IndietroStoichiometry and Chemical Equations: Calculations with Chemical Formulas
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Stoichiometry: Calculations with Chemical Formulas and Equations
Chemical Equations
Chemical equations are symbolic representations of chemical reactions, showing the reactants and products involved. Each reactant and product is represented by its chemical formula, and coefficients indicate the relative number of molecules or moles participating in the reaction. A balanced chemical equation has equal numbers of atoms for each element on both sides, reflecting the law of conservation of mass.
Coefficients: Numbers placed in front of formulas to indicate the quantity of molecules or moles.
Subscripts: Numbers within formulas that indicate the number of atoms of each element in a molecule; changing subscripts alters the identity of the substance.
Balancing Equations: Adjust coefficients to ensure equal numbers of each atom on both sides; never change subscripts when balancing.
Example: The reaction of hydrogen and oxygen to form water:

Types of Chemical Reactions
Chemical reactions can be classified into several types based on the patterns of reactants and products:
Combination (Synthesis) Reactions: Two or more reactants combine to form a single product. General form:
Decomposition Reactions: A single compound breaks down into two or more products. General form:
Single Replacement (Displacement) Reactions: One element replaces another in a compound. General form:
Double Replacement (Metathesis) Reactions: Two compounds exchange ions to form new compounds. General form:
Combustion Reactions: Rapid reactions with oxygen producing a flame, typically forming and from hydrocarbons.
Formula Weights and Percentage Composition
Formula Weight (FW) and Molar Mass (MM)
The formula weight (FW) of a compound is the sum of the atomic weights of all atoms in its chemical formula, measured in atomic mass units (amu). The molar mass (MM) is the mass in grams of one mole of a substance and is numerically equal to the formula weight in amu.
Formula Weight Example:
Molar Mass Example:
Percentage Composition
The percentage composition of a compound is the mass percentage of each element in the compound. It is calculated as:
Example: For : %C = 42.1%, %H = 6.5%, %O = 51.4%

Avogadro’s Number and the Mole
Definition of the Mole
The mole is the fundamental counting unit in chemistry, representing particles (atoms, molecules, ions, etc.). This number is known as Avogadro’s number.
1 mole of atoms: atoms
1 mole of molecules: molecules
1 mole of ions: ions
Mole Relationships
The relationship between formula weight, molar mass, and the number of particles in one mole is summarized in the following table:
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 |

Limiting Reactants and Theoretical Yield
Limiting Reactant Concept
The limiting reactant is the substance that is completely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed. Any reactant left over is called the excess reactant.
Example: If you have 14 slices of bread and 6 slices of cheese, you can make 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 of product actually obtained from the reaction, which is usually less than the theoretical yield. Percent yield is calculated as:
Example: If the theoretical yield of is 1.39 g and the actual yield is 1.12 g, then percent yield =
Empirical and Molecular Formulas
Empirical Formula Calculation
The empirical formula is the simplest whole-number ratio of atoms in a compound. It can be determined from percent composition or mass data by converting masses to moles and finding the smallest whole-number ratio.
Example: For a compound with 2.06 g H, 32.71 g S, and 65.23 g O in 100 g, the empirical formula is .
Molecular Formula Calculation
The molecular formula is a whole-number multiple of the empirical formula. It is determined by dividing the molecular weight by the empirical formula weight:
Example: If the empirical formula is and the molecular weight is 180.12 g/mol, then the molecular formula is .
Stoichiometric Calculations
Using Balanced Equations for Quantitative Analysis
Balanced chemical equations provide the mole ratios needed to convert between amounts of reactants and products. Calculations can be performed using mass-mass, mass-mole, or mole-mole conversions, often using dimensional analysis.
Example: For , the mole ratio between and is 1:1.
Mass-mass calculation: Convert grams to moles, use the mole ratio, then convert moles to grams.
Unit Conversion Map
Unit conversions in stoichiometry often involve converting between grams, moles, molecules, atoms, and ions using Avogadro’s number and molar mass.
Key conversions: Grams → Moles: Moles → Particles: Particles → Moles:
Summary Table: Mole Relationships
Substance | Formula | Formula Weight (amu) | Molar Mass (g/mol) | 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 |
Additional info: These notes expand on the original content by providing definitions, formulas, and examples for each topic, ensuring completeness and academic quality for exam preparation.