Problem 27 - General Chemistry: Principles and Modern Applications

Question

What is the volume, in milliliters, occupied by \(89.2 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) at \(37^{\circ} \mathrm{C}\) and \(737 \mathrm{mmHg} ?\)

Step-by-Step Solution

Answer

By proceeding through these 5 steps, we obtain the volume of CO2 in liters. To get the volume in milliliters, multiply by 1000. Don't forget to use significant figures.

Step 1: Finding Molar Mass of CO2

First, we need to find out the molar mass of CO2. Carbon (C) has a molar mass of 12.01 g/mol and Oxygen (O) has a molar mass of 16.00 g/mol. Given that CO2 has 1 Carbon atom and 2 Oxygen atoms, we can compute the molar mass as follows: \[ MolarMass_{CO2} = 12.01\ g/mol + 2 * 16.00\ g/mol = 44.01\ g/mol \]

Step 2: Converting grams to moles

Next, we convert the mass of CO2 in grams to moles. We need to divide the given mass by the molar mass: \[n =\frac{89.2\ g\ CO2}{44.01\ g/mol} \]

Step 3: Converting Temperature to Kelvin

The ideal gas law requires the temperature to be in Kelvin. To convert from Celsius to Kelvin, we add 273.15: \[ T = 37^{\circ}C + 273.15 = 310.15\ K \]

Step 4: Converting pressure to atm

The pressure needs to be in atmospheres (atm) for the ideal gas law. To convert from mmHg to atm, we use the conversion factor 1 atm = 760 mmHg: \[ P = \frac{737\ mmHg}{760\ mmHg/atm} \]

Step 5: Substitute and Solve

We have the following values ready: number of moles (n), temperature (T), pressure (P), and ideal gas constant (R=0.0821 L.atm/mol.K). Now, inserting these values into the ideal gas law equation PV=nRT, we solve for V: \[ V = \frac{nRT}{P} \]

Key Concepts

Molar Mass Calculation

Understanding molar mass is crucial when working with gases. In this exercise, we calculate the molar mass of carbon dioxide (CO₂). Molar mass represents the mass of one mole of a substance. To find it for CO₂, we add the molar masses of all atoms in the molecule:

• Carbon (C) has a molar mass of 12.01 g/mol.
• Each Oxygen (O) atom has a molar mass of 16.00 g/mol.

Because CO₂ contains one carbon atom and two oxygen atoms, the molar mass is calculated as follows:

\[ \text{MolarMass}_{CO_2} = 12.01\ \, \text{g/mol} + 2 \times 16.00\ \, \text{g/mol} = 44.01\ \, \text{g/mol} \]
Calculating molar mass allows us to understand how many moles are present in a given mass, which is essential for working with the Ideal Gas Law.

Temperature Conversion

When dealing with gas laws, temperature must be in Kelvin. This is because the Kelvin scale starts at absolute zero, ensuring that all temperatures are positive values, which is required for the math to work out.

To convert Celsius to Kelvin, the formula is straightforward. You simply add 273.15 to the Celsius temperature:

\[ T = 37^{\circ}C + 273.15 = 310.15\ \, K \]
Starting with Celsius ensures we have a baseline understood by most, but transitioning to Kelvin makes equations consistent and reliable in the context of the Ideal Gas Law.

Pressure Conversion

Pressure conversion is an essential step in using the Ideal Gas Law. The standard unit for pressure in this context is atmospheres (atm), although it is often recorded in other units like mmHg.

To convert from mmHg to atm, we divide the pressure by the conversion factor, which is 760 mmHg per atm:

\[ P = \frac{737\ \, mmHg}{760\ \, mmHg/atm} \]
This conversion ensures that all the units are compatible when inserted into the Ideal Gas Law equation, allowing for accurate volume calculations.

Volume Calculation

Volume calculation using the Ideal Gas Law is the final step after converting all other units appropriately. The Ideal Gas Law is expressed as \( PV = nRT \), where:

• \( P \) is the pressure in atm.
• \( V \) is the volume in liters.
• \( n \) is the number of moles of gas.
• \( R \) is the ideal gas constant, \( 0.0821\ \, L.atm/mol.K \).
• \( T \) is the temperature in Kelvin.

To find volume \( V \), we rearrange the equation:

\[ V = \frac{nRT}{P} \]
The volume can then be converted to milliliters by multiplying by 1000, offering the solution in a more practical unit for many applications.