Question The formula mass FM of barium phosphite

Question The formula mass (FM) of barium phosphite, Ba 3(PO 3)2 is a) 184. 30 amu. b) 216. 30 amu. c) 295. 27 amu. d) 190. 37 amu. e) 569. 93 amu. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 1

Answer e) 569. 93 amu. The formula mass of a substance is the sum of the atomic weights of all of the atoms in the compound. Be careful to use all of the subscripts when counting the atoms in Ba 3(PO 3)2. Ba: 3 × 137. 33 amu = 411. 99 amu P: 2 × 30. 97 amu = 61. 94 amu O: 2 × 3 × 16. 00 amu = 96. 00 amu 569. 93 amu Copyright © Houghton Mifflin Company. All rights reserved. 3 | 2

Question You have a sample of zinc (Zn) and a sample of aluminum (Al). You have an equal number of atoms in each sample. Which of the following statements concerning the masses of the samples is true? a) The mass of the zinc sample is more than twice as great as the mass of the aluminum sample. b) The mass of the zinc sample is more than the mass of the aluminum sample, but it is not twice as great. c) The mass of the aluminum sample is more than twice as great as the mass of the zinc sample. d) The mass of the aluminum sample is more than the mass of the zinc sample, but it is not twice as great. e) The masses of each sample are equal. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 3

Answer a) The mass of the zinc sample is more than twice as great as the mass of the aluminum sample. Since zinc has an atomic mass more than twice that of aluminum, the sample of zinc with the same number of atoms will have more than twice the mass of the sample of aluminum. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 4

Question Which of the following is the most accurate description of a mole? a) The mass of carbon in a measured sample of carbon b) The number of atoms in any given mass of an element c) The number of sodium ions in 58. 44 g of sodium chloride d) At least two of the above are accurate descriptions of a mole. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 5

Answer c) The number of sodium ions in 58. 44 g of sodium chloride The molar mass of sodium chloride (Na. Cl) is 58. 44 g. Since there is 1 atom of sodium for each unit of sodium chloride, 58. 44 g of Na. Cl will contain 1 mol of sodium ions. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 6

Question Which of the following is closest to the average mass of one atom of copper? a) 63. 55 g b) 52. 00 g c) 58. 93 g d) 65. 38 g e) 1. 055 × 10− 22 g Copyright © Houghton Mifflin Company. All rights reserved. 3 | 7

Answer e) 1. 055 × 10− 22 g 1 atom Cu 1 mole Cu 63. 55 g Cu 6. 022 x 1023 atomc Cu 1 mole Cu = 1. 055 × 10− 22 g Also, one atom must have a very low mass, and choice e) is the only choice with the correct order or magnitude. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 8

Question Which of the following 100. 0 -g samples contains the greatest number of atoms? a) Magnesium b) Zinc c) Silver d) Calcium e) All samples contains the same number of atoms. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 9

Answer a) Magnesium Divide 100. 0 g by each of the atomic masses. Or, since Mg has the lowest atomic mass, the number of atoms must be the greatest. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 10

Question For which of the following compounds does 1. 0 g represent 2. 27

Answer b) CO 2 The answer may be obtained by solving for molar mass: (1. 0 g)(1 mol/molar mass) = 2. 27 10− 2 mol. The molar mass obtained (44 g/mol) is closest to the answer CO 2. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 12

Question The mass of 0. 822 mol of a diatomic molecule is 131. 3

Answer c) Br 2 The answer may be obtained by solving for molar mass: (131. 3 g)(1 mol/molar mass) = 0. 822 mol. The molar mass obtained (160. g/mol) is closest to the answer Br 2 (2 × 79. 90 g/mol). Copyright © Houghton Mifflin Company. All rights reserved. 3 | 14

Question Which of the following 100. 0 -g samples contains the greatest number of oxygen atoms? a) H 2 O b) N 2 O c) C 3 H 6 O 2 d) CO 2 e) All of the sample have the same number of oxygen atoms. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 15

