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 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 \(\mathrm{atm}\) and \(298 \mathrm{K},\) to completely oxidize 50.0 \(\mathrm{g}\) of glucose.

Short answer

The volumes of dry CO2 produced and O2 consumed in the metabolic oxidation of glucose are as follows: (a) When 24.5 grams of glucose is consumed, 21.32 liters of CO2 are produced at body temperature (37°C) and 0.970 atm. (b) To completely oxidize 50.0 grams of glucose, 40.72 liters of oxygen would be needed at 1.00 atm and 298 K.

Step-by-step solution

Convert the mass of glucose to moles

The molar mass of glucose (\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)) is \((6\times12.01)+(12\times1.01)+(6\times16.00)=180.18 \ \mathrm{g/mol}\). Given mass of glucose = 24.5 g To find the moles of glucose, we'll use the formula: moles = mass / molar mass. Moles of glucose = \(\frac{24.5}{180.18} = 0.1359 \ \mathrm{moles}\)

Use stoichiometry to find the moles of CO2 produced

According to the balanced chemical equation, 1 mole of glucose produces 6 moles of CO2. Moles of CO2 = moles of glucose × 6 = 0.1359 × 6 = 0.8154 moles

Use the Ideal Gas Law to calculate the volume

First, convert the temperature to Kelvin: T(K) = T(°C) + 273.15 = 37 + 273.15 = 310.15 K. Now, we'll use the Ideal Gas Law: \(PV=nRT\). R (gas constant) = 0.0821 \(Latm/molK\) V = ? P = 0.970 atm n = 0.8154 moles T = 310.15 K R = 0.0821 \(L\cdot atm / (mol \cdot K)\) Now, plug in the values into the Ideal Gas Law equation and solve for V: V = \(\frac{nRT}{P} = \frac{(0.8154)(0.0821)(310.15)}{0.970} = 21.32\ \mathrm{L}\) So, when 24.5 grams of glucose is consumed, 21.32 liters of CO2 are produced. **(b) Calculate the volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 50.0 g of glucose.**

Convert the mass of glucose to moles

Given mass of glucose = 50.0 g Molar mass of glucose = 180.18 g/mol Moles of glucose = \(\frac{50.0}{180.18} = 0.2774\ \mathrm{moles}\)

Use stoichiometry to find the moles of O2 consumed

According to the balanced equation, 1 mole of glucose consumes 6 moles of O2. Moles of O2 = moles of glucose × 6 = 0.2774 × 6 = 1.6644 moles

Use the Ideal Gas Law to calculate the volume

P = 1.00 atm n = 1.6644 moles T = 298 K R = 0.0821 \(L\cdot atm / (mol \cdot K)\) V = \(\frac{nRT}{P} = \frac{(1.6644)(0.0821)(298)}{1.00} = 40.72\ \mathrm{L}\) So, to completely oxidize 50.0 grams of glucose, 40.72 liters of oxygen would be needed.

Key concepts

Stoichiometry

Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In our glucose metabolism example, stoichiometry comes into play when we deduce how many moles of CO2 are produced from a certain amount of glucose.

For each mole of glucose (C6H12O6), the chemical equation tells us that six moles of carbon dioxide (CO2) are produced. This fixed ratio allows us to convert moles of one substance (glucose) into moles of another substance (CO2), which is essential to solve the given problem. Understanding the stoichiometric coefficients – the numbers in front of compounds in a balanced chemical equation – is crucial, as they dictate the proportions in which substances react. Simple division and multiplication will then translate these ratios into the desired quantity, whether it's moles, mass, or volume.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation that relates the pressure (P), volume (V), number of moles (n), temperature (T), and a constant (R) for an ideal gas. The equation is expressed as PV = nRT. For our glucose metabolism problem, the Ideal Gas Law enables us to calculate the volume of CO2 produced from the moles we already have, factoring in the body's temperature and pressure conditions.

To solve this problem, we convert the temperature to Kelvin and use the pressure in atmospheres. By inputing the number of moles of CO2 and the known values of R (0.0821 Latm/molK) and the conditions of temperature and pressure, we can solve for V, the volume of CO2 produced. With this understanding, it's clear how the Ideal Gas Law allows us to connect the amount of substance in moles to its manifestation in a more tangible form, like volume, under specified conditions of temperature and pressure.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance and is typically expressed in grams per mole (g/mol). This concept is essential when converting between mass of a substance and moles, which is a central step in calculations involving chemical reactions. In the metabolic oxidation of glucose, we are given the mass of glucose and need to find its amount in moles.

The molar mass of glucose is calculated by summing the molar masses of each element within the molecule: carbon (C), hydrogen (H), and oxygen (O). We utilize the periodic table to find the molar masses of individual elements and multiply by the number of atoms of each element present in a glucose molecule (C6H12O6). Once we have the molar mass, dividing the given mass of glucose by this amount gives us the moles needed to proceed with the stoichiometric and gas law calculations.