Study Notes on Stoichiometry of Chemical Reactions

CHAPTER 7: STOICHIOMETRY OF CHEMICAL REACTIONS

The Mole

• Definition: The mole is a fundamental unit of measurement in chemistry, specifically designed to express macroscopic amounts of a chemical substance in terms of fundamental particles. It functions analogously to everyday collective units like a "pair" (2 items), a "dozen" (12 items), or a "gross" (144 items), but for a vastly larger quantity.

• Formal Definition: A mole is formally defined as the amount of a substance that contains as many discrete entities (whether individual atoms, molecules, ions, electrons, or other particles) as there are atoms in exactly 12 grams (g) of pure carbon-12 (carbon with 6 protons and 6 neutrons, denoted as $^{12}C$).

• Purpose: The mole provides a crucial and practical link between the measurable mass of a sample in grams (a macroscopic property) and the actual, uncountably large number of atoms, molecules, or ions present within that sample (a microscopic property). It enables chemists to perform calculations based on chemical equations, which are expressed in terms of moles.

Avogadro's Number

• Definition: Avogadro's number, universally denoted as $N_A$, is the exact number of entities (atoms, molecules, ions, etc.) in one mole of any substance. Its precisely determined value is $6.02214076 \times 10^{23}$. For most calculations, the value is often rounded to $6.022 \times 10^{23}$.

• Significance: Named in honor of the Italian scientist Amedeo Avogadro, this constant is pivotal for converting between the number of moles of a substance and the actual count of individual particles in a given sample. It allows scientists to bridge the gap between measurable laboratory quantities and the microscopic world of atoms and molecules.

• Equation: The commonly used approximation for Avogadro's number is $N_A = 6.022 \times 10^{23}$. This means that one mole of any substance contains $6.022 \times 10^{23}$ particles of that substance.

• Variety in Moles: While one mole always contains the same number of particles ($N_A$), the mass of one mole of different elements or compounds varies significantly. This is because different atoms have different atomic masses; therefore, a mole of one element will not have the same mass as a mole of another element due to the differing weights of their constituent atoms.

Mole of Atoms

• General Formula: For any element, 1 mole of that element always contains $6.022 \times 10^{23}$ atoms of that specific element. This principle applies universally across the periodic table, making the mole a versatile unit for all elements.

• Specific Examples:

  • 1 mole of carbon = $6.022 \times 10^{23}$ atoms of carbon.

  • 1 mole of sodium = $6.022 \times 10^{23}$ atoms of sodium.

  • This also applies to molecules, e.g., 1 mole of water ($H_2O$) = $6.022 \times 10^{23}$ molecules of water.

Converting Moles to Molecules

• Application of Avogadro's Number: To convert a given number of moles of a substance to the equivalent number of particles (molecules or atoms), Avogadro's number serves as the essential conversion factor. This process utilizes dimensional analysis to ensure proper unit cancellation.

  • For CO2, in 0.50 moles, the calculation involves the following steps:

  • Step 1: Identify the given quantity and the desired quantity.

    Given: 0.50 mole of $CO_2$

    Need: Number of molecules of $CO_2$.

  • Step 2: Plan the conversion pathway. This involves using Avogadro's number to convert from moles to individual molecules.

  $\text{moles of CO}<em>2 \rightarrow \text{molecules of CO}</em>2$

  • Step 3: Set up the conversion factor based on the definition of a mole.

  $\text{1 mole of CO}<em>2 = 6.022 \times 10^{23} \text{ molecules of CO}</em>2$

  • Step 4: Perform the calculation using the conversion factor, arranging it so that the undesired unit (moles) cancels out.

  $0.50 \text{ moles CO}<em>2 \times \frac{6.022 \times 10^{23} \text{ molecules CO}</em>2}{1 \text{ mole CO}<em>2} = 3.011 \times 10^{23} \text{ molecules CO}</em>2$

Examples of Mole Calculations

• Finding atoms from moles: To calculate the number of atoms in 2.0 moles of aluminium (Al):

  • Step 1: Identify the given and needed quantities.

