Molarity in Stoichiometric Calculations: 3 Practical Examples

Explore key examples of using molarity in stoichiometric calculations to enhance your chemistry understanding.
By Jamie

Understanding Molarity in Stoichiometric Calculations

Molarity is a fundamental concept in chemistry used to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution (mol/L). This measurement is crucial for stoichiometric calculations, which involve the relationships between reactants and products in chemical reactions. Below are three diverse, practical examples of using molarity in stoichiometric calculations.

Example 1: Determining the Amount of Reactant Needed

Context

In a laboratory setting, a chemist needs to determine how much sodium hydroxide (NaOH) is required to neutralize a given amount of hydrochloric acid (HCl) solution.

To neutralize HCl, the reaction is:

\[ HCl + NaOH \rightarrow NaCl + H_2O \]

The Example

  • Given: A 0.5 M solution of HCl, and we need to neutralize 0.1 L of this solution.
  • Required: The concentration and volume of NaOH required for complete neutralization.

  1. Calculate the moles of HCl:

    • Moles of HCl = Molarity × Volume = 0.5 mol/L × 0.1 L = 0.05 mol

  2. According to the stoichiometry of the reaction, 1 mole of HCl reacts with 1 mole of NaOH. Therefore, 0.05 moles of HCl will react with 0.05 moles of NaOH.

  3. If we have a 0.2 M NaOH solution:

    • Volume of NaOH required = Moles of NaOH / Molarity = 0.05 mol / 0.2 mol/L = 0.25 L or 250 mL.

Notes

  • Ensure that all measurements are in consistent units, particularly volume (liters).
  • This example demonstrates how molarity aids in calculating the required amounts for chemical reactions.

Example 2: Preparing a Dilution

Context

A researcher needs to prepare a diluted solution from a more concentrated stock solution for a series of tests. Here, we will calculate how to dilute a concentrated H2SO4 solution.

The Example

  • Given: A concentrated H2SO4 solution with a molarity of 18 M, and the researcher needs 0.5 L of a 1 M solution.

  1. Use the dilution equation: \[ M_1V_1 = M_2V_2 \]

    • Where:
      • M1 = initial molarity (18 M)
      • V1 = volume of concentrated solution to be found
      • M2 = final molarity (1 M)
      • V2 = final volume (0.5 L)

  2. Substitute the known values into the equation:
    \[ 18 imes V_1 = 1 imes 0.5 \]

  3. Solve for V1:
    \[ V_1 = \frac{1 \times 0.5}{18} = 0.0278 L \text{ or } 27.8 mL \]

  4. Therefore, to prepare 0.5 L of 1 M H2SO4, the researcher needs to measure 27.8 mL of the 18 M solution and then dilute it with distilled water to reach a total volume of 0.5 L.

Notes

  • Always add acid to water, not the other way around, to prevent exothermic reactions.
  • This example highlights the practical use of molarity in preparing solutions for experiments.

Example 3: Calculating Product Yield in a Reaction

Context

In an industrial process, a company wants to determine the expected yield of a product from a chemical reaction. Here, we will use molarity to find the yield of sodium chloride from the reaction of NaOH and HCl.

The Example

  • Given: 250 mL of 0.5 M HCl is reacted with excess NaOH.
  • Required: Calculate the mass of NaCl produced.

  1. Calculate the moles of HCl:

    • Moles of HCl = Molarity × Volume = 0.5 mol/L × 0.25 L = 0.125 mol

  2. Using the stoichiometry of the reaction (1:1 ratio), moles of NaCl produced = moles of HCl = 0.125 mol.

  3. Calculate the mass of NaCl:

    • Molar mass of NaCl = 58.44 g/mol
    • Mass of NaCl = Moles × Molar Mass = 0.125 mol × 58.44 g/mol = 7.305 g.

Notes

  • This example emphasizes the importance of molarity in predicting product yields in chemical reactions, which is essential for industrial applications.
  • Accurate measurements and calculations are crucial for successful outcomes in chemical manufacturing processes.