AS

Chapter 3 Notes: Chemical Reactions and Stoichiometry

Stoichiometry: Definition and Law of Conservation

  • Stoichiometry is the study of the mass relationships in chemistry.

  • Based on the Law of Conservation of Mass (Antoine Lavoisier, 1789):

    • We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.

  • In chemical reactions, mass is conserved; total mass of reactants equals total mass of products.

Chemical Equations: Components and Notation

  • Chemical equations are concise representations of chemical reactions.

  • Reactants appear on the left side; products on the right side.

  • States are written to the right of each compound: (g) = gas; (l) = liquid; (s) = solid; (aq) = in aqueous solution.

  • Coefficients are inserted to balance the equation to follow the law of conservation of mass.

  • Example: 2 H$2$ + O$2$ → 2 H$_2$O

  • Coefficients balance; subscripts are not changed to balance (why we add coefficients instead of changing subscripts is illustrated below).

  • Why coefficients instead of changing subscripts? Changing subscripts can alter the identity of substances (e.g., 2 H$2$ + O$2$ → 2 H$2$O vs H$2$ + O$2$ → H$2$O$_2$).

Why Coefficients and Subscripts

  • Coefficients adjust the amounts while keeping chemical formulas fixed.

  • Example contrast:

    • 2 H$2$(g) + O$2$(g) → 2 H$_2$O(l)

    • H$2$(g) + O$2$(g) → H$2$O$2$(l)

Three Types of Reactions

  • Combination (synthesis) reactions

    • Examples: 2 Mg(s) + O$_2$(g) → 2 MgO(s)

    • N$2$(g) + 3 H$2$(g) → 2 NH$_3$(g)

    • C$3$H$6$(g) + Br$2$(l) → C$3$H$6$Br$2$(l)

    • In combination reactions two or more substances react to form one product.

  • Decomposition reactions

    • Examples: CaCO$3$(s) → CaO(s) + CO$2$(g)

    • 2 KClO$3$(s) → 2 KCl(s) + O$2$(g)

    • 2 NaN$3$(s) → 2 Na(s) + 3 N$2$(g)

    • In a decomposition reaction one substance breaks down into two or more substances.

  • Combustion reactions

    • Examples: CH$4$(g) + 2 O$2$(g) → CO$2$(g) + 2 H$2$O(g)

    • C$3$H$8$(g) + 5 O$2$(g) → 3 CO$2$(g) + 4 H$_2$O(g)

    • Combustion reactions are generally rapid and produce flame; most often involve oxygen from the air as a reactant.

Formula Weight (FW) and Molecular Weight (MW)

  • Formula Weight (FW): the sum of the atomic weights for the atoms in a chemical formula.

  • FW reflects the quantitative significance of a formula.

  • Example: FW of CaCl$_2$ = 1(40.08) + 2(35.453) = 110.99 amu.

  • Molecular Weight (MW): the sum of the atomic weights of the atoms in a molecule.

  • Example: MW of C$2$H$6$ = 2(12.011) + 6(1.00794) ≈ 30.070 amu.

  • Ionic compounds exist in a 3D lattice and do not form discrete molecules; they use empirical formulas and formula weights (not molecular weights).

Avogadro’s Number and the Mole

  • Avogadro’s number: $N_A = 6.02 \times 10^{23}$ particles per mole.

  • A mole is the amount of substance containing $N_A$ particles.

  • 1 mole of $^{12}$C has a mass of 12.000 g.

  • One mole of a substance (element) has a mass equal to its molar mass in g/mol; for elements, this equals the atomic weight on the periodic table; diatomic elements have twice the atomic weight.

Molar Mass and the Link to Formula Weight

  • Molar mass is the mass of 1 mole of a substance (g/mol).

  • The molar mass of an element equals its atomic weight (in amu) converted to g/mol; for diatomic molecules, multiply by 2.

  • The formula weight (in amu) equals the molar mass (in g/mol).

Using Moles: A Bridge from Molecular to Real-World Scale

  • Moles provide a bridge from the microscopic world to real-world quantities.

  • One mole corresponds to $N_A$ particles.

