Chemistry 9th solution manual

CHAPTER CHEMISTRY: THE STUDY OF CHANGE 1.3 (a) Quantitative This statement clearly involves a measurable distance (b) Qualitative This is a value judgment There is no numerical scale of measurement for artistic excellence (c) Qualitative If the numerical values for the densities of ice and water were given, it would be a quantitative statement (d) Qualitative Another value judgment (e) Qualitative Even though numbers are involved, they are not the result of measurement 1.4 (a) hypothesis 1.11 (a) Chemical property Oxygen gas is consumed in a combustion reaction; its composition and identity are changed (b) Chemical property The fertilizer is consumed by the growing plants; it is turned into vegetable matter (different composition) (c) Physical property The measurement of the boiling point of water does not change its identity or composition (d) Physical property The measurement of the densities of lead and aluminum does not change their composition (e) Chemical property When uranium undergoes nuclear decay, the products are chemically different substances (a) Physical change The helium isn't changed in any way by leaking out of the balloon (b) Chemical change in the battery (c) Physical change The orange juice concentrate can be regenerated by evaporation of the water (d) Chemical change Photosynthesis changes water, carbon dioxide, etc., into complex organic matter (e) Physical change The salt can be recovered unchanged by evaporation 1.12 (b) law (c) theory 1.13 Li, lithium; F, fluorine; P, phosphorus; Cu, copper; As, arsenic; Zn, zinc; Cl, chlorine; Pt, platinum; Mg, magnesium; U, uranium; Al, aluminum; Si, silicon; Ne, neon 1.14 (a) (f) K Pu 1.15 (a) element 1.16 (a) (d) (g) homogeneous mixture homogeneous mixture heterogeneous mixture (b) (g) Sn S (b) (c) (h) compound (b) (e) Cr Ar (d) (i) (c) B Hg element element heterogeneous mixture (e) (d) (c) (f) Ba compound compound homogeneous mixture CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE 1.21 density = 1.22 Strategy: We are given the density and volume of a liquid and asked to calculate the mass of the liquid Rearrange the density equation, Equation (1.1) of the text, to solve for mass mass 586 g = = 3.12 g/mL volume 188 mL density = mass volume Solution: mass = density × volume mass of ethanol = 1.23 ? °C = (°F − 32°F) × 0.798 g × 17.4 mL = 13.9 g mL 5°C 9°F 5°C = 35°C 9°F 5°C = − 11°C (12°F − 32°F) × °F 5°C (102°F − 32°F) × = 39°C 9°F 5°C = 1011°C (1852°F − 32°F) × 9°F 9°F ⎞ ⎛ ⎜ °C × 5°C ⎟ + 32°F ⎝ ⎠ (a) ? °C = (95°F − 32°F) × (b) ? °C = (c) ? °C = (d) ? °C = (e) ? °F = 9°F ⎞ ⎛ ? °F = ⎜ −273.15 °C × + 32°F = − 459.67°F °C ⎟⎠ ⎝ 1.24 Strategy: Find the appropriate equations for converting between Fahrenheit and Celsius and between Celsius and Fahrenheit given in Section 1.7 of the text Substitute the temperature values given in the problem into the appropriate equation (a) Conversion from Fahrenheit to Celsius ? °C = (°F − 32°F) × 5°C 9°F ? °C = (105°F − 32°F) × (b) 5°C = 41°C 9°F Conversion from Celsius to Fahrenheit 9°F ⎞ ⎛ ? °F = ⎜ °C × + 32°F 5°C ⎟⎠ ⎝ 9°F ⎞ ⎛ ? °F = ⎜ −11.5 °C × + 32°F = 11.3 °F °C ⎟⎠ ⎝ CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE (c) Conversion from Celsius to Fahrenheit 9°F ⎞ ⎛ ? °F = ⎜ °C × + 32°F °C ⎟⎠ ⎝ 9°F ⎞ ⎛ ? °F = ⎜ 6.3 × 103 °C × + 32°F = 1.1 × 104 °F 5°C ⎟⎠ ⎝ (d) Conversion from Fahrenheit to Celsius ? °C = (°F − 32°F) × 5°C 9°F ? °C = (451°F − 32°F) × 1.25 1.26 K = (°C + 273°C) 1K 1°C (a) K = 113°C + 273°C = 386 K (b) K = 37°C + 273°C = 3.10 × 10 K (c) K = 357°C + 273°C = 6.30 × 10 K (a) 2 1K 1°C °C = K − 273 = 77 K − 273 = −196°C K = (°C + 273°C) (b) °C = 4.2 K − 273 = −269°C (c) °C = 601 K − 273 = 328°C 1.29 (a) 2.7 × 10 1.30 (a) 10 −2 −8 3.56 × 10 (b) (c) 10 −8 (a) (b) (d) −2 = 0.0152 indicates that the decimal point must be moved places to the left 7.78 × 10 1.31 4.7764 × 10 indicates that the decimal point must be moved two places to the left 1.52 × 10 (b) 5°C = 233°C 9°F −8 −1 = 0.0000000778 145.75 + (2.3 × 10 ) = 145.75 + 0.23 = 1.4598 × 10 79500 2.5 × 102 = 7.95 × 104 2.5 × 102 −3 = 3.2 × 102 −4 −3 −3 (c) (7.0 × 10 ) − (8.0 × 10 ) = (7.0 × 10 ) − (0.80 × 10 ) = 6.2 × 10 (d) (1.0 × 10 ) × (9.9 × 10 ) = 9.9 × 10 10 −3 9.6 × 10 −2 1.32 CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE (a) Addition using scientific notation n Strategy: Let's express scientific notation as N × 10 When adding numbers using scientific