Examples of Stoichiometry and Mole Calculations

Explore practical examples of stoichiometry and mole calculations in chemistry.
By Jamie

Understanding Stoichiometry and Mole Calculations

Stoichiometry is a branch of chemistry that deals with the calculation of reactants and products in chemical reactions. It allows chemists to predict the quantities of substances consumed and produced in a given reaction based on the balanced chemical equation. In this article, we’ll explore three diverse examples of stoichiometry and mole calculations to better understand this essential concept in chemistry.

Example 1: Combustion of Methane

Context

The combustion of methane is a fundamental reaction in chemistry that can be used to calculate the amount of carbon dioxide produced from a given amount of methane burned.

To balance the chemical equation, we have:

\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \]

Example

Suppose we burn 16 grams of methane (CH₄). First, we need to calculate the number of moles of methane:

  1. Determine the molar mass of methane:

    • Carbon (C): 12.01 g/mol
    • Hydrogen (H): 1.01 g/mol
    • Molar mass of CH₄ = 12.01 + (4 × 1.01) = 16.05 g/mol

  2. Calculate moles of methane:

    • Moles of CH₄ = mass (g) / molar mass (g/mol) = 16 g / 16.05 g/mol ≈ 0.997 moles

From the balanced equation, 1 mole of CH₄ produces 1 mole of CO₂. Therefore, 0.997 moles of CH₄ will produce approximately 0.997 moles of CO₂. To find the mass of CO₂ produced:

  1. Calculate the molar mass of CO₂:

    • Carbon (C): 12.01 g/mol
    • Oxygen (O): 16.00 g/mol
    • Molar mass of CO₂ = 12.01 + (2 × 16.00) = 44.01 g/mol

  2. Calculate the mass of CO₂ produced:

    • Mass of CO₂ = moles × molar mass = 0.997 moles × 44.01 g/mol ≈ 43.88 g

Notes

  • This example illustrates how stoichiometry allows us to relate reactants and products quantitatively. Variations can include different reactants and products, as well as incomplete combustion scenarios.

Example 2: Synthesis of Water

Context

The synthesis of water is an important chemical reaction that can be analyzed to determine how much water can be produced from a given amount of hydrogen and oxygen.

The balanced equation for the synthesis of water is:

\[ 2H_2 + O_2 \rightarrow 2H_2O \]

Example

If we start with 4 grams of hydrogen (H₂) and 32 grams of oxygen (O₂), we can determine how much water will be produced.

  1. Calculate moles of hydrogen:

    • Molar mass of H₂ = 2 × 1.01 g/mol = 2.02 g/mol
    • Moles of H₂ = 4 g / 2.02 g/mol ≈ 1.98 moles

  2. Calculate moles of oxygen:

    • Molar mass of O₂ = 2 × 16.00 g/mol = 32.00 g/mol
    • Moles of O₂ = 32 g / 32.00 g/mol = 1.00 mole

According to the balanced equation, 2 moles of H₂ react with 1 mole of O₂ to produce 2 moles of H₂O. Therefore, we should check the limiting reactant:

  • H₂: 1.98 moles H₂ requires 0.99 moles O₂ (1:2 ratio).
  • O₂: 1.00 mole O₂ can react with 2.00 moles H₂.

Since hydrogen is the limiting reactant, we will use its amount to calculate water produced:

  1. Calculate moles of water produced:
  • Moles of H₂O = moles of H₂ = 1.98 moles
  • Mass of H₂O = moles × molar mass of H₂O = 1.98 moles × 18.02 g/mol ≈ 35.64 g

Notes

  • This example emphasizes the concept of limiting reactants in stoichiometric calculations. Variations can include different amounts of reactants or examining other products formed under different conditions.

Example 3: Reaction of Calcium Carbonate with Hydrochloric Acid

Context

The reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) is a common experiment in chemistry to produce carbon dioxide. The balanced equation is:

\[ CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2 \]

Example

If we start with 10 grams of calcium carbonate, we can determine how much carbon dioxide is produced.

  1. Calculate moles of calcium carbonate:

    • Molar mass of CaCO₃ = 40.08 (Ca) + 12.01 (C) + (3 × 16.00) (O) = 100.09 g/mol
    • Moles of CaCO₃ = 10 g / 100.09 g/mol ≈ 0.100 moles

According to the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, 0.100 moles of CaCO₃ will produce 0.100 moles of CO₂.

  1. Calculate the mass of CO₂ produced:

    • Molar mass of CO₂ = 44.01 g/mol
    • Mass of CO₂ = moles × molar mass = 0.100 moles × 44.01 g/mol ≈ 4.40 g

Notes

  • This example highlights the practical application of stoichiometry in laboratory settings. Variations can include using different acids or solid carbonates to observe different reaction rates or products.

These examples of stoichiometry and mole calculations illustrate the importance of understanding chemical reactions quantitatively, enabling better predictions and applications in real-world scenarios.