Question

A compound consists of \(37.5 \% \mathrm{C}, 3.15 \% \mathrm{H},\) and \(59.3 \%\) \(\mathrm{F}\) by mass. When \(0.298 \mathrm{~g}\) of the compound is heated to 50\. \({ }^{\circ} \mathrm{C}\) in an evacuated \(125-\mathrm{mL}\) flask, the pressure is observed to be \(750 . \mathrm{mmHg}\). The compound has three isomers. (a) Calculate the molar mass of the compound. (b) Determine the empirical and molecular formulas of the compound. (c) Draw the Lewis structure for each isomer of the compound.

Short answer

The molar mass is 63.52 g/mol, the empirical formula is CHF, and the molecular formula is \(\text{C}_2\text{H}_2\text{F}_2\). The compound has three isomers.

Step-by-step solution

Calculate moles of gas

First, let's find the number of moles of the compound using the ideal gas law, given the pressure, volume, and temperature.The ideal gas law is \( PV = nRT \), where:- \( P \) is the pressure in atmospheres. Convert \(750 \, \text{mmHg}\) to \text{atm}: \(750 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} = 0.9868 \, \text{atm}\).- \( V \) is the volume in liters: \(125 \, \text{mL} = 0.125 \, \text{L}\).- \( R \) is the ideal gas constant \(0.0821 \, \frac{L \, atm}{mol \, K}\).- \( T \) is the temperature in Kelvin: \(50^{\circ}C + 273.15 = 323.15 \, \text{K}\).Putting the values into the equation: \[ n = \frac{PV}{RT} = \frac{0.9868 \, \text{atm} \times 0.125 \, \text{L}}{0.0821 \, \frac{L \, atm}{mol \, K} \times 323.15 \, \text{K}} \approx 0.00469 \, \text{mol} \].

Calculate Molar Mass

Using the moles of the compound and the mass of the sample, we can now calculate its molar mass:\[ \text{Molar Mass} = \frac{\text{mass}}{\text{moles}} = \frac{0.298 \, g}{0.00469 \, mol} \approx 63.52 \, g/mol \].

Determine Empirical Formula

Find moles of each element in 100 g of the compound.- For carbon: \( \frac{37.5}{12.01} = 3.12 \, mol \)- For hydrogen: \( \frac{3.15}{1.01} = 3.12 \, mol \)- For fluorine: \( \frac{59.3}{19.00} = 3.12 \, mol \)The mole ratio is approximately 1:1:1, making the empirical formula \( \text{CHF} \).

Calculate Molecular Formula

Combine the empirical formula with the experimental molar mass to find the molecular formula.The empirical formula mass of \( \text{CHF} \) is \(12.01 + 1.01 + 19.00 = 32.02 \, g/mol \).The molecular formula is a multiple of the empirical formula. The ratio of molar mass to empirical formula mass is \( \frac{63.52}{32.02} \) which is approximately 2.Thus, the molecular formula is \( \text{C}_2\text{H}_2\text{F}_2 \).

Draw Lewis Structures of Isomers

The compound \( \text{C}_2\text{H}_2\text{F}_2 \) can have different arrangements leading to isomers. There are three possible isomers:1. **1,1-Difluoroethene**: Fluorines both attached to the same carbon.2. **1,2-Difluoroethene (cis and trans)**: Fluorines attached to adjacent carbons with potential cis or trans configuration.For each, - Mirror the distribution of electrons around the C-C bonds.- Ensure proper distribution of the other atoms (H, F) around these carbons. - Consider spatial orientation for the cis-trans isomers.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a key concept used in chemistry to relate the pressure, volume, temperature, and number of moles of a gas in a container. This relationship is represented by the equation \( PV = nRT \), where:

• \( P \) is the pressure of the gas, measured in atmospheres (atm).
• \( V \) is the volume of the gas in liters (L).
• \( n \) is the number of moles of gas.
• \( R \) is the ideal gas constant, \( 0.0821 \, \frac{L \, atm}{mol \, K} \).
• \( T \) is the temperature in Kelvin (K), which can be converted from Celsius by adding 273.15.

To find the number of moles \( n \) in a given situation, you rearrange the equation to \( n = \frac{PV}{RT} \). This is particularly useful when you have measurements of pressure, volume, and temperature, as seen in the exercise where the gas law helped determine the molar mass.

Empirical Formula Calculation

The empirical formula of a compound tells you the simplest whole-number ratio of the atoms present. To calculate this, you need to:

• Convert the mass percentages of each element in the compound to moles by dividing by the atomic mass of each element.
• For example, take the percentage mass of carbon, hydrogen, and fluorine and convert them to moles.
• Find a mole ratio by dividing each mole value by the smallest number of moles calculated.
• Simplify the ratio to the smallest whole numbers to find the empirical formula.

In our exercise, the calculations showed a 1:1:1 ratio of carbon to hydrogen to fluorine, thus giving the empirical formula \( \text{CHF} \). Calculating empirical formulas is foundational for understanding the composition of compounds.

Isomer Structures

Isomers are compounds that have the same molecular formula but different structures. This means they have the same number of each type of atom but arranged differently in space. The concept of isomers introduces:

• Structural Diversity: Different physical and chemical properties arise from the same formula but different structures.
• Types of Isomers: In this context, it specifically looks at structural isomers (same atoms connected differently) and geometric isomers (different spatial arrangements of groups due to restricted rotation, such as cis-trans).

In the exercise, three isomers were possible for \( \text{C}_2\text{H}_2\text{F}_2 \). Recognizing isomers also involves understanding that the spatial arrangement can significantly alter the properties of a molecule.

Lewis Structure Drawing

Lewis structures are visual representations that show how atoms are connected in a molecule and where electrons are located around each atom. To draw a Lewis structure, follow these steps:

• Determine the total number of valence electrons available. For the compound \( \text{C}_2\text{H}_2\text{F}_2 \), calculate using the valence electrons of C, H, and F.
• Arrange the skeleton structure, ensuring each atom bonds in appropriate valence conditions.
• Distribute electrons to satisfy the octet rule (or duet for hydrogen), placing lone pairs around atoms as needed.
• For multiple isomers, explore variations in electron distribution and bonding arrangements.

In this exercise, drawing distinct structures such as 1,1-difluoroethene and 1,2-difluoroethene requires understanding how different placements of atoms and electrons create unique isomers.