Question

Carbon monoxide (CO) detectors sound an alarm when peak levels of carbon monoxide reach 100 parts per million (ppm). This level roughly corresponds to a composition of air that contains \(400,000 \mu \mathrm{g}\) carbon monoxide per cubic meter of air \(\left(400,000 \mu \mathrm{g} / \mathrm{m}^{3}\right)\). Assuming the dimensions of a room are \(18 \mathrm{ft} \times\) \(12 \mathrm{ft} \times 8 \mathrm{ft}\), estimate the mass of carbon monoxide in the room that would register \(100 \mathrm{ppm}\) on a carbon monoxide detector.

Short answer

The mass of carbon monoxide in the room that would register 100 ppm on a carbon monoxide detector is approximately \(19.577\,\mathrm{g}\).

Step-by-step solution

Calculate the volume of the room in cubic feet

First, we'll find the volume of the room using the given dimensions. The room's dimensions are \(18\,\text{ft}\times12\,\text{ft}\times8\,\text{ft}\). To find the volume, multiply the three dimensions together: \[Volume = Length \times Width \times Height\] \[Volume = 18\,\text{ft} \times 12\,\text{ft} \times 8\,\text{ft} = 1,728\,\text{ft}^3\]

Convert the volume of the room to cubic meters

Now, we need to convert the room's volume from cubic feet to cubic meters. To do this, you can use the conversion factor: \(1\,\text{m}^3 = 35.3147\,\text{ft}^3\). So, in our case, we have: \[Volume = 1,728\,\text{ft}^3 \times \frac{1\,\text{m}^3}{35.3147\,\text{ft}^3} = 48.9427\,\text{m}^3\]

Determine the mass of CO in the room

First, we'll need to find the mass concentration of CO that corresponds to 100 ppm. We are given that 100 ppm corresponds to \(400,000 \mu\mathrm{g/m^3}\) or \(4\times10^{5} \mu\mathrm{g/m^3}\). To calculate the mass of CO in the room, we'll use the formula: \[Mass = Concentration \times Volume\] \[Mass = 4\times10^{5}\, \mu\mathrm{g/m^3} \times 48.9427\,\mathrm{m}^3\] \[Mass = 19,577,084.66 \mu\mathrm{g}\] Since we want the mass in grams, we will convert from micrograms to grams: \[Mass = 19,577,084.66 \mu\mathrm{g} \times \frac{1\,\mathrm{g}}{1,\!000,\!000\, \mu\mathrm{g}}\] \[Mass = 19.57708466\,\mathrm{g}\]

Final result

The mass of carbon monoxide in the room that would register 100 ppm on a carbon monoxide detector is approximately \(19.577\,\mathrm{g}\).

Key concepts

Volume Conversion

Understanding the conversion of different volume units is essential when dealing with problems related to gases in enclosed spaces, such as rooms or containers. In the given exercise, the conversion from cubic feet to cubic meters was necessary to match the unit used in the mass concentration value (micrograms per cubic meter).

One cubic meter is equivalent to 35.3147 cubic feet. This conversion factor is derived from the fact that one meter equals approximately 3.28084 feet. Therefore, to convert from cubic feet to cubic meters, you divide the volume in cubic feet by 35.3147. Simplified, the conversion equation is: \[ Volume_{\text{m}^3} = Volume_{\text{ft}^3} \times \frac{1}{35.3147} \]
Using this principle, students can tackle various volume conversion problems, ensuring that they are working with the correct units for their calculations.

Mass Concentration

Mass concentration in the context of air pollutants like carbon monoxide (CO) refers to the mass of the substance present in a unit volume of air. It is typically measured in micrograms per cubic meter. Understanding how to work with mass concentration is crucial to express the quantity of pollutants in the air and compare it with safety standards.

The problem provided gives us a mass concentration of CO: \(400,000\, \mu\mathrm{g/m}^3\) when the CO levels reach 100 ppm. To calculate the mass from the concentration, you use the formula: \[Mass = Concentration \times Volume\]
This straightforward calculation can be applied to various contexts, including environmental science and occupational health.

Stoichiometry

Stoichiometry is the aspect of chemistry that relates to the quantitative relationships between the substances as they participate in chemical reactions, based on the balanced equations. Though our problem does not directly involve a chemical reaction, stoichiometry's principles are applied in the step where we equate the parts per million (ppm) measurement, which is a ratio, to a mass concentration, which involves grams per cubic meters.

This exercise uses the concept of stoichiometry to relate the volume of air to the mass of carbon monoxide. It is a fantastic example that illustrates how stoichiometric principles are not limited to chemical reactions but are also deeply ingrained in environmental and analytical chemistry calculations.

Gas Detection

Gas detection is a critical safety measure in various environments to prevent toxic exposure and explosions. In this exercise, carbon monoxide detectors are used as an example of a device that measures gas concentration levels. These detectors are calibrated to alert when certain thresholds are exceeded, in this case, 100 ppm.

It's essential to understand the calibration of these devices since different gases have different permissible exposure limits. For carbon monoxide, exposure to 100 ppm is considered hazardous, which is why detectors are set to this level. Educating on gas detection helps students appreciate the importance of monitoring air quality and the technology's role in preventing potentially fatal situations caused by toxic gases like CO.