Chapter 3: Problem 59
Freon- \(12\left(\mathrm{CCl}_{2} \mathrm{~F}_{2}\right)\) is used as a refrigerant in air conditioners and as a propellant in aerosol cans. Calculate the number of molecules of Freon-12 in \(5.56 \mathrm{mg}\) of Freon-12. What is the mass of chlorine in \(5.56 \mathrm{mg}\) of Freon-12?
Short Answer
Expert verified
There are \( 2.77 \times 10^{19} \) molecules of Freon-12 in 5.56 mg of Freon-12, and the mass of chlorine in 5.56 mg of Freon-12 is 3.26 mg.
Step by step solution
01
Determine the molar mass of Freon-12
The molecular formula of Freon-12 is \( CCl_2F_2 \). To find the molar mass of Freon-12, we can add the molar masses of its constituent elements:
Molar mass (Freon-12) = Molar mass (Carbon) + 2 × Molar mass (Chlorine) + 2 × Molar mass (Fluorine)
= 12.01 g/mol (C) + 2 × 35.45 g/mol (Cl) + 2 × 19.00 g/mol (F)
= 12.01 + 70.90 + 38.00
= 120.91 g/mol
02
Calculate the number of moles of Freon-12
Now, we can use the given mass (5.56 mg) to calculate the number of moles of Freon-12. First, convert the mass to grams:
5.56 mg × (1 g / 1000 mg) = 0.00556 g
Next, find the number of moles using the molar mass:
Number of moles = (Mass of Freon-12) / (Molar mass of Freon-12)
= 0.00556 g / 120.91 g/mol
= 4.6 × 10^{-5} moles
03
Calculate the number of molecules of Freon-12
Now, use Avogadro's number (6.022 × 10^{23} molecules/mol) to find the number of molecules in the given amount of Freon-12:
Number of molecules = (Number of moles) × (Avogadro's number)
= 4.6 × 10^{-5} moles × 6.022 × 10^{23} molecules/mol
= 2.77 × 10^{19} molecules
04
Calculate the mass of chlorine in 5.56 mg of Freon-12
To find the mass of chlorine in the given amount of Freon-12, first calculate the mass percentage of chlorine in Freon-12:
Mass percentage of chlorine = (Mass of 2 Cl atoms) / (Molar mass of Freon-12) × 100
= (70.9 g/mol) / (120.91 g/mol) × 100
= 58.65 %
Now, use the mass percentage to find the mass of chlorine in 5.56 mg of Freon-12:
Mass of chlorine = (5.56 mg) × (58.65%)
= 3.26 mg
So, there are 2.77 × 10^{19} molecules of Freon-12 in 5.56 mg of Freon-12, and the mass of chlorine in 5.56 mg of Freon-12 is 3.26 mg.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Calculating Molar Mass
Understanding the molar mass of a substance is critical in chemistry. It represents the mass of one mole of a substance, typically in grams per mole (g/mol). To calculate the molar mass, you need the atomic weights of each element in the compound, often found on the periodic table. For example, to calculate the molar mass of Freon-12 (\( CCl_2F_2 \)), you'll add the atomic weights of carbon (C), chlorine (Cl), and fluorine (F), taking into account the number of atoms of each in the molecule.
For each carbon atom (12.01 g/mol), along with two atoms of chlorine (35.45 g/mol each), and two atoms of fluorine (19.00 g/mol each), you end up with the molar mass of Freon-12 by summing these amounts: \[ Molar \ mass \ (Freon-12) = 12.01 \ g/mol \ (C) + 2 \times 35.45 \ g/mol \ (Cl) + 2 \times 19.00 \ g/mol \ (F) = 120.91 \ g/mol. \]This step is essential for converting a given mass of the substance to moles, which is a gateway to finding out quantities like the number of molecules, as we'll see in the mole to molecule conversion section.
