Question

In an experiment reported in the scientific literature, male cockroaches were made to run at different speeds on a miniature treadmill while their oxygen consumption was measured. In \(1 \mathrm{hr}\) the average cockroach running at \(0.08 \mathrm{~km} / \mathrm{hr}\) consumed \(0.8 \mathrm{~mL}\) of \(\mathrm{O}_{2}\) at 1 atm pressure and \(24^{\circ} \mathrm{C}\) per gram of insect mass. (a) How many moles of \(\mathrm{O}_{2}\) would be consumed in \(1 \mathrm{hr}\) by a \(5.2-\mathrm{g}\) cockroach moving at this speed? (b) This same cockroach is caught by a child and placed in a 1-qt fruit jar with a tight lid. Assuming the same level of continuous activity as in the research, will the cockroach consume more than \(20 \%\) of the available \(\mathrm{O}_{2}\) in a 48-hr period? (Air is \(21 \mathrm{~mol} \% \mathrm{O}_{2}\).)

Short answer

In this problem, we calculate the moles of O₂ consumed by a 5.2g cockroach moving at 0.08 km/hr in 1 hour to be \(n\), using the Ideal Gas Law. The total moles of O₂ consumed in 48 hours will be \(48 * n\). We also find the moles of O₂ in a 1-qt jar by calculating the total moles of air in the jar and multiplying it by the 21 mol% of O₂ in the air. Then, we compute the percentage of O₂ consumed by the cockroach in 48 hours as \(\frac{(48 * n)}{0.21 * n_{air}} * 100\%\). By comparing this percentage to 20%, we can determine if the cockroach will consume more than 20% of the available O₂ in a closed jar during a 48-hour period.

Step-by-step solution

Calculate moles of O₂ consumed by the cockroach in 1 hour

Given that the cockroach consumes 0.8 mL of O₂ per gram of its mass in 1 hour: Oxygen consumed by a 5.2g cockroach = 5.2 * 0.8 mL = 4.16 mL Now, we will convert the volume of O₂ consumed into moles using the Ideal Gas Law, given the pressure and temperature conditions of 1 atm and 24°C (297 K), respectively: \(PV = nRT\) where P = pressure (1 atm) V = volume (4.16 mL) n = moles of O₂ R = gas constant (0.0821 L atm / K mol) T = temperature (297 K) We first need to convert the volume from mL to L: V = 4.16 mL / 1000 = 0.00416 L Now, we can solve for n: \(n = \frac{PV}{RT}\)

Calculate the total moles of O₂ consumed by the cockroach in 48 hours

Moles of O₂ consumed in 1 hour = \(n\) Moles of O₂ consumed in 48 hours = \(48 * n\)

Calculate the available moles of O₂ in the jar

The jar has a volume of 1 qt, which can be converted to liters as follows: 1 qt = 0.946 L Now, we need to find the moles of O₂ in the jar, given that air contains 21 mol% O₂. The molar volume of a gas at 1 atm and 24°C (297 K) can be calculated using the Ideal Gas Law: \(V = \frac{nRT}{P}\) We will first find the total moles of air in the jar: \(n_{air} = \frac{PV}{RT}\) The moles of O₂ in the jar can then be calculated as follows: Moles of O₂ in the jar = \(0.21 * n_{air}\)

Check if the cockroach consumes more than 20% of the available O₂ in the jar

Now, we can calculate the proportion of O₂ consumed by the cockroach during the 48-hour period: Percentage of O₂ consumed = \(\frac{(48 * n)}{0.21 * n_{air}} * 100\%) Finally, we will check if the percentage of O₂ consumed by the cockroach is greater than 20%. If it is, then the cockroach will consume more than 20% of the available O₂ in the jar during the 48-hour period. If not, it will consume less than 20% of the available O₂.

Key concepts

Molar Volume of a Gas

The molar volume of a gas is essential when solving stoichiometry problems in chemistry. It refers to the volume occupied by one mole of a gas under specified conditions of temperature and pressure. Standard conditions often refer to 0°C (273 K) and 1 atm pressure, where one mole of an ideal gas occupies 22.4 liters. However, the actual molar volume can differ based on the conditions.

Understanding the concept of molar volume is crucial as it serves as a bridge between the macroscopic measurements of gases (like volume in liters) and the microscopic world of molecules (measured in moles). Students can connect the observable behavior of gases to the amount of substance present, thus providing a comprehensive understanding of chemical reactions involving gases.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes the relationship between pressure (P), volume (V), temperature (T), and number of moles (n) of an ideal gas. The Ideal Gas Law equation is:
\( PV = nRT \)
Here, R is the gas constant. In most stoichiometry problems, including the cockroach experiment, we assume gases behave ideally, which allows us to use this law for calculations.

This law is critical for predicting how a gas will behave under different conditions and is particularly useful when dealing with gases at standard temperature and pressure (STP). In stoichiometry, we frequently use the Ideal Gas Law to convert between volumes of gases and moles, as seen in the textbook exercise.

Moles Calculation

Moles calculation is a central part of stoichiometry and involves the conversion of mass, volume, and particle count into moles, the standard unit of amount in chemistry. Calculating moles allows chemists to quantify substances and use the mole as a counting unit to express amounts of a chemical substance. This is done using Avogadro's number for particles, molar mass for weight, and molar volume for gases.

For gases, we often use the Ideal Gas Law to determine the number of moles. As shown in the solution steps for the cockroach's oxygen consumption, we first converted the volume of oxygen from milliliters to liters and then used the Ideal Gas Law to find the number of moles.

Gas Constant (R)

The gas constant, denoted as R, is a crucial value in the Ideal Gas Law, which provides the proportionality between the product of pressure and volume and the product of the number of moles and temperature of an ideal gas. Its value will vary depending on the units used for pressure, volume, and temperature. In the SI system, its value is approximately 8.314 J/(mol K), but for problems involving liters and atmospheres, it's often used as 0.0821 L atm/(mol K), as seen in the cockroach exercise.

Understanding and correctly using the gas constant is critical for accurately calculating the properties of gases and being able to solve stoichiometry problems involving ideal gases.

Percent Composition

Percent composition is a term that describes the percentage by mass of each element in a compound or the percentage by volume or moles of each component in a mixture. In stoichiometry, this concept is often used to calculate the relative amounts of reactants or products. For instance, in the provided exercise, air is 21 mol% O₂, meaning that for every 100 moles of air, 21 moles are oxygen.

This concept can also be applied to determine how much of a particular substance is present in a mixture or reaction, as shown when calculating the percentage of oxygen the cockroach will consume. Comprehending percent composition is vital for understanding the composition of mixtures and predicting the outcome of chemical reactions.