Chapter 3: Chemical Reactions and Reaction Stoichiometry – Study Notes

Chapter 3: Chemical Reactions and Reaction Stoichiometry

3.1 Chemical Equations

Chemical equations are the symbolic representation of chemical reactions, showing the relationship between reactants and products. They are fundamental for communicating chemical changes and for quantitative calculations in chemistry.

• Chemical equations use arrows to separate reactants (starting materials, left side) from products (ending materials, right side).

• The plus sign (+) separates multiple reactants or products.

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

Balancing Equations

• Follows the Law of Conservation of Mass: the number of atoms of each element must be the same on both sides of the equation.

• Balance by changing coefficients (numbers in front of formulas), not subscripts (which would change the substance).

• Start with elements that appear in only one reactant and one product, then proceed to others.

• Check all elements at the end to ensure balance.

Why Use Coefficients?

• Changing subscripts alters the identity of the substance (e.g., $H_2O$ vs. $H_2O_2$).

• Coefficients adjust the quantity without changing the chemical identity.

Required Symbols in Chemical Equations

• States of matter are indicated in parentheses:

  • (s) = solid

  • (l) = liquid

  • (g) = gas

  • (aq) = aqueous (dissolved in water)

• Example: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$

Sample Exercise: Balancing Chemical Equations

• Balance the following:

  1. __Fe(s) + __O2(g) → __Fe2O3(s)

  2. __Al(s) + __HCl(aq) → __AlCl3(aq) + __H2(g)

  3. __CaCO3(s) + __HCl(aq) → __CaCl2(aq) + __CO2(g) + __H2O(l)

3.2 Simple Patterns of Chemical Reactivity

Chemical reactions can be classified into several types based on their patterns. Understanding these patterns helps predict products and write equations.

• Combination reactions: Two or more substances form one product.

• Decomposition reactions: One substance breaks down into two or more substances.

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

Combination Reactions

• General form: $A + B \rightarrow C$

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

• Predicting products: For a metal and a nonmetal, use common charges to determine the formula.

Combination Reactions	Description
$C(s) + O_2(g) \rightarrow CO_2(g)$	Two or more reactants combine to form a single product.
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$	Many elements react with one another in this fashion to form compounds.
$CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq)$	Formation of a hydroxide from an oxide and water.

Decomposition Reactions

• General form: $C \rightarrow A + B$

• Example: $2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$

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

• Example: $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$

Decomposition Reactions	Description
$PbCO_3(s) \rightarrow PbO(s) + CO_2(g)$	Decomposition of a carbonate.
$Cu(OH)_2(s) \rightarrow CuO(s) + H_2O(g)$	Decomposition of a hydroxide.

Combustion Reactions

• Rapid reactions that produce a flame, usually involving oxygen.

• When burning hydrocarbons (compounds with C and H), the products are always $CO_2$ and $H_2O$.

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

3.3 Formula Weight (FW) and Molecular Weight (MW)

Formula weight and molecular weight are measures of the mass of a chemical formula or molecule, respectively, based on atomic weights from the periodic table.

• Formula weight (FW): Sum of atomic weights for all atoms in a chemical formula (amu).

• Molecular weight (MW): Sum of atomic weights for all atoms in a molecule (amu).

• For ionic compounds, use the empirical formula; for molecules, use the molecular formula.

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

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

Percent Composition

• Percent composition gives the percentage by mass of each element in a compound.

• Formula: $\%\,\text{Element} = \left(\frac{\text{number of atoms of element} \times \text{atomic weight of element}}{\text{FW of substance}}\right) \times 100$

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

3.4 Avogadro’s Number and the Mole

The mole is a counting unit for atoms, molecules, or ions, allowing chemists to relate microscopic particles to macroscopic amounts.

• One mole (mol) contains $6.02 \times 10^{23}$ particles (Avogadro’s number).

• The molar mass (g/mol) of an element is numerically equal to its atomic weight (amu).

• For diatomic elements, the molar mass is twice the atomic weight.

