Stoichiometry and the Mole Concept

Table of Contents

Introduction to Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. The word comes from the Greek "stoicheion" (element) and "metron" (measure).

Stoichiometry allows chemists to:

To master stoichiometry, you must first understand the mole concept, which is the foundation for quantitative chemistry.

The Mole Concept

The mole is the SI unit for amount of substance. It represents a specific number of particles (atoms, molecules, ions, or electrons).

1 mole = 6.022 × 1023 particles

This number (6.022 × 1023) is known as Avogadro's number or Avogadro's constant. The mole concept provides a bridge between the microscopic world of atoms and molecules and the macroscopic world that we can measure in the laboratory.

MOLE 6.022×10²³ Mass (g) Volume Atoms Formula

Figure 1: The central role of the mole concept in chemical calculations

Relative Atomic and Molecular Masses

Relative Atomic Mass (Ar)

The relative atomic mass of an element is the average mass of one atom of that element compared to 1/12 of the mass of a carbon-12 atom.

Relative atomic mass is:

Relative Molecular Mass (Mr)

The relative molecular mass of a compound is the sum of the relative atomic masses of all atoms in the molecule.

Example 1: Calculate the relative molecular mass of sulfuric acid (H2SO4).

Solution:

Mr of H2SO4 = (2 × Ar of H) + (1 × Ar of S) + (4 × Ar of O)

Mr of H2SO4 = (2 × 1.0) + (1 × 32.1) + (4 × 16.0)

Mr of H2SO4 = 2.0 + 32.1 + 64.0 = 98.1

Avogadro's Number

Avogadro's number (NA) is defined as the number of atoms in exactly 12 g of carbon-12.

NA = 6.022 × 1023 mol-1

This incredibly large number allows us to work with quantities of atoms and molecules that are measurable in the laboratory.

Avogadro's Number Visualization 1 mole = 6.022×10²³ particles 12g of carbon-12 Avogadro's number of atoms Scale comparison

Figure 2: Visualization of Avogadro's Number

Example 2: Calculate the number of water molecules in 1.5 moles of water.

Solution:

Number of molecules = number of moles × Avogadro's number

Number of molecules = 1.5 mol × 6.022 × 1023 molecules/mol

Number of molecules = 9.033 × 1023 molecules

Molar Mass

The molar mass (M) of a substance is the mass of one mole of that substance. It is numerically equal to the relative atomic or molecular mass expressed in grams per mole (g/mol).

For any substance:

Example 3: Calculate the molar mass of glucose (C6H12O6).

Solution:

Molar mass of C6H12O6 = (6 × molar mass of C) + (12 × molar mass of H) + (6 × molar mass of O)

Molar mass of C6H12O6 = (6 × 12.0 g/mol) + (12 × 1.0 g/mol) + (6 × 16.0 g/mol)

Molar mass of C6H12O6 = 72.0 g/mol + 12.0 g/mol + 96.0 g/mol = 180.0 g/mol

Converting Between Mass, Moles, and Number of Particles

The mole concept allows us to convert between mass, moles, and number of particles using the following relationships:

Moles Mass (g) Particles Molar mass Avogadro's number Molar mass × Avogadro's number n = m/M N = n×NA

Figure 3: Conversion triangle showing relationships between mass, moles, and particles

Number of moles (n) = Mass (g) ÷ Molar mass (g/mol)

Mass (g) = Number of moles (n) × Molar mass (g/mol)

Number of particles = Number of moles (n) × Avogadro's number (NA)

Empirical and Molecular Formulas

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms present in a compound.

Molecular Formula

The molecular formula shows the actual number of atoms of each element in a molecule of the compound.

The molecular formula is a whole-number multiple of the empirical formula:

Molecular formula = (Empirical formula)n

where n is a whole number.

Empirical vs. Molecular Formula Example: Glucose Empirical Formula CH₂O ×6 Molecular Formula C₆H₁₂O₆ Molecular formula = (Empirical formula)ₙ where n = 6 for glucose

Figure 4: Relationship between empirical and molecular formulas

Determining Empirical Formula from Percentage Composition

Example 4: A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.

Solution:

  1. Assume 100 g of the compound (so percentages become grams)
  2. Convert mass to moles:
    • Moles of C = 40.0 g ÷ 12.0 g/mol = 3.33 mol
    • Moles of H = 6.7 g ÷ 1.0 g/mol = 6.7 mol
    • Moles of O = 53.3 g ÷ 16.0 g/mol = 3.33 mol
  3. Divide by the smallest number of moles:
    • C: 3.33 ÷ 3.33 = 1
    • H: 6.7 ÷ 3.33 = 2.01 ≈ 2
    • O: 3.33 ÷ 3.33 = 1
  4. Empirical formula: CH2O

Determining Molecular Formula from Empirical Formula

Example 5: The empirical formula of a compound is CH2O. If its molecular mass is 180 g/mol, what is its molecular formula?

