Problem 110 - Chemistry The Central Science

Question

Assume that an exhaled breath of air consists of \(74.8 \% \mathrm{~N}_{2}\), \(15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2},\) and \(6.2 \%\) water vapor. (a) If the total pressure of the gases is 0.985 atm, calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is \(455 \mathrm{~mL}\) and its temperature is \(37^{\circ} \mathrm{C},\) calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled. (c) How many grams of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would need to be metabolized to produce this quantity of \(\mathrm{CO}_{2}\) ? (The chemical reaction is the same as that for combustion of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} .\) See Section 3.2 and Problem \(\left.10.59 .\right)\)

Step-by-Step Solution

Answer

(a) Using Dalton's law of partial pressures: \(P_{N_{2}} = 0.748 × 0.985 \mathrm{~atm} = 0.736 \mathrm{~atm}\), \(P_{O_{2}} = 0.153 × 0.985 \mathrm{~atm} = 0.151 \mathrm{~atm}\), \(P_{CO_{2}} = 0.037 × 0.985 \mathrm{~atm} = 0.0364 \mathrm{~atm}\), and \(P_{H_{2}O} = 0.062 × 0.985 \mathrm{~atm} = 0.061 \mathrm{~atm}\). (b) Using the ideal gas law: \(n = \frac{PV}{RT} = \frac{0.0364 \mathrm{~atm} \times 0.455 \mathrm{~L}}{0.0821 \mathrm{~L·atm/mol·K} \times 310.15 \mathrm{~K}} = 5.40 \times 10^{-3} \ moles \ CO_{2}\). (c) Moles of glucose: \(\frac{5.40 \times 10^{-3} \ moles \ CO_{2}}{6} = 9.00 \times 10^{-4} \ moles \ C_{6}H_{12}O_{6}\), grams of glucose: \(9.00 \times 10^{-4} \ moles \ C_{6}H_{12}O_{6} \times 180.16 \ g/mol = 0.162 \ g \ C_{6}H_{12}O_{6}\).

Step 1: Calculate partial pressures of each gas

Using the given percentages for each gas and Dalton's law of partial pressures, we can find the partial pressure of each gas component: \(P_{total} = P_{N_{2}} + P_{O_{2}} + P_{CO_{2}} + P_{H_{2}O}\) Given, total pressure: \(P_{total} = 0.985 \mathrm{~atm}\) Calculate the partial pressure of each component by multiplying the total pressure by its percentage: \(P_{N_{2}} = 0.748 × P_{total}\) \(P_{O_{2}} = 0.153 × P_{total}\) \(P_{CO_{2}} = 0.037 × P_{total}\) \(P_{H_{2}O} = 0.062 × P_{total}\)

Step 2: Determine the number of moles of CO2 exhaled

Using the ideal gas law, we will express the number of moles of CO2 as follows: \(PV = nRT\) where \(P\) is the pressure of the gas, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant and \(T\) is the temperature in Kelvin. We can use the partial pressure of CO2 calculated in Step 1 and the given volume of the exhaled gas (0.455 L) and temperature (37°C = 310.15 K). The ideal gas constant, \(R\), has a value of 0.0821 L·atm/mol·K. Now, rearrange the formula to solve for the number of moles of CO2: \(n = \frac{PV}{RT}\)

Step 3: Calculate the grams of glucose needed to produce the CO2

Using the balanced chemical equation for glucose metabolism (C6H12O6 + 6O2 -> 6CO2 + 6H2O), we can find the amount of glucose needed to produce the moles of CO2 calculated in Step 2. 1 mole of glucose (C6H12O6) produces 6 moles of CO2: \(1 \ mole \ C_{6}H_{12}O_{6} \ → \ 6 \ moles \ CO_{2}\) Now, use the stoichiometry of the balanced chemical equation to find the moles of glucose required to produce the moles of CO2 exhaled: \[Moles \ of \ Glucose = \frac{Moles \ of \ CO_{2} \ exhaled}{6}\] Finally, convert moles of glucose to grams using the molar mass of glucose (\(C_{6}H_{12}O_{6} = 180.16 \ g/mol\)): \[Grams \ of \ glucose = Moles \ of \ glucose × Molar \ mass \ of \ glucose\]

Key Concepts

Partial Pressure Calculations

When dealing with a mixture of gases like exhaled air, each component gas exerts a portion of the total pressure. This is known as its partial pressure. Dalton’s Law of Partial Pressures helps us find the partial pressure of each gas in the mixture. It states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas.

• Total Pressure, \( P_{\text{total}} = P_{N_{2}} + P_{O_{2}} + P_{CO_{2}} + P_{H_{2}O} \)
• For each gas, calculate its partial pressure by multiplying the total pressure by the fraction (or percentage as a decimal) of that gas.
• For example, if the total pressure \( P_{\text{total}} = 0.985 \text{ atm} \), for nitrogen (\( N_2 \)), use \( P_{N_{2}} = 0.748 \times 0.985 \text{ atm} \).

This calculation shows how much of the total pressure each gas contributes, providing insight into the behavior of gas mixtures.

Stoichiometry

Stoichiometry involves calculating relationships between the amounts of reactants and products in a chemical reaction. In this case, it's used to determine how much glucose is needed to produce the amount of \( CO_{2} \) exhaled.

• The complete combustion of glucose \(( C_{6}H_{12}O_{6} )\) is described by the equation:
  \[ C_{6}H_{12}O_{6} + 6O_{2} \rightarrow 6CO_{2} + 6H_{2}O \]
• This equation tells us that 1 mole of glucose produces 6 moles of \( CO_{2} \).
• You calculate the moles of glucose required by dividing the number of moles of \( CO_{2} \) by 6.
• Finally, convert this to grams using the molar mass of glucose \((180.16 \text{ g/mol})\).

Stoichiometry is essential for translating the atomic world into the measurable quantities we observe.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation that relates the pressure, volume, temperature, and number of moles of a gas. It's represented by the equation \( PV = nRT \), where:

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

To find the number of moles of \( CO_{2} \) exhaled:

• Convert the temperature from Celsius to Kelvin.
• Use the partial pressure of \( CO_{2} \), volume of gas (in liters), and rearrange the equation: \( n = \frac{PV}{RT} \)

This allows us to solve for the number of moles, showing how gas laws can bridge theoretical concepts and practical applications.