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Unformatted text preview: General Review: Answers to Review Problems Topic A: Atoms, Molecules and Moles 1) Isotopes are atoms with the same number of protons (so they are the same element) but with different numbers of neutrons. This different number of neutrons results in identical elements having different mass numbers. 2) Barium is element number 56. This means that barium has 56 protons. In a neutral atom, the number of electrons is equal to the number of protons, so this species has 56 electrons. The mass number of any element is the sum of the number of protons and the number of neutrons. Therefore, the number of neutrons in this isotope is 135 – 56 = 79. 3) 4) If 69Ga is 60.4%, then what is the percentage of the isotope xGa ? Use the average atomic mass (from the Periodic Table) to determine the atomic mass of xGa, using the equation to determine a weighted average. 5) 6) What does a charge of +1 tell you about the number of electrons? What is the relationship between the number of electrons and the number of protons? The element is Na. (p. A‐4) Atomic Mass of xGa = 70.93, therefore the mass number of xGa is 71. The molar mass of a compound tells us the mass of one mole of the compound. What are the units of molar mass? Can you manipulate these two pieces of data to give you those units? Molar mass = 114.97 gmol—1 a) 1 mole of C2H6O molecules contains 2 moles of C atoms Therefore: 3.20 moles of C2H6O molecules contains 2 x 3.20 = 6.40 moles of C atoms b) Each mole of C2H6O molecules contains 6.02 x 1023 molecules Therefore: 3.20 moles of C2H6O molecules contains 3.20 x 6.02 x 1023 = 1.93 x 1023 molecules c) From a) we know that there are 6.40 moles of C atoms in 3.20 moles of C2H6O molecules Therefore, 6.40 moles of C atoms contains 6.40 x 6.02 x 1023 = 3.85 x 1024 atoms General Review ‐ 1 Each C2H6O molecule contains a total of 9 atoms. Therefore, 3.20 moles of C2H6O molecules contain 3.20 x 9 = 28.8 moles of atoms The total number of atoms is 28.8 moles x 6.02 x 1023 atomsmol―1 = 1.73 x 1025 atoms d) One molecule of C2H6O contains a total of 9 atoms e) Molar Mass of C2H6O = 46.068 gmol―1 f) One mole of C2H6O molecules contains 2 moles of C atoms, 6 moles of H atoms and 1 mole of O atoms 7) 8) Therefore: x molecules of C2H6O contains a total of 100 atoms x = 11.1 molecules (in theory only, in practice you can’t have 11.1 molecules!) Mass Percent of C = 2 x 12.01 x 100 = 52.1% Mass Percent of O = 16.00 x 100 = 34.7% 46.068 46.068 2 = 0.222 9 6 Mole fraction of H = = 0.667 9 1 Mole fraction of O = = 0.111 9 Mole fraction of C = Convert moles of each species to mass. Express mass of SiO2 as a percentage of the total mass of the mixture. Mass % of SiO2 = 40.3% Mole % of SiO2 = 39.6% Mole % of NaCl = 60.4% Use a 100 g sample, therefore: C = 44.77 g, H = 7.52 g, O = 47.71 g Convert to moles and determine the smallest whole number ratio of moles. Actual ratio of moles is 1.25 C : 2.50 H : 1 O Multiply by 4 to get whole numbers: 5 C : 10 H : 4 O Formula is C5H10O4 For empirical formula, same approach as 8). Ratio is 2 C : 5 H : 1 O, so empirical formula is C2H5O (Molar Mass = 45.06 gmol―1) Molecular formula is a multiple of simplest formula; determined using molar mass. Molecular formula is C4H10O2. General Review ‐ 2 9) 10) Same approach as in 8) and 9) 11) We are using 1.00 x 1023 molecules of oxygen; ie, O2! Find moles of O2 needed. (Use Avogadro’s number) 12) Use a variable (x) to determine the molar mass of C5H12Ox; Molar mass = 5(12.01) + 12(1.008) + x(16.00) = 72.146 + 16.00x 13) What is the relationship between moles of OsO4 and moles of Os in OsO4? Atoms Os = 2.37 x 1018 atoms 14) Start with 100 g of Ne and 100 g of Ar. (You could use any mass here.) 15) What is the relationship between moles of C6H14 and moles C in C6H14? 