    Given: 2.0 moles of Al

    Need: Atoms of Al.

  • Step 2: The conversion factor is derived from Avogadro's number, stating that:

  $\text{1 mole of Al} = 6.022 \times 10^{23} \text{ atoms of Al}$

  • Final Calculation: Multiply the given moles by Avogadro's number:

  $2.0 \text{ moles Al} \times \frac{6.022 \times 10^{23} \text{ atoms Al}}{1 \text{ mole Al}} = 1.2044 \times 10^{24} \text{ atoms Al}$

  The answer will be $1.2 \times 10^{24}$ atoms of Al (keeping two significant figures).

• Finding moles from atoms: For $1.8 \times 10^{24}$ atoms of sulfur (S):

  • Given: $1.8 \times 10^{24}$ atoms of S

  • Find: Moles of S:

  Using Avogadro's number: The relationship is:

  $\text{1 mole of S} = 6.022 \times 10^{23} \text{ atoms S}$

  • Conversion calculates the answer by dividing the number of atoms by Avogadro's number:

  $1.8 \times 10^{24} \text{ atoms S} \times \frac{1 \text{ mole S}}{6.022 \times 10^{23} \text{ atoms S}} \approx 2.99 \text{ moles S}$

  The answer, rounded, is approximately $3.0$ moles of S.

Moles of Elements in a Chemical Formula

• Chemical Formula Example: Aspirin, with the molecular formula $C<em>9H</em>8O_4$. This formula provides the exact number of atoms of each element in one molecule.

  • Atoms per molecule of Aspirin:

  • 9 atoms of Carbon (C)

  • 8 atoms of Hydrogen (H)

  • 4 atoms of Oxygen (O)

• Calculating moles in a compound: The subscripts in a chemical formula represent not only the ratio of atoms in one molecule but also the ratio of moles of each element in one mole of the compound. So, in one mole of aspirin ($C<em>9H</em>8O_4$), the quantities of each atom in moles are:

  • 9 moles of C atoms

  • 8 moles of H atoms

  • 4 moles of O atoms

• Subscripts: These subscripts thus provide direct conversion factors to relate the moles of individual elements to the moles of the overall compound. For instance, for $C<em>9H</em>8O_4$, you can say that for every 1 mole of aspirin, there are 9 moles of C. This is crucial for stoichiometric calculations involving compounds.

Molar Mass

• Resource Definition: The molar mass of a substance is precisely defined as the mass in grams of one mole of that substance. Its standard unit of expression is grams per mole (g/mol).

• Relation to Atomic Mass: A key property of the molar mass is its numerical equivalence to its atomic mass (for elements) or molecular/formula mass (for compounds) when expressed in atomic mass units (amu). For example, if an atom has an average atomic mass of 23.0 amu, then one mole of those atoms will have a mass of 23.0 g.

• Example: Carbon-12 ($^{12}C$) has an atomic mass of exactly 12 amu. Consequently, 1 mole of carbon-12 has a molar mass of 12 g/mol. This directly illustrates how the mole concept links the microscopic scale (amu of a single atom) to the macroscopic scale (grams of a mole of atoms).

Molar Mass Calculation: $Li<em>2CO</em>3$

• Calculating molar mass steps: To find the molar mass of lithium carbonate ($Li<em>2CO</em>3$), sum the molar masses of all constituent atoms, taking into account their subscripts in the formula.

  • Step 1: Identify the molar mass of each element from the periodic table:

  • Lithium (Li) = 6.94 g/mol

  • Carbon (C) = 12.01 g/mol

  • Oxygen (O) = 16.00 g/mol

  • Step 2: Multiply each element's molar mass by its subscript in the chemical formula to find its total contribution to the compound's molar mass:

  • $\text{Li: } 2 \times 6.94 \text{ g/mol} = 13.88 \text{ g}$

  • $\text{C: } 1 \times 12.01 \text{ g/mol} = 12.01 \text{ g}$

  • $\text{O: } 3 \times 16.00 \text{ g/mol} = 48.00 \text{ g}$

  • Final Calculation of Molar Mass: Sum these individual contributions to get the total molar mass of $Li<em>2CO</em>3$.

