Chapter 3 Overview

This chapter covers the fundamental concepts of chemical equations, patterns of chemical reactivity, formula weights, Avogadro’s number, the mole, empirical and molecular formulas, and stoichiometry. It also addresses limiting and excess reactants and reaction yields. Combustion analysis is specifically excluded from exam coverage.

Chemical Equations (Reactions)

Reactants, Products, and Balancing Reactions

• Chemical equations represent chemical reactions, showing the transformation of reactants into products.

• Each side of the equation must have the same number of each type of atom (law of conservation of mass).

• Physical states are indicated in parentheses: (s) for solid, (l) for liquid, (g) for gas, (aq) for aqueous solution.

• Example:

Patterns of Chemical Reactivity

Types of Chemical Reactions

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

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

  • (Additional info: Balanced for illustration)

• Combustion Reactions: A hydrocarbon reacts with oxygen to produce carbon dioxide and water. General form: Examples:

Avogadro’s Number and the Mole

Definition and Relationships

• The mole is the SI unit for amount of substance, defined as entities (Avogadro’s number).

• One mole of atoms has a mass of exactly 12.0000000 g.

• Mass of a single atom: g

• 1 amu (atomic mass unit) = g

• 1 g = amu

Molecular Weight vs. Molar Mass

• Molecular weight is the sum of atomic masses (in amu) of all atoms in a molecule.

• Molar mass is the mass (in g) of one mole of a substance; numerically equal to molecular weight in amu.

• Example: For (acetylene):

  • Molecular weight = amu

  • Molar mass = g/mol

Formula Weights

Formula Weight vs. Molecular Weight

• Formula weight refers to the sum of atomic masses in a formula unit (used for ionic compounds).

• Molecular weight is used for covalent (molecular) compounds.

• Example:

  • (molecular compound): molecular weight = 26.036 amu

  • (ionic compound): formula weight = amu

Calculating Atoms and Mass from Moles

• To find the number of atoms in a sample, multiply moles by Avogadro’s number and the number of atoms per molecule.

• Example: How many total atoms are present in a container with:

  • 0.10 mol He: atoms

  • 0.10 mol : H atoms

  • 0.10 mol : atoms (1 C + 4 H per molecule)

• Mass of 0.25 moles of :

Mass of Elements in Compounds

• To find the mass of a specific element in a compound, use the percent composition or the ratio of atoms.

• Example: Mass of C and H in , (g), and (l):

  • : 1 C (12.01 g/mol), 4 H (4.032 g/mol) per mole

  • : 2 C (24.02 g/mol), 2 H (2.016 g/mol) per mole

  • : 6 C (72.06 g/mol), 6 H (6.048 g/mol) per mole

Conversions: Grams, Moles, Particles, Atoms

• Conversions between mass, moles, and number of particles use molar mass and Avogadro’s number.

• General conversion steps:

  1. Grams Moles:

  2. Moles Particles:

  3. Particles Atoms: Use the number of atoms per molecule or formula unit

• Example: In 100.0 g of water ():

  • Moles of water: mol

  • Molecules of water: molecules

  • H atoms: atoms

  • Grams of H: g (Additional info: Calculated for illustration)

Sample Calculation: Atoms in a Compound

• Question: How many nitrogen atoms are in 4.72 g of magnesium nitrate, ?

• Solution steps:

  1. Find molar mass of : g/mol

  2. Moles: mol

  3. Moles of N: mol

  4. Atoms of N: atoms

Formulas from Analyses

Empirical and Molecular Formulas

• Empirical formula: The simplest whole-number ratio of atoms in a compound.

• Molecular formula: The actual number of atoms of each element in a molecule; a multiple of the empirical formula.

• Finding empirical formula:

  1. Convert mass percentages to grams (assume 100 g sample).

  2. Convert grams to moles for each element.

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

• Example: A compound is 84.4% C and 15.6% H. Empirical formula?

  • 84.4 g C: mol

  • 15.6 g H: mol

  • Ratio: (divide by 7.03)

  • Multiply to get whole numbers: ,

  • Empirical formula:

• Finding molecular formula:

  1. Find empirical formula mass.

  2. Divide molecular weight by empirical formula mass to get the multiple.

  3. Multiply subscripts in empirical formula by this multiple.

  Example: If empirical formula is (mass = 14 g/mol) and molecular weight is 28 g/mol, molecular formula is .

Empirical Formula from Percent Composition

• Example: Which formula is consistent with a compound found to be 36.6% Cr, 12.7% C, and the remainder O?

  Formula	% Cr	% C	% O
  Cr2C2O5	36.6	12.7	50.7
  Cr2(CO3)3	23.3	8.7	68.0
  Cr3(CO3)2	36.6	12.7	50.7
  Cr2(CO3)3	23.3	8.7	68.0

  Additional info: The correct answer matches the given percentages.

Stoichiometry

Quantifying Chemical Reactions

• Stoichiometry is the calculation of reactants and products in chemical reactions using balanced equations.

• Coefficients in balanced equations indicate the mole ratios of reactants and products.

• Example:

  • 1 molecule reacts with 3 molecules to form 2 molecules

  • 1 mol reacts with 3 mol to form 2 mol

Stoichiometric Calculations

• General steps:

  1. Convert given mass to moles (using molar mass).

  2. Use mole ratio from balanced equation to find moles of desired substance.

  3. Convert moles to grams if required.

• Example:

  • How much is produced (in grams) when 2.50 g of reacts?

  • How much is required to produce 25.0 g ?

Limiting Reactants and Percent Yield

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

• Excess reactant: The reactant that remains after the reaction is complete.

• Theoretical yield: The maximum amount of product that can be formed from given reactants.

• Actual yield: The amount of product actually obtained from a reaction.

• Percent yield:

• Example: If 10.0 g and 2.50 g react to form :

  • Identify the limiting reactant.

  • Calculate the amount of formed.

  • If 12.2 g is obtained, calculate the percent yield.

  • Determine how much excess reactant remains.

Sample Limiting Reactant Problem

• Example: 2.0 g reacts with an unknown amount of to produce 76.0 g (). How much remains and how much reacted if the reaction proceeds to 100% yield?

• Solution:

  • Calculate moles of and required for 76.0 g .

  • Determine which reactant is limiting and how much is left over.

Visualizing Limiting Reactants

• When two reactants are mixed in a reaction, the one that runs out first is the limiting reactant.

• After the reaction, some of the excess reactant may remain unreacted.

• Example: If 6 molecules of fluorine and 6 molecules of chlorine react to form chlorine trifluoride () with 100% yield, the number of molecules of each reactant remaining depends on the stoichiometry of the reaction.