16) What does % by mass O actually mean? Write the relationship and use a variable for the molar mass of M in that relationship. Solve for M. 17) M = 56 gmol―1; the element is Fe a) How many moles of C8H9NO2 are in a 500 mg tablet? Number of molecules = 1.99 x 1021 b) One trillion = 1 x 1012 Number of molecules accepted by each cell = 1.99 x 108 molecules Ratio is 2 C : 5.5 H : 1 N, so multiply by 2 to get whole numbers; 4 C : 11 H : 2 N Empirical formula is C4H11N2 (Molar Mass = 87.148 gmol―1) As molar mass of empirical formula is the same as the molar mass of molecular formula; the molecular formula is C4H11N2. How many moles of O2 are needed to form 1 mole of MgSO4? Mass of MgSO4 = 0.0830 moles x 120.38 gmol―1 = 9.99 g Use mass percent of H to determine x. x = 3 Then convert to moles and find mole fraction. X Ne = 0.664 Carbon is 1.11% 13C, moles 13C = moles Carbon x 0.0111 = 0.333 moles General Review ‐ 3 18) What is the Average Atomic Mass of Boron (from Periodic Table)? 19) What is the % by mass of Fe in Fe2O3? Any sample of Fe2O3 will be that % of Fe. A 65.2 g sample of Fe2O3 will contain 45.6 g Fe Use this in a weighted average. Let Fraction 10B = x, Fraction 11B = 1 ‐ x Percent Abundance of 10B = 20.0 % Topic B: Solutions 1) 2) 3) 4) 5) Use the definition of (p. B‐2) Mass of NaCl = 0.350 g ppm of KCl = 200 ppm, so ppm of K+ = 200 x % by mass of K+ in KCl = 105 ppm 37% by mass HCl means that 100 g of HCl solution contains 37 g of HCl. The density of the solution is the link between the mass of the solution (100 g) and the volume of the solution (which is required to determine Molarity). Molarity of HCl = 12.08 M What is the definition of “Molarity”? Molarity = 0.0358 molL―1 (or M) Use this molarity to calculate the volume required to provide 0.560 moles of NaOH: Volume soln = 15.64 L Moles of solute do not change on dilution; moles = Molarity x Volume (in L) The relationship M1V1 = M2V2 may be used here; as the moles of solute must remain constant M2 = 0.375 M General Review ‐ 4 Topic C: Stoichiometry 1) If S8 was the Limiting Reagent, then Cl2 must be present in excess. Use the balanced equation and the limiting reagent to determine how much Cl2 was reacted. Then find the amount in excess. Mass of Cl2 in excess = 72.3 g 2) 3) 4) 5) 6) General Review ‐ 5 Be sure the equation is balanced. Find moles of each species and determine Limiting Reagent. (S8 is the LR; 0.2106 moles) Use the balanced equation to relate the moles of limiting reagent to moles of what you are trying to find. (See below) From Equation: 1 mole S8 produces 4 moles S2Cl2 0.2106 moles S8 produces x moles S2Cl2 x = 0.8424 moles S2Cl2 Mass of S2Cl2 = 113.8 g What does percent yield mean? (p. C‐2) Use same approach as in 1) to determine theoretical yield (but there is no limiting reagent). 120 g is the actual yield. Percentage Yield = 75.4 % Same approach as in 1) and 2) a) Mass N2O5 = 9.17 g (This is the theoretical yield.) b) Percentage Yield = 21.4 % First, write a balanced equation. Same approach as in 1) Mass CO2 = 16.5 g Mass Cu = 133 g If the mixture is 55.0% CaCl2 by mass, what % is CaO? What is the mass of CaO in 80.0 g of mixture? Mass CaCO3 = 64.3 g Topic D: Reaction Types 1) 2) Use approach outlined in Example D.1 on p. D‐2 3) 4) What species accounts for the difference in mass between X and X2O3? Find moles of O and then find moles of X in X2O3 using the Ratio of O : X in X2O3 is 3:2 Determine molar mass of X; 27.00 gmol―1 X is Al 5) 6) 7) 8) First determine the molarity of the base, then the [OH―] and find pOH; convert to pH a) [NaOH] = 0.583 M = [OH―] pOH = ‐log [OH―] = 0.234 pH = 14 – 0.234 = 13.766 b) [Ba(OH)2] = 0.180 M *As each Ba(OH)2 contains 2 OH―, [OH―] = 2 x 0.180 = 0.360 M pH = 14 – 0.444 = 13.566 General Review ‐ 6 C2H6O + 3 O2 2 CO2 + 3 H2O Moles H2O formed = 0.03256 moles a) Empirical formula is C3H2O (Molar Mass = 54.046 gmol―1) b) Molecular formula is C9H6O3 Same approach as in 2) a) Empirical formula is C2H3 (Molar Mass = 27.044 gmol―1) b) Molecular formula is C4H6 2 NaCl (aq) + Pb(NO3)2 (aq) PbCl2 (s) + 2 NaNO3 (aq) (NH4)2S (aq) + 2 AgNO3 (aq) Ag2S (s) + 2 NH4NO3 (aq) First determine the