Chapter 3 Chemistry

Chemical Equations and Stoichiometry

Representing Chemical Reactions

Chemical reactions are represented by chemical equations, which show the reactants and products involved. The numbers in front of the formulas are called coefficients and indicate the relative number of molecules or moles of each substance.

• Balanced chemical equations have equal numbers of atoms of each element on both sides of the arrow, reflecting the law of conservation of mass.

• Coefficients change the amount of a substance, not its identity; subscripts define the identity and should never be changed when balancing equations.

Balancing Chemical Equations

Balancing equations ensures that the number of atoms for each element is the same on both sides. This is achieved by adjusting coefficients, not subscripts.

• Example: $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

• Balance hydrogen and oxygen atoms by adjusting coefficients.

Types of Chemical Reactions

Combination Reactions

Two or more reactants combine to form a single product.

• General form: $A + B \rightarrow C$

• Example: $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$

Decomposition Reactions

A single reactant breaks apart to form two or more products.

• General form: $C \rightarrow A + B$

• Example: $2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2$

Single Replacement Reactions

One element replaces another in a compound.

• General form: $A + BX \rightarrow AX + B$

• Example: $\text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu}$

Double Replacement Reactions

Two compounds exchange ions to form two new compounds.

• General form: $AX + BY \rightarrow AY + BX$

• Example: $\text{Na}_2\text{CO}_3 + 2\text{AgNO}_3 \rightarrow 2\text{NaNO}_3 + \text{Ag}_2\text{CO}_3$

Combustion Reactions

Combustion reactions are rapid and produce a flame, typically involving oxygen as a reactant and forming CO2 and H2O.

• Example: $\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$

Formula Weights and Percentage Composition

Formula Weight (FW)

The formula weight is the sum of the atomic weights of all atoms in a chemical formula.

• Example: FW of HNO3 = 1(1.01) + 1(14.01) + 3(15.99) = 62.99 amu

Percentage Composition

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

• Formula: $\text{Percent composition} = \frac{\text{number of atoms} \times \text{atomic weight}}{\text{formula weight}} \times 100$

• Example: In C12H22O11, %C = 42.1%, %H = 6.5%, %O = 51.4%

The Mole and Avogadro’s Number

Definition of the Mole

The mole is the counting unit for atoms, ions, or molecules. One mole contains Avogadro’s number ($6.02 \times 10^{23}$) of particles.

• 1 mol H2O = $6.02 \times 10^{23}$ molecules H2O

• 1 mol NaCl = $6.02 \times 10^{23}$ formula units NaCl

Mole Relationships Table

This table summarizes the relationship between formula weight, molar mass, and the number of particles in one mole for various substances.

Name of Substance	Formula	Formula Weight (amu)	Molar Mass (g/mol)	Number and Kind of Particles in One Mole
Atomic nitrogen	N	14.0	14.0	6.02 × 1023 N atoms
Molecular nitrogen	N2	28.0	28.0	6.02 × 1023 N2 molecules
Silver	Ag	107.9	107.9	6.02 × 1023 Ag atoms
Silver ions	Ag+	107.9	107.9	6.02 × 1023 Ag+ ions
Barium chloride	BaCl2	208.2	208.2	6.02 × 1023 BaCl2 formula units, 6.02 × 1023 Ba2+ ions, 2 × 6.02 × 1023 Cl- ions

Molar Mass and Conversions

Molar Mass

The molar mass (MM) is the mass in grams of one mole of a substance and is numerically equal to its formula weight in atomic mass units.

• Example: MM of C6H12O6 = 180.12 g/mol

Converting Between Mass, Moles, and Particles

Conversions between grams, moles, and number of particles use the molar mass and Avogadro’s number.

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

• To convert moles to particles: $\text{particles} = \text{moles} \times 6.02 \times 10^{23}$

Empirical and Molecular Formulas

Empirical Formula

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

• Steps: Convert mass to moles, divide by the smallest number of moles, adjust to whole numbers if necessary.

• Example: Sulfuric acid (H2SO4) has an empirical formula identical to its molecular formula.

Molecular Formula

The molecular formula is a whole-number multiple of the empirical formula, determined by the ratio of the molecular weight to the empirical formula weight.

• Formula: $\text{Whole number multiple} = \frac{\text{molecular weight}}{\text{empirical formula weight}}$

• Example: Glucose (C6H12O6) has an empirical formula CH2O and a multiple of 6.

Quantitative Information from Balanced Equations

Stoichiometric Calculations

Balanced equations provide mole ratios for converting between reactants and products. These ratios are used in calculations involving mass, moles, and number of particles.

• Example: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$

• For every 2 moles of H2, 1 mole of O2 is consumed, and 2 moles of H2O are produced.

Limiting Reactants and Percent Yield

Limiting Reactant

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

• Any excess reactant will be left over after the reaction.

• Example: In making sandwiches, if you have 14 slices of bread and 6 slices of cheese, you can make only 6 sandwiches; cheese is the limiting reactant.

Theoretical Yield and Percent Yield

The theoretical yield is the maximum amount of product that can be formed from the limiting reactant. The actual yield is the amount actually obtained, and percent yield is calculated as:

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

• Percent yield is always less than or equal to 100%.

Summary Table: Mole Relationships

Name of Substance	Formula	Formula Weight (amu)	Molar Mass (g/mol)	Number and Kind of Particles in One Mole
Atomic nitrogen	N	14.0	14.0	6.02 × 1023 N atoms
Molecular nitrogen	N2	28.0	28.0	6.02 × 1023 N2 molecules
Silver	Ag	107.9	107.9	6.02 × 1023 Ag atoms
Silver ions	Ag+	107.9	107.9	6.02 × 1023 Ag+ ions
Barium chloride	BaCl2	208.2	208.2	6.02 × 1023 BaCl2 formula units, 6.02 × 1023 Ba2+ ions, 2 × 6.02 × 1023 Cl- ions