Chemical Equations and Stoichiometry

Chemical Equations

Overview of Chemical Equations

• Definition of Chemical Equations: A method to represent a chemical reaction.

• Definition of Chemical Reactions: Chemical changes when substances react to create new substances.

• Example: The reaction of hydrogen and oxygen to form water:

  • $2 ext{H}<em>2(g) + ext{O}</em>2(g) <br>ightarrow 2 ext{H}_2 ext{O}(l)$

  • Reactants: Material before the reaction (e.g.,
    \text{H}2, ext{O}2)

  • Products: Result of the reaction (e.g., ext{H}_2 ext{O})

  • States of Matter: (g) for gas, (l) for liquid, (s) for solid, (aq) for aqueous.

Components of Chemical Equations

• Coefficients: Whole numbers indicating the number of molecules (e.g., in $2 ext{H}_2$, the coefficient indicates 2 molecules of hydrogen).

• Subscripts: Numbers written after element symbols indicating the number of atoms in a molecule (e.g., in $ext{H}_2 ext{O}$, the subscript 2 indicates there are 2 hydrogen atoms).

Law of Conservation of Mass

• Principle: Mass is conserved in a chemical reaction; atoms are merely rearranged.

Balancing Chemical Equations

Balancing Process

• Equal numbers of each type of atom are required on both sides of the equation, as per the Law of Conservation of Mass.

• Changes during balancing:

  • Changing COEFFICIENTS: to balance the equation.

  • NEVER CHANGE SUBSCRIPTS! This would alter the substance represented.

Steps to Balance Equations

1. Write the unbalanced chemical equation.

2. Identify the reactants and products.

3. Count the starting number of atoms on each side.

4. Add coefficients to balance each type of atom.

   • Balancing strategy:

     • Start with elements found in only one reactant and product.

     • Prioritize balance for elements with an unequal number of atoms on either side.

Examples of Balancing Chemical Equations

1. For the unbalanced reaction $ext{As} + ext{NaOH} <br>ightarrow ext{Na}<em>3 ext{AsO}</em>3 + ext{H}_2$

2. Write a balanced equation for the reaction of silver (I) iodide with sodium sulfide, resulting in silver (I) sulfide and sodium iodide.

Example Problem

When lead (II) nitrate reacts with sodium iodide to yield sodium nitrate and lead (II) iodide, determine the coefficient for sodium iodide:

• A. 1

• B. 2

• C. 3

• D. 4

More Balancing Examples

1. $ext{Pb(NO}<em>3)</em>2 + ext{NaI} <br>ightarrow ext{PbI}<em>2 + ext{NaNO}</em>3$

2. $ext{CO} + ext{H}<em>2 ightarrow ext{C}</em>8 ext{H}<em>{18} + ext{H}</em>2 ext{O}$

Chemical Reactivity

Types of Chemical Reactions

1. Combination (Synthesis): Two or more reactants combine to form one product.

   • Example: $2 ext{Mg} + ext{O}_2 <br>ightarrow 2 ext{MgO}$

   • Often results in an ionic compound if a metal and nonmetal react.

2. Decomposition: One substance breaks down into two or more products.

   • Example: $ext{PbCO}<em>3 ightarrow ext{PbO} + ext{CO}</em>2$

   • Example: $2 ext{NaN}<em>3 ightarrow 2 ext{Na} + 3 ext{N}</em>2$

3. Combustion: A substance burns in oxygen, usually generating heat.

   • Example: Combustion of hydrocarbons:

     • Reactants: Hydrocarbon (CxHy or CxHyOz) + $O_2$

     • Products: $CO<em>2$ + $H</em>2O$

   • Write a balanced equation for the combustion of hexane ($C<em>6H</em>{14}$).

Multiple Choice Example for Decomposition Reaction

Which of the following represents a decomposition reaction?

• A. $2 ext{H}<em>2 ext{O} ightarrow 2 ext{H}</em>2 + ext{O}_2$

• B. $ext{H}<em>2 ext{CO}</em>3 <br>ightarrow ext{H}<em>2 ext{O} + ext{CO}</em>2$

• C. $2 ext{Na} + ext{Cl}_2 <br>ightarrow 2 ext{NaCl}$

• D. Both A and B are correct.

Atomic and Formula Masses (Weights)

Atomic Mass Unit (amu)

• Definition of amu: Atomic mass unit is used to measure the mass of an atom or compound.

  • 1 atom of $^{12} ext{C} = 12 ext{ amu}$

  • 1/12 of a $^{12} ext{C}$ atom = 1 amu = $1.66054 imes 10^{-24} ext{ g}$

Mass of Subatomic Particles

• Mass (grams) and Mass (amu):

  • Proton: Mass = $1.673 imes 10^{-24} ext{ g}$, Mass (amu) = $1.0073$

  • Neutron: Mass = $1.675 imes 10^{-24} ext{ g}$, Mass (amu) = $1.0087$

  • Electron: Mass = $9.11 imes 10^{-28} ext{ g}$, Mass (amu) = $5.49 imes 10^{-4}$

Average Atomic Mass

• Definition: A weighted average of all naturally occurring isotopes of an element.

• Weighted Average Explanation: Takes into account the relative amounts of each isotope.

Periodic Table Information

• Elements are listed with atomic weights (mean relative mass) which are generally IUPAC 1997 values noted to 5 significant figures.

