Tomlin CFE Chemistry Review 2021 revised

Exam Information

• Test Dates: TBD
• Test Format:
  • Computer-Based Test
  • 50 Multiple-Choice Questions
  • 80 Minutes
  • Periodic Table Provided
• Reminders:
  • Students MUST use a school computer
  • Charge your computer the night before and throughout the day.
  • Bring a SCIENTIFIC CALCULATOR!
    • Graphing or Programmable calculators NOT ALLOWED!
  • If you are absent and/or know you will be absent on the exam date, please coordinate with your teacher to make up the exam.

Matter and Change: Physical and Chemical Properties

• Physical Property: characteristic of matter that can be observed or measured without changing the sample’s composition
  • Relates to the 5 senses
• Chemical Property: ability or inability of a substance to combine with or change into one or more substances.
  • Ability to react

Matter and Change - Physical and Chemical Changes

• Physical Changes: a change that alters a substance without changing its composition
  • Relates to the 5 senses
• Chemical Changes: one or more substances changing into new substances
  • Commonly known as a chemical reaction.
• Examples of Physical and Chemical Changes

Matter and Change - States of Matter

States of Matter - Phase Changes

Matter and Change - Mixtures

• Mixture: is a combination of two or more pure substances in which each substance retains its chemical properties.
  • Homogeneous Mixture: mixture that has a uniform composition throughout. (Also known as solution)
    • Coffee
    • Saltwater
    • Air
  • Heterogeneous Mixture: mixture that is NOT uniform in composition.
    • House Salad
    • Trail mix
    • Oil and Vinegar
    • Oil and Water

Matter and Change - Separation of Mixtures

• Note: Substances can also be separated by physical means such as using a magnet to separate nails from sand.

Atomic Structure - Subatomic Particles

Atomic Structure - Atomic Number and Mass Number

• Atomic Number: number of protons in an atom.
  • Equal to the number of electrons if the atom is NEUTRAL
  • Determines the identity of an element.
• Mass Number: # of protons + # of neutrons
• Chemical Notation Symbols
• Element Name - Mass Number
  • Example: Copper-63
• Mass #

Atomic Structure - Isotopes

• Isotopes: atoms of the same element, with different numbers of neutrons.
  • Will also differ in atomic mass (and mass number)
  • Have the same number of protons

Calculating Average Weighted Mass of an Element:

• Calculate the average atomic mass of iron if its abundance in nature is 15% iron-55 and 85% iron-56.
• Step 1: Calculate the contribution of each isotope:
  • Iron-55: $(55 \text{ amu}) \times (0.15) = 8.25 \text{ amu}$
  • Iron-56: $(56 \text{ amu}) \times (0.85) = 47.6 \text{ amu}$
• Step 2: Add the contribution from each isotope together:
  • $8.25 \text{ amu} + 47.6 \text{ amu} = 55.85 \text{ amu}$

Electrons in Atoms - Electron Configurations

• Order of Filling Orbitals
• Example: Electron Configuration of Aluminum
• Example: Electron Configuration of Calcium

Principles

• Aufbau Principle:
  • Electrons will occupy the lowest energy level orbital that will receive it.
• Pauli Exclusion Principle:
  • No two electrons within an atom may have an identical set of all four quantum numbers.
• Hund's Rule:
  • Electrons will occupy all empty orbitals in a subshell with single electrons having parallel spins before entering half-filled orbitals.

Orbital diagrams:

• Arrows represent electrons.
• Boron 2p¹
• Carbon 2p²
• Nitrogen 2p³
• Oxygen 2p4

Electrons in Atoms - Energy, Frequency, Wavelength

• Relationship(s) between Energy, Frequency, and Wavelength
  • Wavelength and Frequency are INVERSELY related
  • Energy and Frequency are DIRECTLY related
  • Wavelength and Energy are INVERSELY related
• CONSTANTS:
  • $C = 3.00 \times 10^8 \text{ m/s}$
  • Planck’s constant $\text{= }6.626 \times 10^{-34} \text{ Js}$
• Units: wavelength (m), Energy (J), frequency (s^-1)

Electrons in Atoms - Electromagnetic Spectrum

• Electromagnetic Spectrum
  *Finding The Energy of a Photon
  *Some useful rearrangement triangles
  *The Relationship between light and energy
  *Converting frequency to wavelength