notation, we must write each quantity with the same exponent, n We can then add the N parts of the numbers, keeping the exponent, n, the same Solution: Write each quantity with the same exponent, n n Let’s write 0.0095 in such a way that n = −3 We have decreased 10 by 10 , so we must increase N by 10 Move the decimal point places to the right 0.0095 = 9.5 × 10 −3 Add the N parts of the numbers, keeping the exponent, n, the same −3 9.5 × 10 −3 + 8.5 × 10 18.0 × 10 −3 The usual practice is to express N as a number between and 10 Since we must decrease N by a factor of n 10 to express N between and 10 (1.8), we must increase 10 by a factor of 10 The exponent, n, is increased by from −3 to −2 18.0 × 10 (b) −3 = 1.8 × 10 −2 Division using scientific notation n Strategy: Let's express scientific notation as N × 10 When dividing numbers using scientific notation, divide the N parts of the numbers in the usual way To come up with the correct exponent, n, we subtract the exponents Solution: Make sure that all numbers are expressed in scientific notation 653 = 6.53 × 10 Divide the N parts of the numbers in the usual way 6.53 ÷ 5.75 = 1.14 Subtract the exponents, n 1.14 × 10 (c) +2 − (−8) = 1.14 × 10 +2 + = 1.14 × 10 10 Subtraction using scientific notation n Strategy: Let's express scientific notation as N × 10 When subtracting numbers using scientific notation, we must write each quantity with the same exponent, n We can then subtract the N parts of the numbers, keeping the exponent, n, the same Solution: Write each quantity with the same exponent, n Let’s write 850,000 in such a way that n = This means to move the decimal point five places to the left 850,000 = 8.5 × 10 CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE Subtract the N parts of the numbers, keeping the exponent, n, the same 8.5 × 10 − 9.0 × 10 −0.5 × 10 The usual practice is to express N as a number between and 10 Since we must increase N by a factor of 10 n to express N between and 10 (5), we must decrease 10 by a factor of 10 The exponent, n, is decreased by from to −0.5 × 10 = −5 × 10 (d) Multiplication using scientific notation n Strategy: Let's express scientific notation as N × 10 When multiplying numbers using scientific notation, multiply the N parts of the numbers in the usual way To come up with the correct exponent, n, we add the exponents Solution: Multiply the N parts of the numbers in the usual way 3.6 × 3.6 = 13 Add the exponents, n 13 × 10 −4 + (+6) = 13 × 10 The usual practice is to express N as a number between and 10 Since we must decrease N by a factor of n 10 to express N between and 10 (1.3), we must increase 10 by a factor of 10 The exponent, n, is increased by from to 3 13 × 10 = 1.3 × 10 1.33 (a) (e) four three 1.34 (a) (e) one two or three 1.35 (a) 10.6 m 1.36 (a) Division (b) (f) two one (b) (f) (b) three one 0.79 g (c) (c) (g) five one (c) (g) three one or two (d) (h) two, three, or four two (d) four 16.5 cm Strategy: The number of significant figures in the answer is determined by the original number having the smallest number of significant figures Solution: 7.310 km = 1.283 5.70 km The (bolded) is a nonsignificant digit because the original number 5.70 only has three significant digits Therefore, the answer has only three significant digits The correct answer rounded off to the correct number of significant figures is: 1.28 (Why are there no units?) CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE (b) Subtraction Strategy: The number of significant figures to the right of the decimal point in the answer is determined by the lowest number of digits to the right of the decimal point in any of the original numbers Solution: Writing both numbers in decimal notation, we have 0.00326 mg − 0.0000788 mg 0.0031812 mg The bolded numbers are nonsignificant digits because the number 0.00326 has five digits to the right of the decimal point Therefore, we carry five digits to the right of the decimal point in our answer The correct answer rounded off to the correct number of significant figures is: 0.00318 mg = 3.18 × 10 (c) −3 mg Addition Strategy: The number of significant figures to the right of the decimal point in the answer is determined by the lowest number of digits to the right of the decimal point in any of the original numbers Solution: Writing both numbers with exponents = +7, we have 7 (0.402 × 10 dm) + (7.74 × 10 dm) = 8.14 × 10 dm Since 7.74 × 10 has