Chemistry with Mr. Faucher Name_______________________________ Block______

Chemistry Review for Final Exam

The final exam covers topics from this 2nd

semester. All of my chemistry students will take the Senior Final Exam.

Then, remaining Juniors will be better-prepared to take another Junior Final Exam at the end of the school year.

For Juniors, I will average both Final Exams to obtain 1 final exam score. Your final exam score will then be

averaged with your Semester 1 Exam to get 1 exam score covering the whole year, and counting as 20% of your

final course grade. (Think of this score as your imaginary 5th

quarter grade.)

Chemical Equations

Be able to interpret and balance a chemical equation for a reaction, knowing the roles of the coefficients

and subscripts. Also recognize the italicized states of the reaction chemicals in parentheses.

Know all the parts of a chemical equation and how they are useful in interpreting a chemical reaction.

Cu (s) + 2HCl (aq) CuCl2 (aq) + H2 (g)

The italics in the parentheses indicate the “condition” (state, phase) of the preceding chemical: l = liquid, g = gas, s = solid, aq = aqueous (i.e., dissolved in water) The mole ratio of the chemicals is represented by the coefficients in front of each chemical. The coefficient is 1 if there is none listed. Interpreting the equation above, 1 mole of solid copper reacts with 2 moles of hydrochloric acid (aqueous hydrogen chloride) to make 1 mole of aqueous copper (II) chloride and 1 mole of hydrogen gas.

Molarity & Making Molar Solutions

Understand and apply the concept of molarity and how to make a molar solution of a specific

concentration.

Molarity is the aqueous concentration of a dissolved substance in terms of moles per liter.

Example: A 1.4-M solution of AgNO3 consists of 1.4 moles of silver nitrate crystals dissolved per 1 liter of solution.

To make the solution in a 1-L volumetric flask, find out how many grams you need to weigh by first finding the

molar mass of AgNO3. Then you add it to a volumetric flask half-filled with water. Swirl to dissolve. Then finish

adding water to the 1-L mark.

1.4 moles AgNO3 x 169.9 g AgNO3 = 237.9 g/L AgNO3

1 L 1 mole AgNO3

desired molarity of solution molar mass amount you need to weigh on the balance per Liter of volume.

Molality Molality is the aqueous concentration of a dissolved substance in terms of moles per kilogram of water.

Know how to make a molal solution.

Example: How would a chemist prepare a 0.20-m solution of sugar, C6H12O6 ? First, place 1 kg of water in a beaker. Calculate how many grams of C6H12O6 you need to add to the 1 kg of water after finding its molar mass.

0.20 moles C6H12O6

1 kg H2O x

180.0 g C6H12O6

1 mole C6H12CO6 =

36.0 g C6H12O6

1 kg H2O

desired molality molar mass grams of sugar to add to 1 kg of water

Add the 36.0 g of sugar to the 1 kg of water and stir.

Recognize that given identical molal concentrations of a covalently bonded non-electrolyte (e.g., sugar) and an ionic electrolyte salt (e.g., NaCl), the ionic salt will have more particles in solution because of dissociation.

Example: What is the molality of all the particles dissolved in a 0.20-m solution of C6H12O6 ? Since sugar is a covalent substance that does not dissociate into ions, the molality of all the particles in solution is still 0.200 m. Example: What is the molality of the total number of particles in a 0.20-m solution of NaCl?

Since NaCl is an ionic solid that easily dissociates into separated aqueous ions, 1 mole of NaCl makes 1 mole of Na

+

and 1 mole of Cl-, adding up to 2 moles of dissociated particles.

So the molality of the particles from 0.20-m NaCl is twice the molality, 0.40-m Na

+ and Cl

-.

Stoichiometry Understand and apply the concept of molar mass in determining the mass per mole of an atom or

molecule. The Periodic Table lists the molar masses for each type of atom.

Be able to apply the principles of stoichiometry in using a balanced chemical equation for a reaction to

calculate the amount of grams or moles of a substance from a given amount of grams or moles of another

chemical substance.

