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10.2: Intermolecular Forces
1. What type of intermolecular force do all substances have?
2. What is necessary for a molecule to experience dipole-dipole interactions?
3. What is necessary for a molecule to experience hydrogen bonding?
4. How does varying the temperature change the preferred phase of a substance?
5. Identify the strongest intermolecular force present in each substance.
1. He
2. CHCl3
3. HOF
6. Identify the strongest intermolecular force present in each substance.
1. CH3OH
2. (CH3)2CO
3. N2
7. Identify the strongest intermolecular force present in each substance.
1. HBr
2. C6H5NH2
3. CH4
8. Identify the strongest intermolecular force present in each substance.
1. C10H22
2. HF
3. glucose
Answers
1. dispersion force
2. An H atom must be bonded to an N, O, or F atom.
1. dispersion forces
2. dipole-dipole interactions
3. hydrogen bonding
1. dipole-dipole interactions
2. hydrogen bonding
3. dispersion forces
10.3: Phase Transitions - Melting, Boiling, and Subliming
1. What is the difference between melting and solidification?
2. What is the difference between boiling and condensation?
3. Describe the molecular changes when a solid becomes a liquid.
4. Describe the molecular changes when a liquid becomes a gas.
5. What is the energy change when 78.0 g of Hg melt at −38.8°C?
6. What is the energy change when 30.8 g of Al solidify at 660°C?
7. What is the energy change when 111 g of Br2 boil at 59.5°C?
8. What is the energy change when 98.6 g of H2O condense at 100°C?
9. Each of the following statements is incorrect. Rewrite them so they are correct.
1. Temperature changes during a phase change.
2. The process of a liquid becoming a gas is called sublimation.
10. Each of the following statements is incorrect. Rewrite them so they are correct.
1. The volume of a gas contains only about 10% matter, with the rest being empty space.
2. ΔHsub is equal to ΔHvap.
11. Write the chemical equation for the melting of elemental sodium.
12. Write the chemical equation for the solidification of benzene (C6H6).
13. Write the chemical equation for the sublimation of CO2.
14. Write the chemical equation for the boiling of propanol (C3H7OH).
15. What is the ΔHsub of H2O? (Hint: see Table 10.3.1 "Enthalpies of Fusion for Various Substances" and Table 10.3.2 "Enthalpies of Vaporization for Various Substances".)
16. The ΔHsub of I2 is 60.46 kJ/mol, while its ΔHvap is 41.71 kJ/mol. What is the ΔHfus of I2?
Answers
1. Melting is the phase change from a solid to a liquid, whereas solidification is the phase change from a liquid to a solid.
2. The molecules have enough energy to move about each other but not enough to completely separate from each other.
3. 890 J
4. 10.7 kJ
1. Temperature does not change during a phase change.
2. The process of a liquid becoming a gas is called boiling; the process of a solid becoming a gas is called sublimation.
5. Na(s) → Na(ℓ)
6. CO2(s) → CO2(g)
7. 46.69 kJ/mol
10.4: Properties of Liquids
1. What is the difference between evaporation and boiling?
2. What is the difference between a gas and vapor?
3. Define normal boiling point in terms of vapor pressure.
4. Is the boiling point higher or lower at higher environmental pressures? Explain your answer.
5. Referring to Fig. 10.4.3, if the pressure is 400 torr, which liquid boils at the lowest temperature?
6. Referring to Fig. 10.4.3, if the pressure is 100 torr, which liquid boils at the lowest temperature?
7. Referring to Fig. 10.4.3, estimate the boiling point of ethanol at 200 torr.
8. Referring to Fig. 10.4.3, at approximately what pressure is the boiling point of water 40°C?
9. Explain how surface tension works.
10. From what you know of intermolecular forces, which substance do you think might have a higher surface tension—ethyl alcohol or mercury? Why?
11. Under what conditions would a liquid demonstrate a capillary rise?
12. Under what conditions would a liquid demonstrate a capillary depression?
Answers
1. Evaporation occurs when a liquid becomes a gas at temperatures below that liquid’s boiling point, whereas boiling is the conversion of a liquid to a gas at the liquid’s boiling point.
2. the temperature at which the vapor pressure of a liquid is 760 torr
3. diethyl ether
4. 48°C
5. Surface tension is an imbalance of attractive forces between liquid molecules at the surface of a liquid.
6. Adhesion must be greater than cohesion.
10.5: Solids
1. What is the difference between a crystalline solid and an amorphous solid?
2. What two properties do solids have in common? What two properties of solids can vary?
3. Explain how the bonding in an ionic solid explains some of the properties of these solids.
4. Explain how the bonding in a molecular solid explains some of the properties of these solids.
5. Explain how the bonding in a covalent network solid explains some of the properties of these solids.
6. Explain how the bonding in a metallic solid explains some of the properties of these solids.
7. Which type(s) of solid has/have high melting points?
8. Which type(s) of solid conduct(s) electricity in their solid state? In their liquid state?
9. Which type of solid(s) is/are considered relatively soft?
1. Which type of solid(s) is/are considered very hard?
2. Predict the type of solid exhibited by each substance.
1. Hg
2. PH3
3. CaF2
3. Predict the type of solid exhibited by each substance.
1. (CH2)n (polyethylene, a form of plastic)
2. PCl3
3. NH4Cl
4. Predict the type of solid exhibited by each substance.
1. SO3
2. Br2
3. Na2SO3
5. Predict the type of solid exhibited by each substance.
1. BN (boron nitride, a diamond-like compound)
2. B2O3
3. NaBF4
6. Predict the type of solid exhibited by each substance.
1. H2S
2. Si
3. CsF
7. Predict the type of solid exhibited by each substance.
1. Co
2. CO
3. CaCO3
Answers
1. At the atomic level, a crystalline solid has a regular arrangement of atoms, whereas an amorphous solid has a random arrangement of atoms.
2. The oppositely charged ions are very strongly held together, so ionic crystals have high melting points. Ionic crystals are also brittle because any distortion of the crystal moves same-charged ions closer to each other, so they repel.
3. The covalent network solid is essentially one molecule, making it very hard and giving it a very high melting point.
4. ionic solids, covalent network solids
5. molecular solids
1. metallic
2. molecular solid
3. ionic crystal
1. molecular solid
2. molecular solid
3. ionic crystal
1. molecular solid
2. molecular solid
3. ionic crystal | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Ball_et_al._(Beginning_Chemistry)/10%3A_Solids_and_Liquids.txt |
11.2: Definitions
1. Define solute and solvent.
2. Define saturated, unsaturated, and supersaturated.
3. A solution is prepared by combining 2.09 g of CO2 and 35.5 g of H2O. Identify the solute and solvent.
4. A solution is prepared by combining 10.3 g of Hg(ℓ) and 45.0 g of Ag(s). Identify the solute and solvent.
1. Use Table 11.2.1 - Solubilities of Some Ionic Compounds, to decide if a solution containing 45.0 g of NaCl per 100 g of H2O is unsaturated, saturated, or supersaturated.
2. Use Table 11.2.1 - Solubilities of Some Ionic Compounds, to decide if a solution containing 0.000092 g of AgCl per 100 g of H2O is unsaturated, saturated, or supersaturated.
3. Would the solution in Exercise 5 be described as dilute or concentrated? Explain your answer.
4. Would the solution in Exercise 6 be described as dilute or concentrated? Explain your answer.
1. Identify a solute from Table 11.2.1 - Solubilities of Some Ionic Compounds, whose saturated solution can be described as dilute.
2. Identify a solute from Table 11.2.1 - Solubilities of Some Ionic Compounds, whose saturated solution can be described as concentrated.
3. Which solvent is Br2 more likely soluble in—CH3OH or C6H6?
4. Which solvent is NaOH more likely soluble in—CH3OH or C6H6?
5. Compounds with the formula CnH2n + 1OH are soluble in H2O when n is small but not when n is large. Suggest an explanation for this phenomenon.
6. Glucose has the following structure:
What parts of the molecule indicate that this substance is soluble in water?
Answers
1. The solvent is the majority component of a solution, whereas the solute is the minority component of a solution.
2.
3. solute: CO2; solvent: H2O
4.
5. supersaturated
6.
7. concentrated because there is a lot of solute
8.
9. AgCl or CaCO3
10.
11. C6H6
12.
13. The nonpolar end dominates intermolecular forces when n is large.
11.4: Dilutions and Concentrations
1. What is the difference between dilution and concentration?
2. What quantity remains constant when you dilute a solution?
3. A 1.88 M solution of NaCl has an initial volume of 34.5 mL. What is the final concentration of the solution if it is diluted to 134 mL?
4. A 0.664 M solution of NaCl has an initial volume of 2.55 L. What is the final concentration of the solution if it is diluted to 3.88 L?
5. If 1.00 mL of a 2.25 M H2SO4 solution needs to be diluted to 1.00 M, what will be its final volume?
6. If 12.00 L of a 6.00 M HNO3 solution needs to be diluted to 0.750 M, what will be its final volume?
7. If 665 mL of a 0.875 M KBr solution are boiled gently to concentrate the solute to 1.45 M, what will be its final volume?
8. If 1.00 L of an LiOH solution is boiled down to 164 mL and its initial concentration is 0.00555 M, what is its final concentration?
9. How much water must be added to 75.0 mL of 0.332 M FeCl3(aq) to reduce its concentration to 0.250 M?
1. How much water must be added to 1.55 L of 1.65 M Sc(NO3)3(aq) to reduce its concentration to 1.00 M?
Answers
1. Dilution is a decrease in a solution’s concentration, whereas concentration is an increase in a solution’s concentration.
2.
3. 0.484 M
4.
5. 2.25 mL
6.
7. 401 mL
8.
9. 24.6 mL
11.5: Concentrations as Conversion Factors
1. Using concentration as a conversion factor, how many moles of solute are in 3.44 L of 0.753 M CaCl2?
2. Using concentration as a conversion factor, how many moles of solute are in 844 mL of 2.09 M MgSO4?
3. Using concentration as a conversion factor, how many liters are needed to provide 0.822 mol of NaBr from a 0.665 M solution?
4. Using concentration as a conversion factor, how many liters are needed to provide 2.500 mol of (NH2)2CO from a 1.087 M solution?
5. What is the mass of solute in 24.5 mL of 0.755 M CoCl2?
6. What is the mass of solute in 3.81 L of 0.0232 M Zn(NO3)2?
7. What volume of solution is needed to provide 9.04 g of NiF2 from a 0.332 M solution?
8. What volume of solution is needed to provide 0.229 g of CH2O from a 0.00560 M solution?
9. What volume of 3.44 M HCl will react with 5.33 mol of CaCO3?
2HCl + CaCO3 → CaCl2 + H2O + CO2
10. What volume of 0.779 M NaCl will react with 40.8 mol of Pb(NO3)2?
Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3
11. What volume of 0.905 M H2SO4 will react with 26.7 mL of 0.554 M NaOH?
H2SO4 + 2NaOH → Na2SO4 + 2H2O
12. What volume of 1.000 M Na2CO3 will react with 342 mL of 0.733 M H3PO4?
3Na2CO3 + 2H3PO4 → 2Na3PO4 + 3H2O + 3CO2
13. It takes 23.77 mL of 0.1505 M HCl to titrate with 15.00 mL of Ca(OH)2. What is the concentration of Ca(OH)2? You will need to write the balanced chemical equation first.
14. It takes 97.62 mL of 0.0546 M NaOH to titrate a 25.00 mL sample of H2SO4. What is the concentration of H2SO4? You will need to write the balanced chemical equation first.
15. It takes 4.667 mL of 0.0997 M HNO3 to dissolve some solid Cu. What mass of Cu can be dissolved?
Cu + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2 + 2H2O
16. It takes 49.08 mL of 0.877 M NH3 to dissolve some solid AgCl. What mass of AgCl can be dissolved?
AgCl(s) + 4NH3(aq) → Ag(NH3)4Cl(aq)
17. What mass of 3.00% H2O2 is needed to produce 66.3 g of O2(g)?
2H2O2(aq) → 2H2O(ℓ) + O2(g)
1. A 0.75% solution of Na2CO3 is used to precipitate Ca2+ ions from solution. What mass of solution is needed to precipitate 40.7 L of solution with a concentration of 0.0225 M Ca2+(aq)?
Na2CO3(aq) + Ca2+(aq) → CaCO3(s) + 2Na+(aq)
Answers
1. 2.59 mol
2.
3. 1.24 L
4.
5. 2.40 g
6.
7. 0.282 L
8.
9. 3.10 L
10.
11. 8.17 mL
12.
13. 0.1192 M
14.
15. 7.39 mg
16.
17. 4.70 kg
11.6: Colligative Properties of Solutions
1. What are the three colligative properties that involve phase changes?
2. Which colligative property does not involve a phase change? Give an example of its importance.
3. If 45.0 g of C6H6 and 60.0 g of C6H5CH3 are mixed together, what is the mole fraction of each component?
4. If 125 g of N2 are mixed with 175 g of O2, what is the mole fraction of each component?
5. If 36.5 g of NaCl are mixed with 63.5 g of H2O, what is the mole fraction of each component?
6. An alloy of stainless steel is prepared from 75.4 g of Fe, 12.6 g of Cr, and 10.8 g of C. What is the mole fraction of each component?
7. A solution is made by mixing 12.0 g of C10H8 in 45.0 g of C6H6. If the vapor pressure of pure C6H6 is 76.5 torr at a particular temperature, what is the vapor pressure of the solution at the same temperature?
8. A solution is made by mixing 43.9 g of C6H12C6 in 100.0 g of H2O. If the vapor pressure of pure water is 26.5 torr at a particular temperature, what is the vapor pressure of the solution at the same temperature?
9. At 300°C, the vapor pressure of Hg is 32.97 torr. If 0.775 g of Au were dissolved into 3.77 g of Hg, what would be the vapor pressure of the solution?
10. At 300°C, the vapor pressure of Hg is 32.97 torr. What mass of Au would have to be dissolved in 5.00 g of Hg to lower its vapor pressure to 25.00 torr?
11. If 25.0 g of C6H12O6 are dissolved in 100.0 g of H2O, what is the boiling point of this solution?
12. If 123 g of C10H16O are dissolved in 355 g of C6H6, what is the boiling point of this solution?
13. If 1 mol of solid CBr4 is mixed with 2 mol of CCl4, what is the boiling point of this solution?
14. A solution of C2H2O4 in CH3COOH has a boiling point of 123.40°C. What is the molality of the solution?
15. If 123 g of C10H16O are dissolved in 355 g of C6H6, what is the freezing point of this solution?
16. If 25.0 g of C6H12O6 are dissolved in 100.0 g of H2O, what is the freezing point of this solution?
17. C8H17OH is a nonvolatile solid that dissolves in C6H12. If 7.22 g of C8H17OH is dissolved in 45.3 g of C6H12, what is the freezing point of this solution?
18. A solution of C2H2O4 in CH3COOH has a freezing point of 10.00°C. What is the molality of the solution?
19. If 25.0 g of C6H12O6 are dissolved in H2O to make 0.100 L of solution, what is the osmotic pressure of this solution at 25°C?
20. If 2.33 g of C27H46O are dissolved in liquid CS2 to make 50.00 mL of solution, what is the osmotic pressure of this solution at 298 K?
21. At 298 K, what concentration of solution is needed to have an osmotic pressure of 1.00 atm?
1. The osmotic pressure of blood is about 7.65 atm at 37°C. What is the approximate concentration of dissolved solutes in blood? (There are many different solutes in blood, so the answer is indeed an approximation.)
Answers
1. boiling point elevation, freezing point depression, vapor pressure depression
2.
3. mole fraction C6H6: 0.469; mole fraction C6H5CH3: 0.531
4.
5. mole fraction NaCl: 0.157; mole fraction H2O: 0.843
6.
7. 65.8 torr
8.
9. 27.26 torr
10.
11. 100.71°C
12.
13. 92.9°C
14.
15. −5.65°C
16.
17. −18.3°C
18.
19. 33.9 atm
20.
21. 0.0409 M | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Ball_et_al._(Beginning_Chemistry)/11%3A_Solutions.txt |
12.2: Arrhenius Acids and Bases
1. Define Arrhenius acid.
2. Define Arrhenius base.
3. What are some general properties of Arrhenius acids?
1. What are some general properties of Arrhenius bases?
2. Identify each substance as an Arrhenius acid, an Arrhenius base, or neither.
1. NaOH
2. C2H5OH
3. H3PO4
3. Identify each substance as an Arrhenius acid, an Arrhenius base, or neither.
1. C6H12O6
2. HNO2
3. Ba(OH)2
4. Write the balanced chemical equation for the neutralization reaction between KOH and H2C2O4. What is the salt?
5. Write the balanced chemical equation for the neutralization reaction between Sr(OH)2 and H3PO4. What is the salt?
6. Write the balanced chemical equation for the neutralization reaction between HCl and Fe(OH)3. What is the salt?
7. Write the balanced chemical equation for the neutralization reaction between H2SO4 and Cr(OH)3. What is the salt?
8. CaCl2 would be the product of the reaction of what acid and what base?
9. Zn(NO3)2 would be product of the reaction of what acid and what base?
10. BaSO4 would be product of the reaction of what acid and what base?
11. Na3PO4 would be product of the reaction of what acid and what base?
Answers
1. a compound that increases the H+ concentration in water
2.
3. sour taste, react with metals, and turn litmus red
4.
5.
1. Arrhenius base
2. neither
3. Arrhenius acid
6.
7. 2KOH + H2C2O4 → 2H2O + K2C2O4; K2C2O4
8.
9. 3HCl + Fe(OH)3 → 3H2O + FeCl3; FeCl3
10.
11. HCl and Ca(OH)2
12.
13. H2SO4 and Ba(OH)2
12.3: Brønsted-Lowry Acids and Bases
Exercise $1$
1. Define Brønsted-Lowry acid. How does it differ from an Arrhenius acid?
2. Define Brønsted-Lowry base. How does it differ from an Arrhenius base?
3. Write the dissociation of hydrogen bromide in water as a Brønsted-Lowry acid-base reaction and identify the proton donor and proton acceptor.
4. Write the dissociation of nitric acid in water as a Brønsted-Lowry acid-base reaction and identify the proton donor and proton acceptor.
5. Pyridine (C5H5N) acts as a Brønsted-Lowry base in water. Write the hydrolysis reaction for pyridine and identify the Brønsted-Lowry acid and Brønsted-Lowry base.
6. The methoxide ion (CH3O) acts as a Brønsted-Lowry base in water. Write the hydrolysis reaction for the methoxide ion and identify the Brønsted-Lowry acid and Brønsted-Lowry base.
7. Identify the Brønsted-Lowry acid and Brønsted-Lowry base in this chemical equation.
H3PO4 + OH → H2PO4 + H2O
8. Identify the Brønsted-Lowry acid and Brønsted-Lowry base in this chemical equation.
H2C2O4 + 2F → 2HF + C2O42−
9. Predict the products of this reaction, assuming it undergoes a Brønsted-Lowry acid-base reaction.
HC2H3O2 + C5H5N → ?
10. Predict the products of this reaction, assuming it undergoes a Brønsted-Lowry acid-base reaction.
(C2H5)3N + H2O → ?
11. What is the conjugate acid of H2O? of NH3?
12. What is the conjugate acid of H2PO4? of NO3?
13. What is the conjugate base of HSO4? of H2O?
14. What is the conjugate base of H3O+? of H2SO4?
15. Identify the conjugate acid-base pairs in this reaction.
HSO4 + PO43− → SO42− + HPO42−
16. Identify the conjugate acid-base pairs in this reaction.
HClO3 + (C2H5)3N → ClO3 + (C2H5)3NH+
17. Identify the conjugate acid-base pairs in this reaction.
NH3 + C6H5O → C6H5OH + NH2
18. Identify the conjugate acid-base pairs in this reaction.
C5H5NH+ + C2O42− → C5H5N + HC2O4
nswers
1. A Brønsted-Lowry acid is a proton donor. It does not necessarily increase the H+ concentration in water.
2.
3. HBr + H2O → H3O+ + Br; PD: HBr; PA: H2O
4.
5. C5H5N + H2O → C5H5NH+ + OH; PD: H2O; PA: C5H5N
6.
7. BL acid: H3PO4; BL base: OH
8.
9. C2H3O2 and C5H5NH+
10.
11. H3O+; NH4+
12.
13. SO42−; OH
14.
15. HSO4 and SO42−; PO43− and HPO42−
16.
17. NH3 and NH2; C6H5O and C6H5OH
12.4: Acid-Base Titrations
Exercise $1$
1. Define titration.
2. What is the difference between the titrant and the analyte?
3. True or false: An acid is always the titrant. Explain your answer.
4. True or false: An analyte is always dissolved before reaction. Explain your answer.
5. If 55.60 mL of 0.2221 M HCl was needed to titrate a sample of NaOH to its equivalence point, what mass of NaOH was present?
6. If 16.33 mL of 0.6664 M KOH was needed to titrate a sample of HC2H3O2 to its equivalence point, what mass of HC2H3O2 was present?
7. It takes 45.66 mL of 0.1126 M HBr to titrate 25.00 mL of Ca(OH)2 to its equivalence point. What is the original concentration of the Ca(OH)2 solution?
8. It takes 9.77 mL of 0.883 M H2SO4 to titrate 15.00 mL of KOH to its equivalence point. What is the original concentration of the KOH solution?
Answers
1. a chemical reaction performed in a quantitative fashion
2.
3. False; a base can be a titrant, or the reaction being performed may not even be an acid-base reaction.
4.
5. 0.494 g
6.
7. 0.1028 M
12.5: Strong and Weak Acids and Bases and their Salts
Exercise $1$
1. Differentiate between a strong acid and a weak acid.
1. Differentiate between a strong base and a weak base.
2. Identify each as a strong acid or a weak acid. Assume aqueous solutions.
1. HF
2. HCl
3. HC2O4
3. Identify each as a strong base or a weak base. Assume aqueous solutions.
1. NaOH
2. Al(OH)3
3. C4H9NH2
4. Write a chemical equation for the ionization of each acid and indicate whether it proceeds 100% to products or not.
1. HNO3
2. HNO2
3. HI3
5. Write a chemical equation for the ionization of each base and indicate whether it proceeds 100% to products or not.
1. NH3
2. (CH3)3N
3. Mg(OH)2
6. Write the balanced chemical equation for the reaction of each acid and base pair.
1. HCl + C5H5N
2. H2C2O4 + NH3
3. HNO2 + C7H9N
7. Write the balanced chemical equation for the reaction of each acid and base pair.
1. H3C5H5O7 + Mg(OH)2
2. HC3H3O3 + (CH3)3N
3. HBr + Fe(OH)3
8. Identify each salt as neutral, acidic, or basic.
1. NaBr
2. Fe(NO3)2
3. Fe(NO3)3
9. Identify each salt as neutral, acidic, or basic.
1. NH4I
2. C2H5NH3Cl
3. KI
10. Identify each salt as neutral, acidic, or basic.
1. NaNO2
2. NaNO3
3. NH4NO3
11. Identify each salt as neutral, acidic, or basic.
1. KC2H3O2
2. KHSO4
3. KClO3
12. Write the hydrolysis reaction that occurs, if any, when each salt dissolves in water.
1. K2SO3
2. KI
3. NH4ClO3
13. Write the hydrolysis reaction that occurs, if any, when each salt dissolves in water.
1. NaNO3
2. CaC2O4
3. C5H5NHCl
14. When NH4NO2 dissolves in H2O, both ions hydrolyze. Write chemical equations for both reactions. Can you tell if the solution will be acidic or basic overall?
15. When pyridinium acetate (C5H5NHC2H3O2) dissolves in H2O, both ions hydrolyze. Write chemical equations for both reactions. Can you tell if the solution will be acidic or basic overall?
16. A lab technician mixes a solution of 0.015 M Mg(OH)2. Is the resulting OH concentration greater than, equal to, or less than 0.015 M? Explain your answer.
17. A lab technician mixes a solution of 0.55 M HNO3. Is the resulting H+ concentration greater than, equal to, or less than 0.55 M? Explain your answer.
Answers
1. A strong acid is 100% ionized in aqueous solution, whereas a weak acid is not 100% ionized.
2.
3.
1. weak acid
2. strong acid
3. weak acid
4.
5.
1. HNO3(aq) → H+(aq) + NO3(aq); proceeds 100%
2. HNO2(aq) → H+(aq) + NO2(aq); does not proceed 100%
3. HI3(aq) → H+(aq) + I3(aq); does not proceed 100%
6.
7.
1. HCl + C5H5N → Cl + C5H5NH+
2. H2C2O4 + 2NH3 → C2O42− + 2NH4+
3. HNO2 + C7H9N → NO2 + C7H9NH+
8.
9.
1. neutral
2. acidic
3. acidic
10.
11.
1. basic
2. neutral
3. acidic
12.
13.
1. SO32− + H2O → HSO3 + OH
2. no reaction
3. NH4+ + H2O → NH3 + H3O+
14.
15. NH4+ + H2O → NH3 + H3O+; NO2 + H2O → HNO2 + OH; it is not possible to determine whether the solution will be acidic or basic.
16.
17. greater than 0.015 M because there are two OH ions per formula unit of Mg(OH)2
12.6: Autoionization of Water
Exercise $4$
1. Does [H+] remain constant in all aqueous solutions? Why or why not?
2. Does [OH] remain constant in all aqueous solutions? Why or why not?
3. What is the relationship between [H+] and Kw? Write a mathematical expression that relates them.
4. What is the relationship between [OH] and Kw? Write a mathematical expression that relates them.
5. Write the chemical equation for the autoionization of water and label the conjugate acid-base pairs.
6. Write the reverse of the reaction for the autoionization of water. It is still an acid-base reaction? If so, label the acid and base.
7. For a given aqueous solution, if [H+] = 1.0 × 10−3 M, what is [OH]?
8. For a given aqueous solution, if [H+] = 1.0 × 10−9 M, what is [OH]?
9. For a given aqueous solution, if [H+] = 7.92 × 10−5 M, what is [OH]?
10. For a given aqueous solution, if [H+] = 2.07 × 10−11 M, what is [H+]?
11. For a given aqueous solution, if [OH] = 1.0 × 10−5 M, what is [H+]?
12. For a given aqueous solution, if [OH] = 1.0 × 10−12 M, what is [H+]?
13. For a given aqueous solution, if [OH] = 3.77 × 10−4 M, what is [H+]?
14. For a given aqueous solution, if [OH] = 7.11 × 10−10 M, what is [H+]?
15. What are [H+] and [OH] in a 0.344 M solution of HNO3?
16. What are [H+] and [OH] in a 2.86 M solution of HBr?
17. What are [H+] and [OH] in a 0.00338 M solution of KOH?
18. What are [H+] and [OH] in a 6.02 × 10−4 M solution of Ca(OH)2?
19. If HNO2 is dissociated only to an extent of 0.445%, what are [H+] and [OH] in a 0.307 M solution of HNO2?
20. If (C2H5)2NH is dissociated only to an extent of 0.077%, what are [H+] and [OH] in a 0.0955 M solution of (C2H5)2NH?
Answers
1. [H+] varies with the amount of acid or base in a solution.
2.
3. $\left [ H^{+} \right ]=\frac{K_{W}}{\left [ OH^{-} \right ]}$
4.
5. H2O + H2O → H3O+ + OH; H2O/H3O+ and H2O/OH
6.
7. 1.0 × 10−11 M
8.
9. 1.26 × 10−10 M
10.
11. 1.0 × 10−9 M
12.
13. 2.65 × 10−11 M
14.
15. [H+] = 0.344 M; [OH] = 2.91 × 10−14 M
16.
17. [OH] = 0.00338 M; [H+] = 2.96 × 10−12 M
18.
19. [H+] = 0.00137 M; [OH] = 7.32 × 10−12 M
12.7: The pH Scale
Exercise $1$
1. Define pH. How is it related to pOH?
2. Define pOH. How is it related to pH?
3. What is the pH range for an acidic solution?
4. What is the pH range for a basic solution?
5. What is [H+] for a neutral solution?
6. What is [OH] for a neutral solution? Compare your answer to Exercise 6. Does this make sense?
1. Which substances in Table 12.7.1 are acidic?
2. Which substances in Table 12.7.1 are basic?
3. What is the pH of a solution when [H+] is 3.44 × 10−4 M?
4. What is the pH of a solution when [H+] is 9.04 × 10−13 M?
5. What is the pH of a solution when [OH] is 6.22 × 10−7 M?
6. What is the pH of a solution when [OH] is 0.0222 M?
7. What is the pOH of a solution when [H+] is 3.44 × 10−4 M?
8. What is the pOH of a solution when [H+] is 9.04 × 10−13 M?
9. What is the pOH of a solution when [OH] is 6.22 × 10−7 M?
10. What is the pOH of a solution when [OH] is 0.0222 M?
11. If a solution has a pH of 0.77, what is its pOH, [H+], and [OH]?
12. If a solution has a pOH of 13.09, what is its pH, [H+], and [OH]?
Answers
1. pH is the negative logarithm of [H+] and is equal to 14 − pOH.
2.
3. pH < 7
4.
5. 1.0 × 10−7 M
6.
7. Every entry above pure water is acidic.
8.
9. 3.46
10.
11. 7.79
12.
13. 10.54
14.
15. 6.21
16.
17. pOH = 13.23; [H+] = 1.70 × 10−1 M; [OH] = 5.89 × 10−14 M
12.8: Buffers
Exercise $1$
1. Define buffer. What two related chemical components are required to make a buffer?
1. Can a buffer be made by combining a strong acid with a strong base? Why or why not?
2. Which combinations of compounds can make a buffer? Assume aqueous solutions.
1. HCl and NaCl
2. HNO2 and NaNO2
3. NH4NO3 and HNO3
4. NH4NO3 and NH3
3. Which combinations of compounds can make a buffer? Assume aqueous solutions.
1. H3PO4 and Na3PO4
2. NaHCO3 and Na2CO3
3. NaNO3 and Ca(NO3)2
4. HN3 and NH3
4. For each combination in Exercise 3 that is a buffer, write the chemical equations for the reactions of the buffer components when a strong acid and a strong base is added.
5. For each combination in Exercise 4 that is a buffer, write the chemical equations for the reactions of the buffer components when a strong acid and a strong base is added.
6. The complete phosphate buffer system is based on four substances: H3PO4, H2PO4, HPO42−, and PO43−. What different buffer solutions can be made from these substances?
7. Explain why NaBr cannot be a component in either an acidic or a basic buffer.
8. Two solutions are made containing the same concentrations of solutes. One solution is composed of H3PO4 and Na3PO4, while the other is composed of HCN and NaCN. Which solution should have the larger capacity as a buffer?
9. Two solutions are made containing the same concentrations of solutes. One solution is composed of NH3 and NH4NO3, while the other is composed of H2SO4 and Na2SO4. Which solution should have the larger capacity as a buffer?
Answers
1. A buffer is the combination of a weak acid or base and a salt of that weak acid or base.
2.
3.
1. no
2. yes
3. no
4. yes
4.
5. 3b: strong acid: NO2 + H+ → HNO2; strong base: HNO2 + OH → NO2 + H2O; 3d: strong base: NH4+ + OH → NH3 + H2O; strong acid: NH3 + H+ → NH4+
6.
7. Buffers can be made from three combinations: (1) H3PO4 and H2PO4, (2) H2PO4 and HPO42−, and (3) HPO42− and PO43−. (Technically, a buffer can be made from any two components.)
8.
9. The phosphate buffer should have the larger capacity. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Ball_et_al._(Beginning_Chemistry)/12%3A_Acids_and_Bases.txt |
14.2: Oxidation-Reduction Reactions
1. Is this reaction a redox reaction? Explain your answer.2K(s) + Br2(ℓ) → 2KBr(s)
2. Is this reaction a redox reaction? Explain your answer.
2NaCl(aq) + Pb(NO3)2(aq) → 2NaNO3(aq) + PbCl2(s)
3. Which substance loses electrons and which substance gains electrons in this reaction?
2Mg(s) + O2(g) → 2MgO
4. Which substance loses electrons and which substance gains electrons in this reaction?
16Fe(s) + 3S8(s) → 8Fe2S3(s)
5. Which substance is oxidized and which substance is reduced in this reaction?
2Li(s) + O2(g) → Li2O2(s)
6. Which substance is oxidized and which substance is reduced in this reaction?
2Fe(s) + 3I2(s) → 2FeI3(s)
7. What are two different definitions of oxidation?
1. What are two different definitions of reduction?
2. Assign oxidation numbers to the atoms in each substance.
1. P4
2. SO3
3. SO32−
4. Ca3(PO3)2
3. Assign oxidation numbers to the atoms in each substance.
1. PCl5
2. (NH4)2Se
3. Ag
4. Li2O2
4. Assign oxidation numbers to the atoms in each substance.
1. NO
2. NO2
3. CrCl2
4. CrCl3
5. Assign oxidation numbers to the atoms in each substance.
1. NaH
2. N2O3
3. NO2
4. CuNO3
6. Assign oxidation numbers to the atoms in each substance.
1. CH2O
2. NH3
3. Rb2SO4
4. Zn(C2H3O2)2
7. Assign oxidation numbers to the atoms in each substance.
1. C6H6
2. B(OH)3
3. Li2S
4. Au
8. Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.2NO + Cl2 → 2NOCl
9. Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.Sr + SO3 → SrSO3
10. Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.2KrF2 + 2H2O → 2Kr + 4HF + O2
11. Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.SO3 + SCl2 → SOCl2 + SO2
12. Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.2Rb + MgCl2 → 2RbCl + Mg
13. Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.2C8H18 + 25O2 → 16CO2 + 18H2O
Answers
1. yes because oxidation numbers are changing
2. lose: Mg; gain: O
3. oxidized: Li; reduced: O
1. increase in oxidation number; loss of electrons
1. P: 0
2. S: +6; O: −2
3. S: +4; O: −2
4. Ca: +2; P: +3; O: −2
1. N: +2; O: −2
2. N: +4; O: −2
3. Cr: +2; Cl: −1
4. Cr: +3; Cl: −1
1. C: 0; H: +1; O: −2
2. N: −3; H: +1
3. Rb: +1; S: +6; O: −2
4. Zn: +2; C: 0; H: +1; O: −2
2. oxidized: N; reduced: Cl
3. oxidized: O; reduced: Kr
4. oxidized: Rb; reduced: Mg
14.3: Balancing Redox Reactions
1. Balance these redox reactions by inspection.
1. Na + F2 → NaF
2. Al2O3 + H2 → Al + H2O
2. Balance these redox reactions by inspection.
1. Fe2S3 + O2 → Fe2O3 + S
2. Cu2O + H2 → Cu + H2O
3. Balance these redox reactions by inspection.
1. CH4 + O2 → CO2 + H2O
2. P2O5 + Cl2 → PCl3 + O2
4. Balance these redox reactions by inspection.
1. PbCl2 + FeCl3 → PbCl4 + FeCl2
2. SO2 + F2 → SF4 + OF2
5. Balance these redox reactions by the half reaction method.
1. Ca + H+ → Ca2+ + H2
2. Sn2+ → Sn + Sn4+ (Hint: both half reactions will start with the same reactant.)
6. Balance these redox reactions by the half reaction method.
1. Fe3+ + Sn2+ → Fe + Sn4+
2. Pb2+ → Pb + Pb4+ (Hint: both half reactions will start with the same reactant.)
7. Balance these redox reactions by the half reaction method.
1. Na + Hg2Cl2 → NaCl + Hg
2. Al2O3 + C → Al + CO2
8. Balance these redox reactions by the half reaction method.
1. Br + I2 → I + Br2
2. CrCl3 + F2 → CrF3 + Cl2
9. Balance these redox reactions that occur in aqueous solution. Use whatever water-derived species is necessary; there may be more than one correct balanced equation.
1. Cu + NO3 → Cu2+ + NO2
2. Fe + MnO4 → Fe3+ + Mn
10. Balance these redox reactions that occur in aqueous solution. Use whatever water-derived species is necessary; there may be more than one correct balanced equation.
1. CrO3 + Ni2+ → Cr3+ + Ni3+
2. OsO4 + C2H4 → Os + CO2
11. Balance these redox reactions that occur in aqueous solution. Use whatever water-derived species is necessary; there may be more than one correct balanced equation.
1. ClO + Ti2+ → Ti4+ + Cl
2. BrO3 + Ag → Ag+ + BrO2
12. Balance these redox reactions that occur in aqueous solution. Use whatever water-derived species is necessary; there may be more than one correct balanced equation.
1. H2O2 + NO → N2O3 + H2O
2. VO2+ + NO → V3+ + NO2
13. Explain why this chemical equation is not balanced and balance it if it can be balanced: Cr2+ + Cl2 → Cr3+ + 2Cl
14. Explain why this equation is not balanced and balance it if it can be balanced: O2 + 2H2O + Br2 → 4OH + 2Br
Answers
1. 2Na + F2 → 2NaF
2. Al2O3 + 3H2 → 2Al + 3H2O
1. CH4 + 2O2 → CO2 + 2H2O
2. 2P2O5 + 6Cl2 → 4PCl3 + 5O2
1. Ca + 2H+ → Ca2+ + H2
2. 2Sn2+ → Sn + Sn4+
1. 2Na + Hg2Cl2 → 2NaCl + 2Hg
2. 2Al2O3 + 3C → 4Al + 3CO2
1. 4H+ + Cu + 2NO3 → Cu2+ + 2NO2 + 2H2O in acidic solution; 2H2O + Cu + 2NO3 → Cu2+ + 2NO2 + 4OH in basic solution
2. 24H+ + 3MnO4 + 7Fe → 7Fe3+ + 3Mn + 12H2O in acidic solution; 12H2O + 3MnO4 + 7Fe → 7Fe3+ + 3Mn + 24OH in basic solution
1. 2H+ + ClO + Ti2+ → Cl + H2O + Ti4+ in acidic solution; H2O + ClO + Ti2+ → Cl + Ti4+ + 2OH in basic solution
2. 2H+ + BrO3 + Ag → BrO2 + H2O + Ag+ in acidic solution; H2O + BrO3 + Ag → BrO2 + Ag+ + 2OH in basic solution
1. The charges are not properly balanced. The correct balanced equation is 2Cr2+ + Cl2 → 2Cr3+ + 2Cl.
14.4: Applications of Redox Reactions - Voltaic Cells
1. Draw the voltaic cell represented by this reaction and label the cathode, the anode, the salt bridge, the oxidation half cell, the reduction half cell, the positive electrode, and the negative electrode. Use Fig. 14.4.1 as a guide. Zn + 2Ag+ → Zn2+ + 2Ag
2. Draw the voltaic cell represented by this reaction and label the cathode, the anode, the salt bridge, the oxidation half cell, the reduction half cell, the positive electrode, and the negative electrode. Use Fig. 14.4.1 as a guide. 3Mg + 2Cr3+ → 3Mg2+ + 2Cr
3. What is the voltage of this half reaction? 2F → F2 + 2e
4. What is the voltage of this half reaction? Na → Na+ + e
5. What is the voltage of the voltaic cell in Exercise 1? Consult Table 14.4.1.
6. What is the voltage of the voltaic cell in Exercise 2? Consult Table 14.4.1.
7. Balance this redox reaction and determine its voltage. Is it spontaneous? Li+ + Al → Li + Al3+
8. Balance this redox reaction and determine its voltage. Is it spontaneous? Pb2+ + Ni → Pb + Ni2+
9. Balance this redox reaction and determine its voltage. Is it spontaneous? Cu2+ + Ag + Cl → Cu + AgCl
10. Balance this redox reaction and determine its voltage. Is it spontaneous? Mn2+ + Br2 → MnO4 + Br
11. Which reaction represents the cathode reaction in Exercise 7? The anode reaction?
12. Which reaction represents the cathode reaction in Exercise 8? The anode reaction?
13. Which reaction represents the cathode reaction in Exercise 9? The anode reaction?
14. Which reaction represents the cathode reaction in Exercise 10? The anode reaction?
15. A voltaic cell is based on this reaction: Ni + 2Au+ → Ni2+ + 2Au; If the voltage of the cell is 0.33 V, what is the standard reduction potential of the Au+ + e → Au half reaction?
16. A voltaic cell is based on this reaction: 3Pb + 2V3+ → 3Pb2+ + 2V; If the voltage of the cell is −0.72 V, what is the standard reduction potential of the V3+ + 3e → V half reaction?
17. What species is being oxidized and what species is being reduced in a dry cell?
18. What species is being oxidized and what species is being reduced in an alkaline battery?
19. What species is being oxidized and what species is being reduced in a silver oxide button battery?
20. What species is being oxidized and what species is being reduced in a lead storage battery?
21. Based on the data in Table 14.4.1, what is the highest voltage battery you can construct?
22. Based on the data in Table 14.4.1, what is the lowest voltage battery you can construct? (This may be more challenging to answer than Exercise 21.)
Answers
1. −2.87 V
2. 1.56 V
3. 3Li+ + Al → 3Li + Al3+; −1.39 V; not spontaneous
4. Cu2+ + 2Ag + 2Cl → Cu + 2AgCl; 0.12 V; spontaneous
5. cathode reaction: Li+ + e → Li; anode reaction: Al → Al3+ + 3e
6. cathode reaction: Cu2+ + 2e → Cu; anode reaction: Ag + Cl → AgCl + e
7. 0.08 V
8. oxidized: Zn; reduced: Mn
9. oxidized: Zn; reduced: Ag
10. 5.92 V from the reaction of F2 and Li
14.5: Electrolysis
1. Define electrolytic cell.
2. How does the operation of an electrolytic cell differ from a voltaic cell?
3. List at least three elements that are produced by electrolysis.
4. Write the half reactions for the electrolysis of the elements listed in Exercise 3.
5. Based on Table 14.4.1, what voltage must be applied to an electrolytic cell to electroplate copper from Cu2+?
6. Based on Table 14.4.1, what voltage must be applied to an electrolytic cell to electroplate aluminum from Al3+?
Answers
1. an electrochemical cell in which charge is forced through and a nonspontaneous reaction occurs
2. any three of the following: Al, K, Li, Na, Cl2, or Mg
3. 0.34 V
Additional Exercises
1. Oxidation was once defined as chemically adding oxygen to a substance. Use this reaction to argue that this definition is consistent with the modern definition of oxidation: 2Mg + O2 → 2MgO
2. Reduction was once defined as chemically adding hydrogen to a substance. Use this reaction to argue that this definition is consistent with the modern definition of reduction: C2H2 + 2H2 → C2H6
3. Assign oxidation numbers to the atoms in each substance.
1. Kr (krypton)
2. krypton tetrafluoride (KrF4)
3. dioxygen difluoride (O2F2)
4. Assign oxidation numbers to the atoms in each substance.
1. lithium hydride (LiH)
2. potassium peroxide (K2O2)
3. potassium fluoride (KF)
5. N atoms can have a wide range of oxidation numbers. Assign oxidation numbers for the N atom in each compound, all of which are known compounds.
1. N2O5
2. N2O4
3. NO2
4. NO
5. N2H4
6. NH3
6. Cr atoms can have a wide range of oxidation numbers. Assign oxidation numbers for the Cr atom in each compound, all of which are known compounds.
1. Na2CrO4
2. Na2Cr2O7
3. CrF5
4. CrCl3
5. CrCl2
7. Balance this redox reaction by inspection: S8 + O2 → SO2
8. Balance this redox reaction by inspection: C18H38 + O2 → CO2 + H2O
9. Balance this redox reaction by the half reaction method by assuming an acidic solution: Cr2O72− + Fe → Cr3+ + Fe3+
10. Balance the redox reaction in Exercise 9 by the half reaction method by assuming a basic solution.
11. The uranyl ion (UO22+) is a fairly stable ion of uranium that requires strong reducers to reduce the oxidation number of uranium further. Balance this redox reaction using the half reaction method by assuming an acidic solution.UO22+ + HN3 → U + N2
12. Balance the redox reaction in Exercise 11 by the half reaction method by assuming a basic solution.
13. Zinc metal can be dissolved by acid, which contains H+ ions. Demonstrate that this is consistent with the fact that this reaction has a spontaneous voltage: Zn + 2H+ → Zn2+ + H2
14. Copper metal cannot be dissolved by acid, which contains H+ ions. Demonstrate that this is consistent with the fact that this reaction has a nonspontaneous voltage: Cu + 2H+ → Cu2+ + H2
15. A disproportionation reaction occurs when a single reactant is both oxidized and reduced. Balance and determine the voltage of this disproportionation reaction. Use the data in Table 14.4.1 - Standard Reduction Potentials of Half Reactions: Cr2+ → Cr + Cr3+
16. A disproportionation reaction occurs when a single reactant is both oxidized and reduced. Balance and determine the voltage of this disproportionation reaction. Use the data in Table 14.4.1 - Standard Reduction Potentials of Half Reactions: Fe2+ → Fe + Fe3+
17. What would be overall reaction for a fuel cell that uses CH4 as the fuel?
18. What would be overall reaction for a fuel cell that uses gasoline (general formula C8H18) as the fuel?
19. When NaCl undergoes electrolysis, sodium appears at the cathode. Is the definition of cathode the same for an electrolytic cell as it is for a voltaic cell?
20. When NaCl undergoes electrolysis, chlorine appears at the anode. Is the definition of anode the same for an electrolytic cell as it is for a voltaic cell?
21. An award is being plated with pure gold before it is presented to a recipient. If the area of the award is 55.0 cm2 and will be plated with 3.00 µm of Au, what mass of Au will be plated on the award? The density of Au is 19.3 g/cm3.
22. The unit of electrical charge is called the coulomb (C). It takes 96,500 coulombs of charge to reduce 27.0 g of Al from Al3+ to Al metal. At 1,040 cm3, how many coulombs of charge were needed to reduce the aluminum in the cap of the Washington monument, assuming the cap is pure Al? The density of Al is 2.70 g/cm3.
Answer
As oxygen is added to magnesium, it is being oxidized. In modern terms, the Mg atoms are losing electrons and being oxidized, while the electrons are going to the O atoms.
1. Kr: 0
2. Kr: +4; F: −1
3. O: +1; F: −1
1. +5
2. +4
3. +4
4. +2
5. −2
6. −3
S8 + 8O2 → 8SO2
14H+ + Cr2O72− + 2Fe → 2Cr3+ + 7H2O + 2Fe3+
6HN3 + UO22+ → U + 2H2O + 9N2 + 2H+
The voltage of the reaction is +0.76 V, which implies a spontaneous reaction.
3Cr2+ → Cr + 2Cr3+; −0.50 V
CH4 + 2O2 → CO2 + 2H2O
yes because reduction occurs at the cathode
0.318 g | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Ball_et_al._(Beginning_Chemistry)/14%3A_Oxidation_and_Reduction.txt |
15.1: Radioactivity
Q15.1.1
Define radioactivity.
S15.1.1
Radioactivity is the spontaneous emission of particles and electromagnetic radiation from nuclei of unstable atoms.
Q15.1.2
Give an example of a radioactive element. How do you know if it is radioactive?
Q15.1.3
How many protons and neutrons are in each isotope?
1. $_{5}^{11}\textrm{B}$
2. $_{13}^{27}\textrm{Al}$
3. $_{}^{56}\textrm{Fe}$
4. $_{}^{224}\textrm{Rn}$
S15.1.3
1. 5 protons; 6 neutrons
2. 13 protons; 14 neutrons
3. 26 protons; 30 neutrons
4. 86 protons; 138 neutrons
Q15.1.4
How many protons and neutrons are in each isotope?
1. $_{1}^{2}\textrm{H}$
2. $_{48}^{112}\textrm{Cd}$
3. $_{}^{252}\textrm{Es}$
4. $_{}^{40}\textrm{K}$
Q15.1.5
Describe an alpha particle. What nucleus is it equivalent to?
S15.1.5
An alpha particle is a collection of two protons and two neutrons and is equivalent to a helium nucleus.
Q15.1.6
Describe a beta particle. What subatomic particle is it equivalent to?
Q15.1.7
What are gamma rays?
S15.1.7
Gamma rays are high-energy electromagnetic radiation given off in radioactive decay.
Q15.1.8
Why is it inappropriate to refer to gamma rays as “gamma particles”?
Q15.1.9
Plutonium has an atomic number of 94. Write the nuclear equation for the alpha particle emission of plutonium-244. What is the daughter isotope?
S15.1.9
$_{94}^{244}\textrm{Pu}\rightarrow \: _{92}^{240}\textrm{U}+\: _{2}^{4}\textrm{He}$
daughter isotope: $_{}^{240}\textrm{U}$
Q15.1.10
Francium has an atomic number of 87. Write the nuclear equation for the alpha particle emission of francium-212. What is the daughter isotope?
S15.1.11
$_{50}^{121}\textrm{Sn}\rightarrow \: _{51}^{121}\textrm{Sb}+\: _{-1}^{0}\textrm{e}$
daughter isotope: $_{}^{121}\textrm{Sb}$
Q15.1.11
Tin has an atomic number of 50. Write the nuclear equation for the beta particle emission of tin-121. What is the daughter isotope?
Q15.1.12
Technetium has an atomic number of 43. Write the nuclear equation for the beta particle emission of technetium-99. What is the daughter isotope?
Q15.1.13
Energies of gamma rays are typically expressed in units of megaelectron volts (MeV), where 1 MeV = 1.602 × 10−13 J. Using the data provided in the text, calculate the energy in megaelectron volts of the gamma ray emitted when radon-222 decays.
0.51 MeV
Q15.1.14
The gamma ray emitted when oxygen-19 gives off a beta particle is 0.197 MeV. What is its energy in joules? (See Exercise 13 for the definition of a megaelectron volt.)
Q15.1.15
Which penetrates matter more deeply—alpha particles or beta particles? Suggest ways to protect yourself against both particles.
S15.1.15
Beta particles penetrate more. A thick wall of inert matter is sufficient to block both particles.
Q15.1.16
Which penetrates matter more deeply—alpha particles or gamma rays? Suggest ways to protect yourself against both emissions.
Q15.1.17
Define nuclear fission.
S15.1.17
Nuclear fission is the breaking down of large nuclei into smaller nuclei, usually with the release of excess neutrons.
Q15.1.18
What general characteristic is typically necessary for a nucleus to undergo spontaneous fission?
Q15.2.1
Do all isotopes have a half-life? Explain your answer.
S15.2.1
Only radioactive isotopes have a half-life.
Q15.2.2
Which is more radioactive—an isotope with a long half-life or an isotope with a short half-life?
Q15.2.3
How long does it take for 1.00 g of palladium-103 to decay to 0.125 g if its half-life is 17.0 d?
51.0 d
Q15.2.4
How long does it take for 2.00 g of niobium-94 to decay to 0.0625 g if its half-life is 20,000 y?
Q15.2.5
It took 75 y for 10.0 g of a radioactive isotope to decay to 1.25 g. What is the half-life of this isotope?
25 y
Q15.2.6
It took 49.2 s for 3.000 g of a radioactive isotope to decay to 0.1875 g. What is the half-life of this isotope?
Q15.2.7
The half-live of americium-241 is 432 y. If 0.0002 g of americium-241 is present in a smoke detector at the date of manufacture, what mass of americium-241 is present after 100.0 y? After 1,000.0 y?
S15.2.7
0.000170 g; 0.0000402 g
Q15.2.8
If the half-life of tritium (hydrogen-3) is 12.3 y, how much of a 0.00444 g sample of tritium is present after 5.0 y? After 250.0 y?
Q15.2.9
Explain why the amount left after 1,000.0 y in Exercise 7 is not one-tenth of the amount present after 100.0 y, despite the fact that the amount of time elapsed is 10 times as long.
S15.2.9
Radioactive decay is an exponential process, not a linear process.
Q15.2.10
Explain why the amount left after 250.0 y in Exercise 8 is not one-fiftieth of the amount present after 5.0 y, despite the fact that the amount of time elapsed is 50 times as long.
Q15.2.11
An artifact containing carbon-14 contains 8.4 × 10−9 g of carbon-14 in it. If the age of the artifact is 10,670 y, how much carbon-14 did it have originally? The half-life of carbon-14 is 5,730 y.
3.1 × 10−8 g
Q15.2.12
Carbon-11 is a radioactive isotope used in positron emission tomography (PET) scans for medical diagnosis. Positron emission is another, though rare, type of radioactivity. The half-life of carbon-11 is 20.3 min. If 4.23 × 10−6 g of carbon-11 is left in the body after 4.00 h, what mass of carbon-11 was present initially?
Define rad.
S15.3.1
a unit of radioactive exposure equal to 0.01 J of energy per gram of tissue
Define rem.
Q15.3.3
How does a becquerel differ from a curie?
S15.3.3
A becquerel is 1 decay/s, whereas a curie is 3.7 × 1010 decays/s.
Define curie.
Q15.3.5
A sample of radon gas has an activity of 140.0 mCi. If the half-life of radon is 1,500 y, how long before the activity of the sample is 8.75 mCi?
6.0 × 103 y
Q15.3.6
A sample of curium has an activity of 1,600 Bq. If the half-life of curium is 24.0 s, how long before its activity is 25.0 Bq?
Q15.3.7
If a radioactive sample has an activity of 65 µCi, how many disintegrations per second are occurring?
S15.3.7
2.41 × 106 disintegrations per second
Q15.3.8
If a radioactive sample has an activity of 7.55 × 105 Bq, how many disintegrations per second are occurring?
Q15.3.9
A sample of fluorine-20 has an activity of 2.44 mCi. If its half-life is 11.0 s, what is its activity after 50.0 s?
0.104 mCi
Q15.3.10
Strontium-90 has a half-life of 28.1 y. If 66.7 Bq of pure strontium-90 were allowed to decay for 15.0 y, what would the activity of the remaining strontium-90 be?
Q15.3.11
How long does it take 100.0 mCi of fluorine-20 to decay to 10.0 mCi if its half-life is 11.0 s?
Q15.3.12
Technetium-99 is used in medicine as a source of radiation. A typical dose is 25 mCi. How long does it take for the activity to reduce to 0.100 mCi? The half-life of 99Tc is 210,000 y.
Q15.3.13
Describe how a radiation exposure in rems is determined.
S15.3.13
by using a film badge, which is exposed by the radiation, or a Geiger counter
Q15.3.14
Which contributes more to the rems of exposure—alpha or beta particles? Why?
Q15.3.15
Use Table 15.4.2 to determine which sources of radiation exposure are inescapable and which can be avoided. What percentage of radiation is unavoidable?
S15.3.14
Radioactive atoms in the body, most terrestrial sources, cosmic sources, and nuclear energy sources are likely unavoidable, which is about 27% of the total exposure. If exposure to radon gas is added, the total unavoidable exposure increases to 82%.
Q15.3.16
Name two isotopes that contribute to the radioactivity in our bodies.
Q15.3.17
Explain how a film badge works to detect radiation.
S15.3.17
Film is exposed by the radiation. The more radiation film is subjected to, the more exposed it becomes.
Q15.3.18
Explain how a Geiger counter works to detect radiation.
15.4: Uses of Radioactive Isotopes
Q15.4.1
Define tracer and give an example of how tracers work.
S15.4.1
A tracer is a radioactive isotope that can be detected far from its original source to trace the path of certain chemicals. Hydrogen-3 can be used to trace the path of water underground.
Q15.4.2
Name two isotopes that have been used as tracers.
Q15.4.3
Explain how radioactive dating works.
S15.4.3
If the initial amount of a radioactive isotope is known, then by measuring the amount of the isotope remaining, a person can calculate how old that object is since it took up the isotope.
Q15.4.4
Name two isotopes that have been used in radioactive dating.
Q15.4.5
The current disintegration rate for carbon-14 is 14.0 Bq. A sample of burnt wood discovered in an archeological excavation is found to have a carbon-14 disintegration rate of 3.5 Bq. If the half-life of carbon-14 is 5,730 y, approximately how old is the wood sample?
11,500 y
Q15.4.6
A small asteroid crashes to Earth. After chemical analysis, it is found to contain 1 g of technetium-99 to every 3 g of ruthenium-99, its daughter isotope. If the half-life of technetium-99 is 210,000 y, approximately how old is the asteroid?
Q15.4.7
What is a positive aspect of the irradiation of food?
S15.4.7
increased shelf life (answers will vary)
Q15.4.8
What is a negative aspect of the irradiation of food?
Q15.4.9
Describe how iodine-131 is used to both diagnose and treat thyroid problems.
S15.4.9
The thyroid gland absorbs most of the iodine, allowing it to be imaged for diagnostic purposes or preferentially irradiated for treatment purposes.
Q15.4.10
List at least five organs that can be imaged using radioactive isotopes.
Q15.4.11
Which radioactive emissions can be used therapeutically?
gamma rays
Q15.4.12
Which isotope is used in therapeutics primarily for its gamma ray emissions?
15.5: Nuclear Energy
Q15.5.1
According to Einstein’s equation, the conversion of 1.00 g of matter into energy generates how much energy?
9.00 × 1013 J
Q15.5.2
How much matter needs to be converted to energy to supply 400 kJ of energy, the approximate energy of 1 mol of C–H bonds? What conclusion does this suggest about energy changes of chemical reactions?
Q15.5.3
In the spontaneous fission of lead-208, the following reaction occurs: $_{}^{208}\textrm{Pb}\rightarrow \: _{ }^{129}\textrm{I}\: +\: _{ }^{76}\textrm{Cu}\: +\: 3_{ }^{1}\textrm{n}$ For every mole of lead-208 that decays, 0.1002 g of mass is lost. How much energy is given off per mole of lead-208 reacted?
9.02 × 1012 J
Q15.5.4
In the spontaneous fission of radium-226, the following reaction occurs: $_{}^{226}\textrm{Ra}\rightarrow \: _{ }^{156}\textrm{Pm}\: +\: _{ }^{68}\textrm{Co}\: +\: 2_{ }^{1}\textrm{n}$ For every mole of radium-226 that decays, 0.1330 g of mass is lost. How much energy is given off per mole of radium-226 reacted?
Q15.5.5
Recalculate the amount of energy from Exercise 3 in terms of the number of grams of lead-208 reacted.
4.34 × 1010 J/g
Q15.5.6
Recalculate the amount of energy from Exercise 4 in terms of the number of grams of radium-226 reacted.
Q15.5.7
What is the energy change of this fission reaction? Masses in grams are provided. $\underset{241.0569}{_{ }^{241}\textrm{Pu}}\rightarrow \: \underset{139.9106}{_{ }^{140}\textrm{Ba}}\: +\: \underset{89.9077}{_{ }^{90}\textrm{Sr}}\: +\: \underset{11\times 1.0087}{11_{ }^{1}\textrm{n}}$
−1.28 × 1013 J
Q15.5.8
What is the energy change of this fission reaction? Masses in grams are provided. $\underset{247.0704}{_{ }^{247}\textrm{Cm}}\rightarrow \: \underset{106.9099}{_{ }^{107}\textrm{Ru}}\: +\: \underset{130.9085}{_{ }^{131}\textrm{Te}}\: +\: \underset{9\times 1.0087}{9_{ }^{1}\textrm{n}}$
Q15.5.9
The two rarer isotopes of hydrogen—deuterium and tritium—can also be fused to make helium by the following reaction: $^{2}H+^{3}H\rightarrow ^{4}\textrm{He}+^{1}n$ In the course of this reaction, 0.01888 g of mass is lost. How much energy is given off in the reaction of 1 mol of deuterium and tritium?
1.70 × 1012 J
Q15.5.10
A process called helium burning is thought to occur inside older stars, forming carbon: $3^{4}He\rightarrow ^{12}C$ If the reaction proceeds with 0.00781 g of mass lost on a molar basis, how much energy is given off?
Q15.5.11
Briefly describe how a nuclear reactor generates electricity.
S15.5.11
A nuclear reactor controls a nuclear reaction to produce energy in usable amounts. The energy produced generates steam, which is used to turn a turbine that generates electricity for general use.
Q15.5.12
Briefly describe the difference between how a nuclear reactor works and how a nuclear bomb works.
Q15.5.13
What is a chain reaction?
S15.5.13
a process that generates more reaction pathways for each previous reaction
Q15.5.14
Why must uranium be enriched to supply nuclear energy?
Additional Exercises
QExtra.1
Given that many elements are metals, suggest why it would be unsafe to have radioactive materials in contact with acids.
SExtra.1
Acids can dissolve many metals; a spilled acid can lead to contamination.
QExtra.2
Many alpha-emitting radioactive substances are relatively safe to handle, but inhaling radioactive dust can be very dangerous. Why?
QExtra.3
Uranium can be separated from its daughter isotope thorium by dissolving a sample in acid and adding sodium iodide, which precipitates thorium(III) iodide:
$Th^{3+}(aq)+3I^{-}(aq)\rightarrow ThI_{3}(s)$
If 0.567 g of Th3+ were dissolved in solution, how many milliliters of 0.500 M NaI(aq) would have to be added to precipitate all the thorium?
14.7 mL
QExtra.4
Thorium oxide can be dissolved in acidic solution: $ThO_{2}(s)+4H^{+}\rightarrow Th^{4+}(aq)+2H_{2}O(l)$
How many milliliters of 1.55 M HCl(aq) are needed to dissolve 10.65 g of ThO2?
QExtra.5
Radioactive strontium is dangerous because it can chemically replace calcium in the human body. The bones are particularly susceptible to radiation damage. Write the nuclear equation for the beta emission of strontium-90.
SExtra.5
$_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^{ 0}\textrm{e}$
QExtra.6
Write the nuclear equation for the beta emission of iodine-131, the isotope used to diagnose and treat thyroid problems.
QExtra.7
A common uranium compound is uranyl nitrate hexahydrate [UO2(NO3)2 ⋅ 6H2O]. What is the formula mass of this compound?
502.15 g/mol
QExtra.8
Plutonium forms three oxides: PuO, PuO2, and Pu2O3. What are the formula masses of these three compounds?
QExtra.9
A banana contains 600 mg of potassium, 0.0117% of which is radioactive potassium-40. If 1 g of potassium-40 has an activity of 2.626 × 105 Bq, what is the activity of a banana?
about 18 Bq
QExtra.10
Smoke detectors typically contain about 0.25 mg of americium-241 as part of the smoke detection mechanism. If the activity of 1 g of americium-241 is 1.26 × 1011 Bq, what is the activity of americium-241 in the smoke detector?
QExtra.11
Uranium hexafluoride (UF6) reacts with water to make uranyl fluoride (UO2F2) and HF. Balance the following reaction: $UF_{6}+H_{2}O\rightarrow UO_{2}F_{2}+HF$
SExtra.11
$UF_{6}+2H_{2}O\rightarrow UO_{2}F_{2}+4HF$
QExtra.12
The cyclopentadienyl anion (C5H5) is an organic ion that can make ionic compounds with positive ions of radioactive elements, such as Np3+. Balance the following reaction:
$NpCl_{3}+Be(C_{2}H_{5})_{2}\rightarrow Np(C_{2}H_{5})_{3}+BeCl_{2}$
QExtra.13
If the half-life of hydrogen-3 is 12.3 y, how much time does it take for 99.0% of a sample of hydrogen-3 to decay?
81.7 y
QExtra.14
If the half-life of carbon-14 is 5,730 y, how long does it take for 10.0% of a sample of carbon-14 to decay?
QExtra.15
Although bismuth is generally considered stable, its only natural isotope, bismuth-209, is estimated to have a half-life of 1.9 × 1019 y. If the universe is estimated to have a lifetime of 1.38 × 1010 y, what percentage of bismuth-209 has decayed over the lifetime of the universe? (Hint: Be prepared to use a lot of decimal places.)
SExtra.15
about 0.000000005%
QExtra.16
The most common isotope of uranium (uranium-238) has a half-life of 4.5 × 109 y. If the universe is estimated to have a lifetime of 1.38 × 1010 y, what percentage of uranium-238 has decayed over the lifetime of the universe?
QExtra.17
Refer to Table 15.4.1, and separate the sources of radioactive exposure into voluntary and involuntary sources. What percentage of radioactive exposure is involuntary?
SExtra.17
Radioactive atoms in the body, terrestrial sources, and cosmic sources are truly involuntary, which is about 27% of the total. Radon exposure, medical sources, consumer products, and even nuclear energy sources can be avoided.
QExtra.18
With reference to Table 15.4.1, suggest ways that a practical person can minimize exposure to radioactivity. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Ball_et_al._(Beginning_Chemistry)/15%3A_Nuclear_Chemistry.txt |
(22) What type of reaction is necessary to create the products of the initial stage of fatty acid catabolism? Identify the products? What about for the second stage? Can the third stage be summarized in this way? Why or why not?
(23) Explain the process for the body obtaining fatty acids from our diets. What molecules are responsible for delivering fatty acids into the body? Are fatty acids ever consumed directly or only as part of large biomolecules?
(24) If the bloodstream is aqueous, can lipids dissolve into it for transport? If not, what alternative route through the body do lipids use for transport? Give a brief summary of how a triglyceride goes from entering the digestive system to the specific passages in the body they travel through.
(25) Lipoproteins have a membrane structure similar to that of cell membranes. Which portion, the exterior and interior, are polar and which are non-polar? Which lipoproteins have the greater densities, those with more protein in their composition or less protein?
(26) Based on your previous answer, where would you be more likely to find the lipids in a lipoprotein? Where would you be more likely to find the protein?
(27) Cells contain many organelles with many different purposes and functions. Which organelle is responsible for beta-oxidation? What stage of fatty acid catabolism is this?
(28) How many steps are involved in beta-oxidation? What types of reactions does each step entail? Are any of them more frequent than others?
01.E: Elements and Atoms (Exercises)
These are homework exercises to accompany the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health.
Exercises: Chemistry for Allied Health (Soult)
These are homework exercises to accompany Chapter 1 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
Questions
(click here for solutions)
Q1.1.1
Express the following values in scientific notation.
1. 150,000,000
2. 0.000043
3. 332000
4. 0.0293
5. 932
6. 0.1873
7. 78,000
8. 0.0001
9. 4500
10. 0.00290
11. 6281
12. 0.00700
Q1.1.2
Express the following values in decimal notation.
1. 3.8 x 10-3
2. 9.21 x 105
3. 7.91 x 10-2
4. 2.5 x 106
5. 3.42 x 10-8
6. 5.4 x 105
7. 3 x 10-3
8. 7.34 x 102
9. 9.8 x 10-4
10. 6 x 107
11. 4.20 x 10-6
12. 4.20 x 106
Q1.1.3
What SI base unit would be appropriate for each measurement?
1. the length of a room
2. the amount of carbon in a diamond
3. the mass of NaCl in a bottle
Q1.1.4
List the meaning of each abbreviation of the base units.
1. m
2. K
3. kg
4. s
5. mol
Q1.1.5
What is the the derived unit from the SI base units for the relationship of each pair of quantities?
1. mass and volume
2. distance and time
3. amount of substance and volume
4. area
Q1.1.6
Give the meaning and name of each metrix prefix abbreviation.
1. M
2. m
3. n
4. d
Q1.1.7
Give the abbreviation and meaning of each metrix prefix.
1. kilo
2. centi
3. micro
4. giga
Q1.1.8
Name the prefix with the following numerical meaning.
1. 1/10
2. 1,000,000
3. 1/1,000,000
4. 1/100
5. 1
Q1.1.9
Convert each temperature to the missing one between Celsius and Fahrenheit.
1. 77°F
2. 212°F
3. 37°C
4. 22°C
5. 95°F
6. 15°C
7. 0°F
8. 0°C
9. –10°C
10. –10°F
(click here for solutions)
Q1.2.1
Explain the similarities and differences between accuracy and precision.
Q1.2.2
The density of a copper sample was determined by three different students (shown below). Each performed the measurement three times and is reported below (all values in units of g/cm3). The accepted value for the density of copper is 8.92 g/cm3.
1. Determine if each student's data is accurate, precise, neither or both.
2. What is the average density based on Justin's data?
3. What is the average density based on Jane's data?
• Jane: 8.94, 8.89, 8.91
• Justin: 8.32, 8.31, 8.34
• Julia: 8.64, 9.71, and 9.10
Q1.2.3
Determine the number of significant figures in each of the following values.
1. 406
2. 3.00
3. 3.20
4. 0.25
5. 0.0689
6. 0.002910
7. 3941
8. 46.250
9. 30.21
10. 0.10300
Q1.2.4
Write each value with three significant figures, use scientific notation if necessary.
1. 34500
2. 24
3. 0.0345
4. 0.012
5. 612.8
6. 98.22
7. 0.14928
8. 300
Q1.2.5
Give three examples of exact numbers.
Q1.2.6
Find the result of each of the following calculations and report the value with the correct number of significant figures.
1. 0.23 + 12.2 =
2. 13 - 1.03 =
3. 0.839 + 0.28925 =
4. 28 + 34.4 =
5. 0.8 + 2.3 =
6. 34.9 - 0.583 =
7. 21 - 0.132 =
8. 0.840 + 0.9334
Q1.2.7
Find the result of each of the following calculations and report the value with the correct number of significant figures.
1. 34 x 0.12 =
2. 68.2 / 0.78 =
3. 3.29 x 104 x 16.2 =
4. 0.8449 x 29.7 =
5. 5.92 x 103 / 0.628 =
6. 3.00 x 2.6 =
7. 2.50 x 9.331 =
8. 3.20 / 12.75 =
(click here for solutions)
Q1.3.1
What is a conversion factor?
Q1.3.2
What is the conversion factor between each pair of units?
1. feet and inches
2. mL and cm3
3. kg and g
4. cm and m
5. mm and cm
6. inches and centimeters
7. grams and pounds
8. g and µg (mcg)
Q1.3.3
Complete each of the following conversions.
1. 34 cm to m
2. 3.7 ft to in
3. 345 mg to Mg
4. 5.3 km to mm
5. 4.0 L to mL
6. 3.45 x 103 mm to km
7. 78 cm3 to mL
8. 0.85 kg to dg
9. 13 pints to gallon
10. 0.35 L to cm3
Q1.3.4
Complete each of the following conversions.
1. 342 cm3 to dm3
2. 2.70 g/cm3 to kg/L
3. 34 mi/hr to km/min
4. 0.00722 km2 to m2
5. 4.9 x 105 mcm3 to mm3
6. 80. km/hr to mi/hr
(click here for solutions)
Q1.4.1
Solve each of the following.
1. What percent of 35 is 8.2?
2. What percent of 56 is 12?
3. What percent of 923 is 38?
4. What percent of 342 is 118?
Q1.4.2
Solve each of the following?
1. What is 42% of 94?
2. What is 83% of 239?
3. What is 16% of 45?
4. What is 38% of 872?
Q1.4.3
Solve each of the following?
1. 42 is 34% of what number?
2. 73 is 82% of what number?
3. 13 is 57% of what number?
4. 75 is 25% of what number?
5. 25 is 15% of what number?
6. 98 is 76% of what number?
Q1.4.4
A patient originally weighs 182 pounds and loses 15.0% of their body weight. What is their final weight?
Q1.4.5
A patient's original weight was 135 pounds and they lose 12 pounds. What percent of their body weight did they lose?
Q1.4.6
A patient needs to increase their calcium supplement by 25% a week. If they are currently taking a 300. mg supplement, how much more will they need to take?
Q1.4.7
An infant's birth weight is 7 pounds, 1 ounce. Her discharge weight is 6 pounds, 13 ounces. What percent of her birth weight did she lose?
Q1.4.8
A patient needs a 20.% decrease in their medication dosage from 125 mg. What will his dosage be after the decrease?
Answers
1.1: Measurements Matter
Q1.1.1
1. 1.5 × 108
2. 4.3 × 10–5
3. 3.32 × 105
4. 2.93 × 10–2
5. 9.32 × 102
6. 1.873 × 10–1
7. 7.8 × 104
8. 1 × 10–4
9. 4.5 × 103
10. 2.9 × 10–3
11. 6.281 × 103
12. 7 × 10–3
Q1.1.2
1. 0.0038
2. 921000
3. 0.0791
4. 2500000
5. 0.0000000342
6. 540000
7. 0.003
8. 734
9. 0.00098
10. 60000000
11. 0.00000420
12. 4200000
Q1.1.3
1. meter
2. mole
3. kilogram
Q1.1.4
1. meter
2. Kelvin
3. kilogram
4. second
5. mole
Q1.1.5
1. kg/m3
2. m/s
3. mol/m3 is based on SI base units, but mol/L is also acceptable
4. m2
Q1.1.6
1. Mega, 106
2. milli, 10–3
3. nano, 10–9
4. deci, 10–1
Q1.1.7
1. k, 103
2. c, 10–2
3. µ (or mc), 10–6
4. G, 109
Q1.1.8
1. deci
2. mega
3. micro
4. centi
5. none (base unit)
Q1.1.9
1. 77°F = 25°C
2. 212°F = 100°C
3. 37°C = 98.6°F
4. 22°C = 72°F
5. 95°F = 35°C
6. 15°C = 59°F
7. 0°F = –18°C
8. 0°C = 32°F
9. –10°C = 14°F
10. –10°F = –23°C
1.2 Significant Figures
Q1.2.1
Accuracy is a measure of how close the values are close to the correct value while precision is a measure of how close values are to each other.
Q1.2.2
1.
• Jane: 8.94, 8.89, 8.91 - accurate and precise
• Justin: 8.32, 8.31, 8.34 - precise
• Julia: 8.64, 9.71, and 9.10 - neither accurate nor precise
1. 8.32 g/cm3
2. 8.91 g/cm3
Q1.2.3
1. 3
2. 3
3. 3
4. 2
5. 3
6. 4
7. 4
8. 5
9. 4
10. 5
Q1.2.4
1. 3.45 × 104
2. 2.40 × 101
3. 3.45 × 10–2
4. 1.20 × 10–2
5. 613 or 6.13 × 102
6. 9.82 × 101
7. 0.149 or 1.49 × 10–1
8. 300. or 3.00 × 102
Q1.2.5
Answers will vary. 12 eggs, 100 cm = 1 m, 1 inch = 2.54 cm, 4 people
Q1.2.6
1. 0.23 + 12.2 = 12.43 = 12.4
2. 13 - 1.03 = 11.97 = 12
3. 0.839 + 0.28925 = 1.12825 = 1.128
4. 28 + 34.4 = 62.4 = 62
5. 0.8 + 2.3 = 3.1
6. 34.9 - 0.583 = 34.317 = 34.3
7. 21 - 0.132 = 20.868 = 21
8. 0.840 + 0.9334 = 1.7734 = 1.773
Q1.2.7
1. 34 x 0.12 = 4.08 = 4.1
2. 68.2 / 0.78 = 87.4358974 = 87
3. 3.29 x 104 x 16.2 = 5.32980 × 105 = 5.33 × 105
4. 0.8449 x 29.7 = 25.09353 = 25.1
5. 5.92 x 103 / 0.628 = 9.4267515 × 103 = 9.43 × 103
6. 3.00 x 2.6 = 7.8
7. 2.50 x 9.331 = 23.3275 = 23.3
8. 3.20 / 12.75 = 0.25098 = 0.251
1.3 Scientific Dimensional Analysis
Q1.3.1
A conversion factor is a relationship between two units. The value in the numerator has some equivalence to the value in the denominator.
Q1.3.2
1. 1 foot = 12 inches
2. 1 mL = 1 cm3
3. 1 kg = 1000 g or 1 × 10–3 kg = 1 g
4. 100 cm = 1 m or 1 cm = 1 × 10–2 m
5. 10 mm = 1 cm
6. 1 inch = 2.54 cm
7. 454 grams = 1 pound
8. 1 g = 1 × 106 µg (mcg) or 1 × 10–6 g = 1 µg (mcg)
Q1.3.3
1. $34 \; cm \times \frac{1 \; m}{100\;cm} = 0.34\;m$
2. $3.7 \; ft \times \frac{12 \; in}{1\;ft}=44.4\;in=44\;in$
3. $345\;mg \times \frac{1\;g}{1000\;mg} \times \frac{1\;Mg}{1\times {10}^6\;g}=3.45 \times {10}^{-7}\;Mg$
4. $5.3\;km\times\frac{1000\;m}{1\;km}\times\frac{1000\;mm}{1\;m}=5.3\times{10}^6\;mm$
5. $4.0\;L\times\frac{1000\;mL}{1\;L}=4.0\times{10}^3\;mL$
6. $3.45\times{10}^3\;mm\times\frac{1\;m}{1000\;mm}\times\frac{1\;km}{1000\;m}=3.45\times{10}^{-3}\;km$
7. $78\;{cm}^3\times\frac{1\;mL}{{cm}^3}=78\;mL$
8. $0.85\;kg\times\frac{1000\;g}{1\;kg}\times\frac{10\;dg}{1\;g}=8.5\times{10}^3\;dg$
9. $13\;pints\times\frac{1\;quart}{2\;pints}\times\frac{1\;gallon}{4\;quarts}=1.6\;gallons$
10. $0.35\;L\times\frac{1000\;mL}{1\;L}\times\frac{1\;mL}{1\;{cm}^3}=3.5\times{10}^2\;{cm}^3$
Q1.3.4
1. $342\;{cm}^3\times\frac{1\;dm}{10\;cm}\times\frac{1\;dm}{10\;cm}\times\frac{1\;dm}{10\;cm}=0.342\;{dm}^3$ or $342\;{cm}^3\times {\left( \frac{1\;dm}{10\;cm} \right)}^3=342\;{cm}^3\times \frac{{1}^3\;{dm}^3}{{10}^3\;{cm}^3}=0.342\;{dm}^3$
2. $\frac{2.70\;g}{{cm}^3}\times\frac{1\;kg}{1000\;g}\times\frac{1\;{cm}^3}{1\;mL}\times\frac{1000\;mL}{1\;L}=\frac{2.70\;kg}{L}$
3. $\frac{34\;mi}{hr} \times \frac{5280\;ft}{1\;mi} \times \frac{12\;in}{1\;ft}\times \frac{2.54\;cm}{1\;in} \times \frac{1\;m}{100\;cm} \times \frac{1\;km}{1000\;m} \times \frac{1\;hr}{60\;min} = \frac{0.91\;km}{min}$
4. $0.00722\;k{m^2} \times \frac{1000\;m}{1\;km} \times \frac{1000\;m}{1\;km} = 7.22 \times {10^3}\;{m^2}$
5. $4.95 \times {10^5}\;mc{m^3} \times \frac{1\;mm}{1000\;mcm} \times \frac{1\;mm}{1000\;mcm} \times \frac{1\;mm}{1000\;mcm} = 4.95 \times {10^ - }^4\;m{m^3}$
6. $\frac{80.\;km}{hr}\times\frac{1000\;m}{1\;km}\times\frac{100\;cm}{1\;m}\times\frac{1\;in}{2.54\;cm}\times\frac{1\;ft}{12\;in}\times\frac{1\;mi}{5280\;ft}=\frac{50.\;mi}{hr}$
1.4 Percentages
Q1.4.1
1. $\%= \frac{part}{whole} \times 100= \frac{8.2}{35} \times 100= 23\%$
2. $\%= \frac{part}{whole} \times 100= \frac{12}{56} \times 100= 21\%$
3. $\%= \frac{part}{whole} \times 100= \frac{38}{923} \times 100= 4.1\%$
4. $\%= \frac{part}{whole} \times 100= \frac{118}{342} \times 100= 34.5\%$
Q1.4.2
1. $\begin{array}{c} \% = \frac{part}{whole} \times 100\ 42\% = \frac{part}{94} \times 100\ part = 39 \end{array}$
2. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 83\% = \frac{part}{239} \times 100\ part = 198=2.0\times{10}^2 \end{array}$
3. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 16\% = \frac{part}{45} \times 100\ part = 7.2\ \end{array}$
4. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 38\% = \frac{part}{872} \times 100\ part = 3.3\times{10}^2 \end{array}$
Q1.4.3
1. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 34\% = \frac{42}{whole} \times 100\ whole = 1.2\times{10}^2 \end{array}$
2. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 82\% = \frac{73}{whole} \times 100\ whole = 89 \end{array}$
3. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 57\% = \frac{13}{whole} \times 100\ whole = 23 \end{array}$
4. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 25\% = \frac{75}{whole} \times 100\ whole = 3.0\times{10}^2 \end{array}$
5. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 15\% = \frac{25}{whole} \times 100\ whole = 1.7\times{10}^2 \end{array}$
6. $\begin{array}{c} \quad \ \% = \frac{part}{whole} \times 100\ 76\% = \frac{98}{whole} \times 100\ whole = 129 \end{array}$
Q1.4.4
$\begin{array}{l} \% = \frac{part}{whole} \times 100\ 15.0\% = \frac{part}{182\;pounds} \times 100\ part = 27.3\;pounds\;lost\ \ 182\;pounds - 27.3\;pounds = 154.7\;pounds = 155\;pounds \end{array}$
Q1.4.5
$\begin{array}{l} \% = \frac{part}{whole} \times 100\ \% = \frac{12\;pounds}{135\;pounds} \times 100\ \% = 8.9\%\;lost\ \end{array}$
Q1.4.6
$\begin{array}{l} \% = \frac{part}{whole} \times 100\ 25\% = \frac{part}{300.\;mg} \times 100\ part = 75\;mg\;more\ \end{array}$
Q1.4.7
Convert both weights to ounces, find the ounces lost, and then find the percent lost.
Birth weight: $\left( 7\;pounds\times 16 \right) + 1\;ounce=113\;ounces$
Discharge weight: $\left( 6\;pounds\times 16 \right) + 13\;ounces=109\;ounces$
Weight lost: $113\;ounces-109\;ounces=4\;ounces$
Percent lost from original brith weight.
$\begin{array}{l} \% = \frac{part}{whole} \times 100\ \% = \frac{4\;ounces}{113\;ounces} \times 100\ \% = 3.5\% =4\% \end{array}$
Q1.4.8
$\begin{array}{l} \% = \frac{part}{whole} \times 100\ 20\% = \frac{part}{125\;mg} \times 100\ part = 25\;mg\;lost\ \ 125\;mg - 25\;mg = 100.\;mg \end{array}$ | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Chem_309_Bennett_(GOB)/Fatty_Acid_Catabolism.txt |
These are homework exercises to accompany Chapter 2 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
Questions
(click here for solutions)
Q2.1.1
Give the names and symbols of three elements.
Q2.1.2
Describe where protons, neutrons, and electrons are located in an atom.
Q2.1.3
Why are all atoms electrically neutral?
Q2.1.4
How many protons are in the nucleus of each of the following atoms?
1. neon (Ne)
2. gold (Au)
3. strontium (Sr)
4. uranium (U)
Q2.1.5
A certain atom contains 28 protons, 28 electrons, and 31 neutrons. Provide the following:
1. atomic number
2. mass number
3. name of element
Q2.1.6
How many protons, neutrons, and electrons are in an atom of cesium-133?
Q2.1.7
How many protons, neutrons, and electrons are there in the atom $\ce{^{19}_9F}$?
Q2.1.8
How many protons, neutrons, and electrons are there in an atom of lead-207?
Q2.1.9
A certain atom has an atomic number of 36 and a mass number of 84. Write out the designation for this isotope in both nuclide symbol form and in hyphenated form.
Q2.1.10
An atom has a mass number of 59 and contains 32 neutrons in its nucleus. What element is it?
Q2.1.11
Complete the Table below:
Isotope Isotope Symbol Atomic Number Mass Number
sodium-23
$\ce{^{75}_{33}As}$
silver-108
Q2.1.12
Which one is an isotope of $\ce{^{40}_{18}Ar}$? Explain.
1. $\ce{^{40}_{20}Ca}$
2. $\ce{^{39}_{18}Ar}$
3. $\ce{^{40}_{18}Ar}$
Q2.1.13
Fill in Table below:
Isotope Number of Protons Number of Electrons Number of Neutrons Isotope Symbol
hydrogen-1
hydrogen-2
beryllium-9
aluminum-27
Q2.1.14
Fill in Table below:
Element Symbol Atomic Number Mass Number # of Protons # of Electrons # of Neutrons Isotope Symbol
nitrogen 14
B 11
30 35
77 116
$\ce{^{56}_{26}Fe}$
(click here for solutions)
Q2.2.1
Define matter and give three examples of matter.
Q2.2.2
Explain the differences between compounds and mixtures.
Q2.2.3
Explain the differences between pure substances and mixtures.
Q2.2.4
Identify each of the following as a pure substance or a mixture.
1. salt water
2. NaCl
3. brewed coffee
4. air
Q2.2.5
Label each of the following as an element, compound, homogeneous mixture, or heterogeneous mixture.
1. silicon
2. sulfuric acid
3. air
4. soda
5. sugar
6. muddy water
7. chicken noodle soup
8. scoop of sand from the beach
Q2.2.6
Identify each of the following elements as a metal, nonmetal, or metalloid.
1. iron (Fe)
2. gallium (Ga)
3. phosphorus (P)
4. potassium (K)
5. neon (Ne)
6. germanium (Ge)
Q2.2.7
Identify each of the following elements as a metal, nonmetal, or metalloid.
1. chlorine (Cl)
2. hydrogen (H)
3. antimony (Sb)
4. titanium (Ti)
5. nitrogen (N)
6. selenium (Se)
Q2.2.8
Describe how you identify molecular and ionic compounds.
Q2.2.9
Label each as an ionic or molecular compound.
1. H2O2
2. N2O5
3. PF3
4. MgCl2
5. KBr
6. AlCl3
Q2.2.10
Label each as an ionic or molecular compound.
1. CaO
2. Na2S
3. NH3
4. CO2
5. N2H4
6. N2O
(click here for solutions)
Q2.3.1
Define "counting number".
Q2.3.2
What is the value of Avogadro's number?
Q2.3.3
What seven elements exist as diatomic molecules in nature?
Q2.3.4
How many atoms of helium are present in each of the following samples?
1. 1 mole
2. 2 moles
3. 2.5 moles
4. 0.5 moles
5. 0.35 moles
Q2.3.5
How many molecules of water are present in each of the following samples?
1. 1 mole
2. 2 moles
3. 2.5 moles
4. 0.5 moles
5. 0.35 moles
Q2.3.6
Compare the answers for each part of questions 4 and 5 to one another. How do they compare? Explain why.
Q2.3.7
How many moles of silicon is 6.73 x 1025 atoms of silicon?
Q2.3.8
How many moles of sodium is 4.29 x 1022 atoms of sodium?
Q2.3.9
How many atoms of each element are in one unit of each compound?
1. H2O2
2. N2O5
3. PF3
4. MgCl2
5. KBr
6. AlCl3
7. CaO
8. Na2S
9. NH3
10. CO2
11. N2H4
12. N2O
Q2.3.10
How many moles of each element are in one mole of each compound?
1. H2O2
2. N2O5
3. PF3
4. MgCl2
5. KBr
6. AlCl3
7. CaO
8. Na2S
9. NH3
10. CO2
11. N2H4
12. N2O
Q2.3.11
How do the answers in questions 9 and 10 compare to one another? Explain the similarities or differences.
Q2.3.12
How many moles of carbon are in 0.75 moles of CCl4? How many moles of chlorine?
Q2.3.13
How many atoms of carbon are in 0.75 moles of CCl4? How many atoms of chlorine?
Q2.3.14
How many moles of hydrogen are in 2.5 moles of H2O? How many moles of oxygen?
Q2.3.15
How many atoms of hydrogen are in 2.5 moles of H2O? How many atoms of oxygen?
Q2.3.16
A sample of CaNO3 contains 3.87 x 1025 atoms of oxygen. How many molecules of CaNO3 are in the sample?
Q2.3.17
A sample of propane gas (C3H8) contains 5.39 x 1024 atoms of carbon. How many atoms of hydrogen are in the sample?
Q2.3.18
What is the molar mass of each of the following elements (in atomic form)?
1. carbon
2. nitrogen
3. sodium
4. hydrogen
5. potassium
6. phosphorus
Q2.3.19
How many moles of each element listed in the previous question are present in a 25.0 g sample of the element?
Q2.3.20
For question 19, all of the samples have the same mass. Are the moles the same? Why or why not?
Q2.3.21
What is the mass of each of the following samples?
1. 0.35 moles sodium
2. 0.75 moles carbon
3. 1.34 moles potassium
4. 1.21 moles silicon
5. 0.95 moles calcium
6. 2.85 moles helium
Q2.3.22
Determine the molar mass of each of the following compounds?
1. CO2
2. N2H4
3. CaF2
4. C6H12O6
5. CH4
6. C6H6
7. Na2SO4
8. K3PO4
9. Al(NO3)3
10. Mg3(PO4)2
Q2.3.23
Calculate the moles of each of the following samples.
1. 25.0 g CO2
2. 10.0 g N2H4
3. 85.0 g CaF2
4. 15.5 g C6H12O6
5. 20.0 g CH4
6. 100.0 g C6H6
7. 30.0 g Na2SO4
8. 75.0 g K3PO4
9. 50.0 g Al(NO3)3
10. 47.2 g Mg3(PO4)2
Q2.3.24
Calculate the mass of each of the following samples.
1. 3.5 mol CO2
2. 0.45 mol N2H4
3. 2.25 mol CaF2
4. 1.75 mol C6H12O6
5. 4.9 mol CH4
6. 8.75 mol C6H6
7. 2.35 mol Na2SO4
8. 0.672 mol K3PO4
9. 0.95 mol Al(NO3)3
10. 1.15 mol Mg3(PO4)2
(click here for solutions)
Q2.4.1
What is the electron arrangement for each of the elements?
a. Na
b. Ne
c. Be
d. N
e. S
f. Cl
Q2.4.2
How many valence electrons are in each element?
a. K
b. P
c. F
d. S
e. Li
f. B
Q2.4.3
What is the octet rule?
(click here for solutions)
Q2.5.1
Define ion.
Q2.5.2
How are anions and cation the same? Different?
Q2.5.3
What is the most common ion formed from each element?
a. Li
b. Na
c. Ca
d. B
e. P
f. S
g. Cl
h. Br
Q2.5.4
How many protons, neutrons, and electrons are present in the ions indicated in the previous question?
Q2.5.5
Identify the following elements.
a. An ion with a 3+ charge and two electrons.
b. An ion with a 1$-$ charge and 18 electrons.
c. An ion with a 1+ charge and 18 electrons.
d. An ion with a 3$-$ charge and 10 electrons.
Q2.5.6
Describe a polyatomic ion.
Q2.5.7
Which are polyatomic ions?
1. NO3
2. O2–
3. NH4+
4. Mg2+
5. Na+
6. O22–
(click here for solutions)
Q2.6.1
What element is present in all organic compounds?
Q2.6.2
Give three examples of metallic elements.
Q2.6.3
Give three examples of nonmetallic elements.
Q2.6.4
What types of elements form an ionic compound?
Q2.6.5
How do the electrons behave in the formation of an ionic bond?
Q2.6.6
What is the overall charge of an ionic compound?
Q2.6.7
What is the formula for the ionic compound formed from each of the following pairs?
a. potassium and sulfur
b. silver and chlorine (silver has a 1+ charge)
c. calcium and oxygen
d. aluminum and iodine
e. barium and nitrogen
f. sodium and phosphorus
g. lithium and fluorine
h. magnesium and nitrogen
i. calcium and sulfur
j. beryllium and bromine
k. zinc and nitrogen (zinc has a 2+ charge)
l. tin and iodine (tin has a 4+ charge)
Q2.6.8
Write the formula for the compound formed between sodium and each of these polyatomic ions. You can look up the formula and charge for each polyatomic ion.
a. carbonate
b. chlorate
c. chlorite
d. phosphate
e. nitrate
f. sulfate
g. chromate
h. dichromate
Q2.6.9
Write the formula for the compound formed between magnesium and each of the polyatomic ions listed in the previous question.
Q2.6.10
Explain when parentheses should and should not be used in the formulas of ionic compounds.
Answers
2.1: Isotopes and Atomic Mass
Q2.1.1
Answers will vary.
Q2.1.2
Protons and neutrons are in the nucleus and electrons are located outside the nucleus.
Q2.1.3
The sum of the charges on ions in an ionic compound must equal zero.
Q2.1.4
1. 10
2. 79
3. 38
4. 92
Q2.1.5
A certain atom contains 28 protons, 28 electrons, and 31 neutrons. Provide the following:
1. 28
2. 59
3. nickel
Q2.1.6
55 protons, 78 neutrons, 55 electrons
Q2.1.7
9 protons, 10 neutrons, 9 electrons
Q2.1.8
82 protons, 125 neutrons, 82 electrons
Q2.1.9
$\ce{^{84}_{36}Kr}$, krypton-84
Q2.1.10
cobalt
Q2.1.11
Isotope Isotope Symbol Atomic Number Mass Number
sodium-23 $\ce{^{23}_{11}Na}$ 11 23
aresenic-75 $\ce{^{75}_{33}As}$ 33 75
silver-108 $\ce{^{108}_{47}Ag}$ 47 108
Q2.1.12
1. $\ce{^{40}_{20}Ca}$ - not an isotope because it is a different element
2. $\ce{^{39}_{18}Ar}$ - isotope because it has the same atomic number but a different atomic mass
3. $\ce{^{40}_{18}Ar}$ - not an isotope because it has the same atomic number and the same atomic mass
Q2.1.13
Isotope Number of Protons Number of Electrons Number of Neutrons Isotope Symbol
hydrogen-1 1 1 0 $\ce{^{1}_{1}H}$
hydrogen-2 1 1 1 $\ce{^{2}_{1}H}$
beryllium-9 4 4 5 $\ce{^{9}_{4}Be}$
aluminum-27 13 13 14 $\ce{^{27}_{13}Al}$
Q2.1.14
Element Symbol Atomic Number Mass Number # of Protons # of Electrons # of Neutrons Isotope Symbol
nitrogen N 7 14 7 7 7 $\ce{^{14}_{7}N}$
boron B 5 11 5 5 6 $\ce{^{11}_{5}B}$
zinc Zn 30 65 30 30 35 $\ce{^{65}_{30}Zn}$
iridium Ir 77 193 77 77 116 $\ce{^{193}_{77}Ir}$
iron Fe 26 56 26 26 30 $\ce{^{56}_{26}Fe}$
2.2: Matter
Q2.2.1
Matter is anything that has mass and occupies space. Examples of matter will vary and can be any object from an atom to a macroscopic object.
Q2.2.2
A compound is a combination of elements with a fixed composition. The elements in the compound do not retain their individual identity by have the properties of the compound. A mixture does not have a fixed composition and each component of the mixture retains its identity and properties.
Q2.2.3
A pure substance contains only one component, either an element or compound, while a mixture contains multiple pure substances.
Q2.2.4
1. mixture
2. pure substance
3. mixture
4. mixture
Q2.2.5
1. element
2. compound
3. homogeneous mixture
4. heterogeneous mixture
5. compound
6. heterogeneous mixture
7. heterogeneous mixture
8. heterogeneous mixture
Q2.2.6
1. metal
2. metal
3. nonmetal
4. metal
5. nonmetal
6. metalloid
Q2.2.7
1. nonmetal
2. nonmetal
3. metalloid
4. metal
5. nonmetal
6. nonmetal
Q2.2.8
Ionic compounds are generally formed between a metal and nonmetal or between a polyatomic ion and another ion. Molecular compounds are composed of two ore more nonmetals.
Q2.2.9
1. molecular
2. molecular
3. molecular
4. ionic
5. ionic
6. ionic
Q2.2.10
1. ionic
2. ionic
3. molecular
4. molecular
5. molecular
6. molecular
2.3: Mole and Molar Mass
Q2.3.1
A counting number is a word that is associated with a specific number.
Q2.3.2
$6.022\times10^{23}$
Q2.3.3
H2, N2, O2, F2, Cl2, Br2, I2
Q2.3.4
1. $6.022\times10^{23}$ atoms
2. $1.204\times10^{24}$ atoms
3. $1.506\times10^{24}$ atoms
4. $3.011\times10^{23}$ atoms
5. $2.108\times10^{23}$ atoms
Q2.3.5
1. $6.022\times10^{23}$ molecules
2. $1.204\times10^{24}$ molecules
3. $1.506\times10^{24}$ molecules
4. $3.011\times10^{23}$ molecules
5. $2.108\times10^{23}$ molecules
Q2.3.6
The numbers are the same for the same number of moles because moles are a counting number. Regardless of what is being counted, a mole will have the same number of items.
Q2.3.7
$6.73 \times {10^{25}}\;atoms\;{\rm{Si}}\left( {\frac{1\;mole}{6.022 \times {10}^{23}\;atoms}} \right) = 112\;moles\;{\rm{Si}}$
Q2.3.8
$4.29 \times {10^{22}}\;atoms\;{\rm{Na}}\left( {\frac{1\;mole}{6.022 \times {10}^{23}\;atoms}} \right) = 0.0712\;moles\;{\rm{Na}}$
Q2.3.9
1. 2 hydrogen atoms, 2 oxygen atoms
2. 2 nitrogen atoms, 5 oxygen atoms
3. 1 phosphorus atom, 3 fluorine atoms
4. 1 magnesium atom, 2 chlorine atoms
5. 1 potassium atom, 1 bromine atom
6. 1 aluminum atom, 3 chlorine atoms
7. 1 calcium atom, 1 oxygen atom
8. 2 sodium atoms, 1 oxygen atom
9. 1 nitrogen atom, 3 hydrogen atoms
10. 1 carbon atom, 2 oxygen atoms
11. 2 nitrogen atoms, 4 hydrogen atoms
12. 2 nitrogen atoms, 1 oxygen atom
Q2.3.10
1. 2 moles hydrogen, 2 moles oxygen
2. 2 moles nitrogen, 5 moles oxygen
3. 1 mole phosphorus, 3 moles fluorine
4. 1 mole magnesium, 2 moles chlorine
5. 1 mole potassium, 1 mole bromine
6. 1 mole aluminum, 3 moles chlorine
7. 1 mole calcium, 1 mole oxygen
8. 2 moles sodium, 1 mole oxygen
9. 1 mole nitrogen, 3 moles hydrogen
10. 1 mole carbon, 2 moles oxygen
11. 2 moles nitrogen, 4 moles hydrogen
12. 2 moles nitrogen, 1 mole oxygen
Q2.3.11
The numbers are the same because the ratios are the same between atoms and moles. Moles are a counting number so they are a multiple of the number of atoms.
Q2.3.12
$0.75\;mol\;{\rm{CCl}_4}\left ( \frac{1\;mol\;{\rm{C}}}{1\;mol\;{\rm{CCl}_4}} \right ) =0.75\;mol\;\rm{C}$
$0.75\;mol\;{\rm{CCl}_4}\left ( \frac{4\;mol\;{\rm{Cl}}}{1\;mol\;{\rm{CCl}_4}} \right ) =3.0\;mol\;\rm{Cl}$
Q2.3.13
$0.75\;mol\;{\rm{CCl}_4}\left ( \frac{1\;mol\;{\rm{C}}}{1\;mol\;{\rm{CCl}_4}} \right )\left ( \frac{6.02\times 10^{23} \;atoms}{1\;mol\;{\rm{C}} }\right ) =4.52\times10^{23}\;atoms\;\rm{C}$
$0.75\;mol\;{\rm{CCl}_4}\left ( \frac{4\;mol\;{\rm{Cl}}}{1\;mol\;{\rm{CCl}_4}} \right )\left ( \frac{6.02\times 10^{23} \;atoms}{1\;mol\;{\rm{Cl}}} \right ) =1.81\times10^{24}\;atoms\;\rm{Cl}$
Q2.3.14
$2.5\;mol\;{\rm{H}_2\rm{O}}\left ( \frac{2\;mol\;{\rm{H}}}{1\;{mol}\;\rm{H}_2\rm{O}} \right ) =5.0\;mol\;\rm{H}$
$2.5\;mol\;{\rm{H}_2\rm{O}}\left ( \frac{1\;mol\;{\rm{O}}}{1\;{mol}\;\rm{H}_2\rm{O}} \right ) =2.5\;mol\;\rm{O}$
Q2.3.15
$2.5\;mol\;{\rm{H}_2\rm{O}}\left ( \frac{2\;mol\;{\rm{H}}}{1\;{mol}\;\rm{H}_2\rm{O}} \right ) \left ( \frac{6.02\times 10^{23} \;atoms}{1\;mol\;{\rm{H}} }\right )=3.01\times10^{24}\;atoms\;\rm{H}$
$2.5\;mol\;{\rm{H}_2\rm{O}}\left ( \frac{1\;mol\;{\rm{O}}}{1\;{mol}\;\rm{H}_2\rm{O}} \right ) \left ( \frac{6.02\times 10^{23} \;atoms}{1\;mol\;{\rm{O}} }\right )=1.51\times10^{24}\;atoms\;\rm{O}$
Q2.3.16
$3.87\times10^{25}\;atoms\;{\rm{O}}\left ( \frac{1\;molec\;{\rm{CaNO}}_3}{3\;atoms\;\rm{O}} \right )=1.29\times10^{25}\;molec\;\rm{CaNO}_3$
Q2.3.17
$5.39\times10^{24}\;atoms\;{\rm{C}}\left ( \frac{8\;atoms\;{\rm{H}}}{3\;atoms\;\rm{C}} \right )=1.44\times10^{25}\;atoms\;\rm{H}$
Q2.3.18
1. 12.01 g/mol
2. 14.01 g/mol
3. 22.99 g/mol
4. 1.008 g/mol
5. 39.10 g/mol
6. 30.97 g/mol
Q2.3.19
1. $25.0\;g\;{\rm{C}}\left ( \frac{1\;mol\;{\rm{C}}}{12.01\;\frac{g}{mol}} \right )=2.08\;mol\;{\rm{C}}$
2. $25.0\;g\;{\rm{N}}\left ( \frac{1\;mol\;{\rm{N}}}{14.01\;\frac{g}{mol}} \right )=1.78\;mol\;{\rm{N}}$
3. $25.0\;g\;{\rm{Na}}\left ( \frac{1\;mol\;{\rm{Na}}}{22.99\;\frac{g}{mol}} \right )=1.09\;mol\;{\rm{Na}}$
4. $25.0\;g\;{\rm{H}}\left ( \frac{1\;mol\;{\rm{H}}}{1.008\;\frac{g}{mol}} \right )=24.8\;mol\;{\rm{H}}$
5. $25.0\;g\;{\rm{K}}\left ( \frac{1\;mol\;{\rm{K}}}{39.10\;\frac{g}{mol}} \right )=0.639\;mol\;{\rm{K}}$
6. $25.0\;g\;{\rm{P}}\left ( \frac{1\;mol\;{\rm{P}}}{30.97\;\frac{g}{mol}} \right )=0.807\;mol\;{\rm{P}}$
Q2.3.20
The moles are different because a mole of atoms of each element has a different mass. Although they have the same mass, the number of atoms of each sample is different. Compare a ton of feathers to a ton of books. The same total mass but a different quantity of each.
Q2.3.21
What is the mass of each of the following samples?
1. $0.35\;mol\;{\rm{Na}}\left ( \frac{22.99\;g}{mol\;{\rm{Na}}} \right )=8.0\;g\;\rm{Na}$
2. $0.75\;mol\;{\rm{C}}\left ( \frac{12.01\;g}{mol\;{\rm{C}}} \right )=9.0\;g\;\rm{C}$
3. $1.34\;mol\;{\rm{K}}\left ( \frac{39.10\;g}{mol\;{\rm{K}}} \right )=52.4\;g\;\rm{K}$
4. $1.21\;mol\;{\rm{Si}}\left ( \frac{28.09\;g}{mol\;{\rm{Si}}} \right )=34.0\;g\;\rm{Si}$
5. $0.95\;mol\;{\rm{Ca}}\left ( \frac{40.08\;g}{mol\;{\rm{Ca}}} \right )=38\;g\;\rm{Ca}$
6. $2.85\;mol\;{\rm{He}}\left ( \frac{4.003\;g}{mol\;{\rm{He}}} \right )=11.4\;g\;\rm{He}$
Q2.3.22
1. CO2 has 1 mole of carbon and 2 moles of oxygen; $\left ( 12.01\frac{g}{mol}\times1 \right )+\left ( 16.00\frac{g}{mol}\times2 \right )=44.01\frac{g}{mol}$
2. N2H4 has 2 moles of nitrogen and 4 moles of hydrogen; $\left ( 14.01\frac{g}{mol}\times2 \right )+\left ( 1.008\frac{g}{mol}\times4 \right )=32.05\frac{g}{mol}$
3. CaF2 has 1 mole of calcium and 2 moles of fluorine; $\left ( 40.08\frac{g}{mol}\times1 \right )+\left ( 19.00\frac{g}{mol}\times2 \right )=78.08\frac{g}{mol}$
4. C6H12O6 has 6 moles of carbon, 12 moles of hydrogen, and 6 moles of oxygen; $\left ( 12.01\frac{g}{mol}\times6 \right )+\left ( 1.008\frac{g}{mol}\times12 \right )+\left ( 16.00\frac{g}{mol}\times6 \right )=180.16\frac{g}{mol}$
5. CH4 has 1 mole of carbon and 4 moles of hydrogen; $\left ( 12.01\frac{g}{mol}\times1 \right )+\left ( 1.008\frac{g}{mol}\times4 \right )=16.04\frac{g}{mol}$
6. C6H6 has 6 moles of carbon and 6 moles of hydrogen; $\left ( 12.01\frac{g}{mol}\times6 \right )+\left ( 1.008\frac{g}{mol}\times6 \right )=78.11\frac{g}{mol}$
7. Na2SO4 has 2 moles of sodium, 1 mole of sulfur, and 4 moles of oxygen; $\left ( 22.99\frac{g}{mol}\times2 \right )+\left ( 32.06\frac{g}{mol}\times1 \right )+\left ( 16.00\frac{g}{mol}\times4 \right )=142.04\frac{g}{mol}$
8. K3PO4 has 3 moles of potassium, 1 mole of phosphorus, and 4 moles of oxygen; $\left ( 39.10\frac{g}{mol}\times3 \right )+\left ( 30.97\frac{g}{mol}\times1 \right )+\left ( 16.00\frac{g}{mol}\times4 \right )=212.27\frac{g}{mol}$
9. Al(NO3)3 has 1 mole of aluminum, 3 moles of nitrogen, and 9 moles of oxygen; $\left ( 26.98\frac{g}{mol}\times1 \right )+\left ( 14.01\frac{g}{mol}\times3 \right )+\left ( 16.00\frac{g}{mol}\times9 \right )=213.01\frac{g}{mol}$
10. Mg3(PO4)2 has 3 moles of magnesium, 2 moles of phosphorus, and 8 moles of oxygen; $\left ( 24.31\frac{g}{mol}\times3 \right )+\left ( 30.97\frac{g}{mol}\times2 \right )+\left ( 16.00\frac{g}{mol}\times8 \right )=262.87\frac{g}{mol}$
Q2.3.23
1. $25.0\;g\;{\rm{CO_2}}\left ( \frac{1\;mol\;{\rm{CO_2}}}{44.01\;g} \right )=0.568\;mol\;\rm{CO_2}$
2. $10.0\;g\;{\rm{N_2H_4}}\left ( \frac{1\;mol\;{\rm{N_2H_4}}}{32.05\;g} \right )=0.312\;mol\;\rm{N_2H_4}$
3. $85.0\;g\;{\rm{CaF_2}}\left ( \frac{1\;mol\;{\rm{CaF_2}}}{78.08\;g} \right )=1.09\;mol\;\rm{CaF_2}$
4. $15.5\;g\;{\rm{C_6H_{12}O_6}}\left ( \frac{1\;mol\;{\rm{C_6H_{12}O_6}}}{180.16\;g} \right )=0.0860\;mol\;\rm{C_6H_{12}O_6}$
5. $20.0\;g\;{\rm{CH_4}}\left ( \frac{1\;mol\;{\rm{CH_4}}}{16.04\;g} \right )=1.25\;mol\;\rm{CH_4}$
6. $100.0\;g\;{\rm{C_6H_6}}\left ( \frac{1\;mol\;{\rm{C_6H_6}}}{78.11\;g} \right )=1.28\;mol\;\rm{C_6H_6}$
7. $30.0\;g\;{\rm{Na_2SO_4}}\left ( \frac{1\;mol\;{\rm{Na_2SO_4}}}{142.04\;g} \right )=0.211\;mol\;\rm{Na_2SO_4}$
8. $75.0\;g\;{\rm{K_3PO_4}}\left ( \frac{1\;mol\;{\rm{K_3PO_4}}}{212.27\;g} \right )=0.353\;mol\;\rm{K_3PO_4}$
9. $50.0\;g\;{\rm{Al(NO_3)_3}}\left ( \frac{1\;mol\;{\rm{Al(NO_3)_3}}}{213.01\;g} \right )=0.235\;mol\;\rm{Al(NO_3)_3}$
10. $47.2\;g\;{\rm{Mg_3(SO_4)_2}}\left ( \frac{1\;mol\;{\rm{Mg_3(SO_4)_2}}}{262.87\;g} \right )=0.180\;mol\;\rm{Mg_3(SO_4)_2}$
Q2.3.24
1. $3.5\;mol\;{\rm{CO_2}}\left ( \frac{44.01\;g}{mol\;{\rm{CO_2}}} \right )=1.50\times10^2\;g\;\rm{CO_2}$
2. $0.45\;mol\;{\rm{N_2H_4}}\left ( \frac{32.05\;g}{mol\;{\rm{N_2H_4}}} \right )=14\;g\;\rm{N_2H_4}$
3. $2.25\;mol\;{\rm{CaF_2}}\left ( \frac{78.08\;g}{mol\;{\rm{CaF_2}}} \right )=176\;g\;\rm{CaF_2}$
4. $1.75\;mol\;{\rm{C_6H_{12}O_6}}\left ( \frac{180.16\;g}{mol\;{\rm{C_6H_{12}O_6}}} \right )=315\;g\;\rm{C_6H_{12}O_6}$
5. $4.9\;mol\;{\rm{CH_4}}\left ( \frac{16.04\;g}{mol\;{\rm{CH_4}}} \right )=79\;g\;\rm{CH_4}$
6. $8.75\;mol\;{\rm{C_6H_6}}\left ( \frac{78.11\;g}{mol\;{\rm{C_6H_6}}} \right )=683\;g\;\rm{C_6H_6}$
7. $2.35\;mol\;{\rm{Na_2SO_4}}\left ( \frac{142.04\;g}{mol\;{\rm{Na_2SO_4}}} \right )=334\;g\;\rm{Na_2SO_4}$
8. $0.672\;mol\;{\rm{K_3PO_4}}\left ( \frac{212.27\;g}{mol\;{\rm{K_3PO_4}}} \right )=143\;g\;\rm{K_3PO_4}$
9. $0.95\;mol\;{\rm{Al(NO_3)_3}}\left ( \frac{213.01\;g}{mol\;{\rm{Al(NO_3)_3}}} \right )=2.0\times10^2\;g\;\rm{Al(NO_3)_3}$
10. $1.15\;mol\;{\rm{Mg_3(PO_4)_2}}\left ( \frac{262.87\;g}{mol\;{\rm{Mg_3(PO_4)_2}}} \right )=302\;g\;\rm{Mg_3(PO_4)_2}$
2.4: Electron Arrangements
Q2.4.1
a. 2, 8, 1
b. 2, 8
c. 2, 2
d. 2, 5
e. 2, 8, 6
f. 2, 8, 7
Q2.4.2
a. 1
b. 5
c. 7
d. 6
e. 1
f. 3
Q2.4.3
The octet rule predicts the stability of an atom based on having eight electrons in its electron shell.
2.5: Ion Formation
Q2.5.1
An ion is a charged species which results from the gain or loss of one ore more electrons.
Q2.5.2
Anions and cations are both charged species which results from the change in the number of electrons. Anions have a negative charge while cations have a positive charge.
Q2.5.3
1. $\rm{Li}^+$
2. $\rm{Na}^+$
3. $\rm{Ca}^{2+}$
4. $\rm{B}^{3+}$
5. $\rm{P}^{3-}$
6. $\rm{S}^{2-}$
7. $\rm{Cl}^{-}$
8. $\rm{Br}^-$
Q2.5.4
1. 3 protons, 4 neutrons, 2 electrons
2. 11 protons, 12 neutrons, 10 electrons
3. 20 protons, 20 neutrons, 18 electrons
4. 5 protons, 6 neutrons, 2 electrons
5. 15 protons, 16 neutrons, 18 electrons
6. 16 protons, 16 neutrons, 18 electrons
7. 17 protons, 18 neutrons, 18 electrons
8. 35 protons, 45 neutrons, 36 electrons
Q2.5.5 ( $\therefore$ = therefore)
1. ion with 3+ charge and 2 electrons $\therefore$ neutral atom had 5 electrons $\therefore$ atom has 5 protons $\therefore$ boron (B)
2. ion with 1$-$ charge and 18 electrons $\therefore$ neutral atom had 17 electrons $\therefore$ atom has 17 protons $\therefore$ chlorine (Cl)
3. ion with 1+ charge and 18 electrons $\therefore$ neutral atom had 19 electrons $\therefore$ atom has 19 protons $\therefore$ potassium (K)
4. ion with a 3$-$ charge and 10 electrons $\therefore$ neutral atom had 7 electrons $\therefore$ atom has 7 protons $\therefore$ nitrogen (N)
Q2.5.6
A polyatomic ion contains multiple atoms working together as a group and has an overall charge.
Q2.5.7
1. polyatomic
2. monatomic
3. polyatomic
4. monatomic
5. monatomic
6. polyatomic
2.6: Ionic Compounds
Q2.6.1
carbon
Q2.6.2
Answers will vary. Most metals are in the first two columns of the periodic table or in the transition metal block.
Q2.6.3
Answers will vary. Nonmetallic elements are located in the upper right corner of the periodic table (examples include nitrogen, oxygen, phosphorus, chlorine, bromine, etc)
Q2.6.4
Ionic compounds are composed of ions of metals and nonmetals. Ionic compounds can also include a polyatomic ion.
Q2.6.5
To form an ionic bond, electrons are transferred from the metal to the nonmetal.
Q2.6.6
Zero
Q2.6.7
What is the formula for the ionic compound formed from each of the following pairs?
1. K2S
2. AgCl
3. CaO
4. AlI3
5. Ba3N2
6. Na3P
7. LiF
8. Mg3N2
9. CaS
10. BeBr2
11. Zn3N2
12. SnI4
Q2.6.8
1. Na2CO3
2. NaClO3
3. NaClO2
4. Na3PO4
5. NaNO3
6. Na2SO4
7. Na2CrO4
8. Na2Cr2O7
Q2.6.9
1. MgCO3
2. Mg(ClO3)2
3. Mg(ClO2)2
4. Mg3(PO4)2
5. Mg(NO3)2
6. MgSO4
7. MgCrO4
8. MgCr2O7
Q2.6.10
Parentheses are used when there is more than one of a polyatomic ion in the formula of an ionic compound. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Chemistry_for_Allied_Health_(Soult)/02.E%3A_Compounds_%28Exercises%29.txt |
These are homework exercises to accompany Chapter 3 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
(click here for solutions)
Q3.1.1
What types of elements typically form binary molecular compounds?
Q3.1.2
Describe the similarities and differences between ionic and molecular compounds.
Q3.1.3
Give the prefix that would be used in the name a molecular compound for each of the following quantities of atoms.
a. 6
b. 3
c. 9
d. 5
e. 8
Q3.1.4
Write the formula for each compound.
1. diphosphorus pentoxide
2. dinitrogen monoxide
3. trisilicon tetranitride
4. dinitrogen pentoxide
5. tetraphosphorus decasulfide
6. disulfur hexafluoride
7. triboron dicarbide
8. tetraselenium tetranitride
(click here for solutions)
Q3.2.1
What elements are found in a hydrocarbon?
Q3.2.2
Describe an alkane.
Q3.2.3
How many carbon atoms are found in octane? propane? nonane?
Q3.2.4
What is the name for the alkane with six carbons? four carbons? two carbons?
Q3.2.5
What is the difference between an alkane and a cycloalkane?
3.1: Molecular Compounds
Q3.1.1
Binary molecular compounds are composed of two nonmetallic elements.
Q3.1.2
They both form as a result of bonding between atoms. Ionic compounds result from the transfer of electrons from one element to another while molecular compounds form bonds through the sharing of electrons.
Q3.1.3
a. hexa
b. tri
c. nona
d. penta
e. octa
Q3.1.4
1. P2O5
2. N2O
3. Si3N4
4. N2O5
5. P4S10
6. S2F6
7. B3C2
8. Se4N4
3.2: Straight-Chain Alkanes
Q3.2.1
carbon and hydrogen
Q3.2.2
An alkane contains only carbon and hydrogen atoms with the carbons connected by single bonds.
Q3.2.3
octane, 8; propane, 3; nonane, 9
Q3.2.4
6, hexane; 4, butane; 2, ethane
Q3.2.5
An alkane contains a chain of carbon atoms while a cycloalkane contains carbons in a ring structure.
04.E: Structure and Function (Exercises)
These are homework exercises to accompany Chapter 4 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
Questions
(Click here for solutions)
Q4.1.1
Describe a Lewis structure.
Q4.1.2
How many bonds are usually formed by each element?
1. sulfur
2. hydrogen
3. nitrogen
4. phosphorus
Q4.1.3
How many bonds does oxygen tend to form? What combinations of bonds (single, double, triple) can it have?
Q4.1.4
How many bonds does carbon tend to form? What combinations of bonds (single, double, triple) can it have?
Q4.1.5
Using Table 4.1.1, draw the Lewis structure for each molecule.
a. H2S
b. CCl4
c. PF3
d. C2H4
(click here for solutions)
Q4.2.1
Given one type of structure (Lewis, condensed, or skeletal), draw the other two.
a. CH3CH2CH3
b.
c.
d. CH3CHBrCH2CH2CHClCH3
e. CH2(OH)CH2CH2CH3
f.
g.
(click here for solutions)
Q4.3.1
How does the presence of a lone pair affect the shape of the molecule?
Q4.3.2
What is the "ideal" bond angle in a tetrahedral (AB4) molecule? How does that change in trigonal pyramidal (AB3E)? Bent (AB2E2)?
Q4.3.3
What is the ideal bond angle for trigonal planar (AB3) molecules? What would be the expected bond angle for a bent (AB2E) molecule?
Q4.3.4
Determine the electron geometry for each of the following molecules.
a.
b.
c.
d.
e.
f.
(click here for solutions)
Q4.4.1
Circle and label the functional groups in each molecule. There may be more than one functional group in a molecule.
a.
b.
c.
d.
e.
f.
Q4.4.2
Draw a molecule which contains an ester, an alcohol, and an aldehyde.
Q4.4.3
Label each alcohol as primary, secondary, or tertiary.
a.
b.
c.
d.
Q4.4.4
Label each amine as primary, secondary, or tertiary.
a.
b.
c.
d.
e.
Answers
4.1: Lewis Electron Dot Structures
Q4.1.1
A Lewis structure shows the connections between atoms (single, double, or triple bonds) as well as any non-bonding electrons.
Q4.1.2
1. 2
2. 1
3. 3
4. 3
Q4.1.3
2 bonds total; 2 single bonds or 1 double bond
Q4.1.4
4 bonds total; 4 single or 2 single and 1 double or 2 double or 1 single and 1 triple
Q4.1.5
a.
b.
c.
d.
4.2: Representing Structures
Q4.2.1
Given one type of structure (Lewis, condensed, or skeletal), draw the other two.
Lewis Structure Condensed Structure Skeletal Structure
a. CH3CH2CH3
b. CH3CH(OH)C(CH3)(OH)CH2CH3
c. CH3C(O)CH3
d. CH3CHBrCH2CH2CHClCH3
e. CH2(OH)CH2CH2CH3
f. CH3CH(CH3)CH2CH2CH3
g. CH3CCCH3
4.3: Molecular Shapes
Q4.3.1
A lone pair causes the compression of other bond angles in the molecule.
Q4.3.2
AB4: 109.5°
AB3E: 107° (less than AB4)
AB2E2: 104.5° (less than AB3E)
Q4.3.3
AB3: 120°
AB2E: less than 120°
Q4.3.4
1. trigonal planar
2. linear
3. tetrahedral
4. linear
5. trigonal planar
6. tetrahedral
4.4: Functional Groups
Q4.4.1
a.
b.
c.
d.
e.
f.
Q4.4.2
Answers will vary. One example is shown.
Q4.4.3
1. secondary
2. primary
3. secondary
4. tertiary
Q4.4.4
1. primary
2. primary
3. secondary
4. tertiary
5. secondary | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Chemistry_for_Allied_Health_(Soult)/03.E%3A_Structures_of_Compounds_%28Exercises%29.txt |
These are homework exercises to accompany Chapter 5 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
Questions
(click here for solutions)
Q5.1.1
Define isomer.
Q5.1.2
Circle the chiral carbons in each structure.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
Q5.1.3
Determine whether each molecule can have a geometric isomer.
Q5.1.4
For each pair, determine if they are isomers. If they are isomers, identify the type.
a.
b.
c.
d.
e.
f.
g.
h.
Q5.1.5
Draw two isomers with the formula C2H6O
(click here for solutions)
Q5.2.1
A monosaccharide has 4 carbon atoms. What is its chemical formula?
Q5.2.2
What are the differences between monosaccharides, disaccharides, oligosaccharides, and polysaccharides?
Q5.2.3
Compare and contrast simple and complex carbohydrates.
Q5.2.4
Draw each of the following structures.
1. aldoheptose
2. ketopentose
3. aldotriose
4. ketotetrose
Q5.2.5
Draw the Fischer projections of two aldohexoses that are different from the four given in Figure 5.2.3.
Q5.2.6
Draw the enantiomers of D-glucose, D-allose, D-mannose, and D-galactose. (See Figure 5.2.3).
Q5.2.7
Determine if each of the following pairs are diastereomers? Epimers? (Look up the structures in the chapter or online.)
1. D-glucose and D-allose
2. D-glucose and D-galactose
3. D-allose and D-mannose
4. D-allose and D-galactose
Q5.2.8
For each of the given Haworth structures, identify the sugar as being in the α or β form.
a.
b.
c.
d.
e.
Q5.2.9
Refer to Figure 5.2.7 when answering these questions.
1. Which blood type is the universal donor? Why?
2. Which blood type is the universal acceptor? Why?
3. Type A blood can be donated to recipients with which types of blood?
4. Type B recipients can accept which types of blood?
Q5.3.1
Define electronegativity.
Q5.3.2
Describe the periodic trends for electronegativity values.
Q5.3.3
Distinguish between nonpolar and polar covalent bonds.
Q5.3.4
Describe the bond between each pair of elements as ionic, polar covalent, or nonpolar covalent. Refer to Table 5.3.1 when answering this question. For the exam, if you need specific values, a table of electronegativity values will be provided. Note that many questions can be answered without the table by knowing the periodic trends.
1. N and O
2. C and P
3. Si and Cl
4. Al and F
5. Al and I
6. P and S
7. C and N
8. B and Cl
9. Be and Br
10. Si and P
Q5.3.5
Place the following bonds in order from least to most polar. Refer to Table 5.3.1 when answering this question.
1. Fe-N
2. H-Cl
3. Ca-O
4. C-S
Q5.3.6
Place the following bonds in order from least to most polar. Use periodic trends to determine the correct order without looking at electronegativity values.
1. PCl
2. SCl
3. PBr
4. CBr
Q5.3.7
Label each of the molecules as nonpolar or polar covalent.
1. CO2
2. CCl4
3. NH3
Q5.3.8
Describe the types of molecules that have the following types of intermolecular forces.
1. London dispersion forces
2. dipole-dipole forces
3. hydrogen bonding
Q5.3.9
Why are the intermolecular forces in H2O and H2S so different from one another?
Q5.3.10
What type(s) of intermolecular forces are present in each of the molecules in the question 5.3.7?
Q5.3.11
What is the relationship between the strength of intermolecular forces in a molecule and its boiling point?
Q5.3.12
Rank the following in order of increasing (smallest to greatest) boiling point.
1. N2
2. CH3OH
3. PH3
Q5.4.1
Define chromatography.
Q5.4.2
List two types of chromatography.
Q5.4.3
Distinguish between the stationary and mobile phases.
5.1: Isomers
Q5.1.1
Isomers are molecules with the same chemical formula but a different structure or arrangement of atoms.
Q5.1.2
Circle the chiral carbons in each structure.
a.
b.
c. No chiral carbons
d. No chiral carbons
e.
f.
g. No chiral carbons.
h.
i.
j.
Q5.1.3
1. No, because of two H atoms on left side of double bond.
2. Yes, because the chlorine atoms are shown trans but could be drawn cis.
3. Yes, because the two groups could be on the same side of the ring or on opposite sides.
4. No, because you need a double bond or a ring to have a geometric isomer.
5. No, because there is only one non-hydrogen group on the ring.
6. Yes, because each carbon in the double bond has two different groups so it could be described as cis or trans.
Q5.1.4
For each pair, determine if they are isomers. If they are isomers, identify the type.
1. structural
2. conformational
3. not isomers
4. structural
5. structural
6. not isomers
7. enantiomers
8. same molecule
Q5.1.5
5.2: Carbohydrate Structures
Q5.2.1
C4H8O4
Q5.2.2
Monosaccharides are a single carbohydrate molecule. Disaccharides have two carbohydrate molecules bonded together. Polysaccharides have three or more carbohydrate molecules bonded together.
Q5.2.3
Simple carbohydrates are monosaccharides or disaccharides and are easily broken down in the body. Complex carbohydrates are polysaccharides such as starches and fiber and take longer to break down in the body.
Q5.2.4
Answers will vary due to orientation of H and OH at each chiral carbon.
a.
b.
c.
d.
Q5.2.5
Any two of the structures shown here in red. The structures in black are from Figure 5.2.3.