Electrons in Atoms - Absorption/Emission of Light

• The excitation of an atom by EM radiation and the emission of EM radiation from the 'excited' atom!
  1. an electron in an outer shell absorbs a photon of EM energy
  2. electron promoted to a higher shell
  3. promoted electron loses energy, falls to lower energy level by losing (emitting) a photon of EM radiation

Periodic Table - Periodic Trends

• Periodic Trends
  *Atomic Size Increases
  *Electronegativity Increases
  *Ionization Energy Increases

Periodic Table - Groups/Families

• Groups or Families of similar elements are arranged in vertical columns on the Periodic Table
• You need to know the four families indicated below as well as the larger group of transition metals
• Alkali Metals
• Alkaline Earth Metals
• Halogens
  *Noble Gases

Periodic Table - Valence Electrons

• Valence electrons: electrons in the outermost energy level
  • Determined using group numbers 1A-8A
  • Elements in the same family (group) have similar reactivity due to the same # of valence electrons
• Lewis Dot Structures: display the number of valence electrons in an atom.
  • See examples on the periodic table.

Ionic Bonding - Bonding Theory

• Ionic Bonding: bonding between a metal and a non-metal.
  • Valence electrons are transferred from the metal (cation) to the non-metal (anion).
  • Ionic Bonds are held together by the electrostatic attraction between cation and anion

Ionic Bonding - Naming

• Steps to Naming Ionic Compounds
  • Name the metal (cation) as is from the periodic table.
    • If the metal is a transition metal, specify the charge of the metal in parenthesis and roman numerals
  • Change the ending of the nonmetal (anion) to “-ide”
  • NO NUMERICAL PREFIXES WHEN NAMING IONIC COMPOUNDS!!!
• Practice: Naming Ionic Compounds and Writing Formulas

Covalent Bonding - Bonding Theory

• Covalent Bonding: bonding between a non-metal and a non-metal.
  • Valence electrons are SHARED between elements

Covalent Bonding - Naming

• Steps to Naming Covalent Molecules
  • Name the least electronegative element first as is.
    • If the first element has more than one atom, use a numerical prefix.
  • Name the second element and change the ending to “-ide”
    • Second element ALWAYS uses a numerical prefix.

Molecular Geometry Chart

Summary of Intermolecular Forces (from strongest to weakest)

*Comparison of Bonding and Attractive Forces

Chemical Reactions - Types of Chemical Reactions

• Types of Chemical Reactions
  1. Neutralization: Acid$(H)$ + Base$(OH)$ salt + $(H(OH))$
     • $HCl + KOH \rightarrow KCl + H_2O$
  2. Combustion: $AB$+ oxygen → $CO<em>2 + H</em>2O$
     • $CH<em>4 + 2O</em>2 \rightarrow CO<em>2 + 2H</em>2O$
  3. Synthesis: $A + B \rightarrow AB$
     • $Zn(s) + I<em>2(s) \rightarrow ZnI</em>2(s)$
  4. Decomposition: $AB \rightarrow A + B$
     • $H<em>2SO</em>3(aq) \rightarrow H<em>2O (l) + SO</em>2(g)$
  5. Single displacement: $A + BC \rightarrow AC + B$
     • $Zn + CuCl<em>2 \rightarrow ZnCl</em>2 + Cu$
  6. Double displacement: $AB + CD \rightarrow AD + CB$
     • $KBr + AgNO<em>3 \Rightarrow KNO</em>3 + AgBr$

Chemistry - Single Replacement Reactions:

• Single Replacement Reactions: use the activity series
• Metals will replace Metals and Halogens will replace Halogens.
• The higher up on the activity series, the more reactive or “attractive” the element.

Chemical Reactions - Balancing Reactions

• Steps in Balancing
  1. Write the formula of the substances.
  2. Count the number of atoms in each element on each side of the equation.
  3. Adjust the coefficient.
  4. Check if the number of atoms each element is the same on both sides.
• Mini-Lesson: Balancing Reaction
• Practice: Balancing Chemical Reactions

The Mole - Calculations

• Relationships:
  • Particle Town $\leftrightarrow$ Mole $\leftrightarrow$ Gramville Station
  • * by molar mass $\leftrightarrow$ ÷ by molar mass
  • Avogadro’s Number = $6.02 x 10^{23}$
  • * by Avogadro’s Number $\leftrightarrow$ ÷ by Avogadro’s Number

