Test Bank For General Chemistry 11th Edition By Ebbing

1. The molecular formula of a particular solid is C8H4O6. Its molecular mass is a. 288 amu. b. 100 amu. c. 296 amu. d. 196 amu. e. 170 amu. ANSWER: d POINTS: 1 DIFFICULTY: easy REFERENCES: 3.1 LEARNING OBJECTIVES: GENE.EBBI.13.67 - Calculate the formula mass from a formula. (Example 3.1) TOPICS: mass and moles of substance stoichiometry KEYWORDS: molecular mass OTHER: general chemistry 2. The hydrocarbon octane has the structural formula CH3(CH2)6CH3. What is the molecular mass of this hydrocarbon? a. 114.2 amu. b. 0.008754 amu. c. 100.1 amu. d. 0. amu. e. 124.2 amu. ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.1 LEARNING OBJECTIVES: GENE.EBBI.13.68 - Calculate the formula mass from a formula. TOPICS: mass and moles of substance stoichiometry 3. A single molecule of polystryene has the repeating unit -[CH2CH(C6H5)]n-, where n is the number of repeating units. What is the value of n if the molecular mass of a single polymer chain is amu? a. units. b. units. c. units. d. units. e. units. ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.1 LEARNING OBJECTIVES: GENE.EBBI.13.68 - Calculate the formula mass from a formula. TOPICS: stoichiometry mass and moles of substance 4. The formula mass of aluminum oxalate, Al2(C2O4)3, is a. 143 amu. b. 318 amu. c. 290 amu. d. 204 amu. e. 197 amu. ANSWER: b POINTS: 1 DIFFICULTY: easy REFERENCES: 3.1 LEARNING OBJECTIVES: GENE.EBBI.13.67 - Calculate the formula mass from a formula. (Example 3.1) TOPICS: stoichiometry mass and moles of substance KEYWORDS: formula mass OTHER: general chemistry 5. The formula mass of zinc acetate trihydrate, Zn(CH3COO)2 • 3H2O, is a. 356 amu. b. 184 amu. c. 292 amu. d. 238 amu. e. 156 amu. ANSWER: d POINTS: 1 DIFFICULTY: easy REFERENCES: 3.1 LEARNING OBJECTIVES: GENE.EBBI.13.67 - Calculate the formula mass from a formula. (Example 3.1) TOPICS: stoichiometry mass and moles of substance KEYWORDS: formula mass OTHER: general chemistry 6. The fully hydrated form of sodium sulfate is the decahydrate, Na2SO4 • 10H2O. When heated the hydrated salt loses water. How many water molecules are found per formula unit in a partially dehydrated sample of sodium sulfate with a formula mass of 197 amu (i.e. find n for Na2SO4 • nH2O)? a. 3 waters. b. 5 waters. c. 7 waters. d. 9 waters. e. 4 waters. ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.1 LEARNING OBJECTIVES: GENE.EBBI.13.68 - Calculate the formula mass from a formula. TOPICS: stoichiometry mass and moles of substance 7. What is the molecular mass of cycloheptane, C7H14? a. 13.02 amu b. 1191.19 amu c. 85.08 amu d. 98.19 amu e. 26.12 amu ANSWER: d POINTS: 1 DIFFICULTY: easy REFERENCES: 3.1 LEARNING OBJECTIVES: GENE.EBBI.13.67 - Calculate the formula mass from a formula. (Example 3.1) TOPICS: stoichiometry mass and moles of substance KEYWORDS: molecular mass OTHER: general chemistry 8. What is the formula mass of barium phosphate, Ba3(PO4)2? a. 506.95 amu b. 570.95 amu c. 232.30 amu d. 1013.90 amu e. 601.92 amu ANSWER: e POINTS: 1 DIFFICULTY: easy REFERENCES: 3.1 LEARNING OBJECTIVES: GENE.EBBI.13.67 - Calculate the formula mass from a formula. (Example 3.1) TOPICS: stoichiometry mass and moles of substance KEYWORDS: formula mass OTHER: general chemistry 9. What is the molecular mass of the hydrocarbon styrene (shown in the figure)? a. 104.1 amu. b. 91.1 amu. c. 103.1 amu. d. 13.0 amu. e. 78.1 amu. ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.1 LEARNING OBJECTIVES: GENE.EBBI.13.69 - Calculate the formula mass from molecular models. (Example 3.2) TOPICS: mass and moles of substance stoichiometry KEYWORDS: molecular mass OTHER: general chemistry 10. What is the molar mass of ammonium sulfite, (NH4)2SO3? a. 98 g/mol b. 116 g/mol c. 55 g/mol d. 180 g/mol e. 84 g/mol ANSWER: b POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.70 - Understand how the molar mass is related to the formula weight of a substance. TOPICS: mass and moles of substance stoichiometry KEYWORDS: formula weight OTHER: general chemistry 11. What is the molar mass of zinc sulfate hexahydrate, ZnSO4 • 6H2O? a. 180. g/mol b. 270. g/mol c. 366 g/mol d. 162 g/mol e. 564 g/mol ANSWER: b POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.70 - Understand how the molar mass is related to the formula weight of a substance. TOPICS: mass and moles of substance stoichiometry KEYWORDS: formula weight OTHER: general chemistry 12. Plastic wrap can be made from poly(vinylidene chloride). A single poly(vinylidene chloride) strand has the general formula -(CH2CHCl)n-, where n ranges from 10,000 to 100,000. What is the molar mass of a single poly(vinylidene chloride) molecule containing repeating units? a. g/mol b. g/mol c. g/mol d. g/mol e. g/mol ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.70 - Understand how the molar mass is related to the formula weight of a substance. TOPICS: stoichiometry mass and moles of substance 13. The dicarboxylic acid potassium hydrogen pthalate (shown in the figure) is used to standardize solutions of strong base. What is the molar mass of this compound? a. 204.2 g/mol b. 192.2 g/mol c. 248.9 g/mol d. 71.08 g/mol e. 172.2 g/mol ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.70 - Understand how the molar mass is related to the formula weight of a substance. TOPICS: stoichiometry mass and moles of substance 14. What is the molar mass of the solid C6H8N2O4? a. 144 g/mol b. 108 g/mol c. 172 g/mol d. 90 g/mol e. 164 g/mol ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.70 - Understand how the molar mass is related to the formula weight of a substance. TOPICS: stoichiometry mass and moles of substance KEYWORDS: molecular mass OTHER: general chemistry 15. A 1.74 g sample of an element contains 7.887 ×1021atoms. What is the element symbol? a. Cs b. I c. In d. Sb e. Cd ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.70 - Understand how the molar mass is related to the formula weight of a substance. TOPICS: stoichiometry | determining chemical formulas 16. Monodisperse polyacrylonitrile contains molecules with the general formula -(CH2CHCN)n-, where n is typically greater than 10,000. Given that a sample of monodisperse polyacrilonitrile weighs 676.8 g and contains molecules of -(CH2CHCN)n-, calculate n. a. b. c. d. e. ANSWER: a POINTS: 1 DIFFICULTY: difficult REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.70 - Understand how the molar mass is related to the formula