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Chapter 3 Stoichiometry Notes

Chemical Reactions and Chemical Equations

  • Combination reaction: two (or more) substances combine to form one product.
  • Chemical equations describe proportions of reactants and products during a chemical reaction.
  • Reactants: the substances that are consumed during a chemical reaction.
  • Products: the substances that are formed during a chemical reaction.

Evidence of a Chemical Reaction

  • Evidence includes:
    • A color change
    • Formation of a solid (precipitate)
    • Evolution of a gas
    • Absorption or emission of heat

Examples of Chemical Equations

  • Describe changes on the atomic level:
    • \mathrm{SO3(g) + H2O(\ell) \rightarrow H2SO4(\ell)}
    • \mathrm{Fe2O3(s) + 3\,H2SO4(aq) \rightarrow 3\,H2O(\ell) + Fe2(SO4)3(aq)}
  • Physical states of products/reactants:
    • (s) = solid; (l) = liquid; (g) = gas; (aq) = aqueous solution

The Mole

  • A mole is a unit for a specific quantity of substance.
  • Common abbreviations:
    • 1 dozen = 12 items
    • 1 gross = 144 items
    • 1 mole = 6.022 × 10^23 particles (Avogadro’s number) N_A = 6.022 \times 10^{23}.
  • A mole is Avogadro’s number of atoms in exactly 12 g of carbon-12.
  • It is a unit for expressing macroscopic quantities (atoms or molecules) involved in chemical reactions.

Mole as a Conversion Factor

  • To convert between the number of particles and an equivalent number of moles, use the conversion factor N_A = 6.022 \times 10^{23} particles per mole.

Molecular Mass / Formula Mass

  • Molecular mass: mass of one molecule of a molecular compound.
  • Formula mass: mass in atomic mass units (amu) of one formula unit of an ionic compound.
  • Examples:
    • M{\mathrm{CO2}} = 12.01 + 2\times 16.00 = 44.01\ \text{amu}
    • M{\mathrm{H2O}} = 2\times 1.008 + 16.00 = 18.02\ \text{amu}
    • Formula mass for NaCl: M_{\mathrm{NaCl}} = 22.99 + 35.45 = 58.43\ \text{amu}

Molar Mass

  • Molar mass is the mass (in grams) equal to the mass of one mole of the substance (in amu).
  • Example: 1 atom of He = 4.003 amu; 1 mole of He (6.022 × 10^23 atoms) = 4.003 g; The molar mass of He is \mathcal{M}(\mathrm{He}) = 4.003\ \text{g mol}^{-1}.
  • Example: Molar mass of ammonium carbonate: (NH4)2CO_3 ≈ 96.086 g/mol (calculation: N, H, C, O contributions add up to 96.086 g/mol).

Conversions: Moles ↔ Mass

  • To interconvert between moles and mass, use molar mass: n = \frac{m}{M} where n = moles, m = mass (g), M = molar mass (g/mol).
  • Example: 1 mole of He has mass 4.003 g → M_{\mathrm{He}} = 4.003\ \text{g mol}^{-1}.

Conversions: Particles ↔ Moles ↔ Mass

  • Particles to moles: n = \frac{N}{N_A} where N is number of particles.
  • Moles to particles: N = n N_A.
  • Mass to moles: n = \frac{m}{M}.
  • Mass to particles: combine the two steps: N = \frac{m}{M} N_A.

Practice: Converting Mass into Number of Moles

  • Example context: the active ingredient in many antacid tablets is calcium carbonate. Some tablets contain 0.400 g of calcium (as Ca^{2+} ions).
  • Question: How many moles of calcium are in each tablet? (Use Ca atomic mass ~ 40.08 g/mol.)

How many atoms in a given mass

  • Example: How many atoms are in 0.551 g of potassium (K)? (Use K atomic mass ~ 39.10 g/mol; Avogadro’s number.)

Diatomic Elements

  • Reference: diatomic elements appear in their elemental form as diatomic molecules (e.g., H2, N2, O2, F2, Cl2, Br2, I_2).

Moles and Chemical Equations

  • Coefficients in a chemical equation indicate the proportions of reactants and/or products.
  • On the macroscale, coefficients indicate the number of moles of each substance.
  • Stoichiometry: the quantitative relation between reactants and products in a chemical equation; indicated in the equation by coefficients.