Solution:

  1. Calculate the empirical formula mass:
    Mass of CH2O = 12.0 + (2 × 1.0) + 16.0 = 30.0 g/mol
  2. Find n:
    n = Molecular mass ÷ Empirical formula mass
    n = 180 g/mol ÷ 30.0 g/mol = 6
  3. Molecular formula = (Empirical formula)n = (CH2O)6 = C6H12O6

Chemical Equations and Calculations

Balanced Chemical Equations

A balanced chemical equation shows the relative amounts of reactants and products in a chemical reaction, with the same number of atoms of each element on both sides of the equation.

Balanced Chemical Equation 2H₂ + O₂ → 2H₂O + → 4 H atoms 2 O atoms 4 H atoms + 2 O atoms

Figure 5: Balanced chemical equation for the formation of water

In a balanced chemical equation:

Stoichiometric Calculations

Once we have a balanced equation, we can use the mole ratios to calculate quantities of reactants and products.

Example 6: Iron(III) oxide reacts with carbon monoxide to produce iron and carbon dioxide according to the following balanced equation:

Fe2O3 + 3CO → 2Fe + 3CO2

How many grams of iron can be produced from 8.0 g of Fe2O3?

Solution:

  1. Convert mass of Fe2O3 to moles:
    Molar mass of Fe2O3 = (2 × 55.8) + (3 × 16.0) = 159.6 g/mol
    Moles of Fe2O3 = 8.0 g ÷ 159.6 g/mol = 0.050 mol
  2. Use the mole ratio from the balanced equation:
    From the balanced equation: 1 mol Fe2O3 produces 2 mol Fe
    Moles of Fe = 0.050 mol Fe2O3 × (2 mol Fe / 1 mol Fe2O3) = 0.10 mol Fe
  3. Convert moles of Fe to mass:
    Mass of Fe = 0.10 mol × 55.8 g/mol = 5.58 g ≈ 5.6 g

Limiting Reactants

In many reactions, one reactant is completely consumed before the others. This reactant, called the limiting reactant, determines the maximum amount of product that can be formed.

To identify the limiting reactant:

  1. Calculate the moles of each reactant present
  2. Convert these to theoretical moles of product using the mole ratios from the balanced equation
  3. The reactant that produces the least amount of product is the limiting reactant
Limiting Reactant Concept H₂ H₂ H₂ O₂ O₂ H₂O H₂O H₂O H₂O O₂

Example Problem

For the reaction: 2H₂ + O₂ → 2H₂O, if you have 6 moles of H₂ and 3 moles of O₂:

  1. 6 moles H₂ × (2 moles H₂O/2 moles H₂) = 6 moles H₂O possible
  2. 3 moles O₂ × (2 moles H₂O/1 mole O₂) = 6 moles H₂O possible
  3. In this case, there is no limiting reactant (stoichiometric amounts)

Self-Assessment Questions

Q1: For the reaction N₂ + 3H₂ → 2NH₃, if you have 1 mole of N₂ and 4 moles of H₂, which reactant is limiting?

Answer:

  1. 1 mole N₂ × (2 moles NH₃/1 mole N₂) = 2 moles NH₃ possible
  2. 4 moles H₂ × (2 moles NH₃/3 moles H₂) ≈ 2.67 moles NH₃ possible
  3. N₂ is the limiting reactant as it produces less product

Q2: In the reaction 2Al + 3Cl₂ → 2AlCl₃, if you have 5 moles of Al and 9 moles of Cl₂, how many moles of AlCl₃ can be produced?

Answer:

  1. 5 moles Al × (2 moles AlCl₃/2 moles Al) = 5 moles AlCl₃ possible
  2. 9 moles Cl₂ × (2 moles AlCl₃/3 moles Cl₂) = 6 moles AlCl₃ possible
  3. Al is limiting, so maximum 5 moles AlCl₃ can be produced

Q3: Why is the concept of limiting reactants important in industrial chemistry?

Answer: The limiting reactant concept is crucial in industry because:

  • It determines the maximum yield of product
  • Helps optimize reactant quantities to minimize waste
  • Affects cost efficiency of chemical processes
  • Determines which reactant needs to be carefully controlled in the process