Answer a) H 2 O Divide 100. 0 g by the molar mass of each compound and multiply by the number of oxygen atoms in the formula. The greater the number of moles of O atoms, the greater the number of O atoms. H 2 O: N 2 O: C 3 H 6 O 2: CO 2: Copyright © Houghton Mifflin Company. All rights reserved. 3 | 16

Question Which of the following 100. 0 -g samples contains the highest percentage oxygen by mass? a) H 2 O b) N 2 O c) C 3 H 6 O 2 d) CO 2 e) All of the sample have the same number of oxygen atoms. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 17

Answer a) H 2 O The mass of the samples does not matter, nor does it matter that the samples have the same mass. H 2 O: [(16. 00 g)/(2. 016 + 16. 00) g] × 100% = 88. 81% O N 2 O: [(16. 00 g)/(28. 02 + 16. 00) g] × 100% = 36. 35% O C 3 H 6 O 2: [(32. 00 g)/(36. 03 + 6. 048 + 32. 00) g] × 100% = 43. 20% O CO 2: [(32. 00 g)/(12. 01 + 32. 00) g] × 100% = 72. 71% O Copyright © Houghton Mifflin Company. All rights reserved. 3 | 18

Question In combustion analysis, the mass of oxygen, in a sample containing only carbon, hydrogen, and oxygen a) is determined using the mass of carbon dioxide, CO 2, collected in an absorber and converting it to mass of oxygen. b) is determined using the mass of water, H 2 O, collected in an absorber and converting it to mass of oxygen. c) is determined by difference, using the initial sample weight and subtracting the amounts of carbon and hydrogen collected by the absorbers. d) is determined by adding the amounts of oxygen collected by the carbon dioxide and water absorbers. e) cannot be determined in a combustion analysis. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 19

Answer c) is determined by difference, using the initial sample weight and subtracting the amounts of carbon and hydrogen collected by the absorbers. Since the compound contains only carbon, hydrogen, and oxygen, we can calculate the mass of oxygen by difference. The carbon weight in the sample is calculated using the mass of carbon dioxide, CO 2, collected in an absorber. The hydrogen weight in the sample is calculated using the mass of water, H 2 O, collected in an absorber. So the mass of O = sample weight − (carbon weight + hydrogen weight) Copyright © Houghton Mifflin Company. All rights reserved. 3 | 20

Question The empirical formula of styrene is CH; its molar mass is 104. 1. What is the molecular formula of styrene? a) b) c) d) C 2 H 4 C 8 H 8 C 10 H 10 C 6 H 6 Copyright © Houghton Mifflin Company. All rights reserved. 3 | 21

Answer b) C 8 H 8 Since the molecular formula is (CH)n, the n may be obtained from the equation (n)(12. 0 g/mol + 1. 0 g/mol) = 104. 1 g/mol, and so n = 8. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 22

Question Which of the following compounds have a different empirical formula than the other compounds? a) b) c) d) Glucose, C 6 H 12 O 6 Sucrose, C 12 H 22 O 11 Fructose, C 6 H 12 O 6 Ribose, C 5 H 10 O 5 Copyright © Houghton Mifflin Company. All rights reserved. 3 | 23

Answer b) Sucrose, C 12 H 22 O 11 All simple carbohydrates have a molecular formula of Cn. H 2 n. On, and an empirical formula CH 2 O. For glucose n = 6, for fructose n = 6, and for ribose n = 5. Sucrose is a complex carbohydrate, with an empirical formula of C 12 H 22 O 11. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 24

Question In the reaction 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + 6 H 2 O (g) How many grams of H 2 O are generated when 1. 54 moles of NH 3 react ? a) 2. 31 g H 2 O b) 41. 6 g H 2 O c) 18. 5 g H 2 O d) 39. 3 g H 2 O e) 6. 51 g H 2 O Copyright © Houghton Mifflin Company. All rights reserved. 3 | 25

