Chapter 3: Chemical Reactions and Reaction Stoichiometry – Study Notes

Stoichiometry

Introduction to Stoichiometry

Stoichiometry is the area of chemistry that examines the quantitative relationships between the amounts of reactants and products in chemical reactions. It is fundamentally based on the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction.

• Law of Conservation of Mass: The total mass of reactants equals the total mass of products in a chemical reaction.

• Antoine Lavoisier (1789): Credited with establishing the law, stating that "nothing is created; an equal amount of matter exists both before and after the experiment."

Chemical Equations

Representing Chemical Reactions

Chemists use chemical equations to represent chemical reactions on paper. These equations show the transformation of reactants into products.

• Reactants: Substances present at the start of the reaction (left side of the equation).

• Products: Substances formed as a result of the reaction (right side of the equation).

• Plus sign (+): Separates multiple reactants or products.

• Arrow (→): Indicates the direction of the reaction, from reactants to products.

Example:

• $2H_2 + O_2 \rightarrow 2H_2O$

Balancing Chemical Equations

How to Balance Equations

Balancing chemical equations ensures the Law of Conservation of Mass is obeyed. The number of atoms of each element must be the same on both sides of the equation.

• Start with an element that appears in only one reactant and one product.

• Balance by changing coefficients (numbers in front of formulas), not subscripts (numbers within formulas).

• Repeat for other elements, checking all elements at the end.

Example:

• $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

Why Use Coefficients Instead of Subscripts?

Changing coefficients adjusts the number of molecules, while changing subscripts alters the identity of the substance.

• Example: Hydrogen and oxygen can form water or hydrogen peroxide:

• $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$

• $H_2(g) + O_2(g) \rightarrow H_2O_2(l)$

• Do not change the formula unless the product is actually different.

Other Symbols in Chemical Equations

States of Matter and Reaction Conditions

Chemical equations often include symbols to indicate the physical state of each substance and reaction conditions.

• (g): Gas

• (l): Liquid

• (s): Solid

• (aq): Dissolved in aqueous (water) solution

• Δ (delta) over arrow: Indicates heat is required for the reaction

Example:

• $CH_4(g) + 2O_2(g) \xrightarrow{\Delta} CO_2(g) + 2H_2O(g)$

Types of Chemical Reactions

Common Reaction Patterns

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

• Combination Reactions: Two or more substances combine to form one product.

• Decomposition Reactions: One substance breaks down into two or more products.

• Combustion Reactions: Rapid reactions with oxygen that produce a flame.

Combination Reactions

• General form: $A + B \rightarrow C$

• Example: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

• Metal and nonmetal combinations often form ionic compounds.

Decomposition Reactions

• General form: $C \rightarrow A + B$

• Example: $CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)$

• Metal carbonates decompose to metal oxides and carbon dioxide when heated.

Combustion Reactions

• Involve oxygen as a reactant and produce heat and light.

• Hydrocarbon combustion produces carbon dioxide and water.

• Example: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)$

Formula and Molecular Weights

Formula Weight (FW)

The formula weight is the sum of the atomic weights of all atoms in a chemical formula, expressed in atomic mass units (amu).

• Example: $FW(H_2SO_4) = 2(1.0) + 32.1 + 4(16.0) = 98.1$ amu

Molecular Weight (MW)

For molecular substances, the formula weight is also called the molecular weight.

• Example: $MW(C_6H_{12}O_6) = 6(12.0) + 12(1.0) + 6(16.0) = 180.0$ amu

Percent Composition

Calculating Percent Composition

The percent composition of a compound is the percentage by mass of each element in the compound.

• Formula: $\%\,\text{Element} = \frac{\text{(number of atoms of element)} \times \text{atomic weight}}{\text{formula weight of compound}} \times 100$

• Example (Carbon in Glucose): $\%\,C = \frac{6 \times 12.0}{180.0} \times 100 = 40.0\%$

The Mole and Avogadro's Number

Definition of the Mole

The mole (mol) is the amount of substance containing as many entities (atoms, molecules, ions) as there are atoms in exactly 12 g of carbon-12.

• Avogadro's Number: $6.022 \times 10^{23}$ particles per mole

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). For elements, it is the atomic weight in g/mol; for compounds, it is the formula weight in g/mol.

• Example: $\text{Molar mass of H}_2SO_4 = 98.1$ g/mol

Mole Relationships and Conversions

Converting Between Mass, Moles, and Number of Particles

Moles provide a bridge between the molecular scale and the real-world scale. Use the following relationships for conversions:

• Mass to Moles: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

• Moles to Number of Particles: $\text{number of particles} = \text{moles} \times 6.022 \times 10^{23}$

Example: How many atoms in 3 g of copper (Cu)?

• $\text{moles Cu} = \frac{3.00}{63.55} = 0.0472$ mol

• $\text{atoms Cu} = 0.0472 \times 6.022 \times 10^{23} = 2.84 \times 10^{22}$ atoms

Empirical and Molecular Formulas

Determining Empirical Formulas

The empirical formula gives the simplest whole-number ratio of atoms in a compound. It can be determined from percent composition:

1. Assume 100 g of compound; convert mass of each element to moles.

2. Divide each mole value by the smallest number of moles.

3. Round to nearest whole number to get subscripts.

Example: Para-aminobenzoic acid (PABA): C (61.31%), H (5.14%), N (10.21%), O (23.33%)

• C: $\frac{61.31}{12.01} = 5.105$ mol

• H: $\frac{5.14}{1.008} = 5.10$ mol

• N: $\frac{10.21}{14.01} = 0.729$ mol

• O: $\frac{23.33}{16.00} = 1.46$ mol

• Divide by 0.729: C (7), H (7), N (1), O (2) → Empirical formula: C7H7NO2

Determining Molecular Formulas

The molecular formula is a whole-number multiple of the empirical formula. Use the molar mass to find the multiple:

• $\text{Whole-number multiple} = \frac{\text{molar mass}}{\text{empirical formula mass}}$

• Example: Empirical formula CH, molar mass 78 g/mol. $\frac{78}{13} = 6$ → Molecular formula: C6H6

Combustion Analysis

Determining Elemental Composition by Combustion

Combustion analysis is used to determine the amounts of C, H, and O in organic compounds.

• Mass of C determined from mass of CO2 produced.

• Mass of H determined from mass of H2O produced.

• Mass of O determined by difference (total mass minus mass of C and H).

Stoichiometric Calculations

Using Balanced Equations for Quantitative Analysis

Coefficients in balanced equations indicate relative numbers of molecules and moles, which can be converted to mass.

• Use mole ratios from coefficients to relate amounts of different substances.

• Convert grams to moles, use mole ratio, then convert moles to grams of desired substance.

Example: How many grams of water produced from 1.00 g glucose?

• $C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)$

• Step 1: $\text{moles glucose} = \frac{1.00}{180.0} = 0.00556$ mol

• Step 2: $0.00556 \text{ mol glucose} \times \frac{6 \text{ mol } H_2O}{1 \text{ mol glucose}} = 0.0333$ mol $H_2O$

• Step 3: $0.0333 \text{ mol } H_2O \times 18.0 \text{ g/mol} = 0.600$ g $H_2O$

Limiting Reactants and Yield

Limiting Reactant

The limiting reactant is the reactant that is completely consumed first, limiting the amount of product formed.

• Identify by comparing mole ratios of reactants to coefficients in the balanced equation.

• Use limiting reactant for all stoichiometric calculations.

Theoretical and Percent Yield

• Theoretical yield: Maximum amount of product possible, calculated from stoichiometry.

• Actual yield: Amount of product actually obtained in the experiment.

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