CHEM 191: Aqueous Solutions & Stoichiometry

• CHEM 191 Module 1 Lecture 1 covers chemical reactions in aqueous solution and stoichiometry.

Learning Objectives

• Understand the concept of the mole.
• How to write and balance chemical equations.
• Perform calculations using the equations n=m/M and c=n/V.
• Understand limiting reactants.

Stoichiometry

• It concerns the relative amounts of reactants and products in a reaction.
• Allows the determination of bulk quantities of reactants and products instead of individual atoms and molecules.

Chemical Equations

• A chemical equation shows reactants' ratio in a chemical reaction to produce products.
• Chemical equations use numbers.
• For example: N2(g) + 3H2(g) → 2NH3(g) indicates 1 molecule of nitrogen reacts with 3 molecules of hydrogen, yielding 2 molecules of ammonia.
• Balanced chemical equations are required in these examples.
• Chemical equations must be balanced to be valid.

The Mole

• The mole (mol) is the SI unit for the amount of a substance.
• One mole is defined as "the amount of substance that contains the same number of specified entities as there are atoms in exactly 12 g of the carbon isotope ¹²C."
• There are 6.022 × 10²³ atoms in 12 g of carbon-12; thus, 1 mole contains 6.022 × 10²³ entities.

Avogadro Constant

• Determined by the International Avogadro Project using a highly enriched 28Si single crystal.
• The constant is 6.02214078(18) × 10²³ mol⁻¹.
• Redefinition of the kilogram (May 2019) allows the Avogadro constant to be defined exactly.

Molar Mass

• Molar mass (M) is the mass of 1 mole of a specific substance.
• Defined as M=m/n, where m is mass and n is the amount of substance.
• Molar mass is measured in grams/mole (g/mol).
• Molar mass values are found in tables.

Example Calculation

• For a diamond made of pure carbon, carbon's molar mass is approximately 12.01 g/mol.
• To find how many moles are in a 109.13 g diamond:
  • n = 109.13 g / 12.01 g/mol = 9.087 mol.

Mole Relationships

• Balanced chemical equations describe mole relationships between reactants and products.
• Example: 2H₂(g) + O₂(g) → 2H₂O(l).
  • 2 molecules or moles of H₂ react with 1 molecule or mole of O₂ to produce 2 molecules or moles of H₂O.

Reaction Stoichiometry

• To calculate the amount of product from a given amount of reactant:
  • Convert grams of reactant A to moles of A using its molar mass.
  • Use the balanced equation's coefficients to find the moles of substance B.
  • Convert moles of B to grams of B using its molar mass.

Example

• Metallic iron reacts with oxygen to produce iron oxide: 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).
• If 2.0 g of iron reacts with excess oxygen, determining the mass of iron oxide formed requires:
  • n(Fe) = 2.0 g / 55.85 g/mol = 0.0358 mol.
  • From the balanced equation, 4 moles of Fe yield 2 moles of Fe₂O₃; thus, 0.0358 mol Fe yields 0.0179 mol Fe₂O₃.
  • Molar mass of Fe₂O₃ = (2 × 55.85 g/mol) + (3 × 16.00 g/mol) = 159.7 g/mol.
  • Mass of Fe₂O₃ = 0.0179 mol × 159.7 g/mol = 2.859 g.

Limiting Reagents

• Reactants may not always be in stoichiometric amounts.
• The limiting reagent limits the reaction.
• Example: 2H₂(g) + O₂(g) → 2H₂O(l). 5.0 g of H₂ reacts with 21.0 g of O₂.
• To determine which reactant remains after the reaction is complete:.
  • n(H₂) = 5.0 g / 2.0 g/mol = 2.5 mol.
  • n(O₂) = 21.0 g / 32 g/mol = 0.66 mol.
  • From stoichiometry, 2.5 mol H₂ produces 2.5 mol H₂O, and 0.66 mol O₂ produces 1.32 mol H₂O.
  • When 1.32 mol H₂O is produced, all O₂ is used; therefore, O₂ limits the reaction.

Limiting Reagents: Continued

• When 1.32 mol H₂O is produced, all the O₂ is used up.
• 1.32 mol H₂O requires 1.32 mol H₂ to have reacted.
• The remaining H₂ = 2.5 mol - 1.32 mol = 1.18 mol.
• Mass of remaining H₂ = 1.18 mol × 2.0 g/mol = 2.4 g.

Solutions

• Solutions are defined by their concentration, which indicates the amount of solute per volume of solvent.
• Concentration is defined with the equation C=n/V.
• For example, in a solution of 10.0 g of glucose (M = 180.156 g/mol) dissolved in water to make a final volume of 0.500 L:
  • n = 10.0 g / 180.156 g/mol = 5.55 × 10⁻² mol.
  • C = 5.55 × 10⁻² mol / 0.500 L = 1.11 × 10⁻¹ mol/L.

Expressing Concentration

• Concentration can be expressed various ways. Traditionally this is mol/L.
• Molarity (M), is a more modern, alternative unit. Molarity equals the amount of moles per cubic decimeter (mol/dm³).

Conclusions

• To be able to do stoichiometry requires two equations: n=m/M and c=n/V.
• Balancing chemical equations are a necessity for accurate applications of stoichiometry.