weight of a substance. TOPICS: stoichiometry mass and moles of substance 17. Monodisperse polyacrylonitrile contains molecules with the general formula -(CH2CHCN)n-, where n is typically greater than 10,000. Given that a sample of monodisperse polyacrylonitrile weighs 665.1 g and contains molecules of -(CH2CHCN)n-, what is the molar mass of the polymer? a. g/mol b. g/mol c. g/mol d. g/mol e. g/mol ANSWER: e POINTS: 1 DIFFICULTY: difficult REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.70 - Understand how the molar mass is related to the formula weight of a substance. TOPICS: stoichiometry mass and moles of substance 18. An atom of an element weighs 3.38× 10–23 g. What is the atomic mass of this element in atomic mass units? a. 19.6 amu b. 18.6 amu c. 17.9 amu d. 20.3 amu e. 17.2 amu ANSWER: d POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.71 - Calculate the mass of atoms and molecules. (Example 3.3) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 19. What is the mass in grams of one propene, C3H6, molecule? a. 6.99× 10–23 g b. 2.53 × 1025 g c. 44.0 g d. 42.0 g e. 1.99 × 10–23 g ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.71 - Calculate the mass of atoms and molecules. (Example 3.3) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 20. What is the mass of carbon in grams found in one molecule of the compound C7H8O4? a. g b. g c. g d. g e. 156.000000 g ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.71 - Calculate the mass of atoms and molecules. (Example 3.3) TOPICS: stoichiometry mass and moles of substance 21. Which of the following compounds contains the largest number of atoms? a. 4.0 mol of H2S b. 3.0 mol of NO3 c. 2.0 mol of Na2SO4 d. 5.0 mol of HBr e. 1.0 mol of CH3COOH ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.72 - Perform calculations using the mole. TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 22. How many atoms of carbon are there in 0.37 mol of procaine, C13H20N2O2, a “pain killer” used by dentists? a. 4.8 × 1023 b. 3.1 × 1024 c. 2.9 × 1024 d. 4.5 × 1023 e. 3.3 × 1024 ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.72 - Perform calculations using the mole. TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 23. A sample of ammonium phosphate, (NH4)3PO4, contains 0.104 mol of nitrogen atoms. The number of moles of oxygen atoms in the sample is a. 0.139 mol. b. 0.0104 mol. c. 0.41 mol. d. 0.419 mol. e. 4.6 mol. ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.72 - Perform calculations using the mole. TOPICS: stoichiometry mass and moles of substance 24. A sample of thallium(III) peroxide, Tl2(O2)3, contains 2.45 mol of thallium(III) ions. The number of moles of peroxide ions in the sample is a. 3.67 mol. b. 1.63 mol. c. 2.45 mol. d. 7.35 mol. e. 4.9361 mol. ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.72 - Perform calculations using the mole. TOPICS: stoichiometry mass and moles of substance 25. Sorbose, C6H12O6, is used in making vitamin C. A sorbose sample containing 36.0 g of carbon atoms also contains ____ g of hydrogen atoms. a. 12 b. 6.04 c. 4.32 × 102 d. 36 e. 2.99 ANSWER: b POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.72 - Perform calculations using the mole. TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 26. Consider the following three samples. A. A sample containing 180 g glucose (C6H12O6) B. A sample containing 90 g glucose and 90 g mannose (C6H12O6) C. A sample containing 180 g mannose Which statement is correct? a. All three samples have the same number of oxygen atoms. b. Both samples A and C have the same number of oxygen atoms, but more than in sample B. c. Sample B has more oxygen atoms than sample A or sample C. d. Sample C has more oxygen atoms than sample A or sample B e. Sample A has more oxygen atoms than sample B or sample C. ANSWER: a POINTS: 1 DIFFICULTY: difficult REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.72 - Perform calculations using the mole. TOPICS: stoichiometry determining chemical formulas KEYWORDS: mole | mole calculations OTHER: general chemistry 27. Styrene's empirical formula is CH. What mass of styrene contains 2.95 × 1021 atoms of hydrogen? The molar mass of styrene is 104 g/mol. a. 0.0391 g b. 0.0636 g c. 0.509 g d. 0.0587 g e. 0.00489 g ANSWER: b POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.73 - Convert from moles of substance to grams of substance. (Example 3.4) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 28. What is the mass of oxygen atoms in 0.380 mol Fe(CO)5? a. 21.2 g b. 74.4 g c. 30.4 g d. 6.08 g e. 22.8 g ANSWER: c POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.73 - Convert from moles of substance to grams of substance. (Example 3.4) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 29. What is the mass in grams of 0.754 mol of glucose, C6H12O6? a. 0.00419 g b. 72.4 g c. 136 g d. 22.6 g e. 239 g ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.73 - Convert from moles of substance to grams of substance. (Example 3.4) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 30. Which one of the following samples has the greatest mass? a. 0.37 mol of camphor, C10H16O b. 4.2 mol of ammonia, NH3 c. 9.3 mol of krypton, Kr d. 4.0 mol of iodine vapor, I2 e. 1.6 mol of formaldehyde, CH2O ANSWER: d POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.73 - Convert from moles of substance to grams of substance. (Example 3.4) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 31. Which of the following contains the greatest mass of oxygen atoms? a. 0.5 mol CoSO4 • 7H2O b. 2.1 mol KHSO4 c. 1.1 mol K2Cr2O7 d. 2.1 mol H2O2 e. 2.1 mol Na2S2O3 ANSWER: b POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.73 - Convert from moles of substance to grams of substance. (Example 3.4) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 32. Calculate the number of moles of bromine present in 20.5 mL of Br2(l), whose density is 3.12 g/mL. a. 2.50 mol b. 0.257 mol c. 0.801 mol d. 0.128 mol e. 0.400 mol ANSWER: e POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.74 - Convert from grams of substance to moles of substance. (Example 3.5) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 33. How many moles of hexachlorobenzene, C6Cl6, are in 5.44 g of C6Cl6? a. 0.0254 mol b. 1.55 × 103 mol c. 0.0755 mol d. 0.019 mol e. 0.0394 mol ANSWER: d POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.74 - Convert from grams of substance to moles of substance. (Example 