Problem 57 - Chemistry The Central Science In Si Units

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 normal body temperature, \(37^{\circ} \mathrm{C},\) and \(101.33 \mathrm{kPa}\) when \(10.0 \mathrm{~g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at \(100 \mathrm{kPa}\) and \(298 \mathrm{~K},\) to completely oxidize \(15.0 \mathrm{~g}\) of glucose.

Step-by-Step Solution

Answer

(a) 8.51 L of CO2; (b) 12.36 L of O2.

Step 1: Calculate Molar Mass of Glucose

The molecular formula for glucose is \( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \). Calculate the molar mass by using the atomic masses \( (\mathrm{C} = 12.01\, \text{g/mol}, \mathrm{H} = 1.008\, \text{g/mol}, \mathrm{O} = 16.00\, \text{g/mol}) \). This results in: \( 6(12.01) + 12(1.008) + 6(16.00) = 180.16\, \text{g/mol} \).

Step 2: Calculate Moles of Glucose

For \( 10.0 \, \text{g} \) of glucose, use the formula: moles = mass / molar mass. Thus, \( \text{moles of glucose} = \frac{10.0}{180.16} \approx 0.0555 \, \text{mol} \).

Step 3: Determine Moles of \(\mathrm{CO}_2\) Produced

From the balanced equation, 1 mole of glucose produces 6 moles of \(\mathrm{CO}_2\). Therefore, \( 0.0555 \, \text{mol of glucose} \times 6 = 0.333 \, \text{mol of } \mathrm{CO}_2 \).

Step 4: Use Ideal Gas Law for \(\mathrm{CO}_2\) Volume

Use the ideal gas law: \( PV = nRT \). Convert temperature: \( 37\degree \mathrm{C} = 310\,\text{K} \) and pressure is \( 101.33 \, \text{kPa} = 101330 \, \text{Pa} \). \( R = 8.314 \, \text{J/(mol·K)} \). Solve for volume: \( V = \frac{nRT}{P} = \frac{0.333 \times 8.314 \times 310}{101330} \approx 8.51 \, \text{L} \).

Step 5: Calculate Moles of Glucose for Oxygen

For \( 15.0 \, \text{g} \) glucose: \( \text{moles of glucose} = \frac{15.0}{180.16} \approx 0.0833 \, \text{mol} \).

Step 6: Determine Moles of \(\mathrm{O}_2\) Needed

From the balanced equation, 6 moles of \(\mathrm{O}_2\) are needed per mole of glucose: \( 0.0833 \, \text{mol of glucose} \times 6 = 0.500 \, \text{mol of } \mathrm{O}_2 \).

Step 7: Use Ideal Gas Law for \(\mathrm{O}_2\) Volume

Use the ideal gas law: \( V = \frac{nRT}{P} \). Convert temperature: \( 298\,\text{K} \) and pressure \( 100 \, \text{kPa} = 100000 \, \text{Pa} \). Solve for volume: \( V = \frac{0.500 \times 8.314 \times 298}{100000} \approx 12.36 \, \text{L} \).

Key Concepts

Molar Mass Calculation

Understanding molar mass calculation is essential in stoichiometry for converting between mass and moles in a chemical reaction. The molar mass of a compound is the sum of the atomic masses of all atoms present in its molecular formula. For glucose, which has the molecular formula \( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \), the calculation goes as follows:

• Carbon (C): 6 atoms, each with an atomic mass of 12.01 g/mol
• Hydrogen (H): 12 atoms, each with an atomic mass of 1.008 g/mol
• Oxygen (O): 6 atoms, each with an atomic mass of 16.00 g/mol

Thus, the molar mass is \( 6(12.01) + 12(1.008) + 6(16.00) = 180.16 \, \text{g/mol} \). This value allows us to calculate how many moles are present in a given mass of glucose, which is crucial for identifying how reactants and products relate in chemical reactions.

Ideal Gas Law

The ideal gas law is a useful equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. This law is represented by the formula \( PV = nRT \), where:

• \( P \) is the pressure of the gas
• \( V \) is the volume of the gas
• \( n \) is the number of moles of the gas
• \( R \) is the ideal gas constant \( 8.314 \, \text{J/(mol·K)} \)
• \( T \) is the temperature of the gas in Kelvin

To apply the ideal gas law, we must convert temperatures to Kelvin and pressures to the appropriate units (Pa for pressure when \( R \) is in \( \text{J/(mol·K)} \)). For example, in calculating the volume of \( \mathrm{CO}_2 \) produced at body temperature (37°C), we first convert the temperature to 310 K and the pressure from kPa to Pa, as done in the steps of the solution. The calculated volume tells us the space that a specific amount of gas occupies under given conditions.

Chemical Reaction Equations

Chemical reaction equations describe the process of reactants converting into products. In the case of glucose oxidation, the balanced equation \( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 (a\, q)+6 \mathrm{O}_2 (g) \rightarrow 6 \mathrm{CO}_2 (g) + 6 \mathrm{H}_2 \mathrm{O} (l) \) shows the complete reaction of glucose with oxygen to produce carbon dioxide and water.

This equation is essential for determining the stoichiometric relationships between the substances involved. Each coefficient in the equation represents the mole ratio of reactants to products. Here, 1 mole of glucose reacts with 6 moles of oxygen to produce 6 moles of carbon dioxide and 6 moles of water. Understanding these relationships enables us to calculate the exact amount of products formed from a given amount of reactants, as illustrated in the exercise solution. Such calculations form the backbone of stoichiometry in chemistry.