Q5.2.6
L-glucose L-allose L-mannose L-galactose
Q5.2.7
1. Epimers because the orientation differs at a single chiral carbon.
2. Diastereomers because the orientation differs at multiple (but not all) chiral carbons.
3. Diastereomers because the orientation differs at multiple (but not all) chiral carbons.
4. Diastereomers because the orientation differs at multiple (but not all) chiral carbons.
Q5.2.8
1. beta
2. alpha
3. beta
4. alpha
5. beta
Q5.2.9
1. O because it has the fewest types of carbohydrates.
2. AB because it has all carbohydrates found on red blood cells.
3. A can give to A or AB.
4. B can receive B or O.
5.3: Polarity and Intermolecular Forces
Q5.3.1
Electronegativity is an atom's attraction to an electron in a bond.
Q5.3.2
Electronegativity increases from left to right and from bottom to top with a maximum at fluorine.
Q5.3.3
Nonpolar covalent bonds have even sharing of electrons between atoms while polar electrons share electrons unevenly.
Q5.3.4
1. polar
2. nonpolar
3. polar
4. ionic
5. polar
6. nonpolar
7. polar
8. polar
9. polar
10. nonpolar
Q5.3.5
C-S < H-Cl < Fe-N < Ca-O
Calculate the difference in electronegativity for each bond. The smaller the difference, the more nonpolar; the greater the difference, the more polar. If a bond has a large enough difference, it is so polar that it is considered ionic.
Q5.3.6
S-Cl < P-Cl < P-Br < C-Br
• S and Cl are closest together so they will have the smallest difference in electronegativity and be the least polar.
• S-Cl and P-Cl can be compared since they share a common element. P is farther left than S, so the electronegativity of P must be less than that of S. Therefore, the difference between P and Cl must be greater than the difference between S and Cl since we are comparing them both to the same element (Cl).
• Compare P-Cl and P-Br. Since Br is further down the periodic table, it has a lower electronegativity value than Cl. Therefore the difference between P and Cl is less than the difference between P and Br. P-Br is more polar than P-Cl.
• Compare P-Br and C-Br. C is further from Br than P is so the difference in electronegativity is greater for C-Br than P-Br making C-Br the more polar bond.
Q5.3.7
1. nonpolar
2. nonpolar
3. polar
Q5.3.8
1. all molecules
2. polar molecules
3. molecules with an H bonded to F, O, or N which is attracted to the F, O, or N on another molecule
Q5.3.9
H2O has hydrogen bonding while H2S does not.
Q5.3.10
1. dispersion
2. dispersion
3. dispersion, dipole-dipole, and hydrogen bonding
Q5.3.11
The stronger the intermolecular forces, the higher the boiling point.
Q5.3.12
N2 < PH3 < CH3OH
All three of these molecules are comparable in size so the differences in dispersion forces are small. N2 is nonpolar and has only dispersion forces. PH3 is polar, so it has dispersion and dipole-dipole forces but no hydrogen bonding. CH3OH has dispersion, dipole-dipole, and hydrogen bonding forces.
5.4: Chromatography
Q5.4.1
Chromatogaphy is used to separate a mixture into its components when they travel at different rates in the mobile phase.
Q5.4.2
Paper, thin-layer, liquid, and gas are the most well-known types of chromatography
Q5.4.3
The mobile phase is a fluid which moves through the stationary phase. The stationary phase holds the sample until the mobile phase moves it along the stationary phase. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Chemistry_for_Allied_Health_(Soult)/05.E%3A_Properties_of_Compounds_%28Exercises%29.txt |
These are homework exercises to accompany Chapter 6 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions.
Questions
(click here for solutions)
Q6.1.1
Define potential energy and chemical potential energy.
Q6.1.2
What is one potential use for substances that have a large amount of chemical potential energy?
Q6.1.3
Describe what happens when two objects that have different temperatures come into contact with one another.
Q6.1.4
Distinguish between system and surroundings.
Q6.1.5
Distinguish between endothermic and exothermic.
Q6.1.6
Two different reactions are performed in two identical test tubes. In reaction A, the test tube becomes very warm as the reaction occurs. In reaction B, the test tube becomes cold. Which reaction is endothermic and which is exothermic? Explain.
Q6.1.7
What is the sign of q for an endothermic process? For an exothermic process?
Q6.1.8
Classify the following as endothermic or exothermic processes.
1. Boiling water
2. Sweating
3. Burning paper
4. Water freezing
Q6.1.9
Convert each value to the indicated units.
1. 150. kcal to Cal
2. 355 J to cal
3. 200. Cal to J
4. 225 kcal to cal
5. 3450. cal to kcal
6. 450. Cal to kJ
7. 175 kJ to cal
Q6.1.10
Equal amounts of heat are applied to 10.0 g samples of iron and aluminum, both originally at 25°C. Which one will be at the higher temperature?
Q6.1.11
Which sample will require more heat to increase the temperature by 10°C?
1. 25.0 g copper
2. 25.0 g lead
Q6.1.12
How much energy is required to heat 50.0 g of silver from 30°C to 50°C?
Q6.1.13
What is the final temperature when 125 J is applied to 20.0 g of lead, initially at 15°C?
Q6.1.14
How much energy is required to raise the temperature of 13.7 g of aluminum from 25.2°C to 61.9°C?
Q6.1.15
A 274 g sample of air is heated with 2250 J of heat and its temperature rises by 8.11°C. What is the specific heat of air at these conditions?
Q6.1.16
98.3 J of heat is supplied to 12.28 g of a substance, and its temperature rises by 5.42°C. What is the specific heat of the substance?
Q6.1.17
A quantity of ethanol is cooled from 47.9°C to 12.3°C and releases 3.12 kJ of heat. What is the mass of the ethanol sample?
Answers
6.1: Heat Flow
Q6.1.1
Potential energy is usually described as the energy of position. Chemical potential energy is energy stored within the chemical bonds of a substance.
Q6.1.2
Answers will vary. The most common example in every day life is the burning of fossil fuels to generate electricity or to run a vehicle.
Q6.1.3
The temperature of the hot object decreases and the temperature of the cold object increases as heat is transferred from the hot object to the cold object. The change in temperature of each depends on the identity and properties of each substance.
Q6.1.4
The system is the specific portion of matter being observed in an experiment and is designated by the experimenter. The surroundings is everything that is not the system.
Q6.1.5
Endothermic processes result in the gain of heat to the system while exothermic processes are associated with the loss of heat from the system.
Q6.1.6
Reaction A is exothermic because heat is leaving the system making the test tube feel hot. Reaction B is endothermic because heat is being absorbed by the system making the test tube feel cold.
Q6.1.7
q is positive for endothermic processes and q is negative for exothermic processes.
Q6.1.8
Classify the following as endothermic or exothermic processes.
1. Endothermic because heat is being added to the water to get it from the liquid state to the gas state.
2. Endothermic because energy is consumed to evaporate the moisture on your skin which lowers your temperature.
3. Exothermic because burning (also known as combustion) releases heat.
4. Exothermic because energy is exiting the system in order to go from liquid to solid. Another way to look at it is to consider the opposite process of melting. Energy is consumed (endothermic) to melt ice (solid to liquid) so the opposite process (liquid to solid) must be exothermic.
Q6.1.9
Convert each value to the indicated units.
1. $150\; kcal\left ( \frac{1\;Cal}{1\;kcal} \right )=150\;Cal$
2. $355\;J\left ( \frac{1\;cal}{4.184\;J} \right )=84.8\;cal$
3. $200.\;Cal\left ( \frac{1000\;cal}{1\;Cal} \right )\left ( \frac{4.184\;J}{1\;cal} \right )=8.37\times 10^{5}\;J$
4. $225\;kcal\left ( \frac{1000\;cal}{1\;kcal} \right )=2.25\times 10^{5}\;cal$
5. $3450.\;cal\left ( \frac{1 kcal\;kJ}{1000\;cal} \right )=3.450\;kcal$
6. $450.\;Cal\left ( \frac{1000\;cal}{1\;Cal} \right )\left ( \frac{4.184\;J}{1\;cal} \right )\left ( \frac{1\;kJ}{1000\;J} \right )=1.88\times 10^{3}\;kJ$ or $450.\;Cal\left ( \frac{4.184\;kJ}{1\;Cal} \right )=1.88\times 10^{3}\;kJ$
7. $175\;kJ\left ( \frac{1000\;J}{1\;kJ} \right )\left ( \frac{1\;cal}{4.184\;J} \right )=4.18\times 10^{4}\;cal$
Q6.1.10
Iron has a specific heat capacity of 0.449 $\text{J/g} \cdot ^\text{o} \text{C}$ which means it takes 0.449 J of energy to raise 1 gram of iron by 1$^\text{o} \text{C}$. Aluminum has a specific heat capacity of 0.897 $\text{J/g} \cdot ^\text{o} \text{C}$ which means it takes 0.897 J of energy to raise 1 gram of aluminum by 1$^\text{o} \text{C}$. When equal amounts of heat are applied, the temperature of the iron will increase more because it takes less energy (heat) to raise its temperature so iron will be at a higher temperature since they both start at 25$^\text{o} \text{C}$.
Q6.1.11
Both samples are the same mass so a comparison of the specific heat must be compared. Copper has a specific heat of 0.385 $\text{J/g} \cdot ^\text{o} \text{C}$ which means it takes 0.385 J of energy to raise 1 gram of copper by 1^\text{o} \text{C}\). Lead has a specific heat of 0.129 $\text{J/g} \cdot ^\text{o} \text{C}$ which means it takes 0.129 J of energy to raise 1 gram of copper by 1^\text{o} \text{C}\). More energy is needed to raise the temperature of copper so more heat will be needed to increase the temperature of copper by 10 ^\text{o} \text{C}\).
Q6.1.12
$\begin{array}{c} q=m\cdot C_{p}\cdot \Delta T\ q=50.0\;g\cdot 0.233\; \frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (50^\text{o} \text{C}-30^\text{o} \text{C} \right )\ q=50.0\;g\cdot 0.233\;\frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (20^\text{o} \text{C} \right )\ q=233\; J \end{array}$
Q6.1.13
$\begin{array}{c} q=m\cdot C_{p}\cdot \Delta T\ 125\;J=20.0\;g\cdot 0.129\;\frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (T_{f}-15^\text{o} \text{C} \right )\ 48.4^\text{o} \text{C}=T_{f}-15^\text{o} \text{C}\ T_{f}=63^\text{o} \text{C} \end{array}$
Q6.1.14
$\begin{array}{c} q=m\cdot C_{p}\cdot \Delta T\ q=13.7\;g\cdot 0.897\;\frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (61.9^\text{o} \text{C}-25.2^\text{o} \text{C} \right )\ q=13.7\;g\cdot 0.897\;\frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (36.7^\text{o} \text{C} \right )\ q=451\; J \end{array}$
Q6.1.15
$\begin{array}{c} q=m\cdot C_{p}\cdot \Delta T\ 2250\;J=274\;g\cdot C_{p}\ \cdot 8.11^\text{o} \text{C}\ C_{p}=1.01 \frac{J}{g\cdot^\text{o} \text{C} } \end{array}$
Q6.1.16
$\begin{array}{c} q=m\cdot C_{p}\cdot \Delta T\ 98.3\;J=12.28\;g\cdot C_{p}\ \cdot 5.42^\text{o} \text{C}\ C_{p}=1.48 \frac{J}{g\cdot^\text{o} \text{C} } \end{array}$
Q6.1.17
$\begin{array}{c} q=m\cdot C_{p}\cdot \Delta T\ -3.12\;kJ=m\cdot 2.44\frac{J}{g\cdot^\text{o} \text{C} }\left (12.3^\text{o} \text{C}-47.9^\text{o} \text{C} \right )\ -3.12\times10^{3}\;J=m\cdot 2.44\frac{J}{g\cdot^\text{o} \text{C} }\left (-35.6^\text{o} \text{C} \right )\ m=35.9\;g \end{array}$ | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Chemistry_for_Allied_Health_(Soult)/06.E%3A_States_of_Matter_%28Exercises%29.txt |
These are homework exercises to accompany Chapter 7 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Answers are available below the solutions.
Questions
(click here for solutions)
Q7.1.1
In which state(s) of matter are atoms
1. closest together?
2. farthest apart?
3. filling the container?
4. fixed in position relative to one another?
5. moving past one another?
6. taking on the shape of the container?
Q7.1.2
Which of the following statements are true? Correct any false statements.
1. All substances exist as a liquid at room temperature and pressure.
2. Water changes from liquid to solid at 32°C.
3. All substances can exist as solids, liquids, or gases.
Q7.1.3
Which state of matter is most compressible?
Q7.1.4
Use online resources to find the boiling point of ethanol and dimethyl ether. Which one is higher? Why?
Q7.1.5
Describe the relationship between boiling point and altitude.
Q7.1.6
Where is the boiling point of ethanol the highest? The lowest?
1. Lexington, KY
2. New Orleans, LA
3. Salt Lake City, UT
4. Same at all locations.
(click here for solutions)
7.2.1
What phase change is described by each term? Is the process endothermic or exothermic?
1. sublimation
2. vaporization
3. fusion
4. deposition
Q7.2.2
List two phase changes that consume energy.
Q7.2.3
List two phase changes that release energy.
Refer to Table 7.2.1 for enthalpy values.
Q7.2.4
What is the enthalpy of fusion, vaporization, freezing, and condensation for each substance?
1. oxygen, O2
2. ethane, C2H6
3. carbon tetrachloride, CCl4
4. lead, Pb
Q7.2.5
How much energy is needed to vaporize 1.4 moles of ammonia (NH3)?
Q7.2.6
How much energy is needed to melt 3.0 moles of ice (H2O)?
Q7.2.7
What is the change in energy when 2.0 moles of ethanol is condensed?
Q7.2.8
What is the change in energy when 2.2 moles of oxygen is condensed?
Q7.2.9
Using the molar mass of water, convert the molar heats of fusion and vaporization for water from units of kJ/mol to kJ/g.
Q7.2.10
Calculate the quantity of heat that is absorbed or released during each process.
1. 655 g of water vapor condenses at 100°C
2. 8.20 kg of water is frozen
3. 40.0 mL of ethanol is vaporized. The density of ethanol is 0.789 g/mL.
4. 25.0 mL of ethanol condenses. The density of ethanol is 0.789 g/mL.
Q7.2.11
Various systems are each supplied with 9.25 kJ of heat. Calculate the mass of each substance that will undergo the indicated process with this input of heat.
1. melt ice at 0°C
2. vaporize water at 100°C
3. vaporize ethanol at 351 K
Q7.2.12
15.5 kJ of energy is released from each change. What mass of substance is involved?
1. condensation of NH3
2. freezing water
3. condensation of ethanol
Q7.2.13
What is $\Delta H_{vap}$ for benzene (C6H6) if 7.88 kJ of energy is needed to vaporize 20.0 g of benzene?
(click here for solutions)
7.3.1
How are gases different from liquids and solids in terms of the distance between the particles?
Q7.3.2
Under what conditions do gases exhibit the most ideal behavior?
Q7.3.3
Which of the following are behaviors of a gas that can be explained by the kinetic-molecular theory?
a. Gases are compressible.
b. Gases exert pressure.
c. All particles of a gas sample move at the same speed.
d. Gas particles can exchange kinetic energy when they collide.
e. Gas particles move in a curved-line path.
Q7.3.4
What is an elastic collision?
Q7.3.5
Perform the indicated conversions for the following pressure measurements.
1. 1.721 atm to mmHg
2. 559 torr to kPa
3. 91.1 kPa to atm
4. 2320 mmHg to atm
Q7.3.6
1. A typical barometric pressure in Redding, California, is about 755 mmHg. Calculate this pressure in atm and kPa.
2. A typical barometric pressure in Denver, Colorado, is 615 mmHg. What is this pressure in atmospheres and kilopascals?
Q7.3.7
How does the average kinetic energy of an air sample near a campfire compare to the average kinetic energy of a sample of air that is far away from it?
(click here for solutions)
7.4.1
Complete the missing temperature values in the table.
oC oF K
25
99
32
0
300
65
Q7.4.2
What units must temperature be in for gas law calculations?
Q7.4.3
Based on R = 0.08206 $\frac{L\dot atm}{mol \dot K}$, what units should be used in ideal gas law calculations?
Q7.4.4
A 1.00 mol sample of gas is at 300 K and 4.11 atm. What is the volume of the gas under these conditions?
Q7.4.5
What is the pressure in a 2.5 L container with 2.5 moles of gas at 293 K?
Q7.4.6
How many moles of carbon monoxide, CO, are in an 11.2-L sample at 744 torr at 55 °C?
Q7.4.7
A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25 °C at ground level. What is the volume of the balloon under these conditions?
Q7.4.8
A balloon inflated with three breaths of air has a volume of 1.7 L. At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?
Q7.4.9
The volume of an automobile air bag was 66.8 L when inflated at 25 °C with 77.8 g of nitrogen (N2) gas. What was the pressure in the bag in kPa?
Q7.4.10
How many moles of gaseous boron trifluoride, BF3, are contained in a 4.3410-L bulb at 788.0 K if the pressure is 1.220 atm? How many grams of BF3?
Q7.4.11
How is the combined gas law is simplified for each set of conditions?
1. constant V and n
2. constant n
3. constant P and V
4. constant T and n
5. constant V and T
6. constant P and n
7. constant T
Q7.4.12
A nitrogen sample has a pressure of 0.56 atm with a volume of 2.0 L. What is the final pressure if the volume is compressed to a volume of 0.75 L? Assume constant moles and temperature.
Q7.4.13
A 2.50-L volume of hydrogen measured at –196 °C is warmed to 100 °C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.
Q7.4.14
A high altitude balloon is filled with 1.41 × 104 L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is –48 °C and the pressure is 63.1 torr?
Q7.4.15
A cylinder of medical oxygen has a volume of 35.4 L, and contains O2 at a pressure of 151 atm and a temperature of 25 °C. What volume of O2 does this correspond to at normal body conditions, that is, 1 atm and 37 °C?
Q7.4.16
A 0.50 L container of helium expands to 1.50 L. By what factor does the pressure change? Assume constant moles and temperature.
Q7.4.17
A sample of oxygen gas has an initial pressure and volume of 1.0 L and 1.0 atm. What is the final pressure if the volume is compressed to 0.50 L? Assume constant moles and temperature.
Q7.4.18
A sample of gas has a volume of 2.75 L at a temperature of 100 K. What is the volume of the gas when the temperature increases to 200 K? Assume constant pressure and moles.
Q7.4.19
What is the final volume of a gas that was originally at 0.75 L at 25°C and a final temperature of 50°C? Assume constant pressure and moles.
Q7.4.20
A sample of nitrogen is at 45°C with a volume of 2.5 L. What is the final temperature in °C if the volume is compressed to 1.4 L? Assume constant pressure and moles.
Q7.4.21
A 2.00 mole sample of gas is in a 3.50 L container. What happens to the volume when an additional 0.75 moles of gas is added? Assume pressure and temperature are constant.
Q7.4.22
A 1.85 mole sample of helium has a volume of 2.00 L. Additional helium is added at constant pressure and temperature until the volume is 3.25 L. What is the total moles of helium present in the sample? What mass of helium was added?
Q7.4.23
If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?
Q7.4.24
If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?
(click here for solutions)
7.5.1
Describe the solution, solvent, and solute.
Q7.5.2
How do solutions differ from compounds? Are solutions heterogeneous or homogeneous mixtures?
Q7.5.3
When KNO3 is dissolved in water, the resulting solution is significantly colder than the water was originally. Is the dissolution of KNO3 an endothermic or an exothermic process?
Q7.5.4
What are the differences between strong, weak and non-electrolytes.
Q7.5.5
Write dissociation equations for the following strong electrolytes.
1. NaCl(s)
2. CoCl3(s)
3. Li2S(s)
4. MgBr2(s)
5. CaF2(s)
Q7.5.6
Based on the given information, identify each as a strong, weak, or non-electrolyte.
1. C6H12O6(s) $\rightarrow$ C6H12O6(aq)
2. NaCl is added to water and the conductivity increases dramatically.
3. 1.5 moles of HCl are added to a container of water. The resulting solution has 1.5 moles of H+ ions and 1.5 moles of Cl$^{\-}$ ions.
4. Acetic acid (CH3COOH) partially dissociates in water.
5. An HCN solution contains 0.50 moles of HCN molecules and 0.05 moles of H+ ions and 0.05 moles of Cl$^{\-}$ ions.
6. Acetone is added to water and the conductivity does not change.
(click here for solutions)
7.6.1
How can you distinguish between a suspension and a solution?
Q7.6.2
How big are the particles in a colloid compared to those of a suspension and a solution?
Q7.6.3
What is the Tyndall effect? Why don’t solutions demonstrate the Tyndall effect?
Q7.6.4
Explain the difference between the dispersed phase and the dispersing medium of a colloid.
Q7.6.5
Identify each of the following descriptions or examples as being representative of a solution, suspension, or colloid. More than one answer may apply.
a. dispersed particles can be filtered out
b. heterogeneous
c. particles are not visible to the unaided eye
d. paint
e. lemonade with no pulp
f. particle size larger than 1 nm
g. milk
h. particles do not settle upon standing
i. fog
(click here for solutions)
7.7.1
Describe the difference between saturated and unsaturated solutions.
Q7.7.2
What are two things that you could do to change an unsaturated solution into a saturated solution?
Q7.7.3
A given solution is clear and colorless. A single crystal of solute is added to the solution. Describe what happens in each of the following situations.
1. The original solution was saturated.
2. The original solution was unsaturated.
Q7.7.4
List the original states (solid, liquid, or gas) of the solute and solvent that are combined to make each of the following solutions.
1. an alloy
2. salt water
3. carbonated water
4. oil in gasoline
Q7.7.5
Answer the following using the solubility curve diagram.
1. How many grams of NH4Cl are required to make a saturated solution in 100 g of water at 70°C?
2. How many grams of NH4Cl could be dissolved in 200 g of water at 70°C?
3. At what temperature is a solution of 50 grams of KNO3 dissolved in 100 grams of water a saturated solution?
4. Which two substances in the above graph have the same solubility at 85°C?
5. How many grams of NaNO3 can be dissolved in 100 grams of water to make a saturated solution at 25°C?
6. How much KI can be dissolved in 5 grams of water at 20°C to make a saturated solution?
Q7.7.6
An exactly saturated solution of KClO3 is prepared at 90°C using 100 grams of water. If the solution is cooled to 20°C, how many grams of KClO3 will recrystallize (i.e. come out of solution)?
Q7.7.7
Indicate whether the following solutions are unsaturated or saturated.
1. 22 grams of KClO3 is dissolved in 100 g of water at 50°C.
2. 60 grams of KNO3 is dissolved in 100 g of water at 50°C.
3. 50 grams of NaCl is dissolved in 100 g of water at 50°C.
Q.7.7.8
Under which set of conditions is the solubility of a gas in a liquid the greatest?
1. low temperature and low pressure
2. low temperature and high pressure
3. high temperature and low pressure
4. high temperature and high pressure
Answers
7.1: States of Matter
Q7.1.1
1. solid
2. gas
3. gas
4. solid
5. liquid and gas
6. liquid and gas
Q7.1.2
Which of the following statements are true? Correct any false statements.
1. All substances exist as a liquid at room temperature and pressure. at some temperature and pressure.
2. Water changes from liquid to solid at 32°C °F.
3. True (although some states are rarely seen for some substances).
Q7.1.3
gas
Q7.1.4
ethanol 78°C; dimethyl ether $-$24°C
Ethanol has stronger intermolecular forces due to having hydrogen bonding which is not seen in dimethyl ether. The stronger the intermolecular forces, the higher the boiling point.
Q7.1.5
As the altitude increases, the boiling point decreases
Q7.1.6
1. Lexington, KY (altitude = 978 feet)
2. New Orleans, LA (altitude = 2 feet) - HIGHEST
3. Salt Lake City, UT (altitude = 4226 feet) - LOWEST
7.2: Heat and Changes of State
Q7.2.1
1. solid to gas; endothermic
2. liquid to gas; endothermic
3. solid to liquid; endothermic
4. gas to solid; exothermic
Q7.2.2
Any two of fusion, vaporization, or sublimation.
Q7.2.3
Any two of freezing, condensation, deposition.
Q7.2.4
Substance $\Delta H_{fus}$ (kJ/mol) $\Delta H_{vap}$ (kJ/mol) $\Delta H_{freezing}$ (kJ/mol) $\Delta H_{condensation}$ (kJ/mol)
oxygen, O2 0.44 6.82 $-$0.44 $-$6.82
ethane, C2H6 2.85 14.72 $-$2.85 $-$14.72
carbon tetrachloride, CCl4 2.67 30.0 $-$2.67 $-$30.0
lead, Pb 4.77 178 $-$4.77 $-$178
Q7.2.5
$1.4\;mol \; \text{NH}_{3}\left ( \frac{23.35 \;kJ}{mol} \right )=33 \; kJ$
Q7.2.6
$3.0\;mol \; \text{H}_2\text{O}\left ( \frac{6.01 \;kJ}{mol} \right )=18 \; kJ$
Q7.2.7
$2.0\;mol \; \text{CH}_3\text{CH}_2\text{OH}\left ( \frac{-38.56\; kJ}{mol} \right )=-77 \; kJ$
Q7.2.8
$2.2\;mol \; \text{O}_2\left ( \frac{-6.82\; kJ}{mol} \right )=-15 \; kJ$
Q7.2.9
$\frac{6.01\; kJ}{mol} \left ( \frac{1\;mol}{18.02\;g} \right )=\frac{0.334\;kJ}{g}$
$\frac{40.7\; kJ}{mol} \left ( \frac{1\;mol}{18.02\;g} \right )=\frac{2.26\;kJ}{g}$
Q7.2.10
1. $655\;g\; \text{H}_2\text{O}\left ( \frac{1\;mol}{18.02\;g} \right ) \left (\frac{-40.7\; kJ}{mol} \right )=-1.48\times10^3\;kJ$
2. $8.20\;kg\; \text{H}_2\text{O}\left ( \frac{1000\;g}{1\;kg} \right ) \left ( \frac{1\;mol}{18.02\;g} \right ) \left (\frac{-6.01\; kJ}{mol} \right )=-2.73\times10^3\;kJ$
3. $40.0\;mL\; \text{CH}_3\text{CH}_2\text{OH}\left ( \frac{0.789\;g}{1\;mL} \right ) \left ( \frac{1\;mol}{46.07\;g} \right ) \left (\frac{38.56\; kJ}{mol} \right )=26.4\;kJ$
4. $25.0\;mL\; \text{CH}_3\text{CH}_2\text{OH}\left ( \frac{0.789\;g}{1\;mL} \right ) \left ( \frac{1\;mol}{46.07\;g} \right ) \left (\frac{-38.56\; kJ}{mol} \right )=-16.5\;kJ$
Q7.2.11
1. $9.25\;kJ \left ( \frac{mol}{6.01\;kJ} \right )\left ( \frac{18.02\;g}{mol} \right )=27.7\;g\;\text{H}_2\text{O}$
2. $9.25\;kJ \left ( \frac{mol}{40.7\;kJ} \right )\left ( \frac{18.02\;g}{mol} \right )=4.10\;g\;\text{H}_2\text{O}$
3. $9.25\;kJ \left ( \frac{mol}{38.56\;kJ} \right )\left ( \frac{46.07\;g}{mol} \right )=11.1\;g\;\text{CH}_3\text{CH}_2\text{OH}$
Q7.2.12
1. $-15.5\;kJ \left ( \frac{mol}{-23.35\;kJ} \right )\left ( \frac{17.03\;g}{mol} \right )=11.3\;g\;\text{NH}_3$
2. $-15.5\;kJ \left ( \frac{mol}{-6.01\;kJ} \right )\left ( \frac{18.02\;g}{mol} \right )=46.5\;g\;\text{H}_2\text{O}$
3. $-15.5\;kJ \left ( \frac{mol}{-38.56\;kJ} \right )\left ( \frac{46.07\;g}{mol} \right )=18.5\;g\;\text{CH}_3\text{CH}_2\text{OH}$
Q7.2.13
Find the moles of benzene.
$20.0\;g\;\text{C}_6\text{H}_6\left ( \frac{1\;mol}{78.11\;g} \right )=0.256\;mol\;\text{C}_6\text{H}_6$
Combine the energy with the moles to calculate the enthalpy of vaporization.
$\Delta H_{vap}=\frac{7.88\;kJ}{0.256\;mol}=\frac{30.8\;kJ}{mol}$
7.3: Kinetic-Molecular Theory
Q7.3.1
Gas particles are much farther from one another than liquid or solid particles.
Q7.3.2
Gases have the most ideal behavior at high temperatures (molecules moving more quickly than at low temperatures so less time to interact) and at low pressure (molecules are farther apart from one another than at high pressure).
Q7.3.3
1. Molecules are very far apart from one another and are compressible.
2. Gases are in constant random motion so they collide with the walls of the container.
3. False. Molecules of the same substance are moving at a range of speeds.
4. Collisions are elastic. Energy is exchanged but not lost when two particles coll
5. False. Particles move in a straight line.
Q7.3.4
A collision in which no energy is lost.
Q7.3.5
1. $1.721\;atm\left ( \frac{760\;mmHg}{1\;atm} \right )=1308\;mmHg$
2. $559\;torr\left ( \frac{101.3\;kPa}{760\;torr} \right )=74.5\;kPa$
3. $91.1\;kPa\left ( \frac{1\;atm}{101.3\;kPa} \right )=0.899\;atm$
4. $2320\;mmHg\left ( \frac{1\;atm}{760\;mmHg} \right )=3.05\;atm$
Q7.3.6
1. $755\;mmHg\left ( \frac{1\;atm}{760\;mmHg} \right )=0.993\;atm$ $755\;mmHg\left ( \frac{101.3\;kpa}{760\;mmHg} \right )=101\;kPa$
2. $615\;mmHg\left ( \frac{1\;atm}{760\;mmHg} \right )=0.809\;atm$ $615\;mmHg\left ( \frac{101.3\;kpa}{760\;mmHg} \right )=82.0\;kPa$
Q7.3.7
Closer to the fire, it is warmer and the kinetic energy of the particles (and therefore the average speed) will be greater.
7.4:The Ideal Gas Equation
Q7.4.1
oC oF K
25 77 298
37 99 310
32 90 305
$-$273 $-$459 0
27 80 300
18 65 291
Q7.4.2
Kelvin
Q7.4.3
P (atm), V (L), n (mol), T (K)
Q7.4.4
A 1.00 mol sample of gas is at 300 K and 4.11 atm. What is the volume of the gas under these conditions?
$PV=nRT$
$\left ( 4.11\; atm \right )V=\left ( 1.00\;mol \right )\left ( 0.08206\frac{L\cdot atm}{mol \cdot K} \right )\left ( 300\;K \right )$
$V=5.99\;L$
Q7.4.5
$PV=nRT$
$P\left ( 2.5\;L \right )=\left ( 2.5\;mol \right )\left ( 0.08206\frac{L\cdot atm}{mol \cdot K} \right )\left ( 293\;K \right )$
$P=24 \;atm$
Q7.4.6
$744\;torr \left (\frac{1\;atm}{760\;mmHg} \right ) =0.979\; atm$
T = 55°C + 273.15 = 328 K
$PV=nRT$
$\left ( 0.979\;atm \right )\left ( 11.2\;L \right )=n \left ( 0.08206\frac{L\cdot atm}{mol \cdot K} \right )\left ( 328\;K \right )$
$n=0.407 \;mol$
Q7.4.7
T = 25°C + 273.15 = 298 K
$PV=nRT$
$\left ( 0.992\;atm \right )V=\left (8.80\; mol \right ) \left ( 0.08206\frac{L\cdot atm}{mol \cdot K} \right )\left ( 298\;K \right )$
$V=217 \;L$
Q7.4.8
x = volume of one breath of air
$3x=1.7\;L$
$x=0.57\;L$
Balloon will have a total of 8 breaths of air (3 original plus 5 additional)
$V = 8x=8(0.57\;L) = 4.6 L$
Q7.4.9
$77.8\;g\;N_2 \left (\frac{1\;mol}{28.02\;g} \right )=2.78\;mol\;N_2$
T = 25°C + 273.15 = 298 K
$PV=nRT$
$P \left (66.8\;L \right )=\left (2.78\; mol \right ) \left ( 0.08206\frac{L\cdot atm}{mol \cdot K} \right )\left ( 298\;K \right )$
$P=1.02 \;atm$
$1.02\;atm \left (\frac{101.3\;kPa}{1\;atm} \right ) = 103\;kPa$
Q7.4.10
$PV=nRT$
$\left (1.220\;atm \right ) \left (4.3410\;L \right )=mol \left ( 0.08206\frac{L\cdot atm}{mol \cdot K} \right )\left ( 788.0\;K \right )$
$n=0.08190 \;mol\; BF_3$
$0.08190\;mol \left (\frac{67.82\;g}{mol} \right )=5.554\;g\;BF_3$
Q7.4.11
1. $\frac{P_i}{T_i}=\frac{P_f}{T_f}$
2. $\frac{P_iV_i}{T_i}=\frac{P_fV_f}{T_f}$
3. $\frac{1}{n_iT_i}=\frac{1}{n_fT_f}$ or $n_iT_i=n_fT_f$
4. $P_iV_i=P_fV_f$
5. $\frac{P_i}{n_i}=\frac{P_f}{n_f}$
6. $\frac{V_i}{T_i}=\frac{V_f}{T_f}$
7. $\frac{P_iV_i}{n_i}=\frac{P_fV_f}{n_f}$
Q7.4.12
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$P_iV_i=P_fV_f$
$0.56\;atm\cdot2.0L=P_f\cdot0.75\;L$
$P_f=1.5\;atm$
Q7.4.13
$K=-196^{\circ}C+273.15=77\;K$
$K=100^{\circ}C+273.15=373\;K$
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{V_i}{T_i}=\frac{V_f}{T_f}$
$\frac{2.50\;L}{77\;K}=\frac{V_f}{373\;K}$
$V_f=12\;L$
Q7.4.14
$K=21^{\circ}C+273.15=294\;K$
$K=-48^{\circ}C+273.15=225\;K$
$745\;torr\left(\frac{1\;atm}{760\;torr}\right)=0.980\;atm$
$63.1\;torr\left(\frac{1\;atm}{760\;torr}\right)=0.0830\;atm$
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{P_iV_i}{T_i}=\frac{P_fV_f}{T_f}$
$\frac{0.980\;atm\cdot1.4\times10^4\;L}{294\;K}=\frac{P_i\cdot0.0830\;atm}{225\:}$
$P_f=1.27\times10^5\;atm$
Q7.4.15
$K=25^{\circ}C+273.15=298\;K$
$K=-37^{\circ}C+273.15=310\;K$
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{P_iV_i}{T_i}=\frac{P_fV_f}{T_f}$
$\frac{151\;atm\cdot35.4\;L}{298\;K}=\frac{1\;atm\cdot V_f}{310\:}$
$V_f=5.56\times10^3\;L$
Q7.4.16
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$P_iV_i=P_fV_f$
Set the initial pressure = x to calculate the factor of change in terms of x.
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$P_iV_i=P_fV_f$
$x\cdot0.50\;L =P_f\cdot1.50\;L$
$P_f=\frac{1}{3}x$
The final pressure is one third of the original pressure.
Q7.4.17
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$P_iV_i=P_fV_f$
$1.0\;atm\cdot1.0\;L =P_f\cdot0.50\;L$
$P_f=2.0\;atm$
Q7.4.18
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{V_i}{T_i}=\frac{V_f}{T_f}$
$\frac{2.75\;L}{100\;K}=\frac{V_f}{200\;K}$
$V_f=5.50\;L$
Q7.4.19
$K=25^{\circ}C+273.15=298\;K$
$K=-50^{\circ}C+273.15=323\;K$
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{V_i}{T_i}=\frac{V_f}{T_f}$
$\frac{0.75\;L}{298\;K}=\frac{V_f}{323\;K}$
$V_f=0.813\;L$
Q7.4.20
$K=45^{\circ}C+273.15=318\;K$
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{V_i}{T_i}=\frac{V_f}{T_f}$
$\frac{2.5\;L}{318\;K}=\frac{1.4\;L}{T_f}$
$T_f=178\;K$
$^{\circ}C=K-273.15$
$^{\circ}C=178-273.15$
$^{\circ}C=-95\;K$
Q7.4.21
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{V_i}{n_i}=\frac{V_f}{n_f}$
$\frac{3.50\;L}{2.00\;mol}=\frac{V_f}{2.75\;mol}$
$V_f=4.81\;L$
Note the final moles is 2.75 because the problem says that 0.75 moles of gas is added to the original amount of 2.00 moles.
Q7.4.22
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{V_i}{n_i}=\frac{V_f}{n_f}$
$\frac{2.00\;L}{1.85\;mol}=\frac{3.25\;L}{n_f}$
$n_f=3.01\;mol$
$\text{moles added} = 3.01\;mol-1.85\;mol$
$\text{moles added} = 1.16\;mol$
$1.16\;mol\;He\left(\frac{4.003\;g}{mol}\right)=4.64\;g\; \text{He}$
Q7.4.23
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{P_i}{T_i}=\frac{P_f}{T_f}$
The temperature is doubled so $T_f=2\cdot T_i$
Let $P_i=x$ to see the factor the pressure changes.
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$\frac{P_i}{T_i}=\frac{P_f}{T_f}$
$\frac{x}{T_i}=\frac{P_f}{2\cdot T_i}$
$x=\frac{P_f}{2}$
$P_f=2x$
The final pressure is twice the initial pressure.
Q7.4.24
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$P_iV_i=P_fV_f$
The volume is tripled so $V_f=3\cdot V_i$
Let $P_i=x$ to see the factor the pressure changes.
$\frac{P_iV_i}{n_iT_i}=\frac{P_fV_f}{n_fT_f}$
$P_iV_i=P_fV_f$
$x\cdot V_i=P_f\cdot 3V_i$
$x=3P_f$
$P_f=\frac{1}{3}x$
The final pressure is one-third of the initial pressure.
7.5: Aqueous Solutions
Q7.5.1
The solute is present in the smaller amount, the solvent is present in the larger amount, and the solution is the combination of the solute and solvent.
Q7.5.2
Solutions are a homoogeneous mixture of two or more compounds.
Q7.5.3
Endothermic because heat was needed to dissolve the KNO3. Heat present in the solution was consumed by the dissolution process.
Q7.5.4
Strong electrolytes completely dissociate into ions in aqueous solution and are conductors of electricity. Weak electrolytes partially dissociate into ions in aqueous solutions are are weak conductors of electricity. Non-electrolytes do not dissociate into ions in aqueous solution and are poor conductors of electricity.
Q7.5.5
1. NaCl(s) → Na+(aq) + Cl(aq)
2. CoCl3(s) → Co3+(aq) + 3Cl(aq)
3. Li2S(s) → 2Li+(aq) + S2–(aq)
4. MgBr2(s) → Mg2+(aq) + 2Br(aq)
5. CaF2(s) → Ca2+(aq) + 2F(aq)
Q7.5.6
Based on the given information, identify each as a strong, weak, or non-electrolyte.
1. non-electrolyte
2. strong electrolyte
3. strong electrolyte
4. weak electrolyte
5. weak electrolyte
6. non-electrolyte
7.6: Colloids and Suspensions
Q7.6.1
A suspension can be separated from the solvent by filtration while a solution cannot because particles settle out of suspensions but not solutions.
Q7.6.2
Particles in a solution are less than 1 nanometer, colloids have particles from 1-1000 nm, and suspensions have particles over 1000 nm.
Q7.6.3
The Tyndall effect is the scattering of visible light by particles. The particles in colloids are large enough to scatter light while the particles in solutions are too small to scatter light. Solutions are transparent (we can see through them) because the particles are so small.
Q7.6.4
The dispersed phase is present in the smaller amount and the dispersing medium is present in a larger amount.
Q7.6.5
1. suspension
2. colloids and suspensions
3. solution
4. colloid
5. solution
6. colloids and suspensions
7. colloid
8. solutions and colloids
9. colloids
7.7: Solubility
Q7.7.1
A saturated solution has the maximum amount of solute dissolved. An unsaturated solution does not have the maximum amount dissolved; additional solute can be added and will dissolve.
Q7.7.2
1. Addition of solute to the solution until no more dissolves.
2. Removal of solvent such as through evaporation.
Q7.7.3
1. The added solute will not dissolve.
2. The added solute will dissolve.
Q7.7.4
1. The solute and solvent are both solids.
2. The solute is a solid and the solvent is a liquid.
3. The solute is a gas and the solvent is a liquid.
4. The solute and solvent are both liquids.
Q7.7.5
1. 60 g NH4Cl
2. 120 g NH4Cl
3. 31oC
4. HCl and KClO3
5. 90 g
6. 7 g
Q7.7.6
At 90°C, 50 g of KClO3 will dissolve in 100 g of water for a saturated solution. At 20°C, only 10 g of KClO3 is dissolved in 100 g of water for a saturated solution. 40 grams of KClO3 will precipitate out of solution.
Q7.7.7
1. saturated
2. unsaturated
3. saturated (with additional undissolved solute)
Q.7.7.8
The solubility of a gas in a liquid is the greatest at low temperature and high pressure. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Chemistry_for_Allied_Health_(Soult)/07.E%3A_Solutions_%28Exercises%29.txt |
These are homework exercises to accompany Chapter 8 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Answers are below the questions.
Questions
(click here for solutions)
Q8.1.1
How is a concentrated solution different from a dilute solution?
Q8.1.2
What is the molarity of a solution prepared with 0.75 moles NaCl in 250. mL solution?
Q8.1.3
What is the mass percent of an aqueous solution prepared by dissolving 12.0 g of solute into 40.0 g of water?
Q8.1.4
What is the volume percent of a solution prepared by adding enough water to 200. mL of acetone to make a total volume of 1.60 L?
Q8.1.5
What mass of glucose is in 250.0 mL of solution that is 5.00% by mass? Assume the density of the solution is 1.00 g/mL.
Q8.1.6
For a solution that contains 25.0 g of NaCl in 300.0 mL of water, find each of the following. The density of the water is 1.00 g/mL. Assume the NaCl does not contribute to the volume of the solution
1. mass percent
2. mass/volume percent
Q8.1.7
For a solution that contains 15.0 mL of methanol 125 mL of ethanol, find each of the following. The density of methanol is 0.792 g/mL and the density of ethanol is 0.789 g/mL.
1. mass percent
2. mass/volume percent
3. volume percent
Q8.1.8
A saline solution has a mass percent concentration of 10.5%. What mass of NaCl is present in 150.0 mL of the solution? Assume the density of the solution is 1.00 g/mL.
Q8.1.9
Calculate the molarity for each solution.
1. 87.2 g of Na2SO4 in enough water to make 500. mL of solution
2. 61.8 g of NH3 in enough water to make 7.00 L of solution
3. 100. mL of ethanol (C2H5OH) in 500. mL of solution (The density of ethanol is 0.789 g/mL.)
Q8.1.10
How many moles of KF are contained in 180.0 mL of a 0.250 M solution?
Q8.1.11
Calculate how many grams of each solute would be required in order to make the given solution.
1. 3.40 L of a 0.780 M solution of iron(III) chloride, FeCl3
2. 60.0 mL of a 4.10 M solution of calcium acetate, Ca(CH3COO)2
Q8.1.12
What volume of a 0.500 M solution of NaI could be prepared with 113 g of solid NaI?
Q8.1.13
Calculate the molarity of the solutions prepared from the following dilutions.
1. 125 mL of 2.00 M HCl is diluted to a volume of 4.00 L.
2. 1.85 mL of 6.30 M AgNO3 is diluted to a volume of 5.00 mL.
Q8.1.14
What volume of 12 M HCl is required to prepare 6.00 L of a 0.300 M solution?
Q8.1.15
What mass of lead is present in 50.0 mL of solution with a lead concentration of 12 ppm?
Q8.1.16
What mass of mercury is present in 175 mL of solution with a mercury concentration of 25 ppb?
Q8.1.17
What is the concentration, in units of ppm, for a solution that contains 34 g of iron in 365 mL of water?
Q8.1.18
How many equivalents are there in 2.0 moles of the ion of each element below?
1. magnesium
2. aluminum
3. sulfur
4. bromine (Br)
5. cesium (Cs)
6. barium (Ba)
Q8.1.19
How many equivalents are present in 2.50 moles of ions for each of the elements in the previous question?
Q8.1.20
How many moles of Ca2+ are given to a patient if they receive 250.0 mL of a solution with a concentration of 132 mEq/L?
Q8.1.21
How many grams of K+ are given to a patient if they receive 500.0 mL of a solution with a concentration of 98 mEq/L?
Q8.1.22
A solution contains 128 mEq/L of Sr2+. What volume of solution is needed to have a total mass of 3.93 g of strontium ions?
(click here for solutions)
Q8.2.1
What is chemical equilibrium?
Q8.2.2
If the reaction H2 + I2 ⇌ 2HI is at equilibrium, do the concentrations of HI, H2, and I2 have to be equal?
Q8.2.3
Do the concentrations at equilibrium depend upon how the equilibrium was reached?
Q8.2.4
What does the equilibrium constant tell us?
Q8.2.5
What does it mean if the Keq is > 1?
Q8.2.6
What does it mean if the Keq is < 1?
Q8.2.7
Does the equilibrium state depend on the starting concentrations?
(click here for solutions)
Q8.3.1
Define Le Chatelier’s principle.
Q8.3.2
List the three factors types of changes that can disturb the equilibrium of a system.
Q8.3.3
How will each change affect the reaction?
PCl5(g) + heat ⇌ PCl3(g) + Cl2(g)
1. Addition of PCl5
2. Addition of Cl2
3. Removal of PCl3
4. Increasing temperature
5. Decreasing temperature
6. Decreasing volume
Q8.3.4
How will each change affect the reaction?
HNO2(aq) ⇌ H+(aq) + NO2(aq)
1. Removal of HNO2
2. Addition of HCl (i.e. adding more H+)
3. Increasing volume
4. Decreasing volume
5. Removal of NO2
6. Addition of OH (which will react with and remove H+)
Q8.3.5
How will each change affect the reaction?
CO2(g) + C(s) ⇌ 2CO(g) $\Delta \text{H}=172.5\; kJ$
1. Addition of CO2
2. Removal of CO2
3. Increasing temperature
4. Decreasing temperature
5. Increasing volume
6. Addition of CO
Q8.3.6
How will each change affect the reaction?
H2(g) + I2(g) ⇌ 2HI(g) $\Delta \text{H} = -9.48\; kJ$
1. Addition of H2
2. Removal of H2
3. Increasing temperature
4. Decreasing temperature
5. Increasing volume
6. Decreasing volume
(click here for solutions)
Q8.4.1
What are some of the features of a semipermeable membrane?
Q8.4.2
Two solutions are separated by a semipermeable membrane. Solution A contains 25.0 g of NaCl in 100.0 mL of water and solution B contains 35.0 g of NaCl in 100.0 mL of water.
1. Which one has a higher concentration?
2. Which way will water molecules flow?
3. Which volume will increase?
4. Which volume will decrease?
5. What will happen to the concentration of solution A?
6. What will happen to the concentration of solution B?
Q8.4.3
What do the prefixes hyper, hypo, and iso mean?
Q8.4.4
Cells are placed in a solution and the cells then undergo hemolysis. What can be said about the relative concentrations of solute in the cell and the solution?
Q8.4.5
Describe the relative concentrations inside and outside a red blood cell when crenation occurs.
Q8.4.6
A saltwater fish is placed in a freshwater tank. What will happen to the fish? Describe the flow of water molecules to explain the outcome.
Q8.4.7
What makes up the "head" region of a phospholipid? Is it hydrophobic or hyrdrophillic?
Q8.4.8
What makes up the "tail" region of a phospholipid? Is it hydrophobic or hyrdrophillic?
(click here for solutions)
Q8.5.1
Which statement below is true? Explain.
1. All Arrhenius bases are also Brønsted-Lowry bases.
2. All Brønsted-Lowry bases are also Arrhenius bases.
Q8.5.2
Classify each of the following as an acid, base, or neither.
1. LiOH
2. HClO4
3. CH3COOH
4. Sr(OH)2
5. CH4
6. CH4OH
Q8.5.3
What does it mean to say that a substance is amphoteric?
Q8.5.4
Identify each reactant in the following reactions as an acid or a base according to the Brønsted-Lowry theory.
1. HIO3(aq) + H2O(l) ⇌ IO3(aq) + H3O+(aq)
2. F(aq) + HClO(aq) ⇌ HF(aq) + ClO(aq)
3. H2PO4(aq) + OH(aq) ⇌ HPO42(aq) + H2O(l)
4. CO32(aq) + H2O(l) ⇌ HCO3(aq) + OH(aq)
Q8.5.5
Referring to question 4, identify the conjugate acid-base pairs in each reaction.
Q8.5.6
Write the formula of each acid’s conjugate base.
1. HNO3
2. HSO3
3. H3AsO4
4. HCOOH
5. HPO42
6. H2S
7. HS
8. HCO3
9. H2CO3
10. H3PO4
11. NaHSO4
Q8.5.7
Write the formula of each base’s conjugate acid.
1. BrO3
2. NH3
3. CH3COO
4. HCO3
5. CN
6. HPO42
7. HS
8. SO42
9. CO32
10. HCO3
11. PH3
Q8.5.8
Explain why the hydrogen phosphate ion (HPO42) is amphoteric.
(click here for solutions)
Q8.6.1
Describe the process by which water self-ionizes, and explain why pure water is considered to be neutral.
Q8.6.2
Indicate whether solutions with the following pH values are acidic, basic, or neutral.
a. pH = 9.4
b. pH = 7.0
c. pH = 5.0
Q8.6.3
How can the pOH of a solution be determined if its pH is known? (Hint: Write a mathematical expression.)
Q8.6.4
Find pH and pOH of each solution.
1. [H+] = 2.3 × 10−4 M
2. [H+] = 8.7 × 10−10 M
3. [OH] = 1.9 × 10−9 M
4. [OH] = 0.60 M
Q8.6.5
Find pH and pOH of each solution.
1. [H+] = 1.0 × 10−5 M
2. [H+] = 2.8 × 10−11 M
3. [OH] = 1.0 × 10−2 M
4. [OH] = 4.4 × 10−9 M
Q8.6.6
Determine [H+] and [OH] in aqueous solutions with the following pH or pOH values.
1. pH = 1.87
2. pH = 11.15
3. pH = 0.95
4. pOH = 6.21
5. pOH = 13.42
6. pOH = 7.03
Q8.6.7
You have prepared 1.00 L of a solution with a pH of 5.00. What is the pH of the solution if 0.100 L of additional water is added to it? (Hint: Calculate the moles of H+ ions present in the solution.)
Q8.6.8
How much water would need to be added to the original solution in question 8 in order to bring the pH to 6.00?
Answers
8.1: Concentrations of Solutions
Q8.1.1
Concentrated solutions have more solute per unit of solvent or solution.
Q8.1.2
$M=\frac{mol\;solute}{L\;soln}$
$M=\frac{0.75\;mol}{0.250\;L}$
$M=3.0\;M$
Q8.1.3
$mass\; \%=\frac{g\;solute}{g\;soln}\times100$
$mass\; \%=\frac{12.0\;g}{40.0\;g+12.0\;g}\times100$
$mass\; \%=23.1\;\%$
Remember, the mass of the solution includes both the solute and solvent.
Q8.1.4
$volume\; \%=\frac{L\;solute}{L\;soln}\times100$
$volume\; \%=\frac{0.200\;L}{1.60\;L}\times100$
$volume\; \%=12.5\;\%$
Volumes can also be used in mL (or any other unit) as long as both volumes are in the same unit.
Q8.1.5
Write the concentration in "expanded form" which shows the relationship to then be used in dimensional analysis.
$5.00\; \%\;m/m=\frac{5.00\;g\;glucose}{100\;g\;solution}$
$250.0\;mL\;soln\left (\frac{1.00\;g\;soln}{mL\;soln}\right )\left (\frac{5.00\;g\;glucose}{100\;g\;soln}\right )=12.5\;g\;glucose$
Q8.1.6
1. $\%\;m\diagup m=\frac{mass\;solute}{mass\;solution}\times 100$
$\%\;m\diagup m=\frac{25.0\;g}{25.0\;g+300\;g}\times 100$
$\%\;m\diagup m=7.69\;\%$
2. The NaCl does not contribute to the volume of the solution so only the volume of the water is used for the volume fo the solution. Given the density is 1.00 g/mL, the volume of the solution is 300.0 mL.
$\%\;m\diagup v=\frac{mass\;solute}{volume\;solution}\times 100$
$\%\;m\diagup v=\frac{25.0\;g}{25.0\;mL+300\;mL}\times 100$
$\%\;m\diagup v=8.33\;\%$
Q8.1.7
The parts of this problem require both the volume and mass of solute and solvent. The volume of the solute and solvent are given so first, find the mass of the solute and solvent so all the values are present before we start calculating the concentrations.
$15.0\;mL\;\text{methanol}\left (\frac{0.792\;g}{mL\;\text{methanol}}\right)=11.9\;g\;\text{methanol}$
$125.0\;mL\;\text{ethanol}\left (\frac{0.789\;g}{mL\;\text{ethanol}}\right)=98.6\;g\;\text{ethanol}$
1. $\%\;m\diagup m=\frac{mass\;solute}{mass\;solution}\times 100$
$\%\;m\diagup m=\frac{11.9\;g\;\text{methanol}}{11.9\;g+98.6\;g}\times 100$
$\%\;m\diagup m=10.8\;\%$
2. $\%\;m\diagup v=\frac{mass\;solute}{volume\;solution}\times 100$
$\%\;m\diagup v=\frac{11.9\;g\;\text{methanol}}{15.0\;mL+125\;mL}\times 100$
$\%\;m\diagup v=8.50\;\%$
3. $\%\;v\diagup v=\frac{volume\;solute}{volume\;solution}\times 100$
$\%\;v\diagup v=\frac{15.0\;mL\;\text{methanol}}{15.0\;mL+125\;mL}\times 100$
$\%\;v\diagup v=10.7\;%$
Q8.1.8
Write the concentration in "expanded form" which shows the relationship to then be used in dimensional analysis.
$10.5\;\%\;m\diagup m=\frac{10.5\;g\;\text{NaCl}}{100\;g\;soln}$
$150.0\;mL\;soln\left (\frac{1.00\;g\;soln}{mL\;soln}\right )\left (\frac{10.5\;g\;\text{NaCl}}{100\;g\;soln}\right )=15.8\;g\;\text{NaCl}$
Q8.1.9
1. $M=\frac{0.614\;mol\;\text{Na}_2\text{SO}_4}{0.500\;L\;soln}=1.23\;M$
2. $M=\frac{3.63\;mol\;\text{NH}_3}{7.00\;L\;soln}=0.519\;M$
3. $M=\frac{1.71\;mol\;\text{EtOH}}{0.500\;L\;soln}=3.43\;M$
Q8.1.10
$0.250\;M=\frac{0.250\;mol\;\text{KF}}{1\;L\;soln}$
$180.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{0.250\;mol\;\text{KF}}{1\;L\;soln}\right)=0.0450\;mol\;\text{KF}$
Q8.1.11
1. $3.40\;L\left(\frac{0.780\;mol}{1\;L}\right)\left(\frac{162.2\;g}{mol}\right)=430.\;g\;\text{FeCl}_3$
2. $60.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{4.10\;mol}{1\;L}\right)\left(\frac{158.17\;g}{mol}\right)=38.9\;g\;\text{Ca(CH}_3\text{COO)}_2$
Q8.1.12
$0.500\;M=\frac{0.500\;mol\;\text{NaI}}{1\;L\;soln}$
$113\;g\;\text{NaI}\left(\frac{1\;mol}{149.89\;g}\right)\left(\frac{1\;L}{0.500\;mol}\right)=1.51\;L\;soln$
Q8.1.13
1. $C_1V_1=C_2V_2$
$2.00\;M\cdot0.125\;L=C_2\cdot4.00\;L$
$C_2=0.0625\;M$
2. $C_1V_1=C_2V_2$
$6.30\;M\cdot1.85\;mL=C_2\cdot5.00\;mL$
$C_2=2.33\;M$
Q8.1.14
$C_1V_1=C_2V_2$
$0.300\;M\cdot6.00\;L=12\;M\cdotV_2$
$V_2=0.15\;L$
Q8.1.15
$12\;ppm\;\text{Pb}=\frac{12\;mg\;\text{Pb}}{1\;L\;soln}$
$50.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{12\;mg} {1\;L}\right)=0.60\;mg\;\text{Pb}$
Q8.1.16
$25\;ppb\;\text{Hg}=\frac{25\;\mu g\;\text{Hg}}{1\;L\;soln}$
$175\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{25\;\mu g} {1\;L}\right)=4.4\;\mu g\;\text{Hg}$
Q8.1.17
$ppm=\frac{mg}{L}$
$34\;g\;\text{Fe}\left(\frac{1\;mg}{10^{-3}\;g}\right)=3.4\times10^4\;mg\;\text{Fe}$
$365\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)=0.365\;L$
$ppm=\frac{3.4\times10^4\;mg\;\text{Fe}}{0.365\;L}=9.3\times10^4\;ppm\;\text{Fe}$
Q8.1.18
1. $2.0\;mol\;\text{Mg}^{2+}\left(\frac{2\;Eq}{1\;mol}\right)=4.0\;Eq$
2. $2.0\;mol\;\text{Al}^{3+}\left(\frac{3\;Eq}{1\;mol}\right)=6.0\;Eq$
3. $2.0\;mol\;\text{S}^{2-}\left(\frac{2\;Eq}{1\;mol}\right)=4.0\;Eq$
4. $2.0\;mol\;\text{Br}^-\left(\frac{1\;Eq}{1\;mol}\right)=2.0\;Eq$
5. $2.0\;mol\;\text{Cs}^{+}\left(\frac{1\;Eq}{1\;mol}\right)=2.0\;Eq$
6. $2.0\;mol\;\text{Ba}^{2+}\left(\frac{2\;Eq}{1\;mol}\right)=4.0\;Eq$
Q8.1.19
1. $2.50\;mol\;\text{Mg}^{2+}\left(\frac{2\;Eq}{1\;mol}\right)=5.00\;Eq$
2. $2.50\;mol\;\text{Al}^{3+}\left(\frac{3\;Eq}{1\;mol}\right)=7.50\;Eq$
3. $2.50\;mol\;\text{S}^{2-}\left(\frac{2\;Eq}{1\;mol}\right)=5.00\;Eq$
4. $2.50\;mol\;\text{Br}^-\left(\frac{1\;Eq}{1\;mol}\right)=2.50\;Eq$
5. $2.50\;mol\;\text{Cs}^{+}\left(\frac{1\;Eq}{1\;mol}\right)=2.50\;Eq$
6. $2.50\;mol\;\text{Ba}^{2+}\left(\frac{2\;Eq}{1\;mol}\right)=5.00\;Eq$
Q8.1.20
$250.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{132\;mEq}{L}\right)\left(\frac{10^{-3}\;Eq}{1\;mEq}\right)\left(\frac{1\;mol\;\text{Ca}^{2+}}{2\;Eq}\right)=0.0165\;mol\;\text{Ca}^{2+}$
Q8.1.21
$500.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{98\;mEq}{L}\right)\left(\frac{10^{-3}\;Eq}{1\;mEq}\right)\left(\frac{1\;mol\;\text{K}^{+}}{1\;Eq}\right)\left(\frac{39.10\;g}{mol}\right)=1.92\;g\;\text{K}^{+}$
Q8.1.22
$3.93\;g\;\text{Sr}^{2+}\left(\frac{1\;mol}{87.62\;g}\right)\left(\frac{2\;Eq}{1\;mol}\right)\left(\frac{1\;mEq}{10^{-3}\;Eq}\right)\left(\frac{1\;L}{128\;mEq}\right)=0.701\;L\;soln$
8.2: Chemical Equilibrium
Q8.2.1
The rate of the forward reaction equals the rate of the reverse reaction.
Q8.2.2
No, the concentrations are constant but the concentrations do not have to be equal.
Q8.2.3
No.
Q8.2.4
The ratio of products and reactants when the system is at equilibrium.
Q8.2.5
More products than reactants are present at equilibrium.
Q8.2.6
More reactants than products present at equilibrium.
Q8.2.7
No. The equilibrium ratio does not depend on the initial concentrations.
8.3: Le Chatelier's Principle
Q8.3.1
Le Chatelier’s principle states that a system at equilibrium is disturbed, it will respond in a way to minimize te disturbance.
Q8.3.2
temperature, change in amount of substance, change in pressure through change in volume
Q8.3.3
1. shift right
2. shift left
3. shift right
4. shift right
5. shift left
6. shift left
Q8.3.4
1. shift left
2. shift left
3. no effect
4. no effect
5. shift right
6. shift right
Q8.3.5
1. shift right
2. shift left
3. shift right
4. shift left
5. shift right
6. shift left
Q8.3.6
1. shift right
2. shift left
3. shift left
4. shift right
5. no effect
6. no effect
8.4: Osmosis and Diffusion
Q8.4.1
A semipermeable membrane allows some substances to pass through but not others.
Q8.4.2
Two solutions are separated by a semipermeable membrane. Solution A contains 25.0 g of NaCl in 100.0 mL of water and solution B contains 35.0 g of NaCl in 100.0 mL of water.
1. Solution B
2. A $\rightarrow$ B
3. B
4. A
5. increase
6. decrease
Q8.4.3
hyper - higher
hypo - lower
iso - same
Q8.4.4
Cells contain fluid with higher concentration than solution outside the cell.
Q8.4.5
Cells contain fluid with a lower concentration than the solution outside the cell.
Q8.4.6
Water molecules will flow from the tank water into the fish because the fish has a higher concentration of salt. If the fish absorbs too much water, it will die.
Q8.4.7
The "head" region is a phosphate group and it is hydrophillic.
Q8.4.8
The "tail" is a hydrocarbon tail and it is hydrophobic.
8.5: Acid-Base Definitions
Q8.5.1
1. TRUE
2. FALSE - Bronsted-Lowry acid-base definitions are broader.
Q8.5.2
Classify each of the following as an acid, base, or neither.
1. base (contains metal and -OH group)
2. acid (formula starts with H and isn't water)
3. acid (contains -COOH which is carboxylic acid functional group)
4. base (contains metal and -OH group)
5. neither
6. neither (-OH group has to be with metal)
Q8.5.3
Amphoteric substances can act as an acid or base.
Q8.5.4
Identify each reactant in the following reactions as an acid or a base according to the Brønsted-Lowry theory.
1. HIO3(aq) - acid; H2O(l) - base; IO3(aq) - base ; H3O+(aq) - acid
2. F(aq) - base; HClO(aq) - acid; HF(aq) - acid; ClO(aq) - base
3. H2PO4(aq) - acid; OH(aq) - base; HPO42−(aq) - base; H2O(l) - acid
4. CO32−(aq) - base; H2O(l) - acid; HCO3(aq) - acid; OH(aq) - base
Q8.5.5
1. HIO3/IO3 and H3O+/H2O
2. HF/Fand HClO/ClO
3. H2PO4/HPO42− and H2O/OH
4. HCO3/CO32− and H2O/OH
Q8.5.6
1. NO3
2. SO32−
3. H2AsO4
4. HCOO(the H that is removed comes from the carboxylic acid functional group)
5. PO43−
6. HS
7. S2−
8. CO32
9. HCO3
10. H2PO4
11. Na2SOor NaSO4
Q8.5.7
Write the formula of each base’s conjugate acid.
1. HBrO3
2. NH4+
3. CH3COOH
4. H2CO3
5. HCN
6. H2PO4
7. H2S
8. HSO4
9. HCO3
10. H2CO3
11. PH4+
Q8.5.8
HPO42− can act as a base and accept a proton to form H2PO4and it can act as an acid and donate a proton to form PO43−.
8.6: The pH Concept
Q8.6.1
H2O(l) H+(aq) + OH(aq)
It's neutral because there are equal amounts of H+ and OH.
Q8.6.2
Indicate whether solutions with the following pH values are acidic, basic, or neutral.
a. basic
b. neutral
c. acidic
Q8.6.3
pH + pOH = 14
Q8.6.4
1. pH = 3.64; pOH = 10.36
2. pH = 9.06; pOH = 4.94
3. pOH = 8.72; pH = 5.28
4. pOH = 0.22; pH = 13.78
Q8.6.5
1. pH = 5.00; pOH = 9.00
2. pH = 10.55; pOH = 4.94
3. pOH = 8.72; pH = 5.28
4. pOH = 0.22; pH = 13.78
Q8.6.6
1. $[\text{H}^+]=1.3\times 10^{-2}\;M, [\text{OH}^-]=7.4\times10^{-13}\;M$
2. $[\text{H}^+]=7.1\times 10^{-12}\;M, [\text{OH}^-]=1.4\times10^{-3}\;M$
3. $[\text{H}^+]=0.11\;M, [\text{OH}^-]=8.9\times10^{-14}\;M$
4. $[\text{OH}^-]=6.2\times10^{-7}\;M, [\text{H}^+]=1.6\times 10^{-8}\;M$
5. $[\text{OH}^-]=3.8\times10^{-14}\;M, [\text{H}^+]=0.26\;M$
6. $[\text{OH}^-]=9.3\times10^{-8}\;M, [\text{H}^+]=1.1\times 10^{-7}\;M$
Q8.6.7
Given pH = 5.00, we know $[\text{H}^+]=1.0\times 10^{-5}\;M$ which means $M=\frac{1.0\times10^{-5}\;mol\;\text{H}^+}{1.00\;L}$.
If 0.100 L of water is added to 1.00 L, then the volume changes to 1.10 L but the moles of H+ does not change. The molarity can be calclulated with the same number of moles and the new volume.
$M=\frac{1.0\times10^{-5}\;mol\;\text{H}^+}{1.00+0.100\;L}$
$M=9.1\times10^{-6}\;M$
$pH = 5.04$
Q8.6.8
How much water would need to be added to the original solution in question 8 in order to bring the pH to 6.00?
To get to pH = 6.00, we need $[\text{H}^+]=1.0\times 10^{-6}\;M$.
Use the dilution formula to calculate the total volume of solution.
$C_1V_1=C_2V_2$
$1.0\times10^{-5}\;M\cdot 1.00\;L=1.0\times10^{-6}\;M\cdot V_2$$V_2=10.\;L$
The total volume is 10. L so 9 L needs to be added to the original 1 L solution. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Chemistry_for_Allied_Health_(Soult)/08.E%3A_Properties_of_Solutions_%28Exercises%29.txt |
These are homework exercises to accompany Chapter 9 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Answers are below the questions.
Questions
(click here for solutions)
Q9.1.1
Describe the difference between a strong acid and a weak acid.
Q9.1.2
Describe the difference between a strong base and a weak base.
Q9.1.3
Identify each of the following as a strong acid, weak acid, strong base, or weak base.
1. HCl
2. NaOH
3. KOH
4. HNO2
5. HNO3
6. HF
7. NH3
8. Ba(OH)2
9. CH3CH2COOH
Q9.1.4
Write an equation representing the behavior of each substance in question 3. Pay attention to the type of arrow used in the equation.
Q9.1.5
A solution is prepared by dissolving 15.0 grams of NaOH in enough water to make 500.0 mL of solution. Calculate the pH of the solution.
Q9.1.6
A solution is prepared by dissolving 22.0 grams of HCl in enough water to make 300.0 mL of solution. Calculate the pH of the solution.
(click here for solutions)
Q9.2.1
What is a buffer?
Q9.2.2
What is the purpose of a buffer?
Q9.2.3
Determine whether or not each of these pairs can act as a buffer.
1. HCl/Cl
2. HF/F
3. H2SO4/HSO4
4. HSO4/SO42
5. H2O/NaOH
6. HNO2/KNO2
7. HCl/NaOH
Q9.2.4
Write the formula of the conjugate base needed to form a buffer with each of the following weak acids.
1. HClO3
2. H2PO4
3. CH3COOH
Q9.2.5
Write the formula of the conjugate acid needed to form a buffer with each of the following weak bases.
1. NH3
2. CH3NH2
Q9.2.6
Describe buffer capacity.
Answers
9.1: Acid and Base Strength
Q9.1.1
A strong acid completely dissociates into ions and a weak acid doesn't completely dissociate into ions.
Q9.1.2
A strong base is a base (metal with an -OH) group that dissociates completely into ions. A weak base is a proton acceptor but not ll of the molecules will accept a proton.
Q9.1.3
Identify each of the following as a strong acid, weak acid, strong base, or weak base.
1. HCl is a strong acid. See list of 6 strong acids.
2. NaOH is a strong base. It has a metal with an -OH group.
3. KOH is a strong base. It has a metal and an -OH group.
4. HNO2 is a weak acid. The formula starts with H but it's not water so it's an acid. Recognize it is weak because it is not on the list of 6 strong acids.
5. HNO3 is a strong acid. The formula starts with H but it's not water so it's an acid. See the list of 6 strong acids.
6. HF is a weak acid. the formula starts with H but it's not water so it's an acid. Recognize it is weak becauce it is not on the list of 6 strong acids.
7. NH3 is a weak base. Amines are weak bases.
8. Ba(OH)2 is a strong base. It has a metal with an -OH group.
9. CH3CH2COOH is a weak base. It has a carboxylic acid functional group so it's an acid. Carboxylic acids are all weak (also, not on the list of 6 strong acids).
Q9.1.4
Write an equation representing the behavior of each substance in question 3. Pay attention to the type of arrow used in the equation.
1. HCl(aq) $\rightarrow$ H+(aq) + Cl(aq)
2. NaOH(aq) $\rightarrow$ Na+(aq) + OH(aq)
3. KOH(aq) $\rightarrow$ K+(aq) + OH(aq)
4. HNO2(aq) $\rightleftharpoons$ H+(aq) + NO2(aq)
5. HNO3(aq) $\rightarrow$ H+(aq) + NO3(aq)
6. HF(aq) $\rightleftharpoons$ H+(aq) + F(aq)
7. NH3(aq) + H2O(l)$\rightleftharpoons$ NH4+(aq) + OH(aq)
8. Ba(OH)2(aq) $\rightarrow$ Ba2+(aq) + 2OH(aq)
9. CH3CH2COOH(aq) $\rightleftharpoons$ H+(aq) + CH3CH2COO(aq)
Q9.1.5
NaOH is a strong base and completely dissociates (see reaction). Since it completely dissociates, the concentration of NaOH equals the concentration of OH. We need to calculate the concentration of NaOH.
NaOH(aq) $\rightarrow$ Na+(aq) + OH(aq)
$15.0\;g\;\text{NaOH}\left(\frac{1\;mol}{40.00\;g}\right)=0.375\;mol\;\text{NaOH}$
$M=\frac{mol\;solute}{L\;soln}=\frac{0.375\;mol}{0.500\;L}=0.750\;M\;NaOH$
[NaOH] = [OH] = 0.750 M
Use [OH] to find pOH.
$p\text{OH} = -log[\text{OH}^-]=-log[0.750]=0.125$
Now, convert from pOH to pH. Note that 14 is an exact number in this context so it does not affect significant figures.
$p\text{H}+p\text{OH}=14$
$p\text{H}=14-p\text{OH}$
$p\text{H}=14-0.125$
$p\text{H}=13.875$
Q9.1.6
HCl is a strong acid and completely dissociates (see reaction). Since it completely dissociates, the concentration of HCl equals the concentration of H+. We need to calculate the concentration of HCl.
HCl(aq) $\rightarrow$ H+(aq) + Cl(aq)
$22.6\;g\;\text{HCl}\left(\frac{1\;mol}{36.46\;g}\right)=0.603\;mol\;\text{HCl}$
$M=\frac{mol\;solute}{L\;soln}=\frac{0.603\;mol}{0.300\;L}=2.01\;M\;HCl$
[HCl] = [H+] = 2.01 M
$p\text{H} = -log[\text{H}^+]=-log[2.01]=-0.303$ (pH can be less than zero if it is a strong acid with a concentration greater than 1 M)
9.2: Buffers
Q9.2.1
A buffer is a weak acid and its conjugate base (or a weak base and its conjugate acid) that helps maintain the pH of a solution.
Q9.2.2
The purpose of a buffer is to resist change of pH in a solution.
Q9.2.3
1. HCl/Cl cannot because HCl is a strong acid.
2. HF/Fcan because HF is a weak acid and F is its conjugate base. It will be added to mixture as a salt (ionic compound) of F (i.e. NaF, KF, etc)
3. H2SO4/HSO4cannot because H2SO4 is a strong acid.
4. HSO4/SO42can because HSO4 is a weak acid and SO42 is its conjugate base. Both compounds will be added to the mixture as salts (i.e. NaHSO4 and Na2SO4).
5. H2O/NaOH cannot becase NaOH is a strong base and NaOH and H2O are not a conjugate acid-base pair.
6. HNO2/KNO2 can because HNO2 is a weak acid and KNO2 contains its conjugate base. Note that KNO2 is a strong electrolyte so it dissociates into K+ and NO2. HNO2 and NO2 form a conjugate acid-base pair.
7. HCl/NaOH cannot because HCl is a strong acid, NaOH is a strong base, and they do not form a conjugate acid-base pair.
Q9.2.4
Write the formula of the conjugate base needed to form a buffer with each of the following weak acids.
1. ClO3(or a salt such as NaClO3)
2. HPO42(or a salt such as Na2HPO4)
3. CH3COO(or a salt such as CH3COONa)
Q9.2.5
Write the formula of the conjugate acid needed to form a buffer with each of the following weak acids.
1. NH4+ (or a salt such as NH4Cl )
2. CH3NH3+ (or a salt such as CH3NH3Cl)
Q9.2.6
Buffer capacity is the amount of acid or base that can be added to a buffer solution before it can no longer resist significant changes in the pH of the solution. Adding small amounts of acid or base will change the pH of a buffer by a small amount and the buffer continues to be effective. If larger amounts of acid or base are added, the buffer capacity is reached and the solution can no longer resist changes in pH. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Chemistry_for_Allied_Health_(Soult)/09.E%3A_Equilibrium_Processes_%28Exercises%29.txt |
These are homework exercises to accompany Chapter 10 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Answers are below the questions.
Questions
(click here for solutions)
Q10.1.1
Write the symbol for the isotope described.
1. 12 protons, 12 electrons, 13 neutrons
2. 17 protons, 17 electrons, 20 neutrons
3. 53 protons, 53 electrons, 78 neutrons
4. 92 protons, 92 electrons, 146 neutrons
Q10.1.2
Determine the number of protons, neutrons, and electrons in each isotope.
1. $\ce{^{195}_{77}Ir}$
2. $\ce{^{209}_{82}Pb}$
3. $\ce{^{211}_{84}Po}$
4. $\ce{^{237}_{93}Np}$
Q10.1.3
Fill in the missing numbers in each equation.
1. $\ce{^{196}_{82}Pb}$ + $\ce{^0_{-1}e}$ → $_{\text{__}}^{\text{__}}\text{Tl}$
2. $\ce{^{28}_{15}P}$ → $_{\text{__}}^{\text{__}}\text{Si}$ + $\ce{^0_1e}$
3. $\ce{^{226}_{88}Ra}$ → $_{\text{__}}^{\text{__}}\text{Rn}$ + $\ce{^4_2 \alpha}$
4. $\ce{^{73}_{30}Zn}$ → $_{\text{__}}^{\text{__}}\text{Ga}$ + $\ce{^0_{-1}e}$
Q10.1.4
Fill in the blanks for each of the nuclear reactions below. State the type of decay in each case.
1. $\ce{^{198}_{79}Au}$ → _______ + $\ce{^0_{-1}e}$
2. $ce{^{57}_{27}Co}$ + $\ce{^0_{-1}e}$ → _______
3. $\ce{^{230}_{92}U}$ → _______ + $\ce{^4_2He}$
4. $\ce{^{128}_{56}Ba}$ → _______ + $\ce{^0_{1}e}$
5. $\ce{^{131}_{53}I}$ → $\ce{^{131}_{54}Xe}$ + _______
6. $\ce{^{239}_{94}Pu}$ → $\ce{^{235}_{92}U}$ + _______
Q10.1.5
Write balanced nuclear reactions for each of the following.
1. Francium-220 undergoes alpha decay.
2. Arsenic-76 undergoes beta decay.
3. Uranium-231 captures an electron.
4. Promethium-143 emits a positron.
(click here for solutions)
Q10.2.1
Describe the main difference between fission and fusion.
Q10.2.2
What is the difference between the fission reactions used in nuclear power plants and nuclear weapons?
Q10.2.3
How do the doses of radioisotopes used in diagnostic procedures and therapeutic treatment compare to one another?
(click here for solutions)
Q10.3.1
What percent of a sample remains after one half-life? Three half-lives?
Q10.3.2
The half-life of polonium-218 is 3.0 min. How much of a 0.540 mg sample would remain after 9.0 minutes have passed?
Q10.3.3
The half-life of hydrogen-3, commonly known as tritium, is 12.26 years. If 4.48 mg of tritium has decayed to 0.280 mg, how much time has passed?
Q10.3.4
The half-life of protactinium-234 is 6.69 hours. If a 0.812 mg sample of Pa-239 decays for 40.14 hours, what mass of the isotope remains?
Q10.3.5
2.86 g of a certain radioisotope decays to 0.358 g over a period of 22.8 minutes. What is the half-life of the radioisotope?
Q10.3.6
Use Table 10.3.2 above to determine the time it takes for 100. mg of carbon-14 to decay to 6.25 mg.
Q10.3.7
A radioisotope decays from 55.9 g to 6.99 g over a period of 72.5 hours. What is the half-life of the isotope?
Q10.3.8
A sample of a radioisotope with a half-life of 9.0 hours has an activity of 25.4 mCi after 36 hours. What was the original activity of the sample?
Q10.3.9
What volume of a radioisotope should be given if a patient needs 125 mCi of a solution which contains 45 mCi in 5.0 mL?
Q10.3.10
Sodium-24 is used to treat leukemia. A 36-kg patient is prescribed 145 μCi/kg and it is supplied to the hospital in a vial containing 250 μCi/mL. What volume should be given to the patient?
Q10.3.11
Using information from the previous question and knowing the half-life of Na-24 is 15 hours, calculate the total dose in μCi given to the patient. How long will it take for the radioactivity to be approximately 80 μCi?
Q10.3.12
Lead-212 is one of the radioisotopes used in the treatment of breast cancer. A patient needs a 15 μCi dose and it is supplied as a solution with a concentration of 2.5 μCi/mL. What volume does the patient need? Given the half-life of lead is 10.6 hours, what will be the radioactivity of the sample after approximately four days?
(click here for solutions)
Q10.4.1
Identify each of the following as a physical or chemical change.
1. melting ice
2. boiling water
3. cooking eggs
4. dissolving salt in water
5. burning match
6. metal reacting with HCl
7. mixing NaCl and KCl
8. decomposition of hydrogen peroxide
Q10.4.2
Give two signs that indicate a chemical change is occurring.
Q10.4.3
What doesn't change when a substance undergoes a physical change?
(click here for solutions)
Q10.5.1
Identify the reactants and products in each chemical reaction.
1. In photosynthesis, carbon dioxide and water react to form glucose and oxygen.
2. Magnesium oxide forms when magnesium is exposed to oxygen gas.
Q10.5.2
Write grammatically correct sentences that completely describe the chemical reactions shown in each equation. You may need to look up the names of elements or compounds.
1. 2H2O2(l) → 2H2O(l) + O2(g)
2. CuCO3(s) → CuO(s) + CO2(g)
3. 2Cs(s) + 2H2O(l) → 2CsOH(aq) + H2(g)
Q10.5.3
How many atoms of each element are represented by the following combinations of coefficients and chemical formulas?
1. 5Br2
2. 2NH3
3. 4(NH4)2SO4
4. 2CH3COOH
5. 3Fe(NO3)3
6. 2K3PO4
Q10.5.4
Balance the following equations.
1. Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g)
2. Li(s) + N2(g) → Li3N(s)
3. Ca(OH)2 + HBr → CaBr2 + H2O
4. C4H10 + O2 → CO2 + H2O
5. NH3 + CuO → Cu + N2 + H2O
Q10.5.5
Balance the following equations.
1. Fe(s) + Cl2(g) → FeCl3(g)
2. C4H10O + O2 → CO2 + H2O
3. As + NaOH → Na3AsO3 + H2
4. SiO2 + HF → SiF4 + H2O
5. N2 + O2 + H2O → HNO3
Answers
10.1: Nuclear Radiation
Q10.1.1
Write the symbol for the isotope described.
1. $_{12}^{25}\text{Mg}$
2. $_{17}^{37}\text{Cl}$
3. $_{53}^{131}\text{I}$
4. $_{92}^{238}\text{U}$
Q10.1.2
1. 77 protons, 77 electrons, 118 neutrons
2. 82 protons, 82 electrons, 127 neutrons
3. 84 protons, 84 electrons, 127 neutrons
4. 93 protons, 93 electrons, 144 neutrons
Q10.1.3
1. $\ce{^{196}_{82}Pb}$ + $\ce{^0_{-1}e}$ →$_{81}^{196}\text{Tl}$
2. $\ce{^{28}_{15}P}$ → $_{14}^{28}\text{Si}$ + $\ce{^0_1e}$
3. $\ce{^{226}_{88}Ra}$ → $_{86}^{222}\text{Rn}$ + $\ce{^4_2 \alpha}$
4. $\ce{^{73}_{30}Zn}$ → $_{31}^{73}\text{Ga}$ + $\ce{^0_{-1}e}$
Q10.1.4
1. $\ce{^{198}_{79}Au}$ → $_{80}^{198}\text{Hg}$ + $\ce{^0_{-1}e}$, beta
2. $\ce{^{57}_{27}Co}$ + $\ce{^0_{-1}e}$ → $_{26}^{57}\text{Fe}$, electron capture
3. $\ce{^{230}_{92}U}$ → $_{90}^{226}\text{Th}$ + $\ce{^4_2He}$, alpha
4. $\ce{^{128}_{56}Ba}$ → $_{55}^{128}\text{Cs}$ + $\ce{^0_{1}e}$, positron
5. $\ce{^{131}_{53}I}$ → $\ce{^{131}_{54}Xe}$ + $_{-1}^{0}e$, beta
6. $\ce{^{239}_{94}Pu}$ → $\ce{^{235}_{92}U}$ + $_{2}^{4}\alpha$ (or can show as $_{2}^{4}\text{He}$), alpha
Q10.1.5
1. $_{87}^{220}\text{Fr}\;\rightarrow\;_{2}^{4}\text{He}\;+\;_{85}^{216}\text{At}$
2. $_{33}^{76}\text{As}\;\rightarrow\;_{-1}^{0}e\;+\;_{34}^{76}\text{Se}$
3. $_{92}^{231}\text{U}\;+\;_{-1}^{0}e\;\rightarrow\;_{91}^{231}\text{Pa}$
4. $_{61}^{143}\text{Pm}\;\rightarrow\;_{1}^{0}e\;+\;_{60}^{143}\text{Nd}$
10.2: Fission and Fusion
Q10.2.1
During fission, big nuclei split into smaller nuclei. During fusion, nuclei combine to form large nuclei.
Q10.2.2
Fission in nuclear power plants is controlled through limiting the availability of neutrons. Nuclear weapons are uncontrolled once the process initiates.
Q10.2.3
Diagnostic amounts are much smaller than therapeutic amounts.
10.3: Half-Life
Q10.3.1
1 half-life: 50%
3 half-lives: 12.5%
Q10.3.2
Time Half-lives Amount
0 minutes 0.540 mg
3 minutes 1 0.270 mg
6 minutes 2 0.135 mg
9 minutes 3 0.0675 mg
Q10.3.3
Amount Half-lives Time
4.48 mg 0 years
2.24 mg 1 12.26 years
1.12 mg 2 24.52 years
0.560 mg 3 36.78 years
0.280 mg 4 49.04 years
Q10.3.4
Time Half-lives Amount
0 hours 0.812 mg
6.69 hours 1 0.406 mg
13.38 hours 2 0.203 mg
20.07 hours 3 0.102 mg
26.76 hours 4 0.0508 mg
33.45 hours 5 0.0254 mg
40.14 hours 6 0.0127 mg
Q10.3.5
Amount Half-lives
2.86 g
1.43 g 1
0.715 g 2
0.358 g 3
It takes three half-lives to go from 2.86 g to 0.358 g in a total time of 22.8 minutes.
$22.8\;min\;\div\;3\;=7.60 \;min$
One half-life is 7.60 minutes.
Q10.3.6
Amount Half-lives Time
100. mg 0 years
50.0 mg 1 5730 years
25.0 mg 2 11460 years
12.5 mg 3 17190 years
6.25 mg 4 22920 years
Q10.3.7
Amount Half-lives
55.9 g
28.0 g 1
14.0 g 2
6.99 g 3
It takes three half-lives to go from 55.9 g to 6.99 g in a total time of 72.5 hours.
$72.5\;hr\;\div\;3\;=24.2 \;hr$
One half-life is 24.2 hours.
Q10.3.8
Fill in the time and half-lives from top to bottom. Start at the bottom of the amount column to fill it in because we know where we end up but not where we started.
Time Half-lives Activity
0 hours 406 mCi
9.0 hours 1 203 mCi
18 hours 2 102 mCi
27 hours 3 50.8 mCi
36 hours 4 25.4 mCi $\leftarrow$ START HERE
Q10.3.9
$125\;mCi\left(\frac{5.0\;mL}{45\;mCi}\right)=14\;mL$
Q10.3.10
Sodium-24 is used to treat leukemia. A 36-kg patient is prescribed 145 μCi/kg and it is supplied to the hospital in a vial containing 250 μCi/mL. What volume should be given to the patient?
$36\;kg\left(\frac{145\;\mu Ci}{kg}\right)\left(\frac{1\;mL}{250\;\mu Ci}\right)=21\;mL$
Q10.3.11
$21\;mL\left(\frac{250\;\mu Ci}{mL}\right)=5250\;\mu Ci$ is the total dose received
Amount Half-lives Time
5250 μCi 0 hours
2625 μCi 1 15 hours
1313 μCi 2 30 hours
656 μCi 3 45 hours
328 μCi 4 60 hours
164 μCi 5 75 hours
82 μCi 6 90 hours
Q10.3.12
Lead-212 is one of the radioisotopes used in the treatment of breast cancer. A patient needs a 15 μCi dose and it is supplied as a solution with a concentration of 2.5 μCi/mL. What volume does the patient need? Given the half-life of lead is 10.6 hours, what will be the radioactivity of the sample after approximately four days?
Volume given: $15\;\mu Ci\left(\frac{1\;mL}{2.5\;\mu Ci}\right)=6.0\;mL$
Elapsed time in hours: $4\;days\left(\frac{24\;hr}{day}\right)=96\;hr$
Time Half-lives Activity
0 hours 15 μCi
10.6 hours 1 7.5 μCi
21.2 hours 2 3.8 μCi
31.8 hours 3 1.9 μCi
42.4 hours 4 0.94 μCi
53.0 hours 5 0.47 μCi
63.6. hours 6 0.23 μCi
74.2 hours 7 0.12 μCi
84.8 hours 8 0.059 μCi
95.6 hours 9 0.029 μCi
10.4: Physical and Chemical Changes
Q10.4.1
1. physical
2. physical
3. chemical
4. physical
5. chemical
6. chemical
7. physical
8. chemical
Q10.4.2
Any two from change in color, formation of gas (i.e. bubbles), formation of precipitate, odor, change in temperature.
Q10.4.3
chemical composition (i.e. chemical formula is the same)
10.5: Chemical Equations
Q10.5.1
1. reactants: carbon dioxide and water; products: glucose and oxygen
2. reactants: magnesium and oxygen; product: magnesium oxide
Q10.5.2
Descriptions may vary.
1. Two moles of liquid hydrogen peroxide decomposes to form two moles of liquid water and one mole of gaseous hydrogen.
2. One mole of solid copper(II) carbonate decomposes to form one mole each of solid copper(II) oxide and gaseous carbon dioxide.
3. Two moles of solid cesium react with 2 moles of liquid water to form 2 moles of aqueous cesium hydroxide and 1 mole of gaseous hydrogen.
Q10.5.3
1. 10 Br
2. 2 N, 6 H
3. 8 N, 32 H, 4 S, 16 O
4. 4 C, 8 H, 4 O
5. 3 Fe, 9 N, 27 O
6. 6 K, 2 P, 8 O
Q10.5.4
1. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
2. 6 Li(s) + N2(g) → 2 Li3N(s)
3. Ca(OH)2 + 2 HBr → CaBr2 + 2 H2O
4. 2 C4H10 + 13 O28 CO2 + 10 H2O
5. 2 NH3 + 3 CuO → 3 Cu + N2 + 3 H2O
Q10.5.5
1. 2 Fe(s) + 3 Cl2(g) → 2 FeCl3(g)
2. C4H10O + 6 O24 CO2 + 5 H2O
3. 2 As + 6 NaOH → 2 Na3AsO3 + 3 H2
4. SiO2 + 4 HF → SiF4 + 2 H2O
5. 2 N2 + 5 O2 + 2 H2O → 4 HNO3 | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Chemistry_for_Allied_Health_(Soult)/10.E%3A_Nuclear_and_Chemical_Reactions_%28Exercises%29.txt |
These are homework exercises to accompany Chapter 11 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health.
Questions
(click here for solutions)
Q11.1.1
What is a free element, and what is the oxidation number for atoms that exist as a free element?
Q11.1.2
What is the highest oxidation number that sulfur can have? The lowest? Consider the number of valence electrons it has.
Q11.1.3
Determine the oxidation numbers of each of the atoms in the following.
1. KMnO4
2. OCl2
3. H2C2O4
4. Li3PO4
5. NaClO
6. Br2
7. ClF3
8. CaCl2
9. K2O
Q11.1.4
Determine the oxidation number of each of the atoms in the following ions.
1. NO2
2. NO3
3. Cr2O72
4. BrO3
5. ClO3
6. BO33
7. CO32
8. NH4+
9. CrO42
10. S2O32
(click here for solutions)
Q11.2.1
Explain why oxidation and reduction always occur together in a reaction.
Q11.2.2
What happens to the oxidizing agent in a redox reaction?
Q11.2.3
What happens to the reducing agent in a redox reaction?
Q11.2.4
Identify each process as an oxidation or a reduction.
1. Rb → Rb+
2. Te → Te2
3. 2H+ → H2
4. P3– → P
5. 2Cl → Cl2
6. Sn4+ → Sn2+
7. Br2 → Br
8. Fe2+ → Fe3+
Q11.2.5
For each equation, 1) identify the oxidation numbers of each element, 2) determine if it is a redox reaction or not, and for redox reactions, 3) identify the species being oxidized and the species being reduced, and 4) identify the oxidizing and reducing agents.
1. 2KClO3(s) → 2KCl(s) + 3O2(g)
2. H2(g) + CuO(s) → Cu(s) + H2O(l)
3. 2Al(s) + 3Mg(NO3)2(aq) → 2Al(NO3)3(aq) + 3Mg(s)
4. 2HNO3(aq) + 6HI(aq) → 2NO(g) + 3I2(s) + 4H2O(l)
5. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
6. 2FeCl3(aq) + H2S(g) → 2FeCl2(aq) + 2HCl(aq) + S(s)
Q11.2.6
Identify the change occurring with respect to the gain or loss of hydrogen or oxygen.
a.
b.
c.
d.
e.
(click here for solutions)
Q11.3.1
Classify the following reactions as combination, decomposition, single replacement, double replacement, or combustion. For each, indicate if it is a redox reaction or not.
1. Cd(s) + H2SO4(aq) → CdSO4(aq) + H2(g)
2. 2Fe(s) + 3Cl2(g) → 2FeCl3(s)
3. C7H8(l) + 9O2(g) → 7CO2(g) + 4H2O(g)
4. 2NH4NO3(s) → 2N2(g) + O2(g) + 4H2O(g)
5. 2CoCl3(aq) + 3Pb(NO3)2(aq) → 2Co(NO3)3(aq) + 3PbCl2(s)
Q11.3.2
Balance the following equations and classify the following reactions as combination, decomposition, single replacement, double replacement, or combustion. For each, indicate if it is a redox reaction or not.
1. Na + Cl2 → NaCl
2. Na3PO4 + KOH → NaOH + K3PO4
3. P4 + O2 → P2O3
4. N2 + H2 → NH3
5. Al + HCl → H2 + AlCl3
6. H2O2 → H2O + O2
7. NH3 + CuO → Cu + N2 + H2O
8. NH4NO3 → N2O + H2O
Q11.3.3
What do the products of combustion reactions have in common?
Q11.3.4
Write and balance combustion reactions for the following compounds.
1. methane (CH4)
2. propane (C3H8)
3. octane (C8H18)
4. ethanol (CH3CH2OH)
5. sucrose (C12H22O11)
(click here for solutions)
Q11.4.1
In a certain reaction, the energy of the reactants is less than the energy of the products (reaction consumes energy). Is the reaction endothermic or exothermic?
Q11.4.2
What are the two driving forces for all chemical reactions and physical processes?
Q11.4.3
Does entropy determine the spontaneity of a reaction? Does enthalpy determine the spontaneity of a reaction?
Q11.4.4
How does an increase in temperature affect the entropy of a system?
Q11.4.5
Which system has the greater entropy in each of the following?
1. solid sodium chloride or a sodium chloride solution
2. bromine liquid or bromine vapor
3. 25 g of water at 80°C or 25 g of water at 50°C
4. liquid mercury or solid mercury
Q11.4.6
How does the entropy of a system change for each of the following processes?
1. A solid melts.
2. A liquid freezes.
3. A liquid boils.
4. A vapor condenses to a liquid.
5. Sugar dissolves in water.
6. A solid sublimes.
(click here for solutions)
Q11.5.1
What is true about the relative amounts of reactants and products at the end of a spontaneous reaction?
Q11.5.2
Can a proposed reaction be spontaneous and yet still not be observed to occur? Explain.
Q11.5.3
The forward reaction is spontaneous for a particular reversible reaction. What can you conclude about the spontaneity of the reverse reaction?
Q11.5.4
Explain how free energy is used to determine whether or not a reaction is spontaneous.
Q11.5.5
Under what conditions of enthalpy and entropy change is a reaction always spontaneous? Under what conditions is a reaction never spontaneous?
Q11.5.6
If the enthalpy change is unfavorable (endothermic), but the entropy change is favorable (increasing), would a high temperature or a low temperature be more likely to lead to a spontaneous reaction?
Q11.5.7
If the enthalpy change is favorable and the entropy change is favorable, would the reaction be spontaneous at high temperatures, low temperatures, or all temperatures?
(click here for solutions)
Q11.6.1
In what unit is the rate of a chemical reaction typically expressed?
Q11.6.2
A 2.50 M solution undergoes a chemical reaction. After 3.00 minutes, the concentration of the solution is 2.15 M. What is the rate of change in M/s?
Q11.6.3
Substance A disappears at a rate of 0.0250 M/s. If the initial concentration is 4.00 M, what is the concentration after one minute?
Q11.6.4
The concentration of product B increases from 0 to 1.75 M in 45 seconds. What is the rate of formation of B?
Q11.6.5
The concentration of product B increases from 0.50 M to 1.25 M in 2.5 seconds. What is the rate of formation of B?
Q11.6.6
Reactant B goes from 2.25 M to 1.50 M in 0.85 seconds. What is the rate of change of B?
Q11.6.7
Does every collision between reacting particles lead to the formation of products? Explain.
Q11.6.8
What two conditions must be met in order for a collision to be effective?
Q11.6.9
Explain why the activation energy of a reaction is sometimes referred to as a barrier.
Q11.6.10
Why is it difficult to study activated complexes?
Q11.6.11
Explain how reaction rates can be affected by
1. changes in concentration.
2. changes in pressure.
3. increased surface area.
4. changes in temperatures.
Q11.6.12
What is the effect of a catalyst on the rate of a reaction?
Q11.6.13
Explain how the presence of a catalyst affects the activation energy of a reaction.
Q11.6.14
Zinc metal reacts with hydrochloric acid. Which one would result in the highest rate of reaction?
1. A solid piece of zinc in 1 M HCl
2. A solid piece of zinc in 3 M HCl
3. Zinc powder in 1 M HCl
4. Zinc powder in 3 M HCl
Q11.6.15
Use the potential energy diagram below to answer the following questions.
1. What is the potential energy of the reactants?
2. What is the potential energy of the products?
3. What is the heat of reaction (ΔH = Eproducts − Ereactants)?
4. What is the potential energy of the activated complex?
5. What is the activation energy for the reaction?
6. Is the reaction endothermic or exothermic?
7. Which of the values in a-e above would be changed by the use of a catalyst in the reaction?
8. What is the activation energy of the reverse reaction?
9. What is the heat of reaction (ΔH = Eproducts − Ereactants) of the reverse reaction?
Q11.6.16
Use the potential energy diagram below to answer the following questions.
1. What is the potential energy of the reactants?
2. What is the potential energy of the products?
3. What is the heat of reaction (ΔH = Eproducts − Ereactants)?
4. What is the potential energy of the activated complex?
5. What is the activation energy for the reaction?
6. Is the reaction endothermic or exothermic?
7. Which of the values in a-e above would be changed by the use of a catalyst in the reaction?
8. What is the activation energy of the reverse reaction?
9. What is the heat of reaction (ΔH = Eproducts − Ereactants) of the reverse reaction?
Answers
11.1: Oxidation Numbers
Q11.1.1
Any element by itself, either monatomic or diatomic, without a charge. The oxidation number of any free element is zero.
Q11.1.2
Sulfur can have an oxidation number up to +6 or as low as $-2$.
Q11.1.3
1. $\overset{+1}{\text{K}}\overset{+7}{\text{Mn}}\overset{-2}{\text{O}_4}$
• Rule 2: K forms an ion with a 1+ charge so its oxidation number is +1.
• Rule 3: O has an oxidation number of $-2$.
• Rule 6 for Mn: $1+x+4(-2)=0\ 1+x+-8=0\ x-7=0\ x=7$
2. $\overset{-2}{\text{O}}\overset{+1}{\text{Cl}_2}$
• Rule 3: O has an oxidation number of $-2$.
• Rule 6 for Cl: $-2+2x=0\ 2x=-2\ x=1$
3. $\overset{+1}{\text{H}_2}\overset{+3}{\text{C}_2}\overset{-2}{\text{O}_4}$
• Rule 3: O has an oxidation number of $-2$.
• Rule 4: H has an oxidation number of \{+1\).
• Rule 6 for C: $2(+1)+2x+4(-2)=0\ 2+2x-8=0\ 2x-6=0\ 2x=6\x=3$
4. $\overset{+1}{\text{Li}_3}\overset{+5}{\text{P}}\overset{-2}{\text{O}_4}$
• Rule 2: Li forms an ion with a 1+ charge so its oxidation number is +1.
• Rule 3: O has an oxidation number of $-2$.
• Rule 6 for P: $3(+1)+x+4(-2)=0\ 3+x+-8=0\ x-5=0\ x=5$
5. $\overset{+1}{\text{Na}}\overset{+1}{\text{Cl}}\overset{-2}{\text{O}}$
• Rule 2: Na forms an ion with a 1+ charge so its oxidation number is +1.
• Rule 3: O has an oxidation number of $-2$.
• Rule 6 for Cl: $1+x+-2=0\ x-1=0\ x=1$
6. $\overset{0}{\text{Br}_2}$
• Rule 1: Oxidation number of a free element is zero.
7. $\overset{+3}{\text{Cl}} \overset{-1}{\text{F}_3}$
• Rule 5: F has an oxidation number of $-1$.
• Rule 6 for Cl: $x+3(-1)=0\ x-3=0\ x=3$
8. $\overset{+2}{\text{Ca}}\overset{-1}{\text{Cl}_2}$
• Rule 2: Ca forms an ion with a 2+ charge so its oxidation number is +2.
• Rule 6 for Cl: $2+2(x)=0\ 2x=-2\ x=-1$
9. $\overset{+1}{\text{K}_2}\overset{-2}{\text{O}}$
• Rule 2: K forms an ion with a 1+ charge so its oxidation number is +1.
• Rule 3: O has an oxidation number of $-2$.
Q11.1.4
Determine the oxidation number of each of the atoms in the following ions.
1. $\overset{+3}{\text{N}}\overset{-2}{\text{O}^-_2}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for N: $x+2(-2)=-1\ x-4=-1\ x=3$
2. $\overset{+5}{\text{N}}\overset{-2}{\text{O}^-_3}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for N: $x+3(-2)=-1\ x-6=-1\ x=5$
3. $\overset{+6}{\text{Cr}_2}\overset{-2}{\text{O}^{2-}_7}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for Cr: $2x+7(-2)=-2\ 2x-14=-2\ 2x=12\ x=6$
4. $\overset{+5}{\text{Br}}\overset{-2}{\text{O}^-_3}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for Br: $x+3(-2)=-1\ x-6=-1\ x=5$
5. $\overset{+5}{\text{Cl}}\overset{-2}{\text{O}^-_3}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for Cl: $x+3(-2)=-1\ x-6=-1\ x=5$
6. $\overset{+3}{\text{B}}\overset{-2}{\text{O}^-_3}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for B: $x+3(-2)=-3\ x-6=-3\ x=3$
7. $\overset{+4}{\text{C}}\overset{-2}{\text{O}^{2-}_3}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for C: $x+3(-2)=-2\ x-6=-2\ x=4$
8. $\overset{-3}{\text{N}}\overset{+1}{\text{H}^+_4}$
• Rule 2: H has an oxidation number of $+1$.
• Rule 7 for N: $x+4(+1)=+1\ x+4=1\ x=-3$
9. $\overset{+6}{\text{Cr}}\overset{-2}{\text{O}^{2-}_4}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for Cr: $x+4(-2)=-2\ x-8=-2\ x=6$
10. $\overset{+2}{\text{S}_2}\overset{-2}{\text{O}^{2-}_3}$
• Rule 2: O has an oxidation number of $-2$.
• Rule 7 for S: $2x+3(-2)=-2\ 2x-6=-2\ 2x=4\ x=2$
11.2: The Nature of Oxidation and Reduction
Q11.2.1
Oxidation and reduction are opposite processes. Electrons are lost by one substance and gained by another so they have a transfer of electrons.
Q11.2.2
The oxidizing agent is reduced.
Q11.2.3
The reducing agent is oxidized.
Q11.2.4
1. oxidation
2. reduction
3. reduction
4. oxidation
5. oxidation
6. reduction
7. reduction
8. oxidation
Q11.2.5
1. 2KClO3(s) → 2KCl(s) + 3O2(g)
1. $\overset{+1}{2\text{K}}\overset{+5}{\text{Cl}}\overset{-2}{\text{O}_3}\rightarrow \overset{+1}{2\text{K}}\overset{-1}{\text{Cl}}~+~\overset{0}{3\text{O}_2}$
2. redox
3. oxygen is being oxidized, chlorine is being reduced
4. KClO3 is the oxidizing and reducing agent (specifically, the oxygen is the reducing agent and the chlorine is the oxidizing agent)
2. H2(g) + CuO(s) → Cu(s) + H2O(l)
1. $\overset{0}{\text{H}_2}~+~\overset{+2}{\text{Cu}}\overset{-2}{\text{O}}\rightarrow \overset{0}{\text{Cu}}~+~\overset{+1}{\text{H}_2}\overset{-2}{\text{O}}$
2. redox
3. hydrogen is being oxidized, copper is being reduced
4. CuO is the oxidizing agent, H2 is the reducing agent
3. 2Al(s) + 3Mg(NO3)2(aq) → 2Al(NO3)3(aq) + 3Mg(s)
1. $overset{0}{2\text{Al}}~+~\overset{+2}{3\text{Mg}}\overset{+5} {\text{(N}}\overset{-2}{\text{O}_3\text{)}_2} \rightarrow \overset{+3}{2\text{Al}}\overset{+5} {\text{(N}}\overset{-2}{\text{O}_3\text{)}_3}~+~\overset{0}{3\text{Mg}}$
2. redox
3. aluminum is being oxidized, magnesium is being reduced
4. Mg(NO3)2 is the oxidizing agent, Al is the reducing agent
4. 2HNO3(aq) + 6HI(aq) → 2NO(g) + 3I2(s) + 4H2O(l)
1. $\overset{+1}{2\text{H}}\overset{+5}{\text{N}}\overset{-2}{\text{O}_3}~+~\overset{+1}{6\text{H}}\overset{-1}{\text{I}} \rightarrow \overset{+2}{2\text{N}}\overset{-2}{\text{O}}~+~\overset{0} {3\text{I}_2}~+~\overset{+1}{4\text{H}_2}\overset{-2}{\text{O}}$
2. redox
3. iodine is being oxidized, nitrogen is being reduced
4. HNO3 is the oxidizing agent, HI is the reducing agent
5. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
1. $\overset{+1}{\text{Ag}}\overset{+5}{\text{N}}\overset{-2}{\text{O}_3}~+~\overset{+1}{\text{Na}}\overset{-1}{\text{Cl}} \rightarrow \overset{+1}{\text{Ag}}\overset{-1}{\text{Cl}}~+~\overset{+1}{\text{Na}}\overset{+5}{\text{N}}\overset{-2}{\text{O}_3}$
2. not redox
6. 2FeCl3(aq) + H2S(g) → 2FeCl2(aq) + 2HCl(aq) + S(s)
1. $\overset{+3}{2\text{Fe}}\overset{-1}{\text{Cl}_3}~+~\overset{+1}{\text{H}_2}\overset{-2}{\text{S}} \rightarrow \overset{+2}{2\text{Fe}}\overset{-1}{\text{Cl}_2}~+~\overset{+1}{2\text{H}}\overset{-1}{\text{Cl}}~+~\overset{0}{\text{S}}$
2. redox
3. sulfur is being oxidized, iron is being reduced
4. FeCl3 is the oxidizing agent, H2S is the reducing agent
Q11.2.6
1. Oxidation due to loss of hydrogen.
2. Reduction due to loss of oxygen.
3. Reduction due to loss of oxygen or gain of hydrogen.
4. Reduction due to gain of hydrogen.
5. Reduction due to gain of hydrogen.
11.3: Types of Chemical Reactions
Q11.3.1
1. single replacement (single displacement), redox
2. combination (synthesis), redox
3. combustion, redox
4. decomposition, redox
5. double replacement (double displacement), not redox
Q11.3.2
1. 2Na + Cl2 → 2NaCl, combination (synthesis), redox
2. Na3PO4 + 3KOH → 3NaOH + K3PO4, double replacement (double displacement), not redox
3. P4 + 3O2 → 2P2O3, combination (synthesis), redox
4. N2 + 3H2 → 2NH3, combination (synthesis), redox
5. 2Al + 6HCl → 3H2 + 2AlCl3, single replacement (single displacement), redox
6. 2H2O2 → 2H2O + O2, decomposition, redox
7. 2NH3 + 3CuO → 3Cu + N2 + 3H2O, decomposition, redox
8. NH4NO3 → N2O + 2H2O, decomposition, not redox
Q11.3.3
Combustion produces CO2 and H2O.
Q11.3.4
1. CH4 + 2O2 → CO2 + 2H2O
2. C3H8 + 5O2 → 3CO2 + 4H2O
3. 2C8H18 + 25O2 → 16CO2 + 18H2O
4. CH3CH2OH + 3O2 → 2CO2 + 3H2O
5. C12H22O11 + 12O2 → 12CO2 + 11H2O
11.4: Entropy
Q11.4.1
endothermic
Q11.4.2
enthalpy and entropy
Q11.4.3
Neither entropy nor enthalpy determine the spontaneity of a reaction but both contribute to determining the spontaneity.
Q11.4.4
Entropy increases with temperature.
Q11.4.5
Which system has the greater entropy in each of the following?
1. solution
2. vapor
3. 80°C
4. liquid
Q11.4.6
How does the entropy of a system change for each of the following processes?
1. increases
2. decreases
3. increasese
4. decreases
5. increases
6. increases
11.5: Spontaneous Reactions and Free Energy
Q11.5.1
There are more products than reactants.
Q11.5.2
Yes, because it is a very slow reaction.
Q11.5.3
The reverse reaction is not spontaneous.
Q11.5.4
Free energy ($\Delta G$) is negative for spontaneous reactions and positive for non-spontaneous reactions.
Q11.5.5
$\Delta G$ is negative (spontaneous) at all temperatures when $\Delta H$ is negative and $\Delta S$ is positive.
$\Delta G$ is positive (non-spontaneous) at all temperatures when $\Delta H$ is positive and $\Delta S$ is negative.
Q11.5.6
high temperature
Q11.5.7
all temperatures
11.6: Rates of Reactions
Q11.6.1
molarity per second, $\frac{M}{s}$ (may be written as $Ms^{-1}$)
Q11.6.2
$rate=\frac{\Delta [A]}{\Delta t}\=\frac{2.15-2.50\;M}{180. \;s}\ =\frac{-0.35\;M}{180.\;s}\ =-0.0019\;\frac{M}{s}$
Q11.6.3
$60\;s\left(\frac{0.0250\;M}{s}\right)=1.50\;M\;\text{lost}\ initial + change =final\ 4.00\;M+(-1.50\;M)=2.50\;M$
Q11.6.4
$rate=\frac{\Delta [B]}{\Delta t}\=\frac{1.75-0\;M}{45 \;s}\ =0.039\;\frac{M}{s}$
Q11.6.5
$rate=\frac{\Delta [B]}{\Delta t}\=\frac{1.25-0.50\;M}{2.5 \;s}\ =\frac{0.75\;M}{2.5\;s}\ =0.30\;\frac{M}{s}$
Q11.6.6
$rate=\frac{\Delta [B]}{\Delta t}\=\frac{1.50-2.25\;M}{0.85 \;s}\ =\frac{-0.75\;M}{0.85\;s}\ =-0.88\;\frac{M}{s}$
Q11.6.7
No, because collisions have to occur at the correct orientation with sufficient energy.
Q11.6.8
Correct orientation and sufficient energy.
Q11.6.9
Because it is the energy requirement that must be met before the reaction can occur. When there is not enough energy, the reaction is blocked from occurring.
Q11.6.10
They are very short-lived.
Q11.6.11
1. Increasing concentration increases the probability of an effective collision so the reaction rate will increase. The reverse is true for a decrease in concentration.
2. Change in pressure of a substance involved in the reaction will have the same effect as a change in concentration.
3. Increasing the surface area creates more places for reactants to interact which will increase the probability of an effective collision so the reaction rate will increase.
4. Increasing the temperature will increase the energy of collisions so a greater number of collisions will have sufficient energy to overcome the activation barrier which will result in an increase in the reaction rate. Decreasing the temperature will have the reverse effect.
Q11.6.12
Catalysts increase the rate of a reaction by lowering the activation energy.
Q11.6.13
Catalysts provide an alternate mechanism or "path" for the reaction to occur. This new mechanism has a lower activation energy so more collisions have enough energy to overcome the barrier which will increase the reaction rate.
Q11.6.14
d. The powder has a higher surface area than a solid piece of zinc and the higher concentration will result in a higher reaction rate.
Q11.6.15
1. 20 kJ/mol
2. 50 kJ/mol
3. 30 kJ/mol
4. 100 kJ/mol
5. 80 kJ/mol
6. endothermic
7. d (energy of the activated complex) and e (activation energy)
8. 50 kJ/mol
9. $-$30 kJ/mol
Q11.6.16
1. 45 kJ/mol
2. 15 kJ/mol
3. $-$30 kJ/mol
4. 92 kJ/mol
5. 47 kJ/mol
6. exothermic
7. d (energy of the activated complex) and e (activation energy)
8. 77 kJ/mol
9. 30 kJ/mol | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Chemistry_for_Allied_Health_(Soult)/11.E%3A_Properties_of_Reactions_%28Exercises%29.txt |
These are homework exercises to accompany Chapter 12 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Answers are below the questions.
Questions
(click here for solutions)
Q12.1.1
Explain the difference between hydrolysis and condensation reactions.
Q12.1.2
What is being added in a hydrogenation reaction? Hydration? Bromination?
Q12.1.3
What is the product of the oxidation of a primary alcohol? A secondary alcohol?
Q12.1.4
Why is a condensation reaction also called a dehydration reaction?
Q12.1.5
Draw the structure for the primary product of each reaction..
1. CH3CH=CH2 + Br2
2. CH3CH2CH=CH2 + H2
3. CH3CH=CHCH3 + H2O →
4. CH3CH2CH=CH2 + H2O →
5. CH3CH2CH=CHCH3 + H2O →
Q12.1.6
Draw the structure of the compound that would be formed from the single oxidation of each compound.
a.
b.
c.
Q12.1.7
Draw the structure of the compound that would be formed from the complete oxidation of each compound.
a.
b.
c.
Q12.1.8
Draw the structure of the compound that would be formed from a single reduction of each compound.
a.
b.
c.
Q12.1.9
Draw the product(s) for the hydrolysis of the given reactant. The red bond in the reactant is the one that is broken.
Q12.1.10
Draw the product(s) for the condensation of the given reactants. The layout of the molecules indicates where the bond will form between them. If there are three molecules, two condensations will occur to produce one final product.
Q12.1.11
Compare the reactants and products in part d and e of 12.1.10. Are the reactants the same or different? Are the products the same or different? Why?
Answers
12.1: Organic Reactions
Q12.1.1
Hydrolysis is a molecule breaking apart after the addition of water. Condensation is two molecules coming together and producing H2O.
Q12.1.2
hydrogenation = adding hydrogen (H2)
hydration = adding water
bromination = adding bromine (Br2)
Q12.1.3
A primary alcohol oxidizes to an aldehyde and then to a carboxylic acid. A secondary alcohol oxidizes to a ketone.
Q12.1.4
Because water is released during a condensation reaction.
Q12.1.5
a.
Each bromine atom is added to one of the carbons in the double bond.
b.
CH3CH2CH2CH3
Each hydrogen atom is added to one of the carbons in the double bonds.
c.
Water is added as H and OH. The carbons in the double bond each have the same number of hydrogen atoms so Markovnikov's rule does not apply. The molecule is symmetrical so adding the OH group to either carbon results in the same molecule.
d.
Water is added as H and OH. The carbon atoms in the double bond have different number of hydrogen atoms so Markovnikov's rule does apply. The H is added to the carbon with more hydrogen atoms and the OH is added to the carbon with fewer hydrogen atoms.
e.
and
Water is added as H and OH. The carbon atoms in the double bond have the same number of hydrogen atoms on each so Markovnikov's rule does not apply. There are two possible products because adding the OH to the carbon on the left side of the double bond results in a different molecule than adding the OH to the carbon on the right side of the double bond. This differs from c because the molecule is not symmetrical.
Q12.1.6
a.
b.
c.
Q12.1.7
a.
b.
c.
Q12.1.8
a.
b.
c.
Q12.1.9
a.
b.
c.
d.
Q12.1.10
a.
b.
c.
d.
e.
Q12.1.11
The reactants are the same, but the products are different because they are combined in a different order. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Chemistry_for_Allied_Health_(Soult)/12.E%3A_Organic_Reactions.txt |
These are homework exercises to accompany Chapter 13 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health.
Questions
(click here for solutions)
Q13.1.1
Read the material at http://hyperphysics.phy-astr.gsu.edu/hbase/organic/essam.html and answer the following questions:
a. What are essential amino acids?
b. What are nonessential amino acids?
c. What happens if you are deficient in an amino acid?
Q13.1.2
Draw the functional groups present in all amino acids.
Q13.1.3
Complete the following for threonine, lysine, and tyrosine.
a. Draw the amino acid.
b. Circle the side chain.
c. Identify whether it is polar, nonpolar, acidic, or basic.
d. At what pH will it exist as a zwitterion?
e. What is the range of pH values when it will be positively charged?
f. What is the range of pH values when it will be negatively charged?
(click here for solutions)
Q13.2.1
Draw the two dipeptides formed from each pair of amino acids.
a. tyrosine and lysine
b. threonine and gluatmine
c. alanine and histidine
Q13.2.2
Draw and give the full names of the amino acids in the following dipeptides.
Q13.2.3
List of all of the possible polypeptides that can be formed from threonine, alanine, and phenylalanine (use three character abbreviations for each amino acid).
Q13.2.4
Draw the following polypeptides.
a. Ser-Tyr-Gln
b. Lys-Met-Gly
Q13.2.5
Identify each of the amino acids in the polypeptide and then name it using the three character abbreviations.
(click here for solutions)
Q13.3.1
Describe the four levels of protein structure.
Q13.3.2
What levels of structure involve hydrogen bonding?
Q13.3.3
What types of structure is the result of interactions between amino acids that are far apart in the primary structure?
Q13.3.4
What types of interactions hold the secondary structure together?
Q13.3.5
What types of interactions hold the tertiary structure together?
Q13.3.6
What levels of structure are affected by denaturation?
Q13.3.7
A protein has one subunit. Would it have a quaternary structure?
Answers
13.1: Amino Acids
Q13.1.1
a. Essential amino acids are those you get from your diet.
b. Nonessential amino acids are produced in the body.
c. Illness and/or degradation of body's proteins.
Q13.1.2
amine and carboxylic acid
Q13.1.3
Complete the following for threonine, lysine, and tyrosine.
threonine
1. polar
2. 5.60
3. < 5.60
4. > 5.60
lysine
1. basic
2. 9.47
3. < 9.47
4. > 9.47
tyrosine
1. polar
2. 5.63
3. < 5.63
4. > 5.63
13.2: Peptides
Q13.2.1
Draw the two dipeptides formed from each pair of amino acids.
a.
b.
c.
Q13.2.2
a. alanine glycine
b. proline phenylalanine
c. tryptophan lysine
Q13.2.3
Thr-Ala-Phe
Thr-Phe-Ala
Ala-Thr-Phe
Ala-Phe-Thr
Phe-Ala-Thr
Phe-Thr-Ala
Q13.2.4
a.
b.
Q13.2.5
Arg-His-Thr-Glu-Ser
13.3: Protein Structure
Q13.3.1
Primary - sequence of amino acids
Secondary - alpha helix and Beta-pleated sheets held together by hydrogen bonds
Tertiary - third level of structure of protein often forming globular or fibrous structure, held together by variety of attractive forces
Quaternary - complex of multiple proteins held together to function as one, held together by variety of attractive forces (same as tertiary)
Q13.3.2
secondary, tertiary, and quaternary structures
Q13.3.3
tertiary structures
Q13.3.4
hydrogen bonds
Q13.3.5
London dispersion forces, hydrogen bonds, dipole-dipole forces, ion-dipole interactions, salt bridges, and disulfide bonds
Q13.3.6
secondary, tertiary, and quaternary
Q13.3.7
No, a quaternary structure must have multiple subunits. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Chemistry_for_Allied_Health_(Soult)/13.E%3A_Amino_Acids_and_Proteins_%28Exercises%29.txt |
These are homework exercises to accompany Chapter 14 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Answers are below the questions.
Questions
(click here for solutions)
Q14.1.1
What do enzymes do in the body?
Q14.1.2
Describe each of the following.
1. active site
2. substrate
3. allosteric site
4. inhibitor
Q14.1.3
How are competitive and noncompetitive inhibitors similar and different from one another?
Q14.1.4
An inhibitor interacts with the enzyme at an allosteric site. Is it competitive or noncompetitive?
Q14.1.5
Describe each of the following enzyme-substrate pairs as using the lock-and-key model or the induced fit model.
Q14.1.6
Each of the following behaves as a cofactor or a coenzyme. Identify each.
1. Zn2+
2. Vitamin B12
3. biotin
4. Fe3+
Vitamin B12
biotin
Q14.1.7
Identify each statement as true or false. Correct the false statement(s).
1. Enzyme activity increases with temperature.
2. Enzyme activity depends on pH.
3. Enzymes are consumed in a chemical reaction.
4. Increasing the concentration will increase the rate of a reaction until all of the enzyme is "occupied".
5. Enzymes can only function at their ideal temperature and pH.
Q14.1.8
The enzyme activity is graphed with respect to the pH of the mixture. Determine the pH at which each enzyme is most effective.
(click here for solutions)
Q14.2.1
What is the functional group in a fatty acid?
Q14.2.2
What is the difference between a fat and an oil?
Q14.2.3
Butter is a fat that is a solid at room temperature. What type of fatty acid does butter contain? How do you know?
Q14.2.4
Describe the difference between saturated and unsaturated fatty acids.
Q14.2.5
Explain why molecules of saturated and unsaturated fatty acids have different shapes.
Q14.2.6
Draw each structure.
1. Saturated fatty acid with 18 carbon atoms.
2. Monounsaturated fatty acid with 14 carbon atoms.
3. Polyunsaturated fatty acid with 14 carbon atoms.
4. Monounsaturated with 16 carbon atoms.
5. Polyunsaturated fatty acid with 18 carbon atoms and three double bonds.
Q14.2.7
Where does the body get essential fatty acids?
Q14.2.8
What molecules react to form a triglyceride?
Q14.2.9
Draw a triglyceride formed from three identical fatty acids.
Q14.2.10
Draw a triglyceride formed from three different fatty acids.
(click here for solutions)
Q14.3.1
What is a phospholipid?
Q14.3.2
Which part of the phospholipid molecule is water-soluble?
Q14.3.3
Which part is not water-soluble?
Q14.3.4
What is the purpose of a semipermeable membrane like the cell membrane?
Answers
14.1: Enzymes
Q14.1.1
Enzymes act as catalysts for biological reactions.
Q14.1.2
Describe each of the following.
1. The active site is where the reaction occurs in the enzyme.
2. The substrate is the molecule that interacts with the enzyme.
3. An allosteric site is on the enzyme away from the active site. Inhibitors can interactive with the enzyme at the allosteric site.
4. An inhibitor is a molecule that interacts with the molecule to slow or stop a reaction.
Q14.1.3
Both competitive and noncompetitive inhibitors slow or stop a reaction. Competitive inhibitors bind with the active site and non-competitive inhibitors bind with an allosteric site.
Q14.1.4
noncompeititve
Q14.1.5
1. lock & key
2. lock & key
3. induced fit
Q14.1.6
1. cofactor
2. coenzyme
3. coenzyme
4. cofactor
Q14.1.7
Identify each statement as true or false. Correct the false statement(s).
1. Enzyme activity increases with temperature and then decreases with increasing temperature beyond a peak temperature.
2. True
3. Enzymes are not consumed in a chemical reaction.
4. True
5. Enzymes
6. can only function most effectively at their ideal temperature and pH, but can function at a range of temperature and pH values.
Q14.1.8
1. pH = 8
2. pH = 6
14.2: Lipids and Triglycerides
Q14.2.1
carboxylic acid
Q14.2.2
Fats are solids at room temperatures while oils are liquids.
Q14.2.3
Saturated fatty acid.
Q14.2.4
Saturated fatty acids contain only single bonds between carbon atoms while unsaturated fatty acids contain one or more double bonds.
Q14.2.5
Saturated fatty acids have a straight chain of carbon atoms while unsaturated fatty acids have a bend at every double bond.
Q14.2.6
a.
b.
Answers will vary. The double bond may be located between any two carbon atoms.
c.
Answers will vary. There should be multiple double bonds and can be located between any pairs of carbon atoms.
d.
Answers will vary. The double bond may be located between any two carbon atoms.
e.
Answers will vary. There should be three double bonds and can be located between any pairs of carbon atoms.
Q14.2.7
Essential fatty acids come from the food we eat.
Q14.2.8
Glycerol and three fatty acids from a triglyceride.
Q14.2.9
Answers will vary but each "tail" should have the same number of carbon atoms.
Q14.2.10
Answers will vary but each "tail" should have a different number of carbon atoms.
14.3: Phospholipids in Cell Membranes
Q14.3.1
A phospholipid is a triglyceride where one of the fatty acid tails is replaced by a phosphate group.
Q14.3.2
The phosphate end is water-soluble.
Q14.3.3
The hydrocarbon tails are not water-soluble.
Q14.3.4
To control the flow of substances in and out of the cell.
15.E: Metabolic Cycles (Exercises)
These are homework exercises to accompany Chapter 15 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health.
15.1: Glycolysis
Q15.1.1
What are the three stages of cellular respiration?
Q15.1.2
What is the purpose of glycolysis?
Q15.1.3
What is the output of glycolysis from a single glucose molecule?
Q15.1.4
How many molecules of ATP are "invested" in glycolysis? How many are produced?
Q15.1.5
Define aerobic and anaerobic.
15.2: The Citric Acid Cycle
Q15.2.1
Where does the Krebs cycle occur in the cell?
Q15.2.2
What happens to the pyruvate produced during glycolysis?
Q15.2.3
How many reactions does it take to complete the cycle?
Q15.2.4
How many "turns" of the citric acid cycle must occur for each molecule of glucose entering glycolysis?
Q15.2.5
What is he output of the citric acid cycle?
Q15.2.6
Trace the six carbon atoms originally from acetyl-CoA through the Krebs Cycle. Trace the flow of energy from the pyruvates produced in glycolysis through the Krebs Cycle.
Q15.2.7
How many energy carriers are produced during the Krebs cycle per acetyl-CoA?
15.3: Lactic Acid Fermentation
Q15.3.1
What is fermentation?
Q15.3.2
Define lactic acid fermentation.
Q15.3.3
Identify yourself as a “sprinter” or an “endurance runner” and predict the type of muscle fiber (red or white) which predominates in your body. Explain your reasoning.
Q15.3.4
What is the chemical equation of lactic acid fermentation?
15.4: The Electron Transport Chain
Q15.4.1
What molecules "feed" the electron transport chain?
Q15.4.2
What is the primary product of the electron transport chain?
Q15.4.3
Where do the reactions of the electron transport chain occur? | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_Chemistry_for_Allied_Health_(Soult)/14.E%3A_Biological_Molecules_%28Exercises%29.txt |
These are homework exercises to accompany the Ball et al. "The Basics of GOB Chemistry" Textmap.
Exercises: General Organic and Biological Chemistry (Ball et al.)
