Stoichiometry

Stoichiometry Notes with Examples

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1. Introduction to Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction.

Key concepts:

Mole ratio

Limiting and excess reactants

Theoretical, actual, and percent yield

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2. Mole Concept in Stoichiometry

In stoichiometry, calculations are based on moles, which help relate masses of substances in a reaction.

1 mole of any substance contains Avogadro’s number (6.022 × 10²³) of particles.

Molar mass (g/mol) is the mass of one mole of a substance.

Example:

Molar mass of H₂O = (2 × 1) + (1 × 16) = 18 g/mol

This means 1 mole of water weighs 18 grams.

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3. Balanced Chemical Equations

A balanced chemical equation ensures that the number of atoms of each element is the same on both sides.

Example: Combustion of methane

CH₄ + 2O₂ → CO₂ + 2H₂O

Interpretation: 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.

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4. Mole-to-Mole Conversions

Using the balanced equation, we can determine how many moles of a product form from a given amount of reactant.

Example:
How many moles of CO₂ are produced when 3 moles of CH₄ react?

Solution:
From the balanced equation:

1 \text{ mole } CH₄ \rightarrow 1 \text{ mole } CO₂

3 \times \left(\frac{1 \text{ mole } CO₂}{1 \text{ mole } CH₄}\right) = 3 \text{ moles of CO₂}

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5. Mass-to-Mass Conversions

Stoichiometry also allows us to calculate the mass of reactants or products.

Example:
How many grams of CO₂ are produced from 16 g of CH₄?

Solution:

Molar mass of CH₄ = 16 g/mol

Molar mass of CO₂ = 44 g/mol

From the equation:

1 \text{ mole } CH₄ \rightarrow 1 \text{ mole } CO₂

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6. Limiting and Excess Reactants

Limiting reactant: The reactant that gets completely used up first, limiting the amount of product formed.

Excess reactant: The reactant that remains after the reaction is complete.

Example:
Given the reaction:

2H₂ + O₂ → 2H₂O

Solution:

2 moles of H₂ react with 1 mole of O₂.

5 moles of H₂ would require:

5 \times \left(\frac{1 \text{ mole } O₂}{2 \text{ moles } H₂}\right) = 2.5 \text{ moles of O₂}

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7. Percent Yield

Percent yield compares the actual yield (from experiment) to the theoretical yield (calculated).

\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100

Example:
If the theoretical yield of a reaction is 10 g and the actual yield obtained is 8 g:

\left(\frac{8}{10}\right) \times 100 = 80\% \text{ yield}