Problem 51 - Chemistry The Central Science In Si Units

Question

(a) Calculate the density of dinitrogen tetroxide gas \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\) at \(111.5 \mathrm{kPa}\) and \(0{ }^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a gas if \(2.70 \mathrm{~g}\) occupies \(0.97 \mathrm{~L}\) at \(134.7 \mathrm{~Pa}\) and \(100^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The density of \(\text{N}_2\text{O}_4\) is \(4.52 \text{ g/L}\); the molar mass of the gas is \(60940.86 \text{ g/mol}\).

Step 1: Convert Pressure and Temperature for Part (a)

First, convert the given pressure from kilopascals to pascals for consistency in units. We have \( 111.5 \text{ kPa} = 111500 \text{ Pa}\). Additionally, convert the temperature from Celsius to Kelvin: \( T = 0^\circ \text{C} + 273.15 = 273.15 \text{ K} \).

Step 2: Use the Ideal Gas Law for Density Calculation

The Ideal Gas Law \( PV = nRT \) can be rearranged to \( P = \frac{n}{V} RT \). Knowing that \( n/V \) is the molar volume and \( n = \frac{m}{M} \) (where \(m\) is mass and \(M\) is molar mass), we substitute \( n \) and rearrange to find density \( \rho = \frac{m}{V} = \frac{PM}{RT} \). Substitute \( P = 111500 \text{ Pa} \), \( R = 8.314 \text{ J/mol K} \), \( T = 273.15 \space \text{K} \), and \( M = 92.02 \text{ g/mol} \) (molar mass of \(\text{N}_2\text{O}_4\)) to find density.

Step 3: Calculate Density for Part (a)

Substituting the values gives: \( \rho = \frac{111500 \times 92.02}{8.314 \times 273.15} = 4.52 \text{ g/L} \).

Step 4: Convert Pressure and Temperature for Part (b)

For the second part, convert the pressure from pascals to kilopascals for consistency: \( 134.7 \text{ Pa} = 0.1347 \text{ kPa} \). Similarly, convert the temperature from Celsius to Kelvin: \( T = 100^\circ \text{C} + 273.15 = 373.15 \text{ K} \).

Step 5: Calculate Molar Volume at STP

Using the Ideal Gas Law \( PV = nRT \), rearrange to \( n = \frac{PV}{RT} \). Substituting \( P = 0.1347 \text{ kPa} \), \( V = 0.97 \text{ L} \), \( R = 8.314 \text{ J/mol K} \), and \( T = 373.15 \text{ K} \), calculate \( n \) to determine moles present: \( n = \frac{0.1347 \times 0.97}{8.314 \times 373.15} = 4.43 \times 10^{-5} \text{ mol} \).

Step 6: Calculate Molar Mass for Part (b)

To find the molar mass \( M \), use the relation \( M = \frac{m}{n} \) where \( m = 2.70 \text{ g} \). Thus, \( M = \frac{2.70}{4.43 \times 10^{-5}} = 60940.86 \text{ g/mol} \).

Key Concepts

Gas Density Calculation

To understand how to calculate gas density, we first need to utilize the Ideal Gas Law which is expressed as the equation \( PV = nRT \). In this equation, \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant \( 8.314 \, \text{J/mol K} \), and \( T \) is the temperature in Kelvin.
When rearranging this equation to find density, we realize that \( n/V \) is essentially the molar volume of the gas and can be rewritten in terms of molar mass \( M \) and mass \( m \) as \( n = \frac{m}{M} \). This allows us to derive the formula for density \( \rho = \frac{m}{V} = \frac{PM}{RT} \). This relation points out that density depends on pressure, temperature, and the type of gas (via molar mass).
For example, if we calculate the density of dinitrogen tetroxide \( \text{N}_2\text{O}_4 \) at \( 111500 \, \text{Pa} \) and \( 273.15 \, \text{K} \), substituting \( P = 111500 \, \text{Pa} \), \( T = 273.15 \, \text{K} \), and \( M = 92.02 \, \text{g/mol} \), the resulting density is approximately \( 4.52 \, \text{g/L} \). This demonstrates how alterations in conditions affect gaseous substances.

Molar Mass Determination

Determining the molar mass of a gas involves using the Ideal Gas Law in conjunction with experimental measurements. Molar mass aids us in linking the mass of a gas to the amount of substance present. By employing the relationship \( M = \frac{m}{n} \), where \( m \) is the sample mass and \( n \) is the number of moles, we can effectively find the molar mass.
In a scenario where \( 2.70 \, \text{g} \) of gas occupies \( 0.97 \, \text{L} \) at \( 134.7 \, \text{Pa} \) and \( 373.15 \, \text{K} \), the moles \( n \) is calculated using \( n = \frac{PV}{RT} \). Here, \( P = 0.1347 \, \text{kPa} \), \( V = 0.97 \, \text{L} \), and \( R = 8.314 \, \text{J/mol K} \), resulting in \( n = 4.43 \times 10^{-5} \, \text{mol} \). The molar mass \( M \) is then \( \frac{2.70}{4.43 \times 10^{-5}} = 60940.86 \, \text{g/mol} \), illustrating the procedure of using mass and volume data to deduce molar mass.

Unit Conversion in Chemistry

Unit conversion is pivotal in chemistry, particularly when utilizing equations like the Ideal Gas Law. Standard consistency in units allows for accurate calculations and avoids errors.

• Convert pressure: If pressure is given in kilopascals (kPa), convert to pascals (Pa) by multiplying by 1000. For example, \( 111.5 \, \text{kPa} = 111500 \, \text{Pa} \).
• Convert temperature: Always use Kelvin when substituting into gas laws. To change degrees Celsius to Kelvin, add 273.15. For instance, \( 0^\circ \text{C} \) becomes \( 273.15 \, \text{K} \).

These conversions ensure that each component of the gas law fits together correctly.
Practicing unit conversions prepares students to handle diverse chemical problems effectively, be it in lab settings or theoretical equations.