  • The number of particles in a molecule depends on the number of atoms of each element; multiplication by $N_A$ scales to macroscopic amounts.

Mole Relationships

  • One mole of atoms, ions, or molecules contains Avogadro’s number of those particles: $N_A$.

  • One mole of molecules or formula units contains $N_A$ times the number of atoms or ions of each element in the compound.

Determining Empirical Formulas

  • Use percent composition to determine the simplest whole-number ratio of atoms.

  • Steps:
    1) Convert percent to moles: moles of each element = (percent by mass) / (atomic weight).
    2) Divide each by the smallest number of moles.
    3) If necessary, multiply by a small integer to obtain whole numbers; assign subscripts accordingly.

Empirical Formula Example: PABA

  • Para-aminobenzoic acid (PABA) composition: C 61.31%, H 5.14%, N 10.21%, O 23.33%.

  • Basis: 100.00 g of compound.

Empirical Formula Example: Calculations

  • Convert to moles (use standard atomic weights):

    • C: 61.31 g / 12.01 g/mol = 5.105 mol

    • H: 5.14 g / 1.00794 g/mol ≈ 5.09 mol

    • N: 10.21 g / 14.01 g/mol ≈ 0.7288 mol

    • O: 23.33 g / 16.00 g/mol ≈ 1.456 mol

Empirical Formula Example: Mole Ratios

  • Divide by smallest number of moles (0.7288):

    • C: 5.105 / 0.7288 ≈ 7.005

    • H: 5.09 / 0.7288 ≈ 6.984

    • N: 0.7288 / 0.7288 = 1.000

    • O: 1.456 / 0.7288 ≈ 2.001

  • Round to nearest whole numbers:

    • C ≈ 7, H ≈ 7, N ≈ 1, O ≈ 2

  • Empirical formula: C$7$H$7$NO$2$ (often written as C$7$H$7$NO$2$)

Determining a Molecular Formula

  • The molecular formula is a whole-number multiple of the empirical formula.

  • If the molar mass (molecular weight) is known, multiplier $n$ = (molar mass) / (empirical formula weight).

  • Example: If empirical formula is CH (FW = 13.0 g/mol) and molar mass is 78 g/mol, then $n = 78 / 13 = 6$; molecular formula is C$6$H$6$.

Combustion Analysis

  • Compounds containing C, H, and O are analyzed via combustion.

  • C is determined from the mass of CO$2$ produced; H from the mass of H$2$O produced; O determined by difference after C and H are determined.

Quantitative Relationships in Stoichiometry

  • The coefficients in a balanced equation show the relative numbers of molecules (or moles) of reactants and products.

  • These relative numbers can be converted to masses using molar masses.

Stoichiometric Calculations

  • We have learned to convert between grams and moles.

  • The new calculation is how to compare two different materials using the mole ratio from the balanced equation.

  • Example: How many grams of water can be produced from 1.00 g of glucose?

    • Reaction: C$6$H${12}$O$6$(s) + 6 O$2$(g) → 6 CO$2$(g) + 6 H$2$O(l)

    • Steps: convert starting material to moles; use the mole ratio from the balanced equation; convert to grams.

Limiting Reactants

  • The limiting reactant is the reactant present in the smallest stoichiometric amount; it determines the maximum amount of product that can be formed.

  • Example: In the shown scenario, H$2$ is the limiting reactant; O$2$ is the excess.

  • Before reaction: 10 H$2$ and 7 O$2$; After reaction: 10 H$2$O and 2 O$2$ (no H$_2$ left).

  • The limiting reactant is used in all stoichiometry calculations to determine amounts of products and amounts of any other reactants used.

Theoretical Yield and Actual Yield

  • Theoretical yield: maximum amount of product that can be made based on stoichiometry.

  • Actual yield: amount actually produced in a reaction.

  • Percent yield = (actual yield / theoretical yield) × 100.

Percent Yield

  • Percent yield = \frac{Actual\,Yield}{Theoretical\,Yield} \times 100.

  • This metric assesses the efficiency of a reaction and can be affected by side reactions, losses, and experimental conditions.