Chapter 3: Chemical Reactions and Reaction Stoichiometry – Study Notes

Chapter 3: Chemical Reactions and Reaction Stoichiometry

Objectives

• Write and interpret chemical equations.

• Recognize simple patterns of chemical activity.

• Calculate formula weights and percentage composition.

• Relate masses, moles, and numbers of particles.

• Determine empirical formulas from reaction analysis.

• Draw mathematical conclusions about stoichiometry in chemical reactions.

Conservation of Mass

Law of Conservation of Mass

The Law of Conservation of Mass states that matter is neither created nor destroyed in a chemical reaction. The total mass of reactants equals the total mass of products.

• First articulated by Antoine Lavoisier.

• All atoms present in the reactants are accounted for among the products.

Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in chemical reactions.

Anatomy of a Chemical Equation

Structure of Chemical Equations

• Reactants appear on the left side of the equation.

• Products appear on the right side of the equation.

• The states of substances are indicated in parentheses: (s) for solid, (l) for liquid, (g) for gas, (aq) for aqueous solution.

• Chemical equations must be balanced so that the number of atoms of each element is the same on both sides.

• Chemical formulas use subscripts to indicate the number of atoms in a molecule (e.g., H2O has 2 H and 1 O atom).

• Coefficients in front of formulas indicate the number of molecules or moles (e.g., 2 H2O means 2 molecules or 2 moles of water).

Example: The balanced equation for the combustion of methane:

$\mathrm{CH}_4(g) + 2\ \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) + 2\ \mathrm{H}_2\mathrm{O}(g)$

• 1 molecule of methane reacts with 2 molecules of oxygen to produce 1 molecule of carbon dioxide and 2 molecules of water.

Balancing Chemical Equations

• Write correct formulas for all reactants and products.

• Balance elements one at a time using coefficients.

• Balance metals first, then nonmetals, and finally hydrogen and oxygen.

• Check that the total number of atoms of each element is the same on both sides.

Example: Balancing the combustion of propane:

$\mathrm{C}_3\mathrm{H}_8 + 5\ \mathrm{O}_2 \rightarrow 3\ \mathrm{CO}_2 + 4\ \mathrm{H}_2\mathrm{O}$

Types of Chemical Reactions

Combination (Synthesis) Reactions

Two or more substances combine to form a single product.

• General form: $A + B \rightarrow AB$

• Example: $2\ \mathrm{Mg}(s) + \mathrm{O}_2(g) \rightarrow 2\ \mathrm{MgO}(s)$

Decomposition Reactions

A single compound breaks down into two or more simpler substances.

• General form: $AB \rightarrow A + B$

• Example: $2\ \mathrm{KClO}_3(s) \rightarrow 2\ \mathrm{KCl}(s) + 3\ \mathrm{O}_2(g)$

Combustion Reactions

Rapid reactions that produce a flame, usually involving oxygen as a reactant and producing CO2 and H2O.

• Example: $\mathrm{CH}_4(g) + 2\ \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) + 2\ \mathrm{H}_2\mathrm{O}(g)$

Formula Weights and Percentage Composition

Formula Weight (FW)

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

• Example: For CaCl2:

$\text{FW} = 1 \times 40.1\ \text{amu (Ca)} + 2 \times 35.5\ \text{amu (Cl)} = 111.1\ \text{amu}$

Molecular Weight (MW)

If the substance is a molecule, the formula weight is called the molecular weight.

• Example: For ethane (C2H6):

$\text{MW} = 2 \times 12.0\ \text{amu (C)} + 6 \times 1.0\ \text{amu (H)} = 30.0\ \text{amu}$

Percent Composition

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

Formula:

$\%\ \text{Element} = \frac{(\text{number of atoms})(\text{atomic weight})}{\text{FW of compound}} \times 100$

• Example: Percent C in C2H6:

$\%\ \text{C} = \frac{2 \times 12.0}{30.0} \times 100 = 80.0\%$

The Mole and Avogadro's Number

Definition of the Mole

A mole (mol) is the amount of substance that contains as many entities (atoms, molecules, ions) as there are atoms in exactly 12 g of carbon-12. This number is called Avogadro's number ($N_A$):

$N_A = 6.022 \times 10^{23}$ particles/mol

Molar Mass

The molar mass of a substance is the mass in grams of one mole of its entities. It is numerically equal to the formula weight in amu, but expressed in g/mol.

