UPCAT Chemistry — Stoichiometry & Chemical ReactionsStudy Notes

Chemical reactions can be classified into several main types based on their patterns and characteristics. Understanding these patterns helps predict products and balance equations more easily. **1. Synthesis (Composition) Reactions** Pattern: A + B → AB In synthesis reactions, two or more simple substances combine to form a more complex compound. For example: Zn + I₂ → ZnI₂. This is like combining ingredients to make a dish - simpler components come together to form something more complex. **2. Decomposition Reactions** Pattern: AB → A + B Decomposition reactions break down complex compounds into simpler substances. For example: 2Al₂O₃ → 4Al + 3O₂. Think of this as breaking down a compound into its basic building blocks, like taking apart a machine to see its individual parts. **3. Single Replacement (Displacement) Reactions** Pattern: AB + X → AX + B In these reactions, one element replaces another in a compound. For example: Br₂ + 2KI → 2KBr + I₂. Imagine substituting one player for another in a team - the more active element takes the place of the less active one. **4. Double Displacement Reactions** Pattern: AB + XY → AY + XB Two compounds exchange ions to form two new compounds. For example: K₂Cr₂O₇ + Ca(OH)₂ → CaCr₂O₇ + 2KOH. This is like two couples switching dance partners. **5. Combustion Reactions** Pattern: Compound + O₂ → CO₂ + H₂O Combustion involves rapid reaction with oxygen, usually producing heat and light. For example: C₆H₆ + O₂ → CO₂ + H₂O (when balanced). Think of burning fuel in a car engine or wood in a fireplace. **6. Acid-Base Reactions** Pattern: Acid + Base → Salt + H₂O These neutralization reactions occur when acids and bases react to form salt and water. For example: Sr(OH)₂ + H₂SO₄ → SrSO₄ + 2H₂O.

A chemical equation represents a chemical reaction using chemical formulas and symbols. Understanding the components of chemical equations is crucial for solving stoichiometry problems. **Components of Chemical Equations:** - **Reactants**: Substances that undergo chemical change (left side of arrow) - **Products**: Substances formed as a result of the reaction (right side of arrow) - **Coefficients**: Numbers placed in front of formulas to balance the equation - **Arrow (→)**: Indicates the direction of reaction, meaning 'yields' or 'produces' **Law of Conservation of Mass**: This fundamental principle states that mass cannot be created or destroyed in chemical reactions. The total mass of reactants equals the total mass of products. This means atoms are neither created nor destroyed, only rearranged. **Balancing Chemical Equations:** To balance equations, follow these systematic steps: 1. **Identify reactants and products**: Write the unbalanced equation 2. **Count atoms of each element**: List the number of atoms on both sides 3. **Add coefficients**: Use whole numbers to balance atoms (never change subscripts) 4. **Check your work**: Ensure equal numbers of each type of atom on both sides Example: Balancing Fe + O₂ → Fe₂O₃ Step 1: Fe + O₂ → Fe₂O₃ (unbalanced) Step 2: Left side - 1 Fe, 2 O; Right side - 2 Fe, 3 O Step 3: 4Fe + 3O₂ → 2Fe₂O₃ (balanced) Step 4: Left side - 4 Fe, 6 O; Right side - 4 Fe, 6 O ✓

The mole is the fundamental unit for counting particles in chemistry. Understanding moles is essential for all stoichiometric calculations. **The Mole Defined:** A mole (mol) is the amount of substance that contains 6.022 × 10²³ elementary entities (atoms, molecules, ions, etc.). This number is called Avogadro's number (Nₐ). Think of a mole like a dozen - just as a dozen always means 12, a mole always means 6.022 × 10²³ particles. **Molar Mass:** Molar mass (M) is the mass of one mole of a substance, expressed in grams per mole (g/mol). For elements, the molar mass equals the atomic mass from the periodic table. For compounds, add up the atomic masses of all atoms in the formula. **Calculating Molar Mass:** For H₂O (water): - H: 1.00 g/mol × 2 atoms = 2.00 g/mol - O: 16.00 g/mol × 1 atom = 16.00 g/mol - Total molar mass = 18.00 g/mol **Mass-Mole-Number Conversions:** These three quantities are related through mathematical relationships: - Mass (g) ↔ Moles: Use molar mass as conversion factor - Moles ↔ Number of particles: Use Avogadro's number - Mass ↔ Number of particles: Convert through moles **Practical Example:** How many molecules are in 5g of water? 5g H₂O × (1 mol/18.00g) × (6.022×10²³ molecules/1 mol) = 1.67×10²³ molecules This step-by-step approach ensures accurate calculations in stoichiometry problems.