These are homework exercises to accompany Chapter 1 of the Ball et al. "The Basics of GOB Chemistry" Textmap.
1.1: What Is Chemistry?
1. Based on what you know, which fields are branches of science?
1. meteorology (the study of weather)
2. astrophysics (the physics of planets and stars)
3. economics (the study of money and monetary systems)
4. astrology (the prediction of human events based on planetary and star positions)
5. political science (the study of politics)
2. Based on what you know, which fields are a branches of science?
1. history (the study of past events)
2. ornithology (the study of birds)
3. paleontology (the study of fossils)
4. zoology (the study of animals)
5. phrenology (using the shape of the head to determine personal characteristics)
3. Which of the following are examples of matter?
1. a baby
2. an idea
3. the Empire State Building
4. an emotion
5. the air
6. Alpha Centauri, the closest known star (excluding the sun) to our solar system
4. Which of the following are examples of matter?
1. your textbook
2. brain cells
3. love
4. a can of soda
5. breakfast cereal
5. Suggest a name for the science that studies the physics of rocks and the earth.
6. Suggest a name for the study of the physics of living organisms.
7. Engineering is the practical application of scientific principles and discoveries to develop things that make our lives easier. Is medicine science or engineering? Justify your answer.
8. Based on the definition of engineering in Exercise 7, would building a bridge over a river or road be considered science or engineering? Justify your answer.
9. When someone says, “I have a theory that excess salt causes high blood pressure,” does that person really have a theory? If it is not a theory, what is it?
10. When a person says, “My hypothesis is that excess calcium in the diet causes kidney stones,” what does the person need to do to determine if the hypothesis is correct?
11. Some people argue that many scientists accept many scientific principles on faith. Using what you know about the scientific method, how might you argue against that assertion?
12. Most students take multiple English classes in school. Does the study of English use the scientific method?
Answers
1.
1. science
2. science
3. not science
4. not science
5. not science
2.
1. not science
2. science
3. science
4. science
5. not science
1.
1. matter
2. not matter
3. matter
4. not matter
5. matter
6. matter
4.
• matter
• matter
• not matter
• matter
• matter
1. geophysics
6. biophysics
1. Medicine is probably closer to a field of engineering than a field of science, but this may be arguable. Ask your doctor.
8. Engineering
1. In scientific terms, this person has a hypothesis.
2. Conduct experiments to determine if kidney stones contain calcium.
3. Science is based on reproducible facts, not blind belief.
4. No.
Exercises
1. Does each statement refer to a chemical property or a physical property?
1. Balsa is a very light wood.
2. If held in a flame, magnesium metal burns in air.
3. Mercury has a density of 13.6 g/mL.
4. Human blood is red.
2. Does each statement refer to a chemical property or a physical property?
1. The elements sodium and chlorine can combine to make table salt.
2. The metal tungsten does not melt until its temperature exceeds 3,000°C.
3. The ingestion of ethyl alcohol can lead to disorientation and confusion.
4. The boiling point of isopropyl alcohol, which is used to sterilize cuts and scrapes, is lower than the boiling point of water.
3. Define element. How does it differ from a compound?
4. Define compound. How does it differ from an element?
5. Give two examples of a heterogeneous mixture.
6. Give two examples of a homogeneous mixture.
7. Identify each substance as an element, a compound, a heterogeneous mixture, or a solution.
1. xenon, a substance that cannot be broken down into chemically simpler components
2. blood, a substance composed of several types of cells suspended in a salty solution called plasma
3. water, a substance composed of hydrogen and oxygen
8. Identify each substance as an element, a compound, a heterogeneous mixture, or a solution.
1. sugar, a substance composed of carbon, hydrogen, and oxygen
2. hydrogen, the simplest chemical substance
3. dirt, a combination of rocks and decaying plant matter
9. Identify each substance as an element, a compound, a heterogeneous mixture, or a solution.
1. air, primarily a mixture of nitrogen and oxygen
2. ringer’s lactate, a standard fluid used in medicine that contains salt, potassium, and lactate compounds all dissolved in sterile water
3. tartaric acid, a substance composed of carbon, hydrogen, and oxygen
10. Identify each material as an element, a compound, a heterogeneous mixture, or a solution.
1. equal portions of salt and sand placed in a beaker and shaken up
2. a combination of beeswax dissolved in liquid hexane
3. hydrogen peroxide, a substance composed of hydrogen and oxygen
11. What word describes each phase change?
1. solid to liquid
2. liquid to gas
3. solid to gas
12. What word describes each phase change?
1. liquid to solid
2. gas to liquid
3. gas to solid
Answers
1.
1. physical property
2. chemical property
3. physical property
4. physical property
2.
1. chemical property
2. physical property
3. chemical property
4. physical property
1. An element is a substance that cannot be broken down into chemically simpler components. Compounds can be broken down into simpler substances.
4. A compound is composed of two or more elements combined in a fixed ratio. An element is the simplest chemical substance.
1. a salt and pepper mix and a bowl of cereal (answers will vary)
6. vinegar and rubbing alcohol (answers will vary)
1.
1. element
2. heterogeneous mixture
3. compound
8.
1. compound
2. element
3. heterogeneous mixture
1.
1. solution
2. solution
3. compound
10.
1. heterogeneous mixture
2. solution
3. compound
11.
• melting or fusion
• boiling or evaporation
• sublimation
12.
1. freezing
2. condensation
3. deposition
Exercises
1. Why are both parts of a quantity important when describing it?
2. Why are measurements an important part of any branch of science, such as chemistry?
3. You ask a classmate how much homework your chemistry professor assigned. Your classmate answers, “twenty.” Is that a proper answer? Why or why not?
4. Identify the number and the unit in each quantity.
1. five grandchildren
2. 16 candles
3. four score and seven years
4. 40 days and 40 nights
5. 12.01 grams
6. 9.8 meters per second squared
7. 55 miles per hour
8. 98.6 degrees Fahrenheit
Answers
1. The number states how much, and the unit states of what. Without the number and the unit, a quantity cannot be properly communicated.
2. Measurements are needed to carry out experiments.
3. No, it is not a proper answer; you do not know whether the professor meant homework problem number 20 or 20 homework problems.
4.
1. The number is 5, and the unit is grandchildren.
2. The number is 16, and the unit is candles.
3. The number is 4 score and 7 (= 87), and the unit is years.
4. The number is 40, and the units are days and nights.
5. The number is 12.01, and the unit is grams.
6. The number is 9.8, and the unit is meters per second squared.
7. The number is 55, and the unit is miles per hour.
8. The number is 98.6, and the unit is degrees Fahrenheit.
Exercises
1. Why is scientific notation useful in expressing numbers?
2. What is the relationship between the power and the number of places a decimal point is moved when going from standard to scientific notation?
3. Express each number in scientific notation.
1. 0.00064
2. 5,230,000
3. −56,200
4. 0.000000000220
5. 1.0
4. Express each number in scientific notation.
1. 678
2. −1,061
3. 0.000560
4. 0.0000003003
5. 100,000,000
5. Express each number in standard form.
1. 6.72 × 104
2. 2.088 × 10−4
3. −3 × 106
4. 9.98 × 10−7
6. Express each number in standard form.
1. 9.05 × 105
2. 1.0 × 10−3
3. 6.022 × 1023
4. 8.834 × 10−12
7. Complete the following table:
7. Complete the following table
Incorrect Scientific Notation Correct Scientific Notation
54.7 × 104
0.0066 × 103
3,078 × 100
8. Complete the following table:
8. Complete the following table
Incorrect Scientific Notation Correct Scientific Notation
234.0 × 101
36 × 10−4
0.993 × 105
Answers
1. Scientific notation is more convenient than listing a large number of zeros.
2. The power is the number of places a decimal point is moved when going from standard to scientific notation. It is positive if the decimal point is moved to the left; negative if moved to the right.
1.
1. 6.4 × 10−4
2. 5.23 × 106
3. −5.62 × 104
4. 2.20 × 10−10
5. 1.0 × 100
4.
1. 6.78 × 102
2. −1.061 × 103
3. 5.60 × 10−4
4. 3.003 × 10−7
5. 1 × 108
1.
1. 67,200
2. 0.0002088
3. −3,000,000
4. 0.000000998
6. Express each number in standard form.
1. 905,000
2. 0.0010
3. 602,200,000,000,000,000,000,000
4. 0.000000000008834
1. Answers to question 7: Complete the table
Incorrect Scientific Notation Correct Scientific Notation
54.7 × 104 5.47 × 105
0.0066 × 103 6.6 × 100
3,078 × 100 3.078 × 103
8.
Answers to question 8: Complete the table
Incorrect Scientific Notation Correct Scientific Notation
234.0 × 101 2.340 × 103
36 × 10−4 3.6 × 10−3
0.993 × 105 9.93 × 104
Exercises
1. Define significant figures. Why are they important?
2. Define the different types of zeros found in a number and explain whether or not they are significant.
3. How many significant figures are in each number?
1. 140
2. 0.009830
3. 15,050
4. 221,560,000
5. 5.67 × 103
6. 2.9600 × 10−5
4. How many significant figures are in each number?
1. 1.05
2. 9,500
3. 0.0004505
4. 0.00045050
5. 7.210 × 106
6. 5.00 × 10−6
5. Round each number to three significant figures.
1. 34,705
2. 34,750
3. 34,570
6. Round each number to three significant figures.
1. 34,705
2. 34,750
3. 34,570
7. Perform each operation and express the answer to the correct number of significant figures.
1. 467.88 + 23.0 + 1,306 = ?
2. 10,075 + 5,822.09 − 34.0 = ?
3. 0.00565 + 0.002333 + 0.0991 = ?
8. Perform each operation and express the answer to the correct number of significant figures.
1. 0.9812 + 1.660 + 8.6502 = ?
2. 189 + 3,201.8 − 1,100 = ?
3. 675.0 − 24 + 1,190 = ?
9. Perform each operation and express the answer to the correct number of significant figures.
1. 439 × 8,767 = ?
2. 23.09 ÷ 13.009 = ?
3. 1.009 × 876 = ?
10. Perform each operation and express the answer to the correct number of significant figures.
1. 3.00 ÷ 1.9979 = ?
2. 2,300 × 185 = ?
3. 16.00 × 4.0 = ?
11. Use your calculator to solve each equation. Express each answer in proper scientific notation and with the proper number of significant figures. If you do not get the correct answers, you may not be entering scientific notation into your calculator properly, so ask your instructor for assistance.
1. (5.6 × 103) × (9.04 × 10−7) = ?
2. (8.331 × 10−2) × (2.45 × 105) = ?
3. 983.09 ÷ (5.390 × 105) = ?
4. 0.00432 ÷ (3.9001 × 103) = ?
12. Use your calculator to solve each equation. Express each answer in proper scientific notation and with the proper number of significant figures. If you do not get the correct answers, you may not be entering scientific notation into your calculator properly, so ask your instructor for assistance.
1. (5.2 × 106) × (3.33 × 10−2) = ?
2. (7.108 × 103) × (9.994 × 10−5) = ?
3. (6.022 × 107) ÷ (1.381 × 10−8) = ?
4. (2.997 × 108) ÷ (1.58 × 1034) = ?
Answers
1. Significant figures represent all the known digits plus the first estimated digit of a measurement; they are the only values worth reporting in a measurement.
2. Leading zeros serve as decimal place holders only, hence are not significant. Sandwiched zeros are significant. Trailing zeros following non-zero numbers without a decimal point is not significant. Trailing zeros following non-zero digits with a decimal point are significant.
1.
1. two
2. four
3. four
4. five
5. three
6. five
4.
1. three
2. two
3. four
4. five
5. four
6. three
1.
1. 34,700
2. 34,800
3. 34,600
6.
1. 34,710
2. 0.005411
3. 8.904 × 108
1.
1. 1,797
2. 15,863
3. 0.1071
8.
1. 11.291
2. 2,291
3. 1,841
9.
1. 3,850,000
2. 1.775
3. 884
10.
1. 1.50
2. 430,000
3. 64
11.
1. 5.1 × 10−3
2. 2.04 × 104
3. 1.824 × 10−3
4. 1.11 × 10−6
12.
1. 1.7 × 105
2. 7.104 × 10-1
3. 4.361 × 1015
4. 1.90 × 10-26
Exercises
1. List four base units.
2. List four derived units.
3. How many meters are in 1 km? How many centimeters are in 1 m?
4. How many grams are in 1 Mg? How many microliters are in 1 L?
5. Complete the following table:
5. Complete the following table
Unit Abbreviation
centiliter
ms
cm
kL
micrometer
6. Complete the following table:
6. Complete the following table
Unit Abbreviation
microliter
kilosecond
dL
ns
millimeter
7. What are some appropriate units for density?
8. A derived unit for velocity, which is the change of position with respect to time, is meters per second (m/s). Give three other derived units for velocity.
Answers
1. second, meter, kilogram, and kelvin (answers will vary)
2. square meter (m2), cubic meter (m3), grams per milliliter (g/mL), milliliters per second (mL/s) answers will vary
1. 1,000; 100
4. 1,000,000; 1,000,000
1. Answer to question 5: Complete the following table
Unit Abbreviation
centiliter cL
millisecond ms
centimeter cm
kiloliter kL
micrometer µm
6.
Answers to question 6: Complete the following table
Unit Abbreviation
microliter µL
kilosecond ks
deciliter dL
nanosecond ns
millimeter mm
7. grams per liter, grams per milliliter, and kilograms per liter (answers will vary)
8. kilometers per hour; meters per minute; millimeters per second (answers will vary)
Exercises
1. Give the two conversion factors you can construct using each pair of units.
1. meters and kilometers
2. liters and microliters
3. seconds and milliseconds
2. Give the two conversion factors you can construct using each pair of units.
1. grams and centigrams
2. millimeters and meters
3. liters and megaliters
3. How many meters are in 56.2 km?
4. How many seconds are in 209.7 ms?
5. How many microliters are in 44.1 L?
6. How many megagrams are in 90.532 g?
7. Convert 109.6 kg into micrograms. Express your final answer in scientific notation.
8. Convert 3.8 × 105 mm into kilometers. Express your final answer in scientific notation.
9. Convert 3.009 × 10−5 ML into centiliters. Express your final answer in scientific notation.
10. Convert 99.04 dm into micrometers. Express your final answer in scientific notation.
11. The density of ethyl alcohol is 0.79 g/mL. What is the mass of 340 mL of ethyl alcohol in kilograms? Do a 2-step conversion.
12. The density of a certain fraction of crude oil is 1.209 g/mL. What is the mass of 13,500 mL of this fraction?
13. The density of ethyl alcohol is 0.79 g/mL. What is the volume of 340 g of ethyl alcohol in liters? Do a 2-step conversion.
14. The density of a certain component of crude oil is 1.209 g/mL. What is the volume of 13,500 g of this component?
Answers
1.
1. $\mathrm{\dfrac{1,000\:m}{1\:km}\,;\:\dfrac{1\:km}{1,000\:m}}$
2. $\mathrm{\dfrac{1,000,000\:\mu L}{1\:L}\,;\:\dfrac{1\:L}{1,000,000\:\mu L}}$
3. $\mathrm{\dfrac{1,000\:ms}{1\:s}\,;\:\dfrac{1\:s}{1,000\:ms}}$
2.
1. $\mathrm{\dfrac{1\:g}{100\:cg}\,;\:\dfrac{100\:cg}{1\:g}}$
2. $\mathrm{\dfrac{1,000\:mm}{1\:m}\,;\:\dfrac{1\:m}{1,000\:mm}}$
3. $\mathrm{\dfrac{1,000,000\:L}{1\:ML}\,;\:\dfrac{1\:ML}{1,000,000\:L}}$
1. 5.62 × 104 m
4. 2.097 × 10-1 s
1. 4.41 × 107 µL
6. 9.0532 × 10-5 g
1. 1.096 × 1011 µg
8. 0.38 km; 3.8 × 10-1 km
1. 3.009 × 103 cL
10. 9.904 × 106 µm
11. 0.27 kg
12. 16,300 mL
13. 0.43 L
14. 11,200 mL
Exercises
1. Vitamin C tablets can come in 500 mg tablets. How many of these tablets are needed to obtain 10 g of vitamin C?
2. Vitamin C tablets can come in 500 mg tablets. How many of these tablets are needed to obtain 10 g of vitamin C?
3. The recommended daily allowance (RDA) for magnesium for 19-30 yrs old+ men is 400 mg. Magnesium supplements come in 200 mg capsules. How many capsules need to be taken to meet RDA?
4. A 175 lb patient is to undergo surgery and will be given an intravenous anesthetic. The safe dosage of anesthetic is up 12 mg/kg of body weight. Determine the maximum dose of anesthetic that should be used. (Hint: 2.2 lbs = 1 kg)
5. The safe dosage of an IV antibiotic for children weighing more than 2.0kg is 60.mg per kilogram of body weight. How many mg should be administered to a child weighing 16 kg?
6. A drug dose of 1.5 mg/kg is ordered for a child weighing 70.4 lbs. How many mg of the drug should be administered? If the drug is available as 60 mg/2 mL. How many mL must the nurse administer?
Answers
1. 20 tablets
2. 11,000 mg; 11 g
3. 2 tablets
4. 955 mg
5. 960 mg
6. 48 mg; 1.6 mL
Exercises
1. A sample of urine has a density of 1.105 g/cm3. What is the mass of 0.255 L of this urine?
2. The hardest bone in the body is tooth enamel, which has a density of 2.91 g/cm3. What is the volume, in liters, of 75.9 g of tooth enamel?
3. Some brands of aspirin have 81 mg of aspirin in each tablet. If a person takes 8 tablets per day, how many grams of aspirin is that person consuming every day?
4. The US government has a recommended daily intake (RDI) of 5 µg of vitamin D per day. (The name recommended daily allowance was changed to RDI in 1997.) If milk contains 1.2 µg per 8 oz glass, how many ounces of milk are needed to supply the RDI of vitamin D?
5. The population of the United States, according to the 2000 census, was 281.4 million people.
1. How many significant figures does this number have?
2. What is the unit in this quantity?
3. Express this quantity in proper scientific notation.
6. The United States produces 34,800,000,000 lb of sugar each year, and much of it is exported to other countries.
1. How many significant figures does this number have?
2. What is the unit in this quantity?
3. Express this quantity in proper scientific notation.
7. Construct a conversion factor that can convert from one unit to the other in each pair of units.
1. from millimeters to kilometers
2. from kilograms to micrograms
3. from centimeters to micrometers
8. Construct a conversion factor that can convert from one unit to the other in each pair of units.
1. from kilometers to micrometers
2. from decaliters to milliliters
3. from megagrams to milligrams
9. What is the density of a dextrose solution if 355 mL of the solution has a mass of 406.9 g?
10. What is the density of a dental amalgam (an alloy used to fill cavities) if 1.005 kg of the material has a volume of 433 mL? Express your final answer in grams per milliliter.
For Exercises 11–16, see the accompanying table for the relationships between English and SI units.
For Exercises 11–16, see the accompanying table for the relationships between English and SI units.
1 m ≈ 39.36 in. ≈ 3.28 ft ≈ 1.09 yd
1 in. ≈ 2.54 cm
1 km ≈ 0.62 mi
1 kg ≈ 2.20 lb
1 lb ≈ 454 g
1 L ≈ 1.06 qt
1 qt ≈ 0.946 L
1. Approximately how many inches are in 4.76 m?
2. Approximately how many liters are in 1 gal, which is exactly 4 qt?
3. Approximately how many kilograms are in a person who weighs 170 lb?
4. The average distance between Earth and the sun is 9.3 × 107 mi. How many kilometers is that?
5. Show mathematically that 1 L equals 1 dm3.
6. Show mathematically that 1 L equals 1,000 cm3.
Answers
1. 282 g
2. 26.1 cm3; 0.0261 L
1. 650 mg
4. 30 oz
1.
1. four significant figures
2. people
3. 2.841 × 108 people
6.
1. three significant figures
2. pound (lb)
3. 3.48 x 1010 lbs
7
1. $\mathrm{\dfrac{1\:km}{10^6\:mm}}$
2. $\mathrm{\dfrac{10^9\:\mu g}{1\:kg}}$
3. $\mathrm{\dfrac{10^4\:\mu m}{1\:cm}}$
8. Construct a conversion factor that can convert from one unit to the other in each pair of units.
1. $\mathrm{\dfrac{10^9\:\mu m}{1\:km}}$
2. $\mathrm{\dfrac{100\:mL}{1\:dL}}$
3. $\mathrm{\dfrac{10^9\:mg}{1\:Mg}}$
9. 1.15 g/mL
10. 2.321 g/mL
11. 187 in.
12. 3.784 L
13. 77 kg
14. 1.5 × 108 km
15. $\mathrm{1\:L=0.001\:m^3\times\left(\dfrac{1\:dm}{0.1\:m}\right)^3=1\:dm^3}$
16. $\mathrm{1\:L=0.001\:m^3\times\left(\dfrac{1\:cm}{0.01\:m}\right)^3=1000\:cm^3}$ | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/01.E%3A_Exercises.txt |
These are homework exercises to accompany Chapter 2 of the Ball et al. "The Basics of GOB Chemistry" Textmap.
Exercises
1. Which of the following substances are elements?
1. sodium
2. milk
3. gold
4. water
5. air
6. liquefied nitrogen
2. Which of the following substances are elements?
1. paper
2. electricity
3. neon
4. carbon
5. wood
6. concrete
3. Write the chemical symbol for each element.
1. silver
2. sulfur
3. nitrogen
4. neon
4. Write the chemical symbol for each element.
1. bromine
2. oxygen
3. lithium
4. boron
5. Explain why it is improper to write CO as the chemical symbol for cobalt.
6. Explain why it is improper to write NO as the chemical symbol for nobelium.
7. Complete the following table.
Element Symbol Element Name
F
Fe
I
Cr
C
P
8. Complete the following table.
Element Symbol Element Name
Mg
Mn
Ca
Cl
K
Pt
Answers
1.
1. element
2. not an element
3. element
4. not an element
5. not an element
6. element
2.
1. not an element
2. not an element
3. element
4. element
5. not an element
6. not an element
1.
1. Ag
2. S
3. N
4. Ne
4.
1. Br
2. O
3. Li
4. B
1. By convention, the second letter in an element’s symbol is always lowercase.
6. By convention, the second letter in an element’s symbol is always lowercase. Additionally, NO represents a compound.
7.
Element Symbol Element Name
F fluorine
Fe iron
I iodine
Cr chromium
C carbon
P phosphorus
8.
Element Symbol Element Name
Mg magnesium
Mn manganese
Ca calcium
Cl chlorine
K potassium
Pt platinum
Exercises
1. Which of the following elements exist as diatomic molecules?
1. helium
2. hydrogen
3. iodine
4. gold
2. Which of the following elements exist as diatomic molecules?
1. chlorine
2. potassium
3. silver
4. oxygen
3. Why is it proper to represent the elemental form of helium as He but improper to represent the elemental form of hydrogen as H?
4. Why is it proper to represent the elemental form of chlorine as Cl2 but improper to represent the elemental form of calcium as Ca2?
Answers
1.
1. no
2. yes
3. yes
4. no
2.
1. yes
2. no
3. no
4. yes
3. Hydrogen exists as a diatomic molecule in its elemental form; helium does not exist as a diatomic molecule.
4. Chlorine exists as a diatomic molecule in its elemental form; calcium does not exist as a diatomic molecule.
Exercises
1. Which is smaller—an electron or a helium atom?
2. Which is smaller—an electron or a helium atom?
3. Which subatomic particle has a positive charge? Which subatomic particle has a negative charge?
4. Which subatomic particle is electrically neutral? Does it exist inside or outside the nucleus?
5. Protons are among the (most, least) massive subatomic particles, and they are found (inside, outside) the nucleus.
6. Electrons are among the (most, least) massive subatomic particles, and they are found (inside, outside) the nucleus.
7. Describe why Rutherford used the term planetary model to describe his model of atomic structure.
8. Why is the planetary model not an appropriate way to describe the structure of an atom?
9. What happened to most of the alpha particles in Rutherford’s experiment? Explain why that happened.
10. Electrons account for the (majority, minority) of the (mass, volume) of an atom.
Answers
1. An electron is smaller.
2. An atom of lead is larger.
1. proton; electron
4. neutron; inside the nucleus
1. most; inside
6. least; outside
1. Electrons are in orbit about the nucleus.
8. Electrons do not have specific circular orbits about the nucleus.
9. Most of the alpha particles went through the metal sheet because atoms are mostly empty space.
10. minority; mass
Exercises
1. How many protons are in the nucleus of each element?
1. radon
2. tungsten
3. chromium
4. beryllium
2. How many protons are in the nucleus of each element?
1. sulfur
2. uranium
3. calcium
4. lithium
3. What are the atomic numbers of the elements in Exercise 1?
4. What are the atomic numbers of the elements in Exercise 2?
5. How many electrons are in neutral atoms of the elements in Exercise 1?
6. How many electrons are in neutral atoms of the elements in Exercise 2?
7. Complete the following table.
Number of Protons Number of Neutrons Element Name Isotope Symbol
80 120
$\mathrm{^{55}_{26}Fe}$
2 hydrogen
8. Complete the following table.
Number of Protons Number of Neutrons Element Name Isotope Symbol
$\mathrm{^{3}_{2}He}$
95 153
21 potassium
9. State the number of protons, neutrons, and electrons in neutral atoms of each isotope.
1. 131I
2. 40K
3. 201Hg
4. 19F
10. State the number of protons, neutrons, and electrons in neutral atoms of each isotope.
1. 3H
2. 133Cs
3. 56Fe
4. 207Pb
11. What is the mass number of a gallium atom that has 38 neutrons in it?
12. What is the mass number of a uranium atom that has 143 neutrons in it?
13. Complete each sentence.
1. 48Ti has _____ neutrons.
2. 40Ar has _____ neutrons.
3. 3H has _____ neutrons.
14. Complete each sentence.
1. 18O has _____ neutrons.
2. 60Ni has _____ neutrons.
3. 127I has _____ neutrons.
Answers
1.
1. 86
2. 74
3. 24
4. 4
2.
1. 16
2. 92
3. 20
4. 3
1. 86, 74, 24, and 4
4. 16, 92, 20, 3
1. 86, 74, 24, and 4
6. 16, 92, 20, 3
1. Number of Protons Number of Neutrons Element Name Isotope Symbol
80 120 mercury $\mathrm{^{200}_{80}Hg}$
26 29 iron $\mathrm{^{55}_{26}Fe}$
1 2 hydrogen $\mathrm{^{3}_{1}H}$
8.
Number of Protons Number of Neutrons Element Name Isotope Symbol
2 1 helium $\mathrm{^{3}_{2}He}$
95 153 americium $\mathrm{^{248}_{95}Am}$
19 21 potassium $\mathrm{^{40}_{19}K}$
1.
1. protons: 53; neutrons: 78; electrons: 53
2. protons: 19; neutrons: 21; electrons: 19
3. protons: 80; neutrons: 121; electrons: 80
4. protons: 9; neutrons: 10; electrons: 9
10.
1. protons: 1; neutrons: 2; electrons: 1
2. protons: 55; neutrons: 78; electrons: 55
3. protons: 26; neutrons: 30; electrons: 26
4. protons: 82; neutrons: 125; electrons: 82
11. 69
12. 235
13.
1. 26
2. 22
3. 2
14.
1. 10
2. 32
3. 74
Exercises
1. What is the atomic mass of zinc in atomic mass units?
2. What is the atomic mass of barium in atomic mass units?
3. What is the average mass of a single magnesium atom in grams?
4. What is the average mass of a single calcium atom in grams?
5. What is the mass of 1.000 × 1024 aluminum atoms in grams?
6. What is the mass of 5.000 × 1023 carbon atoms in grams?
7. Which has more mass—1 tungsten atom or 11 oxygen atoms?
8.Which has more mass—1 tungsten atom or 11 oxygen atoms?
9. Determine the atomic mass of lithium, given the isotopic composition: 92.4% lithium-7 (mass 7.016 u) and 7.60% lithium-6 (mass 6.015 u).
10. Determine the atomic mass of neon, given the isotopic composition: 90.48% neon-20 (mass 19.992 u), 0.27% neon-21 (mass 20.994 u), and 9.25% neon-22 (mass 21.991 u).
11. Determine the atomic mass of magnesium, given the isotopic composition: 78.70% magnesium-24 (mass 23.98 u), 10.13% magnesium-25 (mass 24.99 u) and 11.17% magnesium-26 (mass 25.98).
12. Determine the atomic mass of bromine, given the isotopic composition: 50.69% bromine-79 (mass 78.9183 u), 49.31% and bromine-81 (mass 80.9163 u).
Answers
1. 65.4 u
2. 137.33 u
1. 4.038 × 10−23 g
4. 6.657 × 10−23 g
1. 44.81 g
6. 9.974 g
7. 1 tungsten atom
8. 1 magnesium atom
9. 6.94 u
10. 20.18 u
11. 24.30 u
12. 79.90 u
Exercises
1. What is the maximum number of electrons that can fit in an s subshell? Does it matter what shell the s subshell is in?
2. What is the maximum number of electrons that can fit in a p subshell? Does it matter what shell the p subshell is in?
3. What is the maximum number of electrons that can fit in a d subshell? Does it matter what shell the d subshell is in?
4. What is the maximum number of electrons that can fit in an f subshell? Does it matter what shell the f subshell is in?
5. What is the electron configuration of a carbon atom?
6. What is the electron configuration of a sulfur atom?
7. What is the valence shell electron configuration of a calcium atom?
8. What is the valence shell electron configuration of a selenium atom?
9. What atom has the electron configuration 1s22s22p5?
10. What atom has the electron configuration 1s22s22p63s23p3?
11. Draw a representation of the electronic structure of an oxygen atom.
12. Draw a representation of the electronic structure of a phosphorus atom.
13. A potassium atom has ____ core electrons and ____ valence electrons.
14. A silicon atom has ____ core electrons and ____ valence electrons.
Answers
1. 2; no
2. 6; no
1. 10; no
4. 14; no
1. 1s22s22p2
6. 1s22s22p63s23p4
1. 4s2
8. 4s24p4
1. fluorine
10. phosphorus
11.
12.
13. 18; 1
14. 10; 4
Exercises
1. Which elements have chemical properties similar to those of magnesium?
1. sodium
2. fluorine
3. calcium
4. barium
5. selenium
2. Which elements have chemical properties similar to those of lithium?
1. sodium
2. calcium
3. beryllium
4. barium
5. potassium
3. Which elements have chemical properties similar to those of chlorine?
1. sodium
2. fluorine
3. calcium
4. iodine
5. sulfur
4. Which elements have chemical properties similar to those of carbon?
1. silicon
2. oxygen
3. germanium
4. barium
5. argon
5. Which elements are alkali metals?
1. sodium
2. magnesium
3. aluminum
4. potassium
5. calcium
6. Which elements are alkaline earth metals?
1. sodium
2. magnesium
3. aluminum
4. potassium
5. calcium
7. Which elements are halogens?
1. oxygen
2. fluorine
3. chlorine
4. sulfur
5. carbon
8. Which elements are noble gases?
1. helium
2. hydrogen
3. oxygen
4. neon
5. chlorine
9. Which pairs of elements are located in the same period?
1. H and Li
2. H and He
3. Na and S
4. Na and Rb
10. Which pairs of elements are located in the same period?
1. V and Nb
2. K and Br
3. Na and P
4. Li and Mg
11. In each pair of atoms, which atom has the greater atomic radius?
1. H and Li
2. N and P
3. Cl and Ar
4. Al and Cl
12. In each pair of atoms, which atom has the greater atomic radius?
1. H and He
2. N and F
3. Cl and Br
4. Al and B
13. Scandium is a (metal, nonmetal, semimetal) and is a member of the (main group elements, transition metals).
14. Silicon is a (metal, nonmetal, semimetal) and is a member of the (main group elements, transition metals).
Answers
1.
1. no
2. no
3. yes
4. yes
5. no
2.
1. yes
2. no
3. no
4. no
5. yes
1.
1. no
2. yes
3. no
4. yes
5. no
4.
1. yes
2. no
3. yes
4. no
5. no
1.
1. yes
2. no
3. no
4. yes
5. no
6.
1. no
2. yes
3. no
4. no
5. yes
1.
1. no
2. yes
3. yes
4. no
5. no
8.
1. yes
2. no
3. no
4. yes
5. no
1.
1. no
2. yes
3. yes
4. no
10.
1. no
2. yes
3. yes
4. no
11.
1. Li
2. P
3. Cl
4. Al
12.
1. H
2. N
3. Br
4. Al
13. metal; transition metals
14. semimetal; main group elements
2.8: Elements, Atoms, and the Periodic Table (Exercises)
Q2.1.1
Which of the following substances are elements?
1. sodium
2. milk
3. gold
4. water
5. air
6. liquefied nitrogen
Hint: what is the definition of an element?
Answer:
1. element
2. not an element
3. element
4. not an element
5. not an element
6. element
Solution:
At first we need to know the definition of an element: a chemical element is a PURE chemical substance consisting of a SINGLE type of atom. Also, we need to know the compositions of these substances. The substances that can be represented with a SINGLE chemical symbol (maybe with subscript or superscript).
1. Sodium is Na an element
2. Milk is an organic solution containing lipids, proteins, salts minerals, vitamins, carbohydrates and other miscellaneous compounds not an element
3. Gold is Au an element
4. Water is H2O → not an element
5. Air is composed of many gases such O2, N2 → not an element
6. Liquefied nitrogen is still N2 → an element.
Q2.1.2
Which of the following substances are elements?
1. paper
2. electricity
3. neon
4. carbon
5. wood
6. concrete
Hint: what is the definition of an element?
Answer:
1. not an element
2. not an element
3. element
4. element
5. not an element
6. not an element element
Solution:
At first we need to know the definition of an element: a chemical element is a PURE chemical substance consisting of a SINGLE type of atom. Also, we need to know the compositions of these substances. The substances that can be represented with a SINGLE chemical symbol (can have subscript or superscript).
1. Paper is mostly cellulose composed of carbon, hydrogen and oxygen [(C6H10O5)n] → not an element
2. Electricity is a form of energy not atoms not an element
3. Neon is a noble gas with chemical symbol Ne an element
4. Carbon is C → an element
5. Wood is majorly composed with cellulose and lignin → not an element
6. Concrete is composed of water and coarse granular material → not an element.
Q2.1.3
Write the chemical symbol for each element.
1. silver
2. sulfur
3. nitrogen
4. neon
Hint: what is the first or first two letters of the name of the element? Sometimes it's the Latin name of the element. However, sometimes it's not the first 2 letters.
Answer:
1. Ag
2. S
3. N
4. Ne
Solution:
Usually, the chemical symbol of an element is the first or first 2 letters of the English name or Latin name of the element with the first letter ALWAYS in uppercase and the second ALWAYS letter in lowercase. However, that's not always the case, such as Tin (Stannum in Latin) is Sn. This is the time a Periodic Table comes in handy.
Q2.1.4
Write the chemical symbol for each element.
1. bromine
2. oxygen
3. lithium
4. boron
Hint: what is the first or first two letters of the name of the element? Sometimes it's the Latin name of the element. However, sometimes it's not the first 2 letters.
Answer:
1. Br
2. O
3. Li
4. B
Solution:
Usually, the chemical symbol of an element is the first or first 2 letters of the English name or Latin name of the element with the first letter ALWAYS in uppercase and the second letter ALWAYS in lowercase. However, that's not always the case, such as Tin (Stannum in Latin) is Sn. This is the time a Periodic Table comes in handy.
Q2.1.5
Explain why it is improper to write CO as the chemical symbol for cobalt.
Hint: How would this notation confuse the readers?
Answer: If "CO" was written as the chemical symbol for cobalt, it would be the same as the chemical formula for carbon monoxide. This ambiguity would create a lot problems such as in labeling and might lead to life-threatening danger.
Solution:
The chemical symbol is ALWAYS written with the first letter in uppercase and the second letter in lowercase.
Q2.1.6
Explain why it is improper to write NO as the chemical symbol for nobelium.
Hint: How would this notation confuse the readers?
Answer: If "NO" was written as the chemical symbol for cobalt, it would be the same as the chemical formula for nitrogen monoxide. This ambiguity would create a lot problems such as in labeling and might lead to life-threatening danger.
Solution:
The chemical symbol is ALWAYS written with the first letter in uppercase and the second letter in lowercase.
Q2.1.7
Complete the following table.
Element Symbol Element Name
F
Fe
I
Cr
C
P
Hint: The element symbols are usually the first or first two letters of the element names.
Answer
Element Symbol Element Name
F fluorine
Fe iron
I iodine
Cr chromium
C carbon
P phosphorus
Solution:
Usually, the chemical symbol of an element is the first or first 2 letters of the English name or Latin name of the element. Also, metals tend to end in -ium; halogens gases (group 17 gases) end in -ine; noble gases except helium end in -on; gases other than halogens and noble gases tend to end in -gen. However, that's not always the case, such as Tin (Stannum in Latin) is Sn. This is the time a Periodic Table comes in handy.
Q2.1.8
Complete the following table.
Element Symbol Element Name
Mg
Mn
Ca
Cl
K
Pt
Hint: The element symbols are usually the first or first two letters of the element names.
Answer
Element Symbol Element Name
Mg Magnesium
Mn Manganese
Ca Calcium
Cl Chlorine
K Potassium
Pt
Platinum
Solution:
Usually, the chemical symbol of an element is the first or first 2 letters of the English name or Latin name of the element. Also, metals tend to end in -ium; halogens gases (group 17 gases) end in -ine; noble gases except helium end in -on; gases other than halogens and noble gases tend to end in -gen. However, that's not always the case, such as Tin (Stannum in Latin) is Sn. This is the time a Periodic Table comes in handy.
Q2.2.1
Which of the following elements exist as diatomic molecules?
1. helium
2. hydrogen
3. iodine
4. gold
Hint: Which atoms don't have an octet?
Answer:
1. No
2. Yes
3. Yes
4. No
Solution:
Diatomic molecules are molecules composed of only two atoms. If an atom has full valence shell, the atom is very stable and hardly react with other atoms. When an atom doesn't have a complete valence shell, it wants to pair up the unpaired electrons in valence shell to reach a stabler status. So, if an atom has unpaired valence electrons, it will form one or more covalent bonds with another atom which also has unpaired valence electrons to form an octet. Metal atoms form metallic bonds and exist as large continuous lattice structure.
1. Helium is in the first period and has 2 valence electrons, so its valence shell is full→ monatomic
2. Hydrogen is also in the first period but has only 1 valence electrons, so it has an unpaired electron in the valence shell. 2 Hydrogen atoms form 1 covalent bond diatomic
3. Iodine is in group VII so it has 7 valence electrons leaving one valence electron unpaired. 2 Iodine atoms form 1 covalent bond diatomic
4. First of all, gold is metal. Second, gold is a transition metal so it has more than one unpaired valence electron → not diatomic
Q2.2.2
Which of the following elements exist as diatomic molecules?
1. chlorine
2. potassium
3. silver
4. oxygen
Hint: Which atoms don't have an octet?
Answer:
1. Yes
2. No
3. No
4. Yes
Solution
Diatomic molecules are molecules composed of only two atoms. If an atom has full valence shell, the atom is very stable and hardly react with other atoms. When an atom doesn't have a complete valence shell, it wants to pair up the unpaired electrons in valence shell to reach a stabler status. So, if an atom has unpaired valence electrons, it will form one or more covalent bonds with another atom which also has unpaired valence electrons to form an octet. Metal atoms form metallic bonds and exist as large continuous lattice structure.
1. Chlorine is in group VII so it has 7 valence electrons leaving one valence electron unpaired. 2 Chlorine atoms form 1 covalent bond diatomic
2. Potassium is in group I so it has 1 unpaired valence electron. However, even 2 potassium atoms pair up, they cannot form an octet not diatomic
3. Silver is a transition metal so it has more than one unpaired valence electron → not diatomic
4. Oxygen is in group VI so it has 6 valence electrons so it has 2 unpaired valence electron. 2 Chlorine atoms form 2 covalent bonds diatomic
Q2.1.3
Why is it proper to represent the elemental form of helium as He but improper to represent the elemental form of hydrogen as H?
Hint: How many atoms are there in these 2 gas molecules?
Answer: Because Hydrogen molecules are diatomic while Helium molecules are monatomic.
Solution: If an atom has unpaired valence electrons, it will form one or more covalent bonds with another atom which also has unpaired valence electrons to form an octet. Hydrogen is a group I element, so it has 1 valence electron. Therefore, it wants to form one covalent bond with, in this case, another hydrogen atom to complete its duet (since hydrogen only has the first shell). On the other hand, Helium already has a complete valence shell so it is stable and exist in a monatomic form.
Q2.1.4
Why is it proper to represent the elemental form of chlorine as Cl2 but improper to represent the elemental form of calcium as Ca2?
Hint: How many atoms are there in these 2 gas molecules?
Answer: Because Chlorine molecules are diatomic while Calcium molecules exist as large continuous lattice structure.
Solution: If an atom has unpaired valence electrons, it will form one or more covalent bonds with another atom which also has unpaired valence electrons to form an octet. Chlorine is a group VII element, so it has 7 valence electrons. Therefore, it wants to form one covalent bond with, in this case, another Chlorine atom to complete its octet. Calcium is a group II element, but it's metal. So, calcium atoms form metallic bonds and exist in a vast network structure called lattice system.
Additional Exercises
1. If the atomic radius of sodium atoms is 1.86 × 10−10 m, how many sodium atoms are needed to make a line that is 1.00 cm in length?
2. If the atomic radius of osmium atoms is 1.34 × 10−10 m, how many osmium atoms are needed to make a line that is 5.85 m in length?
3. What might be the electron configuration of K+, an atom that has lost an electron?
4. What might be the electron configuration of Cl, an atom that has gained an additional electron?
5. The electron configuration of the Ti atom is 1s22s22p63s23p64s23d2. What is the valence shell electron configuration of Ti?
6. The electron configuration of the Ge atom is 1s22s22p63s23p64s23d104p2. What is the valence shell electron configuration of Ge?
7. What is the mass of an electron in atomic mass units?
8. In a footnote in this chapter, an alpha particle was defined as a particle with 2 protons and 2 neutrons. What is the mass, in grams, of an alpha particle? (Hint: what element does an alpha particle resemble?)
9. A sample of the nonexistent element mythium consists of 25.59% of an isotope with mass number 580, 32.74% of an isotope with mass number 582, and 41.67% of an isotope with mass number 581. What is the atomic mass of mythium?
10. Because the distribution of isotopes is different on different planets in the solar system, the average atomic mass of any element differs from planet to planet. Assume that on Mercury, a rather hot planet, there is more deuterium left in the atmosphere than on Earth, so that 92.55% of the hydrogen on Mercury is 1H, while the remainder is 2H. What is the atomic mass of hydrogen on Mercury?
11. The compound that sodium makes with chlorine has sodium and chlorine atoms in a 1:1 ratio. Name two other elements that should make a compound having a 1:1 ratio of atoms with sodium.
12. The compound that magnesium makes with oxygen has magnesium to oxygen atoms in a 1:1 ratio. Name two other elements that should make a compound having a 1:1 ratio of atoms with magnesium. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/02.E%3A_Exercises.txt |
These are homework exercises to accompany Chapter 3 of the Ball et al. "The Basics of GOB Chemistry" Textmap.
3.1: Two Types of Bonding
Concept Review Exercises
1. What is the octet rule?
2. How are ionic bonds formed?
Answers
1. The octet rule is the concept that atoms tend to have eight electrons in their valence electron shell.
2. Ionic bonds are formed by the attraction between oppositely charged ions.
Exercises
1. Why is an ionic compound unlikely to consist of two positively charged ions?
2. Why is an ionic compound unlikely to consist of two negatively charged ions?
3. A calcium atom has two valence electrons. Do you think it will lose two electrons or gain six electrons to obtain an octet in its outermost electron shell?
4. An aluminum atom has three valence electrons. Do you think it will lose three electrons or gain five electrons to obtain an octet in its outermost electron shell?
5. A selenium atom has six valence electrons. Do you think it will lose six electrons or gain two electrons to obtain an octet in its outermost electron shell?
6. An iodine atom has seven valence electrons. Do you think it will lose seven electrons or gain one electron to obtain an octet in its outermost electron shell?
Answers
1. Positive charges repel each other, so an ionic compound is not likely between two positively charged ions.
3. It is more likely to lose two electrons.
5. It is more likely to gain two electrons.
3.2: Ions
Concept Review Exercises
1. What are the two types of ions?
2. Use Lewis diagrams to illustrate the formation of an ionic compound from a potassium atom and an iodine atom.
3. When the following atoms become ions, what charges do they acquire?
1. Li
2. S
3. Ca
4. F
Answers
1. Cations have positive charges, and anions have negative charges.
2.
3.
1. 1+
2. 2−
3. 2+
4. 1−
Key Takeaways
• Ions can be positively charged or negatively charged.
• A Lewis diagram is used to show how electrons are transferred to make ions and ionic compounds.
Exercises
1. Identify each as a cation, an anion, or neither.
1. H+
2. Cl
3. O2
4. Ba2+
5. CH4
6. CS2
2. Identify each as a cation, an anion, or neither.
1. NH3
2. Br
3. H
4. Hg2+
5. CCl4
6. SO3
3. Write the electron configuration for each ion.
1. Li+
2. Mg2+
3. F
4. S2−
4. Write the electron configuration for each ion.
1. Na+
2. Be2+
3. Cl
4. O2−
5. Draw Lewis diagrams for the ions listed in Exercise 3. Also include Lewis diagrams for the respective neutral atoms as a comparison.
6. Draw Lewis diagrams for the ions listed in Exercise 4. Also include Lewis diagrams for the respective neutral atoms as a comparison.
7. Using Lewis diagrams, show the electron transfer for the formation of LiF.
8. Using Lewis diagrams, show the electron transfer for the formation of MgO.
9. Using Lewis diagrams, show the electron transfer for the formation of Li2O.
10. Using Lewis diagrams, show the electron transfer for the formation of CaF2.
11. What characteristic charge do atoms in the first column of the periodic table have when they become ions?
12. What characteristic charge do atoms in the second column of the periodic table have when they become ions?
13. What characteristic charge do atoms in the third-to-last column of the periodic table have when they become ions?
14. What characteristic charge do atoms in the next-to-last column of the periodic table have when they become ions?
Answers
1.
1. cation
2. anion
3. neither
4. cation
5. neither
6. neither
3.
1. 1s2
2. 1s22s22p6
3. 1s22s22p6
4. 1s22s22p63s23p6
5.
a.
b.
c.
d.
1. 1+
1. 2−
3.3: Formulas for Ionic Compounds
Concept Review Exercises
1. What information is contained in the formula of an ionic compound?
2. Why do the chemical formulas for some ionic compounds contain subscripts, while others do not?
3. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Mg2+ and I
2. Na+ and O2−
Answers
1. the ratio of each kind of ion in the compound
2. Sometimes more than one ion is needed to balance the charge on the other ion in an ionic compound.
3.
1. MgI2
2. Na2O
Key Takeaways
• Proper chemical formulas for ionic compounds balance the total positive charge with the total negative charge.
• Groups of atoms with an overall charge, called polyatomic ions, also exist.
Exercises
1. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Na+ and Br
2. Mg2+ and Br
3. Mg2+ and S2−
2. Write the chemical formula for the ionic compound formed by each pair of ions.
1. K+ and Cl
2. Mg2+ and Cl
3. Mg2+ and Se2
3. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Na+ and N3−
2. Mg2+ and N3−
3. Al3+ and S2−
4. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Li+ and N3−
2. Mg2+ and P3−
3. Li+ and P3−
5. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Fe3+ and Br
2. Fe2+ and Br
3. Au3+ and S2−
4. Au+ and S2−
6. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Cr3+ and O2−
2. Cr2+ and O2−
3. Pb2+ and Cl
4. Pb4+ and Cl
7. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Cr3+ and NO3
2. Fe2+ and PO43
3. Ca2+ and CrO42
4. Al3+ and OH
8. Write the chemical formula for the ionic compound formed by each pair of ions.
1. NH4+ and NO3
2. H+ and Cr2O72
3. Cu+ and CO32
4. Na+ and HCO3
9. For each pair of elements, determine the charge for their ions and write the proper formula for the resulting ionic compound between them.
1. Ba and S
2. Cs and I
10. For each pair of elements, determine the charge for their ions and write the proper formula for the resulting ionic compound between them.
1. K and S
2. Sc and Br
11. Which compounds would you predict to be ionic?
1. Li2O
2. (NH4)2O
3. CO2
4. FeSO3
5. C6H6
6. C2H6O
12. Which compounds would you predict to be ionic?
1. Ba(OH)2
2. CH2O
3. NH2CONH2
4. (NH4)2CrO4
5. C8H18
6. NH3
Answers
1.
1. NaBr
2. MgBr2
3. MgS
1.
1. Na3N
2. Mg3N2
3. Al2S3
1.
1. FeBr3
2. FeBr2
3. Au2S3
4. Au2S
1.
1. Cr(NO3)3
2. Fe3 (PO4)2
3. CaCrO4
4. Al(OH)3
1.
1. Ba2+, S2−, BaS
2. Cs+, I, CsI
1.
1. ionic
2. ionic
3. not ionic
4. ionic
5. not ionic
6. not ionic
3.4: Ionic Nomenclature
Concept Review Exercises
1. Briefly describe the process for naming an ionic compound.
2. In what order do the names of ions appear in the names of ionic compounds?
3. Which ionic compounds can be named using two different systems? Give an example.
Answers
1. Name the cation and then the anion but don’t use numerical prefixes.
2. the cation name followed by the anion name
3. Ionic compounds in which the cation can have more than one possible charge have two naming systems. FeCl3 is either iron(III) chloride or ferric chloride (answers will vary).
Key Takeaways
• Each ionic compound has its own unique name that comes from the names of the ions.
Exercises
1. Name each ion.
1. Ra2+
2. P3−
3. H2PO4
4. Sn4+
2. Name each ion.
1. Cs+
2. As3
3. HSO4
4. Sn2+
3. Name the ionic compound formed by each pair of ions.
1. Na+ and Br
2. Mg2+ and Br
3. Mg2+ and S2−
4. Name the ionic compound formed by each pair of ions.
1. K+ and Cl
2. Mg2+ and Cl
3. Mg2+ and Se2
5. Name the ionic compound formed by each pair of ions.
1. Na+ and N3−
2. Mg2+ and N3−
3. Al3+ and S2−
6. Name the ionic compound formed by each pair of ions.
1. Li+ and N3−
2. Mg2+ and P3−
3. Li+ and P3−
7. Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate.
1. Fe3+ and Br
2. Fe2+ and Br
3. Au3+ and S2−
4. Au+ and S2−
8. Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate.
1. Cr3+ and O2−
2. Cr2+ and O2−
3. Pb2+ and Cl
4. Pb4+ and Cl
9. Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate.
1. Cr3+ and NO3
2. Fe2+ and PO43
3. Ca2+ and CrO42
4. Al3+ and OH
10. Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate.
1. NH4+ and NO3
2. H+ and Cr2O72
3. Cu+ and CO32
4. Na+ and HCO3
11. Give two names for each compound.
1. Al(HSO4)3
2. Mg(HSO4)2
12. Give two names for each compound.
1. Co(HCO3)2
2. LiHCO3
Answers
1.
1. the radium ion
2. the phosphide ion
3. the dihydrogen phosphate ion
4. the tin(IV) ion or the stannic ion
1.
1. sodium bromide
2. magnesium bromide
3. magnesium sulfide
1.
1. sodium nitride
2. magnesium nitride
3. aluminum sulfide
1.
1. iron(III) bromide or ferric bromide
2. iron(II) bromide or ferrous bromide
3. gold(III) sulfide or auric sulfide
4. gold(I) sulfide or aurous sulfide
1.
1. chromium(III) nitrate or chromic nitrate
2. iron(II) phosphate or ferrous phosphate
3. calcium chromate
4. aluminum hydroxide
1.
1. aluminum hydrogen sulfate or aluminum bisulfate
2. magnesium hydrogen sulfate or magnesium bisulfate
3.5: Formula Mass
Concept Review Exercises
1. What is the relationship between atomic mass and formula mass?
2. How are subscripts used to determine a formula mass when more than one polyatomic ion is present in a chemical formula?
Answers
1. The formula mass is the sum of the atomic masses of the atoms in the formula.
2. The subscript is distributed throughout the parentheses to determine the total number of atoms in the formula.
Key Takeaways
• Formula masses of ionic compounds can be determined from the masses of the atoms in their formulas.
Exercises
1. What is the formula mass for the ionic compound formed by each pair of ions?
1. Na+ and Br
2. Mg2+ and Br
3. Mg2+ and S2−
2. What is the formula mass for the ionic compound formed by each pair of ions?
1. K+ and Cl
2. Mg2+ and Cl
3. Mg2+ and Se2
3. What is the formula mass for the ionic compound formed by each pair of ions?
1. Na+ and N3−
2. Mg2+ and N3−
3. Al3+ and S2−
4. What is the formula mass for the ionic compound formed by each pair of ions?
1. Li+ and N3−
2. Mg2+ and P3−
3. Li+ and P3−
5. What is the formula mass for each compound?
1. FeBr3
2. FeBr2
3. Au2S3
4. Au2S
6. What is the formula mass for each compound?
1. Cr2O3
2. CrO
3. PbCl2
4. PbCl4
7. What is the formula mass for each compound?
1. Cr(NO3)3
2. Fe3(PO4)2
3. CaCrO4
4. Al(OH)3
8. What is the formula mass for each compound?
1. NH4NO3
2. H2Cr2O7
3. Cu2CO3
4. NaHCO3
9. What is the formula mass for each compound?
1. Al(HSO4)3
2. Mg(HSO4)2
10. What is the formula mass for each compound?
1. Co(HCO3)2
2. LiHCO3
Answers
1.
1. 102.90 amu
2. 184.11 amu
3. 56.38 amu
1.
1. 83.00 amu
2. 100.93 amu
3. 150.17 amu
1.
1. 295.50 amu
2. 215.60 amu
3. 490.30 amu
4. 426.10 amu
1.
1. 238.00 amu
2. 357.49 amu
3. 156.08 amu
4. 78.01 amu
1.
1. 318.22 amu
2. 218.47 amu | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/03.E%3A_Exercises.txt |
Additional Exercises
1. What number shell is the valence electron shell of a sodium atom? What number shell is the valence shell of a sodium ion? Explain the difference.
2. What number shell is the valence electron shell of a bromine atom? What number shell is the valence shell of a bromide ion? Explain the difference between these answers and the answers to Exercise 1.
3. What is the electron configuration of each ion?
1. K+
2. Mg2+
3. F
4. S2−
4. What is the electron configuration of each ion?
1. Li+
2. Ca2+
3. Cl
4. O2−
1. If a sodium atom were to lose two electrons, what would be the electron configuration of the resulting cation?
2. Considering that electron shells are typically separated by large amounts of energy, use your answer to Exercise 5a to suggest why sodium atoms do not form a 2+ cation.
1. If a chlorine atom were to gain two electrons, what would be the electron configuration of the resulting anion?
2. Considering that electron shells are typically separated by large amounts of energy, use your answer to Exercise 6a to suggest why chlorine atoms do not form a 2− anion.
5. Use Lewis diagrams and arrows to show the electron transfer that occurs during the formation of an ionic compound among Mg atoms and F atoms. (Hint: how many atoms of each will you need?)
6. Use Lewis diagrams and arrows to show the electron transfer that occurs during the formation of an ionic compound among K atoms and O atoms. (Hint: how many atoms of each will you need?)
7. Mercury forms two possible cations—Hg2+ and Hg22+, the second of which is actually a two-atom cation with a 2+ charge.
1. Using common names, give the probable names of these ions.
2. What are the chemical formulas of the ionic compounds these ions make with the oxide ion, O2−?
8. The uranyl ion (UO22+) is a common water-soluble form of uranium. What is the chemical formula of the ionic compound uranyl nitrate? What is the chemical formula of the ionic compound uranyl phosphate?
9. The formal chemical name of the mineral strengite is iron(III) phosphate dihydrate. What is the chemical formula of strengite? What is the formula mass of strengite?
10. What is the formula mass of MgSO4·7H2O?
11. What is the formula mass of CaSO4·½H2O?
12. What mass does 20 formula units of NaCl have?
13. What mass does 75 formula units of K2SO4 have?
14. If an atomic mass unit equals 1.66 × 10−24 g, what is the mass in grams of one formula unit of NaCl?
15. If an atomic mass unit equals 1.66 × 10−24 g, what is the mass in grams of 5.00 × 1022 formula units of NaOH?
16. If an atomic mass unit equals 1.66 × 10−24 g, what is the mass in grams of 3.96 × 1023 formula units of (NH4)2SO4?
17. Both tin and lead acquire 2+ or 4+ charges when they become ions. Use the periodic table to explain why this should not surprise you.
18. Which ion would you expect to be larger in size—In3+ or Tl3+? Explain.
19. Which ion would you expect to be smaller in size—I or Br? Explain.
20. Which ion with a 2+ charge has the following electron configuration? 1s22s22p6
21. Which ion with a 3− charge has the following electron configuration? 1s22s22p6
Answers
1. For sodium, the valence shell is the third shell; for the sodium ion, the valence shell is the second shell because it has lost all its third shell electrons.
1. 1s22s22p63s23p6
2. 1s22s22p6
3. 1s22s22p6
4. 1s22s22p63s23p6
1. 1s22s22p5
2. It probably requires too much energy to form.
1. mercuric and mercurous, respectively
2. HgO and Hg2O, respectively
1. FePO4·2H2O; 186.86 u
1. 145.16 u
1. 13,070.25 u
1. 3.32 g
1. Both tin and lead have two p electrons and two s electrons in their valence shells.
1. Br because it is higher up on the periodic table
1. N3− | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/03E%3A_Exercises_II.txt |
Concept Review Exercises
1. How is a covalent bond formed between two atoms?
2. How does covalent bonding allow atoms in group 6A to satisfy the octet rule?
Answers
1. Covalent bonds are formed by two atoms sharing electrons.
2. The atoms in group 6A make two covalent bonds.
Exercises
1. Define covalent bond.
2. What is electron sharing?
3. Draw the Lewis diagram for the covalent bond in the H2 molecule.
4. Draw the Lewis diagram for the covalent bond in the Br2 molecule.
5. Draw the Lewis diagram for the covalent bond in the HCl molecule.
6. What is the difference between a molecule and a formula unit?
7. Why do hydrogen atoms not follow the octet rule when they form covalent bonds?
8. Draw the Lewis diagram for the covalent bonding in H2S. How many bonding electrons and nonbonding electrons are in the molecule?
9. Draw the Lewis diagram for the covalent bonding in CF4. How many bonding electrons and nonbonding electrons are in the molecule?
10. Draw the Lewis diagram for the covalent bonding in PCl3. How many bonding electrons and nonbonding electrons are in the molecule?
11. How many covalent bonds does a hydrogen atom typically form? Why?
12. How many covalent bonds does an oxygen atom typically form? Why?
13. Tellurium atoms make covalent bonds. How many covalent bonds would a tellurium atom make? Predict the formula of a compound between tellurium and hydrogen.
14. Tin atoms make covalent bonds. How many covalent bonds would a tin atom make? Predict the formula of a compound between tin and hydrogen.
15. Astatine is a synthetic element, made one atom at a time in huge “atom-smasher” machines. It is in the halogen group on the periodic table. How many covalent bonds would an atom of this element form?
16. There have been reports that atoms of element 116 (Lv) were made by smashing smaller atoms together. Using the periodic table, determine what column element 116 would be in and suggest how many covalent bonds an atom of this element would form.
Answers
1. A covalent bond is formed when two atoms share electrons.
2. Electron sharing joins two atoms in a covalent bond. This is a more stable arrangement than 2 individual atoms.
3.
4.
5.
6. A molecule is a discrete combination of atoms; a formula unit is the lowest ratio of ions in a crystal. 7. Hydrogen atoms follow the duet rule (not the octet rule). This is because it has only one shell and this shell can only hold 2 electrons.
8.
bonding electrons: 4; nonbonding electrons: 4
9.
bonding electrons: 8; nonbonding electrons: 24
10.
bonding electrons: 6; nonbonding electrons: 20
11. Hydrogen atoms form only one covalent bond because they have only one valence electron to pair.
12. Oxygen atoms form 2 covalent bonds because oxygen atoms have 6 valence electrons (2 lone pairs plus 2 unpaired electrons that are shared to achieve octet).
13. two; H2Te
14. four: SnH4
15. one
16. two
Concept Review Exercises
1. How do you recognize a covalent compound?
2. What are the rules for writing the molecular formula of a simple covalent compound?
3. What are the rules for naming a simple covalent compound?
Answers
1. A covalent compound is usually composed of two or more nonmetal elements.
2. It is just like an ionic compound except that the element further down and to the left on the periodic table is listed first and is named with the element name.
3. Name the first element first and then the second element by using the stem of the element name plus the suffix -ide. Use numerical prefixes if there is more than one atom of the first element; always use numerical prefixes for the number of atoms of the second element.
Exercises
1. Identify whether each compound has covalent bonds.
1. NaI
2. Na2CO3
3. N2O
4. SiO2
2. Identify whether each compound has covalent bonds.
1. C2H6
2. C6H5Cl
3. KC2H3O2
4. Ca(OH)2
3. Identify whether each compound has ionic bonds, covalent bonds, or both.
1. Na3PO4
2. K2O
3. COCl2
4. CoCl2
4. Identify whether each compound has ionic bonds, covalent bonds, or both.
1. FeCl3
2. Fe(NO3)3
3. (NH2)2CO
4. SO3
5. Which is the correct molecular formula—H4Si or SiH4? Explain.
6. Which is the correct molecular formula—SF6 or F6S? Explain.
7. Write the name for each covalent compound.
1. SiF4
2. NO2
3. CS2
4. P2O5
8. Write the name for each covalent compound.
1. CO
2. S2O3
3. BF3
4. GeS2
9. Write the formula for each covalent compound.
1. iodine trichloride
2. disulfur dibromide
3. arsenic trioxide
4. xenon hexafluoride
10. Write the formula for each covalent compound.
1. boron trichloride
2. carbon dioxide
3. tetraphosphorus decoxide
4. germanium dichloride
11. Write two covalent compounds that have common rather than systematic names.
12. What is the name of the simplest organic compound? What would its name be if it followed the nomenclature for binary covalent compounds?
Answers
1. no
2. yes
3. yes
4. yes
2.
1. yes
2. yes
3. yes
4. yes
1. both
2. ionic
3. covalent
4. ionic
4.
1. ionic
2. both
3. covalent
4. covalent
1. SiH4; except for water, hydrogen is almost never listed first in a covalent compound.
6. SF6; the less electronegative atom (S) is written first
1. silicon tetrafluoride
2. nitrogen dioxide
3. carbon disulfide
4. diphosphorus pentoxide
8.
1. carbon monoxide
2. disulfur trioxide
3. boron trifluoride
4. germanium disulfide
1. ICl3
2. S2Br2
3. AsO3
4. XeF6
10.
1. BCl3
2. CO2
3. P4O10
4. GeCl2
1. H2O and NH3 (water and ammonia) (answers will vary)
2. CH4; carbon tetrahydride
Exercises
1. What is one clue that a molecule has a multiple bond?
2. Draw the Lewis diagram for each of the following.
a. H2O
b. NH3
c. C2H6
d. CCl4
3. Each molecule contains double bonds. Draw the Lewis diagram for each. The first element is the central atom.
1. CS2
2. C2F4
3. COCl2
4. Each molecule contains multiple bonds. Draw the Lewis diagram for each. Assume that the first element is the central atom, unless otherwise noted.
1. N2
2. CO
3. HCN (The carbon atom is the central atom.)
4. POCl (The phosphorus atom is the central atom.)
5. Explain why hydrogen atoms do not form double bonds.
6. Why is it incorrect to draw a double bond in the Lewis diagram for MgO?
Answers
1. If single bonds between all atoms do not give all atoms (except hydrogen) an octet, multiple covalent bonds may be present.
2. a.
b.
c.
d.
3. a.
b.
c.
4. a.
b.
c.
d.
5. Hydrogen can accept only one more electron; multiple bonds require more than one electron pair to be shared.
6. MgO is an ionic compound (Mg transfers two electrons to O). The electrons are not shared hence it's incorrect to draw a double bond.
This is the Lewis dot structure of MgO.
Concept Review Exercises
1. What is the name for the distance between two atoms in a covalent bond?
2. What does the electronegativity of an atom indicate?
3. What type of bond is formed between two atoms if the difference in electronegativities is small? Medium? Large?
Answers
1. bond length
2. Electronegativity is a qualitative measure of how much an atom attracts electrons in a covalent bond.
3. nonpolar; polar; ionic
Exercises
1. Which is longer—a C–H bond or a C–O bond? (Refer to Table \(1\).)
2. Which is shorter—an N–H bond or a C–H bond? (Refer to Table \(1\).)
3. A nanometer is 10−9 m. Using the data in Table \(1\) and Table \(2\), determine the length of each bond in nanometers.
1. a C–O bond
2. a C=O bond
3. an H–N bond
4. a C≡N bond
4. An angstrom (Å) is defined as 10−10 m. Using Table \(1\) and Table \(2\), determine the length of each bond in angstroms.
1. a C–C bond
2. a C=C bond
3. an N≡N bond
4. an H–O bond
5. Refer to Exercise 3. Why is the nanometer unit useful as a unit for expressing bond lengths?
6. Refer to Exercise 4. Why is the angstrom unit useful as a unit for expressing bond lengths?
7. Using Figure \(3\), determine which atom in each pair has the higher electronegativity.
1. H or C
2. O or Br
3. Na or Rb
4. I or Cl
8. Using Figure \(3\), determine which atom in each pair has the lower electronegativity.
1. Mg or O
2. S or F
3. Al or Ga
4. O or I
9. Will the electrons be shared equally or unequally across each covalent bond? If unequally, to which atom are the electrons more strongly drawn?
1. a C–O bond
2. an F–F bond
3. an S–N bond
4. an I–Cl bond
10. Will the electrons be shared equally or unequally across each covalent bond? If unequally, to which atom are the electrons more strongly drawn?
• a C–C bond
• a S–Cl bond
• an O–H bond
• an H–H bond
11. Arrange the following bonds from least polar to most polar: H-F, H-N, H-O, H-C
12. Arrange the following bonds from least polar to most polar: C-F, C-N, C-O, C-C
Answers
1. A C–O bond is longer.
2. An H-N bond is shorter than an H-C bond.
1. 0.143 nm
2. 0.120 nm
3. 0.100 nm
4. 0.116 nm
4.
1. 1.54 Å
2. 1.34 Å
3. 1.10 Å
4. 0.97 Å
1. Actual bond lengths are very small, so the nanometer unit makes the expression of length easier to understand.
6. Actual bond lengths are very small, so the angstrom unit makes the expression of length easier to understand.
1. C
2. O
3. Na
4. Cl
8.
1. Mg
2. S
3. Al
4. I
9.
1. unequally toward the O
2. equally
3. unequally toward the N
4. unequally toward the Cl
10.
• equally
• unequally toward the Cl
• unequally toward the O
• equally
11. The electronegativity difference increases from 0.4; 0.9; 1.4; 1.9. Hence, the least to most polar: H-C, H-N, H-O, H-F
12. The electronegativity difference increases from 0; 0.5; 1.0; 1.5. Hence, the least to most polar: C-C, C-N, C-O, C-F
Concept Review Exercises
1. How do you determine the molecular mass of a covalent compound?
2. How do you determine the shape of a molecule?
3. How do you determine whether a molecule is polar or nonpolar?
Answers
1. The molecular mass is the sum of the masses of the atoms in the formula.
2. The shape of a molecule is determined by the position of the atoms, which in turn is determined by the repulsion of the bonded and lone electron pairs around the central atom.
3. If all the bonds in a molecule are nonpolar, the molecule is nonpolar. If it contains identical polar bonds that are oriented symmetrically opposite each other (linear, trigonal planar or tetrahedral) then the molecule is nonpolar. If it contains polar bonds that don't cancel each other's effects, the molecule is polar.
Exercises
1. What is the molecular mass of each compound?
1. H2S
2. N2O4
3. ICl3
4. HCl
2. What is the molecular mass of each compound?
1. O2F2
2. CCl4
3. C6H6
4. SO3
3. Aspirin (C9H8O4) is a covalent compound. What is its molecular mass?
4. Cholesterol (C27H46O) is a biologically important compound. What is its molecular mass?
5. What is the shape of each molecule? State whether it is polar or nonpolar.
1. H2S
2. COCl2
3. SO2
6. What is the shape of each molecule? State whether it is polar or nonpolar.
1. NBr3
2. SF2
3. SiH4
7. Predict the shape of nitrous oxide (N2O), which is used as an anesthetic. A nitrogen atom is in the center of this three-atom molecule. Is this polar?
8. Predict the shape of acetylene (C2H2), which has the two carbon atoms in the middle of the molecule with a triple bond. What generalization can you make about the shapes of molecules that have more than one central atom?
Answers
1. 34.08 amu
2. 92.02 amu
3. 233.25 amu
4. 36.46 amu
2. What is the molecular mass of each compound?
1. 70.00 amu
2. 153.81 amu
3. 78.12 amu
4. 80.06 amu
1. 180.17 amu
4. 386.73 amu
1. bent; polar
2. trigonal planar; nonpolar
3. bent; polar
6.
1. pyramidal; polar
2. bent; polar
3. tetrahedral; nonpolar
7. linear; polar
8. linear; in a molecule with more than one central atom, the geometry around each central atom needs to be examined.
Concept Review Exercises
1. What is organic chemistry?
2. What is a functional group? Give at least two examples of functional groups.
Answers
1. Organic chemistry is the study of the chemistry of carbon compounds.
2. A functional group is a specific structural arrangement of atoms or bonds that imparts a characteristic chemical reactivity to the molecule; alcohol group and carboxylic group (answers will vary).
Exercises
1. Give three reasons why carbon is the central element in organic chemistry.
2. Are organic compounds based more on ionic bonding or covalent bonding? Explain.
3. Identify the type of hydrocarbon in each structure.
4. Identify the type of hydrocarbon in each structure.
5. Identify the functional group(s) in each molecule.
6. Identify the functional group(s) in each molecule.
7. How many functional groups described in this section contain carbon and hydrogen atoms only? Name them.
8. What is the difference in the ways the two oxygen atoms in the carboxyl group are bonded to the carbon atom?
Answers
1. Carbon atoms bond reasonably strongly with other carbon atoms. Carbon atoms bond reasonably strongly with atoms of other elements. Carbon atoms make a large number of covalent bonds (four).
2. Organic compounds are based on covalent bonding or electron sharing. The atoms C, H, O, N that make up organic compounds are all nonmetals.
3.
a. alkane
b. alkene
c. alkene
d. alkyne
4.
a. alkene
b. alkane
c. alkyne
d. alkene
5.
a. alcohol
b. carboxyl
c. alcohol
d. alkene and alkyne
6.
1. a carbon-carbon double bond and alcohol
2. carboxyl group
3. carbon-carbon double bond and alcohol
4. carbon-carbon double bond; alcohol and carboxyl group
7. two; carbon-carbon double bonds and carbon-carbon triple bonds
8. There are two oxygen atoms in a carboxyl group: one is double-bonded while the other is OH, single bonded to the same carbon atom.
Additional Exercises
Use the atomic masses found in Figure 2.7.1
1. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of each molecule of (a) H2S (b) N2O4 (c) ICl3 (d) NCl3?
2. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of (a) O2F2 (b) CCl4 (c) C6H6 (d) SO3?
3. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of 5.00 × 1022 molecules of C9H8O4?
4. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of 1.885 × 1020 molecules of C27H46O?
5. Acetic acid has the following structure:
This molecule can lose a hydrogen ion (H+) and the resulting anion can combine with other cations, such as Na+:
Name this ionic compound.
6. Formic acid (HCOOH) loses a hydrogen ion to make the formate ion (HCOO). Write the formula for each ionic compound: potassium formate, calcium formate, and ferric formate.
7. Cyanogen has the formula C2N2. Propose a bonding scheme that gives each atom the correct number of covalent bonds. (Hint: the two carbon atoms are in the center of a linear molecule.)
8. How many carbon–carbon single bonds, linked together, are needed to make a carbon chain that is 1.000 cm long?
9. How many carbon–carbon double bonds, linked together, are needed to make a carbon chain that is 1.000 cm long?
10. In addition to themselves, what other atoms can carbon atoms bond with and make covalent bonds that are nonpolar (or as nonpolar as possible)?
11. What is the greatest possible electronegativity difference between any two atoms? Use Figure 4.4 to find the answer.
12. Acetaminophen, a popular painkiller, has the following structure:
Name the recognizable functional groups in this molecule. Do you think there are other groups of atoms in this molecule that might qualify as functional groups?
13. Glutamic acid is the parent compound of monosodium glutamate (known as MSG), which is used as a flavor enhancer. Glutamic acid has the following structure:
Name the functional groups you recognize in this molecule. Do you think there are other groups of atoms in this molecule that might qualify as functional groups?
Answers
1.
a: 5.661 × 10−23 g
b: 1.528 × 10−22 g
c: 3.874 × 10−22 g
d: 1.999 × 10−22 g
2.
a: 1.163 × 10−22 g
b: 2.555 × 10−22 g
c: 1.298 × 10−22 g
d: 1.330 × 10−22 g
3. 14.96 g
4. 0.1211 g
5. sodium acetate
6.
a. KHCOO
b. Ca(HCOO)2
c. Fe(HCOO)3
1. :N≡C–C≡N:
2. 6.49 × 107 C-C bonds
3. 7.46 × 107 C=C bonds
4. Hydrogen atoms make relatively nonpolar bonds with carbon atoms.
5. The greatest electronegativity difference is 3.2, between F and Rb.
6. alcohol; the ring with double bonds, and the O=C-NH are also likely functional groups.
7. carboxyl and -NH2 functional groups
Additional Questions
1. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of each molecule of (a) H2S (b) N2O4 (c) ICl3 (d) NCl3?
2. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of (a) O2F2 (b) CCl4 (c) C6H6 (d) SO3?
3. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of 5.00 × 1022 molecules of C9H8O4?
4. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of 1.885 × 1020 molecules of C27H46O?
5. Acetic acid has the following structure:
This molecule can lose a hydrogen ion (H+) and the resulting anion can combine with other cations, such as Na+:
Name this ionic compound.
6. Formic acid (HCOOH) loses a hydrogen ion to make the formate ion (HCOO). Write the formula for each ionic compound: potassium formate, calcium formate, and ferric formate.
7. Cyanogen has the formula C2N2. Propose a bonding scheme that gives each atom the correct number of covalent bonds. (Hint: the two carbon atoms are in the center of a linear molecule.)
8. The molecular formula C3H6 represents not only propene, a compound with a carbon–carbon double bond, but also a molecule that has all single bonds. Draw the molecule with formula C3H6 that has all single bonds.
9. How many carbon–carbon single bonds, linked together, are needed to make a carbon chain that is 1.000 cm long?
10. How many carbon–carbon double bonds, linked together, are needed to make a carbon chain that is 1.000 cm long?
11. In addition to themselves, what other atoms can carbon atoms bond with and make covalent bonds that are nonpolar (or as nonpolar as possible)?
12. What is the greatest possible electronegativity difference between any two atoms? Use Figure 4.4 to find the answer.
13. Acetaminophen, a popular painkiller, has the following structure:
Name the recognizable functional groups in this molecule. Do you think there are other groups of atoms in this molecule that might qualify as functional groups?
14. Glutamic acid is the parent compound of monosodium glutamate (known as MSG), which is used as a flavor enhancer. Glutamic acid has the following structure:
Name the functional groups you recognize in this molecule. Do you think there are other groups of atoms in this molecule that might qualify as functional groups?
Answers
1. 1a: 5.75 × 10−23 g; 1b: 1.53 × 10−22 g; 1c: 3.88 × 10−22 g; 1d: 6.06 × 10−23 g
1. 14.96 g
1. sodium acetate
1. N≡C–C≡N
1. 6.49 × 107 bonds
1. Hydrogen atoms make relatively nonpolar bonds with carbon atoms.
1. alcohol; the N–H group, the ring with double bonds, and the C=O are also likely functional groups. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/04E%3A_Exercises.txt |
5.1: The Law of Conservation of Matter
1. What is the law of conservation of matter?
2. How does the law of conservation of matter apply to chemistry?
ANSWERS
1. The law of conservation of matter states that in any given system that is closed to the transfer of matter, the amount of matter in the system stays constant
2. The law of conservation of matter says that in chemical reactions, the total mass of the products must equal the total mass of the reactants.
Exercises
1. Express the law of conservation of matter in your own words.
2. Explain why the concept of conservation of matter is considered a scientific law.
3. Potassium hydroxide ($\ce{KOH}$) readily reacts with carbon dioxide ($\ce{CO2}$) to produce potassium carbonate ($\ce{K2CO3}$) and water ($\ce{H2O}$). How many grams of potassium carbonate is produced if 224.4 g of $\ce{KOH}$ reacted with 88.0 g of $\ce{CO2}$. The reaction also produced 36.0 g of water.
Answers
1. Matter may not be created or destroyed.
2. The concept is a scientific law because it is based on experimentation.
3. 276.4 g
Concept Review Exercises
1. What are the parts of a chemical equation?
2. Explain why chemical equations need to be balanced.
Answers
a. reactants and products
b. Chemical equations need to be balanced to satisfy the law of conservation of matter.
Exercises
1. Write a chemical equation to express the fact that hydrogen gas and solid iodine react to make gaseous hydrogen iodide. Make sure the equation satisfies the law of conservation of matter.
2. Write a chemical equation to express the fact that sodium metal and chlorine gas react to make solid sodium chloride. Make sure the equation satisfies the law of conservation of matter.
3. Write an equation expressing the fact that hydrogen gas and fluorine gas react to make gaseous hydrogen fluoride. Make sure the equation satisfies the law of conservation of matter.
4. Write an equation expressing the fact that solid potassium and fluorine gas react to make solid potassium fluoride. Make sure the equation satisfies the law of conservation of matter.
5. Mercury reacts with oxygen to make mercury(II) oxide. Write a balanced chemical equation that summarizes this reaction.
6. Octane (C8H18) reacts with oxygen to make carbon dioxide and water. Write a balanced chemical equation that summarizes this reaction.
7. Propyl alcohol (C3H7OH) reacts with oxygen to make carbon dioxide and water. Write a balanced chemical equation that summarizes this reaction.
8. Sulfuric acid (H2SO4) reacts with iron metal to make iron(III) sulfate and hydrogen gas. Write a balanced chemical equation that summarizes this reaction.
9. Balance each equation.
1. MgCl2 + K → KCl + Mg
2. C6H12O6 + O2 → CO2 + H2O
3. NaN3 → Na + N2 (This is the reaction used to inflate airbags in cars.)
10. Balance each equation.
1. NH4NO3 → N2O + H2O
2. TiBr4 + H2O → TiO2 + HBr
3. C3H5N3O9 → CO2 + N2 + O2 + H2O (This reaction represents the decomposition of nitroglycerine.)
11. Balance each equation.
1. NH3 + O2 → NO + H2O
2. Li + N2 → Li3N
3. AuCl → Au + AuCl3
12. Balance each equation.
1. NaOH + H3PO4 → Na3PO4 + H2O
2. N2H4 + Cl2 → N2 + HCl
3. Na2S + H2S → NaSH
13. Chromium(III) oxide reacts with carbon tetrachloride to make chromium(III) chloride and phosgene (COCl2). Write the balanced chemical equation for this reaction.
14. The reaction that occurs when an Alka-Seltzer tablet is dropped into a glass of water has sodium bicarbonate reacting with citric acid (H3C6H5O7) to make carbon dioxide, water, and sodium citrate (Na3C6H5O7). Write the balanced chemical equation for this reaction.
15. When sodium hydrogen carbonate is used to extinguish a kitchen fire, it decomposes into sodium carbonate, carbon dioxide and water. Write a balanced chemical equation for this reaction.
16. Elemental bromine gas can be generated by reacting sodium bromide with elemental chlorine. The other product is sodium chloride. Write a balanced chemical equation for this reaction.
Answers
1. H2(g) + I2(s) → 2HI(g)
2. 2Na(s) + Cl2(g) → 2NaCl(s)
1. H2(g) + F2(g) → 2HF(g)
4. 2K(s) + F2(g) → 2KF(s)
1. 2Hg + O2 → 2HgO
6. 2C8H18 + 25O2 → 16CO2 + 18H2O
1. 2C3H7OH + 9O2 → 6CO2 + 8H2O
8. 3H2SO4 + 2Fe → Fe2(SO4)3 + 3H2
1. MgCl2 + 2K → 2KCl + Mg
2. C6H12O6 + 6O2 → 6CO2 + 6H2O
3. 2NaN3 → 2Na + 3N2
10.
1. NH4NO3 → N2O + 2H2O
2. TiBr4 + 2H2O → TiO2 + 4HBr
3. 4C3H5N3O9 → 12CO2 + 6N2 + O2 + 10H2O
11.
1. 4NH3 + 5O2 → 4NO + 6H2O
2. 6Li + N2 → 2Li3N
3. 3AuCl → 2Au + AuCl3
12.
1. 3NaOH + H3PO4 → Na3PO4 + 3H2O
2. N2H4 + 2Cl2 → N2 + 4HCl
3. Na2S + H2S → 2NaSH
13. Cr2O3 + 3CCl4 → 2CrCl3 + 3COCl2
14. 3NaHCO3 + H3C6H5O7 → 3CO2 + 3H2O + Na3C6H5O7
15. 2NaHCO3 → Na2CO3 + CO2 + H2O
16. 2NaBr + Cl2 → Br2 + 2NaCl
Concept Review Exercises
1. Explain how stoichiometric ratios are constructed from a chemical equation.
2. Why is it necessary for a chemical equation to be balanced before it can be used to construct conversion factors?
Answers
1. Stoichiometric ratios are made using the coefficients of the substances in the balanced chemical equation.
2. A balanced chemical equation is necessary so one can construct the proper stoichiometric ratios.
Exercises
1. Balance this equation and write every stoichiometric ratio you can from it.
NH4NO3 → N2O + H2O
2. Balance this equation.
N2 + H2 → NH3
3. Balance this equation and write every stoichiometric ratio you can from it.
Fe2O3 + C → Fe + CO2
4. Balance this equation.
Fe2O3 + CO → Fe + CO2
5. Balance this equation and determine how many molecules of CO2 are formed if 15 molecules of C6H6 are reacted.
C6H6 + O2 → CO2 + H2O
6. Balance this equation and determine how many formula units of Ag2CO3(s) are produced if 20 formula units of Na2CO3 are reacted.
Na2CO3(aq) + AgNO3(aq) → NaNO3(aq) + Ag2CO3(s)
7. Copper metal reacts with nitric acid according to this equation:
3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(ℓ)
1. Verify that this equation is balanced.
2. How many Cu atoms will react if 488 molecules of aqueous HNO3 are reacted?
8. Gold metal reacts with a combination of nitric acid and hydrochloric acid according to this equation:
Au(s) + 3HNO3(aq) + 4HCl(aq) → HAuCl4(aq) + 3NO2(g) + 3H2O(ℓ)
1. Verify that this equation is balanced.
2. How many Au atoms react with 639 molecules of aqueous HNO3?
9. Sulfur can be formed by reacting sulfur dioxide with hydrogen sulfide at high temperatures according to this equation:
SO2(g) + 2H2S(g) → 3S(g) + 2H2O(g)
1. Verify that this equation is balanced.
2. How many S atoms will be formed from by reacting 1,078 molecules of H2S?
10. Nitric acid is made by reacting nitrogen dioxide with water:
3NO2(g) + H2O(ℓ) → 2HNO3(aq) + NO(g)
1. Verify that this equation is balanced.
2. How many molecules of NO will be formed by reacting 2,268 molecules of NO2?
Answers
1. NH4NO3 → N2O + 2H2O; the stoichiometric ratios are $\mathrm{\dfrac{1NH_4NO_3}{1N_2O}\:,\: \dfrac{1NH_4NO_3}{2H_2O}\:,\: \dfrac{1N_2O}{2H_2O}\:,}$ and their reciprocals.
2. N2 + 3H2 → 2NH3
1. 2Fe2O3 + 3C → 4Fe + 3CO2; the stoichiometric ratios are $\mathrm{\dfrac{2Fe_2O_3}{3C}\:,\: \dfrac{2Fe_2O_3}{4Fe}\:,\: \dfrac{2Fe_2O_3}{3CO_2}\:,\: \dfrac{3C}{4Fe}\:,\: \dfrac{3C}{3CO_2}\:,\: \dfrac{4Fe}{3CO_2}\:,\: }$ and their reciprocals.
4. Fe2O3 + 3CO → 2Fe + 3CO2
1. 2C6H6 + 15O2 → 12CO2 + 6H2O; 90 molecules
6. Na2CO3(aq) + 2AgNO3(aq) → 2NaNO3(aq) + Ag2CO3(s); 20 formula units
7.
1. It is balanced.
2. 183 atoms
8.
1. It is balanced.
2. 213 atoms
9.
1. It is balanced.
2. 1,617 atoms
10.
1. It is balanced.
2. 756 molecules
Concept Review Exercises
1. What is the difference between a combination reaction and a combustion reaction?
2. Give the distinguishing characteristic(s) of a decomposition reaction.
3. How do we recognize a combustion reaction?
Answers
1. A combination reaction produces a certain substance; a combustion reaction is a vigorous reaction, usually a combination with oxygen, that is accompanied by the production of light and/or heat.
2. In a decomposition reaction, a single substance reacts to make multiple substances as products.
3. A combustion reaction is typically a vigorous reaction accompanied by light and/or heat, usually because of reaction with oxygen.
Exercises
1. Identify each type of reaction.
1. C6H5CH3 + 9O2 → 7CO2 + 4H2O
2. 2NaHCO3 → Na2CO3 + H2O + CO2
3. C + 2H2 → CH4
2. Identify each type of reaction.
1. P4O10 + 6H2O → 4H3PO4
2. FeO + SO3 → FeSO4
3. CaCO3(s) → CO2(g) + CaO(s)
3. Identify each type of reaction.
1. 2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
2. Hg(ℓ) + ½O2 (g) → HgO(s)
3. CH2CH2(g) + Br2(ℓ) → CH2BrCH2Br
4. Identify each type of reaction.
1. Ti(s) + O2(g) → TiO2(s)
2. H2SO3(aq) → H2O(ℓ) + SO2(g)
3. 3O2(g) → 2O3(g)
Answers
1.
a. combustion
b. decomposition
c. combination
2.
a. combination
b. combination
c. decomposition
3.
a. decomposition
b. combustion (also combination)
c. combination
4.
a. combination
b. decomposition
c. combination
Concept Review Exercises
1. Give two different definitions for oxidation and reduction.
2. Give an example of each definition of oxidation and reduction.
Answers
1. Oxidation is the loss of electrons or the addition of oxygen; reduction is the gain of electrons or the addition of hydrogen.
2. Zn → Zn2+ +2e (oxidation); C2H4 + H2 → C2H6 (reduction) (answers will vary)
Exercises
1. Which reactions are redox reactions? For those that are redox reactions, identify the oxidizing and reducing agents.
1. NaOH + HCl → H2O + NaCl
2. 3Mg + 2AlCl3 → 2Al + 3MgCl2
3. H2O2 + H2 → 2H2O
4. KCl + AgNO3 → AgCl + KNO3
2. Which reactions are redox reactions? For those that are redox reactions, identify the oxidizing and reducing agents.
1. 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O
2. 2C2H6 + 7O2 → 4CO2 + 6H2O
3. 2NaHCO3 → Na2CO3 + CO2 + H2O
4. 2K + 2H2O → 2KOH + H2
3. Balance each redox reaction by writing appropriate half reactions and combining them to cancel the electrons.
1. Ca(s) + H+(aq) → Ca2+(aq) + H2(g)
2. I(aq) + Br2(ℓ) → Br(aq) + I2(s)
4. Balance each redox reaction by writing appropriate half reactions and combining them to cancel the electrons.
1. Fe(s) + Sn4+(aq) → Fe3+(aq) + Sn2+(aq)
2. Pb(s) + Pb4+(aq) → Pb2+(aq) (Hint: both half reactions will yield the same product.)
Answers
1.
a. no
b. yes; oxidizing agent: AlCl3; reducing agent: Mg
c. yes; oxidizing agent: H2O2; reducing agent: H2
d. no
2.
a. yes; oxidizing agent: HNO3; reducing agent: Cu
b. yes; oxidizing agent: O2; reducing agent: C2H6
c. no
d. yes; oxidizing agent: H2O; reducing agent: K
3.
1. Combined: Ca + 2H+ → Ca2+ + H2
2. Combined: Br2 + 2I → 2Br + I2
4.
a. (Fe → Fe3+ + 3e) x 2
(Sn4++ 2e→ Sn2+) x 3
Combined: 2Fe + 3Sn4+ → 2Fe3+ + 3Sn2+
b. Pb → Pb2+ + 2e
Pb4++ 2e→ Pb2+
Combined: Pb + Pb4+ → 2Pb2+
Concept Review Exercise
1. Give some biochemical examples of oxidation and reduction reactions.
Answer
1. photosynthesis and antioxidants in foods (answers will vary)
Exercises
1. A typical respiratory reaction discussed in the text is the oxidation of glucose (C6H12O6):
C6H12O6 + 6O2 → 6CO2 + 6H2O
Is this a redox reaction? If so, what are the oxidizing and reducing agents?
2. The major net reaction in photosynthesis is as follows:
6CO2 + 6H2O → C6H12O6 + 6O2
Is this a redox reaction? If so, what are the oxidizing and reducing agents?
3. What would be the ultimate organic product if CH3CH2CH2OH were to react with a solution of K2Cr2O7?
4. What would be the ultimate organic product if CH3CH2CH2CH2OH were to react with a solution of K2Cr2O7?
5. What would be the final organic product if CH3CH2CHOHCH3 were to react with a solution of K2Cr2O7?
6. What would be the major organic product if CH3CH2CHOHCH2CH3 were to react with a solution of K2Cr2O7?
7. What alcohol is produced in the reduction of acetone [(CH3)2CO]?
8. What alcohol is produced in the reduction of propanal (CH3CH2CHO)?
Answers
1. yes; oxidizing agent: O2; reducing agent: C6H12O6
2. yes; oxidizing agent: CO2; reducing agent: H2O
1. CH3CH2COOH
4. CH3CH2CH2COOH
1. CH3CH2C(O)CH3, where the carbon is double bonded to the oxygen
6. CH3CH2C(O)CH2CH3, carbon #3 is double bonded to the oxygen
7. CH3CHOHCH3, or isopropyl alcohol
8. CH3CH2CH2OH
1.
Additional Exercises
1. Isooctane (C8H18) is used as a standard for comparing gasoline performance. Write a balanced chemical equation for the combustion of isooctane.
2. Heptane (C7H16), like isooctane (see Exercise 1), is also used as a standard for determining gasoline performance. Write a balanced chemical equation for the combustion of heptane.
3. What is the difference between a combination reaction and a redox reaction? Are all combination reactions also redox reactions? Are all redox reactions also combination reactions?
4. Are combustion reactions always redox reactions as well? Explain.
5. A friend argues that the equation
Fe2+ + Na → Fe + Na+
is balanced because each side has one iron atom and one sodium atom. Explain why your friend is incorrect.
6. Some antacids contain aluminum hydroxide [Al(OH)3]. This compound reacts with excess hydrochloric acid (HCl) in the stomach to neutralize it. If the products of this reaction are water and aluminum chloride, what is the balanced chemical equation for this reaction? What is the stoichiometric ratio between the number of HCl molecules made to the number of H2O molecules made?
7. Sulfuric acid is made in a three-step process: (1) the combustion of elemental sulfur to produce sulfur dioxide, (2) the continued reaction of sulfur dioxide with oxygen to produce sulfur trioxide, and (3) the reaction of sulfur trioxide with water to make sulfuric acid (H2SO4). Write balanced chemical equations for all three reactions.
8. If the products of glucose metabolism are carbon dioxide and water, what is the balanced chemical equation for the overall process? What is the stoichiometric ratio between the number of CO2 molecules made to the number of H2O molecules made?
9. Historically, the first true battery was the Leclanché cell, named after its discoverer, Georges Leclanché. It was based on the following reaction:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Identify what is being oxidized, what is being reduced, and the respective reducing and oxidizing agents.
Answers
1. 2C8H18 + 25O2 → 16CO2 + 18H2O
2. C7H16 + 11O2→ 7CO2+ 8H2O
1. A combination reaction makes a new substance from more than one reactant; a redox reaction rearranges electrons. Most (not all) combination reactions are redox reactions. Not all redox reactions are combination reactions.
4. All combustion reactions are redox reactions. In combustion a chemical is combining with oxygen and that chemical is being oxidized. Oxygen, on the other hand, is being reduced.
1. Your friend is incorrect because the number of electrons transferring is not balanced. A balanced equation must not only have the same number of atoms of each element on each side of the equation but must also have the same charge on both sides.
6. Al(OH)3 + 3HCl → 3H2O + AlCl3; 1:1
1. (1) S + O2 → SO2; (2) 2SO2 + O2 → 2SO3; (3) SO3 + H2O → H2SO4
8. C6H12O6 + 6O2→ 6CO2+ 6H2O; 1:1
1. oxidized and reducing agent: Zn; reduced and oxidizing agent: Cu2+ | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/05E%3A_Exercises.txt |
Additional Exercises
1. If the average male has a body mass of 70 kg, of which 60% is water, how many moles of water are in an average male?
2. If the average female is 60.0 kg and contains 0.00174% iron, how many moles of iron are in an average female?
3. How many moles of each element are present in 2.67 mol of each compound?
1. HCl
2. H2SO4
3. Al(NO3)3
4. Ga2(SO4)3
4. How many moles of each element are present in 0.00445 mol of each compound?
1. HCl
2. H2SO4
3. Al2(CO3)3
4. Ga2(SO4)3
5. What is the mass of one hydrogen atom in grams? What is the mass of one oxygen atom in grams? Do these masses have a 1:16 ratio, as expected?
6. What is the mass of one sodium atom in grams?
7. If 6.63 × 10−6 mol of a compound has a mass of 2.151 mg, what is the molar mass of the compound?
8. Hemoglobin (molar mass is approximately 64,000 g/mol) is the major component of red blood cells that transports oxygen and carbon dioxide in the body. How many moles are in 0.034 g of hemoglobin?
Answers
1. 2,330 mol
1. 2.67 mol of H and 2.67 mol of Cl
2. 5.34 mol of H, 2.67 mol of S, and 10.68 mol of O
3. 2.67 mol of Al, 8.01 mol of N, and 24.03 mol of O
4. 5.34 mol of Ga, 8.01 mol of S, and 32.04 mol of O
1. H = 1.66 × 10−24 g and O = 2.66 × 10−23 g; yes, they are in a 1:16 ratio.
1. 324 g/mol
07.E: Exercises
Concept Review Exercises
1. What is the relationship between energy and heat?
2. What units are used to express energy and heat?
Answers
1. Heat is the exchange of energy from one part of the universe to another. Heat and energy have the same units.
2. Joules and calories are the units of energy and heat.
Exercises
1. Define energy.
2. What is heat?
3. What is the relationship between a calorie and a joule? Which unit is larger?
4. What is the relationship between a calorie and a kilocalorie? Which unit is larger?
5. Express 1,265 cal in kilocalories and in joules.
6. Express 9,043.3 J in calories and in kilocalories.
7. One kilocalorie equals how many kilojoules?
8. One kilojoule equals how many kilocalories?
9. Many nutrition experts say that an average person needs 2,000 Cal per day from his or her diet. How many joules is this?
10. Baby formula typically has 20.0 Cal per ounce. How many ounces of formula should a baby drink per day if the RDI is 850 Cal?
Answers
1. Energy is the ability to do work.
2. Heat is a form of enery (thermal) that can be transferred from one object to another.
1. 1 cal = 4.184 J; the calorie is larger.
4. 1 kilocalorie(1 Cal) = 1000 cal; the kcal is larger.
1. 1.265 kcal; 5,293 J
6. 2161.4 cal; 2.1614 kcal
1. 1 kcal = 4.184 kJ
8. 1 kJ = 0.239 kcal
9. 8.4 × 106 J
10. 42.5 oz
Concept Review Exercise
1. Describe the relationship between heat transfer and the temperature change of an object.
2. Describe what happens when two objects that have different temperatures come into contact with one another.
Answer
1. Heat is equal to the product of the mass, the change in temperature, and a proportionality constant called the specific heat.
2. The temperature of the hot object decreases and the temperature of the cold object increases as heat is transferred from the hot object to the cold object. The change in temperature of each depends on the identity and properties of each substance.
Exercises
1. The melting point of mercury is −38.84oC. Convert this value to degrees Fahrenheit and Kelvin.
2. A pot of water is set on a hot burner of a stove. What is the direction of heat flow?
3. Some uncooked macaroni is added to a pot of boiling water. What is the direction of heat flow?
4. How much energy in calories is required to heat 150 g of H2O from 0°C to 100°C?
5. How much energy in calories is required to heat 125 g of Fe from 25°C to 150°C?
6. If 250 cal of heat were added to 43.8 g of Al at 22.5°C, what is the final temperature of the aluminum?
7. If 195 cal of heat were added to 33.2 g of Hg at 56.2°C, what is the final temperature of the mercury?
8. A sample of copper absorbs 145 cal of energy, and its temperature rises from 37.8°C to 41.7°C. What is the mass of the copper?
9. A large, single crystal of sodium chloride absorbs 98.0 cal of heat. If its temperature rises from 22.0°C to 29.7°C, what is the mass of the NaCl crystal?
10. If 1.00 g of each substance in Table 7.3 were to absorb 100 cal of heat, which substance would experience the largest temperature change?
11. If 1.00 g of each substance in Table 7.3 were to absorb 100 cal of heat, which substance would experience the smallest temperature change?
12. Determine the heat capacity of a substance if 23.6 g of the substance gives off 199 cal of heat when its temperature changes from 37.9°C to 20.9°C.
13. What is the heat capacity of gold if a 250 g sample needs 133 cal of energy to increase its temperature from 23.0°C to 40.1°C?
Answers
1. -37.910F and 234.31 K
2. Heat flows into the pot of water.
3. Heat flows to the macaroni.
4. 15,000 cal
5. 1,690 cal
6. 49.0°C
7. 234°C
8. 404 g
9. 61 g
10. Mercury would experience the largest temperature change.
11. hydrogen (H2)
12. 0.496 cal/g•°C
13. 0.031 cal/g•°C
Concept Review Exercises
1. Explain what happens when heat flows into or out of a substance at its melting point or boiling point.
2. How does the amount of heat required for a phase change relate to the mass of the substance?
3. What is the direction of heat transfer in boiling water?
4. What is the direction of heat transfer in freezing water?
5. What is the direction of heat transfer in sweating?
Answers
1. The energy goes into changing the phase, not the temperature.
2. The amount of heat is a constant per gram of substance.
3. Boiling. Heat is being added to the water to get it from the liquid state to the gas state.
4. Freezing. Heat is exiting the system in order to go from liquid to solid. Another way to look at it is to consider the opposite process of melting. Energy is consumed (endothermic) to melt ice (solid to liquid) so the opposite process (liquid to solid) must be exothermic.
5. Sweating. Heat is consumed to evaporate the moisture on your skin which lowers your temperature.
Exercises
1. How much energy is needed to melt 43.8 g of Au at its melting point of 1,064°C?
2. How much energy is given off when 563.8 g of NaCl solidifies at its freezing point of 801°C?
3. What mass of ice can be melted by 558 cal of energy?
4. How much ethanol (C2H5OH) in grams can freeze at its freezing point if 1,225 cal of heat are removed?
5. What is the heat of vaporization of a substance if 10,776 cal are required to vaporize 5.05 g? Express your final answer in joules per gram.
6. If 1,650 cal of heat are required to vaporize a sample that has a heat of vaporization of 137 cal/g, what is the mass of the sample?
7. What is the heat of fusion of water in calories per mole?
8. What is the heat of vaporization of benzene (C6H6) in calories per mole?
9. What is the heat of vaporization of gold in calories per mole?
10. What is the heat of fusion of iron in calories per mole?
Answers
1. 670 cal
2. 69,630 cal
3. 6.98 g
4. 27.10 g
1. 8,930 J/g
6. 12.0 g
1. 1,440 cal/mol
8. 7,350 cal/mol
9. 80,600 cal/mol
10. 3,530 cal/mol
Concept Review Exercises
1. What is the connection between energy and chemical bonds?
2. Why does energy change during the course of a chemical reaction?
3. Two different reactions are performed in two identical test tubes. In reaction A, the test tube becomes very warm as the reaction occurs. In reaction B, the test tube becomes cold. Which reaction is endothermic and which is exothermic? Explain.
4. Classify "burning paper" as endothermic or exothermic processes.
Answers
1. Chemical bonds have a certain energy that is dependent on the elements in the bond and the number of bonds between the atoms.
2. Energy changes because bonds rearrange to make new bonds with different energies.
3. Reaction A is exothermic because heat is leaving the system making the test tube feel hot. Reaction B is endothermic because heat is being absorbed by the system making the test tube feel cold.
4. "Burning paper" is exothermic because burning (also known as combustion) releases heat
Exercises
1. Using the data in Table 7.4.1, calculate the energy of one C–H bond (as opposed to 1 mol of C–H bonds). Recall that 1 mol = 6.022 x 1023 C–H bonds
2. Using the data in Table 7.4.1, calculate the energy of one C=C bond (as opposed to 1 mol of C=C bonds). Recall that 1 mol = 6.022 x 1023 C=C bonds
3. Is a bond-breaking process exothermic or endothermic?
4. Is a bond-making process exothermic or endothermic?
5. Is each chemical reaction exothermic or endothermic?
1. 2SnCl2(s) + 33 kcal → Sn(s) + SnCl4(s)
2. CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ) + 213 kcal
6. Is each chemical reaction exothermic or endothermic?
1. C2H4(g) + H2(g) → C2H6(g) + 137 kJ
2. C(s, graphite) + 1.9 kJ → C(s, diamond)
Answers
1. 1.661 × 10−19 cal
2. 2.408 x 10-19 cal
3. endothermic
4. exothermic
5.
a. endothermic
b. exothermic
6.
a. exothermic
b. endothermic
Concept Review Exercise
1. What is the energy content per gram of proteins, carbohydrates, and fats?
Answer
1. proteins and carbohydrates: 4 kcal/g; fats: 9 kcal/g
Exercises
1. An 8 oz serving of whole milk has 8.0 g of fat, 8.0 g of protein, and 13 g of carbohydrates. Approximately how many kilocalories does it contain?
2. A serving of potato chips has 160 kcal. If the chips have 15 g of carbohydrates and 2.0 g of protein, about how many grams of fat are in a serving of potato chips?
3. The average body temperature of a person is 37°C, while the average surrounding temperature is 22°C. Is overall human metabolism exothermic or endothermic?
4. Cold-blooded animals absorb heat from the environment for part of the energy they need to survive. Is this an exothermic or an endothermic process?
5. If the reaction ATP → ADP gives off 7.5 kcal/mol, then the reverse process, ADP → ATP requires 7.5 kcal/mol to proceed. How many moles of ADP can be converted to ATP using the energy from 1 serving of potato chips (see Exercise 2)?
6. If the oxidation of glucose yields 670 kcal of energy per mole of glucose oxidized, how many servings of potato chips (see Exercise 2) are needed to provide the same amount of energy?
1. 156 kcal
2. 10.2 g
3. exothermic
4. endothermic
5. 21.3 mol
6. 4.2 servings
Additional Exercises
1. Sulfur dioxide (SO2) is a pollutant gas that is one cause of acid rain. It is oxidized in the atmosphere to sulfur trioxide (SO3), which then combines with water to make sulfuric acid (H2SO4).
1. Write the balanced reaction for the oxidation of SO2 to make SO3. (The other reactant is diatomic oxygen.)
2. When 1 mol of SO2 reacts to make SO3, 23.6 kcal of energy are given off. If 100 lb (1 lb = 454 g) of SO2 were converted to SO3, what would be the total energy change?
2. Ammonia (NH3) is made by the direct combination of H2 and N2 gases according to this reaction:
N2(g) + 3H2(g) → 2NH3(g) + 22.0 kcal
1. Is this reaction endothermic or exothermic?
2. What is the overall energy change if 1,500 g of N2 are reacted to make ammonia?
3. A 5.69 g sample of iron metal was heated in boiling water to 99.8°C. Then it was dropped into a beaker containing 100.0 g of H2O at 22.6°C. Assuming that the water gained all the heat lost by the iron, what is the final temperature of the H2O and Fe?
4. A 5.69 g sample of copper metal was heated in boiling water to 99.8°C. Then it was dropped into a beaker containing 100.0 g of H2O at 22.6°C. Assuming that the water gained all the heat lost by the copper, what is the final temperature of the H2O and Cu?
5. When 1 g of steam condenses, 540 cal of energy is released. How many grams of ice can be melted with 540 cal?
6. When 1 g of water freezes, 79.9 cal of energy is released. How many grams of water can be boiled with 79.9 cal?
7. The change in energy is +65.3 kJ for each mole of calcium hydroxide [Ca(OH)2] according to the following reaction:
Ca(OH)2(s) → CaO(s) + H2O(g)
How many grams of Ca(OH)2 could be reacted if 575 kJ of energy were available?
8. The thermite reaction gives off so much energy that the elemental iron formed as a product is typically produced in the liquid state:
2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(ℓ) + 204 kcal
How much heat will be given off if 250 g of Fe are to be produced?
9. A normal adult male requires 2,500 kcal per day to maintain his metabolism.
1. Nutritionists recommend that no more than 30% of the calories in a person’s diet come from fat. At 9 kcal/g, what is the maximum mass of fat an adult male should consume daily?
2. At 4 kcal/g each, how many grams of protein and carbohydrates should an adult male consume daily?
10. A normal adult male requires 2,500 kcal per day to maintain his metabolism.
1. At 9 kcal/g, what mass of fat would provide that many kilocalories if the diet was composed of nothing but fats?
2. At 4 kcal/g each, what mass of protein and/or carbohydrates is needed to provide that many kilocalories?
11. The volume of the world’s oceans is approximately 1.34 × 1024 cm3.
1. How much energy would be needed to increase the temperature of the world’s oceans by 1°C? Assume that the heat capacity of the oceans is the same as pure water.
2. If Earth receives 6.0 × 1022 J of energy per day from the sun, how many days would it take to warm the oceans by 1°C, assuming all the energy went into warming the water?
12. Does a substance that has a small specific heat require a small or large amount of energy to change temperature? Explain.
13. Some biology textbooks represent the conversion of adenosine triphosphate (ATP) to adenosine diphosphate (ADP) and phosphate ions as follows:
ATP → ADP + phosphate + energy
What is wrong with this reaction?
14. Assuming that energy changes are additive, how much energy is required to change 15.0 g of ice at −15°C to 15.0 g of steam at 115°C? (Hint: you will have five processes to consider.)
Answers
1.
1. 2SO2 + O2 → 2SO3
2. 16,700 kcal
2.
1. exothermic
2. 1177 kcal
1. about 23.1°C
4. about 23.0°C
5. 6.76 g
6. 0.148 g
1. 652 g
8. 457 kcal
1.
1. 83.3 g
2. 438 g
10.
a. 278 g
b. 625 g
11.
1. 1.34 × 1024 cal
2. 93 days
12. A substance with smaller specific heat requires less energy per unit of mass to raise its temperature,
13. A reactant is missing: H2O is missing.
14. Total energy = 11,019 cal | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/06.E%3A_Exercises.txt |
Concept Review Exercise
1. What types of intermolecular interactions can exist in compounds?
2. What is the difference between covalent network and covalent molecular compounds?
Answer
1. ionic bonding, network covalent, dispersion forces, dipole-dipole interactions, and hydrogen bonding.
2. Covalent network compounds contain atoms that are covalently bonded to other individual atoms in a giant 3-dimensional network. Covalent molecular compounds contain individual molecules that are attracted to one another through dispersion, dipole-dipole or hydrogen bonding.
Exercises
1. List the three common phases in the order you are likely to find them—from lowest temperature to highest temperature.
2. List the three common phases in the order they exist from lowest energy to highest energy.
3. List these intermolecular interactions from weakest to strongest: London forces, hydrogen bonding, and ionic interactions.
4. List these intermolecular interactions from weakest to strongest: covalent network bonding, dipole-dipole interactions, and dispersion forces.
5. What type of intermolecular interaction is predominate in each substance?
1. water (H2O)
2. sodium sulfate (Na2SO4)
3. decane (C10H22)
6. What type of intermolecular interaction is predominate in each substance?
1. diamond (C, crystal)
2. helium (He)
3. ammonia (NH3)
7. Explain how a molecule like carbon dioxide (CO2) can have polar covalent bonds but be nonpolar overall.
8. Sulfur dioxide (SO2) has a formula similar to that of carbon dioxide (see Exercise 7) but is a polar molecule overall. What can you conclude about the shape of the SO2 molecule?
9. What are some of the physical properties of substances that experience covalent network bonding?
10. What are some of the physical properties of substances that experience only dispersion forces?
Answers
1. solid, liquid, and gas
2. solid, liquid, and gas
1. London forces, hydrogen bonding, and ionic interactions
4. dispersion, dipole-dipole, network covalent
1. hydrogen bonding
2. ionic interactions
3. dispersion forces
6. a. network covalent
b. dispersion
c. hydrogen bonding
1. The two covalent bonds are oriented in such a way that their dipoles cancel out.
8. SO2 is not a linear molecule. It has a bent or V-shape.
9. very hard, high melting point
10. very soft, very low melting point
Concept Review Exercise
1. How do the strengths of intermolecular interactions in solids and liquids differ?
Answer
1. Solids have stronger intermolecular interactions than liquids do.
Exercises
1. What are the general properties of solids?
2. What are the general properties of liquids
3. What are the general properties of gases?
4. What phase or phases have a definite volume? What phase or phases do not have a definite volume?
5. Name a common substance that forms a crystal in its solid state.
6. Name a common substance that forms an amorphous solid in its solid state.
7. Are substances with strong intermolecular interactions likely to be solids at higher or lower temperatures? Explain.
8. Are substances with weak intermolecular interactions likely to be liquids at higher or lower temperatures? Explain.
9. State two similarities between the solid and liquid states.
10. State two differences between the solid and liquid states.
11. If individual particles are moving around with respect to each other, a substance may be in either the _______ or ________ state but probably not in the _______ state.
12. If individual particles are in contact with each other, a substance may be in either the ______ or _______ state but probably not in the ______ state.
Answers
1. hard, specific volume and shape, high density, cannot be compressed
2. fixed volume, no definite shape, high density, individual molecules touch each other but in a random way
1. variable volume and shape, low density, compressible
4. solid and liquid have definite volume; gas has no definite volume
1. sodium chloride (answers will vary)
6. glass
1. At higher temperatures, their intermolecular interactions are strong enough to hold the particles in place.
8. Substances with weak intermolecular interactions are likely to be liquids at lower temperatures. Their attractive forces are more easily broken hence they melt more readily.
1. high density; definite volume
10. Solids have definite shape while liquids don't. In solids, molecules occupy fixed positions in a pattern, while in liquids, the molecules are moving in a random arrangement.
11. liquid; gas; solid
12. solid; liquid; gas
Concept Review Exercise
1. What is pressure, and what units do we use to express it?
Answer
1. Pressure is the force per unit area; its units can be pascals, torr, millimeters of mercury, or atmospheres.
Exercises
1. What is the kinetic theory of gases?
2. According to the kinetic theory of gases, the individual gas particles are (always, frequently, never) moving.
3. Why does a gas exert pressure?
4. Why does the kinetic theory of gases allow us to presume that all gases will show similar behavior?
5. Arrange the following pressure quantities in order from smallest to largest: 1 mmHg, 1 Pa, and 1 atm.
6. Which unit of pressure is larger—the torr or the atmosphere?
7. How many torr are there in 1.56 atm?
8. Convert 760 torr into pascals.
9. Blood pressures are expressed in millimeters of mercury. What would be the blood pressure in atmospheres if a patient’s systolic blood pressure is 120 mmHg and the diastolic blood pressure is 82 mmHg? (In medicine, such a blood pressure would be reported as “120/82,” spoken as “one hundred twenty over eighty-two.”)
10. In weather forecasting, barometric pressure is expressed in inches of mercury (in. Hg), where there are exactly 25.4 mmHg in every 1 in. Hg. What is the barometric pressure in millimeters of mercury if the barometric pressure is reported as 30.21 in. Hg?
Answers
1. Gases are composed of tiny particles that are separated by large distances. Gas particles are constantly moving, experiencing collisions with other gas particles and the walls of their container. The velocity of gas particles is related to the temperature of a gas. Gas particles do not experience any force of attraction or repulsion with each other.
2. always
1. A gas exerts pressure as its particles rebound off the walls of its container.
4. Because the molecules are far apart and don't have attractive forces between them
1. 1 Pa, 1 mmHg, and 1 atm
6. atm
1. 1,190 torr
8. 98,700 Pa
9. 0.158 atm; 0.108 atm
10. 767.3 mm Hg
Concept Review Exercises
1. What properties do the gas laws help us predict?
2. What makes the ideal gas law different from the other gas laws?
Answers
1. Gas laws relate four properties: pressure, volume, temperature, and number of moles.
2. The ideal gas law does not require that the properties of a gas change.
Exercises
1. What conditions of a gas sample should remain constant for Boyle’s law to be used?
2. What conditions of a gas sample should remain constant for Charles’s law to be used?
3. Does the identity of a gas matter when using Boyle’s law? Why or why not?
4. Does the identity of a gas matter when using Charles’s law? Why or why not?
5. A sample of nitrogen gas is confined to a balloon that has a volume of 1.88 L and a pressure of 1.334 atm. What will be the volume of the balloon if the pressure is changed to 0.662 atm? Assume that the temperature and the amount of the gas remain constant.
6. A sample of helium gas in a piston has a volume of 86.4 mL under a pressure of 447 torr. What will be the volume of the helium if the pressure on the piston is increased to 1,240 torr? Assume that the temperature and the amount of the gas remain constant.
7. If a gas has an initial pressure of 24,650 Pa and an initial volume of 376 mL, what is the final volume if the pressure of the gas is changed to 775 torr? Assume that the amount and the temperature of the gas remain constant.
8. A gas sample has an initial volume of 0.9550 L and an initial pressure of 564.5 torr. What would the final pressure of the gas be if the volume is changed to 587.0 mL? Assume that the amount and the temperature of the gas remain constant.
9. A person draws a normal breath of about 1.00 L. If the initial temperature of the air is 18°C and the air warms to 37°C, what is the new volume of the air? Assume that the pressure and amount of the gas remain constant.
10. A person draws a normal breath of about 1.00 L. If the initial temperature of the air is −10°C and the air warms to 37°C, what is the new volume of the air? Assume that the pressure and the amount of the gas remain constant.
11. An air/gas vapor mix in an automobile cylinder has an initial temperature of 450 K and a volume of 12.7 cm3. The gas mix is heated to 565°C. If pressure and amount are held constant, what is the final volume of the gas in cubic centimeters?
12. Given the following conditions for a gas: Vi = 0.665 L, Ti = 23.6°C, Vf = 1.034 L. What is Tf in degrees Celsius and kelvins?
13. Assuming the amount remains the same, what must be the final volume of a gas that has an initial volume of 387 mL, an initial pressure of 456 torr, an initial temperature of 65.0°C, a final pressure of 1.00 atm, and a final temperature of 300 K?
14. When the nozzle of a spray can is depressed, 0.15 mL of gas expands to 0.44 mL, and its pressure drops from 788 torr to 1.00 atm. If the initial temperature of the gas is 22.0°C, what is the final temperature of the gas?
15. Use the ideal gas law to show that 1 mol of a gas at STP has a volume of about 22.4 L.
16. Use a standard conversion factor to determine a value of the ideal gas law constant R that has units of L•torr/mol•K.
17. How many moles of gas are there in a 27.6 L sample at 298 K and a pressure of 1.44 atm?
18. How many moles of gas are there in a 0.066 L sample at 298 K and a pressure of 0.154 atm?
19. A 0.334 mol sample of carbon dioxide gas is confined to a volume of 20.0 L and has a pressure of 0.555 atm. What is the temperature of the carbon dioxide in kelvins and degrees Celsius?
20. What must V be for a gas sample if n = 4.55 mol, P = 7.32 atm, and T = 285 K?
21. What is the pressure of 0.0456 mol of Ne gas contained in a 7.50 L volume at 29°C?
22. What is the pressure of 1.00 mol of Ar gas that has a volume of 843.0 mL and a temperature of −86.0°C?
23. A mixture of the gases $N_2$, $O_2$, and $Ar$ has a total pressure of 760 mm Hg. If the partial pressure of $N_2$ is 220 mm Hg and of $O_2$ is 470 mm Hg, What is the partial pressure of $Ar$?
24. What percent of the gas above is Ar?
25. Apply Henry’s Law to the diagram below to explain:
why oxygen diffuses from the alveoli of the lungs into the blood and from the blood into the tissues of the body. why carbon dioxide diffuses from the tissues into the blood and from the blood into the alveoli and then finally out into the atmosphere.
Answers
1. temperature and amount of the gas
2. pressure and amount of the gas
3. The identity does not matter because the variables of Boyle’s law do not identify the gas.
4. The identity does not matter because the variables of Charles law do not identify the gas.
5. 3.79 L
6. 31.1 mL
7. 92.1 mL
8. 918.4 torr
9. 1.07 L
10. 1.18 L
11. 23.7 cm3
12. 461 K; 1880C
13. 206 mL
14. 835 K; 5620C
15. The ideal gas law confirms that 22.4 L equals 1 mol.
16. $\dfrac{760\: torr}{1\: atm}$
17. 1.63 mol
18. 4.2 x 10-4 mol
19. 405 K; 132°C
20. 14.5 L
21. 0.151 atm
22. 18.2 atm
23. 70 mm Hg
24. 9.2%
25. Gases diffuse from high concentration to low concentration (Henry's Law). The partial pressure of oxygen is high in the alveoli and low in the blood of the pulmonary capillaries. As a result, oxygen diffuses across the respiratory membrane from the alveoli into the blood. It's also higher partial pressure in the blood than in the tissues, hence it transfers to the tissues. On the other hand, carbon dioxide diffuses from the tissues (highest CO2 partial pressure) and across the respiratory membrane from the blood into the alveoli and out to the atmosphere.
Additional Exercises
1. How many grams of oxygen gas are needed to fill a 25.0 L container at 0.966 atm and 22°C?
2. A breath of air is about 1.00 L in volume. If the pressure is 1.00 atm and the temperature is 37°C, what mass of air is contained in each breath? Use an average molar mass of 28.8 g/mol for air.
3. The balanced chemical equation for the combustion of propane is as follows:
$C_3H_{8(g)} + 5O_{2(g)} \rightarrow 3CO_{2(g)} + 4H_2O_{(ℓ)}$
1. If 100.0 g of propane are combusted, how many moles of oxygen gas are necessary for the reaction to occur?
2. At STP, how many liters of oxygen gas would that be?
4. The equation for the formation of ammonia gas (NH3) is as follows:
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$
At 500°C and 1.00 atm, 10.0 L of N2 gas are reacted to make ammonia.
1. If the pressures and temperatures of H2 and NH3 were the same as those of N2, what volume of H2 would be needed to react with N2, and what volume of NH3 gas would be produced?
2. Compare your answers to the balanced chemical equation. Can you devise a “shortcut” method to answer Exercise 4a?
5. At 20°C, 1 g of liquid H2O has a volume of 1.002 mL. What volume will 1 g of water vapor occupy at 20°C if its pressure is 17.54 mmHg? By what factor has the water expanded in going from the liquid phase to the gas phase?
6. At 100°C, 1 g of liquid H2O has a volume of 1.043 mL. What volume will 1 g of steam occupy at 100°C if its pressure is 760.0 mmHg? By what factor has the water expanded in going from the liquid phase to the gas phase?
7. Predict whether NaCl or NaI will have the higher melting point. Explain. (Hint: consider the relative strengths of the intermolecular interactions of the two compounds.)
8. Predict whether CH4 or CH3OH will have the lower boiling point. Explain. (Hint: consider the relative strengths of the intermolecular interactions of the two compounds.)
9. A standard automobile tire has a volume of about 3.2 ft3 (where 1 ft3 equals 28.32 L). Tires are typically inflated to an absolute pressure of 45.0 pounds per square inch (psi), where 1 atm equals 14.7 psi. Using this information with the ideal gas law, determine the number of moles of air needed to fill a tire if the air temperature is 18.0°C.
10. Another gas law, Amontons’s law, relates pressure and temperature under conditions of constant amount and volume:
$\mathrm{\dfrac{P_i}{T_i}=\dfrac{P_f}{T_f}}$
If an automobile tire (see Exercise 9) is inflated to 45.0 psi at 18.0°C, what will be its pressure if the operating temperature (i.e., the temperature the tire reaches when the automobile is on the road) is 45.0°C? Assume that the volume and the amount of the gas remain constant.
Answers
1. 31.9 g
2. 1.13 g
3.
1. 11.4 mol
2. 255 L
4.
a. 30.0 L H2 and 20.0 L NH3
b. the mole ratio in the balanced equation is the same as the volume ratio.
$\mathrm{10.0\: L\: N_2\times\dfrac{3\: L\ H_2}{1\: L\ N_2}=30.0\: L\ H_2}$
$\mathrm{10.0\: L\: N_2\times\dfrac{2\: L\ NH_3}{1\: L\ N_2}=20.0\: L\ NH_3}$
1. 57.75 L; an expansion of 57,600 times
6. 1.698 L; an expansion of 1,628 times
1. NaCl; with smaller anions, NaCl likely experiences stronger ionic bonding.
8. CH4 will have the lower boiling point because its intermolecular force (London dispersion force only) is weaker than those in CH3OH. Aside from London dispersion, CH3OH has dipole-dipole and hydrogen bonding.
9. 11.6 mol
10. 49.2 psi | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/08.E%3A_Exercises.txt |
Concept Review Exercises
1. What causes a solution to form?
2. How does the phrase like dissolves like relate to solutions?
Answers
1. Solutions form because a solute and a solvent have similar intermolecular interactions.
2. It means that substances with similar intermolecular interactions will dissolve in each other.
Exercises
1. Define solution.
2. Give several examples of solutions.
3. What is the difference between a solvent and a solute?
4. Can a solution have more than one solute in it? Can you give an example?
5. Does a solution have to be a liquid? Give several examples to support your answer.
6. Give at least two examples of solutions found in the human body.
7. Which substances will probably be soluble in water, a very polar solvent?
1. sodium nitrate (NaNO3)
2. hexane (C6H14)
3. isopropyl alcohol [(CH3)2CHOH]
4. benzene (C6H6)
8. Which substances will probably be soluble in toluene (C6H5CH3), a nonpolar solvent?
1. sodium nitrate (NaNO3)
2. hexane (C6H14)
3. isopropyl alcohol [(CH3)2CHOH]
4. benzene (C6H6)
9. The solubility of alcohols in water varies with the length of carbon chain. For example, ethanol (CH3CH2OH) is soluble in water in any ratio, while only 0.0008 mL of heptanol (CH3CH2CH2CH2CH2CH2CH2OH) will dissolve in 100 mL of water. Propose an explanation for this behavior.
10. Dimethyl sulfoxide [(CH3)2SO] is a polar liquid. Based on the information in Exercise 9, which do you think will be more soluble in it—ethanol or heptanol?
Answers
1. a homogeneous mixture
2. vinegar, dextrose IV, saline IV, coffee, tea, wine
1. A solvent is the majority component of a solution; a solute is the minority component of a solution.
4. yes. Coke or Pepsi has sugar, caffeine and carbon dioxide as solutes.
1. A solution does not have to be liquid; air is a gaseous solution, while some alloys are solid solutions (answers will vary).
6. Urine, plasma
1. probably soluble
2. probably not soluble
3. probably soluble
4. probably not soluble
8.
1. probably not soluble
2. probably soluble
3. probably not soluble
4. probably soluble
9. Small alcohol molecules have strong polar intermolecular interactions, so they dissolve in water. In large alcohol molecules, the nonpolar end overwhelms the polar end, so they do not dissolve very well in water.
10. Ethanol is a smaller molecule. It will be more soluble in water than heptanol.
Concept Review Exercises
1. What are some of the units used to express concentration?
2. Distinguish between the terms solubility and concentration.
Answers
1. % m/m, % m/v, ppm, ppb, molarity, and Eq/L (answers will vary)
2. Solubility is typically a limit to how much solute can dissolve in a given amount of solvent. Concentration is the quantitative amount of solute dissolved at any concentration in a solvent.
Exercises
1. Define solubility. Do all solutes have the same solubility?
2. Explain why the terms dilute or concentrated are of limited usefulness in describing the concentration of solutions.
3. If the solubility of sodium chloride (NaCl) is 30.6 g/100 mL of H2O at a given temperature, how many grams of NaCl can be dissolved in 250.0 mL of H2O?
4. If the solubility of glucose (C6H12O6) is 120.3 g/100 mL of H2O at a given temperature, how many grams of C6H12O6 can be dissolved in 75.0 mL of H2O?
5. How many grams of sodium bicarbonate (NaHCO3) can a 25.0°C saturated solution have if 150.0 mL of H2O is used as the solvent?
6. If 75.0 g of potassium bromide (KBr) are dissolved in 125 mL of H2O, is the solution saturated, unsaturated, or supersaturated?