3.5) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 34. How many moles of iron atoms are contained in 4.16 g of iron? a. 232 mol b. 0.0548 mol c. 0.0745 mol d. 0.116 mol e. 0.160 mol ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.74 - Convert from grams of substance to moles of substance. (Example 3.5) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 35. How many moles of pentane, C5H12, are contained in a 31-g sample? a. 0.52 mol b. 0.43 mol c. 0.74 mol d. 3.9 mol e. 3.1 mol ANSWER: b POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.74 - Convert from grams of substance to moles of substance. (Example 3.5) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 36. Sodium cyclamate, C6H11NHSO3Na, was used at one time as an artificial sweetener. C6H11NHSO3Na has a molecular mass of 201.2 g/mol. How many moles of sodium cyclamate are contained in a 56.0-g sample? a. 0.175 mol b. 0.160 mol c. 0.278 mol d. 0.255 mol e. %1,0f mol ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.74 - Convert from grams of substance to moles of substance. (Example 3.5) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 37. How many moles of silver are contained in 7.00 kg of silver? a. 64.9 mol b. 64.9 × 101 mol c. 64.9 × 10–3 mol d. 64.9 × 103 mol e. 64.9 × 10–1 mol ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.74 - Convert from grams of substance to moles of substance. (Example 3.5) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 38. A 0.0106-mol sample of urea, NH2CONH2, contains a. 6.02 × 1023 molecules. b. 2.55 × 1022 molecules. c. 5.10 × 1022 atoms. d. 2.55 × 1023 atoms. e. 1.06 × 1024 atoms. ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 39. How many molecules are there in 104 g of pentylene glycol, HO(CH2)5OH? a. 1 b. (6.02 × 1023)/104 c. 104 d. 104 × (6.02 × 1023) e. 6.02 × 1023 ANSWER: e POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 40. How many molecules are there in 60 g of acetic acid, C2H4O2? a. (6.02 × 1023) / 60 b. 60 c. 6.02 × 1023 d. 30 e. 60 × (6.02 × 1023) ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 41. In 0.266 mol of trimellitic acid, C6H3(COOH)3, there are a. 1.60 × 1022 molecules. b. 6.41 × 1024 molecules. c. 4.80 × 1023 oxygen atoms. d. 1.44 × 1024 carbon atoms. e. 2.67 × 1023 hydrogen atoms. ANSWER: d POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 42. How many molecules are there in 1.54 kg of hydrazine, N2H4? a. 4.81 × 1023 b. 2.90 × 1025 c. 1.88 × 1022 d. 1.54 × 1026 e. 2.56 × 1021 ANSWER: b POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 43. How many molecules are there in 2.10 mg of mannose, C6H12O6, which is a sweet-tasting sugar that has a bitter aftertaste? a. 5.16 × 1021 b. 2.87 × 1021 c. 6.32 × 1018 d. 3.49 × 1024 e. 7.02 × 1018 ANSWER: e POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 44. How many atoms are present in 495 g of KPF6 (MM = 184.1 g/mol)? a. 2.38 × 1025 b. 1.62 × 1021 c. 2.69 × 1021 d. 1.21 × 1026 e. 1.29 × 1025 ANSWER: e POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 45. Styrene's empirical formula is CH. When it is heated to 200°C, it is converted into a polymer, polystyrene, which has excellent insulating properties. What mass of styrene contains 1.77 × 1021 molecules of styrene? The molar mass of styrene is 104 g/mol. a. 0.00293 g b. 0.305 g c. 0.587 g d. 0.0382 g e. 0.0235 g ANSWER: b POINTS: 1 DIFFICULTY: difficult REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 46. In 0.300 mol of dimethylhydrazine, (CH3)2N2H2, there are a. 1.81 × 1024 molecules. b. 2.51 × 1023 atoms. c. 1.81 × 1022 molecules. d. 2.17 × 1024 atoms. e. 1.08 × 1024 atoms. ANSWER: d POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 47. Which is a reasonable mass corresponding to 1026 molecules of a substance? a. 100 ng b. 100 g c. 100 mg d. 100 kg e. 100 µg ANSWER: d POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 48. How many aluminum atoms are there in 52 g of Al2S3? a. 4.2 × 1023 b. 1.6 × 1021 c. 2.1 × 1023 d. 1.1 × 1021 e. 6.3 × 1023 ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 49. Which one of the following contains 4.21 × 1024 atoms? a. 91 g C2H2 b. 168 g O2 c. 56 g CH4 d. 294 g N2 e. 28 g He ANSWER: e POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 50. A sample of 496 g of white phosphorus, P4, contains the same number of atoms as a. 192 g of ozone (O3). b. 56 g of nitrogen (N2). c. 92 g of sodium. d. 120 g of formaldehyde (CH2O). e. 128 g of oxygen (O2). ANSWER: d POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 51. The total number of oxygen atoms in 1.93 g of CaCO3 (MM = 100.0 g/mol) is a. 2.24 × 1023. b. 4.65 × 1022. c. 3.49 × 1022. d. 1.92 × 1023. e. 5.81 × 1022. ANSWER: c POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 52. A sample of 96 g of ozone, O3, contains the same number of atoms as a. 96 g of oxygen (O2). b. 8.06 g of hydrogen (H2). c. 76 g of fluorine (F2). d. 54 g of aluminum (Al). e. 117 g of nickel (Ni). ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 53. Which of the following samples contains the smallest number of molecules? a. 5.00 g of TNT, C7H5N3O6 b. 5.00 g of benzene, C6H6 c. 5.00 g of glucose, C6H12O6 d. 5.00 g of naphthalene, C10H8 e. 5.00 g of formaldehyde, CH2O ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 54. Which of the following samples contains the largest number of atoms? a. 1 g N2 b. 1 g Be c. 1 g Br2 d. 1 g P4 e. 1 g Si ANSWER: b POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 55. Which of the following samples contains the largest number of molecules? a. 10. g Ca b. 10. g Br2 c. 10. g Ar d. 10. g O2 e. 10. g S8 ANSWER: d POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.2 LEARNING OBJECTIVES: GENE.EBBI.13.75 - Calculate the number of molecules in a given mass of substance. (Example 3.6) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 56. In 1928, 1.0 g of rhenium, Re, was isolated from 660 kg of the ore molybenite. The percent by mass of this element in the molybenite was a. 0.66 %. b. 0.15 %. c. 3.5 × 10–4 %. d. 6.6 × 10–3 %. e. 1.5 × 10–4 %. ANSWER: e POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.76 - Calculate the percentage composition of the elements in a compound. (Example 3.7) TOPICS: stoichiometry determining chemical formulas KEYWORDS: mass percentage OTHER: general chemistry 57. An ore sample is found to contain 10.500000 g of cobalt and 87.3 g waste rock (gangue). What is the percent by mass of cobalt in the ore? a. 