Lecture 10_2025

Key Concepts in Stoichiometry and Solutions

• Stoichiometry Basics

  • The process of calculating the quantities of reactants and products in a chemical reaction.

  • Steps in Stoichiometry:

    1. Balance the atoms and charges in the chemical equation.

    2. Convert mass to moles using the formula:[ n = \frac{\text{mass}}{\text{molar mass}} ]

    3. Identify the limiting reagent, which is the reactant that is completely consumed first and limits the amount of product formed.

    4. Use moles and coefficients from the balanced equation to calculate the amount of desired product.

    5. Convert moles of desired product back to mass (if required).

    6. Calculate the percentage yield of the reaction using the formula:[ % \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100% ]

• Definitions

  • Solvent: The substance that dissolves the solute; typically present in greater quantity.

  • Solute: The substance that is dissolved in the solvent.

  • Solution: A homogeneous mixture consisting of solute and solvent.

  • Concentration (c): A measure of the amount of solute dissolved per unit volume of solution.

    • Molarity (M) is a common unit of concentration, defined as:[ c = \frac{n}{V} ]where ( n ) = amount of substance in moles, ( V ) = volume of solution in liters.

Molarity and Concentration Calculations

• Calculating Molarity

  • Example 1:

    • For a saline solution with ( 0.10 , ext{M} ) of NaCl:

      • Volume = 10 L

      • Using the formula ( n = cV ):[ n = 0.10 , ext{mol L}^{-1} \times 10 , ext{L} = 1.0 , ext{mol} ]

      • Mass of NaCl needed:[ m = n \times M = 1.0 , ext{mol} \times 58.5 , \text{g mol}^{-1} = 58.5 , ext{g} ]

• Other Concentration Forms

  • Weight/Volume (%)

    • Example 2:

      • A 3% w/v bleach solution contains 3 g of NaOCl in 100 mL.

      • Molar mass of NaOCl = 74.5 g/mol.

      • Calculate moles:[ n = \frac{3.00 , ext{g}}{74.5 , ext{g mol}^{-1}} = 0.0403 , ext{mol} ]

      • Molarity:[ c = \frac{0.0403 , ext{mol}}{0.1 , ext{L}} = 0.403 , ext{M} ]

Reaction Involvement in Solutions

• Neutralization Reaction Example

  • Example 3:

    • For a reaction between NaOH and HCl:

      • Reaction: ( \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} )

      • Given volumes: 25.00 mL of NaOH and 22.40 mL of 0.1040 M HCl.

      • Calculate moles of HCl:[ n (HCl) = V \times c = 0.02240 , \text{L} \times 0.1040 , ext{mol L}^{-1} = 0.002330 , ext{mol} ]

      • Since ( 1 , \text{mol HCl} ) reacts with ( 1 , \text{mol NaOH} ), we find ( n (NaOH) = 0.002330 , ext{mol} ):

      • Molarity of NaOH: [ [NaOH] = \frac{n}{V} = \frac{0.002330 , ext{mol}}{0.02500 , ext{L}} = 0.09318 , ext{M} ]

Additional Learning Outcomes

• Be familiar with key terms associated with solutions and concentration, including solute, solvent, and molarity.

• Ability to perform calculations involving concentrations, volume, and moles.

• Prepare for further exercises involving dissolving compounds and calculating their concentrations in different forms, to be presented in the next lecture.

Questions for Review

1. Consider dissolving 23.4 g of sodium sulfate (Na2SO4) in enough water to form 125 mL of solution:

   • Write the equation for dissolving sodium sulfate.

   • Calculate molar mass of sodium sulfate.

   • Determine moles in 23.4 g of sodium sulfate.

   • Calculate the concentration of Na2SO4 in solution.

   • Find concentrations of sodium ions [Na+] and sulfate ions [SO4^2-] in the solution.