CHM1045 Latturner Final Exam (1-7, 9)

Potassium permanganate (KMnO4) is a chemical commonly used as a disinfectant. Calculate the mass percent of each element in this compound.

> calculate molar mass of KMnO4 KMnO4 = 158.032 g/mol > find molar mass of each element in compound 1 mole K = 39.098 g/mol, 1 mole Mn = 54.938 g/mol, 4 moles O = 63.996 g/mol > (molar mass of element/total molar mass) x 100 = mass % (39.098/158.032) x 100 = *24.741% K* (54.938/158.032) x 100 = *34.764% Mn* (63.996/158.032) x 100 = *40.496% O*

Give the number of protons, neutrons, and electrons for the given isotope, or give the isotope symbol (with X being the element symbol, A being the mass number, and q being the charge), for the following: a) 84Sr2+ b) 235U0 c) 31P3- d) protons = 36, neutrons = 48, electrons = 36

> look at periodic table atomic # for protons, # of neutrons = atomic mass - protons, # electrons = protons + charge on isotope a) protons = 38, neutrons = 46, electrons = 36 b) protons = 92, neutrons = 143, electrons = 92 c) protons = 15, neutrons = 16, electrons = 18 d) 84Kr0

Titration of a 20.0 mL sample of a sulfuric acid solution is carried out to determine its molarity. It requires the addition of 1.7 mL of 0.0811 M NaOH to reach the end point of the titration. What is the concentration of the sulfuric acid solution?

> write balanced equation H2SO4(aq) + 2 NaOH(aq) -> Na2SO4(aq) + 2 H2O(l) > convert molarity to mol (0.0017 L)(0.0811 mol/L) = 0.0001379 moles NaOH >use stoichiometry to find H2SO4 moles (0.0001379 moles NaOH)(1 mole H2SO4 / 2 moles NaOH) = 0.000068935 moles H2SO4 > divide H2SO4 moles by L of sulfuric acid solution to find concentration Molarity = (0.000068935 moles H2SO4)/(0.0200 L) = *0.0034 M*

An aqueous solution of Na2CO3 is mixed with an aqueous solution of Cr(NO3)3. Write the balanced molecular, total ionic, and net ionic equation and include the states of reactants and products.

Molecular: 3 Na2CO3 (aq) + 2 Cr(NO3)3 (aq) -> 6 NaNO3 (aq) + Cr2(CO3)3 (s) Total ionic: 6 Na+(aq) + 3 CO32-(aq) + 2 Cr3+(aq) + 6 NO3−(aq) -> 6 Na+(aq)+ 6 NO3−(aq) + Cr2(CO3)3 (s) Net ionic: 3 CO32- (aq) + 2 Cr3+ (aq) -> Cr2(CO3)3 (s)

20.3 mL of 1.025 M AgNO3 aqueous solution is added to 21.9 mL of a 0.571 M aqueous solution of MgCl2. a) Write the balanced molecular equation for the reaction. Include states of reagents. b) Which reactant is limiting? c) What is the mass of solid that will precipitate?

a) 2 AgNO3(aq) + MgCl2(aq) -> 2 AgCl(s) + Mg(NO3)2 (aq) b) find mols of each reactant using M AgNO3 solution: (0.0203 L)(1.025 mol/L) = 0.020808 moles AgNO3 MgCl2 solution: (0.0219 L)(0.571 mol/L) = 0.012505 moles MgCl2 > use stoichometry to find mols of AgCl (0.020808 moles AgNO3)(2 moles AgCl / 2 moles AgNO3) = 0.020808 moles AgCl (0.012505 moles MgCl2)(2 moles AgCl / 1 mole MgCl2) = 0.025010 moles AgCl *AgNO3 is limiting* because it produces less moles c) convert moles of AgCl w limiting reactant to g using molar mass of AgCl (0.020808 moles AgCl)(143.32 g/mol) = *2.98 g of AgCl*

Sodium metal reacts violently with excess liquid water to produce hydrogen gas (H2) and aqueous sodium hydroxide. a) Write the balanced reaction, identify oxidation states of all the atoms, and label the reducing agent. b) If you start with 125 grams of sodium metal, how many grams of H2 should be produced?

