71% carbon by mass, this means you have 93.71g of carbon and the remaining 6.29g is hydrogen (100% - 93.71%).
Convert these masses to moles by dividing each by the element's molar mass: for carbon it's 12.01 g/mol, and for hydrogen, it’s 1.01 g/mol. This calculation gives 7.80 moles of carbon and 6.23 moles of hydrogen.
Next, derive the simplest ratio by dividing each mole value by the smallest number of moles calculated. Here, 6.23 is the baseline, leading to a ratio of approximately 1.25:1. However, chemical formulas use whole numbers, so multiply these ratios to get a whole number. Multiplying by 4 gives a ratio of C₅H₄, but real data shows C₁₀H₈ is more accurate in practice.

Ideal Gas Law

The Ideal Gas Law connects pressure, volume, temperature, and the number of moles of a gas in a single equation: \( PV = nRT \). This equation is particularly useful for calculating properties of gases under certain conditions.
- **P** stands for pressure (in atmospheres, atm)- **V** is the volume (in liters, L)- **n** is the number of moles- **R** is the ideal gas constant, 0.0821 L·atm/mol·K- **T** is the temperature (in Kelvin, K)
For example, suppose you want to find the initial pressure of naphthalene gas. Convert 3.00g of naphthalene to moles using its molar mass, divide by volume (5.00L) and multiply by the gas constant and temperature (298K).
The answer provides the initial pressure in atm, which is crucial for further equilibrium calculations.

Percent Sublimated

Percent sublimated refers to how much of a solid substance has turned into gas. This is important in understanding the behavior of compounds like naphthalene, which sublimates readily at room temperature.
To calculate the percentage sublimated, you need the amount of naphthalene that has escaped into the gas phase at equilibrium. This mass is determined using both the Ideal Gas Law and the equilibrium pressure. Subtract this mass from the initial mass and divide by the initial mass, then multiply by 100 to express as a percentage.
In our problem, 0.962 grams has sublimated from the initial 3.00 grams, leading to a 32.1% sublimation. This percentage helps illustrate how much of the naphthalene has transformed from solid to gas under given conditions.

Equilibrium Constant

The equilibrium constant \( K \) for a reaction represents the ratio of the concentrations of products to reactants at equilibrium. For naphthalene's sublimation, the constant is given as \( 4.29 \times 10^{-6} \) at 298K. This low value signifies that at equilibrium, very little naphthalene is in the gas phase under these conditions.
When dealing with gaseous equilibria, this constant lets you determine the pressures of the gases involved once equilibrium is reached. This is pivotal when calculating how much naphthalene sublimates into a closed space.
Having such a low equilibrium constant implies that the pressure of the gas is also low, confirming that naphthalene prefers remaining in solid form at room temperature unless specific conditions drive a significant phase change.