For each carbon atom (12.01 g/mol), along with two atoms of chlorine (35.45 g/mol each), and two atoms of fluorine (19.00 g/mol each), you end up with the molar mass of Freon-12 by summing these amounts: \[ Molar \ mass \ (Freon-12) = 12.01 \ g/mol \ (C) + 2 \times 35.45 \ g/mol \ (Cl) + 2 \times 19.00 \ g/mol \ (F) = 120.91 \ g/mol. \]This step is essential for converting a given mass of the substance to moles, which is a gateway to finding out quantities like the number of molecules, as we'll see in the mole to molecule conversion section.
Avogadro's Number
When delving into chemistry problems, you'll frequently encounter Avogadro's number, which is a cornerstone of the mole concept. Avogadro's number, \( 6.022 \times 10^{23} \), is the quantity of entities (like atoms or molecules) in one mole of a substance. This constant enables the conversion between microscopic and macroscopic scales in chemistry, acting as a bridge between the number of entities and the amount of substance in moles.
To illustrate, if you have a mole of any substance, you essentially have \( 6.022 \times 10^{23} \) particles of that substance. Using Avogadro's number is crucial when converting moles to the number of particles, whether you're talking about atoms in an element or molecules in a compound.
To illustrate, if you have a mole of any substance, you essentially have \( 6.022 \times 10^{23} \) particles of that substance. Using Avogadro's number is crucial when converting moles to the number of particles, whether you're talking about atoms in an element or molecules in a compound.
Mole to Molecule Conversion
Once you have the number of moles of a substance, you might need to find out how many molecules it contains. This is where Avogadro's number becomes valuable. Multiply the number of moles by Avogadro's number, and voilà—you have the number of molecules.
For instance, with the 0.00556 g of Freon-12 from our exercise, we calculated the number of moles to be \( 4.6 \times 10^{-5} \) moles. By multiplying this by Avogadro's number: \[Number \ of \ molecules = (Number \ of \ moles) \times (Avogadro's \ number) = 4.6 \times 10^{-5} \ moles \times 6.022 \times 10^{23} \ molecules/mol = 2.77 \times 10^{19} \ molecules.\]
This result indicates the exact count of Freon-12 molecules present in 5.56 mg of the substance. This conversion is invaluable in quantitative analyses and synthetic processes.
For instance, with the 0.00556 g of Freon-12 from our exercise, we calculated the number of moles to be \( 4.6 \times 10^{-5} \) moles. By multiplying this by Avogadro's number: \[Number \ of \ molecules = (Number \ of \ moles) \times (Avogadro's \ number) = 4.6 \times 10^{-5} \ moles \times 6.022 \times 10^{23} \ molecules/mol = 2.77 \times 10^{19} \ molecules.\]
This result indicates the exact count of Freon-12 molecules present in 5.56 mg of the substance. This conversion is invaluable in quantitative analyses and synthetic processes.
Mass Percentage Calculation
The mass percentage of an element in a compound is a way of expressing its concentration. Essentially, it is the mass of the element in a compound compared to the total mass of the compound, multiplied by 100 to get a percentage. This calculation is key when you want to find the actual mass of a particular element in a sample of the compound.
Here's how we applied this in our exercise: \[Mass \ percentage \ of \ chlorine = (Mass \ of \ 2 \ Cl \ atoms) / (Molar \ mass \ of \ Freon-12) \times 100.\] With the mass percentage known, we then determined the mass of chlorine in our sample. By multiplying the mass percentage by the sample's total mass, we obtained the chlorine's mass: \[Mass \ of \ chlorine = (5.56 \ mg) \times (58.65\%) = 3.26 \ mg.\]
Mastering mass percentage calculations is essential for chemists, especially when assessing the purity of a sample or carrying out stoichiometric computations in reactions.
Here's how we applied this in our exercise: \[Mass \ percentage \ of \ chlorine = (Mass \ of \ 2 \ Cl \ atoms) / (Molar \ mass \ of \ Freon-12) \times 100.\] With the mass percentage known, we then determined the mass of chlorine in our sample. By multiplying the mass percentage by the sample's total mass, we obtained the chlorine's mass: \[Mass \ of \ chlorine = (5.56 \ mg) \times (58.65\%) = 3.26 \ mg.\]
Mastering mass percentage calculations is essential for chemists, especially when assessing the purity of a sample or carrying out stoichiometric computations in reactions.