Law of Conservation of Mass

  • The sum of the masses of the reactants in a chemical equation is equal to the sum of the masses of the products.

Balancing Chemical Equations (1 of 4)

  • Balanced equations follow the Law of Conservation of Mass (unbalanced shown for reference).

Balancing Chemical Equations (2 of 4)

  • Four-step method:
    1) Write correct formulas for reactants and products, including physical states.
    2) Check whether the expression is balanced by counting atoms on each side of the reaction.
    3) Balance an element that appears in only one reactant and one product first.
    4) Choose coefficients to balance other elements as needed.

Balancing Chemical Equations (3 of 4)

  • (Illustrative problems and balanced equations typically shown in class.)

Balancing Chemical Equations (4 of 4)

  • (Further practice problems and examples.)

Example Balancing Problems

  • Aluminum metal reacts with oxygen gas to form aluminum oxide. Write the balanced chemical equation.
  • Magnesium metal reacts with iodine gas to form magnesium iodide. Write the balanced chemical equation.

Combustion Reactions

  • Combustion reactions are heat-producing reactions between oxygen (O_2) and another element or compound.
  • Example: SO2(g) + 2 O2(g) → SO_3(g)
  • Combustion of hydrocarbons produces CO2 and H2O.
    • Example: CH4(g) + O2(g) → CO2(g) + H2O(g)

Balancing Chemical Equations (Combustion Examples)

  • Typical combustion balancing problems include reactions like:
    • N2 + H2 → NH_3
    • CH4 + O2 → CO2 + H2O
    • CaH2 + H2O → Ca(OH)2 + H2
    • C6H{12}O6 + O2 → CO2 + H2O

Respiration and Photosynthesis

  • Photosynthesis (plants):
    • CO2(g) + 6 H2O(ℓ) → C6H{12}O6(aq) + 6 O2(g)
  • Respiration (reverse of photosynthesis):
    • C6H{12}O6(aq) + 6 O2(g) → CO2(g) + 6 H2O(ℓ)

Stoichiometry: Key Concepts and Analogy

  • The word stoichiometry comes from Greek stoicheon (element) and metron (to measure).
  • Stoichiometry is the quantitative relationship between reactants and products; it’s analogous to following a baking recipe.
  • Example: How much of each reactant is needed to make a certain amount of product (e.g., 2 cakes in a cheesecake recipe with given ingredient blocks).
  • Cheesecake analogy: 3 blocks cream cheese + 1 cup sugar + 5 eggs → 1 cheesecake; useful for understanding limiting vs excess reactants.

Practice: Stoichiometry Problems

  • Example 1: How much CO2 would be formed if 10.0 g of C5H_{12} were completely burned in oxygen?
    • Balanced equation: \mathrm{C5H{12} + 8 O2 \rightarrow 5 CO2 + 6 H_2O}
    • Solve using stoichiometric ratios to find g CO_2 produced.
  • Example 2: Methanol combustion:
    • \mathrm{2 CH3OH + 3 O2 \rightarrow 2 CO2 + 4 H2O}
    • If 209 g of CH3OH are used, calculate the mass of H2O produced.
  • Example 3: H2 + I2 → 2 HI
    • Given masses and mole relationships, determine unknown quantities (illustrative setup shown in class materials).
  • Example 4: Determine grams of O2 required to burn 10.0 g of glucose (C6H{12}O6):
    • Balanced form: \mathrm{C6H{12}O6 + 6 O2 \rightarrow 6 CO2 + 6 H2O}
    • Compute g O_2 needed.

Practice: Always Start with a Balanced Equation

  • Example: Sn + HF → SnF2 + H2 (theoretical yields shown in class)
    • If 100.0 g HF is used, determine the grams of SnF_2 that could be produced (theoretical yield).
    • Steps shown: Grams -> Moles -> Grams using molar masses and stoichiometric coefficients.

Percent Yield

  • Definitions:
    • Theoretical yield: the maximum amount of product possible from given quantities of reactants.
    • Actual yield: the amount of product actually formed in a reaction.
  • Percent yield formula: \text{Percent Yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\%.