10.7 % b. 12.114 % c. 0.107 % d. 0.12228 % e. 1.2879 % ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.76 - Calculate the percentage composition of the elements in a compound. (Example 3.7) TOPICS: stoichiometry determining chemical formulas 58. An ore sample with a mass of 92.3 g is found to contain 17.7% by mass iron. What mass of iron is contained in the ore? a. 16.300000 g b. 78.4 g c. 1.77 g d. 82.2 g e. 464 g ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.76 - Calculate the percentage composition of the elements in a compound. (Example 3.7) TOPICS: stoichiometry determining chemical formulas 59. What is the percent by mass nitrogen in (NH4)2SO3? a. 24.1 % b. 12.961 % c. 64.6 % d. 39.3 % e. 4.73 % ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.76 - Calculate the percentage composition of the elements in a compound. (Example 3.7) TOPICS: stoichiometry determining chemical formulas 60. What is the percentage by mass of chlorine in the insecticide Lindane, C6H6Cl6? a. 20.6 % b. 1.00 % c. 69.6 % d. 8.80 % e. 73.1 % ANSWER: e POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.76 - Calculate the percentage composition of the elements in a compound. (Example 3.7) TOPICS: stoichiometry determining chemical formulas 61. The mineral leadhillite, which is essentially Pb4(SO4)(CO3)2(OH)2 (FW = 1079 g/mol), contains ____% hydrogen by mass. a. 76.81 b. 0.1868 c. 2.226 d. 17.79 e. 2.971 ANSWER: b POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.76 - Calculate the percentage composition of the elements in a compound. (Example 3.7) TOPICS: stoichiometry determining chemical formulas KEYWORDS: mass percentage OTHER: general chemistry 62. Which of the following compounds has the highest percentage of hydrogen atoms by mass? a. CH3COOH b. C2H5OH c. CH3OH d. H2CO3 e. H2C2O4 ANSWER: b POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.76 - Calculate the percentage composition of the elements in a compound. (Example 3.7) TOPICS: stoichiometry determining chemical formulas KEYWORDS: mass percentage OTHER: general chemistry 63. Which of the following compounds has the highest percentage of nitrogen by mass? a. (NH4)2SO3 b. NaNO3 c. N2F4 d. NH4NO2 e. HNO3 ANSWER: d POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.76 - Calculate the percentage composition of the elements in a compound. (Example 3.7) TOPICS: stoichiometry determining chemical formulas KEYWORDS: mass percentage OTHER: general chemistry 64. Which of the following compounds has the same percentage of carbon and hydrogen by mass as cyclohexane, C6H12? a. C6H14, hexane b. C4H8, butylene c. C6H10, cyclohexene d. C6H6, benzene e. C6H12O6, glucose ANSWER: b POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.76 - Calculate the percentage composition of the elements in a compound. (Example 3.7) TOPICS: stoichiometry determining chemical formulas KEYWORDS: mass percentage OTHER: general chemistry 65. What is the mass percentage of carbon in the compound C6H6O3? a. 4.8 % b. 61.9 % c. 57.1 % d. 38.1 % e. 20.0 % ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.76 - Calculate the percentage composition of the elements in a compound. (Example 3.7) TOPICS: stoichiometry determining chemical formulas KEYWORDS: mass percentage OTHER: general chemistry 66. What is the percentage by mass of nitrogen in ammonium phosphate, (NH4)3PO4? a. 28.1 % b. 14.15 % c. 52.4 % d. 39.9 % e. 2.27 % ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.76 - Calculate the percentage composition of the elements in a compound. (Example 3.7) TOPICS: stoichiometry determining chemical formulas 67. A crystal of the mineral troegerite, (UO2)3(AsO4)2 • 12H2O (FM = 1304 amu), contains ____% uranium by mass. a. 18.6 b. 47.2 c. 62.8 d. 30.0 e. 54.8 ANSWER: e POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.76 - Calculate the percentage composition of the elements in a compound. (Example 3.7) TOPICS: stoichiometry determining chemical formulas KEYWORDS: mass percentage OTHER: general chemistry 68. How many grams of hydrogen atoms are present in 16.4 g of water? a. 33.1 g b. 0.91 g c. 1.84 g d. 2.20 g e. 10.9 g ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.77 - Calculate the mass of an element in a given mass of compound. (Example 3.8) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 69. How many grams of potassium are present in 28.7 g of K2Cr2O7? a. 7.63 g b. 1.468 g c. 3.81 g d. 78.2 g e. 14.4 g ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.77 - Calculate the mass of an element in a given mass of compound. (Example 3.8) TOPICS: stoichiometry determining chemical formulas KEYWORDS: mass percentage OTHER: general chemistry 70. NaHCO3 is the active ingredient in baking soda. How many grams of oxygen are present in 0.25 g of NaHCO3? a. 0.009 g b. 0.048 g c. 2.98 × 103 g d. 0.016 g e. 0.14g ANSWER: e POINTS: 1 DIFFICULTY: easy REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.77 - Calculate the mass of an element in a given mass of compound. (Example 3.8) TOPICS: stoichiometry determining chemical formulas KEYWORDS: mass percentage OTHER: general chemistry 71. Which of the following contains the greatest mass of bromine atoms? a. 15.0 g of KBr b. 29.0 g of Br2 c. 0.096 mol of KBr d. 0.078 mol of Br2 e. 25.0 g of NaBrO3 ANSWER: b POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.77 - Calculate the mass of an element in a given mass of compound. (Example 3.8) TOPICS: stoichiometry determining chemical formulas KEYWORDS: mass percentage OTHER: general chemistry 72. The amount of calcium in a 15.0-g sample was determined by converting the calcium to calcium oxalate, CaC2O4. The CaC2O4 weighed 10.3 g. What is the percent of calcium in the original sample? a. 8.8 % b. 21.5 % c. 12.1 % d. 27.5 % e. 68.7 % ANSWER: b POINTS: 1 DIFFICULTY: difficult REFERENCES: 3.3 LEARNING OBJECTIVES: GENE.EBBI.13.77 - Calculate the mass of an element in a given mass of compound. (Example 3.8) TOPICS: stoichiometry mass and moles of substance KEYWORDS: mole | mole calculations OTHER: general chemistry 73. A compound containing only carbon, hydrogen, and oxygen is subjected to elemental analysis. Upon complete combustion, a 0.7916-g sample of the compound produced 1.581 g of CO2 and 0.6474 g of H2O. What is the empirical formula of the compound? a. C3H6O3 b. C3H3O c. C2H4O d. C2H2O e. CH2O3 ANSWER: c POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.4 LEARNING OBJECTIVES: GENE.EBBI.13.78 - Calculate the percentage of C, H, and O from combustion data. (Example 3.9) TOPICS: stoichiometry determining chemical formulas KEYWORDS: elemental analysis OTHER: general chemistry 74. A 3.391 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 6.477 g CO2 and 3.978 g H2O. What mass of oxygen is contained in the original sample? a. 1.178 g b. 1.4899 g c. 3.086 g d. 2.499 g e. 0.5874 g ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.4 LEARNING OBJECTIVES: GENE.EBBI.13.78 - Calculate the percentage of C, H, and O from combustion data. (Example 3.9) TOPICS: stoichiometry determining chemical formulas 75. A 4.215 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 9.582 g CO2 and 3.922 g H2O. What percent by mass of oxygen is contained in the original sample? a. 27.54 % b. 32.75 % c. 12.73 % d. 13.42 % e. 6.939 % ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.4 LEARNING OBJECTIVES: GENE.EBBI.13.79 - Calculate the percentage of C, H, and O from combustion data. TOPICS: stoichiometry determining chemical formulas 76. A 4.957 g sample of a hydrocarbon is burned in an excess of dioxygen, producing 9.284 g CO2 and water. What mass of hydrogen is contained in the original sample? a. 2.423 g b. 4.327 g c. 14.24 g d. 1.794 g e. 2.478 g ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.4 LEARNING OBJECTIVES: GENE.EBBI.13.78 - Calculate the percentage of C, H, and O from combustion data. (Example 3.9) TOPICS: stoichiometry determining chemical formulas 77. A 3.903 g sample of a hydrocarbon is burned in an excess of dioxygen, producing 7.31177 g CO2 and 4.488 g H2O. What is the empirical formula of the hydrocarbon? a. CH3 b. CH2 c. C2H3 d. CH4 e. CH ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.80 - Determine the empirical formula of a binary compound. TOPICS: stoichiometry determining chemical formulas 78. A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in dioxygen to yield 0.01962 mol of CO2 and 0.01961 mol of H2O. What is the empirical formula of the compound? a. CHO b. C3H3O2 c. C2H2O d. C3H6O2 e. C6H3O2 ANSWER: d POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.4 LEARNING OBJECTIVES: GENE.EBBI.13.78 - Calculate the percentage of C, H, and O from combustion data. (Example 3.9) TOPICS: stoichiometry determining chemical formulas 79. A sample containing only carbon, hydrogen, phosphorus, and oxygen is subjected to elemental analysis. After complete combustion, a 0.3584-g sample of the compound yields 0.5139 g of CO2, 0.3155 g of H2O, and 0.2762 g of P4O10. What is the empirical formula of the compound? a. CH3PO b. C2H3PO c. C2H6P2O4 d. C3H9PO e. CH2P4O13 ANSWER: d POINTS: 1 DIFFICULTY: difficult REFERENCES: 3.4 LEARNING OBJECTIVES: GENE.EBBI.13.78 - Calculate the percentage of C, H, and O from combustion data. (Example 3.9) TOPICS: stoichiometry determining chemical formulas KEYWORDS: elemental analysis OTHER: general chemistry 80. A sample containing only carbon, hydrogen, and silicon is subjected to elemental analysis. After complete combustion, a 0.7020-g sample of the compound yields 1.4 g of CO2, 0.86 g of H2O, and 0.478 g of SiO2. What is the empirical formula of the compound? a. CH3Si b. C2H4Si c. C4H12Si d. C6H12Si2 e. CH2Si ANSWER: c POINTS: 1 DIFFICULTY: difficult REFERENCES: 3.4 LEARNING OBJECTIVES: GENE.EBBI.13.78 - Calculate the percentage of C, H, and O from combustion data. (Example 3.9) TOPICS: stoichiometry determining chemical formulas KEYWORDS: elemental analysis OTHER: general chemistry 81. Of the following, the only empirical formula is a. C4H10. b. C4H8. c. C5H14. d. H2O2. e. O3. ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.81 - Define empirical formula. TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 82. Which of the following is the empirical formula for the molecule below? a. CHO b. CH3COOH c. C2H4O2 d. CH2O e. none of the above. ANSWER: d POINTS: 1 DIFFICULTY: easy REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.81 - Define empirical formula. TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 83. Analysis of a compound showed that it contained 76.0 % fluorine atoms and 24.0 % carbon atoms by mass. What is its empirical formula? a. CF2 b. C2F3 c. CF3 d. C2F5 e. CF ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.82 - Determine the empirical formula of a binary compound from the masses of its elements. (Example 3.10) TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 84. A sample of an oxide of antimony (Sb) contained 48.5 g of antimony combined with 15.9 g of oxygen. What is the simplest formula for the oxide? a. SbO2 b. SbO c. Sb2O3 d. Sb3O e. Sb2O5 ANSWER: e POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.82 - Determine the empirical formula of a binary compound from the masses of its elements. (Example 3.10) TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 85. Chlorine was passed over 1.41 g of heated titanium, and 4.54 g of a chloride-containing compound of Ti was obtained. What is the empirical formula of the chloride-containing compound? a. TiCl2 b. TiCl4 c. TiCl d. TiCl3 e. Ti2Cl3 ANSWER: d POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.82 - Determine the empirical formula of a binary compound from the masses of its elements. (Example 3.10) TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 86. A 4.94-g sample of an oxide of chromium contains 3.06 g of chromium. Calculate the simplest formula for the compound. a. CrO5 b. Cr2O c. CrO2 d. CrO e. Cr2O3 ANSWER: c POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.82 - Determine the empirical formula of a binary compound from the masses of its elements. (Example 3.10) TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 87. A compound is composed of only C and H. It contains 92.26 % C. What is its empirical formula? a. C2H5 b. C2H3 c. C3H4 d. CH e. CH2 ANSWER: d POINTS: 1 DIFFICULTY: easy REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.83 - Determine the empirical formula from the percentage composition. (Example 3.11) TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 88. A compound composed of only C and H contains 79.88 % C by mass. What is its empirical formula? a. CH3 b. CH c. C2H d. CH4 e. CH2 ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.83 - Determine the empirical formula from the percentage composition. (Example 3.11) TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 89. A hydrocarbon, subjected to elemental analysis, was found to contain 79.89 % carbon