Example: In our silver lab, we had the following equation for the reaction of a copper coil dipped into a silver

nitrate solution:

Cu (s) + 2AgNO3 (aq) 2Ag (s) + Cu(NO3)2 (aq)

If 3.00 grams of copper reacted, how many grams of silver were produced?

(Hint: Use “situation circles” to focus your thinking on the specific parts of the equation you need to think about. Ignore

everything else outside the circles. Draw a “mole bridge” to give you a plan to follow for the chemical hop.)

3.00 g Cu x 1 mole Cu

63.5 g Cu x

2 moles Ag

1 mole Cu x

107.9 g Ag

1 mole Ag = 10.2 g Ag Notice all units cancel except g Ag.

given molar mass mole ratio molar mass answer (with proper unit and chemical identity!)

Understand that determining the number of atoms is not the same thing as determining the amount of grams. It does not involve molar mass, just mole ratios. (This is because the number of atoms is a “count” of atoms following a simple “mole” ratio.)

Cu (s) + 2AgNO3 (aq) 2Ag (s) + Cu(NO3)2 (aq)

Example: If 3,000 atoms of copper reacted, how many atoms of silver were produced?

3,000 atoms Cu x 2 atoms Ag

1 atom Cu = 6,000 atoms Ag Notice all units cancel except atoms Ag.

given mole ratio answer (with proper unit and chemical identity!)

Oxidation/Reduction

Oxidation happens when the oxidation number increases (to a more positive value) due to a loss of electrons.

Reduction occurs when the oxidation number is reduced (to a less positive value).

Spectator ions do not actively participate in a reaction and therefore keep their charge.

Know how to identify what metal is oxidized and what metal ion is reduced in a redox reaction.

Know how to identify the spectator ion(s) in a chemical reaction.

Example: Cu (s) + 2AgNO3 (aq) 2Ag (s) + Cu(NO3)2 (aq)

In the equation above, from our silver lab, solid Cu

0 atoms are oxidized to aqueous Cu

2+ ions, the aqueous Ag

1+ ion

is reduced to solid Ag0 atoms. The oxidation number of the NO3

1- aqueous ions remains the same, making them

spectator ions. All of this happens in one beaker.

Chemical Equilibrium

Chemical equilibrium is achieved when the rate of products being formed in a chemical reaction equals the rate of

the reactants being re-formed. This makes the concentration of the products constant. The concentrations of the

reactants are also necessarily constant.

Using LeChatelier’s Principle, predict how a chemical system will respond to a stress placed on that

system when a reactant or product is removed from the chemical system.

Example: Cu (s) + 2AgNO3 (aq) dynamic

equilibrium

↔ 2Ag (s) + Cu(NO3)2 (aq)

After enough time elapses, dynamic equilibrium is reached. Then, for every molecule of Cu(NO3)2 being formed,

another one “unforms” as a molecule of AgNO3 is being re-made. This keeps the average concentration of each

aqueous ion constant.

If copper (II) nitrate product is removed from the reaction mixture, a stress is created and the chemical system will

respond and relieve that stress (caused by the lack of that chemical) by increasing the forward rate of the reaction

to make more Cu(NO3)2 .

Recognize and understand the formula used to determine the equilibrium constant for a chemical

reaction. Know what is going on with subscripts, superscripts, and concentration brackets in the formula.

Since the equilibrium constant for a particular chemical reaction does not change, the Keq expression is

an excellent visual tool we can use to help determine the effect of the LeChatelier response to a

removed/added chemical reactant or product.

The Keq expression numerator has the product concentrations. The denominator has the reactant

concentrations.. Therefore, when Keq > 1, products are favored. When Keq < 1, reactants are favored.

Pure liquids and solids are not listed in the Keq expression

Example: What is the equilibrium constant equation for the chemical process below?

2A + B2 A2B2

Notice that the molar concentration of the product is on top of the fraction bar, and the molar concentrations of

the reactants are below the fraction bar. Also notice that the coefficient for A in the reaction becomes its

superscript in the expression. This means that the molar concentration for A must be squared.

You can see that when K > 1, it means that the concentration of the product (A2B2 ) is higher than the

concentration of the reactants, and the reaction wants to proceed to favor the products at equilibrium.