Practice: Limiting Reactants and Related Concepts

  • The Cheese Burger Analogy illustrates limiting reactant vs excess reactant.
    • If one reactant is consumed first, it is the limiting reactant, and it determines the amount of product formed.
  • Example: In a reaction represented by the equation, identify the limiting reagent and the amount of product formed.
  • Practice: Limiting Reactants problems:
    • If 10.0 g CH4 is burned with 20.0 g O2 to form CO2 and H2O, identify the limiting reactant.

Limiting Reactant: Worked Example

  • SO3(g) + H2O(g) → H2SO4(l)
    • Given 20.00 g SO3 and 10.00 g H2O, determine the limiting reactant.

Practice: More Limiting Reactants Problems

  • Example: In a reaction, 124 g Al reacted with 601 g Fe2O3 to form Al2O3 and Fe; calculate mass of Al2O3 formed.
  • Conclusion: The substance that limits the amount of product is the limiting reagent; the amount of product is the amount obtained from the limiting reagent.

Percent Yield Practice Problems

  • Example: Aluminum burns in bromine liquid to form aluminum bromide. In one experiment, 6.0 g of Al reacts with excess Br_2 to yield 50.3 g of aluminum bromide. Calculate the theoretical and percent yields.

Stoichiometry: Percent Composition and Empirical Formulas

  • Percent composition (mass percent) formula:
    • \%
      = \left(\frac{\text{mass of element}}{\text{mass of compound}}\right) \times 100\%
  • Example: %C in CH_4 is calculated as: \% \, C = \frac{12.01 \times 1}{(12.01 + 4 \times 1.008)} \times 100\% = 74.88\%.
  • Example: %C in C{10}H{16} is 74.88% C; %H is 25.12%; those values illustrate the method.

Empirical Formulas from Percent Composition

  • Approach:
    1) Assume 100 g of substance.
    2) Convert masses of each element to moles.
    3) Compute mole ratios.
    4) If necessary, convert to smallest whole-number ratios.
  • Example: Methane (CH4) with 74.88% C and 25.12% H yields empirical formula CH2 (smallest whole-number ratio).

Empirical vs Molecular Formulas

  • Empirical formula: simplest whole-number mole ratio of elements in a compound.
  • Molecular formula: actual mole ratio of elements in a compound.
  • Relationship: Molecular formula = (empirical formula) × n, where n is a whole-number multiplier.
  • To determine the molecular formula, you need the empirical formula and the molar (molecular) mass of the compound.
  • Example: If empirical formula is CH2 and molar mass is 56 g/mol, the molecular formula could be C4H_8 (since empirical mass is 14 g/mol and 56/14 = 4).
  • In some class materials, glucose is used as an example with empirical formula CH2O, molecular mass 180 g/mol, yielding C6H{12}O6 as the molecular formula.

Empirical Formulas from Percent Composition: Practice Problems

  • Example: Glucose has 40.03% C, 6.667% H, 53.33% O and a molar mass of 180 g/mol. The empirical formula is CH2O, leading to the molecular formula C6H{12}O6.
  • Example: A compound with 54.55% C, 9.09% H, and 36.36% O has molar mass 132 g/mol. The empirical formula is C2H4O, and the molecular formula is C6H{12}O_3.

Ascorbic Acid (Vitamin C)

  • Contains C, H, and O.
  • Percent composition: C ≈ 40.9%, H ≈ 4.55%, O ≈ 54.6%.
  • Task: Determine empirical formula from percent composition and, if given molecular mass, determine the molecular formula.

Combustion Analysis

  • Combustion analysis determines percent C and H in a hydrocarbon from the masses of CO2 and H2O produced:
    • CaHb + O2(g) → CO2(g) + H_2O(g)
  • From the masses of CO2 and H2O produced, compute moles of C and H, then deduce the empirical formula.

Practice: Combustion Analysis Problems

  • Limonene (a hydrocarbon) combustion: 0.671 g of limonene yields 2.168 g CO2 and 0.710 g H2O. Determine the empirical formula of limonene.
  • Another combustion problem: A compound containing C, H, and O, upon combustion, yields 6.60 g CO2 and 2.70 g H2O from 3.70 g of compound. Determine its empirical formula.