Problem 108 - 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 \(0.985 \mathrm{~atm}\), 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 10.57.)

Step-by-Step Solution

Answer

\(P_{N2} = 0.985 \mathrm{~atm} \times 0.748 = 0.737 \mathrm{~atm}\) \(P_{O2} = 0.985 \mathrm{~atm} \times 0.153 = 0.151 \mathrm{~atm}\) \(P_{CO2} = 0.985 \mathrm{~atm} \times 0.037 = 0.036 \mathrm{~atm}\) \(P_{H2O} = 0.985 \mathrm{~atm} \times 0.062 = 0.061 \mathrm{~atm}\) \(n_{CO2} = \frac{0.036 \mathrm{~atm} \times 0.455 \mathrm{L}}{0.0821\mathrm{~L\cdot{atm}}\mathrm{/(\mathsf{mol\cdot{K}})} \times 310.15 \mathrm{K}} = 6.03 \times 10^{-4}\; \mathrm{mol}\) Moles of glucose: \(\frac{6.03 \times 10^{-4}\; \mathrm{mol \; CO2}}{6} = 1.01 \times 10^{-4} \; \mathrm{mol \; glucose}\) Grams of glucose: \(1.01 \times 10^{-4} \; \mathrm{mol \; glucose} \times 180 \frac{\mathrm{g}}{\mathrm{mol}} = 1.81 \times 10^{-2} \; \mathrm{g}\)

Step 1: Calculate the partial pressure of each component

Use Dalton's law of partial pressures to calculate the partial pressure of each component in the exhaled breath, multiplying the total pressure (0.985 atm) by the percentage composition of each gas. For nitrogen (N2): \(P_{N2} = 0.985 \mathrm{~atm} \times 0.748\) For oxygen (O2): \(P_{O2} = 0.985 \mathrm{~atm} \times 0.153\) For carbon dioxide (CO2): \(P_{CO2} = 0.985 \mathrm{~atm} \times 0.037\) For water vapor: \(P_{H2O} = 0.985 \mathrm{~atm} \times 0.062\) Now, calculate the partial pressures for each gas.

Step 2: Calculate the number of moles of CO2 exhaled

Use the ideal gas law, which states that PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, convert the temperature from Celsius to Kelvin: \(T = 37^{\circ} \mathrm{C} + 273.15 = 310.15 \mathrm{K}\) Now, use the ideal gas law to find the number of moles of CO2 exhaled: \(n_{CO2} = \frac{P_{CO2} \times V}{R \times T}\) We have already found \(P_{CO2}\) in step 1, and the volume (V) is given as 455 mL. Make sure to convert the volume to liters to ensure consistent units: \(V = 455 \times 10^{-3} \mathrm{L} = 0.455 \mathrm{L}\) The gas constant (R) is 0.0821 L atm/(mol K). Calculate the number of moles of CO2 exhaled.

Step 3: Calculate the amount of glucose metabolized to produce CO2

To find the amount of glucose metabolized to produce the CO2, we need to know the stoichiometric coefficients in the balanced combustion reaction for glucose: \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} + 6 \mathrm{O}_{2} \rightarrow 6 \mathrm{CO}_{2} + 6 \mathrm{H}_{2}\mathrm{O}\) For every mole of glucose, 6 moles of CO2 are produced. So, the relation between glucose and CO2 is 1:6. Find the moles of glucose required by dividing the moles of CO2 obtained in step 2 by 6. Now, we need to calculate the grams of glucose required. The molar mass of glucose is: \(\textrm{C}_{6}\textrm{H}_{12}\textrm{O}_{6} = 6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \textrm{ g/mol}\) Finally, multiply the moles of glucose by the molar mass to obtain the grams of glucose metabolized to produce the given amount of CO2 in the exhaled breath.

Key Concepts

Understanding the Ideal Gas Law

The ideal gas law is a fundamental principle in chemistry that helps us understand how gases behave under different conditions. It is represented by the equation \( PV = nRT \). Here, \( P \) stands for pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. This law combines several gas laws and simplifies calculations related to gases.
To solve problems using the ideal gas law, follow these steps:

• Ensure all units are compatible. Common units include atmospheres for pressure, liters for volume, and Kelvin for temperature.
• Use the ideal gas constant suitable for the units, often \( R = 0.0821 \) L atm/mol K.
• Rearrange the formula to find the unknown quantity, such as moles (\( n \)) or volume (\( V \)).

Understanding how to manipulate the variables is crucial for solving any gas-related problem.

Exploring Stoichiometry

Stoichiometry is the study of quantitative relationships in chemical reactions. It’s fundamental for determining how much of each substance is involved in a reaction. It uses balanced chemical equations to show these relationships.
For instance, in the combustion reaction of glucose: \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} + 6 \mathrm{O}_{2} \rightarrow 6 \mathrm{CO}_{2} + 6 \mathrm{H}_{2}\mathrm{O} \), stoichiometry tells us that one molecule of glucose produces six molecules of \( \mathrm{CO}_{2} \).

• First, ensure the chemical equation is balanced.
• Use coefficients to convert between moles of reactants and products.
• Utilize these relationships to calculate unknown quantities, such as the amount of a reactant needed or product formed.

By understanding stoichiometry, you can predict the outcomes of chemical reactions and calculate quantities needed or produced.

Introduction to Combustion Reactions

Combustion reactions are chemical processes where substances react with oxygen to release energy. These reactions are exothermic, as they release heat and light. Combustion is crucial in everyday life, from burning fuels to driving engines.
The reaction of glucose with oxygen is a classic combustion reaction: \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} + 6 \mathrm{O}_{2} \rightarrow 6 \mathrm{CO}_{2} + 6 \mathrm{H}_{2}\mathrm{O} \). In this reaction:

• Glucose breaks down to form carbon dioxide and water.
• Oxygen acts as an oxidizing agent.
• The reaction releases energy, vital for biological processes in organisms.

Understanding combustion helps us appreciate how energy is generated and consumed, informing industries and environmental management.

Molar Mass Calculation Made Easy

Calculating molar mass is essential for translating between the atomic scale and macroscopic quantities. Molar mass is the mass of one mole of a substance and is usually expressed in g/mol.
To find the molar mass of a compound like glucose \( (\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}) \):

• List the elements and their quantities: Carbon (C), Hydrogen (H), and Oxygen (O).
• Multiply the number of atoms of each element by their respective atomic masses (C = 12 g/mol, H = 1 g/mol, O = 16 g/mol).
• Add the results to get the overall molar mass: \( 6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \) g/mol.

Understanding how to calculate molar mass is pivotal for converting between moles and grams, crucial for laboratory work and real-world applications alike.