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Chemistry And Life In The Laboratory: Experiments, 6th Edition Solution Manual

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    Experiment 1 - “Thinking Metric” Reagents - None Special Equipment - meter sticks 8½" x 11" pieces of paper string and scissors rectangular objects such as boxes or blocks pieces of chalk nickel coins 100 tablets (aspirin, vitamin) or other small objects pebbles, rubber stoppers, or metal shot Suggestions and Precautions - The key is to have the students “think metric” rather than to stress conversions. Prelaboratory Questions - 1. For: allow for easier importation and exportation of trade products. Against: expense of conversion; resistance to re-education. 2. mile, kilometer, meter, foot, inch, centimeter, micron, Angstrom. 3. barrel, cubic foot, gallon, liter, milliliter, cubic milliliter. 4. The density is less than that of water. 5. Overall density, including air cavity, is less; otherwise greater. Bodies can be made to float or sink depending on chest expansion. Observations and Results - Part 3- Source of discrepancy - balance is off; personal error. A gram is about a fifth of a nickel’s mass. Imagine a nickel that is cut into five pieces. Average weight. Part 4- Wood samples of same kind? - yes; samples of different wood? - no. Relative densities of wood and water - density of wood is less than that of water. Wood would float on mercury because wood’s density is less than 13.6 g/ml.

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    Questions and Problems - 1. 3.7851 24 gal • 90.841 1 gal = 2. 110 km 1 mi • 68.4 mi/hr hr 1.609 km = 3. 2.5 cm • 12 cm • 20 cm = 600 cu cm; 600 g water 4. 4 years • 186,300 mi/sec • 60 sec/min • 24 hr/da • 365 da/yr = 23,500,000,000,000 miles 5. wt (lb) ÷ 2.2 lb/kg 6. 60 min/hr x 24 hr/da x da/10 cda = 144 min/cda 365 day/yr x 1 kda/1000 da = 0.365 yr/kda 7. (a) Probably safe (b) 0.5 L/sec x 60 sec/min x 60 min/hr x 8 hr = 14,400 L (c) 14,400 L x 1.2 g/L = 17,280 g; 12,200 g = 12.2 kg (d) 17,280 g x 1/106 = 0.017 g of X; 17 mg of X (e) Move information needed

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    Experiment 2 - “Molecular Motion - The Particulate Nature of Matter” Reagents - Each Student 100 Students methyl salicylate, C 8 H 8 O 3 need 0.5 ml per demonstration hydrochloric acid, HCl, 12 M 500 ml ammonium hydroxide, NH 4 OH, 15 M 500 ml acetone, C 3 H 6 O 1 ml 100 ml sodium chloride, NaCl 4 g 400 g p-dichlorobenzene, C 6 H 4 Cl 2 0.25 g 25 g iodine, I 2 0.3 g 30 g potassium permanganate, KMnO 4 0.2 g 20 g Special Equipment - watch glass (for group experiment) hot plate (for group experiment) time piece with second hand (for group experiment) chalk string and meter stick 60 cm x 2 cm glass tube (for demonstration) cotton plugs (for demonstration) thermometer, °C lined, white paper Suggestions and Precautions - To avoid pre-release, bring the correct amount of oil of wintergreen in a closed vial and drop it directly onto the watch glass on a heated hotplate. Have no drafts in the room (open windows, hood drafts, etc.). Use only small crystals of iodine or the vapor will cause problems in the room. Define the size of a small pea. Prelaboratory Questions - 1. Molecules of perfume vaporize and travel across the room to sensory cells in the nose. 2. The intermolecular distance in a gas is much greater. 3. Solids: touching each other in a regular three-dimensional patter. Liquids: random arrangement with molecules in close contact with each other. 4. Faster. More energy is being added to the system. 5. Slower. 6. Although their speed is great, because of collisions they travel by a circuitous route. Observations and Results - Part 1- Explain how the odor came: Vaporized molecules traveled by means of a zig-zag course due to molecular collisions. Difficult if you assume matter is continuous; could say matter travels in a continuous ray. You would expect ammonia to travel faster because its

