Problem 6 - Chemistry and Chemical Reactivity

Question

Suppose you find the CO concentration in your home is \(10 .\) ppm by volume at 1.00 atm pressure and \(25^{\circ} \mathrm{C} .\) What is the CO concentration in \(\mathrm{mg} / \mathrm{L}\) and in ppm by mass. (The average molar mass for dry air is \(28.96 \mathrm{g} / \mathrm{mol}\) at 1.00 atm pressure and \(\left.25^{\circ} \mathrm{C} .\right)\)

Step-by-Step Solution

Answer

The CO concentration is 0.0114 mg/L and 9.63 ppm by mass.

Step 1: Convert ppm by Volume to Mole Fraction

To find the mole fraction of CO in the air, remember that 10 ppm by volume means there are 10 parts of CO per 1,000,000 parts of air. Therefore, the mole fraction of CO, denoted as \( x_{CO} \), is \( x_{CO} = \frac{10}{1,000,000} = 1 \times 10^{-5} \).

Step 2: Calculate the Molar Mass of CO

The molar mass of carbon (C) is approximately 12.01 g/mol, and the molar mass of oxygen (O) is approximately 16.00 g/mol. Thus, the molar mass of carbon monoxide (CO) is \( M_{CO} = 12.01 + 16.00 = 28.01 \) g/mol.

Step 3: Use Ideal Gas Law to Find CO Concentration in mg/L

At standard conditions of 1 atm and 25°C, we use the ideal gas law \( PV = nRT \) to find the molar concentration. Use \( P = 1 \text{ atm} \), \( R = 0.0821 \text{ L atm mol}^{-1} \text{K}^{-1} \), \( T = 298 \text{ K} \). From the ideal gas law, the concentration in moles/L is \( c = \frac{P}{RT} = \frac{1}{0.0821 \times 298} = 0.0408 \text{ mol/L} \). Now, multiply this concentration by the mole fraction of CO to find its concentration in moles/L: \( c_{CO} = 0.0408 \times 1 \times 10^{-5} = 4.08 \times 10^{-7} \text{ mol/L} \). To convert this to mg/L, multiply by the molar mass of CO: \( \text{Concentration in mg/L} = 4.08 \times 10^{-7} \times 28.01 \times 1000 = 0.0114 \text{ mg/L} \).

Step 4: Find the Density of Dry Air

First, calculate the density of air using the ideal gas law. The molar volume of an ideal gas at 25°C and 1 atm is \( V_m = \frac{RT}{P} = \frac{0.0821 \times 298}{1} = 24.46 \text{ L/mol} \). The density of air is then \( \rho_{\text{air}} = \frac{M_{\text{air}}}{V_m} = \frac{28.96}{24.46} = 1.184 \text{ g/L} \).

Step 5: Calculate ppm by Mass of CO in Air

To find ppm by mass, first calculate the mass of CO per volume: 0.0114 mg/L. Then compare it to the mass of the air in the same volume: 1.184 g/L. The ppm by mass is \( \text{ppm by mass} = \frac{\text{mass of CO}}{\text{mass of air}} \times 10^6 = \frac{0.0114}{1184} \times 10^6 = 9.63 \text{ ppm} \).

Key Concepts

Mole Fraction

The concept of mole fraction is a way to express the concentration of a component in a mixture, such as gases in a room. It's simple to understand and calculate because it deals with the ratio of moles of a particular gas to the total moles of all gases in the mixture. For example, if we talk about the mole fraction of carbon monoxide (CO) in air, we denote it as \( x_{CO} \). This fraction is calculated based on the number of moles of CO compared to the total moles within the air.

In our exercise, we were given a concentration of 10 ppm by volume, which translates to having 10 parts of CO in a million parts of air. Thus, the mole fraction of CO is calculated as \( x_{CO} = \frac{10}{1,000,000} = 1 \times 10^{-5} \). This system helps us understand the proportion of CO in the air without needing to convert to complex units.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry linking the pressure, volume, and temperature of a gas to its number of moles. The formula is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is temperature. This equation allows us to calculate the unknown parameter if others are known.

In the context of our problem, we could use the ideal gas law to determine the molar concentration \( c \), which is the number of moles per liter. By rearranging to \( c = \frac{P}{RT} \), we compute the concentration at standard conditions (1 atm and 25°C) as \( 0.0408 \text{ mol/L} \). This step is pivotal because it translates the mole fraction of CO into an actual concentration value. Understanding this law enables us to predict the behavior and characteristics of gases under various conditions.

PPM Conversion

PPM, or parts per million, is a unit of concentration often used for measuring the amount of one substance in another. It is quite helpful for very dilute concentrations, like with gases in the air. The conversion from ppm by volume to ppm by mass involves comparing the actual mass of a gas to the surrounding medium.

In this problem, after finding the concentration of CO in mg/L (0.0114 mg/L), we compare it to the density of dry air, calculated using its molar mass. Dry air's molar mass is \( 28.96 \text{ g/mol} \). We find the density as \( 1.184 \text{ g/L} \) and express the CO concentration relative to air's mass, which results in a ppm by mass value. The ppm by mass for CO turns out to be approximately 9.63 ppm. This comparison gives a sense of how much CO is present relative to the whole air volume, which is important for assessing air quality.

Molar Mass Calculation

The molar mass of a compound is the mass of one mole of its molecules and is vital for converting between moles and grams. Calculating the molar mass involves summing the masses of all atoms in the molecule.

For carbon monoxide (CO), we sum the molar masses of carbon (C) and oxygen (O), which are about 12.01 g/mol and 16.00 g/mol, respectively. Therefore, the molar mass of CO is calculated as \( M_{CO} = 12.01 + 16.00 = 28.01 \text{ g/mol} \). This value enables us to convert the moles of CO determined from using the ideal gas law into a mass concentration in mg/L. Understanding how to calculate molar mass is crucial whenever you're converting between concentrations in chemical reactions or understanding how much of a substance is present in any given situation.