FINAL EXAM REVIEW Fall Semester

FINAL FORMAT 35 multiple choice questions 2 free response Electrochemistry Chemical equilibrium

FREE RESPONSE 2000

FREE RESPONSE 2000

FREE RESPONSE 2000

FREE RESPONSE 2000

FREE RESPONSE 2000

FREE RESPONSE 2000

Topics you wanted to cover Compound Formulas - in packet Periodicity - in packet Electrochemistry – difference between galvanic and electrolytic cell calculations and drawings; calculating plating Chemical equilibrium: ICE problems, Partial pressure, Kp Solution Stoichiometry Calculating average atomic mass Basic stoichiometry – percent composition, empirical and molecular formulas. Redox reactions – half reactions, esp. acidic vs basic

ELECTROLYTIC VERSUS VOLTAIC

Stoichiometry of Electrolytic Processes Step 1 – convert current and time to quantity of charge in coulombs (amps)(time) = total charge transferred in coulombs (Coulomb/sec)(sec) = coulombs Step 2 – convert quantity of charge in coulombs to moles of electrons coulombs ÷ (96,485 coulomb/mol e-) = mol e- Step 3 – Convert moles of electrons to moles of substance (mol e-) (mole substance/mol e-) = mol substance Step 4 – Convert moles of substance to grams of substance (mol substance)(formula mass of substance) = mass of substance

CALCULATING TIME, MASS, OR CURRENT Basic Structure: Coulombs = time (s) x current Take Coulombs and determine moles of electrons using Faradays’ constant. Determine the ratio of moles of electrons to the solid (based on the charge). Convert moles of solid to mass of solid plated. If they give you the mass, work backwards to time or current.

FINAL EXAM REVIEW = 1930.9736 = 1931C 1931C = 3.2183 = 3.22amps 600.s 24. A constant current was passed through a solution of AuCl4-1 ions between gold electrodes. After a period of 10.0min, the cathode increased in mass by 1.314g. What was the current? 10.0min AuCl4- 1.314gAu I = ? 1.314g Au 1 mol Au 1 mol AuCl4-1 3 mole e- 96485F 196.97g Au 1 mol Au 1 mol AuCl4- 1 mol e- = 1930.9736 = 1931C 1931C = 3.2183 = 3.22amps 600.s

CHEMICAL EQUILIBRIUM Equilibrium Expressions involving Pressure KP = K(RT)∆n Reverse the reaction: K = 1/K = K′ If the original is multiplied by some factor, n, then K = Kn

CHEMICAL EQUILIBRIUM Doing an ICE problem Determine the molarities of the given substances. Fill out the ice chart with what you know. If solving for K, they have to give you a way to solve for x. If solving for x, they have to give you K. Then solve for x (and then determine concentrations) or determine K.

ICE At a particular temperature, 12.0mol SO3 is placed in a 3.0L rigid container and the SO3 dissociates by the reaction 2SO3(g) ↔2SO2(g) + O2(g) At equilibrium, 3.0mol of SO2 is present. Calculate K for this rxn. ?MSO3 = 12.0mol/3.0L = 4.0mol SO3 2SO3 ↔ 2SO2 + O2 I 4.0 0 0 C -2x +2x +x E 4.0-2x 2x x 2x = 1.0M SO2 (from 3.0mol/3.0L), so x = 0.50M

ICE K = [O2][SO2]2 = (0.50)(1.0)2 = 0.056 [SO3]2 (3.0)2

SOLUTION STOICHIOMETRY Write the balanced equation and the net ionic equation. Figure out what is the precipitate. Convert molarity to moles: mol = MV Convert each mole to moles of the precipitate. Detemrine LIMITING REACTANT – smallest number. Find the mass produced from the limiting moles GOING FURTHER: Need to find the ions?

3NaOH(aq) + Fe(NO3)3(aq)  Fe(OH)3(s) + 3NaNO3(aq) FINAL EXAM REVIEW What mass of solid iron(III) hydroxide can be produced when 150.0mL of a 0.400M Fe(NO3)3 is added to 250.0mL of a 0.500M NaOH? 3NaOH(aq) + Fe(NO3)3(aq)  Fe(OH)3(s) + 3NaNO3(aq) Fe(NO3)3 mol = MV = (0.1500L)(0.400M) = 0.0600mol NaOH mol = MV = (0.2500L)(0.500M) = 0.125mol 0.0600mol Fe(NO3)3 1 mol Fe(OH)3 = 0.0600mol Fe(OH)3 1 mol Fe(NO3)3 0.125mol NaOH 1 mol Fe(OH)3 = 0.0417mol Fe(OH)3 1 mol NaOH 0.0417mol Fe(OH)3 106.87g Fe(OH)3 = 4.56g Fe(OH)3 1 mole Fe(OH)3 LR

DETERMINING ION CONCENTRATION 15mL of a 1.0M potassium iodide solution reacts with 20.0mL of 6.0M lead(IV) nitrate. Calculate the concentration of potassium and lead(IV) ions in the supernatant (the clear liquid above the precipitate). Write the balanced equation. 4KI + Pb(NO3)4  PbI4 + 4KNO3 Determine if there is a change in moles of ions from reactant to product. 4K+  4K+ no change

DETERMINING ION CONCENTRATION Convert molarity to moles 15mL 1L 1.0mol KI 1 mol K+ = 0.015mol K+ 1000mL 1L 1 mol KI Convert back to molarity by putting moles over whole volume. 0.015mol K+ = 0.43M K+ (0.020L + 0.015L) Back at #2, if there is a change in the number of ions Lead goes from 1 ion (reactant) to 0 ions (product) So, you must do stoichiometry to determine what is limiting.

