Stoichiometry and Stoichiometric Calculations

🔍 What is Stoichiometry?

Stoichiometry deals with calculating masses (and sometimes volumes) of reactants and products in chemical reactions. It comes from Greek words: stoicheion (element) and metron (measure).

⚖️ Balanced Chemical Equations

A balanced equation has the same number of atoms of each element on both sides. Example (combustion of methane):

\[ \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)} \]

Key points:

  • Reactants → Products (CH₄ and O₂ are reactants; CO₂ and H₂O are products).
  • Stoichiometric coefficients (numbers in front) show mole ratios: 1 mol CH₄ reacts with 2 mol O₂ to give 1 mol CO₂ and 2 mol H₂O.
  • Relationships:
    • Moles ⇄ Mass ⇄ Number of molecules
    • Volume ⇄ Moles (using density if needed)

🧪 Finding Empirical & Molecular Formulas

Empirical formula (simplest whole-number ratio of atoms):

  1. Find moles of each element.
  2. Divide by the smallest mole value.
  3. Convert ratios to whole numbers (multiply if needed).
  4. Write the formula (e.g., CH₂Cl).

Molecular formula (actual atom count):

  1. Calculate empirical formula mass.
  2. Divide molar mass by empirical mass: \( n = \frac{\text{Molar mass}}{\text{Empirical mass}} \).
  3. Multiply empirical formula by \( n \) (e.g., CH₂Cl × 2 = C₂H₄Cl₂).

⚠️ Limiting Reagent

The reactant that gets used up first and stops the reaction. It determines how much product forms. Always identify it when reactant amounts aren’t balanced!

Example: If 50.0 kg N₂ and 10.0 kg H₂ make NH₃:

  • N₂ moles = \( 17.86 \times 10^2 \) mol
  • H₂ moles = \( 4.96 \times 10^3 \) mol
  • H₂ is limiting (needed: \( 5.36 \times 10^3 \) mol; available: \( 4.96 \times 10^3 \) mol).

🧪 Reactions in Solutions

Concentration units:

  1. Mass Percent (w/w%): \[ \text{Mass %} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100 \] Example: 2 g A + 18 g water → Mass% of A = \( \frac{2}{20} \times 100 = 10\% \).
  2. Mole Fraction (x): \[ x_{\text{A}} = \frac{n_{\text{A}}}{n_{\text{A}} + n_{\text{B}}}, \quad x_{\text{B}} = \frac{n_{\text{B}}}{n_{\text{A}} + n_{\text{B}}} \] (n = moles of component)
  3. Molarity (M): \[ M = \frac{\text{Moles of solute}}{\text{Volume (L) of solution}} \] Example: 4 g NaOH in 250 mL solution → Molarity = \( \frac{4/40}{0.250} = 0.4 \) M.
    Tip: Use \( M_1V_1 = M_2V_2 \) for dilution!
  4. Molality (m): \[ m = \frac{\text{Moles of solute}}{\text{Mass (kg) of solvent}} \] Example: 3 M NaCl (density = 1.25 g/mL) → Molality = \( \frac{3}{1.0745} = 2.79 \) m.
    Note: Molality doesn’t change with temperature! 🌡️

📚 Important Concepts for NEET

  1. Limiting Reagent Calculations (e.g., identifying it and finding product yield).
  2. Molarity vs. Molality (definitions, units, temperature dependence).
  3. Stoichiometric Conversions (mass → moles → molecules using balanced equations).
  4. Empirical & Molecular Formulas (from % composition or combustion data).
  5. Concentration Units (mass%, mole fraction, dilution problems).

💡 Solved Problems

Problem 1: Calculate water produced by combusting 16 g methane.
Solution: \[ \ce{CH4 + 2O2 -> CO2 + 2H2O} \] 16 g CH₄ = 1 mol → Gives 2 mol H₂O = 36 g water. 💧

Problem 2: Moles of methane to produce 22 g CO₂.
Solution: 44 g CO₂ requires 16 g CH₄ → 22 g CO₂ requires \( \frac{22}{44} \times 1 \) mol CH₄ = 0.5 mol.

Problem 3: Molarity of 4 g NaOH in 250 mL solution.
Solution: Moles of NaOH = \( \frac{4}{40} = 0.1 \) mol → Molarity = \( \frac{0.1}{0.250} = 0.4 \) M.