only two digits to the right of the decimal point, two digits are carried to the right of the decimal point in the final answer 1.37 1.38 dm = 226 dm 0.1 m (a) ? dm = 22.6 m × (b) ? kg = 25.4 mg × (c) ? L = 556 mL × (d) ? g cm = 10.6 kg m3 0.001 g kg × = 2.54 × 10−5 kg mg 1000 g × 10−3 L = 0.556 L mL × 1000 g ⎛ × 10−2 m ⎞ ×⎜ ⎟ = 0.0106 g/cm ⎜ cm ⎟ kg ⎝ ⎠ (a) Strategy: The problem may be stated as ? mg = 242 lb A relationship between pounds and grams is given on the end sheet of your text (1 lb = 453.6 g) This relationship will allow conversion from pounds to grams A metric conversion is then needed to convert −3 grams to milligrams (1 mg = × 10 g) Arrange the appropriate conversion factors so that pounds and grams cancel, and the unit milligrams is obtained in your answer CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE Solution: The sequence of conversions is lb → grams → mg Using the following conversion factors, mg 453.6 g lb × 10−3 g we obtain the answer in one step: ? mg = 242 lb × 453.6 g mg × = 1.10 × 108 mg lb × 10−3 g Check: Does your answer seem reasonable? Should 242 lb be equivalent to 110 million mg? How many mg are in lb? There are 453,600 mg in lb (b) Strategy: The problem may be stated as 3 ? m = 68.3 cm Recall that cm = × 10 −2 3 m We need to set up a conversion factor to convert from cm to m Solution: We need the following conversion factor so that centimeters cancel and we end up with meters × 10−2 m cm Since this conversion factor deals with length and we want volume, it must therefore be cubed to give ⎛ × 10−2 m ⎞ × 10−2 m × 10−2 m × 10−2 m × × = ⎜ ⎟ ⎜ cm ⎟ cm cm cm ⎝ ⎠ We can write ⎛ × 10−2 m ⎞ ? m = 68.3 cm × ⎜ ⎟ = 6.83 × 10−5 m ⎜ cm ⎟ ⎝ ⎠ 3 Check: We know that cm = × 10 −6 −5 × 10 gives 6.83 × 10 −6 3 m We started with 6.83 × 10 cm Multiplying this quantity by (c) Strategy: The problem may be stated as ? L = 7.2 m 3 In Chapter of the text, a conversion is given between liters and cm (1 L = 1000 cm ) If we can convert 3 −2 m to cm , we can then convert to liters Recall that cm = × 10 m We need to set up two conversion 3 factors to convert from m to L Arrange the appropriate conversion factors so that m and cm cancel, and the unit liters is obtained in your answer CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE Solution: The sequence of conversions is 3 m → cm → L Using the following conversion factors, ⎛ cm ⎞ ⎜ ⎟ ⎜ × 10−2 m ⎟ ⎝ ⎠ 1L 1000 cm3 the answer is obtained in one step: ⎛ cm ⎞ 1L = 7.2 × 103 L ? L = 7.2 m3 × ⎜ ⎟ × ⎜ × 10−2 m ⎟ 1000 cm3 ⎝ ⎠ 3 Check: From the above conversion factors you can show that m = × 10 L Therefore, m would equal × 10 L, which is close to the answer (d) Strategy: The problem may be stated as ? lb = 28.3 μg A relationship between pounds and grams is given on the end sheet of your text (1 lb = 453.6 g) This relationship will allow conversion from grams to pounds If we can convert from μg to grams, we can then −6 convert from grams to pounds Recall that μg = × 10 g Arrange the appropriate conversion factors so that μg and grams cancel, and the unit pounds is obtained in your answer Solution: The sequence of conversions is μg → g → lb Using the following conversion factors, × 10−6 g μg lb 453.6 g we can write ? lb = 28.3 μg × × 10−6 g lb × = 6.24 × 10−8 lb μg 453.6 g Check: Does the answer seem reasonable? What number does the prefix μ represent? Should 28.3 μg be a very small mass? 1.39 1255 m mi 3600 s × × = 2808 mi/h 1s 1609 m 1h 1.40 Strategy: The problem may be stated as ? s = 365.24 days You should know conversion factors that will allow you to convert between days and hours, between hours and minutes, and between minutes and seconds Make sure to arrange the conversion factors so that days, hours, and minutes cancel, leaving units of seconds for the answer CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE Solution: The sequence of conversions is days → hours → minutes → seconds Using the following conversion factors, 24 h day 60 1h 60 s we can write ? s = 365.24 day × 24 h 60 60 s × × = 3.1557 × 107 s day 1h Check: Does your answer seem reasonable? Should there be a very large number of seconds in year? 1.609 km 1000 m 1s × × × = 8.3 mi km 60 s 3.00 × 10 m 1.41 (93 × 106 mi) × 1.42 (a) ? in/s = (b) ? m/min = (c) ? km/h = 1.43 6.0 ft × 168 lb × mi 5280 ft 12 in × × × = 81 in/s 13 mi ft 60 s mi 1609 m × = 1.2 × 102 m/min 13 mi mi 1609 m km 60 × × × = 7.4 km/h 13 mi 1000 m 1h 1m = 1.8 m 3.28 