and 20.11 % hydrogen by mass. What is the empirical formula of the hydrocarbon? a. CH4 b. C2H6 c. C4H d. C10H e. CH3 ANSWER: e POINTS: 1 DIFFICULTY: easy REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.83 - Determine the empirical formula from the percentage composition. (Example 3.11) TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 90. A particular compound contains, by mass, 25.5 % carbon, 40.4 % fluorine, and 34.1 % oxygen. A 0.050-mol sample of this compound weighs 5.80 g. The molecular formula of this compound is a. C2F2O2. b. C3F3O. c. CFO. d. C4F4O4. e. C5F5O5. ANSWER: d POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.83 - Determine the empirical formula from the percentage composition. (Example 3.11) TOPICS: stoichiometry determining chemical formulas KEYWORDS: molecular formula OTHER: general chemistry 91. What is the empirical formula of an oxide of nitrogen that contains 25.93 % nitrogen by mass? a. NO2 b. N2O c. N2O3 d. NO e. N2O5 ANSWER: e POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.83 - Determine the empirical formula from the percentage composition. (Example 3.11) TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 92. The analysis of an organic compound showed that it contained 0.630 mol of C, 0.0300 mol of H, 0.420 mol of O, and 0.210 mol of N. How many nitrogen atoms are there in the empirical formula for this compound? a. 10 b. 7 c. 2 d. 4 e. 3 ANSWER: b POINTS: 1 DIFFICULTY: easy REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.83 - Determine the empirical formula from the percentage composition. (Example 3.11) TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 93. A sample containing 0.400 mol of a compound is composed of 2.41 × 1023 atoms of sodium, 14.19 g of chlorine atoms, and 19.22 g of oxygen atoms. The formula of the compound is a. NaClO3. b. NaClO5. c. NaClO. d. NaClO4. e. NaClO2. ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.83 - Determine the empirical formula from the percentage composition. (Example 3.11) TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 94. The analysis of an organic compound showed that it contained 0.105 mol of C, 0.315 mol of H, and 0.0350 mol of N. Its molecular mass is 59 amu. How many atoms of carbon are there in the empirical formula for the compound, and how many are in the molecular formula? a. empirical = 2, molecular = 3 b. empirical = 3, molecular = 9 c. empirical = 3, molecular = 3 d. empirical = 2, molecular = 4 e. empirical = 5, molecular = 10 ANSWER: c POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.83 - Determine the empirical formula from the percentage composition. (Example 3.11) TOPICS: stoichiometry determining chemical formulas KEYWORDS: molecular formula OTHER: general chemistry 95. A given hydrocarbon is burned in the presence of oxygen gas and is converted completely to water and carbon dioxide. The mole ratio of H2O to CO2 is 1.125:1.00. The hydrocarbon could be a. C2H2. b. C2H6. c. CH4. d. C3H4. e. C4H9. ANSWER: e POINTS: 1 DIFFICULTY: difficult REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.83 - Determine the empirical formula from the percentage composition. (Example 3.11) TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 96. A given hydrocarbon is burned in the presence of oxygen gas and is converted completely to carbon dioxide and water. Equal numbers of moles of CO2 and H2O are produced. The hydrocarbon could be a. C3H4. b. C5H10. c. C2H3. d. CH3. e. C3H5. ANSWER: b POINTS: 1 DIFFICULTY: difficult REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.83 - Determine the empirical formula from the percentage composition. (Example 3.11) TOPICS: stoichiometry determining chemical formulas KEYWORDS: empirical formula OTHER: general chemistry 97. An organic compound that has the empirical formula CHO has a molecular mass of 174 amu. Its molecular formula is a. C7H7O7. b. C4H4O4. c. C2H2O2. d. C6H6O6. e. C10H10O10. ANSWER: d POINTS: 1 DIFFICULTY: easy REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.84 - Understand the relationship between the molecular weight of a substance and its empirical formula weight. TOPICS: determining chemical formulas stoichiometry KEYWORDS: molecular formula OTHER: general chemistry 98. A certain compound has a molar mass of 270 g/mol. Which is a possible empirical formula for this compound? a. CH2O b. CHO c. C2H2O2 d. C2HO e. C2H2O ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.84 - Understand the relationship between the molecular weight of a substance and its empirical formula weight. TOPICS: determining chemical formulas stoichiometry KEYWORDS: empirical formula OTHER: general chemistry 99. The empirical formula for a group of compounds is CHCl. Lindane, a powerful insecticide, is a member of this group. The molar mass of lindane is 290.8. How many atoms of carbon does a molecule of lindane contain? a. 3 b. 2 c. 4 d. 6 e. 8 ANSWER: d POINTS: 1 DIFFICULTY: easy REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.84 - Understand the relationship between the molecular weight of a substance and its empirical formula weight. TOPICS: stoichiometry determining chemical formulas KEYWORDS: molecular formula OTHER: general chemistry 100. The empirical formula of styrene is CH; its molar mass is 104.1 g/mol. What is the molecular formula of styrene? a. C6H6 b. C2H4 c. C8H8 d. C10H12 e. none of these ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.84 - Understand the relationship between the molecular weight of a substance and its empirical formula weight. TOPICS: stoichiometry determining chemical formulas KEYWORDS: molecular formula OTHER: general chemistry 101. The empirical formula of butadiene is C2H3. An experimental determination of the molar mass of butadiene by a student yields the value of 54 g/mol. What is the molecular formula of butadiene? a. C5H10 b. C2H3 c. C4H6 d. C3H8 e. C6H9 ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.84 - Understand the relationship between the molecular weight of a substance and its empirical formula weight. TOPICS: stoichiometry determining chemical formulas KEYWORDS: molecular formula OTHER: general chemistry 102. An organic compound has a molecular mass of 324.2 amu and contains 66.63% carbon by mass. How many carbon atoms are in each molecule of this compound? a. 17 b. 21 c. 26 d. 29 e. 18 ANSWER: e POINTS: 1 DIFFICULTY: easy REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.85 - Determine the