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 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 hr by a 5.2 -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 part (a), a 5.2-g cockroach consumes \(0.000191\;mol\) of O2 in 1 hour. For part (b), the cockroach consumes more than 20% of the available O2 in a 1-qt fruit jar within a 48-hour period (0.009168 mol consumed > 0.001597 mol which is 20% of available O2).

Step-by-step solution

1. Calculate the amount of O2 consumed by a 5.2-g cockroach in 1 hour

Since the average cockroach consumes 0.8 mL of O2 per gram of insect mass in 1 hour, and the mass of the cockroach under consideration is 5.2 g, we can calculate the amount of O2 consumed by the 5.2-g cockroach using the given relation: \(O_2\)consumed \(= \) mass of cockroach \(\times\) \(O_2\) consumed per gram \(\times\) 1 hour \(= 5.2 \times 0.8\) \(=4.16\;mL\) To find the number of moles of O2 consumed, we need to convert mL to moles using the ideal gas law: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. Given: P = 1 atm, V = 4.16 mL (converted into L), R = 0.0821 L.atm/mol.K, T = 24°C (converted into K) Next, we will rearrange the ideal gas law formula to solve for n. n = PV/RT

2. Calculate the number of moles of O2 consumed by a 5.2-g cockroach in 1 hour

We will use the temperature in Kelvin (T = 273 + 24 = 297 K) and convert volume into litres: V = 4.16 mL = 0.00416 L. Now we can plug in the values into our formula: n = (1 atm * 0.00416 L) / (0.0821 L.atm/mol.K * 297 K) n = 0.000191 mol In part (a), the cockroach consumes 0.000191 mol of O2 in 1 hour.

3. Calculate the total moles of O2 in the jar

The jar has a volume of 1 qt. We need to convert this volume into litres. 1 qt is approximately 0.946 L. Now using the ideal gas law formula, we can calculate the total number of moles of O2 inside the jar: Total moles of O2 in air = PV/RT = (1 atm * 0.946 L) / (0.0821 L.atm/mol.K * 297 K) = 0.03803 mol As the air is 21 mol% O2, the total moles of O2 in the jar = 0.03803 * 0.21 = 0.007986 mol

4. Calculate the amount of O2 consumed in 48 hours

The cockroach consumes 0.000191 mol of O2 in 1 hour. So, the amount of O2 consumed in 48 hours = 0.000191 mol * 48 hr = 0.009168 mol

5. Check if the cockroach consumes more than 20% of the available O2

O2 consumed in 48 hours = 0.009168 mol Now, we need to calculate the 20% of the total O2 present in the jar: 20% of 0.007986 mol = 0.001597 mol Since 0.009168 mol > 0.001597 mol, the cockroach consumes more than 20% of available O2 in a 48-hour period. So, the answer to part (b) is yes.

Key concepts

Moles and Molar Volume

Understanding moles and molar volume is crucial for solving stoichiometry problems, as these concepts are fundamental to quantifying substances in chemistry. A mole is a unit that represents a specific number of particles, such as atoms, molecules, or ions. This number is Avogadro's number, which is approximately 6.022 x 10^23 particles.

Molar volume, specifically for gases, is the volume one mole of gas occupies at a given temperature and pressure. At standard temperature and pressure (STP), one mole of any ideal gas occupies 22.4 liters. However, when conditions differ from STP, we must use the ideal gas law to calculate the molar volume. In the cockroach exercise, to calculate the amount of O2 consumed, we had to convert the volume of O2 from milliliters to moles. Since the volume was not at STP, using the ideal gas law was necessary to find the accurate molar volume under the experimental conditions.

Ideal Gas Law

The ideal gas law, represented by the equation PV = nRT, is a powerful equation in chemistry that relates the pressure (P), volume (V), number of moles (n), temperature (T), and the ideal gas constant (R). It describes the behavior of an ideal gas, providing a way to calculate the amount of substance when the other variables are known.

When we applied the ideal gas law to the cockroach exercise, we used it in two distinct scenarios: first, to calculate the moles of O2 the cockroach consumed and second, to determine the total moles of O2 in the jar. For both scenarios, we needed to ensure units were consistent, and temperatures were converted to Kelvin. Remember, the ideal gas law is only accurate under conditions where gases behave ideally, meaning high temperature and low pressure.

Percent Composition

Percent composition is another fundamental concept in chemistry that represents the fraction of a specific substance within a mixture, expressed as a percentage. For gases, this is often described as the mole percent, which indicates the proportion of the gas by moles in a mixture.

In the case of air, it is composed of various gases, but about 21% of its volume is oxygen, which is referred to as its mole percent composition. Understanding percent composition is essential, as it helped us to determine the moles of O2 in the jar. With this information, we could calculate whether the cockroach would consume more than 20% of the available O2 in the given time period. Accurate knowledge of percent composition is necessary when dealing with chemical reactions and environmental studies, like in our example of predicting the survival scenario of the cockroach.