Electrochemical Cells

Dividing the Cu (s) + 2AgNO3 (aq) 2Ag (s) + Cu(NO3)2 (aq) reaction mixture into 2 separate beakers, connected with a salt bridge and a wire to allow electron flow, still allows oxidation/reduction to occur. Now we have created an electrochemical cell, acting as a battery when electrons from the reducing agent, Cu

0 , move along

the wire to the oxidizing agent , Ag1+

. As a result, the oxidizing agent is reduced to a lower oxidation number, as it precipitates as solid Ag

0 atoms onto the strip.

Cu (s) | Cu

2+ (1.0 M) || Ag

1+ (1.0 M) | Ag (s)

Anode Cathode Oxidation Reduction Reducing Agent Oxidizing Agent

2Ag

1+(aq) + 2e

- 2Ag (s) higher on Table 16-4 E

o reduction = +0.80 V

Cu (s) Cu2+

(aq) + 2e-

lower on Table 16-4 Eooxidation = - 0.34 V

____________________________________________________________________________

2Ag1+

(aq) + Cu (s) 2Ag (s) + Cu2+

(aq) Eocell = +0.46 V

net ionic equation electrochemical cell voltage

Example: If the copper metal strip (anode) was replaced with an aluminum strip as the new anode,

how would the new electrochemical cell voltage (Eocell ) compare with the one made of copper?

(Hint: Refer to Table 16-4 on the next page.)

Since aluminum is lower on the table than copper, it is a stronger reducing agent than copper. (Its vertical distance from the oxidizing agent, Ag1+

, is farther in this arrangement, so the generated voltage should be greater.)

3Ag

1+(aq) + 3e

- 3Ag (s) E

o reduction = +0.80 V

Al (s) Al3+

(aq) + 3e-

Eooxidation = + 1.66 V (Notice the sign change. Why?)

____________________________________________________________________________

3Ag1+

(aq) + Al (s) 3Ag (s) + Al3+

(aq) Eocell = +2.46 V (This is 2 volts higher!)

net ionic equation electrochemical cell voltage

VSEPR Theory and Bonding

Re-watch the two VSEPR videos from the Resources Links on my teacher website.

Be able to predict the geometrical shape of a molecule based on its number of “things” around the

central atom and whether or not there is a lone-pair of electrons around that central atom. Know how

the geometrical shape of a molecule and its electron distributions influence whether the molecule is

polar or nonpolar.

Example: Is CO2 a polar or nonpolar molecule? Explain your reasoning.

This linear molecule has oxygens pushing away from each other because of lone-pair electron repulsion. Both ends

of the molecule are negative, making it non-polar. Polar molecules have a positive end and a negative end.

Understand the degree of sharing or “stealing” of electrons associated with each type of chemical bond.

Covalent bonds involve the sharing of electrons. Ionic bonds involve the “stealing” of electrons.

Know how to use the concept that central atoms usually have a stable octet of electrons in predicting if

there will be lone-pair electrons affecting the geometrical shape of the molecule.

Be able to remember and apply the HONC1234 Rule in determining the number of chemical bonds

hydrogen, oxygen, nitrogen and carbon atoms can have.

Example: Notice that the HONC1234 Rule tells us that the central N atom will have three bonds (contributing 6

electrons), leaving a lone pair of electrons to complete a stable octet on N for a total of 8 electrons around it. This

electron pair repels the hydrogens so that the molecule cannot be planar (flat). Because the hydrogen atoms are

pushed down, the resulting shape is trigonal pyramidal.

Be able to understand and apply the expression “Like dissolves like”.

Example: Polar molecules like the bent H2O molecule are attracted to other polar molecules like NH3. That’s why

ammonia dissolves so well in water. Non-polar molecules like Styrofoam dissolve very well in non-polar solvents

like gasoline. So pouring gasoline into a Styrofoam cup will make it go right through the cup!

Each line between two atoms represents one bond, consisting of two electrons.

Example: How many electrons are around the central carbon of CO2? (answer: 8 e-)

Gas Laws

Know the relationships between pressure, volume, and temperature. The combined gas law equation is 𝑃1 𝑉1 𝑇1

= 𝑃2 𝑉2 𝑇2

. This equation relates the first set of conditions for P,V,T to a second set of conditions.

