Stock Solution Preparation And Dilution Calculations In Chemistry

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Hey guys! Today, we're diving into a classic chemistry problem: stock solution preparation and dilution. This is a fundamental concept in chemistry, especially in lab settings, where we often need solutions of specific concentrations. We'll break down the steps involved in calculating concentrations and dilutions, making it super easy to understand. So, let's get started!

Problem Overview: Preparing and Diluting Ammonium Sulfate Solution

Let's consider this problem. We're making a stock solution by dissolving 66.05 g of ammonium sulfate, $(NH_4)_2SO_4$, in enough water to make 250 mL of solution. Then, we take a 10.0 mL sample of this solution and dilute it to 50.0 mL. Our goal is to figure out the concentration of the final diluted solution. To solve this, we need to understand a few key concepts: molar mass, molarity, and the dilution equation. Let's dive deep into the process, so you can nail similar problems every time!

Step 1: Calculating the Molar Mass of $(NH_4)_2SO_4$

The first crucial step in solving this problem is to calculate the molar mass of ammonium sulfate, $(NH_4)_2SO_4$. The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). To find the molar mass of a compound, we add up the atomic masses of each element in the compound, considering the number of atoms of each element.

Ammonium sulfate, $(NH_4)_2SO_4$, consists of the following elements:

  • Nitrogen (N)
  • Hydrogen (H)
  • Sulfur (S)
  • Oxygen (O)

The chemical formula tells us that there are:

  • 2 Nitrogen atoms (from the $(NH_4)_2$ part)
  • 8 Hydrogen atoms (from the $(NH_4)_2$ part, 2 x 4 = 8)
  • 1 Sulfur atom
  • 4 Oxygen atoms

Now, let's look up the atomic masses of these elements from the periodic table:

  • Nitrogen (N): Approximately 14.01 g/mol
  • Hydrogen (H): Approximately 1.01 g/mol
  • Sulfur (S): Approximately 32.07 g/mol
  • Oxygen (O): Approximately 16.00 g/mol

To calculate the molar mass of $(NH_4)_2SO_4$, we multiply the atomic mass of each element by the number of atoms of that element in the compound and then add them all together:

(2 × 14.01 g/mol) + (8 × 1.01 g/mol) + (1 × 32.07 g/mol) + (4 × 16.00 g/mol)

= 28.02 g/mol (for N) + 8.08 g/mol (for H) + 32.07 g/mol (for S) + 64.00 g/mol (for O)

Adding these values together:

  1. 02 + 8.08 + 32.07 + 64.00 = 132.17 g/mol

So, the molar mass of $(NH_4)_2SO_4$ is approximately 132.17 g/mol. Knowing the molar mass is essential because it allows us to convert between mass (grams) and moles, which is a key step in calculating molarity.

In summary, we've broken down the process of calculating the molar mass step by step. This foundational calculation is crucial for determining the concentration of solutions, and without it, we can't proceed accurately with the rest of the problem. Take your time with this step, double-check your work, and you'll be well on your way to mastering solution chemistry!

Step 2: Calculating the Molarity of the Stock Solution

Now that we've determined the molar mass of $(NH_4)_2SO_4$, which is 132.17 g/mol, we can move on to calculating the molarity of the stock solution. Molarity, represented by the symbol M, is a measure of the concentration of a solution. It tells us how many moles of solute (the substance being dissolved) are present in one liter of solution. The formula for molarity is:

Molarity (M) = Moles of solute / Liters of solution

In our problem, we dissolved 66.05 g of $(NH_4)_2SO_4$ in enough water to make 250 mL of solution. To calculate the molarity, we first need to convert the mass of $(NH_4)_2SO_4$ to moles and the volume of the solution from milliliters to liters.

Converting Grams to Moles

To convert grams to moles, we use the molar mass we calculated earlier. We divide the mass of the substance by its molar mass:

Moles of $(NH_4)_2SO_4$ = Mass of $(NH_4)_2SO_4$ / Molar mass of $(NH_4)_2SO_4$

Moles of $(NH_4)_2SO_4$ = 66.05 g / 132.17 g/mol

Moles of $(NH_4)_2SO_4$ ≈ 0.500 moles

Converting Milliliters to Liters

To convert milliliters to liters, we divide the volume in mL by 1000, since there are 1000 mL in 1 L:

Liters of solution = Volume in mL / 1000

Liters of solution = 250 mL / 1000

Liters of solution = 0.250 L

Calculating Molarity

Now that we have the moles of $(NH_4)_2SO_4$ and the volume of the solution in liters, we can calculate the molarity:

Molarity (M) = Moles of solute / Liters of solution

Molarity (M) = 0.500 moles / 0.250 L

Molarity (M) = 2.00 M

So, the molarity of the stock solution is 2.00 M. This means that there are 2.00 moles of $(NH_4)_2SO_4$ in every liter of the stock solution.

