Stoichiometry and Chemical Calculations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Stoichiometry — Key Facts

Stoichiometry is the calculation of reactants and products in chemical reactions based on the law of conservation of mass. The word comes from Greek: “stoicheion” (element) + “metron” (measure).

Key Definitions:

• Mole (mol): Amount of substance containing Avogadro’s number of particles
• Avogadro’s Number: $N_A = 6.022 \times 10^{23}$ particles per mole
• Molar Mass: Mass of one mole of a substance (g/mol)
• Molar Volume: Volume occupied by one mole of gas at STP = 22.4 L

** Fundamental Stoichiometric Calculations:**

Quantity	Formula	Units
Number of moles	$n = \frac{m}{M}$	mol
Number of particles	$N = n \times N_A$	particles
Molar volume (gas)	$V = n \times 22.4$	L at STP
Concentration	$M = \frac{n}{V}$	mol/L

where m = mass, M = molar mass, V = volume.

Balancing Chemical Equations:

A balanced equation shows:

• Conservation of atoms (atoms in = atoms out)
• Conservation of charge
• Conservation of mass

Example: $2H_2 + O_2 \rightarrow 2H_2O$

• H atoms: 4 on both sides ✓
• O atoms: 2 on both sides ✓

⚡ ECAT Exam Tip: Always balance equations before doing stoichiometric calculations. The coefficients give the mole ratio. In limiting reagent problems, convert all to moles first.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding…

Types of Stoichiometric Problems:

1. Mole-Mole Calculations: Given: $N_2 + 3H_2 \rightarrow 2NH_3$ If 2 moles of N₂ react, moles of NH₃ produced = 2 × 2 = 4 mol

2. Mass-Mass Calculations: Step 1: Convert given mass to moles Step 2: Use mole ratio from balanced equation Step 3: Convert back to mass

Example: Calculate mass of H₂O produced from 36 g of H₂ $2H_2 + O_2 \rightarrow 2H_2O$ Moles of H₂ = 36/2 = 18 mol Moles of H₂O = 18 mol H₂ × (2 mol H₂O / 2 mol H₂) = 18 mol Mass of H₂O = 18 × 18 = 324 g

3. Limiting Reagent Problems:

The limiting reagent is completely consumed and determines the maximum product possible.

Example: 10 g H₂ + 10 g O₂ → H₂O Moles H₂ = 10/2 = 5 mol Moles O₂ = 10/32 = 0.3125 mol O₂ is limiting (needs 1.5 mol H₂ per mol O₂, but we have excess H₂)

4. Percent Yield:

$$Percent\ Yield = \frac{Actual\ Yield}{Theoretical\ Yield} \times 100%$$

5. Empirical and Molecular Formula:

Empirical formula = simplest whole number ratio Molecular formula = (Empirical formula)_n

Example: Compound contains 40% C, 6.67% H, 53.33% O

• C: 40/12 = 3.33 mol
• H: 6.67/1 = 6.67 mol
• O: 53.33/16 = 3.33 mol
• Divide by smallest (3.33): C₁H₂O₁ → Empirical = CH₂O
• If molar mass = 180 g/mol: n = 180/30 = 6
• Molecular formula = (CH₂O)₆ = C₆H₁₂O₆

⚡ ECAT Exam Tip: In percent composition problems, always use atomic masses: C=12, H=1, O=16, N=14, Cl=35.5, S=32, Na=23, Fe=56, Ca=40 g/mol.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Advanced Stoichiometric Concepts:

1. Equivalent Weight:

$$Equivalent\ Weight = \frac{Molar\ Mass}{n-factor}$$

The n-factor depends on the reaction type:

• Acid-base: n = number of H⁺ or OH⁻ donated/accepted
• Redox: n = change in oxidation state per atom
• For Na₂CO₃ in acid-base: n = 2 (2 H⁺ consumed)
• For KMnO₄ in acidic medium: n = 5 (Mn goes from +7 to +2)

2. Normality (N):

$$N = \frac{equivalents\ of\ solute}{volume\ of\ solution\ in\ L}$$

Relationship with Molarity: $N = M \times n$

3. Titration Calculations:

For acid-base titration: $N_1 V_1 = N_2 V_2$ (at equivalence point)

For redox titration: $n_1 N_1 V_1 = n_2 N_2 V_2$

4. Volume Relationships (Gay-Lussac’s Law):

“At constant temperature and pressure, the ratio of volumes of gaseous reactants and products is a simple whole number ratio.”

Example: $2H_2 + O_2 \rightarrow 2H_2O$ Volume ratio: 2 : 1 : 2

5. Molarity by Dilution:

$$M_1 V_1 = M_2 V_2$$

This works because moles don’t change: $n_1 = n_2$

6. Stoichiometry of Solutions:

7. Oxidation Number Method for Balancing Redox:

Step 1: Assign oxidation numbers Step 2: Identify oxidation and reduction half-reactions Step 3: Balance atoms other than O and H Step 4: Balance O by adding H₂O and H by adding H⁺ (acidic) or OH⁻ (basic) Step 5: Balance charge with electrons Step 6: Multiply half-reactions and add

Example — Balancing Fe²⁺ + Cr₂O₇²⁻ (acidic):

Oxidation: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ Reduction: $Cr_2O_7^{2-} + 6e^- \rightarrow 2Cr^{3+}$

Multiply Fe by 6: $6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-$ Add: $6Fe^{2+} + Cr_2O_7^{2-} \rightarrow 6Fe^{3+} + 2Cr^{3+}$

Balance O with H₂O: $Cr_2O_7^{2-}$ has 7 O → add 7 H₂O on right Balance H with H⁺: 14 H⁺ on left

Final: $14H^+ + 6Fe^{2+} + Cr_2O_7^{2-} \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O$

⚡ ECAT 2023 Question Analysis: Questions on limiting reagent, percent yield, and oxidation number calculations appeared repeatedly. The equivalent weight concept (n-factor method) is frequently tested in ECAT. Remember: 1 L of 1 N solution contains 1 equivalent of solute.

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