ft 453.6 g kg × = 76.2 kg lb 1000 g 55 mi 1.609 km × = 88 km/h 1h mi 1.44 ? km/h = 1.45 62 m mi 3600 s × × = 1.4 × 102 mph 1s 1609 m 1h 1.46 0.62 ppm Pb = 0.62 g Pb × 106 g blood 0.62 g Pb 6.0 × 103 g of blood × = 3.7 × 10−3 g Pb × 10 g blood 10 CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE 1.47 (a) 1.42 yr × 365 day 24 h 3600 s 3.00 × 108 m mi × × × × = 8.35 × 1012 mi yr day 1h 1s 1609 m (b) 32.4 yd × 36 in 2.54 cm × = 2.96 × 103 cm yd in (c) 3.0 × 1010 cm in ft × × = 9.8 × 108 ft/s 1s 2.54 cm 12 in (a) ? m = 185 nm × (b) ? s = (4.5 × 109 yr) × (c) ⎛ 0.01 m ⎞ −5 ? m = 71.2 cm3 × ⎜ ⎟ = 7.12 × 10 m cm ⎝ ⎠ (d) ⎛ cm ⎞ 1L ? L = 88.6 m × ⎜ = 8.86 × 104 L ⎟ × ⎜ × 10−2 m ⎟ 1000 cm3 ⎝ ⎠ 1.48 × 10−9 m = 1.85 × 10−7 m nm 365 day 24 h 3600 s × × = 1.4 × 1017 s yr day 1h 3 2.70 g density = 1.50 density = 1.51 Substance (a) water (b) carbon (c) iron (d) hydrogen gas (e) sucrose (f) table salt (g) mercury (h) gold (i) air 1.52 See Section 1.6 of your text for a discussion of these terms cm3 × ⎛ cm ⎞ kg 3 ×⎜ ⎟ = 2.70 × 10 kg/m 1000 g ⎝ 0.01 m ⎠ 1.49 0.625 g 1L mL × × = 6.25 × 10−4 g/cm 1L 1000 mL cm3 Qualitative Statement colorless liquid black solid (graphite) rusts easily colorless gas tastes sweet tastes salty liquid at room temperature a precious metal a mixture of gases Quantitative Statement freezes at 0°C density = 2.26 g/cm density = 7.86 g/cm melts at −255.3°C at 0°C, 179 g of sucrose dissolves in 100 g of H2O melts at 801°C boils at 357°C density = 19.3 g/cm contains 20% oxygen by volume (a) Chemical property Iron has changed its composition and identity by chemically combining with oxygen and water (b) Chemical property The water reacts with chemicals in the air (such as sulfur dioxide) to produce acids, thus changing the composition and identity of the water (c) Physical property The color of the hemoglobin can be observed and measured without changing its composition or identity CHAPTER 24: ORGANIC CHEMISTRY (e) 659 This is an eight carbon chain with methyl groups on the 3, 4, and carbons CH CH CH CH 24.28 CH CH CH CH CH CH CH The hydrogen atoms have been omitted from the skeletal structure for simplicity H3C (a) H C H C C (b) H CH2CH3 H C CH2CH3 CH2CH3 H H3C HC H C H C CH3 C CH2CH2CH3 HC CH2CH3 (c) C (d) 24.31 CH Br Cl (b) (a) CH CH CH CH CH 24.32 (c) H 3C CH Strategy: We follow the IUPAC rules and use the information in Table 24.2 of the text When a benzene ring has more than two substituents, you must specify the location of the substituents with numbers Remember to number the ring so that you end up with the lowest numbering scheme as possible, giving preference to alphabetical order Solution: (a) Since a chloro group comes alphabetically before a methyl group, let’s start by numbering the top carbon of the ring as If we number clockwise, this places the second chloro group on carbon and a methyl group on carbon This compound is 1,3−dichloro−4−methylbenzene (b) If we start numbering counterclockwise from the bottom carbon of the ring, the name is 2−ethyl−1,4−dinitrobenzene Numbering clockwise from the top carbon gives 3−ethyl−1,4−dinitrobenzene Numbering as low as possible, the correct name is 2−ethyl−1,4−dinitrobenzene (c) Again, keeping the numbers as low as possible, the correct name for this compound is 1,2,4,5−tetramethylbenzene You should number clockwise from the top carbon of the ring 660 CHAPTER 24: ORGANIC CHEMISTRY 24.35 (a) There is only one isomer: CH3OH (b) There are two structures with this molecular formula: CH3−CH2−OH and CH3−O−CH3 (c) The cyclic di-alcohol has geometric isomers O CH3CH2 H C O OH CH3 H H H C H (d) O O C C H H C H H O CH3 OH H C H C C O O H H C C H H OH C H There are two possible alcohols and one ether CH3CH2CH2OH CH3CH2 CH3CHCH3 O CH3 OH 24.36 Strategy: Learning to recognize functional groups requires memorization of their structural formulas Table 24.4 of the text shows a number of the important functional groups Solution: (a) H3C−O−CH2−CH3 contains a C−O−C group and is therefore an ether 24.37 (b) This molecule contains an RNH2 group and is therefore an amine (c) This molecule is an aldehyde It contains a carbonyl group in which one of the atoms bonded to the carbonyl carbon is a hydrogen atom (d) This molecule also contains a carbonyl group However, in this case there are no hydrogen atoms bonded to the carbonyl carbon This molecule is a ketone (e) This molecule contains a carboxyl group It is a carboxylic