Molarity And Mass Calculations Using The Periodic Table

Hey guys! Today, we're diving into some awesome chemistry problems involving molarity and mass calculations. We'll be using the periodic table as our trusty sidekick to help us figure things out. So, grab your calculators, and let's get started!

Question 1: Moles of $AgNO_3$ in Solution

Moles of silver nitrate ($AgNO_3$) is our focus here. Imagine you're in the lab, and you need to prepare a solution of silver nitrate ($AgNO_3$). You have 1.50 liters of a solution with a concentration of 0.050 M. The big question is: how many moles of $AgNO_3$ are actually dissolved in that solution? Molarity, as you might recall, is the measure of the concentration of a solution and is defined as the number of moles of solute per liter of solution. So, understanding molarity is crucial for accurately preparing solutions in chemistry. To solve this, we'll use the fundamental relationship between molarity, moles, and volume.

Understanding Molarity

Molarity (M) is defined as the number of moles of solute per liter of solution. Mathematically, it’s expressed as:

$$Molarity (M) = \frac{Moles ext{ of solute}}{Liters ext{ of solution}}$$

In this scenario, we're given the molarity (0.050 M) and the volume (1.50 L). Our mission is to find the number of moles of $AgNO_3$. We can rearrange the formula to solve for moles:

$$Moles ext{ of solute} = Molarity (M) \times Liters ext{ of solution}$$

Calculation Steps

Let’s plug in the values we have:

$$Moles ext{ of } AgNO_3 = 0.050 ext{ M} \times 1.50 ext{ L}$$

Performing the calculation:

$$Moles ext{ of } AgNO_3 = 0.075 ext{ moles}$$

So, we've found that there are 0.075 moles of $AgNO_3$ present in 1.50 L of a 0.050 M solution. This calculation highlights the direct relationship between molarity and the number of moles in a given volume of solution. Understanding and applying this relationship is fundamental in chemistry for preparing solutions of specific concentrations. Whether you're titrating acids and bases, performing precipitation reactions, or conducting spectrophotometry, knowing the exact number of moles in your solution is critical for accurate and reproducible results.

Why This Matters

This type of calculation is super important in the lab. If you're doing an experiment, you need to know exactly how much of a chemical you're using. Too much or too little can throw off your results. Imagine you're synthesizing a new compound or running a titration; precision is key. The molarity calculation helps ensure that you're adding the correct amount of reactants, leading to successful and accurate experiments. It’s not just about getting the right answer in a textbook; it’s about making sure your experiments work in the real world.

Question 2: Mass of $(NH_4)_2S$ in Solution

Next up, let's tackle a problem where we need to find the mass of ammonium sulfide $(NH_4)_2S$ in a solution. Suppose you have 3.00 L of a 0.0200 M solution of $(NH_4)_2S$, and you're asked to calculate the mass of $(NH_4)_2S$ present. This question takes us a step further, combining molarity with the concept of molar mass to find the mass of a compound. To solve this, we'll need to use the molarity to find the number of moles, and then use the molar mass to convert moles to grams. This is a common type of calculation in chemistry, bridging the gap between solution concentration and the actual mass of the chemical compound.

The Strategy

Here's our game plan:

1. Calculate moles: Use the molarity and volume to find the number of moles of $(NH_4)_2S$.
2. Find molar mass: Use the periodic table to calculate the molar mass of $(NH_4)_2S$.
3. Convert to mass: Use the moles and molar mass to calculate the mass of $(NH_4)_2S$ in grams.

Step 1: Calculate Moles

We'll use the same molarity formula as before:

$$Moles ext{ of solute} = Molarity (M) \times Liters ext{ of solution}$$

Plugging in the given values:

$$Moles ext{ of } (NH_4)_2S = 0.0200 ext{ M} \times 3.00 ext{ L}$$

$$Moles ext{ of } (NH_4)_2S = 0.0600 ext{ moles}$$

So, we have 0.0600 moles of $(NH_4)_2S$ in the solution. This step is a straightforward application of the molarity formula, showing how molarity provides a direct link to the amount of substance present in the solution. Knowing the number of moles is a crucial intermediate step towards finding the mass, as it connects the concentration of the solution to the actual quantity of the compound.

Step 2: Find the Molar Mass of $(NH_4)_2S$

This is where the periodic table becomes our best friend. The molar mass is the sum of the atomic masses of all the atoms in the compound. For $(NH_4)_2S$, we have:

• 2 Nitrogen (N) atoms
• 8 Hydrogen (H) atoms
• 1 Sulfur (S) atom

Let's look up the atomic masses on the periodic table:

• N: Approximately 14.01 g/mol
• H: Approximately 1.01 g/mol
• S: Approximately 32.07 g/mol

Now, let's calculate the molar mass of $(NH_4)_2S$:

$$ext{Molar mass of } (NH_4)_2S = (2 imes 14.01) + (8 imes 1.01) + (1 imes 32.07)$$

$$ext{Molar mass of } (NH_4)_2S = 28.02 + 8.08 + 32.07$$

$$ext{Molar mass of } (NH_4)_2S = 68.17 ext{ g/mol}$$

The molar mass of $(NH_4)_2S$ is approximately 68.17 g/mol. This calculation showcases how the periodic table is an indispensable tool in chemistry. By providing the atomic masses of elements, it allows us to calculate the molar masses of compounds, which is essential for converting between moles and grams. This conversion is a fundamental skill in quantitative chemistry, used in various applications from stoichiometry to preparing solutions.

Step 3: Convert Moles to Mass

Now that we have the moles and the molar mass, we can calculate the mass of $(NH_4)_2S$:

$$Mass = Moles imes Molar ext{ mass}$$

Plugging in our values:

$$Mass ext{ of } (NH_4)_2S = 0.0600 ext{ moles} imes 68.17 ext{ g/mol}$$

$$Mass ext{ of } (NH_4)_2S = 4.0902 ext{ g}$$

So, there are approximately 4.09 grams of $(NH_4)_2S$ in 3.00 L of a 0.0200 M solution. This final calculation brings everything together, demonstrating how molarity, molar mass, and the periodic table work in concert to determine the mass of a compound in a solution. This skill is incredibly valuable in the lab, where you often need to weigh out specific amounts of chemicals for reactions or experiments. Accurate mass determination is critical for ensuring the success and reproducibility of chemical processes.

Real-World Importance

Why go through all this trouble? Well, imagine you're working in a pharmaceutical lab. You need to prepare a medication with a very specific amount of the active ingredient. If you mess up the mass calculation, you could end up with a drug that's ineffective or even dangerous. Precision is crucial. These calculations ensure that you're accurately measuring out chemicals, whether it's for synthesizing new materials, analyzing environmental samples, or developing new medicines. It’s a foundational skill that underpins much of the work in chemistry and related fields.

Wrapping Up

So, we've tackled two important types of chemistry problems today, all with the help of the periodic table. Remember, understanding molarity, moles, and molar mass is key to success in chemistry. Keep practicing, and you'll be a pro in no time! Chemistry can seem daunting, but breaking down problems into manageable steps and understanding the underlying concepts makes it much more approachable. Plus, the satisfaction of solving a challenging problem is pretty awesome. Keep exploring, keep learning, and who knows? Maybe you'll be the one discovering the next breakthrough in chemistry!

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