Compounds, Mole, and Stoichiometry

Introduction

This study guide covers essential topics from General Chemistry, focusing on the relationships between compounds, the mole concept, and stoichiometry. These foundational concepts are critical for understanding chemical reactions, quantitative calculations, and the composition of substances.

Atomic Mass and Isotopes

Atomic Mass

The atomic mass of an atom is determined by the number of protons and neutrons in its nucleus. By international agreement, atomic mass is measured in atomic mass units (amu), where 1 amu is defined as exactly one-twelfth the mass of a carbon-12 atom.

• Atomic mass unit (amu):

• Isotopes: Atoms of the same element with different numbers of neutrons.

• Example: Chlorine has two main isotopes: chlorine-35 (75.77%) and chlorine-37 (24.23%).

The atomic mass of chlorine is calculated as:

Calculating Atomic Weight from Isotopic Abundance

To find the atomic weight of an element with multiple isotopes, use the weighted average based on natural abundance.

• Example: Magnesium has three isotopes: 23.99 amu (78.99%), 24.99 amu (10.00%), 25.98 amu (11.01%).

• Calculation:

The Mole Concept

Definition of the Mole

The mole is the amount of substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 g of carbon-12. This number is known as Avogadro's number ().

• Avogadro's number:

• One mole of any element contains atoms, molecules, or formula units.

Molar Mass

The molar mass of a substance is the mass of one mole of its entities (atoms or molecules), expressed in grams per mole (g/mol).

• Example: Sodium (Na) has a molar mass of 22.99 g/mol.

• Example: Phosphorus (P) has a molar mass of 30.97 g/mol.

Important Conversion Factors

• Number of moles:

• Number of atoms:

Example Calculations:

• How many moles of magnesium are in 87.3 g of Mg?

• Calculate the number of grams of lead (Pb) in 12.4 moles of Pb.

• Calculate the number of atoms in 0.551 g of potassium (K).

Molecular Mass and Percent Composition

Molecular Mass

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

• Example: Methanol (CH3OH): amu

Percent Composition by Mass

The percent composition of a compound is the percentage by mass of each element in the compound.

• Example: Water (H2O): g/mol

• Percent H:

• Percent O:

Example: Sulfuric acid (H2SO4):

• Molar mass = g/mol

• Percent H:

• Percent S:

• Percent O:

Empirical and Molecular Formulas

The empirical formula gives the simplest whole-number ratio of atoms in a compound. The molecular formula gives the actual number of atoms of each element in a molecule.

• To determine the empirical formula from percent composition, assume 100 g of compound and convert mass to moles for each element.

• To determine the molecular formula, use the empirical formula and the compound's molar mass.

Example: A compound contains 2.52 g B and 1.803 g H. If the molecular mass is known, the molecular formula can be determined.

Chemical Reactions and Chemical Equations

Chemical Equations

A chemical reaction is a process in which substances are transformed into new substances. A chemical equation uses chemical symbols to represent the reactants and products.

• General format: Reactants → Products

• Example:

Balancing Chemical Equations

Balancing ensures the law of conservation of mass is obeyed. The number of atoms of each element must be the same on both sides of the equation.

• Change coefficients, not subscripts.

• Example:

Stoichiometry

Stoichiometric Calculations

Stoichiometry is the quantitative study of reactants and products in a chemical reaction. The coefficients in a balanced equation represent the number of moles of each substance.

• Convert mass to moles:

• Use mole ratios from the balanced equation to relate quantities.

• Convert moles back to mass if needed.

Example:

• How many moles of NO2 are formed from 1 mole of O2?

• How many grams of NO2 are formed from 2 moles of NO?

Limiting Reagent and Excess Reagent

The limiting reagent is the reactant that is completely consumed first, limiting the amount of product formed. The excess reagent is the reactant that remains after the reaction is complete.

• Identify the limiting reagent by comparing the mole ratios of reactants.

• Calculate the amount of product formed and the amount of excess reagent left.

Example: 124 g of Al reacts with 601 g of Fe2O3. Calculate the mass of Al2O3 formed and the excess reagent remaining.

Theoretical Yield, Actual Yield, and Percent Yield

Theoretical yield is the maximum amount of product that can be formed from the limiting reagent. Actual yield is the amount of product actually obtained. Percent yield measures the efficiency of a reaction.

• Percent yield is always less than or equal to 100%.

Example: If 1.54 × 103 g of V2O5 reacts with 1.96 × 103 g of Ca, and 803 g of V is obtained, calculate the percent yield.

Summary Table: Key Stoichiometric Relationships

Additional info: Some example problems and images were referenced in the original materials. For full mastery, students should practice calculations using the formulas and concepts provided above.