TEST 2

Chapter 3: Stoichiometry: Calculations with Chemical Formulas and Equations

Chemical Equations

• Definition: Represents chemical reactions through chemical equations.

  • Example: When hydrogen gas burns, it reacts with oxygen gas to form water.

• Coefficients: Numbers in front of formulas, indicating the relative number of molecules involved in the reaction.

• Balanced Chemical Reaction:

  • Has an equal number of atoms of each element on both sides of the reaction arrow.

  • Principle: Matter is neither created nor destroyed.

Difference Between Coefficients and Subscripts

• Coefficients:

  • Placed in front of a formula; it changes the amount of the substance without altering its identity.

  • Example:

  • H2O: 1 molecule of water (2 H, 1 O)

  • 2 H2O: 2 molecules of water (4 H, 2 O)

• Subscripts:

  • Changing a subscript changes the identity of the substance.

  • Example:

  • H2O (water) vs H2O2 (hydrogen peroxide).

  • Rule: NEVER change subscripts when balancing an equation.

Balancing Chemical Equations

• Balancing process for the example: CH4 + O2 → CO2 + H2O.

  • Initial state:

  • C: 1, H: 4, O: 2 (reactants)

  • C: 1, H: 2, O: 3 (products)

  • Observation: Not balanced; need equal numbers of H atoms.

• Adjust by placing coefficients:

  • New equation: CH4 + 2 O2 → CO2 + 2 H2O.

  • Confirm balanced state:

  • C: 1, H: 4, O: 4 (both reactants and products).

Practice Balancing Equations

• Practice problems:

  • (a) Li (s) + N2 (g) → Li3N2 (s)

  • (b) TiCl4 (l) + H2O (l) → TiO2 (s) + HCl (aq)

  • (c) NH4NO3 (s) + N2 (g) → N2 (g) + O2 (g) + H2O (g)

  • (d) O2 + NO → NO2.

Chemical Reactivity Types

1. Combination Reactions:

   • Definition: Two or more reactants combine to form a single product.

   • General Reaction: A + B → C.

   • Examples:

     • C (s) + O2 (g) → CO2 (g)

     • CaO (s) + H2O (l) → Ca(OH)2 (s).

2. Decomposition Reactions:

   • Definition: A single reactant breaks apart to form two or more substances.

   • General Reaction: C → A + B.

   • Examples:

     • 2 KClO3 (s) → 2 KCl (s) + 3 O2 (g).

3. Single Replacement (Displacement):

   • Definition: One similar substance replaces another in a reaction.

   • General Reaction: A + BX → AX + B.

   • Example:

     • Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s).

4. Double Replacement (Displacement):

   • Definition: Two similar substances replace each other; occurs between ionic substances.

   • General Reaction: AX + BY → AY + BX.

   • Example:

     • Na2CO3(aq) + 2 AgNO3(aq) → 2 NaNO3(aq) + Ag2CO3(s).

5. Combustion Reactions:

   • Definition: Rapid reactions that produce a flame, typically involving O2.

   • Hydrocarbons produce CO2 and H2O when burned in oxygen.

   • Example:

     • C3H8 (g) + 5 O2 → 3 CO2 (g) + 4 H2O (g).

   • Notes on balancing: Balance oxygen atoms last and possibly use half-coefficients.

Formula Weights

• Definition: Formula weight (FW) is the sum of the atomic weights (AW) of the atoms in a chemical formula.

• Atomic weight is listed under the chemical symbol in the periodic table.

• Example Calculation for HNO3:

  • FW = 1(AW of H) + 1(AW of N) + 3(AW of O)

  • FW = 1(1.01 amu) + 1(14.01 amu) + 3(15.99 amu) = 62.99 amu.

• Query: Calculate FW of C12H22O11.

Percentage Composition

• Definition: Percentage by mass contributed by each element in a substance.

• Formula to calculate % composition of an element:
  $ext{Percent Composition} = rac{ ext{(FW of the compound)}}{ ext{(number of atoms) * (atomic weight of element)}} × 100$.

• Example for C12H22O11:

  • Carbon: 12 * 12.01 amu = 144.1 amu.

