Problem 28 - Chemistry and Chemical Reactivity

Question

A 0.0125 -g sample of a gas with an empirical formula of 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?

Step-by-Step Solution

Answer

The molecular formula is \(C_2H_2F_4\).

Step 1: Convert Pressure to Atmospheres

The gas pressure is given in mm Hg. To work with the ideal gas law, we need to convert the pressure to atmospheres (atm). We use the conversion factor: 1 atm = 760 mm Hg.\[ P = \frac{13.7 \text{ mm Hg}}{760 \text{ mm Hg/atm}} \approx 0.01803 \text{ atm} \]

Step 2: Convert Temperature to Kelvin

The temperature is given in degrees Celsius. To use the ideal gas law, we need the temperature in Kelvin. The conversion formula is: \( T(K) = T(°C) + 273.15 \).\[ T = 22.5 + 273.15 = 295.65 \text{ K} \]

Step 3: Convert Volume to Liters

The volume is provided in milliliters. We need to convert it to liters for use in the ideal gas law. There are 1000 mL in 1 L.\[ V = \frac{165}{1000} = 0.165 \text{ L} \]

Step 4: Use the Ideal Gas Law to Calculate Moles

Now use the ideal gas law equation \(PV = nRT\), where \(R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}\). Solve for \(n\), the number of moles.\[ n = \frac{PV}{RT} = \frac{0.01803 \times 0.165}{0.0821 \times 295.65} \approx 1.226 \times 10^{-4} \text{ moles} \]

Step 5: Calculate the Molar Mass of the Compound

We know the mass of the sample is 0.0125 g. Use the moles calculated in the previous step to determine the molar mass.\[ \text{Molar Mass} = \frac{\text{mass}}{\text{moles}} = \frac{0.0125}{1.226 \times 10^{-4}} \approx 101.98 \text{ g/mol} \]

Step 6: Determine the Molecular Formula

The molar mass of the empirical formula of CHF\( _{2} \) is calculated based on the atomic weights: C (12.01), H (1.01), F (19.00) x 2.\[ \text{Molar Mass of CHF}_2 = 12.01 + 1.01 + 2 \times 19.00 = 50.02 \text{ g/mol} \]Divide the molar mass calculated in Step 5 by the molar mass of the empirical formula:\[ \text{Factor} = \frac{101.98}{50.02} \approx 2 \]Thus, the molecular formula is \(C_2H_2F_4\).

Key Concepts

Ideal Gas Law

The ideal gas law is a cornerstone concept in chemistry. It correlates pressure, volume, temperature, and number of moles of a gas. The formula is expressed as \( PV = nRT \), where \( P \) denotes pressure, \( V \) represents volume, \( n \) is the number of moles, \( R \) stands for the ideal gas constant, and \( T \) indicates temperature in Kelvin.

To effectively use the ideal gas law,

• Convert all units properly. For this exercise, pressure must be in atmospheres, volume in liters, and temperature in Kelvin.
• Adjust for standard conditions with \( R \) often taken as \( 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \).

Applying these adjustments simplify the calculation of unknown variables like moles (\( n \)), when other variables are known.

Conversion Factors

Conversion factors play a critical role in chemistry to ensure consistency and correctness in unit measurements. This exercise requires conversions across several units for the ideal gas law to be applicable.

Let's explore the essential conversions used:

• Pressure: Convert from mm Hg to atm using \( 1 \text{ atm} = 760 \text{ mm Hg} \).
• Temperature: Convert from Celsius to Kelvin using \( T(K) = T(°C) + 273.15 \).
• Volume: Convert from milliliters to liters, knowing \( 1 \text{ L} = 1000 \text{ mL} \).

These conversions ensure all quantities are in standard units, allowing accurate applications in calculations using the ideal gas law.

Empirical Formula

An empirical formula represents the simplest whole-number ratio of elements in a compound. This exercise starts with CHF\( _{2} \) as the given empirical formula.

To determine the molecular formula from the empirical formula, compare:

• The molar mass obtained from experiment (in this case, around \( 101.98 \text{ g/mol} \)).
• The calculated molar mass of the empirical formula (for CHF\( _{2} \), \( 50.02 \text{ g/mol} \)).

Dividing the experimental molar mass by the empirical formula's molar mass gives a factor. This factor tells how many empirical formula units make up the molecular formula, leading to the discovery that \( C_2H_2F_4 \) is the molecular formula for this sample.

Moles Calculation

Calculating moles is pivotal in chemical equations and reactions since it quantifies substances. Utilizing the ideal gas law, moles (\( n \)) are derived using \( n = \frac{PV}{RT} \). This exercise required finding the moles of the given gas under certain conditions.

Here's a simplified look into the steps:

• Determine the values for pressure, volume, and temperature in their appropriate units.
• Apply these values into the equation along with the ideal gas constant to solve for \( n \), the number of moles.

In the exercise, solving yielded \( 1.226 \times 10^{-4} \text{ moles} \). This is used in conjunction with mass to find molar mass, aiding in molecular formula determination.