Problem 34 - Chemistry Matter and Change

Question

Solve How much ammonium chloride \(\left(\mathrm{NH}_{4} \mathrm{Cl}\right),\) in grams, is needed to produce 2.5 \(\mathrm{L}\) of a 0.5 \(\mathrm{M}\) aqueous solution?

Step-by-Step Solution

Answer

66.875 grams of ammonium chloride are needed.

Step 1: Understand Molarity

Molarity (M) is defined as the number of moles of solute per liter of solution. In this problem, we are given a solution with a molarity of 0.5 M, meaning there are 0.5 moles of ammonium chloride in every liter of solution.

Step 2: Calculate Moles of Solute

To find the total moles of ammonium chloride needed, multiply the molarity by the volume of the solution in liters. Since the solution's volume is 2.5 liters and its molarity is 0.5 M, we calculate the moles of ammonium chloride as follows: \( \text{moles} = 0.5 \text{ M} \times 2.5 \text{ L} = 1.25 \text{ moles} \).

Step 3: Find Molar Mass of NH4Cl

The molar mass of ammonium chloride (\( \text{NH}_4\text{Cl} \)) is found by adding the atomic masses of its elements: Nitrogen (N) is 14 g/mol, each Hydrogen (H) is 1 g/mol, and Chlorine (Cl) is 35.5 g/mol. Thus, molar mass = 14 + (4 \times 1) + 35.5 = 53.5 g/mol.

Step 4: Convert Moles to Grams

Now, use the molar mass to convert the moles of \( \text{NH}_4\text{Cl} \) into grams. Multiply the number of moles by the molar mass: \( 1.25 \text{ moles} \times 53.5 \text{ g/mol} = 66.875 \text{ grams} \).

Key Concepts

Ammonium Chloride

Ammonium chloride is a white crystalline solid that is often used in both industrial and laboratory settings. Its chemical formula is \( \text{NH}_4\text{Cl} \), which means it contains nitrogen, hydrogen, and chlorine in its structure.

It easily dissolves in water, creating an acidic solution. This property makes it useful in applications like metalwork, dyeing of textiles, and in the manufacture of dry cell batteries.
Furthermore, ammonium chloride is commonly used in the medical field as a treatment to acidify urine. Being able to calculate its amounts in chemical reactions is important for its effective use.
Understanding the role of ammonium chloride in solution reactions is crucial for students studying chemistry. This compound serves as an excellent example to practice concepts like molarity and stoichiometry due to its frequent use in illustrations of these chemical principles.

Moles Calculation

Calculating moles is an essential skill in chemistry that helps us understand the amount of a substance in a given sample.

To find this, we often use the formula relating molarity to volume. Molarity, which is moles of solute per liter of solution, allows us to determine how many moles are present.

• For instance, if we need to find how many moles of ammonium chloride are necessary for a 2.5-liter solution with a molarity of 0.5 M, we simply multiply the volume by molarity: \( 0.5 \text{ M} \times 2.5 \text{ L} = 1.25 \text{ moles} \).

This calculation is foundational in chemistry, as it provides the basis to further determine the mass or volume of different substances involved in a chemical reaction.

Molar Mass

The molar mass of a compound is a critical concept that tells us the weight of one mole of a substance. This is expressed in grams per mole (g/mol).

To find the molar mass of ammonium chloride \( \text{NH}_4\text{Cl} \), we sum up the atomic weights of its constituent atoms:

• Nitrogen (N) has an atomic mass of 14 g/mol.
• Each of the four hydrogens (H) contributes 1 g/mol.
• Chlorine (Cl) adds 35.5 g/mol.

Adding these, we get a molar mass of 53.5 g/mol for \( \text{NH}_4\text{Cl} \).
Knowing the molar mass is crucial when converting moles to grams or vice-versa, and helps bridge the gap between the molecular and tangible world in chemistry.

Stoichiometry

Stoichiometry is a powerful concept in chemistry that deals with the quantitative relationships in chemical reactions, helping us predict how much of each substance is involved.

In the example of ammonium chloride, using stoichiometry involves leveraging the relationship between molarity, volume, and molar mass to determine the amount of needed substance.

• Here, knowing 1.25 moles of \( \text{NH}_4\text{Cl} \) are required and having calculated a molar mass of 53.5 g/mol, stoichiometry allows converting moles to grams:
  \( 1.25 \text{ moles} \times 53.5 \text{ g/mol} = 66.875 \text{ grams} \).

This process exemplifies stoichiometry’s utility in both lab settings and theoretical problems, ensuring chemists can precisely predict and measure the substances they work with.