The Mole Concept, Percent Composition, and Stoichiometry in General Chemistry

Chapter 3: Mole Concept and Stoichiometry

Introduction

This section covers foundational concepts in General Chemistry, including the mole, percent composition, empirical and molecular formulas, and stoichiometry. Mastery of these topics is essential for quantitative chemical analysis and problem-solving in laboratory and industrial settings.

The Mole and Avogadro's Number

Definition and Importance

• Mole (mol): The SI unit for the amount of substance, defined as containing exactly $6.022 \times 10^{23}$ elementary entities (Avogadro's Number, $N_A$).

• Avogadro's Number ($N_A$): $6.022 \times 10^{23}$, the number of atoms in exactly 12 grams of carbon-12.

• Purpose: The mole allows chemists to count atoms, molecules, or ions by weighing macroscopic amounts of material.

Example: 1 mole of carbon-12 atoms has a mass of 12.00 g and contains $6.022 \times 10^{23}$ atoms.

Counting Numbers in Chemistry

• Pair: 2 items

• Dozen: 12 items

• Mole: $6.022 \times 10^{23}$ items

Atomic Mass Unit (amu) and Molar Mass

Definitions

• Atomic Mass Unit (amu): A unit of mass used to express atomic and molecular weights; 1 amu is defined as one twelfth the mass of a carbon-12 atom.

• Molar Mass: The mass of one mole of a substance, expressed in grams per mole (g/mol). For elements, the molar mass in g/mol is numerically equal to the atomic mass in amu.

Example: 1 atom of boron = 10.81 amu; 1 mole of boron atoms = 10.81 g.

Percent Composition by Mass

Definition and Calculation

Percent composition expresses the mass percentage of each element in a compound.

• Formula:

$\text{Percent by mass of element} = \left( \frac{n \times \text{atomic weight of element}}{\text{formula weight of compound}} \right) \times 100\%$

• n: Number of atoms of the element in the formula unit.

• Formula weight: Sum of atomic weights of all atoms in the compound.

Example: For water ($\mathrm{H_2O}$):

• Hydrogen: 2 atoms × 1.01 g/mol = 2.02 g/mol

• Oxygen: 1 atom × 16.00 g/mol = 16.00 g/mol

• Formula weight = 18.02 g/mol

• Percent H = $(2.02/18.02) \times 100\% = 11.2\%$

• Percent O = $(16.00/18.02) \times 100\% = 88.8\%$

Applications:

• Purity analysis in industry and pharmaceuticals

• Forensic analysis (e.g., identifying unknown powders)

• Determining empirical formulas

Empirical and Molecular Formulas

Empirical Formula

• Definition: The simplest whole-number ratio of atoms of each element in a compound.

• Determination: Use mass or percent composition data to calculate moles of each element, then divide by the smallest number of moles to get the ratio.

Example: A compound contains 8.56 g C and 1.44 g H. Calculate moles of each, find the ratio, and write the empirical formula.

Molecular Formula

• Definition: The actual number of atoms of each element in a molecule of the compound.

• Relationship: $\text{Molecular formula} = n \times \text{Empirical formula}$, where $n$ is a whole number.

• Calculation: $n = \frac{\text{Molar mass of compound}}{\text{Molar mass of empirical formula}}$

Example: If empirical formula is CH and molar mass is 78 g/mol, empirical formula mass is 13 g/mol, so $n = 78/13 = 6$, molecular formula is $C_6H_6$.

Stoichiometry

Definition and Importance

Stoichiometry is the study of quantitative relationships between reactants and products in a chemical reaction, based on the balanced chemical equation.

• Balanced Equation: Shows the relative number of moles of reactants and products.

• Stoichiometric Coefficients: Used as conversion factors between substances.

Stoichiometric Calculations

1. Ensure the equation is balanced.

2. Convert given mass to moles (using molar mass).

3. Use mole ratios from the balanced equation to find moles of desired substance.

4. Convert moles back to mass if required.

Example: $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$

• How many moles of water are formed from 3.65 g of $H_2$?

• How many grams of $O_2$ are consumed?

Limiting Reactant and Theoretical Yield

Limiting Reactant

• Definition: The reactant that is completely consumed first, thus limiting the amount of product formed.

• Identification: Calculate the amount of product each reactant could produce; the smallest amount indicates the limiting reactant.

Example: $3H_2(g) + N_2(g) \rightarrow 2NH_3(g)$ If 3.0 mol $N_2$ and 6.0 mol $H_2$ react, determine the limiting reactant and amount of $NH_3$ formed.

Theoretical and Percent Yield

• Theoretical Yield: The maximum amount of product that can be formed from the limiting reactant.

• Actual Yield: The amount of product actually obtained from a reaction.

• Percent Yield:

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

Note: Percent yield cannot exceed 100%.

Summary Table: Key Concepts and Formulas

Concept	Definition	Key Formula
Mole	Amount of substance containing $6.022 \times 10^{23}$ entities	1 mol = $6.022 \times 10^{23}$ particles
Molar Mass	Mass of 1 mole of substance (g/mol)	Sum of atomic masses in formula
Percent Composition	Mass percent of each element in compound	$\frac{n \times \text{atomic mass}}{\text{formula mass}} \times 100\%$
Empirical Formula	Simplest whole-number ratio of atoms	From mass or % composition
Molecular Formula	Actual number of atoms in molecule	$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$
Percent Yield	Efficiency of reaction	$\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%$

Additional info:

• Significant figures are important in all calculations; use the correct number as specified in problems.

• Knowledge of periodic table groups, element symbols, and naming conventions is foundational for these topics.