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	$6.02 \times 10^{23}$ N atoms
Molecular nitrogen	N2	28.0	28.0	$6.02 \times 10^{23}$ N2 molecules
Silver	Ag	107.9	107.9	$6.02 \times 10^{23}$ Ag atoms
Barium chloride	BaCl2	208.2	208.2	$6.02 \times 10^{23}$ BaCl2 units

Converting Amounts Using Moles

• Moles provide a bridge between the molecular and real-world scales.

• Conversions use molar mass and Avogadro’s number.

• Example: How many atoms in 3 g of Cu?

$3\,\text{g}\,Cu \times \frac{1\,\text{mol}\,Cu}{63.5\,\text{g}\,Cu} \times \frac{6.02 \times 10^{23}\,\text{atoms}}{1\,\text{mol}\,Cu} = 3 \times 10^{22}\,\text{atoms}$

3.5 Empirical Formulas from Analysis

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

• Steps:

  1. Convert mass % to grams (assume 100 g sample).

  2. Convert grams to moles for each element.

  3. Divide by the smallest number of moles to get the ratio.

Example: Para-aminobenzoic Acid (PABA)

• Given: C (61.31%), H (5.14%), N (10.21%), O (23.33%)

• Convert to grams (per 100 g): C = 61.31 g, H = 5.14 g, N = 10.21 g, O = 23.33 g

• Convert to moles:

  • C: $61.31 \div 12.01 = 5.105$ mol

  • H: $5.14 \div 1.01 = 5.09$ mol

  • N: $10.21 \div 14.01 = 0.7288$ mol

  • O: $23.33 \div 16.00 = 1.456$ mol

• Divide by smallest (0.7288):

  • C: $5.105 \div 0.7288 = 7$

  • H: $5.09 \div 0.7288 = 7$

  • N: $0.7288 \div 0.7288 = 1$

  • O: $1.456 \div 0.7288 = 2$

• Empirical formula: $C_7H_7NO_2$

Molecular Formulas from Empirical Formulas

• The molecular formula is a whole-number multiple of the empirical formula.

• Formula: $\text{Whole number multiple} = \frac{\text{Molecular weight (MW)}}{\text{Empirical formula weight (FW)}}$

• Example: Empirical formula CH, MW = 78 g/mol. $13 \times 6 = 78$, so molecular formula is $C_6H_6$.

Combustion Analysis

• Used to determine empirical formulas of compounds containing C, H, and O.

• Mass of C is determined from mass of $CO_2$ produced.

• Mass of H is determined from mass of $H_2O$ produced.

• Mass of O is found by subtracting the mass of C and H from the total mass of the sample.

3.6 Quantitative Information from Balanced Equations

Balanced equations provide the mole ratios needed for quantitative calculations in chemical reactions (stoichiometry).

• Coefficients indicate relative numbers of molecules and moles of reactants and products.

• Stoichiometric calculations use these ratios to convert between masses of different substances in a reaction.

Stoichiometric Calculations

• Convert grams of substance A to moles using molar mass.

• Use the mole ratio from the balanced equation to find moles of substance B.

• Convert moles of B to grams if needed.

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

$\text{Moles } C_6H_{12}O_6 = \frac{1.00\,g}{180.0\,g/mol} = 0.00556\,mol$

$\text{Moles } H_2O = 0.00556\,mol \times \frac{6\,mol\,H_2O}{1\,mol\,C_6H_{12}O_6} = 0.0333\,mol$

$\text{Mass } H_2O = 0.0333\,mol \times 18.02\,g/mol = 0.600\,g$

3.7 Limiting Reactants, Theoretical Yield, and Percent Yield

In reactions with more than one reactant, the limiting reactant is the one that is completely consumed first, limiting the amount of product formed.

• Limiting reactant: Reactant present in the smallest stoichiometric amount; determines the maximum amount of product.

• Excess reactant: Reactant(s) left over after the reaction is complete.

• Theoretical yield: Maximum amount of product possible, calculated from the limiting reactant.

• Actual yield: Amount of product actually obtained from the reaction.

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

Example: If 2.6 g of $C_6H_6$ reacts with excess $O_2$ to produce 1.25 g of water, percent yield is:

$\text{Percent yield} = \frac{1.25\,g}{1.799\,g} \times 100 = 69.5\%$

Additional info: Actual yield is always less than theoretical yield due to losses and inefficiencies in real reactions.