7. Calculate the mass/mass percent of a saturated solution of NaCl. Use the data from Table $1$ "Solubilities of Various Solutes in Water at 25°C (Except as Noted)", assume that masses of the solute and the solvent are additive, and use the density of H2O (1.00 g/mL) as a conversion factor.
8. Calculate the mass/mass percent of a saturated solution of MgCO3 Use the data from Table $1$ "Solubilities of Various Solutes in Water at 25°C (Except as Noted)", assume that masses of the solute and the solvent are additive, and use the density of H2O (1.00 g/mL) as a conversion factor.
9. Only 0.203 mL of C6H6 will dissolve in 100.000 mL of H2O. Assuming that the volumes are additive, find the volume/volume percent of a saturated solution of benzene in water.
10. Only 35 mL of aniline (C6H5NH2) will dissolve in 1,000 mL of H2O. Assuming that the volumes are additive, find the volume/volume percent of a saturated solution of aniline in water.
11. A solution of ethyl alcohol (C2H5OH) in water has a concentration of 20.56% v/v. What volume of C2H5OH is present in 255 mL of solution?
12. What mass of KCl is present in 475 mL of a 1.09% m/v aqueous solution?
13. The average human body contains 5,830 g of blood. What mass of arsenic is present in the body if the amount in blood is 0.55 ppm?
14. The Occupational Safety and Health Administration has set a limit of 200 ppm as the maximum safe exposure level for carbon monoxide (CO). If an average breath has a mass of 1.286 g, what is the maximum mass of CO that can be inhaled at that maximum safe exposure level?
15. Which concentration is greater—15 ppm or 1,500 ppb?
16. Express the concentration 7,580 ppm in parts per billion.
17. What is the molarity of 0.500 L of a potassium chromate solution containing 0.0650 mol of K2CrO4?
18. What is the molarity of 4.50 L of a solution containing 0.206 mol of urea [(NH2)2CO]?
19. What is the molarity of a 2.66 L aqueous solution containing 56.9 g of NaBr?
20. If 3.08 g of Ca(OH)2 is dissolved in enough water to make 0.875 L of solution, what is the molarity of the Ca(OH)2?
21. What mass of HCl is present in 825 mL of a 1.25 M solution?
22. What mass of isopropyl alcohol (C3H8O) is dissolved in 2.050 L of a 4.45 M aqueous C3H8O solution?
23. What volume of 0.345 M NaCl solution is needed to obtain 10.0 g of NaCl?
24. How many milliliters of a 0.0015 M cocaine hydrochloride (C17H22ClNO4) solution is needed to obtain 0.010 g of the solute?
25. Aqueous calcium chloride reacts with aqueous silver nitrate according to the following balanced chemical equation:
CaCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + Ca(NO3)2(aq)
How many moles of AgCl(s) are made if 0.557 L of 0.235 M CaCl2 react with excess AgNO3? How many grams of AgCl are made?
26. Sodium bicarbonate (NaHCO3) is used to react with acid spills. The reaction with sulfuric acid (H2SO4) is as follows:
2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2H2O(ℓ) + 2CO2(g)
If 27.6 mL of a 6.25 M H2SO4 solution were spilled, how many moles of NaHCO3 would be needed to react with the acid? How many grams of NaHCO3 is this?
27. The fermentation of glucose to make ethanol and carbon dioxide has the following overall chemical equation:
C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)
If 1.00 L of a 0.567 M solution of C6H12O6 were completely fermented, what would be the resulting concentration of the C2H5OH solution? How many moles of CO2 would be formed? How many grams is this? If each mole of CO2 had a volume of 24.5 L, what volume of CO2 is produced?
28. Aqueous sodium bisulfite gives off sulfur dioxide gas when heated:
2NaHSO3(aq) → Na2SO3(aq) + H2O(ℓ) + SO2(g)
If 567 mL of a 1.005 M NaHSO3 solution were heated until all the NaHSO3 had reacted, what would be the resulting concentration of the Na2SO3 solution? How many moles of SO2 would be formed? How many grams of SO2 would be formed? If each mole of SO2 had a volume of 25.78 L, what volume of SO2 would be produced?
29. What is the concentration of a 1.0 M solution of K+(aq) ions in equivalents/liter?
30. What is the concentration of a 1.0 M solution of SO42(aq) ions in equivalents/liter?
31. A solution having initial concentration of 0.445 M and initial volume of 45.0 mL is diluted to 100.0 mL. What is its final concentration?
32. A 50.0 mL sample of saltwater that is 3.0% m/v is diluted to 950 mL. What is its final mass/volume percent?
Answers
1. Solubility is the amount of a solute that can dissolve in a given amount of solute, typically 100 mL. The solubility of solutes varies widely.
2. The term dilute means relatively less solute and the term concentrated implies relatively more solute. Both are of limited usefulness because these are not accurate.
3. 76.5 g
4. 90.2 g
5. 12.6 g
6. unsaturated
7. 26.5%
8. 2.15%
9. 0.203%
10. 3.4%
11. 52.4 mL
12. 5.18 g
13. 0.00321 g
14. 2.57 x 10-4 g
15. 15 ppm
16. 7,580,000 ppb
17. 0.130 M
18. 0.0458 M
19. 0.208 M
20. 0.0475 M
21. 37.6 g
22. 548 g
23. 0.496 L
24. 20 mL
25. 0.262 mol; 37.5 g
26. 0.345 mol; 29.0 g
27. 1.13 M C2H5OH; 1.13 mol of CO2; 49.7 g of CO2; 27.7 L of CO2
28. 0.503 M Na2SO3; 0.285 mol SO2; 18.3 g SO2; 471 L SO2
29. 1.0 Eq/L
30. 2.0 Eq/L
31. 0.200 M
32. 0.16 % m/v
Concept Review Exercise
1. Explain how the solvation process describes the dissolution of a solute in a solvent.
Answer
1. Each particle of the solute is surrounded by particles of the solvent, carrying the solute from its original phase.
Exercises
1. Describe what happens when an ionic solute like Na2SO4 dissolves in a polar solvent.
2. Describe what happens when a molecular solute like sucrose (C12H22O11) dissolves in a polar solvent.
3. Classify each substance as an electrolyte or a nonelectrolyte. Each substance dissolves in H2O to some extent.
1. NH4NO3
2. CO2
3. NH2CONH2
4. HCl
4. Classify each substance as an electrolyte or a nonelectrolyte. Each substance dissolves in H2O to some extent.
1. CH3CH2CH2OH
2. Ca(CH3CO2)2
3. I2
4. KOH
5. Will solutions of each solute conduct electricity when dissolved?
1. AgNO3
2. CHCl3
3. BaCl2
4. Li2O
6. Will solutions of each solute conduct electricity when dissolved?
1. CH3COCH3
2. N(CH3)3
3. CH3CO2C2H5
4. FeCl2
Answers
1. Each ion of the ionic solute is surrounded by particles of solvent, carrying the ion from its associated crystal.
2. Each sucrose molecule is surrounded by solvent molecules (attracted to each other via intermolecular forces of attraction).
1. electrolyte
2. nonelectrolyte
3. nonelectrolyte
4. electrolyte
4.
• nonelectrolyte
• electrolyte
• nonelectrolyte
• electrolyte
5.
1. yes
2. no
3. yes
4. yes
6.
a. no
b. no
c. no
d. yes
Concept Review Exercises
1. What are the colligative properties of solutions?
2. Explain how the following properties of solutions differ from those of the pure solvent: vapor pressure, boiling point, freezing point, and osmotic pressure.
Answers
1. Colligative properties are characteristics that a solution has that depend on the number, not the identity, of solute particles.
2. In solutions, the vapor pressure is lower, the boiling point is higher, the freezing point is lower, and the osmotic pressure is higher.
Exercises
1. In each pair of aqueous systems, which will have the lower vapor pressure?
1. pure water or 1.0 M NaCl
2. 1.0 M NaCl or 1.0 M C6H12O6
3. 1.0 M $\ce{CaCl2}$ or 1.0 M (NH4)3PO4
2. In each pair of aqueous systems, which will have the lower vapor pressure?
1. 0.50 M Ca(NO3)2 or 1.0 M KBr
2. 1.5 M C12H22O11 or 0.75 M Ca(OH)2
3. 0.10 M Cu(NO3)2 or pure water
3. In each pair of aqueous systems, which will have the higher boiling point?
1. pure water or a 1.0 M NaCl
2. 1.0 M NaCl or 1.0 M C6H12O6
3. 1.0 M $\ce{CaCl2}$ or 1.0 M (NH4)3PO4
4. In each pair of aqueous systems, which will have the higher boiling point?
1. 0.50 M Ca(NO3)2 or 1.0 M KBr
2. 1.5 M C12H22O11 or 0.75 M Ca(OH)2
3. 0.10 M Cu(NO3)2 or pure water
5. Estimate the boiling point of each aqueous solution. The boiling point of pure water is 100.0°C.
1. 0.50 M NaCl
2. 1.5 M Na2SO4
3. 2.0 M C6H12O6
6. Estimate the freezing point of each aqueous solution. The freezing point of pure water is 0.0°C.
1. 0.50 M NaCl
2. 1.5 M Na2SO4
3. 2.0 M C6H12O6
7. Explain why salt (NaCl) is spread on roads and sidewalks to inhibit ice formation in cold weather.
8. Salt (NaCl) and calcium chloride ($\ce{CaCl2}$) are used widely in some areas to minimize the formation of ice on sidewalks and roads. One of these ionic compounds is better, mole for mole, at inhibiting ice formation. Which is that likely to be? Why?
9. What is the osmolarity of each aqueous solution?
1. 0.500 M NH2CONH2
2. 0.500 M NaBr
3. 0.500 M Ca(NO3)2
10. What is the osmolarity of each aqueous solution?
1. 0.150 M KCl
2. 0.450 M (CH3)2CHOH
3. 0.500 M Ca3(PO4)2
11. A 1.0 M solution of an unknown soluble salt has an osmolarity of 3.0 osmol. What can you conclude about the salt?
12. A 1.5 M NaCl solution and a 0.75 M Al(NO3)3 solution exist on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and the direction of solvent flow, if any, across the membrane.
Answers
1. 1.0 M NaCl
2. 1.0 M NaCl
3. 1.0 M (NH4)3PO4
2.
• 1.0 M KBr
• 0.75 M Ca(OH)2
• 0.10 M Cu(NO3)2
3.
1. 1.0 M NaCl
2. 1.0 M NaCl
3. 1.0 M (NH4)3PO4
4.
• 1.0 M KBr
• 0.75 M Ca(OH)2
• 0.10 M Cu(NO3)2
5.
1. 100.5°C
2. 102.3°C
3. 101°C
6.
1. -1.9°C
2. -8.6°C
3. -3.8°C
7. NaCl lowers the freezing point of water, so it needs to be colder for the water to freeze.
8. $\ce{CaCl2}$ splits up into 3 ions while NaCl splits up into 2 ions only. $\ce{CaCl2}$ will be more effective.
9.
1. 0.500 osmol
2. 1.000 osmol
3. 1.500 osmol
10.
1. 0.300 osmol
2. 0.450 osmol
3. 2.50 osmol
11. It must separate into three ions when it dissolves.
12. Both NaCl and Al(NO3)3 have 3.0 osmol. There will be no net difference in the solvent flow.
Concept Review Exercises
1. What is chemical equilibrium?
2. What does the equilibrium constant tell us?
Answers
1. The rate of the forward reaction equals the rate of the reverse reaction.
2. The ratio of products and reactants when the system is at equilibrium.
EXERCISES
1. If the reaction H2 + I2 ⇌2HI is at equilibrium, do the concentrations of HI, H2, and I2 have to be equal?
2. Do the concentrations at equilibrium depend upon how the equilibrium was reached?
3. What does it mean if the Keq is > 1?
4. What does it mean if the Keq is < 1?
5. Does the equilibrium state depend on the starting concentrations?
6. Write an expression for the equilibrium constant K equation.
a. PCl5 (g) ⇄ PCl3 (g) + Cl2 (g)
b. $2\,O_{3}\,(g) \rightleftharpoons 3\,O_{2}\,(g)$
7. Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: 3C2H2(g)⟶C6H6(g). Which value of K would make this reaction most useful commercially? Explain your answer.
a. K ≈ 0.01
b. K ≈ 1
c. K ≈ 10.
8. Tell whether the reactants or the products are favored at equilibrium:
a. 2NH3(g) ⇌ N2(g) + 3H2(g) K = 172
b. 2SO3(g) ⇌ 2SO2(g) + O2(g) K = 0.230
c. 2NO(g) + Cl2(g) ⇌ 2NOCl(g) K = 4.6×104
d. N2(g) + O2(g) ⇌ 2NO(g) K = 0.050
AnswerS
1. No, the concentrations are constant but the concentrations do not have to be equal.
2. No.
3. More products than reactants are present at equilibrium.
4. More reactants than products present at equilibrium.
5. No. The equilibrium ratio does not depend on the initial concentrations.
6.
a. $K=\dfrac{[PCl_3][Cl_2]}{[PCl_5]}$
b. $K=\dfrac{[O_2]^3}{[O_3]^2}$
7. The answer is c. K ≈ 10.
Since $K=\dfrac{[C_6H_6]^2}{[C_2H_2]^3}$ (K≈10), this means that C6H6 predominates over C2H2. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.
8.
a. products
b. reactants
c. products
d. reactants
Concept Review Exercises
1. Define Le Chatelier’s principle.
2. List the three factors types of changes that can disturb the equilibrium of a system.
Answers
1. Le Chatelier’s principle states that a system at equilibrium is disturbed, it will respond in a way to minimize the disturbance.
2. temperature, change in amount of substance, change in pressure through change in volume
EXERCISES
1. How will each change affect the reaction?
PCl5(g) + heat ⇌PCl3(g) + Cl2(g)
1. Addition of PCl5
2. Addition of Cl2
3. Removal of PCl3
4. Increasing temperature
5. Decreasing temperature
6. Decreasing volume
2. How will each change affect the reaction?
HNO2(aq) ⇌H+(aq) + NO2(aq)
1. Removal of HNO2
2. Addition of HCl (i.e. adding more H+)
3. Increasing volume
4. Decreasing volume
5. Removal of NO2
6. Addition of OH (which will react with and remove H+)
3. How will each change affect the reaction?
CO2(g) + C(s) ⇌2CO(g) ΔH=172.5kJ
1. Addition of CO2
2. Removal of CO2
3. Increasing temperature
4. Decreasing temperature
5. Increasing volume
6. Addition of CO
4. How will each change affect the reaction?
H2(g) + I2(g) ⇌2HI(g) ΔH=−9.48kJ
1. Addition of H2
2. Removal of H2
3. Increasing temperature
4. Decreasing temperature
5. Increasing volume
6. Decreasing volume
1.
1. shift right
2. shift left
3. shift right
4. shift right
5. shift left
6. shift left
2.
1. shift left
2. shift left
3. no effect
4. no effect
5. shift right
6. shift right
3.
1. shift right
2. shift left
3. shift right
4. shift left
5. shift right
6. shift left
4.
1. shift right
2. shift left
3. shift left
4. shift right
5. no effect
6. no effect
Concept Review Exercises
1. What are some of the features of a semipermeable membrane?
2. What do the prefixes hyper, hypo, and iso mean?
Answers
1. A semipermeable membrane allows some substances to pass through but not others.
2. hyper – higher; hypo – lower; iso - same
EXERCISES
1. Two solutions are separated by a semipermeable membrane. Solution A contains 25.0 g of NaCl in 100.0 mL of water and solution B contains 35.0 g of NaCl in 100.0 mL of water.
1. Which one has a higher concentration?
2. Which way will water molecules flow?
3. Which volume will increase?
4. Which volume will decrease?
5. What will happen to the concentration of solution A?
6. What will happen to the concentration of solution B?
2. Two solutions with different concentrations and compositions are separated by a semipermeable membrane. The left-hand solution is a .50 M solution of MgSO4, while the right-hand solution contains CaCl2 at a concentration of .40 M. Determine the direction of the flow of solvent, left or right.
3. Given the following situations, wherein two tanks of different solutions are separated by a semipermeable membrane, determine the direction of the flow of solvent (water).
a. Solution A contains a 0.40 M concentration of CaCl2, while Solution B contains a 0.45 M concentration of KI
b. Solution A contains a 1.00 M concentration of NH4Cl, while Solution B contains a 1.00 M concentration of CH2O
4. Cells are placed in a solution and the cells then undergo hemolysis. What can be said about the relative concentrations of solute in the cell and the solution?
5. Describe the relative concentrations inside and outside a red blood cell when crenation occurs.
6. A saltwater fish is placed in a freshwater tank. What will happen to the fish? Describe the flow of water molecules to explain the outcome.
7. What makes up the "head" region of a phospholipid? Is it hydrophobic or hydrophilic?
8. What makes up the "tail" region of a phospholipid? Is it hydrophobic or hydrophilic?
Answers
1. Two solutions are separated by a semipermeable membrane. Solution A contains 25.0 g of NaCl in 100.0 mL of water and solution B contains 35.0 g of NaCl in 100.0 mL of water.
1. Solution B
2. A →→ B
3. B
4. A
5. increase
6. decrease
2. Water (solvent) flows from left to right.
3.
a. Water flows from Solution B to Solution A.
b. Water flows from Solution B to Solution A.
4. Cells contain fluid with higher concentration than solution outside the cell.
5. Cells contain fluid with a lower concentration than the solution outside the cell.
6. Water molecules will flow from the tank water into the fish because the fish has a higher concentration of salt. If the fish absorbs too much water, it will die.
7. The "head" region is a phosphate group and it is hydrophilic.
8. The "tail" is a hydrocarbon tail and it is hydrophobic.
Contributors
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
Additional Exercises
1. Calcium nitrate reacts with sodium carbonate to precipitate solid calcium carbonate:
$Ca(NO_3)_{2(aq)} + Na_2CO_{3(aq)} \rightarrow CaCO_{3(s)} + NaNO_{3(aq)}$
1. Balance the chemical equation.
2. How many grams of Na2CO3 are needed to react with 50.0 mL of 0.450 M Ca(NO3)2?
3. Assuming that the Na2CO3 has a negligible effect on the volume of the solution, find the osmolarity of the NaNO3 solution remaining after the CaCO3 precipitates from solution.
2. The compound HCl reacts with sodium carbonate to generate carbon dioxide gas:
$HCl_{(aq)} + Na_2CO_{3(aq)} \rightarrow H_2O_{(ℓ)} + CO_{2(g)} + NaCl_{(aq)}$
1. Balance the chemical equation.
2. How many grams of Na2CO3 are needed to react with 250.0 mL of 0.755 M HCl?
3. Assuming that the Na2CO3 has a negligible effect on the volume of the solution, find the osmolarity of the NaCl solution remaining after the reaction is complete.
3. Estimate the freezing point of concentrated aqueous HCl, which is usually sold as a 12 M solution. Assume complete ionization into H+ and Cl ions.
4. Estimate the boiling point of concentrated aqueous H2SO4, which is usually sold as an 18 M solution. Assume complete ionization into H+ and HSO4 ions.
5. Seawater can be approximated by a 3.0% m/m solution of NaCl in water. Determine the molarity and osmolarity of seawater. Assume a density of 1.0 g/mL.
6. Human blood can be approximated by a 0.90% m/m solution of NaCl in water. Determine the molarity and osmolarity of blood. Assume a density of 1.0 g/mL.
7. How much water must be added to 25.0 mL of a 1.00 M NaCl solution to make a resulting solution that has a concentration of 0.250 M?
8. Sports drinks like Gatorade are advertised as capable of resupplying the body with electrolytes lost by vigorous exercise. Find a label from a sports drink container and identify the electrolytes it contains. You should be able to identify several simple ionic compounds in the ingredients list.
9. Occasionally we hear a sensational news story about people stranded in a lifeboat on the ocean who had to drink their own urine to survive. While distasteful, this act was probably necessary for survival. Why not simply drink the ocean water? (Hint: See Exercise 5 and Exercise 6 above. What would happen if the two solutions in these exercises were on opposite sides of a semipermeable membrane, as we would find in our cell walls?)
Answers
1.
1. Ca(NO3)2(aq) + Na2CO3(aq) → CaCO3(s) + 2NaNO3(aq)
2. 2.39 g
3. 1.80 osmol
2.
a. 2HCl (aq) + Na2CO3(aq) → H2O(ℓ) + CO2(g) + 2NaCl (aq)
b. 10.0 g
c. 1.51 M
1. −45.6°C
4. 118°C
1. 0.513 M; 1.026 osmol
6. molarity = 0.15 M; osmolarity = 0.31 M
7. 75.0 mL
8. magnesium chloride, calcium chloride (answers may vary)
1. The osmotic pressure of seawater is too high. Drinking seawater would cause water to go from inside our cells into the more concentrated seawater, ultimately killing the cells. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09.E%3A_Exercises.txt |
Concept Review Exercises
1. Give the Arrhenius definitions of an acid and a base.
2. What is neutralization?
Answers
1. Arrhenius acid: a compound that increases the concentration of hydrogen ion (H+) in aqueous solution; Arrhenius base: a compound that increases the concentration of hydroxide ion (OH) in aqueous solution.
2. the reaction of an acid and a base
Exercises
1. Give two examples of Arrhenius acids.
2. Give two examples of Arrhenius bases.
3. List the general properties of acids.
4. List the general properties of bases.
5. Name each compound. (For acids, look up the name in Table 10.1.1. For bases, use the rules for naming ionic compounds from Chapter 3.)
a. HBr(aq)
b. Ca(OH)2(aq)
c. HNO3(aq)
d. Fe(OH)3(aq)
6. Name each compound.
a. HI(aq)
b. Cu(OH)2(aq)
c. H3PO4(aq)
d. CsOH(aq)
7. Write a balanced chemical equation for the neutralization of Ba(OH)2(aq) with HNO3(aq).
8. Write a balanced chemical equation for the neutralization of H2SO4(aq) with Cr(OH)3(aq).
9. Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction.
10. Identify the salt produced in each acid-base reaction below. Then, balance the equation.
a. 2HCl + Sr(OH)2 2H2O + ??
b. KNO3; HNO3 + KOH ?? + H2O
c. HF + Ca(OH)2 ---> ?? + H2O
11. How many moles of sodium hydroxide (NaOH) are needed to neutralize 0.844 mol of acetic acid (HC2H3O2)? (Hint: begin by writing a balanced chemical equation for the process.)
12. How many moles of perchloric acid (HClO4) are needed to neutralize 0.052 mol of calcium hydroxide [Ca(OH)2]? (Hint: begin by writing a balanced chemical equation for the process
13. Hydrazoic acid (HN3) can be neutralized by a base.
a. Write the balanced chemical equation for the reaction between hydrazoic acid and calcium hydroxide.
b. How many milliliters of 0.0245 M Ca(OH)2 are needed to neutralize 0.564 g of HN3?
14. Citric acid (H3C6H5O7) has three hydrogen atoms that can form hydrogen ions in solution.
a. Write the balanced chemical equation for the reaction between citric acid and sodium hydroxide.
b. If an orange contains 0.0675 g of H3C6H5O7, how many milliliters of 0.00332 M NaOH solution are needed to neutralize the acid?
15. Magnesium hydroxide [Mg(OH)2] is an ingredient in some antacids. How many grams of Mg(OH)2 are needed to neutralize the acid in 158 mL of 0.106 M HCl(aq)? It might help to write the balanced chemical equation first.
16. Aluminum hydroxide [Al(OH)3] is an ingredient in some antacids. How many grams of Al(OH)3 are needed to neutralize the acid in 96.5 mL of 0.556 M H2SO4(aq)? It might help to write the balanced chemical equation first.
17. Write the balanced chemical equation for the reaction between HBr and Ca(OH)2. What volume of 0.010 M HBr solution is be required to neutralize 25 mL of a 0.0100M Ca(OH)2 solution?
18. Write the balanced chemical equation for the reaction between HNO3 and KOH. What volume of 0.5M HNO3 is required to neutralize 60 mL of 0.4M KOH solution?
Answers
1. HCl and HNO3 (answers will vary)
2. NaOH and Ca(OH)2 (answers will vary)
1. sour taste, react with metals, react with bases, and turn litmus red
2. bitter taste, feels slippery, react with acids and turn litmus blue
1. a. hydrobromic acid
b. calcium hydroxide
c. nitric acid
d. iron(III) hydroxide
6. a. hydroiodic acid
b. cupric hydroxide
c. phosphoric acid
d. cesium hydroxide
7. 2HNO3(aq) + Ba(OH)2(aq) → Ba(NO3)2(aq) + 2H2O
8. 3H2SO4(aq) + 2Cr(OH)3(aq) → Cr2(SO4)3(aq) + 6H2O
9. Mg(OH)2 + 2HCl --> MgCl2 + 2H2O
10. a. SrCl2; 2HCl + Sr(OH)2 2H2O + SrCl2
b. KNO3; HNO3 + KOH KNO3 + H2O
c. CaF2; 2HF + Ca(OH)2 CaF2 + 2H2O
11. 0.844 mol
12. 0.104 mol
13. Part 1: 2HN3(aq) + Ca(OH)2 → Ca(N3)2 + 2H2O
Part 2: 268 mL
14. Part 1: H3C6H5O7(aq) + 3NaOH(aq) → Na3C6H5O7(aq) + 3H2O
Part 2: 317.5 mL
15. 0.488 g
16. 2.79 g
17. 2HBr + Ca(OH)2 CaBr2 + 2H2O; 50 mL HBr
18. HNO3 + KOHKNO3 + H2O; 48 mL HNO3
Concept Review Exercise
1. Give the definitions of a Brønsted-Lowry acid and a Brønsted-Lowry base.
Answer
1. A Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor.
Exercises
1. Label each reactant as a Brønsted-Lowry acid or a Brønsted-Lowry base.
HCl(aq) + NH3(aq) → NH4+(aq) + Cl(aq)
2. Label each reactant as a Brønsted-Lowry acid or a Brønsted-Lowry base.
H2O(ℓ) + N2H4(aq) → N2H5+(aq) + OH(aq)
3. Explain why a Brønsted-Lowry acid can be called a proton donor.
4. Explain why a Brønsted-Lowry base can be called a proton acceptor.
5. Write the chemical equation of the reaction of ammonia in water and label the Brønsted-Lowry acid and base.
6. Write the chemical equation of the reaction of methylamine (CH3NH2) in water and label the Brønsted-Lowry acid and base.
7. Demonstrate that the dissolution of HNO3 in water is actually a Brønsted-Lowry acid-base reaction by describing it with a chemical equation and labeling the Brønsted-Lowry acid and base.
8. Identify the Brønsted-Lowry acid and base in the following chemical equation:
C3H7NH2(aq) + H3O+(aq) → C3H7NH3+(aq) + H2O(ℓ)
9. Identify the Brønsted-Lowry acid and the Brønsted-Lowry base in each of the following equations
1. \(\ce{NO2- + H2O ⟶ HNO2 + OH-}\)
2. \(\ce{HBr + H2O ⟶ H3O+ + Br-}\)
3. \(\ce{HS- + H2O ⟶ H2S + OH-}\)
4. \(\ce{H2PO4- + OH- ⟶HPO4^2- + H2O}\)
5. \(\ce{H2PO4- + HCl ⟶ H3PO4 + Cl-}\)
10. Write the chemical equation for the reaction that occurs when cocaine hydrochloride (C17H22ClNO4) dissolves in water and donates a proton to a water molecule. (When hydrochlorides dissolve in water, they separate into chloride ions and the appropriate cation.)
11. If codeine hydrobromide has the formula C18H22BrNO3, what is the formula of the parent compound codeine?
Answers
1. HCl: Brønsted-Lowry acid; NH3: Brønsted-Lowry base
2. H2O: Brønsted-Lowry acid; N2H4: Brønsted-Lowry base
3. A Brønsted-Lowry acid gives away an H+ ion—nominally, a proton—in an acid-base reaction.
4. A Brønsted-Lowry base accepts an H+ ion (a proton) in an acid-base reaction.
5. NH3 + H2O → NH4+ + OH (here NH3 = Brønsted-Lowry base; H2O = Brønsted-Lowry acid)
6. CH3NH2 + H2O → CH3NH3+ + OH (here CH3NH2 = Brønsted-Lowry base; H2O = Brønsted-Lowry acid)
7. HNO3 + H2O → H3O+ + NO3(here HNO3 = Brønsted-Lowry acid; H2O = Brønsted-Lowry base)
8. C3H7NH2(aq) + H3O+(aq) → C3H7NH3+(aq) + H2O(ℓ) (here H3O+ = Brønsted-Lowry acid; C3H7NH2 = Brønsted-Lowry base)
Concept Review Exercises
1. Explain how water can act as an acid.
2. Explain how water can act as a base.
Answers
1. Under the right conditions, H2O can donate a proton, making it a Brønsted-Lowry acid.
2. Under the right conditions, H2O can accept a proton, making it a Brønsted-Lowry base.
Exercises
1. Is H2O(ℓ) acting as an acid or a base?
H2O(ℓ) + NH4+(aq) → H3O+(aq) + NH3(aq)
2. Is H2O(ℓ) acting as an acid or a base?
CH3(aq) + H2O(ℓ) → CH4(aq) + OH(aq)
3. In the aqueous solutions of some salts, one of the ions from the salt can react with water molecules. In some C2H3O2 solutions, the following reaction can occur:
C2H3O2(aq) + H2O(ℓ) → HC2H3O2(aq) + OH(aq)
Is H2O acting as an acid or a base in this reaction?
4. In the aqueous solutions of some salts, one of the ions from the salt can react with water molecules. In some NH4+ solutions, the following reaction can occur:
NH4+(aq) + H2O → NH3(aq) + H3O+(aq)
Is H2O acting as an acid or a base in this reaction?
5. Why is pure water considered neutral?
Answers
1. base
2. acid
1. acid
2. base
5. When water ionizes, equal amounts of H+ (acid) and OH(base) are formed, so the solution is neither acidic nor basic: H2O(ℓ) → H+(aq) + OH(aq)
Concept Review Exercises
1. Explain the difference between a strong acid or base and a weak acid or base.
2. Explain what is occurring when a chemical reaction reaches equilibrium.
3. Define pH.
AnswerS
1. A strong acid or base is 100% ionized in aqueous solution; a weak acid or base is less than 100% ionized.
2. The overall reaction progress stops because the reverse process balances out the forward process.
3. pH is a measure of the hydrogen ion concentration.
Exercises
1. Name a strong acid and a weak acid. (Hint: use Table 10.4.1.)
2. Name a strong base and a weak base. (Hint: use Table 10.4.1.)
3. Is each compound a strong acid or a weak acid? Assume all are in aqueous solution. (Hint: use Table 10.4.1.)
1. HF
2. HC2H3O2
3. HCl
4. HClO4
4. Is each compound a strong acid or a weak acid? Assume all are in aqueous solution. (Hint: use Table 10.4.1.)
1. H2SO4
2. HSO4
3. HPO42
4. HNO3
5. Is each compound a strong base or a weak base? Assume all are in aqueous solution. (Hint: use Table 10.4.1.)
1. NH3
2. NaOH
3. Mg(OH)2
4. Cu(OH)2
6. Is each compound a strong base or a weak base? Assume all are in aqueous solution. (Hint: use Table 10.4.1.)
1. KOH
2. H2O
3. Fe(OH)2
4. Fe(OH)3
7. Write the chemical equation for the equilibrium process for each weak acid in Exercise 3.
8. Write the chemical equation for the equilibrium process for each weak acid in Exercise 4.
9. Write the chemical equation for the equilibrium process for each weak base in Exercise 5.
10. Write the chemical equation for the equilibrium process for each weak base in Exercise 6.
11. Which is the stronger acid—HCl(aq) or HF(aq)?
12. Which is the stronger base—KOH(aq) or Ni(OH)2(aq)?
13. Consider the two acids in Exercise 11. For solutions that have the same concentration, which one would you expect to have a lower pH?
14. Consider the two bases in Exercise 12. For solutions that have the same concentration, which one would you expect to have a higher pH?
15. Consider the list of substances in Table \PageIndex3\PageIndex3.2"The pH Values of Some Common Solutions". What is the most acidic substance on the list that you have encountered recently?
16. Consider the list of substances in Table \PageIndex3\PageIndex3.2"The pH Values of Some Common Solutions". What is the most basic substance on the list that you have encountered recently?
17. Indicate whether solutions with the following pH values are acidic, basic, or neutral:
1. pH = 9.4
2. pH = 7.0
3. pH = 1.2
4. pH = 6.5
Answers
1. strong acid: HCl; weak acid: HC2H3O2 (answers will vary)
2. strong base: NaOH; weak base: NH3 (answers will vary)
1. weak
2. weak
3. strong
4. strong
1. strong
2. weak
3. weak
4. strong
1. weak
2. strong
3. strong
4. weak
1. strong
2. weak
3. weak
4. weak
7. 3a: HF(aq) ⇆ H+(aq) + F(aq); 3b: HC2H3O2(aq) ⇆ H+(aq) + C2H3O2(aq)
8. 4b: HSO4(aq) ⇆ H+(aq) + SO42(aq); 4c: HPO42(aq) ⇆ H+(aq) + PO43(aq)
9. 5a: NH3(aq) + H2O ⇆ NH4+(aq) + OH(aq); 5d: Cu(OH)2(aq) ⇆ Cu2+(aq) + 2OH(aq)
10. 6b: H2O + H2O ⇆ H3O+(aq) + OH(aq); 6c: Fe(OH)2(aq) ⇆ Fe2+(aq) + 2OH(aq); 6d: Fe(OH)3(aq) ⇆ Fe3+(aq) + 3OH(aq)
11. HCl(aq)
12. KOH(aq)
13. HCl(aq)
14. KOH(aq)
15. (answers will vary)
16. (answers will vary)
17. 1. basic
2. neutral
3. acidic (strongly)
4. acidic (mildly)
Concept Review Exercise
1. Explain how a buffer prevents large changes in pH.
Answer
1. A buffer has components that react with both strong acids and strong bases to resist sudden changes in pH.
Exercises
1. Describe a buffer. What two related chemical components are required to make a buffer?
2. Can a buffer be made by combining a strong acid with a strong base? Why or why not?
3. Which solute combinations can make a buffer? Assume all are aqueous solutions.
1. HCl and NaCl
2. HNO2 and NaNO2
3. NH4NO3 and HNO3
4. NH4NO3 and NH3
4. Which solute combinations can make a buffer? Assume all are aqueous solutions.
1. H3PO4 and Na3PO4
2. NaHCO3 and Na2CO3
3. NaNO3 and Ca(NO3)2
4. HN3 and NH3
5. For each combination in Exercise 3 that is a buffer, write the chemical equations for the reactions of the buffer components when a strong acid and a strong base is added.
6. For each combination in Exercise 4 that is a buffer, write the chemical equations for the reaction of the buffer components when a strong acid and a strong base is added.
7. The complete phosphate buffer system is based on four substances: H3PO4, H2PO4, HPO42, and PO43. What different buffer solutions can be made from these substances?
8. Explain why NaBr cannot be a component in either an acidic or a basic buffer.
9. Explain why Mg(NO3)2 cannot be a component in either an acidic or a basic buffer.
Answers
1. A buffer resists sudden changes in pH. It has a weak acid or base and a salt of that weak acid or base.
2. No. Combining a strong acid and a strong base will produce salt and water. Excess strong acid or strong base will not act as a buffer.
1. not a buffer
2. buffer
3. not a buffer
4. buffer
4. 1. not a buffer
2. buffer
3. not a buffer
4. not buffer
1. 3b: strong acid: H+ + NO2 → HNO2; strong base: OH + HNO2 → H2O + NO2; 3d: strong acid: H+ + NH3 → NH4+; strong base: OH + NH4+ → H2O + NH3
2. 4b: strong acid: H+ + CO32 → HCO3; strong base: OH + HCO3 → H2O + CO32;
1. Buffers can be made by combining H3PO4 and H2PO4, H2PO4 and HPO42, and HPO42 and PO43.
2. NaBr splits up into two ions in solution, Na+ and Br. Na+ will not react with any added base knowing that NaOH is a strong base. Br- will not react with any added acid knowing that HBr is a strong acid. Because NaBr will not react with any added base or acid, it does not resist change in pH and is not a buffer.
1. Mg(NO3)2 includes two types of ions, Mg2+ and NO3. Mg(OH)2 is strong base and completely dissociates (100% falls apart), so Mg2+ will not react with any added base (0% combines with OH). HNO3 is strong acid and completely dissociates (100% falls apart), so NO3 will not react with any added acid (0% combines with H+). Because Mg(NO3)2 will not react with any added base or acid, it does not resist change in pH and is not a buffer.
Additional Exercises
1. The properties of a 1.0 M HCl solution and a 1.0 M HC2H3O2 solution are compared. Measurements show that the hydrochloric acid solution has a higher osmotic pressure than the acetic acid solution. Explain why.
2. Of a 0.50 M HNO3 solution and a 0.50 M HC2H3O2 solution, which should have the higher boiling point? Explain why.
3. The reaction of sulfuric acid [H2SO4(aq)] with sodium hydroxide [NaOH(aq)] can be represented by two separate steps, with only one hydrogen ion reacting in each step. Write the chemical equation for each step.
4. The reaction of aluminum hydroxide [Al(OH)3(aq)] with hydrochloric acid [HCl(aq)] can be represented by three separate steps, with only one hydroxide ion reacting in each step. Write the chemical equation for each step.
5. A friend brings you a small sample of an unknown chemical. Assuming that the chemical is soluble in water, how would you determine if the chemical is an acid or a base?
6. A neutral solution has a hydrogen ion concentration of about 1 × 10−7 M. What is the concentration of the hydroxide ion in a neutral solution?
7. The Lewis definitions of an acid and a base are based on electron pairs, not protons. A Lewis acid is an electron pair acceptor, while a Lewis base is an electron pair donor. Use Lewis diagrams to show that
H+(aq) + OH(aq) → H2O(ℓ)
is an acid-base reaction in the Lewis sense as well as in the Arrhenius and Brønsted-Lowry senses.
8. Given the chemical reaction
NH3(g) + BF3(g) → NH3—BF3(s)
show that the reaction illustrated by this equation is an acid-base reaction if we use the Lewis definitions of an acid and a base (see Exercise 7). The product contains a bond between the N and B atoms.
Answers
1. HCl is a strong acid and yields more ions in solution. HC2H3O2 is a weak acid and undergoes partial ionization in solution.
2. HNO3 is a strong acid while HC2H3O2 is a weak acid. HNO3 dissociates 100% and its solution contains more ions. The more ions the solution contains the lower is its vapor pressure; the higher temperature is required for it to boil.
1. H2SO4 + NaOH → NaHSO4 + H2O; NaHSO4 + NaOH → Na2SO4 + H2O
2. Al(OH)3 + HCl → Al(OH)2Cl + H2O; Al(OH)2Cl + HCl → Al(OH)Cl2 + H2O; Al(OH)Cl2 + HCl → AlCl3 + H2O
1. One way is to add it to NaHCO3; if it bubbles, it is an acid. Alternatively, add the sample to litmus and look for a characteristic color change (red for acid, blue for base).
2. In a neutral solution, [OH-] = [H+] = 1.0 x 10-7 M
1. The O atom is donating an electron pair to the H+ ion, making the base an electron pair donor and the acid an electron pair acceptor.
2. The N atom is donating a lone pair to B in BF3, Hence NH3 is the Lewis base and BF3 is the Lewis acid. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/10.E%3A_Acids_and_Bases_%28Exercises%29.txt |
11.1 Radioactivity
Concept Review Exercise
1. What are the major types of radioactivity? Write chemical equations demonstrating each type.
Answer
1. The major types of radioactivity are alpha decay, beta decay, and gamma ray emission; alpha decay with gamma emission: $\mathrm{_{86}^{222}Rn \rightarrow \, _{84}^{218}Po + \, ^4_2He + \gamma}$; beta decay: $\ce{_6^{14}C \rightarrow _7^{14}N + ^0_{-1}e}$ (answers will vary)
Exercises
1. Define radioactivity.
2. Give an example of a radioactive isotope.
3. How many protons and neutrons are in each isotope?
1. $\mathrm{^{11}_5B}$
2. $\mathrm{^{27}_{13}Al}$
3. 56Fe
4. 224Rn
4. How many protons and neutrons are in each isotope?
1. $\mathrm{^{2}_{1}H}$
2. $\mathrm{^{112}_{48}Cd}$
3. 252Es
4. 40K
5. Describe an alpha particle. What nucleus is it equivalent to?
6. Describe a beta particle. What subatomic particle is it equivalent to?
7. Explain what gamma rays are.
8. Explain why it is inappropriate to refer to gamma rays as gamma “particles.”
9. Plutonium has an atomic number of 94. Write the chemical equation for the alpha particle emission of 244Pu. What is the daughter isotope?
10. Francium has an atomic number of 87. Write the chemical equation for the alpha particle emission of 212Fr. What is the daughter isotope?
11. Tin has an atomic number of 50. Write the chemical equation for the beta particle emission of 121Sn. What is the daughter isotope?
12. Technetium has an atomic number of 43. Write the chemical equation for the beta particle emission of 99Tc. What is the daughter isotope?
13. Energies of gamma rays are typically expressed in units of megaelectron volts (MeV), where 1 MeV = 1.602 × 10−13 J. Using data provided in the text, calculate the energy, in megaelectron volts, of the gamma ray emitted when radon-222 decays.
14. The gamma ray emitted when oxygen-19 gives off a beta particle is 0.197 MeV. What is its energy in joules? (See Exercise 13 for the definition of a megaelectron volt.)
15. Which penetrates matter more deeply—alpha particles or beta particles? Suggest ways to protect yourself against both particles.
16. Which penetrates matter more deeply—alpha particles or gamma rays? Suggest ways to protect yourself against both emissions.
17. Define nuclear fission.
18. What general characteristic is typically necessary for a nucleus to undergo spontaneous fission?
Answers
1. Radioactivity is the spontaneous emission of particles and radiation from atomic nuclei.
2. C-14 or 14C is an example of radioactive isotope (answers may vary).
1.
1. 5 protons; 6 neutrons
2. 13 protons; 14 neutrons
3. 26 protons; 30 neutrons
4. 86 protons; 138 neutrons
4.
1. 1 proton; 1 neutron
2. 48 protons; 64 neutrons
3. 99 protons;153 neutrons
4. 19 protons; 21 neutrons
1. An alpha particle is a combination of two protons and two neutrons and is equivalent to a helium nucleus.
6. A beta particle is an electron.
1. Gamma rays are high-energy electromagnetic radiation given off in radioactive decay.
8. Gamma rays have no mass. Hence not a particle.
1. $\mathrm{^{244}_{94}Pu\rightarrow \, _2^4He +\, ^{240}_{92}U}$; the daughter isotope is $\mathrm{^{240}_{92}U}$, an atom of uranium.
10. $\mathrm{^{212}_{87}Fr\rightarrow \, _2^4He +\, ^{208}_{85}At}$; the daughter isotope is $\mathrm{^{208}_{85}At}$, an atom of astatine.
11. $\mathrm{_{50}^{121}Sn \rightarrow \, _{-1}^0e + \, _{51}^{121}Sb}$; the daughter isotope is $\mathrm{_{51}^{121}Sb}$, an atom of antimony.
12. $\mathrm{_{43}^{99}Tc \rightarrow \, _{-1}^0e + \, _{44}^{99}Mo}$; the daughter isotope is $\mathrm{_{44}^{99}Mo}$, an atom of antimony.
13. 0.512 MeV
14. 3.16 x 10-14 J
15. Beta particles; shielding of the appropriate thickness can protect against both alpha and beta particles.
16. Gamma rays; can be shielded by thick, dense material such as lead (Pb). Alpha particles has low energy; can shielded by a piece of paper.
17. Nuclear fission is when large nuclei break down into smaller nuclei.
18. A nucleus must be very large. Examples are Th-232 and U-235.
11.2 Half-Life
Concept Review Exercises
1. Define half-life.
2. Describe a way to determine the amount of radioactive isotope remaining after a given number of half-lives.
Answers
1. Half-life is the amount of time needed for half of a radioactive material to decay.
2. take half of the initial amount for each half-life of time elapsed
Exercises
1. Do all isotopes have a half-life? Explain.
2. Which is more radioactive—an isotope with a long half-life or an isotope with a short half-life?
3. What percent of a sample remains after one half-life? Three half-lives?
4. The half-life of polonium-218 is 3.0 min. How much of a 0.540 mg sample would remain after 9.0 minutes have passed?
5. The half-life of protactinium-234 is 6.69 hours. If a 0.812 mg sample of Pa-239 decays for 40.14 hours, what mass of the isotope remains?
6. How long does it take for 1.00 g of 103Pd to decay to 0.125 g if its half-life is 17.0 d?
7. How long does it take for 2.00 g of 94Nb to decay to 0.0625 g if its half-life is 20,000 y?
8. It took 75 y for 10.0 g of a radioactive isotope to decay to 1.25 g. What is the half-life of this isotope?
9. It took 49.2 s for 3.000 g of a radioactive isotope to decay to 0.1875 g. What is the half-life of this isotope?
Answers
1. Only radioactive isotopes have half-lives.
2. An isotope with a shorter half-life decay more rapidly is more radioactive.
3. 1 half-life: 50%; 3 half-lives: 12.5%
4. 9.0 min = 3 half-lives (make 3 arrows): 0.540 mg --> 0.270 mg --> 0.135 mg --> 0.0675 mg
5. 0.0127 mg
6. 51.0 d
7. 100 000 y
8. 25 y
9. 12.3 s
11.3 Units of Radioactivity
Concept Review Exercise
1. What units are used to quantify radioactivity?
Answer
1. the curie, the becquerel, the rad, the gray, the sievert, and the rem
Exercises
1. Define rad.
2. Define rem.
3. How does a becquerel differ from a curie?
4. How is the curie defined?
5. A sample of radon gas has an activity of 140.0 mCi. If the half-life of radon is 1,500 y, how long before the activity of the sample is 8.75 mCi?
6. A sample of curium has an activity of 1,600 Bq. If the half-life of curium is 24.0 s, how long before its activity is 25.0 Bq?
7. If a radioactive sample has an activity of 65 µCi, how many disintegrations per second are occurring?
8. If a radioactive sample has an activity of 7.55 × 105 Bq, how many disintegrations per second are occurring?
9. Describe how a radiation exposure in rems is determined.
10. Which contributes more to the rems of exposure—alpha or beta particles? Why?
11. Use Table 11.3.2 to determine which sources of radiation exposure are inescapable and which can be avoided. What percentage of radiation is unavoidable?
12. What percentage of the approximate annual radiation exposure comes from radioactive atoms that are in the body naturally?
13. Explain how a film badge works to detect radiation.
14. Explain how a Geiger counter works to detect radiation.
Answers
1. Known as the radiation absorbed dose, a rad is the absorption of 0.01 J/g of tissue.
2. Known as roentgen equivalent man, a rem is an absorption of one rad times a factor. The factor is variable depending on the type of emission and the type of irradiated tissue.
1. A becquerel is smaller and equals 1 decay per second. A curie is 3.7 × 1010 Bq.
4. A curie is defined as 3.7 × 1010 decays per second.
1. 6000 y
6. 144 s
1. 2.41 × 106 disintegrations per second
8. 7.55 × 105 disintegrations per second
1. The radiation exposure is determined by the number of rads times the quality factor of the radiation.
10. Alpha contributes more than beta because of its bigger size and electrical charge.
11. At least 16% (terrestrial and cosmic sources) of radioactivity is unavoidable; the rest depends on what else a person is exposed to.
12. About 11% come from radioactive atoms that are in the body naturally.
13. A film badge uses film, which is exposed as it is subjected to radiation.
14. The Geiger counter consists of a tube with electrodes and is filled with an inert (argon) gas. Radiation entering the tube ionizes the gas, and the ions are attracted to the electrodes and produce an electric pulse (clicking sound).
11.4 Uses of Radioactive Isotopes
Concept Review Exercise
1. Describe some of the different ways that amounts of radioactivity are applied in society.
Answer
1. Radioactive isotopes are used in dating, as tracers, and in medicine as diagnostic and treatment tools.
Exercises
1. Define tracer is and give an example of how tracers work.
2. Name two isotopes that have been used as tracers.
3. Explain how radioactive dating works.
4. Name an isotope that has been used in radioactive dating.
5. The current disintegration rate for carbon-14 is 14.0 Bq. A sample of burnt wood discovered in an archaeological excavation is found to have a carbon-14 decay rate of 3.5 Bq. If the half-life of carbon-14 is 5,700 y, approximately how old is the wood sample?
6. A small asteroid crashes to Earth. After chemical analysis, it is found to contain 1 g of technetium-99 to every 3 g of ruthenium-99, its daughter isotope. If the half-life of technetium-99 is 210,000 y, approximately how old is the asteroid?
7. What do you think are some of the positive aspects of irradiation of food?
8. What do you think are some of the negative aspects of irradiation of food?
9. Describe how iodine-131 is used to both diagnose and treat thyroid problems.
10. List at least five organs that can be imaged using radioactive isotopes.
11. Which radioactive emissions can be used therapeutically?
12. Which isotope is used in therapeutics primarily for its gamma ray emissions?
13. What volume of a radioisotope should be given if a patient needs 125 mCi of a solution which contains 45 mCi in 5.0 mL?
14. Sodium-24 is used to treat leukemia. A 36-kg patient is prescribed 145 μCi/kg and it is supplied to the hospital in a vial containing 250 μCi/mL. What volume should be given to the patient?
15. Lead-212 is one of the radioisotopes used in the treatment of breast cancer. A patient needs a 15 μCi dose and it is supplied as a solution with a concentration of 2.5 μCi/mL. What volume does the patient need? Given the half-life of lead is 10.6 hours, what will be the radioactivity of the sample after approximately four days?
Answers
1. A tracer follows the path of a chemical or a physical process. One of the uses of a tracer is following the path of water underground (answers will vary).
2. Tritium (3H) and Carbon-14 (14C) (answers will vary)
1. Radioactive dating works by comparing the amounts of parent and daughter isotopes and calculating back to how long ago all of the material was just the parent isotope.
4. Carbon-14 (14C) and Uranium-235 (235U) (answers will vary)
1. about 11,400 y
6. about 420,000 y
1. increased shelf life (answers will vary)
8. reduction in the food's vitamin content and cost
1. Iodine-131 is preferentially absorbed by the thyroid gland and can be used to measure the gland’s activity or destroy bad cells in the gland.
10. brain, bone, heart, thyroid, lung (answers will vary)
11. gamma rays, beta particles, or alpha particles
12. cobalt-60
13. 125mCi x (5.0mL/45mCi)=14mL
14. 36kg x (145μCi/kg) x (1mL/250μCi)=21mL
15. Volume given: 15μCi x (1mL/2.5μCi) = 6.0mL
Elapsed time in hours: 4 days x (24 hr/day) = 96 hr
Number of half-lives: 96 hrs/10.6 hours = 9
Radioactivity remaining after 9 half-lives: 0.029 μCi
11.5 Nuclear Energy
Concept Review Exercises
1. How is nuclear energy produced?
2. What is the difference between fission and fusion?
Answers
1. Nuclear energy is produced by carefully controlling the speed of a fission reaction.
2. In fission, large nuclei break down into small ones; in fusion, small nuclei combine to make larger ones. In both cases, a lot of energy is emitted.
Exercises
1. In the spontaneous fission of uranium-233, the following reaction occurs:
233U + 1n → 142Ce + 82Se + 101n
For every mole of 233U that decays, 0.1355 g of mass is lost. How much energy is given off per mole of 233U reacted?
2. In the spontaneous fission of plutonium-241, the following reaction occurs:
241Pu + 1n → 104Ru + 124Sn + 141n
For every mole of 241Pu that decays, 0.1326 g of mass is lost. How much energy is given off per mole of 241Pu reacted?
3. The two rarer isotopes of hydrogen—deuterium and tritium—can also be fused to make helium by the following reaction:
2H + 3H → 4He + 1n
In the course of this reaction, 0.01888 g of mass is lost. How much energy is emitted in the reaction of 1 mol of deuterium and tritium?
4. A process called helium burning is thought to occur inside older stars, forming carbon:
34He → 12C
If the reaction proceeds with 0.00781 g of mass lost on a molar basis, how much energy is given off?
5. Briefly describe how a nuclear reactor generates electricity.
6. Briefly describe the difference between how a nuclear reactor works and how a nuclear bomb works.
7. What is a chain reaction?
8. Why must uranium be enriched to supply nuclear energy?
Answers
1. 1.22 × 1013 J
2. 1.19 × 1013 J
1. 1.70 × 1012 J
4. 7.03 × 1011 J
1. A nuclear reactor generates heat, which is used to generate steam that turns a turbine to generate electricity.
6. Both nuclear reactor and nuclear bomb are powered by fission reaction however, in a nuclear reactor, the fission is monitored and controlled to occur continuously for a much longer time. In a nuclear bomb, the reaction is uncontrolled to explode in one event.
7. A chain reaction is an ever-expanding series of processes that, if left unchecked, can cause a runaway reaction and possibly an explosion.
8. Natural uranium ores contain only 0.7% U-235. Most nuclear reactors require enriched U-235 for their fuel.
11.6: Chapter Summary
Additional Exercises
1. Given that many elements are metals, suggest why it would be unsafe to have radioactive materials in contact with acids.
2. Many alpha-emitting radioactive substances are relatively safe to handle, but inhaling radioactive dust can be very dangerous. Why?
3. Uranium can be separated from its daughter isotope thorium by dissolving a sample in acid and adding sodium iodide, which precipitates thorium(III) iodide:
Th3+(aq) + 3I(aq) → ThI3(s)
If 0.567 g of Th3+ were dissolved in solution, how many milliliters of 0.500 M NaI(aq) would have to be added to precipitate all the thorium?
4. Thorium oxide can be dissolved in an acidic solution:
ThO2(s) + 4H+ → Th4+(aq) + 2H2O(ℓ)
How many milliliters of 1.55 M HCl(aq) are needed to dissolve 10.65 g of ThO2?
5. Radioactive strontium is dangerous because it can chemically replace calcium in the human body. The bones are particularly susceptible to radiation damage. Write the nuclear equation for the beta emission of strontium-90.
6. Write the nuclear equation for the beta emission of iodine-131, the isotope used to diagnose and treat thyroid problems.
7. A common uranium compound is uranyl nitrate hexahydrate [UO2(NO3)2_6H2O]. What is the formula mass of this compound?
8. Plutonium forms three oxides: PuO, PuO2, and Pu2O3. What are the formula masses of these three compounds?
9. A banana contains 600 mg of potassium, 0.0117% of which is radioactive potassium-40. If 1 g of potassium-40 has an activity of 2.626 × 105 Bq, what is the activity of a banana?
10. Smoke detectors typically contain about 0.25 mg of americium-241 as part of the smoke detection mechanism. If the activity of 1 g of americium-241 is 1.26 × 1011 Bq, what is the activity of americium-241 in the smoke detector?
11. Uranium hexafluoride (UF6) reacts with water to make uranyl fluoride (UO2F2) and hydrogen fluoride (HF). Balance the following chemical equation:
UF6 + H2O → UO2F2 + HF
12. The cyclopentadienyl anion (C5H5) is an organic ion that can make ionic compounds with positive ions of radioactive elements, such as Np3+. Balance the following chemical equation:
NpCl3 + Be(C5H5)2 → Np(C5H5)3 + BeCl2
Answers
1. Acids can dissolve metals, making aqueous solutions.
2. Alpha rays are dangerous only when the alpha emitter is in direct contact with tissue cells inside the body.
1. 14.7 mL
4. 104 mL
1. $\mathrm{^{90}_{38}Sr\rightarrow \, _{-1}^{0}e + \, _{39}^{90}Y}$
6. $\mathrm{^{131}_{53}I\rightarrow \, _{-1}^{0}e + \, _{54}^{131}Xe}$
1. 502 g/mol
8. PuO = 260.06 g/mol; PuO2 = 276.06 g/mol; Pu2O3 = 536.12 g/mol
9. about 18 Bq
10. 3.15 x 107 Bq
11. UF6 + 2H2O → UO2F2 + 4HF
12. 2NpCl3 + 3Be(C5H5)2 → 2Np(C5H5)3 + 3BeCl2 | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/11.E%3A_Nuclear_Chemistry_%28Exercises%29.txt |
12.1: Organic Chemistry
Concept Review Exercises
1. Classify each compound as organic or inorganic.
1. C3H8O
2. CaCl2
3. Cr(NH3)3Cl3
4. C30H48O3N
2. Which compound is likely organic and which is likely inorganic?
1. a flammable compound that boils at 80°C and is insoluble in water
2. a compound that does not burn, melts at 630°C, and is soluble in water
Answers
1.
1. organic
2. inorganic
3. inorganic
4. organic
2.
1. organic
2. inorganic
1. Classify each compound as organic or inorganic.
1. C6H10
2. CoCl2
3. C12H22O11
2. Classify each compound as organic or inorganic.
1. CH3NH2
2. NaNH2
3. Cu(NH3)6Cl2
3. Which member of each pair has a higher melting point?
1. CH3OH and NaOH
2. CH3Cl and KCl
4. Which member of each pair has a higher melting point?
1. C2H6 and CoCl2
2. CH4 and LiH
1.
1. organic
2. inorganic
3. organic
1.
1. NaOH
2. KCl
Concept Review Exercises
1. In the homologous series of alkanes, what is the molecular formula for the member just above C8H18?
2. Use the general formula for alkanes to write the molecular formula of the alkane with 12 carbon atoms.
1. C9H20
2. C12H26
Exercises
1. What compounds contain fewer carbon atoms than C3H8 and are its homologs?
2. What compounds contain five to eight carbon atoms and are homologs of C4H10?
Answer
1. CH4 and C2H6
Concept Review Exercises
1. In alkanes, can there be a two-carbon branch off the second carbon atom of a four-carbon chain? Explain.
2. A student is asked to write structural formulas for two different hydrocarbons having the molecular formula C5H12. She writes one formula with all five carbon atoms in a horizontal line and the other with four carbon atoms in a line, with a CH3 group extending down from the first attached to the third carbon atom. Do these structural formulas represent different molecular formulas? Explain why or why not.
Answers
1. No; the branch would make the longest continuous chain of five carbon atoms.
2. No; both are five-carbon continuous chains.
Key Takeaway
• Alkanes with four or more carbon atoms can exist in isomeric forms.
Exercises
1. Briefly identify the important distinctions between a straight-chain alkane and a branched-chain alkane.
2. How are butane and isobutane related? How do they differ?
3. Name each compound.
4. Write the structural formula for each compound.
1. hexane
2. octane
5. Indicate whether the structures in each set represent the same compound or isomers.
1. CH3CH2CH2CH3 and
2. CH3CH2CH2CH2CH3 and
Answers
1. Straight-chain alkanes and branched-chain alkanes have different properties as well as different structures.
1.
1. pentane
2. heptane
1.
1. no
2. yes
Exercises
1. Write the condensed structural formula for each structural formula.
2. A condensed structural formula for isohexane can be written as (CH3)2CHCH2CH2CH3. Draw the line-angle formula for isohexane.
3. Draw a line-angle formula for the compound CH3CH2CH(CH3)CH2CH2CH3.
4. Give the structural formula for the compound represented by this line-angle formula:
Answers
1.
1. CH3CH3
2. CH3CH2CH3
3. CH3CH2CH2CH2CH3
Concept Review Exercises
1. What is a CH3 group called when it is attached to a chain of carbon atoms—a substituent or a functional group?
2. Which type of name uses numbers to locate substituents—common names or IUPAC names?
1. substituent
2. IUPAC names
Exercises
1. Briefly identify the important distinctions between an alkane and an alkyl group.
2. How many carbon atoms are present in each molecule?
1. 2-methylbutane
2. 3-ethylpentane
3. How many carbon atoms are present in each molecule?
1. 2,3-dimethylbutane
2. 3-ethyl-2-methylheptane
4. Draw the structure for each compound.
1. 3-methylpentane
2. 2,2,5-trimethylhexane
3. 4-ethyl-3-methyloctane
5. Draw the structure for each compound.
1. 2-methylpentane
2. 4-ethyl-2-methylhexane
3. 2,2,3,3-tetramethylbutane
6. Name each compound according to the IUPAC system.
7. Name each compound according to the IUPAC system.
8. What is a substituent? How is the location of a substituent indicated in the IUPAC system?
9. Briefly identify the important distinctions between a common name and an IUPAC name.
Answers
1. An alkane is a molecule; an alkyl group is not an independent molecule but rather a part of a molecule that we consider as a unit.
1.
1. 6
2. 10
1.
1.
1. 2,2,4,4-tetramethylpentane
2. 3-ethylhexane
1. Common names are widely used but not very systematic; IUPAC names identify a parent compound and name other groups as substituents.
Concept Review Exercises
1. Without referring to a table, predict which has a higher boiling point—hexane or octane. Explain.
2. If 25 mL of hexane were added to 100 mL of water in a beaker, which of the following would you expect to happen? Explain.
1. Hexane would dissolve in water.
2. Hexane would not dissolve in water and would float on top.
3. Hexane would not dissolve in water and would sink to the bottom of the container.
Answers
1. octane because of its greater molar mass
2. b; hexane is insoluble in water and less dense than water.
Exercises
1. Without referring to a table or other reference, predict which member of each pair has the higher boiling point.
1. pentane or butane
2. heptane or nonane
2. For which member of each pair is hexane a good solvent?
1. pentane or water
2. sodium chloride or soybean oil
1. pentane
2. nonane
Concept Review Exercises
1. Why are alkanes sometimes called paraffins?
2. Which halogen reacts most readily with alkanes? Which reacts least readily?
Answers
1. Alkanes do not react with many common chemicals. They are sometimes called paraffins, from the Latin parum affinis, meaning “little affinity.”
2. most readily: $F_2$; least readily: $I_2$
Exercises
1. Why do alkanes usually not react with ionic compounds such as most laboratory acids, bases, oxidizing agents, or reducing agents?
2. Write an equation for the complete combustion of methane ($CH_4$), the main component of natural gas).
3. What is the most important reaction of alkanes?
4. Name some substances other than oxygen that react readily with alkanes.
Answers
1. Alkanes are nonpolar; they do not attract ions.
Concept Review Exercises
1. What is the IUPAC name for the HFC that has the formula CH2FCF3? (Hint: you must use a number to indicate the location of each substituent F atom.)
2. What is the IUPAC name for the HCFC that has the formula CHCl2CF3?
Answers
1. 1,1,1,2-tetrafluoroethane
2. 1,1,1-trifluoro-2,2-dichloroethane
Exercises
1. Write the condensed structural formula for each compound.
1. methyl chloride
2. chloroform
2. Write the condensed structural formula for each compound.
1. ethyl bromide
2. carbon tetrachloride
3. Write the condensed structural formulas for the two isomers that have the molecular formula C3H7Br. Give the common name and the IUPAC name of each.
4. Write the condensed structural formulas for the four isomers that have the molecular formula C4H9Br. Give the IUPAC name of each.
5. What is a CFC? How are CFCs involved in the destruction of the ozone layer?
6. Explain why each compound is less destructive to the ozone layer than are CFCs.
1. fluorocarbons
2. HCFCs
Answers
1. CH3Cl
2. CHCl3
1. CH3CH2CH2Br, propyl bromide, 1-bromopropane; CH3CHBrCH3, isopropyl bromide, 2-bromopropane
1. compounds containing Cl, F, and C; by releasing Cl atoms in the stratosphere
Concept Review Exercises
1. What is the molecular formula of cyclooctane?
2. What is the IUPAC name for this compound?
Answers
1. C8H16
2. ethylcyclopropane
Exercises
1. Draw the structure for each compound.
1. ethylcyclobutane
2. propylcyclopropane
2. Draw the structure for each compound.
1. methylcyclohexane
2. butylcyclobutane
3. Cycloalkyl groups can be derived from cycloalkanes in the same way that alkyl groups are derived from alkanes. These groups are named as cyclopropyl, cyclobutyl, and so on. Name each cycloalkyl halide.
4. Halogenated cycloalkanes can be named by the IUPAC system. As with alkyl derivatives, monosubstituted derivatives need no number to indicate the position of the halogen. To name disubstituted derivatives, the carbon atoms are numbered starting at the position of one substituent (C1) and proceeding to the second substituted atom by the shortest route. Name each compound.
Answers
1. cyclopentyl bromide
2. cyclohexyl chloride
Additional Exercises
1. You find an unlabeled jar containing a solid that melts at 48°C. It ignites readily and burns readily. The substance is insoluble in water and floats on the surface. Is the substance likely to be organic or inorganic?
2. Give the molecular formulas for methylcyclopentane, 2-methylpentane, and cyclohexane. Which are isomers?
3. What is wrong with each name? (Hint: first write the structure as if it were correct.) Give the correct name for each compound.
1. 2-dimethylpropane
2. 2,3,3-trimethylbutane
3. 2,4-diethylpentane
4. 3,4-dimethyl-5-propylhexane
4. What is the danger in swallowing a liquid alkane?
5. Distinguish between lighter and heavier liquid alkanes in terms of their effects on the skin.
6. Following is the line formula for an alkane. Draw its structure and give its name.
7. Write equations for the complete combustion of each compound.
1. propane (a bottled gas fuel)
2. octane (a typical hydrocarbon in gasoline).
8. The density of a gasoline sample is 0.690 g/mL. On the basis of the complete combustion of octane, calculate the amount in grams of carbon dioxide (CO2) and water (H2O) formed per gallon (3.78 L) of the gasoline when used in an automobile.
9. Draw the structures for the five isomeric hexanes (C6H14). Name each by the IUPAC system.
10. Indicate whether the structures in each set represent the same compound or isomers.
11. Consider the line-angle formulas shown here and answer the questions.
1. Which pair of formulas represents isomers? Draw each structure.
2. Which formula represents an alkyl halide? Name the compound and write its condensed structural formula.
3. Which formula represents a cyclic alkane? Name the compound and draw its structure.
4. What is the molecular formula of the compound represented by (i)?
Answers
1. organic
1.
1. Two numbers are needed to indicate two substituents; 2,2-dimethylpropane.
2. The lowest possible numbers were not used; 2,2,3-trimethylbutane.
3. An ethyl substituent is not possible on the second carbon atom; 3,5-dimethylheptane.
4. A propyl substituent is not possible on the fifth carbon atom; 3,4,5-trimethyloctane.
1. Lighter alkanes wash away protective skin oils; heavier alkanes form a protective layer.
1.
1. C3H8 + 5O2 → 3CO2 + 4H2O
2. 2C8H18 + 25O2 → 16CO2 + 18H2O
1. CH3CH2CH2CH2CH2CH3; hexane
1.
1. ii and iii; CH3CH2CH2CH2CH2CH2CH3 and
2. iv; 3-chloropentane; CH3CH2CHClCH2CH3
3. i; ethylcyclopentane;
4. C7H14 | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/12.E%3A_Organic_Chemistry%3A_Alkanes_and_Halogenated_Hydrocarbons_%28Exercise.txt |
Concept Review Exercises
1. Briefly identify the important distinctions between a saturated hydrocarbon and an unsaturated hydrocarbon.
2. Briefly identify the important distinctions between an alkene and an alkane.
3. Classify each compound as saturated or unsaturated. Identify each as an alkane, an alkene, or an alkyne.
1. CH3CH2C≡CCH3
Answers
1. Unsaturated hydrocarbons have double or triple bonds and are quite reactive; saturated hydrocarbons have only single bonds and are rather unreactive.
2. An alkene has a double bond; an alkane has single bonds only.
1. saturated; alkane
2. unsaturated; alkyne
3. unsaturated; alkene
Exercises
1. Draw the structure for each compound.
1. 2-methyl-2-pentene
2. 2,3-dimethyl-1-butene
3. cyclohexene
2. Draw the structure for each compound.
1. 5-methyl-1-hexene
2. 3-ethyl-2-pentene
3. 4-methyl-2-hexene
3. Name each compound according to the IUPAC system.
4. Name each compound according to the IUPAC system.
Answers
1. 2-methyl-1-pentene
2. 2-methyl-2-pentene
3. 2,5-dimethyl-2-hexene
Concept Review Exercises
1. What are cis-trans (geometric) isomers? What two types of compounds can exhibit cis-trans isomerism?
2. Classify each compound as a cis isomer, a trans isomer, or neither.
Answers
1. Cis-trans isomers are compounds that have different configurations (groups permanently in different places in space) because of the presence of a rigid structure in their molecule. Alkenes and cyclic compounds can exhibit cis-trans isomerism.
1. trans (the two hydrogen atoms are on opposite sides)
2. cis (the two hydrogen atoms are on the same side, as are the two ethyl groups)
3. cis (the two ethyl groups are on the same side)
4. neither (flipping the bond does not change the molecule. There are no isomers for this molecule)
Exercises
1. Draw the structures of the cis-trans isomers for each compound. Label them cis and trans. If no cis-trans isomers exist, write none.
1. 2-bromo-2-pentene
2. 3-hexene
3. 4-methyl-2-pentene
4. 1,1-dibromo-1-butene
5. 2-butenoic acid (CH3CH=CHCOOH)
2. Draw the structures of the cis-trans isomers for each compound. Label them cis and trans. If no cis-trans isomers exist, write none.
1. 2,3-dimethyl-2-pentene
2. 1,1-dimethyl-2-ethylcyclopropane
3. 1,2-dimethylcyclohexane
4. 5-methyl-2-hexene
5. 1,2,3-trimethylcyclopropane
Answer
1. a: none. There are two distinct geometric isomers, but since there are there are four different groups off the double bond, these are both cis/trans isomers (they are technically E/Z isomers discussed elsewhere).
b:
c:
d:
e:
Concept Review Exercises
1. Briefly describe the physical properties of alkenes. How do these properties compare to those of the alkanes?
2. Without consulting tables, arrange the following alkenes in order of increasing boiling point: 1-butene, ethene, 1-hexene, and propene.
Answers
1. Alkenes have physical properties (low boiling points, insoluble in water) quite similar to those of their corresponding alkanes.
2. ethene < propene < 1-butene < 1-hexene
Exercises
1. Without referring to a table or other reference, predict which member of each pair has the higher boiling point.
1. 1-pentene or 1-butene
2. 3-heptene or 3-nonene
2. Which is a good solvent for cyclohexene, pentane or water?
1. 1-pentene
2. 3-nonene
Concept Review Exercises
1. What is the principal difference in properties between alkenes and alkanes? How are they alike?
2. If C12H24 reacts with HBr in an addition reaction, what is the molecular formula of the product?
Answers
1. Alkenes undergo addition reactions; alkanes do not. Both burn.
2. C12H24Br2
Exercises
1. Complete each equation.
1. (CH3) 2C=CH2 + Br2
2. $\mathrm{CH_2\textrm{=C}(CH_3)CH_2CH_3 + H_2 \xrightarrow{Ni}}$
2. Complete each equation.
1. $\mathrm{CH_2\textrm{=CHCH=C}H_2 + 2H_2\xrightarrow{Ni}}$
2. $\mathrm{(CH_3)_2\textrm{C=C}(CH_3)_2 + H_2O \xrightarrow{H_2SO_4}}$
Answer
1. (CH3)2CBrCH2Br
2. CH3CH(CH3)CH2CH3
Concept Review Exercises
1. What is a monomer? What is a polymer? How do polymer molecules differ from the molecules we have discussed in earlier sections of this chapter?
2. What is addition polymerization? What structural feature usually characterizes molecules used as monomers in addition polymerization?
3. What is the molecular formula of a polymer molecule formed by the addition polymerization of 175 molecules of vinyl chloride (CH2=CHCl)?
Answers
1. Monomers are small molecules that can be assembled into giant molecules referred to as polymers, which are much larger than the molecules we discussed earlier in this chapter.
2. In addition polymerization, the monomers add to one another in such a way that the polymer contains all the atoms of the starting monomers.
3. C350H525Cl175
Exercises
1. Write the condensed structural formula of the monomer from which Saran is formed. A segment of the Saran molecule has the following structure: CH2CCl2CH2CCl2CH2CCl2CH2CCl2.
2. Write the condensed structural formula for the section of a molecule formed from four units of the monomer CH2=CHF.
1. H2C=CCl2
Concept Review Exercises
1. Briefly identify the important differences between an alkene and an alkyne. How are they similar?
2. The alkene (CH3)2CHCH2CH=CH2 is named 4-methyl-1-pentene. What is the name of (CH3)2CHCH2C≡CH?
3. Do alkynes show cis-trans isomerism? Explain.
Answers
1. Alkenes have double bonds; alkynes have triple bonds. Both undergo addition reactions.
2. 4-methyl-1-pentyne
3. No; a triply bonded carbon atom can form only one other bond. It would have to have two groups attached to show cis-trans isomerism.
Exercises
1. Draw the structure for each compound.
1. acetylene
2. 3-methyl-1-hexyne
2. Draw the structure for each compound.
1. 4-methyl-2-hexyne
2. 3-octyne
3. Name each alkyne.
1. CH3CH2CH2C≡CH
2. CH3CH2CH2C≡CCH3
1. H–C≡C–H
1. 1-pentyne
2. 2-hexyne
Concept Review Exercises
1. How do the typical reactions of benzene differ from those of the alkenes?
2. Briefly describe the bonding in benzene.
3. What does the circle mean in the chemist’s representation of benzene?
Answers
1. Benzene is rather unreactive toward addition reactions compared to an alkene.
2. Valence electrons are shared equally by all six carbon atoms (that is, the electrons are delocalized).
3. The six electrons are shared equally by all six carbon atoms.
Exercises
1. Draw the structure of benzene as if it had alternate single and double bonds.
2. Draw the structure of benzene as chemists usually represent it today.
Concept Review Exercises
1. Briefly identify the important characteristics of an aromatic compound.
2. What is meant by the prefixes meta, ortho, or para? Give the name and draw the structure for a compound that illustrates each.
3. What is a phenyl group? Give the structure for 3-phenyloctane.
Answers
1. An aromatic compound is any compound that contains a benzene ring or has certain benzene-like properties.
2. meta = 1,3 disubstitution; (answers will vary)
ortho = 1,2 disubstitution
para = 1,4 disubstitution or 1-bromo-4-chlorobenzene
3. phenyl group: C6H5 or
3-phenyloctane:
Exercises
1. Is each compound aromatic?
2. Is each compound aromatic?
3. Draw the structure for each compound.
1. toluene
2. m-diethylbenzene
3. 3,5-dinitrotoluene
4. Draw the structure for each compound.
1. p-dichlorobenzene
2. naphthalene
3. 1,2,4-trimethylbenzene
5. Name each compound with its IUPAC name.
6. Name each compound with its IUPAC name.
Answers
1. yes
2. no
1. ethylbenzene
2. isopropylbenzene
3. o-bromotoluene
4. 3,5-dichlorotoluene
Additional Exercises
1. Classify each compound as saturated or unsaturated.
1. CH3C≡CCH3
2. Classify each compound as saturated or unsaturated.
3. Give the molecular formula for each compound.
4. When three isomeric pentenes—X, Y, and Z—are hydrogenated, all three form 2-methylbutane. The addition of Cl2 to Y gives 1,2-dichloro-3-methylbutane, and the addition of Cl2 to Z gives 1,2-dichloro-2-methylbutane. Draw the original structures for X, Y, and Z.
5. Pentane and 1-pentene are both colorless, low-boiling liquids. Describe a simple test that distinguishes the two compounds. Indicate what you would observe.
6. Draw and name all the alkene cis-trans isomers corresponding to the molecular formula C5H10. (Hint: there are only two.)
7. The complete combustion of benzene forms carbon dioxide and water:
C6H6 + O2 → CO2 + H2O
Balance the equation. What mass, in grams, of carbon dioxide is formed by the complete combustion of 39.0 g of benzene?
8. Describe a physiological effect of some PAHs.
9. What are some of the hazards associated with the use of benzene?
10. What is wrong with each name? Draw the structure and give the correct name for each compound.
1. 2-methyl-4-heptene
2. 2-ethyl-2-hexene
3. 2,2-dimethyl-3-pentene
11. What is wrong with each name?
1. 2-bromobenzene
2. 3,3-dichlorotoluene
3. 1,4-dimethylnitrobenzene
12. Following are line-angle formulas for three compounds. Draw the structure and give the name for each.
13. Following are ball-and-stick molecular models for three compounds (blue balls represent H atoms; red balls are C atoms). Write the condensed structural formula and give the name for each.
Answers
1.
1. unsaturated
2. unsaturated
1.
1. C6H10
2. C4H8
1. Add bromine solution (reddish-brown) to each. Pentane will not react, and the reddish-brown color persists; 1-pentene will react, leaving a colorless solution.
1. 2C6H6 + 15O2 → 12CO2 + 6H2O; 132 g
1. carcinogenic, flammable
1.
1. number not needed
2. can’t have two groups on one carbon atom on a benzene ring
3. can’t have a substituent on the same carbon atom as the nitro group
1.
1. CH3CH=CHCH2CH2CH3; 2-hexene | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/13.E%3A_Unsaturated_and_Aromatic_Hydrocarbons_%28Exercises%29.txt |
Concept Review Exercises
1. What is the functional group of an alkene? An alkyne?
2. Does CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3 have a functional group? Explain.
Answers
1. carbon-to-carbon double bond; carbon-to-carbon triple bond
2. No; it has nothing but carbon and hydrogen atoms and all single bonds.
Exercises
1. What is the functional group of 1-butanol (CH3CH2CH2CH2OH)?
2. What is the functional group of butyl bromide, CH3CH2CH2CH2Br?
1. OH
Concept Review Exercises
1. Is isobutyl alcohol primary, secondary, or tertiary? Explain.
2. What is the LCC in 2-ethyl-1-hexanol? What is taken as the LCC in naming the compound? Explain.
Answers
1. primary; the carbon atom bearing the OH group is attached to only one other carbon atom
2. 7 carbon atoms; the 6-atom chain includes the carbon atom bearing the OH group
Exercises
1. Name each alcohol and classify it as primary, secondary, or tertiary.
1. CH3CH2CH2CH2CH2CH2OH
2. Name each alcohol and classify it as primary, secondary, or tertiary.
3. Draw the structure for each alcohol.
1. 3-hexanol
2. 3,3-dimethyl-2-butanol
3. cyclobutanol
4. Draw the structure for each alcohol.
1. cyclopentanol
2. 4-methyl-2-hexanol
3. 4,5-dimethyl-3-heptanol
Answers
1. 1-hexanol; primary
2. 3-hexanol; secondary
3. 3,3-dibromo-2-methyl-2-butanol; tertiary
Concept Review Exercises
1. Why is ethanol more soluble in water than 1-hexanol?
2. Why does 1-butanol have a lower boiling point than 1-hexanol?
Answers
1. Ethanol has an OH group and only 2 carbon atoms; 1-hexanol has one OH group for 6 carbon atoms and is thus more like a (nonpolar) hydrocarbon than ethanol is.
2. The molar mass of 1-hexanol is greater than that of 1-butanol.
Exercises
Answer the following exercises without consulting tables in the text.
1. Arrange these alcohols in order of increasing boiling point: ethanol, methanol, and 1-propanol.
2. Which has the higher boiling point—butane or 1-propanol?
3. Arrange these alcohols in order of increasing solubility in water: 1-butanol, methanol, and 1-octanol.
4. Arrange these compounds in order of increasing solubility in water: 1-butanol, ethanol, and pentane.
Answers
1. methanol < ethanol < 1-propanol
1. 1-octanol < 1-butanol < methanol
Concept Review Exercises
1. Why is methanol more toxic than ethanol?
2. How does rubbing alcohol cool a feverish patient?
Answers
1. Methanol is oxidized to formaldehyde, which destroys tissue; ethanol is oxidized to acetaldehyde and then acetic acid, a normal metabolite.
2. Evaporation removes heat.
Exercises
1. From what alkene is ethanol made? Draw its condensed structural formula.
2. Can methanol be made from an alkene? Explain.
Answer
1. ethylene; CH2=CH2
14.5: Reactions of Alcohols
Conceptual Questions
1. In a reaction, compound W with the molecular formula C4H10O is converted to compound X with the formula C4H8O. Is W oxidized, reduced, dehydrated, or none of these? Explain.
2. In a reaction, 2 mol of compound Y with the molecular formula C4H10O is converted to 1 mol of compound Z with the formula C8H18O. Is Y oxidized, reduced, or neither? Explain.
Answers
1. oxidized; H is removed
2. neither; water is removed
Exercises
1. Name the three major types of chemical reactions of alcohols.
2. Why do tertiary alcohols not undergo oxidation? Can a tertiary alcohol undergo dehydration?
3. Draw the structure of the product for each reaction.
4. Draw the structure of the product for each reaction.
5. Write an equation for the dehydration of 2-propanol to yield each compound type.
1. an alkene
2. an ether
6. Draw the structure of the alkene formed by the dehydration of cyclohexanol.
Answers
1. dehydration, oxidation, and esterification
1. $\mathrm{CH_3CHOHCH_3\underset{180^\circ C,\: excess\: acid}{\xrightarrow{conc\: H_2SO_4}} CH_3COCH_3+H_2O}$
2. $\mathrm{2CH_3CHOHCH_3 \underset{180^\circ C,\: excess\: acid}{\xrightarrow{conc\: H_2SO_4}}(CH_3)_2CHOCH(CH_3)_2+H_2O}$
Concept Review Exercises
1. In the oxidation of propylene glycol to pyruvic acid, what functional groups in the reactant are involved? What new functional groups appear in the product?
2. Oxalate ion is formed by the oxidation of ethylene glycol. In what kind of reaction is the oxalate ion involved?
Answers
1. two OH groups; a ketone group and a carboxylic acid group
2. precipitation
Exercises
1. What is a glycol?
2. Why is ethylene glycol so much more toxic to humans than propylene glycol?
3. Draw the structure for each compound.
1. 1,5-pentanediol
2. propylene glycol
4. Draw the structure for each compound.
1. 1,3-hexandiol
2. glycerol
Answers
1. an alcohol with two OH groups on adjacent carbon atoms
1. HOCH2CH2CH2CH2CH2OH
Concept Review Exercises
1. How do phenols differ from alcohols in terms of structure and properties?
2. How do phenols differ in properties from aromatic hydrocarbons?
Answers
1. Phenols have an OH group attached directly to an aromatic ring. Phenols are weakly acidic.
2. Phenols have an OH group and are somewhat soluble in water.
Exercises
1. Name each compound.
2. Name each compound.
3. Draw the structure for each compound.
1. m-iodophenol
2. p-methylphenol (p-cresol)
4. Draw the structure for each compound.
1. 2,4,6-trinitrophenol (picric acid)
2. 3,5-diethylphenol
Answers
1. o-nitrophenol
2. p-bromophenol
Concept Review Exercises
1. Why does diethyl ether (CH3CH2OCH2CH3) have a much lower boiling point than 1-butanol (CH3CH2CH2CH2OH)?
2. Which is more soluble in water—ethyl methyl ether (CH3CH2OCH3) or 1-butanol (CH3CH2CH2CH2OH)? Explain.
Answers
1. Diethyl ether has no intermolecular hydrogen bonding because there is no OH group; 1-butanol has an OH and engages in intermolecular hydrogen bonding.
2. Ethyl methyl ether (three carbon atoms, one oxygen atom) is more soluble in water than 1-butanol (four carbon atoms, one oxygen atom), even though both can engage in hydrogen bonding with water.
Exercises
1. How can ethanol give two different products when heated with sulfuric acid? Name these products.
2. Which of these ethers is isomeric with ethanol—CH3CH2OCH2CH3, CH3OCH2CH3, or CH3OCH3?
3. Name each compound.
1. CH3OCH2CH2CH3
4. Name each compound.
1. CH3CH2CH2CH2OCH3
2. CH3CH2OCH2CH2CH3
5. Draw the structure for each compound.
1. methyl ethyl ether
2. tert-butyl ethyl ether
6. Draw the structure for each compound.
1. diisopropyl ether
2. cyclopropyl propyl ether
Answers
1. Intramolecular (both the H and the OH come from the same molecule) dehydration gives ethylene; intermolecular (the H comes from one molecule and the OH comes from another molecule) dehydration gives diethyl ether.
1. methyl propyl ether
2. ethyl isopropyl ether
1. CH3OCH2CH3
Concept Review Exercises
1. Give the structure and IUPAC name for the compound that has the common name m-bromobenzaldehyde.
2. Give the IUPAC name for glyceraldehyde, (HOCH2CHOHCHO). (Hint: as a substituent, the OH group is named hydroxy.)
Answers
1. 3-bromobenzaldehyde
2. 2,3-dihydroxypropanal
Answers
1. The stem names of aldehydes and ketones are derived from those of the parent alkanes, defined by the longest continuous chain (LCC) of carbon atoms that contains the functional group.
2. For an aldehyde, drop the -e from the alkane name and add the ending -al. Methanal is the IUPAC name for formaldehyde, and ethanal is the name for acetaldehyde.
3. For a ketone, drop the -e from the alkane name and add the ending -one. Propanone is the IUPAC name for acetone, and butanone is the name for ethyl methyl ketone.
4. To indicate the position of a substituent on an aldehyde, the carbonyl carbon atom is always considered to be C1; it is unnecessary to designate this group by number.
5. To indicate the position of a substituent on a ketone, number the chain in the manner that gives the carbonyl carbon atom the lowest possible number. In cyclic ketones, it is understood that the carbonyl carbon atom is C1.
6. Give the structure and IUPAC name for the compound that has the common name m-bromobenzaldehyde.
7. Give the IUPAC name for glyceraldehyde, (HOCH2CHOHCHO). (Hint: as a substituent, the OH group is named hydroxy.)
8. 3-bromobenzaldehyde
9. 2,3-dihydroxypropanal
10. Name each compound.
11. Name each compound.
1. CH3CH2CH2CH2CH2CHO
12. Draw the structure for each compound.
1. butyraldehyde
2. 2-hexanone
3. p-nitrobenzaldehyde
13. Draw the structure for each compound.
1. 5-ethyloctanal
2. 2-chloropropanal
3. 2-hydroxy-3-pentanone
1. propanal or propionaldehyde
2. butanal or butyraldehyde
3. 3-pentanone or diethyl ketone
4. benzaldehyde
1. CH3CH2CH2CHO
Concept Review Exercises
1. What feature of their structure makes aldehydes easier to oxidize than ketones?
2. How does the carbon-to-oxygen bond of aldehydes and ketones differ from the carbon-to-carbon bond of alkenes?
Answers
1. the H on the carbonyl carbon atom
2. The carbon-to-oxygen double bond is polar; the carbon-to-carbon double bond is nonpolar.
Exercises
1. Which compound in each pair has the higher boiling point?
1. acetone or 2-propanol
2. dimethyl ether or acetaldehyde
2. Which compound in each pair has the higher boiling point?
1. butanal or 1-butanol
2. acetone or isobutane
3. Draw the structure of the alcohol that could be oxidized to each compound.
1. cyclohexanone
2. 2-methyl-1-propanal
4. Draw the structure of the alcohol that could be oxidized to each compound.
1. 2-pentanone
2. o-methylbenzaldehyde
5. Acetaldehyde is treated with each substance.
1. Ag+(aq)—What inorganic product, if any, is formed?
2. K2Cr2O7 in an acid solution—What organic product, if any, is formed?
6. Acetone is treated with each substance.
1. Ag+(aq) —What inorganic product, if any, is formed?
2. K2Cr2O7 in an acid solution—What organic product, if any, is formed?
Answers
1. 2-propanol
2. acetaldehyde
1. silver metal (Ag)
2. acetic acid (CH3COOH)
Concept Review Exercises
1. What is the functional group of a thiol? Write the condensed structural formula for ethanethiol (ethyl mercaptan).
2. What is the functional group of a disulfide? Write the condensed structural formula for dipropyl disulfide.
Answers
1. SH; CH3CH2SH
2. –S–S–; CH3CH2CH2SSCH2CH2CH3
Exercises
1. A common natural gas odorant is tert-butyl mercaptan. What is its condensed structural formula?
2. Write the equation for the oxidation of ethanethiol to diethyl disulfide.
1. (CH3)3CSH
Extra Exercises
1. Describe two ways that ethanol can be prepared. Which method is used to produce alcoholic beverages?
2. Give the structure of the alkene from which isopropyl alcohol is made by reaction with water in an acidic solution.
3. Ethanol is used as a solvent for some drugs that are not soluble in water. Why is methanol not used in medicines?
4. Give the structure of the alkene that is made from tert-butyl alcohol [(CH3)3COH] by reaction with water in an acidic solution.
5. Classify each conversion as oxidation, dehydration, or hydration (only the organic starting material and product are shown):
1. CH3OH → HCHO
2. CH3CHOHCH3 → CH3CH=CH2
3. CH2=CHCH2CH3 → CH3CHOHCH2CH3
6. Classify each conversion as oxidation, dehydration, or hydration (only the organic starting material and product are shown.):
1. CH3CHOHCH3 → CH3COCH3
2. HOOCCH=CHCOOH → HOOCCH2CHOHCOOH
3. 2 CH3OH → CH3OCH3
7. Why is methanol so much more toxic to humans than ethanol?
8. Each of the four isomeric butyl alcohols is treated with potassium dichromate (K2Cr2O7) in acid. Give the product (if any) expected from each reaction.
9. Draw the structures and give IUPAC names for the four isomeric aldehydes having the formula C5H10O.
10. Write an equation for the reaction of phenol with aqueous NaOH.
11. Write an equation for the ionization of phenol in water.
12. Draw the structures and give the common and IUPAC names for the three isomeric ketones having the formula C5H10O.
13. As we shall see in Chapter 16 "Carbohydrates", 2,3-dihydroxypropanal and 1,3-dihydroxyacetone are important carbohydrates. Draw their structures.
14. Glutaraldehyde (pentanedial) is a germicide that is replacing formaldehyde as a sterilizing agent. It is less irritating to the eyes, the nose, and skin. Write the condensed structural formula of glutaraldehyde.
15. Why does the oxidation of isopropyl alcohol give a ketone, whereas the oxidation of isobutyl alcohol gives an aldehyde?
16. Identify each compound as an alcohol, a phenol, or an ether. Classify any alcohols as primary (1°), secondary (2°), or tertiary (3°).
1. CH3CH2CH2OH
17. Identify each compound as an alcohol, a phenol, or an ether. Classify any alcohols as primary, secondary, or tertiary.
1. CH3CH2OCH2CH3
18. Tell whether each compound forms an acidic, a basic, or a neutral solution in water.
19. When water is added to ethylene in the presence of an acid catalyst, only one product—ethanol—is possible. However, when water is added to propylene, two products are possible—1-propanol and 2-propanol—but only 2-propanol is formed. In 1870, the Russian chemist Vladimir V. Markovnikov proposed a rule to predict the products of such reactions: Considering water to be HOH, the hydrogen atom of water goes on the carbon atom (of the two involved in the double bond) that has the most hydrogen atoms already bonded to it. The OH group goes on the carbon atom with fewer hydrogen atoms. Use Markovnikov’s rule to predict the product of the addition of water to each compound.
1. 2-methylpropene
2. 1-butene
3. 2-methyl-1-pentene
4. 2-methyl-2-pentene
20. Ethyl alcohol, like rubbing alcohol (isopropyl alcohol), is often used for sponge baths. What property of alcohols makes them useful for this purpose?
21. In addition to ethanol, the fermentation of grain produces other organic compounds collectively called fusel oils (FO). The four principal FO components are 1-propanol, isobutyl alcohol, 3-methyl-1-butanol, and 2-methyl-1-butanol. Draw a structure for each. (FO is quite toxic and accounts in part for hangovers.)
22. Draw and name the isomeric ethers that have the formula C5H12O.
23. Menthol is an ingredient in mentholated cough drops and nasal sprays. It produces a cooling, refreshing sensation when rubbed on the skin and so is used in shaving lotions and cosmetics. Thymol, the aromatic equivalent of menthol, is the flavoring constituent of thyme.
1. To what class of compounds does each belong?
2. Give an alternate name for thymol.
24. Write the equation for the production of ethanol by the addition of water to ethylene. How much ethanol can be made from 14.0 kg of ethylene?
25. Methanol is not particularly toxic to rats. If methanol were newly discovered and tested for toxicity in laboratory animals, what would you conclude about its safety for human consumption?
26. The amino acid cysteine has the formula HSCH2CH(NH2)COOH. What is the sulfur-containing functional group in the cysteine molecule?
27. The amino acid methionine has the formula CH3SCH2CH2CH(NH2)COOH. What functional groups are in methionine?
28. Tetrahydrocannabinol is the principal active ingredient in marijuana. What functional groups are present in this molecule?
Answers
1. addition of water to ethylene; fermentation (for beverages)
1. Methanol is too toxic.
1.
1. oxidation
2. dehydration
3. hydration
1. Methanol is oxidized in the body to toxic formaldehyde; ethanol is oxidized to the less toxic acetaldehyde.
1. C6H5OH + H2O → C6H5O + H3O+
1. Isopropyl alcohol is a secondary alcohol, whereas isobutyl alcohol is a primary alcohol.
1.
1. ether
2. tertiary alcohol
3. phenol
4. secondary alcohol
1.
1. tert-butyl alcohol
2. 2-butanol
3. 2-methyl-2-pentanol
4. 2-methyl-2-pentanol
1.
1. menthol: alcohol; thymol: phenol
2. 2-isopropyl-5-methylphenol
1. It might be ruled safe until tested on humans.
1. sulfide, amino, and carboxylic acid | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/14.E%3A_Organic_Compounds_of_Oxygen_%28Exercises%29.txt |
Concept Review Exercises
1. What is the IUPAC name for the straight-chain carboxylic acid with six carbon atoms?
2. The straight-chain aldehyde with five carbon atoms has the common name valeraldehyde. What is the common name of the corresponding straight-chain carboxylic acid?
Answers
1. hexanoic acid
2. valeric acid
Exercises
1. Draw the structure for each compound.
1. heptanoic acid
2. 3-methylbutanoic acid
3. 2,3-dibromobenzoic acid
4. β-hydroxybutyric acid
2. Draw the structure for each compound.
1. o-nitrobenzoic acid
2. p-chlorobenzoic acid
3. 3-chloropentanoic acid
4. α-chloropropionic acid
3. Name each compound with either the IUPAC name, the common name, or both.
1. (CH3)2CHCH2COOH
2. (CH3)3CCH(CH3)CH2COOH
3. CH2OHCH2CH2COOH
4. Name each compound with its IUPAC name.
1. CH3(CH2)8COOH
2. (CH3)2CHCCl2CH2CH2COOH
3. CH3CHOHCH(CH2CH3)CHICOOH
Answers
1. CH3CH2CH2CH2CH2CH2COOH
1. 3-methylbutanoic acid; β-methylbutyric acid
2. 3,4,4-trimethylpentanoic acid
3. 4-hydroxybutanoic acid; γ- hydroxybutyric acid
Concept Review Exercises
1. Caproic acid (hexanoic acid) can be prepared in an oxidation reaction from
1. what alcohol?
2. what aldehyde?
2. Give the structures of the aldehyde and the carboxylic acid formed by the oxidation of isobutyl alcohol [(CH3)2CHCH2OH].
Answers
1. CH3CH2CH2CH2CH2CH2OH
2. CH3CH2CH2CH2CH2CHO
Exercises
1. Caprylic acid (octanoic acid) can be prepared in an oxidation reaction from
1. what alcohol?
2. what aldehyde?
2. Give the structures of the aldehyde and the carboxylic acid formed by the oxidation of 1,4-butanediol (HOCH2CH2CH2CH2OH).
Answer
1. CH3CH2CH2CH2CH2CH2CH2CH2OH
2. CH3CH2CH2CH2CH2CH2CH2CHO
Concept Review Exercises
1. Which compound has the higher boiling point—butanoic acid (molar mass 88) or 2-pentanone (molar mass 86)? Explain.
2. Would you expect butyric acid (butanoic acid) to be more or less soluble than 1-butanol in water? Explain.
Answers
1. butyric acid because of hydrogen bonding (There is no intermolecular hydrogen bonding in 2-pentanone.)
2. more soluble because there is more extensive hydrogen bonding
Exercises
1. Which compound has the higher boiling point—CH3CH2CH2OCH2CH3 or CH3CH2CH2COOH? Explain.
2. Which compound has the higher boiling point—CH3CH2CH2CH2CH2OH or CH3CH2CH2COOH? Explain.
3. Which compound is more soluble in water—CH3COOH or CH3CH2CH2CH3? Explain.
4. Which compound is more soluble in water—CH3CH2COOH or CH3CH2CH2CH2CH2COOH? Explain.
Answers
1. CH3CH2CH2COOH because of hydrogen bonding (There is no intermolecular hydrogen bonding with CH3CH2CH2OCH2CH3.)
1. CH3COOH because it engages in hydrogen bonding with water (There is no intermolecular hydrogen bonding with CH3CH2CH2CH3.)
Concept Review Exercises
1. How does the neutralization of a carboxylic acid differ from that of an inorganic acid? How are they similar?
2. What products are formed when a carboxylic acid is neutralized with a strong base? What additional product is formed when a carboxylic acid is neutralized with a carbonate or a bicarbonate?
Answers
1. Insoluble carboxylic acids often form soluble carboxylate salts. Both form a salt and water.
2. a carboxylate salt and water; carbon dioxide
Exercises
1. Write the equation for the ionization of CH3CH2CH2COOH in water.
2. Write the equation for the neutralization of CH3CH2CH2COOH with sodium hydroxide [NaOH(aq)].
3. Write the equation for the reaction of CH3COOH with sodium carbonate [Na2CO3(aq)].
4. Write the equation for the reaction of CH3CH2COOH with sodium bicarbonate [NaHCO3(aq)].
5. Write the equation for the ionization of propionic acid in water.
6. Write the equation for the ionization of γ-chloropentanoic acid in water.
7. Write an equation for the reaction of butyric acid with each compound.
1. aqueous NaOH
2. aqueous NaHCO3
8. Write the condensed structural formula for each compound.
1. potassium acetate
2. calcium propanoate
9. Name each compound.
1. CH3CH2CH2COOLi+
2. CH3CH2CH2COONH4+
Answers
1. CH3CH2CH2COOH(aq) + H2O(ℓ) → CH3CH2CH2COO(aq) + H3O+(aq)
1. 2CH3COOH + Na2CO3(aq) → 2CH3COONa+(aq) + H2O(ℓ) + CO2(g)
1. CH3CH2COOH(aq) + H2O(ℓ) → CH3CH2COO(aq) + H3O+(aq)
1. CH3CH2CH2COOH(aq) + NaOH(aq) → CH3CH2CH2COONa+(aq) + H2O(ℓ)
2. CH3(CH2)2COOH + NaHCO3(aq) → CH3(CH2)COONa+(aq) + H2O(ℓ) + CO2(g)
1. lithium butyrate (lithium butanoate)
2. ammonium butanoate or ammonium butyrate
Concept Review Exercises
1. From what carboxylic acid and what alcohol can isopropyl hexanoate be made?
2. From what carboxylic acid and what alcohol can cyclobutyl butyrate be made?
Answers
1. hexanoic acid and isopropyl alcohol
2. butyric acid and cyclobutyl alcohol
Exercises
1. Draw the structure for each compound.
1. methyl acetate
2. ethyl pentanoate
3. phenyl acetate
4. isopropyl propionate
2. Draw the structure for each compound.
1. ethyl hexanoate
2. ethyl benzoate
3. phenyl benzoate
4. ethyl 3-methylhexanoate
3. Name each compound with both the common name and the IUPAC name.
4. Name each compound with both the common name and the IUPAC name.
Answers
1. methyl formate; methyl methanoate
2. ethyl propionate; ethyl propanoate
Concept Review Exercises
1. Which compound has the higher boiling point—CH3CH2CH2CH2OH or CH3COOCH3? Explain.
2. Which compound is more soluble in water—methyl butyrate or butyric acid? Explain.
Answers
1. CH3CH2CH2CH2OH because there is intermolecular hydrogen bonding (There is no intermolecular hydrogen bonding in CH3COOCH3.)
2. butyric acid because of hydrogen bonding with water
Exercises
1. Which compound has the higher boiling point—CH3CH2CH2COOH or CH3CH2CH2COOCH3? Explain.
2. Which compound is more soluble in water—methyl acetate or octyl acetate? Explain.
Answer
1. CH3CH2CH2COOH because there is intermolecular hydrogen bonding (There is no intermolecular hydrogen bonding in CH3CH2COOCH3.)
Concept Review Exercises
1. From what carboxylic acid and what alcohol can the ester isopropyl nonanoate be made?
2. From what carboxylic acid and what alcohol can the ester cyclobutyl butyrate be made?
Answers
1. nonanoic acid and isopropyl alcohol
2. butyric acid and cyclobutyl alcohol
Exercises
1. Write the equation for the reaction of acetic acid with each compound.
1. ethanol
2. 1-butanol in the presence of a mineral acid catalyst
2. Write the equation for the reaction of benzoic acid with each compound.
1. methanol
2. 1-propanol in the presence of a mineral acid catalyst
Concept Review Exercises
1. How do acidic hydrolysis and basic hydrolysis of an ester differ in terms of
1. products obtained?
2. the extent of reaction?
2. What is saponification?
Answers
1. acidic hydrolysis: carboxylic acid + alcohol; basic hydrolysis: carboxylate salt + alcohol
2. basic hydrolysis: completion; acidic hydrolysis: incomplete reaction
1. the basic hydrolysis of an ester
Exercises
1. Write an equation for the acid-catalyzed hydrolysis of ethyl acetate.
2. Write an equation for the base-catalyzed hydrolysis of ethyl acetate.
3. Complete each equation.
4. Complete each equation.
1. $\mathrm{(CH_3)_2CHCOOCH_2CH_3 + H_2O \overset{H^+}{\rightleftharpoons}}$
2. CH3COOCH(CH3)2 + KOH(aq) →
Answers
1. $\mathrm{CH_3COOCH_2CH_3 + H_2O \xrightarrow{H^+} CH_3COOH + CH_3CH_2OH}$
1. CH3COONa(aq) + CH3CH2CH2OH
2. CH3CH2CH2COOH + CH3CH2OH
Exercises
1. Draw the structure for each compound.
1. diethyl hydrogen phosphate
2. methyl dihydrogen phosphate
3. 1-glycerol phosphate
2. Name each compound.
Concept Review Exercises
1. To what inorganic compound are the amines related?
2. How are amines classified?
Answers
1. ammonia
2. by the number of hydrocarbon groups on the nitrogen atom: primary amine, one group; secondary amine, two groups; tertiary amine, three groups
Exercises
1. Draw the structure for each compound and classify the amine as primary, secondary, or tertiary.
1. dimethylamine
2. diethylmethylamine
3. 2-aminoethanol
2. Draw the structure for each compound and classify the amine as primary, secondary, or tertiary.
1. 3-aminopentane
2. 1,6-diaminohexane
3. ethylphenylamine
3. Draw the structure for each compound.
1. aniline
2. m-bromoaniline
4. Draw the structure for each compound.
1. 2-chloroaniline
2. 3,5-dichloroaniline
5. Name each compound.
1. CH3CH2CH2NH2
6. Name each compound.
1. (CH3CH2)3N
2. (CH3CH2)2NCH3
7. Draw the structure for each compound.
1. dimethylammonium chloride
2. anilinium chloride
8. Draw the structure for each compound.
1. ethylmethylammonium chloride
2. anilinium nitrate
9. Name each compound.
1. [CH3CH2NH2CH2CH3]+Br
2. [(CH3CH2)3NH]+I
10. Name each compound.
1. [(CH3)3NH]+NO3
2. [(CH3CH2)2NH2]+Cl
Answers
1. CH3NHCH3; secondary
2. tertiary
3. HOCH2CH2NH2; primary
1. propylamine
2. isopropylmethylamine
3. 2-aminopentane
1. [(CH3)2NH2+]Cl
1. diethylammonium bromide
2. triethylammonium iodide
Concept Review Exercises
1. Which compound has the higher boiling point, CH3CH2CH2CH2CH2NH2 or CH3CH2CH2CH2CH2CH3? Explain.
2. Which compound is more soluble in water, CH3CH2CH2CH2CH3 or CH3CH2NHCH2CH3? Explain.
Answers
1. CH3CH2CH2CH2CH2NH2 because the nitrogen-to-hydrogen (N–H) bonds can engage in hydrogen bonding; CH3CH2CH2CH2CH2CH3 cannot engage in hydrogen bonding
2. CH3CH2NHCH2CH3 because amines can engage in hydrogen bonding with water; alkanes cannot engage in hydrogen bonding
Exercises
1. Which compound of each pair has the higher boiling point? Explain.
1. butylamine or pentane
2. CH3NH2 or CH3CH2CH2CH2CH2NH2
2. Which compound of each pair has the higher boiling point? Explain.
1. butylamine or butyl alcohol
2. trimethylamine or propylamine
3. Which compound is more soluble in water—CH3CH2CH3 or CH3CH2NH2? Explain.
4. Which compound is more soluble in water—CH3CH2CH2NH2 or CH3CH2CH2CH2CH2CH2NH2? Explain.
Answers
1. butylamine because the N–H bonds can engage in hydrogen bonding; pentane cannot engage in hydrogen bonding
2. CH3CH2CH2CH2CH2NH2 because it has a greater molar mass than CH3NH2
1. CH3CH2NH2 because amines can engage in hydrogen bonding with water; alkanes cannot engage in hydrogen bonding
Concept Review Exercises
1. Explain the basicity of amines.
2. Contrast the physical properties of amines with those of alcohols and alkanes.
3. What is a heterocyclic compound?
Answers
1. Amines have a lone pair of electrons on the nitrogen atom and can thus act as proton acceptors (bases).
2. The solubilities of amines are similar to those of alcohols; the boiling points of primary and secondary amines are similar to those of alcohols; the boiling points of tertiary amines, which cannot engage in hydrogen bonding because they do not have a hydrogen atom on the nitrogen atom, are comparable to those of alkanes.
3. Heterocyclic compounds are ring compounds with atoms other than carbon atoms in the ring.
Exercises
1. What salt is formed in each reaction? Write its condensed structural formula.
1. CH3NH2(aq) + HBr(aq) →
2. CH3NHCH3(aq) + HNO3(aq) →
2. What salt is formed in each reaction? Draw its structure.
Answer
1. CH3NH3+Br(aq)
2. [CH3NH2CH3]+NO3(aq)
Concept Review Exercises
1. Name this compound with the common name and the IUPAC name.
2. Draw a the structural formulae for pentanamide.
Answers
1. β-bromobutyramide (3-bromobutanamide)
Exercises
1. Draw the structure for each compound.
1. formamide
2. hexanamide
2. Draw the structure for each compound.
1. propionamide
2. butanamide
3. Name each compound with the common name, the IUPAC name, or both.
4. Name the compound.
Answers
1. propionamide (propanamide)
2. α-methylbutyramide (2-methylbutanamide)
Concept Review Exercises
1. Which compound has the higher boiling point—pentanamide (CH3CH2CH2CH2CONH2) or propyl acetate (CH3COOCH2CH2CH3)? Explain.
2. Which compound is more soluble in water—propanamide (CH3CH2CONH2) or 1-pentene (CH2=CHCH2CH2CH3)? Explain.
Answers
1. pentanamide because the nitrogen-to-hydrogen (N–H) and the carbon-to-oxygen double (C=O) bonds can engage in hydrogen bonding; propyl acetate cannot engage in hydrogen bonding
2. propanamide because the N–H and C=O bonds can engage in hydrogen bonding with water; 1-pentene cannot engage in hydrogen bonding with water
Exercises
1. Which compound has the higher boiling point—butyramide (CH3CH2CH2CONH2) or ethyl acetate (CH3COOCH2CH3)? Explain.
2. Which compound has the higher boiling point—butyramide or dimethylacetamide [CH3CON(CH3)2]? Explain.
3. Which compound is more soluble in water—acetamide (CH3CONH2) or 1-butene (CH2=CHCH2CH3)? Explain.
4. Which compound is more soluble in water—CH3CONHCH3 or 2-methylbutane [CH3CH(CH3)CH2CH3)]? Explain.
Answers
1. butyramide because the nitrogen-to-hydrogen (N–H) and the carbon-to-oxygen double (C=O) bonds can engage in hydrogen bonding; ethyl acetate cannot engage in hydrogen bonding
1. acetamide because the N–H and C=O bonds can engage in hydrogen bonding with water; 1-butene cannot engage in hydrogen bonding with water
Concept Review Exercises
1. Write the condensed structural formulas and give names of the two compounds from which butanamide (CH3CH2CH2CONH2) is formed.
2. Write the condensed structural formulas and names of the two compounds from which CH3CH2CH2CH2CH2CONHCH2CH2CH3 is formed.
Answers
1. CH3CH2CH2COOH (butanoic acid) and NH3 (ammonia)
2. CH3CH2CH2CH2CH2COOH (hexanoic acid) and CH3CH2CH2NH2 (propylamine)
Exercises
1. Write the condensed structural formulas and names of the two compounds from which pentanamide (CH3CH2CH2CH2CONH2) is formed.
2. Write the condensed structural formulas and names of the two compounds from which CH3CONHCH3 is formed.
Answer
1. CH3CH2CH2CH2COOH (pentanoic acid) and NH3 (ammonia)
Concept Review Exercises
1. What are the products of the hydrolysis of an amide?
2. When the amide CH3CH2CH2CH2CONH2 is hydrolyzed in an NaOH solution, the products are CH3CH2CH2CH2COONa+ and NH3. What products are obtained when CH3CH2CH2CH2CONH2 is hydrolyzed in an hydrochloric acid solution?
Answers
1. a carboxylic acid and ammonia or an amine
2. CH3CH2CH2CH2COOH and NH4Cl
Exercises
1. Complete each equation.
2. Complete each equation.
Answer
1. CH3COOH + NH3
Concept Review Exercises
1. What are the products of the hydrolysis of an amide?
2. When the amide CH3CH2CH2CH2CONH2 is hydrolyzed in an NaOH solution, the products are CH3CH2CH2CH2COONa+ and NH3. What products are obtained when CH3CH2CH2CH2CONH2 is hydrolyzed in an hydrochloric acid solution?
ADDITIONAL EXERCISES
1. Of the families of organic compounds discussed in this chapter, which are known for their typically unpleasant odors? Which for their characteristically pleasant aromas?
2. What is esterification of a carboxylic acid? How does it differ from neutralization?
3. Like alcohols, phenols form esters with carboxylic acids. The hydrocarbon group from phenol is called phenyl. Draw the structure of phenyl acetate.
4. Describe the hydrogen bonding in carboxylic acids, both acid-acid and acid-water. How does this influence their physical properties?
5. Which compound is more soluble in water—benzoic acid or sodium benzoate? Explain.
6. Dicarboxylic acids have two carboxyl groups and are named with the ending -dioic acid. Give the equation for the reaction of 1,5-pentanedioic acid (HOOCCH2CH2CH2COOH; common name, glutaric acid) with each of the following:
1. 1 mol of NaOH
2. 2 mol of NaOH
7. Without consulting tables, arrange the following compounds in order of increasing boiling point: butyl alcohol, methyl acetate, pentane, and propionic acid.
8. From which alcohol might each acid be prepared via oxidation with acidic dichromate?
1. CH3CH2COOH
2. HCOOH
3. HOOCH2COOH
4. (CH3)2CHCH2COOH
9. The distinctive aroma and flavor of oranges are due in part to octyl acetate, an ester formed from 1-octanol (octyl alcohol) and acetic acid. Write the condensed structural formula for octyl acetate.
10. A lactone is a cyclic ester. What product is formed in following reaction?
11. A lactam is a cyclic amide. What product is formed in the following reaction?
12. Draw the structures for the eight isomeric amines that have the molecular formula C4H11N. Give each a common name and classify it as primary, secondary, or tertiary.
13. Draw the structures for the five isomeric amines that have the molecular formula C7H9N and contain a benzene ring. Classify each compound as primary, secondary, or tertiary.
14. Cocaine is usually used in the form of the salt cocaine hydrochloride and sniffed up the nose. Some prefer to ingest their cocaine by smoking it (mixed with tobacco, for example). Before smoking, the cocaine hydrochloride must be converted back to the free base (that is, to the molecular form). Explain the choice of dosage form for each route of administration.
15. Draw the structures all the isomeric amides that have the molecular formula C4H9NO.
16. An ester with the molecular formula C6H12O2 was hydrolyzed in aqueous acid to yield an acid Y and an alcohol Z. Oxidation of the alcohol with potassium dichromate (K2Cr2O7) gave the identical acid Y. What is the condensed structural formula of the ester?
17. The neutralization of 125 mL of a 0.400 M NaOH solution requires 5.10 g of a monocarboxylic acid. Draw all the possible structures for the acid.
18. If 3.00 g of acetic acid reacts with excess methanol, how many grams of methyl acetate are formed?
19. How many milliliters of a 0.100 M barium hydroxide solution are required to neutralize 0.500 g of dichloroacetic acid?
ANSWERS
1. unpleasant: carboxylic acids; pleasant: esters
2.
3.
4. sodium benzoate because it is ionic and forms ion-dipole forces with water; benzoic acid can engage in hydrogen bonding only with water
5.
6. pentane < methyl acetate < butyl alcohol < propionic acid
7.
8. CH3COOCH2CH2CH2CH2CH2CH2CH2CH3
9.
10. H3N+CH2CH2COOH
11.
12.
13.
14.
15. 20.0 mL | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/15.E%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives_%28Exercises%29.txt |
Concept Review Exercises
1. Why is photosynthesis important?
2. Identify the differences among monosaccharides, disaccharides, and polysaccharides.
Answers
1. Photosynthesis is the process by which solar energy is used to reduce carbon dioxide to carbohydrates, which are needed for energy by plants and other living organisms that eat plants.
2. A monosaccharide is the simplest carbohydrate and cannot be hydrolyzed to produce a smaller carbohydrate; a disaccharide is composed of two monosaccharide units; and a polysaccharide contains many saccharide units.
Exercises
1. When an aqueous solution of trehalose is heated, two molecules of glucose are produced for each molecule of trehalose. Is trehalose a monosaccharide, a disaccharide, or a polysaccharide?
2. When an aqueous solution of arabinose is heated, no other molecules are produced. Is arabinose a monosaccharide, a disaccharide, or a polysaccharide?
Answer
1. Trehalose is a disaccharide because it is hydrolyzed into two molecules of glucose (a monosaccharide).
Concept Review Exercises
1. What is a chiral carbon?
2. Describe how enantiomers differ.
Answers
1. A chiral carbon is a carbon atom with four different groups attached to it.
2. Enantiomers are mirror images of each other; they differ in the arrangements of atoms around a chiral carbon.
Exercises
1. Identify each sugar as an aldose or a ketose and then as a triose, tetrose, pentose, or hexose.
1. D-glucose
2. L-ribulose
3. D-glyceraldehyde
2. Identify each sugar as an aldose or a ketose and then as a triose, tetrose, pentose, or hexose.
1. dihydroxyacetone
2. D-ribose
3. D-galactose
3. Identify each sugar as an aldose or a ketose and then as a D sugar or an L sugar.
4. Identify each sugar as an aldose or a ketose and then as a D sugar or an L sugar.
Answers
1. aldose; hexose
2. ketose; pentose
3. aldose; triose
1. aldose; D sugar
2. ketose; L sugar
Concept Review Exercises
1. Describe the similarities and differences in the structures of D-glucose and D-galactose.
2. Describe similarities and differences in the structures of D-glucose and D-fructose.
Answers
1. Both monosaccharides are aldohexoses. The two monosaccharides differ in the configuration around the fourth carbon atom.
2. Both monosaccharides are hexoses. D-glucose is an aldohexose, while D-fructose is a ketohexose.
Exercises
1. Identify each sugar by its common chemical name.
1. blood sugar
2. levulose
2. Identify each sugar by its common chemical name.
1. dextrose
2. brain sugar
3. Identify each sugar as an aldohexose or a ketohexose.
1. glucose
2. galactose
3. fructose
4. What hexose would you expect to be most abundant in each food?
1. honey
2. milk
3. cornstarch
Answers
1. D-glucose
2. D-fructose
1. aldohexose
2. aldohexose
3. ketohexose
Concept Review Exercises
1. Define each term.
1. mutarotation
2. anomer
3. anomeric carbon
2. How can you prove that a solution of α-D-glucose exhibits mutarotation?
Answers
1. the ongoing interconversion between anomers of a particular carbohydrate to form an equilibrium mixture
2. a stereoisomer that differs in structure around what was the carbonyl carbon atom in the straight-chain form of a monosaccharide
3. the carbon atom that was the carbonyl carbon atom in the straight-chain form of a monosaccharide
1. Place a sample of pure α-D-glucose in a polarimeter and measure its observed rotation. This value will change as mutarotation occurs.
Exercises
1. Draw the cyclic structure for β-D-glucose. Identify the anomeric carbon.
2. Draw the cyclic structure for α-D-fructose. Identify the anomeric carbon.
3. Given that the aldohexose D-mannose differs from D-glucose only in the configuration at the second carbon atom, draw the cyclic structure for α-D-mannose.
4. Given that the aldohexose D-allose differs from D-glucose only in the configuration at the third carbon atom, draw the cyclic structure for β-D-allose.
Concept Review Exercises
1. Why are monosaccharides soluble in water?
2. What is a reducing sugar?
Answers
1. Monosaccharides are quite soluble in water because of the numerous OH groups that readily engage in hydrogen bonding with water.
2. any carbohydrate capable of reducing a mild oxidizing agent, such as Tollens’ or Benedict’s reagents, without first undergoing hydrolysis
Exercises
1. Which gives a positive Benedict’s test—L-galactose, levulose, or D-glucose?
2. Which gives a positive Benedict’s test—D-glyceraldehyde, corn sugar, or L-fructose?
3. D-Galactose can be oxidized at the sixth carbon atom to yield D-galacturonic acid and at both the first and sixth carbon atoms to yield D-galactaric acid. Draw the Fischer projection for each oxidation product.
4. D-Glucose can be oxidized at the first carbon atom to form D-gluconic acid, at the sixth carbon atom to yield D-glucuronic acid, and at both the first and sixth carbon atoms to yield D-glucaric acid. Draw the Fischer projection for each oxidation product.
Answers
1. All three will give a positive Benedict’s test because they are all monosaccharides.
Concept Review Exercise
1. What monosaccharides are obtained by the hydrolysis of each disaccharide?
1. sucrose
2. maltose
3. lactose
Answer
1. D-glucose and D-fructose
2. two molecules of D-glucose
3. D-glucose and D-galactose
Exercises
1. Identify each sugar by its common chemical name.
1. milk sugar
2. table sugar
2. Identify each sugar by its common chemical name.
1. cane sugar
2. malt sugar
3. For each disaccharide, indicate whether the glycosidic linkage is α or β.
4. For each disaccharide, indicate whether the glycosidic linkage is α or β.
5. Identify each disaccharide in Exercise 3 as a reducing or nonreducing sugar. If it is a reducing sugar, draw its structure and circle the anomeric carbon. State if the OH group at the anomeric carbon is in the α or the β position.
6. Identify each disaccharide in Exercise 4 as a reducing or nonreducing sugar. If it is a reducing sugar, draw its structure and circle the anomeric carbon. State if the OH group at the anomeric carbon is in the α or β position.
7. Melibiose is a disaccharide that occurs in some plant juices. Its structure is as follows:
1. What monosaccharide units are incorporated into melibiose?
2. What type of linkage (α or β) joins the two monosaccharide units of melibiose?
3. Melibiose has a free anomeric carbon and is thus a reducing sugar. Circle the anomeric carbon and indicate whether the OH group is α or β.
8. Cellobiose is a disaccharide composed of two glucose units joined by a β-1,4-glycosidic linkage.
1. Draw the structure of cellobiose.
2. Is cellobiose a reducing or nonreducing sugar? Justify your answer.
Answers
1. lactose
2. sucrose
1. 3a: nonreducing; 3b: reducing
1. galactose and glucose
2. α-glycosidic linkage
Concept Review Exercises
1. What purposes do starch and cellulose serve in plants?
2. What purpose does glycogen serve in animals?
Answers
1. Starch is the storage form of glucose (energy) in plants, while cellulose is a structural component of the plant cell wall.
2. Glycogen is the storage form of glucose (energy) in animals.
Exercises
1. What monosaccharide is obtained from the hydrolysis of each carbohydrate?
1. starch
2. cellulose
3. glycogen
2. For each carbohydrate listed in Exercise 1, indicate whether it is found in plants or mammals.
3. Describe the similarities and differences between amylose and cellulose.
4. Describe the similarities and differences between amylopectin and glycogen.
Answers
1. glucose
2. glucose
3. glucose
1. Amylose and cellulose are both linear polymers of glucose units, but the glycosidic linkages between the glucose units differ. The linkages in amylose are α-1,4-glycosidic linkages, while the linkages in cellulose they are β-1,4-glycosidic linkages.
ADDITIONAL EXERCISES
1. Draw the Fischer projections for D-glucose and D-ribose. Identify all the functional groups in each structure.
2. Draw the Fischer projections for D-galactose and D-fructose. Identify all the functional groups in each structure.
3. L-Fucose is an aldohexose that is often incorporated into oligosaccharides attached to cell membranes. It is also known as 6-deoxy-L-galactose. Draw the structure of L-fucose.
4. D-glucitol, also known as sorbitol, is added to shredded coconut to keep it soft and to pharmaceutical products to increase the absorption of nutrients. It is prepared industrially by the reduction of D-glucose. Propose a structure for D-glucitol.
5. Which would give a positive Benedict’s test—lactose, amylopectin, D-ribose, sucrose, D-glyceraldehyde, or amylose?
6. Which enzyme hydrolyzes each carbohydrate?
1. maltose
2. lactose
3. cellulose
4. sucrose
7. What structural characteristics are necessary if a disaccharide is to be a reducing sugar? Draw the structure of a hypothetical reducing disaccharide composed of two aldohexoses.
8. Raffinose, a trisaccharide found in beans and sugar beets, contains D-galactose, D-glucose, and D-fructose. The enzyme α-galactase catalyzes the hydrolysis of raffinose to galactose and sucrose. Draw the structure of raffinose. (The linkage from galactose to the glucose unit is α-1,6).
9. What reagent(s) could be used to carry out each conversion?
10. What reagents are necessary to carry out each conversion?
11. The structure of lactulose is shown here. What monosaccharide units compose this disaccharide?
12. N-acetylglucosamine is synthesized from D-glucosamine, which in turn is obtained from D-glucose. What reagents are needed for the conversion of D-glucosamine to N-acetylglucosamine?
13. Hyaluronic acid is a heteropolymer that acts as a lubricating agent in the fluids of joints and the eyes. Its structure consists of repeating disaccharide units containing glucuronic acid and N-acetylglucosamine connected by a β-1,3-linkage. Draw the structure of the disaccharide unit found in hyaluronic acid.
14. Several artificial sweeteners are discussed in this chapter.
1. Which are currently approved for use in the United States?
2. Which has (or have) a bitter, metallic aftertaste?
3. Which was (or were) most recently approved for use in the United States?
4. Which contain(s) potassium?
15. If 3.0 mmol (3.0 × 10−3 mol) samples of saccharin, cyclamate, aspartame, and acesulfame K were each dissolved in separate beakers containing 500 mL of pure water, which solution would have the sweetest taste? Which solution would have the least sweet taste? Justify your answers.
16. Identify two functional groups found in aspartame, acesulfame K, and sucralose.
17. Why does a deficiency of lactase lead to cramps and diarrhea?
18. How does galactosemia differ from lactose intolerance in terms of the cause of the disease and its symptoms and severity?
ANSWERS
1.
2.
3. Lactose, D-ribose, and D-glyceraldehyde would give a positive Benedict’s test.
4.
5. To be a reducing sugar, a disaccharide must contain an anomeric carbon atom that can open up to form an aldehyde functional group, as shown in this disaccharide (answers will vary).
6.
7.
1. The carbohydrate is being oxidized; Tollens’ or Benedict’s reagent could be used.
2. To form the compound shown, an aldehyde must react with methanol (CH3OH) and an acid catalyst.
8.
9. galactose and fructose
10.
11.
12. Sucralose would be expected to have the sweetest taste because its relative sweetness is the highest. Lactose would have the least sweet taste because it has the lowest relative sweetness.
13.
14. Intestinal bacteria can act on the lactose present in the intestine to produce organic acids and gases. The buildup of water and bacterial decay products leads to cramps and diarrhea. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/16.E%3A_Carbohydrates_%28Exercises%29.txt |
Additional Exercises
1. The melting point of elaidic acid is 52°C.
1. What trend is observed when comparing the melting points of elaidic acid, oleic acid, and stearic acid? Explain.
2. Would you expect the melting point of palmitelaidic acid to be lower or higher than that of elaidic acid? Explain.
2. Examine the labels on two brands of margarine and two brands of shortening and list the oils used in the various brands.
3. Draw a typical lecithin molecule that incorporates glycerol, palmitic acid, oleic acid, phosphoric acid, and choline. Circle all the ester bonds.
4. In cerebrosides, is the linkage between the fatty acid and sphingosine an amide bond or an ester bond? Justify your answer.
5. Serine is an amino acid that has the following structure. Draw the structure for a phosphatidylserine that contains a palmitic acid and a palmitoleic acid unit.
6. Explain whether each compound would be expected to diffuse through the lipid bilayer of a cell membrane.
1. potassium chloride
2. CH3CH2CH2CH2CH2CH3
3. fructose
7. Identify the role of each steroid hormone in the body.
1. progesterone
2. aldosterone
3. testosterone
4. cortisol
8. How does the structure of cholic acid differ from that of cholesterol? Which compound would you expect to be more polar? Why?
