Stoichiometry Notes - Chapter 3: Chemical Reactions and Reaction Stoichiometry

Stoichiometry: Core Concepts

Stoichiometry is the study of mass relationships in chemistry, based on the Law of Conservation of Mass (Antoine Lavoisier, 1789):
- "We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment." Thus, the total mass participating in a chemical change equals the total mass of products formed.
Purpose: relate quantities of reactants and products in chemical reactions via quantitative relationships.

Chemical Equations and Balancing

Chemical equations are concise representations of chemical reactions with:
- Reactants on the left, products on the right.
- States of matter shown in parentheses after each species (g) = gas, (l) = liquid, (s) = solid, (aq) = aqueous.
- Coefficients balance the equation to obey the law of conservation of mass; subscripts in formulas are not changed to balance.
Example (balanced):
- ext{CH}4(g) + 2 ext{O}2(g)
  ightarrow ext{CO}2(g) + 2 ext{H}2 ext{O}(g)
Balancing emphasizes using coefficients, not changing subscripts, to ensure atoms are conserved across reactants and products.
Important distinction: Changing subscripts can alter identity of substances (e.g., 2 H2 + O2 → 2 H2O does not equal H2 + O2 → H2O2); the latter would produce a different compound (hydrogen peroxide) and is not the way to balance standard equations.

Types of Reactions

Combination (synthesis) reactions: two or more substances react to form one product.
- Examples:
- 2\,\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\,\text{MgO}(s)
- \text{N}2(g) + 3\,\text{H}2(g) \rightarrow 2\,\text{NH}_3(g)
- \text{C}3\text{H}6(g) + \text{Br}2(l) \rightarrow \text{C}3\text{H}6\text{Br}2(l)
Decomposition reactions: a single substance breaks down into two or more substances.
- Examples:
- \text{CaCO}3(s) \rightarrow \text{CaO}(s) + \text{CO}2(g)
- 2\,\text{KClO}3(s) \rightarrow 2\,\text{KCl}(s) + \text{O}2(g)
- 2\,\text{NaN}3(s) \rightarrow 2\,\text{Na}(s) + 3\,\text{N}2(g)
Combustion reactions: rapid reactions typically producing flame, often involving oxygen.
- Examples:
- \text{CH}4(g) + 2\,\text{O}2(g) \rightarrow \text{CO}2(g) + 2\,\text{H}2\text{O}(g)
- \text{C}3\text{H}8(g) + 5\,\text{O}2(g) \rightarrow 3\,\text{CO}2(g) + 4\,\text{H}_2\text{O}(g)
- Notes: Combustion reactions are generally rapid, flame-producing, and often involve atmospheric oxygen as a reactant.

Formula Weights, Molecular Weights, and Ionic Compounds

Formula Weight (FW): sum of atomic weights in a formula; used for ionic compounds as well as covalent formulas.
- Example: FW(CaCl₂) = 1(40.08) + 2(35.453) = 110.99 amu
Molecular Weight (MW): sum of atomic weights for the atoms in a molecule; used for molecular compounds.
- Example: MW(C₂H₆) = 2(12.011) + 6(1.00794) ≈ 30.0696 + 6.04764 ≈ 36.1173 amu (often rounded to 36.12 amu)
Ionic compounds use empirical formulas and formula weights rather than molecular weights because they do not consist of discrete molecules.

Percent Composition

Formula to find the percentage by mass contributed by each element:
- \%\text{Element} = \frac{(\text{number of atoms})(\text{atomic weight})}{\text{FW of the compound}} \times 100
Example: For ethane, C₂H₆, FW ≈ 30.070 amu; %C = \frac{(2)(12.011)}{30.070} \times 100 ≈ 79.887\%
This shows how mass is distributed among elements in a compound.

Avogadro’s Number and Molar Mass

Avogadro’s number: 6.02 × 10^{23} particles per mole; 1 mole of anything contains NA particles.
1 mol of ¹²C has mass exactly 12.000 g.
Molar Mass: mass of 1 mole of a substance (g/mol). For elements, the molar mass equals the atomic weight (rounded). For diatomic elements, multiply by 2 if applicable.
The formula weight in amu (atomic mass units) equals the molar mass in g/mol for a given substance.

Using Moles as a Bridge

Moles provide a bridge from the molecular scale to macroscopic (real-world) quantities.
Concept: convert between grams, moles, and particles using molar mass, Avogadro’s number, and balancing coefficients.

Mole Relationships

1 mole of atoms, ions, or molecules contains NA particles: N_A = 6.02\times 10^{23}
1 mole of molecules or formula units contains NA particles times the number of atoms/ions per formula unit.

Determining Empirical Formulas

Goal: determine the simplest whole-number ratio of elements in a compound from percent composition.
Three steps:
1) Convert percent composition to moles of each element (using molar masses).
2) Divide all mole values by the smallest number of moles to obtain the simplest ratio.
3) If needed, multiply by the smallest common factor to obtain whole numbers.