The Mole - Percent Composition

• Percent Composition by Mass
  • $\% \text{ by mass} = \frac{\text{mass of element}}{\text{mass of compound}} \times 100$
• Steps to Solve for Percent Composition (with example)
  • $PCl_5$ (Phosphorus Pentachloride)
    1. Find the molar mass of all elements in the compound:
       • $P = 30.974 \text{g}$
       • $Cl = 5(35.453 \text{g}) = 177.265 \text{g}$
    2. Find the molecular mass:
       • $PCl_5 = 30.974 \text{g} + 177.265 \text{g} = 208.239 \text{g}$
    3. Divide each molar mass by the molecular mass and multiply by 100:
       • $P = \frac{30.974 \text{g}}{208.239 \text{g}} \times 100 = 14.87\%$
       • $Cl = \frac{177.265 \text{g}}{208.239 \text{g}} \times 100 = 85.13\%$
• Therefore, Phosphorus Pentachloride is 14.87% P and 85.13% Cl by mass.

The Mole - Empirical Formula

• Empirical Formula: a chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms.

Stoichiometry - Mole to Mole

• All stoichiometry problems MUST have a balanced chemical reaction.
• Mole - to- Mole calculation will be in every stoichiometry calculation
• Most important calculation because it allows us to change substances due to their mole ratios

Stoichiometry - Mole to Mass

• If a reaction starts with 10.0 moles of propane, how many grams of carbon dioxide will it yield?
  • $C<em>3H</em>8 + 5O<em>2 \longrightarrow 3CO</em>2 + 4H_2O$
  • $\text{? g } CO<em>2 = 10 \text{ mol } C</em>3H<em>8 \times \frac{3.0 \text{ mol } CO</em>2}{1 \text{ mol } C<em>3H</em>8} \times \frac{44 \text{ g } CO<em>2}{1 \text{ mol } CO</em>2} = 1320 \text{ g } CO_2$
• Switch Step #'s come from Equation
• $2 H<em>2 + O</em>2 \rightarrow 2 H_2O$
  • If I have 8 moles of $O<em>2$ and excess $H</em>2$, how many grams of $H_2O$ can I make?
  • $\frac{8 \text{ mol } O<em>2}{1} \times \frac{2 \text{ mol } H</em>2O}{1 \text{ mol } O<em>2} \times \frac{18.016 \text{ g } H</em>2O}{1 \text{ mol } H<em>2O} = 288.256 \text{ g } H</em>2O$
  • *Switch Step!

Stoichiometry - Mass to Mole

• Mass-Mole Practice
  • $3 CuSO<em>4 + 2 Al \rightarrow Al</em>2(SO<em>4)</em>3 + 3 Cu$
  • How many moles of $CuSO_4$ are required to react with 13.5 g of Al?
    • $\frac{13.5 \text{ g } Al}{1} \times \frac{1 \text{ mol } Al}{26.98 \text{ g } Al} \times \frac{3 \text{ mol } CuSO<em>4}{2 \text{ mol } Al} = 0.751 \text{ mol } CuSO</em>4$
  • How many moles of $Al<em>2(SO</em>4)_3$ are produced when 13.5g of Al react?
    • $\frac{13.5 \text{ g } Al}{1} \times \frac{1 \text{ mol } Al}{26.98 \text{ g } Al} \times \frac{1 \text{ mol } Al<em>2(SO</em>4)<em>3}{2 \text{ mol } Al} = 0.250 \text{ mol } Al</em>2(SO<em>4)</em>3$

Stoichiometry - Mass to Mass

• Consider the reaction of photosynthesis:
  • $6CO<em>2 + 6H</em>2O \rightarrow C<em>6H</em>{12}O<em>6 + 6O</em>2$
• molar mass of $C<em>6H</em>{12}O_6 = 180 \text{ g/mol}$
• How many grams of water would be needed to produce 500 grams of sugar?
  • $\frac{500 \text{ g } C<em>6H</em>{12}O<em>6}{1} \times \frac{1 \text{ mol } C</em>6H<em>{12}O</em>6}{180 \text{ g } C<em>6H</em>{12}O<em>6} \times \frac{6 \text{ mol } H</em>2O}{1 \text{ mol } C<em>6H</em>{12}O<em>6} \times \frac{18.016 \text{ g } H</em>2O}{1 \text{ mol } H<em>2O} = 300.27 \text{ g } H</em>2O$
• $N<em>2(g) + 3 H</em>2(g) \rightarrow 2 NH_3 (g)$
  • How many grams of nitrogen are needed to produce the 38.5 g of $NH_3$ produced in the previous example?
  • $\frac{38.5 \text{ g } NH<em>3}{1} \times \frac{1 \text{ mol } NH</em>3}{17.03 \text{ g } NH<em>3} \times \frac{1 \text{ mol } N</em>2}{2 \text{ mol } NH<em>3} \times \frac{28.0 \text{ g } N</em>2}{1 \text{ mol } N<em>2} = 31.7 \text{ g } N</em>2$

