Problem 100 - 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 \(0.980 \mathrm{~atm}\), 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.)

Step-by-Step Solution

Answer

The partial pressures of the gas mixture are \(P_{N_2} = 0.733 \, atm, P_{O_2} = 0.150 \, atm, P_{CO_2} = 0.036 \, atm\), and \(P_{H_2O} = 0.061 \, atm\). The number of moles of \(CO_2\) exhaled is \(0.000644 \, mol\), and the mass of glucose that would need to be metabolized to produce this quantity of \(CO_2\) is \(0.0193 \, g\).

Step 1: Apply Dalton's Law of Partial Pressures

According to Dalton's Law, the total pressure of the gas mixture is equal to the sum of the partial pressures of all the components present in the mixture. It can be represented as: \[P_{total} = P_{N_2} + P_{O_2} + P_{CO_2} + P_{H_2O}\] Here, \(P_{total}\) is the total pressure of the gases, and \(P_{N_2}, P_{O_2}, P_{CO_2}, P_{H_2O}\) are the partial pressures of the individual components. Since we know the total pressure and percentages of each component, we can calculate the partial pressures.

Step 2: Calculate partial pressures

Given that the total pressure is \(0.980 \, atm\), we can now find the partial pressures of each component: \[P_{N_2} = 0.980 \, atm \times 74.8\% = 0.980 \, atm \times 0.748 = 0.733 \, atm\] \[P_{O_2} = 0.980 \, atm \times 15.3\% = 0.980 \, atm \times 0.153 = 0.150 \, atm\] \[P_{CO_2} = 0.980 \, atm \times 3.7\% = 0.980 \, atm \times 0.037 = 0.036 \, atm\] \[P_{H_2O} = 0.980 \, atm \times 6.2\% = 0.980 \, atm \times 0.062 = 0.061 \, atm\] (b) Calculate the number of moles of CO₂ exhaled

Step 3: Use the ideal gas law

To find the number of moles of CO₂, we can use the ideal gas law equation: \[PV=nRT,\] where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles of the gas, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin. First, let's convert the temperature to Kelvin: \[37^{\circ} C = 37 + 273.15 = 310.15 K\] Next, we can use the ideal gas law equation to find the number of moles of CO₂ in the exhaled gas sample: \[n_{CO_2} = \frac{P_{CO_2}V}{RT}\] \[n_{CO_2} = \frac{0.036 \, atm \, \times 0.455 \, L}{0.0821 \, \frac{L \, atm}{mol \, K} \, \times 310.15 \, K}\] \[n_{CO_2} = \frac{0.01638 \, atm \, L}{25.44 \, \frac{L \, atm}{mol}}\] \[n_{CO_2} = 0.000644 \, mol\] (c) Calculate the mass of glucose required

Step 4: Use stoichiometry

The glucose undergoes combustion to produce CO₂ according to the following equation: \[C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 +6H_2O.\] By using stoichiometry, we can calculate the mass of glucose required to produce the CO₂ present in the exhaled air sample: \[1 \, mol \, of \, C_6H_{12}O_6 \, \rightarrow 6 \, mol \, of \, CO_{2}.\] Hence, we can calculate the mass of glucose corresponding to the \(0.000644 \, mol\) of CO₂: \[mass \, of \, glucose = \frac{0.000644 \, mol \, of \, CO_{2}}{6 \, mol \, CO_{2}} \, \times 1 \, mol \, C_6H_{12}O_6 \, \times 180.18 \, g/mol\] \[mass \, of \, glucose = 0.000644 \times 30.03 \, g\] \[mass \, of \, glucose = 0.0193 \, g.\]

Key Concepts

Dalton's law of partial pressures

Dalton's Law of Partial Pressures is a key principle in chemistry, especially when dealing with gas mixtures. It provides a straightforward way to understand how different gases behave when mixed. According to this law, the total pressure of a gas mixture is the sum of the partial pressures of each component gas.

This means that each gas in a mixture contributes to the total pressure as if it were alone in the same volume. Mathematically, this can be expressed as:\[P_{\text{total}} = P_{A} + P_{B} + P_{C} + \ldots \]
where \(P_{A}\), \(P_{B}\), and \(P_{C}\) are the partial pressures of components A, B, and C, respectively. This concept is particularly helpful for calculating partial pressures when you know the percentage composition of each gas and the total pressure, as demonstrated in the solution.

Just multiply the total pressure by the fraction (or percentage) of each individual gas to find their respective partial pressures, as shown:\[P_{\text{gas}} = P_{\text{total}} \times \text{fraction of gas}\]

stoichiometry

Stoichiometry is an essential concept in chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the principle of the conservation of mass, where the amount of reactants equals the amount of products.

This exercise demonstrates stoichiometry by using the chemical equation for glucose combustion:\[C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O\]

Through stoichiometry, you can determine how much of a reactant is needed or how much product can be formed from a specific amount of another substance. It's like a recipe in cooking, where precise amounts of ingredients make a specific dish.

To calculate the amount of glucose needed to produce a certain amount of \(CO_2\), you take the ratio from the balanced equation. For every mole of glucose, six moles of \(CO_2\) are produced. This relationship helps you determine the grams of glucose required based on the exhaled \(CO_2\), using the molar mass of glucose for conversion.

chemical reactions

Chemical reactions are processes in which substances, known as reactants, are transformed into different substances, called products. This transformation involves breaking and forming chemical bonds, changing their properties and energies.

The combustion of glucose is a clear example of a chemical reaction, where \(C_6H_{12}O_6\) glucose combines with \(O_2\) oxygen to produce \(CO_2\) carbon dioxide and \(H_2O\) water. The reaction releases energy, which the body uses for various functions.

Understanding chemical reactions, especially in the context of metabolism, helps explain everyday processes like breathing and energy production. Glucose combustion in the body is analogous to burning fuel in an engine, providing energy necessary for life processes.

The balanced chemical equation provides a concise way to understand the reaction stoichiometry, showing the proportional relationships between reactants and products. It guides calculations for estimating product yields or reactant requirements.