  $\text{1 mole of } Li<em>2CO</em>3 = 13.88 \text{ g} + 12.01 \text{ g} + 48.00 \text{ g} = 73.89 \text{ g/mol}$

Example: Molar Mass of $C<em>2H</em>6O$

• Formula: The compound is ethanol, $C<em>2H</em>6O$ (or $CH<em>3CH</em>2OH$).

• Elements and Their Molar Mass (from periodic table):

  • C = 12.01 g/mol

  • H = 1.008 g/mol

  • O = 16.00 g/mol

• Calculation Steps: Determine the mass contribution of each element and sum them.

  • $\text{2 moles C: } 2 \times 12.01 \text{ g/mol} = 24.02 \text{ g}$

  • $\text{6 moles H: } 6 \times 1.008 \text{ g/mol} = 6.048 \text{ g}$

  • $\text{1 mole O: } 1 \times 16.00 \text{ g/mol} = 16.00 \text{ g}$

• Final Molar Mass: Add the masses of all atoms to find the total molar mass of $C<em>2H</em>6O$.
  $(24.02 + 6.048 + 16.00 = 46.068 \text{ g/mol})$
  Rounded to two decimal places, the molar mass is $46.07 \text{ g/mol}$.

Converting Grams to Moles

• Example: To convert 737 g of sodium chloride (NaCl) to moles:

  • Step 1: Identify the given quantity and the desired quantity.

  • Given: 737 g of NaCl

  • Need: Moles of NaCl.

  • Step 2: Write a plan: Convert grams (g) of NaCl to moles (mol) using the molar mass of NaCl as the conversion factor.

  • Conversion factor: First, calculate the molar mass of NaCl:

    • Na: 22.99 g/mol

    • Cl: 35.45 g/mol

    • Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol.

  • Final Setup: Arrange the conversion factor so that grams cancel out, leaving moles.
    $\text{moles NaCl} = 737 \text{ g NaCl} \times \frac{1 \text{ mol NaCl}}{58.44 \text{ g NaCl}} = 12.61 \text{ moles NaCl}$

Practical Applications of Stoichiometry

• Broad Utility: Stoichiometry is the quantitative study of reactants and products in chemical reactions. It is essential for predicting the amounts of substances involved in reactions and plays a crucial role in various fields.

• Examples: Determine grams of elemental compounds after reactions or calculate the amount of product formed from a given amount of reactant.

  • Limiting Reactants: Identifying which reactant will be completely consumed first, thereby limiting the amount of product that can be formed.

  • Theoretical Yield: Calculating the maximum amount of product that could be formed from given amounts of reactants, assuming 100% efficiency.

  • Percentage Yield: Comparing the actual amount of product obtained in an experiment to the theoretical yield, expressed as a percentage, to evaluate reaction efficiency:
    $\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

• Apply stoichiometric ratios derived from balanced chemical equations to precisely track conversions between reactants and products, ensuring that no matter is unaccounted for.

Reaction Stoichiometry

• Law of Conservation of Mass: A foundational principle in chemistry, stating that in any closed system, matter cannot be created or destroyed. This means that the total mass of the reactants before a chemical reaction must equal the total mass of the products after the reaction.

• Mass Relationship: This law is directly applied in balancing chemical equations. A balanced chemical equation ensures that the number of atoms of each element is the same on both the reactant and product sides, guaranteeing that the mass of products equals the mass of the reactants. This allows for quantitative predictions about reaction outcomes.

Types of Chemical Reactions

• Combustion reactions: These are high-temperature reactions that typically involve the rapid oxidation of a fuel (often a hydrocarbon or other organic compound) in the presence of an oxidant, usually oxygen ($O<em>2$). They typically produce carbon dioxide ($CO</em>2$) and water ($H_2O$) as products, along with significant amounts of heat and light.