Stoichiometry - Limiting and Excess Reactants

• Limiting Reactants
  • To find the limiting reactant, you must do a separate stoichiometry problem for each reactant. Go from amount of reactant amount of product. The reactant that makes the least amount of product will be your limiting reactant.
  • Used up in a reaction
  • Determines the amount of product
• Excess Reactant
  • Added to ensure that the other reactant is completely used up
  • Cheaper and easier to recycle
• $2 H<em>2 + O</em>2 \rightarrow 2 H_2O$
  • What is your limiting and excess reactants if you have 5.43 grams of hydrogen and 17.03 grams of oxygen?
  • $\frac{5.43 \text{ g } H<em>2}{1} \times \frac{1 \text{ mol } H</em>2}{2.02 \text{ g } H<em>2} \times \frac{2 \text{ mol } H</em>2O}{1 \text{ mol } H<em>2} \times \frac{18.02 \text{ g } H</em>2O}{1 \text{ mol } H<em>2O} = 48.44 \text{ g } H</em>2O$
  • $\frac{17.03 \text{ g } O<em>2}{1} \times \frac{1 \text{ mol } O</em>2}{32.00 \text{ g } O<em>2} \times \frac{2 \text{ mol } H</em>2O}{1 \text{ mol } O<em>2} \times \frac{18.02 \text{ g } H</em>2O}{1 \text{ mol } H<em>2O} = 19.18 \text{ g } H</em>2O$

Stoichiometry - Percent Yield

• Formula for Percent Yield
  • $\text{Percent yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$
• Example 1: In a particular experiment 10.0 g of sugar should be produced but only 0.664 g is produced. What is the percentage yield?
  • $\% \text{ yield} = \frac{0.664 \text{ g}}{10.0 \text{ g}} \times 100 = 6.64 \%$
• Problem 2 continued
  • If the actual yield from the above reaction was 81 grams, what is the percent yield?
  • Actual yield = 81 grams
  • Theoretical yield = 93.95 g (from last slide)
  • $\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$
  • $\text{Percent yield} = \frac{81}{93.95} \times 100$
  • Percent yield = 86.22%

Solutions - Molarity

*Molarity!
*$\text{Molarity} = \frac{\text{Number of Moles}}{\text{Liters of Solution}}$
*$\text{M} = \frac{n}{V} = \frac{\text{Number of Moles}}{\text{Volume}}$
*What is the molarity of a solution containing 0.32 moles of NaCl in 3.4 liters?
*$\text{molarity} = \frac{0.32 \text{ moles } NaCl}{3.4 \text{ L}} = 0.094 \text{ M } NaCl$
*Practice: Calculate the molarity for a solution where you dissolve 50.0 g of calcium nitrate in enough water to make 2.00 L of solution.
*$\text{M} = \frac{\text{moles solute}}{\text{L solution}} = \frac{0.3047 \text{ mol } Ca(NO<em>3)</em>2}{2.00 \text{ L solution}} = 0.152 \text{ M } Ca(NO<em>3)</em>2$
*$50.0 \text{ g } Ca(NO<em>3)</em>2 \times \frac{1 \text{ mol } Ca(NO<em>3)</em>2}{164.1 \text{ g } Ca(NO<em>3)</em>2} = 0.3047 \text{ mol } Ca(NO<em>3)</em>2$

Acids/Base - Arrhenius Acid/Base

*ARRHENIUS ACIDS & BASES
*ARRHENIUS ACID
*An Arrhenius acid is any substance that provides hydrogen ions, H+, when dissolved in water.
*ARRHENIUS BASE
*An Arrhenius base is any substance that provides hydroxide ions, OH, when dissolved in water.
*EXAMPLES OF AN ARRHENIUS ACID AND BASE
*$HNO<em>3$ is an acid: $HNO</em>3(aq) \longrightarrow H^+ (aq) + NO<em>3^- (aq)$ *$KOH$ is a base: $KOH(aq) \longrightarrow K^+ (aq) + OH^- (aq)$ *Arrhenius Acids: *$HCI \longrightarrow H^+ + CI$ *$HNO</em>3 \longrightarrow H^+ + NO<em>3^-$ *Arrhenius Bases: *$NaOH \longrightarrow Na^+ + OH^-$ *$Ca(OH)</em>2 \longrightarrow Ca^{2+} + 2OH^-$

