Solutions and Molarity

Definition and Calculation of Molarity

Molarity (M) is a measure of concentration, defined as the number of moles of solute per liter of solution. It is a fundamental concept in solution chemistry, used to quantify the amount of a substance dissolved in a given volume.

• Formula:

• Key Terms:

  • Solute: The substance being dissolved.

  • Solvent: The substance doing the dissolving (often water).

Example: Calculating Volume from Mass and Molarity

To find the volume of solution containing a given mass of solute:

• Given: 75.6 g of acetic acid (CH3COOH), molar mass = 60.05 g/mol, solution molarity = 0.839 mol/L.

• Calculate moles:

• Calculate volume:

Example: Finding Volume Needed for a Given Mass of Solute

• Given: 3.81 g glucose (C6H12O6), molar mass = 180.2 g/mol, solution molarity = 2.53 M.

• Calculate moles:

• Calculate volume:

Calculating Molarity from Mass and Volume

• Given: 1.77 g ethanol (CH3CH2OH) in 85.0 mL solution.

• Molar mass:

• Moles:

• Molarity:

Calculating Volume from Mass and Molarity

• Given: 6.22 g NaOH, molar mass = 40.0 g/mol, solution molarity = 0.315 M.

• Moles:

• Volume:

Ion Concentration in Solution

• For ionic compounds, the concentration of ions can be determined from the formula and the molarity of the compound.

• Example: Ammonium sulfate ((NH4)2SO4), 0.0236 M solution.

• Each formula unit yields 2 NH4+ ions: ions

Calculating Molarity with Density

When given mass of solute and solvent, and the density of the solution, follow these steps:

• Find total mass of solution: solute mass + solvent mass

• Use density to find volume:

• Calculate moles of solute:

• Calculate molarity:

Example:

• 45.0 g glycerol (C3H8O3), 50.0 g water, density = 1.095 g/mL

• Total mass: 95.0 g

• Volume:

• Moles:

• Molarity:

Stoichiometry and Chemical Equations

Balancing Chemical Equations

Balancing chemical equations ensures the conservation of mass and atoms. Each side of the equation must have the same number of each type of atom.

• Coefficients are used to balance equations.

• Fractional coefficients are acceptable, but typically equations are multiplied to yield whole numbers.

Example:

• Unbalanced: N2(g) + O2(g) → N2O5(g)

• Balanced: 2 N2(g) + 5 O2(g) → 2 N2O5(g)

Types of Stoichiometry

• Formula Stoichiometry: Based on the chemical formula.

• Reaction Stoichiometry: Based on the coefficients in the balanced chemical equation.

Stoichiometric Calculations

Stoichiometry allows conversion between mass, moles, and molecules of reactants and products using the balanced chemical equation.

• Convert known quantities to moles.

• Use mole ratios from the balanced equation.

• Convert moles of desired substance to mass or molecules as needed.

Example: Molecules Produced

• Combustion of propane: C3H8 + 5 O2 → 3 CO2 + 4 H2O

• 0.75 mol propane produces:

• Number of molecules: molecules

Limiting Reactant and Yield Calculations

Limiting Reactant Concept

The limiting reactant is the reactant that is completely consumed first, thus limiting the amount of product formed. The other reactants are in excess.

• Identify the limiting reactant by calculating the amount of product each reactant can produce.

• The reactant that produces the least amount of product is the limiting reactant.

Limiting Reactant Calculation Steps

1. Balance the chemical equation.

2. Convert all reactant amounts to moles.

3. Choose a product and calculate how much each reactant can produce.

4. The reactant that produces the least product is the limiting reactant.

5. Use the limiting reactant to calculate the amount of product and excess reactant remaining.

Example: Synthesis of Silicon Nitride

• Reaction: 3 Si + 2 N2 → Si3N4

• Given: 2.00 g Si, 1.50 g N2

• Moles Si: mol

• Moles N2: mol

• Calculate product from each reactant:

  • Si:

  • N2:

• Si is limiting reactant.

Theoretical Yield and Percent Yield

• Theoretical yield: Maximum amount of product possible, calculated from the limiting reactant.

• Actual yield: Amount of product actually obtained from the experiment.

• Percent yield:

Example: Iron Production

• Reaction: Fe3O4 + 3 CO → 2 Fe + 3 CO2

• Given: 245.0 g Fe3O4, 120.2 g CO

• Moles Fe3O4: mol

• Moles CO: mol

• Fe3O4 can produce mol Fe

• CO can produce mol Fe

• CO is limiting reactant; theoretical yield: g Fe

• If actual yield is 87.9 g, percent yield:

Summary Table: Key Solution and Stoichiometry Calculations

Additional info: These notes expand on the provided examples and calculations, adding definitions, formulas, and stepwise procedures for clarity and completeness.