• Single and Double Replacement reactions:

  • Single Replacement (or Single Displacement): An element reacts with a compound, displacing another element from the compound. The general form is $A + BC \rightarrow AC + B$. This depends on the relative reactivity of the elements.

  • Double Replacement (or Double Displacement): Two compounds react, exchanging ions to form two new compounds. The general form is $AB + CD \rightarrow AD + CB$. These reactions often lead to the formation of a precipitate, a gas, or water.

• Formation of Precipitates: When two aqueous solutions are mixed, and an insoluble solid (a precipitate) forms, it is a common type of double replacement reaction. These reactions are used to illustrate solubility principles and to identify the presence of specific ions in solutions, as well as to produce reaction byproducts with specific physical properties.

Percent Composition

• Definition: Percent composition is defined as the mass fraction of each element in a compound, expressed as a percentage. It indicates the relative amount of each element by mass within a compound.

• Formula: The formula to calculate the percent composition by mass for an element in a compound is:
  $\text{Percent by mass of element} = \frac{\text{mass of element in 1 mole of compound}}{\text{molar mass of compound}} \times 100\%$

• Application: Percent composition is a key piece of empirical data. It is crucial in determining the empirical and subsequently the molecular formula of an unknown compound based on experimental analysis of its constituent elements' mass ratios.

Empirical vs Molecular Formulas

• Empirical Formula: This is the simplest whole number ratio of atoms for each element present in a compound. It represents the lowest possible integer values of the subscripts in a chemical formula. For example, the empirical formula for glucose ($C<em>6H</em>{12}O<em>6$) is $CH</em>2O$.

• Molecular Formula: This represents the actual number of atoms of each element present in a single molecule of the compound. It is always a whole number multiple of the empirical formula. The relationship can be expressed as: $\text{Molecular Formula} = n \times \text{Empirical Formula}$, where $n$ is a whole number integer ($n = \frac{\text{Molar Mass of Molecular Formula}}{\text{Molar Mass of Empirical Formula}}$).

Oxidation-Reduction Reactions (Redox)

• Definition: Redox reactions are chemical reactions that involve the transfer of electrons between reactants. These are fundamental to many chemical and biochemical processes.

• Oxidation: Defined as the loss of one or more electrons by an atom, ion, or molecule. An increase in the oxidation state (or oxidation number) of an element indicates oxidation.

• Reduction: Defined as the gain of one or more electrons by an atom, ion, or molecule. A decrease in the oxidation state (or oxidation number) of an element indicates reduction.

• Oxidizing and Reducing Agents: These reactions always occur simultaneously.

  • Oxidizing Agent (or Oxidant): The substance that causes another substance to be oxidized (by accepting electrons from it) and is itself reduced.

  • Reducing Agent (or Reductant): The substance that causes another substance to be reduced (by donating electrons to it) and is itself oxidized.

Acid-Base Reactions

• Definition: Acid-base reactions involve the transfer of protons ($H^+$ ions) or the formation of coordinate covalent bonds.

• Acids: According to the Arrhenius definition, acids are substances that increase the concentration of hydronium ions ($H_3O^+$) when dissolved in water. According to the Brønsted-Lowry definition, acids are proton ($H^+$) donors.

• Bases: According to the Arrhenius definition, bases are substances that increase hydroxide ion ($OH^-$) concentrations in solutions. According to the Brønsted-Lowry definition, bases are proton ($H^+$) acceptors.

• Neutralization: These are typically exothermic double-replacement reactions that occur between an acid and a base. The products of a complete neutralization reaction are a salt and water. For example, hydrochloric acid ($HCl$) and sodium hydroxide ($NaOH$) react to produce sodium chloride ($NaCl$) and water ($H<em>2O$): $HCl</em>{(aq)} + NaOH<em>{(aq)} \rightarrow NaCl</em>{(aq)} + H<em>2O</em>{(l)}$. The "salt" often refers to an ionic compound formed from the cation of the base and the anion of the acid.