1. What fatty acid is the precursor for the prostaglandins?
2. Identify three biological effects of prostaglandins.
9. Why is it important to determine the ratio of LDLs to HDLs, rather than just the concentration of serum cholesterol?
Answers
1. Stearic acid has the highest melting point, followed by elaidic acid, and then oleic acid with the lowest melting point. Elaidic acid is a trans fatty acid, and the carbon chains can pack together almost as tightly as those of the saturated stearic acid. Oleic acid is a cis fatty acid, and the bend in the hydrocarbon chain keeps these carbon chains from packing as closely together; fewer interactions lead to a much lower melting point.
2. The melting point of palmitelaidic acid should be lower than that of elaidic acid because it has a shorter carbon chain (16, as compared to 18 for elaidic acid). The shorter the carbon chain, the lower the melting point due to a decrease in intermolecular interactions.
1. regulates the menstrual cycle and maintains pregnancy
2. regulates salt metabolism by stimulating the kidneys to retain sodium and excrete potassium
3. stimulates and maintains male sex characteristics
4. stimulates the conversion of proteins to carbohydrates
1. arachidonic acid
2. induce smooth muscle contraction, lower blood pressure, and contribute to the inflammatory response
17E: Lipids
Concept Review Exercises
1. Give an example of each compound.
f
1. saturated fatty acid
2. polyunsaturated fatty acid
3. monounsaturated fatty acid
2. Why do unsaturated fatty acids have lower melting points than saturated fatty acids?
Answers
1. stearic acid (answers will vary)
2. linoleic acid (answers will vary)
3. palmitoleic acid (answers will vary)
1. Unsaturated fatty acids cannot pack as tightly together as saturated fatty acids due to the presence of the cis double bond that puts a “kink” or bend in the hydrocarbon chain.
Exercises
Classify each fatty acid as saturated or unsaturated and indicate the number of carbon atoms in each molecule.
1. palmitoleic acid
2. myristic acid
3. linoleic acid
Classify each fatty acid as saturated or unsaturated and indicate the number of carbon atoms in each molecule.
1. stearic acid
2. oleic acid
3. palmitic acid
• Write the condensed structural formula for each fatty acid.
1. lauric acid
2. palmitoleic acid
3. linoleic acid
• Write the condensed structural formulas for each fatty acid.
1. oleic acid
2. α-linolenic acid
3. palmitic acid
• Arrange these fatty acids (all contain 18 carbon atoms) in order of increasing melting point. Justify your arrangement.
• Arrange these fatty acids (all contain 16 carbon atoms) in order of increasing melting point. Justify your arrangement.
1. CH3(CH2)14COOH
Answers
1. unsaturated; 16 carbon atoms
2. saturated; 14 carbon atoms
3. unsaturated; 18 carbon atoms
1. CH3(CH2)10COOH
2. CH3(CH2)5CH=CH(CH2)7COOH
3. CH3(CH2)3(CH2CH=CH)2(CH2)7COOH
1. c < a < b; an increase in the number of double bonds will lower the melting point because it is more difficult to closely pack the fatty acids together.
Concept Review Exercises
1. What functions does fat serve in the body?
2. Which of these triglycerides would you expect to find in higher amounts in oils? In fats? Justify your choice.
Answers
1. Fats provide energy for living organisms. They also provide insulation for body organs and transport fat-soluble vitamins.
2. The triglyceride on the left is expected to be present in higher amounts in fats because it is composed of a greater number of saturated fatty acids. The triglyceride on the right is expected to be present in higher amounts in oils because it is composed of a greater number of unsaturated fatty acids.
Exercises
1. Draw the structure for each compound.
1. trimyristin
2. a triglyceride likely to be found in peanut oil
2. Draw the structure for each compound.
1. tripalmitin
2. a triglyceride likely to be found in butter
3. Draw structures to write the reaction for the complete hydrogenation of tripalmitolein (Table \(1\) for the condensed structure of palmitoleic acid). Name the product formed.
4. Draw structures to write the reaction for the complete hydrogenation of trilinolein (Table \(1\) for the condensed structure of linoleic acid). Name the product formed.
5. Draw structures to write the reaction for the hydrolysis of trilaurin in a basic solution (Table \(1\) for the condensed structure of lauric acid).
6. Draw structures to write the reaction for the hydrolysis of tristearin in a basic solution (Table \(1\) for the condensed structure of stearic acid).
1. What compounds with a disagreeable odor are formed when butter becomes rancid?
2. How are these compounds formed?
3. How can rancidity be prevented?
1. What compound with a disagreeable odor is formed when unsaturated fatty acids react with oxygen in the atmosphere?
2. How can this process be prevented?
Answers
1
1. smaller carboxylic acids, such as butyric, caprylic, and capric acids
2. These compounds are formed by the hydrolysis of the triglycerides found in butter.
3. Rancidity can be prevented by covering the butter (to keep out moisture) and storing it in a refrigerator. (Cold temperatures slow down hydrolysis reactions.)
Concept Review Exercises
1. Name the structural unit that must be present for a molecule to be classified as a
1. phospholipid.
2. glycolipid.
3. sphingolipid.
2. Why is it important that membrane lipids have dual character—part of the molecule is hydrophilic and part of the molecule is hydrophobic?
3. Why do you suppose lecithins (phosphatidylcholines) are often added to processed foods such as hot cocoa mix?
Answers
1. a phosphate group
2. a saccharide unit (monosaccharide or more complex)
3. sphingosine
1. The dual character is critical for the formation of the lipid bilayer. The hydrophilic portions of the molecule are in contact with the aqueous environment of the cell, while the hydrophobic portion of the lipids is in the interior of the bilayer and provides a barrier to the passive diffusion of most molecules.
2. Lecithin acts as an emulsifying agent that aids in the mixing of the hot cocoa mix with water and keeps the cocoa mix evenly distributed after stirring.
Exercises
1. Classify each as a phospholipid, a glycolipid, and/or a sphingolipid. (Some lipids can be given more than one classification.)
2. Classify each as a phospholipid, a glycolipid, and/or a sphingolipid. (Some lipids can be given more than one classification.)
3. Draw the structure of the sphingomyelin that has lauric acid as its fatty acid and ethanolamine as its amino alcohol.
4. Draw the structure of the cerebroside that has myristic acid as its fatty acid and galactose as its sugar.
1. Distinguish between an integral protein and a peripheral protein.
2. What is one key function of integral proteins?
Answers
1. phospholipid
2. sphingolipid and glycolipid
1. Integral proteins span the lipid bilayer, while peripheral proteins associate with the surfaces of the lipid bilayer.
2. aid in the movement of charged and polar species across the membrane
Concept Review Exercises
1. Distinguish between a saponifiable lipid and a nonsaponifiable lipid.
2. Identify a key function for each steroid.
1. bile salt
2. cholesterol
3. estradiol
Answers
1. A saponifiable lipid reacts with aqueous alkali to yield simpler components, while a nonsaponifiable lipid does not react with alkali to yield simpler components.
1. acts as an emulsifying agent to break down large fat globules and keep these globules suspended in the aqueous digestive environment
2. a key component of mammalian cell membranes (answers will vary)
3. stimulates female sex characteristics and regulates changes during the menstrual cycle
Exercises
1. Which of these compounds are steroids—tripalmitin, cephalin, or cholesterol?
2. Which of these compounds are steroids—vitamin D, cholic acid, or lecithin?
3. Draw the basic steroid skeleton and label each ring with the appropriate letter designation.
4. Identify each compound as an adrenocortical hormone, a female sex hormone, or a male sex hormone.
1. progesterone
2. aldosterone
3. testosterone
4. cortisol
1. cholesterol
1. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/17.E%3A_Lipids_%28Exercises%29.txt |
Concept Review Exercises
1. What is the general structure of an α-amino acid?
2. Identify the amino acid that fits each description.
1. also known as aspartate
2. almost as strong a base as sodium hydroxide
3. does not have a chiral carbon
1. aspartic acid
2. arginine
3. glycine
3. Write the side chain of each amino acid.
1. serine
2. arginine
3. phenylalanine
4. Write the side chain of each amino acid.
1. aspartic acid
2. methionine
3. valine
5. Draw the structure for each amino acid.
1. alanine
2. cysteine
3. histidine
6. Draw the structure for each amino acid.
1. threonine
2. glutamic acid
3. leucine
7. Identify an amino acid whose side chain contains a(n)
1. amide functional group.
2. aromatic ring.
3. carboxyl group.
8. Identify an amino acid whose side chain contains a(n)
1. OH group
2. branched chain
3. amino group
1. CH2OH
1.
2.
1. asparagine or glutamine
2. phenylalanine, tyrosine, or tryptophan
3. aspartic acid or glutamic acid
Concept Review Exercises
1. Define each term.
1. zwitterion
2. isoelectric point
2. Draw the structure for the anion formed when alanine (at neutral pH) reacts with a base.
3. Draw the structure for the cation formed when alanine (at neutral pH) reacts with an acid.
Answers
1. an electrically neutral compound that contains both negatively and positively charged groups
2. the pH at which a given amino acid exists in solution as a zwitterion
Exercises
1. Draw the structure of leucine and determine the charge on the molecule in a(n)
1. acidic solution (pH = 1).
2. neutral solution (pH = 7).
3. a basic solution (pH = 11)
2. Draw the structure of isoleucine and determine the charge on the molecule in a(n)
1. acidic solution (pH = 1).
2. neutral solution (pH = 7).
3. basic solution (pH = 11).
Concept Review Exercises
1. Distinguish between the N-terminal amino acid and the C-terminal amino acid of a peptide or protein.
2. Describe the difference between an amino acid and a peptide.
3. Amino acid units in a protein are connected by peptide bonds. What is another name for the functional group linking the amino acids?
Answers
1. The N-terminal end is the end of a peptide or protein whose amino group is free (not involved in the formation of a peptide bond), while the C-terminal end has a free carboxyl group.
2. A peptide is composed of two or more amino acids. Amino acids are the building blocks of peptides.
3. amide bond
Exercises
1. Draw the structure for each peptide.
1. gly-val
2. val-gly
2. Draw the structure for cys-val-ala.
3. Identify the C- and N-terminal amino acids for the peptide lys-val-phe-gly-arg-cys.
4. Identify the C- and N-terminal amino acids for the peptide asp-arg-val-tyr-ile-his-pro-phe.
Answers
1. C-terminal amino acid: cys; N-terminal amino acid: lys
Concept Review Exercises
1. What is the predominant attractive force that stabilizes the formation of secondary structure in proteins?
2. Distinguish between the tertiary and quaternary levels of protein structure.
3. Briefly describe four ways in which a protein could be denatured.
Answers
1. hydrogen bonding
2. Tertiary structure refers to the unique three-dimensional shape of a single polypeptide chain, while quaternary structure describes the interaction between multiple polypeptide chains for proteins that have more than one polypeptide chain.
3. (1) heat a protein above 50°C or expose it to UV radiation; (2) add organic solvents, such as ethyl alcohol, to a protein solution; (3) add salts of heavy metal ions, such as mercury, silver, or lead; and (4) add alkaloid reagents such as tannic acid
Exercises
1. Classify each protein as fibrous or globular.
1. albumin
2. myosin
3. fibroin
2. Classify each protein as fibrous or globular.
1. hemoglobin
2. keratin
3. myoglobin
3. What name is given to the predominant secondary structure found in silk?
4. What name is given to the predominant secondary structure found in wool protein?
5. A protein has a tertiary structure formed by interactions between the side chains of the following pairs of amino acids. For each pair, identify the strongest type of interaction between these amino acids.
1. aspartic acid and lysine
2. phenylalanine and alanine
3. serine and lysine
4. two cysteines
6. A protein has a tertiary structure formed by interactions between the side chains of the following pairs of amino acids. For each pair, identify the strongest type of interaction between these amino acids.
1. valine and isoleucine
2. asparagine and serine
3. glutamic acid and arginine
4. tryptophan and methionine
7. What level(s) of protein structure is(are) ordinarily disrupted in denaturation? What level(s) is(are) not?
8. Which class of proteins is more easily denatured—fibrous or globular?
Answers
1. globular
2. fibrous
3. fibrous
1. β-pleated sheet
1. ionic bonding
2. dispersion forces
3. dispersion forces
4. disulfide linkage
1. Protein denaturation disrupts the secondary, tertiary, and quaternary levels of structure. Only primary structure is unaffected by denaturation.
Concept Review Exercise
In the small intestine, sucrose is hydrolyzed to form glucose and fructose in a reaction catalyzed by sucrase.
1. Identify the substrate in this reaction.
2. Name the enzyme.
1. sucrose
2. sucrase
Exercises
1. Identify the substrate catalyzed by each enzyme.
1. lactase
2. cellulase
3. peptidase
2. Identify the substrate catalyzed by each enzyme.
1. lipase
2. amylase
3. maltase
3. Identify each type of enzyme.
1. decarboxylase
2. protease
3. transaminase
4. Identify each type of enzyme.
1. dehydrogenase
2. isomerase
3. lipase
Answers
1. lactose
2. cellulose
3. peptides
1. lyase
2. hydrolase
3. transferase
Concept Review Exercises
1. Distinguish between the lock-and-key model and induced-fit model of enzyme action.
2. Which enzyme has greater specificity—urease or carboxypeptidase? Explain.
Answers
1. The lock-and-key model portrays an enzyme as conformationally rigid and able to bond only to substrates that exactly fit the active site. The induced fit model portrays the enzyme structure as more flexible and is complementary to the substrate only after the substrate is bound.
2. Urease has the greater specificity because it can bind only to a single substrate. Carboxypeptidase, on the other hand, can catalyze the removal of nearly any amino acid from the carboxyl end of a peptide or protein.
Exercises
1. What type of interaction would occur between each group present on a substrate molecule and a functional group of the active site in an enzyme?
1. COOH
2. NH3+
3. OH
4. CH(CH3)2
2. What type of interaction would occur between each group present on a substrate molecule and a functional group of the active site in an enzyme?
1. SH
2. NH2
3. C6H5
4. COO
3. For each functional group in Exercise 1, suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you identified.
4. For each functional group in Exercise 2, suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you identified.
Answers
1. hydrogen bonding
2. ionic bonding
3. hydrogen bonding
4. dispersion forces
1. The amino acid has a polar side chain capable of engaging in hydrogen bonding; serine (answers will vary).
2. The amino acid has a negatively charged side chain; aspartic acid (answers will vary).
3. The amino acid has a polar side chain capable of engaging in hydrogen bonding; asparagine (answers will vary).
4. The amino acid has a nonpolar side chain; isoleucine (answers will vary).
Concept Review Exercises
1. The concentration of substrate X is low. What happens to the rate of the enzyme-catalyzed reaction if the concentration of X is doubled?
2. What effect does an increase in the enzyme concentration have on the rate of an enzyme-catalyzed reaction?
Answers
1. If the concentration of the substrate is low, increasing its concentration will increase the rate of the reaction.
2. An increase in the amount of enzyme will increase the rate of the reaction (provided sufficient substrate is present).
Exercises
1. In non-enzyme-catalyzed reactions, the reaction rate increases as the concentration of reactant is increased. In an enzyme-catalyzed reaction, the reaction rate initially increases as the substrate concentration is increased but then begins to level off, so that the increase in reaction rate becomes less and less as the substrate concentration increases. Explain this difference.
2. Why do enzymes become inactive at very high temperatures?
3. An enzyme has an optimum pH of 7.4. What is most likely to happen to the activity of the enzyme if the pH drops to 6.3? Explain.
4. An enzyme has an optimum pH of 7.2. What is most likely to happen to the activity of the enzyme if the pH increases to 8.5? Explain.
Answers
1. In an enzyme-catalyzed reaction, the substrate binds to the enzyme to form an enzyme-substrate complex. If more substrate is present than enzyme, all of the enzyme binding sites will have substrate bound, and further increases in substrate concentration cannot increase the rate.
1. The activity will decrease; a pH of 6.3 is more acidic than 7.4, and one or more key groups in the active site may bind a hydrogen ion, changing the charge on that group.
Concept Review Exercises
1. What is the difference between a cofactor and a coenzyme?
2. How are vitamins related to coenzymes?
Answers
1. A coenzyme is one type of cofactor. Coenzymes are organic molecules required by some enzymes for activity. A cofactor can be either a coenzyme or an inorganic ion.
2. Coenzymes are synthesized from vitamins.
Exercises
1. Identify each vitamin as water soluble or fat soluble.
1. vitamin D
2. vitamin C
3. vitamin B12
2. Identify each vitamin as water soluble or fat soluble.
1. niacin
2. cholecalciferol
3. biotin
3. What vitamin is needed to form each coenzyme?
1. pyridoxal phosphate
2. flavin adenine dinucleotide
3. coenzyme A
4. nicotinamide adenine dinucleotide
4. What coenzyme is formed from each vitamin?
1. niacin
2. thiamine
3. cyanocobalamin
4. pantothenic acid
5. What is the function of each vitamin or coenzyme?
1. flavin adenine dinucleotide
2. vitamin A
3. biotin
6. What is the function of each vitamin or coenzyme?
1. vitamin K
2. pyridoxal phosphate
3. tetrahydrofolate
Answers
1. fat soluble
2. water soluble
3. water soluble
1. vitamin B6 or pyridoxine
2. vitamin B2 or riboflavin
3. pantothenic acid
4. vitamin B3 or niacin
1. needed by enzymes that catalyze oxidation-reduction reactions in which two hydrogen atoms are transferred
2. needed for the formation of vision pigments
3. needed by enzymes that catalyze carboxylation reactions
Concept Review Exercises
1. What are the characteristics of an irreversible inhibitor?
2. In what ways does a competitive inhibitor differ from a noncompetitive inhibitor?
Answers
1. It inactivates an enzyme by bonding covalently to a particular group at the active site.
2. A competitive inhibitor structurally resembles the substrate for a given enzyme and competes with the substrate for binding at the active site of the enzyme. A noncompetitive inhibitor binds at a site distinct from the active site and can bind to either the free enzyme or the enzyme-substrate complex.
Exercises
1. What amino acid is present in the active site of all enzymes that are irreversibly inhibited by nerve gases such as DIFP?
2. Oxaloacetate (OOCCH2COCOO) inhibits succinate dehydrogenase. Would you expect oxaloacetate to be a competitive or noncompetitive inhibitor? Explain.
1. serine
Additional Exercises
1. Draw the structure of the amino acid γ-aminobutyric acid (GABA). Would you expect to find GABA in the amino acid sequence of a protein? Explain.
2. Draw the structure of the amino acid homocysteine (R group = CH2CH2SH). Would you expect to find homocysteine in the amino acid sequence of a protein? Justify your answer.
3. Write equations to show how leucine can act as a buffer (that is, how it can neutralize added acid or base).
4. Write equations to show how isoleucine can act as a buffer (that is, how it can neutralize added acid or base).
5. Glutathione (γ-glutamylcysteinylglycine) is a tripeptide found in all cells of higher animals. It contains glutamic acid joined in an unusual peptide linkage involving the carboxyl group of the R group (known as γ-carboxyl group), rather than the usual carboxyl group (the α-carboxyl group). Draw the structure of glutathione.
6. Draw the structure of the pentapeptide whose sequence is arg-his-gly-leu-asp. Identify which of the amino acids have R groups that can donate or gain hydrogen ions.
7. Bradykinin is a peptide hormone composed of nine amino acids that lowers blood pressure. Its primary structure is arg-pro-pro-gly-phe-ser-pro-phe-arg. Would you expect bradykinin to be positively charged, negatively charged, or neutral at a pH of 6.0? Justify your answer.
8. One of the neurotransmitters involved in pain sensation is a peptide called substance P, which is composed of 11 amino acids and is released by nerve-cell terminals in response to pain. Its primary structure is arg-pro-lys-pro-gln-gln-phe-phe-gly-leu-met. Would you expect this peptide to be positively charged, negatively charged, or neutral at a pH of 6.0? Justify your answer.
9. Carbohydrates are incorporated into glycoproteins. Would you expect the incorporation of sugar units to increase or decrease the solubility of a protein? Justify your answer.
10. Some proteins have phosphate groups attached through an ester linkage to the OH groups of serine, threonine, or tyrosine residues to form phosphoproteins. Would you expect the incorporation of a phosphate group to increase or decrease the solubility of a protein? Justify your answer.
11. Refer to Table 18.5 and determine how each enzyme would be classified.
1. the enzyme that catalyzes the conversion of ethanol to acetaldehyde
2. the enzyme that catalyzes the breakdown of glucose 6-phosphate to glucose and inorganic phosphate ion (water is also a reactant in this reaction)
12. Refer to Table 18.5 and determine how each enzyme would be classified.
1. the enzyme that catalyzes the removal of a carboxyl group from pyruvate to form acetate
2. the enzyme that catalyzes the rearrangement of 3-phosphoglycerate to form 2-phosphoglycerate
13. The enzyme lysozyme has an aspartic acid residue in the active site. In acidic solution, the enzyme is inactive, but activity increases as the pH rises to around 6. Explain why.
14. The enzyme lysozyme has a glutamic acid residue in the active site. At neutral pH (6–7), the enzyme is active, but activity decreases as the pH rises. Explain why.
15. The activity of a purified enzyme is measured at a substrate concentration of 1.0 μM and found to convert 49 μmol of substrate to product in 1 min. The activity is measured at 2.0 μM substrate and found to convert 98 μmol of substrate to product/minute.
1. When the substrate concentration is 100 μM, how much substrate would you predict is converted to product in 1 min? What if the substrate concentration were increased to 1,000 μM (1.0 mM)?
2. The activities actually measured are 676 μmol product formed/minute at a substrate concentration of 100 μM and 698 μmol product formed/minute at 1,000 μM (1.0 mM) substrate. Is there any discrepancy between these values and those you predicted in Exercise 15a? Explain.
16. A patient has a fever of 39°C. Would you expect the activity of enzymes in the body to increase or decrease relative to their activity at normal body temperature (37°C)?
17. Using your knowledge of factors that influence enzyme activity, describe what happens when milk is pasteurized.
Answers
1. This amino acid would not be found in proteins because it is not an α-amino acid.
1. Bradykinin would be positively charged; all of the amino acids, except for arginine, have R groups that do not become either positively or negatively charged. The two arginines are R groups that are positively charged at neutral pH, so the peptide would have an overall positive charge.
1. Carbohydrates have many OH groups attached, which can engage in hydrogen bonding with water, which increases the solubility of the proteins.
1.
1. oxidoreductase
2. hydrolase
1. The enzyme is active when the carboxyl group in the R group of aspartic acid does not have the hydrogen attached (forming COO); the hydrogen is removed when the pH of the solution is around pH 6 or higher.
1.
1. at 100 μM, you would predict that the rate would increase 100 times to 4,900 μmol of substrate to product in 1 min; at 1.0 mM, you would predict an increase to 49,000 μmol of substrate to product in 1 min.
2. There is a great discrepancy between the predicted rates and actual rates; this occurs because the enzyme becomes saturated with substrate, preventing a further increase in the rate of the reaction (the reaction is no longer linear with respect to substrate concentration because it is at very low concentrations).
1. When milk is pasteurized, it is heated to high temperatures. These high temperatures denature the proteins in bacteria, so they cannot carry out needed functions to grow and multiply. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/18.E%3A_Amino_Acids_Proteins_and_Enzymes_%28Exercise%29.txt |
Concept Review Exercises
1. Identify the three molecules needed to form the nucleotides in each nucleic acid.
1. DNA
2. RNA
2. Classify each compound as a pentose sugar, a purine, or a pyrimidine.
1. adenine
2. guanine
3. deoxyribose
4. thymine
5. ribose
6. cytosine
Answers
1. nitrogenous base (adenine, guanine, cytosine, and thymine), 2-deoxyribose, and H3PO4
2. nitrogenous base (adenine, guanine, cytosine, and uracil), ribose, and H3PO4
1. purine
2. purine
3. pentose sugar
4. pyrimidine
5. pentose sugar
6. pyrimidine
Exercises
1. What is the sugar unit in each nucleic acid?
1. RNA
2. DNA
2. Identify the major nitrogenous bases in each nucleic acid.
1. DNA
2. RNA
3. For each structure, circle the sugar unit and identify the nucleotide as a ribonucleotide or a deoxyribonucleotide.
4. For each structure, circle the sugar unit and identify the nucleotide as a ribonucleotide or a deoxyribonucleotide.
5. For each structure, circle the nitrogenous base and identify it as a purine or pyrimidine.
6. For each structure, circle the nitrogenous base and identify it as a purine or pyrimidine.
Answers
1. ribose
2. deoxyribose
Concept Review Exercises
1. Name the two kinds of nucleic acids.
2. Which type of nucleic acid stores genetic information in the cell?
1. What are complementary bases?
2. Why is it structurally important that a purine base always pair with a pyrimidine base in the DNA double helix?
Answers
1. deoxyribonucleic acid (DNA) and ribonucleic acid (RNA)
2. DNA
1. the specific base pairings in the DNA double helix in which guanine is paired with cytosine and adenine is paired with thymine
2. The width of the DNA double helix is kept at a constant width, rather than narrowing (if two pyrimidines were across from each other) or widening (if two purines were across from each other).
Exercises
1. For this short RNA segment,
1. identify the 5′ end and the 3′ end of the molecule.
2. circle the atoms that comprise the backbone of the nucleic acid chain.
3. write the nucleotide sequence of this RNA segment.
2. For this short DNA segment,
1. identify the 5′ end and the 3′ end of the molecule.
2. circle the atoms that comprise the backbone of the nucleic acid chain.
3. write the nucleotide sequence of this DNA segment.
3. Which nitrogenous base in DNA pairs with each nitrogenous base?
1. cytosine
2. adenine
3. guanine
4. thymine
4. Which nitrogenous base in RNA pairs with each nitrogenous base?
1. cytosine
2. adenine
3. guanine
4. thymine
5. How many hydrogen bonds can form between the two strands in the short DNA segment shown below?
5′ ATGCGACTA 3′ 3′ TACGCTGAT 5′
6. How many hydrogen bonds can form between the two strands in the short DNA segment shown below?
5′ CGATGAGCC 3′ 3′ GCTACTCGG 5′
Answers
1. c. ACU
1. guanine
2. thymine
3. cytosine
4. adenine
1. 22 (2 between each AT base pair and 3 between each GC base pair)
Concept Review Exercises
1. In DNA replication, a parent DNA molecule produces two daughter molecules. What is the fate of each strand of the parent DNA double helix?
2. What is the role of DNA in transcription? What is produced in transcription?
3. Which type of RNA contains the codon? Which type of RNA contains the anticodon?
Answers
1. Each strand of the parent DNA double helix remains associated with the newly synthesized DNA strand.
2. DNA serves as a template for the synthesis of an RNA strand (the product of transcription).
3. codon: mRNA; anticodon: tRNA
Exercises
1. Describe how replication and transcription are similar.
2. Describe how replication and transcription differ.
3. A portion of the coding strand for a given gene has the sequence 5′‑ATGAGCGACTTTGCGGGATTA‑3′.
1. What is the sequence of complementary template strand?
2. What is the sequence of the mRNA that would be produced during transcription from this segment of DNA?
4. A portion of the coding strand for a given gene has the sequence 5′‑ATGGCAATCCTCAAACGCTGT‑3′.
1. What is the sequence of complementary template strand?
2. What is the sequence of the mRNA that would be produced during transcription from this segment of DNA?
Answers
1. Both processes require a template from which a complementary strand is synthesized.
3.
1. 3′‑TACTCGCTGAAACGCCCTAAT‑5′
2. 5′‑AUGAGCGACUUUGCGGGAUUA‑3′
Concept Review Exercises
1. What are the roles of mRNA and tRNA in protein synthesis?
2. What is the initiation codon?
3. What are the termination codons and how are they recognized?
Answers
1. mRNA provides the code that determines the order of amino acids in the protein; tRNA transports the amino acids to the ribosome to incorporate into the growing protein chain.
2. AUG
3. UAA, UAG, and UGA; they are recognized by special proteins called release factors, which signal the end of the translation process.
Exercises
1. Write the anticodon on tRNA that would pair with each mRNA codon.
1. 5′‑UUU‑3′
2. 5′‑CAU‑3′
3. 5′‑AGC‑3′
4. 5′‑CCG‑3′
2. Write the codon on mRNA that would pair with each tRNA anticodon.
1. 5′‑UUG‑3′
2. 5′‑GAA‑3′
3. 5′‑UCC‑3′
4. 5′‑CAC‑3′
3. The peptide hormone oxytocin contains 9 amino acid units. What is the minimum number of nucleotides needed to code for this peptide?
4. Myoglobin, a protein that stores oxygen in muscle cells, has been purified from a number of organisms. The protein from a sperm whale is composed of 153 amino acid units. What is the minimum number of nucleotides that must be present in the mRNA that codes for this protein?
5. Use Figure \(3\) to identify the amino acids carried by each tRNA molecule in Exercise 1.
6. Use Figure \(3\) to identify the amino acids carried by each tRNA molecule in Exercise 2.
7. Use Figure \(3\) to determine the amino acid sequence produced from this mRNA sequence: 5′‑AUGAGCGACUUUGCGGGAUUA‑3′.
8. Use Figure \(3\) to determine the amino acid sequence produced from this mRNA sequence: 5′‑AUGGCAAUCCUCAAACGCUGU‑3′
Answers
1. 3′‑AAA‑5′
2. 3′‑GUA‑5′
3. 3′‑UCG‑5′
4. 3′‑GGC‑5′
1. 27 nucleotides (3 nucleotides/codon)
1. 1a: phenyalanine; 1b: histidine; 1c: serine; 1d: proline
1. met-ser-asp-phe-ala-gly-leu
Concept Review Exercises
1. What effect can UV radiation have on DNA?
2. Is UV radiation an example of a physical mutagen or a chemical mutagen?
1. What causes PKU?
2. How is PKU detected and treated?
Answers
1. It can lead to the formation of a covalent bond between two adjacent thymines on a DNA strand, producing a thymine dimer.
2. physical mutagen
1. the absence of the enzyme phenylalanine hydroxylase
2. PKU is diagnosed by assaying a sample of blood or urine for phenylalanine or one of its metabolites; treatment calls for an individual to be placed on a diet containing little or no phenylalanine.
Exercises
1. A portion of the coding strand of a gene was found to have the sequence 5′‑ATGAGCGACTTTCGCCCATTA‑3′. A mutation occurred in the gene, making the sequence 5′‑ATGAGCGACCTTCGCCCATTA‑3′.
1. Identify the mutation as a substitution, an insertion, or a deletion.
2. What effect would the mutation have on the amino acid sequence of the protein obtained from this mutated gene (use Figure 19.14)?
2. A portion of the coding strand of a gene was found to have the sequence 5′‑ATGGCAATCCTCAAACGCTGT‑3′. A mutation occurred in the gene, making the sequence 5′‑ATGGCAATCCTCAACGCTGT‑3′.
1. Identify the mutation as a substitution, an insertion, or a deletion.
2. What effect would the mutation have on the amino acid sequence of the protein obtained from this mutated gene (use Figure 19.14)?
1. What is a mutagen?
2. Give two examples of mutagens.
3. For each genetic disease, indicate which enzyme is lacking or defective and the characteristic symptoms of the disease.
1. PKU
2. Tay-Sachs disease
Answers
1. substitution
2. Phenylalanine (UUU) would be replaced with leucine (CUU).
1. a chemical or physical agent that can cause a mutation
2. UV radiation and gamma radiation (answers will vary)
Questions
1. Describe the general structure of a virus.
2. How does a DNA virus differ from an RNA virus?
3. Why is HIV known as a retrovirus?
4. Describe how a DNA virus invades and destroys a cell.
1. Describe how an RNA virus invades and destroys a cell.
2. How does this differ from a DNA virus?
5. What HIV enzyme does AZT inhibit?
6. What HIV enzyme does raltegravir inhibit?
Answers
1. A virus consists of a central core of nucleic acid enclosed in a protective shell of proteins. There may be lipid or carbohydrate molecules on the surface.
2. A DNA virus has DNA as its genetic material, while an RNA virus has RNA as its genetic material.
3. In a cell, a retrovirus synthesizes a DNA copy of its RNA genetic material.
4. The DNA virus enters a host cell and induces the cell to replicate the viral DNA and produce viral proteins. These proteins and DNA assemble into new viruses that are released by the host cell, which may die in the process.
5. -
6. reverse transcriptase
7. -
Additional Exercises
1. For this nucleic acid segment,
1. classify this segment as RNA or DNA and justify your choice.
2. determine the sequence of this segment, labeling the 5′ and 3′ ends.
2. For this nucleic acid segment,
1. classify this segment as RNA or DNA and justify your choice.
2. determine the sequence of this segment, labeling the 5′ and 3′ ends.
3. One of the key pieces of information that Watson and Crick used in determining the secondary structure of DNA came from experiments done by E. Chargaff, in which he studied the nucleotide composition of DNA from many different species. Chargaff noted that the molar quantity of A was always approximately equal to the molar quantity of T, and the molar quantity of C was always approximately equal to the molar quantity of G. How were Chargaff’s results explained by the structural model of DNA proposed by Watson and Crick?
4. Suppose Chargaff (see Exercise 3) had used RNA instead of DNA. Would his results have been the same; that is, would the molar quantity of A approximately equal the molar quantity of T? Explain.
5. In the DNA segment
5′‑ATGAGGCATGAGACG‑3′ (coding strand) 3′‑TACTCCGTACTCTGC‑5′ (template strand)
1. What products would be formed from the segment’s replication?
2. Write the mRNA sequence that would be obtained from the segment’s transcription.
3. What is the amino acid sequence of the peptide produced from the mRNA in Exercise 5b?
6. In the DNA segment
5′‑ATGACGGTTTACTAAGCC‑3′ (coding strand) 3′‑TACTGCCAAATGATTCGG‑5′ (template strand)
1. What products would be formed from the segment’s replication?
2. Write the mRNA sequence that would be obtained from the segment’s transcription.
3. What is the amino acid sequence of the peptide produced from the mRNA in Exercise 6b?
7. A hypothetical protein has a molar mass of 23,300 Da. Assume that the average molar mass of an amino acid is 120.
1. How many amino acids are present in this hypothetical protein?
2. What is the minimum number of codons present in the mRNA that codes for this protein?
3. What is the minimum number of nucleotides needed to code for this protein?
8. Bradykinin is a potent peptide hormone composed of nine amino acids that lowers blood pressure.
1. The amino acid sequence for bradykinin is arg-pro-pro-gly-phe-ser-pro-phe-arg. Postulate a base sequence in the mRNA that would direct the synthesis of this hormone. Include an initiation codon and a termination codon.
2. What is the nucleotide sequence of the DNA that codes for this mRNA?
9. A particular DNA coding segment is ACGTTAGCCCCAGCT.
1. Write the sequence of nucleotides in the corresponding mRNA.
2. Determine the amino acid sequence formed from the mRNA in Exercise 9a during translation.
3. What amino acid sequence results from each of the following mutations?
1. replacement of the underlined guanine by adenine
2. insertion of thymine immediately after the underlined guanine
3. deletion of the underlined guanine
10. A particular DNA coding segment is TACGACGTAACAAGC.
1. Write the sequence of nucleotides in the corresponding mRNA.
2. Determine the amino acid sequence formed from the mRNA in Exercise 10a during translation.
3. What amino acid sequence results from each of the following mutations?
1. replacement of the underlined guanine by adenine
2. replacement of the underlined adenine by thymine
11. Two possible point mutations are the substitution of lysine for leucine or the substitution of serine for threonine. Which is likely to be more serious and why?
12. Two possible point mutations are the substitution of valine for leucine or the substitution of glutamic acid for histidine. Which is likely to be more serious and why?
Answers
1.
1. RNA; the sugar is ribose, rather than deoxyribose
2. 5′‑GUA‑3′
1. In the DNA structure, because guanine (G) is always paired with cytosine (C) and adenine (A) is always paired with thymine (T), you would expect to have equal amounts of each.
1.
1. Each strand would be replicated, resulting in two double-stranded segments.
2. 5′‑AUGAGGCAUGAGACG‑3′
3. met-arg-his-glu-thr
1.
1. 194
2. 194
3. 582
1.
1. 5′‑ACGUUAGCCCCAGCU‑3′
2. thr-leu-ala-pro-ala
1. thr-leu-thr-pro-ala
2. thr-leu-val-pro-ser
3. thr-leu-pro-gin
1. substitution of lysine for leucine because you are changing from an amino acid with a nonpolar side chain to one that has a positively charged side chain; both serine and threonine, on the other hand, have polar side chains containing the OH group. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/19.E%3A_Nucleic_Acids_%28Exercises%29.txt |
Concept Review Exercise
1. Why is ATP referred to as the energy currency of the cell?
Answer
1. ATP is the principal molecule involved in energy exchange reactions in biological systems.
Exercises
1. How do ATP and ADP differ in structure?
2. Why does the hydrolysis of ATP to ADP involve the release of energy?
3. Identify whether each compound would be classified as a high-energy phosphate compound.
1. ATP
2. glucose 6-phosphate
3. creatine phosphate
4. Identify whether each compound would be classified as a high-energy phosphate compound.
1. ADP
2. AMP
3. glucose 1-phosphate
Answers
1. ATP has a triphosphate group attached, while ADP has only a diphosphate group attached.
1. yes
2. no
3. yes
Concept Review Exercises
1. Distinguish between each pair of compounds.
1. pepsin and pepsinogen
2. chymotrypsin and trypsin
3. aminopeptidase and carboxypeptidase
2. What are the primary end products of each form of digestion?
1. carbohydrate digestion
2. lipid digestion
3. protein digestion
3. In what section of the digestive tract does most of the carbohydrate, lipid, and protein digestion take place?
Answers
1. Pepsinogen is an inactive form of pepsin; pepsin is the active form of the enzyme.
2. Both enzymes catalyze the hydrolysis of peptide bonds. Chymotrypsin catalyzes the hydrolysis of peptide bonds following aromatic amino acids, while trypsin catalyzes the hydrolysis of peptide bonds following lysine and arginine.
3. Aminopeptidase catalyzes the hydrolysis of amino acids from the N-terminal end of a protein, while carboxypeptidase catalyzes the hydrolysis of amino acids from the C-terminal end of a protein.
1. glucose, fructose, and galactose
2. monoglycerides and fatty acids
3. amino acids
1. the small intestine
Exercises
1. What are the products of digestion (or stage I of catabolism)?
2. What is the general type of reaction used in digestion?
3. Give the site of action and the function of each enzyme.
1. chymotrypsin
2. lactase
3. pepsin
4. maltase
4. Give the site of action and the function of each enzyme.
1. α-amylase
2. trypsin
3. sucrase
4. aminopeptidase
1. What is the meaning of the following statement? “Bile salts act to emulsify lipids in the small intestine.”
2. Why is emulsification important?
5. Using chemical equations, describe the chemical changes that triglycerides undergo during digestion.
6. What are the expected products from the enzymatic action of chymotrypsin on each amino acid segment?
1. gly-ala-phe-thr-leu
2. ala-ile-tyr-ser-arg
3. val-trp-arg-leu-cys
7. What are the expected products from the enzymatic action of trypsin on each amino acid segment?
1. leu-thr-glu-lys-ala
2. phe-arg-ala-leu-val
3. ala-arg-glu-trp-lys
Answers
1. proteins: amino acids; carbohydrates: monosaccharides; fats: fatty acids and glycerol
1. Chymotrypsin is found in the small intestine and catalyzes the hydrolysis of peptide bonds following aromatic amino acids.
2. Lactase is found in the small intestine and catalyzes the hydrolysis of lactose.
3. Pepsin is found in the stomach and catalyzes the hydrolysis of peptide bonds, primarily those that occur after aromatic amino acids.
4. Maltase is found in the small intestine and catalyzes the hydrolysis of maltose.
1. Bile salts aid in digestion by dispersing lipids throughout the aqueous solution in the small intestine.
2. Emulsification is important because lipids are not soluble in water; it breaks lipids up into smaller particles that can be more readily hydrolyzed by lipases.
1. gly-ala-phe and thr-leu
2. ala-ile-tyr and ser-arg
3. val-trp and arg-leu-cys
Concept Review Exercises
1. What is a metabolic pathway?
2. What vitamin is required to make coenzyme A?
3. What is the net yield of Glycolysis as far as ATP?
4. Name the enzymes that are key regulatory sites in Glycolysis.
5. Why are the enzymes in the previous question targets for regulation?
6. Why is the priming phase necessary?
7. Draw the entire pathway for glycolysis including enzymes, reactants and products for each step.
8. Where does beta-oxidation occur?
9. What is the average net yield of ATP per carbon?
10. Where exactly is water formed during the process of fatty acid degradation? (Hint: H2O is formed when when the one of the products of beta-oxidation is passed through another of the metabolic pathways)
11. During the process of beta-oxidation, why is it that [FAD] is used to oxidize an alkane to an alkene while NAD+ is used to oxidize an alchol to a carbonyl
12. Draw out the entire process of the degradation of a triglyceride, include enzymes and products and reactants for each step.
Answers
1. A metabolic pathway is a series of biochemical reactions by which an organism converts a given reactant to a specific end product.
2. pantothenic acid
Concept Review Exercises
1. What is the main function of the citric acid cycle?
2. Two carbon atoms are fed into the citric acid cycle as acetyl-CoA. In what form are two carbon atoms removed from the cycle?
3. What are mitochondria and what is their function in the cell?
Answers
1. the complete oxidation of carbon atoms to carbon dioxide and the formation of a high-energy phosphate compound, energy rich reduced coenzymes (NADH and FADH2), and metabolic intermediates for the synthesis of other compounds
2. as carbon dioxide
3. Mitochondria are small organelles with a double membrane that contain the enzymes and other molecules needed for the production of most of the ATP needed by the body.
Exercises
1. Replace each question mark with the correct compound.
1. $\mathrm{?\xrightarrow{aconitase}isocitrate}$
2. $\mathrm{?\, +\, ? \xrightarrow{citrate\: synthase} citrate + coenzyme\: A}$
3. $\mathrm{fumarate \xrightarrow{fumarase}\, ?}$
4. $\mathrm{isocitrate + NAD^+ \xrightarrow{?} \alpha\textrm{-ketoglurate} + NADH + CO_2}$
2. Replace each question mark with the correct compound.
1. $\mathrm{malate + NAD^+ \xrightarrow{?} oxaloacetate + NADH}$
2. $\mathrm{?\, +\, ? \xrightarrow{nucleoside\: diphosphokinase} GDP + ATP}$
3. $\mathrm{\textrm{succinyl-CoA} \xrightarrow{\textrm{succinyl-CoA synthetase}} \,?\, +\, ?}$
4. $\mathrm{succinate + FAD \xrightarrow{succinate\: dehydrogenase}\, ? + FADH_2}$
3. From the reactions in Exercises 1 and 2, select the equation(s) by number and letter in which each type of reaction occurs.
1. isomerization
2. hydration
3. synthesis
4. From the reactions in Exercises 1 and 2, select the equation(s) by number and letter in which each type of reaction occurs.
1. oxidation
2. decarboxylation
3. phosphorylation
5. What similar role do coenzyme Q and cytochrome c serve in the electron transport chain?
6. What is the electron acceptor at the end of the electron transport chain? To what product is this compound reduced?
7. What is the function of the cytochromes in the electron transport chain?
1. What is meant by this statement? “Electron transport is tightly coupled to oxidative phosphorylation.”
2. How are electron transport and oxidative phosphorylation coupled or linked?
Answers
1. citrate
2. oxaloacetate + acetyl-CoA
3. malate
4. α-ketoglutarate hydrogenase complex
1. reaction in 1a
2. reaction in 1c
3. reaction in 1b
1. Both molecules serve as electron shuttles between the complexes of the electron transport chain.
1. Cytochromes are proteins in the electron transport chain and serve as one-electron carriers.
• Concept Review Exercises
1. In glycolysis, how many molecules of pyruvate are produced from one molecule of glucose?
2. In vertebrates, what happens to pyruvate when
1. plenty of oxygen is available?
2. oxygen supplies are limited?
3. In anaerobic glycolysis, how many molecules of ATP are produced from one molecule of glucose?
Answers
1. two
1. Pyruvate is completely oxidized to carbon dioxide.
2. Pyruvate is reduced to lactate, allowing for the reoxidation of NADH to NAD+.
2. There is a net production of two molecules of ATP.
Exercises
1. Replace each question mark with the correct compound.
1. $\mathrm{fructose\: 1,6\textrm{-bisphosphate} \xrightarrow{aldolase}\, ?\, +\, ?}$
2. $\mathrm{? + ADP \xrightarrow{pyruvate\: kinase} pyruvate + ATP}$
3. $\mathrm{dihydroxyacetone\: phosphate \xrightarrow{?} glyceraldehyde\: 3\textrm{-phosphate}}$
4. $\mathrm{glucose + ATP \xrightarrow{hexokinase} \, ? + ADP}$
2. Replace each question mark with the correct compound.
1. $\mathrm{fructose\: 6\textrm{-phosphate} + ATP \xrightarrow{?} fructose\: 1,6\textrm{-bisphosphate} + ADP}$
2. $\mathrm{? \xrightarrow{phosphoglucose\: isomerase} fructose\: 6\textrm{-phosphate}}$
3. $\mathrm{glyceraldehyde\: 3\textrm{-phosphate} + NAD^+ + P_i \xrightarrow{?} 1,3\textrm{-bisphosphoglycerate} + NADH}$
4. $\mathrm{3\textrm{-phosphoglycerate} \xrightarrow{phosphoglyceromutase} \, ?}$
3. From the reactions in Exercises 1 and 2, select the equation(s) by number and letter in which each type of reaction occurs.
1. hydrolysis of a high-energy phosphate compound
2. synthesis of ATP
4. From the reactions in Exercises 1 and 2, select the equation(s) by number and letter in which each type of reaction occurs.
1. isomerization
2. oxidation
5. What coenzyme is needed as an oxidizing agent in glycolysis?
6. Calculate
1. the total number of molecules of ATP produced for each molecule of glucose converted to pyruvate in glycolysis.
2. the number of molecules of ATP hydrolyzed in phase I of glycolysis.
3. the net ATP production from glycolysis alone.
7. How is the NADH produced in glycolysis reoxidized when oxygen supplies are limited in
1. muscle cells?
2. yeast?
1. Calculate the number of moles of ATP produced by the aerobic oxidation of 1 mol of glucose in a liver cell.
2. Of the total calculated in Exercise 9a, determine the number of moles of ATP produced in each process.
1. glycolysis alone
2. the citric acid cycle
3. the electron transport chain and oxidative phosphorylation
Answers
1. glyceraldehyde 3-phosphate + dihydroxyacetone phosphate
2. phosphoenolpyruvate
3. triose phosphate isomerase
4. glucose 6-phosphate
1. reactions 1b, 1d, and 2a
2. reaction 1b
1. NAD+
1. Pyruvate is reduced to lactate, and NADH is reoxidized to NAD+.
2. Pyruvate is converted to ethanol and carbon dioxide, and NADH is reoxidized to NAD+.
• Concept Review Exercises
1. How are fatty acids activated prior to being transported into the mitochondria and oxidized?
2. Draw the structure of hexanoic (caproic) acid and identify the α-carbon and the β-carbon.
Answers
1. They react with CoA to form fatty acyl-CoA molecules.
Key Takeaways
• Fatty acids, obtained from the breakdown of triglycerides and other lipids, are oxidized through a series of reactions known as β-oxidation.
• In each round of β-oxidation, 1 molecule of acetyl-CoA, 1 molecule of NADH, and 1 molecule of FADH2 are produced.
• The acetyl-CoA, NADH, and FADH2 are used in the citric acid cycle, the electron transport chain, and oxidative phosphorylation to produce ATP.
Exercises
1. For each reaction found in β-oxidation, identify the enzyme that catalyzes the reaction and classify the reaction as oxidation-reduction, hydration, or cleavage.
2. What are the products of β-oxidation?
3. How many rounds of β-oxidation are necessary to metabolize lauric acid (a saturated fatty acid with 12 carbon atoms)?
4. How many rounds of β-oxidation are necessary to metabolize arachidic acid (a saturated fatty acid with 20 carbon atoms)?
5. When myristic acid (a saturated fatty acid with 14 carbon atoms) is completely oxidized by β-oxidation, how many molecules of each are formed?
1. acetyl-CoA
2. FADH2
3. NADH
6. When stearic acid (a saturated fatty acid with 18 carbon atoms) is completely oxidized by β-oxidation, how many molecules of each are formed?
1. acetyl-CoA
2. FADH2
3. NADH
7. What is the net yield of ATP from the complete oxidation, in a liver cell, of one molecule of myristic acid?
8. What is the net yield of ATP from the complete oxidation, in a liver cell, of one molecule of stearic acid?
Answers
1. enoyl-CoA hydratase; hydration
2. thiolase; cleavage
3. acyl-CoA dehydrogenase; oxidation-reduction
1. five rounds
1. 7 molecules
2. 6 molecules
3. 6 molecules
1. 112 molecules
20.7: Stage II of Protein Catabolism
Concept Review Exercises
1. Write the equation for the transamination reaction between alanine and oxaloacetate.
2. Name the two products that are formed.
1. What is the purpose of oxidative deamination?
Answers
1. pyruvate and aspartate
1. Oxidative deamination provides a reaction in which the amino group [as the ammonium (NH4+) ion] is removed from a molecule, not simply transferred from one molecule to another. Most of the NH4+ ion is converted to urea and excreted from the body.
Exercises
1. Write the equation for the transamination reaction between valine and pyruvate.
2. Write the equation for the transamination reaction between phenylalanine and oxaloacetate.
3. What products are formed in the oxidative deamination of glutamate?
4. Determine if each amino acid is glucogenic, ketogenic, or both.
1. phenylalanine
2. leucine
3. serine
5. Determine if each amino acid is glucogenic, ketogenic, or both.
1. asparagine
2. tyrosine
3. valine
Answers
1. α-ketoglutarate, NADH, and NH4+
1. glucogenic
2. both
3. glucogenic
Additional Exercises
1. Hydrolysis of which compound—arginine phosphate or glucose 6-phosphate—would provide enough energy for the phosphorylation of ATP? Why?
2. If a cracker, which is rich in starch, is chewed for a long time, it begins to develop a sweet, sugary taste. Why?
3. Indicate where each enzymes would cleave the short peptide ala-ser-met-val-phe-gly-cys-lys-asp-leu.
1. aminopeptidase
2. chymotrypsin
4. Indicate where each enzymes would cleave the short peptide ala-ser-met-val-phe-gly-cys-lys-asp-leu.
1. trypsin
2. carboxypeptidase
5. If the methyl carbon atom of acetyl-CoA is labeled, where does the label appear after the acetyl-CoA goes through one round of the citric acid cycle?
6. If the carbonyl carbon atom of acetyl-CoA is labeled, where does the label appear after the acetyl-CoA goes through one round of the citric acid cycle?
7. The average adult consumes about 65 g of fructose daily (either as the free sugar or from the breakdown of sucrose). In the liver, fructose is first phosphorylated to fructose 1-phosphate, which is then split into dihydroxyacetone phosphate and glyceraldehyde. Glyceraldehyde is then phosphorylated to glyceraldehyde 3-phosphate, with ATP as the phosphate group donor. Write the equations (using structural formulas) for these three steps. Indicate the type of enzyme that catalyzes each step.
8. What critical role is played by both BPG and PEP in glycolysis?
9. How is the NADH produced in glycolysis reoxidized when oxygen supplies are abundant?
10. When a triglyceride is hydrolyzed to form three fatty acids and glycerol, the glycerol can be converted to glycerol 3-phosphate and then oxidized to form dihydroxyacetone phosphate, an intermediate of glycolysis. (In this reaction, NAD+ is reduced to NADH.) If you assume that there is sufficient oxygen to completely oxidize the pyruvate formed from dihydroxyacetone phosphate, what is the maximum amount of ATP formed from the complete oxidation of 1 mol of glycerol?
11. How is the FADH2 from β-oxidation converted back to FAD?
12. If 1 mol of alanine is converted to pyruvate in a muscle cell (through transamination) and the pyruvate is then metabolized via the citric acid cycle, the electron transport chain, and oxidative phosphorylation, how many moles of ATP are produced?
13. If the essential amino acid leucine (2-amino-4-methylpentanoic acid) is lacking in the diet, an α-keto acid can substitute for it. Give the structure of the α-keto acid and the probable reaction used to form leucine from this α-keto acid.
Answers
1. The hydrolysis of arginine phosphate releases more energy than is needed for the synthesis of ATP, while hydrolysis of glucose 6-phosphate does not.
1.
1. The enzyme will cleave off amino acids one at a time beginning with alanine (the N-terminal end).
2. following phenylalanine
1. Half of the label will be on the second carbon atom of oxaloacetate, while the other half will be on the third carbon atom.
1. When oxygen is abundant, NADH is reoxidized through the reactions of the electron transport chain.
1. FADH2 is reoxidized back to FAD via the electron transport chain. | textbooks/chem/Introductory_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry/Exercises%3A_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20.E%3A_Energy_Metabolism_%28Exercises%29.txt |
• 1.1: Matter and energy
Matter, i.e., anything that occupies space and has mass, and energy, i.e., the ability to do work are their forms and basic properties are described.
• 1.2: What is chemistry?
Chemistry, i.e., a study of matter's composition, properties, and transformations are described. The scientific method comprising of observations, hypothesis, experiments, law, and theory is also described.
• 1.3: Measurements
Measurements representing properties quantitatively, Units, i.e., quantities defined by the standard that peoples agree to use as a reference, systems of units, i.e., a set of units used for measurements, and basic mathematics involved in handling the units are described.
• 1.4: Significant Figures
Significant figures, i.e., all the digits reported in the measured value, including the estimated digit is described. How to determine the significant figures and how to handle them in calculations are also presented.
• 1.5: Unit conversions
Conversion of values in one unit to the same value in another unit using conversion factors derived from the equality between the two units is described along with the rules for handling significant figures in these calculations.
• 1.6: Equations and graphs
Manipulating the equations commonly used in chemistry, graphs for a visual presentation of trends,s and interpreting graphs are described.
• 1.7: Density and specific gravity measurements
Density, i.e., mass-to-volume ratio, which is a physical characteristic of a substance is described. Specific gravity, i.e., a ratio of the density of a substance-to-the density of water which is a unitless number is also described.
• 1.8: Heat and its measurements
Measurements of heat, i.e., energy in transfer, are described. The heat released from food and its calculations are also presented.
• 1.9: Heat and changes in physical states of matter
Effects of heat on a substance, i.e., temperature change or change of physical state of the substance are described.
01: Matter energy and their measurements
Learning Objectives
• Define matter, element, molecule, compound, molecule, mixture, homogeneous and heterogeneous mixtures, the law of constant composition, states of matter, i.e., solid, liquid, gas, and plasma, energy and its types kinetic and potential energy, endothermic and exothermic and exothermic processes, and the law of conservation of energy.
• Be able to write and interpret symbols of elements, molecules, and compounds.
What is the matter?
Matter is anything that has mass and occupies space.
Matter is a natural material that makes up the universe. The matter comprises tiny particles called atoms held together by forces called bonds. The matter is classified as pure if it has a constant and a non-variable composition of the type of atoms. Pure matter is either an element or a compound.
Element
An element is pure matter that is composed of only one type of atom
For example, carbon, shown in Fig. 1.1.1, is an element. Elements can not convert to simpler matter by physical or chemical methods. There are around a hundred different elements known at this time. For example, hydrogen, oxygen, carbon, nitrogen, sodium, chlorine, iron, cobalt, gold, and silver are a few elements.
Symbol of an element
Symbols represent elements, the first alphabet of their English or non-English name, written in capital letters. For example, C for carbon, O for oxygen, and H for hydrogen. Usually, another alphabet is also chosen from the element's name and written as a small letter, e.g., He for helium, Co for cobalt. Some element symbols are derived from non-English names, e.g., Fe for iron is from its Latin name Ferrum, and Au for gold is from its Latin name Aurum.
Two consecutive capital letters do not represent an element; they may combine two elements.
CO is not a symbol of an element; it is a pure matter, a combination of carbon and oxygen bonded together in a 1:1 atom ratio.
Molecule
A molecule is a group of two or more atoms held together by forces called chemical bonds. The molecule is the smallest particle of matter that can exist freely. A single atom of some elements can exist freely and is also considered a molecule. For example, He, O2, P4, and S8 are examples of elements having molecules composed of one, two, four, and eight atoms of the same element, respectively, as illustrated in Fig. 1.1.2. Although metal elements exist as a vast number of atoms bonded together by a special type of bond called metallic bonds, their symbol is that of a single atom, e.g., Fe for iron and Au for gold.
Compound
A compound is a pure matter composed of atoms of two or more different elements in a constant whole-number ratio held together by chemical bonds.
The symbol of a compound is a combination of its constituent elements with a subscript to the right of the element symbol representing the whole number ratio of the atoms of the element in the compound. For example, H2O symbolizes a compound called water composed of hydrogen (H) and oxygen (O) atoms in a 2:1 ratio. Similarly, NaCl symbolizes a table salt compound composed of sodium (Na) and chlorine (Cl) atoms in a 1:1 ratio. If the symbol of a compound also represents a molecule of the compound, it is called a molecular formula. For example, H2O is a molecule formula of water. On the other hand, table salt is another class of compound composed of a vast number of atoms of its constituent elements arranged in a specific arrangement in 3D space called a crystal lattice, as illustrated in Fig. 1.1.3. When the compound symbol does not represent a molecule, it only means the simple whole-number ratio of the constituent elements; it is called the formula unit. For example, NaCl is a chemical formula of a compound called table salt.
Law of definite proportion or law of constant composition
A chemical compound always contains the number of atoms of its component elements in a constant ratio, or the ratio of masses of the constituent elements is a constant in a given compound.
Pure matter, i.e., elements or compounds, has a fixed proportion of atoms of element/s independent of the source or method of their preparation. It is also called a chemical or a substance. The atoms in a compound are held together by attractive forces called chemical bonds. Constituent elements in a compound can be separated only by breaking chemical bonds and making new chemical bonds which is a chemical reaction.
Mixture
Two or more pure substances mixed such that ratio of the constituents is variable is called a mixture.
For example, table salt can be mixed with water, but the salt (NaCl) to water (H2O) can be varied to give less salty or more salty water in a mixture. A mixture in which the components are thoroughly mixed, and the composition is constant throughout a given sample, is a homogeneous mixture or a solution. For example, table salt dissolved in water is an example of a homogeneous mixture or a solution. Other homogeneous mixtures include sugar dissolved in water; the air, a mixture of nitrogen, oxygen, carbon dioxide, and other gases; and metal alloys like brass, a mix of copper and zinc metals, etc.
It is a heterogeneous mixture if the miitsnents are not thoroughly mixed and the composition varies within different regions of a given sample. For example, sulfur (S) powder mixed with iron (Fe) filling is a heterogeneous mixture shown in Fig. 1.1.4. Other examples of heterogeneous mixtures include smoke which is a mixture of air and carbon particles; smog which is a mixture of liquid water droplets suspended in air; and orange juice which is a mixture of sugar, water, fiber particles, etc. Fig. 1.1.5 illustrates the classification of the matter described above.
States of matter
Matter exists in one of the four physical states or phases, i.e., solid (s), liquid (l), gas (g), or plasma. Fig. 1.1.6 illustrates the four states of matter at the molecular level, and Fig. 1.1.7 shows examples of the four states of matter.
Solid-state
In the solid state, the particles, i.e., atoms or molecules, are very close to each other and held strongly by intermolecular forces. The particles can vibrate around their mean positions, but they cannot slide past each other. Expansion and contraction in a solid state are negligible. The solid has a fixed shape and a fixed volume.
Liquid state
In the liquid state, the particles are close enough to experience strong intermolecular interactions that usually do not let the particles cross the liquid boundary. Still, the particles can move around within the liquid. Consequently, the particles in a liquid can flow and acquire the shape of the container but have a fixed volume. Expansion and contraction are negligible in the liquid state.
Gas state
In the gas state, the particles are far apart. The intermolecular interactions are negligible in the gas phase due to the large distances between the particles. The gas molecules move in straight lines in random directions until they collide with other molecules or the walls of the container. The collisions are elastic; that is, the molecules bounce off like elastic balls, and the total kinetic energy of the system is conserved. If exposed to space, the particles keep moving into space. In other words, the particles in a gas can flow, acquire the shape of the container, and expand or contract to fill up the available space. The gases do not have a fixed shape and do not have a fixed volume.
Plasma state
In the plasma state, the particles are far apart like gases, and a portion of the negative charge of the particles, i.e., electrons, are separated from the positive charge potion, i.e., the nucleus. In other words, the atoms in the plasma state are ionized. The plasma state is not common on Earth but is the universe's most common state of matter. For example, the matter in the sun and stars is in the plasma state. Examples of the plasma state on Earth include the matter in the lightning bolts and electrical sparks.
What is energy
Energy is a quantitative property transferred to an object and recognizable as performing work or as heat or light. In simple words: energy is the ability to do work.
What we commonly encounter other than matter is energy. There are two basic types of energies, i.e., the kinetic energy of moving objects and potential energy stored by the position of an object in a force field, e.g., gravitation potential energy under the gravity force and chemical potential energy under the electrical fields in the bonds, as illustrated in Figure $8$.
Kinetic energy
The energy of moving objects is kinetic energy. The mathematical form of kinetic energy (KE) is $K E=\frac{1}{2} m v^{2}$, where m is the mass and v is the velocity of the moving object. Examples of kinetic energy include the energy of all moving things that we see around, like moving vehicles or a moving turbine that generates electricity. Thermal energy is also the kinetic energy of the atoms and molecules in matter.
Potential energy
Potential energy is due to the position of an object in a force field. Examples of force fields responsible for the potential energy include electric, magnetic, gravitational, and elastic forces. Examples of potential energies are electrical energy and gravitational energy. Light is potential energy due to moving electric and magnetic fields. Chemical energy is potential energy stored in chemical bonds in electrostatic potential energy.
Gravitation potential energy is due to position or height relative to the earth. The earth attracts other objects with force $F = mg$, where F is the force, m is the object's mass, and g is the acceleration due to gravity. When an object falls, the potential energy changes to several forms of energy, including kinetic energy, work against friction from air, sound, and work done in deformations when it hits the ground.
Chemical potential energy is due to chemical bonds that hold the atoms together in a molecule or compound through electrical forces between negative charge electrons and positive charge nuclei. Bond-forming always releases energy, and bond-breaking absorbs the same energy. Each bond has different bond energy. In chemical reactions, some bonds break, and some bonds form.
A combination of kinetic and potential energy is also possible, e.g., in mechanical waves. A sound wave is a mechanical wave that combines kinetic and potential energy –kinetic because particles move and potential due to the elasticity of the material in which the deformation (sound) is propagating.
Exothermic and endothermic reactions
The balance between the energy needed to break the bonds and the energy released from forming new bonds determines whether the chemical reaction releases or absorbs energy. Chemical reactions that release energy are called exothermic reactions. For example, the combustion of methane gas in kitchen burners releases energy. A reaction that absorbs energy is called an endothermic reaction. For example, photosynthesis converts carbon dioxide and water to glucose by absorbing sunlight, as illustrated in Fig. 1.1.9.
Law of conservation of energy
The energy conservation law states that energy can transform from one form to another, but the total energy of an isolated system remains the same.
Energy can transform from one form to another through work or heat. An oscillating pendulum is an example of kinetic and gravitational potential energy periodically transforming into each other through work. Radiation energy in sunlight transfers to chemical energy during photosynthesis, heat, and work during glucose metabolism, but the total energy remains the same, as illustrated in Figure $10$. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/01%3A_Matter_energy_and_their_measurements/1.01%3A_Matter_and_energy.txt |
Learning Objectives
• Define chemistry, physical properties, physical process, intensive and extensive properties, chemical properties, chemical process, and the law of multiple proportions.
• Be able to write and interpret chemical equations.
• Understand the scientific method and its steps, including observation, hypothesis, experiment, conclusion, law, theory, and how theory evolves knowledge.
Chemistry is the study of matter. More specifically, chemistry studies matter's composition, properties, and transformations.
The properties of matter are of two types, physical properties, and chemical properties, as illustrated in Fig. 1.3.1.
Physical property and physical process
Physical property
Physical property is the property that, when observed, does not change the elemental composition of the matter.
Examples include color, mass, volume, electrical conductivity, and heat conductivities.
Intensive properties
Physical properties that do not depend on the amount of matter are called intensive properties, e.g., color, density, and heat conductivity.
Extensive properties
Physical properties that depend on the amount of the substance, like, mass and volume, are called extensive properties.
Physical process
Any process that changes the matter somehow but does not change the elemental composition is called a physical process.
For example, melting solid to liquid or boiling liquid to a gas state are physical processes.
Mixtures can be separated using physical processes based on the differences in the physical properties of the constituents. Fig. 1.3.2 demonstrates that a magnetic material like iron can be separated from a nonmagnetic material like sulfur using a magnet.
The filtration process can separate a heterogeneous mixture of liquid and solid, like sand in water. Water passes through the filter paper leaving behind the sand particles on the filter.
The distillation process can separate homogeneous mixtures of solids in liquids or liquids in liquids based on the difference in the boiling points of the components. For example, salt dissolved in water separates by distilling off the water, leaving behind the solid salt. Distillation can also separate a mixture of two or more liquids if their boiling points differ; e.g., a distillation of crude oil separates the components based on their boiling points.
Chromatography is another technique often used to separate mixtures. For example, a mixture of inks is adsorbed on a porous paper and separated by ascending through the capillaries in the paper. The component of the ink mixture separate because some components have more ability to stay adsorbed in the solid phase and less ability to solubilize in the liquid phase than the other components. Fig. 1.3.2 Illustrates the physical separation processes described.
Chemical property and chemical process
Chemical property
Chemical properties relate to the change in the elemental composition of the matter.
For example, methane ($\ce{CH4}$) in natural gas is combustible -this is a chemical property. It means methane (\ce{CH4}\)) and oxygen (\ce{O2}\)) change their elemental composition to become carbon dioxide (\ce{CO2}\))and water (\ce{H2O}\)) and release heat after ignition.
Chemical process
A process that changes the elemental composition is called a chemical process or chemical reaction.
For example, photosynthesis is a chemical process that converts carbon dioxide and water to glucose using energy from sunlight.
Chemical equation
A chemical equation represents a chemical reaction in the form of symbols of elements and compounds involved.
Substances consumed in a chemical reaction are reactants; the substances formed are products. The reactants are written on the left side, separated by a plus sign, followed by an arrow, and products are on the right side of the arrow, as illustrated in Fig. 1.3.3. For example, the following chemical equation represents the combustion of methane.
\mathrm{CH}_{4}+2 \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\nonumber
Note that the chemical formula without any preceding number, e.g., O2 and CO2, represents one molecule or a unit amount of the chemical. The number preceding the formula is called the coefficient, and it represents the number of particles or the number of units involved. For example, the coefficient of 2 in 2H2O in the above chemical equation represents two molecules of water formed or two moles of water formed, where the mole is the unit amount. Note that the chemical composition has changed in the chemical reaction. Before, one substance was carbon and hydrogen atoms in a 1:4 ratio, and the other substance was oxygen atoms. After the reaction, one substance is carbon and oxygen atoms in a 1:2 ratio, and the other is hydrogen and oxygen in a 2:1 proportion.
The physical state of matter is sometimes shown in a chemical equation by the following symbols: (s) for solid, (l) for liquid, (g) for gas, and (aq) for a substance dissolved in water, as illustrated in Fig. 1.3.3.
Scientific method -how does science make progress?
Chemistry is one branch of science. Science knowledge is gathered systematically from generation to generation through the scientific method.
The scientific method starts with making observations, giving a tentative explanation, i.e., hypotheses, testing the hypothesis, i.e., experiment, and deducing a conclusion from the investigation. A truth found through repeated experiments becomes a law, and a comprehensive explanation of related findings gathered over time becomes a theory.
Figure $4$ illustrates these scientific method steps and is described below.
Observation
Observation is the active acquisition of information from a primary source. For example, you fill the air in a car tire and notice that the pressure reading on the gauge increases. This is an observation.
Hypothesis
A hypothesis is a tentative explanation of the observation or a law based on available scientific knowledge. For example, John visits a friend and starts sneezing. The friend says I have a cat, and you might be allergic to cats. This tentative explanation of John's sneezing is a hypothesis.
Experiment
Experiments test the hypothesis. For example, John visits another friend with a cat to determine whether he is allergic to cats. If he sneezes in this experiment, it supports the hypothesis. If he does not sneeze, the experiment disproves the hypothesis.
Conclusion
A hypothesis proven true becomes a conclusion. The hypothesis is rejected or revised if the experiment results do not support it. For example, scientists worldwide and in different periods attempted to convert other metals to gold and failed every time. It concluded that elements do not transform into the more simple matter by any physical or chemical reaction.
Law
It becomes law if an observation is universally true in repeated experiments. Examples of law are the following.
1. The pressure of any gas is directly proportional to the amount of gas if temperature and volume are kept constant, is Avogadro's law.
2. The proportion of atoms of different elements in a compound is always the same, a law of constant proportion.
3. Mass before any chemical reaction is the same as after the chemical reaction, i.e., mass is conserved in any chemical reaction or process, a law of conservation of mass.
Theory
A theory is a comprehensive explanation based on scientific principles to explain several laws and conclusions on a related topic. For example, the knowledge gathered over time on the properties of matter led Dalton to put forward Dalton's atomic theory.
Dalton's atomic theory
Postulates of Dalton's atomic theory are:
1. elements are composed of tiny indivisible particles called atoms;
2. atoms of any one element are identical to each other but different from atoms of any other element;
3. Atoms of different elements react with each other in a constant whole-number ratio to produce a compound;
4. atoms in a compound separate and recombine to give new material. Still, the atoms are neither created nor destroyed in the reaction.
These postulates explain the properties of the matter described in the previous sections. For example, elements can not convert to simpler substances by any physical or chemical process because they are composed of one type of atom, and atoms are indivisible according to the first postulate. Compounds can convert to elements by the chemical reaction because the atoms in the compounds can separate and recombine according to the fourth postulate.
What happens after a theory is accepted?
The theory goes through the test of time. If it keeps explaining the results of future experiments, it remains valid. It is either rejected or revised if it is disproved or cannot explain some observations of future investigations. For example, "atoms are not divisible" is no more considered valid. According to current knowledge, atoms can divide into subatomic particles like electrons, protons, and neutrons. However, the subatomic particles do not represent the element anymore.
"Atoms of the same element are the same" has been revised because isotopes are atoms of the same element that are different in some respects. The statement "atoms are neither created nor destroyed" is still valid for chemical reactions but does not hold in nuclear reactions where atoms of one element can convert to atoms of other elements. Similarly, matter can be converted into energy and vice versa in nuclear reactions, following Einstein's famous equation: E = mc2, where E is energy, m is mass, and c is the speed of light. It means the law of conservation of mass and the law of conservation of energy are not valid individually in a nuclear reaction. Still, the mass and energy together are conserved in nuclear reactions. These are examples of revisions made in theory over time.
The theory is a basic knowledge that allows the prediction of new laws and leads to new ideas on related concepts. For example, Dalton predicted the law of multiple proportions, also known as Dalton's law, i.e., if atoms of two elements can combine in one whole number ratio to give a specific compound, they may also mix in another whole number ratio to give another compound.
Law of multiple proportions
When two elements form more than one compound, the proportions of the atoms of elements in those chemical compounds can be expressed in small whole-number ratios, or the ratio of the masses of the second element in the two compounds that combines with a fixed mass of the first, is a small whole number ratio.
For example, hydrogen and oxygen atoms can mix in a 2:1 ratio to provide water (H2O). Still, they can also combine in a 2:2 ratio to give hydrogen peroxide (H2O2), a different compound. Similarly, carbon and hydrogen combine in a 1:4 ratio to make methane (CH4); they can combine in a 2:6 ratio to make another ethane compound (C2H6). | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/01%3A_Matter_energy_and_their_measurements/1.02%3A_What_is_chemistry.txt |
Measurements are an essential part of making observations needed to develop science. Several measurements are commonly done in everyday life, as illustrated in Fig. 1.3.1. The measured values have two components: a number and a unit.
Numbers
The numbers are composed of digits. The digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The digits are written in a row in the number, e.g., 123, which means one hundred and twenty-three. The numbers include a decimal point. If the decimal point is not marked, it is assumed to be present at the right of the number. For example, 123 is 123. by conversion, with the decimal point shown in red font.
Place value of a number
The digits have place values. The place values are relative to the decimal point, i.e., the 1st digit to the left of the decimal point is ones, the 2nd is tens, 3rd is hundreds, and so on. The digit 1st to the right of the decimal point is tenth, 2nd is a hundredth, 3rd is thousandth, and so, as shown in Fig. 1.3.2. For example, in 231.45, the digit 2 is hundreds, i.e., 200, 3 is tens, i.e., thirty, 1 is ones, that is one, 4 is tenth, i.e., four divided by ten, and 5 is hundredth, that is five divided by hundred.
Sign of a number
Numbers have signs, either +ve or -ve, to the left of a number, e.g., -23.4 and +430. By conversion, no sign means +ve. The signs are relative to zero; the -ve sign means the number is less than zero, and the +ve sign means the number is more than zero.
The number and the sign in calculations
The rules for the sign in a calculated answer are the following.
1. When two positive numbers add, the answer has a +ve sign, e.g., 3+2 = 5.
2. When two negative numbers add, the answer has –ve sign, e.g., -4 + (-2) = -6.
3. When two numbers having opposite signs add, subtract the smaller number from the larger number, and the answer has the sign of the larger number. For example, -5 +3 = -2.
4. In subtraction, change the sign of the subtracted number and then follow the addition rules. For example, subtract 3 from 5: 5-(+3) = 5-3 = 2. Note that 3 is subtracted, and its sign changed before operating addition. Another example: subtract -6 from 2: 2-(-6) = 2+6 = 8
5. When two positive numbers multiply, the answer has a +ve sign, e.g., 2x3 = 6.
6. When two negative numbers multiply, the answer has a +ve sign, e.g., (-4) x (-3) = 12.
7. When the two numbers multiplied have opposite signs, the answer has a –ve sign, e.g., (-3) x 2 = -6 and 4 x (-4) = -16.
8. When a number is divided by another number, it follows multiplication rules for the sign. For example, $\frac{-4}{-2}=2, \quad \frac{4}{2}=2, \quad\frac{-4}{2}=-2, \text { and } \frac{9}{-3}=-3$.
Percentage calculations
The percentage (%) is the part out of a hundred, as illustrated in Fig. 1.3.3. The percentage is calculated as part divided by the total and then multiplied by a hundred, i.e.:
\text { percentage } \%=\frac{\text { Part amount }}{\text { Total amount }} \times 100\nonumber
Example $1$
Calculate the percentage of aspirin if there is 81 mg aspirin in a 325 mg tablet?
Solution
part = 81 mg aspirin, Total = 325 mg tablet
Formula:
\text { percentage }=\frac{\text { Part amount }}{\text { Total amount }} \times 100\nonumber
Plug in values in the formula and calculate:
\text { Percentage }=\frac{81 \mathrm{mg} \text { aspirin }}{325 \mathrm{mg} \text { tablet }} \times 100=25 \% \text { aspirin }\nonumber
Example $2$
A piece of 18K green color gold jewelry has 7.5 g gold, 2.0 g silver, and 0.5 g copper. Calculate the percentage of gold in the jewelry?
Solution
Part = 7.5 g gold, Total = 7.5 g gold + 2.0 g silver + 0.5 g copper = 10 g jewelry.
Plug in values in the formula and calculate:
\text { percentage }=\frac{\text { Part amount }}{\text { Total amount }} \times 100=\frac{7.5 \mathrm{ggold}}{10 \mathrm{~g}} \times 100=75 \% \mathrm{gold}\nonumber
Writing numbers in scientific notation
Sometimes the given number is too large or too small to be easily written, read, and grasped. Scientific notation is one approach to changing a too large or a too-small number into an easily readable and writable number. The following steps convert a given number to scientific notation.
1. Move the decimal point to the right or the left side, one digit at a time, till the largest non-zero digit becomes one's place. For example, move the decimal in 12,700,000 seven times to the right to obtain 1.27, and move the decimal in 0.000,006 six-time to the left to get 6. The numbers 1.27 and 6 obtained are the coefficients of the scientific notation.
2. The coefficient is multiplied by 10x, where x is a power of ten. The power of ten equals the number of times the decimal moved. The sign of the power is +ve if the decimal moved to the left and -ve if the decimal moved to the right. For example, 12,700,000 in scientific notation is 1.27 x 107, and 0.000,006 is 6 x 10-6.
Units
Physical properties like mass, length, and temperature are measured. The measured value is a combination of a number and a unit, as illustrated in Fig. 1.3.4. For example, a person's height is 1.83 meters, where 1.83 is a number and a meter is a unit.
Units are quantities defined by the standard that peoples agree to use as a reference.
For example, the meter is defined as the distance light travels in a vacuum in $\frac{1}{299,792,458}$ of a second.
Systems of units
There are different sets of units used in different systems of units. For example, 1.83 meters and 6.00 feet show the same length value but using a unit of ‘meter’ from the international system of units (SI) and ‘foot’ from the English system of units. The international system of units (SI) is universally used in scientific work. There are seven base units in SI, as listed in Table 1.
Table 1: Base units in the International System of Units (SI).
Measurement Unit Abbreviation
Time second s
Length meter m
Mass kilogram kg
Temperature kelvin K
Amount of substance mole mol
Electric current ampere A
Note
The following section is based on the 2019 redefinition of the SI base units: https://en.Wikipedia.org/wiki/2019_redefinition _of_the_SI_base_units#Kilogram, accessed on May 2nd, 2020
Time
Time is the progress of existence and events that occur in succession from the past through the present to the future.
In old times, the time measuring device was the hour sandglass shown in Fig. 1.3.5. The basic unit of time is second (s), a standard unit of time in all the measurements systems. Other units of time are minute (min) which is equal to 60 s, and hour which is equal to 60 min or 3600 s.
Definition: Second
The duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atoms at a temperature of 0 K.
Length
Length is a measure of distance, i.e., a numerical measurement of how far apart the objects or points are.
Fig. 1.3.6. illustrated the concept of length. The SI unit of length is a meter (m).
Meter
The meter is defined as the distance that light travels in a vacuum in $\frac{1}{299,792,458}$ of a second.
Mass
The mass of an object is a measure of its inertia.
Inertia is the resistance of any physical object to any change in its velocity. Mass determines the strength of the gravitational attraction of an object to another object -a property commonly used in modern balances for mass measurements, as shown in Fig. 1.3.7. SI unit of mass is the kilogram (kg).
Kilogram
Earlier definition: The mass of one cubic decimeter of water at the melting point of ice.
Current definition: Kilogram (kg) is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit kg⋅m2⋅s1.
Temperature
Temperature is a physical property of matter that expresses hotness or coldness, as illustrated in Fig. 1.3.8.
Temperature is a manifestation of the thermal energy of the matter, which is a source of the flow of energy in the form of heat from a hot object to a cold object when they are in contact with each other.
The SI unit of temperature is Kelvin (K).
Kelvin
Kelvin (K) is defined by taking the fixed numerical value of the Boltzmann constant k to be 1.380649×10−23 when expressed in the unit kg⋅m2⋅s2⋅K1.
A 0 K, also called absolute zero, is the temperature of a matter at which no energy can be removed as heat from the matter. The freezing point of water is 273.15 K, and the boiling point of water is 373.15 K.
Amount of substance
In chemistry, the amount of substance (n) measures the number of specified elementary entities. The elementary particles in chemistry are usually atoms in the case of elements and molecules or formula units in the case of compounds. SI unit of the amount of a substance is a mole (mol).
Mole (mol)
Mole is exactly 6.02214076×1023 elementary entities.
Fig. 1.3.9 illustrates one mole of aluminum, copper, and carbon. The molar mass is the mass in grams of one mole of that substance, i.e., the mass of 6.02214076×1023 atoms or molecules. Usually, the number of particles is shown with four significant figures, i.e., 6.022 x 1023 atoms or molecules in one mole of the substance.
Electric current
Electric current is the flow rate of electric charge past a point or a region.
It could be the flow of electrons in electric wires or the flow of cations and anions in opposite directions as in electrolytes, as illustrated in Fig. 1.3.10. The SI unit of current is ampere (A).
Definition: Ampere (A)
Ampere (A) is defined by taking charge of an electron (e) to be 1.602176634×10−19 coulomb (C), where C is equivalent to ampere-second (A.s).
Luminous intensity
Luminous intensity measures the wavelength-weighted power emitted by a light source in a particular direction per unit solid angle.
The solid angle is measured in steradian (sr), analogous to the radian. The radian is a planar angel that gives the length of the circumference of a circle, and the steradian is a 3D angle, like a cone, that gives an area on the surface of a sphere, as shown in Fig. 1.3.11.
SI unit of luminous intensity in a given direction is Candela (cd).
Candela (cd)
Candela (cd) is defined by taking the fixed numerical value of the luminous efficacy of monochromatic radiation of frequency 540×1012 Hz, Kcd, to be 683 cd⋅sr⋅W1, or cd⋅sr⋅kg1⋅m2⋅s3, where W is watt –a SI unit of power described by kg.m2.s-3.
Prefixes in SI
In several situations, the measured number with the base unit is either too large or too small. For example, a person's height is comfortable to represent in the meter as the height is usually in a 1 m to 2 m range. However, the diameter of the earth, i.e., 12,700,000 m, and the diameter of red blood cells, i.e., 0.000,006 m, are too large and too small, respectively. The unit needs to be revised so that the number with it is easy to read and write.
Prefixes are used in SI to increase or decrease the base unit by order of tens.
For example, kilo (k) means a thousand times, i.e., 1 km means 1000 m and 1 kg means 1000 g. Similarly, micro (µ) means one-millionth time, i.e., 1 µm is 10-6 m, and 1 µg is 10-6 g. Table 2 lists commonly used prefixes in SI.
Table 2: Commonly used prefixes in SI (note: m means “meter” in SI units, but as a prefix, it means “mili”)
Prefix Means Abbreviation
Gega 109 G
Mega 106 M
Kilo 103 k
Deci 10-1 d
Centi 10-2 c
Mili 10-3 m
Micro 10-6 µ
Nano 10-9 n
Pico 10-12 p
A new unit may be defined and used if there is no appropriate prefix available in SI for some specific type of measurement. For example, the diameter of atoms varies in the range of 1 x 10-10 m to 5 x 10-10 m, where the prefix pico (p, 10-12) is too small, and the prefix nano (n, 10-9) is large. A new unit called angstrom (Å) is defined as 1Å = 10-10 m for reporting atomic diameter and inter-atomic distances.
Derived units
The units in SI other than the seven base units are Derived units obtained by combining the base units.
For example, The SI unit of volume is meter-cube (m3), equal to the space occupied by a cube of 1m on each edge, as illustrated in Fig. 1.3.12.
Usually, the volume is reported in decimeter-cube (dm3), commonly known as liter (L). One liter is a volume occupied by a cube that is one dm on each edge. Another commonly used unit of volume is the centimeter-cube (cm3), which is also called cc or mL. One mL is the volume occupied by a cube that is one cm on each edge. The dm3 is a thousandth of m3, and cm3 is the thousandth of dm3, i.e., 1000 dm3 = 1 m3, and 1000 cm3 = 1 dm3, as illustrated in Fig. 1.3.11.
Relationship of SI units with metric and English system of units
SI was developed from the metric system. Some basic units are different, but both systems have much in common, using the same prefixes. The English system of units uses a different set of units except for the common unit of time. Table 3 compares the standard measuring units in the three systems of measurement.
Table 3: Common measurement units in three conventional systems of measurements
Quantity English unit Metric unit SI unit Relationships
Mass Pound (lb) Gram (g) Kilogram (kg)
1 kg = 2.205 lb
1 kg = 1000 g
Length Foot (ft) Meter (m) Meter (m) 1 m = 3.281 ft
Volume Quart (qt) Liter (L) Cubic meter (m3)
0.946 L = 1 qt
1 m3 = 1000 L
Energy calorie (cal) calorie (cal) Joule (J) 4.184 J = 1 cal
Temperature Degree Fahrenheit (°F) Degree Celsius (°C) Kelvin (K)
°F = (1.8 × °C) + 32
K = °C + 273.15 | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/01%3A_Matter_energy_and_their_measurements/1.03%3A_Measurements.txt |
Significant figures are related to errors associated with the measured numbers. It is important to understand the significant figures because when calculations are made using numbers with errors, the answer cannot have less error than the error in any original number. The answer needs correction for the significant figures.
Precision and accuracy
The measured numbers have two types of errors, systematic and random errors, that determine the accuracy and precision of the measured number, as illustrated in Fig. 1.4.1.
Systematic errors
Systematic errors are constant, i.e., they have the same value in every measurement. For example, meter rod is a little short or a little long than a meter, it will introduce a systematic error. Systematic errors usually happen due to inaccurate calibration of the measuring instrument. The systematic errors determine how much the measured value differs from the actual value.
Random errors
Random errors are the statistical variability of the measured number. Random errors vary from one observation to another. Random errors cancel out if many measurements are taken and averaged. Scientific measurements are usually taken at least in triplicate and averaged to minimize random errors. The random errors determine how close the repeat measured numbers are to each other.
Accuracy
Accuracy or trueness of the measurement is defined as how close the average value is to the actual value.
The closer the average is to the actual value, the more accurate or true it is, as illustrated in Fig. 1.4.1. The trueness depends on systematic errors, i.e., less systematic error, more accurate the average.
Precision
Precision is defined as how close the individual measurements are to each other.
The closer the individual values are to each other, the more precise the measurement is, irrespective of whether it is accurate or not, as illustrated in Fig. 1.4.1. Precision depends on random errors, i.e., more substantial random errors mean less precision.
Exact and inexact number
There are two types of numbers, count numbers that are exact and measured numbers that are inexact.
If the value is a counted number, it is an exact number.
That is, there is no error in it. For example, a purchase of one dozen oranges contains exactly 12 nos of oranges; it can not be 11.5 or 12.5.
Inexact numbers and error range
When a value is measured, it comes with an error of measurement.
A measured number with an error is called an inexact number.
For example, when the same one dozen oranges are purchased by mass, the balance may read it 1572.6 g, or 1573 g, or 1570 g, depending on whether the smallest digit that the balance displays is 0.1g, 1 g, or 10 g. Suppose the balance is accurate to 1 g and reports the mass 1573 g; the actual mass may be anywhere in the range of 1572.5g-to-1573.4g. The smallest measured digit, i.e., the number in one's place, in this case, is an estimated number associated with an error. By convention, the estimated digit has ±1 errors associated with it. For example, the above-mentioned measured numbers are reported in science as 1572.6 g ± 0.1 g, 1573 g ± 1 g, or 1570 g ± 10 g, respectively. The estimated digits are shown in bold fonts in the examples.
The smallest digit in the display of digital instruments is an estimated number. In measurement using instruments that do not have a digital display, the smallest digit marked on the instrument plus one digit less than the minimum marked digit is added to the reported value. The smallest reported digit is an estimated digit. For example, the length of the pencil in Fig 1.4.2 is reported as 17.7 cm using the ruler on the bottom, where 17 includes the smallest digit marked on the ruler, and the last digit, i.e., 0.7 is an estimated digit. By convention, the error range in this value is shown as 17.7 ± 0.1. The same length is 17.70 cm using the ruler on the top in Fig. 1.4.2l, where 17.7 includes the smallest digit marked on the ruler, and the last reported digit, i.e., 0, is an estimated digit. By convention, the error range in this value is shown as 17.70 ± 0.01. The estimated digits are marked in bold fonts.
Rules to determine the significant figures in measured numbers
Significant figures
All the digits reported in the measured value, including the estimated digit, are significant figures (SF).
For example, 1572.6 g, 1573 g, and 1570 g have significant figures of 5, 4, and 3, respectively.
Caution
Note that zero in the last reading 1570g is not significant; it is a placeholder zero that is needed to place the estimated digit 7 at tens place.
It is crucial to find significant figures in measured numbers because, when they are used in calculations, the answer cannot have less error than the maximum error in any measured number used in the calculation. The rules to determine the significant numbers in a measured number are the following.
1. All non-zero digits are significant, e.g., 1572 has 4 SFs. The zeros may or may not be significant. In the following examples, the zeros in bold fonts are nonsignificant.
2. Zeros between non-zero digits are significant, e.g., 1305.6 has 5 SFs.
3. Leading zeros are not significant, e.g., 0.0134 has 3 SFs.
4. Trailing zeros are not significant if there is no decimal point present, e.g., 1570 has 3 SFs. Trailing zeros are significant if the decimal point is present, e.g., 1570. has 4 SFs because the decimal point is present. Similarly, 0.0124 has 3 SFs, but 0.01240 has 4 SFs because the decimal point is present.
5. Confusion arises when more than one trailing zeros and the decimal point is absent. For example, 1500 g has 2 SFs by convention, but if the balance was accurate to 10 g, one of the zero was an estimated digit and was significant. Converting the number to a scientific notation resolves this issue. The coefficient part of the scientific notation shows all the significant figures in the measurement. For example, the number 1500 g, if shown in scientific notation as 1.5 x 103 has 2 SFs, but the same number shown as 1.50 x 103 has 3 SFs.
Rounding the calculated answer involving inexact numbers
When inexact numbers are used in calculations, the answer needs to be rounded to an appropriate number of significant figures, determined by the following rules.
Rules of rounding
1. A number is rounded by keeping the larger digits equal to significant figures and dropping or replacing the remaining smaller digits with placeholder zeros. The placeholder zeros are in bold fonts in the following examples. For example, 13543 becomes 13500 when rounded to three significant.
2. If the largest digit dropped is 4 or less than 4, it is simply dropped. For example, 23145 becomes 23100, when rounded to three significant figures.
3. If the largest digit dropped is 5 or more than 5, then the smallest digit retained is increased by one. For example, 13543 becomes 14000 when rounded into two significant figures.
Rules for determining the significant figures in a calculated answer
In the following rules, the track of significant figures that dictate the significant figures in the answer is kept by using bold fonts.
1. In addition and subtraction, the answer has the same number of decimal places as the number with the smallest number of decimal places in the original numbers. For example, 13.2 + 12.252 = 25.452 is rounded to 25.5 to keep one decimal place.
2. In multiplication and division, the answer has the same number of significant figures as the original number with the smallest number of significant figures. For example, 1.35 x 2.1 = 2.835 is rounded to 2.8.
Note
If mathematical operations are performed in a series of steps, keep track of the significant figures but do not round off intermediate answers. Carry as many digits as possible from the intermediate answers to the next calculation step. Round off the final answer following the above rules. For example, (13.2 + 12.252) x (1.35 x 2.1) = 25.452 x 2.835 = 72.15642 is round to 72 in agreement with 2.835 that should have been rounded to 2.8 in a one-step calculation. Rounding the intermediate answers will lead to incorrect final answer of 71 instead of more correct 72, i.e. (13.2 + 12.252) x (1.35 x 2.1) = 25.5 x 2.8 = 71
Caution
Exact numbers have an unlimited number of significant figures, which means they do not restrict the significant figures in the calculated answer.
Example $1$
If 12 oranges weigh 1572.6 g calculate the mass of 1 orange in grams?
Solution
1572.6 \mathrm{~g} / 12=131.05 \mathrm{~g}\nonumber
Explanation: The answer has 5 SFs because 12 is a counted number and exact. The only inexact number in the calculation that dictates the significant figures in the solution is 1572.6, which has 5 SFs.
Example $2$
One dozen oranges were were sold 11 times. Calculate the total oranges sold?
Solution
12 \times 11=132 \text { oranges }\nonumber
Explanation: The answer is not rounded because both the numbers in the calculation are exact, so the answer is also exact with unlimited significant figures. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/01%3A_Matter_energy_and_their_measurements/1.04%3A_Significant_Figures.txt |
Conversion of values in one unit to the same value in another unit, as illustrated in Figure $1$ is often needed in scientific calculations. The unit conversion includes the following.
• Conversion of the same type of measurement in the same system of measurement, e.g., conversion of a measured value of the length in meters to kilometers in SI;
• conversion of the same type of measurement in different systems of measurements, e.g., conversion of a measured value of the length in kilometers from SI to miles in the English system; and
• conversion of one type of measurement to another type of measurement, e.g., conversion of a measured value of the mass in g to volume in mL of a substance using the density of the substance.
Conversion factors
The conversion factors are derived from equality between the given unit and the desired unit. For example, 1 cm = 10-2 m is equality between centimeter and meter. The conversion factors are derived from the equality by the following steps.
Both sides of the equality are divided by one side to get one conversion factor. For example, $1 = \frac{10^{-2} ~m}{1 ~cm}$, which is a conversion factor for cm to m.
Then both sides are divided by the other side of the equality to get the second conversion factor. For example, $1 = \frac{1 ~cm}{10^{-2} ~m}$, which is a conversion factor for m to cm.
Since both the conversion factors are equal to one, multiplying a value with a conversion factor changes the number and the unit, such that the new number and the new unit together represent the same value. The unit of the given number should be opposite to the same unit in the conversion factor, i.e., numerator versus denominator or denominator versus numerator, to cancel them out and leave the desired unit in the answer. For example, 1.83m is converted to cm unit using the conversion factor $\frac{1 ~cm}{10^{-2} ~m}$ as:
1.83 \cancel{\mathrm{~m}} \times \frac{1 \mathrm{~cm}}{10^{-2} \cancel{\mathrm{~m}}}=183 \mathrm{~cm}\nonumber
Keep track of the units that cancel out and the unit left in the answer. If all the units cancel out, leaving only the desired unit means the chosen conversion factor is correct. An incorrect conversion factor leads to an unwanted unit in the answer, e.g., in the above calculation if incorrect conversion factor is chose, it will lead to:$1.83 ~m \times \frac{10^{-2} ~m}{1 \mathrm{~cm}}=0.0183 \frac{\mathrm{~m}^{2}}{\mathrm{~cm}}\nonumber$, where no unit is canceled and the answer has units that are not the desired. It means an incorrect conversion factor was employed.
Converting the same type of measurement in the same system
The prefixes in the SI, listed in section 1.3 Table 2, establish equality between the base and the prefixed units. For example, centi (c) means 10-2. Therefore $1 \mathrm{~cm}=10^{-2} \mathrm{~m}\nonumber$1cm = 10-2 m is an equality that gives two conversion factors:
$\frac{1~cm}{10^{-2}~m}\quad\text{ and } \quad\frac{10^{-2}~m}{1~cm}\nonumber$
Some of the common qualities in SI are listed in Table 1 and the English system in Table 2.
Note
The prefixes are exact numbers. The equalities within the same system of measurement are exact numbers. Therefore the equalities and the conversion factors derived from them are exact numbers. Significant figures in the answers involving exact and inexact numbers are dictated by inexact numbers only.
Example $1$
Convert 325 cm to m units?
Solution
$325 \cancel{~cm}\times\frac{10^{-2}~m}{1 \cancel{~cm}}=3.25 ~m\nonumber$
Note that in this example, the answer has the significant figures the same as in the given number because the conversion factor within the same system of measurement is numbers.
Table 1: Some of the common equalities in SI
Length Mass Volume
1 km = 1,000 m 1 kg = 1,000 g 1 L = 10 dL
1 m = 100 cm 1 g = 1,000 mg 1 L = 1,000 mL
1 m = 1,000 mm 1 mg = 1,000 μg 1 L = 1,000,000 μL
1 cm = 10 mm 10-1 1 dL = 100 mL
1 mL = 1 cm3 = 1 cc
Table 2: Some of the common equalities in the English System
Length Mass Volume
ft = 12 in. 1 lb = 16 oz 1 qt = 4 cups
1 yd = 3 ft 1 ton = 2,000 lb 1 qt = 2 pints
1 mi = 5,280 ft 1 qt = 32 fl oz
1 gal = 4 qt
Conversion of the same type of measurement in different systems
Table 3 lists some common qualities between SI and the English systems.
These equalities between different systems usually have one side in the equality, which is the number 1, as exact, while the other side is considered an inexact number.
For example, 1 kg = 2.205 pounds (lb) has an exact number (1 kg) on the left side but an inexact number (2.205 lb) with 4 SFs on the right. Remember that only the inexact numbers dictate the significant figures in the answer.
There are some exceptions to the above general rule. Some equalities between units in different systems are defined and considered exact. They are stated to be exact in the reference tables.
For example, 1 inch = 2.54 cm is defined, which means both sides are the exact number.
Table 3: some of the common SI-to-English system equalities
Length Mass Volume
2.54 cm = 1 in. defined and exact 1 kg = 2.20 lb 946 mL = 1 qt
1 m = 39.4 in. 454 g = 1 lb 1 L = 1.06 qt
1 km = 0.621 mi 28.4 g = 1 oz 29.6 mL = 1 fl oz
Conversion of one type of measurement to another
Sometimes, equality between two different units is known under specific conditions. For example, density (d) which is mass (m) per unit volume (V), is a relationship between mass and volume of a given substance, i.e., $d= \frac{m}{v}$. The density is a conversion factor used to convert the volume to the mass of a substance. Reciprocal density, i.e., $\frac{1}{d}=\frac{v}{m}$, is the second conversion factor that converts the mass to the volume of the substance.
Example $2$
The density of ethanol at 20 oC is 0.7893 g/mL; what is the mass of 10.0 mL of ethanol?
Solution
Multiply the given volume with the conversion factor that has volume in the denumerator, i.e., $frac{m}{v} to get the mass desired. 10.0 \cancel{\mathrm{~m L}} \times \frac{0.7893 \mathrm{~g}}{1 \cancel{\mathrm{~mL}}}=7.89 \mathrm{~g}\nonumber Note that 1mL is exact, and 10.0 and 0.7893 are inexact numbers with 3SF and 4 SF, respectively. The answer has 3 SFs. Example \(3$
The density of gold is 19.30 g/mL; what is the volume of 10.123 g of gold?
Solution
Multiply the given mass with the conversion factor that has a mass in the denumerator, i.e., $frac{v}{m}$ to get the mass desired.
10.123 \cancel{\mathrm{~g}} \times \frac{1 \mathrm{~mL}}{19.30 \cancel{\mathrm{~g}}}=0.5245 \mathrm{~mL}
Conversion factors derived from chemical equations
Chemical equations show relationships or equalities between reactants and products measured in moles. Mole is a SI unit for the amount of substance. Mole equals 6.02 x 1023 particles (atoms or molecules) of the substance. Like dozen means 12 of something, mole means 6.02 x 1023 atoms of element or molecules/formula units for compounds in a chemical equation.
For calculations in chemistry, the number of moles of a substance is considered equal to its coefficient in a balanced chemical equation.
For example:
$\ce{2H2 + O2 -> 2H2O}\nonumber$
gives the following equalities and their corresponding conversion factors:
1. equality: 1 mole O2 = 2 moles H2, conversion factors: $\ce{\frac{1 ~mol ~O2}{2 ~mol ~H2}}$ and $\ce{\frac{2 ~mol ~H2}{1 ~mol ~O2}}$
2. equality: 1 mole O2 = 2 mole H2O, conversion factors: $\ce{\frac{1 ~mol ~O2}{2 ~mol ~H2O}}$ and $\ce{\frac{2 ~mol ~H2O}{1 ~mol ~O2}}$ and
3. equality: 2 moles H2 = 2 moles H2O, conversion factors: $\ce{\frac{2 ~mol ~H2}{2 ~mol ~H2O}}$ and $\ce{\frac{2 ~mol ~H2O}{2 ~mol ~H2}}$.
The use of these conversion factors in the calculation is explained in the following example.
Example $4$
If 5 moles of oxygen (O2) is consumed, how many moles of water are produced by the chemical equation mentioned above?Solution
Given: 5 mol O2, Desired: ? mol H2O
Multiply the given quantity with the conversion factor that has the given unit in denumerator and the desired unit in the numerator:
5.0 \cancel{\text { mole } \mathrm{O}_{2}} \times \frac{2 \text { mole } \mathrm{H}_{2} \mathrm{O}}{1 \cancel{\text { mole } \mathrm{O}_{2}}}=10 \text { mole } \mathrm{H}_{2} \mathrm{O}\nonumber
Unit conversion involving more than one conversion factor
Often, there is no direct conversion factor between the given unit and the desired unit. In this situation, convert the given unit to another unit that can, later on, be linked with the desired unit, as explained in the following examples.
Example $5$
How many µg are in 10.0 mg?
Solution
Problems like this can be solved in two steps: i) by converting the given unit to the base unit and then, ii) converting the base unit to the desired unit:
10.0 \cancel{~m g} \times \frac{10^{-3} ~g}{1 \cancel{~m g}}=1.00 \times 10^{-2}~ g\nonumber
1.00 \times 10^{-2} \cancel{~g}\times \frac{1 ~µ g}{10^{-6} \cancel{~g}}=1.00 \times 10^{4} ~µ g\nonumber
Note that the first conversion factor converts mg to g, and then the second conversion factor converts g to the desired unit µg. The same calculation can be done using two conversion factors in a rwo:
10.0 \cancel{\mathrm{~mg}} \times \frac{10^{-3} \cancel{\mathrm{~g}}}{10 \cancel{\mathrm{mg}}} \times \frac{1 ~µ g}{10^{-6} \cancel{\mathrm{~g}}}=1.00 \times 10^{4} ~µ g\nonumber
Example $6$
What is 100. km/h speed in m/s, where h represents hours?
Solution
This problem asks to convert two units, i.e., km to m and h to s. First convert one unit and then follow on to convert the second unit as:
\frac{100 . \cancel{\mathrm{~km}}}{\cancel{\mathrm{h}}} \times \frac{10^{3} \mathrm{~m}}{1 \cancel{\mathrm{~km}}} \times \frac{1 \cancel{\mathrm{~h}}}{60 \cancel{\mathrm{~min}}} \times \frac{1 \cancel{\mathrm{~min}}}{60 \mathrm{~s}}=27.8 \mathrm{~m} / \mathrm{s}\nonumber
Note that the first conversion factor converts km to m, and then two conversion factors are needed to convert h to s via min.
Example $7$
A prescription says a dosage of 0.225 mg of Synthroid to be taken once a day. If tablets in stock contain 75 µg of Synthroid, how many tablets are needed per day?
Solution
The given mass is in mg, while the equality "1 tablet = 75 µg" takes mass in µg. First convert mg to base unit, i.e., g, then from g to needed unit, i.e., µg, and finally take appropriate conversion factor from the two given by the equality to convert µg to tablet, i.e., three conversion factors in a row:
0.225 \cancel{\mathrm{~mg}} \times \frac{10^{-3} \cancel{\mathrm{~g}}}{1 \cancel{\mathrm{mg}}} \times \frac{1 \cancel{\mathrm{~µg}}}{10^{-6} \cancel{\mathrm{~g}}} \times \frac{1 \text { tablet }}{75 \cancel{\mathrm{µg}}}=3.0 \text { tablet }\nonumber
Example $8$
A healthy person has 16% body fat by mass. Calculate the mass of fat in kg of a person who weighs 180. lb?
Solution
Given: mass of a person = 180. lb, Desired: mass of body fat in kg.
16% body fat by mass means: 16 lb body fat = 100 lb body mass, and the equality kg and lb is: 1 kg = 2.20 lb. Take one conversion factor from each equality such that the units cancel out leaving the desired unit in the answer:
180. \cancel{\text{lb body mass }} \times \frac{16 \cancel{\mathrm{~lb} \text { body fat }}}{100 \cancel{\mathrm{~lb} \text { body mass }}} \times \frac{1 \mathrm{~kg} \text { body fat }}{2.20 \cancel{\mathrm{~lb}\text { body fat }}}=16 \mathrm{~kg} \text { body fat }\nonumber
Note that there are three inexact numbers in the calculation, i.e., 180, 16, and 2.20, and the answer as two significant figures in agreement with the smallest significant figure among the inexact numbers. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/01%3A_Matter_energy_and_their_measurements/1.05%3A_Unit_conversions.txt |
Equalities where both sides have a single term, i.e., monomial, lead to the conversion factors. If one or both sides of the equality have more than one term, i.e., polynomial, it leads to a formula that does the same job, i.e., to convert units. Temperature conversion equations and ideal gas equations and the manipulation procedure of these equations are described below as examples of how to manipulate the questions needed for making calculations.
Temperature conversion equations
There are three temperature scales in common use: Celsius (oC), Kelvin (K), and Fahrenheit (oF), illustrated in Fig. 1.6.1.
Celsius (oC)
The Celsius scale has O oC at the freezing point of water and 100 oC at the boiling point of water. Celsius is the base unit of temperature in the metric system.
The Kelvin scale
Kelvin is the base unit of temperature in the SI. The freezing point of water is 273.15 K, and the boiling point of water is 373.15 K. For most practical purposes, the freezing point of water is reported as 273 K and the boiling point 373 K, i.e., accurate to three significant figures. Celsius scale units are the same size but shifted up by 273 compared to the Kelvin scale. So, the relationship between Kelvin and Celsius is:
T_{K}= T_{C}+273,\nonumber
where TK is the temperature in Kelvin, and TC is the temperature in degrees Celsius. This equation converts temperature in Kelvin to temperature in Celsius.
A 0 K, also called absolute zero, is the temperature of a matter at which no energy can be removed as heat from the matter. There is no negative temperature on the Kelvin scale.
Fahrenheit (oF)
Fahrenheit is the base unit of the English system, with 32 oF at the freezing point of water and 212 oF at the boiling point of water. Fahrenheit is $\frac{5}{9}$ times shorter and shifted up by 32 than Celsius. So the relationship between the two is:
T_{F}=\frac{9}{5} \times T_{C}+32,\nonumber
where TF is the temperature in Fahrenheit, and TC is the temperature in degrees Celsius. This equation converts temperature in Celsius to temperature in Fahrenheit.
Manipulating temperature conversion equations
The equation for converting Celsius to Fahrenheit is:
T_{F}=\frac{9}{5} \times T_{C}+32,\nonumber
Addition or subtraction of the same number on the two sides of an equation does not change the equality. Subtracting 32 from both sides of the above equation leads to:
T_{F} -32=\frac{9}{5} \times T_{C}\cancel{+32}\cancel{-32},\nonumber
T_{F} -32=\frac{9}{5} \times T_{C}.\nonumber
Multiplication or division by the same number on both sides of an equation does not change equality. Remember that multiplication or division should apply to every term on either side of the equality. Enclose the side with more than one term in small brackets and then do the multiplication of division operation so that it applies to each term in the bracket. Multiplying both sides of the above equation with $\frac{5}{9}$ leads to:
\frac{5}{9} \times\left(T_{F}-32\right)=\cancel{\frac{5}{9}} \times \cancel{\frac{9}{5}} \times T_{C}\nonumber
\frac{5}{9} \times\left(T_{F}-32\right)=T_{C}\nonumber
Swapping the sides of an equation does not change equality. Swapping the sides in the above equation to bring TC to the left:
T_{C}=\frac{5}{9} \times\left(T_{F}-32\right)\nonumber
This is the equation for Fahrenheit to Celsius conversion.
The procedure of rearranging an equation described above applies to all algebraic equations. For example, start with a relationship that converts Celsius to Kelvin:
T_{K}= T_{C}+273,\nonumber
subtract 273 from both sides:
T_{K}-273= T_{C}\cancel{+273}\cancel{-273},\nonumber
T_{K}-273= T_{C},\nonumber
and finally swap the left and right side to bring TC to the left:
T_{C}=T_{K}-273.\nonumber
This is the equation for Kelvin to Celsius conversion.
Ideal gas equation
The ideal gas equation relates more than two variables:
$PV=nRT,\nonumber$
where P is pressure, V is volume, n is the amount of gas in moles, T is the temperature (in K), and R is the proportionality constant called an ideal gas constant. Dividing both sides of the equation with V leads to:
$P\times\frac{V}{V}=\frac{nRT}{V},\nonumber$
$P=\frac{n R T}{V}.\nonumber$
It allows calculating the pressure of a gas sample if the amount in moles, temperature in kelvin, and volume of the gas sample are known, along with the value of the constant R in the consistent units. Similarly, rearranging the equation leads to formulas for calculating, V, T or n of a gas sample:
V=\frac{n R T}{P}, \quad T=\frac{P V}{n R}, \quad \text { and } \quad n=\frac{P V}{R T}\nonumber
Graphs
The graph is a visual presentation of a relationship between two variables. Fig. 1.6.2 shows a graph that presents a relationship between the volume and pressure of a given amount of gas at a constant temperature, known as Boyle’s law.
The typical components of a graph are the following.
1. A title that tells what the graph is about, e.g., “Boyle’s Law: ” in the graph of Fig. 1.6.2.
2. Axes: the x-axis is a horizontal line, and the y-axis is a vertical line. Axes usually have an evenly distributed scale starting from zero. The x-axis represents the independent variable, and the y-axis represents the dependent variable. For example, volume is independent, and pressure is the dependent variable in Fig. 1.6.2.
3. Axes labels that tell the variable's name and the units of measurement. For example, volume (in3) and pressure (in Hg), where in3 and in Hg are the units of the variables.
4. Symbols representing experimental points. For example, ∆ symbols in Fig 1.6.2, at the crossing of a vertical line starting from an experimental value independent variable on the x-axis and a horizontal line starting from the corresponding value of the dependent variable on the y-axis.
5. A curve connects the experimental points and shows the trend in the relationship. For example, the cure in Fig. 1.6.2 tells that the pressure decreases as the volume increases.
Interpretation of a graph
Interpretation is reading the trend or relationship between the variables plotted. For example, Fig. 1.6.2 shows that the pressure decreases as the volume of gas increases. The curve also allows reading the value of one variable from the value of the other. For example, if the volume is 30 in3, the pressure would be ~50 in Hg. To read: draw a vertical line from the given value on the x-axis and a horizontal line from the point where the vertical line crosses the curve. Then, read the value where the horizontal line meets the y-axis. The process is reversed when the given value is of the dependent variable on the y-axis, and the desired is the corresponding value of the independent variable on the x-axis. For example, if the pressure is 40 in Hg, the volume is about 35 in3. Fig. 1.6.3 is another example of a graph that shows a relationship between the molar mass and density of the gases at a constant temperature. The cure in this graph tells that the density of gases increases as the molar mass increases. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/01%3A_Matter_energy_and_their_measurements/1.06%3A_Equations_and_graphs.txt |
Density
Density is the mass-to-volume ratio of a substance.
Density is a physical characteristic of matter. Each substance has a characteristic density that can be used as one hint in identifying a substance.
Gases have very low density, usually expressed in g/L. For example, air density is around 1.224 g/L at sea level and 15 oC. The density of liquids and solids is usually expressed in g/mL. For example, the density of water at 4 oC is 1.00 g/mL.
Objects that are less dense than water float, and the denser objects than water sink in the water. For example, oil is less dense than water and floats on water. Metals are denser than water and sink in water. The density of some common substances is listed in Table 1
Table 1: Densities of some common substances
Substance Density (g/mL)
hydrogen 0.000089
carbon dioxide 0.0019
ethyl alcohol 0.7893
water 1.00
magnesium 1.74
table salt 2.16
aluminum 2.70
iron 7.86
copper 8.92
silver 10.50
lead 11.34
mercury 13.59
gold 19.30
Density measurement
Density (d) is calculated from the mass (m) and volume (V) of a substance by the formula:
$d = \frac{m}{V}\nonumber$
Mass is usually measured using an analytical balance. The volume of liquids can be measured using a graduated cylinder, pipet, or density bottle. The volume of regular solids can be calculated from the geometric parameters. For example, the volume of a rectangle is equal to length x width x height. The volume of a cube is equal to the edge length cubed.
The volume of an irregular shaped sold is usually measured because the substances that are denser than water sink and displace an equal amount of water. Fig. 1.7.1 illustrates the density measurement of an irregular-shaped solid object that sinks in water, as explained in the following example.
Example $1$
Wwhat the density of the object in Fig. 1.7.1?
Soution
the mass (m) of the object on balance is 1000.00 g. The volume of the object is equal to the volume of the water displaced by the object, which is 225.0 mL – 100.0 mL = 125.0 mL.
d=\frac{m}{V}=\frac{1000.00 \mathrm{~g}}{125.0 \mathrm{~mL}}=8.000 \frac{\mathrm{g}}{\mathrm{ml}}\nonumber
Caution
Most of the time, Calculators give more significant numbers and sometimes less than needed; both need correction. In example 1.7.1, the calculator displays 8, i.e., one significant figure, but three zeros are added to make four significant figures.
Bone density and osteoporosis
Osteoporosis is a bone disease associated with decreased bone density, particularly in older adults. Bones always lose and gain calcium, magnesium, and phosphate. In childhood, the bones build faster than decay, but in old age, the process reverses, and the bones start to thin, loos strength, and become more prone to fracture, as illustrated in Fig. 1.7.2. Hormonal changes, diseases, and some medications contribute to bone thinning. Severe loss of bone density is called osteoporosis.
Specific gravity
Specific gravity is the ratio of the object's density to the density of water, i.e.:
\text { Specific gravity }=\frac{\text { Density of an object }}{\text { Density of water }}\nonumber
Specific gravity is the ratio of the object's density to the density of water, i.e.:
\text { Specific gravity }=\frac{\text { Density of an object }}{\text { Density of water }}\nonumber
The units cancel out in the ratio. Therefore, the specific gravity is a unitless number. The density of water is 1.0 g/mL at room temperature, so the specific gravity is equal to the density of the object expressed without a unit.
When substances dissolve in water, the density of the solution is usually different from pure water. For example, the density of whole blood for humans is ~1.060 g/mL. The density of urine varies in the range of 1.0050 g/mL to 1.030 g/mL. Both the blood and urine have dissolved substances in water that increase the density from that of pure water. Both high and low density or specific gravity than the normal range of urine indicates medical problems. An increase in the specific gravity of urine indicates that it is due to an increase in the solutes caused by dehydration, diarrhea, or infection. Similarly, a decrease in solute concentration decreases the specific gravity of urine, which indicates medical problems like renal failure.
Specific gravity measurement
The specific gravity is usually measured using an instrument called a hydrometer. The hydrometer partially submerges in the liquid sample, and the reading on the scale at the air-water junction point is recorded, as illustrated in Fig. 1.7.3. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/01%3A_Matter_energy_and_their_measurements/1.07%3A_Density_and_specific_gravity_measurements.txt |
Heat
Heat is energy in transfer to and from a thermodynamic system by a mechanism other than the work or the transfer of matter.
Energy exists in different forms, but energy transforms from one form to another through work or heat. In chemical reactions, a significant portion of energy transfer happens through heat. Green leaves in plants trap the energy from light and store it as chemical energy in the form of glucose synthesized from water and carbon dioxide, as illustrated in Fig. 1.8.1.
A reverse process happens when we eat food, i.e., the food substances are converted to carbon dioxide, water, and energy, as illustrated in Fig. 1.8.2. Some energy is used to maintain our body temperature at an average level. Another portion is used to drive chemical reactions that consume energy and synthesize substances we need.
Body Temperature
The human body continually loses energy as heat to the environment. The heat released by exothermic reactions in the food digestion process supplies the heat needed to maintain the body temperature at an average level. Human body temperature varies over a small range, as illustrated in Fig. 1.8.3. Hypothermia is <35.0 °C (95.0 °F), normal body temperate is 36.5–37.5 °C (97.7–99.5 °F), fever and hyperthermia is >37.5 or 38.3 °C (99.5 or 100.9 °F), Hyperpyrexia is >40.0 or 41.0 °C (104.0 or 105.8 °F).
Measurement of heat
Temperature is a manifestation of the thermal energy of an object, but the temperature and the energy are not the same. Temperature is a measure of hotness or coldness, and it is intensive, and energy is an extensive physical property of matter. For example, 1 g of water has some energy, but 2 g of water at the same temperature has twice the energy.
Specific heat
The calory (c or cal) is a non-SI unit of heat, work, and energy.
The calory (c or cal)
The energy needed to raise the temperature of 1 g of water by 1 oC is 1 calory
The heat needed to raise the temperature of 2 g of water by 1 oC is 2 calory, and to raise the temperature of 2 g of water by 2 oC is 4 calory. That is, heat energy (q) is directly proportional to both the mass (m) and change in temperature (∆T) of an object:
q \propto m \Delta T\nonumber
Introducing a constant of proportionality, i.e., specific heat (Cs) changes the proportionality to equality, that is known as the heat equation:
q=C_{s} m \Delta T\nonumber
, where Cs is Specific heat.
Specific heat of a substance
The heat energy needed to raise the temperature of 1g of a substance by 1 oC (or 1K) is specific heat if the substance.
The mathematical form of the definition of specific heat is:
C_{S}=\frac{q}{m \Delta T}\nonumber
The units of specific heat are $\frac{c a l}{g .^{\circ} \mathrm{C}}$, where cal is a non-SI unit of energy. The SI unit of energy is the joule (J). The following is an exact relationship between calory and joule:
1 \mathrm{~cal}=4.184 \mathrm{~J} \text { (exact) }\nonumber
So, the specific heat (Cs) in SI units is: $\frac{J}{g \cdot^{o} C}$
The energy unit in nutrition and food is food calory or Calory (C) that is written with capital C, and it is equal to 1000 cal, i.e.,
\mathrm{C}=1000 \mathrm{~cal}=1 \mathrm{~kcal}=4184 \mathrm{~J}\nonumber
, where the capital letter C is food calory.
The specific heat is a characteristic physical property of a material. It varies because the energies of vibrational motions of bonds and other motions that contribute to the thermal energy content differ from material to material. Table 1 lists specific heats of some familiar materials.
Water has higher specific heat than other materials, i.e., it absorbs or releases more heat for the same change in temperature. The environmental temperature variation remains moderate in coastal areas because the temperature change of water is less for the same amount of energy absorbed or released. Note that the specific heat of liquid water is different from solid water (ice) and gaseous water (steam).
Table 1: Specific heat of some common substances
Substance Specific heat ($\frac{J}{g \cdot^{o} C}$)
Al 0.902
C (graphite) 0.720
Fe 0.451
Cu 0.385
Au 0.128
NH3 (ammonia) 4.70
H2O (liquid) 4.184
H2O (ice) 2.06
H2O(steam) 2.00
C2H5OH (l) (ethanol) 2.46
(CH2OH)2 (l) (ethylene glycol, antifreeze) 2.42
CCl4 (carbon tetrachloride) 0.861
Wood 1.76
Concrete 0.88
Glass 0.84
Calculations using specific heat
The specific heat values are used in the heat equation ($q=C_{s} m \Delta T$) to calculate the temperature change for a given amount of heat or the heat energy needed for a given temperature change of a material. A coffee cup calorimeter illustrated in Fig. 1.8.4 is commonly used. It consists of a coffee cup with a lid. It contains water in which the hot or cold object is immersed, or a chemical reaction is performed for heat exchange with the water. A wire loop is used to stir the water. A thermometer records the change in the temperature of the water. The heat absorbed or released by the water is calculated from the data of mass, specific heat, and temperature change of the water.
Example $1$
Immersing a hot object in 50.0 grams of water in a coffee cup calorimeter increased the water temperature from 22.0 oC to 28.8 oC. How much heat is gained or lost by water?
Solution
Given: m = 50.0 g water, Cs = 4.184 (\frac{J}{g \cdot^{o} C}\), ∆T = Tf – Ti = 28.8 oC – 22.0 oC = 6.8 oC, Desired: heat gained or lost by water, i.e, qwater = ?
Calculations:
\mathrm{q}_{\text {water }}=\mathrm{C}_{\mathrm{s}} \mathrm{m} \Delta \mathrm{T}=4.184 \frac{\mathrm{J}}{\mathrm{g}{ }^{\circ} \mathrm{C}} \times 50.0 \mathrm{~g} \times 6.8^{\circ} \mathrm{C}=+1400 \mathrm{~J}\nonumber
The +ve sign tells water gained 1400 J heat.
Example $2$
Ammonium nitrate (NH4NO3) is used as a cold pack in hospitals to decrease the temperature of a targeted skin area. When 3.21 g of NH4NO3 is dissolved in 50.0 g of water in a coffee cup calorimeter at 25.0 oC, the temperature decreases to 20.4 oC. What is the amount of heat absorbed or released by water?
Solution
Given: m of solution = 50.0g H2O + 3.21 g NH3NO3 = 53.2 g, Cs = 4.184 (\frac{J}{g \cdot^{o} C}\), ∆T = Tf – Ti =20.4 oC - 25.0 oC = -4.6 oC, Desired: heat gained or lost by the solution, i.e, qsolution = ?
Calculations:
\mathrm{q}_{\text {solution }}=\mathrm{C}_{\mathrm{s}} \mathrm{m} \Delta \mathrm{T}=4.184 \frac{\mathrm{J}}{g^{\circ} \mathrm{C}} \times 53.2 \mathrm{~g} \times\left(-4.6^{\circ} \mathrm{C}\right)=-1.0 \times 10^{3} \mathrm{~J}\nonumber
The –ve sign tells water lost heat
Note
In the problems involving the heating of solution, the mass of solution includes the mass of water and the solute. It is assumed that the specific heat and density of the dilute solution are the same as that of pure water.
Example $3$
Calcium chloride (CaCl2) is used as a hot pack, i.e., to raise the temperature of a targeted portion of the skin. When CaCl2 dissolves in water, heat is released by the process and absorbed by the surrounding water, increasing the temperature of the water.
When 5.00 g of CaCl2 is dissolved in 50.0 g of water at 23oC in a coffee cup calorimeter, the temperature rises to 39.2oC. What is the amount of heat absorbed or released by water?
Solution
Given: m of solution = 50.0g H2O + 5.00 g CaCl2 = 55.0 g, Cs = 4.184 (\frac{J}{g \cdot^{o} C}\), ∆T = Tf – Ti =39.2 oC - 23.0 oC = 16.2 oC, Desired: heat gained or lost by the solution, i.e, qsolution = ?
Calculations:
\mathrm{q}_{\text {solution }}=\mathrm{C}_{\mathrm{s}} \mathrm{m} \Delta \mathrm{T}=4.184 \frac{\mathrm{J}}{g .^{\circ} \mathrm{C}} \times 55.0 \mathrm{~g} \times\left(16.2^{\circ} \mathrm{C}\right)=3.7 \times 10^{3} \mathrm{~J}\nonumber
The +ve sign tells water gained heat.
The energy released from foods
Foods and fossil fuels are carbon and hydrogen compounds and may also contain some other elements like oxygen, nitrogen, phosphorous, and sulfur. Fuels are used in combustion reactions to release heat for different purposes. The same combustion reactions happen in living systems in several steps to get energy from the foods. For example, the chemical equation for the combustion of glucose is:
$\ce{C6H12O6 (s) + 6O2 (g) -> 6CO2 (g) + 6H2O (l)}\quad\Delta \mathrm{H}^{\circ}=-2803 \mathrm{~kJ}\nonumber$
, where ∆Ho is the enthalpy of reaction, i.e., the heat evolved or absorbed at constant pressure conditions. This reaction releases 2803 kJ of energy as heat per mole of glucose. The –ve sign of ∆Ho tells that the reaction releases the energy, i.e., the reaction is exothermic. The same amount of energy is released when glucose is digested and converted to the same products in the living system under the same conditions.
Measurement of energy released from food
The heat released from the combustion of a substance is measured using a bomb calorimeter illustrated in Fig. 1.8.5. The substance is placed in a particular cell surrounded by an excess of oxygen. An electric spark ignites the substance, and the heat released causes a temperature increase in the water and the other materials in the meter. The heat capacity (Ccp), i.e., the product of the meter's average specific heat and mass and the water components, is calibrated before the experiment. The heat released from the given amount of the substance is calculated from the heat capacity and the measured temperate increase.
Fuel value
The heat of combustion per g of a food is called the fuel value of the food
Glucose is one of the carbohydrates. Carbohydrates are food components that are compounds of carbon, hydrogen, and oxygen in which the ratio of hydrogen-to-oxygen is 2-to-1. The average fuel value of carbohydrates is 17 kJ/g.
The second major class of food is fat. For example, the chemical equation of tristearin C57H110O6, which is a fat, is:
$\ce{2C57H110O6 (s) + 16O2 (g) -> 114CO2 (g) + 110H2O (l)}\quad\Delta \mathrm{H}^{\circ}=-71609 \mathrm{~kJ}\nonumber$
The –ve sign with energy value tells the energy is released, or the reaction is exothermic. The average fuel value of fats is 38 kJ/g.
The third primary type of food is protein. Protein is mainly used as a building material in living systems, but it is also used as a source of energy. The average fuel value of protein is 17 kJ/g.
The average fuel values of the three major food components are listed in Table 2. These values are used to calculate the fule value of food servings, as explained in the following examples.
Table 2: Average fuel value of food components
Food Fuel value kJ/g
Carbohydrates 17
Proteins 17
Fats 38
Example $1$
Calculate the energy released from one cup of orange juice that contains 26 g of carbohydrates, no fat, and 2 g of protein.
Solution
Carbohydrate: $26 \mathrm{~g} \times 17 \mathrm{~kJ} / \mathrm{g}=442 \mathrm{~kJ}$
Protein: $\quad\ 2 \mathrm{~g} \times 17 \mathrm{~kJ} / \mathrm{g}=34 \mathrm{~kJ}$
Total energy released $\quad\quad=476 \mathrm{~kJ}$
The energy value in kJ can be converted to food calories (C) by using the conversion factor based on the following relations.
\text { Food calory C }=1000 \mathrm{~cal}=1 \mathrm{~kcal}=4184 \mathrm{~J}\nonumber | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/01%3A_Matter_energy_and_their_measurements/1.08%3A_Heat_and_its_measurements.txt |
Among the four physical states of matter, solid has the lowest thermal energy. Intermolecular forces in solids are strong and do not let the molecules slide past each other. The molecules and the bonds in them can still have vibrational motions that account for the thermal energy contents of the material.