DETERMINING ION CONCENTRATION Determine limiting reactant 15mL 1L 1.0mol KI 1 mol PbI4 = 0.0038mol PbI4 1000mL 1L 4 mol KI 20mL 1L 6.0mol Pb(NO3)4 1 mol PbI4 1000mL 1L 1 mol Pb(NO3)4 = 0.12mol PbI4 Determine moles of the ions in the LR: 0.038mol Pb+4 Determine how much it should make 20mL 1L 6.0mol Pb(NO3)4 1 mol Pb+4 1000mL 1L 1 mol Pb(NO3)4 = 0.12mol Pb+4 LR

DETERMINING ION CONCENTRATION Take was you started with and subtract what you used to make the solid to get the amt of ion left over. 0.12mol Pb+4 - 0.038mol Pb+4 = 0.12mol Pb+4 0.12mol Pb+4 = 3.42857 = 3.4M Pb+4 0.035L

AVERAGE ATOMIC MASS The average atomic mass on the periodic table comes from the weighted averages of the isotopes. To determine the mass, you need the following: The masses of the isotopes Their percent abundance Multiple those two together, add all the numbers to get the average atomic mass.

AVERAGE ATOMIC MASS The atomic masses of iridium-191 is an 191.0 amu and iridium-193 is 193.0 amu. The percentage abundance for each is 37.58% ( iridium-191) and 62.42% ( iridium-193). Calculate the average atomic mass.   191Ir 191.0 amu x 0.3758 = 71.78 amu 193Ir 193.0 amu x 0.6262 = 120.9 amu 192.68 = 192.7 amu

STOICHIOMETRY Percent composition: g part x 100 = answer g whole Empirical Formula: If they give you a percentage, change it to mass out of 100. Convert to moles Do mole ratios, dividing all by the smallest number Determine formula from whole numbers

STOICHIOMETRY Molecular Formula: Once you have the empirical formula, determine its mass from the periodic table. Divide that mass into the given molecular mass to get the ratio. That number is what all elements in the empirical formula get multiplied by.

EMPIRICAL FORMULA While trace impurities of iron and chromium in natural corundum form the gemstones ruby and sapphire, they are basically a binary compound of aluminum and oxygen, with 52.9% Al and 47.1% O. Find the empirical formula and give the chemical name for corundum.   52.9g Al 1 mol Al = 1.96071 = 1.96 mol Al 26.98g Al 47.1g O 1 mol O = 2.94375 = 2.94 mol O 16.00g O 1.96mol Al = 1 2.94 mol O = 1.5 1.96 mol Al 1.96 mol Al Al2O3 aluminum oxide

Molecular formula The simplest formula for vitamin C is C3H4O3. Experimental data indicates that the molecular mass of vitamin C is 180 amu.   C 3 x 12.01 = 36.03 H 4 x 1.01 = 4.04 formula mass O 3 x 16.00 = 48.00 emp. C3H4O3 88.07g 88.07 molec. ? 180 180 / 88.07 = 2 C3H4O3 x 2  C6H8O6

REDOX REACTIONS The fundamental principle in balancing redox equations is that the number of electrons lost in an oxidation process (increase in oxidation number) must equal the number of electrons gained in the reduction process (decrease in oxidation number). Write the oxidation numbers and find out what two elements changes charge.

REDOX REACTIONS STEPS TO FOLLOW: Write out the equation. Then change it to a net ionic equation if it is not one already, omit the spectator ions, and assign oxidation numbers to all atoms in the equation. Write them above the element.   +1 -2 +5 -2 -1 0 +1 -2 HS-1 + IO3-1  I-1 + S + H2O Write the separate half-reactions. -2 0 +5 -1 HS-1  S IO3-1  I-1

REDOX REACTIONS Balance all the elements except O and H (already balanced in this one). HS-1  S IO3-1  I-1 If the oxygen is unbalanced, add enough water (H2O) to the side deficient in oxygen.   HS-1  S IO3-1  I-1 + 3H2O

REDOX REACTIONS Add sufficient hydrogen ions (H+) to the side deficient in hydrogen to balance the hydrogen.   HS-1  S + H+1 6H+1 + IO3-1  I-1 + 3H2O

REDOX REACTIONS Write the electrons in each half reaction. -2 0 -2 0 HS-1  S + H+1 + 2e- +5 -1 6e- + 6H+1 + IO3-1  I-1 + 3H2O Determine the least common multiple and multiply each to get it so that the number of electrons gained equals the number of electrons lost. (x3) 3HS-1  3S + H+1 + 6e- (x1) 6e- + 6H+1 + IO3-1  I-1 + 3H2O

REDOX REACTIONS Add the two half reactions together and return the spectator atoms. Delete anything that exactly occurs on both sides. (Notice how the water showed back up!) 3HS-1 + 6e- + 6H+1 + IO3-1  3S + 3H+1 + 6e- + I-1 + 3H2O  becomes 3H+1 3HS-1 + 3H+1 + IO3-1  3S + I-1 + 3H2O

REDOX REACTIONS If acidic, you: If Basic, you: Add water to balance the oxygens Add H+ to balance the hydrogens If Basic, you: Add OH- to get rid of all the H+ by making water. Add the same number of OH- to both sides.