Acids/Base - Bronsted-Lowry Acid/Base

*Bronsted-Lowry Acid-Base Definition
*An acid is a proton donor, any species that donates an H+ ion.
*A base is a proton acceptor, any species that accepts an H+ ion.
*Conjugate Acid-Base Pair
*An acid-base reaction is a proton transfer process.
*Acids donate a proton to water
*Bases accept a proton from water
*Conjugate Acid-Base Pair
*$HCI$ donates a proton to water:
*$HCI + H<em>2O \longrightarrow H</em>3O^+ + CI$
*$NH<em>3$ accepts a proton from water: *$NH</em>3 + H<em>2O \longrightarrow NH</em>4^+ + OH^-$

Acids/Bases - pH Scale

*Explain and use the pH scale
*The pH Scale
*pH is a measure of how acidic or basic a solution is, which means it is a measure of the concentration of hydrogen ions [H+] in a solution.
*The pH scale ranges from 0 to 14.
*Acidic solutions have pH values below 7
*A solution with a pH of 7 is neutral (pure water)
*Basic/alkali solutions have pH values above 7.

Acids/Bases - Calculating pH and pOH

*pH & pOH Calculations
*$pH = -log [H<em>3O^+]$ *$pOH = -log[OH^-]$ *[H3O^+] = 10^{-pH}
*[OH^-] = 10^{-pOH}
*[H3O^+][OH^-] = 1 \times 10^{-14} *pH + pOH = 14 *pH Calculations *pH = -log{10}[H^+]
*[H+] = 10^{-pH}
*[H^+][OH^-] = 1 \times 10^{-14}
*POH
*$pOH = -log_{10}[OH^-]$
*[OH^-] = 10^{-pOH}

Acids/Bases - Strong/Weak Acids & Bases

• Strong Acids: produce the greatest concentration of H+ ions.
• Strong Bases: produce the greatest concentration of OH- ions.

Thermochemistry - Calculating “q”

• Equation for Calculating Heat
  • $q=c \times m \times \Delta T$
    • q represents the heat absorbed or released.
    • c represents the specific heat of the substance.
    • m represents the mass of the sample in grams.
    • $\Delta T$ is the change in temperature in °C, or $T<em>f - T</em>i$
• The quantity of heat absorbed or released by a substance is equal to the product of its specific heat, the mass of the substance, and the change in its temperature.
• Practice: q=mcΔT
• An empty tin can has a mass of 14.9 grams. If the specific heat of tin is 0.228 J/g°C, how much heat is absorbed if the temperature increases by 20.0°C?

Thermochemistry - Heating Curves

• Heating Curve of Water

Thermochemistry - Flow of Heat

• Heat will flow from an area of higher temperature to an area of lower temperature until temperatures are equal (thermal equilibrium).

Rates/Equilibrium - Reaction Coordinates

• Reactant energy > Product energy the reaction is EXOTHERMIC
  • Exothermic reaction results in the loss of heat so, H is negative (“-“)
• Reactant energy < Product energy the reaction is ENDOTHERMIC
  • Endothermic reaction results in the absorption of heat so, H is positive (“+“)

Rates/Equilibrium - Catalysts

• Catalysts work by lowering the activation energy needed for a chemical reaction to proceed.
  • Activation Energy: minimum amount of energy that reactant particles must have to form the activated complex and lead to a reaction.

Gases - Kinetic Molecular Theory

• Postulates of the Kinetic Molecular Theory of Gases
  1. Gases consist of tiny particles (atoms or molecules)
  2. These particles are so small, compared with the distances between them, that the volume (size) of the individual particles can be assumed to be negligible (zero).
  3. The particles are in constant random motion, colliding with the walls of the container. These collisions with the walls cause the pressure exerted by the gas.
  4. The particles are assumed not to attract or to repel each other.
  5. The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature of the gas

Gases - Simple Gas Laws

Gases - Simple Gas Laws (Problems)

*Boyle's Law Example Problems
*Charles' Law Example Problems
*Gay-Lussac Example Problems
*Combined Gas Law Example Problems

Gases - Ideal Gas Law

• $PV=NRT$
• R = The Ideal Gas Constant
  • $R=0.0821 \frac{(L \cdot atm)}{(mol \cdot K)}$
  • $R= 8.31 \frac{(L \cdot kPa)}{(mol \cdot K)}$
• V has to be in