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    molecular mass is less, thus its velocity will be faster. Suggest several explanations - different sensitivities of noses; drafts or different air circulation patterns. How does this experiment prove: The product (NH 4 Cl) was formed in the air above the liquid levels of the reactants - therefore, the reactant molecules had to travel through the air. Part 2- How do you know - because acetone evaporates faster (disappears faster). What do relative rates of evaporation - the attractive forces between water molecules is greater than the forces between acetone molecules. Explain this observation - as those molecules with higher energies vaporize first, molecules with lower energy (the cooler ones) are left behind. Suggest a reason why molecules - in a concentrated solution there are many particles which get in the way of the water molecules escaping, thus increasing the amount of energy (and increasing the boiling point) needed to vaporize the solvent. Part 3- Solid mothballs activated your nose because molecules vaporized from the solid and then diffused through the air to the nose. Part 4- Typical pathway for a particle: Questions and Problems - 1. colorless 2. bluish purple 3. yes - sound is the displacement of air molecules and it travels in waves from the point where the sound originated to the eardrum - the displaced molecules displace other molecules, which eventually lead to collision with the ear drum. This sets off nerve impulses which the brain interprets as a particular sound. 4. (a) yes; no change in matter (b) no; odor is transfer of molecules – matter lost 5. Molecules in one cup, 7.5 x 10 24 . Volume of oceans, 2.8 x 10 20 L . Fraction of molecules from original cup times molecules in cup: 0.25 L 2.8 x 10 20 L x 7.5 x 10 24 molecules = about 7,000 molecules original 6. Each received, 7 x 10 23 atoms / 7 x 10 9 persons = 1 x 10 14 atoms/person $.01/1 x 10 4 atoms = $1 x 10 -6 /atom 1 x 10 14 atoms/person x 1 x 10 -6 $/atom = $10 8 = 100 million dollars

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    Experiment 3 - “What Is a Chemical Reaction?” Reagents - Each Student 100 Students iron dust, Fe (reagent, free of sulfur) 1 g 100 g sulfur, S 1 g 100 g carbon disulfide, CS 2 (demonstration) 5 ml hydrochloric acid, HCl, 6 M 0.5 ml 50 ml camphor, C 10 H 16 O 0.2 g 20 g copper sulfate, CuSO 4 , solution, 0.1 M 1 ml 100 ml potassium chromate, K 2 CrO 4 , solution, 0.1 M 0.5 ml 50 ml lead acetate, Pb(C 2 H 3 O 2 ) 2 , solution, 0.1 M 0.5 ml 50 ml calcium carbonate, CaCO 3 0.1 g 10 g acetic acid, HC 2 H 3 O 2 , 1 M 1 ml 100 ml saturated calcium hydroxide, Ca(OH) 2 , solution 2 ml 200 ml iron nail 1/student 100 nails Special Equipment - magnet mortar and pestle plastic straw Suggestions and Precautions - Use only reagent iron dust which is essentially free of sulfur. Inform the students that a faint odor of H 2 S from the reaction of iron with HCl is due to contamination with sulfide. The camphor should be heated sufficiently to melt it but not enough to make it burn or physical and chemical changes will be confused. Prelaboratory Questions - 1. Change in appearance, change in odor. 2. (a) fermenting sugar; (c) burning wood; (e) cooking meat; (h) rusting iron. 3. They get used up (consumed) and changed into different substances. 4. Convenience, clarity and brevity; a chemical reaction is what actually takes place - an equation is a written statement of that occurrence. 5. endothermic: absorbs heat (energy); exothermic: evolves heat. 6. Attractive (electrostatic) forces which hold the atoms in a molecule together. Bonds are broken and new bonds formed. Observations and Results - Part 1- This proves that the change which occurred with CS 2 was physical because the sulfur was recoverable in its original form. Two simple tests are magnet for iron and solubility in CS 2 for sulfur. Fe + S → FeS. Formula for this gas: H 2 S. The fact that the gas formed when treating the heated iron-sulfur mixture with acid had a different odor than the gas