a) 2 Na(s) + 2 H2O (l) -> H2 (g) + 2 NaOH (aq) Na oxidation state = 0 H2O oxidation state = +1 for H, -2 for O H2 oxidation state = 0 NaOH oxidation state = +1 for Na, -2 for O, +1 for H *2 Na (s) is the reducing agent* because its going from 0 to +1 b) convert g to mol (125 g Na)(1 mol / 22.990 g Na) = 5.437 moles Na > use stoichiometry to find moles H2 produced (5.437 moles Na) (1 mole H2 produced / 2 moles Na used) = 2.7186 moles H2 > convert moles to g (2.7186 moles H2)(2.0158 g/mol H2) = *5.48 grams of H2*

Potassium perchlorate, KClO4, decomposes when it is heated, forming potassium chloride and oxygen gas. a) Write the balanced reaction. b) Identify the oxidation numbers of all the atoms in the reactant and products. c) If you start with 2.58 grams of KClO4, how many molecules of oxygen should be produced? d) If the oxygen gas is collected over water at 30.0 °C at a pressure of 757.2 mmHg, the volume of gas is 667 mL. What was your percent yield? (The vapor pressure of water at 30.0 °C is 31.8 mmHg.)

a) KClO4 -> KCl (aq) + 2 O2 (g) b) KClO4: *K = +1, O = -2*, *Cl =* 0=(1*1)+(1*x)-(2*4) -> *7* KCl: *K = +1, Cl = -1* O2: *O = 0* c) convert g to mol, use stoichiometry to find mol of O2, convert mol to molecules (2.58 g) (1 mol / 138.55 g) = 0.01862 moles KClO4 (0.01862 moles KClO4)(2 moles O2 / 1 mole KClO4) = 0.03724 moles O2 (0.03724 moles O2)(6.022 x 1023 molecules /mol) = *2.24 x 10^22 molecules of O2* d) find P(O2) using Ptotal = PO2 + PH2O 757.2 mmHg = PO2 + 31.8 mmHg PO2 = 725.4 mmHg > use n=PV/RT to find actual yield P = 725.4 mmHg (1 atm / 760 mmHg) = 0.9545 atm V = 667 mL = 0.667 L R = 0.082057 L·atm/mol·K T = 30.0C + 273.15 = 303.15 K n = (0.9545 atm)(0.667 L) / (0.082057 L·atm/mol·K)(303.15 K) n = 0.0256 moles O2 > divide actual yield by theoretical yield of O2 (found in part c) theoretical yield = 0.03724 moles O2 % yield = [(0.0256 moles)/(0.03724 moles)] x 100 = *68.7 % yield*

Naturally occurring vanadium contains a mixture of 50V (atomic mass 49.9471585 amu) and 51V (atomic mass 50.9439595 amu). a) Without doing any math, estimate which isotope will be the most abundant in a natural sample of vanadium; explain your answer in one sentence. b) Given the mass listed on the periodic table, calculate the percent abundances of the two isotopes.

a) The average mass of vanadium (on periodic table) is 50.942 amu; this is VERY close to the isotopic mass of 51V, so most vanadium atoms will be 51V b) mass on vanadium on periodic table = 50.942 > the percent fractions for 50V and 51V must add to 1 (or 100%) so do x for 50V and 1-x for 51V 50.942 amu = [(X)(49.9471585 amu)] + [(1-X)(50.9439595 amu)] *with ti-84 plus calculator: you can solve equation above by going to MATH, ^ , SOLVER, and entering equation = 0, then hover over x= and press ALPHA ENTER* x = 0.0019658, 1-x = 0.9980342 > convert to % *% abundance of 50V = 0.19658%, % abundance of 51V = 99.80342%*

Acetone is C3H6O and is a liquid with a density of 0.784 g/mL, used as nail polish remover. a) A bottle of acetone contains 155 mL of acetone. How many moles of acetone are in the bottle? b) How many molecules of acetone are in the bottle? c) Calculate the number of C, H, and O atoms in the bottle.

a) convert density (g/mL) to g (0.784g/mL)(155mL) = 121.52g > convert g to moles acetone molar mass = 58.079g (121.52g)(1mol/58.079g) = *2.09 moles* b) convert moles to molecules (2.09 moles)(6.022x10^23 molecules/1 mole) = *1.26x10^24 molecules* c) find amount of atoms in molecule 3 C, 6 H, and 1 O > (molecules)(atoms of E) = atoms in molecule for that element (1.26 x 1024 molecules)(3 carbon atoms / 1 molecule) = *3.78 x 10^24 atoms of C* (1.26 x 1024 molecules)(6 hydrogen atoms / 1 molecule) = *7.56 x 10^24 atoms of H* (1.26 x 1024 molecules)(1 oxygen atom / 1 molecule) = *1.26 x 10^24 atoms of O*