molecular formula from the percentage composition and molecular weight. (Example 3.12) TOPICS: determining chemical formulas stoichiometry KEYWORDS: molecular formula OTHER: general chemistry 103. An unknown organic compound contains 41.4 % carbon, 3.47 % hydrogen, and 55.1 % oxygen by mass. A 0.030-mol sample of this compound weighs 3.48 g. What is the molecular formula of the organic compound? a. C3H3O b. C2H2O2 c. C4H4O4 d. C8H8O8 e. CHO ANSWER: c POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.85 - Determine the molecular formula from the percentage composition and molecular weight. (Example 3.12) TOPICS: stoichiometry determining chemical formulas KEYWORDS: molecular formula OTHER: general chemistry 104. A compound is composed of only C and Cl. It contains 18.42 % C by mass and has a molar mass of 260.76 g/mol. What is its molecular formula? a. CCl b. C4Cl6 c. C3Cl4 d. C2Cl2 e. C2Cl3 ANSWER: b POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.85 - Determine the molecular formula from the percentage composition and molecular weight. (Example 3.12) TOPICS: stoichiometry determining chemical formulas KEYWORDS: molecular formula OTHER: general chemistry 105. An organic compound has a molar mass of 169.3 g/mol and contains 10.63 % hydrogen atoms by mass. How many hydrogen atoms are in each molecule of this compound? a. 18 b. 7 c. 22 d. 29 e. 9 ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.85 - Determine the molecular formula from the percentage composition and molecular weight. (Example 3.12) TOPICS: stoichiometry determining chemical formulas KEYWORDS: molecular formula OTHER: general chemistry 106. A compound has a molar mass of 319.9g/mol and contains 30.01 % oxygen atoms by mass. How many oxygen atoms are in each molecule of this compound? a. 6 b. 4 c. 10 d. 3 e. 7 ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.85 - Determine the molecular formula from the percentage composition and molecular weight. (Example 3.12) TOPICS: stoichiometry determining chemical formulas KEYWORDS: molecular formula OTHER: general chemistry 107. A compound contains 43.84 % carbon atoms, 3.65% hydrogen atoms, and 8.68 % fluorine atoms by mass. Each molecule of this compound contains one fluorine atom. What is the total number of carbon, hydrogen, and fluorine atoms in one molecule of this compound? a. 17 b. 11 c. 10 d. 7 e. 14 ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.85 - Determine the molecular formula from the percentage composition and molecular weight. (Example 3.12) TOPICS: stoichiometry determining chemical formulas KEYWORDS: molecular formula OTHER: general chemistry 108. A molecular compound contains 92.3 % carbon and 7.7 % hydrogen by mass. If 0.142 mol of the compound weighs 11.08 g, what is its molecular formula? a. C8H8 b. C8H10 c. C4H8 d. C6H6 e. CH ANSWER: d POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.85 - Determine the molecular formula from the percentage composition and molecular weight. (Example 3.12) TOPICS: stoichiometry determining chemical formulas KEYWORDS: molecular formula OTHER: general chemistry 109. Analysis of a compound containing only C and Cl revealed that it contains 33.33 % C atoms by number and has a molar mass of 248.75 g/mol. What is the molecular formula of this compound? a. CCl2 b. C2Cl6 c. C2Cl4 d. CCl3 e. C3Cl6 ANSWER: e POINTS: 1 DIFFICULTY: difficult REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.85 - Determine the molecular formula from the percentage composition and molecular weight. (Example 3.12) TOPICS: stoichiometry determining chemical formulas KEYWORDS: molecular formula OTHER: general chemistry 110. Complete combustion of a 0.40-mol sample of a hydrocarbon, CxHy, gives 2.40 mol of CO2 and 1.60 mol of H2O. The molecular formula of the original hydrocarbon is a. C2H8. b. C6H8. c. C9H16. d. C5H5. e. C4H7. ANSWER: b POINTS: 1 DIFFICULTY: difficult REFERENCES: 3.5 LEARNING OBJECTIVES: GENE.EBBI.13.85 - Determine the molecular formula from the percentage composition and molecular weight. (Example 3.12) TOPICS: stoichiometry determining chemical formulas KEYWORDS: molecular formula OTHER: general chemistry 111. A chemical reaction has the equation: 2A + B → C. Which of the following figures best illustrates a stoichiometric ratio of A and B? I. II. III. IV. a. I only b. III only c. II only d. Both I and IV e. IV only ANSWER: d POINTS: 1 DIFFICULTY: easy REFERENCES: 3.6 LEARNING OBJECTIVES: GENE.EBBI.13.86 - Relate the coefficients in a balanced chemical equation to the number of molecules or moles (molar interpretation). TOPICS: stoichiometry stoichiometry calculation KEYWORDS: molar interpretation OTHER: general chemistry 112. The balanced chemical equation for the combustion of methane is: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) Which of the following statements concerning this chemical equation is/are correct? 1. One gram of methane gas reacts with two grams of dioxygen gas, producing one gram of carbon dioxide gas and two grams of gaseous water. 2. One mole of methane gas reacts with two moles of dioxygen gas, producing one mole of carbon dioxide gas and two moles of gaseous water. 3. One molecule of methane gas reacts with two molecules of dioxygen gas, producing one molecule of carbon dioxide gas and two molecules of gaseous water. a. 1 only b. 2 only c. 2 and 3 d. 1 and 3 e. 1,2 and 3 ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.6 LEARNING OBJECTIVES: GENE.EBBI.13.86 - Relate the coefficients in a balanced chemical equation to the number of molecules or moles (molar interpretation). TOPICS: stoichiometry stoichiometry calculation KEYWORDS: molar interpretation OTHER: general chemistry NOTES: REVISED 113. Balance the following expression: __ CH3CH2COOH + __ O2 → __ CO2 + __ H2O How many moles of O2 are required for the complete combustion of 6 mol of propanoic acid? a. 6 mol b. 33 mol c. 21 mol d. 38 mol e. 4 mol ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.6 LEARNING OBJECTIVES: GENE.EBBI.13.86 - Relate the coefficients in a balanced chemical equation to the number of molecules or moles (molar interpretation). TOPICS: stoichiometry stoichiometry calculation KEYWORDS: molar interpretation OTHER: general chemistry 114. The products of the combustion of acetone with oxygen are shown in the following equation: __ CH3COCH3 + __ O2 → __ CO2 + __ H2O When properly balanced, the equation indicates that ____ mol of O2 are required for each mole of CH3COCH3. a. 4 b. 6 c. 3.5 d. 7 e. 2 ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.6 LEARNING OBJECTIVES: GENE.EBBI.13.86 - Relate the coefficients in a balanced chemical equation to the number of molecules or moles (molar interpretation). TOPICS: stoichiometry stoichiometry calculation KEYWORDS: molar interpretation OTHER: general chemistry 115. Ammonia, NH3, and oxygen can be reacted together in the presence of a catalyst to form only nitrogen monoxide and water. The number of moles of oxygen consumed for every 1.00 mole of NO produced is . a. 1.25 b. 5 c. 3.75 d. 0.625 e. 2.5 ANSWER: a POINTS: 1 DIFFICULTY: difficult REFERENCES: 3.6 LEARNING OBJECTIVES: GENE.EBBI.13.86 - Relate the coefficients in a balanced chemical equation to the number of molecules or moles (molar interpretation). TOPICS: stoichiometry stoichiometry calculation KEYWORDS: molar interpretation OTHER: general chemistry 116. 