acid (f) This molecule contains a hydroxyl group (−OH) It is an alcohol (g) This molecule has both an RNH2 group and a carboxyl group It is therefore both an amine and a carboxylic acid, commonly called an amino acid Aldehydes can be oxidized easily to carboxylic acids The oxidation reaction is: O O CH C H O2 CH C OH CHAPTER 24: ORGANIC CHEMISTRY 661 Oxidation of a ketone requires that the carbon chain be broken: O CH 24.38 C CH O2 H 2O + CO2 Alcohols react with carboxylic acids to form esters The reaction is: HCOOH + CH3OH ⎯⎯ → HCOOCH3 + H2O The structure of the product is: O H C O CH 24.39 (methyl formate) Alcohols can be oxidized to ketones under controlled conditions The possible starting compounds are: OH OH CH CH CH CHCH OH CH CH CHCH CH (CH )2 CHCHCH The corresponding products are: O O CH CH CH CCH O CH CH CCH CH (CH )2 CHCCH Why isn't the alcohol CH3CH2CH2CH2CH2OH a possible starting compound? 24.40 The fact that the compound does not react with sodium metal eliminates the possibility that the substance is an alcohol The only other possibility is the ether functional group There are three ethers possible with this molecular formula: CH3−CH2−O−CH2−CH3 CH3−CH2−CH2−O−CH3 (CH3)2CH−O−CH3 Light−induced reaction with chlorine results in substitution of a chlorine atom for a hydrogen atom (the other product is HCl) For the first ether there are only two possible chloro derivatives: ClCH2−CH2−O−CH2−CH3 CH3−CHCl−O−CH2−CH3 For the second there are four possible chloro derivatives Three are shown below Can you draw the fourth? CH3−CHCl−CH2−O−CH3 CH3−CH2−CHCl−O−CH3 CH2Cl−CH2−CH2−O−CH3 For the third there are three possible chloro derivatives: CH3 CH2Cl CH3 CH O CH3 CH3 CH Cl O The (CH3)2CH−O−CH3 choice is the original compound CH2Cl (CH3)2 CH O CH3 662 CHAPTER 24: ORGANIC CHEMISTRY 24.41 (a) The product is similar to that in Problem 24.38 O (b) H C CH 3CH 2O Addition of hydrogen to an alkyne gives an alkene H C C CH + H H 2C CH CH CH CH The alkene can also add hydrogen to form an alkane H 2C (c) CH CH + H CH HBr will add to the alkene as shown (Note: the carbon atoms at the double bond have been omitted for simplicity) C2H H + H Br C2H CH Br CH H H How you know that the hydrogen adds to the CH2 end of the alkene? 24.42 (a) ketone (b) 24.43 The four isomers are: CH ester (c) ether CH 2Cl CH CH Cl Cl Cl 24.44 This is a Hess's Law problem See Chapter If we rearrange the equations given and multiply times the necessary factors, we have: 2CO2(g) + 2H2O(l) ⎯⎯ → C2H4(g) + 3O2(g) C2H2(g) + O2(g) ⎯⎯ → 2CO2(g) + H2O(l) ΔH° = 1411 kJ/mol ΔH° = −1299.5 kJ/mol H2(g) + 12 O2(g) ⎯⎯ → H2O(l) ΔH° = −285.8 kJ/mol C2H2(g) + H2(g) ⎯⎯ → C2H4(g) ΔH° = −174 kJ/mol The heat of hydrogenation for acetylene is −174 kJ/mol 24.45 (a) Cyclopropane because of the strained bond angles (The C−C−C angle is 60° instead of 109.5°) (b) Ethylene because of the C=C bond (c) Acetaldehyde (susceptible to oxidation) CHAPTER 24: ORGANIC CHEMISTRY 24.46 24.47 To form a hydrogen bond with water a molecule must have at least one H−F, H−O, or H−N bond, or must contain an O, N, or F atom The following can form hydrogen bonds with water: (a) carboxylic acids (c) (a) The empirical formula is: ethers (d) aldehydes (f) amines mol H = 3.17 mol H 1.008 g H H: 3.2 g H × C: 37.5 g C × mol C = 3.12 mol C 12.01 g C F: 59.3 g F × mol F = 3.12 mol F 19.00 g F This gives the formula, H3.17C3.12F3.12 Dividing by 3.12 gives the empirical formula, HCF (b) When temperature and amount of gas are constant, the product of pressure times volume is constant (Boyle's law) (2.00 atm)(0.322 L) (1.50 atm)(0.409 L) (1.00 atm)(0.564 L) (0.50 atm)(1.028 L) = = = = 0.664 atm⋅L 0.614 atm⋅L 0.564 atm⋅L 0.514 atm⋅L The substance does not obey the ideal gas law (c) Since the gas does not obey the ideal gas equation exactly, the molar mass will only be approximate Gases obey the ideal gas law best at lowest pressures We use the 0.50 atm data n = PV (0.50 atm)(1.028 L) = = 0.0172 mol RT (0.0821 L ⋅ atm/K ⋅ mol)(363 K) Molar mass = 1.00 g = 58.1 g/mol 0.0172 mol This is reasonably close to C2H2F2 (64 g/mol) (d) The C2H2F2 formula is that of difluoroethylene Three isomers are possible The carbon atoms are omitted for simplicity (see Problem 24.17) H F F F H F H F H H F H Only the third isomer has no dipole moment 24.48 (e) The name is trans−difluoroethylene (a) rubbing alcohol (b) vinegar (c) moth balls (d) organic synthesis (e) organic synthesis (f) antifreeze (g) fuel (natural gas) (h) synthetic polymers 663 664 CHAPTER 24: ORGANIC CHEMISTRY 24.49 In any stoichiometry problem, you must start with a balanced equation The balanced equation for the combustion reaction is: 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l) To find the number of moles of octane in one liter, use density as a conversion factor to find grams of octane, then use the molar mass of octane to convert to moles of octane The strategy is: L octane → mL octane → g octane → mol octane 1.0 L × mol C8 H18 1000 mL 0.70 g C8 H18 × × = 6.13 mol C8 H18 1L mL C8 H18 114.2 g C8 H18 Using the mole ratio from the balanced equation, the number of moles of oxygen used is: 6.13 mol C8 H18 × 25 mol O2 = 76.6 mol O 2 mol C8 H18 From the ideal gas equation, we can calculate the volume of oxygen VO2 = nO2 RT P = (76.6 mol)(293 K) 0.0821 L ⋅ atm × = 1.84 × 103 L 1.00 atm mol ⋅ K Air is only 22% O2 by volume Thus, 100 L of air will contain 22 L of O2 Setting up the appropriate conversion factor, we find that the volume of air is: ? vol of air = (1.84 × 103 L O2 ) × 24.50 100 L air = 8.4 × 103 L air 22 L O (a) 2−butyne has three C−C sigma bonds (b) Anthracene is: There are sixteen C−C sigma bonds (c) C C C C C There are six C−C sigma bonds 24.51 (a) A benzene ring has six carbon-carbon bonds; hence, benzene has six C−C sigma bonds (b) Cyclobutane has four carbon-carbon bonds; hence, cyclobutane has four sigma bonds CHAPTER 24: ORGANIC CHEMISTRY (c) 665 Looking at the carbon skeleton of 3−ethyl−2−methylpentane, you should find seven C−C sigma bonds C C C C C C C C 24.52 (a) The easiest way to calculate the mg of C in CO2 is by mass ratio There are 12.01 g of C in 44.01 g CO2 or 12.01 mg C in 44.01 mg CO2 ? mg C = 57.94 mg CO2 × 12.01 mg C = 15.81 mg C 44.01 mg CO Similarly, ? mg H = 11.85 mg H O × 2.016 mg H = 1.326 mg H 18.02 mg H O The mg of oxygen can be found by difference ? mg O = 20.63 mg Y − 15.81 mg C − 1.326 mg H = 3.49 mg O (b) Step 1: Calculate the number of moles of each element present in the sample Use molar mass as a conversion factor mol C = 1.316 × 10−3 mol C 12.01 g C ? mol C = (15.81 × 10−3 g C) × Similarly, ? mol H = (1.326 × 10−3 g H) × ? mol O = (3.49 × 10−3 g O) × mol H = 1.315 × 10−3 mol H 1.008 g H mol O = 2.18 × 10−4 mol O 16.00 g O Thus, we arrive at the formula C1.316 × 10−3 H1.315 × 10−3 O2.18 × 10−4 , which gives the identity and the ratios of atoms present However, chemical formulas are written with whole numbers Step 2: Try to convert to whole numbers by dividing all the subscripts by the smallest subscript C: 1.316 × 10−3 2.18 × 10−4 = 6.04 ≈ H: 1.315 × 10−3 2.18 × 10−4 = 6.03 ≈ O: 2.18 × 10−4 2.18 × 10−4 = 1.00 This gives us the empirical formula, C6H6O (c) The presence of six carbons and a corresponding number of hydrogens suggests a benzene derivative A plausible structure is shown below OH 666 CHAPTER 24: ORGANIC CHEMISTRY 24.53 The structural isomers are: 1,2−dichlorobutane CH 1,3−dichlorobutane * CH Cl CH CH CH 2Cl 2,3−dichlorobutane *CHCl CH CH 2Cl 1,1−dichlorobutane CH CH 2Cl CH CH CH 2Cl 2,2−dichlorobutane CH CH CH CH Cl2 1,3−dichloro−2−methylpropane CH 2Cl CH 1,4−dichlorobutane *CH Cl CH * CHCl CH CH CCl2 CH 1,2−dichloro−2−methylpropane CH CH 2Cl CH CCl CH 2Cl CH 1,1−dichloro−2−methylpropane CH CH CH Cl2 CH The asterisk identifies the asymmetric carbon atom 24.54 First, calculate the moles of each element C: (9.708 × 10−3 g CO2 ) × mol CO mol C × = 2.206 × 10−4 mol C 44.01 g CO mol CO H: (3.969 × 10−3 g H O) × mol H O mol H × = 4.405 × 10−4 mol H 18.02 g H O mol H O The mass of oxygen is found by difference: 3.795 mg compound − (2.649 mg C + 0.445 mg H) = 0.701 mg O O: (0.701 × 10−3 g O) × mol O = 4.38 × 10−5 mol O 16.00 g O This gives the formula is C2.206 × 10−4 H 4.405 × 10−4 O 4.38 × 10−5 Dividing by the smallest number of moles gives the empirical formula, C5H10O We calculate moles using the ideal gas equation, and then calculate the molar mass n = PV (1.00 atm)(0.0898 L) = = 0.00231 mol RT (0.0821 L ⋅ atm/K ⋅ mol)(473 K) molar mass = g of substance 0.205 g = = 88.7 g/mol mol of substance 0.00231 mol CHAPTER 24: ORGANIC CHEMISTRY 667 The formula mass of C5H10O is 86.13 g, so this is also the molecular formula Three possible structures are: CH2 H2C H2C CH2 H2C CH2 CH2 H2C O 24.55 (a) CH O CH3 H2C CH CH2 O CH2 In comparing the compound in part (a) with the starting alkyne, it is clear that a molecule of