Module 1 Lecture 1 2025 2 per page

Learning Objectives

• Understand the concept of the mole.

• Write balanced chemical equations.

• Carry out calculations involving the mole and molar mass.

• Understand the concept of a limiting reactant.

Stoichiometry

• Definition: Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions.

• Origin: The term comes from Greek roots meaning "element" and "to measure".

• Purpose: It allows the weighing of bulk quantities of substances instead of measuring individual molecules.

Chemical Equations

• Chemical equations represent the ratios of reactants reacting to produce products.

• Example: The equation N2(g) + 3H2(g) → 2NH3(g) indicates that one molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia.

• Importance: Chemical equations must be balanced to reflect the conservation of mass.

The Mole

• The mole (mol) is the SI unit for the amount of substance.

• Definition: One mole corresponds to 6.022 × 10^23 entities, equivalent to the number of atoms in 12 g of 12C.

• Significance: It helps chemists relate mass to the number of particles in a substance.

Molar Mass

• Definition: Molar mass (M) is the mass of one mole of a substance.

• Units: Measured in g/mol (grams per mole).

• Calculation: M is defined by the formula M = m/n, where m is mass and n is the number of moles.

• Example: The molar mass of carbon is 12.01 g/mol. If a diamond weighs 109.13 g, the amount of carbon in it can be calculated using the molar mass.

Mole Relationships

• Balanced equations provide mole ratios between reactants and products, essential for stoichiometric calculations.

Limiting Reagents

• Concept: In reactions with an imbalanced ratio of reactants, one may be exhausted first (the limiting reagent).

• Example: For the reaction 2H2(g) + O2(g) → 2H2O(l), if given 5.0 g of H2 and 21.0 g of O2, the reaction yields a limited amount of water based on the reactants' proportions.

• Calculation: Determining remaining excess reactants based on the limiting reagent is crucial.

Solutions

• Solutions are defined based on the concentration of solute in solvent.

• Example: Dissolving 10.0 g of glucose in water to create a specific volume solution. Calculations involve concentration expressed in mol/L.

Conclusions

• Mastering stoichiometry requires understanding and practicing two main equations: M = m/n and c = n/V.

• Balancing chemical equations is vital for successful stoichiometric calculations.

• Practice is essential for proficiency in stoichiometry.