2KHCO3(s) → K2CO3(s) + CO2(g) + H2O(l) How many moles of potassium carbonate will be produced if 487 g of potassium hydrogen carbonate are heated? a. 2.43 mol b. 3.52 mol c. 17.4 mol d. 244 mol e. 4.87 mol ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.7 LEARNING OBJECTIVES: GENE.EBBI.13.87 - Relate the quantities of reactant to the quantity of product. (Example 3.13) TOPICS: stoichiometry stoichiometry calculation KEYWORDS: amounts of substances OTHER: general chemistry 117. Calculate the number of moles of O2 required to react with phosphorus to produce 3.56 g of P4O6. (Molar mass P4O6 = 219.9 g/mol) a. 0.0162 mol b. 0.111 mol c. 0.0486 mol d. 0.0324 mol e. 0.097 mol ANSWER: c POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.7 LEARNING OBJECTIVES: GENE.EBBI.13.87 - Relate the quantities of reactant to the quantity of product. (Example 3.13) TOPICS: stoichiometry stoichiometry calculation KEYWORDS: amounts of substances OTHER: general chemistry 118. Elemental sulfur can be converted to sulfur dioxide by combustion in air. Sulfur dioxide will react with water to form sulfurous acid (see balanced equation below). SO2(g) + H2O(l) → H2SO3(l) What mass of sulfur dioxide is needed to prepare 29.36 g of H2SO3(l)? a. 22.91g b. 37.61 g c. 0.4582 g d. 0.3576 g e. 29.36 g ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.7 LEARNING OBJECTIVES: GENE.EBBI.13.87 - Relate the quantities of reactant to the quantity of product. (Example 3.13) TOPICS: stoichiometry stoichiometry calculation 119. One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III) hydroxide from a solution containing rhodium(III) sulfate according to the following balanced chemical equation: Rh2(SO4)3(aq) + 6NaOH(aq) → 2Rh(OH)3(s) + 3Na2SO4(aq) If 3.20 g of rhodium(III) sulfate reacts with excess sodium hydroxide, what mass of rhodium(III) hydroxide may be produced? a. 1.99 g b. 6.4 g c. 3.2 g d. 0.498 g e. 3.98 g ANSWER: a POINTS: 1 DIFFICULTY: easy REFERENCES: 3.7 LEARNING OBJECTIVES: GENE.EBBI.13.87 - Relate the quantities of reactant to the quantity of product. (Example 3.13) TOPICS: stoichiometry stoichiometry calculation KEYWORDS: amounts of substances OTHER: general chemistry 120. Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows: Cu2S(s) + O2(g) → 2Cu(s) + SO2(g) What mass of copper(I) sulfide is required in order to prepare 0.100 kg of copper metal? a. 0.1 kg b. 0.05 kg c. 0.0752 kg d. 0.25 kg e. 0.125 kg ANSWER: e POINTS: 1 DIFFICULTY: easy REFERENCES: 3.7 LEARNING OBJECTIVES: GENE.EBBI.13.87 - Relate the quantities of reactant to the quantity of product. (Example 3.13) TOPICS: stoichiometry stoichiometry calculation KEYWORDS: amounts of substances OTHER: general chemistry 121. One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III) hydroxide from a solution containing rhodium(III) sulfate according to the following balanced chemical equation: Rh2(SO4)3(aq) + 6NaOH(aq) → 2Rh(OH)3(s) + 3Na2SO4(aq) What mass of sodium hydroxide is required to precipitate 68.0 g of rhodium(III) hydroxide from a solution containing excess rhodium(III) sulfate? a. 5.88 g b. 53 g c. 17.6 g d. 204 g e. 68 g ANSWER: b POINTS: 1 DIFFICULTY: easy REFERENCES: 3.7 LEARNING OBJECTIVES: GENE.EBBI.13.87 - Relate the quantities of reactant to the quantity of product. (Example 3.13) TOPICS: stoichiometry stoichiometry calculation KEYWORDS: amounts of substances OTHER: general chemistry 122. Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows: Cu2S(s) + O2(g) → 2Cu(s) + SO2(g) If 0.190 kg of copper(I) sulfide reacts with excess oxygen, what mass of copper metal may be produced? a. 0.19 kg b. 0.0379 kg c. 0.0758 kg d. 0.151 kg e. 0.38 kg ANSWER: d POINTS: 1 DIFFICULTY: easy REFERENCES: 3.7 LEARNING OBJECTIVES: GENE.EBBI.13.88 - Relate the quantities of two reactants or two products. (Example 3.14) TOPICS: stoichiometry stoichiometry calculation KEYWORDS: amounts of substances OTHER: general chemistry 123. One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III) hydroxide from a solution containing rhodium(III) sulfate according to the following balanced chemical equation: Rh2(SO4)3(aq) + 6NaOH(aq) → 2Rh(OH)3(s) + 3Na2SO4(aq) If 0.730 g of rhodium(III) hydroxide is produced, what mass of sodium sulfate is also produced? a. 0.673 g b. 1.09 g c. 1.01 g d. 0.73 g e. 0.449 g ANSWER: c POINTS: 1 DIFFICULTY: easy REFERENCES: 3.7 LEARNING OBJECTIVES: GENE.EBBI.13.88 - Relate the quantities of two reactants or two products. (Example 3.14) TOPICS: stoichiometry stoichiometry calculation KEYWORDS: amounts of substances OTHER: general chemistry 124. The balanced equation for the combustion of ethanol is 2C2H5OH(g) + 7O2(g) → 4CO2(g) + 6H2O(g) How many grams of dioxygen are required to burn 3.4 g of C2H5OH? a. 8.2 g b. 11 g c. 2.3 g d. 38 g e. 57 g ANSWER: a POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.7 LEARNING OBJECTIVES: GENE.EBBI.13.88 - Relate the quantities of two reactants or two products. (Example 3.14) TOPICS: stoichiometry stoichiometry calculation 125. 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g) According to the equation above, how many grams of aluminum are needed to completely react with 4.32 mol of hydrochloric acid? a. 350 g b. 52.6 g c. 38.9 g d. 4.32 g e. 116.6 g ANSWER: c POINTS: 1 DIFFICULTY: moderate REFERENCES: 3.7 LEARNING OBJECTIVES: GENE.EBBI.13.88 - Relate the quantities of two reactants or two products. (Example 3.14) TOPICS: stoichiometry stoichiometry calculation KEYWORDS: amounts of substances OTHER: general chemistry 126. A chemical reaction has the e