Answer b) 41. 6 g H 2 O Using the balanced chemical equation, 4

Question In the reaction 2 A + 3 B C 4. 0 moles of A react with 4. 0 moles of B. Which of the following choices best answers the question: “Which reactant is limiting? ” a) Neither is limiting because equal amounts (4. 0 mol) of each reactant are used. b) A is limiting because two is smaller than three (the coefficients in the balanced equation). c) A is limiting because there are 2 moles but 4. 0 moles are needed. d) B is limiting because three is larger than two (the coefficients in the balanced equation). e) B is limiting because there are 4. 0 moles but 6. 0 moles are needed. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 27

Answer e) B is limiting because there are 4. 0 moles but 6. 0 moles are needed. Using the balanced chemical equation, 4. 0 mol of A would require (4. 0 mol A)(3 mol B/2 mol A) or 6. 0 mol of B, but there are only 4. 0 mol of B, so B is the limiting reactant. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 28

Question The limiting reactant in a reaction a) has the lowest coefficient in a balanced equation. b) is the reactant for which you have the fewest number of moles. c) has the lowest ratio of moles available/coefficient in the balanced equation. d) has the lowest ratio of coefficient in the balanced equation/moles available. e) none of these Copyright © Houghton Mifflin Company. All rights reserved. 3 | 29

Answer c) has the lowest ratio of moles available/coefficient in the balanced equation. Using the moles of materials available and the balanced chemical equation, the ratio of the moles of a material available to its corresponding coefficient in the balanced equation gives the quantity available per quantity needed. By comparing these ratios, the smallest will be the limiting reactant. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 30

Question Consider the following balanced equation: A + 5 B 3 C + 4 D Which of the following choices best answers the question, When equal masses of A and B are reacted, which is limiting? a) If the molar mass of A is greater than the molar mass of B, then A must be limiting. b) If the molar mass of A is less than the molar mass of B, then A must be limiting. c) If the molar mass of A is greater than the molar mass of B, then B must be limiting. d) If the molar mass of A is less than the molar mass of B, then B must be limiting. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 31

Answer d) If the molar mass of A is less than the molar mass of B, then B must be limiting. Since the masses are equal, if the molar mass of A is less than that of B, the number if moles of A will be greater than the number of moles of B. Since more moles of B are required (a 1: 5 ratio of A: B according to the balanced equation), B must be limiting. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 32

Question Which of the following reaction mixtures would produce the greatest amount of product according to the following chemical equation? 2 H 2 + O 2 2 H 2 O a) 2 mol H 2 and 2 mol O 2 b) 2 mol H 2 and 3 mol O 2 c) 2 mol H 2 and 1 mol O 2 d) 3 mol H 2 and 1 mol O 2 e) Each of the choices above would produce the same amount of product. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 33

Answer e) Each of the choices above would produce the same amount of product. Each will produce 2 mol of H 2 O: For choice a, H 2 is limiting; 2. 0 mol H 2 will produce 2. 0 mol H 2 O. For choice b, H 2 is limiting; 2. 0 mol H 2 will produce 2. 0 mol H 2 O. For choice c, neither is limiting; 2. 0 mol H 2 will produce 2. 0 mol H 2 O; 1. 0 mol O 2 will produce 2. 0 mol H 2 O. For choice d, O 2 is limiting; 1. 0 mol O 2 will produce 2. 0 mol H 2 O. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 34

Question Methyl salicylate (oil of wintergreen) can be prepared in the lab by heating salicylic acid, C 7 H 6 O 3, with methanol, CH 3 OH, according to the following reaction: C 7 H 6 O 3 + CH 3 OH C 8 H 8 O 3 + H 2 O If the student starts with 2. 53 g of salicylic acid an excess of methanol and collects 2. 62 g of methyl salicylate, what is the student’s percentage yield? (Hint: molecular weight [MW] C 7 H 6 O 3= 138. 12 g/mol; MW CH 3 OH = 32. 04 g/mol; MW C 8 H 8 O 3 = 152. 14 g/mol) a) 94. 0% b) 106% c) 96. 7% d) 87. 7% Copyright © Houghton Mifflin Company. All rights reserved. 3 | 35