The temperature reflects the thermal energy content of the material—the addition of heat increase the vibrational motions, and temperature increases. Ultimately, the solid changes to a liquid and the liquid changes to a gas phase as more heat is added, as illustrated in Figure 1.9.1.
Melting and freezing
When the temperature reaches the melting point of the solid upon heating, the temperature does not increase further, but the sold changes gradually to the liquid phase. The heat added at the melting point is used to change the particles from a well-arranged form in the solid to an irregular arrangement in the liquid phase. This process is called the melting of solid.
Heat of fusion ($\Delta$Hfus)
The energy needed to melt a unit amount of the substance is the heat of fusion (∆Hfus).
The heat of fusion is usually expressed in the units of joules per gram ($\frac{J}{g}$) for the unit amount in grams or in joules per mole ($\frac{J}{mol}$) for the unit amount in moles.
If heat is removed from a substance at its melting point, the reverse of melting, i.e., freezing, happens, i.e., the liquid gradually changes from liquid to solid phase. The energy equal to the heat of fusion is released during the freezing process. Fig. 1.9.2 shows ice and water at 0 oC –an example of melting and freezing.
Vaporization and condensation
After melting, the heat addition causes an increase in the temperature of the liquid until the boiling point is reached. Some of the molecules in a liquid have high enough kinetic energy to cross the liquid-gas boundary and become gas phase. This process is called vaporization.
Heat of vaporization ($\Delta$Hvap
The energy needed to evaporate a unit amount of a liquid is called the heat of vaporization (∆Hvap).
The heat of vaporization is usually expressed in the units of joules per gram ($\frac{J}{g}$) for the unit amount in grams or joules per mole ($\frac{J}{mol}$) for the unit amount in moles.
The reverse of evaporation is called condensation, which releases heat equal to the heat of vaporization. Fig. 1.9.3 demonstrates the co-existence of liquid and gas-phase bromine at room temperate through the simultaneous evaporation and condensation processes.
When the temperature reaches the boiling point of the liquid, the temperature does not increase further, but the added heat is used to evaporate the liquid. Heating increases the temperature of the gas phase after all of the liquid has changed to the gas phase.
Sublimation and deposition
The solid can change directly to the gas phase without going through the liquid phase. This process is called sublimation.
The energy required in sublimation (∆Hsub) is the addition of the heat of fusion and the heat of vaporization, i.e.,:
$\Delta H_{\text {sub }}=\Delta H_{\text {fus}}+\Delta H_{\text {vap }}\nonumber$
The reverse of the sublimation is called deposition, i.e., the gas phase changes directly to the solid phase. Fig. 1.8.4 shows the sublimation of iodine crystals on a hot plate and deposition of iodine gas on an ice-cold watch glass.
The sublimation is responsible for drying clothes below 0 oC conditions in cold areas. Sublimation is also used in freeze-drying vegetables and other foods. Bacteria can not grow on dried foods because they need some moisture to grow. Fig. 1.9.5 shows the terminologies related to the phase changes described in the previous paragraphs.
Heating curve
A graphical presentation of the relationship of heat added versus the temperature change and phase changes of a matter is called a heating curve.
Fig. 1.9.6 shows the heating curve of water. The curve shows the heating of ice initially, followed by co-existing of solid and liquid at the freeing point, then hating of liquid water, then co-existing of liquid and gas phases at the boiling point, and finally the heating of steam –the gas phase of water. The reverse of the heating curve is called the cooling curve.
Heat calculations on heating or cooling curves
The heat required or released can be calculated by using the specific heat of the substance's solid, liquid, and gas phases. The heat of fusion is needed at the freezing point, and the heat of evaporation is needed at the substance's boiling point. The heat calculation is explained in the following example.
Example $1$
Calculate the energy required to heat 10.0 g of ice from -20.0 oC to steam (water vapor) at 110 oC?
Solution
1st step –heating the ice from -20.0 oC to the melting point of the ice, i.e., 0.00 oC:
m = 10.0 g, Cs of ice = 2.06 $\frac{J}{g \cdot{ }^{\circ} \mathrm{C}}$, ∆T = 0.00 oC – (-20.0 oC) = 20.0 oC
$\mathrm{q}_{1}=\mathrm{C}_{\mathrm{s}} \mathrm{m} \Delta \mathrm{T}=2.06 \frac{J}{g \cdot{ }^{\circ} \mathrm{C}}\times 10.0 \mathrm{~g} \times 20.0^{\circ} \mathrm{C}=412 \mathrm{~J}\nonumber$
2nd step – melting of ice, multiply the heat of fusion with the amout of substance:
m = 10.0 g, ∆Hfus = 334 $\frac{J}{g}$.
$\mathrm{q}_{2}=\Delta \mathrm{H}_{\text {fus }} \times \mathrm{m}=334\frac{J}{g}\ \times 10.0 \mathrm{~g}=3340 \mathrm{~J}\nonumber$
3rd step –hating of the water from 0.00 oC to the boiling point of water, i.e., 100 oC:
m = 10.0 g, Cs of liquid water = 4.184 $\frac{J}{g \cdot{ }^{\circ} \mathrm{C}}$, ∆T = 100 oC – 0.00 oC = 100 oC
$\mathrm{q}_{3}=\mathrm{C}_{\mathrm{s}} \mathrm{m} \Delta \mathrm{T}=4.184 \frac{J}{g \cdot{ }^{\circ} \mathrm{C}}\times 10.0 \mathrm{~g} \times 100^{\circ} \mathrm{C}=4180 \mathrm{~J}\nonumber$
4th step – boiling of liquid water, multiply the heat of vaporization with the amount of the substance:
m = 10.0 g, ∆Hvap = 2260 $\frac{J}{g}$
$\mathrm{q}_{4}=\Delta \mathrm{H}_{\text {vap }} \times \mathrm{m}=2260 \frac{J}{g}\ \times 10.0 \mathrm{~g}=22600 \mathrm{~J}\nonumber$
5th step –hating of the stem from 100 oC to 110 oC:
m = 10.0 g, Cs of steam = 2.00 $\frac{J}{g \cdot{ }^{\circ} \mathrm{C}}$, ∆T = 110 oC – 100 oC = 10.0 oC
$\mathrm{q}_{5}=\mathrm{C}_{\mathrm{s}} \mathrm{m} \Delta \mathrm{T}=2.00\frac{J}{g \cdot{ }^{\circ} \mathrm{C}} \times 10.0 \mathrm{~g} \times 10.0^{\circ} \mathrm{C}=200 \mathrm{~J}\nonumber$
Total heat needed = $q_{1}+q_{2}+q_{3}+q_{4}+q_{5}=412 \mathrm{~J}+3340 \mathrm{~J}+4180 \mathrm{~J}+22600 \mathrm{~J}+200 \mathrm{~J}=30700 \mathrm{~J}$
$30700 \mathrm{~J} \times \frac{1 \mathrm{~kJ}}{1000 \mathrm{~J}}=30.7 \mathrm{~kJ}\nonumber$
Note
The most significant portion of the heat is consumed in boiling the water to steam, i.e., 22.6 kJ out of 30.7 kJ total. The same amount of heat is released when the steam condenses. That is why the steam burn is much more severe than the burn by hot water. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/01%3A_Matter_energy_and_their_measurements/1.09%3A_Heat_and_changes_in_physical_states_of_matter.txt |
• 2.1: Dalton’s atomic theory
Dalton's atomic theory describing an atom of an element is presented.
• 2.2: Subatomic particles and a modern view of an atom
Subatomic particles, i.e., electrons, protons, and neutrons, along with the modern view of atom, i.e., who the subatomic particles are arranged in an atom are described.
• 2.3: Atoms of elements
Symbols of elements and the terms atomic number, mass number, atomic mass, and isotopes are introduced.
• 2.4: The periodic table
The periodic table, i.e., the arrangement of elements in columns and rows based on their atomic number is described. The classification of elements into metals, metalloids, and nonmetals, the classification based on their physical states are also described.
• 2.5: Electrons in atoms
Quantization of electron energies, the arrangement of electrons in an atom, and the rules for determining the electron configuration of 1st twenty elements are described.
• 2.6: The periodic trends in properties of the elements
Periodic trends in atomic size, ionization energy, and electronegativity are described. The trend of valence electrons that determine the chemical characteristics of an element and their representation in Lewis dot symbols is also presented.
02: Elements
Atom
Atom is the smallest particle of an element that retains the element's characteristics.
For example, gold is an element. Magnifying a section of the gold surface large enough would look like a packing of atoms, as illustrated in Fig. 2.1.1.
Dalton's atomic theory was the first significant attempt to explain the basic knowledge of atoms gained over time. Postulates of Dalton's atomic theory are the following.
Postulates of Dalton's atomic theory
1. Elements are composed of tiny indivisible particles called atoms.
2. Atoms of an element are identical but different from atoms of any other element.
3. Atoms of different elements react with each other in a fixed whole numbers proportion to produce a compound.
4. Atoms in a compound can separate and recombine to give new substances. Still, the atoms are neither created nor destroyed in the reaction.
Dalton's atomic theory is the basis of the current atomic theory, though the atoms are no more considered ‘indivisible.’ According to the current knowledge, subatomic particles like electrons, protons, and neutrons compose the atoms. However, the subatomic particles do not represent the element.
2.02: Subatomic particles and a modern view of an atom
Atoms are composed of fundamental subatomic particles: electrons, protons, and neutrons.
Electron was the first subatomic particle discovered. The discovery of electrons is related to the study of cathode rays and the basic knowledge of charges, i.e., there are two types of charges +ve and –ve; like charges repel each other; opposite charges attract each other as illustrated in Fig. 2.2.1; electric and magnetic field deflects the moving charges.
The discovery of the electron
Cathode rays
Cathode rays are a type of radiation emitted from a cathode (negatively charged electrode) when a high electric field is applied across a pair of electrodes under reduced pressure conditions, as illustrated in Fig. 2.2.2. The cathode rays travel in the inter-electrode space. If there is a hole in the anode, the cathode rays can pass through the hole and keep moving in a straight path. The cathode rays emit light when they strike a fluorescent screen.
J.J Thomson's experiments -the discovery of electrons
J.J. Thomson studied the cathode rays and found that an electric field deflects the cathode rays towards the positive electrode. This observation indicates that the cathode rays were negative charges. Changing the cathode material did not change the properties of the cathode rays. J.J. Thomson concluded from these observations that the cathode rays were streams of particles, called electrons, that are present in the atoms of all elements.
Further discoveries revealed that the charge on the electrons is 1.602 x 10-19 C, and their mass is 9.10 x 10-28 g. The electrons are incredibly light, about two thousand times lighter than the lightest atom.
Plum Pudding Model of an atom
Based on the information from cathode ray experiments, J.J. Thomson concluded that there are –ve charge electrons and a +ve matter representing almost all the atom's mass. He proposed the plum-pudding model of atoms, i.e., the positive matter is like a diffused cloud or jelly that occupies the atom's space, and electrons are embedded in it like fruits in jelly in the case of the plum-pudding dessert dish as shown in Fig. 2.2.3.
The discovery of the nucleus of an atom
${\alpha}$-Rays
${\alpha}$-Rays, pronounced as alpha-rays, are high-energy radiations emitted from some radioactive sources. The ${\alpha}$-rays are composed of ${\alpha}$-particles that are helium atoms without any electrons.
Rutherford's gold foil experiment -the discovery of the nucleus
Rutherford tested the plum pudding model of the atom by bombarding ${\alpha}$-rays on a thin gold foil. He expected that ${\alpha}$-particles to pass through the gold foil un-deflected like bullets fired through a Styrofoam sheet. He observed that although most of the ${\alpha}$-particles passed through the gold-foil un-deflected, one in ~20,000 deflected at larger angles, as illustrated in Fig. 2.2.4.
Since the plum-pudding model of an atom could not explain the deflection of the ${\alpha}$-particles, Rutherford concluded that there was a tiny but very dense region in the center of an atom, now called the nucleus, that deflected the ${\alpha}$-particles. Rutherford’s gold foil experiment led to the discovery of the nucleus.
The discovery of proton
Rutherford predicted that a positively charged fundamental particle of atoms should reside in the nucleus. Later on, Rutherford observed that shining ${\alpha}$ -rays on nitrogen gas produced positively charged particles called protons. The protons are about 20,000 times heavier than electrons but carry a +ve charge equal in magnitude to the –ve charge on an electron.
The discovery of the neutron
The mass of protons and electrons did not account for an atom's overall mass, which led to a search for another subatomic particle. James Chadwick discovered that bombarding ${\alpha}$ -rays on a beryllium target produced a highly penetrating radiation consisting of a beam of neutral particles, now called neutrons. The presence of neutrons in the nucleus accounts for the missing mass of atoms.
Other sub-atomic particles have been discovered, e.g., quarks that constitute protons and neutrons, but their knowledge is not critical for understanding basic chemistry.
Modern view of an atom
Protons and neutrons reside in the nucleus with a diameter of 10-15 m. Electrons occupy the region outside the nucleus with a diameter of 10-10 m, as illustrated in Fig. 2.2.5. If the nucleus is about the size of a marble, the atom would be about the size of a soccer field.
Properties of the subatomic particles
The basic SI units of mass, electric charge, and distance are too big for atomic-scale measurements. New units are defined for this purpose, i.e.:
Atomic mass unit (amu)
Atomic mass unit (amu) which is 1/12 of the mass of a single carbon atom that has 6 protons and 6 neutrons in it.
$1 \mathrm{~amu}=1.660539606660(50) \times 10^{-27} \mathrm{~kg}\nonumber$
Electron charge (e)
Charge on one electrons is electron charge (e). $1 ~e=1.602176634 \times 10^{-19} \mathrm{~C}\nonumber$
Angstrom(Å)
1 Å = 10-10 m
These units are usually used for masses, charges, and diameters of atoms. Table 1 lists the basic properties of the subatomic particles.
Table 1: Basic properties of subatomic particles
Particle Charge (e) Mass (amu)
Proton +1 1.0073
Neutron 0 1.0078
Electron -1 5.486 x 10-4
How the subatomic particles are held in the atom
The gravitational force is negligible in the case of the behavior of subatomic particles in the atoms. Although electrons repel other electrons, they stay in a tiny space due to attraction towards the nucleus. Similarly, protons repel each other, but the electrical force is small compared to the strong nuclear force that holds protons and neutrons together in the nucleus. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/02%3A_Elements/2.01%3A_Daltons_atomic_theory.txt |
Symbol of an element
Elements are represented by symbols, the first alphabet of their English or non-English name, written in capital letters. For example, C for carbon, N for nitrogen, and I for iodine. Usually, another alphabet is also chosen from the element's name and written as a small letter, e.g., Al for aluminum, Zn for zinc, and Ca for calcium. Some element symbols are derived from non-English names, e.g., Na for sodium from the Latin name natium, Cu for copper from Latin cupurum, and Ag for silver from Latin argentum.
Atoms of the same element have the same number of protons, and atoms of different elements have different numbers of protons. In other words, the number of protons in an atom defines the element. There are 118 elements known at this time; the number of protons in atoms varies from 1 for hydrogen to 118 for Oganesson (Og).
Atomic number
The number of protons in an atom is called the atomic number.
The atomic number defines the element. A subscript to the left of the symbol of an element represents the atomic number. For example, $\ce{ _1H}$ shows one proton in a hydrogen atom, and $\ce{_6C}$ shows 6 protons in a carbon atom.
Mass number
The number of protons plus the number of neutrons in an atom is the mass number.
A superscript to the left of the symbol of an element represents the mass number. For example, $\ce{^1_1H}$ is a hydrogen atom with atomic number 1, mass number 1, and no neutrons, while $\ce{^19_9F}$ is a fluorine atom with 9 protons and 10 neutrons.
Number of electrons
The number of electrons in an atom equals the number of protons minus the charge on the atom.
The number of electrons is equal to the number of protons in the case of a neutral atom, as there is no charge on a neutral atom.
Cation
A neutral atom can lose some electrons and become a positively charged particle, called a cation.
The charge is represented as a superscript on the right side of the element symbol, e.g. $\ce{^1_1H^+}$ is hydrogen without any electron, i.e., 1 proton, 0 neutrons, and 0 electrons. $\ce{^20_40Ca^2+}$ is calcium with two fewer electrons than protons, i.e., 20 protons, 20 neutrons, and 18 electrons.
Anion
An atom can gain electrons and become a negatively charged particle, called an anion.
For example, $\ce{^16_8O^2-}$ is oxygen with two more electrons than protons on it, i.e., 8 protons, 8 neutrons, and 10 electrons. A $\ce{^19_9F^-}$ is fluorine with one more electron than protons on it, i.e., 9 protons, 10 neutrons, and 10 electrons. Fig. 2.3.1 illustrates the gain or loss of electrons from neutral atoms.
Calculating the number of protons, neutrons, and electrons
In general, an atom of a hypothetical element X is represented as ${ }_{Z}^{A} X^{I ~charge}$ where Z is the atomic number, A is the mass number, and I is an integer equal to charge number and charge the sign of the charge number: + or -.The number of protons, neutrons, and electrons is calculated by using the following formulas:
Number of protons = A,
Number of neutrons = AZ, and
Number of electrons = Z – (charge I),
where a charge is a + or - sign of the charge number.
Example $1$
Calculate the number of protons, neutrons, and electrons in $\ce{^16_8O}$, $\ce{^16_8O^2-}$, and $\ce{^16_8O^+}$ ?
Solution
$\ce{^16_8O}$: number of protons = Z = 8, number of neutrons = AZ = 16-8 = 8,
and number of electrons = Z – (charge I) = 8 – 0 = 8.
$\ce{^16_8O^2-}$: number of protons = Z = 8, number of neutrons = AZ = 16-8 = 8,
and number of electrons = Z – (charge I) = 8 – (-2) = 8 + 2 = 10.
$\ce{^16_8O^1+}$: number of protons = Z = 8, number of neutrons = AZ = 16-8 = 8,
and number of electrons = Z – (charge I) = 8 – (+1) = 8 – 1 = 9.
Note
If charge number I is 1 in ${ }_{Z}^{A} X^{I ~charge}$, it is usually not written, but a number more than one is written. For example, $\ce{^16_8O^-}$ has change = -1, $\ce{^16_8O^2-}$ has charge charge -2, and $\ce{^16_8O^+}$ has charge = +1.
Isotopes
All atoms of the same element have the same number of protons but can have a different number of neutrons. For example, $\ce{^1_1H}$, $\ce{^2_1H}$, and $\ce{^3_1H}$ have neutrons equal to 0, 1, and 2, respectively.
Atoms of the same element that have a different number of neutrons are called isotopes.
For examples, $\ce{^1_1H}$, $\ce{^2_1H}$, and $\ce{^3_1H}$ are isotopes of hydrogen illustrated in Fig. 2.3.2. Another example is $\ce{^6_3Li}$, and $\ce{^7_3Li}$ are two isotopes of lithium. Natural samples of elements usually have almost constant ratios of isotopes. Table 1 lists some Isotopes of elements and their percent abundance in typical natural samples.
Table 1: Some of the important isotopes of elements with abundance in typical natural samples
Element Isotopes Abundance (%)
Hydrogen
$\ce{^1_1H}$
$\ce{^2_1H}$
$\ce{^3_1H}$
99.99
0.01
Negligible
Lithium
$\ce{^6_3Li}$
$\ce{^7_3Li}$
7.6
92.4
Carbon
$\ce{^12_6C}$
$\ce{^13_6C}$
$\ce{^14_6C}$
98.93
1.07
Negligible
Chlorine
$\ce{^35_17Cl}$
$\ce{^37_17Cl}$
75.78
24.22
Bromine
$\ce{^79_35Br}$
$\ce{^81_35Br}$
50.69
49.31
Uranium
$\ce{^235_92U}$
$\ce{^238_92U}$
0.72
99.28
Atomic mass
The atomic mass listed in the periodic table is the weighted average of the masses of the isotopes present in a natural sample of the element.
The following formula calculates the atomic mass:
$\text { Atomic mass }=\sum[(\text { mass of isotope }) \times(\text { fractional abuncance of the isotope })\nonumber$
, where $\sum$ means summation over all isotopes of the element, the fractional abundance of the isotope is the % abundance divided by 100. Fig. 2.3.3. illustrates how the atomic mass is listed in a periodic table.
Example $2$
Calculate the atomic mass of chlorine with two isotopes in nature samples, i.e., $\ce{^35_17Cl}$ of mass 34.969 amu and % abundance 75.78% and $\ce{^37_17Cl}$ of mass 36.996 amu and % abundance 24.22%.
Solution
Formula: $\text { Atomic mass }=\sum[(\text { mass of isotope }) \times(\text { fractional abuncance of the isotope })$
Plug in the given values in the formula and calculate:
$\left(34.969 \mathrm{~amu} \times \frac{75.78}{100}\right)+\left(36.996 \mathrm{~amu} \times \frac{24.22}{100}\right)=35.45 \mathrm{~amu}\nonumber$
Note
1. Atomic masses of isotopes are close to but not the same as their mass numbers. For example, $\ce{^35_17Cl}$ has mass number mass = 35, but the atomic mass of this isotope is 34.969 amu as shown in the above example.
2. The weighted average atomic mass is usually closer to the mass number of the most abundant isotope, e.g., 35.45 amu in the above example is close to the mass number 35 of $\ce{^35_17Cl}$ isotope, which is the most abundant isotope.
3. The periodic table reports the atomic mass as calculated in the above example, i.e., the weighted average of the masses of the isotopes present in the natural sample of the element. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/02%3A_Elements/2.03%3A_Atoms_of_elements.txt |
Early developments
The discoveries of elements happened over a long time. As the list of known elements grew, scientists tried to arrange them based on their properties. Mendeleev arranged elements in a table based on atomic masses. It resulted in elements having similar properties placed next to each other in most cases. There were gaps intentionally left in the table for the elements that were predicted based on the knowledge from the periodic table but not yet discovered.
Few exceptions were there observed where the properties of the elements did not agree with the group in which they were placed based on their atomic masses. Mosely developed a method to measure atomic numbers based on X-ray spectroscopy. The arrangement of the elements based on the atomic number instead of atomic masses removed the discrepancies in Mendeleev’s periodic table.
The modern periodic table of elements
• The modern periodic table of elements arranges the elements according to the increasing order of the atomic number starting from atomic number 1 for H and ending with atomic number 118 for Og, as shown in Fig. 2.4.1.
• The elements are arranged in horizontal rows called periods and vertical columns called groups.
Periods
The periodic table has seven horizontal rows called periods. The periods are numbered: 1 at the top to 7 at the bottom.
1. The 1st period has only two elements: hydrogen in group 1 and helium in group 18, with a gap from group 2 to group 17.
2. The 2nd and the 3rd periods have eight elements each, filling groups 1 and 2 followed by groups 13 to 18, leaving a gap from group 3 to group 12.
3. The periods 4th and 5th periods have eighteen elements that are filled successively from group 1 to group 18.
4. The periods 6th and 7th have 32 elements, each: the first two in groups 1 and 2, the next fourteen elements in separate rows below the table. These two rows of 14 elements each are called Actinides and Lanthanoids, respectively. Then the next sixteen elements fill the groups 3 to 18.
Groups
The periodic table has 18 vertical columns called groups or families. The groups are numbed starting from 1 on the leftmost and going through to 18 at the rightmost.
Alkali metals
The 1st group is called alkali metals. The alkali metals include lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr), shown in Fig. 2.4.2. The alkali metals are the most reactive among the metals in the periodic table. They react vigorously with water, as shown in Fig. 2.4.3.
Figure \(2\): Alkali metals, from left to right: lithium, sodium, potassium, rubidium, and cesium. Source: Tomihahndorf at German Wikipedia, Dnn87 Contact email: [email protected], and http://images-of-elements.com/potassium.php.
Hydrogen
Hydrogen is in group 1 but is not included in alkaline earth metals. Hydrogen is a nonmetal and has properties quite different from alkali metals or any other group of elements.
Alkaline earth metals
The 2nd group is called alkaline earth metals. It includes beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). They are reactive metals but less reactive than alkali metals. Alkaline earth metals impart characteristic color to a flame. Salts of alkali metals are used in firework formulation to give distinctive colors to the firework, as shown in Fig. 2.4.4.
Transition metals
Groups 3 to 12 are called transition metals. They include precious metals like gold, silver, platinum, and construction metals like iron. Some make catalysts and are found in enzymes and other bio-molecules, like hemoglobin and chlorophyll.
Group 13 to 16
Group 13 to group 16 does not have a unique name. They comprise nonmetals at the top and metals at the bottom of each group called post-transition metals. Important nonmetals include carbon, nitrogen, oxygen, phosphorous, and sulfur.
Halogens
Group 17 elements are called halogens. The halogens include fluorine (F), chlorine (Cl), bromine, iodine (I), and astatine (At). The halogens are highly reactive nonmetals. Chlorine is gas, bromine is liquid, and iodine is solid at room temperature, as shown in Fig. 2.4.5.
Noble gases
Group 18 is called noble gases. They include helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). They are the least reactive of all the elements. Noble gases are used to create an inert atmosphere for chemical reactions. Noble gases are also used in the lighting system because of their chemically inert nature , as illustrated in Fig. 2.4.6.
Metals, metalloids, and nonmetals
Metals
The elements towards the right-bottom corner of the periodic table are metals except for hydrogen, which is a nonmetal. Metals have common characteristics, including:
1. they are shiny,
2. solid at room temperature (except mercury which is liquid),
3. malleable (can be hammered into sheets) and ductile (can be drawn into wires),
4. good conductors of heat and electricity, and
5. tend to lose electrons and form ionic compounds when they react with nonmetals.
The metals of group 1 are called alkali metals; group 2 are called alkaline earth metals, group 3 to group 13 are called transition metals, the two-row below transition metals in the periodic table are called lanthanoids, and actinides or inner transition metals, and metals in the group 13 to group 16 are called post-transition metals. Elements other than transition or inner-transition metals, i.e., the elements of groups 1 and 2 and groups 13 to 18, are collectively called main group elements or representative elements.
Metalloids
The dividing line between metals and nonmetals is a staircase line starting from \(\ce{_5B}\) and ending at \(\ce{_85At}\). The elements on the staircase line are metalloids except for aluminum and polonium, which are considered metals. Metalloids have properties in-between metals and nonmetals; e.g., they have moderate heat and electrical conductivity.
Nonmetals
Elements towards the top-right corner of the periodic table and hydrogen are called nonmetals.
1. The nonmetals usually have properties opposite to metals, e.g., they are not typically shiny, brittle if solid, and poor conductors of heat and electricity.
2. Nonmetals tend to make ionic compounds by accepting electrons from metals and making molecular compounds by reacting with each other.
Two groups in nonmetals also have unique names, i.e., group 17 is called halogens, and group 18 is called Noble gases.
The most reactive and nonreactive elements
Generally, alkali metals are the most reactive, followed by alkaline earth metals, and halogens are the most reactive nonmetals. Noble gases are the least reactive nonmetals, also called inert gases.
Fig. 2.4.7 shows examples of a metal, a metalloid, and a nonmetal.
Elements essential for life
The elements that are the main constituents of humans and other living organisms are oxygen, carbon, hydrogen, nitrogen, and sulfur. Phosphorous is present in bone, teeth, and DNA. Calcium and magnesium are the main constituents of bones and teeth and perform some other body functions. Sodium and potassium cations are the main electrolytes in body fluids, and chloride anion balances the charge. Iron is present in hemoglobin that carries oxygen to the cells. These elements are essential to life, and they are macronutrients. Besides these, several other elements are needed in a small amount. They are essential for life and called micronutrients. Fig. 2.4.8 shows the macronutrients in pink and micronutrients in blue color in a periodic table.
Physical and chemical states of elements
Elements in gases state at ambient conditions
The term molecule is generally used for an electrically neutral group of two or more atoms held together by chemical bonds. In the kinetic molecular theory of gases, a molecule is the smallest particle of an element or compound with a stable and independent existence. Atoms of noble gases exist as independent species in the gas phase at room temperature, i.e., as monoatomic molecules like He, Ne, Ar, Kr, Xe, and Rn. Other elements that are gases at room temperature or diatomic molecules, i.e., H2, N2, O2, F2, and Cl2, are called dihydrogen, dinitrogen, dioxygen, and dichlorine, respectively. Note that usually prefix di is not used for the name of the element, i.e., these are usually called hydrogen, nitrogen, oxygen, fluorine, and chlorine, respectively. Monoatomic species of these elements, i.e., H, N, O, F, and Cl, exist, but they are very reactive species called free radicals, and they do not survive for a long time.
Elements in the liquid state at ambient conditions
Two elements exist as liquids at room temperature, i.e., mercury (Hg), a metal, and bromine (Br2), a no metal that exists as diatomic molecules. Four elements, francium, cesium, gallium, and rubidium, are solid metals at 25 oC, but become liquid when the temperature is slightly warmer.
Elements in the solid-state at ambient conditions
All elements not mentioned in the previous two sections are solid at room temperature. They may be diatomic or polyatomic molecules, e.g., I2, O3, P4, S8, diiodine, ozone, tetraphosphorous, and octasulfur, with two or three, four, and eight atoms, respectively, in a molecule.
Allotropes
Different forms of the same element in the same physical state are called allotropes.
The allotropes are different structural modifications of the element. For example, carbon exists in several allotropic forms; two of them are shown in Fig. 2.4.9. Another example is O2, and O3 are gaseous forms of oxygen.
Elements that exist as giant molecules
Some elements exist as giant molecules, i.e., a collection of many atoms bonded with each other through a 3D network of bonds. For example, carbon is a giant molecule in several allotropic forms, including diamond, graphite, carbon nanotubes, and fullerenes. Fig. 2.4.9 shows the bonding in diamond and graphite allotropes. The whole piece of diamond is one molecule with carbon atoms interconnected in a 3D network of bonds.
Metals also exist as a collection of many atoms bonded together by metallic bonds. The metal atoms exist as +ions arranged in a well-defined 3D arrangement called crystal lattice with some of the outermost electrons roaming around in the whole piece of the metal, forming a sea of electrons around the metal atoms, as illustrated in Fig. 2.4.10.
The elements in giant molecules and metals are represented by element symbols without any subscript, e.g., C is carbon, Fe is iron, Au is gold, etc. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/02%3A_Elements/2.04%3A_The_periodic_table.txt |
The electronic structure of an atom determines the properties of the element. Knowledge of electromagnetic radiation is described first as it plays an essential role in understanding the electronic structure of atoms.
Electromagnetic radiations
Electromagnetic radiations are waves that are oscillating electric and magnetic fields. The wave propagates in one direction, e.g., along the x-axis. The electric field oscillates perpendicular to it, e.g., along the y-axis. The magnetic field oscillates perpendicular to both, e.g., along the z-axis, as illustrated in Fig. 2.5.1. The distance between two consecutive maxima or between any two consecutive same phase points along the wave is called wavelength ($\lambda$, pronounced ‘lambda’). The number of waves that pass a reference point in one second is called frequency ($\nu$, pronounced ‘nu’). The speed of electromagnetic radiation is called the speed of light (c). The speed of light is the product wavelength and frequency, i.e.,
$c = \lambda\nu\nonumber$
The speed of light (c) is 3.00 x 108 m/s in a vacuum. The energy (E) of electromagnetic radiation is directly proportional to frequency, i.e.,
$E = h\nu\nonumber$
, where h is a constant, called plank’s constant. Replacing $\nu$ with $\frac{c}{\lambda}$ shows that the energy is inversely proportional to the wavelength.
$E = \frac{h c}{\lambda}\nonumber$
Fig. 2.5.2 illustrates the range of electromagnetic radiations that differ from each other concerning wavelength, frequency, or energy. The separation of the radiations based on their wavelength gives a spectrum. Visible light is a small portion of the spectrum of electromagnetic radiations, as illustrated in Fig. 2.5.2.
Continuous and line spectrum
The spectrum of sunlight contains radiations of all wavelengths within the visible range. The spectrum that contains all wavelengths in the range is called a continuous spectrum. Spectrum from sodium lamps or emissions from other atoms contains only some discrete wavelengths. The spectrum that contains discrete wavelengths is called the line spectrum. For example, as illustrated in Fig. 2.5.3, the emission from the hydrogen atom is a line spectrum.
Energy levels of electrons in an atom
The question is why the atomic emission has a discrete wavelength or discrete energy. The answer to this question came from the discoveries that concluded that the electron in an atom does not have continuous energy values; they have discrete energy values called shells and subshells.
The shell
Quantum numbers determine the allowed energy values of an electron in an atom.
• Principal quantum number (n) can have any integer value starting from 1, i.e., 1, 2, 3, 4, and so on.
• The smaller the n, the lower is the energy state, and the closer the electron is to the nucleus, the more tightly held the electron is by the nucleus.
• The value of n defines the shell, i.e., 1st shell has n = 1, 2nd shell has n = 2, 3rd shell has n =3, and so on.
Bohr introduced this concept of quantization of electronic energy levels. Fig. 2.5.4 illustrates Bohr’s atomic model.
• When an electron jumps from a lower shell to a higher shell, it absorbs electromagnetic radiation of energy equal to the energy gap between the initial and the final shell.
• When an electron jumps from a higher shell to a lower shell, it emits radiation equal to the energy gap between the initial and the final shell.
Fig. 2.5.3 illustrates the emission of radiation from atoms –it is a line spectrum because only discrete energy levels, called shells, are allowed to electrons in an atom.
The subshell
A second quantum number, called Azimuthal quantum number (l) defines subshells within a shell.
The subshells are usually designated as s, p, d, f, ...
Each shell has subshells equal to the shell number. For example, 1st shall have only one subshell, i.e., s. It is designated 1s, where the number is the principal quantum number, and the letter is the subshell. The 2nd shell has two subshells 2s and 2p; the 3rd shell has three subshells 3s, 3p, and 3d; and the 4th shell has four subshells 4s, 4p, 4d, and 4f.
The energy order of subshells is 1s<2s<2p<3s<3p<4s and so on.
Fig. 2.5.6 helps in remembering the order. This figure is drawn by placing the orbitals in columns and shell numbers in rows in increasing order of n from top to bottom, starting from 1s orbitals in the first column and first row, p orbitals in the second, d in the third, and f in the fourth. The filling of electrons follows arrows going from corner to corner, starting from the top left corner of the topmost cell, as described in more detail in a later section.
Orbital
The orbital is the region in space around the nucleus of an atom where electrons are most likely found.
Each subshell has a certain number of orbitals in them. The orbitals have a specific shape and orientation. The s subshell has only one orbital spherically symmetrical, like a ball with a nucleus at the center. The 1s orbital is smaller than 2s, and 2s is smaller than 3s, but they all have a spherically symmetrical shape, as illustrated in Fig. 2.5.5. The p subshell has three orbitals. Each p orbital is a dumbbell shape with two lobes, i.e., px oriented along the x-axis, py along the y-axis, and pz along the z-axis, as illustrated in Fig. 2.5.5.
Degenerate orbitals
A set of orbitals having the same energy is called degenerate orbitals.
All the three p orbitals in the same shell have the same energy, i,e, 2px, 2py, and 2pz is a set of degenerate orbitals. The d subshell has five degenerate orbitals, and f subshell has seven degenerate orbitals, as shown in Fig. 2.5.7. Their shapes and orientations are more complex and not shown here.
The electron configuration of atoms
The electron configuration is the distribution of electrons in the atom's orbitals.
Rules for distributing electrons in atom's orbitals
1. Each orbital can take a maximum of two electrons.
2. The orbitals are filled in the order of increasing energy: the lowest energy orbital is filled with electrons before the next higher energy orbital starts filling up.
The energy order of orbitals is 1s<2s<2p<3s<3p<4s and so on, as shown in Fig. 2.5.6 and Fig. 2.5.7.
The electron configuration of 1st-row elements
The hydrogen atom has one electron, and it occupies 1s orbital, i.e., 1s1 represents the electron configuration of the hydrogen atom. The superscript in 1s1 shows the number of electrons in the sub-shell. A helium atom has two electrons, and both occupy 1s orbital, giving helium the electron configuration of 1s2. Hydrogen and helium are in the 1st-row of the periodic table and have the 1st-shell containing electrons.
The electron configuration of 2nd-row elements
The 2nd row starts from lithium with 3 electrons: the first two occupy 1s, and the third occupy 2s giving lithium the electron configuration 1s2 2s1. The next element is beryllium with 4 electrons with the configuration 1s2 2s2. The next element is a boron with 5 electrons and the electron configuration 1s2 2s2 2p1. Carbon has 6 electrons with the electron configuration 1s2 2s2 2p2. Remember that the s subshell has one orbital and can take a maximum of two electrons, but the p subshell has three orbitals and can take a maximum of six electrons, i.e., two per orbital. Nitrogen, oxygen, fluorine and neon have 7, 8, 9, and 10 electrons and have the electron configuration 1s22s22p3, 1s22s22p4, 1s22s22p5, and 1s22s22p6, respectively. Atomic number 3 to 10, i.e., lithium to neon, completes the 2nd-row.
Valence electrons and core electrons
The outermost shell is called the valence shell, and electrons in the valence shell are called valence electrons. 1st shell is the valence shell for 1st-row elements hydrogen and helium. 2nd-row elements have an inner shell with configuration 1s2, and a valence shell containing 2s and 2p being filled. The inner shell is also called the core-shell, and the electrons in the core-shell are called the core electrons.
The electron configuration of 3rd-row and 4th-row elements
The 3rd-row starts with sodium atomic number 11 and ends with argon atomic number 18. The electron configuration of the 3rd-row elements repeats the pattern of the 2nd-row, i.e., a set of full core-shells 1s22s22p6, and a valence shell having 3s or 3p being filled. The first two elements of the 4th-row are potassium and calcium with atomic numbers 19 and 20 having configurations 1s22s22p63s23p64s1 and 1s22s22p63s23p64s2, respectively. The electron configuration of elements with atomic numbers beyond 20 is more complicated, involving d and f subshell, and they are out of the scope of this book. Fig. 2.5.8 shows the electron configuration of the first twenty elements described above. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/02%3A_Elements/2.05%3A_Electrons_in_atoms.txt |
Properties of elements generally show a periodic trend that correlates with their position in the periodic table. The properties and their periodic trends are described below.
Valence electrons
Electrons in the outermost shell are the valence electrons. Fig. 2.5.8 shows the valence electrons of the first twenty elements in red fonts.
Periodic trend of valence electrons
All elements in a group have the same number of valence electrons equal to the first digit of their group number. For example, 1st group of hydrogen and alkali metals has one valence electron, 2nd group of alkali metals has two valence electrons, halogens in 17th group have seven valence electrons, and noble gases in 18th group have eight valence electrons.
Caution
The transition metals in groups 3 to 12, and inner-transition metals, i.e., lanthanoids and actinoids -the two rows of elements placed below the periodic table, are the exception to the general trend of valence electrons described above. Valence electron configurations of transition metals and inner-transition metals are not described here; it is beyond the scope of this book.
The valence electrons mainly determine the chemical properties of the elements. The elements in the same group have similar chemical properties because they have the same valence shell electron configuration. The elements in a row show a gradual change in chemical properties because their valence shell electron configuration changes gradually along the row.
Lewis symbols
1. Lewis symbols show the valence electrons as dots around the symbol of an element. One dot represents one valence electron, e.g., .
2. The dots are shown on any of the four sides of the symbol.
3. A single dot on the top, bottom, left, or right is shown four valence electrons. Then start pairing the dots beyond four valence electrons, as shown in Fig. 2.6.1 for the first twenty elements.
4. Helium is an exception that has only two valence electrons, but they are shown paired.
The electron dots in the Lewis structure are a convenient way to determine how many bonds an atom of an element can make.
Generally, each unpaired dot can make one bond.
For example, a hydrogen atom with one unaired dot can make one bond as in H-H. A bond is represented by a line between the bonded atoms. A bond is formed by sharing unpaired valence electrons. It is called a covalent bond. Carbon, nitrogen, oxygen, and fluorine with 4, 3, 2, and 1 unpaired dot can make 4, 3, 2, and 1 bond, e.g., in the following molecules: , , , and .
• Each line in these molecules represents a bonding electron pair, and
• the pair of dots represent valence electrons that are not involved in bonding, called lone pair of electrons.
Atomic size
Electrons exist around the nucleus in a cloud-like appearance with no clearly defined boundaries. Therefore, the atomic size generally refers to the covalent radius of an atom that is one-half of the distance between the nuclei covalently bonded in a homonuclear molecule, like Cl2, I2, H2, as illustrated in Fig. 2.6.2.
Periodic trend of atomic sizes
1. The atomic size generally increases from top to bottom in a group because the valence electrons add in the higher shell in each consecutive member of a group down the column.
2. The atomic size generally decreases from left to right in a row because the valence electrons are in the same shell while more protons add to the nucleus, increasing the pull on the valence electrons, as illustrated in Fig. 2.6.3.
Ionization energy
The positively charged nucleus attracts the negatively charged electrons. Therefore removal of an electron from the atom requires energy. The ionization produces a cation with fewer electrons than the parent neutral atom, i.e., cations.
Ionization energy
Ionization energy is the energy needed to remove an electron from a neutral atom, as in the following reaction.
$\mathrm{Na}+\text { ionization energy } \rightarrow \mathrm{Na}^{1+}+\mathrm{e}^{-}\nonumber$
Periodic trend in ionization energy
1. The ionization energy generally decreases from top to bottom in a column because the valence electrons are further away and experience less pull to the nucleus down the column.
2. The ionization energy generally increases from left to right in a row because the valence electrons are in the same shell while more protons add to the nucleus, which increases the pull on the valence electrons.
Fig. 2.6.4 illustrates the periodic trend in the ionization energy.
Electronegativity
Definition: Electronegativity
Electronegativity is the ability of an atom in a compound to attract the bonding electron pair to itself. Electropositivity is the opposite of electronegativity. Electronegativity is a property of an atom in a compound, i.e., a bonded atom, not a feature of an individual atom.
There are several electronegativity scales. The most commonly used is the Pauling electronegativity scale. Fig. 2.6.5 shows the electronegativities values on Pauling’s electronegativity scale.
Periodic trend in electronegativity
1. The electronegativity generally decreases from top to bottom in a column because the atomic size increases down the column making the nucleus less effective in pulling the bonding electrons.
2. The electronegative generally increases from left to right in a row because the atomic size decrease from left to right in a row, making the nucleus more effective in pulling the bonding electron pair to itself.
Metallic character
The metallic character relates to the ease of losing an electron in a chemical reaction. The metallic character trend is opposite to the trend of ionization energy.
Periodic trend in metallic character
1. The metallic character generally increases from top to bottom in a column because the atomic size increases down the column, making the valence electrons less tightly held and easier to remove.
2. The metallic character generally decreases from left to right in a row because the atomic size decreases from left to right, making the valence electrons more tightly held and difficult to remove.
Summary of the periodic trends
The ionization energy and the electronegativity generally increase from left to right in a row and from bottom to top in a column. The atomic size and the metallic character are opposite, i.e., they increase from right to left in a row and from top to bottom in a column. Fig. 2.6.6 summarizes the periodic trend in the properties of the elements. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/02%3A_Elements/2.06%3A_The_periodic_trends_in_properties_of_the_elements.txt |
Chemical bonds
What is a compound?
Compounds are a pure form of matter formed by atoms of more than one element combined in a constant whole number ratio.
The bonds connect the atoms in the compounds. Sharing or transferring some valance electrons from one atom to the other makes the bonds. Noble gases have a full valence shell of eight valence electrons, except helium which has a full valence shell of two valence electrons. The noble gases are the least reactive, i.e., the most inert group of elements.
Octet rule
The octet rule states that atoms of all elements other than noble gases tend to share, lose, or gain valence electrons to acquire the electron configuration of the nearest noble gas having eight valence electrons.
Covalent bonds
A bond formed by sharing electrons is a covalent bond.
When a nonmetal atom combines with another nonmetal atom, they usually make a covalent bond. A covalent bond is a pair of shared electrons, called a bonding pair of electrons, where each bonded atom contributes one electron. For example, chlorine has seven valence electrons and needs one more to complete its octet. Hydrogen has one valence electron and requires one more to acquire the electron configuration of helium, i.e., duet instead of the octet. Hydrogen and chlorine combine by sharing one electron to make the compound HCl. Similarly, oxygen, nitrogen, and carbon make 2, 3, and 4 covalent bonds with hydrogen to complete their octet and make compounds H2O, NH3, and CH4, respectively, as illustrated in Fig. 3.1.1.
Two atoms can share one, two, or three electrons to make a single, a double, or a triple covalent bond. For example, H2 has a single bond (H-H), O2 has a double bond (O=O), and N2 has a triple bond (N≡N), where each line between the atoms represent one covalent bond. Fig. 3.1.2 illustrates the formation of three covalent bonds in N2. A valence electron pair that is not involved in bonding is called a nonbonding pair. One nonbonding pair and three bonding pairs complete the octet of each nitrogen atom in the N2 molecule, as shown in Fig. 3.1.2.
Note
The bonding pair of electrons counts towards the total valence electrons of each bonded atom., i.e., in H-H each hydrogen atom has two valence electrons, and in :N≡N: each nitrogen has eight valence electrons; two in the nonbonding pair and six in three bonding pairs.
The formula of a covalent compound
A compound is represented by a chemical formula that combines the symbols of its constituent elements. More electropositive elements are usually written first, e.g., HF, NO, CO. Some exceptions to this rule, e.g., NH3 and CH4, have more electronegative elements written first. The majority of the covalent compounds exist as discrete molecules. A subscript to the right of the element’s symbol represents the number of atoms of the component of the molecule. For example, H2O has two hydrogen atoms and one oxygen atom in a water molecule. Note that subscript 1 is not written, i.e., the symbol of an element alone represents one atom.
A few covalently bonded compounds are giant molecules where the atoms are held together by a 3D network of bonds. The formula of these compounds shows the simplest whole-number ratio of elements in the compound. For example, Fig. 3.1.3 shows SiO2 present in high-purity sand and quartz, a giant molecule.
Ionic bond
A bond formed by the transfer of electrons from one atom to the other atoms is an ionic bond.
A compound that has ionic bonds is an ionic compound. Usually, metal atoms lose electrons and become cations, and nonmetal atoms gain electrons to become anions. The electrostatic attraction between the opposite charges holds the ions together in the ionic compound. For example, sodium (Na) loses one electron, and fluorine (F) gains one electron to make a compound sodium fluoride (NaF), as illustrated in Fig. 3.1.4.
Ionic compounds
Table salt, i.e., NaCl is an example of an ionic compound. The Na completes its octet by losing one electron and becoming Na+ cation. Losing electrons reduces the electron-electron repulsion, but the electron-nucleus attraction remains the same. Consequently, the electron cloud around the nucleus shrinks. Similarly, the chlorine atom has seven valence electrons. After gaining one electron, it becomes Cl- anion with its octet complete. Gaining electrons increases the electron-electron repulsion, but the electron-nucleus attraction remains the same. Consequently, the electron cloud around the nucleus expands. Fig. 3.1.5 illustrates the formation of an ionic bond and the accompanying changes in the total number of electrons and sizes relative to the parent neutral atoms in the case of NaCl formation.
The ionic bond is not localized or unidirectional. The electrostatic force is all around the ions. Therefore, the cations surround the anions, and the anion surrounds the cations in a regular array in a 3D crystal lattice. Fig 3.1.6 illustrates the structure of NaCl.
Note
1. The formula of the ionic compounds represents the simplest whole-number ratio of the atoms of the constituent elements.
2. A cation is always smaller in size than its parent neutral atom.
3. An anion is always larger than its parent neutral atom.
Criteria to distinguish between ionic and covalent bond
Usually, a metal and a nonmetal bond is ionic, and the bond between two nonmetals is covalent. Better criteria are based on the difference in electronegativities of the bonded atoms. Electronegativity is the ability of an atom to attract a pair of bonded electrons to itself. If the electronegativity difference is significant, the bonding electrons completely transfer to the more electronegative atom, and the bond is ionic. There is no single value of electronegativity difference to separate ionic and covalent bonds, but usually, the electronegativity difference of more than 1.8 results in an ionic bond. Otherwise, a covalent bond, but the bonding electrons are more towards the more electronegative atom, making it a polar covalent bond. An electronegativity difference less than 0.5 is considered a noncovalent bond, but a true noncovalent bond forms when the bonded atoms are the same element.
Properties of compounds
The properties of the compounds are usually altogether different from the properties of their constituent elements. For example, hydrogen (H2) is a gas that burns in oxygen, oxygen (O2) is a gas that assists combustion, but water (H2O) is a liquid that extinguishes fire. Similarly, sodium (Na) is a soft metal that melts at 97.79 oC, chlorine (Cl2) is yellowish color gas, but sodium chloride (NaCl) is a transparent crystal that melts at 801 oC.
The intermolecular interactions in covalent molecules are weak to moderate relative to the strength of covalent bonds or ionic bonds. Therefore, the covalent molecules are usually gases like O2, NH3, CH4, liquids like H2O, or soft and low melting solids like waxes, glucose (C6H12O6, mp 146oC).
Ions in the ionic compounds are held together by strong ionic bonds in a 3D array of crystal lattices. Therefore, ionic compounds are usually hard solids with high melting points. For example, NaCl melts at 801 oC. Covalent compounds that exist as a 3D network of covalent bonds, i.e., as giant molecules, are usually hard materials having a higher melting point than ionic compounds. For example, SiO2 present in sand and quartz is a hard solid that melts at 1,710 oC. Diamond –the hardest known substance, is a giant molecule of carbon atoms held together in a 3D network of covalent bonds that melts around 4,027 oC. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/03%3A_Compounds/3.01%3A_Bonding_in_compounds.txt |
Binary ionic compounds are compounds composed of monoatomic cations and monoatomic anions. For example, NaCl is a binary ionic compound composed of monoatomic cations Na+ and monoatomic anions Cl-. Another example is CaCl2 composed of monoatomic cations Ca2+ and monoatomic anions Cl-.
Charge on monoatomic ions
Nonmetals and metalloids of group 14 to group 17 usually form monoatomic anions. The charge on the anions is equal to the group number minus eighteen. For example, halogens in group 17 have charge: 17-18 = -1, oxygen in group 16 has charge: 16-18 = -2, and nitrogen in group 15 has charge: 15-18 = -3. Metals usually form cations: metals of group 1 from +1, metals of group 2 form +2, and aluminum of group 13 form +3 charge on cations, as shown in Fig. 3.2.1. Other metals have variable charges in compounds. The charge of the metals having variable charge can be calculated from the compound's chemical formula because the total –ve charge should be equal to the total +ve charge to make the compound neutral.
Example \(1\)
calculate the charge of iron ion in FeCl2?
Solution
Three are two chloride anions, each with a -1 charge, making a total of -2. So the charge on cation has to be +2 to balance the negative charge. Answer: Fe2+.
Example \(2\)
Calculate the charge on an iron ion in Fe2O3?
Solution
There are three oxygen anions, each with a -2 charge, making -6. So the total charge on two iron atoms should be +6, i.e., the charge on iron atoms is +3. Answer: Fe3+.
Names of monoatomic ions
Name of a monoatomic anion
The name of a monoatomic anion is the element's name with the last syllable replaced with –ide ion. For example, Cl- is a chloride ion, O2- is an oxide ion, N3- is a nitride ion, S2- is sulfide ion, and C4- is a carbide ion, derived from the element names chlorine, oxygen, nitrogen, sulfur, and carbon, respectively.
Name of a monoatomic cation having a fixed charge
Alkali metals have +1, alkaline earth metals have +2, and aluminum has +3 charge. Their name is the name of the element ending with ion. For example, Na+ is a sodium ion, Ca2+ is a calcium ion, and Al3+ is an aluminum ion.
Name of a monoatomic cation having a variable charge
The names of the cations that have a variable charge is the name of the element followed by charge in roman numeral enclosed in small bracts and ending with ion. For example, Fe2+ is an iron(II) ion, and Fe3+ is an iron(III) ion, Cu+ is a copper(I) ion, and Cu2+ is a copper(II) ion.
Ions in body fluids
The ions that are important in body fluids include sodium ion (Na+), potassium ion (K+), calcium ion (Ca2+), magnesium ion (Mg2+), and chloride ion (Cl-), as shown in Fig. 3.2.2. Na+ is present in fluids inside the cells. It regulates and controls body fluids. K+ is present in fluids outside the cells and regulates body fluids and cell functions. Ca2+ and Ma2+ are present in the body fluids outside the cells, where Ca2+ is needed for muscle contraction, Mg2+ is needed for muscle contraction, nerve control, and enzymes. Cl- is primarily present to balance the charge of the cations in the body fluids.
Writing the formula of a binary ionic compounds
The formula of an ionic compound is the symbol of the cation element with a subscript number followed by the symbol of the anion element with a subscript number. The formula shows the simplest whole-number ratio of the constituent elements in the subscripts, such that the total positive charge is equal to the total negative charge.script on the right of the symbols tell the number if they're more than one atom of the element.
Rules of writing the formula of a binary ionic compound
The rules are illustrated in Fig. 3.2.3:
1. Write cation followed by anion with charges,
2. swap the charge as a subscript of the opposite ion,
3. simplify the subscript to the simplest whole-number ratio,
4. use the simplified subscript in the final formula, and
5. do not write the subscript if it is one.
Writing the names of an ionic compound from the formula
Writing the name of ionic compounds of cations with fixed charge
If the cation has a fixed charge in compounds, the name starts with the name of the element of the cation, followed by the name of the anion without the world ion at the end. For example, KI is potassium iodide, and CaCl2 is calcium chloride.
Writing the name of ionic compounds of cations with variable charge
Write the name of the cation, including the charge in roman numerals enclosed in small brackets but without the word ion at the end, followed by the name of the anion, without ion at the end. For example, FeCl2 is iron(II) chloride, and Fe2O3 is iron(III) oxide. Additional examples are given in Table 1.
Transition metals with fixed charge
Silver, zinc, and cadmium cations have fixed changes: Ag+, Zn2+, and Cd2+. Names of these cations are the names of the element with or without charge shown in roman numerals, both ways it is correct.
Table 1: Examples of writing names from the formulae of binary ionic compounds
Example# Formula Name of the cation Name of the anion Name of the compound
1 NaCl Sodium ion Chloride ion Sodium chloride
2 Al2O3 Aluminum ion Oxide ion Aluminum oxide
3 FeCl3 Iron(III) ion Chloride ion Iron(III) chloride
4 CuO Copper(II) ion Oxide ion Copper(II) oxide
5 AgCl
Silver(I) ion, or
Silver ion
Chloride ion
Silver(I) chloride, or
Silver chloride | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/03%3A_Compounds/3.02%3A_Naming_binary_ionic_compounds.txt |
Polyatomic ions
Polyatomic ions are molecular ions composed of two or more atoms bonded by covalent bonds and acting as a single unit, but unlike molecules, they have a net charge on them.
The examples include cations like ammonium ion (\(\ce{NH4^+}\)), and hydronium ion (\(\ce{H3O^+}\)); and anions like hydroxide ion (\(\ce{OH^-}\)), and cyanide ion (\(\ce{CN^-}\)). Usually, the name of polyatomic cations ends with –ium, and the name of polyatomic anions end with –ide, except for oxyanions that have separate rules for their nomenclature.
Oxyanion
The oxyanions are oxides of nonmetals that are molecular ions. Examples include carbonate (\(\ce{CO3^2-}\)) , nitrate (\(\ce{NO3^2-}\)), phosphate (\(\ce{PO4^3-}\)), and sulfate (\(\ce{SO4^2-}\)). The following guidelines will help remember the names and charges of oxyanions in most cases.
Oxyanions having elements from groups 14, 15, and 16
1. the 2nd-row elements C, and N have three oxygen, and 3-row elements P, and S have four oxygen, i.e., one oxygen more than the row number,
2. the name of the polyatomic anion is the name of the atom other than oxygen with the last syllable replaced with –ate, and
3. the charge on these oxyanions equals the number of valence electrons on the non-oxygen atom minus twice the number of oxygen atoms. For example: in (\(\ce{PO4^3-}\)), phosphorous has 5 valence electrons and there are 4 oxygen atoms, so the charge = valence elections in the central atom -2 x number of oxygen atoms = 5-(2x4) = -3; in (\(\ce{SO4^2-}\)), sulfur has 6 valence electrons and there are 4 oxygen atoms, so the charge = 6-(2x4) = -2; in (\(\ce{CO3^2-}\)), carbon has 4 valence electrons and there are 3 oxygen atoms, so the charge = 4-(2x3) = -2; and in (\(\ce{NO3^-}\)), nitrogen 6 valence electrons and there are 3 oxygen atoms, so the charge = 6-(2x3) = -1.
4. There is a 2nd set of oxyanions of the elements mentioned above with one less oxygen but the same charge and the last syllable of the name changed from -ate to -ite. For example: (\(\ce{NO3^-}\)) is nitrate and (\(\ce{NO2^-}\)) is nitrite; (\(\ce{PO4^3-}\)) is phosphate and (\(\ce{PO3^3-}\)) is phosphite; and (\(\ce{SO4^2-}\)) sulfate and (\(\ce{SO3^2-}\)) is sulfite.
Oxyanions of halogens
Oxyanins of chlorine, bromine, and iodine are also common oxyanions with the following in common.
1. They have -1 charge,
2. a halogen with four oxygen is named by adding prefix "per-" to the name of the halogen with last syllable replaced with -ate, for example; (\(\ce{ClO4^-}\)) is perchlorate, (\(\ce{BrO4^-}\)) is perbromate, and (\(\ce{IO4^-}\)) is periodate.
3. a halogen with three oxygen is named as name of the halogen with last syllable replaced with -ate, for example; (\(\ce{ClO3^-}\)) is chlorate, (\(\ce{BrO3^-}\)) is bromate, and (\(\ce{IO3^-}\)) is iodate,
4. a halogen with two oxygen is named as name of the halogen with last syllable replaced with -ite; for example, (\(\ce{ClO2^-}\)) is chlorite,(\(\ce{BrO2^-}\)) is bromite, and (\(\ce{IO2^-}\)) is iodite,
5. a halogen with one oxygen is named by adding prefix "hypo-" to the name of the halogen with last syllable replaced with -ite; for example, (\(\ce{ClO^-}\)) is hypochlorite, (\(\ce{BrO^-}\)) is hypobromite, and (\(\ce{IO3^-}\)) is hypoiodaite.
Acids of oxyanions
Oxyanions are acids when their charge in neutralized with protons (\(\ce{H^+}\)). Names of the acids are the names of oxyanions with -ate replaced with -ic acid and -ite replaced with -ous acid. For example: (\(\ce{HNO3}\)) is nitric acid and (\(\ce{HNO2}\)) is nitrous acid; (\(\ce{H3PO4}\)) is phosphoric acid and (\(\ce{H3PO3}\)) is phosphorous acid; and (\(\ce{H2SO4}\)) sulfuric acid and (\(\ce{H2SO3}\)) is sulfurus acid.
The prefixes "per-" and "hypo-" in the cases of oxyanions of halogens remain in the acid name. For example: (\(\ce{HClO4}\)) is perchloric acid; (\(\ce{HClO3}\)) is chloric acid; (\(\ce{HClO2}\)) is chlorous acid; and (\(\ce{HClO}\)) is hypochlorous acid.
Oxyanions with one proton attached but charge one them not fully neutralized, i.e., they are still polyatomic anion are named beginning with hydrogen and ending with the name of the oxyanion. For example: (\(\ce{HSO4^-}\)) is hydrogen sulfate; (\(\ce{HSO3^-}\)) is hydrogen sulfite; and (\(\ce{HPO4^2-}\)) is hydrogen phosphate.
Oxyanions with two protons attached but charge one them not fully neutralized, i.e., they are still polyatomic anion are named beginning with dihydrogen and ending with the name of the oxyanion. For example, (\(\ce{H2PO4^-}\)) is dihydrogen phosphate
fTwo oxyanions containing a transition metal as the central atom in common use as reagents are in chemistry are chromate (\(\ce{CrO4^2-}\)) and permanganate (\(\ce{MnO4^-}\)). Table 3.3.1 lists the formulas and names of some of the common polyatomic ions.
Table 1: Names of some of the common polyatomic ions
Formula Name Formula Name
(\(\ce{NH4^+}\)) Ammonium (\(\ce{MnO4^-}\)) Permanganate
(\(\ce{H3O^+}\)) Hydronium (\(\ce{BrO4^-}\)) Perbromate
(\(\ce{HO^-}\)) Hydorxide (\(\ce{IO4^-}\)) Periodate
(\(\ce{CN^-}\)) Cynide (\(\ce{CrO4^2-}\)) Chromate
(\(\ce{CO3^2-}\)) Carbonate (\(\ce{CO2^2-}\)) Carbonite
(\(\ce{NO3^-}\)) Nitrate (\(\ce{NO2^-}\)) Nitrite
(\(\ce{PO4^3-}\)) Phosphate (\(\ce{PO3^3-}\)) Phosphite
(\(\ce{SO4^2-}\)) Sulfate (\(\ce{SO3^2-}\)) Sulfite
(\(\ce{HCO3^-}\)) Hydrogen carbonate (\(\ce{ClO4^-}\)) Perchlorate
(\(\ce{HSO4^-}\)) Hydrogne sulfate (\(\ce{ClO3^-}\)) Chlorate
(\(\ce{HPO4^2-}\)) Hydrogenphosphate (\(\ce{ClO2^-}\)) Chlorite
(\(\ce{H2PO4^-}\)) Dihydrogenphosphate (\(\ce{ClO^-}\)) Hypochlorite
Names of compounds containing polyatomic ions
Rules for naming ionic compounds containing polyatomic ions are the same as binary ionic compounds. That is, write the name of the cation followed by the name of the anion. For cations with variable charge, keep the roman numeral in the compound's name. For example, NaNO3 is sodium nitrate, CaCO3 is calcium carbonate, FeCO3 is iron(II) carbonate, NH4Cl is ammonium chloride.
Writing formulae of compounds containing polyatomic ions
The polyatomic ion acts as a single unit, i.e., they are molecular ions. The writing formula of compounds containing polyatomic ions is the same as writing the formula of a binary ionic compound, except that the polyatomic ions must remain intact as a unit. If a subscript is needed for the anion, place the polyatomic ion within small brackets and write the subscript outside the bracket. Just like the subscript to the right of the monoatomic anion tells how many atoms of the anions are there, the subscript to the right of the small bracket around a polyatomic anion tells how many polyatomic anions are there in the compound. For example, iron(III) nitrate is Fe(NO3)3; sodium carbonate is Na2CO3; ammonium phosphate is (NH4)3PO4; potassium permanganate is KMnO4; and calcium phosphate is Ca3(PO4)2. Note that in Fe(NO3)3, there are three nitrate ions, i.e., one iron atom, three nitrogen atoms, and nine oxygen atoms, in the formula unit of the compound. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/03%3A_Compounds/3.03%3A_Polyatomic_ions_and_their_compounds.txt |
Acids donate produce protons and anions when dissolved in water. The naming is different for acids that have oxyanions and the others.
Naming acid that does not contain oxyanion
There are two nomenclatures, first as ionic compounds and second as acids.
Naming acids as ionic compounds
Wright the name of the cation element, i.e., hydrogen, followed by the name of anion without ion at the end. For example, HCl is hydrogen chloride, and HCN is hydrogen cyanide. The ionic name is preferred when the compound is not acting as an acid, e.g., pure HCl in the gas phase.
Naming as acids
Add prefix hydro- to the name of anion and replace the last syllable from –ide to –ic acid. For example, HCl is hydrochloric acid, HCN is hydrocyanic acid, HI is hydroiodic acid. The acid name is preferred when the compound acts as an acid, particularly when it is in solution form in water.
Naming acid that contains oxyanion
Begin the name with the name of oxyanion and change the last syllable from -ate to -ic acid or from -ite to -ous acid. If there is prefix per- or hypo to the name of the oxyanion, it stays in the acid name. For example: (\(\ce{NO3^-}\)) is nitrate and (\(\ce{HNO3}\)) is nitric acid ; (\(\ce{NO2^-}\)) is nitrite and (\(\ce{HNO2}\)) is nitrous acid; (\(\ce{SO4^2-}\)) is sulfate and {H2SO4}\)) sulfuric acid; (\(\ce{ClO4^-}\)) is perchlorate and (\(\ce{HClO4}\)) is perchloric acid; and (\(\ce{ClO^-}\)) is hypochlorite and (\(\ce{HClO}\)) is hypochlorous acid.
One of the commonly encountered oxyacid is acetic acid. Acetic acid is an organic acid with the formula CH3COOH, where the last hydrogen attached with oxygen is the acidic proton, and the other three oxygen attached with carbon are not acidic. The anion from acetic acid is called acetate ion that has a formula CH3COO- which may also be written as C2H3O2-.
3.05: Naming binary covalent compounds
What is a binary covalent compound?
Binary covalent compounds are atoms of two different elements held together by covalent bonds. Usually, they are composed of nonmetals elements, e.g., laughing gas NO, acid rain causing gas SO2, etc.
Writing formulae of binary covalent compounds.
The molecular formula of a binary covalent compound shows the symbols of constituent elements, followed by a subscript showing how many atoms of the element are in the molecule. Usually, the symbol of an element closer to metals in a period or a group is written first, followed by the symbol of the other element. For example, CO2, NO, P2O5, where carbon, nitrogen, and phosphorous are nearer to the metals in the periodic table than the other element. An exception occurs when the compound contains oxygen combined with chlorine, bromine, or iodine, where oxygen is closer to metals but is second in the formula, e.g., ClO2.
Writing the names of binary covalent compounds
The name of binary covalent compounds contains prefixes, listed in Table 1, to indicate the number of atoms followed by the name of the elements according to the following rules:
1. name of the first element in the formula with a prefix showing the number of atoms, followed by,
2. the name of the second element with a prefix showing the number of atoms and its last syllable replaced with –ide.
3. Do not write mono- if it applies to the first element in the formula, but write mono- if it applies to the second element. Write all other prefixes.
4. If the prefix ends with a vowel and the element name begins with a vowel, drop the ending vowel of the prefix.
Table 3.5.1 lists the prefixes used to represent the number of atoms from 1 to 10. Examples of the names are: NO is nitrogen monoxide, CO2 is carbon dioxide, PCl3 is phosphorous trichloride, P2O5 is diphosphorus pentoxide, SiO2 is silicon dioxide. Trivial names are well known for some molecular compounds and they are often used in the place of systematic names, e.g., water for H2O and ammonia for NH3.
Table 1: Prefixes used in naming binary covalent compounds
Prefix Means
Mono- 1
di- 2
Tri- 3
tetra- 4
penta- 5
hexa- 6
hepta- 7
octa- 8
nona- 9
deca- 10 | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/03%3A_Compounds/3.04%3A_Naming_acids.txt |
Lewis symbols of elements
Lewis symbol of element shows the symbol of element with valence electrons shown as dots placed on top, bottom, left, and right sides of the symbol. Valence electrons up to four are shown as a single dot on either side of the symbol. The 5th, 6th, 7th, and 8th valence electron dots are paired with any of the first four dots. For example, represent hydrogen, beryllium, boron, carbon, nitrogen, fluorine, and neon with 1, 2, 3, 4, 5, 6, 7, and 10 valence electrons, respectively. Lewis symbols of first twenty elements are shown in section 2.6, Table 1.
Lewis structures of simple molecules
An unpaired dot in the Lewis symbol of an element can make one bond by sharing it with an unpaired dot of another atom. The shared pair of dots represents a pair of bonding electrons, a covalent bond. For example, as shown below, a Hydrogen atom has one unpaired valence electron and makes a covalent bond with another hydrogen atom.
The covalent bond is usually represented by a single line between the bonded atoms, e.g., the H2 molecule shown in the above equation is generally shown as H-H. An example is a reaction between hydrogen having one valence electron and carbon having four valence electrons react to form CH4 molecule.
Similarly, hydrogen reacts with nitrogen, oxygen, and fluorine to form the following molecules: , , and . Each line in these molecules represents a bonding electron pair, and the pair of dots represent valence electrons that are not involved in bonding, called lone pair of electrons. The lone pair is usually omitted from the Lewis structure unless it is needed to emphasize their presence for some reason.
Procedure for writing Lewis structures of molecules
A systematic approach for writing the Lewis structure of molecules is explained with the help of the following example.
Example \(1\)
Draw the Lewis structures of CH4, PCl3, CO2, and HCN
Solution
Step 1: Add the valence electrons of all the molecules' atoms:
• CH4 has 4 valence electrons in C, and 1 in each of the four H: = 4 + 1x4 = 8 valence electrons
• PCl3 has 5 valence electros in P and 7 in each of the three Cl: = 5 + 7x3 = 26 valence electrons
• CO2 has 4 valence electrons in C and 6 in each of the two O: = 4 + 6x2 = 16 valence electrons
• HCN has 1 valence electron in H, 4 in C, and 5 in N: = 1 + 4 + 5 = 10 valence electrons
Step 2: Place the element symbol with more valances, i.e., having more unpaired dots in its Lewis structure, in the center and the rest of the atoms on four sides:
Step 3: Draw a line between the outer atom and the central atom to represent a single covalent bond:
Step 4: Every single bond consumes two valence electrons. Subtract the total number of valence electrons consumed in all the bonds from the total valence electrons initially present in step 1:
• CH4: 4 bond = 8 electrons consumed. 8 – 8 = 0 electrons left
• PCl3: 3 bond = 6 electrons consumed. 26 – 6 = 20 electrons left
• CO2: 2 bond = 4 electrons consumed. 16 – 4 = 12 electrons left
• HCN: 2 bond = 4 electrons consumed. 10 – 4 = 6 electrons left
Step 5: Distribute the remaining electrons as lone pairs, first to the outer atoms to complete their octet (duet in the case of hydrogen) and then to the central atom to complete its octet
Step 6: Check that the octet of each atom is complete (duet for hydrogen). If yes, the Lewis structure is complete, e.g., as in the cases of CH4 and PCl3 in the present examples. If not, move one of the lone pair of electrons from a neighboring atom to make a double bond, as shown by the red color arrows in the figure in the previous step. If the octet is still not complete, move one more lone pair of electrons from a neighboring atom: from the same atom to make a triple bond, as in the case of HCN above, or from another neighboring atom to make two double bonds, as in the case of CO2 above. The result is Lewis structures shown below.
Note
To draw the Lewis structure of the most stable form, try to keep covalent bonds with an atom equal to the number of unpaired dots in the Lewis symbol of the atoms. For example, have 1, 1, 2, 3, and 4 unpaired electrons. So, hydrogen makes one, fluorine one, oxygen two, nitrogen 3, and carbon four covalent bonds in the stable molecules. When pulling a lone pair from neighboring atoms is needed to result in a double or triple bond, it is preferable to keep the resulting number of covalent bonds equal to the number of unpaired electrons in the Lewis symbol of the atom. For example, in the case of CO2 molecule, if both lone pairs were pulled from the same oxygen in step 6 above, the resulting Lewis structure would have been , which is technically correct Lewis structure but not the most stable form of this molecule. Note that the latter structure has one covalent bond in one oxygen and three covalent bonds in the other instead of two covalent bonds needed for oxygen.
Exceptions to the octet rule
The octet rule generally applies in most cases, but there are exceptions in a few cases:
1. Atoms of hydrogen, lithium, and beryllium tend to share, lose, or gain valence electrons to acquire the electron configuration of the nearest noble gas helium having two valence electrons. This is called the duet rule.
2. Sometimes the atoms may end up in compounds with less than eight valence electrons. This often happens in the case of boron compounds and aluminum compounds of group 13. For example, boron has three valence electrons resulting in compounds like BF3 with three covalent bonds and six instead of eight valence electrons around boron.
3. Atoms of elements in period three and beyond sometimes end up in compounds with more than eight valence electrons. For example, PCl3 has octet complete, but PCl5 has 10 valence electrons. Similarly, sulfur has its octet complete in H2S but has 12 valence electrons in H2SO4. This happens because atoms in period thee and beyond have larger sizes and they have valence electrons in d or f orbitals in addition to the valence s and p orbitals. Table 1 lists some examples of exceptions to the octet rule.
Table 1: Some examples of exceptions to the octet rule.
Structure H:H
Name Hydrogen Beryllium dichloride Boron trifluoride Phosphorous pentafluoride Sulfuric acid
Valence electrons on the central atom 2 4 6 10 12 | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/03%3A_Compounds/3.06%3A_Lewis_structures_of_molecules.txt |
The Lewis structure tells the connection between atoms and any lone pair present, but it does not tell the exact angles of bonds around the central atom or the actual shape of the molecule. The conventional way of presenting a Lewis structure of a molecule shows it as a planer, e.g., CH4 as which implies the molecule to be planar and H-C-H bond angles to be either 90o or 180o. The actual CH4 molecule is nonplanar as with all H-C-H bond angles 109.5o. the following theory helps explain the actual shapes of molecules.
Valence shell electron pair repulsion (VSERP) theory
The valence shell electron pair repulsion theory predicts the shape of the molecule and bond angles based on the fact that valence shell electrons around the central atom in a molecule are grouped. The electron groups repel each other and go as far apart from each other as possible.
Electron groups
A lone pair, a single bond, a double bond, and a triple bond, each of these is one electron group. This is because two elections of single bond, four electrons of a double bond, and six electrons of a triple bond are located in the region along the axis of the bond, i.e., they are grouped together. Similarly, a lone pair is located in a defined space around the atom. For example, carbon in methane () has four electron groups that are the four single bonds (C-H bonds) around the carbon. Carbon in carbon dioxide (O=C=O) has two electron groups that are the two double bonds around the carbon, and in H-C≡N has two electron groups that are a single bond (C-H bond) and a triple bond (C≡N bond).
Molecular shapes and bond angles based on VSEPR theory
One electron group
One electron group between two atoms is always a linear molecule. For example, H-H ,O=O, N≡N, and H-Cl , are linear molecules, where hydrogen is white and chlorine is green in the H-Cl model.
Two-electron groups
Two-electron groups are farthest apart in a linear geometry with the central atom in the middle of the line and the bond angles of 180o around the central atom. The examples include CO2 and HCN , where the central carbon atom is gray, hydrogen is white, nitrogen is blue, and oxygen is red.
Three-electron groups
Three-electron groups are farthest apart when they are at the corners of a triangle in a planar trigonal geometry with the central atom in the middle of the triangle and the bond angles of 120o around the central atom. Examples include BF3 , and H2CO , where boron is pink, F is green, carbon is gray, oxygen is red, and hydrogen is white.
If one of the electron domains is a lone pair, the electron domain geometry remains the same, but the geometry of the atoms in the molecule, i.e., molecule geometry, is bent. For example, has three electron domains and trigonal planar electron domain geometry, but there is one lone pair. So, the molecule geometry is bent as , where sulfur is yellow, and oxygen is red (lone pair in not shown).
Four-electron groups
Four-electron groups are farthest apart when they are at the corners of a tetrahedron in a tetrahedral geometry with the central atom at the center of the tetrahedron and the bond angles of 109.5o around the central atom as: . An example is methane CH4 , where carbon is gray, and hydrogens are white.
If one of the electron domains is a lone pair, the electron domain geometry is still tetrahedral, but the molecule geometry is trigonal pyramidal as with three pereferal atoms at the corners of the triangel and the central atom raised to the top of the pyramid. An example is ammonia (:NH3) , where nitrogen is blue, and hydrogens are white.
If two electron domains are lone pairs, the electron domains geometry is still tetrahedral, but the molecule geometry is bent. An example is water ( ), where oxygen is red, and hydrogens are white.
Table 1 is the summary of the electron domain geometries and the corresponding molecular geometries.
Table 1: Common molecular geometries around the central atom
Electron domain Lone pair Electron domain geometry Molecule geometry Bond angles Examples
1 0 Linear Linear - HCl
2 0 Linear Linear 180o
CO2
HCN
3 0 Trigonal pyramidal Trigonal pyramidal 120o H2CO
3 2 Trigonal pyramidal Bent 120o
4 0 Tetrahedral Tetrahedral 109.5o CH4
4 1 Tetrahedral Trigonal pyramidal 109.5o :NH3
4 2 Tetrahedral Bent 109.5o
3.08: Polarity of molecules
The polarity of an atom
The negative charge of electrons balances the positive charge of protons in an atom. The electrons symmetrically distributed around the nucleus leave no negative or positive end. The atoms are nonpolar. Fig. 3.8.1 illustrates the polarity of a hydrogen atom with color codes.
The polarity of a covalent bond
When atoms of one element combine to make a covalent bond, e.g., H-H and F-F, the positive and negative charges are still symmetrical, and the bond is nonpolar, i.e., no negative end separated from a positive end. However, when atoms of different elements combine to make a covalent bond, the more electronegative atom attracts the bonding electron pair towards itself stronger than the other atom. The separation of positive and negative charges happens as the electrons shift more towards the electronegative atom. The bond becomes polar with a partially positive (δ+) end on the electropositive atom and a partially negative (δ-) end on the electronegative atom. For example, fluorine is more electronegative than hydrogen. Consequently, fluorine pulls the bonding electron pair towards itself in the H-F molecule, creating a partial negative charge (δ-) on fluorine and a partial positive charge (δ+) on hydrogen. The H-F bond is polar. Fig. 3.8.2 illustrates the polarity in H-F with color codes.
Bond polarity is a vector that has a magnitude and direction and can be represented by an arrow, like other vectors, as shown in Fig. 3.8.3 for the case of a water molecule.
A bond is categorized as a nonpolar covalent, polar covalent, or an ionic bond based on the following convention: nonpolar covalent if the electronegative difference of the bonded atom is less than 0.5, a polar covalent if the electronegativity difference is between 0.5 to 1.9, and an ionic if the electronegativity difference is more than 1.9.
The polarity of a Molecule
The molecules fall into the following categories concerning molecular polarity.
The molecule is nonpolar if there is no polar bond in it, e.g., H-H, F-F, and CH4 are nonpolar molecules. Fig. 3.8.4 illustrates CH4 molecules with green color electron clouds that represent a nonpolar molecule.
If there is only one polar bond in a molecule, then the molecule is polar, e.g., the H-F molecule shown in Fig. 3.8.2.
If there is more than one polar bond in a molecule, the molecule may be polar or may be nonpolar, depending on the symmetry of the molecule: a) Polarity vector of individual bonds cancel out in symmetric molecules making the molecule nonpolar. For example, symmetric molecules, like CO2, BF3, and CCl4, are nonpolar, although each bond in them is polar. Fig. 3.8.5 illustrates the symmetric molecules that have polar bonds, but the polarity of bonds cancels each other, making the molecule nonpolar. b) If a molecule has polar bonds and it is not symmetric, the polarity vectors do not cancel out, and the molecule is polar. Examples of polar molecules include CHCl3, NH3, and H2O, as illustrated in Fig. 3.8.6. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/03%3A_Compounds/3.07%3A_Molecular_shapes_Valence_shell_electron_pair_repulsion_%28VSEPR%29_theory.txt |
There are electrostatic interaction between charges or partial charges, i.e., the same charges attract each other, and opposite charges repel each other, as illustrated in Fig. 3.9.1.There are two types of electrostatic forces in compounds or molecules, intramolecular forces that exist between the bonded atoms of a compound or a molecule, and intermolecular forces that exist between molecules as described below.
Intramolecular forces
Intramolecular forces are the chemical bonds holding the atoms together in the molecules. The three major types of chemical bonds are the metallic bond, the ionic bond, and the covalent bond.
Metallic bond
Metals exist as a collection of many atoms as +ions arranged in a well-defined 3D arrangement called crystal lattice with some of the outermost electrons roaming around in the whole piece of the metal, forming a sea of electrons around the metal atoms, as illustrated in Fig. 3.9.2. The attraction between +ions and the sea of free moving electrons is the metallic bond that holds the atoms together in a piece of metal. The metallic bond is usually the strongest type of chemical bond.
Ionic bond
When the electronegativity difference between bonded atoms is large, i.e., more than 1.9 in most cases, the bonding electrons completely transfer from a more electropositive atom to a more electronegative atom creating a cation and an anion, respectively. There is the electrostatic interaction between cation and anion, i.e., the same charges attract each other, and opposite charges repel each other, as illustrated in Fig. 3.9.1. The cations and anions orient themselves in a 3D crystal lattice in such a way that attractive interactions maximize and the repulsive interactions minimize, as illustrated in Fig. 3.9.3. Ionic bonds are usually weaker than metallic bonds but stronger there the other types of bonds.
Covalent bond
When the electronegativity difference between bonded atoms is moderate to zero, i.e., usually less than 1.9, the bonding electrons are shared between the bonded atoms, as illustrated in Fig. 3.9.4. The attractive force between the bonding electrons and the nuclei is the covalent bond that holds the atoms together in the molecules. The covalent bond is usually weaker than the metallic and the ionic bonds but much stronger than the intermolecular forces.
Criteria to predict the type of chemical bond
Metals tend to have lower electronegativity and nonmetals have higher electronegativity. When the electronegativity difference between the bonded atoms is large, usually more than 1.9, the bond is ionic. Generally, a bond between a metal and a nonmetal is ionic. When the electronegativity difference is low, usually less than 1.9, the bond is either metallic or covalent. Nonmetals tend to make a covalent bond with each other. Nonmetals also have higher electronegativities. So, when the average electronegativity of the bonded atom is high and the electronegativity difference between them is low, they tend to make a covalent bond. Metals tend to make the metallic bond with each other. Metals also tend to have lower electronegativity values. So, when the average electronegativity of the bonded atom is low and the electronegativity difference between them is also low, they tend to make a metallic bond. Fig. 3.9.5 illustrates the criteria to predict the type of chemical bond based on the electronegativity difference. Keep in mind that there is no sharp boundary between metallic, ionic, and covalent bonds based on the electronegativity differences or the average electronegativity values.
Intermolecular forces
Intermolecular forces are the electrostatic interactions between molecules. The intermolecular forces are usually much weaker than the intramolecular forces, but still, they play important role in determining the properties of the compounds. The major intermolecular forces include dipole-dipole interaction, hydrogen bonding, and London dispersion forces.
Dipole-dipole interactions
Polar molecules have permanent dipoles, one end of the molecule is partial positive (δ+) and the other is partial negative (δ-). The polar molecules have electrostatic interactions with each other through their δ+ and δ- ends called dipole-dipole interactions, though these interactions are weaker than ionic bonds. The polar molecules orient in a way to maximize the attractive forces between the opposite charges and minimize the repulsive forces between the same charges, as illustrated in Fig. 3.9.6.
Hydrogen bonds
Hydrogen bonding is a dipole-dipole interaction when the dipole is a hydrogen bond to O, N, or F, e.g. in water molecules as illustrated in Fig. 3.9.7. Although hydrogen bond is a dipole-dipole interaction, it is distinguished from the usual dipole-dipole interactions because of the following special features.
1. The electronegativity difference between H and O, N, or F is usually more than other polar bonds.
2. The charge density on hydrogen is higher than the δ+ ends of the rest of the dipoles because of the smaller size of hydrogen.
3. The δ+ Hydrogen can penetrate in less accessible spaces to interact with the δ- O, N, or F of the other molecule because of its small size.
A hydrogen bond is usually stronger than the usual dipole-dipole interactions. Hydrogen bonding is the most common and essential intermolecular interaction in biomolecules. For example, two strands of DNA molecules are held together through hydrogen bonding, as illustrated in Fig. 3.9.8. Proteins also acquire structural features needed for their functions mainly through hydrogen bonding.
London dispersion forces
It may appear that the nonpolar molecules should not have intermolecular interactions. Practically, there are intermolecular interactions called London dispersion forces, in all the molecules, including the nonpolar molecules. The electron cloud around atoms is not all the time symmetrical around the nuclei. It temporarily sways to one side or the other, generating a transient dipole. The transient dipole induces a dipole in the neighboring. A transient dipole-induced dipole interaction, called London dispersion force or wander Wall’s force, is established between the neighboring molecules as illustrated in Fig. 3.9.9. Although London dispersion forces are transient, they keep re-appearing randomly distributed in space and time. London dispersion forces are not unique to nonpolar molecules, they are present in all types of molecules, but these are the only intramolecular forces present in the nonpolar molecules. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/03%3A_Compounds/3.09%3A_Intramolecular_forces_and_intermolecular_forces.txt |
• 4.1: Stoichiometry
The terms stoichiometry, i.e., a study of quantities of mass and energy consumed or produced in a chemical reaction, and atomic mass, i.e., the weighted average of the masses of isotopes of an element present in a natural sample, are described.
• 4.2: The mole
Mole plays a central part in stoichiometry relating to the number of particles through Avogadro's number, to the mass of substance in grams through the molar mass, and to moles of other substances through a balanced chemical equation. These stoichiometric calculations are explained.
• 4.3: Chemical reaction
answers to what is a chemical reaction equation, how to write it, and how to balance it are given.
• 4.4: Patterns of chemical reactions
General patterns of chemical reaction, i.e., combination, decomposition, single displacement, double displacement, and combustion, and general classification of chemical reactions, i.e., oxidation-reduction, acid-base, and precipitation reactions are described.
• 4.5: Oxidation-reduction reactions
Oxidation-reduction or redox reactions involving the transfer of electrons, splitting the redox reaction into oxidation-half and reduction-half, combining the two halves to write an overall redox reaction equation, and recognizing oxidation or reduction in biological reactions are described.
• 4.6: Energetics of chemical reactions
The energies evolved or absorbed during a reaction, the energy barrier that the reactants must surpass before the reactions can happen, the effects of the energy barrier on the rate of reactions, and the factors that affect the rate of reaction are introduced.
• 4.7: Stoichiometric calculations
Stoichiometric calculations involving the use of reciprocal of molar mass as a conversion factor from grams to moles of a substance, mole to mole conversion factor from a chemical equation to find the moles of one substance from the know moles of another, and finally using molar mass as a conversion factor to calculate the mass of a substance from its know moles are described.
04: Stoichiometry the quantification of chemical reactions
What is Stoichiometry
Stoichiometry (stoi·chi·om·e·try /ˌstɔɪkiˈɒmɪtri/) is the study of the quantities of substances and energy consumed and produced in chemical reactions.
The basis of the stoichiometric calculations is the law of conservation of mass which states that the mass is neither created nor destroyed in a chemical reaction. Another form of the law states that atoms are neither created nor destroyed in a chemical reaction. It is the basis of stoichiometric calculations that are described in this chapter.
It can be concluded from the law of mass action that atoms of each element and their masses are the same in reactants and products. A balanced chemical equation shows atoms of each element and the total mass of reactants equal to that in the product, as illustrated in Fig. 4.1.1. The number of atoms and molecules is related to their quantity in moles through Avogadro's number. The mole, in turn, is related to the mass of the substance through molar mass in grams. These relationships are described in the next sections.
Atomic mass
Atomic mass is the weighted average of the masses of the isotopes present in a natural sample of the element, as explained with an example calculation in section 2.3. The mass of a single atom or molecule is expressed in the atomic mass unit (amu), which is equal to $\frac{1}{2}$th of the mass of $\ce{^12_6C}$ isotope of carbon that is unbound.
Atomic mass is listed in a periodic table as a number below the symbol and name of the element, as illustrated in Fig. 4.1.2. Atomic mass is listed as a number without a unit because it is the mass of an atom in amu and it is also the mass of one mole (molar mass) of the atom in grams. Molar mass is often used in stoichiometry calculation as explained in the next sections. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/04%3A_Stoichiometry_the_quantification_of_chemical_reactions/4.01%3A_Stoichiometry.txt |
Avogadro’s number and mole
The Avogadro's number
The Avogadro's number is equal to 6.02214076 x 1023 exact.
Just like the dozen is equal to 12, the Avogadro's number 6.02214076 x 1032 is exact by definition, but usually, 6.022 x 1023 is used in calculations with 4 significant figures.
Mole (mol)
The mole is a SI unit of the amount of a substance that is equal to 6.02214076 x 1023 particles of the substance.
The particle of a substance are usually atoms, ions, or molecules. For example, 6.02214076 x 1023 atoms of $\ce{^12_6C}$ isotope is one mole of $\ce{^12_6C}$. The number 6.02214076 x 1032 is exact by definition, but usually, 6.022 x 1023 is used with 4 significant figures.
One mole of a substance is equal to one Avogadro’s number of atoms, molecules, or formula units of the substance. i.e.,
$1 \text { Avogadro's number of particles }=1 \mathrm{~mol}=6.022 \times 10^{23} \text { particles }\nonumber$
, where particles are atoms, molecules, or formula units in chemistry.
The equality between Avogadro's number and mole gives two conversion factor:
$\frac{6.022 \times 10^{23} \text { particles }}{1 \text { mol }} \quad\text { and }\quad \frac{1 \mathrm{~mol}}{6.022 \times 10^{23} \text { particles }}\nonumber$
, where the first factor is used to convert a number of moles to a number of particles and the second for a number of particles to number moles conversions, as explained in the following examples.
Example $1$
How many aspirin ($\ce{C9H8O4}$ molecules are in 0.0139 mol of aspirin?
Solution
Step 1. Write the given quantity and the desired quantity.
Given: 0.0139 mol ($\ce{C9H8O4}$, Desired: ? molecules of ($\ce{C9H8O4}$
Step 2. Write the two conversion factors from equality between the given and the desired quantity.
$\frac{6.022 \times 10^{23} \text { particles }}{1 \text { mol }} \quad\text { and}\quad ~\frac{1 \mathrm{~mol}}{6.022 \times 10^{23} \text { particles }}\nonumber$
Step 3. Multiply the given quantity with the conversion factor that cancels the given unit and leaves the desired unit in the answer.
$0.0139 \cancel{\text { mol } \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}}{1 \cancel{\mathrm{~mol} \mathrm{~C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}}}=8.37 \times 10^{21} \text { molecules } \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\nonumber$
Example $2$
How many moles of aspirin (($\ce{C9H8O4}$ ) are in 9.50 x 1025 molecules of aspirin?
Solution
Step 1. Write the given quantity and the desired quantity.
Given: 9.50 x 1025 molecules of aspirin, Desired: ? mol of aspirin
Step 2. Write the two conversion factors from equality between the given and the desired quantity.
$\frac{6.022 \times 10^{23} \text { particles }}{1 \text { mol }} \quad\text { and } \quad\frac{1 \mathrm{~mol}}{6.022 \times 10^{23} \text { particles }}\nonumber$
Step 3. Multiply the given quantity with the conversion factor that cancels the given unit and leaves the desired unit in the answer.
$9.50 \times 10^{25} \cancel{\text { molecules } \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}} \times \frac{1 \mathrm{~mol} \mathrm{~C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}}{6.022 \times 10^{23}\cancel{\text { molecules } \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}}}=158 \mathrm{~mol} \mathrm{} \mathrm{~C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\nonumber$
Moles of elements in a mole of a compound
Moles of an element in a mole of a compound is equivalent to atoms of the element in a molecule or formula unit of the compound. For example in 1 mole of glucose ($\ce{C6H12O6}$) there are 6 moles of carbon, 12 moles of hydrogen, and 6 moles of oxygen. Each of these equalities between a mole of a substance and the moles of the element in it gives two conversion factors for the calculations.
Example $3$
How many moles of hydrogen are in 3.0 moles of glucose ($\ce{C6H12O6}$)?
Solution
Step 1. Write the given quantity and the desired quantity.
Given: 3.0 moles of ($\ce{C6H12O6}$), Desired: ? mol of H
Step 2. Write the two conversion factors from equality between the given and the desired quantity.
$\frac{12 {~mol} ~H}{1 {~m o l} ~C_{6} H_{12} O_{6}} \quad\text { and } \quad\frac{12 {~mol} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{1 {~mol} ~H}\nonumber$
Step 3. Multiply the given quantity with the conversion factor that cancels the given unit and leaves the desired unit in the answer.
$3.0 \cancel{\mathrm{~mol}\mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}} \times \frac{12 \mathrm{~mol} \mathrm{~H}}{1 \cancel{\mathrm{~mol} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}}=36 {~mol} ~H\nonumber$
Example $4$
Calculate the moles of oxygen are in 0.50 mole of $\ce{Ca(NO3)2}$?
Solution
Step 1. Write the given quantity and the desired quantity.
Given: 0.50 moles of $\ce{Ca(NO3)2}$, Desired: ? mol of O
Step 2. Write the two conversion factors from equality between the given and the desired quantity. Note that $\ce{NO3^-}$ is polyatomic ion that has three oxygen atoms in it. There are two $\ce{NO3^-}$ in the formula unit as shown by subscript 2 outside the bracket enclosing the polyatomic anion (\ce{NO3^-}\). So, the equality is:
1 mole $\ce{Ca(NO3)2}$ = 6 mole O,
and the two conversion factors from the equality are:
$\frac{1 ~mol ~\ce{Ca(NO3)2}}{6 ~m o l ~O} \quad\text { and }\quad \frac{6 ~mol ~O}{1 ~mol ~\ce{Ca(NO3)2}}\nonumber$
Step 3. Multiply the given quantity with the conversion factor that cancels the given unit and leaves the desired unit in the answer.
$0.50 \cancel{~mol~\ce{Ca(NO3)2}}\times \frac{6 ~mol ~O}{1 \cancel{~mol ~\ce{Ca(NO3)2}}}=3.0 ~mol ~O\nonumber$
Molar mass
The mass of one mole of a substance, in $\frac{g}{mol}$, is called the molar mass.
Recall that one mole = 1 Avogadro's number, i.e., 6.022×1023 atoms, molecules, or formula units of a substance. Fig. 4.2.1 helps in visualizing the molar masses of aluminum, copper, and carbon. The molar mass of an element or a compound is a reasonable quantity to be measured on an analytical balance commonly available in laboratories, while the mass of an individual atom or molecule is too small to be easily measured. That is why the mole is commonly used in stoichiometric calculations.
The molar mass of atoms of an element
The atomic mass listed in the periodic table is the molar mass of atoms of the element in $\frac{g}{mol}$. For example, the atomic mass of H listed in the periodic table is 1.008, so the molar mass of H is 1.008 $\frac{g}{mol}$. Similarly, the atomic mass of O listed in the periodic table is 15.999, and the molar mass of O is 15.999 $\frac{g}{mol}$.
The molar mass of molecules of an element
The molar mass of molecules of an element is the sum of atomic mass of atoms in the molecule expressed in $\frac{g}{mol}$. For example, the molar mass of H2 is 1.008 $\frac{g}{mol}$ + 1.008 $\frac{g}{mol}$ = 2.016 $\frac{g}{mol}$.
The molar mass of a compound
The molar mass of a compound is the sum of the atomic masses of all the atoms in the molecular formula or formula unit of the compound. For example, the molar mass of water (H2O) is the sum of the molar mass of two hydrogen atoms + the molar mass of one oxygen atom, i.e., 2x1.008 $\frac{g}{mol}$ H + 15.999 $\frac{g}{mol}$ O = 18.02 $\frac{g}{mol}$ H2O. In other words, to calculate the molar mass of a compound, take the atomic masses of the constituent elements from a periodic table, multiply them with the number of atoms of the element in the formula of the compound, and then add these numbers, as explained in the following examples. Note that the unit $\frac{g}{mol}$ can also be written as: g.mol-1.
Example $5$
Calculate the molar mass of NaOH?
Solution
Step 1. Find the atomic masses of the constituent elements from the periodic table.
Na = 22.990 g.mol-1, O = 15.999 g.mol-1, H = 1.008 g.mol-1.
Step 2. Multiply the atomic masses with the number of atoms in the fourmula.
1x22.990 g.mol-1 Na, 1x15.999 g.mol-1O, 1x1.008 g.mol-1 H.
Step 3. Add all the numbers from step 2.
1x22.990 g.mol-1 Na + 1x15.999 g.mol-1O + 1x1.008 g.mol-1 H = 39.997 g.mol-1 NaOH
Example $6$
Calculate the molar mass of Ca(NO3)2?
Solution
Step 1. Find the atomic masses of the constituent elements from the periodic table.
Ca = 40.078 g.mol-1, N = 14.007 g.mol-1, O = 15.999 g.mol-1.
Step 2. Multiply the atomic masses with the number of atoms in the formula. (Note that there are two NO3 units in (NO3)2?, To get the total number of atoms, multiply the subscript outside the bracket with the subscript to the element symbol inside the bracket to get the total number of atoms. That is, there are 1x2 = 2 N and 3x2 = 6 O in this compound.
1x40.078 g.mol-1 Ca, 2x14.007 g.mol-1 N, 6x15.999 g.mol-1 O.
Step 3. Add all the numbers from step 2.
1x40.078 g.mol-1 Ca + 2x14.007 g.mol-1N + 6x15.999 g.mol-1 O = 164.008 g.mol-1 Ca(NO3)2.
Conversion from grams to moles and moles to grams of a substance
The molar mass in $\frac{g}{mol}$ is a conversion factor converting the amount of a substance in moles to the mass of the substance in grams. Reciprocal of the molar mass in $\frac{mol}{g}$ is a conversion factor converting the mass of a substance in grams to the amount of the substance in moles. The conversions are explained in the following examples.
Example $7$
How many grams of water are in 2.50 moles of water?
Solution
Step 1. Write the given quantity and the desired quantity.
Given: 2.50 moles H2O, Desired: ? g H2O
Step 2. Calculate the molar mass of the substance.
2x1.008 g.mol-1 H + 15.999 g.mol-1 O = 18.02 g.mol-1 H2O.
Step 3. Write the molar mass of the substance, and it's reciprocal as the two conversion factors.
$\frac{18.02 ~g ~\mathrm{H}_{2} \mathrm{O}}{1 \mathrm{~mol} \mathrm{} ~\mathrm{H}_{2} \mathrm{O}}\quad\text { and} \quad~\frac{1 \mathrm{~mol} \mathrm{} ~\mathrm{H}_{2} \mathrm{O}}{18.02 ~g ~\mathrm{H}_{2} \mathrm{O}}\nonumber$
Step 4. Multiply the given quantity with the conversion factor that cancels the given unit and leaves the desired unit in the answer.
$2.50 \cancel{\mathrm{~mol} \mathrm{~} \mathrm{H}_{2} \mathrm{O}} \times \frac{18.02 \mathrm{~g} ~\mathrm{H}_{2} \mathrm{O}}{1 \cancel{\mathrm{~mol} ~\mathrm{H}_{2} \mathrm{O}}}=45.1 \mathrm{~g} \mathrm{~} \mathrm{H}_{2} \mathrm{O}\nonumber$
Example $8$
How many moles are present in 2.50 g of aspirin? (Aspirin $\ce{C9H8O4}$, molar mass 180.2 g.mol-1 )
Solution
Step 1. Write the given quantity and the desired quantity.
Given: 2.50 g $\ce{C9H8O4}$, Molar mass of $\ce{C9H8O4}$ = 180 g.mol-1, Desired: ? g H2O
Step 2. Calculate the molar mass of the substance.: given 180.2 g.mol-1
Step 3. Write the molar mass of the substance, and it's reciprocal as the two conversion factors.
$\frac{180 ~g~\ce{C9H8O4}}{1~mol ~\ce{C9H8O4}}\quad\text { and}\quad ~\frac{1~mol~\ce{C9H8O4}}{180~g~\ce{C9H8O4}}\nonumber$
Step 4. Multiply the given quantity with the conversion factor that cancels the given unit and leaves the desired unit in the answer.
$2.5 \cancel{~g~\ce{C9H8O4}}\times\frac{1~mol~\ce{C9H8O4}}{180\cancel{~g~\ce{C9H8O4}}}\text { = 0.0193}~mol~\ce{C9H8O4}\nonumber$
Example $9$
How many moles of NaOH are in 10.0 g of NaOH?
Solution
Step 1. Write the given quantity and the desired quantity.
Given: 10.0 g NaOH, Desired: ?mol NaOH
Step 2. Calculate the molar mass of the substance.
22.99 g.mol-1 Na + 16.00 g.mol-1 O + 1.01 g.mol-1 H = 40.00 g.mol-1 NaOH.
Step 3. Write the molar mass of the substance and it's reciprocal as the two conversion factors.
$\frac{40.00 ~g~NaOH}{1~mol ~NaOH}\quad\text { and}\quad ~\frac{1~mol~NaOH}{40.00~g~NaOH}\nonumber$
Step 4. Multiply the given quantity with the conversion factor that cancels the given unit and leaves the desired unit in the answer.
$10.0 \cancel{~g~NaOH}\times\frac{1~mol~NaOH}{40.00\cancel{~g~NaOH}}\text { = 0.250 mol NaOH}\nonumber$ | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/04%3A_Stoichiometry_the_quantification_of_chemical_reactions/4.02%3A_The_mole.txt |
What is a chemical reaction?
A chemical reaction is a combination, separation, or rearrangement of atoms in a substance.
Atoms are healed together in a substance by chemical bonds. The combination makes bonds, separation breaks bonds, and rearrangement breaks some of the old bonds and makes some new bonds in the substances during a chemical reaction. It results in new substances with a different composition of elements than the starting substances. The starting substances are called reactants and the new substances formed are called products.
When a chemical bond is broken energy is required and when a chemical bond is formed energy is released. The amount of energy depended on the chemical bond, but for the same bond, the energy that needs to break is the same energy released to make the bond. Therefore, energy is released in some chemical reactions and absorbed in others.
Examples of chemical reactions are photosynthesis which converts carbon dioxide and water into glucose in the green leaves of plants using energy from sunlight. Digestion of food is a chemical reaction that releases the energy needed for the functioning of living things. The burning of a candle is a chemical reaction that converts the organic compounds in the fule to carbon dioxide, water, and heat energy. Rusting iron is another chemical reaction that converts the element iron to compound iron oxide.
Indications of a chemical reaction
A chemical reaction is usually accompanied by some physical changes that can be observed. The changes include color change, flame, heat, light, the evolution of a gas, formation of a precipitate, etc., as illustrated in Fig. 4.3.1
Representing a chemical reaction- a chemical equation
Chemical equaitron
A chemical equation represents a chemical reaction.
1. The formulas of the reactants are written on the left side separated by a + sign.
2. Formulas of the products are written on the right-side separated by a + sign.
3. An arrow pointing in the direction of products separates the reactants from the products.
Fig. 4.3.2 illustrates the photosynthesis reaction between carbon dioxide ($\ce{CO2}$ and water ($\ce{H2O}$ producing glucose ($\ce{C6H12O6}$ and oxygen ($\ce{O2}$:
$\ce{6CO2 + 6H2O ->[sun light] C6H12O6 + 6O2}\nonumber$
Writing a chemical equation
1. First correct formulas of reactants and products are written, separated by plus signs and arrows.
2. Then the number of species is adjusted by adding a number, called a coefficient, at the beginning of a formula. The coefficients are needed to make atoms of each element equal on both sides of the equation because atoms are neither created nor destroyed in a chemical reaction.
3. Subscripts within the formulas can not be changed as they represent the composition of the substance that is constant.
4. The coefficient is not written if it is one.
For example, Fig. 4.3.2 shows coefficients 6 for carbon dioxide, water, and oxygen, but no coefficient for glucose means the coefficient is actually one for glucose in a balanced chemical equation for photosynthesis reaction.
Symbols in a chemical equation
Sometimes the physical state of the substances is shown by symbols in small brackets next to the formula.
1. The symbols for the physical state are: (s) for solid, (l) for liquid, (g) for gas, and (aq) for a substance dissolved in water. The symbol (aq) stands for aqueous, i.e., water.
2. If a gas evolves, it can be shown by an up-arrow (↑), and if a precipitate forms, it is shown by a down-arrow (↓).
For example, combustion of butane ($ce{C4H10}$ in lighter shown in Fig. 4.3.1. can be represented as:
$\ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H26O(g)}\nonumber$
, where all the reactants and products are in the gas phase. Reaction of NaCl and AgNO3 in the water that results in a precipitate, as shown in Fig. 4.3.1 is:
$\ce{NaCl(aq) + AgNO3(aq) -> AgCl(s) (v) + NaNO3(aq)}\nonumber$
, where NaCl, AgNO3, and NaNO3 are in the water, and AgCl precipitates out as white solid. The reaction of limestone with HCl, shown in Fig. 4.3.1 is:
$\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g) ^}\nonumber$
Reaction conditions, catalysts, or heat can be written above or below the arrow. Heat can also be represented by ∆ symbol, e.g.,
$\ce{CaCO3(s) + CaO(s) ->[\Delta] CO2(g)^}\nonumber$
A two way arrows $\ce{<=>}$ or $\leftrightarrows$ represents a two ways reaction, e.g.,
$\ce{2NO2(g) <=> N2O4(g)}\nonumber$
Balancing a chemical equation
The first step is wright correct formulas of reactants and products separated by plus signs and an arrow. For example, the initial equation for burning hydrogen burns in oxygen and producing water would be:
$\ce{H2 + O2 -> H2O}\nonumber$
Note that hydrogen is written as H2 (not H) and oxygen as O2 (not O) because these reactants usually exist as molecules, not as atoms.
The next step is to add coefficients to balance atoms of each element on the two sides of the equation. For example, in the above equation hydrogen is balanced but oxygen is not. Balance oxygen by changing the coefficient of water from 1 to 2:
$\ce{H2 + O2 -> 2H2O}\nonumber$
Caution
Subscripts in the formulae can not be changed, as they are constant. For example, if O2 is changed to O in the above equation to balance the oxygen, it is incorrect as O2 is molecular oxygen which is a different chemical than atomic oxygen O.
The coefficient is a multiplier of each subscript in the formula, i.e., in 2H2O there are 2x2 = 4 hydrogen and 2x1 =2 oxygen. Now look for the other elements again: note that hydrogen atoms have changed to 4 on the right side. To balance hydrogen, change the coefficient of H2 from 1 to 2:
$\ce{2H2 + O2 -> 2H2O}\nonumber$
Check again: Now, the atoms of each element are the same on both sides, i.e., the equation is balanced, as illustrated in Fig. 4.3.3.
Rules to balance a chemical equation
The method described above to balance a chemical equation is called the hit and trial method. There are no hard and fast rules in this method. The general guidelines are:
1. if an element occurs only in one reactant and one product, balance it first,
2. if there is a free element in the reactants or products, balanced it last,
3. if there is a polyatomic ion in reactants and products, balance the number of polyatomic ions as a unit, i.e., add coefficient to but do not change the subscripts in the polyatomic ion or of any formula.
4. in some cases, a fractional coefficient is needed to balance the equation, in that case first use the fractional coefficient and when the equation is balanced, remove the fraction by multiplying all the coefficients with a common multiplier,
5. if the set of coefficients in the balanced equation is not in the simplest whole-number ratio, the equation is still balanced, but it is recommended to convert the set of coefficients to the simplest whole-number ratio.
The following examples explain the guidelines.
Example $1$
Combustion of propane (C3H8) is used in gas welding, as shown in Fig. 4.3.4. The propane reacts with oxygen and produces carbon dioxide and water. Write and balance the chemical equation of the reaction?
Solution
Start with writing the correct formulae of reactants and products in a chemical equation form:
$\ce{C3H8 + O2 -> CO2 + H2O}\nonumber$
According to rule 1, look at carbon or hydrogen first. Both are not balanced. Balance the carbon by changing the coefficient of CO2 from 1 to 3:
$\ce{C3H8 + O2 -> 3CO2 + H2O}\nonumber$
Balance the hydrogen by changing the coefficient of H2O from 1 to 4:
$\ce{C3H8 + O2 -> 3CO2 + 4H2O}\nonumber$
Now, look at oxygen (rule 2) and balance it by changing the coefficient of O2 from 1 to 5:
$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}\nonumber$
The last equation is balanced, as illustrated in Fig. 4.3.5.
Example $2$
The reaction between Aluminium metal powder and iron oxide, called thermite reaction, is highly exothermic that melts the ion. The thermite reaction is used for railroad welding as shown in Fig. 4.3.6. Wright and balance the reaction equation.
Solution
Start with writing the correct formulae of reactants and products in a chemical equation form:
$\ce{Al + Fe2O3 -> Fe + Al2O3}\nonumber$
Rule 1 does not apply as all elements are in one reactant and one product. According to rule 2, look at either one of the compounds. Balance aluminum by changing the coefficient in the eactant from 1 to 2:
$\ce{2Al + Fe2O3 -> Fe + Al2O3}\nonumber$
Now balance iron by change its coefficient from 1 to 2 in the product:
$\ce{2Al + Fe2O3 -> 2Fe + Al2O3}\nonumber$
All atoms are balanced at this stage and coefficients are in the simplest whole-number ratio. So the equation is balanced.
Example $3$
Methanol (CH4O) reacts with oxygen and produces carbon dioxide and water. Write and balance the chemical reaction equation?
Solution
The starting equation with correct formulas is:
$\ce{CH4O + O2 -> CO2 + H2O}\nonumber$
According to rule 1, look at carbon and hydrogen first. Carbon is already balanced. There are 4 hydrogens on the left but only two on the right side, so balance them by changing the coefficient of water from 1 to 2:
$\ce{CH4O + O2 -> CO2 + 2H2O}\nonumber$
Now, look at oxygen (rule 2). There are three oxygen atoms on the right (one in methanol and two in free element oxygen) but four oxygen on the right, it is not balanced. Changing the coefficient of CH4O from 1 to 2 balances the oxygen, but carbon and hydrogen go out of balance. Changing the coefficient of O2 from 1 to 2 does not balance the oxygen. A fractional coefficient 3/2 for O2 works at this stage (rule 4):
$\ce{CH4O + 3/2 O2 -> CO2 + 2H2O}\nonumber$
All elements are balanced, i.e., the equation is balanced, but, according to rule 5, it is better to remove the fraction by multiplying each coefficient in the equation with 2:
$\ce{2CH4O + 3O2 -> 2CO2 + 4H2O}\nonumber$
The set of coefficients in the balanced equation is already in the simplest whole-number ratio, so no further step is needed in this case.
Example $4$
Aluminum reacts sulfuric acid to produce aluminum sulfate and hydrogen gas. Wright a balanced chemical equation for the reaction.
Solution
The starting equation with correct formulas of reactants and products is:
$\ce{Al + H2SO4 -> Al2(SO4)3 + H2}\nonumber$
All elements occur in one reactant and one product, so rule 1 do not apply. According to rule 2, leave aluminum till the end and, according to rule 3, balance the polyatomic ion (SO42-) as a unit by changing the coefficient of sulfuric acid from 1 to 3:
$\ce{Al + 3H2SO4 -> Al2(SO4)3 + H2}\nonumber$
Now balance the hydrogen by changing it coefficient from 1 to 3:
$\ce{Al + 3H2SO4 -> Al2(SO4)3 + 3H2}\nonumber$
Finally balance aluminum by changing its coefficient from 1 to 2:
$\ce{2Al + 3H2SO4 -> Al2(SO4)3 + 3H2}\nonumber$
Double-check at the end –All elements are balanced, and all the coefficients are in the simplest whole-number ratio. No further step is needed. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/04%3A_Stoichiometry_the_quantification_of_chemical_reactions/4.03%3A_Chemical_reaction.txt |
General types of chemical reactions
There are several ways to classify chemical reactions. The general types of chemical reactions fall in the categories of combination, decomposition, replacement, and combustion reactions, as illustrated in Fig. 4.4.1.
Combination reactions
A compound is synthesized or formed from two or more substances, e.g.:
$\ce{C + O2 -> CO2}\nonumber$
$\ce{2H2 + O2 -> 2H2O}\nonumber$
$\ce{2Mg + O2 -> 2MgO}\nonumber$
$\ce{2Na + Cl2 -> 2NaCl}\nonumber$
$\ce{CaO + CO2 -> CaCO3}\nonumber$
Fig. 4.4.2 shows an example of a hydrogen with oxygen combination reaction that is being developed for use as a fuel in the future.
Decomposition reactions
The decomposition reactions are the reverse of the combination reaction, i.e., one compound splits apart into two or more substances, usually by heating, e.g.:
$\ce{H2CO3 -> H2O + CO2}\nonumber$
$\ce{CaCO3 ->[\Delta] CaO + CO2}\nonumber$
$\ce{2KClO3 ->[\Delta] 2KCl + 3O2}\nonumber$
$\ce{2H2O (l) ->[Electrolyisis] 2H2(g) + O2(g)}\nonumber$
Fig. 4.4.3 illustrates the last reaction, i.e., decomposition of water through electrolysis.
Replacement or substitution reactions
There are two sub-classes of this category of reactions, i.e., single replacement and double replacement reaction.
Single replacement reactions involve one substance replacing a part of another, e.g.:
$\ce{Zn(s) + CuCl2(aq) -> ZnCl2(aq) + Cu(s)}\nonumber$
$\ce{2Al(s) + 6HCl(aq) -> AlCl3(aq) + 3H2(g)}\nonumber$
Fig. 4.4.4 shows an example of a single replacement reaction of magnesium resulting in hydrogen gas formation.
Double replacement reactions or metathesis involve the mutual exchange of partners between two substances, e.g. the following precipitation reactions:
$\ce{NaCl(aq) + AgNO3(aq) -> AgCl(s)(v) + NaNO3(aq)}\nonumber$
$\ce{Na2CO3(aq) + CaCl2(aq) -> CaCO3(s)(v) + 2NaCl(aq)}\nonumber$
Combustion reactions
Combustion is a reaction of a substance with oxygen, often with the formation of flame and release of much heat, e.g.:
$\ce{C8H16 + 12O2 -> 8CO2 + 8H2O + Heat}\nonumber$
$\ce{C + O2 -> CO2 + Heat}\nonumber$
$\ce{2H2 + O2 -> 2H2O + Heat}\nonumber$
$\ce{2Mg + O2 -> 2MgO + Heat}\nonumber$
Fig. 4.4.5 shows the above reaction, i.e., combustion of Mg in air.
Usually, combustion is considered as the reaction of a substance containing carbon and hydrogen with oxygen resulting in carbon dioxide, water, flame, and heat, e.g., burning methane on a kitchen stove:
$\ce{CH4 + 3O2 -> CO2 + 2H2O + Heat}\nonumber$
Classification of chemical reaction
The chemical reactions are generally classified based on what exchanges during the reaction. These include;
1. the transfer of electrons in oxidation-reduction reactions,
2. transfer of protons in acid-base reactions, and
3. a part of reactants mutually exchanges in precipitation reactions, as described below.
Oxidation-reduction reactions
The oxidation-reduction or redox reaction involves the exchange of electrons. For example, reactions between a metal and nonmetal involve the transfer of electrons from the metal to the nonmetal forming an ionic bond, as shown in Fig. 4.4.6.
Acid-base reactions
The acid-base reactions involve the transfer of protons from an acid to a base, as shown in Fig. 4.4.7.
Precipitation reactions
These are double displacement reactions in water that results in the precipitation of one of the products, as shown in Fig. 4.4.8.
The precipitation reactions and the acid-base reactions are described in the later chapters. The oxidation-reduction reactions are discussed in the following section. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/04%3A_Stoichiometry_the_quantification_of_chemical_reactions/4.04%3A_Patterns_of_chemical_reactions.txt |
What are Oxidation and Reduction?
The oxidation-reduction is defined in three ways:
1. Oxidation is the loss of electrons and reduction is the gain of electrons. The word OIL RIG helps in remembering this definition.
2. The addition of oxygen is oxidation, and the removal of oxygen is a reduction.
3. The removal of hydrogen is oxidation, and the addition of hydrogen is a reduction.
The oxidation-reduction or in short redox reaction is one of the most common types of chemical reactions happening in and around us. For example, rusting of metals, photosynthesis, digestion of food, and combustion of fuels are redox reactions.
Oxidation-half and reduction-half reactions
The oxidation-reduction reactions happen in a pair because one thing loses electrons and is oxidized; the other thing gains the electrons and is reduced. That is why it is commonly called redox reaction, where red- represents reduction and –ox represents oxidation.
The green patina on the statue of liberty, shown in Fig. 4.5.1, is a result of the oxidation of copper:
$\ce{2Cu(s) + O2(g) -> CuO(s)}\nonumber$
The transfer of electrons becomes apparent when the reaction is split into the oxidation-half and reduction-half. The oxidation-half is:
$\ce{Cu -> Cu^2+ + 2e^-}\nonumber$
copper lost electrtons, so copper is oxides. The reduction half is:
$\ce{O2 + 4e^- -> 2O^2-}\nonumber$
Oxygen gained electrons, so oxygen is reduced.
Oxidizing agent and reducing agent
• The substance that oxidizes another substance is an oxidizing agent. Oxygen is an oxidizing agent in the above reaction, as it oxidizes copper.
• The substance that reduces another substance is a reducing agent. Copper is a reducing agent in the above reaction, as it reduces oxygen.
The electrons lost in an oxidation-half must be equal to the electrons gained by the accompanying reduction-half. Multiplying the oxidation-half with 2 makes the electrons lost equal to the electrons gained in the reduction half in the above reaction. Then adding the oxidation and the reduction half gives the overall reaction:
$\ce{2Cu -> 2Cu^2+ + \cancel{4e^-}}\nonumber$
$\ce{O2 + \cancel{4e^-} -> 2O^2-}\nonumber$
$\text {Overall reaction: } \ce{2Cu(s) + O2(g) -> CuO(s)}\nonumber$
Manipulating a chemical equation
The chemical equations can be manipulated like algebraic equations, i.e., they can be multiplied or divided by a constant, added, and subtracted, as demonstrated in the example of the copper redox-half reactions above. Note that electrons on one side of the equation canceled the electrons on the other side of the equation during the addition operation.
A silver color zinc strip dipped in copper nitrate solution becomes coated with a layer of reddish color copper, as shown in Fig. 4.5.2. The molecular equation of the reaction, that shows formula units of compounds in the reactants and products, is:
$\ce{Zn(s) + Cu(NO3)2(aq) -> Cu(s) + Zn(NO3)2(aq)}\nonumber$
The complete ionic equation for the reaction is obtained by showing the dissolved ionic compounds as ions, e.g.:
$\ce{Zn(s) + Cu^2+ (aq) + 2NO3^{-} (aq) -> Cu(s) + Zn^2+ (aq) + 2NO3^- (aq)}\nonumber$
The oxidation-half of the reaction is:
$\ce{Zn(s) -> Zn^2+ (aq) + 2e^-}\nonumber$
The reduction-half of the reaction is:
$\ce{Cu^2+ (aq) + 2e{-} -> Cu(s)}\nonumber$
The net ionic equation that is the addition of the oxidation-half and the reduction-half is:
$\ce{Zn(s) + Cu^2+ (aq) -> Cu(s) + Zn^2+ (aq)}\nonumber$
Note that NO3- was on both sides of the complete ionic equation and has been canceled out in the net ionic equation:
$\ce{Zn(s) + Cu^2+ (aq) + \cancel{2NO3^{-} (aq)} -> Cu(s) + Zn^2+ (aq) + \cancel{2NO3^- (aq)}}\nonumber$
Ions that do not participate in the chemical reaction are called spectator ions, and they appear on both sides of the molecular equation, like NO3- in this case.
Biological oxidation and reduction reactions
In the redox reactions involving metal species, the transfer of electrons is usually evident through the oxidation-half and the reduction-half reactions, as in the above examples. In the organic and biochemical redox reactions, the transfer of electrons is usually not so obvious, but the transfer of hydrogen or oxygen is usually apparent. For example, the metabolism of methanol $\ce{H3C-OH}$ starts with oxidation through the loss of hydrogen:
$\ce{H3C-OH -> H2C=O + 2H}\nonumber$
Metanol is oxidized to formaldehyde $\ce{H2C=O}$. Formaldehyde oxidizes further by gaining oxygen:
$\ce{H2C=O -> HCOOH}\nonumber$
Finally, the formic acid (HCOOH) is oxidized by gaining oxygen and forming carbon dioxide and water:
$\ce{2HCOOH -> 2O=C=O + 2H2O}\nonumber$
The increase in the C-O bond from a single bond in methanol to two double bonds (four C-O bonds) in carbon dioxide is a clear indication of the oxidation of the carbon.
Removal of hydrogen is also oxidation, e.g., hydroxymalonate is oxidized to oxomalonate by an enzyme hydroxymalonate dehydrogenase:
The reverse of these, i.e., removal of oxygen and addition of hydrogen is reduction. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/04%3A_Stoichiometry_the_quantification_of_chemical_reactions/4.05%3A_Oxidation-reduction_reactions.txt |
Energetics deals with the energies involved in chemical reactions. There are two types of energies: the first is related to thermodynamics which is the energy released or absorbed when the reactants convert to the products, and the second is related to the kinetics of the reaction, i.e., the energy that the reacting molecules must possess to surpass the energy barrier for conversion to the products.
The progress of the reaction versus energy graph is shown in Fig. 4.7.1.
The starting point shows the energy level of reactants. Remember that every substance has internal energy which is manifested as temperature. The molecules are moving around due to the internal energy and collied with other molecules. The collision between the reactants is usually the first requirement for the reaction to proceed.
The F- ion in this example collides with the target H3C-I from the side opposite to the C-I bond. The collision at an angle other than 180o to the C-I bond is less effective or ineffective in initiating the C-I bond breaking. So, the proper orientation of the reacting molecules at the time of the collision is the second requirement.
After a properly oriented collision, the C-I bond starts breaking, the F-C bond starts making, and the energy of the system rises to a maximum level where the bond breaking and making are about halfway. The specie at the maxima of the energy curve is called activated complex or a transition state. The difference in energy between the reactants and the transition state is called activation energy (∆G). The reactants must have enough kinetic energy to surpass the activation energy barrier, which is the third requirement for the reaction to happen. A summary of the basic requirements of a chemical reaction to happen is the following.
Requirement of a chemical reaction
1. The reacting molecules collide with each other –the more frequent the collision, the faster the reaction.
2. The molecules should have proper orientation at the time of the collision –this factor varies from reaction to reaction.
3. The colliding molecules must have energy, in addition to the potential chemical energy, equal to or higher than the energy of activation –the lower the activation energy, the faster the reaction.
The activated complex forms if the basic requirements are fulfilled. The activated complex rolls down the hill on the energy scale and settles at the energy level of the products where the old bonds are entirely broken, and the new bonds are fully formed. The graph of the reaction progress versus the energy, as in Fig. 4.7.1 is called the reaction coordinate diagram.
Exothermic and endothermic reactions
The energy absorbed or released in the form of heat at constant pressure conditions is called enthalpy change (∆H), which is almost the same as the internal energy absorbed or released (∆G) in most of the reactions.
A chemical reaction that releases heat is exothermic. A chemical reaction that absorbs heat is an endothermic reaction.
Bond-forming is always exothermic, and bond-breaking is the opposite, i.e., endothermic, as illustrated in Fig. 4.7.2.
In a chemical reaction, some bonds break, and some bonds form. If the bond-breaking absorbs less heat than the heat released in the new bond making, the reaction is exothermic. The products are lower in energy than the reactants, and the ∆H is negative for exothermic reactions. The opposite is true for an endothermic reaction, i.e., the bond-breaking absorbs more heat than the bond-forming; the products are higher in energy than the reactants, and ∆H is positive, as illustrated in Fig. 4.8.3.
An example of an exothermic reaction is the combustion of methane that releases heat used for cooking.
$\ce{CH4 + 2O2 -> CO2 + 2H2O \quad\Delta H^{o} = -891 kJ}\nonumber$
An example of an endothermic reaction is photosynthesis.
$\ce{6CO2 + 6H2O -> C6H12O6 + 6O2 \quad\Delta H^{o} = +2800 kJ}\nonumber$
Note that the sign of ∆H is –ve for exothermic and +ve for an endothermic reaction. Sometimes, the energy is shown as a product for exothermic reaction, and as reactant for edothermic reaction. For example, the above two creations may be written as:
$\ce{CH4 + 2O2 -> CO2 + 2H2O + 891 kJ}\nonumber$
$\ce{6CO2 + 6H2O + 2800 kJ -> C6H12O6 + 6O2}\nonumber$
The instant hot or cold packs used in hospitals are based on the heat of dissolution of salts in water, as shown in Fig. 4.7.4. The salt and the water are in separate pockets in the pack. When the seal is ruptured, the salt dissolves in water and releases or absorbs heat. The dissolution of CaCl2 salt in water is an exothermic process that is the source of heat in a hot pack:
$\ce{CaCl2(s) -> CaCl2(aq) \quad\Delta H^{o} = -82 kJ}\nonumber$
The dissolution of NH4NO3 salt in water is an endothermic process that is utilized to absorb heat in a cold-pack.
$\ce{NH4NO3(s) -> NH4NO3(aq) \quad\Delta H^{o} = -26 kJ}\nonumber$
Factors that affect the rate of a chemical reaction
The rate of a chemical reaction, i.e., the amount of a reactant consumed or the amount of a product formed per unit time, is dependent on the energy of activation of the reaction —the higher the activation energy, the slower the rate of a reaction and vice versa. A reaction requires collision between the reactant molecules, with proper orientation, and sufficient energy to surpass the activation energy barrier.
Any factor that increases the rate of collisions, enhances the proper orientation or increases the kinetic energy of the molecules causes an increase in the rate of reaction. The factors include concentration, temperature, and catalysts.
Effect of concentration of reactants
The higher the concentration, the more frequent the collisions, and the faster the reaction, as illustrated in Fig. 4.7.5.
In the breathing process, oxygen (O2) binds with hemoglobin (Hb) in the lungs.
Hb(aq) + O2(g) → HbO2(aq)
Patients having breathing problems are given breathing masks with a higher concentration of oxygen than in the atmosphere to increase the rate of oxygen binding with the hemoglobin.
Effect of temperature
The kinetic energy of molecules at a given temperature follows Boltzman distribution, as illustrated in Fig. 4.8.6. Increased temperature increases the average kinetic energy of the molecules which increases the fraction of molecules with more than the activation energy.
