Question

A \(0.0125-\mathrm{g}\) sample of a gas with an empirical formula of \(\mathrm{CHF}_{2}\) is placed in a 165 -mL. flask. It has a pressure of \(13.7 \mathrm{mm}\) Hg at \(22.5^{\circ} \mathrm{C} .\) What is the molecular formula of the compound?

Short answer

The molecular formula is \( \text{C}_2 \text{H}_2 \text{F}_4 \).

Step-by-step solution

Convert Pressure to Atmospheres

Convert the given pressure from mm Hg to atmospheres using the conversion factor 1 atm = 760 mm Hg. \( P = \frac{13.7 \text{ mm Hg}}{760} = 0.01803 \text{ atm} \).

Convert Volume to Liters

Convert the volume from mL to L. Since 1000 mL = 1 L, \( V = \frac{165 \text{ mL}}{1000} = 0.165 \text{ L} \).

Convert Temperature to Kelvin

Convert the temperature from Celsius to Kelvin using the formula \( T = 22.5 + 273.15 = 295.65 \text{ K} \).

Apply the Ideal Gas Law to Find Moles

Use the ideal gas law \( PV = nRT \) to find the number of moles (n) of the gas. Use \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \). Substitute the known values: \( (0.01803 \text{ atm})(0.165 \text{ L}) = n (0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1})(295.65 \text{ K}) \). Solving for n gives \( n = 1.235 \times 10^{-4} \text{ mol} \).

Calculate Molar Mass

Use the mass of the gas and the number of moles to calculate the molar mass. \( \text{Molar Mass} = \frac{0.0125 \text{ g}}{1.235 \times 10^{-4} \text{ mol}} \approx 101.2 \text{ g/mol} \).

Determine Molecular Formula

The empirical formula \( \text{CHF}_2 \) has an empirical formula mass: \( 12.01 + 1.01 + 2(19.00) = 50.02 \text{ g/mol} \). The ratio of molar mass to empirical formula mass is \( \frac{101.2}{50.02} \approx 2 \). Therefore, the molecular formula is \( \text{C}_2 \text{H}_2 \text{F}_4 \).

Key concepts

Empirical Formula

The empirical formula is a simplified representation of the relative number of each type of atom in a compound. It shows the smallest whole number ratio of the elements present in the compound. In the exercise you were given, the empirical formula provided was \( \text{CHF}_2 \). This means that in the simplest form, each molecule of the compound consists of one carbon (C) atom, one hydrogen (H) atom, and two fluorine (F) atoms.

Empirical formulas are incredibly useful, as they provide a basic understanding of the composition of a compound without providing the exact number of atoms. For example, the empirical formula for glucose is \( \text{CH}_2\text{O} \), indicating that the ratio of carbon, hydrogen, and oxygen is 1:2:1.

• Empirical formulas do not reflect the actual amount but represent the smallest ratio.
• Knowing the empirical formula is key when transitioning to the molecular formula, as seen in the exercise.

Molecular Formula

A molecular formula provides a detailed view of the actual number of atoms in a single molecule of a compound. It is essentially an expanded version of the empirical formula, giving more specific information.

In the exercise, the empirical formula was \( \text{CHF}_2 \), but through calculations, the molecular formula \( \text{C}_2 \text{H}_2 \text{F}_4 \) was determined. This illustrates that while the elemental composition remains the same, the number of each element in a molecule can increase proportionally.

How to Determine the Molecular Formula:

• Find the molar mass of the compound using the Ideal Gas Law, or another method.
• Calculate the empirical formula mass.
• Divide the molar mass by the empirical formula mass to get a multiple. In this case, it was approximately 2 (\( \frac{101.2}{50.02} \approx 2 \)).
• Multiply the subscripts in the empirical formula by this multiple to get the molecular formula.

This example solidifies understanding by providing insight into the relationship between the two formula types.

Gas Laws

Gas laws describe how gases behave in different conditions, such as changes in temperature, pressure, and volume. In the exercise, the Ideal Gas Law was used to solve the problem, which is one of the fundamental gas laws.

The Ideal Gas Law formula is:\[PV = nRT\]Where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume of the gas.
• \( n \) is the number of moles of the gas.
• \( R \) is the gas constant (0.0821 L atm K\(^{-1}\) mol\(^{-1}\)).
• \( T \) is the temperature in Kelvin.

In our exercise, we used this law to find the number of moles \( n \), which was a crucial step in determining the molecular formula. It started by converting all given units to the ones suitable for the Ideal Gas Law, allowing precise calculations.

Key Points about Gas Laws:

• They help predict how a change in one condition affects others in a gas system.
• The Ideal Gas Law is used extensively in chemistry when dealing with gases under ideal conditions.
• Understanding the conditions under which these laws apply is crucial for applying them accurately to scientific problems.