Problem 57 - Chemistry: The Central Science

Question

The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), in our bodies produces \(\mathrm{CO}_{2}\), which is expelled from our lungs as a gas: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and \(0.970 \mathrm{~atm}\) when \(24.5 \mathrm{~g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at \(1.00\) atm and \(298 \mathrm{~K}\), to completely oxidize \(50.0 \mathrm{~g}\) of glucose.

Step-by-Step Solution

Answer

(a) The volume of dry CO2 produced at body temperature (37°C) and 0.970 atm when 24.5 g of glucose is consumed in this reaction is approximately 21.03 L. (b) The volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 50.0 g of glucose is approximately 40.63 L.

Step 1: Convert the mass of glucose into moles

First, we need to determine the molar mass of glucose, C6H12O6. To do that, look up the atomic masses of carbon, hydrogen, and oxygen and multiply them by their respective numbers in the molecule: Molar mass of glucose = (6 × 12.01) + (12 × 1.01) + (6 × 16.00) = 72.06 + 12.12 + 96.00 = 180.18 g/mol Now, convert the given mass of glucose into moles: Part (a): For 24.5 g of glucose: moles of glucose = mass of glucose / molar mass of glucose = 24.5 g / 180.18 g/mol = 0.1359 mol Part (b): For 50.0 g of glucose: moles of glucose = mass of glucose / molar mass of glucose = 50.0 g / 180.18 g/mol = 0.2774 mol

Step 2: Use stoichiometry to find moles of gases

The balanced equation of the process is: C6H12O6(aq) + 6 O2(g) -> 6 CO2(g) + 6 H2O(l) Using stoichiometry, we can calculate the moles of CO2 and O2 involved: - For part (a), 1 mole of glucose produces 6 moles of CO2. - For part (b), 1 mole of glucose requires 6 moles of O2. Part (a): Calculate moles of CO2 produced: moles of CO2 = 0.1359 mol × (6 moles CO2 / 1 mole glucose) = 0.8154 mol (approximately) Part (b): Calculate moles of O2 required: moles of O2 = 0.2774 mol × (6 moles O2 / 1 mole glucose) = 1.6644 mol (approximately)

Step 3: Calculate the volume of gases using the ideal gas law

The ideal gas law states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin. Part (a): Calculate the volume of CO2 at 37°C (310.15 K) and 0.970 atm: V = nRT / P V = (0.8154 mol)(0.0821 L·atm/mol·K)(310.15 K) / (0.970 atm) ≈ 21.03 L Part (b): Calculate the volume of O2 at 298 K and 1.00 atm: V = nRT / P V = (1.6644 mol)(0.0821 L·atm/mol·K)(298 K) / (1.00 atm) ≈ 40.63 L

Step 4: Results

(a) The volume of dry CO2 produced at body temperature (37°C) and 0.970 atm when 24.5 g of glucose is consumed in this reaction is approximately 21.03 L. (b) The volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 50.0 g of glucose is approximately 40.63 L.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that describes how gases behave under various conditions of pressure, volume, temperature, and quantity. The formula is represented as \( PV = nRT \), where:

• \( P \) stands for pressure, measured in atmospheres (atm).
• \( V \) is the volume, usually measured in liters (L).
• \( n \) represents the number of moles of the gas.
• \( R \) is the ideal gas constant, which is \( 0.0821 \text{ L·atm/mol·K} \).
• \( T \) is the temperature in Kelvin (K).

To use the Ideal Gas Law effectively, always ensure your units are consistent: - Pressure should be in atm, volume in liters, - and temperature converted from Celsius to Kelvin by adding 273.15. In our problem, we used this law to find the volume of gases like \( CO_2 \) and \( O_2 \), ensuring accurate measurements by manipulating the formula to solve for \( V \), i.e., \( V = \frac{nRT}{P} \). Understanding this principle helps in calculating how much gas you obtain from reactions, crucial for processes like metabolic oxidation.

Molar Mass Calculation

Calculating the molar mass of a compound is an essential skill for solving stoichiometry problems. Molar mass is the weight of one mole of a substance, measured in grams per mole (g/mol). To find it, sum up the atomic masses of all the atoms in a compound.

• For glucose (\( \text{C}_6\text{H}_{12}\text{O}_6 \)), add: \( 6 \times 12.01 \) of carbon, \( 12 \times 1.01 \) of hydrogen, and \( 6 \times 16.00 \) of oxygen.
• The calculation gives \( 72.06 + 12.12 + 96.00 = 180.18 \text{ g/mol} \).

Once you have the molar mass, convert a given mass of a substance into moles using the formula:\[\text{moles} = \frac{\text{mass}}{\text{molar mass}}\]For instance, with glucose, if you have \( 24.5 \text{ g} \), using its molar mass \( 180.18 \text{ g/mol} \), you'd have \( \frac{24.5}{180.18} \) moles. Accurately determining moles is vital for stoichiometric calculations, which predict the amounts of reactants and products in chemical reactions.

Metabolic Oxidation

Metabolic oxidation is a process by which organisms convert substances into energy. During the oxidation of glucose, your body produces carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)), releasing energy in the process. The chemical equation for glucose oxidation is:\[\text{C}_6\text{H}_{12}\text{O}_6 + 6 \text{O}_2 \rightarrow 6 \text{CO}_2 + 6 \text{H}_2\text{O}\]This equation indicates that to convert one mole of glucose, six moles of oxygen are required, producing six moles of \( \text{CO}_2 \) and \( \text{H}_2\text{O} \). For each mole of glucose metabolized, a proportional transformation of reactants to products occurs. Understanding how metabolic reactions like this work can help us not only understand the impacts of dietary choices on energy production but also their implications on respiratory processes like oxygen intake and \( \text{CO}_2 \) expulsion. This concept is foundational in biochemistry and helps to bridge interactions between chemistry and biology, particularly in an organism's energy management.