Introduction to Analytical Chemistry - Notes

1. Meaning & Scope:

• Analytical chemistry deals with separation, identification, and quantitative determination of chemical substances.
• Types of Analysis:
  • Qualitative: Detects presence/absence of elements or ions.
  • Quantitative: Measures the amount of each component.

2. Types of Chemical Analysis:

• Chemical Methods:
  • Dry tests: Preliminary tests without dissolving the sample.
  • Wet tests: Sample dissolved in solvent for detailed analysis.
• Qualitative analysis detects cations and anions.
• Quantitative analysis involves:
  • Gravimetric analysis: Based on mass measurement.
  • Volumetric (Titrimetric) analysis: Based on volume of solutions reacting.

3. Importance of Analytical Chemistry:

Used in industry, medicine, agriculture, and forensic science to monitor raw materials, products, air, soil, fertilizers, and drugs.

4. Mathematical Operations & Error Analysis:

• Scientific notation: Expressing large/small numbers as N × 10ⁿ.
• Accuracy: Closeness to true value.
• Precision: Reproducibility of results.
• Errors:
  • Absolute Error = Observed – True value
  • Relative Error:
    Relative Error = 

    Absolute error True value
• Deviation: Difference between observed value and mean value.

5. Significant Figures:

• Indicate certainty in measurement.
• Rules:
  1. Non-zero digits are significant.
  2. Zeros between non-zero digits are significant.
  3. Leading zeros are not significant.
  4. Trailing zeros are significant only after a decimal point.
  5. All digits in scientific notation are significant.
• Rounding off: If the next digit is less than 5, leave it; if it is 5 or greater, increase by 1.

6. Determination of Molecular Formula:

• Empirical formula: Simplest whole-number ratio of atoms in a compound.
• Molecular formula: Actual number of atoms in a molecule.
  Molecular formula = n × Empirical formula
  where:
  n = 

  Molar mass Empirical formula mass

Example: Finding the Empirical Formula

A compound contains: C = 40.0 %, H = 6.7 %, O = 53.3 %.

Solution:

1. Step 1: Assume 100 g of the compound.
   This means we have: C = 40.0 g, H = 6.7 g, O = 53.3 g.
2. Step 2: Convert mass of each element to moles.
   Moles of C = 40.0 g / 12.01 g·mol^-1 = 3.33 mol
   Moles of H = 6.7 g / 1.008 g·mol^-1 = 6.65 mol
   Moles of O = 53.3 g / 16.00 g·mol^-1 = 3.33 mol
3. Step 3: Find the simplest whole-number ratio.
   Divide each mole value by the smallest number of moles (3.33):
   C: 3.33 / 3.33 = 1
   H: 6.65 / 3.33 ≈ 2
   O: 3.33 / 3.33 = 1
   The simplest ratio is 1 : 2 : 1.

Answer: The empirical formula is CH₂O.

7. Stoichiometric Calculations:

Based on balanced chemical equations.

General Steps:

1. Write and balance the chemical equation.
2. Convert the given data (mass, volume, etc.) into moles.
3. Use the mole ratio from the balanced equation to find moles of the desired substance.
4. Convert the moles back into the required units (mass, volume, etc.).

Mass-Mass Relationships

Example 1: CaCO₃ → CaO

Reaction: CaCO₃(s) → CaO(s) + CO₂(g)

Given 5.00 g of CaCO₃, find the mass of CaO produced.

Solution:

1. Moles of CaCO₃ = 5.00 g / 100 g·mol^-1 = 0.0500 mol
2. From the equation, 1 mol CaCO₃ produces 1 mol CaO. So, moles of CaO = 0.0500 mol.
3. Mass of CaO = 0.0500 mol × 56 g·mol^-1 = 2.80 g

Answer: 2.80 g of CaO.

Example 2: Combustion of methane

Reaction: CH₄ + 2 O₂ → CO₂ + 2 H₂O

Given 24.0 g of CH₄, find the mass of CO₂ and H₂O formed.