Stoichiometric calculations use balanced equations to determine quantities of reactants and products in chemical reactions. These calculations are the heart of quantitative chemistry. **Basic Stoichiometry Steps:** 1. Write and balance the chemical equation 2. Convert given quantity to moles 3. Use mole ratios from balanced equation 4. Convert to desired units **Mole-to-Mole Calculations:** The coefficients in balanced equations give mole ratios. For the reaction 2H₂ + O₂ → 2H₂O, the mole ratios are: - 2 mol H₂ : 1 mol O₂ : 2 mol H₂O - This means 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. **Mass-to-Mass Calculations:** Example: How many grams of water form when 4g of hydrogen burns? 2H₂ + O₂ → 2H₂O Step 1: Convert mass to moles 4g H₂ × (1 mol H₂/2.02g H₂) = 1.98 mol H₂ Step 2: Use mole ratio 1.98 mol H₂ × (2 mol H₂O/2 mol H₂) = 1.98 mol H₂O Step 3: Convert to mass 1.98 mol H₂O × (18.02g H₂O/1 mol H₂O) = 35.7g H₂O **Limiting Reactant Problems:** When reactants are not in stoichiometric proportions, one reactant will be completely consumed first. This limiting reactant determines the maximum amount of product that can form. The other reactant(s) are in excess. To identify the limiting reactant: 1. Calculate moles of each reactant 2. Use stoichiometry to determine how much product each could make 3. The reactant that produces the least product is limiting **Percent Yield:** Actual yield is often less than theoretical yield due to side reactions, incomplete reactions, or losses during isolation. Percent yield = (Actual yield/Theoretical yield) × 100% This measures the efficiency of a chemical reaction.

Many chemical reactions occur in solution, making it essential to understand concentration measurements and calculations. **Molarity (M):** Molarity is the most common concentration unit in chemistry, defined as moles of solute per litre of solution. Molarity = moles of solute / liters of solution M = n / V **Preparing Solutions:** To prepare a solution of known molarity: 1. Calculate moles of solute needed 2. Measure the required mass of solute 3. Dissolve in less than the final volume of solvent 4. Dilute to the exact final volume Example: Prepare 250 mL of 0.100 M NaCl solution Moles needed = 0.100 M × 0.250 L = 0.0250 mol NaCl Mass needed = 0.0250 mol × 58.44 g/mol = 1.46 g NaCl **Dilution Calculations:** When diluting concentrated solutions, use the dilution formula: M₁V₁ = M₂V₂ Where: - M₁ = initial molarity - V₁ = initial volume - M₂ = final molarity - V₂ = final volume **Molality (m):** Molality is moles of solute per kilogram of solvent (not solution). Unlike molarity, molality doesn't change with temperature since it's based on mass, not volume. Molality = moles of solute / kg of solvent **Solution Stoichiometry:** When reactions occur in solution, use molarity to convert between volume and moles: Volume (L) × Molarity (M) = moles Example: How many moles of HCl are in 25.0 mL of 0.150 M HCl? moles = 0.0250 L × 0.150 M = 0.00375 mol HCl

Chemical formulas represent the composition of compounds. Understanding the relationship between empirical and molecular formulas is crucial for determining compound structures. **Empirical Formula:** The empirical formula shows the simplest whole-number ratio of elements in a compound. It represents the lowest terms ratio, like reducing a fraction to its simplest form. **Molecular Formula:** The molecular formula shows the actual number of atoms of each element in a molecule. It may be the same as the empirical formula or a whole-number multiple of it. **Determining Empirical Formulas:** From percent composition data: 1. Assume 100g sample 2. Convert percentages to grams 3. Convert grams to moles 4. Divide by smallest number of moles 5. Multiply by integers if needed to get whole numbers Example: A compound is 40.0% C, 6.7% H, and 53.3% O Step 1: 40.0g C, 6.7g H, 53.3g O Step 2: - C: 40.0g ÷ 12.01g/mol = 3.33 mol - H: 6.7g ÷ 1.008g/mol = 6.6 mol - O: 53.3g ÷ 16.00g/mol = 3.33 mol Step 3: Divide by 3.33: C₁H₂O₁ = CH₂O (empirical formula) **Determining Molecular Formulas:** If you know the molecular mass: Molecular formula = (Empirical formula)ₙ where n = molecular mass ÷ empirical formula mass Using the previous example, if molecular mass = 180 g/mol: Empirical formula mass of CH₂O = 30.0 g/mol n = 180 ÷ 30.0 = 6 Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ **Combustion Analysis:** For compounds containing C, H, and O, combustion analysis determines composition: - All carbon forms CO₂ - All hydrogen forms H₂O - Oxygen is calculated by difference This method is particularly useful for organic compounds where direct analysis might be difficult.