Know how to recognize and interpret the graphs of each of the 3 gas laws. P and V are the only variables

that are inversely (oppositely) related. When P↑ , V↓ .

Know the concept behind the operation of a mercury barometer. What pushes the mercury in the tube to

its height?

Example: If the atmospheric pressure were to decrease, what would happen to the level of mercury in the tube?

Know how to use Dalton’s Law of Partial Pressures to adjust for the water vapor pressure when

determining the pressure of a collected generated gas using the water-displacement method. (Also

remember that the amount of trapped water vapor depends on temperature.)

When the liquid level inside the eudiometer (gas-collection tube) matches the surrounding water level in the water jar, the total

pressure of the gases inside the tube equals the outside barometric “atmospheric” pressure of the classroom.

Patm = Ptotal = Phydrogen gas + P water vapor Therefore, Phydrogen gas = Patm - P water vapor

Acids & Bases

Know the definition of pH.

pH is defined as the negative log of the hydrogen ion concentration. Mathematically, pH = -log [H+]

Know which numbers on the pH scale correspond to acids, neutrality, and bases.

pH values range from 1 to 14, with 7 being neutral. Acids can be between pH 0-7, bases between pH 7-14.

Know what kinds of information we can obtain by knowing the pH of a solution.

We know, for example, that if pH = 3, that means that we have an acid and [H+] = 1 x 10

-3 M. This is a stronger

concentration than the concentration of the hydroxide base ions, [OH-] = 1 x 10-11

M, needed to reach [H+] x [OH-]

= 1 x 10-14

M. An acid simply has more H+ ions than OH

- ions.

Know exactly what it means in terms of ion interactions when we say that an acid is neutralizing a base,

or vice versa.

Example: A 50-mL sample of HCl of unknown molarity was titrated with 0.15-M NaOH. Neutralization occurs at pH=7. This means that the number of acid H

+ ions equals the number of base OH

- ions, cancelling each other out.

What is the molarity of the acid sample?

Answer: Since it took and equal volume amount of base to neutralize the sample of acid, and this means that they both have the same number of moles of acid and base ions, at the same volume, then they both must be equal in molarity (0.15M).

Hypothetically, if it taken only 25 mL of base to neutralize the 50 mL sample of acid, that would mean the molarity of the acid would be twice as weak, at 0.075 M HCl.

At neutrality, an acid plus a base makes a salt plus water: HCl + NaOH NaCl + HOH

0 10 20 30 40 50 60 70 80 90 1000

2

4

6

8

10

12

14pH of HCl vs. Volume of NaOH Added

pH

Volume_NaOH_added

mL

Colligative Properties

Colligative properties, like freezing point depression, Tfp, only depend on the number of particles, not their nature

(identity).

Example: The molar mass of barium chloride is 208 g/mole. The beaker to the left contains 1 kg of water. When 208 g of BaCl2 are added to that beaker, what is the molality of BaCl2? Since BaCl2 is a salt, it quickly dissociates (breaks up) into its electrolyte ions, as shown in the diagram to the right. What is the total molality of the ion particles in that beaker?

BaCl2 dissociation→ Ba

2+ + Cl

- + Cl

-

A 1-molal solution of barium (II) chloride is made in the left-hand beaker. This salt dissociates and forms the

particles shown in the right-hand beaker. The mole ratio is 1BaCl2 : 1Ba+

: 2Cl-. This means that 1 mole of BaCl2

makes a total of 3 moles of ion particles. The molality of the ion particles in the right-hand beaker is therefore 3 m.

Know how to interpret a freezing-point depression graph.

Example: The slope of the graph shown below is k = -1.86°C/m. Calculate the freezing point depression of a 3.0-m solution of glucose, a covalently-bonded non-electrolyte.

(Hint: If you want an equation, you can obtain it from what you know about the definition of slope.)

Answer: -5.58° C

.

0 1 2 3 4

8

6

4

2

0

Freezing Point Depression

Moles of Solute Particles in 1 kg Water

Dro

p in F

reezin

g P

oin

t T

em

per

atu

re (

°C )

Tfp

molalityglucose