In this step, we've converted the given mass of the solute into moles and the volume of the solution into liters, and then used these values to calculate the molarity of the stock solution. This is a critical step because it allows us to express the concentration of the solution in a standard unit (moles per liter), which is essential for further calculations, especially when we dilute the solution.

Understanding how to calculate molarity is fundamental in chemistry. It's used in various applications, from preparing solutions in the lab to performing stoichiometric calculations. By breaking down the process into smaller steps, we've made it easier to grasp. Remember, practice makes perfect, so try working through similar problems to solidify your understanding!

Step 3: Applying the Dilution Equation

After calculating the molarity of the stock solution, which we found to be 2.00 M, the next step is to determine the concentration of the diluted solution. For this, we'll use the dilution equation. The dilution equation is a simple yet powerful tool that helps us calculate the new concentration of a solution after it has been diluted. It's based on the principle that the number of moles of solute remains constant during dilution; only the volume of the solution changes. The dilution equation is expressed as:

M1V1=M2V2M_1V_1 = M_2V_2

Where:

  • M1M_1 is the molarity of the stock solution (initial concentration)
  • V1V_1 is the volume of the stock solution used for dilution (initial volume)
  • M2M_2 is the molarity of the diluted solution (final concentration)
  • V2V_2 is the final volume of the diluted solution

In our problem, we took a 10.0 mL sample of the 2.00 M stock solution and diluted it to 50.0 mL. Let's identify the values we know:

  • M1M_1 (Molarity of stock solution) = 2.00 M
  • V1V_1 (Volume of stock solution) = 10.0 mL
  • V2V_2 (Final volume of diluted solution) = 50.0 mL

We need to find M2M_2, the molarity of the diluted solution. Plugging the known values into the dilution equation:

(2. 00 M) × (10.0 mL) = M2M_2 × (50.0 mL)

Now, we solve for M2M_2:

M2M_2 = (2.00 M × 10.0 mL) / 50.0 mL

M2M_2 = 20.0 M mL / 50.0 mL

M2M_2 = 0.400 M

So, the molarity of the diluted solution is 0.400 M. This means that after diluting the stock solution, the concentration of $(NH_4)_2SO_4$ in the final solution is 0.400 moles per liter.

The dilution equation is a fundamental concept in chemistry labs because it allows us to accurately prepare solutions of desired concentrations from stock solutions. Understanding how to apply this equation correctly is essential for experimental work and other chemical calculations. By keeping the number of moles of solute constant, we can easily calculate the new concentration when the volume changes.

In this step, we've walked through the application of the dilution equation, highlighting the importance of identifying the known values and solving for the unknown. Remember, this equation is a cornerstone in solution chemistry, and mastering it will significantly enhance your problem-solving skills in this area.

Summary: Putting It All Together

Alright guys, let's recap what we've done to solve this problem! We started with 66.05 g of $(NH_4)_2SO_4$ dissolved in 250 mL of water to create a stock solution. Then, we diluted a 10.0 mL sample of this stock solution to 50.0 mL. Our mission was to find the concentration of the final diluted solution.

Here’s a quick rundown of the steps we took:

  1. Calculated the molar mass of $(NH_4)_2SO_4$: We added up the atomic masses of each element in the compound to get a molar mass of 132.17 g/mol. This step is crucial for converting grams to moles.
  2. Determined the molarity of the stock solution: We converted the mass of $(NH_4)_2SO_4$ to moles (0.500 moles) and the volume of the solution to liters (0.250 L). Then, we used the molarity formula to find that the stock solution had a concentration of 2.00 M.
  3. Applied the dilution equation: We used the equation $M_1V_1 = M_2V_2$ to calculate the molarity of the diluted solution. By plugging in the known values (2.00 M and 10.0 mL for the stock solution, and 50.0 mL for the final volume), we found that the diluted solution has a molarity of 0.400 M.

So, the final answer is that the concentration of the diluted solution is 0.400 M. Woohoo!

This problem illustrates a fundamental concept in chemistry: preparing solutions and diluting them to the desired concentrations. Understanding these steps is essential for anyone working in a lab or studying chemistry. You’ll use these calculations all the time, so it’s great to get comfortable with them now. Remember, practice makes perfect! Try working through similar problems to really nail these concepts.

If you found this breakdown helpful, awesome! Keep practicing, and you'll be a solution-making pro in no time. Chemistry can seem daunting at first, but breaking it down into manageable steps makes it much easier. Keep up the great work, guys! You've got this!

This comprehensive guide should help anyone tackling similar problems. By breaking down the process into manageable steps and explaining the underlying concepts, we've made it easier to understand and apply these calculations. Remember, chemistry is all about understanding the fundamentals, and with a solid grasp of these principles, you'll be well-equipped to tackle more complex problems in the future.