Problem 114
Question
A Freon leak in the air-conditioning system of a large building releases \(12 \mathrm{~kg}\) of \(\mathrm{CHF}_{2} \mathrm{Cl}\) per month. If the leak is allowed to continue, how many kilograms of Cl will be emitted into the atmosphere each year?
Step-by-Step Solution
Verified Answer
The yearly emission of Cl is approximately 57 kg.
1Step 1: Understand the chemical composition of CHF2Cl
The chemical formula for the Freon compound mentioned is CHF2Cl, which indicates that each molecule contains 1 atom of chlorine (Cl).
2Step 2: Calculate the monthly mass of Cl in the leak
We need to find the mass of Cl in one molecule of CHF2Cl and use this to determine the mass of Cl leaked per month. The atomic mass of H (hydrogen) is approximately 1, F (fluorine) is approximately 19, and Cl (chlorine) is approximately 35.5. The molecular mass of CHF2Cl is roughly 1(1) + 2(19) + 35.5 = 74.5. The proportion of Cl in CHF2Cl by mass is 35.5/74.5. Multiplying the total monthly leak of CHF2Cl (12 kg) by this ratio will give us the mass of Cl leaked each month.
3Step 3: Multiply by the number of months in a year
To find the yearly emission of Cl, multiply the monthly emission of Cl by 12, as there are 12 months in a year.
Key Concepts
Molecular Mass CalculationStoichiometryEnvironmental Impact of Chemical Leaks
Molecular Mass Calculation
To comprehend chemical reactions and substance quantities, one must understand the concept of molecular mass. The molecular mass of a compound is the sum of the atomic masses of all the atoms in its molecular formula. For instance, the Freon compound CHF2Cl consists of carbon (C), hydrogen (H), fluorine (F), and chlorine (Cl) atoms.
When calculating the molecular mass, we start by identifying the atomic mass of each element. Let's consider the weights of these atoms in atomic mass units (amu): C is approximately 12 amu, H is about 1 amu, F is around 19 amu, and Cl has a mass of around 35.5 amu. We then multiply the atomic mass of each element by the number of times the element appears in the compound. For CHF2Cl, the calculation would be as follows: molecular mass of CHF2Cl = 12(1) + 1(1) + 2(19) + 35.5 = 86.5 amu.
Understanding this concept allows students to relate the microscopic mass of molecules to macroscopic quantities used in laboratories. Incorporating this guidance into calculations ensures that students accurately determine the chemical quantities involved in reactions or environmental releases.
When calculating the molecular mass, we start by identifying the atomic mass of each element. Let's consider the weights of these atoms in atomic mass units (amu): C is approximately 12 amu, H is about 1 amu, F is around 19 amu, and Cl has a mass of around 35.5 amu. We then multiply the atomic mass of each element by the number of times the element appears in the compound. For CHF2Cl, the calculation would be as follows: molecular mass of CHF2Cl = 12(1) + 1(1) + 2(19) + 35.5 = 86.5 amu.
Understanding this concept allows students to relate the microscopic mass of molecules to macroscopic quantities used in laboratories. Incorporating this guidance into calculations ensures that students accurately determine the chemical quantities involved in reactions or environmental releases.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It involves calculations based on the laws of conservation of mass and balanced chemical equations.
Using stoichiometry, we can understand the proportions of substances involved in reactions and their conversions from one substance to another. This concept is crucial in determining the amount of Cl emitted in the Freon leak scenario mentioned. We use the stoichiometric ratio derived from the molecular mass calculation to find out the mass of chlorine in the leaked Freon.
Using stoichiometry, we can understand the proportions of substances involved in reactions and their conversions from one substance to another. This concept is crucial in determining the amount of Cl emitted in the Freon leak scenario mentioned. We use the stoichiometric ratio derived from the molecular mass calculation to find out the mass of chlorine in the leaked Freon.
Applying Stoichiometry to Freon Leak
For CHF2Cl, with a molecular mass of 86.5 amu and chlorine's mass of 35.5 amu, the stoichiometric proportion of Cl is calculated as the ratio of the mass of Cl to the total mass of CHF2Cl. This ratio is subsequently applied to the total mass of the compound leaked to determine the quantity of Cl specifically. By following the steps provided in the exercise, we arrive at the precise measure of Cl that would be released into the atmosphere each month and, multiplying by 12, each year.Environmental Impact of Chemical Leaks
Chemical leaks, such as the given example of a Freon (CHF2Cl) leak, can have significant environmental repercussions. The release of chlorine gas into the atmosphere doesn't just deplete precious air quality but can also contribute to broader environmental issues, such as ozone layer depletion.
Considering the environmental impact, it becomes imperative to calculate and understand the potential damage from such leaks. When we quantify the amount of Cl emitted over a year, as outlined in the exercise, we not only get a scope of the issue but also gain insight necessary to develop mitigation strategies.
Considering the environmental impact, it becomes imperative to calculate and understand the potential damage from such leaks. When we quantify the amount of Cl emitted over a year, as outlined in the exercise, we not only get a scope of the issue but also gain insight necessary to develop mitigation strategies.
Impact of Chlorine on the Ozone Layer
The presence of Cl in the atmosphere is particularly concerning since chlorine atoms can catalyze the breakdown of ozone molecules, leading to thinning of the protective ozone layer. This example ties stoichiometry to environmental chemistry, highlighting the importance of careful chemical analysis and regulation to protect our planet.Other exercises in this chapter
Problem 111
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