HBr has been added to the triple bond The reaction is: Br CH CH H C C CH CH + HBr H C C CH CH H (b) This compound can be made from the product formed in part (a) by addition of bromine to the double bond Br Br CH Br CH H C C CH CH + Br H C C CH CH Br (c) This compound can be made from the product of part (a) by addition of hydrogen to the double bond Br CH H Br CH H C C CH CH + H H C C CH CH H 24.56 A carbon atom is asymmetric if it is bonded to four different atoms or groups In the given structures the asymmetric carbons are marked with an asterisk (*) H CH (a) CH CH CH * O * CH NH C H NH (b) H * Br H * Br CH3 668 CHAPTER 24: ORGANIC CHEMISTRY 24.57 The isomers are: Cl Cl Cl Cl Cl Cl Cl Cl Did you have more isomers? Remember that benzene is a planar molecule; "turning over" a structure does not create a new isomer 24.58 Acetone is a ketone with the formula, CH3COCH3 We must write the structure of an aldehyde that has the same number and types of atoms (C3H6O) Removing the aldehyde functional group (−CHO) from the formula leaves C2H5 This is the formula of an ethyl group The aldehyde that is a structural isomer of acetone is: O CH CH C H 24.59 The structures are: H2 C (a) (b) H2C H3C C CH2 C H3C OH H H H C C C C H H H H H H H2C (c) CH3 CH2 (e) (d) Br Br alcohol CH 24.61 Ethanol has a melting point of −117.3°C, a boiling point of +78.5°C, and is miscible with water Dimethyl ether has a melting point of −138.5°C, a boiling point of −25°C (it is a gas at room temperature), and dissolves in water to the extent of 37 volumes of gas to one volume of water 24.62 In Chapter 11, we found that salts with their electrostatic intermolecular attractions had low vapor pressures and thus high boiling points Ammonia and its derivatives (amines) are molecules with dipole−dipole attractions If the nitrogen has one direct N−H bond, the molecule will have hydrogen bonding Even so, these molecules will have much weaker intermolecular attractions than ionic species and hence higher vapor pressures Thus, if we could convert the neutral ammonia−type molecules into salts, their vapor pressures, and thus associated odors, would decrease Lemon juice contains acids which can react with ammonia−type (amine) molecules to form ammonium salts ⎯⎯ → NH4 (c) C (a) + ether C 24.60 NH3 + H (b) CH + aldehyde RNH2 + H (d) + carboxylic acid ⎯⎯ → RNH3 + (e) amine CH3 CHAPTER 24: ORGANIC CHEMISTRY 24.63 Cyclohexane readily undergoes halogenation; for example, its reaction with bromine can be monitored by seeing the red color of bromine fading Benzene does not react with halogens unless a catalyst is present 24.64 Marsh gas (methane, CH4); grain alcohol (ethanol, C2H5OH); wood alcohol (methanol, CH3OH); rubbing alcohol (isopropyl alcohol, (CH3)2CHOH); antifreeze (ethylene glycol, CH2OHCH2OH); mothballs (naphthalene, C10H8); vinegar (acetic acid, CH3COOH) 24.65 A mixture of cis and trans isomers would imply some sort of random addition mechanism in which one hydrogen atom at a time adds to the molecule 669 The formation of pure cis or pure trans isomer indicates a more specific mechanism For example, a pure cis product suggests simultaneous addition of both hydrogen atoms in the form of a hydrogen molecule to one side of the alkyne In practice, the cis isomer is formed 24.66 The asymmetric carbons are shown by asterisks: (a) H H H H C *C C Cl (b) CH H H H Cl H 24.67 OH CH *C *C CH OH (c) All of the carbon atoms in the ring are asymmetric Therefore there are five asymmetric carbon atoms (a) Sulfuric acid ionizes as follows: + − H2SO4(aq) ⎯⎯ → H (aq) + HSO4 (aq) + − The cation (H ) and anion (HSO4 ) add to the double bond in propene according to Markovnikov’s rule: OSO3 H CH CH CH + H + + HSO4 − CH C CH H Reaction of the intermediate with water yields isopropanol: OSO3 H CH C CH + H O H OH CH C CH + H SO4 H Since sulfuric acid is regenerated, it plays the role of a catalyst (b) The other structure containing the −OH group is CH3−CH2−CH2−OH propanol (c) From the structure of isopropanol shown above, we see that the molecule does not have an asymmetric carbon atom Therefore, isopropanol is achiral 670 CHAPTER 24: ORGANIC CHEMISTRY (d) 24.68 Isopropanol is fairly volatile (b.p = 82.5°C), and the −OH group allows it to form hydrogen bonds with water molecules Thus, as it evaporates, it produces a cooling and soothing effect on the skin It is also less toxic than methanol and less expensive than ethanol The