Problem 14 - Chemistry: An Atoms-Focused Approach

Question

Calculate the molarity of each of the following solutions: a. \(31.76 \mathrm{g} \mathrm{LiClO}_{4}\) in \(475.0 \mathrm{mL}\) of solution b. \(6.37 \mathrm{g}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in \(250.0 \mathrm{mL}\) of solution c. \(2.97 \mathrm{g} \mathrm{KBr}\) in \(75.0 \mathrm{mL}\) of solution d. \(0.773 \mathrm{g} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) in \(100.0 \mathrm{mL}\) of solution

Step-by-Step Solution

Answer

Question: Calculate the molarity of the following solutions: a. 31.76 g of LiClO₄ in 475.0 mL of solution b. 6.37 g of (NH₄)₂SO₄ in 250.0 mL of solution c. 2.97 g of KBr in 75.0 mL of solution d. 0.773 g of Pb(NO₃)₂ in 100.0 mL of solution Answer: a. The molarity of LiClO₄ solution is 0.628 M. b. The molarity of (NH₄)₂SO₄ solution is 0.194 M. c. The molarity of KBr solution is 0.333 M. d. The molarity of Pb(NO₃)₂ solution is 0.0233 M.

Step 1: Convert mass of solute to moles using molar mass

We take the mass of the solutes given and divide it by the respective molar mass to find the number of moles. a. For LiClO₄, the molar mass = 28.9 g/mol (Li) + 35.5 g/mol (Cl) + 4 * 16.0 g/mol (O) = 106.4 g/mol Moles of LiClO₄ = (31.76 g) / (106.4 g/mol) = 0.2983 mol b. For (NH₄)₂SO₄, the molar mass = 2*(14.0 g/mol (N) + 4 * 1.0 g/mol (H)) + 32.1 g/mol (S) + 4 * 16.0 g/mol (O) = 132.1 g/mol Moles of (NH₄)₂SO₄ = (6.37 g) / (132.1 g/mol) = 0.0482 mol c. For KBr, the molar mass = 39.1 g/mol (K) + 79.9 g/mol (Br) = 119.0 g/mol Moles of KBr = (2.97 g) / (119.0 g/mol) = 0.0250 mol d. For Pb(NO₃)₂, the molar mass = 207.2 g/mol (Pb) + 2*(14.0 g/mol (N) + 3 * 16.0 g/mol (O)) = 331.2 g/mol Moles of Pb(NO₃)₂ = (0.773 g) / (331.2 g/mol) = 0.00233 mol

Step 2: Convert volume of solution to liters

The volume of each solution is given in milliliters (mL). We will convert these volumes to liters (L) by dividing by 1000. a. Volume of LiClO₄ solution = 475.0 mL = 0.475 L b. Volume of (NH₄)₂SO₄ solution = 250.0 mL = 0.250 L c. Volume of KBr solution = 75.0 mL = 0.075 L d. Volume of Pb(NO₃)₂ solution = 100.0 mL = 0.100 L

Step 3: Calculate molarity (M)

Now, we'll calculate the molarity of each solution using the moles of solute calculated in Step 1 and the volumes of the solution in liters converted in Step 2. Molarity (M) = (moles of solute) / (volume of solution in L) a. Molarity of LiClO₄ solution: (0.2983 mol) / (0.475 L) = 0.628 M b. Molarity of (NH₄)₂SO₄ solution: (0.0482 mol) / (0.250 L) = 0.194 M c. Molarity of KBr solution: (0.0250 mol) / (0.075 L) = 0.333 M d. Molarity of Pb(NO₃)₂ solution: (0.00233 mol) / (0.100 L) = 0.0233 M

Key Concepts

Stoichiometry

Stoichiometry is like a recipe for chemistry. It helps us determine how much of each ingredient, or reactant, we'll need to make a desired amount of product. When working with chemical reactions, stoichiometry involves calculations that describe the quantitative relationship between reactants and products using a balanced chemical equation. This is crucial for making predictions about the outcomes of chemical processes.

• Reactants are the starting substances.
• Products are what you get after the reaction.
• The coefficients in a balanced chemical equation tell you the relative amounts of each substance involved in the reaction.

Using stoichiometry, you can calculate how many grams of a reactant are needed to produce a specific quantity of product. These calculations are essential in industrial settings where efficiency and cost-effectiveness are critical.

Moles and Molar Mass

Understanding moles and molar mass is key to solving many chemistry problems. A mole is simply a counting unit, much like a dozen for eggs, but immensely larger and more suited for counting tiny particles like atoms and molecules. One mole of any substance contains exactly Avogadro's number of particles, which is \(6.022 \times 10^{23}\).

• The molar mass of a substance is the mass of one mole of that substance and is typically expressed in grams per mole (g/mol).
• To find the molar mass, add up the atomic masses of each atom in the chemical formula using the periodic table.

In the exercise, the molar mass helps us convert the mass of a solute in grams into moles, allowing for further calculations like molarity. By doing this, you can handle the quantitative aspects of chemical reactions more effectively.

Solution Concentration

Solution concentration refers to the amount of solute dissolved in a given volume of solvent. Molarity is a common way to express this concentration, defined as moles of solute per liter of solution, denoted as \(M\).

• To calculate molarity, you need two pieces of information: the number of moles of solute and the volume of the solution in liters.
• The formula for molarity is \[M = \frac{\text{moles of solute}}{\text{liters of solution}}\]

In the provided exercise, once we have converted both mass to moles and volume to liters, calculating solution concentration becomes a straightforward application of the molarity formula. Understanding solution concentration helps in preparing solutions of precise strength for various chemical applications.

Chemical Formulas

Chemical formulas are a way to represent chemical compounds using symbols and numbers to convey key information about the atoms involved and their ratios. This shorthand makes it easier to understand complex chemical compositions.

• Compositions like \(\mathrm{LiClO}_4\) or \(\mathrm{KBr}\) are chemical formulas describing specific compounds.
• The subscript numbers in a chemical formula tell you how many of each type of atom are in a molecule of the compound.
• Chemical formulas not only identify the elements in a compound but also help in calculating properties like molar mass.

Understanding chemical formulas is crucial since it is the first step in many calculations, like determining moles and stoichiometry. Knowing how to read and use them will make tackling chemistry problems much more approachable and less intimidating.