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    which was formed from iron and acid (sulfur doesn’t react in acid). FeS + 2HCl → H 2 S + FeCl 2 . Part 2- Burning of methane- bonds broken: C-H, O ═ O; bonds formed: C ═ O, H-O. CuSO 4 + Fe → FeSO 4 + Cu. Pb(C 2 H 3 O 2 ) 2 + K 2 CrO 4 → 2KC 2 H 3 O 2 + PbCrO 4 . Calcium carbonate is insoluble in water. A physical property of acetic acid is its odor. The evidence for the disappearance of the reactants is that CaCO 3 dissolves. The evidence for the new products is the evolution of CO 2 . CaCO 3 + 2HC 2 H 3 O 2 → Ca(C 2 H 3 O 2 ) 2 + CO 2 + H 2 O Questions and Problems - 1. internal combustion engine; gas stove; food metabolism. 2. bonds broken: H-H, O ═ O; bonds formed: O-H. Formed bonds are stronger. 3. add acid to the substance - if effervescence occurs, it is CaCO 3 . 4. use vinegar (or other household acid) to dissolve. 5. H 2 , hydrogen gas, 2 atoms/molecules. 6. internal combustion engine (a spark); 2H 2 + O 2 → 2H 2 O; cooking of food. 7. 6 bonds broken, 6 bonds formed.

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    Experiment 4 - “The Calorie: A Unit of Energy” Reagents - Each Student 100 Students ammonium chloride, NH 4 Cl 10 g 1000 g calcium chloride, CaCl 2 10 g 1000 g chipped ice 100 g 10 kg isopropyl alcohol 1 ml 100 ml Special Equipment - thermometer, °C styrofoam cup, 6 oz. cups work nicely; 4 oz. size is o.k. Suggestions and Precautions - In Part 3, the volume of water must be determined immediately when the temperature no longer drops (approximately 2°C). All of Part 3 must be done rapidly to avoid heat loss to the surroundings. Prelaboratory Questions - 1. calorie, erg, joule, BTU, foot-pound. 2. Calorie: The amount of heat (energy) required to raise the temperature of one gram of water one degree centigrade (from 14.5 to 15.5°C). Specific heat: The amount of energy required to raise one gram of a substance one degree centigrade. Heat of fusion: The energy required for converting a solid to a liquid (per gram or per mole). 3. It is used to change the liquid to a vapor (steam) by supplying the energy needed to overcome the attractive forces between the molecules. Heat of vaporization is the amount of energy required to change a liquid to a gas (per gram or per mole). 4. (40 g) (50°C - 25°C) (1 cal/g - °C) = 1000 calories. 5. (100 g) (x - 20°C) (1 cal/g - °C) = 1000 calories; x = 30°C. Observations and Results - None Questions and Problems - 1. 100 ml 1 g • (70º - 30º ) (1 cal/g-ºC) 5001 11 ml 20, 000 kilocalories 1000 cal/kilocalorie = i 2. (5 g) (80 cal/g) + (5 g) (100° - 0°C) (1 cal/g - °C) + (5 g) (540 cal/g) = 3600 calories 3. (80 g) (80 cal/g) = 6400 calories used up in melting ice. (20 g) (540 cal/g) = 10,800 calories liberated in condensing steam. (10,800 - 6400 calories) = (80 g) (x° - 0°C) (1 cal/g-°C); x = 55°C; mixing 80 g water at 55°C with 20 g water at 100°C = 64°C.