An increase In temperature increases the rate of chemical reactions—generally, a 10 oC increase in temperature doubles the rate of a chemical reaction.
Some practical examples of the use of this principle are the following. Storing foods in refrigerators at lower temperatures decreases the rate of reactions resulting in longer life of the foods. Cooking foods in a pressure cooker increases the temperature resulting in faster cooking than in an open pan. In some cardiac surgeries, the body temperature is lowered to 28 oC to decrease the rate of metabolism which decreases the oxygen demand so that the heart may be stopped temporarily for the surgery.
Effect of catalysts and enzymes
The catalysts and enzymes increase the rate of reaction by decreasing the energy of activation of the reaction through an alternate route, as illustrated in Fig. 4.7.7.
The catalysts and the enzymes do not consume in the reaction -they regenerate and repeat the action. Enzymes are the catalysts in biochemical reactions. Enzymes also increase the rate of reaction by binding with the reactants and properly orienting them for the reaction. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/04%3A_Stoichiometry_the_quantification_of_chemical_reactions/4.06%3A_Energetics_of_chemical_reactions.txt |
Conversion factors from a chemical equation
Stoichiometry pronounced as “stɔɪkiˈɒmɪtri” is the calculation of the amount of reactants and products in a chemical reaction. It is based on the fact that a balanced chemical equation is also a set of mole-to-mole equalities between the reactants and the products. Each equality gives two conversion factors that allow calculating the mole of one substance from the given mole of any other substance in the equation.
Fig. 4.6.1 lists the chemical equation for photosynthesis reaction, the mole-to-mole equalities from the equation, and the two conversion factors from each of the equality, as an example. The conversion factors are used to calculate the unknown quantity in the mole from the known quantity in the mole of any other reactant or product in the same chemical equation, as explained in the following examples.
Mole-to-mole conversion from a chemical equation
Calculation of the mole of the desired substance from the given mole of any reactant or product needs one conversion factor. The steps are:
1. write the given quantity and the desired quantity,
2. write the balanced chemical equation,
3. write the equality between the given and the desired substances,
4. right the conversion factor that has the given substance in the denominator and the desired substance in the numerator,
5. multiply the given quantity with the conversion factor. Double-check to make sure that it cancels the given substance and leaves the desired substance in the answer.
Example $1$
Calculate the moles of glucose produced from 3.0 moles of carbon dioxide in the photosynthesis reaction?
Solution
i. Given: 3.0 mole CO2, Desired: ? moles of C6H12O6
ii. Chemical equation: $\ce{6CO2 + 6H2O -> C6H12O6 + 6O2}$
iii. The desired equality: 6 mol CO2 = 1 mol C6H12O6,
iv. Desired conversion factor: $\frac{1 \mathrm{~mol} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{6 \mathrm{~mol} \mathrm{~CO_2}}$
v. Calculations:$3.0 \cancel{\mathrm {~mol~CO_2}} \times \frac{1 \mathrm{~mol} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{6 \cancel{\mathrm{~mol} \mathrm{~CO_2}}}=0.50 \mathrm{~mol} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\nonumber$
Example $2$
Magnesium reacts with HCl by this reaction: $\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)^}$. Calculate the moles of Mg needed to produce 3 moles of H2?
Solution
i. Given: 3 mol of H2. Desired: ? Moles of Mg.
ii. The chemical equation is given.
iii. The desired equality: 1 mol Mg = 1 mol H2.
iv. The desired conversion factor: $\frac{1 ~mol~Mg}{1 ~mol{~H}_{2}}$
v. Calculation: $3 \cancel{\mathrm{~mol} \mathrm{~H}_{2}} \times \frac{1 \mathrm{~mol} \mathrm{~Mg}}{1 \cancel{\mathrm{~mol} \mathrm{~H}_{2}}}=3 \operatorname{~mol} ~M g\nonumber$
Mole-to-mass conversion from a chemical equation
Calculations described in the previous two examples show calculating moles of the desired substance from the moles of the given substance using a conversion factor from a chemical equation. The molar mass is a conversion factor from a mole-to-gram of the substance. So, add a molar mass of the desired substance as a second conversion factor, as explained in the following examples. Double-check that all the units cancel out, leaving the mass unit of the desired substance in the final answer.
Example $3$
Calculate grams of AgCl precipitate formed from 2.0 moles of CaCl2 consumed in the following reaction: $\ce{2AgNO3 + CaCl2 -> Ca(NO3)2 + 2AgCl(s)(v)}\nonumber$
Solution
i. Given: 2.0 moles of CaCl2. Desired: ? g AgCl
ii. Molar mass of AgCl = 1x107.87 g Ag.mol-1 + 1x35.45 g C.mol-1 = 143.3 g AgCl.mol-1
iii. The chemical equation is given.
iv. The desired equality: 1 mol CaCl2 = 2 mol AgCl, and 1 mol AgCl =143.3 g AgCl.
v. The desired conversion factors: $\frac{2 \mathrm{~mol} \mathrm{~AgCl}}{1 \mathrm{~mol} \mathrm{~CaCl}_{2}} \quad\text { and }\quad \frac{143.3 \mathrm{~g} \mathrm{~AgCl}}{1 \mathrm{~mol} \mathrm{~AgCl}}\nonumber$
vi. Calculation: $2.0\cancel{\mathrm{~mol}\mathrm{~CaCl}_{2}}\times\frac{2 \cancel{\mathrm{~mol} \mathrm{~AgCl}}}{1 \cancel{\mathrm{~mol} \mathrm{~CaCl}_{2}}} \times\frac{143.3 \mathrm{~g} \mathrm{~AgCl}}{1 \cancel{\mathrm{~mol} \mathrm{~AgCl}}}=573.3 \mathrm{~g} \mathrm{~AgCl}\nonumber$
Example $4$
How many grams of carbon dioxide are needed to react with 2 moles of water in the photosynthesis reaction?
Solution
i. Given: 2.0 moles of H2O Desired: ? g CO2.
ii. Molar mass of CO2 = 1x12.011 g C.mol-1 + 2x15.999 g O.mol-1 = 44.009 g CO2.mol-1
iii. The chemical equation: : 6CO2 + 6 H2O → C6H12O6 + 6O2.
iv. The desired equality: 6 mol CO2 = 6 mol H2O, and 1 mol CO2 = = 44.009 g CO2.
v. The desired conversion factors: $\frac{6 \mathrm{~mol} \mathrm{~CO}_{2}}{6 \mathrm{~mol} \mathrm{~H}_{2}}\mathrm{O}\quad \text { and }\quad \frac{44.009 \mathrm{~g} \mathrm{~CO}_{2}}{1 \mathrm{~mol} \mathrm{~CO}_{2}}\nonumber$
vi. Calculation: $2.0 \cancel{\mathrm{~mol} \mathrm{~} \mathrm{H}_{2} \mathrm{O}} \times \frac{6 \cancel{\mathrm{~mol} \mathrm{~CO}_{2}}}{6 \cancel{\mathrm{~mol} \mathrm{~H}_{2} \mathrm{O}}} \times \frac{44.009 \mathrm{~g} \mathrm{~CO}_{2}}{1 \cancel{\mathrm{~mol} \mathrm{~CO}_{2}}}=88 \mathrm{~g} \mathrm{~CO}{ }_{2}\nonumber$
Mass-to-mass conversion from a chemical equation
A chemical equation gives a mole-to-mole conversion factor. If the given substance is in grams and the desired substance is also in grams, then two additional conversion factors based on the molar masses are needed. That is, the following conversions are needed:
Mass of given substance⇒mole of the given substance⇒mole of the desired substance⇒grams of the desired substance.
The reciprocal molar mass of the given substance is the first conversion factor, the mole to mole conversion factor from the chemical equation is the second conversion factor, and the molar mass of the desired substance is the third conversion factor needed. Make sure that each conversion factor cancels the denominator unit of its multiplier to the right, and the desired unit is left in the answer. The following examples explain these calculations.
Example $5$
How many grams of Mg are needed to produce 1.01 g of H2 gas in this reaction: $\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)^}\nonumber$
Solution
i. Given: 1.01 g H2 Desired: ? g Mg.
ii. Molar mass of H2 = 2 x1.008 g H.mol-1 = 2.016 g H2.mol-1, and molar mass = 24.305 g Mg.mol-1.
iii. The chemical equation is given in the problem.
iv. The desired equalities : 1 mol Mg = 1 mol H2, 1 mol H2 = 2.016g H2, 1 mol Mg = 24.305 g Mg
v. Calculate by multiplying the given quantity consecutively with the three desired conversion factors from the equalities:
$1.01 \cancel{\mathrm{~g ~H_{2}}} \times \frac{1 \cancel{\mathrm{~mol} \mathrm{~H}_{2}}}{2.016 \cancel{\mathrm{~g} \mathrm{~H}_{2}}} \times \frac{1 \cancel{\mathrm{~mol} \mathrm{~Mg}}}{1 \cancel{\mathrm{~mol} \mathrm{~H}_{2}}} \times \frac{24.305 \mathrm{~g} \mathrm{~Mg}}{1 \cancel{\mathrm{~mol} \mathrm{~Mg}}}=11.9 \mathrm{~g~Mg}\nonumber$
Example $6$
How many grams of glucose are produced if 22.0 g of carbon dioxide is consumed in the photosynthesis reaction?
Solution
i. Given: 22.0 g CO2 Desired: ? g C6H12O6.
ii. Molar masses: of CO2 = 1x12.011 g C.mol-1 + 2x15.999 g O.mol-1 = 44.009 g CO2.mol-1, and molar mass of C6H12O6 = 6x12.011 g C.mol-1 + 12x1.008 g H.mol-1 + 6x15.999 g O.mol-1 = 180.156 g C6H12O6.mol-1 .
iii. The chemical equation: $\ce{6CO2 + 6H2O -> C6H12O6 + 6O2}$
iv. The desired equalities: 6 mol CO2 = 1 mol C6H12O6, 1 mol CO2 = 44.009 g CO2, 1 mol C6H12O6 = 180.156 g C6H12O6.
v. Calculate by multiplying the given quantity consecutively with the three desired conversion factors from the equalities:
$22.0 \cancel{\mathrm{~g} \mathrm{~CO}_{2}} \times \frac{1 \cancel{\mathrm{~mol} \mathrm{~CO}_{2}}}{44.009 \cancel{\mathrm{~g} \mathrm{~CO}_{2}}} \times \frac{1 \cancel{\mathrm{~mol} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}}{6 \cancel{\mathrm{~mol} \mathrm{~CO}_{2}}} \times \frac{180.156 \mathrm{~g} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{1 \cancel{\mathrm{~mol} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}}=15.0 \mathrm{~g} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\nonumber$
Fig. 4.6.2 illustrates what each conversion factor does in the case of the above example number 4.6.5 | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/04%3A_Stoichiometry_the_quantification_of_chemical_reactions/4.07%3A_Stoichiometric_calculations.txt |
• 5.1: Introduction to solution
The solution, i.e., a homogeneous mixture is described along with the concepts of solvent, solute, and types of solution. Water as a solvent and its dissolution mechanism are described.
• 5.2: Solubility
Solubility, i.e., the maximum amount of solute that can dissolve in a given amount of solvent, is described along with the terminologies. Solubility guidelines that help in predicting the solubility of ionic compounds in water are listed.
• 5.3: Electrolytes
Electrolytes, i.e., the compounds that dissolve in water and dissociated partially or fully into ions are described. Equivalent i.e., a mole of charge is also introduced along with body fluids and their concentration.
• 5.4: Concentration of solutions
Concentration, i.e., the amount of solute in a given amount of solution, is described along with its different units. Preparation of and dilution of the solution are also presented.
• 5.5: Osmosis
Osmosis, i.e., the passage of water and small molecules across a semipermeable member with a net flow towards a more concentrated solution is described. The role of osmosis in water purification, in living systems like excretion of wastes by kidneys, are also described.
05: Solutions
Solutions are all around us, e.g., air, seawater, body fluids, metal alloys are solutions. Fig. 5.1.1 illustrates that air is a mixture of nitrogen, oxygen, carbon dioxide, and some other gases; Fig. 5.1.2 illustrates that seawater is a mixture of water, chloride, sodium, sulfate, magnesium, and some other ions, and Fig. 5.1.3. illustrates that about 60% of the human body is composed of solutions called body fluids.
What is a solution?
• A solution is a homogeneous mixture of two or more pure substances.
• The substance that is in a large amount in the solution is called the solvent.
• The substance that is in smaller amounts in a solution is called the solute.
For example, the air is a solution in which nitrogen is the solvent, and water is the solvent in seawater and body fluids. Oxygen, carbon dioxide, and water vapors are solutes in the air; and sodium, chloride, sulfate, magnesium, and some other ions are solutes in seawater.
Types of solution
The solutions are generally classified in two ways: i) based on the physical state of the solution and the solute, and ii) based on the particle size of the solute.
Types of solution based on the physical state of the solution and the solute
The solutions can be classified based on the physical state of the solution, solvent, and solute. For example, the air is gas in a gas solution; carbonated water is a gas in a liquid solution; vinegar is a liquid in a liquid solution; metal alloys are solid in solid solutions. Table 5.1.1 lists the major types of solutions, solvents, and primary solutes in them.
Table 1: Examples of main types of solutions and solvent and major solute in them.
Type Example solvent Primary solute
Gas in gas Air Nitrogen Oxygen
Gas in liquid Carbonated water Water Carbon dioxide
Liquid in liquid Vinegar Water Acetic acid
Solid in liquid Seawater Water Sodium chloride
Solid in solid Brass Copper Zinc
Types of solutions based on the particle size of the solute
Solution
A solution is a homogeneous mixture comprising smaller component/s called solute/s of small molecules or ions comparable in size to the molecules of a larger component called the solvent.
For example, NaCl dissolved in water is a solution. The solute is almost uniformly distributed in the solvent, making a homogeneous mixture. The solute does not separate by filtration or by a semipermeable membrane but can be separated by some other physical process. For example, the distillation process separates a solid in a liquid or a liquid into a liquid solution. The solution is transparent, though it may be colored. A light passing through a solution is not visible, as shown in Fig. 5.1.4.
Suspension
A suspension is a heterogeneous mixture of solvent and solute particles of larger than 10,000 Å.
For example, muddy water is a suspension. If the suspension is allowed to stand, the suspended particles settle down and separate. The suspended particles can be filtered out. Some medicines, e.g., milk of magnesia, are suspensions. It is instructed to shake just before administering medicine to re-suspend the settled suspension.
Colloid
A colloid falls between a solution and a suspension. The colloidal particles are larger molecules like proteins or groups of molecules or ions.
Unlike a suspension, the colloids usually do not settle if allowed to stand. The colloidal particles can not be filtered but can be separated by a semipermeable membrane. When a light beam passes through a colloid, it scatters by the colloid particles, called the Tyndall effect, and becomes visible, as shown in Fig. 5.1.4.
Examples of colloids include:
1. fog and clouds that are liquid water droplets dispersed in air;
2. smoke that is solid carbon particles dispersed in air;
3. whipped cream that is air dispersed in a liquid;
4. styrofoam is a gas dispersed in a solid; and
5. ager medium that is liquid dispersed in a solid medium.
Water –a universal solvent
Water (H2O) is an essential substance for life. It covers more than 70% of the earth’s surface (Fig. 5.1.5), and it comprises more than 60% of the human body (5.1.3). In addition to being the most abundant solvent, water is a universal solvent because it is a polar molecule with a partial negative charge on oxygen and a partial positive charge on hydrogen atoms as shown in Fig. 5.1.6. The polarity of water molecules allows them to interact with other water molecules as well as with other polar compounds through dipole-dipole interactions and with other ions through ion-dipole interactions. These interactions help to dissolve a lot of polar and ionic compounds that are in and around us.
How does water dissolves polar and ionic compounds?
Water molecules establish electrostatic interaction, called hydrogen bonding, through the partial +ve end of one molecule with a partial –ve end of a neighboring molecule. These interactions impart unique properties to water, like its relatively higher boiling point and melting point compared to other substances of similar molecular weight. Other polar substances have similar interactions, e.g., ethanol has hydrogen bonding similar to water as illustrated in Fig. 5.1.7.
Dissolution of polar substance in water
When water mixes with other polar substances, like ethanol, some of the hydrogen bonding between water molecules replace with similar hydrogen bonding with ethanol molecules. Since the electrostatic potential energy is similar, the natural tendency to go towards more dispersion drives the dispersion of ethanol molecules uniformly in water resulting in the solution.
Ionic compounds are held together by electrostatic forces between opposite ions, i.e., ionic bonds. When an ionic compound is added to water, water molecules surround the cation and establish ion-dipole interaction by orienting their partial -ve end to the cation. Similarly, water molecules establish ion-dipole interaction with anions by orienting their partial +ev end towards the anion, as illustrated in Fig. 5.1.8
Dissolution of ionic compounds in water
The ion-dipole interactions, along with nature’s tendency to disperse the particles, are usually strong enough to overcome the ionic bonds, dissociate the compounds into ions, and disperse them almost uniformly in the water.
The separation of the cations from the anions of the ionic compound is called dissociation.
The formation of a layer of water molecules around ions, driven by ion-dipole interactions, is called hydration.
Non-polar substances, like vegetable oil or gasoline, do not dissolve in water. The molecules in non-polar substances have only London dispersion forces. They easily dissolve in non-polar solvents like hexane or carbon tetrachloride that have similar London dispersion forces among their molecules.
The fact that ionic and polar substances dissolve in polar solvents and non-polar substances dissolve in non-polar solvents of similar intermolecular interactions is called “like dissolves like.” | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/05%3A_Solutions/5.01%3A_Introduction_to_solution.txt |
Solubility and its related terminologies
The majority of solutes do not dissolve in water or other solvents in all proportions.
What is solubility
The maximum proportion of the solute that can dissolve in a given amount of the solvent, usually expressed in grams of solute in 100 grams of solvent, is called the solubility of the solute in the solvent.
1. Substances that make a solution when mixed in any proportion are called miscible. For example, ethanol and water are miscible.
2. Some substances make a solution when mixed in some proportion but not in all proportions; these are called partially miscible. For example, n-butanol can mix in water up to 7.3 g n-butanol/100 ml water.
3. Substances that almost do not dissolve in each other are called immiscible. For example, n-Hexane is immiscible in water.
4. The solution that has not yet reached its solubility limits and can dissolve more solute added is called an unsaturated solution.
5. The solution that has reached its solubility limits and can not dissolve if more solute is added to it is called a saturated solution.
When a solute is added to a saturated solution, it does dissolve, but, at the same time, the dissociated components recombine to form the crystals of the solute, i.e., recrystallize, at the same rate, so that there is no net dissolution, as illustrated in Fig. 5.2.1.
Effect of temperature on solubility
The solubility of most of the solids and liquids increases as the temperature increases.
For example, when a saturated solution of sugar in water is heated, it can dissolve more sugar. Fig 5.2.2 show the solubility vs temperature curves for some compounds.
When a hot saturated solution is cooled, the concentration of solute in the solution becomes above the solubility limits, making a supersaturated solution.
The supersaturated solution is unstable and ultimately crystallizes out the excess solute leaving any impurities in the solution, as illustrated in Fig. 5.2.3. This process is called re-crystallization, which is used to purify the solutes.
Medical issues related with solubility
The crystallization of excess solute from a supersaturated solution is responsible for some medical problems like gout and kidney stone. Gout is the crystallization of uric acid in the cartilage, tendons, and soft tissues when the concentration of the uric acid in blood plasms exceeds its solubility limit of ~7 mg/100 mL at 37 oC. It causes redness, swelling, and pain in the affected area, as illustrated in Fig. 5.2.4. Kidney stones are solid materials formed in the urinary tubes, as illustrated in Fig. 5.2.5. Kidney stones are the result of the crystallization of excess calcium phosphate, calcium oxalate, or uric acid in the urine.
Opposite to the solids and liquids, the solubility of gases generally decreases with an increase in temperature.
That is why the carbonated water releases dissolved gas when heated, causing pressure increase, which, in turn, causes the bursting of the soda can
Effect of pressure on solubility
The pressure has almost no effect on the solubility of solids and liquids but has a strong effect on the solubility of gases.
Henry's law
The solubility of gases in liquids is directly proportional to the pressure of the gas above the liquid.
An increase in pressure causes a decrease in the gas volume that increases the gas concentration. More frequent collision of the gas molecules with the gas-liquid boundary in a concentrated solution causes an increase in the rate of dissolution of the gas in the liquid, as illustrated in Fig 5.2.6. The opposite happens when the gas pressure decreases. For example, carbon dioxide starts bubbling out when a soda can is open because the gas escapes resulting in a decrease in the gas pressure above the liquid and a decrease in the solubility of the gas in water.
Solubility guidelines for dissolution of ionic compounds in water
If the solubility of a compound is less than 0.01 mol/L, it is considered insoluble.
The solubility of ionic compounds in water depends on the nature of the compound. For example, lead(II)iodide (PbI2) and silver chloride (AgCl) are insoluble in water because the solubility of PbI2 is 0.0016 mol/L of the solution and the solubility of AgCl is about 1.3 x 10-5 mol/L of solution. Potassium iodide (KI) and Pb(NO3)2 are soluble in water. When aqueous solutions of KI and Pb(NO3)2 are mixed, the concentration of PbI2 in the mixture goes above its solubility limits, and it precipitates out, as illustrated in Fig. 5.2.7.
There are no fail-proof guidelines for predicting the solubility of ionic compounds in water. However, the following guideline can predict the solubility of most ionic compounds.
Soluble ions
1. Salts of alkali metals (Li1+, Na1+, K1+, Rb1+, Cs1+ ) and ammonia (NH41+) are soluble. No exceptions.
2. Salts of nitrate (NO31-), acetate (CH3COO1-), and perchlorate (ClO41-) are soluble. No exceptions.
3. Salts of chloride (Cl1-), bromide (Br1-), and Iodide (l1-) are soluble, except when the cation is Pb2+, Hg22-, or Ag1+. (Remember the acronym “LMS” based on the first letter of the element name, or phrase ‘Let Me See” to recall Lead, Mercury, and Silver.)
4. Sulfates (SO42-) are soluble except when the cation Pb2+, Hg22-, Ag1+, or a heavy alkaline earth metal ions: calcium (Ca2+), barium (Ba2+), or strontium (Sr2+). ((recall “Let Me See” for Lead, Mercury, and Silver. Remember the acronym “CBS” based on the first letter of the element name, or the phrase “Come By Soon” to recall calcium, barium, and strontium.)
Insoluble ions
1. Hydroxide (OH1-) and sulfides (S2-) are insoluble except when the cation is an alkali metal, ammonia, or a heavy alkaline earth metal ions: Ca2+, Ba2+, and Sr2+. (Recall the phrase “Come By Soon” to recall calcium, barium, and strontium.)
2. Carbonates (CO32-), phosphates (PO43-), and oxide (O2-) are insoluble except when the cation is an alkali metal, ammonia
3. If there is a conflict between the two guidelines, then the guideline listed first has priority. For example, the salts of insoluble ions become soluble when the cation is an alkali metal, ammonia (rule#1).
Fig. 5.2.8 shows precipitates of some insoluble ionic compounds formed by mixing aqueous solutions of appropriate soluble ionic compounds. The precipitation can be predicted, as illustrated in Fig. 5.2.9. List the ions of the soluble ionic compounds and then cross-combine the cations of one with the anion of the other to make the potential products. If any of the potential products is an insoluble ionic compound, it precipitates out. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/05%3A_Solutions/5.02%3A_Solubility.txt |
What is an electrolyte?
Electric current is defined as the movement of electric charges. The substances through which an electric current can flow are called electrical conductors, and the others are electrical nonconductors. Metals are electrical conductors because valence electrons of metal atoms can move around in a piece of metal. Ionic compounds are composed of cations and anions, but the ions in a solid can not move around. Therefore, solid ionic compounds are electrical nonconductors. Pure water does not have a sufficient concentration of ions in it and is an electrical nonconductor. The ionic compounds dissociate into ions when dissolved in water. The solution of ionic compounds in water is an electrical conductor because the ions can move around in the solution, as illustrated in Fig. 5.3.1.
Substances that produce electrically conducting solution when dissolved in water (or in another polar solvent) are called electrolytes.
All ionic compounds, acids, and bases produce ions in water and are classified are electrolytes.
Substances that produce an electrically nonconducting solution when dissolved in water are called nonelectrolytes.
Molecular compounds other than acids and bases, such as methanol, acetone, sugar, and glucose, remain neutral molecules when dissolved in water. The molecular solutes, other than acids and bases, are nonelectrolytes.
Strong and weak electrolytes
Substances that almost 100% dissociate into ions when dissolved in water are strong electrolytes.
1. All ionic compounds that are soluble in water are strong electrolytes.
2. Strong acids are strong electrolytes.
3. Strong bases are ionic compounds and strong electrolytes.
A strong electrolyte does not mean that it is necessarily highly soluble in water. It means that the portion of the solute that dissolves, it also dissociates 100% into ions in water, e.g., all ionic compounds. The solubility Ca(OH)2 is only 0.16 g Ca(OH)2/100 g water at 20 oC, but all the dissolved Ca(OH)2 dissociates into Ca2+ and OH- ions.
Strong bases are hydroxides of alkali metals, i.e., LiOH, NaOH, KOH, RbOH, and CsOH, and hydroxides of heavy alkaline earth metals, i.e., Ca(OH)2, Sr(OH)2 and Ba(OH)2, which are strong electrolytes. Strong acids, i.e., HCl, HBr, HI, HClO4, HNO3, and H2SO4, are molecular compounds but are strong electrolytes because they dissociate almost 100% into ions when dissolved in water. For example, HCl almost wholly dissociates into ions when dissolved in water.
\mathrm{HCl}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{3} \mathrm{O}^{1+}(\mathrm{aq})+\mathrm{Cl}^{1-}(\mathrm{aq})\nonumber
Substances that partially dissociate into ions when dissolved in water are weak electrolytes. Weak acids and weak bases are weak electrolytes.
Week acids and week bases, like acetic acid (CH3COOH) and ammonia (NH3), are soluble in water, but partially dissociate into ions. For example, if 1-mole acetic acid or 1-mole of ammonia is dissolved in 1 liter of water at room temperature, they establish the following equilibrium between dissolved molecules and dissociated ions:
\begin{aligned}
&\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{l}) \stackrel{\text { Water }}{\longrightarrow} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \stackrel{\rightarrow}{\longleftarrow} \mathrm{CH}_{3} \mathrm{COO}^{1-}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{1+}(\mathrm{aq}), \
&\mathrm{NH}_{3}(\mathrm{~g}) \quad \stackrel{\text { Water }}{\longrightarrow} \mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \stackrel{\rightarrow}{\longleftarrow} \mathrm{NH}_{4}{ }^{1+}(\mathrm{aq})+\mathrm{OH}^{1-}(\mathrm{aq}),
\end{aligned}\nonumber
where only about 0.4% of the dissolved molecules dissociate into ions, the remaining about 99.6% of molecules remain neutral. Weak acids and weak bases are weak electrolytes.
Fig. 5.3.2 illustrates the difference between nonelectrolytes, strong electrolytes, and weak electrolytes.
Figure $2$: Illustration of a nonelectrolyte that does not conduct electricity, a strong electrolyte that has high electrical conductivity, and a weak electrolyte that has low electrical conductivity. Source: Karishma50/ (https;//creativecommons.org/licences/by-sa/4.0)
Equivalent
The amount of molecules and atoms is usually measured in moles. Ionic compounds are composed of ions but are overall neutral because the +ve charge is balanced by the –ve charge. Therefore, the mole ratio of cations to anions is not always one to one. For example, NaCl has a one-to-one mole ratio of Na+ and Cl-, but CaCl2 has a one-to-two ratio of Ca2+ and Cl-. A new unit, called equivalent (officially abbreviated as Equiv but commonly abbreviated as Eq), is introduced, to differentiate between a mole of ion and a mole of charge on the ion.
The equivalent is the amount of a substance needed to react with or supply one mole of hydrogen ions (H+) in an acid-base reaction, or react with or supply one mole of electrons in a redox reaction. In other words, the number of equivalents of a given ion in a solution is equal to the number of moles of that ion multiplied by its valence.
$1 \mathrm{~mol} \mathrm{~Na}^{+}=1 \mathrm{~Eq} \mathrm{~Na}^{+} \text {, but } \quad 1 \mathrm{~mol} \mathrm{~Ca}^{2+}=2 \mathrm{~Eq} \mathrm{~Ca}^{2+} \text {, and } 1 \mathrm{~mol} \mathrm{} \mathrm{~Al}^{3+}=3 \mathrm{~Ea} \mathrm{~Al}^{3+}\nonumber$
Similarly,
$1 \mathrm{~mol} \mathrm{~Cl}^{-}=1 \mathrm{~Eq} \mathrm{~Cl}^{-}\text{, but} \quad 1 \mathrm{~mol} \mathrm{~CO}_{3}{ }^{2-}=2 \mathrm{~Eq} \mathrm{~CO}{ }_{3}{ }^{2-}\text{ and } 1 \mathrm{~mol} \mathrm{~PO}_{4}{ }^{3-}=3 \mathrm{~Ea} \mathrm{~PO}_{4}{ }^{3-}\nonumber$
The solutions of electrolytes are overall electrically neutral, i.e., the number of equivalents of cations is equal to the number of equivalents of anions in the solution.
For example, if 1 mole of NaCl is dissolved in water, there is 1 Eq of Na+ ions and 1 Eq of Cl- ions in the solution. If one mole of CaCl2 is dissolved in the water there is 2 Eq of Ca2+ ions and 2 Eq of Cl- ions in the solution. Similarly, if 1 mole of NaCl and 1 mole of AlCl3 are dissolved in the water there is 1 Eq of Na+ ions, 3 Eq of Al3+ ions, and 4 Eq of Cl- ions to balance cations in the water.
Equality gives two conversion factors. For example, the equality 1 mol Ca2+ = 2 Eq Ca2+ gives the following two conversion factors:
$\frac{1 \operatorname{~mol~} C a^{2+}}{2 ~E q ~C a^{2+}}\nonumber$
and
$\frac{2 ~E a ~C a^{2+}}{1 ~m o l ~C a^{2+}}\nonumber$
The first conversion factor converts the given amount in equivalents to moles, and the second converts the given amount in moles to equivalents of the ion as explained in the following examples.
Example $1$
a) Calculate mEq of Fe3+ in 0.0200 mol of Fe3+? b) if chloride ion is the only anion in the solution, how many mEq of Cl- are present in the solution?
Solution
a) Given 0.0200 mol Fe3+. Desired mEq Fe3+.
The equality: 1 mol Fe3+ = 3 Eq Fe3+ gives the conversion factor $\frac{3 ~E q ~F e^{3+}}{1 ~m o l ~F e^{3+}}$ that converts the given amount in moles to Eq of Fe3+. Another equality: 1 Eq Fe3+ = 1000 mEq Fe3+ gives the conversion factor $\frac{1000 ~m E q ~F e^{3+}}{1 ~E q ~F e^{3+}}$, that converts the Eq of Fe3+ to mEq of Fe3+.
Calculations: $0.0200 \cancel{\text { mol } ~F e^{3+}} \times \frac{3 \cancel{~E q ~F e^{3+}}}{1 \cancel{~m o l ~F e^{3+}}} \times \frac{1000 ~m E q ~F e^{3+}}{1 \cancel{~E q ~F e^{3+}}}=60 ~mE q ~F e^{3+}\nonumber$
Example $2$
An intervenous saline solution contains 145 mEq/L of Na+. How many moles of Na+ are in 0.500 L of the solution?
Solution
a) Given: 145 mEq Na+ /L solution and 0.500 L solution, Desired: ? mol Na+
Conversion factors needed: The concentration in mEq/L is the first factor for L to mEq conversion, the equality 1 mEq Na+ = 1 mmol Na+ givens the second conversion needed for mEq to mmol conversion, and the equality 1000 mmol Na+ = 1 mol Na+ given the third conversion factor needed needed for mEq to mol conversion. The three conversion factors are applied one after the other in a single row in the following calculation:
$0.500 \cancel{\text { L solution }} \times \frac{145 \cancel{\mathrm{~mEq} \mathrm{~Na}}}{1 \cancel{\text {L solution }}} \times \frac{1 \cancel{\mathrm{~mmol} \mathrm{~Na}^{+}}}{1 \cancel{\mathrm{~mEq} \mathrm{~Na}^{+}}} \times \frac{1 \mathrm{~mol} \mathrm{~Na}^{+}}{1000 \cancel{\mathrm{~mmol} \mathrm{~Na}^{+}}}=0.0725 \mathrm{~mol} \mathrm{~Na}^{+}\nonumber$
Note how the units of the numerator in one fraction are canceled by the units of the denumerator of the following fraction leaving only the desired units uncancelled that become the units of the answer number.
Electrolytes in body fluids
Fig. 5.3.3 shows the electrolytes and their concentrations commonly found in body fluids. They play an essential role in cell and body functions. For example, sodium ions regulate the water content and play a role in electrical impulse transmission in the nervous system. Potassium ions play a role in maintaining a regular heartbeat and also play a part in electrical impulse transmission. Chloride ions are there to balance the charge. Carbonate ions are involved in maintaining the pH of the blood.
Concentrations of electrolytes in body fluids are not high enough to be reported in a mole or equivalent units. The electrolytes in the body fluids are usually reported in millimoles (mmol) or milliequivalent (mEq) units, where: $1000 \mathrm{~mmol}=1 \mathrm{~mol}$ and $1000 \mathrm{~mEq}=1 \mathrm{~Eq}$.
The overall concentration of electrolytes in intravenous fluids given to patients is about the same as of electrolytes in the body fluids. For example, Ringer’s lactate solution contains about: 130 mEq/L Na+, 4 mEq/L K+, 3 mEq/L Ca2+, 109 mEq/L Cl- , and 28 mEq/L lactate-. Note that the overall +ve charge (130+4+3= 137 mEq/L) is equal to the overall –ve charge (109+28 = 137 mEq/L). | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/05%3A_Solutions/5.03%3A_Electrolytes.txt |
The concentration of a solution tells the amount of solute dissolved in a given amount of solution.
Making a solution of know concentration
A measured amount of solute is dissolved in enough solvent to make the desired volume of the solution, as illustrated in Fig. 5.4.1. The concentration of the solution can be expressed in different ways using mass, volume, or mole units, as explained in the following.
Concentration in percent (%)
What is percent (%)
Percentage (%) is a number or ratio that represent a fraction of 100.
For example, 5% means 5:100, where 5 is the part, and 100 is the total. The percentage is calculated as a hundred times of part by total, i.e.,
\text { Percentage }(\%)=\frac{\text { part }}{\text { Total }} \times 100 .\nonumber
Example $1$
A 50.0 g NaCl is dissolved in water to make a 500 g solution. What is the percentage of NaCl in the solution?
Solution
Given part = 50.0 g NaCl, and total = 500 g solution. Desired: %NaCl in the solution?
Formula: $\text { Percentage }(\%)=\frac{\text { part }}{\text { Total }} \times 100 \text {. }\nonumber$
Calculations: $\text { Percentage }(\%)=\frac{50.0 cancel{\mathrm{~g}}}{500 cancel{\mathrm{~g}}} \times 100=10.0 ~\% \mathrm{NaCl}$
The units cancel in the fraction calculation part, and a % sign is added to the answer to tell that it is a fraction out of a hundred.
Mass percent (m/m)% concentration
The mass percent concentration expresses the mass units of solute in a hundred mass units of the solution.
\operatorname{Mass}(\%)=\frac{\text { mass of solute }(g)}{\text { mass of solute }(g)+\text { mass of solvent }(g)} \times 100=\frac{\text { mass of solute }(g)}{\text { mass of solution }(g)} \times 100\nonumber
Note that the total is solute and solvent added together, i.e., solution.
Example $2$
What is the mass % of NaOH in a solution prepared by dissolving 10.0 g NaOH in 100 g water?
Solution
Given solute = 10.0 g, and solvent = 100 g, Desired: Mass% NaOH?
Formula: $\text { Mass }(\%)=\frac{\text { mass of solute }(g)}{\text { mass of solute }(g)+\text { mass of solvent }(g)} \times 100$
Calculations: $\text { Mass }(\%)=\frac{10.0 \mathrm{~g} \mathrm{NaOH}}{(10.0 \mathrm{~g}+100 \mathrm{~g}) \text { solution }} \times 100=9.09 \% \mathrm{NaOH}$
Note that the mass% concentration and its reciprocal are two conversion factor: $\frac{\text { given } g \text { solute }}{100 g \text { solution }}$ and $\frac{100 g \text { solution }}{\text { given g solute }}$
Example $3$
Neosporin antibiotic is a 3.5% m/m neomycin solution. How many grams of neomycin are in 50 g of ointment?
Solution
Given: $3.5 \% \text { neomycin }=\frac{3.5 \mathrm{~g} \text { neomycin }}{100 \mathrm{~g} \text { solution }}$, and Solution amount = 50 g, Desired: ? g neomycin
Calculations: $50 \cancel{\mathrm{~g} \text { solution }} \times \frac{3.5 \mathrm{~g} \text { neomycin }}{100 \cancel{\mathrm{~g} \text { solution }}}=1.8 \mathrm{~g} \text { neomycin. }\nonumber$
Volume percent (v/v)% concentration
The volume percent concentration expresses the volume units of solute in a hundred volume units of the solution.
The mathematical form of the v/v % is:
\text { Volume }(\%)=\frac{\text { volume of solute }(\mathrm{mL})}{\text { volume of solution }(\mathrm{mL})} \times 100\nonumber
Fig. 5.4.2 shows the volume percent concentration ranges of different classes of fragrances.
Example $4$
What is the volume % of rose extract in a solution prepared by dissolving 14.0 mL rose extract in a solvent to make 200 mL of solution?
Solution
Given: Solute = 14.0 g, and solution = 200 g, Desired: Volume% rose solution?
Formula:
\text { Volume }(\%)=\frac{\text { volume of solute }(m L)}{\text { volume of solution }(m L)} \times 100\nonumber
Calculations:
\text { Volume }(\%)=\frac{14 \mathrm{~mL} \text { rose extract }}{200 \mathrm{~mL} \text { solution }} \times 100=7.0 \% \text { rose solution }\nonumber
The two conversion factors for v/v % concentration are:
\frac{\text { given mL solute }}{100 \mathrm{~mL} \text { solution }} \quad\text {, and }\quad \frac{100 \mathrm{~mL} \text { solution }}{\text { given } \mathrm{~mL} \text { solute }}\nonumber
Example $5$
What is the volume of bromine (Br2) in 250 mL of 4.8% v/v of Br2 solution in carbon tetrachloride?
Solution
Given: Concentration 4.8% v/v bromine $=\frac{4.8 ~m L \text { bromine }}{100 ~m L \text { solution }}$, volume of solution = 250 mL, Desired: Volume of solute, i.e., ? mL bromine.
Calculations: $250 \mathrm{~mL} \text { solution } \times \frac{4.8 \mathrm{~mL} \text { bromine }}{100 \mathrm{~mL} \text { solution} .}=12 \mathrm{~mL ~bromine} \text {. }\nonumber$
Mass/volume percent (m/v)% concentration
The mass/volume percent concentration expresses the mass units of solute in a hundred volume units of solution.
Mathematical form of m/v % is:
\text { Mass/volume }(\%)=\frac{\text { mass of solute }(g)}{\text { volume of solution }(m L)} \times 100\nonumber
Example $6$
What is the mass/volume % of glucose solution prepared by dissolving 50 g glucose in enough water to make 1000 mL of solution?
Solution
Given: Solute = 50.0 g, and Solution = 1000 mL, Desired: Mass/volume % glucose solution?
Formula:
\text { Mass/volume }(\%)=\frac{\text { mass of solute }(g)}{\text { volume of solution }(m L)} \times 100\nonumber
Calculations:
\frac{\text { Mass }}{\text { volume }}(\%)=\frac{50 \mathrm{~g} \text { glucose }}{1000 \mathrm{~mL} \text { solution }} \times 100=5.0 \% \text { glucose solution by } \frac{\mathrm{m}}{\mathrm{v}} .\nonumber
The two conversion factors from ms/v % concentration are:
\frac{\text { given } g \text { solute }}{100 \mathrm{~mL} \text { solution }}\quad \text { and }\quad \frac{100 \mathrm{~mL} \text { solution }}{\text { given } \mathrm{g} \text { solute }}\nonumber
Example $7$
How many grams of clindamycin antibiotics are in a 45 mL capsule of the 1.0% (m/v) clindamycin?
Solution
Given: % m/v concentration: 1.08 % m/v clindamycin $=\frac{1.0 \mathrm{~g} \text { clindamycin }}{100 \mathrm{~mL} \text { solution }}$, and volume of solution = 45 mL, Desired: ? g clindamycin?
Calculations:
45 \text { mL solution } \times \frac{1.0 \mathrm{~g} \text { clindamycin }}{100 \mathrm{~mL} \text { solution }}=4.5 \mathrm{~g} \text { clindamycin. }\nonumber
Parts per million (ppm) and parts per billion (ppb) concentration
Parts per million (ppm) is a number or ratio expressed as a fraction of a million (106).
For example, 2 ppm means $\frac{2}{1,000,000}$ or 2:1,000,000, where 2 is the part, and 1,000,000 is the total. The concentration in ppm is calculated as a million times of part by total, i.e.:
\text { Concentration in ppm }=\frac{p a r t}{\text { Total }} \times 10^{6}\nonumber
Parts per billion (ppb) is a number or ratio expressed as a fraction of billion (109).
That is:
\text { Concentration in ppb }=\frac{\text { part }}{\text { Total }} \times 10^{9}\nonumber
Like percentage concentration, the ppm and ppb can be mass/mass (m/m), volume/volume (v/v) or mass/volume (m/v).
Example $8$
EPA’s action limit for copper is 1.3 mg/L in drinking water. What is this limit in ppm of copper m/v in the drinking water?
Solution
Given: 1.3 mg copper in 1L solution, Desired: ? ppm m/v of Copper in water
Formula: $\text { Concentration in ppm }=\frac{\operatorname{solute}(\mathrm{~g})}{\operatorname{solution}(\mathrm{~mL})} \times 10^{6}$
Calculations: First, convert the given units of mass and volume into the corresponding units that the formula takes, then plug the values in the formula and calculate.
$\text { Solute }=1.3 \mathrm{~mg} \times \frac{1 \mathrm{~g}}{1000 \mathrm{~mg}}=0.0013 \mathrm{~g}\nonumber$
$\text { Solution }=1 \mathrm{~L} \times \frac{1000 \mathrm{~mL}}{1 \mathrm{~L}}=1000 \mathrm{~mL}\nonumber$
$\text { Concentration in ppm }=\frac{0.0013 \mathrm{~g}}{1000 \mathrm{~mL}} \times 10^{6}=1.3 \mathrm{~ppm} \text { copper } v / \mathrm{m}\nonumber$
Example $9$
EPA’s action limit for lead is 0.015 mg/L in drinking water. What is this limit in ppb of lead m/v in the drinking water?
Solution
Given: 0.015 mg in 1L solution, Desired: ? ppb m/v of Lead in water
Formula: $\text { Concentration in ppb }=\frac{\operatorname{solute}(g)}{\operatorname{solution}(\mathrm{mL})} \times 10^{9}$
Calculations: First, convert the given units of mass and volume into the corresponding units that the formula takes, then plug the values in the formula and calculate.
$\text { Solute }=0.015 \mathrm{~mg} \times \frac{1 \mathrm{~g}}{1000 \mathrm{~mg}}=0.000015 \mathrm{~g}\nonumber$
$\text { Solution }=1 \mathrm{~L} \times \frac{1000 \mathrm{~mL}}{1 \mathrm{~L}}=1000 \mathrm{~mL}\nonumber$
$\text { Concentration in ppb }=\frac{0.000015 \mathrm{~g}}{1000 \mathrm{~mL}} \times 10^{9}=15 \mathrm{~ppb} \text { lead } \mathrm{v} / \mathrm{m}\nonumber$
Molarity
Molarity (M) expresses the moles of solute in a liter of solution.
The most common solution concentration unit used in chemistry is molarity (M):
$\operatorname{Molarity}~({M})=\frac{n~(\text {moles of solute })}{V ~\text { (Litters of solution })}\nonumber$
Example $10$
What is the molarity (M) of a solution prepared by dissolving 50.0 g NaOH in enough water to make 250 mL solution?
Solution
Given: Solute = 50.0 g NaOH, Volume of solution = 250 mL, Desired: ? M NaOH solution?
Formula: $\text { Molarity }(M)=\frac{n~(\text {moles of solute })}{V~(\text {Litters of solution })}$
Calculations: First, convert the given units of mass and volume into the corresponding units that the formula takes, then plug the values in the formula and calculate.
$\text { Solute }=50.0 \mathrm{~g} \mathrm{~NaOH} \times \frac{1 \mathrm{~mol} \mathrm{~NaOH}}{40.00 \mathrm{~g} \mathrm{~NaOH}}=1.25 \mathrm{~mol} \mathrm{~NaOH}\nonumber$
$\text { Solution }=250 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}=0.250 \mathrm{~L}\nonumber$
$\text { Molarity }(M)=\frac{1.25 \mathrm{~mol} \mathrm{~NaOH}}{0.250 \mathrm{~L} \mathrm{~solution}}=5.00 \mathrm{~M} \mathrm{~NaOH}\nonumber$
The two conversion factor from molarity are the following:
$\frac{n \text { (moles of solute) }}{V \text { (Litters of solution) }} \quad\text { and }\quad \frac{V \text { (Litters of solution) }}{n \text { (moles of solute })}\nonumber$
Example $11$
How many litters of 0.211 M HCl solution are needed to provide 0.400 mol of HCl?
Solution
Given: amount of solute = 0.400 mol HCl, Concentration of solute = 0.211M = $\frac{0.211 \text { mol HCl}}{1 L \text { solution }}$
Calculations: $\text { Liters of solution needed }=0.400 \mathrm{~mol} \mathrm{~HCl} \times \frac{1 \mathrm{~L} \text { soluton }}{0.211 \mathrm{~mol} \mathrm{~HCl}}=1.90 \mathrm{~L} \text { solution }\nonumber$
Example $12$
A 2.50 L of 1.12 M NaOH solution contains how many moles of NaOH?
Solution
Given: Volume of solution = 2.50 L solution, concentration of solution = $1.12 \mathrm{~M} \mathrm{~NaOH}=\frac{1.12 \mathrm{~mol} \mathrm{~NaOH}}{1 \mathrm{~L} \text { solution }}$, Desired: ? moles of NaOH?
Calculations: $\text { Moles of } \mathrm{~NaOH} \text { in the solution }=2.50 \text { L solution } \times \frac{1.12 \mathrm{~mol} \mathrm{~NaOH}}{1 \mathrm{~L} \text { solution }}=2.80 \mathrm{~mol} \mathrm{} \mathrm{~NaOH}\nonumber$
Dilution of solutions
Dilution of a solution is the addition of a solvent to decrease the solute concentration of the solute in the solution.
The product of concentration (C) and volume (V) is the amount of solute, i.e.,
\text { Amount of solute }=C \text { in } \frac{\text { amount of solute }}{\text { volume of solution }} \times V \text { in volume of solution. }\nonumber
The amount of solute does not change by adding solvent. Therefore, the product of concentration and volume, i.e., CV, which is the amount of solute, is a constant, i.e.,
C_{1} V_{1}=C_{2} V_{2}=\text { amount of solute }\nonumber
Fig. 5.4.3 shows that if the initial concentration is C1, the initial volume is V1, and after dilution, the final concentration is C2, the final volume is V2, then C1V1 = C2V2 = amount of solute that is constant. If three of the four variables in this equation are known, the missing one can be calculated, as explained in the following example.
Caution
Keep in mind that the concentrations and volumes should be in the same units on both sides of the equation: C1V1 = C2V2. If they are not in the same units, convert them to the same units before plunging them in the formula.
Example $13$
how much volume of 11.3 M HCl is needed to prepare 250 mL of 2.00 M HCl?
Solution
Given: C1 = 11.3 M HCl, C2 = 2.00 M HCl, V2 = 250 mL solution, Desired V1 = ?
Formula: C1V1 = C2V2, rearrange it to isolate the desired parameter: $V_{1}=\frac{C_{2} V_{2}}{C_{1}}$
Calculations: $V_{1}=\frac{C_{2} V_{2}}{C_{1}}=\frac{2.00 \mathrm{~M~HCl} \times 250 \mathrm{~mL} \text { solution }}{11.3 \mathrm{~M} \mathrm{~HCl}}=44.2 \mathrm{~mL} \text { solution }\nonumber$
Example $14$
what is the molarity of the NaOH solution prepared by diluting 100 mL of 0.521 M NaOH solution to 500 mL?
Solution
Given C1 = 0.521 M NaOH, V1 = 100 mL solution, V2 = 500 mL solution, Desired: Concentration of the final solution C2 = ? M NaOH
Formula: C1V1 = C2V2, rearrange to isolate the desired parameter: $C_{2}=\frac{C_{1} V_{1}}{V_{2}}$
Calculations: $C_{2}=\frac{C_{1} V_{1}}{V_{2}} = \frac{0.521 \mathrm{~M~NaOH} \times 100 \mathrm{~mL} \mathrm{~NaOH}}{500 \mathrm{~mL} \mathrm{~NaOH}}=0.104 \mathrm{~M~NaOH}\nonumber$
Example $15$
Dopamine is administered intravenously to a patient to increase blood pressure. How many milliliters (mL) of a 4.0% (m/v) dopamine solution is needed to prepare 250 mL of a 0.030% m/v) solution?
Solution
Given: C1 = 4.0% (m/v), C2 = 0.030% (m/v), V2 = 250 mL solution, Desired V1= ?
Formula: C1V1 = C2V2, rearrange to isolate the desired parameter: $V_{1}=\frac{C_{2} V_{2}}{C_{1}}$
Calculations: $V_{1}=\frac{C_{2} V_{2}}{C_{1}}=\frac{0.030 \% \times 250 \mathrm{~mL} \text { solution }}{4.0 \%}=5.0 \mathrm{~mL} \text { of solution }\nonumber$
Logarithmic dilution
Take a unit volume of a given solution and add enough solvent to increase the volume of the solution 10 times for a logarithmic diluiton.
A logarithmic dilution is ten times dilution, i.e., proven by the following formula: $C_{2}=\frac{C_{1} V_{1}}{V_{2}}=\frac{1 \mathrm{~mL} \times C_{1}}{10 \mathrm{~mL}}=0.1 \times C_{1}\nonumber$
Repeating the above step with the diluted solution results in 10x10 = 100-time dilution, and repeating third-time results in 10x10x10 = 1000-time dilution. This dilution of 10 times in each step is called logarithmic dilution. Fig. 5.4.4 shows that five steps of logarithmic dilution on a 10% initial solution results in a concentration of 10 ppm in the final solution. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/05%3A_Solutions/5.04%3A_Concentration_of_solutions.txt |
Semipermeable membranes surround living cells and organelles in the cells. The semipermeable membranes allow water and small molecules but do not allow the passage of large molecules and ions.
Osmosis is the passage of water and small molecules across a semipermeable membrane with a net flow from a less concentrated solution to a more concentrated solution.
Osmosis helps in the absorption, retention, and flow of water, nutrients, and other molecules required in the biological systems. Fig. 5.5.1 illustrates the process of osmosis.
Osmotic pressure
Osmosis causes net water flow from the less concentrated to the more concentrated solution across the semipermeable membrane. Consequently, the water level rises in the less concentrated solution compartment, as illustrated in Fig. 5.5.2. The difference in the height of water increases and applies pressure, pumping water back to the more concentrated side until the flow of water is equal on the two sides.
The pressure that prevents additional water flow to the more concentrated solution side of the semipermeable membrane is called osmotic pressure.
The osmotic pressure is proportional to the overall concentration of the solute particles. For example, 0.1 molar NaCl has osmotic pressure about twice that of 0.1 molar glucose because each mole of glucose adds one mole of solute particles, while each mole of NaCl produces two moles of particles, i.e., one-mole Na+ and one-mole Cl- ions in the solution.
Reverse Osmosis
Reverse osmosis is the net flow of water to the less concentrated or pure water across the semipermeable membrane by applying external pressure more than the osmotic pressure on the more concentrated solution side.
Reverse osmosis is the net flow of water to the less concentrated or pure water across the semipermeable membrane by applying external pressure more than the osmotic pressure. Note that solvent flow in reverse osmosis driven by external pressure is the opposite of regular osmosis. Reverse osmosis is used to produce drinking water from seawater sources, as illustrated in Fig. 5.5.3.
Isotonic, hypertonic, and hypotonic solutions
Cell membranes are semipermeable membranes that separate intracellular and extracellular fluids. The concentration difference across the membrane and the resulting osmotic pressure plays an essential role in cell functions. Intravenous solutions injected into patients must have the same osmotic pressure as the body fluids.
Solutions with the same solute particle concentration and osmotic pressure are called isotonic. If the two solutions across a semipermeable membrane do not have the same solute particle concentration, the solution with higher solute particle concentration and higher osmotic pressure is hypertonic, and the other has lower solute particle concentration and lower osmotic pressure is hypotonic.
Remember that hyper- means more and hypo-means less, concerning solute particle concentration in the case of osmosis.
Cells placed in an external solution may retain their size, shrink, or swell depending on the relative osmotic pressure of fluid inside and outside of the cell, as illustrated in Fig. 5.5.4. For example, red blood cells placed in an isotonic solution retain their size because the flow of water into and out of the cell is the same.
• Typical isotonic solutions are 0.9% m/v NaCl solution in water or 5% m/v glucose solution in water.
• Red blood cells placed in a hypertonic solution shrink in size due to more flow of water out than into the cell –a process called crenation.
• Red blood cells placed in a hypotonic solution swell and burst due to more water flow into than out of the cells –a process called hemolysis.
A similar situation happens in plant cells that are placed in different environments concerning osmotic pressure, as illustrated In Fig. 5.5.5.
Dialysis
Dialysis separates colloids from water, dissolved ions, and molecules of small dimention
Dialysis is similar to osmosis with the difference that in dialysis water, small molecules and ions can pass through a dialyzing membrane leaving behind collide particles like proteins and starch molecules, as illustrated in Fig. 5.5.6.
Role of dialysis in the human body
Kidneys filter the blood by dialysis process. There are over a million tubular structures, called nephrons, in the kidney surrounded by dialyzing membranes. The nephrons filter the water, small molecules like glucose, amino acids, urea, and ions from the blood. Useful products and most of the water reabsorb later on, but urea and other waste products excrete through urine, as illustrated in Fig. 5.5.7.
Hemodialysis
Hemodialysis is used to extract urea and other waste products from the blood when a person's kidney fails to remove them, as illustrated in Fig. 5.5.8. A hemodialysis system is a kind of artificial kidney in which the blood flows through long cellophane tubes placed in an isotonic solution containing NaCl, KCl, NaHCO3, and glucose. Cellophane is a dialyzing membrane that does not let proteins, other large molecules, and blood pass through it, but urea excretes. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/05%3A_Solutions/5.05%3A_Osmosis.txt |
• 6.1: What is an Acid and a Base?
General properties of acids and bases are described. Arrhenius's definitions and nomenclature of Arrhenius's acids and bases are also presented.
• 6.2: Brønsted–Lowry acids and bases
Brønsted–Lowry's definition, i.e., acids are proton donors, and bases are proton acceptors are described. The related terms: acid/conjugate bases pair, mono- di-, tri-protic acids, and amphoteric substances are also described.
• 6.3: Strength of acids and bases
The strength of acids and bases, i.e., the extent of dissociation of the dissolved acid or base into ions in water is described. The relative strength of the acid/conjugate base pair is also explained.
• 6.4: Acid-base equilibrium
The establishment of an equilibrium between an acids-base mixture and its conjugates is described. Ways to move the equilibrium forward or reverse, by varying concentrations, pressure in the cases involving gases, and temperature is explained based on Le Châtelier’s principle.
• 6.5: Dissociation of water
About one in half a billion water molecules act as an acid by donating a proton to another water molecule that acts as a base, producing hydroxide ion and hydronium ion, respectively. The product of hydroxide ion and hydronium ion concentration is constant, but the acidic solution has more hydronium ions, the basic solution has more hydroxide ions, and the neutral solution has equal.
• 6.6: The pH
pH, i.e., log of reciprocal of hydronium ion concentration, its and measurement using pH paper and pH indicator is described. The importance of and pH of body fluid, acid rain, and its effects are introduced.
• 6.7: Acid-base reactions
Reactions of acids with metals, carbonate and bicarbonate salts, and Arrhenius bases are introduced. Ways of representing acid-base reaction by molecular equation, complete ionic equations, net ionic equations, and determination of acid or base concentration by titration method are described.
• 6.8: pH Buffers
pH Buffer, i.e., a mixture of a weak acid and its conjugate base or vice versa, is used to regulate the pH within a narrow range. Mechanisms of buffer action and buffer action in the blood are explained.
06: Acids and bases
General Properties of Acids and Bases
We commonly encounter acids and bases in our foods –some foods are acidic, and others are basic (alkaline) as illustrated in Fig. 6.1.1.
The general properties of acids and bases are the following.
1. Acids taste sour, e.g., citrus fruits taste source because of citrus acid and ascorbic acid, i.e., vitamin C, in them. Basic (alkaline) substances, on the other hand, taste bitter.
2. Basic (alkaline) substances feel soupy, while acidic substances may sting.
3. The acids turn blue litmus paper to read but do not change the color of red litmus paper. Bases turn red litmus paper blue but do not change the color of blue litmus paper, as illustrated in Fig. 6.1.2.
4. Phenolphthalein indicator turns colorless in acid and turns pink in basic solution, as illustrated in Fig. 6.1.3.
5. Acids and bases neutralize each other. Hydrochloric acid is found in the stomach that helps digestion. Excess hydrochloric acid may cause acid burns—antacids like milk of magnesia are bases that help by neutralizing excess acid in the stomach.
Arrhenius's Definition of Acids and Bases
The earliest definition of acids and bases is Arrhenius's definition which states that:
• An acid is a substance that forms hydrogen ions H+ when dissolved in water, and
• A base is a substance that forms hydroxide ions OH- when dissolved in water.
For example, hydrochloric acid ($\ce{HCl}$) is an acid because it forms $\ce{H^{+}}$ when it dissolves in water.
$\mathrm{HCl}(\mathrm{g}) \stackrel{\text { Water }}{\longrightarrow} \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\nonumber$
Similarly, NaOH is a base because it forms OH- when it dissolves in water.
$\mathrm{NaOH}(\mathrm{s}) \stackrel{\text { Water }}{\longrightarrow} \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\nonumber$
Note that hydrogen ion H+ does not exist in reality. It bonds with water molecules and exists as hydronium ion H3O+(aq).
$\mathrm{H}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\nonumber$
However, $\ce{H^{+}(aq)}$ is often written in the place of (H3O+(aq).
Naming Arrhenius acids and bases
Table 1 lists the names and formulas of some of the common acids and their anions.
Rules for naming acids
1. The names end with the word “acid.”
2. If the anion is not an oxyanion, then add the prefix hydro- to the name of the anion and change the last syllable of the anion name to –ic. For example, Cl- is a chloride ion, and HCl is hydrochloric acid.
3. If the anion is an oxyanion with the last syllable –ate, change the last syllable with –ic. Do not use the prefix hydro-, but add the last word “acid.” If there is a prefix per- in the name of the oxyanion, retain the prefix in the acid name. For example, NO3- is a nitrate, and HNO3 is nitric acid. Another example, ClO4- is a perchlorate, and HClO3 is perchloric acid.
4. If the anion is an oxyanion with the last syllable –ite, change the last syllable with –ous. Do not use the prefix hydro-, but add the last word “acid.” If there is a prefix hypo- in the name of the oxyanion, retain the prefix in the acid name. For example, NO2- is nitrite, and HNO2 is nitrous acid. Another example, ClO- is hypochlorite, and HClO is hypochlorous acid.
Table 1: Names of some common acids and their anions.
Acid formula Acid name Anion Anion name
HCl Hydrochloric acid Cl- Chloride
HBr Hydrobromic acid Br- Bromide
HI Hydroiodic acid I- Iodide
HCN Hydrocyanic acid CN- Cyanide
HNO3 Nitric acid NO3- Nitrate
HNO2 Nitrous acid NO2- Nitrite
H2SO4 Sulfuric acid SO42- Sulfate
H2SO3 Sulfurous acid SO32- Sulfite
H2CO3 Carbonic acid CO3- Carbonate
CH3COOH Acetic acid CH3COO- Acetate
H3PO4 Phosphoric acid PO43- Phosphate
H3PO3 Phosphorous acid PO33- Phosphite
HClO4 Perchloric acid ClO4- Perchlorate
HClO3 Chloric acid ClO3- Chlorate
HClO2 Chlorous acid ClO2- Chlorite
HClO Hypoclorous acid ClO- Hypochlorite
Table 2 lists the names and formulas of some of the common Arrhenius bases.
Naming Arrhenius Bases
The Arrhenius bases are ionic compounds of metal and hydroxide ion, and their name starts with the name of the metal element followed by the name of the anion, i.e., hydroxide. For example, NaOH is sodium hydroxide.
Table 2: Names of some of the common Arrhenius bases
Formula Name
LiOH Lithium hydroxide
NaOH Sodium hydroxide
KOH Potassium hydroxide
Ca(OH)2 Calcium hydroxide
Sr(OH)2 Strontium hydroxide
Ba(OH)2 Barium hydroxide | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/06%3A_Acids_and_bases/6.01%3A_What_is_an_acid_and_a_base.txt |
Some bases do not have hydroxide ions in their formula, yet they act as bases and neutralize acids. For example, ammonia (NH3) and calcium carbonate (CaCO) do not contain hydroxide ions, but they neutralize acids. Further, Arrhenius's definition limits the acid-base reactions in the water medium. The acid-base reactions can take place in other mediums also, e.g., HCl –an acid, and NH3 –a base can react with and neutralize each other in the gas phase also. The Brønsted–Lowry bordered the definition of acids and bases by including the bases mentioned above and also by including acid-base reactions in a non-aqueous medium.
Brønsted–Lowry's definition of acids and bases
Brønsted–Lowry's definition states that:
1. An acid is a proton donor, and
2. A base is a proton acceptor.
or example, HCl is an acid because it donates a proton to the water solvent.
$\mathrm{HCl}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\nonumber$
Water is a base in the above reaction because it accepts a proton from the acid. In a reaction between HCl and NH3:
$\mathrm{HCl}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\nonumber$
HCl is an acid because it donates its proton to NH3, and the NH3 is a base because it accepts a proton, as shown in Fig. 6.2.1.
Brønsted–Lowry's acids have ionizable protons that they donate to bases. Therefore, Brønsted–Lowry's acid is generally written as HA, where H+ is the donatable proton, and A- is the anion of the acid. Examples of acids are HCl, H2SO4, HNO3, and CH3COOH. Note that acetic acid has only one acidic proton that is attached to the O atom in the carboxylic acid group (–COOH). The rest of the protons attached to carbon atoms are not acidic. All organic acids have a carboxylic acid group (–COOH). Brønsted–Lowry's acid may have net +ve charge, no charge, or net –ve charge on it. For example, H3O+, HCl, and HSO4- are all acids because they can donate a proton to a base.
Mono-, di- and tri-protic acids
Acids that have only one acidic proton are mono-protic, e.g., HCl, HNO3, CH3COOH, are mono-protic acids where the acid proton is shown in bold font. Some acids have two acidic protons –they are di-protic, e.g., H2SO4 and H2CO3 are di-protic. The acids with three acidic protons are tri-protic, e.g., H3PO3 is a tri-protic acid. For example, phosphoric acid (H3PO4) can dissociate and donate three protons, as shown in chemical reactions below:
\begin{aligned}
&\mathrm{H}_{3} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{H}_{2} \mathrm{PO}_{4}{ }^{-} \
&\mathrm{H}_{2} \mathrm{PO}_{4}{ }^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HPO}_{4}{ }^{2-} \
&\mathrm{HPO}_{4}{ }^{2-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{PO}_{4}{ }^{3-}
\end{aligned}
The base accepts proton by making a bond with it. The bond is a pair of bonded electrons. Since the proton is a hydrogen atom without an electron, both electrons in the bond come from the base. The base must have a lone pair of electrons on it. The base is usually represented as to emphasize a lone pair of electrons on it that is shown as a pair of dots. For example, ammonia, water, and hydroxide ion (, , and ) are Brønsted–Lowry's bases, because each of these has an atom with lone pair or lone pairs of electrons on them.
Conjugate acid-base pairs
The acid-base reactions described above are one-way reactions, i.e., reactants go-to products almost 100%. However, the majority of the acid-base reactions are two ways, i.e., reactants form the products and the products react with each other and re-form the reactants. Double arrows between reactants and products represent the two ways reactions. For example, hydrofluoric acid (HF) is a weak electrolyte; it partially dissociates in water to form F- and H3O+, and the products react to re-form the reactants, as shown in Fig. 6.2.2. In the reverse reaction, H3O+ is acting as an acid, and F- is acting as a base. The acid and the base in the products are called conjugate acid and conjugate base, respectively. The acid HF becomes conjugate base F- after removal of a proton, and the base H2O becomes conjugate acid H3O+ after accepting a proton.
The conjugate acid-base pair
The conjugate acid-base pair is related to the loss and gain of H+. For example, HF/F- is a conjugate acid-base pair, and H3O+/H2O is also a conjugate acid-base pair.
In other words, remove the acidic proton from an acid to get its conjugate base and add a proton to a base to get its conjugate acid.
Another example is ammonia NH3 which dissolves in water and accepts a proton to form its conjugate acid NH4+, as shown in Fig. 6.2.3. Water H2O acts as an acid by donating a proton and forming its conjugate base OH-. The two conjugate acid-base pairs in this reaction are NH4+ /NH3 and H2O/OH- that are related by loss and gain or an H+.
Example $1$
Identify the conjugate acid-base pairs in the following reaction?
$\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq}) \rightleftarrows \mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{NH}_{4}^{+}\nonumber$
Solution
1. Identify the substance that has donated a proton in the reactants –it is an acid.
2. Remove a proton from the acid to form its conjugate base: H3PO4/H2PO4-.
3. Identify the substance that has accepted a proton in the reactants –it is a base.
4. Add a proton to the base to form its conjugate acid: NH3/NH4+.
Note
Note that loss of a proton from an acid forms its conjugated base with the charge decreased by one, e.g., H3PO4/H2PO4-, HSO4-/SO42-, and NH4+/NH3. Similarly, the gain of a proton by a base forms its conjugate acid with the charge increased by one, e.g., HPO42-/H2PO4-, HCO3-/H2CO3, and NH3/NH4+.
Amphoteric substances
Water acts as a base in some reactions, e.g., with HF, and as an acid in some reactions, e.g., with NH3.
Substances like water that can act as an acid and also as a base are called amphoteric substances.
Other examples of amphoteric substances include HSO4-, HCO3-, and NH3, as illustrated in Fig. 6.2.4. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/06%3A_Acids_and_bases/6.02%3A_BrnstedLowry_acids_and_bases.txt |
The strength of acid HA is the extent to which the acid dissociates into H+ and A- ions, as illustrated in Fig. 6.3.1.
Strong acids
Strong acids, like HCl, almost 100% dissociate into ions when they dissolve in water.
\mathrm{HCl}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\nonumber
One arrow is used to indicate that the reaction is nearly 100% complete.
Strong acids include HClO4, H2SO4, HI, HBr, HCl, and HNO3
Weak acids
Weak acids dissolve in water but partially dissociate into ions.
For example, acetic acid (CH3COOH) is a weak acid, 1 M acetic acid dissolves in water, but only 0.4% of the dissolved molecules dissociate into ions, the remaining 99.6% remain undissociated, as illustrated in Fig. 6.3.2. and equation of the dissociation equilibrium below.
\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \stackrel{\rightarrow}{\longleftarrow} \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq})\nonumber
Two arrows pointing in opposite directions are used for the dissociation of weak acids to indicate that the reaction is an equilibrium, i.e., two ways.
Often the arrows are not equal in size -the longer arrow points to acid-base pair that is weaker and present in a larger concentration at equilibrium than their conjugate pair.
Strong bases
Strong bases almost %100 dissociate into ions when dissolved in water. For example, NaOH is a strong base, and it dissociates almost 100% into ions in water.
Strong bases almost %100 dissociate into ions when dissolved in water. For example, NaOH is a strong base, and it dissociates almost 100% into ions in water.
\mathrm{NaOH}(\mathrm{s}) \stackrel{\text { Water }}{\longrightarrow} \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\nonumber
One arrow is used for the dissolution of strong bases to indicate that the reaction is almost complete.
Strong bases include hydroxides of alkali metals, i.e., LiOH, NaOH, KOH, RbOH, CsOH, and hydroxides of heavy alkaline earth metals, i.e., Ca(OH)2, Sr(OH)2, and Ba(OH)2.
The last three, i.e., the hydroxides of heavy alkaline earth metals, have low solubility in water, but the dissolved fraction exists as ions.
Weak bases
Weak bases partially dissociate into ions when dissolved in water.
For example, ammonia is a weak base –only 0.42% of the dissolved ammonia molecules dissociate into ammonium ions and hydroxide ions in water from a 1 M solution of ammonia.
\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \stackrel{\rightarrow}{\longleftarrow} \mathrm{NH}_{4}{ }^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\nonumber\
Weak bases in household use include ammonia (NH3) in window cleaners, NaClO in bleach, Na2CO3 and Na3PO4 in laundry detergent, NaHCO3 in tooth past, Na2CO3 in baking powder, CaCO3 for use in lawns, Mg(OH)2 and Al(OH)3 in antacids and laxatives.
The weak bases mentioned above are all ionic compounds except ammonia. Ionic compounds are strong electrolytes, i.e., they dissociate into ions almost 100% upon dissolution in water. It appears to contradict the fact that these ionic compounds are weak bases. It does not actually contradict, because the base properties do not refer to these ionic compounds, the base properties refer to the reactions of their polyatomic anions, i.e., ClO-, CO32-, and PO43- with water, as shown in the reactions below:
\begin{aligned}
&\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O} \stackrel{\rightarrow}{\longleftarrow} \mathrm{HClO}+\mathrm{OH}^{-} \
&\mathrm{CO}_{3}^{2-}+2 \mathrm{H}_{2} \mathrm{O} \stackrel{\rightarrow}{\longleftarrow} \mathrm{H}_{2} \mathrm{CO}_{3}+2 \mathrm{OH}^{-} \text {, and } \
&\mathrm{PO}_{4}{ }^{3-}+3 \mathrm{H}_{2} \mathrm{O} \stackrel{\rightarrow}{\longleftarrow} \mathrm{H}_{3} \mathrm{PO}_{4}+3 \mathrm{OH}^{-}
\end{aligned}\nonumber
The above reactions are equilibrium reactions that are more favored in the revers than the forward direction, producing a small number of OH- ions compared to the anion on the reactant sides. The last two examples, i.e., Mg(OH)2 and Al(OH)2 are classified as weak bases because they are considered insoluble in water. The solubility of Mg(OH)2 is 0.00064 g/100 mL (25 °C), and the solubility of Al(OH)3 is 0.0001 g/100 mL, which are in the range of insoluble ionic compounds.
The solubility and the strength of acids and bases are two different things. A strong base may be less soluble, and a weak base may be more soluble or vice versa, but a dissolved strong base exists as ions only, and a dissolved weak base exists both as molecules and ions.
The relative strength of acid-conjugate base pair
A general rule is that the stronger the acid, the weaker the conjugate base, and vice versa.
The conjugate bases of strong acids have negligible base strength, and the conjugate acids of strong basses have negligible acid strength. Fig. 6.3.3. illustrates the relative strengths of some acids and their conjugated bases.
Direction of acid-base equilibrium
In any Brønsted–Lowry acid-base reaction, the general rule is that a stronger acid and a stronger base tend to form a weaker acid and a weaker base.
For example, a dissociation reaction between HCl and H2O is almost 100% complete because HCl is a stronger acid than H3O+ and H2O is a stronger base than Cl-:
\mathrm{HCl}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Cl}^{-}\nonumber
The dissolution of acetic acid (CH3COOH) and ammonia (NH3) are equilibrium reactions because all the acids, bases, and their conjugates are in the weak acids or weak bases category. However, acetic acid and water dominate over their conjugates H3O+ and CH3COO- by 99.6:0.4 ratio (in 1 M acetic acid solution) because the conjugate acid H3O+ is a stronger acid than CH3COOH, and conjugate base CH3COO- is a stronger base than H2O.
\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \stackrel{\rightarrow}{\longleftarrow} \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq})\nonumber
The longer arrow, in the unbalanced equilibrium arrows, points to the acid-base pair in the reaction that exists in a higher concentration relative to their conjugates.
Similarly, ammonia (NH3) and water (H2O) dominate over their conjugates NH4+ and OH- by ~99.6:0.4 ratio (1M ammonia solution) because the conjugate acid NH4+ is a stronger acid than H2O and conjugate base OH- is a stronger base than NH3.
\left.\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})\right) \stackrel{\rightarrow}{\longleftarrow} \mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\nonumber | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/06%3A_Acids_and_bases/6.03%3A_Strength_of_acids_and_bases.txt |
Most of the acid-base reactions are equilibrium reactions, i.e., the reactants form the products, and the products react to re-form the reactants. The reaction is shown with double arrows to indicate that both the forward and the reverse reactions happen simultaneously.
Pre-equilibrium and equilibrium
Initially, there are only reactants present. The concentration of the reactants decreases over time as they convert to the products. The rate of the reaction is proportional to the concentration of the reactants. So, the rate of the forward reaction decreases over time. The products build up over time. The products react with each other to re-form the reactants, i.e., the reverse reaction. The rate of the reverse reaction increases over time as the concentration of the products increases, until the rate of reverse reaction becomes equal to the rate of the forward reaction, as illustrated in Fig. 6.4.1
• At the beginning of the equilibrium when the rate of the reverse reaction is slower than the rate of forward reaction is the pre-equilibrium phase.
• The equilibrium phaes starts at the point when the rate of reverse reaction becomes equal to the rate of the forward reaction
At the point when the rate of reverse reaction becomes equal to the rate of the forward reaction, equilibrium has reached. The concentration of the reactants and products do not change at equilibrium because they are consumed and re-formed at the same rate –it is a dynamic equilibrium.
The equilibrium reactions are not limited to acid-base reactions; they are common in all types of chemical reactions. Fig. 6.4.2 illustrates the concentrations and rates of reactions changes in pre-equilibrium and equilibrium phases with the help of an actual chemical equilibrium reaction between a decomposition reaction of a colorless gas N2O4 and its reverse reaction, i.e., a combination reaction of brown color gas NO2.
What happens when a chemical equilibrium is disturbed?
The concentration of the reactants and products does not change in the mixture when the reaction is at equilibrium because the rate of the forward reaction is equal to the rate of the reverse reaction. Any change made at the equilibrium adds stress to the equilibrium. The system moves in a way to relieve the stress. For example, if one of the reactants is added to the mixture, the rate of forward reaction increases removing the added reactant until a new equilibrium establishes, i.e., the system relieves the stress by removing the added reactant.
Le Chatelier's principle
If a chemical equilibrium is disturbed, the rates of forward and reverse reactions change to relieve the stress and re-establish the equilibrium.
The stresses can be changes in the concentration, pressure, or temperature, as explained in the following section.
Effect of concentration on a chemical equilibrium
The following are the consequences of concentration changes on a chemical reaction at equilibrium.
1. If a reactant is added, the forward reaction increases to remove the reactant.
2. If a reactant is removed, the forward reaction decreases; consequently, the reverse reaction adds the reactant.
3. If a product is added, the reverse reaction increases to remove the product.
4. If a product is removed, the reverse reaction decreases; consequently, the forward reaction adds the product.
Fig. 6.4.3 illustrates the effects with the help of water level in two tanks connected through a conduit in the left tank representing reactants and in the right tank representing products.
Fig 6.4.4 demonstrates the effect of concentration change on the chemical equilibrium between brown dichromate ions and yellow chromate ions:
\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}, \text { brown })+\mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightleftarrows 2 \mathrm{CrO}_{4}^{2-}(\mathrm{aq} \text {, yellow })+2 \mathrm{H}^{+}(\mathrm{aq})\nonumber
The addition of acid to the test tube on the left increases H+ in the system, increasing reverse reaction, which can be observed by the increased brown color in the middle test tube. Then, the addition of a base removes H+ from the system by an acid-base reaction: $\mathrm{H}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}$. The decrease in H+ shifts the equilibrium to the product side, which can be observed by the increased yellow color in the test tube on the right.
A practical example of the effect of concentration on a chemical equilibrium is the binding of oxygen (O2) with hemoglobin (Hb) during the breathing process.
\mathrm{Hb}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftarrows \mathrm{HbO}_{2}(\mathrm{aq})\nonumber
The concentration of oxygen is higher in the lunges that shift the equilibrium to the product side, binding more oxygen with the hemoglobin. When the blood arrives in tissues, the concentration of oxygen is lower in tissues, causing the equilibrium to shift to the reactant side, releasing the oxygen.
The oxygen concentration decreases as the altitude increases. The mountain climbers may experience hypoxia, i.e., inadequate supply of oxygen to the body because the lower level of oxygen at high altitude may shift the equilibrium to the right, resulting in less binding of oxygen with the hemoglobin in the lungs. The body reacts by producing more hemoglobin, but it takes about 10 days for the body to re-adjust the hemoglobin level in the blood. Peoples living at higher altitudes usually have a higher level of hemoglobin in their blood for the reasons described above.
Effect of pressure on a chemical equilibrium
Changes in pressure do not affect the concentration of solids and liquid. However, an increase in pressure decreases the volume, and, consequently, increases the concentration of gases, as illustrated in Fig. 6.4.5.
If a chemical equilibrium involves gases, an increase in pressure has the same effect as increases in the gaseous reactant or product—the equilibrium shifts in the direction where there are fewer moles of gases.
Fig. 6.4.6 demonstrated this effect for the equilibrium between colorless N2O4 gas and brown color NO2 gas:
\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}, \text { colorless }) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{~g}, \text { brown })\nonumber
The color becomes light in going from a syringe on the right to the syringe in the middle. It can be explained based on the fact that the concentration of gases decreased upon an increase in volume as a result of a decrease in pressure. Then the color becomes darker over time, as shown by the syringe on the right. The darker color indicates that the equilibrium has shifted to the product side where there are more moles of gas, to relieve the stress.
Effect of temperature on a chemical equilibrium
Exothermic reaction releases heat, i.e., heat is one of the products. If the forward reaction is exothermic, the reverse reaction must be endothermic by the same amount and vice versa.
When a reversible reaction at equilibrium is disturbed by increasing temperature, the equilibrium shifts in the endothermic direction of the reaction to remove the heat and vice versa.
In the equilibrium between colorless N2O4 and brown color NO2 gases, the reaction is endothermic in the forward direction:
\mathrm{N}_{2} \mathrm{O}_{4}(g, \text { colorless })+\text { Heat } \rightleftarrows 2 \mathrm{NO}_{2}(g, \text { brown })\nonumber
An increase in temperature shifts the equilibrium in the endothermic direction to relieve the stress, as demonstrated in Fig. 6.4.7. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/06%3A_Acids_and_bases/6.04%3A_Acid-base_equilibrium.txt |
How do water molecules dissociate?
Water is an amphoteric substance, which means water can accept a proton acting as a base, and it can also donate A proton acting as an acid.
About one water molecule in half a billion dissociates into an OH- ion by losing a proton to another water molecule. The molecule that receives a proton becomes H3O+.
The dissociation of water is an equilibrium reaction in which one water molecule donates its proton to another water molecule.
The water molecule that receives proton is acting as a base, and it converts to conjugate acid H3O+. The other water molecule that donates a proton is acting as an acid, and it converts to conjugate base OH-. The arrows in the reaction show that the base uses one of its lone pairs of electrons to make a bond with proton, and the previous bond pair of electrons turns into a third lone pair of electrons on the oxygen atom of the base. The reaction is reversible, i.e., the conjugate acid (H3O+) and the conjugate base (OH-) react to re-form the two water molecules.
Water dissociation constant
The dissociation of water is an equilibrium reaction. It means the rate of the forward reaction is equal to the rate of the reverse reaction and the concentration of the reactants and products do not change at equilibrium. The molar concentration of H3O+ represented as [H3O+] is equal to 10-7 M in a pure water sample at 25 oC, where M is in moles/Liter. The molar concentration of OH- represented as [OH-] is equal to the molar concentration of H3O+ in pure water, i.e., [H3O+] = [OH-] = 10-7 M.
The product of the molar concentration of H3O+ and OH- in water is a constant called water dissociation constant Kw equal to 10-14 at 25 oC, i.e.:
$\mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=\left(10^{-7}\right)\left(10^{-7}\right)=10^{-14}\nonumber\nonumber$
Neutral, acidic, or basic aqueous solutions
A solution that has an equal concentration of H3O+ and OH-, each equal to 10-7 M, is a neutral solution.
An acidic solution has an acid dissolved in water. When an acid dissolves in water it dissociates adding more H3O+. The [OH-] must decrease to keep the Kw constant.
A solution that has [H3O+] more than 10-7, and [OH-] less than 10-7 is an acidic solution.
A basic solution has a base dissolved in water. When a base dissolves in water it dissociates adding more OH-. The [H3O+] must decrease to keep the Kw constant.
A solution that has [H3O+] less than 10-7, and [OH-] more than 10-7 is a basic solution.
Calculations of [H3O+] and [OH-] based on Kw
The water dissociation constant remains the same whether the aqueous solution is neutral, acidic, or basic, i.e.:
\mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}_{-}^{+}\right]\left[\mathrm{OH}^{-}\right]=\left(10^{-7}\right)\left(10^{-7}\right)=10^{-14} \text { at } 25^{\circ} \mathrm{C}\nonumber
Therefore, if the molar concentration of hydronium ions [H3O+] is known, the molar concentration of hydroxide ions [OH-] can be calculated using the following formula:
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{\mathrm{K}_{w}}{[\mathrm{OH}^{-}]}=\frac{10^{-14}}{[\mathrm{OH}^{-}]}\nonumber$
Similarly, if the molar concentration of hydroxide ions [OH-] is known, the molar concentration of hydronium ions [OH-] can be calculated using the following formula:
$\left[\mathrm{OH}^{-}\right]=\frac{K_{w}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}=\frac{10^{-14}}{\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]}\nonumber$
When a strong acid like HCl dissolves in water, it dissociates ~100% into ions. Therefore, the [H3O+] is equal to the molar concentration of the acid. The amount H3O+ added by dissociation of water molecules is very small compared to that coming from the dissociation of a strong acid and can be neglected. Similarly, when a strong base like NaOH dissolves in water, it dissociates ~100% into ions. Therefore, the [OH-] is equal to the molar concentration of the base.
Note
When a weak acid or a weak base dissolves in water, it partially dissociates into ions. Therefore, the [H3O+] or the [OH-] in the cases of weak acids and weak bases has to be determined experimentally for the calculations.
Example $1$
Calculate the concentration of OH- ions in a 0.10 M HNO3 solution?
Solution
The HNO3 is a strong acid. Therefore, [HNO3] = 0.10 M = [H3O+]. Desired [OH-] = ?
Formula: $\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]}$
Plug in values an calculate: $\left[0 \mathrm{H}^{-}\right]=\frac{10^{-14}}{0.10}=10^{-13}\mathrm{~M}$
Example $2$
A vinegar solution has [H3O+] = 2.0 x 10-3. a) What is the hydroxide ion concentration in the vinegar solution? b) is the solution acidic, basic, or neutral?
Solution
a) Given [H3O+] = 2.0 x 10-3. Desired [OH-] = ?
Formula: $\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]}$
Plug in values and calculate: $\left[0 H^{-}\right]=\frac{10^{-14}}{2.0 \times 10^{-3}}=5.0 \times 10^{-12} \mathrm{M}$
b) The solution is acidic because [H3O+] > [OH-].
Example $3$
Calculate the value of [H3O+] and [OH-] in a 0.010 M NaOH solution?
Solution
The NaOH is a strong base. Therefore [NaOH] = 0.010 M = [OH-]. Desired [H3O+] = ?
Formula: $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\mathrm{K}_{\mathrm{w}} /\left[\mathrm{OH}^{-}\right]=10^{-14} /\left[\mathrm{OH}^{-}\right]$
Calculations: $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=10^{-14} / 0.010=10^{-12} \mathrm{M}$
Example $4$
a) Calculate the [H3O+] in an ammonia solution that has [OH-] = 4.0 x 10-4 M? b) Is the
solution acidic, basic, or neutral?
Solution
a) Given [OH-] = 4.0 x 10-4. Desired [H3O+] = ?
Formula:$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\mathrm{K}_{\mathrm{w}} /\left[\mathrm{OH}^{-}\right]=10^{-14} /\left[\mathrm{OH}^{-}\right]$
Calculations: $\left[0 H^{-}\right]=\frac{10^{-14}}{4.0 \times 10^{-4}}=2.5 \times 10^{-11} \mathrm{M}$
b) The solution is basic because [H3O+] < [OH-]. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/06%3A_Acids_and_bases/6.05%3A_Dissociation_of_water.txt |
What is pH?
The acidity of an aqueous solution relates to H3O+ ion concentration, but there are two problems with it. The first problem is that the molar concentration of hydronium ion [H3O+] varies over a wide range, usually from 1-to-0.00000000000001 which is not easy to comprehend. Log scale solves the problem, e.g., if a number increases from 1 to 1000, it increases from 0 to 3 on a log scale to the base 10. The second problem is that the [H3O+] is usually a small number between 0 and 1. The log of a number that is between 0 and 1 is a negative number that is not easy to grasp mentally. If a number is between 0 and 1, its reciprocal is more than 1, and its log is a +ve number.
The pH is defined as a log of reciprocal of the molar concentration of hydronium ions [H3O+]. It can also be stated that the pH is a negative log of the molar concentration of hydronium ion [H3O+]. The mathematical form of the pH is the following.
p H=\log \frac{1}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\nonumber
A pH is usually a positive number between 0 to 14.
Calculating pH
First, determine the molar concentration of hydronium ion [H3O+]. Than take log base 10 of [H3O+] and change the sign of the answer, i.e.:
p H=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\nonumber
Example $1$
Calculate the pH of neutral water that has [H3O+] = 10-7M?
Solution
Given [H3O+] = 10-7 M, desired pH = ?
Formula: $p H=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$
Plug in values and calculate: $p H=-\log \left(10^{-7}\right)=7.0$
The keys-sequence for the calculation on a Microsoft Window’s scientific calculator is (the last box is the output of the keys-sequence):
Significant figures in pH value
The given number in scientific notation is 1 x 10-7 which has one significant figure shown by the coefficient part of the number in bold font. The number of decimal places in the pH number shows the significant figures, i.e., 7.0 has one decimal place shown in bold font, which means it has one significant figure.
Example $2$
Calculate the pH of the 0.010 M HCl solution?
Solution
Given [HCl] = 0.010 M = [H3O+], desired pH = ?
Formula: $p H=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$
Plug in values and calculate: $p H=-\log (0.010)=2.00$
The keys-sequence for the calculation on a Microsoft Window’s scientific calculator is (the last box is the output of the keys-sequence):
Note that the given number 0.010 M has two significant figures shown in bold fonts. So, the answer pH 2.00 also has two significant figures shown in bold fonts.
Example $3$
Calculate the pH of 0.10 M HCl solution?
Solution
Given [HCl] = 0.10 M = [H3O+], desired pH = ?
Formula: $p H=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$
Calculations: $p H=-\log (0.10)=1.00$
As the acid 0.01M and 0.10M is gradually added to increase the acid from neutral water in example 1 to 0.01M in example 2 to 0.10 M in example 3, , the pH gradually decreased from 7 to 2 to 1.
The pH of the acidic solution is less than 7. The more acidic the solution, the lower the pH.
The pH decreases as the [H3O+] increases because the reciprocal of [H3O+] is used in the pH calculation. The larger the given number, the smaller the reciprocal; it translates to “the more acidic the solution, the lower the pH.
Example $4$
Calculate the pH of 0.010 M NaOH solution?
Solution
Given [NaOH] = 0.010 M = [OH-], desired pH = ?
First, calculate [H3O+] from the give [OH-]: $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=10^{-14} /\left[\mathrm{OH}^{-}\right]=10^{-14} / 0.010=1.0 \times 10^{-12}$
Than calculate the pH: $p H=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log \left(1.0 \times 10^{-12}\right)=12.00$
Note that the given number 0.010 M has two significant figures shown in bold fonts. So, the answer pH 12.00 also has two significant figures shown in bold fonts.
Example $5$
Calculate the pH of 0.10 M NaOH solution?
Solution
Given [NaOH] = 0.10 M = [OH-], desired pH = ?
First, calculate [H3O+] from the give [OH-]: $10^{-14} /\left[\mathrm{OH}^{-}\right]=10^{-14} / 0.10=1.0 \times 10^{-13}$
Formula: [H3O+] = 10-14/[OH-] = 10-14/0.10 = 1.0 x 10-13.
Than calculate the pH: $p H=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log \left(1.0 \times 10^{-13}\right)=13.00$
As the base, 0.010M and 0.10M are gradually added in the previous two exams to the neutral water at pH 7, the pH increases from 7 to 12 to 13.
The pH of the basic solution is more than 7. The more basic the solution, the higher the pH.
pH measurement
The pH is usually measured in laboratories by digital pH meters. The electrode of the pH meter is first calibrated with solutions of know pH values, and then the electrode is dipped in the test solution to read its pH value. Universal pH indicator papers are available that turn to a specific color when placed in the solution. The pH is read by matching the color of the test paper with the color on the chart. Fig. 6.6.1 shows a digital pH meter and two universal pH indicating papers commonly used for pH measurements in laboratories.
The pH indicators are weak acids or weak bases that change color in a specific pH range. A few drops of the solution of a pH indicator paper are added to the test solution in which pH is changed by adding acid or base. The change in the color of the solution indicates the pH range in which the indicator changes the color. Figure 6.6.2 shows the colors and the color transition ranges of some of the common pH indicators.
Calculating hydronium ion concentration from pH
The formula for calculating the molar concentration of hydronium ions [H3O+] is obtained by rearranging the pH formula. $p H=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$, re-arranges to: $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\operatorname{antilog}(-p H)=10^{-p H}$.
Example $6$
Calculate [H3O+] of a urine sample that has pH 7.56?
Solution
Given: pH = 7.56, Desired: [H3O+] = ?
Formula: $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=10^{-p H}$.
Plug in the values and calculate: $\left[H_{3} O^{+}\right]=10^{-p H}=10^{-(7.56)}=2.8 \times 10^{-8} M$
The keys-sequence for the calculation on a Microsoft Window’s scientific calculator is (the last box is the output of the keys-sequence):
The calculated answer in the above example is a lengthy number, but it is rounded to two significant figures shown in bold font in 2.8$\times 10^{-8} M$, because the given number pH 7.56 has two significant figures shown in bold font.
Significance of pH Scale
The pH scale varies from ~0 to ~14. The pH of 7 is neutral, pH more than 7 is basic, and pH less than 7 is acidic, as illustrated in Fig. 6.6.3. Different foods have different pH values, as shown in Fig. 6.1.1. Similarly, several of the acids and bases in household use have a specific pH range, as shown in Fig. 6.6.3.
The control of pH is essential in the proper functioning of biological systems. Plants thrive if the soil does not have too acidic or too basic a pH. Lacks and revers have a specific range of pH in which the aquatic life can survive. US natural water has a pH in the range of 6.5 to 8.5 range. Seawater has a specific narrow range of pH 7.5 to pH 8.4 in which the life in the sea can function correctly.
The pH of body fluids
Human body fluids vary in pH, as shown in Fig. 6.6.4. Saliva in the mouth is slightly acidic, but the stomach has the lowest pH in the body. The strongly acidic pH in the stomach helps in digesting some foods, and it also helps in killing bacteria that may enter the stomach through the foods. When the food enters the large intestine, the pH changes to basic which helps in digesting the foods that could not be digested in the acidic environment of the stomach. The pH changes to more basic in the small intestine.
The blood has a pH of 7.4, and it can vary in a small range of 7.35 to 7.45. If the blood pH goes outside of the 7.35 to 7.45 range it can result in medical problems. Enzymes in the body need a specific pH range because hydrogen bonding plays a vital role in the structures needed for proper functioning. pH changes affect the hydrogen bonding and can make the enzymes less active or may inactivate them. The pH of blood is maintained by a complex action of buffers that are described in the later sections. The pH of urine can vary over a broad range from 4.6 to 8, depending on the recent diet and exercises.
Acid rain
Acidic gases like NO, NO2, N2O4, SO2 are released into the environment primarily during the combustion of fossil fuels. These gases dissolve in the rainwater and make the rain acidic. For example, sulfur dioxide dissolves in water and makes sulfurous acid:
$\mathrm{SO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightleftarrows \mathrm{H}_{2} \mathrm{SO}_{3}(\mathrm{aq})\nonumber$
The acid rain damages the environment by making the soil, river water, and lack-water acidic. The acidic soil and water, in turn, affect the plants and aquatic life. The acid water also reacts with calcium carbonate and corrodes metals that are responsible for damage to the sculpture and other structures, as illustrated in Fig. 6.6.5, Fig. 6.6.6, and Fig. 6.6.7. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/06%3A_Acids_and_bases/6.06%3A_The_pH.txt |
Acid-base reactions are chemical reactions that involve the transfer of a proton (H+). Examples include the reactions of acids with metals, carbonates, and Arrhenius bases, described in the following.
Reactions of acids with metals
Metals tend to give out an electron and become cations. The majority of the metals, called reactive metals, give out electrons to protons in the acids and release H2 gas. For example, Fig. 6.7.1 shows magnesium reacting with hydrochloric acid by the following reaction:
$\mathrm{Mg}(\mathrm{s})+2 \mathrm{HCl} \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g}) \uparrow\nonumber$
Alkali metals like Na give out one electron, alkaline earth metals like Mg give out two electrons, and aluminum gives out three electrons. A proton can accept one electron. Therefore, the number of acidic protons should be equal to the number of electrons given out to balance the electron transfer in these reactions.
Note
Noble metals or jewelry metals like gold, silver, platinum, copper are exceptions –they do not react with acids.
Example $1$
write a balanced equation for the reaction of aluminum with HBr?
Solution
Step 1) write the equation with the correct formulas of reactants and products.
$\mathrm{Al}(\mathrm{s})+\mathrm{HBr}(\mathrm{aq}) \rightarrow \mathrm{AlBr}_{3}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})\nonumber$
Note that Al gives out three electrons to make Al3+; that is why three Br- are attached to it in the product to balance the charge.
Step 2) balance the electrons lost by the metal with the number of acidic protons.
$\mathrm{Al}(\mathrm{s})+3 \mathrm{HBr}(\mathrm{aq}) \rightarrow \mathrm{AlBr}_{3}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})\nonumber$
Step 3) Balance the rest of the elements by the hit and trial method. There are 3 hydrogen atoms on the left, balance them by adding 3/2 coefficient to the hydrogen on the right.
$\mathrm{Al}(\mathrm{s})+3 \mathrm{HBr}(\mathrm{aq}) \rightarrow \mathrm{AlBr}_{3}(\mathrm{aq})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g})\nonumber$
Step 4. Although the equation is balanced, it is recommended to remove the fraction by multiplying the coefficients in the whole equation with the highest common factor to obtain the final balanced equation.
$2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HBr}(\mathrm{aq}) \rightarrow 2 \mathrm{AlBr}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{~g})\nonumber$
Reactions of acids with carbonates and hydrogen carbonates
Carbonic acid (H2CO3) is a weak acid found in carbonated water. The H2CO3 is a product of carbon dioxide (CO2) and water (H2O) by the following equilibrium reaction.
$\ce{CO2(aq) + H2O(l) <=> H2CO3(aq)}\nonumber$
Hydrogen carbonate (HCO3-) and carbonate (CO32-) are one and two hydrogen less, respectively than carbonic acid. The salts containing HCO3- and CO32- accept one and two protons, respectively, from acids to make H2CO3. The H2CO3 decomposes to carbon dioxide and water by the reverse reaction shown above. Fig. 6.7.2 shows the reaction of sodium hydrogen carbonate with hydrochloric acid.
$\mathrm{NaHCO}_{3}(\mathrm{~s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})+\mathrm{CO}_{2}(\mathrm{~g}) \uparrow\nonumber$
Similarly, calcium carbonate found in limestone reacts with acids. For example, sulfuric acid is one of the components in acid rain that reacts with calcium carbonate and damages sculptures made of stone.
$\mathrm{CaCO}_{3}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow \mathrm{CaSO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{~g}) \uparrow\nonumber$
Reactions of acids with Arrhenius bases
Acids release proton (H+) and Arrhenius bases release hydroxide ions (OH-) in solution. When an acid mix with the Arrhenius base, H+ and OH- ions react with each other and produce water molecules.
$\mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber$
This is the net ionic equation of a reaction of an acid with an Arrhenius base. The cation of the base and the anion of the acid do not react –they are spectator ions. The reaction equation for an acid-base reaction is written in various ways explained below.
Molecular equation
The molecular equation shows the formulas of the substances. For example, the molecular equation of the reaction of HCl with NaOH is the following.
$\mathrm{HCl}(\mathrm{aq})+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber$
Complete ionic equation
The complete ionic equation shows the electrolytes dissociated into ions as they actually exist in the water phase, i.e., as aqueous ions. The complete ionic equation of the reaction of HCl with NaOH is the following.
$\mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Na}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{Na}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\nonumber$
It is evident that Na+ and Cl- do not react –they are the spectator ions.
Net ionic equation
The spectator ions are present on both sides of the equation in equal numbers. The spectator ions can be canceled out, like the terms in an algebraic equation.
$\mathrm{H}^{+}(\mathrm{aq})+\cancel{\mathrm{Cl}^{-}(\mathrm{aq})}+\cancel{\mathrm{Na}^{+}(\mathrm{aq})}+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\cancel{\mathrm{Na}^{+}(\mathrm{aq})}+\cancel{\mathrm{Cl}^{-}(\mathrm{aq})}\nonumber$
The net ionic equation shows the substances that are not spectator ions after canceling out the spectator ions from the complete ionic equation. The net ionic equation of the reaction of HCl with NaOH is the following.
$\mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber$
Note
All of the reactions of the strong acids with strong Arrhenius bases have the same net ionic equation.
Writing a balanced chemical equation of the reaction of acids with Arrhenius bases
The following example shows the steps needed to write the balanced equation of the reaction of an acid with an Arrhenius base.
Example $2$
Write a balanced chemical equation of a reaction between HCl and Ca(OH)2?
Solution
Step 1) Write the formula of acid and base in the reactants and salt and water in the products. All the strong electrolytes dissolve in water, so use (aq) to represent their state.
$\mathrm{HCl}(\mathrm{aq})+\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \text { Salt(aq) }+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber$
Step 2) Balance the H+ in the acid with the OH- in the base.
$2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \mathrm{Salt}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber$
Step 3) Balance the H2O with the H+ in the acid, or with the OH- in the base.
$2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \text { Salt }(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{I})\nonumber$
Step 4) Write the salt by combing the cations from the base and the anions from the acid. Make sure the charges are balanced in the salt to make it a neutral substance.
$2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{I})$
Example $3$
Write a balanced chemical equation of a reaction between H2SO4 and Ba(OH)2?
Solution
Step 1) Write the formula of acid and base in the reactants and salt and water in the products. All the strong electrolytes dissolve in water, so use (aq) to represent their state.
$\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \mathrm{Salt}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber$
Step 2) Balance the H+ in the acid with the OH- in the base. They are already balanced in the above equation.
Step 3) Balance the H2O with the H+ in the acid, or with the OH- in the base.
$\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \mathrm{Salt}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber$
Step 4) Write the salt by combing the cations from the base and the anions from the acid. Make sure the charges are balanced in the salt to make it a neutral substance.
$\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \mathrm{BaSO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber$
Antacids
The stomach sometimes produces excess HCl that may cause heartburn.
Antacids are the substances used to neutralize the excess HCl in the stomach.
Fig. 6.7.3 shows antacid tablets. The antacids include Arrhenius bases with very low solubility in water, like Al(OH)3, and Mg(OH)2, or weak bases like CaCO2, and NaHCO3. that react with and neutralize the acid:
\begin{aligned} &\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{~s})+3 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \ &\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \ &\mathrm{CaCO}_{3}(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})+\mathrm{CO}_{2}(\mathrm{~g}) \ &\mathrm{NaHCO}_{3}(\mathrm{~s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})+\mathrm{CO}_{2}(g) \end{aligned}\nonumber
Side effects of the antacids
The antacids have side effects, e.g., aluminum and calcium-containing antacids may cause constipation. Magnesium-containing antacids have a laxative effect. Some formulations use a mixture of aluminum and magnesium-containing antacids to cancel out the side effect of each other. The calcium-containing antacid is not recommended for persons who tend to develop kidney stones, because the kidney stone is usually a salt of calcium. Sodium-containing antacids may increase the sodium level in the body fluid and may raise the blood pH.
Acid-base titration
The acid-base titration is an analytical process of determining the concentration of acid, called analyte, by neutralizing it with a base of know concentration, called the standard, or vice versa.
Usually, the base is in a burette and added drop by drop to a known volume of the acid in an Erlenmeyer flask, as illustrated in Fig. 6.7.4. A few drops of an acid-base indicator are mixed with the acid. The indicator changes color at a specific narrow pH range.
• The point when the stoichiometric amount of the base has been added to the acid is called the equivalence point.
• The point when the indicator changes color is the end-point.
Usually, the equivalence point is almost the same as the end-point.
The calculation steps after the titration are usually the following.
Steps in acid-base titration calculations
1. The amount, in moles, of the standard is calculated by multiplying the amount in liters with the molarity of the standard.
2. Then, the amount, in moles, of the analyte, is calculated by using the mol-to-mol conversion factor from the balanced chemical equation of the reaction.
3. Finally, the molarity of the analyte is calculated by dividing moles by the volume in liters of the analyte.
Example $4$
What is the molarity of the HCl solution if 50.0 mL of the HCl solution requires 42.0 mL of 0.123 M NaOH solution in the titration?
Solution
Step 1) Given: $[\mathrm{NaOH}]=0.123 \mathrm{~M}=\frac{0.123 \operatorname{~mol} ~N a O H}{1 \mathrm{~NaOH}}\nonumber$
$\text{Volume of standard}=42.0 \mathrm{~mL} .=42.0 \mathrm{~mL} \mathrm{~NaOH} \times \frac{1 \mathrm{~L~NaOH}}{1000 \mathrm{~mL} \mathrm{~NaOH}}=0.0420 ~L \mathrm{~NaOH}\nonumber$
$\text{Vol. analyte}=50.0 \mathrm{~mL}=50.0 \mathrm{~mL} \mathrm{~HCl} \times \frac{1 \mathrm{~L~HCl}}{1000 \mathrm{~mL} \mathrm{~HCl}}=0.0500 \mathrm{~L~HCl}\nonumber$
Step 2) Calculate the moles of the standard from the volume and the molarity product.
\text { Moles of } \mathrm{~NaOH} \text {~consumed }=0.0420 \mathrm{~L} \mathrm{~NaOH} \times \frac{0.123 \mathrm{~mol} \mathrm{~NaOH}}{1 \mathrm{~L} \mathrm{~NaOH}}=0.00517 \mathrm{~mol}{ } \mathrm{~NaOH}\nonumber
Step 3) write the balanced chemical equation of the reaction.
\mathrm{HCl}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber\
Step 4) write the conversion factor for mole of standard to mole of analyte calculation from the equation.
\frac{1 \text {mol } \mathrm{~HCl}}{1 \mathrm{~mol } \mathrm{~NaOH}}\nonumber
Step 5) Calculate the moles of the analyte by multiplying the moles of the standard and the conversion factor.
0.00517 \mathrm{~mol} \mathrm{~NaOH} \times \frac{1 \mathrm{~mol} \mathrm{~HCl}}{1 \mathrm{~mol} \mathrm{~NaOH}}=0.00517 \mathrm{~mol} { }\mathrm{~HCl}\nonumber
Step 6) Calculate the molarity of the analyte by dividing the moles with the volume in liters of the analyte.
0.00517 \mathrm{~mol} \mathrm{~NaOH} \times \frac{1 \mathrm{~mol} \mathrm{~HCl}}{1 \mathrm{~mol} \mathrm{~NaOH}}=0.00517 \mathrm{~mol}{ } \mathrm{~HCl}\nonumber
After learning the steps thoroughly, the calculations can be done in a few steps, as shown in the following.
Calculations: $\quad 42.0 \mathrm{~mL} \mathrm{~NaOH} \times \frac{1 \mathrm{~L} \mathrm{~NaOH}}{1000 \mathrm{~mL} \mathrm{~NaOH}} \times \frac{0.123 \mathrm{~mol} \mathrm{~NaOH}}{1 \mathrm{~L} \mathrm{~NaOH}} \times \frac{1 \mathrm{~mol} \mathrm{~HCl}}{1 \mathrm{~mol} \mathrm{~NaOH}}=0.00517 \mathrm{~mol}{ } \mathrm{~HCl}\nonumber$
Molarity of HCl: $\quad M=\frac{n(\mathrm{mol})}{V(L)}=\frac{0.00517 \mathrm{~mol} \mathrm{~HCl}}{50.0 \mathrm{~mL} \mathrm{~HCl}} \times \frac{1000 \mathrm{~mL} \mathrm{~HCl}}{1 \mathrm{~L~HCl}}=0.103 \mathrm{~M} { }\mathrm{~HCl}\nonumber$
Example $5$
What is the molarity of the H2SO4 solution if 50.0 mL of the H2SO4 solution requires 32.3 mL of 0.201 M NaOH solution in the titration?
Solution
Step 1) Given:
\begin{aligned}
&{[\mathrm{NaOH}]=0.201 \mathrm{~M}=\frac{0.201 \mathrm{~mol} \mathrm{~NaOH}}{1 \mathrm{~L} \mathrm{~NaOH}}} \
&\text { Vol. standard }=42.0 \mathrm{~mL}=32.3 \mathrm{~mL} \mathrm{~NaOH} \times \frac{1 \mathrm{~L} \mathrm{~NaOH}}{1000 \mathrm{~mL} \mathrm{~NaOH}}=0.0323 \mathrm{~L} \mathrm{~NaOH} \
&\text { Vol. analyte }=50.0 \mathrm{~mL}=50.0 \mathrm{~mL~HCl} \times \frac{1 \mathrm{~L} \mathrm{~HCl}}{1000 \mathrm{~mL} \mathrm{~HCl}}=0.0500 \mathrm{~L} \mathrm{~HCl}
\end{aligned}\nonumber
Step 2) Calculate the moles of the standard from the volume and the molarity product.
\text { Moles of } \mathrm{~NaOH} \text { consumed }=0.0323 \mathrm{~N} \mathrm{~NaOH} \times \frac{0.201 \mathrm{~mol} \mathrm{~NaOH}}{1 \mathrm{~L} \mathrm{~NaOH}}=0.00649 \text {~mol } {~NaOH }\nonumber
Step 3) write the balanced chemical equation of the reaction.
\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber
Step 4) write the conversion factor for the moles of standard to mole of analyte calculation from the equation.
\frac{1 \mathrm{~mol}\mathrm{~H}_{2} \mathrm{SO}_{4}}{2 \mathrm{~mol} \mathrm{~NaOH}}\nonumber
Step 5) Calculate the moles of the analyte by multiplying the moles of the standard and the conversion factor.
0.00649 \text { mol } \mathrm{~NaOH} \times \frac{1 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{SO}_{4}}{2 \mathrm{~mol} \mathrm{~NaOH}}=0.00325 \mathrm{~mol} \mathrm{~H}_{2} {SO}_{4}\nonumber
Step 6) Calculate the molarity of the analyte by dividing the moles with the volume in liters of the analyte.
\text { Concentration of } \mathrm{~H}_{2} \mathrm{SO}_{4}=\frac{0.00325 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{SO}_{4}}{0.0500 \mathrm{~L} \mathrm{~H}_{2} \mathrm{SO}_{4}}=0.0649 \mathrm{~M} {~H_{2}} SO_{4}\nonumber
The same calculation in a summary form is in the following.
$\text {Concentration of} \mathrm{~H}_{2} \mathrm{SO}_{4}=\frac{0.00325 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{SO}_{4}}{0.0500 \mathrm{~L~H}_{2} \mathrm{SO}_{4}}=0.0649 \mathrm{~M} \mathrm{~H}_{2} \mathrm{SO}_{4}\nonumber$
$\text{Moles of} \mathrm{~H}_{2} \mathrm{SO}_{4} =\quad 32.3 \mathrm{~mL} \mathrm{~NaOH} \times \frac{1 \mathrm{~L} \mathrm{~NaOH}}{1000 \mathrm{~mL} \mathrm{~NaOH}} \times \frac{0.201 \mathrm{~mol} \mathrm{~NaOH}}{1 \mathrm{~L} \mathrm{~NaOH}} \times \frac{1 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{SO}_{4}}{2 \mathrm{~mol} \mathrm{~NaOH}}=0.00326 \mathrm{~mol} \mathrm{} \mathrm{~H}_{2} \mathrm{SO}_{4}\nonumber$
$\text{Molarity of} \mathrm{~H}_{2} \mathrm{SO}_{4}: \quad M=\frac{n(\mathrm{~mol})}{V(L)}=\frac{0.00326 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{SO}_{4}}{50.0 \mathrm{~mL} \mathrm{~H}_{2} \mathrm{SO}_{4}} \times \frac{1000 \mathrm{~mL} \mathrm{~H}_{2} \mathrm{SO}_{4}}{1 \mathrm{~L~H}_{2} \mathrm{SO}_{4}}=0.0649 \mathrm{~M} \mathrm{~H}_{2} \mathrm{SO}_{4}\nonumber$ | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/06%3A_Acids_and_bases/6.07%3A_Acid-base_reactions.txt |
What is a pH buffer?
A pH buffer is an aqueous solution consisting of a weak acid and its conjugate base or vice versa that lets the pH change to be minimal when a small amount of a strong acid or a strong base adds to it.
For example, the addition of 0.020 mol HCl into 1 L of water changes pH from 7 to 1.7, i.e., about an 80% change in pH. The addition of 0.020 mol NaOH to the same water changes pH from 7 to 12.3, i.e., again, about an 80% change in pH. In contrast to pure water, 1 L of buffer solution containing 0.50 mol acetic acid (CH3COOH) and 0.50 mol CH3COO- -the conjugate of the acetic acid, changes pH from 4.74 to 4.70 by the addition of the same 0.020 mol HCl and from 4.74 to 4.77 by the addition of 0.020 mol NaOH, i.e., about 1% change in pH, as illustrated in Fig. 6.8.1.
Buffers of different initial pH values can be prepared, e.g., by varying the ratio of the weak acid to its conjugate base or by using a different set of a weak acid and its conjugate base. One example is a buffer of initial pH 4.74 comprising 0.5 M acetic acid and 0.5 M sodium acetate shown in Fig. 6.8.1. Another example is a buffer comprising 0.1M dihydrogen phosphate and 0.1M hydrogen phosphate that has an initial pH of 7.21, as shown in Fig. 6.8.2.
Buffer capacity
Buffer capacity refers to how much amount of a strong acid or a strong base the buffer can handle before a drastic change in pH happens.
Higher the amount of weak acid/conjugate base higher the buffer capacity. The buffer has an equal amount of weak acid and its conjugate base has a higher buffer capacity than the same buffer that has an unequal ratio of the acid and its conjugate base.
Caution
Strong acid and its conjugate base or a strong base and its conjugate acid do not make a buffer solution.
Mechanism of buffer action
The buffer contains a weak acid and its conjugate base in equilibrium. For example, acetic acid/sodium acetate buffer has the following equilibrium.
\mathrm{CH}_{3} \mathrm{COOH} \rightleftarrows \mathrm{H}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-}\nonumber
The molar concentration of hydrogen ions [H+] defines the pH of the solution. The conjugate base consumes any strong acid added.
\mathrm{HA}+\mathrm{CH}_{3} \mathrm{COO}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{A}^{-}\nonumber
, where HA is any strong acid and A- is its conjugate base. The concentration of CH3COOH increases and CH3COO- decrease, but pH decreases small because [H+] increases only a little. Similarly, the weak acid consumes any strong base added.
\mathrm{MOH}+\mathrm{CH}_{3} \mathrm{COOH} \rightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{M}^{+}+\mathrm{H}_{2} \mathrm{O}\nonumber
Where MOH is any strong base, and M+ is its conjugate acid. Fig. 6.8.3 illustrates the mechanism of buffer action.
Note
Strong acid and its conjugate base mixture have no buffer action because the acid dissociates almost 100%, leaving no acid behind for the buffer action. The conjugate base of a strong acid is a very weak base that does not react with the added acids. The same explanation applies to a strong base and its conjugate acid mixture having no buffer action.
Buffers in the blood
The blood maintains its pH of ~7.4 primarily by the carbonic acid/hydrogen carbonate buffer system. The blood pH in the range of 7.45 to 7.35 is considered healthy, but outside of this range causes medical problems. If the blood pH decreases to 6.8 or increases to 8.0, death may occur. It is critical to maintain the blood pH in a narrow range for the cells to function correctly. Specifically, the enzymes and other proteins have secondary, tertiary, and quaternary structures needed for their proper functions. Hydrogen bonding plays a crucial role in defining the protein’s structure. pH changes alter the hydrogen bonding making the proteins less effective or ineffective in their functions.
Buffer systems regulate the blood pH. The primary blood buffer system is carbonic acid/hydrogen carbonate, as illustrated in Fig. 6.8.4. Metabolic processes in the cells produce carbon dioxide (CO2) that enters the bloodstream and produces carbonic acid (H2CO3) by reacting with water. Kidneys supply the hydrogen carbonate (HCO3-) –the conjugate base of the carbonic acid and also keep a reservoir of the HCO3-. The H2CO3 consumes any base added and the HCO3- consumes any acid added, thus minimizing the pH change due to the added acids and bases.
The primary mechanism for pH regulation by the H2CO3/HCO3- buffer in the blood is through the lungs. When the blood pH is acidic compared to the average, the breathing rate increases exhaling more CO2 that decreases the concentration of H+, following the blue arrows in Fig. 6.8.4, increasing the pH. The decrease in the breathing rate has the opposite effect. Kidneys also regulate the blood pH by adding or removing HCO3-, but the kidney’s response is delayed compared to the response of the lungs. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/06%3A_Acids_and_bases/6.08%3A_pH_Buffers.txt |
• 7.1: Characteristics of gases
Four parameters, i.e., amount, volume, temperature, and pressure that define the properties of any gas are explained. Measurement of gas pressure, blood pressure, and the kinetic molecular theory of gas are explained.
• 7.2: The pressure-volume relationship
Boyle's law is described, i.e., the volume of a given amount of gas varies in inverse proportion to the pressure of the gas at a constant temperature. The breathing process, i.e., inhalation and exhalation are explained based on the pressure-volume relationship of the gases.
• 7.3: The temperature-volume relationship
Charles's law, i.e., the volume of a given amount of gas is directly proportional to absolute temperature at constant pressure is described.
• 7.4: The pressure-temperature relationship
Gay-Lussac’s law, i.e., the pressure of a gas is directly proportional to the temperature in the Kelvin scale provided the volume and amount of gas is not changed, is described.
• 7.5: The combined gas law
Combined gas law, i.e., the relationship between pressure, volume, and temperature of a given amount of gas are described.
• 7.6: The volume-amount relationship
Avogadro’s law, i.e., the volume of a gas is directly proportional to the amount of gas in moles provided the temperature and pressure of the gas are not changed is described.
• 7.7: Ideal gas law
Combined relationships between all four parameters that together define the properties of gases, i.e., volume, temperature, pressure, and amount of gas are described with examples. The molar volume of an ideal gas at standard temperature and pressure is described and compared with the molar volume of real gases.
• 7.8: Dalton’s law of partial pressure
Dalton’s law of partial pressure, i.e., “the total pressure of a mixture of gases equals the sum of the pressure that each component gas in the mixture would exert if it was present alone” is described. The hyperbaric chamber used for some medical treatments is also described.
07: Gases
What is a gas
Gas is one of the four states of matter that falls between the liquid and the plasma state. The air around us is a gas composed of ~78% nitrogen (N2), ~21% oxygen, and the remaining ~1% are other gases, including carbon dioxide, water vapors, argon, etc., as illustrated in Fig. 7.1.1.
Elements that exist as gases at room temperature and atmospheric pressure include noble gases which are monoatomic molecules including helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn), and diatomic elemental molecules that include hydrogen (H2), nitrogen (N2), oxygen (O2), and two halogens: fluorine (F2) and chlorine (Cl2). Several molecular compounds exist as gases at room temperature, e.g., carbon dioxide (CO2), carbon monoxide (CO), methane (CH4), nitrogen dioxide (NO2), sulfur dioxide (SO2), hydrogen sulfide (H2S), ammonia (NH3).
The properties of gases are defined by their amount in moles (n), volume (V), temperature (T), and pressure (P). These parameters are described in the following.
The amount of a gas
The amount of gas is measured in moles (n), i.e., 6.022 x 1023 molecules of a gas is one mole of the gas. The gas has mass and can be measured in mass units like grams or kilograms, but the mass needs to be divided by the molar mass of the gas to obtain the amount in moles for calculations.
Note
The amount of gas in moles is needed in gas laws because the properties of gases are proportional to the number of molecules, not to the masses of the molecules.
Example $1$
How many moles are in 10.0 g of oxygen gas?
Solution
Step 1. Given: Mass = 10.0 g O2, Desired: ? moles of O2.
Step 2. Equality between the given and the desired units: 1 mole of O2 = 32.00 g O2.
Step 3. The conversion factors from the equality: $\frac{1~mol~O_2}{32.00~g~O_2}$ and $\frac{32.00~g~O_2}{1~mol~O_2}$
Step 4. Multiply the given quantity with the conversion factor that cancels the given and leaves the desired unit in the answer: $10.0 \cancel{~g~O_2)}\times\frac{1~mol~O_2}{32.00\cancel{~g~O_2}} = 0.312 ~g~O_2\nonumber$
The volume of a gas
Volume (V) is the space a substance occupies. The SI unit of volume is a meter cube (m3), i.e., the space inside of a 1m x 1m x 1m cube.
Usually, the volume is reported in liters (L), which is dm3, or milliliter (mL), which is cm3. The relationship between the volume units is the following.
$1 \mathrm{~m}^{3}=1000 \mathrm{~L}\nonumber$
$1 \mathrm{~L}=1000 \mathrm{~mL}\nonumber$
The gases do not have a fixed shape or volume. The gases acquire the shape of the container. The gases expand or contract to fill the available space in a container.
The temperature of a gas
The temperature (T) is a measure of how hot or cold an object is.
The temperature is a manifestation of the thermal energy of a substance. Thermal energy is a source of heat. Heat is the flow of energy from a hot to a cold object. The SI unit of temperature is Kelvin (K) which has a value of 273.15 K at the freezing point of water and 375.15 K at the boiling point of water. The zero of the kelvin scale is called the absolute zero at which no more energy can flow out of a substance as heat. There is no negative number on the kelvin scale of temperature.
Other commonly used units of temperature are Celsius (oC) which has a value of 0 oC at the freezing point of water and 100 oC at the boiling point of water. The relationship between temperature in kelvin (TK) and the temperature in Celsius (TC) is the following.
$T_{K}=T_{C}+273.15\nonumber$
Fahrenheit is a temperature scale used in the English system of measurement that has a value of 32 oF at the freezing point of water and 212 oF at the boiling point of water. The relationship between the temperature in Fahrenheit (TF) and the temperature in Celsius (TC) is the following.
$T_{F}=\frac{9}{5} T_{C}+32\nonumber$
Note
The temperature in Kelvin (K) is used in the gas law calculations.
The pressure of a gas
Pressure (P) is the force (F) per unit area (A), i.e., $P = \frac{F}{A}$.
The matter has mass and applies a force, which is the weight due to gravitational pull. For example, a column 1 m x 1m at sea level and height equal to the earth’s atmosphere, as illustrated in Fig. 7.1.2 has a mass of ~10,000 kg. Mercury (Hg) filled in a column of 1 m x 1m and height 0.760 m has the same mass as the mass of air of 1 m x 1 m column extending from the sea level to the entire atmosphere of the earth.
A column of air extending from sea level to the entire atmosphere applied pressure equal to one atmosphere (atm), where atm is the pressure unit. One atmosphere (atm) is equal to 760 millimeter of mercury (mmHg), where mmHg is another unit of pressure. One mmHg is also called Torr. The relationship between atm, mmHg, and Torr is the following.
$1\mathrm{~atm}=760 \mathrm{~mmHg} = 760 \mathrm{~Torr}\nonumber$
Pascal (Pa)
The SI unit of pressure is pascal (Pa), which is pressure (P) applied by a force (F) of one newton (N) on an area of one meter square (m2). Mathematical form of the unit is: $1 ~Pascal = \frac{1~Newton}{1 ~meter-square}$ or $Pa = \frac{N}{m^2}$
Kilopascal is 1000 Pa, and kilopascal is also called the bar. The English system unit of pressure is pound-force per square inch (psi). The relationship between different units of pressure mentioned above is the following.
$1 \mathrm{~atm}=760 \mathrm{~mm} \mathrm{Hg}=760 \mathrm{~Torr}=1.01325 \times 10^{5} \mathrm{~Pa}=1.01325 \times 10^{2} \mathrm{~kP}=1.01325 \mathrm{~bar}=14.7 \mathrm{~psi}\nonumber$
These relationships are changed to conversion factors that allow conversion from one to another pressure unit, as explained in the following examples.
Example $2$
Express 51 mm Hg in a) atm and b) in Pa.
Solution
Given: Pressure = 51 mm Hg. Required: pressure in atm and Pa
Relationship between mm Hg and atm: $1\mathrm{~atm}=760 \mathrm{~mm Hg}\nonumber$
The conversion factors from the equality: $\frac{1 \mathrm{~atm}}{760 \mathrm{~mm} \mathrm{Hg}}\nonumber$ and $\frac{760 \mathrm{~mm} \mathrm{Hg}}{1 \mathrm{~atm}}\nonumber$
Calculations: multiply the given quantity with the conversion factor that cancels the given unit and leaves the desired unit in the answer.
$51 \cancel{\mathrm{~mm} \mathrm{Hg}} \times \frac{1 \mathrm{~atm}}{760 \cancel{\mathrm{~mm} \mathrm{Hg}}}=6.7 \times 10^{-2} \mathrm{~atm}\nonumber$
Similarly, the relationship between mm Hg and Pa: $760 \mathrm{~mm Hg}=1.01325 \times 10^{5} \mathrm{~Pa}\nonumber$
The conversion factors from the equality: $\frac{760 \mathrm{~mm Hg}}{1.01325 \times 10^{5} \mathrm{~Pa}}\nonumber$ and $\frac{1.01325 \times 10^{5} \mathrm{~Pa}}{760 \mathrm{~mm Hg}}\nonumber$
Calculations: multiply the given quantity with the conversion factor that cancels the given unit and leaves the desired unit in the answer.
$51 \cancel{\mathrm{~mm} \mathrm{Hg}} \times \frac{1.01325 \times 10^{5} \mathrm{~Pa}}{760 \cancel{\mathrm{~mm} \mathrm{Hg}}}=6.8 \times 10^{3} \mathrm{~Pa}\nonumber$
Measurement of gas pressure
The instrument used to measure the atmospheric pressure is called a barometer.
The barometer, shown in Fig. 7.1.3, was invented by Evangelista Torricelli. It is a glass tube that is more than 760 mm long and closed at one end. The glass tube is filled with mercury and inverted to dip its open end in a dish that contains a layer of mercury. The mercury falls from the tube due to its weight but stops falling when the column of mercury is 760 mm in height at sea level due to the atmospheric pressure pushing it back into the tube. Remember 760 mmHg = 1 atm. Above 760 mm is a vacuum in the tube.
The pressure of gases in a laboratory is usually measured using an instrument called a manometer.
A manometer is a U-shaped glass tube partially filled with mercury, as shown in Fig. 7.1.4. One end of the U-tube is connected to a gas chamber in which pressure is being measured, and the other end is either closed (with a vacuum on the closed end), called a closed-ended manometer, or it is open to the atmosphere, called an open-ended manometer.
In the case of a closed-ended manometer, the gas pressure is equal to the difference in the height of mercury in the two arms of the U-tube in mmHg, i.e., height (h) from point N to point B in mm units as shown in Fig. 7.1.4.
Example $3$
Calculate the pressure in atm for the gas shown in Fig. 7.1.4 in the closed-ended manometer if h = 40 mm?
Solution
In a closed-ended manometer, the gas pressure is equal to the height h in mm Hg: P = 40 mm Hg.
Pressure in atm:
$40 \cancel{\mathrm{~mm} \mathrm{Hg}} \times \frac{1 \mathrm{~atm}}{760 \cancel{\mathrm{~mm} \mathrm{Hg}}}=5.3 \times 10^{-2} \mathrm{~atm}\nonumber$
In the case of an open-ended manometer, the gas pressure is equal to the difference in the height of mercury in the two arms of the U-tube in mm of Hg plus the atmospheric pressure outside expressed in mm of Hg.
Variation of air pressure with altitude
The air pressure decreases, and the gas concentration decreases as the altitude increases, as illustrated in Fig. 7.1.5. The oxygen concentration in the air also decreases as the altitude increases. That is why mountain climbers experience an inadequate supply of oxygen to the body at high altitudes -a medical condition called hypoxia.