Solution:

1. Moles of CH₄ = 24.0 g / 16 g·mol^-1 = 1.50 mol
2. From the equation: 1 mol CH₄ → 1 mol CO₂ and 2 mol H₂O.
3. Moles of CO₂ = 1.50 mol. Mass of CO₂ = 1.50 mol × 44 g·mol^-1 = 66.0 g.
4. Moles of H₂O = 1.50 × 2 = 3.00 mol. Mass of H₂O = 3.00 mol × 18 g·mol^-1 = 54.0 g.

Answer: 66.0 g CO₂ and 54.0 g H₂O.

Mass-Volume Relationships

Example 3: Propane combustion

Reaction: C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Given 2.20 g C₃H₈, find the volume of O₂ (at STP) required.

Solution:

1. Moles of C₃H₈ = 2.20 g / 44 g·mol^-1 = 0.0500 mol
2. From the equation: 1 mol C₃H₈ requires 5 mol O₂. Moles of O₂ = 0.0500 × 5 = 0.250 mol.
3. Volume O₂ at STP = 0.250 mol × 22.4 L·mol^-1 = 5.60 L.

Answer: 5.60 L O₂ at STP.

8. Limiting Reagent:

The reactant that is completely consumed first in a chemical reaction and therefore limits the amount of product that can be formed.

Example 4: Limiting Reagent (NO and O₂ → NO₂)

Reaction: 2 NO + O₂ → 2 NO₂

Given 8.0 mol NO and 7.0 mol O₂. Which is the limiting reagent? How many moles of NO₂ are formed?

Solution:

1. Calculate moles of NO₂ from each reactant.
2. From 8.0 mol NO: (8.0 mol NO) × (2 mol NO₂ / 2 mol NO) = 8.0 mol NO₂.
3. From 7.0 mol O₂: (7.0 mol O₂) × (2 mol NO₂ / 1 mol O₂) = 14.0 mol NO₂.
4. The smaller amount of product (8.0 mol NO₂) is what can be formed.

Answer: NO is the limiting reagent, and 8.0 mol of NO₂ will be formed.

9. Concentration of Solutions:

• Mass Percent (w/w %):
  Mass % = 

  Mass of solute Mass of solution

   × 100%

  Worked Example:

  You add 5.0 g of NaCl to 95.0 g of water.

  Mass of solution = 5.0 g + 95.0 g = 100.0 g

  Mass % NaCl = (5.0 g / 100.0 g) × 100% = 5.0%

• Mole Fraction (X):
  X_A = 

  Moles of A Total moles of all components

  Worked Example:

  Dissolve 0.50 mol of glucose in 1.00 mol of water.

  Total moles = 0.50 mol + 1.00 mol = 1.50 mol

  X_glucose = 0.50 mol / 1.50 mol = 0.333

  X_water = 1.00 mol / 1.50 mol = 0.667

• Molarity (M):
  M = 

  Moles of solute Volume of solution in Liters

  Note: Molarity is temperature-dependent.

  Worked Example:

  Dissolve 4.0 g of NaOH (Molar Mass = 40.00 g/mol) and make up to a 250 mL solution.

  Moles of NaOH = 4.0 g / 40.00 g·mol^-1 = 0.100 mol

  Volume = 250 mL = 0.250 L

  Molarity (M) = 0.100 mol / 0.250 L = 0.400 M

• Molality (m):
  m = 

  Moles of solute Mass of solvent in kg

  Note: Molality is temperature-independent.

  Worked Example:

  Dissolve 0.50 mol of a solute in 200.0 g of solvent.

  Mass of solvent = 200.0 g = 0.200 kg

  Molality (m) = 0.50 mol / 0.200 kg = 2.50 m

10. Graphical Analysis:

• Data is plotted on graphs to visualize relationships between variables (e.g., Volume ∝ Temperature).
• An average smooth curve drawn through plotted points should have an equal distribution of points above and below the line, representing the trend while minimizing random errors.