red bromine vapor absorbs photons of blue light and dissociates to form bromine atoms Br2 → 2Br• The bromine atoms collide with methane molecules and abstract hydrogen atoms Br• + CH4 → HBr + •CH3 The methyl radical then reacts with Br2, giving the observed product and regenerating a bromine atom to start the process over again: •CH3 + Br2 → CH3Br + Br• Br• + CH4 → HBr + •CH3 24.69 and so on From the molar mass of the alkene, we deduce that there can only be carbon atoms Therefore, the alkene is CH2=CH−CH3 (propene) The reactions are: H2C CH CH3 H2SO4 H2O OH H3C CH CH3 K2Cr2O7 H+ O H3C C CH3 Markovnikov's rule 24.70 2−butanone is O H3C C CH2 CH3 CH2 CH3 Reduction with LiAlH4 produces 2-butanol OH H3C C H This molecule possesses an asymmetric carbon atom and should be chiral However, the reduction produces an equimolar d and l isomers; that is, a racemic mixture (see Section 22.4 of the text) Therefore, the optical rotation as measured in a polarimeter is zero CHAPTER 24: ORGANIC CHEMISTRY 24.71 671 The structures of three alkenes that yield 2-methylbutane CH3 CH3CHCH2CH3 on hydrogenation are: CH3 CH3 H2C 24.72 C CH2 CH3 H3C C CH3 CH CH3 H3C CH CH CH2 To help determine the molecular formula of the alcohol, we can calculate the molar mass of the carboxylic acid, and then determine the molar mass of the alcohol from the molar mass of the acid Grams of carboxylic acid are given (4.46 g), so we need to determine the moles of acid to calculate its molar mass The number of moles in 50.0 mL of 2.27 M NaOH is 2.27 mol NaOH × 50.0 mL = 0.1135 mol NaOH 1000 mL soln The number of moles in 28.7 mL of 1.86 M HCl is 1.86 mol HCl × 28.7 mL = 0.05338 mol HCl 1000 mL soln The difference between the above two numbers is the number of moles of NaOH reacted with the carboxylic acid 0.1135 mol − 0.05338 mol = 0.06012 mol This is the number of moles present in 4.46 g of the carboxylic acid The molar mass is M = 4.46 g = 74.18 g/mol 0.06012 mol A carboxylic acid contains a −COOH group and an alcohol has an −OH group When the alcohol is oxidized to a carboxylic acid, the change is from −CH2OH to −COOH Therefore, the molar mass of the alcohol is 74.18 g − 16 g + (2)(1.008 g) = 60.2 g/mol With a molar mass of 60.2 g/mol for the alcohol, there can only be oxygen atom and carbon atoms in the molecule, so the formula must be C3H8O The alcohol has one of the following two molecular formulas OH CH3CH2CH2OH H3C CH CH3 672 CHAPTER 24: ORGANIC CHEMISTRY 24.73 There are 18 structural isomers and 10 of them are chiral The asymmetric carbon atoms are marked with an asterisk C C C C C C OH OH OH C C C C *C C C C C C C C C C OH C C C C *C C OH C *C C C C *C C C *C C C C OH OH C C OH C C C C C C C C *C C C C C C C C OH C C C C OH C C C OH C C *C C C C *C C C OH C C C C C C C C OH C C C C C C C C *C OH C C OH C C C C C C C OH C OH C *C C C C C OH C C C CHAPTER 24: ORGANIC CHEMISTRY 24.74 (a) 673 Reaction between glycerol and carboxylic acid (formation of an ester) O CH O C R CH O (b) CH O C R' NaOH H 2O CH O CH O C CH OH OH OH O + R Glycerol C − O Na + Fatty acid salts (soap) R'' A fat or oil (c) Molecules having more C=C bonds are harder to pack tightly together Consequently, the compound has a lower melting point (d) H2 gas with either a heterogeneous or homogeneous catalyst would be used See Section 13.6 of the text (e) Number of moles of Na2S2O3 reacted is: 20.6 mL × 0.142 mol Na 2S2 O3 1L × = 2.93 × 10−3 mol Na 2S2 O3 1000 mL 1L The mole ratio between I2 and Na2S2O3 is 1:2 The number of grams of I2 left over is: (2.93 × 10−3 mol Na 2S2 O3 ) × mol I2 253.8 g I2 × = 0.372 g I2 mol Na 2S2 O3 mol I Number of grams of I2 reacted is: (43.8 − 0.372)γ = 43.4 g I2 The iodine number is the number of grams of iodine that react with 100 g of corn oil iodine number = 43.4 g I × 100 g corn oil = 123 35.3 g corn oil ... = 75.8 g chlorine 18 CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE The chlorine solution is only percent chlorine by mass We can now calculate the volume of chlorine solution that must be added to... the mixed solution should be based on the percentage of each liquid and its density Because the solid object is suspended in the mixed solution, it should have the same density as this solution. .. When the carbon dioxide gas is released, the mass of the solution will decrease If we know the starting mass of the solution and the mass of solution after the reaction is complete (given in the