Chapter 3 – Chemical Reactions and Reaction Stoichiometry: Study Notes

Chapter 3 – Chemical Reactions and Reaction Stoichiometry

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction. It is 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: Developed by Antoine Lavoisier in 1789; the total mass of reactants equals the total mass of products.

• Stoichiometric Calculations: Allow chemists to predict the amounts of substances consumed and produced in a reaction.

Chemical Equations & Balancing

Chemical Equations

Chemical equations use chemical formulas to represent the substances involved in a reaction. Reactants are written on the left, products on the right, separated by an arrow.

• Reactants: Starting materials (left side).

• Products: Ending materials (right side).

• Arrow (→): Indicates the direction of the reaction.

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

Balancing Chemical Equations

Balancing ensures the same number of each type of atom on both sides of the equation, in accordance with the Law of Conservation of Mass.

• Only coefficients (numbers in front of formulas) can be changed, not subscripts (numbers within formulas).

• Balance elements that appear only once on each side first, then balance others.

• Balance hydrogen and oxygen atoms last.

• Check your work to ensure the smallest whole-number coefficients.

Example:

• Unbalanced: $\mathrm{CH_4 + O_2 \rightarrow CO_2 + H_2O}$

• Balanced: $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$

Other Symbols in Chemical Equations

• (s): Solid

• (l): Liquid

• (g): Gas

• (aq): Aqueous solution (dissolved in water)

• Conditions such as heat may be indicated above the arrow (e.g., $\Delta$).

Example: $\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)}$

Aqueous Solutions

An aqueous solution is formed when a substance dissolves in water. Many reactions in chemistry occur in aqueous solution.

• Example: $\mathrm{NaCl}$ dissolves in water to form $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ ions.

Types of Chemical Reactions

Simple Patterns of Chemical Reactivity

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

• Combination (Synthesis) Reaction: Two or more substances combine to form one product.

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

• Combustion Reaction: A substance reacts rapidly with oxygen, often producing heat and light.

Combination Reactions

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

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

• When a metal reacts with a nonmetal, the product is an ionic compound.

Decomposition Reactions

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

• Example: $2\mathrm{NaN}_3(s) \rightarrow 2\mathrm{Na}(s) + 3\mathrm{N}_2(g)$

• Heating a metal carbonate produces a metal oxide and carbon dioxide:

• $\mathrm{CaCO}_3(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO}_2(g)$

Combustion Reactions

• Rapid reactions with oxygen that produce a flame.

• Hydrocarbons combust to form $\mathrm{CO}_2$ and $\mathrm{H}_2O$.

• Example: $\mathrm{C_3H_8}(g) + 5\mathrm{O}_2(g) \rightarrow 3\mathrm{CO}_2(g) + 4\mathrm{H}_2O(g)$

Formula Weight (FW) and Molecular Weight (MW)

Formula Weight (FW)

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

• For ionic compounds, use the empirical formula.

• Example: $\mathrm{NaCl}$: FW = (AW of Na) + (AW of Cl) = 22.99 + 35.45 = 58.44 amu

Molecular Weight (MW)

The molecular weight is the sum of the atomic weights of all atoms in a molecule (used for molecular compounds).

• Example: $\mathrm{CO_2}$: MW = (2 × 16.00) + 12.01 = 44.01 amu

Percent Composition

Percent composition is the percentage by mass contributed by each element in a substance.

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

• Example: In glucose ($\mathrm{C_6H_{12}O_6}$), calculate the percent of carbon.

Avogadro's Number & The Mole

Avogadro's Number

Avogadro's number ($6.022 \times 10^{23}$) is the number of particles (atoms, molecules, ions) in one mole of a substance.

• 1 mole = $6.022 \times 10^{23}$ entities

• Allows conversion between atomic/molecular scale and macroscopic scale.

Molar Mass

The molar mass is the mass of one mole of a substance, numerically equal to the formula weight in grams per mole (g/mol).

• Example: $\mathrm{NaCl}$ has a formula weight of 58.5 amu; its molar mass is 58.5 g/mol.

Mole Relationships

• One mole of atoms, ions, or molecules contains Avogadro's number of particles.

• The number of atoms of each element in a formula is multiplied by Avogadro's number for one mole of the compound.

Converting Amounts

Moles act as a bridge between the molecular scale and the real-world scale. Use conversion factors to go from mass to moles to atoms (or vice versa).

• Example: How many atoms are in 7 g of copper ($\mathrm{Cu}$)?

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 data.

1. Assume 100 g of the compound (so percentages become grams).

2. Convert grams to moles using atomic masses.

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

4. Round to nearest whole number for subscripts.

Example: For a compound with 61.31% C, 5.14% H, 10.21% N, 23.33% O, the empirical formula is $\mathrm{C_7H_7NO_2}$.

Determining Molecular Formulas

The molecular formula is a multiple of the empirical formula. If the molar mass is known, divide the molar mass by the empirical formula mass to find the multiple.

• Formula: $\text{Molecular formula} = n \times \text{Empirical formula}$, where $n = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$

Combustion Analysis

Combustion analysis is used to determine the empirical formula of compounds containing C, H, and O by burning the compound and measuring the amounts of $\mathrm{CO_2}$ and $\mathrm{H_2O}$ produced.

• Mass of C is determined from the mass of $\mathrm{CO_2}$ produced.

• Mass of H is determined from the mass of $\mathrm{H_2O}$ produced.

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

Stoichiometry in Chemical Reactions

Stoichiometric Calculations

Coefficients in balanced equations give the ratios of reactants and products in moles. These ratios are used as conversion factors in calculations.

• Example: How many grams of water can be produced from 1.00 g of glucose in the reaction $\mathrm{C_6H_{12}O_6} + 6O_2 \rightarrow 6CO_2 + 6H_2O$?

Limiting Reactant and Excess Reagent

The limiting reactant is the reactant that is completely consumed first, thus limiting the amount of product formed. The excess reagent is the reactant that remains after the reaction is complete.

• Identify the limiting reactant by comparing the mole ratios of reactants used to those required by the balanced equation.

Theoretical Yield, Actual Yield, and Percent Yield

• Theoretical Yield: The maximum amount of product that can be formed from the given amounts of reactants (calculated value).

• Actual Yield: The amount of product actually obtained from a reaction (measured value).

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

Stoichiometry Example

Given: $\mathrm{NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)}$

• Identify the limiting reactant if starting with 50.0 g NaOH and 50.0 g HCl.

• Calculate the theoretical yield of NaCl.

• Calculate the percent yield if 47.8 g NaCl is obtained.

Term	Definition	Formula/Example
Formula Weight (FW)	Sum of atomic weights in a formula	FW of NaCl = 22.99 + 35.45 = 58.44 amu
Molecular Weight (MW)	Sum of atomic weights in a molecule	MW of CO2 = 12.01 + 2×16.00 = 44.01 amu
Mole	6.022 × 1023 particles	1 mol H2O = 6.022 × 1023 molecules
Percent Composition	Percent by mass of each element	%C in C6H12O6 = (6×12.01/180.16)×100% ≈ 40%
Theoretical Yield	Maximum possible product	Calculated from stoichiometry
Percent Yield	Actual yield as percent of theoretical	Percent yield = (actual/theoretical) × 100%

Additional info: These notes expand on the provided slides by including definitions, formulas, and examples for each key concept, as well as a summary table for quick reference.