A sample of an unknown compound is analyzed and it is found to consist of the following mass percentages: 40.00 % carbon, 6.71 % hydrogen, and 53.29 % oxygen. a) What is the empirical formula of the compound? b) In a separate experiment, the molar mass of this compound is found to be 90.1 g/mol. What is the molecular formula of the compound?

a) convert mass % to g then moles pretend sample is 100g: 40.00% C = 40.00g C, 6.71% H = 6.71g H, and 53.29% O = 53.29g O (40.00 g)(1 mol / 12.011g) = 3.33 mol C (6.71 g)(1 mol / 1.0079 g) = 6.66 mol H (53.29 g)(1 mol / 15.999 g) = 3.33 mol O > divide all mols by smallest # of mols calculated 3.33 mol C /3.33 = 1 mol C 6.66 mol H /3.33 = 2 mole H 3.33 mol O /3.33 = 1 mol O > write as empirical formula *empirical formula is CH2O* b) find molar mass of empirical formula CH2O 12.011 + 2(1.0079) + 15.999 = 30 g/mol > molecular molar mass/empirical molar mass = # of times greater molecular form is 90.1 / 30 = 3, so *molecular formula is C3H6O3*

Lithium carbonate, Li2CO3, is used in the manufacture of lithium ion batteries, and it is also used for the treatment of mood disorders. a) What mass (in grams) of lithium carbonate should be used to prepare 350.0 mL of a 1.125 M solution of lithium carbonate? b) What volume (in mL) of the concentrated solution from part (a) is needed to prepare 100.0 mL of a 0.0625 M solution of lithium carbonate?

a) find mols of Li2CO3 using molarity, convert to grams using molar mass of Li2CO3 (0.3500 L)(1.125 mol/L) = 0.39375 moles (0.39375 moles of Li2CO3)(73.89 g/mol) = *29.09 grams of Li2CO3* b) use M1V1 = M2V2 (1.125 mol/L)(V1) = (0.0625 mol/L)(0.1000 L) V1 = 0.005555 L = *5.56 mL*

Permanganate ion MnO4− (aq) reacts with iodide anion I− (aq) in acidic solution to form I2 (aq) and MnO2 (s). a) Identify the reducing agent. b) What is the oxidation state of manganese in MnO4− ? c) Balance the reaction in acidic solution using the half reaction method.

a) write balanced equation (except O and H) w/ oxidation states MnO4- + 2I- -> I2 + MnO2 MnO4-: O = -2, Mn = -1=(-2*4)+(1*x) = 7 2I-: I- = -1 I2: I = 0 MnO2: O = -2, Mn = 0 = (-2*2)+(1*x) = 4 *I−* (iodide, gets oxidized from -1 to 0, is therefore the reducing agent) b) Mn in MnO4- = *+7*, Mn in MnO2 = *+4* c) write out half reactions + balance except O and H MnO4− -> MnO2 ; 2I− -> I2 > balance O by adding H2O MnO4− -> MnO2 + 2 H2O ; 2I− -> I2 > balance H by adding H+ 4 H+ + MnO4− -> MnO2 + 2 H2O ; 2I− -> I2 > balance charges by adding e- find total charge on each side of arrow 4H+ = 4, MnO4- = -1; 4 + -1 = 3 MnO2 = 0, 2H2O = 0; 0+0=0 adding 3e- to left side makes total charge 0, matching right side 3e- + 4 H+ + MnO4− -> MnO2 + 2 H2O find total charge on each side of arrow 2I- = 2-, I2 = 0 adding 2e- to right side makes total charge -2, matching left side 2 I− -> I2 + 2 e- > make e-s on each half reaction same # (multiply entire equation) 6 e- + 8 H+ + 2 MnO4− -> 2 MnO2 + 4 H2O ; 6 I− -> 3 I2 + 6 e- > add reactions together and cross out extras (cross out bolded) *6 e-* + 8 H+ + 2 MnO4− + 6 I− -> 2 MnO2 + 4 H2O + 3 I2 + *6 e-* *8 H+ + 2 MnO4− + 6 I− -> 2 MnO2 + 4 H2O + 3 I2*