Stoichiometry, Chemical Reactions, and Composition Analysis

Stoichiometry and Chemical Reactions

Introduction to Stoichiometry

Stoichiometry is the quantitative study of reactants and products in chemical reactions. It allows chemists to predict the amounts of substances consumed and produced in a given reaction.

• Mole Concept: The mole is a fundamental unit in chemistry representing $6.022 \times 10^{23}$ entities (Avogadro's number).

• Molar Mass (M): The mass of one mole of a substance, expressed in grams per mole (g/mol).

• Formula Mass: The sum of atomic masses in a chemical formula.

• Example: For $H_2O$, molar mass = $2 \times 1.01$ (H) $+ 16.00$ (O) $= 18.02$ g/mol.

Types of Chemical Reactions

Chemical reactions can be classified based on the changes occurring in reactants and products.

• Combination (Synthesis) Reaction: Two or more substances combine to form a single product. Example: $2CaO(s) + 2SO_2(g) \rightarrow 2CaSO_3(s)$

• Decomposition Reaction: A single compound breaks down into two or more simpler substances. Example: $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$

• Combustion Reaction: A substance reacts with oxygen, releasing energy as heat and light. Example: $C_xH_y + O_2 \rightarrow CO_2 + H_2O$ (Complete combustion)

• Incomplete Combustion: Occurs when oxygen is limited, producing CO and/or C (soot) in addition to $CO_2$ and $H_2O$.

Oxidation and Reduction

Oxidation and reduction (redox) reactions involve the transfer of electrons between substances.

• Oxidation: Loss of electrons by a substance.

• Reduction: Gain of electrons by a substance.

• Example: Oxidation of glucose: $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O$

Working with Moles and Mass

Mole Calculations

Mole calculations are essential for converting between mass, moles, and number of particles.

• Number of Moles ($n$): $n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

• Example: For 33.6 g of $NH_3$ ($M = 17.03$ g/mol): $n = \frac{33.6}{17.03} = 1.97$ mol

• Stoichiometric Ratios: Use coefficients from balanced equations to relate moles of reactants and products.

Percent Yield and Limiting Reactant

Percent yield measures the efficiency of a reaction, while the limiting reactant determines the maximum amount of product formed.

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

• Actual Yield: Amount of product actually obtained from the experiment.

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

• Limiting Reactant: The reactant that is completely consumed first, limiting the amount of product.

Composition Analysis of Compounds

Empirical and Molecular Formulas

Empirical formulas show the simplest whole-number ratio of atoms in a compound, while molecular formulas show the actual number of atoms.

• Empirical Formula: Simplest ratio of elements.

• Molecular Formula: Actual number of atoms in a molecule.

• Example: $C_2H_4O_2$ (molecular) has an empirical formula of $CH_2O$.

Combustion Analysis

Combustion analysis is used to determine the composition of organic compounds by burning a sample and measuring the products.

• Process: Burn sample in excess $O_2$, measure mass of $CO_2$ and $H_2O$ produced.

• Calculation: Use masses of $CO_2$ and $H_2O$ to determine moles of C and H in the original sample.

• Example: If 2.2 g sample produces 3.7 g $CO_2$ and 1.8 g $H_2O$, calculate moles of C and H.

Sample Calculations

• Mass Increase: Used to determine the amount of oxygen added during combustion.

• Example: $\text{Mass Increase} = \text{Final Mass} - \text{Initial Mass}$

• Percent Composition: $\text{Percent Composition} = \frac{\text{Mass of Element}}{\text{Total Mass}} \times 100\%$

Summary Table: Types of Chemical Reactions

Type of Reaction	General Form	Example	Key Features
Combination	$A + B \rightarrow AB$	$2CaO + 2SO_2 \rightarrow 2CaSO_3$	Two or more reactants form one product
Decomposition	$AB \rightarrow A + B$	$CaCO_3 \rightarrow CaO + CO_2$	One reactant breaks into two or more products
Combustion	$C_xH_y + O_2 \rightarrow CO_2 + H_2O$	$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O$	Reactant burns in oxygen, produces energy
Redox	Electron transfer	Oxidation of glucose	Involves oxidation and reduction

Key Equations

• $n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

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

• $\text{Percent Composition} = \frac{\text{Mass of Element}}{\text{Total Mass}} \times 100\%$

Additional info:

• Some notes referenced "cycle" calculations, which typically refer to iterative steps in empirical formula determination.

• Combustion analysis is a standard method for determining the empirical formula of organic compounds.