Stoichiometry and Chemical Reactions: Study Notes for General Chemistry

Stoichiometry: Calculations with Chemical Formulas and Equations

Balancing Chemical Equations

Balancing chemical equations ensures that the number of atoms for each element is the same on both sides of the equation, reflecting the law of conservation of mass.

• Diatomic Elements: Elements that naturally exist as molecules of two atoms: H2, O2, F2, Cl2, Br2, I2.

• Writing Chemical Formulas: Use the charge criss-cross method to write formulas for ionic compounds (e.g., magnesium nitride).

• Balancing Steps:

  • Adjust coefficients (numbers in front of formulas) to balance atoms.

  • Never change subscripts in chemical formulas to balance equations.

• Reference: See Section 3.1 in your textbook for more examples.

Simple Patterns of Chemical Reactivity

Chemical reactions can be classified into several types based on their patterns. The three most common types are:

• Combination (Synthesis) Reactions: Two or more substances combine to form a single product.

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

• Decomposition Reactions: A single compound breaks down into two or more simpler substances.

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

• Combustion Reactions: A substance (often containing C and H) reacts with O2 to produce CO2 and H2O, usually with the release of heat and light (flame).

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

Formula Weights and Percentage Composition

The formula weight (or molecular weight for covalent compounds) is the sum of the atomic weights of all atoms in a chemical formula.

• Formula Weight Example: For $Ca(NO_2)_3$, the formula weight is 164.1 amu.

• Smallest Units:

  • Covalent compounds: molecules

  • Ionic compounds: formula units

• Percentage Composition: The percent by mass of each element in a compound.

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

Avogadro’s Number and the Mole

The mole is a fundamental unit in chemistry representing 6.02 × 1023 particles (Avogadro’s number). It allows chemists to count atoms, molecules, or ions by weighing them.

• Definition: One mole of a substance has a mass in grams equal to its atomic or formula weight in amu.

• Avogadro’s Number: $6.02 \times 10^{23}$ particles/mole.

• Molar Mass: The mass of one mole of a substance (g/mol), numerically equal to the formula weight in amu.

• Conversions: Grams ↔ Moles ↔ Particles (cannot convert directly between grams and particles without going through moles).

Conversion Relationships

Given	Conversion Factor	Find
Mass (g)	Molar Mass (g/mol)	Moles
Moles	Avogadro’s Number ($6.02 \times 10^{23}$)	Particles

Additional info: Always use the correct molar mass for the substance in question; do not mix up substances when converting.

Empirical Formula and Its Determination

The empirical formula shows the simplest whole-number ratio of elements in a compound. It can be determined from percent composition or combustion analysis data.

• From Percent Composition:

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

  2. Convert grams of each element to moles: $\text{moles} = \frac{\text{grams}}{\text{atomic mass}}$

  3. Divide all mole values by the smallest number of moles to get the simplest ratio (subscripts).

• From Combustion Data:

  1. Find moles of C from moles of CO2 produced.

  2. Find moles of H from moles of H2O produced (multiply by 2 for H atoms).

  3. Find mass of O by subtracting masses of C and H from total mass of compound.

Example: If a compound contains 40% C, 6.7% H, and 53.3% O, the empirical formula is CH2O.

Quantitative Information from Balanced Equations

Balanced chemical equations provide the mole ratios of reactants and products, which are essential for stoichiometric calculations.

• Coefficients: Indicate the number of moles (or molecules) of each substance involved.

• Mass Conservation: Total mass of reactants equals total mass of products, but the number of moles may differ.

• Stoichiometric Calculations: Use the coefficients to relate moles of one substance to moles of another.

Stoichiometry Calculation Steps

Step	Description
1. Mass of given (g)	Convert to moles using molar mass
2. Moles of given	Use mole ratio from balanced equation
3. Moles of unknown	Convert to grams using molar mass

Important: Never use the molar mass of one substance with the grams of another.

Limiting Reactants and Theoretical Yield

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

• Limiting Reactant: The reactant that runs out first and determines the maximum amount of product.

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

• Procedure:

  1. Calculate the moles of one reactant needed to react with the other.

  2. Compare with the amount available to identify the limiting reactant.

• Theoretical Yield: The maximum amount of product that can be formed from the limiting reactant, calculated using stoichiometry.

• Actual Yield: The amount of product actually obtained from the reaction (usually less than theoretical yield).

• Percent Yield:

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

Yield Calculation Table

Term	Definition
Theoretical Yield	Predicted amount of product from limiting reactant
Actual Yield	Amount of product obtained in the lab
Percent Yield	$(\text{Actual Yield} / \text{Theoretical Yield}) \times 100\%$

Example: If the theoretical yield is 10.0 g and the actual yield is 8.5 g, then $\%\ \text{Yield} = \frac{8.5}{10.0} \times 100\% = 85\%$.