Answer a) 94. 0% The theoretical yield is So the percentage yield is Note

Question A student adds 85. 92 g of KCl to a 500. 00 -m. L volumetric flask. He then adds water, dissolves the solids, and dilutes the flask to the mark. What is the molar concentration of KCl in the flask? a) 172 M KCl b) 1. 72 M KCl c) 4. 610 M KCl d) 2. 305 M KCl Copyright © Houghton Mifflin Company. All rights reserved. 3 | 37

Answer d) 2. 305 M KCl Molarity is defined as moles of solute per liter of solution. 85. 92 g KCl 1 mole KCl 500. 0 m. L 74. 55 g KCl Copyright © Houghton Mifflin Company. All rights reserved. 1000 m. L 1 L = 2. 305 M KCl 3 | 38

Question Consider five solutions that each have the same mass of solute in 100. 0 m. L of solution. Which has the highest concentration as measured in molarity? a) KCl b) Na. Cl c) Na 2 SO 4 d) Na. F e) Ca. Cl 2 Copyright © Houghton Mifflin Company. All rights reserved. 3 | 39

Answer d) Na. F The highest concentration as measured in molarity for the same volume of solution will be determined by the solute with the smallest molar mass. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 40

Question Which of the following solutions contains the greatest number of ions? a) 400. 0 m. L of 0. 10 M Na. Cl b) 300. 0 m. L of 0. 10 M Ca. Cl 2 c) 200. 0 m. L of 0. 10 M Fe. Cl 3 d) 800. 0 m. L of 0. 10 M sucrose Copyright © Houghton Mifflin Company. All rights reserved. 3 | 41

Answer b) 300. 0 m. L of 0. 10 M Ca. Cl 2 The concentrations are all the same, so we must look at the volumes and number of ions per compound. While Fe. Cl 3 dissolves into four ions, and Ca. Cl 2 dissolves into only three, the greater volume of the Ca. Cl 2 solution more than makes up for this. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 42

Question Which of the following solutions contains the greatest total ion concentration? a) One mole of potassium chloride dissolved in 1. 0 L of solution b) One mole of iron(II) nitrate dissolved in 1. 0 L of solution c) One mole of potassium hydroxide dissolved in 1. 0 L of solution d) One mole of sodium phosphate dissolved in 1. 0 L of solution Copyright © Houghton Mifflin Company. All rights reserved. 3 | 43

Answer d) One mole of sodium phosphate dissolved in 1. 0 L of solution All of the possible answers are strong electrolytes, so the one that dissociates into the most ions (KCl into 2 mol ions, Fe(NO 3)2 into 3 -mol ions, KOH into 2 -mol ions, Na 3 PO 4 into 4 -mol ions) is the answer. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 44

Question You have separate aqueous solutions of Na. OH and Ca(OH)2 with the same concentrations. You wish to neutralize an aqueous solution of HCl. Which basic solution would require more volume to neutralize the acid? a) The Na. OH solution b) The Ca(OH)2 solution c) More information is needed to answer this question. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 45

Answer a) The Na. OH solution The basic solution requiring the greater volume to neutralize the acid is the one with the fewer hydroxide ions available to react with the hydrogen ions. 2 HCl (aq) + Ca(OH)2 (aq) Ca. Cl 2 (aq) + H 2 O (l) HCl (aq) + Na. OH (aq) Na. Cl (aq) + H 2 O (l) Copyright © Houghton Mifflin Company. All rights reserved. 3 | 46

Question A student needs to prepare 250. 0 m. L of a 0. 300 M Ba. Cl 2 and has a stock solution of 1. 50 M Ba. Cl 2. What volume of the concentrated stock solution does the student need to prepare the dilute solution? a) 50. 0 m. L b) 1250 m. L c) 250 m. L d) 112. 5 m. L e) 200. m. L Copyright © Houghton Mifflin Company. All rights reserved. 3 | 47

Answer a) 50. 0 m. L Because the moles of solute do not change during dilution, Mi × Vi = Mf × Vf. This equation can be rearranged to the concentration of the dilute solution is one-fifth of the concentrated solutions, so the volume needed is onefifth of the dilute solution volume. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 48