  • Hydrogen: 22 * 1.008 amu = 22.18 amu.

  • Oxygen: 11 * 16.00 amu = 176.0 amu.

  • Total mass = 342.3 amu.

  • % C = (144.1/342.3)*100 = 42.1% (similar calculations for H and O).

• Example for Fe(OH)3:

  • % Fe = 52.26%, % O = 44.91%, % H = 2.83%.

Avogadro’s Number and the Mole

• Definition: The mole (abbreviated mol) is the counting unit for numbers of atoms, ions, or molecules in a sample.

• Avogadro's Number: $6.02 × 10^{23}$. All entities per mole contain this number.

  • Examples:

  • 1 mol of 12C atoms = $6.02 × 10^{23}$ atoms.

  • 1 mol of H2O = $6.02 × 10^{23}$ molecules.

  • 1 mol of NO3- ions = $6.02 × 10^{23}$ ions.

  • 1 mol of NaCl = $6.02 × 10^{23}$ formula units.

Calculating Moles and Molar Mass

• Molar mass (MM): the mass (in grams) of one molecule of a substance, numerically equal to its formula weight (FW).

• Example Calculation for C6H12O6:

  • $FW = 6(12.01) + 12(1.01) + 6(15.99) = 180.12 ext{ amu}$

• To find: How many molecules are in 1 mole of C6H12O6?

• Calculate the MM: MM of C6H12O6 = 180.12 g/1 mol.

Empirical Formulas from Analyses

• Definition: Determine the simplest whole-number ratio of elements in a compound.

• Example for sulfuric acid (H2SO4) from % composition:

  • % H = 2.06 g, % S = 32.71 g, % O = 65.23 g in 100 g total.

  • Number of moles:

  • H: $2.06 g / 1.01 g/mol = 2.04 mol$

  • S: $32.71 g / 32.07 g/mol = 1.02 mol$

  • O: $65.23 g / 15.99 g/mol = 4.08 mol$.

• Simplest ratio calculation: H: 2.04 = 2, S: 1.02 = 1, O: 4.08 = 4.

  • Empirical formula: H2SO4.

Molecular Formulas from Empirical Formulas

• The molecular formula can be derived from the empirical formula if either the molecular weight or molar mass is known.

• Example calculation with hydrogen peroxide (H2O2):

  • MW of H2O2 = 34.0 g/mol, empirical formula weight = 17.0 g/mol → Multiples by 2.

• Example with glucose (C6H12O6):

  • MW = 180.12 g/mol, empirical formula weight = 30.03 g/mol → Multiples by 6.

Quantitative Information from Balanced Equations

• Coefficients in a balanced equation indicate the relative numbers of molecules and moles while allowing conversion of quantities of reactants and products.

• Example with hydrogen reaction:

  • 2 H2 + O2 → 2 H2O.

  • 2 molecules of H2 for every 1 molecule of O2 implies a mole ratio of:
    $2 mol H2 ext{ and } 1 mol O2.$

• Conversion setup example:

  • If given 1.57 mol O2, find moles of H2O produced using mole ratios.

Limiting Reactants

• Definition: The limiting reactant is the reactant that is completely consumed in the reaction and limits the amount of product formed.

• Example: 14 slices of bread and 6 slices of cheese ➔ You can make 6 sandwiches, cheese is limiting.

Calculating Product Quantities from Limiting Reactant

• When calculating the product formed using limiting reactants, the stoichiometric calculations must consider the amount available for reaction and the ratios from the balanced equation.

• Example: How many moles of NH3 from given amounts of N2 and H2.

Theoretical Yields

• Definition: The theoretical yield is the calculated quantity of product when all limiting reactant is exhausted; usually higher than the actual yield obtained from the reaction.

• Percent yield:
  $ext{Percent yield} = rac{ ext{actual yield}}{ ext{theoretical yield}} × 100.$

• Example calculation; using silver nitrate and produced zinc nitrate findings using limiting reactant approach.

Practice Problems

• Various exercises available to calculate moles, mass from grams, and conversion involving density and reaction coefficients.

• Students are encouraged to use dimensional analysis and other conversion factors to solidify their understanding of stoichiometry.