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    Experiment 5 - “Reliability of Data: The Composition of a Compound” Reagents - Each Student 100 Students copper sulfate pentahydrate, CuSO 4 • 5H 2 O 1 g 100 g Special Equipment – None Suggestions and Precautions - Heat the CuSO 4 • 5H 2 O with a low flame only as long as a blue color is present. Prelaboratory Questions - 1. compounds - a definite percentage composition by weight, whereas, in mixtures, it is variable. Examples: compounds - salt (NaCl), sugar (C 6 H 12 O 6 ), ammonia (NH 3 ); mixtures - alloys, sand and salt, air. 2. 15.12 g - 14.72 g • 100 2.72% 14.72 g = 3. 10.231 10.278 10.256 10.363 10.282 g 4 + + + = 10.282 - 10.231 = 0.051 10.278 - 10.282 = 0.004 average deviation = 0.027 10.256 - 10.282 = 0.026 10.363 - 10.282 = 0.081 10.363 is not reliable because its deviation from the mean is at least three times the average of the other three values. Therefore, take the mean of the first three values, 10.255. 4. a. Na-2; S-1; O-14; H-20; total 37 atoms b. 14 atoms • 100 37.8% 37 atoms = c. 10 water molecules 5. add together (0.2464 • 2.53) = 0.579 g of sand (0.0423 • 2.35) = 0.099 g of clay (0.7113 • 2.35) = 1.672 g of ash; Grind together with mortar and pestle. Observation and Results - Sources of error: balance poorly adjusted - either way experimenter’s technique - either way not heating CuSO 4 enough - low picking up moisture - low losses due to spattering – high

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    CuSO 4 + 5H 2 O → CuSO 4 • 5H 2 O The data obtained support the law of constant composition because they give fairly constant percentages of water in this hydrate, thus showing that the components are present in the compound in a definite ratio. Questions and Problems - 1. 2 (18 g/mole) 100 20.9% 172 g/mole = i 2. Range = 24.45% - 24.27% = 0.18% 24.40 24.32 (24.45) 2 24.27 Mean = 24.38% 5 + + + = Average deviation = 24.40 - 24.38 = 0.02 24.38 - 24.32 = 0.06 24.45 - 24.38 = 0.07 24.45 - 24.38 = 0.07 24.38 - 24.27 = 0.11 0.33 = 0.066% 5 3. 63.5 24.45 - 24.38 100 25.45% 100 4.2% error 249.5 25.45 = = i i 4. 324.0 100 48.62% 666.43 = i 48.70 48.65 48.62 - 48.71 100 0.18% 48.62 49.50 47.99 194.84 48.71% experimental 4 = = i 48.70 48.70 - 48.45 0.25 48.64 48.65 - 48.45 0.20 47.99 48.45 - 47.99 0.46 145.34 0.91 48.45 0.30 3 3 = = = = = 5. Cyclamates, saccharine, various pesticides, tris; opinions vary; thalidomide.

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    Experiment 6 - “Getting Acquainted with Metals” Reagents - Each Student 100 Students iron, Fe, strips or pieces 0.5 g 50 g tin, Sn, strips or pieces 0.5 g 50 g lead, Pb, strips or pieces 0.5 g 50 g copper, Cu, strips or pieces 1 g 100 g magnesium, Mg, strips or pieces 0.5 g 50 g aluminum, Al, strips or pieces 0.5 g 50 g zinc, Zn, strips or pieces 0.5 g 50 g sodium, Na, strips or pieces 0.5 g 50 g iron nail 2/student 200 nails steel wool 1 g 100 g hydrochloric acid, HCl, 1 M 10 ml 1000 ml hydrochloric acid, HCl, 6 M 10 ml 1000 ml hydrochloric acid, HCl, 12 M 5 ml 500 ml copper sulfate, CuSO 4 , solution, 0.1 M 10 ml 1000 ml ferrous sulfate, FeSO 4 , solution, 0.1 M 10 ml 1000 ml silver nitrate, AgNO 3 , solution, 0.1 M 10 ml 1000 ml Special Equipment- knife (for demonstration) Suggestions and Precautions- It is very worthwhile to let students observe the instructor cutting sodium metal so they can see how lustrous it is at first and how rapidly it tarnishes. Let them understand why it is stored under a liquid hydrocarbon. Note how the exothermic reaction causes sodium pieces to heat up and melt as they float on a cushion of hydrogen over the water. At the end, there is usually a small ignition of the hydrogen.