• Elements with no stable isotopes have atomic weights listed in square brackets, indicating representation by the longest-lived isotope.

Example Calculation

• Given isotopes of silicon:

  • $^{28} ext{Si}$ – 27.97693 amu (92.23%)

  • $^{29} ext{Si}$ – 28.97649 amu (4.68%)

  • $^{30} ext{Si}$ – 29.97377 amu (3.09%)

Difference in Terms

1. Atoms: Have atomic masses (weights).

2. Compounds: Have formula masses (weights).

   • To find a compound's formula mass, sum the atomic masses of each atom in the compound.

3. Molecular Mass: Term used interchangeably with formula mass for molecular compounds.

   • Use of the term “formula mass” is preferred for ionic compounds due to their non-molecular nature.

Example Calculation of Molecular and Formula Masses

• Calculate molecular mass for water (H$_2$O).

• Formula mass for calcium phosphate $ext{Ca}<em>3( ext{PO}</em>4)_2.$

Percent Composition

Definition of Percent Composition

• Percent Composition: The percentage by mass of each element in a compound.

  • Formula:
    $ext{Percent of Element} = rac{ ext{(mass of element in formula)} imes ext{(atomic mass of element)}}{ ext{formula mass of compound}} imes 100$

Example Calculation

• Determine the percentage of each element in acetone (C$3$H$6$O):

  • Total percentage = rac{ ext{mass of C}}{ ext{mass of C}3 ext{H}6 ext{O}} imes 100

Avogadro’s Number and the Mole

Definition of a Mole

• Mole: The SI unit of measurement for the amount of substance.

  • Defined as the number of atoms in 12 grams of carbon-12 (12C).

  • Avogadro's Number: $6.022 imes 10^{23}$ particles (atoms, molecules, etc.) are equal to 1 mole.

Understanding the Size of a Mole

• 1 mole of marbles covers the Earth in a layer about 3 miles thick.

• 1 mole of pennies distributed among the world’s population gives approximately $8 trillion to each person.

• 1 mole of water molecules is equivalent to about 18 mL.

Relationships in Moles

• The greater the mass of a particle, the greater the mass of one mole of that substance.

  • Example distinction:

  • 12 pennies vs. 12 bowling balls.

  • 1 mol of pennies vs. 1 mol of bowling balls.

Molar Mass

• Definition of Molar Mass: The mass in grams of one mole of a substance.

• The mass of an atom/molecule expressed in amu corresponds to its molar mass in grams/mole (i.e., 1 amu = 1 g/mol).

• Molar mass from the periodic table can be interpreted as:

  1. Atomic mass in amu.

  2. Molar mass in grams/mole.

Example Calculations

1. Calculate the number of molecules in 2.47 moles of ammonia (NH$_3$).

2. Determine the conversion factors necessary to solve for moles of sodium from a given number of atoms.

Stoichiometry

Definition of Stoichiometry

• Stoichiometry: The study of the relative amounts of elements or compounds in a reaction.

• It employs mole ratios obtained from balanced chemical equations to relate quantities of reactants/products.

Example Mole Ratios

• Derived from the balanced equation: $2 ext{Mg} + ext{O}_2 <br>ightarrow 2 ext{MgO}$

• Coefficients provide mole ratios that serve as conversion factors between different reactants and products.

Example Calculations in Stoichiometry

1. In the production of ammonia via the Haber-Bosch process, if 21.4 moles of NH$3$ are produced, calculate the moles of N$2$ required.

2. If complete combustion of glucose (C$6H{12}O6$) occurs, find the grams of CO$2$ exhaled after eating candy containing 30.5g of glucose.

3. Reaction between lithium hydroxide and carbon dioxide. Calculate how many grams of carbon dioxide can be absorbed by 1.00g of LiOH.

Limiting Reactants

Concept of Limiting Reactants

• Limiting Reactant: The reactant that runs out first and limits product formation.

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

• Analogy: Like making s’mores, if marshmallows run out, production ceases despite excess chocolate and graham crackers.

Example Problems with Limiting Reactants

1. For the reaction $2 ext{Al} + 3 ext{I}<em>2 ightarrow 2 ext{AlI}</em>3$, if beginning with 1.20 mol of Al and 2.40 mol of I$2$, determine the limiting reactant and max amount of AlI$3$ formed.

2. For $2 ext{Al} + 6 ext{HBr} <br>ightarrow 2 ext{AlBr}<em>3 + 3 ext{H}</em>2$, begin with 3.22 mol Al and 4.96 mol HBr, identify limiting and excess reactants and calculate leftover moles.

Theoretical Yield and Percent Yield

Definitions

• Theoretical Yield: The maximum amount of product obtainable from the limiting reactant in a chemical reaction.

• Actual Yield: The quantity of product that is actually produced in a reaction.

• Percent Yield: A comparison of actual yield to theoretical yield, defined as:
  $ext{Percent Yield} = rac{ ext{theoretical yield}}{ ext{actual yield}} imes 100$

Example Calculations

1. For the decomposition reaction of potassium chlorate generating oxygen gas, if you start with 20.0 grams of KClO$3$ and produce 6.85 grams of O$2$, determine the percent yield.

2. For the reaction of chlorinated methane and chlorine gas, if 25.0g of CHCl$3$ and 25.0g of Cl$2$ yield 5.20g of HCl, calculate the percent yield of HCl produced in the reaction.