Determining Empirical Formulas: Example (PABA)

Given composition (for para-aminobenzoic acid, PABA): C 61.31%, H 5.14%, N 10.21%, O 23.33%
Assume 100.00 g sample for convenience.
- C: 61.31 g × (1 mol / 12.01 g) ≈ 5.105 mol C
- H: 5.14 g × (1 mol / 1.008 g) ≈ 5.09 mol H
- N: 10.21 g × (1 mol / 14.01 g) ≈ 0.7288 mol N
- O: 23.33 g × (1 mol / 16.00 g) ≈ 1.456 mol O
Mole ratios by dividing by the smallest amount (0.7288):
- C: 5.105 / 0.7288 ≈ 7.005 ≈ 7
- H: 5.09 / 0.7288 ≈ 6.984 ≈ 7
- N: 0.7288 / 0.7288 = 1.000 ≈ 1
- O: 1.456 / 0.7288 ≈ 2.001 ≈ 2
Empirical formula from subscripts: ext{C}7\text{H}7\text{N}\text{O}_2

Determining a Molecular Formula

The molecular formula is a whole-number multiple of the empirical formula.
If the molar mass (M) of the compound is known, compute the multiple n = M / MW(empirical).
Then molecular formula = (empirical formula) × n.
Example: if empirical formula is CH and molar mass is 78 g/mol, then n = 78 / 13 = 6, and the molecular formula is ext{C}6\text{H}6.

Combustion Analysis (C, H, O determination)

Compounds containing C, H, and O can be analyzed by combustion:
- C is determined from CO₂ produced.
- H is determined from H₂O produced.
- O is determined by difference after C and H have been determined.
This yields the amounts of C, H, and O in the original sample.

Quantitative Relationships from Balanced Equations

The coefficients in a balanced equation specify the relative numbers of:
- molecules (or formula units) of reactants and products,
- or moles of reactants and products.
These can be converted to masses using molar masses and to volumes using gas molar volumes (e.g., at STP).
At STP, 1 mole occupies 22.4 L; at other conditions, gas volumes follow respective equations (not detailed here).

Stoichiometric Calculations: Overview

You can convert between grams and moles for reactants and products, but now you also use the MOLE RATIO from the balanced equation to relate different substances.
Core steps:
1) Convert known mass of a reactant to moles using its molar mass.
2) Use the balanced equation to convert moles of the known substance to moles of the desired substance via the mole ratio.
3) Convert moles of the desired substance to grams using its molar mass.
4) Optionally, convert to liters of a gas at STP using 22.4 L/mol.
5) If counting particles, use Avogadro’s number 6.02 × 10^23 particles/mol.
Example reaction: \text{C}6\text{H}{12}\text{O}6(s) + 6\,\text{O}2(g) \rightarrow 6\,\text{CO}2(g) + 6\,\text{H}2\text{O}(l)
Problem outline: How many grams of water can be produced from 1.00 g of glucose? First convert glucose to moles, then use the mole ratio to find moles of H₂O, then convert to grams.

Limiting Reactants and Yields

The limiting reactant is the reactant present in the smallest stoichiometric amount; it limits how much product can be formed.
- In a common example, H₂ may be the limiting reactant while O₂ is present in excess.
The limiting reactant determines the theoretical yield of the reaction.
Theoretical yield: maximum amount of product that could be formed from the given amounts of reactants, calculated via stoichiometry.
Actual yield: amount of product actually obtained from a reaction.
Percent yield: how much of the theoretical yield was actually obtained:
- \%\text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100
Note: In many real experiments, actual yield is less than theoretical yield due to side reactions, incomplete reactions, etc.

Quick Reference: Common Quantities and Conversions

Coefficients in balanced equations provide mole ratios between substances.
Masses are related to moles via molar masses; gases relate to moles via 22.4 L per mole at STP.
Particles relate to moles via Avogadro’s number: N_A = 6.02\times 10^{23} \text{ particles/mol}.
The mass of one mole of a substance (molar mass) is in g/mol; the formula weight in amu matches the molar mass in g/mol for a given substance.

Summary of Key Equations (LaTeX)

Balanced equation: aA + bB \rightarrow cC + dD
Atomic/molecular weights (example):
- FW(\text{CaCl}_2) = 40.08 + 2(35.453) = 110.99\ \text{amu}
- MW(\text{C}2\text{H}6) = 2(12.011) + 6(1.00794) \approx 36.12\ \text{amu}
Percent composition: \%\text{Element} = \frac{(\text{no. atoms})(\text{atomic weight})}{\text{FW}} \times 100
Avogadro’s number: N_A = 6.02 \times 10^{23} \text{ mol}^{-1}
Empirical formula from percent composition: (ratio of moles / smallest mole value) gives subscripts, e.g. \text{C}7\text{H}7\text{NO}_2 for PABA.
Molecular formula from empirical formula: ext{Molecular formula} = ( ext{empirical formula}) \times n, \quad n = \frac{\text{molar mass}}{\text{MW(empirical)}}
Combustion analysis (C, H, O): C from CO₂, H from H₂O, O by difference.
Stoichiometric use of mole ratios: \text{moles of product} = \text{(moles of reactant A)} \times \left( \frac{\text{coeff}{\text{product}}}{\text{coeff}{\text{reactant A}}} \right)
Theoretical vs Actual yields: \%\text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100