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    Prelaboratory Questions - 1. active moderately active inactive magnesium lead copper potassium iron gold sodium tin silver calcium aluminum platinum nickel 2. Cr - chrome plating, stainless steel Ta - synthetic bone parts in humans Ni - coinage, stainless steel Hg - thermometers, barometers Zn - galvanizing coating Sn - coating for steel cans Ra - radiation for cancer treatment Ag - jewelry, coinage, instrumentation 3. Cu, Cu 2+ ; Cu metal is red, lustrous; Cu 2+ solutions are blue. H 2 , H 3 O + ; H 2 is a gas; H 3 O + is an ion in solution having acidic properties. 4. a. Beneficial Mg +2 , Ca +2 , Fe +2 b. Harmful Hg +2 , Pb +2 , Cd +2 Observations and Results Part 1-3- 4Na + O 2 → 2Na 2 O. Also, 2Na + 2H 2 O → 2NaOH + H 2 . Sodium is stored under a hydrocarbon to protect it from oxidation by the air. Kerosene, toluene, or mineral oil are selected because they have low vapor pressures and convenient liquid ranges. 2Zn + O 2 → 2ZnO 2Al + 3/2O 2 → Al 2 O 3 Magnesium heated in air produces MgO. Make a sketch: Mg lost electrons; other atoms accepted electrons from Mg, such as oxygen. Can you explain your observation - increasing temperature increases the energy of the reactants, allowing more frequent and more effective collisions. Magnesium would make a good night flare. Would any metals react differently? - All those which reacted with air should react faster and more vigorously since the concentration of oxygen is increased significantly. Steel wool has a greater surface area.

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    2Fe + 3/2O 2 → Fe 2 O 3 Al + 3H + → Al 3+ + 3/2 H 2 (g) (Cu - no reaction) Silver is used in jewelry; sodium would be least suitable. All are true, but just the first four are based on the statement only. Part 4- a. Fe + CuSO 4 → FeSO 4 + Cu b. n.r. c. Cu + 2AgNO 3 → Cu(NO 3 ) 2 + 2Ag Which is more active: Fe; Fe; Cu Arrange the 3 metals: Fe, Cu, Ag Arrange all metals: Na, Mg, Al, Zn, Fe, Sn, Pb, H, Cu, Ag Questions and Problems- 1. K, Ba, Ca, Na, Mg, Al, Sn, Pb. 2. The most active metals are located toward the bottom left portion of the chart; active nonmetals are top right; most active metal: Cs; most active non-metal: F. 3. NaOH, Mg(OH) 2 , Al(OH) 3 , Zn(OH) 2 , Fe(OH) 2 , Sn(OH) 4 , Pb(OH) 2 , Cu(OH) 2 . 4. Ag 2 S. 5. 8 × 10 7 tons of gold; density, 1.2 × 10 3 lb/ft 3 ; so, 67,000 cubic ft; e.g., building 10 × 80 × 80.