• Example: H2O has a formula weight of 18.016 amu, so its molar mass is 18.016 g/mol.

Conversions Using the Mole

• To convert grams to moles: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

• To convert moles to number of particles: $\text{number of particles} = \text{moles} \times N_A$

Example: How many moles are in 123.2 g of Cl2?

$\text{moles} = \frac{123.2\ \text{g}}{70.90\ \text{g/mol}} = 1.738\ \text{mol}$

Example: How many atoms are in 3.91 mol of Xe?

$3.91\ \text{mol} \times 6.022 \times 10^{23}\ \text{atoms/mol} = 2.35 \times 10^{24}\ \text{atoms}$

Empirical and Molecular Formulas

Definitions

• Molecular formula: Shows the exact number of atoms of each element in a molecule (e.g., C4H10).

• Empirical formula: Shows the simplest whole-number ratio of atoms in a compound (e.g., C2H5 for C4H10).

Calculating Empirical Formulas

1. Obtain the mass (or percent composition) of each element in the compound.

2. Convert masses to moles using atomic weights.

3. Divide each mole value by the smallest number of moles to get the simplest ratio.

4. If necessary, multiply ratios by a whole number to get whole-number subscripts.

Example: A compound contains 61.31% C, 5.14% H, 10.21% N, and 23.33% O. Find the empirical formula.

• Convert to moles: 61.31 g C × (1 mol/12.01 g) = 5.105 mol C, etc.

• Divide by smallest: C: 5.105/0.7288 ≈ 7, H: 5.09/0.7288 ≈ 7, N: 0.7288/0.7288 = 1, O: 1.456/0.7288 ≈ 2

• Empirical formula: C7H7NO2

Stoichiometric Calculations

Using Balanced Equations

Coefficients in a balanced equation indicate the ratios of moles of reactants and products.

• Example: $2\ \mathrm{H}_2(g) + \mathrm{O}_2(g) \rightarrow 2\ \mathrm{H}_2\mathrm{O}(g)$

• 2 mol H2 react with 1 mol O2 to produce 2 mol H2O.

General Stoichiometric Problem Steps

1. Convert given mass to moles.

2. Use mole ratios from the balanced equation to find moles of desired substance.

3. Convert moles to grams if required.

Example: How many grams of H2O are produced from 1.00 g of C6H12O6?

• Step 1: $1.00\ \text{g} \div 180.0\ \text{g/mol} = 0.00556\ \text{mol}$

• Step 2: $0.00556\ \text{mol} \times \frac{6\ \text{mol H}_2\text{O}}{1\ \text{mol C}_6\text{H}_{12}\text{O}_6} = 0.0334\ \text{mol H}_2\text{O}$

• Step 3: $0.0334\ \text{mol} \times 18.02\ \text{g/mol} = 0.602\ \text{g H}_2\text{O}$

Limiting Reactants and Yields

Limiting Reactant

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

1. Calculate moles of each reactant.

2. Use stoichiometry to determine which reactant produces less product; this is the limiting reactant.

Example: If 2 g of A (molar mass 20 g/mol) and 3 g of B (molar mass 30 g/mol) react according to $A + 2B \rightarrow BA$, find the limiting reactant and amount of product.

• Moles A: $2/20 = 0.1$ mol; Moles B: $3/30 = 0.1$ mol

• 0.1 mol A needs 0.2 mol B, but only 0.1 mol B is available, so B is limiting.

• Product formed: $0.05$ mol BA ($0.1$ mol B $\div$ 2), mass: $0.05 \times 80 = 4$ g BA

Theoretical, Actual, and Percent Yield

• Theoretical yield: Maximum amount of product predicted by stoichiometry.

• Actual yield: Amount of product actually obtained from an experiment.

• Percent yield: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$