Moles And Mass Calculation For 2.15 X 10^23 NH3 Molecules

Hey guys! Let's dive into a fascinating chemistry problem where we'll calculate the number of moles and the corresponding mass for a given number of ammonia (NH3) molecules. This is a classic stoichiometry problem, and we're going to break it down step by step so it’s super easy to understand. So, grab your calculators, and let’s get started!

Understanding Moles, Molecules, and Avogadro's Number

Before we jump into the calculations, let's quickly recap the key concepts. The mole is a fundamental unit in chemistry that helps us quantify the amount of a substance. Think of it like a chemist's dozen, but instead of 12, we’re dealing with a massive number: 6.022 x 10^23, also known as Avogadro's number. This number represents the number of atoms, molecules, ions, or other entities in one mole of a substance. So, when we say one mole of NH3, we're talking about 6.022 x 10^23 molecules of NH3.

Avogadro's number is the cornerstone of these calculations, linking the microscopic world of atoms and molecules to the macroscopic world we can measure in grams. It allows us to convert between the number of particles (molecules) and the amount of substance (moles). To really grasp this, imagine you have a bag filled with marbles. If you know the number of marbles in the bag, and you know that a “mole” of marbles is a specific (very large!) number, you can figure out how many “moles” of marbles you have. This is precisely what we’re doing with NH3 molecules.

Now, why is this important? Because in chemical reactions, we don’t deal with single molecules; we deal with billions upon billions of them. Moles provide a convenient way to express these enormous quantities. Furthermore, the concept of molar mass comes into play. The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). For example, the molar mass of NH3 is the sum of the atomic masses of one nitrogen atom and three hydrogen atoms. We'll use the molar mass to convert from moles to mass, giving us the weight of our sample of NH3.

In essence, understanding moles, Avogadro's number, and molar mass is crucial for solving stoichiometry problems. It's the foundation upon which we build our calculations. Once you understand these concepts, converting between numbers of molecules, moles, and grams becomes a straightforward process. So, let's move on and apply these concepts to our problem involving 2.15 x 10^23 molecules of NH3.

Step 1: Converting Molecules to Moles

The first part of our problem is to determine how many moles are equivalent to 2.15 x 10^23 molecules of NH3. We'll use Avogadro's number as our conversion factor. Remember, 1 mole of any substance contains 6.022 x 10^23 entities (in this case, molecules). To convert from molecules to moles, we'll divide the given number of molecules by Avogadro's number. Here's the formula we'll use:

Moles = (Number of Molecules) / (Avogadro's Number)

Let's plug in the values:

Moles of NH3 = (2.15 x 10^23 molecules) / (6.022 x 10^23 molecules/mol)

Now, we just need to do the math. When you divide 2.15 x 10^23 by 6.022 x 10^23, you'll notice that the 10^23 terms cancel out, which simplifies our calculation significantly. The result is approximately 0.357 moles. So, 2.15 x 10^23 molecules of NH3 is equivalent to about 0.357 moles.

This step is crucial because it bridges the gap between the number of individual molecules (a microscopic perspective) and the amount of substance in moles (a macroscopic perspective). By using Avogadro's number, we can easily convert from a count of molecules to a more chemically relevant unit, the mole. This conversion is the cornerstone for the next step, where we'll calculate the mass corresponding to these moles. Knowing the number of moles allows us to use the molar mass, a characteristic property of NH3, to find the mass. So, we've taken the first step in unraveling the problem and are well on our way to finding the final answer. Let's move on to calculating the mass!

Step 2: Calculating the Molar Mass of NH3

Now that we know the number of moles, the next step is to find out what mass this corresponds to. To do this, we need to calculate the molar mass of NH3. Remember, the molar mass is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). We can calculate the molar mass by adding up the atomic masses of all the atoms in the molecule. You can find these atomic masses on the periodic table.

NH3 consists of one nitrogen atom (N) and three hydrogen atoms (H). Let's look up their atomic masses:

• Nitrogen (N): Approximately 14.01 g/mol
• Hydrogen (H): Approximately 1.01 g/mol

Now, we'll add up the masses to get the molar mass of NH3:

Molar mass of NH3 = (1 x Atomic mass of N) + (3 x Atomic mass of H) Molar mass of NH3 = (1 x 14.01 g/mol) + (3 x 1.01 g/mol) Molar mass of NH3 = 14.01 g/mol + 3.03 g/mol Molar mass of NH3 ≈ 17.04 g/mol

So, the molar mass of NH3 is approximately 17.04 g/mol. This means that one mole of NH3 weighs about 17.04 grams. This value is the bridge between moles and mass. It’s a constant for NH3, just like Avogadro’s number is a constant for the number of entities in a mole. By knowing the molar mass, we can convert any amount of NH3 in moles to its corresponding mass in grams, or vice versa.

Understanding how to calculate molar mass is fundamental in chemistry. It's a skill that you'll use repeatedly in stoichiometry and other types of calculations. It connects the microscopic world of atomic masses to the macroscopic world of measurable masses. Now that we have the molar mass of NH3, we are ready to take the final step: calculating the mass of our 0.357 moles of NH3. Let's move on to that final calculation!

Step 3: Converting Moles to Mass

We're in the home stretch now! We've already calculated the number of moles of NH3 (0.357 moles) and the molar mass of NH3 (17.04 g/mol). Now, we just need to convert moles to mass. To do this, we'll use the following formula:

Mass = (Number of Moles) x (Molar Mass)

This formula is a direct application of the definition of molar mass: the mass of one mole of a substance. If we have a certain number of moles, we simply multiply that number by the mass of one mole to find the total mass. Let's plug in our values:

Mass of NH3 = (0.357 moles) x (17.04 g/mol)

Now, let's do the multiplication:

Mass of NH3 ≈ 6.08 grams

So, 2.15 x 10^23 molecules of NH3 correspond to approximately 6.08 grams. This is our final answer! We started with a number of molecules, converted it to moles using Avogadro's number, and then converted moles to mass using the molar mass. This process is a classic example of how stoichiometry allows us to relate quantities at the molecular level to measurable quantities in the lab.

This final step beautifully illustrates the power of the mole concept. It allows us to take an abstract number of molecules and translate it into a tangible mass that we can weigh on a balance. The ability to convert between these units is essential in chemistry because it allows us to accurately measure and predict the outcomes of chemical reactions. So, with this calculation, we've successfully completed the problem and gained a deeper understanding of the relationship between moles, molecules, and mass.

Final Answer

To summarize, 2.15 x 10^23 molecules of NH3 are equivalent to approximately 0.357 moles, and this number of molecules corresponds to a mass of approximately 6.08 grams. We did it! We successfully navigated this stoichiometry problem. By breaking it down into manageable steps—converting molecules to moles, calculating molar mass, and converting moles to mass—we made the problem much easier to tackle.

Key Takeaways

• Moles are your friend: The mole is the central unit for converting between numbers of particles and mass. Get comfortable using it!
• Avogadro's number is the bridge: Use Avogadro's number to convert between molecules (or atoms, ions, etc.) and moles.
• Molar mass links moles and grams: The molar mass is your key to converting between moles and mass. Make sure you know how to calculate it.

I hope this breakdown has made the problem clear and understandable. Remember, practice makes perfect, so keep working on these types of problems, and you'll become a stoichiometry pro in no time! Keep up the great work, guys!