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    Experiment 11 - “Weights of Reactants and Products: Preparation of a Metal” Reagents - Each Student 100 Students lead carbonate, PbCO 3 2 g 200 g wood charcoal 1 g 100 g Special Equipment- None Suggestions and Precautions - The type of flame is critical to the success of the experiment. The flame must be cool and yellow - a hot flame produces extremely hard material which cannot be removed from the crucible. The yellow flame is easily disturbed by drafts in the lab. Students must be instructed to keep the flame close to the bottom of the crucible. Stirring to obtain a large globule of lead is important; otherwise, lead “sand” will be obtained, which is difficult to separate from the carbon. Prelaboratory Questions 1. 4 + 6 = 10. 2. 123.9 g/mole P 219.9 g/mole P 4 O 6 • 100 = 56.3%P; 96 g/mole O 219.9 g/mole P 4 O 6 • 100 = 43.7% O. 3. 7.5 g PbCO 3 267.9 g/mole PbCO 3 = 0.028 mole PbCO 3 ; 0.028 mole PbO; 0.028 mole Pb; 0.014 mole C; Yes In the experiment: 4 g PbCO 3 267.9 g/mole PbCO 3 = 0.0149 mole PbCO 3 = 0.0149 mole PbO - will require 0.0149 2 mole C = 0.00747 mole C = 0.090 g C and 0300 g - 0.090 g = 0.21 g excess. 4. 2ZnS + 3O 2 → 2ZnO + 2SO 2 2ZnO + C → 2Zn + CO 2 5. A hard, fused lead oxide would form which could not be removed from the crucible. PbCO 3 would not be heated sufficiently hot to give PbO.

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    Observations and Results - If the weight of PbO is too high, the PbCO 3 may not have been heated either hot enough or long enough. A low value suggests mechanical loss. Questions and Problems - 1. Let x = weight of ore required to yield one ton of lead. Then: x • 0.147(% lead in ore) • 0.95 (% recovery) = 1 ton. x = 7.16 tons of ore. 2. M. Wt. PbCO 3 (Cerussite) = 267 g/mole. 207.2 g/mole 267.9 g/mole • 100 = 77.3% Pb M. Wt. PbS (Galena) = 239.2 g/mole. 207.2 g/mole 267.9 g/mole • 100 = 86.6% Pb 3. 2PbS + 3O 2 → 2PbO + 2SO 2 4. grams of lead/yr = 3 gal/day • 365 days/yr • 0.4 g/day = 438 g/yr . 5. fishing sinkers; storage batteries. Discontinued: lead in paint; tetraethyl lead in gasoline; lead shot in some shotgun shells.

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    Experiment 12 - “The pH Scale and the Use of Indicators” Reagents - Each Pair 100 Students distilled water 100 ml 10,000 ml vinegar, 10% solution 35 ml 1750 ml grapefruit juice 35 ml 1750 ml 7-Up, 50% solution 35 ml 1750 ml low alkalinity shampoo, 10% solution 35 ml 1750 ml ordinary shampoo, 10% solution 35 ml 1750 ml liquid laundry detergent, 5% solution 25 ml 1250 ml household ammonia 25 ml 1250 ml baking soda, NaHCO 3 , solution, 5% 25 ml 1250 ml aspirin tablet (in 20 ml water) 1 tab/pair 50 tablets Bufferin (in 20 ml water) 1 tab/pair 50 tablets buffers - pH 4 25 ml for demonstration pH 9 25 ml for demonstration hydrochloric acid, HCl, 0.01 M 5 ml 250 ml - be sure to use boiled distilled water sodium hydroxide, NaOH, 0.01 M 5 ml 250 ml - be sure to use boiled distilled water methyl orange 3 ml 150 ml - 1 g methyl orange per liter water phenolphthalein 3 ml 150 ml - 0.05 g phenolphthalein per 50 ml reagent grade ethanol and 50 ml water bromthymol blue 3 ml 150 ml - 0.01 g bromthyrnol blue per 16 ml 0.01 N NaOH and 234 ml water (this can also be obtained commercially prepared) alizarin yellow 3 ml 150 ml - 0.1% in methanol or 1 g alizarin yellow per 1 liter methanol methyl red 3 ml 150 ml - 1 g methyl red per 600 ml reagent grade ethanol, diluted with 400 ml water Special Equipment - pH meter

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