Chemical Reactions and Stoichiometry: Study Notes

Chemical Reactions and Stoichiometry

Writing and Balancing Chemical Reactions

Chemical equations represent the transformation of reactants into products. Writing and balancing these equations is fundamental to understanding chemical processes.

• Identify reactants and products: Place reactants on the left and products on the right of the arrow.

• Assign correct chemical formulas and states: Use (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous.

• Balance the equation: Adjust coefficients to ensure the same number of each atom on both sides. Balance hydrogen and oxygen last, and treat polyatomic ions as units if they appear unchanged on both sides.

• Check your work: Count atoms of each element on both sides to confirm balance.

Example: Balancing the synthesis of ammonia:

• Unbalanced: N2(g) + H2(g) → NH3(g)

• Balanced: N2(g) + 3H2(g) → 2NH3(g)

Classifying Chemical Reactions

Chemical reactions can be classified into several types based on their general patterns:

• Combination (Synthesis): Two or more substances combine to form one product. Generic: A + B → C

• Decomposition: A single compound breaks down into two or more simpler substances. Generic: C → A + B

• Combustion: A hydrocarbon reacts with oxygen to produce carbon dioxide and water. Generic: hydrocarbon + O2(g) → CO2(g) + H2O(l)

Example: Combustion of methane: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

Calculating Formula Weights and Molecular Weights (Molar Mass)

The formula weight (for ionic compounds) and molecular weight (for molecular compounds) are calculated by summing the atomic masses of all atoms in the formula.

• Count the number of each atom in the formula.

• Multiply by the atomic mass (from the periodic table).

• Sum the total mass.

Example: For NaCl: Na (22.99 amu) + Cl (35.45 amu) = 58.44 amu

Example: For Ca(OH)2: Ca (40.08 amu) + 2 × [O (16.00 amu) + H (1.008 amu)] = 74.10 amu

Example: For N2O4: 2 × N (14.01 amu) + 4 × O (16.00 amu) = 92.02 amu

Avogadro’s Number and the Mole Concept

Avogadro’s number (6.022 × 1023) is the number of particles (atoms, molecules, or formula units) in one mole of a substance. It allows conversion between microscopic (particles) and macroscopic (grams, moles) scales.

• One mole of any substance contains 6.022 × 1023 particles.

• Use the molar mass (g/mol) to convert between grams and moles.

Example: 2.3 g NaCl × (1 mol / 58.44 g) × (6.022 × 1023 units / 1 mol) = number of formula units

Empirical and Molecular Formulas

The empirical formula gives the simplest whole-number ratio of elements in a compound. The molecular formula gives the actual number of atoms of each element in a molecule.

• To find the empirical formula from percent composition:

  1. Assume a 100 g sample (percentages become grams).

  2. Convert grams to moles using atomic masses.

  3. Divide by the smallest number of moles to get ratios.

  4. Assign subscripts to write the empirical formula.

• To find the molecular formula:

  1. Calculate the empirical formula mass.

  2. Divide the molecular mass by the empirical formula mass to get a whole number.

  3. Multiply the empirical formula subscripts by this number.

Example: Empirical formula for ascorbic acid (C3H4O3), molecular mass = 176.0 amu, empirical mass = 88.0 amu. Molecular formula: C6H8O6.

Percent Composition by Mass

The percent by mass of an element in a compound is calculated as:

$\text{Percent by mass} = \frac{\text{mass of element in 1 mol of compound}}{\text{molar mass of compound}} \times 100\%$

Example: Oxygen in KMnO4: (4 × 16.00 g) / (39.10 + 54.94 + 4 × 16.00) × 100%

Stoichiometry: Mole and Mass Relationships

Stoichiometry involves using balanced chemical equations to relate quantities of reactants and products.

• Convert grams to moles: Use molar mass.

• Use mole ratios: From the balanced equation, relate moles of one substance to another.

• Convert moles to grams: Use molar mass.

Example: How many grams of NH3 are produced from 1.00 g N2?

1. Convert grams N2 to moles N2: $1.00\,\text{g} \times \frac{1\,\text{mol}}{28.0\,\text{g}}$

2. Use mole ratio: $1\,\text{mol N}_2 \rightarrow 2\,\text{mol NH}_3$

3. Convert moles NH3 to grams: $\text{mol NH}_3 \times 17.0\,\text{g/mol}$

Limiting Reactant and Percent Yield

The limiting reactant is the reactant that is completely consumed first, limiting the amount of product formed. Percent yield compares the actual yield to the theoretical yield.

• Identify the limiting reactant by comparing the mole ratios of reactants used and required.

• Calculate the maximum amount of product (theoretical yield) from the limiting reactant.

• Percent yield: $\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

Example: If 2.5 g NH3 is made but the theoretical yield is 3.0 g, percent yield = (2.5/3.0) × 100% = 83.3%

Important Constants and Equations

• Avogadro’s number: $6.022 \times 10^{23}$ particles/mol

• Molar mass: $\text{g/mol}$ (from periodic table)

• Percent by mass: $\frac{\text{mass of element}}{\text{molar mass}} \times 100\%$

Additional info:

• For all calculations, always use the periodic table for atomic masses.

• Dimensional analysis is a systematic approach to unit conversions in stoichiometry.