molarity of the acid, then the [H+] and find pH. a) [HCl] = 0.0625 M = [H+] pH = ‐ log [H+] = 1.204 b) [H2SO4] = 0.400 M *As each H2SO4 contains 2 H+, [H+] = 2 x 0.400 = 0.800 M pH = 0.097 pOH = 0.444, 9) pH = 2.50; [H+] = 10―pH = 10―2.50 = 0.00316 M Convert molL―1 to gL―1 using the molar mass: Concentration in gL―1 = 0.115 gL―1 10) There are two sources of NO3―; find moles of each 12) 13) Same approach as in 12). Be sure to write the balanced equation for each reaction. a) pH = 12.574 b) pH = 12.95 c) pH = 2.365 d) pH = 11.699 Moles NaNO3 = 0.01600 moles = moles NO3― Moles Mg(NO3)2 = 0.006998 moles (Each Mg(NO3)2 contains 2 NO3―, therefore moles NO3― = 2 x 0.006998 = 0.01400 moles) Total moles NO3― = 0.01600 + 0.01400 = 0.03000 moles [NO3―] = 0.0200 M [OH―] = 10―pOH = 10―1.75 = 0.01778 M Each Ca(OH)2 contains 2 OH― and 1 Ca2+ [Ca2+] = [OH―] ÷ 2 = 0.01778 ÷ 2 = 0.00889 M H2SO4 + 2 NaOH 2 H2O + Na2SO4 Find moles of each species and find limiting reagent. (It is NaOH.) Determine amount of H2SO4 in excess; excess strong species controls the pH. To find pH, we need [H+], which can be determined from [H2SO4]xs = 0.1375 M (Remember to use total volume when calculating this molarity!) Each H2SO4 contains 2 H+, so [H+] = 0.1375 M x 2 = 0.275 M pH = 0.561 11) If pH = 12.25, the solution is basic, so convert to pOH. pOH = 14.00 – 12.25 = 1.75 14) Write a balanced equation for the reaction. In a complete reaction, both reagents are completely reacted, so there is no limiting reagent. Determine moles NaOH and relate to moles of H2SO4 reacted using balance equation: moles of H2SO4 reacted = moles NaOH ÷ 2 [H2SO4] = 0.06973 M General Review ‐ 7 15) General formula for a diprotic acid is H2A; Reaction is: H2A + 2 NaOH 2 H2O + Na2A This is a complete reaction; there is no limiting reagent and neither reactant is present in excess. Find moles of NaOH and determine moles of H2A from balanced equation. Moles NaOH = 0.100 M x 0.0250 L = 0.00250 moles Molar Mass of H2A = 124.8 gmol―1 16) An oxidizing agent is itself reduced; therefore it gains electrons. Those electrons must come from the species which is being oxidized. 17) Begin by assigning oxidation numbers to the elements that are defined in the Rules (p. D‐11, 12) a) Correct; a species which is being reduced accepts electrons from another species. b) Incorrect; if the species is being reduced, it does not donate electrons to any other species. c) Incorrect; an element in an intermediate oxidation state (somewhere between the highest and the lowest) is capable of accepting electrons and therefore changing to a lower oxidation state. d) Incorrect; an element in its lowest oxidation state cannot accept any more electrons; it can only donate them and thereby undergo oxidation. e) Incorrect; an element in its highest oxidation state may accept electrons and thereby undergo reduction. Only a) is correct. H3PO4 H2S NaBr 3 H @ +1 = +3 4 O @ 2 = 8 1 P @ x = x Overall Charge = 0 = +3 8 + x S is = 2 Br is = 1 x = +5 MnO4 Mn is = 7 S is = 6 H is = 1 SF6 MgH2 General Review ‐ 8 18) Determine the oxidation number for Oxygen in each compound using the Rules (p. D – 11, 12) 19) Determine the oxidation number for Phosphorus in each compound using the Rules (p. D–11, 12) 20) a) The Li has been oxidized, losing 1 electron. Each H in H2 has been reduced, gaining 1 electron. b) The Mg has been oxidized, losing 2 electrons. Each H in HCl has been reduced, gaining 1 electron. The Cl in HCl has been neither oxidized or reduced. The phosphorus in P4O6 has an oxidation number of +3, all the compounds have phosphorus in the +5 oxidation state. (Answer d) a) PCl5: b H3PO4: c) Na4P2O7: d) P4O6: e NaH2PO4: +5 +5 +5 +3 +5 a) H2O: 2 b) OF2: 2 c O2: 0 e) Cl2O: 2 * correct * incorrect * correct * correct d) H2O2 : 1 * correct General Review ‐ 9 ...
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This note was uploaded on 04/12/2011 for the course BIO 1222 taught by Professor Maxwell during the Spring '08 term at UWO.
- Spring '08