Blood pressure
The heart pumps the blood into the circulatory system. When the heart contracts, it applies the pressure on the blood in it, and the blood pumps out of the heart into the circulatory system, as illustrated in Fig. 7.1.6. The blood pressure in the circulatory system is highest at this point, and it is called systolic pressure. It can be in the range of 100 to 200 mm Hg. A desirable systolic pressure range is 100 to 120 mm Hg. When heart muscles relax, the cavity in the heart expands and more blood fills in the heart. The blood pressure in the circulatory system is minimum at this point, and it is called diastolic pressure. The diastolic pressure may vary in the range of 60 to 110 mm Hg. A desirable diastolic pressure is less than 80 mm Hg.
The device used to measure blood pressure is called a sphygmometer. It consists of a cuff that wraps around the upper arm, a pump to inflate air in the cuff, and a stethoscope to hear the sound of blood flow, as illustrated in Fig. 7.1.7. The cuff is inflated with air to a pressure above the systolic pressure and it results in cutting off the blood flow through the brachial artery in the upper arm. A stethoscope is used to hear the sound of blood flow. No sound is heard at this point. The pressure in the cuff is slowly reduced. When the pressure in the cuff is equal to systolic pressure, the blood begins to spurt into the artery, and a tapping sound is heard through the stethoscope. At the point when the pressure is equal to the diastolic pressure, the blood flows freely through the artery, and the tapping sound disappears. The blood pressure reading is reported as a set of two numbers, e.g., 120/80, where the higher number is systolic, and the lower number is the diastolic pressure.
Kinetic molecular theory of gases
The physical characteristics of the gases are explainable by a model called the kinetic molecular theory of gases, illustrated in Fig. 7.1.8.
Postulates of kinetic molecular theory of gases
1. Gases consist of particles called molecules that are far apart from each other. As a consequence of the vast distances, the intermolecular forces are negligible, and the volume occupied by the gas molecules is negligible.
2. The gas molecules are moving in straight lines in random directions until they collide with another molecule or the wall of the container.
3. The collisions are elastic, i.e., the total energy is conserved, but the molecules change direction and change speed after the collision. It results in the distribution of molecular speeds over a wide range.
4. The average kinetic energy of the gas molecules is directly proportional to the absolute temperature. The average kinetic energy of the molecules of different gases is the same at the same temperature.
Explanation of the characteristics of the gases based on the kinetic molecular theory of gases
1. Upon exposing gas to a vacuum, the gas molecules keep moving in a straight path in the vacuum until they collide with a container boundary and bounce back. It explains gases are expandable. Collisions of gas molecules with the wall exert outward pressure on the wall and, as a reaction, the wall exerts an equal inward pressure on the gas. Applying more pressure to the gas through any portion of the gas boundary disturbs the above-mentioned balance of action and reaction forces, resulting in an inward movement of the portion of the boundary surface. In other words, the gases are compressible.
2. If a boundary separating two gases is removed, the random molecular motions ultimately mix the two gases making a homogeneous mixture, i.e., gases are entirely miscible with other gases.
3. When temperature increases the gas molecules move faster and colloid more frequently, which explains the fact that gases exert more pressure upon heating and expand if the boundary is not rigid. The more frequent bursting of tires in summer is due to the pressure increase due to heating.
4. When more gas fills in the tire, the pressure increases due to more frequent collisions. Compressing the gas into smaller volume have the same effect, i.e., the molecular collisions increase increasing the pressure.
The physical characteristics of the gases are independent of the nature of the gas molecules due to negligible intermolecular interactions but depend on four physical properties, i.e., pressure, volume, temperature, and amount of gas. The following sections describe the relationships between the physical properties of the gases. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/07%3A_Gases/7.01%3A_Characteristics_of_gases.txt |
Pressure-volume relationship
Consider a gas in a cylinder with a piston pushing it down due to massive objects placed on it and due to outside air pressure, as illustrated in Fig. 7.2.1, The gas molecules strike the piston surface, applying pressure upward equal to the downward pressure applied by the piston that keeps the piston stationary. Increasing the pressure on the piston, e.g., by adding more weight to it, causes the piston to move down, reducing the gas volume. The gas molecules have less distance to travel before striking the piston surface which increases the collision frequency and causes the gas pressure to increase until becomes equal to the outside pressure.
When a change in one parameter causes a change in another, the parameters are related. When an increase in one parameter causes a decrease in another, the two are inversely proportional to each other. Robert Boyle studied the quantitative relationship between the volume and pressure of the gas, keeping the quantity of gas and the temperature constant. The research concluded in a law called Boyle’s law, which states that:
Boyle's law
The volume of a gas is inversely proportional to the pressure of the gas provided the temperature and the amount of the gas remain constant.
Fig. 7.2.2 illustrates the results of Boyle’s experiment. The mathematical form of Boyle’s law is: $V \propto \frac{1}{P}\nonumber$, or $V=\frac{\mathrm{k}}{P}\nonumber$, or $PV=\mathrm{k}\nonumber$, where k is a consant. The graph of volume versus pressure is curvilinear, but the graph of volume versus the reciprocal of pressure is a linear graph showing the inverse proportionality between the volume and the pressure. Since the product PV is a constant, it implies that: $P_{1} V_{1}=P_{2} V_{2}={k}\nonumber$, i.e., a product of initial pressure (P1) and initial volume (V1) is equal to the product of final pressure (P2) and final volume (V2) of gas provided the quantity of the gas and temperature does not change.
Example $1$
The pressure of a 1.32 L sample of SO2 gas at 0.532 atm is increased to 1.231 atm. Calculate the new volume of the gas if the temperature and the quantity of the gas remain the same?
Solution
Given: P1 = 0.532 atm, P2 = 1.231 atm, V1 = 1.32 L V2 = ?
Formula: $P_{1} V_{1}=P_{2} V_{2}={k}$, rearrange to isolate the desired variable: $V_{2}=\frac{P_{1} V_{1}}{P_{2}}$.
Plug in the values in the rearranged formula and calculate: $V_{2}=\frac{0.532 \mathrm{~atm} \times 1.32 \mathrm{~L}}{1.231 \mathrm{~atm}}=0.570 \mathrm{~L}$
Example $2$
An oxygen tank holds 20.0 L of oxygen at a pressure of 10.0 atm. What is the final pressure when the gas is released and occupies a volume of 200 L?
Solution
Given: V1 = 20.0 L, V2 = 200 L, P1 = 10.0 atm P2 = ?
Formula: $P_{1} V_{1}=P_{2} V_{2}={k}$, rearrange to isolate the desired variable: $P_{2}=\frac{P_{1} V_{1}}{V_{2}}$.
Plug in the values in the rearranged formula and calculate: $P_{2}=\frac{10.0 \mathrm{~atm} \times 20.0 \mathrm{~L}}{200 \mathrm{~L}}=1.00 \mathrm{~atm}$.
Breathing process
Boyle’s law explains the mechanism of the breathing process. Lungs are elastic structures like balloons placed in the thoracic cavity, as illustrated in Fig. 7.2.3. The diaphragm muscle makes a flexible floor and ribs surround the cavity.
Inhalation
The inhalation or the inspiration process starts when the diaphragm contract and move down and the rib muscles contract, expanding the thoracic cavity. Volume increases, the air pressure decreases inside the inside thoracic cavity and the atmospheric air flows into the lungs until the pressure in the lungs is equal to the outside pressure.
Exhalation
The exhalation or the expiration process starts when the diaphragm expands and moves upwards, and the rib muscles relax, contracting the thoracic cavity. Volume decreases and the air pressure increases inside the thoracic cavity that pumps the air out of the lungs into the atmosphere. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/07%3A_Gases/7.02%3A_The_pressure-volume_relationship.txt |
Consider a gas in a cylinder with a piston in Fig. 7.3.1. Increasing temperature increases the average kinetic energy (KE) of the gas molecules. The kinetic energy (KE) is directly proportional to the velocity of the molecules, i.e., $KE=\frac{1}{2}mv^2\nonumber$, where m is the mass and v is the velocity. So, increasing temperature increases the velocity resulting in more frequent and more forceful collisions resulting in increased gas pressure inside the chamber. The gas volume starts to increase causing the pressure to decrease until the pressure inside the chamber is equal to the pressure outside. In other words, increasing temperature increases the volume of the gas if the pressure and amount of gas are not changed.
If two related parameters increase or decrease together, they are directly proportional to each other.
Charles's law
Charles’s law states that the volume of a given amount of gas is directly proportional to the temperature in the Kelvin scale at constant pressure.
Fig. 7.3.2 demonstrates that the volume of a gas decreases when the gas is cooled down.
The mathematical forms of Charles’s law are the following.
$V\propto{T}\nonumber$, or $V=\mathrm{k}T\nonumber$, or $\frac{V}{T}=\mathrm{k}\nonumber$
, where k is a constant, V is volume, and T is the temperature (in kelvin scale) of the gas. Since $\frac{V}{T}$ is a constant, it implies that
$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}=\mathrm{k}\nonumber$
where V1 is the initial volume, T1 is the initial temperature in Kelvin, V2 is the final volume, and T2 is the final temperature in Kelvin, provided the amount of gas and pressure do not change. Note that the kelvin scale is used in Charles’s law because the kelvin scale does not have negative numbers which means the linear curve starts from the origin without any y-intercept. If the given temperature is not in the kelvin scale, first convert the temperature to the Kelvin scale and then use the gas laws for the calculations.
Example $1$
A sample of CO2 occupies 3.23 L volume at 25.0 oC. Calculate the volume of the gas at 50.0 oC if pressure and amount of gas do not change?
Solution
Given: T1 = 25.0 oC + 273 = 298 K, T2 = 50.0oC + 273 = 323 K, V1 = 3.23 L, V2 = ?
Formula: $\frac{V_1}{T_1}=\frac{V_2}{T_2}$, rearrange the formula to isolate the desired variable: $V_{2}=\frac{V_{1} T_{2}}{T_{1}}$
Plug in the values in the rearranged formula and calculate: $V_{2}=\frac{3.23 \mathrm{~L} \times 323 \mathrm{~K}}{298 \mathrm{~K}}=3.50 \mathrm{~L}$
Charles’s law explains the drifting of warm air upward in the atmosphere. As the gas is wormed, its volume increases and its density decreases which makes the gas drift upward. A hot air balloon, shown in Fig. 7.3.3 operates using hot air.
7.04: The pressure-temperature relationship
Consider a gas in a cylinder with a piston in Figure $1$. Increasing temperature increases the average kinetic energy (KE) and the average velocity of the gas molecules resulting in more frequent and more forceful collisions which result in increased gas pressure applied on the piston or the walls of the gas container.
Gay-Lussac's law
Gay-Lussac’s law states that the pressure of a gas is directly proportional to the absolute temperature provided the volume and amount of gas are not changed.
The mathematical forms of Gay-Lussac’s law are the following.
$P\propto{T}\nonumber$, or $P=\mathrm{k}T\nonumber$, or $\frac{P}{T}=\mathrm{k}, \nonumber$
where $k$ is a constant, $P$ is pressure, and $T$ is the temperature (in kelvin scale) of the gas. Since Since $\frac{P}{T}$ is a constant, it implies that
$\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}=\mathrm{k}, \nonumber$
where $P_1$ is the initial pressure, $T_1$ is the initial temperature in Kelvin, $P_2$ is the final pressure, and $T_2$ is the final temperature in Kelvin, provided the amount of gas and volume do not change.
Example $1$
The pressure of an oxygen tank containing 15.0 L oxygen is 965 Torr at 55 oC. What will be the pressure when the tank is cooled to 16 oC.
Solution
First, convert the temperatures to the Kelvin scale before applying gas laws.
Given: T1 = 55 oC + 273.15 = 328.15 K, T2 = 16 oC + 273.15 = 289.15 K, P1 = 965 Torr, P2 = ?
Formula:
$\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}, \nonumber$
rearrange the formula to isolate the desired variable:
$P_{2}=\frac{P_{1} T_{2}}{T_{1}} \nonumber$
Plug in the values in the rearranged formula and calculate:
$P_{2}=\frac{965 \mathrm{~Torr} \times 298.15\mathrm{~K}}{328.15 \mathrm{~K}}=850 \mathrm{~Torr} \nonumber$
7.05: The combined gas law
The laws relating to pressure $P$, volume $V$, and temperature $T$ for a constant amount $n$ of a gas are the following:
1. If $n$ and $T$ are constant: $P_{1} V_{1}=P_{2} V_{2}$, that is Boyle's law.
2. If $P$ and $n$ are constant: $\dfrac{V_{1}}{T_{1}}=\dfrac{V_{2}}{T_{2}}$, that is Charles's law.
3. If $\mathrm{V}$ and $\mathrm{n}$ are constant: $\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}$, that is Gay Lussac's law.
All three relationships are combined in the following law.
Combined gas law
If $\mathrm{n}$ is constant: $\dfrac{P_{1} V_{1}}{T_{1}}=\dfrac{P_{2} V_{2}}{T_{2}}\nonumber$, that is the combined gas law.
The combined gas law allows calculating the effect of varying two parameters on the third.
Example $1$
A weather balloon contains $212 \mathrm{~L}$ of helium at $25^{\circ} \mathrm{C}$ and $750 \mathrm{~mm} \mathrm{Hg}$. What is the volume of the balloon when it ascends to an altitude where the temperature is $-40{ }^{\circ} \mathrm{C}$ and $540 \mathrm{~mm} \mathrm{Hg}$, assuming the quantity of gas remains the same?
Solution
Given and desired parameters (temperatures must be converted to Kelvin scale):
$\begin{array}{lll} \mathrm{P}_{1}=750 \mathrm{~mm} \mathrm{Hg}, & \mathrm{V}_{1}=212 \mathrm{~L}, & \mathrm{~T}_{1}=25^{\circ} \mathrm{C}+273.15=298.15 \mathrm{~K} \ \mathrm{P}_{2}=540 \mathrm{~mm} \mathrm{Hg}, & \mathrm{V}_{2}=? & \mathrm{~T}_{2}=-40^{\circ} \mathrm{C}+273.15=233.15 \mathrm{~K} \end{array}\nonumber$
Formula:
$\dfrac{P_{1} V_{1}}{T_{1}}=\dfrac{P_{2} V_{2}}{T_{2}}, \nonumber$
rearrange the formula to isolate the desired parameter:
$V_{2}=\dfrac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}. \nonumber$
Calculations:
$V_{2}=\dfrac{750 \cancel{\mathrm{~mm} \mathrm{Hg}} \times 212 \mathrm{~L} \times 233.15 \cancel{\mathrm{~K}}}{298.15 \cancel{\mathrm{~K}} \times 540 \cancel{\mathrm{~mm} \mathrm{Hg}}}=230 \mathrm{~L}. \nonumber$
7.06: The volume-amount relationship
Figure $1$ illustrates the effect of the amount of gas on the volume. Adding more gas molecules increases the collision frequency of the molecules with the walls increasing the gas pressure. The gas expands to reduce pressure until the pressure of gas in the chamber is equal to the outside pressure.
Avogadro's law
Avogadro’s law states that the volume of a gas is directly proportional to the amount of gas in moles provided the temperature and pressure of the gas are not changed.
The mathematical forms of Avogadro's law are the following.
$V\propto {n}\nonumber$
or
$V=\mathrm{k}n\nonumber$
or
$\dfrac{V}{n}=\mathrm{k}\nonumber$
where $V$ is the volume, $n$ is the number of moles, and $k$ is the constant for the gas under the conditions of constant temperature and pressure. Since $\mathrm{V} / \mathrm{n}$ is a constant, it implies that:
$\dfrac{V_{1}}{n_{1}}=\dfrac{V_{2}}{n_{2}}=\mathrm{k}\nonumber$
where $V_{1}$ and $n_{1}$ is initial volume and the initial amount of gas, respectively, and $V_{2}$ and $n_{2}$ is final volume and the final amount of gas in mole, provided the temperature and pressure are not changed.
Example $1$
A weather balloon containing $3.0$ moles of helium has a volume of $66 \mathrm{~L}$. What is the final volume if $2.0$ moles of helium is added to it. The pressure and temperature of the gas do not change?
Solution
Given $\quad V_{1}=66 L, \quad n_{1}=3.0 \mathrm{~mol}, \quad V_{2}=?, \quad n_{2}=3.0+2.0=5.0 \mathrm{~mol}$
Formula:
$\dfrac{V_{1}}{n_{1}}=\dfrac{V_{2}}{n_{2}}, \nonumber$
rearrange the formula to isolate the desired variable:
$V_{2}=\dfrac{V_{1} n_{2}}{n_{1}}. \nonumber$
Plug in the values and calculate:
$V_{2}=\dfrac{66 \mathrm{~L} \times \times 5.0 \mathrm{~mol}}{3.0 \mathrm{~mol}}=110 \mathrm{~L}. \nonumber$ | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/07%3A_Gases/7.03%3A_The_temperature-volume_relationship.txt |
Combined relationships between pressure, volume, temperature, and amount of gases
The gas laws described above give the following relationships between volume $V$, pressure $P$, the temperature in kelvin $T$, and the amount of gas in moles $n$:
$V \propto 1 / {P}$, at constant $T$ and $n$,
$V\propto T$, at constant $P$ and $n$, and
$V \propto n$, at constat $T$ and $P$.
The three proportionalities combine to give the following ideal gas relationship:
$V \propto \frac{nT}{P}\nonumber$
The proportionality changes to an equation by introducing a constant of proportionality:
$V=\frac{n R T}{P}\nonumber$, that rearranges to $P V=n R T \nonumber$
where $R$ is the proportionality constant called the gas constant.
Ideal gas law
The equation: $PV=n R T$ is called the ideal gas law.
The value of $R$ can be calculated by: $R=\frac{P V}{n T}$, where $n$ is the quantity of gas in a mole, $T$ is the temperature in kelvin, $P$ is the pressure that can be in various units, and $V$ is the volume that can be in various units. The value of $R$ in different units of $P$, $V$, and $PV$ products are given in Table 1. If values of any three among the $P, V, n$, and $T$ are known, the value of the fourth one can be calculated by using the ideal gas law.
Caution
In these calculations, the units of $R$ should be in agreement with the units of $P, V, n$, and $T$. If they are not in agreement, the given unit of $P, V, n$, and $T$ must be converted to agree with the units of $R$.
Table 1: The numerical values of gas constant R in various units
Value Units
0.08206 L-atm/mol-K
8.314 J/mol-K
1.987 cal/mol-K
8.314 m3-Pa/mol-K
62.36 L-torr/mol-K
The ideal gas equation in a rearranged form is $\frac{P V}{n T}=R$ is a constant, that implies that:
$\frac{P_{1} V_{1}}{n_{1} T_{1}}=\frac{P_{2} V_{2}}{n_{2} T_{2}}=R\nonumber$
If one or two parameters in the ideal gas equation $\frac{P_{1} V_{1}}{n_{1} T_{1}}=\frac{P_{2} V_{2}}{n_{2} T_{2}}$ are constant; they cancel out, leaving the relationship between the remaining parameters, e.g.
if $n$ and $T$ are constant: $P_{1} V_{1}=P_{2} V_{2}$, that is Boyle's law,
if $P$ and $n$ are constant: $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$, that is Charles's law,
if $P$ and $T$ are constant: $\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}$, that is Avogadro's law,
if $V$ and $n$ are constant: $\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}$, that is Gay Lussac's law,
if $V$ and $T$ are constant: $\frac{P_{1}}{n_{1}}=\frac{P_{2}}{n_{2}}$, that is pressure-mole relationship,
and if $n$ is constant: $\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$, that is combined gas law.
Example $1$
Calculate the volume of 1.000 mole of a gas in liters (L) at 0.000 oC and 1.000 atm?
Solution: Given $n$=1.000 mol, $T$= 0 oC+273.15=273.15 K, $P$=1.000 atm, and $R$=0.08206 L-atm/mol-K
Solution
Given $n$=1.000 mol, $T$= 0 oC+273.15=273.15 K, $P$=1.000 atm, and $R$=0.08206 L-atm/mol-K
Formula: $PV=nRT$, rearrange the formula to: $V=\frac{n R T}{P}$.
Plug in the values and calculate: $V=\frac{1.000 \mathrm{~mol} \times 0.08206 \frac{\mathrm{L}-\mathrm{atm}}{\mathrm{mol}-\mathrm{K}} \times 273.15 \mathrm{~K}}{1.000 \mathrm{~atm}}=22.41 \mathrm{~L}$
Note
The volume of 1 mol of an ideal gas is 22.41 L at 0 oC and 1 atm pressure, as calculated in the above example
Example $2$
Calculate the volume of a container that has 1.50 mol of He gas at 7.2 atm and 25 oC?
Solution
$P$=7.2 atm, $V$=? $n$=1.50 mol, $T$=25 oC}+273=298 K, R=0.08206 $\mathrm{~L}-\mathrm{atm} / \mathrm{mol}-\mathrm{K}$
Formula: $PV=nRT$, rearrange the formula to: $V=\frac{n R T}{P}$
Plug in the values and calculate: $V=\frac{1.50 \mathrm{~mol} \times 0.08206 \frac{\mathrm{L}-\mathrm{atm}}{\mathrm{mol}-K} \times 298 \mathrm{~K}}{7.2 \mathrm{~atm}}=5.1 \mathrm{~L}$
Example $3$
Calculate the pressure in a 5.1 L container that has 0.60 mol of He at 25 oC?
Solution
$P$ = ?, $V$=5.1 L, $n$ = 0.60 mol, $T$=25 oC+273=298 K, $R=0.08206 \frac{\mathrm{L}-\mathrm{atm}}{\mathrm{mol}-K}$
Formula: $PV=nRT$, rearrange the formula, plug in the values and calculate: $P=\frac{n R T}{V}=\frac{0.60 \mathrm{~mol} \times 0.08206 \frac{\mathrm{L-atm}}{\mathrm{mol}-K} \times 298 \mathrm{~K}}{5.1 \mathrm{~L}}=2.9 \mathrm{~atm}$
Example $4$
Calculate the pressure of 0.60 mol of He in example 3 mixed with 1.50 mol of He in example 2 in a container of 5.1 L volume at 25 oC?Solution
$P_{total}$ = ?, $n_{total}$ = 1.5 mol + 0.60 mol = 2.1 mol, $V$ = 5.1 L, T = 25 oC + 273 = 298 K, and $R=0.08206 \frac{\mathrm{L}-\mathrm{atm}}{\mathrm{mol}-K}$
Formula: $P_{\text {total }} V=n_{\text {total }} R T$, rearrange, plug in the values and calculate: $P_{\text {total }}=\frac{n_{\text {total }} R T}{V}=\frac{2.1 \mathrm{~mol} \times 0.08206 \frac{L-a t m}{m o l-K} \times 298 \mathrm{~K}}{5.1 \mathrm{~L}}=10.1 \mathrm{~atm}$
Note
The $P_{\text {total }}$ of 2.1 mol He in example 4 is equal to $P$ of 1.50 mole He in example 2+$P$ of 0.6 mol He in example 3, i.e., 7.2 atm +2.9 atm =10.1 atm. This calculation demonstrates that when gases are mixed, the total pressure is the sum of the pressures that each fraction will exert if it was alone in that space. It is demonstrated by mixing the same gas, i.e., He with He, but it remains true when different gases are mixed, as long as all the gases involved obey the ideal gas law.
Molar volume of gases at standard temperature and pressure
The temperature of 0oC and pressure of 1 atm is called standard temperature and pressure (STP) for gases.
Standard temperature and pressure
Currently accepted STP is 0oC and 1 bar. The molar volume of an ideal gas at 0oC and 1 bar is 22.71 L, but for most practical purposes, the older definition of STP of 0oC and 1 atm is used.
The calculations in example 1 of the previous section show that the molar volume of an ideal gas is 22.41 L at STP. Fig. 7.7.1 illustrates the molar volume of an ideal gas at STP.
Fig. 7.7.2 shows that the molar volume of real gases is very close to that of the ideal gas. The small differences between the molar volume of real gases and the ideal gas are because ideal gas molecules are assumed to have negligible volume and negligible intermolecular interactions. The real gas molecules do have some volume and some intermolecular interactions that cause deviations of real gases from the ideal behavior. However, for practical purposes, the calculations based on ideal gas law remain applicable for the majority of real gases under ambient conditions.
The molar volume of gases at STP is an equality between the number of moles and the volume of gas at STP, i.e.:
$1 \text { mol gas }=22.4 \mathrm{~L} \text { gas at STP }\nonumber$
The equality gives two conversion factors, i.e.,
$\frac{1 \text { mol gas }}{22.4 \mathrm{~L} \text { gas }}\text {, and }\frac{22.4 \mathrm{~L} \text { gas }}{1 \text { mol gas }} \text {.}\nonumber$
The conversion factors are used to convert volume to moles and mol to volume of gas, respectively, at STP.
Example $5$
Calculate the volume of 64.0 g of oxygen at STP?
Solution
Given: mass of oxygen = 55.2 g. Required: volume of the oxygen =?
Strategy: 1st convert the grams of oxygen to moles of oxygen by using reciprocal of molar mass as a conversion factor and then convert the moles of oxygen to volume of oxygen by using 2nd conversion factor described above:
$\text { volume of oxygen }=64.0 \cancel{\mathrm{~g} \text { oxygen }} \times \frac{1 \cancel{\text { mol oxygen }}}{32 \cancel{\mathrm{~g} \text { of oxygen }}} \times \frac{22.41 \mathrm{~L}}{1 \cancel{\text { mol oxygen }}}=44.82 \mathrm{~L} \text { oxygen }\nonumber$ | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/07%3A_Gases/7.07%3A_Ideal_gas_law.txt |
Calculations in examples 2 and 3 of section 7.7 prove that 1.50 mol He in 5.1 L chamber exerts a pressure of 7.2 atm; 0.6 mol He in the same chamber exerts a pressure of 2.9 atm, and a mixture of the two in the same chamber at the same temperature exerts a pressure equal to the sum of the pressures that each fraction exerts if it is alone in the chamber, i.e., $P_{total} = 7.2 { atm} + 2.9 { atm} = 10.1 { atm}$. What if one of the gas was hydrogen, and the other helium? The answer is: that the calculations using the ideal gas law remain the same because it is the number of molecules, not the type of molecules that are involved in the calculations. Properties of gases depend on the number of moles of gas $n$, and not on the nature of the gas, as illustrated in Fig. 7.8.1.
Dalton's law of partial pressure
The total pressure of a mixture of gases equals the sum of the pressure that each component gas in the mixture would exert if it was present alone.
The mathematical form of Dalton’s law is:
$P_{\text {tatal }}=P_{\text {He }}+P_{\text {exygen }}=9.3 \mathrm{~atm}+2.4 \mathrm{~atm}=11.5 \mathrm{~atm}\nonumber$
, where $P_{1}$, $P_{2}$, $P_{3}$ are partial pressures of individual gas#1, #2, #3 in the mixture.
Atmospheric air is a mixture of nitrogen, oxygen, argon, carbon dioxide, water vapors, and trance amount of some other gases. The atmospheric pressure is the sum of the partial pressures of components of air, as illustrated in Fig. 7.8.2.
Example $1$
A 46 L He and 12 L O2 sample, both at 1.0 atm and 25 oC were pumped into a 5.0 L scuba diving tank at 25 oC. Calculate the partial pressure of each gas and the total pressure in the tank?
Solution
For He before mixing: $V$ = 46 L, $T$ = 25 oC + 273 = 298 K, $P$ = 1.0 atm, $n_{He}$=?, $R=0.08206 \frac{L-\mathrm{atm}}{\mathrm{mol}-K}$
Formula and calculations: $n_{\mathrm{He}}=\mathrm{PV} / \mathrm{RT}=\frac{1.0 \mathrm{~atm} \times 46 \mathrm{~L}}{0.08206 \frac{\mathrm{Latm}}{\mathrm{mol}-\mathrm{K}} \times 298 \mathrm{~K}}=1.9 \mathrm{~mol}$
For $O_{2}$ before mixing: $V$=12 L, $T$ =25 oC + 273 =298 K, $P$=1.0 atm, $n_{O_{2}}$ = ?, $R=0.08206 \frac{\mathrm{L}-\mathrm{atm}}{\mathrm{mol}-\mathrm{K}}$
Formula and calculations: $n_{H_{2}}=\mathrm{PV} / \mathrm{RT}=\frac{1.0 \mathrm{~atm} \times 12 \mathrm{~L}}{0.08206 \frac{\mathrm{L}-\mathrm{atm}}{\mathrm{mol}-K} \times 298 \mathrm{~K}}=0.49 \mathrm{~mol}$
After mixing: $V$=5 L, $T$ =25 oC+273 = 298 K, $n_{He}$=1.9 mol, $n_{O_{2}}$ =0.49 mol, $R=0.08206 \frac{L-\mathrm{atm}}{\mathrm{mol}-K}$.
Requried: $P_{He }$ = ?, $P_{O_{2}}$=?, and $P_{\text {total }}$=?
Formula: $P_{H e}=\frac{n_{H e} R T}{V}$ = $\frac{1.9 \mathrm{~mol} \times 0.08206 \frac{L-a t m}{\mathrm{~mol}-K} \times 298 \mathrm{~K}}{5.0 \mathrm{~L}}=9.3 \mathrm{~atm}$, $P_{O_{2}}=\frac{n_{\text {oxygen } R T}}{V}$ = $\frac{0.49 \mathrm{~mol} \times 0.08206 \frac{\mathrm{L}-\mathrm{atm}}{\mathrm{mol}-K} \times 298 \mathrm{~K}}{5.0 \mathrm{~L}}=2.4 \mathrm{~atm}$,
$P_{\text {total }}=P_{He }+P_{O_{2}}=9.3 \mathrm{~atm}+2.4 \mathrm{~atm}=11.5 \mathrm{~atm.}\nonumber$
Hyperbaric chamber -a medical tool
The hyperbaric chamber is an air chamber that is at two to three atmospheric pressure, as shown in Fig. 7.8.3. The solubility of gases increases with an increase in pressure. A patient placed in a hyperbaric chamber has a higher concentration of oxygen dissolved in blood because the partial pressure of oxygen is two to three times higher than in the atmospheric air. The higher concentration of oxygen is toxic to many strains of bacteria. Therefore the hyperbaric chambers are used to treat burn patients, in surgeries, and to treat some cancers.
The hyperbaric chambers are also used to treat carbon monoxide (CO) poisoning because the higher concentration of oxygen in the chamber can displace the CO bound with hemoglobin faster than atmospheric oxygen does. Another use of hyperbaric chambers is to treat scuba divers suffering from the bends. If a diver ascends too quickly, the nitrogen dissolved in blood makes bubbles in the vessels that block the blood flow –a condition called bends. The divers suffering from the bends are placed in a hyperbaric chamber at high pressure, and then the pressure is slowly decreased to atmospheric pressure. The nitrogen dissolves in blood under higher pressure and slowly diffuses out through the lungs as the pressure is gradually decreased. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/07%3A_Gases/7.08%3A_Daltons_law_of_partial_pressure.txt |
• 8.1: Introduction to nuclear chemistry
The nuclear reactions that involve changes in the nucleus of an atom, radioactivity, and the related terminologies are introduced.
• 8.2: Radioactivity
Radioactive nucleoids disintegrate and emit smaller particles and energy, and the decay modes like alpha-, beta-, gamma-, or positron-decay modes are described.
• 8.3: Half-life of radioisotopes
Half-life, i.e., the time it takes for half of a radioactive sample to decay, characteristics of the half-life and its use in dating ancient objects are described.
• 8.4: Radiation measurements
Measurements of radioactivity, i.e., disintegration per unit time, absorbed dose, i.e., energy deposited by radiation per unit mass, equivalent dose, and effective dose that take account of the relative damaging effects of each type of radiation and sensitivity of the organ exposed to the radiation are described.
• 8.5: Ionizing radiation exposures
Interaction of radiation with matter, its effects on humans, radiation exposure from natural radiation sources in the environment, and protective measure against the radiation exposures are described.
• 8.6: Medical uses of radioisotopes
Applications of nuclear radiation in medical treatment, like medical imaging and killing of cancer cells using external or internal radiation sources, are introduced.
• 8.7: Making radioisotopes for medical uses
Production of radionuclides commonly used in medical therapies is introduced.
• 8.8: Nuclear fusion and fission
The nuclear fusion process is a source of energy from the sun and stars and thermonuclear weapons, and nuclear fission is a source of energy in the nuclear power plants and in conventional nuclear weapons are introduced.
08: Nuclear chemistry
What is a nuclear reaction?
Unlike chemical reactions that involve valence electrons, nuclear reactions involve changes in the nucleus of an atom, as shown in Fig. 8.1.1.
Nuclear reaction
Nuclear reactions involve changes in the nucleus of an atom. A nuclear reaction may result in one or more of the following: i) conversion of an atom to its isotope or an atom of another element, ii) conversion of mass into energy or vice versa, and iii) release of nuclear radiations.
Although nuclear reactions are less numerous than chemical reactions, they are essential in many aspects, e.g., they are the source of energy in the sun and stars and the synthesis of elements in the universe. Nuclear reactions are becoming essential in human life in the form of electricity production from nuclear power plants, a source of radioisotopes for medical imaging to visualize organs and diagnose diseases, to treat tumors, and cancerous cells, as shown in Fig. 8.1.2.
Nuclear nomenclature and symbols
Nucleoid
Nucleoid is another name for the nucleus of an atom, that is often used in nuclear chemistry
The composition of a nucleoid is represented by the same symbol that represents the isotopes of elements, as $\ce{^{A}_{Z}X}$, where $\ce{X}$ is the element symbol, Z is the number of protons, and A is the number of protons and neutron in the nucleus. For example, carbon exists as a mixture of $\ce{^12_6C}$, and $\ce{^13_6C}$ isotopes. The name of the element, followed by the number of nucleons separated by a hyphen, is another way of representing a nucleoid. For example, carbon-12, carbon-13, and carbon-14 represent the carbon nucleoids having 6 protons each, but 6, 7, and 8 neutrons, respectively. Similarly, hydrogen exists as a mixture of $\ce{^1_1H}$, $\ce{^2_1H}$, and $\ce{^3_1H}$, that can also be represented as hydrogen-1, hydrogen-2, and hydrogen-3, respectively.
Nucleons
The protons and neutrons are also called nucleons.
Nuclear reaction
A nuclear reaction is a process in which two nuclei, or a nucleus and an external subatomic particle, collide to produce one or more new nuclides.
The nuclear reaction is a reaction that involves nucleoids. The reactant nucleoid, called the parent nucleoid, usually transforms into a different nucleoid called the daughter nucleoid. The daughter nucleoid may be an isotope of the parent nucleoid, or it may be a different element. The conversion of an isotope to another isotope of the same or a different element is a nuclear reaction that is called transmutation or a nuclear transformation, as shown in Fig. 8.1.3.
Nuclear radiations
Nuclear radiation or radioactivity is the particles and energy emitted by the nucleus during a nuclear reaction.
The nuclear reaction is accompanied by the emission of nuclear radiations including high-energy electromagnetic traditions called gamma-rays ( $\ce{\gamma}$-rays), subatomic particles like electrons, positrons, protons, neutrons, or a small nucleus, like $\ce{^4_2He^2+}$ called alpha-particles ( $\ce{\alpha}$-particles). Nuclear radiations are ionizing radiations, i.e., they can knock off electrons from the atoms they come in contact with.
Radioactive
The nucleoids that are capable of spontaneous disintegration, causing the emission of nuclear radiation, are called radioactive.
The process of emission of nuclear radiation by a spontaneous disintegration of radioactive nucleoids is called radioactivity, as illustrated in Fig. 8.1.4.
The nuclear radiations include gamma-rays ( $\ce{\gamma}$-rays), alpha-particles ( $\ce{\alpha}$-particles), beta-particle ( $\ce{\beta}$-particles), neutrons (n), and positron ($\ce{\beta^+}$-particles).
Gamma-rays
The gamma-rays are electromagnetic radiations that have no mass and have energy higher than that of X-rays. The symbol $\ce{\gamma}$, $\ce{_{0}^{0}\gamma}$, or $\ce{\gamma}$-ray represents a gamma-ray.
Alpha-particles
The alpha-particles ($\ce{\alpha}$-particles) are helium nuclei with two protons, two neutrons, and without electrons, i.e., $\ce{^4_2He^2+}$. The $\ce{\alpha}$-particles are also represented as $\ce{^4_2He}$ or Helium-4.
Beta-particles
The beta-particles ($\ce{\beta}$-particles) are fast-moving electrons that have atomic number -1, charge -1, and negligible mass. The symbol $\ce{\beta}$, $\ce{\beta^{-}}$, $\ce{_{-1}^{0}\beta}$, or $\ce{_{-1}^{0}{e}}$ also represents a β-particle.
Positrons
Positrons are anti-particle of electron, i.e., they have the same mass but opposite charge than that of an electron. The symbol $\ce{+\beta}$, $\ce{\beta^{+}}$, $\ce{_{+1}^{0}\beta}$, or $\ce{_{+1}^{0}{e}}$ represents a positron. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/08%3A_Nuclear_chemistry/8.01%3A_Introduction_to_nuclear_chemistry.txt |
The cause of radioactivity
Only a particular combination of protons and neutrons forms stable nucleoids; the rest are unstable nucleoids, as illustrated in Fig. 8.2.1. The observations on the stable nucleoids are the following.
Observations on the stable nucleoids
1. Hydrogen with one proton and no neutrons is a stable nucleoid.
2. Other light nucleoids, up to atomic number 20 are usually stable when the number of protons is equal to the number of neutrons
3. The medium nucleoids from atomic numbers 20 to 82 are generally stable when the number of neutrons is more than the number of protons.
4. The heavier nucleoids with atomic number 84 or more are unstable.
The nucleoids shown as black dots in Fig. 8.2.1 are naturally occurring stable nucleoids. All nucleoids shown in colored dots other than black are unstable.
Note
The unstable nucleoids are radioactive -they spontaneously disintegrate, i.e., they tend to re-arrange the nucleon composition in the nucleus to become a more stable nucleoid.
The radioactivity releases energy and particles, i.e., nuclear radiation along with the re-arrangement of the nucleus, as explained in the following sections.
Alpha-decay
The $\ce{\alpha}$-emission is one of the important processes for stabilizing heavy unstable nucleoids shown with yellow dots in Fig. 8.2.1. Since the $\ce{\alpha}$-particle has two protons and two neutrons, the daughter nucleoid produced along with the $\ce{\alpha}$-decay has two fewer protons and four fewer mass numbers in it than the parent nucleoid, as illustrated in Fig. 8.2.2.
For example:
$\ce{_92^238U -> _90^234Th + _2^4He}\nonumber$
Some times the $\ce{\alpha}$-decay is accompanied by $\ce{\gamma}$-emission, e.g.:
$\ce{_84^210Po -> _82^206Pb + _2^4He + \gamma}\nonumber$
Note that the $\ce{\gamma}$-rays have zero mass, so they do not change the atomic number and mass number of the parent nucleoid.
Steps of writing a nuclear reaction equation
1. Write the symbols of the know nucleoids, particles, and radiations in the reactants and products, separated by an arrow. Leave a question mark for the unknown pieces of information.
2. Balance the mass number on the two sides of the equation.
3. Balance the atomic number on the two sides of the equation.
4. Wright the symbols of the unknown nucleoid or particles by finding them in the periodic table, based on the atomic numbers.
$\ce{\alpha}$-decay in smoke detectors
Smoke detectors used in homes need $\ce{\alpha}$-particles for their function. Americium-241 is the $\ce{\alpha}$-decay-emitter used in the smoke detectors.
Example $1$
Write the nuclear reaction equation for the $\ce{\alpha}$-decay of americium-241.
Solution
Step 1. The symbol and the atomic number of americium in the periodic table are Am and 95, respectively. So the initial equation is:
$\ce{_95^241Am -> _{?}^{?}{?} + _2^4He}\nonumber$
Step 2. Balance the mass number on the two sides of the equation, i.e., the mass number of unknown nucleoid is 241-4 = 237:
$\ce{_95^241Am -> _{?}^{237}{?} + _2^4He}\nonumber$
Step 3. Balance the atomic number on the two sides of the equation, i.e., the atomic number of the unknown nucleoid is 95-2 = 93:
$\ce{_95^241Am -> _{93}^{237}{?} + _2^4He}\nonumber$
Step 4. Find the symbol of the unknown nucleoid from the periodic table of elements, i.e., the element at atomic number 93 is neptunium symbol Np:
$\ce{_95^241Am -> _{93}^{237}{Np} + _2^4He}\nonumber$
This is the balanced nuclear equation for the $\ce{\alpha}$-decay of americium-241 in smoke detectors.
Radium-226, present in many types of rocks and soils, is an $\ce{\alpha}$-emitter producing radon-226 in the process. The radon-226 is also an $\ce{\alpha}$-emitter that can diffuse into houses from the rocks and soil underneath the buildings. Radon is an environmental health issue in the buildings when its concentration becomes above a certain level. The nuclear equation for the $\ce{\alpha}$-decay of radon-222 is the following.
$\ce{_86^222Rn -> _{84}^{218}{Po} + _2^4He}\nonumber$
Example $2$
Write the nuclear equation for the α-decay of radium-226.
Solution
Step 1. The initial equation is:
$\ce{_88^226Rn -> _{?}^{?}{?} + _2^4He}\nonumber$
Step 2. Balance the mass number on the two sides of the equation, i.e., is 226-4 = 222:
$\ce{_88^226Rn -> _{?}^{222}{?} + _2^4He}\nonumber$
Step 3. Balance the atomic number on the two sides of the equation, i.e., 88-2 = 86:
$\ce{_88^226Rn -> _{86?}^{222}{?} + _2^4He}\nonumber$
Step 4. Find the symbol of the unknown nucleoid from the periodic table of elements, i.e., the element at atomic number 86 is radon symbol Rn:
$\ce{_88^226Rn -> _{?}^{222}{Rn} + _2^4He}\nonumber$
This is the balanced nuclear equation for the $\ce{\alpha}$-decay of radium-226 in rocks and soil.
Beta-decay
The nucleoids marked blue in Fig. 8.2.1. have more neutrons than needed for stability. They usually stabilize them by converting one of the neutrons (n) into a proton (p) and an electron (e) by the following nuclear process:
$\ce{_{0}^{1}{n} -> _{1}^{1}{p} + _{-1}^{0}{e}}\nonumber$
The proton stays in the nucleus, but the electron emits from the nucleus, as illustrated in Fig. 8.2.3. The emitted electron is called $\ce{\beta}$-particle. The process of the $\ce{\beta}$-particle emission is called beta-decay. Note that the neutron has zero atomic number as there is no proton in it, and the electron has a -1 atomic number to balance the +1 atomic number of the proton. Proton has +1 and electron has -1 charge, which is also balanced. The mass number of an electron is zero as it has negligible mass compared to the mass of a proton or a neutron. The electrical charges and the emission of another particle called neutrino are ignored in this equation.
An example of β-decay is the transformation of nitrogen-16 to oxygen-17:
$\ce{_{7}^{16}{N} -> _{8}^{16}{O} + _{-1}^{0}{e}}\nonumber$
Note that in the $\ce{\beta}$-decay process, the mass number remains the same, but the atomic number increases by one in the daughter nucleus. The nuclear equation is balanced because the mass number is the same (16 = 16+0), and the atomic number is also the same (7 = 8-1) on the two sides of the equation.
Uses of some beta-emitters
Iodine-131 is used for radiation therapy of an overactive thyroid gland.
Yttrium-90 is used to treat cancer and is also injected into large joints to relieve the pain due to arthritis.
Phosphorous-32 is used to treat leukemia and other blood disorders.
Carbon-14 is used to determine the age of a fossil or an old object.
Example $3$
Write the nuclear equation for the $\ce{\beta}$-decay of iodine-131.
Solution
Step 1. The symbol and the atomic number of iodine in the periodic table are I and 53, respectively. So the initial equation is:
$\ce{_{53}^{131}{I} -> _{?}^{?}{?} + _{-1}^{0}{e}}\nonumber$
Step 2. Balance the mass number on the two sides of the equation, i.e., the mass number of unknown nucleoid is 131-0 = 131:
$\ce{_{53}^{131}{I} -> _{?}^{131}{?} + _{-1}^{0}{e}}\nonumber$
Step 3. Balance the atomic number on the two sides of the equation, i.e., the atomic number of the unknown nucleoid is 53-(-1) = 54:
$\ce{_{53}^{131}{I} -> _{54}^{131}{?} + _{-1}^{0}{e}}\nonumber$
Step 4. Find the symbol of the unknown nucleoid from the periodic table of elements, i.e., the element at atomic number 54 is xenon symbol Xe:
$\ce{_{53}^{131}{I} -> _{54}^{131}{Xe} + _{-1}^{0}{e}}\nonumber$
This is the balanced nuclear equation for the α-decay of iodine-131 that is used to treat over-active thyroid glands.
Example $4$
Write the nuclear equation for the β-decay of yttrium-90.
Solution
Step 1. The symbol and the atomic number of yttrium in the periodic table are Y and 39, respectively. So the initial equation is:
$\ce{_{39}^{90}{Y} -> _{?}^{?}{?} + _{-1}^{0}{e}}\nonumber$
Step 2. Balance the mass number on the two sides of the equation, i.e., the mass number of unknown nucleoid is 90-0 = 90:
$\ce{_{39}^{90}{Y} -> _{?}^{90}{?} + _{-1}^{0}{e}}\nonumber$
Step 3. Balance the atomic number on the two sides of the equation, i.e., the atomic number of the unknown nucleoid is 39-(-1) = 40:
$\ce{_{39}^{90}{Y} -> _{40}^{90}{?} + _{-1}^{0}{e}}\nonumber$
Step 4. Find the symbol of the unknown nucleoid from the periodic table of elements, i.e., the element at atomic number 40 is zirconium symbol Z:
$\ce{_{39}^{90}{Y} -> _{40}^{90}{Z} + _{-1}^{0}{e}}\nonumber$
This is the balanced nuclear equation for the β-decay of yttrium-90.
Positron emission
The nucleoids marked orange in Fig. 8.2.1. have more protons than needed for stability. They usually stabilize them by converting one of the protons (p) into a neutron (n) and a positron $\ce{\beta^+}$ by the following nuclear process:
$\ce{_{1}^{1}{p} -> _{0}^{1}{n} + _{1}^{0}{e}}\nonumber$
The neutron stays in the nucleus, but the positron emits from the nucleus, as illustrated in Fig. 8.2.4. Note that the positron has a +1 mass number that balances the +1 atomic number of the proton on the other side of the equation. The positron has a +1 charge that also balances the +1 charge of the proton on the other side of the equation. The mass number of a positron is zero as it has negligible mass compared to the mass of a proton or a neutron. The electrical charges are not shown in the nuclear equation.
Carbon-11 is an example of positron-emitter:
$\ce{_{6}^{11}{I} -> _{5}^{11}{B} + _{1}^{0}{e}}\nonumber$
Note that in the positron-emission process, the mass number remains the same, but the atomic number decreases by one in the daughter nucleus. The nuclear equation is balanced because the mass number is the same (11 = 11+0), and the atomic number is also the same (7 = 5+1) on the two sides of the equation.
Uses of some positron emittors
Positron emission is used in positron emission tomography (PET) which is a medical imaging technique. Short-lived positron-emitting isotopes 11C, 13N, 15O, and 18F used for positron emission tomography are typically produced by proton irradiation of natural or enriched targets described in a later section.
Fluorine-18 in fluorodeoxyglucose, abbreviated as [18F]FDG is a positron emitter commonly used to detect cancer, and in [18F]NaF is widely used for detecting bone formation. Other examples are oxygen-15 in [15O]H2O used to measure blood flow and nitrogen-13 used to tag ammonia molecules for myocardial perfusion imaging.
$\ce{_{8}^{15}{O} -> _{7}^{15}{N} + _{1}^{0}{e}}\nonumber$
$\ce{_{7}^{13}{N} -> _{6}^{13}{C} + _{1}^{0}{e}}\nonumber$
Example $5$
Write the nuclear equation for the positron-emission of fluorine-18?
Solution
Step 1. The symbol and the atomic number of fluorine in the periodic table are F and 9, respectively. So the initial equation is:
$\ce{_{9}^{18}{F} -> _{?}^{?}{?} + _{1}^{0}{e}}\nonumber$
Step 2. Balance the mass number on the two sides of the equation, i.e., the mass number of unknown nucleoid is 18-0 = 131:
$\ce{_{9}^{18}{F} -> _{?}^{131}{?} + _{1}^{0}{e}}\nonumber$
Step 3. Balance the atomic number on the two sides of the equation, i.e., the atomic number of the unknown nucleoid is 9-(+1) = 8:
$\ce{_{9}^{18}{F} -> _{8}^{131}{?} + _{1}^{0}{e}}\nonumber$
Step 4. Find the symbol of the unknown nucleoid from the periodic table of elements, i.e., the element at atomic number 8 is oxygen symbol O:
$\ce{_{9}^{18}{F} -> _{8}^{18}{?} + _{1}^{0}{e}}\nonumber$
This is the balanced nuclear equation for the positron-emission of fluorine-18.
Gamma-emission
The gamma-rays are high-energy electromagnetic radiations that do not have mass or charge. So, pure $\ce{\gamma}$-emission happens from the nucleus, but it does not result in transmutation, simply the nucleoid changes from a more unstable state, called a metastable state, to a relatively stable state, as illustrated in Fig. 8.2.5.
A symbol m or * next to the mass number as a superscript to the right indicates the metastable state of the parent nucleoid. For example, technetium-99m is a $\ce{\gamma}$-emitter widely used in medical imaging:
$\ce{_{43}^{99m}{Tc} -> _{43}^{199}{Tc } + \gamma}\nonumber$
Similarly, boron-11m is a $\ce{\gamma}$-emitter:
$\ce{_{5}^{11m}{B} -> _{5}^{11}{B } + \gamma}\nonumber$
Note that the nucleoid remains the same after $\ce{\gamma}$-emission, except for the change form metastable to a more stable state indicated by m.
Often, the $\ce{\gamma}$-emission accompanies $\ce{\alpha}$-emission or $\ce{\beta}$-emission. For example, polonium-210 decays by a simultaneous $\ce{\alpha}$-emission and $\ce{\gamma}$-emissions.
$\ce{_{84}^{210}{Po} -> _{82}^{206}{Pb } + _2^4He + \gamma}\nonumber$
Similarly, iridium-192 used in implants to treat breast cancer, and cobalt-60 used as an external radiation source for cancer treatment, simultaneously emit $\ce{\beta}$ and $\ce{\gamma}$-rays.
$\ce{_{77}^{192}{Ir} -> _{78}^{192}{Pt } + _{-1}^{0}{e} + \gamma}\nonumber$
$\ce{_{27}^{60}{Co} -> _{28}^{60}{Ni } + _{-1}^{0}{e} + \gamma}\nonumber$
Iodine-131 decays to $\ce{\beta}$-particle and xenon-131m that is rapidly followed by a $\ce{\gamma}$-decay of xenon-131m.
$\ce{_{53}^{131}{I} -> _{54}^{131m}{Xe } + _{-1}^{0}{e}}\nonumber$
$\ce{_{54}^{131m}{Xe} -> _{54}^{131}{Xe} + \gamma}\nonumber$
Less common forms of radioactivity
Several relatively less common forms of radioactivity are known. Some examples are the following.
1. Neutron-emission is a mode of radioactive decay in which one or more neutrons are ejected from a nucleus.
2. Proton-emission is a rare form of radioactivity in which a proton emits from a nucleoid.
3. Spontaneous fission is a radioactive process in which a more massive nucleoid breaks into smaller nucleoids, often along with the emission of smaller nuclear particles.
4. In electron-capture, an external electron is captured to react with proton and produce a neutron in the nucleus.
For example, beryllium-7 decays by electron capture, as shown in the following equation.
$\ce{_{4}^{7}{Be} + _{-1}^{0}{e} -> _{3}^{7}{Li} + \gamma}\nonumber$
Note that the mass number remains the same, but the atomic number decreases by one in the electron-capture process.
Chromium-51, which is used for imaging the spleen, decays by the electron capture and $\ce{\gamma}$-emission.
$\ce{_{24}^{51}{Cr} + _{-1}^{0}{e} -> _{23}^{51}{V} + \gamma}\nonumber$
Summary of decay mode of radioactive nucleoids
Fig. 8.2.6 summarizes the changes in the nucleoid composition upon different radioactive events described above. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/08%3A_Nuclear_chemistry/8.02%3A_Radioactivity.txt |
The half-life (t1/2) of a radioisotope is the time it takes for half of the sample to decay.
It tells the rate of decay of the radioisotope – the faster the rate of decay, the shorter the half-life.
General features of the half-life
1. The half-life is different for different nucleoids, as shown in Fig. 8.3.1, and Table 1. It varies from a fraction of a second to more than 1020 s, i.e., more than 3 trillion years.
2. The farther a nucleoid is away from the stable nucleoid (shown by black dots in Fig. 8.3.1), the less stable it is, and the faster it decays.
3. The half-life is independent of concentration, temperature, and pressure, i.e., the t1/2 is a characteristic constant of a radioisotope.
4. Natural radioactive isotopes usually have a longer half-life, e.g., t1/2 carbon-14 is 5730 years and uranium-235 is 7.0 x 108 years.
5. The radioisotopes used for imaging and treatment in medical sciences are usually synthesized and have a short half-life so that they may not persist in the body for an unnecessarily long time. For example, phosphorous-32, iodine-131, and technetium-99m have half-lives of 14.3 days, 8.1 days, and 6.0 hours, respectively.
Table 1: Half-lives (t1/2) of some common radioisotopes
Radioisotope Symbol Half-life Use
Carbon-14 $\ce{_6^14C}$ 5730 years Radioisotope dating
Hydrogen-3 $\ce{_1^3H}$ 12.3 years Radioisotope dating
Potassium-40 $\ce{_19^40K}$ 1.3 x 109 years Radioisotope dating
Rhenium-187 $\ce{_75^187Re}$ 4.3 x 1010 years Radioisotope dating
Uranium-238 $\ce{_92^238U}$ 4.5 x 109 years Radioisotope dating
Uranium-235 $\ce{_92^235U}$ 7.0 x 108 years Nuclear reactor fuel
Cobalt-60 $\ce{_27^60Co}$ 5.3 years Medical (external radiation source)
Iodine-131 $\ce{_53^131I}$ 8.1 days Medical
Iron-59 $\ce{_26^59Fe}$ 45 days Medical
Molybdenum-99 $\ce{_42^99Mo}$ 67 hours Medical
Sodium-24 $\ce{_11^24Na}$ 15 hours Medical
Technetium-99m $\ce{_43^{99m}Te}$ 6 hours Medical
Phosphoros-32 $\ce{_15^32P}$ 14.3 dsys Medical
Decay curve of radioisotopes
During each successive half-life, half of the initial amount will radioactively decay, as illustrated in Fig. 8.3.2. for the case of phosphorous-32 that decays with a half-life of 14.3 days by the following nuclear reaction
$\ce{_15^32P -> _16^32S + _{-1}^{0}{e}}\nonumber$
Suppose there is 100 mg of the phosphorous-32 in the beginning; 50 mg will be left behind after 14.3 days, i.e., after 1 half-life; and 25 mg will be left after 28.6 days, i.e., after 2 half-lives. A negligible amount of the parent isotope phosphorous-32 is left after 9 half-lives.
The amount of a radioisotope remaining after the given time can be calculated from the known initial amount and time spent, by the following formula:
$m_{f}=m_{i}(0.5)^{n}\nonumber$
where mi is the initial amount, mf is the final amount, and n is the number of half-lives passed. The formula works even if the number of half-lives is not a whole number.
Example $1$
If 50.0 mg of iodine-131 was injected for medical treatment, how many milligrams will be left after 40.5 days? (Half-life of iodine-131 is 8.1 days)
Solution
Given: mi = 50.0 mg, Time = 40.5 days, Desired ? mf
The equality: 1 half-live = 8.1 days, gives the following conversion factors. $\frac{1 \text { half-life }}{8.1 \text { days }} \quad\text{and}\quad \frac{8.1 \text { days }}{1 \text { half-life }}\nonumber$
For calculating the half-lives, multiple the given time with the conversion factor that cancels the time:
$n=40.5 \cancel{\text { days }} \times \frac{1 \text { half-life }}{8.1 \cancel{\text { days }}}=5 \text { half-lives }\nonumber$
For calculating the amount left, plug in the values in the formula:
$m_{f}=m_{i}(0.5)^{n}=50.0 \mathrm{mg}(0.5)^{5}=1.56 \mathrm{~mg}\nonumber$
Radioisotope dating
Natural radioactivity is used to establish the age of objects of archeological, anthropological, or historic interest. All living objects have carbon in their composition. Carbon-14 is a radioactive isotope of carbon with a half-life of 5730 years. Carbon-14 is produced by the transmutation of nitrogen-14 upon neutron bombardment by cosmic rays, as illustrated in Fig. 8.3.3. Its concentration in a carbon source for the living organism remains almost constant because its decay counterbalances its production by cosmic rays. Living organisms continuously replenish carbon, so the carbon-14 concentration remains almost constant as long as the object is alive. After the object dies, the carbon-14 decreases with time, reducing to half after one half-life. The carbon-12 isotope is not radioactive, so its concentration remains constant. Measurement of the carbon-14/carbon-12 ratio allows calculating the age of the object after its death. The age of early civilizations, like the Indus valley civilization examples shown in Fig. 8.3.4 were determined by the carbon-14 dating method.
Because the carbon-14 decreases with time, the object older than 20,000 years of age does not have sufficient carbon-14 left to determine their age accurately. Other radioisotopes with a longer half-life, e.g., uranium-238 with a half-life of 4.5 x 109 years, are used to determine the age of ancient objects. For example, the age of rock samples from the moon, as shown in Fig. 8.3.5 was determined by uranium-238 radioisotope dating. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/08%3A_Nuclear_chemistry/8.03%3A_Half-life_of_radioisotopes.txt |
Radioactivity measurement
Radioactivity is measured in terms of the rate of radioactive events. Nuclear radiations are ionizing radiations, i.e., they knock off electrons from atoms or molecules that come in their path, leaving behind cations. Geiger Muller counter is one of the radiations measuring instruments that counts the disintegration of radionucleotide per second by registering the current produced by the ionization action of the radiation, as illustrated in Fig. 8.4.1. It is not just one ionization event; the nuclear particle keeps ionizing the atoms in its track until its energy is exhausted, as illustrated in Fig. 8.4.2. The instrument records the flash of electric current produced by the ionization of each radioactive disintegration.
Becquerel (Bq)
The SI unit of radioactivity is Becquerel (Bq), i.e., the number of nuclei that disintegrate per second.
The common unit of radiation intensity is Curie (Ci), i.e., 3.7 x 1010 disintegrations per second. The relationship between Becquerel and Curie is the following.
$1 \mathrm{Ci}=3.7 \times 10^{10} \mathrm{~Bq}=3.7 \times 10^{10} \text { disintegrations }\nonumber$
Often the radioisotope for medical use has the information of millicurie per milliliter (mCi/mL) from which the volume for the desired dose can be calculated.
Example $1$
A patient must be given a 5.0 mCi dose of iodine-131 that is available as Na131I solution containing 3.8 mCi/mL. What volume of the solution should be administered?
Solution
Use the reciprocal of 3.8 mCi/mL as a conversion factor:
$5.0 \cancel{\mathrm{~mCi}} \times \frac{1 \mathrm{~mL}}{3.8 \cancel{\mathrm{~mCi}}}=1.3 \mathrm{~mL} \text { dose }\nonumber$
Radiation exposure measurements
Absorbed dose
The ionizing radiation dose or called the absorbed dose is measured in terms of energy deposited by ionizing radiation in a unit mass of matter being irradiated, as illustrated in Fig. 8.4.3
Gray (Gy)
The SI unit of absorbed dose is gray (Gy) which is defined as the absorption of one joule of radiation energy per kilogram of matter (J/kg).
The common unit of absorbed dose is rad, which stands for radiation absorbed dose. The rad is one-hundredth of a gray, i.e.:
$1 \mathrm{~Gy}=100 \mathrm{~rad}\nonumber$
Equivalent dose
The same amount of energy deposited in tissues by different types of radiation carries different levels of health risks in terms of causing cancer and genetic damage, expressed as a radiation weighting factor (WR), as illustrated in Fig. 8.4.4, and listed in Fig. 8.4.5. For example, 1 Gy of $\ce{beta}$-particles carries a risk of 5.5% chances of eventually developing cancer, while 1 Gy of $\ce{alpha}$-particles has 20 times more risk compared to the β-particle (ref.: https://en.Wikipedia.org/wiki/Sievert, accessed on 07/15/2020). The health risk of the ionizing radiation is measured in the units of equivalent dose. Sievert (Sv) is an SI unit of an equivalent dose of ionizing radiation that measures the health effects of low levels of ionizing radiation on the human body.
Sievert (Sv)
The equivalent doze in Sievert (Sv) is equal to the product of absorbed dose in grays (Gy) multiplied the radiation weighting factor (WR), i.e., $\text { The equivalent dose in } \mathrm{Sv}=\text { Absorbed dose in } \mathrm{Gy} \times \mathrm{W}_{\mathrm{R}}$
The common unit of equivalent dose is rem (rem stands for roentgen equivalent man), which is:
$1 \mathrm{~Sv}=100 \mathrm{rem}\nonumber$
The personnel working in a radiation environment are required to wear film badges or electronic personal dosimeters, as shown in Fig. 8.4.6, that record the dose received. A record of each person's dose is usually maintained by the radiation facilities to comply with the allowed radiation exposure limits.
Effective dose
The equivalent dose is equal to the effective dose in sievert (Sv) when the whole human body is exposed equally to the radiation. If part of the body is exposed, then an effective dose in sievert (Sv) is calculated by the summation of the product of equivalent dose in Sv with tissue weighting factor (WT) for each tissue exposed to the radiation, as illustrated in Fig. 8.4.4 and calculated with example in Fig. 8.4.7. The reason for this calculation is that the effect of the same equivalent dose is different in different tissues. The tissue weighting factors (WT) are listed in Table 1.
An effective dose takes the absorbed dose and adjusts it for radiation type and organ sensitivity, i.e.,:
$\text { The effective dose in } \mathrm{Sv}=\text { Equivalent dose in } \mathrm{Sv} \times \mathrm{W}_{\mathrm{T}}\nonumber$
Table 1: Tissue weighting factors (WT) ICRP103 (2007), source of data: Source: https://en.Wikipedia.org/wiki/Sievert, accessed on 07/16/2020
Organs WT
Gonads 0.08
Red bone marrow 0.12
Colon 0.12
Lung 0.12
Stomach 0.12
Breasts 0.12
Bladder 0.04
Liver 0.04
Oesophagus 0.04
Thyroid 0.04
Skin 0.01
Bone surface 0.01
Salivary glands 0.01
Brain 0.01
Remainder of body 0.12
Total 1.00 | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/08%3A_Nuclear_chemistry/8.04%3A_Radiation_measurements.txt |
Interaction of radiation with matter
The ionizing radiations knock off electrons from atoms and molecules. Water composes about 60% of the human body. The oxygen atom in water has eight valence electrons, four in the two bonding pairs, and four in the two lone pairs, as shown in Fig. 8.5.1. If one of them is knocked off by radiation, the result is a radical-cation. It is radical because it has one unpair electron and its octet is incomplete, and cation because an electron has been lost. The radical cation can then release a proton to become a hydroxyl radical. The radicals are very reactive species due to the incomplete octet. The radicals tend to react with any material around that causes damage to the tissues. Notably, the damage to DNA is the most dangerous, causing mutation, cancer, and hereditary problems.
The cells that are proliferating are more susceptible to the harmful effects of radiation exposure, including bone marrow, skin, reproductive organs, and intestinal lining, as well as all the cells of growing children. If the bone marrow cells are damaged, the red blood cells may not be produced. Damage to the reproductive cells or the cells of a fetus may cause congenital disabilities. Fortunately, the cancer cells are multiplying and affected by the radiation exposure much more than the surrounding healthy cells, which allows for selectively killing the cancerous cells by radiation treatment. The use of ionizing radiations in cancer treatment is described in a later section.
Effects of radiation exposure on humans
Exposure of humans to less than 0.25 Sv usually does not have any noticeable effect. Whole-body exposure of 1 Sv results in a temporary decrease in the white blood cell counts. More than 1 Sv exposure may cause nausea, vomiting, fatigue, and a reduction in white blood cell count. More than 3 Sv doses to the whole body can decrease the white blood cell count to zero and cause diarrhea, hair loss, and infection. Exposure to 5 Sv can cause death in 50% of the people receiving the dose –it is called a lethal dose for one-half of the population (LD50). Whole-body exposure of 6 Sv or higher is fatal to all humans within a few weeks. Fig. 8.5.2 illustrates the effects.
Background radiation exposures
Humans are exposed to radiation in the cases of nuclear accidents, during nuclear radiation treatments, particularly for treating different forms of cancers, and during medical imaging for medical diagnostic purposes. Besides these human-made radiation exposures, humans are regularly exposed to natural sources of radiation, called background radiation. The background radiation can be in foods, e.g., potassium-40 is a naturally occurring radioactive isotope present in potassium-containing foods. Carbon-14, radon-222, strontium-90, and iodine-131 are other radioisotopes present in the air and foods around us. Radioactive isotopes of uranium and thorium and their decay products are the source of radiation in soil. The extraterrestrial sources of radiation are cosmic rays that are stopped in the upper atmosphere, but some may reach the ground and expose the people. The average annual radiation dose per person in the U.S. is 6.2 millisieverts (620 millirem).
Radiation protection
Different types of radiation have different penetration depths in the air and the human body. The $\ce{\alpha}$-particles are heavy with two protons and two neutrons; they cause much ionization but travel a few centimeters in air. A sheet of paper can stop the $\ce{\alpha}$-particles, as illustrated in Fig. 8.5.3. Exposure to $\ce{\alpha}$-particles from an external source affects only the outer layer of the skin. However, an internal source can cause significant damage to the nearby tissues, as illustrated in Fig. 8.5.4
The $\ce{\beta}$-particles are fast-moving electrons that can travel several meters in the air. A 2-4 mm thick aluminum plate can stop them. The $\ce{\beta}$-particles can penetrate 4 to 5 centimeters into the tissue. External exposure to the $\ce{\beta}$-particles can burn the skin, but internal organs remain safe. Internal exposure to the $\ce{\beta}$-particle is more dangerous than external exposure.
The $\ce{\gamma}$-rays and X-rays can travel long distances in air, up to half a kilometer, and are not easily stopped by common material. A thick layer of lead or concrete shield is required to stop the $\ce{\gamma}$-rays and X-rays. External exposure to the $\ce{\gamma}$-rays and X-rays is the most dangerous because these rays can penetrate deep and damage the organs.
Neutrons are high-energy neutral particles that also have high penetrating power. They lose energy by colliding with atoms of substances. The most effective shielding is the water or concrete that has moisture in it to stop the neutrons effectively.
The workers in a radiation environment wear heavy clothes, gloves, and lab coats to provide additional protection. The radioactive materials are usually stored in shielded containers, even the syringes containing radioactive materials for injection are shielded. The general rules for protection against the radiation are:
Radiation protection measures
1. keep minimum possible time in the radiation environment -less the time means less exposure,
2. keep as much distance from the radiation source as possible -the radiation intensity drops inversely proportional to the square of the distance, and
3. keep shielding between you and the source of radiation as much as possible -the more the shielding the less the exposure. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/08%3A_Nuclear_chemistry/8.05%3A_Ionizing_radiation_exposures.txt |
The use of nuclear chemistry in medical technologies is increasing over time. The medical uses can be divided into two categories :
1. medical imaging of organs or diagnosing any malfunction and
2. therapeutic use, mainly for killing cancerous cells.
Radioisotopes in medical imaging
Criteria of radioisotope selection for medical imaging
1. those that emit $\ce{\gamma}$-rays, directly or indirectly, during their decay,
2. can be delivered to the organ of interest in pure form or tagged in a compound,
3. are short-lived or can be excreted from the body after use.
Table 8.6.1 lists some radioisotopes commonly used in medical imaging. A low dose of the radioisotope is administered to a patient. The $\ce{\gamma}$-rays cross over the body and are recorded like X-rays. A computer finally converts the recording into a useful image. The image is compared with an image of a healthy organ to diagnose any malfunction.
Table 8.6.1: Some of the commonly used radioisotopes in medical imaging
Radioisotope Symbol Mode of decay Half-life Use in medical imaging
Carbon-11 $\ce{_6^11C}$ β+, g 20.3 m Brain scan to trace glucose
Fluorine-19 $\ce{_9^18F}$ β+, g 109m Brain scan to trace glucose
Chromium-51 $\ce{_24^51Cr}$ E.C., g 27.3 d Diagnose albinism; image the spleen and gastrointestinal track
Gallium-67 $\ce{_31^67Ga}$ E.C., g 78.3 h Whole-body scan for tumors
Selenium-75 $\ce{_34^75Se}$ E.C., g 118 d Pancreas scan
Krypton-81m $\ce{_36^{81m}Kr}$ g 13.3 s Lung ventilation scan
Xenon-133 $\ce{_54^133Xe}$ β 5.24 d Lung ventilation scan
Strontium-81 $\ce{_38^81Sr}$ β 22.2 m Scan for bone cancer and bone diseases
Mercury-197 $\ce{_80^197Hg}$ E.C., g 64.1 h Kidney scan
Iron-59 $\ce{_26^59Fe}$ β, g 44.5 d Bone marrow function and anemias
Iodine-131 $\ce{_53^131I}$ β, g 8.04 d Diagnosis of thyroid malfunction
Iodine-123 $\ce{_53^123I}$ E.C., g 13.2 h Diagnosis of thyroid malfunction
Technetium-99m $\ce{_43^{99m}Te}$ g 6.01 h Brain, liver, kidney, bone scans, and diagnosis of the damaged heart muscle
Phosphorous-32 $\ce{_15^32P}$ β 14.3 d Detect eye tumors
Thallium-201 $\ce{_81^201Tl}$ E.C., g 3.05 d Heart scan and exercise stress test
An example of medical imaging is the thyroid gland in the neck that produces the hormone thyroxin, which controls the overall rate of metabolism in the body. Each thyroxin molecule contains four iodine atoms. Administration of radioactive Na131I or Na123I salt accumulates the iodine in the thyroid gland in a few hours. Decay of 131I and 123I involves $\ce{\gamma}$-emission.
$\ce{_53^131I -> _54^131Xe + _{-1}^{0}{e}} + \gamma\nonumber$
$\ce{_53^123I + _{-1}^{0}{e} -> _52^123Xe} + \gamma\nonumber$
The $\ce{\gamma}$-emission from the iodine localized in the thyroid gland is recorded, as shown in Fig. 8.6.1. An overactive thyroid (hyperthyroidism) cumulates more and underactive thyroid (hypothyroidism) cumulates less iodine than a healthy thyroid.
Another example is positron emission tomography (PET). Positron emitters like carbon-11 and fluorine-18 incorporated in a suitable compound like glucose allow following the metabolic path of the compound. For example, 18-fluorodeoxyglucose )18-FDG) is a glucose molecule in which one of the oxygen is replaced with 18F. Intravenous injection of the 18-FDG ultimately results in the cumulation of 18-FDG in the brain and other body organs where glucose is used in the metabolic process. The 18F emits a positron, which, being an anti-particle of the electron, reacts with the electron and produces two g $\ce{\gamma}$-rays.
$\ce{_9^18F -> _8^18O + _{1}^{0}{e}} + \gamma\nonumber$
$\ce{_{1}^{0}{e} + _{1}^{0}{e} ->} 2\gamma\nonumber$
The $\ce{\gamma}$-rays are used to obtain an image of the organ. The image reveals problem areas in the form of an abnormal concentration of glucose in the part of the organ. For example, Fig. 8.6.2 compares the PET image of a healthy brain versus a brain with Alzheimer's disease.
Since glucose metabolism happens in all organs, whole-body PET scans can be used to diagnose lung, colorectal, head and neck, and esophageal cancers as well as other diseases that involve abnormal glucose metabolism. For example, tumors have high metabolic rates; the PET scans using 18-FDG are used to detect them, as shown in Fig. 8.6.3.
Medical imaging without using radioisotopes
Major examples of medical imaging using external radiation sources include the following.
1. X-rays, which are external radiation source, is commonly used for medical imaging of organs and bones.
2. Computed tomography (CT) scan uses computer processing of many X-ray measurements, from different angles, to produce a cross-sectional view (virtual slices) of the organ.
3. Magnetic resonance image (MRI) is another powerful medical imaging technique that is based on the fact that hydrogen atoms splint into two two energy states when placed in a strong magnetic field. When illuminated with infrared (IR) radiation of the energy matching with the energy gap between the two groups, the hydrogen atoms are excited from the lower to higher energy state. The decay of the excited state emits the IR radiations that are recorded to obtain the image of soft tissues that contain many hydrogen atoms in the form of water molecules. Fig. 8.6.4 shows some examples.
Radiation therapy
The purpose of radiation therapy is to selective kill the diseased cells or tissues by exposing them to radiation. Higher radiation doses are required for therapy than for imaging. The radiation source can be external or internal.
External radiation therapy
In the external irradiation, the radiation from the source, such as coba which are often used, other radiation sources are being developed, such as proton beam from the cyclotron, which have been used to kill inoperable tumors near or in the eye, skull base, or spine. Proton therapy uses a beam of proton to deliver radiation directly to the tumor.
Internal radiation therapy
Internal radiation therapy is used when a short-lived radioisotope can be made to selectively concentrate in the organ or tissue of interest. For example, iodine-131 is $\ce{\beta}$ and $\ce{\gamma}$-emitter, is administered to a patient, is picked up by the thyroid gland, and is used to treat hyperthyroidism. Another example is actinium-225, which is an $\ce{\alpha}$-emitter with a half-life of 10 days. Actinium-225 installs in a monoclonal antibody that is attached to a prostate-specific antigen and delivered selectively to treat prostate tumors.
Brachytherapy
Brachytherapy or seed implantation is a form of internal radiation treatment that delivers a high dose for a short period compared to the external radiation treatment. Fig. 8.6.6 shows the sites in the body where brachytherapy can be used to treat cancer. For example, 40 or more titanium capsules, about the size of a rice grain (Fig. 8.6.7), are implanted to treat prostate cancer. The seed contains $\ce{\gamma}$-emitter like iodine-125 (half-life 60 days), palladium-103 (half-life 17 days), or cesium-131 (half-life 10 days). The seed may be left in the body because, due to the short half-life, they are no more significantly radioactive after the treatment.
Another option is temporary brachytherapy, e.g., iridium-192 needles that deliver higher radiation doses are inserted to treat prostate cancer and removed after 5 to 10 min. The iridium-192 wires are also used as a follow-up treatment after breast cancer surgery to kill any residual or recurring cancer cells. The iridium-192 wires are inserted through a catheter implanted in the space from where the tumor was removed. The wires are removed after delivering the required radiation dose. The process is repeated twice a day for five days. The catheter is removed, and no radioactive material is left in the body after the treatment.
8.07: Making radioisotopes for medical uses
Natural radioisotopes usually have a long half-life and are not best suited for medical applications. The medical application usually requires short-lived radioisotopes. The radioisotopes are usually produced in nuclear reactors where particles, like $\ce{\alpha}$-particles, $\ce{\beta}$-particles, and neutrons, are abundant. Particle accelerators, such as the one shown in Fig. 8.7.1 also accelerate and direct the nuclear particles at the targets. The high-energy nuclear particles may be absorbed by and transmute the target nuclei to radioisotopes in a nuclear reaction.
Radioisotopes in medical applications are usually produced by the particle bombardment method. One reaction that happens naturally by neutron bombardment from cosmic rays on nitrogen-14 is the following.
$\ce{_7^14N + _{0}^{1}{n} -> _6^14C + _{1}^{1}{p}}\nonumber$
An example of an artificial nuclear reaction initiated by $\ce{\alpha}$-particle bombardment on nitrogen, observed by Rutherford that lead to the discovery of proton, is illustrated in Fig. 8.7.2.
$\ce{_2^4He + _{7}^{14}{N} -> _8^17O + _{1}^{1}{p}}\nonumber$
Another example is the nuclear reaction initiated by $\ce{\alpha}$-particles on beryllium, observed by James Chadwick, which lead to the discovery of the neutron.
$\ce{_2^4He + _{4}^{9}{Be} -> _6^12C + _{0}^{1}{n}}\nonumber$
An example of radioisotope production for medical uses is the following. Gold-198, used as a tracer in the liver, is produced by neutron bombardment on gold-197.
$\ce{_79^197Au + _{0}^{1}{n} -> _79^198Au}\nonumber$
Similarly, gallium-67 used in medical diagnostics is produced by proton bombardment on zinc-66.
$\ce{_30^66Zn + _{1}^{1}{p} -> _31^67Ga}\nonumber$
Molybdenum-99 is a radioactive isotope produced in a nuclear reactor by neutron bombardment of molybdenum-98.
$\ce{_42^98Mo + _{0}^{1}{n} -> _42^99Mo} + \gamma\nonumber$
Molybdenum-99 is also produced as a fission product of uranium-235. Molybdenum-99 decays to technetium-99m that has several uses in nuclear medical imaging and treatment.
$\ce{_42^99Mo -> _43^{99m}{Tc} + _{-1}^{0}{e}}\nonumber$
Technetium-99m is short-lived (half-life 6 h), and needs to be produced in the hospital to minimize its decay during the transport. Its parent molybdenum-99 has a half-life of 66h and can be transported without significant decay during the transport. Molybdenum-99/technetium-99m generators are supplied to the hospitals in a shielded container. Fig. 8.7.3 illustrates the first Molybdenum-99/technetium-99m generator developed at Brookhaven National Laboratory. Molybdate (MoO42) ion is adsorbed onto alumina adsorbent in a column. When molybdenum-99 decays to technetium-99m, the ion change to pertechnetate (TcO4), which is less tightly bound to the alumina. Pouring a saline solution through the column elutes the technetium-99m as TcO4 ion, which is then used for medical purposes in the hospitals.
Destruction of an inoperable tumor has also been tested by $\ce{\alpha}$-emission from boron-10 upon neutron bombardment.
$\ce{_{0}^{1}{n} + _5^10B -> _4^7Li + _2^4He}\nonumber$ | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/08%3A_Nuclear_chemistry/8.06%3A_Medical_uses_of_radioisotopes.txt |
Conversion of matter into energy in nuclear reactions
When free nucleons come together to make a nucleoid, they release energy equal to the nuclear binding energy. The energy is produced at the expense of the mass of the nucleons following the famous Einstein equation:
$E = mc^2$
, where E is the energy, m is the mass, and c is the speed of light. In other words, the mass of nucleons in nucleoids is slightly less than the mass of the free nucleons, as some of their mass is released as nuclear binding energy.
The nuclear binding energy per nucleon is different for different nucleoids, as shown in Fig. 8.8.1. The composition of nucleoids changes during the nuclear reaction, and the difference in the nuclear binding energy is released as energy during the process, which is the source of energy in the sun, stars, nuclear power plants, and nuclear weapons.
The most stable nucleoid with the highest average binding energy per nucleon is iron-56.
Nuclear fusion
The lighter nucleoids tend to combine and make heavier nucleoids that are more stable and lease a tremendous amount of energy –a process called nuclear fusions.
Nuclear fission
The heavier nucleoids, particularly those having a mass number higher than 92, tend to split into two smaller nucleoids that are more stable and lease a large amount of energy –a process called nuclear fission.
Both the fusion and fission processes release a tremendous amount of energy, called nuclear energy or atomic energy, which is the difference between the nuclear binding energy of the parent and the daughter nucleoid.
Nuclear fusion -a source of energy in the sun and stars
The universe is composed of about 98% hydrogen and helium. Sun is composed of about 74% hydrogen, 25% helium, and 1% all other elements. The light nucleoid like hydrogen and helium fuse to make heavier nucleoids and subatomic particles like neutrons in a process called nuclear fusion. Nuclear fusion releases energy that is the difference in the nuclear binding energy of their nucleons in the daughter and parent nucleoids. Fusion energy is the source of energy for the sun and stars. Some of the nuclear fusion reactions, illustrated in Fig. 8.1.1 are the following.
$\ce{_1^1H + _1^1H -> _2^2H + _{1}^{0}{e}}\nonumber$
$\ce{_1^1H + _1^2H -> _2^3H} \nonumber$
$\ce{_2^3H + _2^3H -> _2^4H + 2_{1}^{1}{H}}\nonumber$
$\ce{_2^3H + _1^1H -> _2^4H + _{1}^{0}{e}}\nonumber$
Although hydrogen is abundant on earth and a potential source of energy, the controlled fusion reaction is not yet economically feasible. The main difficulty is that the fusion of nuclei requires extremely high temperature, greater than a hundred thousand Celsius, and extremely high pressure, greater than a hundred thousand atmospheres, to overcome the repulsive forces of the like charges of the nuclei before they can fuse. For this reason, nuclear fusion reactions are also called thermonuclear reactions. Such conditions exist in sun and stars where these reactions routinely happen, but there are many technical problems yet to overcome before the energy can be harnessed economically from the nuclear fusion reactions on earth. A nuclear explosion can create the conditions needed for the thermonuclear reactions and are used to carry out uncontrolled thermonuclear reactions to boost the explosive power of nuclear weapons. Such a nuclear weapon is called a thermonuclear or hydrogen bomb.
One of the potential candidate reactions for nuclear fusion reactors is the fusion of deuterium with tritium, illustrated in Fig. 8.8.2, which requires a little less harsh conditions. Most of the research is focused on the device called Tokamak, which uses strong magnetic fields to contain and heat the materials for the reaction, but so far it is not economically feasible to use it as a source of energy for commercial uses.
Nuclear fission -the source of energy in nuclear power plants and in nuclear weapons
Nuclear fission is a radioactive decay process in which a heavier nucleoid splits into two or lighter nucleoids and releases a tremendous amount of energy that is the difference in the binding energy of nucleons in the daughter and the parent nucleoids. The nucleoids that do spontaneous fission are marked green in Fig. 8.2.1.
Nuclear fuel
Induced fission is fission that occurs upon bombardment of a nucleoid with a nuclear particle, like a neutron. For example, the bombardment of uranium-235 or plutonium-239 by a neutron causes the fission and releases about three neutrons along with the smaller nucleoids, as illustrated in Fig. 8.8.3. The neutrons released by the fission cause fission of another nucleoid starting a nuclear chain reaction. The nuclei like uranium-235 or plutonium-239 that produce neutrons in the product that are capable of continuing the fission in a chain reaction are called fissile. The fissile nucleoids are nuclear fuels used in nuclear power plants to produce nuclear energy.
Fate of neutron released in a fission event
The neutrons released in a fission event may be lost, absorbed by another nucleus like uranium-238 that does not fission, or may be absorbed by another fissile nucleoid and repeat the fission, as illustrated in Fig. 8.8.4. Three situations may arise, If, on the average;
1. less than one neutron released causes new fission, the process dyes,
2. one neutron released causes new fission, the fission continues in a controlled way as in nuclear power plants,
3. more than one neutron released causes new fission, the fission increases exponentially resulting in a nuclear explosion, as illustrated in Fig. 8.8.3.
Note that each fission event does not necessarily split the parent nucleoid into exact same daughter nucleoids, it can split into different combinations of daughter nucleoids resulting in a mixture of products that are usually radioactive, as illustrated in Fig. 8.8.4.
Atomic bomb
A certain minimum mass of the fissile material, called critical mass, is necessary for a fission event to grow exponentially and cause a nuclear explosion. The critical mass is less for a more enriched fissile material. Therefore, usually enriched uranium-235 or plutonium 239 is used as an explosive in an atomic bomb. The mass more than the critical mass, called supercritical mass, is split into sub-critical portions which are kept separate in the weapon to avoid a nuclear explosion. When needed, the conventional chemical explosion is used to force the sub-critical portions to combine and make a supercritical mass for a nuclear explosion to happen, as illustrated in Fig. 8.8.5.
Nuclear power plant
Nuclear power plants, like the one illustrated in Fig. 8.8.6, harness the energy released during fission for electricity production. The fissile material is assembled in the form of fuel rods in the core of a nuclear reactor. It is important to sustain the nuclear chain reaction but it is critical to keep it under control so that the chain reaction may not grow exponentially and cause a nuclear explosion. Rods of material, like palladium or boron that are efficient neutron absorbers, are included in the design of the reactor core. The neutron absorber rods can be lowered or raised to control the neutron flux available to sustain the fission chain reaction.
The reactor core is surrounded by a coolant and moderator, which may be regular water (H2O) or heavy water (D2O), or some other material. It is called coolant because it takes away the heat produced by the fission reaction and moderator because it also slows down the fast neutrons released in fission. The slow neutrons are more effective for the fission reaction. The coolant transfers its heat to a steam generator. All of these components are housed in a containment building to keep the radioactivity contained, even if a nuclear accident happens. The steam is used to drive a turbine for electricity production, and the condensed water is cooled using water from the cooling tower and returned for reuse, as illustrated in Fig. 8.8.6.
Nuclear power plants produce about 20% of the electricity in the USA. Fig. 8.8.7 shows a nuclear power plant in Arkansas. In some countries, the contribution of nuclear power to the total electricity is much higher, e.g., about 70% of electricity is produced by nuclear power plants in France. Nuclear power is considered an energy source of the future, or at least for a transition from fossil fuel to the next major energy source when the fossil fuel reserves will exhaust.
The fission products in the nuclear power plants become radioactive waste that needs to be stored for at least 10 half-lives to reach an acceptable radioactivity level. Based on the 28.8 years half-life of strontium-90 which is the long-lived and dangerous product in nuclear waste, a storage time of 300 years is needed. Handling radioactive waste is an issue of concern associated with nuclear power production. Plutonium-239 is another radioactive isotope produced from neutron absorption by uranium-238. Uranium-238 is a major constituent (99.28%) in the natural uranium along with fissile uranium-235 (0.71%). Plutonium-239 has a long half-life (24,000 Years) but it can be extracted in a fuel reprocessing plant and used as a nuclear fuel. | textbooks/chem/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/08%3A_Nuclear_chemistry/8.08%3A_Nuclear_fusion_and_fission.txt |
• 1.1: What is organic chemistry?
Organic chemistry is the study of the properties and reactions of organic compounds.
• 1.2: What is a chemical bond
The chemical bonds connect the atoms in the compounds. Transferring or sharing some valance electrons from one atom to another makes the bonds. Intermolecular forces are kind of bonds through which molecules interact with each other using transient or permanent dipoles.
• 1.3: Hybridization of orbitals and 3D structures of simple organic compounds
Atomes mix valence s and p orbitals in different proportions to make hybrid orbitals to make bonds in molecules. The geometry hybrid orbitals of the central atom determine the molecule's geometry around the central atom.
• 1.4: Representing organic compounds
Organic molecules contain chains and rings of carbons with single, double, or triple bonds connecting the carbons. As described in this chapter, the organic molecules are represented in multiple ways, including Lewis, condensed, and skeletal formulae.
• 1.5: Formal Charge
A formal charge is assigned to an atom in a molecule based on the assumption that bonding electrons are shared equally. Calculating formal charges and characteristics of some organic species having formal charges are described.
• 1.6: Resonance
If more than one Lewis structure can be drawn for a compound, these structures are resonance contributing structures, and the actual structure is a resonance hybrid. Resonance hybrid is always more stable than the contributing structures. Factors stabilizing the resonance hybrid are described.
01: Bonding in organic compounds
Learning Objectives
• Recognize organic compounds from their formula.
• Understand some logical reason for why nature has selected \(\ce{C}\) and \(\ce{H}\) as the main constituent of organic compounds.
What are organic compounds?
The compounds usually synthesized in living things are called organic compounds. The organic compounds are primarily composed of carbon (\(\ce{C}\)) and hydrogen (\(\ce{H}\)). For example, methane (\(\ce{CH4}\)) produced by decaying plant materials is composed of one \(\ce{C}\) and four \(\ce{H's}\). Often one or more atoms of elements other than \(\ce{C}\) and \(\ce{H}\) are also present in organic compounds, like oxygen (\(\ce{O}\)), nitrogen (\(\ce{N}\)), phosphorous (\(\ce{P}\)), sulfur (\(\ce{S}\)), etc. For example, \(\ce{O}\) atoms are present in glucose (\(\ce{C6H12O6}\)). The atoms other than \(\ce{C}\) and \(\ce{H}\), e.g., \(\ce{O}\) in \(\ce{C6H12O6}\), are called heteroatoms. Figure \(1\) illustrates fruits and vegetables composed of organic compounds.
Why has nature chosen \(\ce{C}\) and \(\ce{H}\) as primary constituents of organic compounds?
Nature has chosen \(\ce{C}\) and \(\ce{H}\) as primary constituents composing the organic compounds because of several reasons, some of which are the following.
• \(\ce{C}\) is a member of second-row elements in the periodic table that usually make stronger and more stable bonds than the elements in the higher rows.
• \(\ce{C}\) makes four bonds in neutral molecules, which is higher than any other element of the second row can make. For example, \(\ce{C}\) has for bonds in a methane molecule represent as: \(\ce{\scriptsize{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}\), where each line represents a bond.
• \(\ce{C}\) can make chains and rings, e.g., ethane: \(\ce{\scriptsize{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}\), and a propane: \(\ce{\scriptsize{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}\) are chains of two and three \(\ce{C's}\). The ability of \(\ce{C}\) to make a chain of atoms is called catenation.
• A \(\ce{C}\) atom can make bonds with more than two \(\ce{C's}\) resulting in branched compounds that increases the number of compounds possible, e.g. four \(\ce{C's}\) molecule can be in a straight chain as in n-butane: \(\ce{\scriptsize{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}\) or in a branched chain as in isobutane: \(\ce{\scriptsize{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\!\overset{\overset{\Large{H-\overset{\overset{\Large{H}}|}{C}\!-H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}\).
• Two \(\ce{C's}\) can make single, double, or triple bonds with each other allowing more variations. For example, ethane (\(\ce{\scriptsize{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}\)) has all single bonds; ethene ( \(\ce{\scriptsize{H}-{\underset{\underset{\Large{H}} |}{C}}={\underset{\underset{\Large{H}} |}{C}}\!-H}\) ) as double bond; and acetylene ( \(\ce{\scriptsize{H-C≡C-H}}\)) has a triple bond between \(\ce{C's}\).
• \(\ce{H}\) is the lightest monovalent atom that can occupy the valencies of \(\ce{C'}\) not used in \(\ce{C}\) to \(\ce{C}\) bonds.
• \(\ce{H}\) bonded with a strongly electronegative atom like \(\ce{O}\) or \(\ce{N}\) can interact with a \(\ce{O}\) or \(\ce{N}\) atom of a neighboring molecule through hydrogen bonding that plays a vital role in the functioning of organic molecules in living things.
• \(\ce{C}\) or \(\ce{H}\) in an organic compound can be replaced with a heteroatom that tremendously increases the variety of organic compounds available to living things.
Some other elements have the desired characteristics of \(\ce{C}\) and \(\ce{H}\) but are associated with significant disadvantages. For example, silicon (\(\ce{Si}\)), like \(\ce{C}\), makes i) four bonds, ii) straight and branched chains, and iii) a single bond with hydrogen. The drawbacks of \(\ce{Si}\) are i) it is two times heavier than \(\ce{C}\), ii) it makes weaker unstable \(\ce{Si-Si}\), and \(\ce{Si-H}\) bonds compared to \(\ce{C-C}\) and \(\ce{C-H}\) bonds, and iii) its oxidation product is solid silicon dioxide (\(\ce{SiO2}\)) which is insoluble in water and would have been difficult to excrete than the gaseous \(\ce{CO2}\) from the \(\ce{C}\) compounds which are easier to exhale. Similarly, halogens are monovalent like hydrogen, but halogens are significantly heavier than \(\ce{H}\) and make weaker bonds with \(\ce{C}\) than \(\ce{C-H}\) bonds.
What is organic chemistry?
Organic chemistry is the study of the properties and reactions of organic compounds.
The following section describes chemical bonding that determines the compound's physical properties and chemical reactivities. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/01%3A_Bonding_in_organic_compounds/1.01%3A_What_is_organic_chemistry.txt |
Learning Objectives
• Describe the nature of ionic, covalent, and coordinate covalent bonds.
• Understand the nature of a bond's polarity and molecules' polarity.
• Describe the nature of intermolecular forces and their effect on some properties of the compounds.
• Compare the properties of organic and inorganic compounds.
Chemical bonds were introduced in general chemistry. This is a review of chemical bonds for understanding the differences between inorganic and organic compounds. The chemical bonds connect the atoms in the compounds. Transferring or sharing some valance electrons from one atom to another makes the bonds. There are three major types of chemical bonds, ionic, covalent, and coordinate covalent bonds, as described below.
Ionic bond
An ionic bond is formed by transferring some valence electrons from one atom (usually from a metal atom) to another (usually to a nonmetal atom). The atom that loses an electron becomes a positively charged specie called a cation. For example, sodium atom $\ce{Na}$ loses one electron and become $\ce{Na^{+}}$, or calcium $\ce{Ca}$ loses two electrons and becomes $\ce{Ca^{2+}}$. The atom that gains an electron becomes a negatively charged specie called an anion. For example, chlorine atom $\ce{Cl}$ gains one electron and become $\ce{Cl^{-}}$, or oxygen $\ce{O}$ gains two electrons and becomes $\ce{O^{2-}}$.
The same charges repel each other, and opposite charges attract each other. The cations and the anions come together in the structure of the compound in such a way that the distances between the centers of opposite charges are minimized to increase attraction. The spaces between the centers of the same charges are maximized to decrease repulsion, as illustrated in Figure $1$ for the case of $\ce{NaCl}$. An ionic bond is the net attractive force holding the ions together in an ionic compound. ionic compounds are hard and have high melting and boiling points due to the strong electrostatic attractive forces. Ionic compounds are brittle because a slight displacement of one layer of atoms relative to the other layer can place similar charges next to each other and split due to repulsive forces between similar charges.
Covalent bond
What is a covalent bond?
A bond formed by sharing valence electrons is a covalent bond. Each bonded atom contributes one electron in a shared pair of electrons called a covalent bond, e.g., $\ce{H-Cl}$ bond illustrated in Figure $2$. The shared pair of electrons is called a bonding pair, and it is usually represented as a single line between the bonded atoms as in $\ce{H-Cl}$. Note that there are three unshared pairs of electrons on $\ce{Cl}$ atom in $\ce{H-Cl}$ molecule as illustrated in Figure $2$. The unshared electron pairs are called nonbonding or lone pairs of electrons. Atoms may share two electrons each to make a double bond, e.g., in $\ce{O=O}$, or three electrons each to create a triple bond, e.g., in $\ce{N≡N}$, as illustrated in Figure $2$.
How is a covalent bond formed?
Isolated atoms have potential energy due to several factors, like movements of atoms, movements of and electrostatic interactions among subatomic particles, etc. When two atoms approach each other, attractive and repulsive forces develop. The attractive forces are the attraction between opposite charges, i.e., between the nucleus of one atom and the electrons of the other. The repulsive forces are the repulsion between the same charges. When the two atoms can place more electrons between the nuclei by overlapping their atomic orbitals, the attractive forces between nuclei and electrons become stronger than the repulsive forces. Figure $3$ illustrates this situation for the case of $\ce{H-H}$ covalent bond formation. The two atoms move towards each other due to the attractive force resulting in a decrease in their potential energy until the minimum potential energy is reached. Decreasing the internuclear distance to less than the bond distance results in repulsive forces increasing more than attractive forces, and the overall potential energy increases.
Bond length, bond energy, and the bond strength
• Bond lenght is the distance between the centers of the two atoms at the potential energy minimum. For example, the bond distance $\ce{H-H}$ is 0.74 Å.
• The difference in the energy of isolated atoms and the energy state at the bonding distance is the bond energy. For example, the bond energy of $\ce{H-H}$ is 4.52 eV as shown in Figure $3$
• Bond energy is always released when the bond is formed, and the same amount of energy is absorbed when the same bond is broken. Stronger bonds have higher bond energy, and weaker bonds have lower bond energy.
The polarity of a covalent bond
The polarity was introduced in general chemistry. Here is a quick review. When the bonded atoms are the same, the bonding electron pair equally shared between them and the bond is nonpolar. For example, $\ce{H-H}$, $\ce{F-F}$, and $\ce{C-C}$ bonds are nonpolar. When the bonded atoms differ, one atom attracts the bonding election more than the other. The ability of an atom to attract the bonding electrons to itself is called electronegativity. If the difference in the electronegativity values of the bonded atoms is less than 0.5, the bond is still considered nonpolar. However, if the difference in the electronegativity values of the bonded atoms is between 0.5 to 1.9, the bond is a polar covalent. When the electronegativity difference is more than 1.9, it is usually an ionic bond.
In a polar covalent bond, the more electronegative atom has a partial negative ($\delta$-) charge because the electronegative atom has more share of negatively charged electrons, and the other atom has partial positive ($\delta$+) charge, e.g., $\ce{\overset{\delta{+}}{H}{-}\overset{\delta{-}}{Cl}}$, $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{O}}$, and $\ce{\overset{\delta{-}}{O}{-}\overset{\delta{+}}{H}}$.
The polarity of a molecule
The polarity of molecules is described in general chemistry. Recall that polarity is a vector quantity. It is also shown as an arrow over the bond with the arrowhead pointing to $\delta$- end and the tail with a plus sign starting from $\delta$+ end of the polar bond, e.g., $\ce{\overset{\delta{+}}{H}{-}\overset{\delta{-}}{Cl}}$ can be represented in vector form as: $\ce{\overset{\Large{+\!->}}{H-Cl}}$. Suppose there is more than one polar bond in a molecule. In that case, the polarity vectors may cancel out each other, e.g., two equal and opposite polarity vectors in a carbon dioxide molecule ($\ce{\overset{\large{<-\!++\!->}}{O=C=O}}$) cancel each other resulting in a nonpolar molecule. Therefore, the molecule with more than one polar bond may or may not be polar, depending on whether the polarities cancel each other. The following rules determine the polarity of a molecule:
1. A molecule is polar if there is only one polar bond in it.
2. A molecule is polar if there are polar bonds, but the molecule is not symmetric. In this case, the polarities do not entirely cancel out.
3. A molecule is nonpolar if there are polar bonds, but the molecule is symmetric, e.g., i) linear with two equal bonds, ii) trigonal planar with three equal bonds, iii) tetrahedral with four equal bonds. In these cases, the polarities cancel out.
4. A molecule is nonpolar if there is no polar bond in it.
Coordinate covalent bond
The coordinate covalent bond is the same as the covalent bond, except that both bonding electrons are donated by one of the two bonded atoms. The coordinate covalent bond is also called a dative bond. For example, the bond formed between the boron atom of boron trifluoride ($\ce{BF3}$) and the nitrogen atom of ammonia ($\ce{ _{\bullet}^{\bullet}\! NH3}$) in the following reaction: $\ce{F3B + _{\bullet}^{\bullet}\! NH3 -> F3 \overset{-}{B}- \overset{+}{N}H3}$, where $\ce{N}$ donates its lone pair to make the bond. The charges developed on the $\ce{N}$ and $\ce{B}$ after bonding is explained later in the formal charge section.
Intermolecular interactions
Although a covalent bond is almost as strong as an ionic bond, the covalent bond forces operate within a molecule. The question is, what holds the molecules together in solid molecular compounds? These are electrostatic interactions called intermolecular forces that were introduced in general chemistry. Intermolecular forces are much weaker than ionic, covalent, or coordinate covalent bonds. A line shows a covalent bond, while dotted lines usually show intermolecular forces. Three major intermolecular forces exist: i) London dispersion forces, ii) dipole-dipole interaction, and iii) hydrogen bonding.
London dispersion
Atoms have +ve protons in the nucleus and equal -ve electrons outside the nucleus. The atom is nonpolar because the electrons and protons are equal, and the -ve electrons are symmetrically distributed around the +ve nucleus. Electrons can be considered like clouds that can temporarily sway to one side or the other. If the electron cloud sways to one side, the atom is no more nonpolar, the side of the nucleus becomes a partial positive ($\delta$+) pole, and the side where electrons move to becomes a partial negative ($\delta$-) pole. This transient dipole in one atom induces a temporary dipole in the neighboring atom due to repulsion between the same charges or attraction between the opposite charges. The dipoles orient themselves to maximize attractive force between opposite charges and minimize repulsion between the same changes resulting in a net attractive force. This attraction between transient dipole-induced dipole is called the London dispersion force. The same phenomenon happens in the molecules.
The transient dipole appears and disappears randomly. London dispersion force is proportional to molecular mass because more mass means more electrons and a higher probability of temporary dipole. That is why smaller nonpolar organic compounds, like methane ($\ce{CH4}$, boiling point ~-160 oC) are gases, large like decane ($\ce{C10H22}$, boiling point ~+174 oC) are liquid, and even larger like eicosane ($\ce{C20H42}$, melting point ~37 oC, boiling point ~+343 oC) are solid.
Dipole-dipole interactions
Polar molecules have a permanent dipole in addition to the London dispersions force (transient dipoles). Therefore, polar molecules have higher intermolecular interaction and, consequently, higher melting and boiling points than nonpolar molecules of comparable molecular masses. For example, acetone ($\ce{C2H6O}$), a polar molecule, is liquid (melting point ~-95 oC and boiling point ~+56 oC), and butane ($\ce{C4H10}$), a nonpolar molecule of the same molecular mass (58 g/mole), is a gas (melting point ~-137 oC and boiling point ~0 oC).
Hydrogen bonding
Hydrogen bonding is a particular class of dipole-dipole interactions that involve $\ce{\overset{\delta{-}}{O}{-}\overset{\delta{+}}{H}}$, $\ce{\overset{\delta{-}}{N}{-}\overset{\delta{+}}{H}}$, or $\ce{\overset{\delta{-}}{F}{-}\overset{\delta{+}}{H}}$ dipole. Hydrogen bonding is usually stronger than dipole-dipole interactions because i) the dipoles in hydrogen bonding are usually stronger than the other dipoles, ii) $\delta{+}$ charge is denser on a small $\ce{H}$ than the same charge on larger atoms, and iii) being small, $\ce{H}$ can penetrate even in tight spaces to establish hydrogen bonding. For example, methanol ($\ce{CH3OH}$), capable of hydrogen bonding, is a liquid (boiling point ~+65 oC), formaldehyde ($\ce{H2C=O}$), a polar molecule without hydrogen bonding capability, is a gas (boiling point ~-19 oC), and ethane ($\ce{C2H2}$), a nonpolar molecule, is a gas with much lower boiling point (~-89 oC), all three having comparable molar masses 32 g/mole, 30 g/mole, and 30 g/mole, respectively. Further, hydrogen bonding is more important in living things because $\ce{H}$ is one of the primary constituents of organic compounds.
The intermolecular forces play an essential role in the chemistry of living things, e.g., as illustrated in Figure $4$ for defining the shape of a protein molecule.
Comparison of bonding and properties of organic and inorganic compounds
Organic compounds
1. Organic compounds are usually covalently bonded molecules, e.g., methane ($\ce{CH4}$), glucose ($\ce{C6H12O6}$), etc.
2. Organic compound are primarily composed of $\ce{C}$ and $\ce{H}$. Other elements, like $\ce{O}$, $\ce{N}$, $\ce{P}$, $\ce{S}$, etc., may also be present in small amounts.
3. The polarity of organic compounds varies from nonpolar to highly polar, e.g., decane ($\ce{C10H22}$) is nonpolar, and glucose ($\ce{C6H12O6}$) is polar.
4. Organic compounds are usually insoluble in water, e.g., decane ($\ce{C10H22}$) is insoluble in water, except a few highly-polar compounds that are soluble, e.g., glucose ($\ce{C6H12O6}$) is soluble in water.
5. Organic compounds usually have lower melting and boiling points, e.g., decane ($\ce{C10H22}$) melts at ~-30 oC and boils at ~174 oC.
6. Organic compounds are usually soft solids, liquids, or sometimes gases, e.g., glucose ($\ce{C6H12O6}$) is a soft solid, decane ($\ce{C10H22}$) is liquid, and methane ($\ce{CH4}$) is a gas.
Inorganic compounds
1. Inorganic compounds are usually ionic or highly polar covalent, e.g., table salt ($\ce{NaCl}$) is ionic and ammonium phosphate ($\ce{(NH4)3PO4}$ -a fertilizer, is a combination of covalent and ionic.
2. Inorganic compounds are usually composed of a combination of metals and nonmetals. e.g., calcium carbonate ($\ce{CaCO3}$ is composed of metal ($\ce{Ca}$) and nonmetals ($\ce{C}$) and ($\ce{O}$).
3. Inorganic compounds are usually soluble in water, e.g., $\ce{NaCl}$ and $\ce{(NH4)3PO4}$ are soluble in water. Some exceptions exist, e.g., $\ce{CaCO3}$ is not soluble in water.
4. Inorganic compounds usually have high melting and boiling points, e.g., $\ce{NaCl}$ melts at ~-801 oC and boils at ~1465 oC.
5. Inorganic compounds are usually hard and brittle solids, with few liquids and gases, e.g., $\ce{NaCl}$ and $\ce{CaCO3}$ are hard and brittle solids. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/01%3A_Bonding_in_organic_compounds/1.02%3A_What_is_a_chemical_bond.txt |
Learning Objectives
• Understand sp3, sp2, and sp hybridization of $\ce{C}$, $\ce{N}$, $\ce{O}$, and halogens in organic compounds.
• Understand bond angels and geometries in simple molecules having sp3, sp2, and sp hybridization of $\ce{C}$, $\ce{N}$, $\ce{O}$.
• Read an electrostatic potential map of a simple organic molecule and recognize the partially charged regions and the lone pairs.
• Understand free rotation around single and triple bonds but not around double bonds.
Prelude to the hybridization of atomic orbitals
Electron configuration of $\ce{C}$ atom is: 1s2, 2s2, 2px1, 2py1, 2pz0, where 1s is the core-shell and 2s and 2p are in the valence shell. Since two unpaired electrons are in the valence shell in 2px1, 2py1 orbitals, it implies that $\ce{C}$ should have two bonds in its compound. In reality, $\ce{C}$ has four bonds, e.g., there are four $\ce{C-H}$ covalent bonds in methane ($\ce{CH4}$). It can be explained by assuming that one electron is promoted from 2s to 2p orbital of $\ce{C}$, resulting in the electron configuration of 1s2, 2s1, 2px1, 2py1, 2pz0, which allows four bonds. The rationale is that making tow addition bonds releases more energy that more than compensates the energy consumed in promoting an electron from 2s to 2p orbital.
The molecular formula of methane is ($\ce{CH4}$), which shows the elemental composition but does not show bonds. The Lewis structure of methane is $\ce{ {H}-\overset{\overset{\huge{H}}|}{\underset{\underset{\huge{H}} |}{C}}\!-H}$ which shows covalent bonds as lines but does not show the actual geometry of the molecule. The geometry of the methane molecule is tetrahedral. $\ce{C}$ is at the center of the tetrahedron, and four $\ce{H}$ atoms at the corners of the tetrahedron as illustrated in Figure $1$ a and b. The four bonds are equal and at 109.5o from each other.
The geometry of the methane molecule, shown in Figure $1$ a and b, disagrees with the geometry of valence orbitals of $\ce{C}$ illustrated in Figure $1$ c. The s orbital is a spherical shape with a single phase shown by a single shade, while p orbitals have two lobes of opposite phases along the axis indicated by two different shades. The three p orbitals are perpendicular to each other, each lying along one of the three axes: px along the x-axis, py along the y-axis, and pz along the z-axis. Suppose the valence s and the three p orbitals of $\ce{C}$ were involved in the bonding. In that case, there should have been three bonds from the p orbitals perpendicular to each other and one bond from the s orbital different from the other three at any angle around the $\ce{C}$ nucleus. This difference between the actual and the expected geometry of $\ce{CH4}$ molecule is explained by the orbital hybridization concept.
According to the orbital hybridization concept, atomic orbitals can combine and produce an equal number of hybrid orbitals of energy and orientation different from the constituent orbitals. The s and all three, any two, or any one of the p valence orbitals of $\ce{C}$ can mix and produce four sp3, three sp2, or two sp hybrid orbitals.
The geometry around $\ce{C}$ atoms in organic molecules agrees with the geometry of the hybrid orbitals, as described in the following sections.
sp3 Hybridization of $\ce{C}$
sp3 Hybridization process
The s-orbital has a spherical shape and has a single phase as illustrated in $2$ a. Each p orbital has two lobes lying on a straight line with the nucleus at the center. The two lobes have opposite phases, represented by shades as illustrated in $2$ a. The term phase is about the wave nature of electrons. Waves have crests and troughs that are opposite phases.
When one s and three p orbitals mix, a set of four equal hybrid orbitals, called sp3 orbitals, is formed as illustrated in $2$ b. An sp3 orbital has two lobes of opposite phases, a more prominent lobe and a smaller lobe, lying along a straight line with the nucleus between the two lobes. Each sp3 orbital has one-fourth of s and three-fourths of p orbital character. The sp3 orbitals are arranged in a tetrahedral geometry around the nucleus, as illustrated in Figure $2$ b. In the case of $\ce{C}$, the four valence electrons distribute one each in the four sp3 hybrid orbitals.
Formation of methane ($\ce{CH4}$) molecule
Each of the four sp3 orbitals of $\ce{C}$ overlaps with one 1s orbital of $\ce{H}$, making a set of four covalent bonds as illustrated in Figure $2$ c. The bond formed by a head-on overlap of atomic orbitals is called a sigma-bond ($\sigma$-bond), which in the present case is called $\sigma$sp3-s bond. The result is a $\ce{CH4}$ molecule with all bonds equal in a tetrahedral geometry around $\ce{C}$, which agrees with the observed geometry, as illustrated in Figure $1$: a. All bonds are equal of 108.70 pm bond length and 109.5o bond angles in a $\ce{CH4}$ molecule. Figure $3$ presents an interactive model of $\ce{CH4}$ molecule for lea
Methane
Figure $3$: Interactive model of $\ce{CH4}$ molecule. The model can be enlarged by two fingers gesturing the touchpad or right-clicking and moving and rotated by swiping one finger in the image area or left click the mouse and moving.
Formation of ethane ($\ce{CH3CH3}$) molecule
Carbon can form a chain of carbons by making covalent bonds between carbon atoms. For example, a $\ce{C}$ overlaps a bigger lobe of one of its sp3 hybrid orbitals with the bigger lobe of the sp3 orbital of the second carbon along the axis of the sp3 orbitals to make a $\sigma$sp3-sp3 bond as illustrated in Figure $4$ a. The remaining sp3 orbitals on each $\ce{C}$ make $\sigma$sp3-s bond by overlapping with s orbital of $\ce{H}$ resulting in an ethane $\ce{C2H6}$ molecule.
Rotation around a single bond is possible under ambient conditions. Rotation around a $\ce{C-H}$ bond does not change the shape or orientation of any atom in the molecule. However, rotation around $\ce{C-C}$ in $\ce{CH3CH3}$ molecule results in different orientations of $\ce{H}$ atoms on one $\ce{C}$ relative to the $\ce{H}$ atoms on the other $\ce{C}$, as illustrated in Figure $4$ c. The structures with different orientations of atoms in the same molecule, as a result of rotation around a single bond, are called conformers of each other.
sp2 Hybridization of $\ce{C}$
sp2 Hybridization process
One s and two p orbitals in a valence shell can mix and produce a set of three equal hybrid orbitals, called sp2 orbitals, as illustrated in Figure $5$. An sp2 orbital has one-third s-orbital and two-third p-orbital character. An sp2 orbital comprises two lobes, a more prominent lobe and a smaller lobe of opposite polarity along a straight line with the nucleus between the two lobes, as in the case of the sp3 orbital. The three sp2 are in a plane at 120o from each other, i.e., a trigonal planer geometry with the nucleus in the middle. One p orbital left out from the hybridization lies perpendicular to the plane of the three sp2 orbitals.
Formation of ethene ($\ce{H2C=CH2}$) molecule
A $\ce{C}$ can form multiple bonds with a $\ce{C}$ or a heteroatom. For example, a $\ce{C}$ overlaps a bigger lobe of one of its sp2 hybrid orbitals with the bigger lobe of the sp2 orbital of the second carbon along the axis of the sp2 orbital to make a $\sigma$sp2-sp2 bond as illustrated in Figure $5$b. The remaining sp2 orbitals on each $\ce{C}$ make $\sigma$sp2-s bond by overlapping with s orbital of $\ce{H}$. Two p orbitals on adjacent carbons orient themselves parallel and overlap sideways, forming a pi-bond ($\pi$-bond). This results in the ethene molecule illustrated in Figure $5$c.
The $\pi$-bond is weaker than a $\sigma$-bond because the sideways overall of two parallel p-orbitals making a $\pi$-bond is less than the head-one overlap of orbitals making a $\sigma$-bond. Therefore, a $\pi$-bond is formed only when a $\sigma$-bond can not be created. A single bond is always a $\sigma$-bond. A $\pi$-bond and a $\sigma$-bond together, i.e., $\ce{C=C}$ is called a double bond. A double bond is stronger than a single bond. The $\pi$-bond becomes more weak if the p-orbitals constituting it are not parallel because the overly is less between nonparllel p-orbitals than between parallel p-orbitals. Ther is no overall lap between p-orbitals perpendicular to each other, i.e., there is no $\pi$-bond when p-orbitals are at 90o (perpendicular). Therefore, rotation around a double does not happen until the $\pi$-bond is broken. There is no free rotation around $\ce{C=C}$ bond.
Geometry around each carbon in the ethene molecule is trigonal planar and bond angles close to the predicted value of 120o, as illustrated in Figure $5$d. All six atoms in the ethene ( $\ce{CH2CH2}$) molecule are in the same plane and are locked in this geometry due to no free rotation around the $\ce{C=C}$ bond.
sp Hybridization of $\ce{C}$
sp Hybridization process
One s, and one p orbitals in a valence shell can mix and produce a set of two equal hybrid orbitals, called sp orbitals, as illustrated in Figure $6$a. An sp-orbital has 50% s-orbital and 50% p-orbital character. Like sp3 and sp2 orbitals, an sp orbital comprises two lobes, a more prominent lobe and a smaller lobe of opposite polarity along a straight line with the nucleus between the two lobes. The two sp orbitals are in a line at 180o from each other, i.e., a linear geometry with the nucleus in the middle. Two p orbitals left out from the hybridization lie perpendicular to the axis of the sp orbitals and perpendicular to each other, as illustrated in Figure $6$a.
Formation of ethyne ($\ce{HC#CH}$) molecule
An sp-hybridized $\ce{C}$ overlaps a more prominent lobe of one of its sp orbitals with the bigger lobe of sp orbital of the second carbon along the axis of the sp orbitals to make a $\sigma$sp-sp bond as illustrated in Figure $6$b. The remaining sp orbitals on each $\ce{C}$ make $\sigma$sp-s bond by overlapping with s orbital of $\ce{H}$. One p orbitals on a $\ce{C}$ orient itself parallel to one p-orbital on the other and overlap sideways, forming a $\pi$-bond. The other two p orbitals are also parallel to each other and overlap sideways to form the second $\pi$-bond as illustrated in Figure $6$c. This results in an ethyne molecule with a triple ($\ce{C#C}$) bond, i.e., a $\sigma$-bond and two $\pi$-bonds between the two $\ce{C}$ atoms, as illustrated in Figure $6$d.
A triple-bond is stronger than a double-bond, and a double-bond is stronger than a single-bond. Unlike a double bond, rotation around a triple bond happens because as the $\pi$-bond starts breaking due to p-orbitals going away from parallel orientation, a new $\pi$-bond starts making with the alternate p-orbital on the neighboring $\ce{C}$. This fact becomes apparent by comparing the $\pi$-bond electron clouds in the cases of a double bond of ethene and a triple bond of ethyne illustrated in Figure $7$. However, the rotation around a triple-bond means less as its geometry is linear.
Hybridization of $\ce{N}$, $\ce{O}$, and halogens
Heteroatoms like $\ce{N}$, $\ce{O}$, and halogen atoms (represented as: $\ce{X}$) can also mix all or some of their valence shell s and p orbitals to have sp3, sp2, or sp hybridization like $\ce{C}$. The differences are the following.
Number of covalent bonds and lone pairs in compounds
Table 1 shows the number of covalent bonds and the number of lone pairs on atoms of elements commonly found in organic compounds. $\ce{C}$ has four valence electrons that distribute one each in the hybrid orbitals and the remaining p orbitals. Lewis symbols of elements represent valence electrons around the symbol of the element. For example, Lewis symbol of $\ce{C}$ is $\ce{.\overset{\Large{\cdot}}{\underset{\Large{\cdot}}{C}}.}$ that shows four unpaired electrons in the valence shell. Each unpaired valence electron can make one covalent bond. So, $\ce{C}$ makes four covalent bonds in organic compounds.
$\ce{H}$ has one valence electron, represented as $\ce{\!\overset{\Large{\cdot}}{H}}$ that shows one unpaired electron that makes one bond and no lone pair on $\ce{H}$ in their compounds.
$\ce{N}$ has five valence electrons represented in Lewis symbol as $\ce{.\overset{\Large{\cdot}}{\underset{\Large{\cdot\cdot}}{N}}.}$ that shows three unpaired electrons and one electron pair. The unpaired electrons make one bond each while the paired electrons remain as a lone pair on $\ce{N}$ in its compounds.
$\ce{O}$ has six valence electrons represented in Lewis symbol as $\ce{.\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot\cdot}}{O}}.}$ that shows two unpaired electrons and to electron pairs. The unpaired electrons make one bond each while the paired electrons remain as two lone pairs on $\ce{O}$ in its compounds.
Halogen $\ce{X}$ ( where $\ce{X}$ can be $\ce{F}$, $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$) has seven valence electrons. For example, $\ce{Cl}$ represented as $\ce{:\!\overset{\Large{\cdot}}{\underset{\Large{\cdot\cdot}}{Cl}}\!:}$ that shows one unpaired electron that makes one bond while the paired electrons remain as three lone pairs on $\ce{X}$ in their compounds.
Table 1: Number of covalent bonds and lone pairs of electrons around elements commonly found in organic compounds.
Atom (Lewis symbol) Number of Covalent bonds Number of lone pairs
$\ce{.\overset{\Large{\cdot}}{\underset{\Large{\cdot}}{C}}.}$ 4 0
$\ce{\!\overset{\Large{\cdot}}{H}}$ 1 0
$\ce{.\overset{\Large{\cdot}}{\underset{\Large{\cdot\cdot}}{N}}.}$ 3 1
$\ce{.\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot\cdot}}{O}}.}$ 2 2
Halogen $\ce{:\!\overset{\Large{\cdot}}{\underset{\Large{\cdot\cdot}}{X}}\!:}$ ( where $\ce{X}$ can be $\ce{F}$, $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$) 1 3
Geometry around the central atom
sp3 Hybridization
An sp3 hybridized $\ce{C}$ makes four $\sigma$-bonds in a tetrahedral geometry around $\ce{C}$ where bonds are around 109o from eachother, as illustrated Figure $8$a for the case of methane ($\ce{CH4}$). The electrostatic potential map of methane is also shown.
An electrostatic potential map shows electron cloud in colors: green means neutral or nonpolar region, red means $\delta{-}$ region, blue means $\delta {+}$ region, yellow is between neutral and $\delta{-}$ and light blue is between neutral and $\delta {+}$.
Methane is nonpolar, shown by green in the electrostatic potential map.
The sp3 orbitals of $\ce{N}$ have tetrahedral geometry where three orbitals make $\sigma$-bonds while a lone pair occupy the fourth. The three peripheral atoms are at the base forming a triangle while $\ce{N}$ is raised from the middle of the triangle resulting in a trigonal-pyramidal geometry, as illustrated in Figure $8$b for $\ce{NH3}$ molecule. The trigonal-pyramidal geometry is a modified form of tetrahedral geometry where the fourth corner of the tetrahedron is missing due to being occupied by a lone pair. The bond angles are close to 109o as predicted by the tetrahedral geometry. The molecule is not symmetric with polar bonds, so, it is polar, which is evident from its electrostatic potential map. The red region is due to the polarity of $\ce{N-H}$ bonds and also due to the lone pair region showing red because it is in an sp3-orbital which, unlike a p-orbital, is not symmetric with most of the electron located in the more prominent lobe on one side of the nucleus.
The sp3 orbitals of $\ce{O}$ have tetrahedral geometry where two orbitals make $\sigma$-bonds while lone pairs occupy the other two. The two corners of a tetrahedron which are peripheral atoms and the $\ce{O}$ in the middle of the tetrahedron shape, results in a bent shape with the other two corners of the tetrahedron missing due to being occupied by lone pairs as illustrated in Figure $8$c for $\ce{H2O}$ molecule. The bent shape is a modified form of tetrahedral geometry where two corners of the tetrahedron are missing due to being occupied by lone pairs. The bond angle is close to 109o as predicted by the tetrahedral geometry. The electrostatic potential map reflects the polarity of two $\ce{O-H}$ bonds and the two lone pairs in sp3-orbitals.
The sp3 orbitals of halogen atoms have tetrahedral geometry where one orbital makes a $\sigma$-bond while lone pairs occupy the other three. One bond can make only a linear geometry as illustrated in Figure $8$d for $\ce{HCl}$ molecule. The other three corners of the tetrahedron are missing due to being occupied by lone pairs. The electrostatic potential map reflects the polarity of the bond and the three lone pairs in sp3-orbitals.
sp2 Hybridization
An sp2 hybridized $\ce{C}$ makes three $\sigma$-bonds using its sp2 orbitals in a trigonal planer geometry around $\ce{C}$ and a $\pi$-bond using its p-orbital. The $\sigma$-bonds are around 120o from each other, as illustrated in Figure $9$a for ethene ($\ce{H2C=CH2}$) molecule. The electrostatic potential map of $\ce{H2C=CH2}$ in Figure $9$a shows red region around the axis of the $\ce{C=C}$ bond which is due to the $\pi$-bondin electrons placed above and below the $\sigma$-bond. Further the $\ce{H's}$ are bluish compared to those in $\ce{CH4}$ molecule indicating that the $\ce{C-H}$ is slightly polar with $\delta {+}$ charge on the $\ce{H}$.
The sp2 orbitals of $\ce{N}$ have trigonal planer geometry where two of the sp2 orbitals make $\sigma$-bonds, and the lone pair occupies the third sp2 orbital. The p-orbital makes a $\pi)-bond. The geometry around an sp2 hybridized \(\ce{N}$ is bent with a bond angle around 120o as the trigonal planer geometry predicted. The bent geometry of sp2 hybridized $\ce{N}$ is illustrated in Figure $9$b for methanimine ($\ce{H2C=NH}$) molecule. The electrostatic potential map reflects the bond polarity and the lone pair of electrons in one sp2-orbital as a red region.
The sp2 orbitals of $\ce{O}$ have trigonal planer geometry where one sp2 orbital makes a $\sigma$-bond, and a lone pair occupies the other two sp2 orbitals. The p-orbital makes a $\pi)-bond. An sp2-hybridized \(\ce{O}$ has a linear geometry around $\ce{O}$ atom, as illustrated in Figure $9$c for formaldehyde ($\ce{H2C=O}$) molecule. The electrostatic potential map reflects the bond polarity and the lone pair of electrons in two sp2-orbitals as a red region.
The sp2 or hybridization of halogens is not observed except in a few cases, e.g., when one of the lone pairs is involved in resonance which will be described in a later section.
sp Hybridization
An sp hybridized $\ce{C}$ makes two $\sigma$-bonds using its sp-orbitals that are at 180o from each other, i.e., in a linear geometry. The remaining two p-orbitals make two $\pi$-bonds perpendicular to each other and around the sigma bond resulting in a linear geometry around the sp hybridized carbon, as illustrated in Figure $10$a for ethyne ($\ce{HC#CH}$) molecule. Electrostatic map shows red region around the $\ce{C#C}$ bond axis which is due to $\pi$-electrons and it also shows that the $\ce{C-H}$ bond in ethyne is polar with $\delta {+}$ charge on $\ce{H}$
The sp orbitals of $\ce{N}$ have linear geometry. One of the sp orbitals makes a $\sigma$-bond, and the other is occupied by a lone pair, and the two p-orbital make two $\pi$-bonds resulting in a triple bond. The geometry around an sp hybridized $\ce{N}$ is linear, as illustrated in Figure $10$b for hydrogen cyanide ($\ce{HCN}$) molecule. The electrostatic potential map reflects the bond polarity and shows the red region where the lone pair of electrons are located in an sp-orbital of nitrogen.
The sp hybridization of $\ce{O}$ is rare in organic compounds, but it does exist, e.g., in carbon monoxide molecule ($\ce{CO}$) molecule. It involves formal charges on the atoms, which will be described in a later section. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/01%3A_Bonding_in_organic_compounds/1.03%3A_Hybridization_of_orbitals_and_3D_structures_of_simple_organic_compounds.txt |
Learning Objectives
• Read and draw molecular, Lewis, condensed, and structural formulae of simple organic compounds.
• Understand the mixed versions of and slight variations in drawing the formulae of simple organic compounds.
The molecular formula, Lewis formula, condensed formula, skeletal formula, or a combination of these can represent an organic compound. These ways of representing organic compound and what they mean is described below.
Molecular formula
The molecular formula tells the symbols of the elements that compose the compound, and the subscript to the element symbol denotes how many atoms of that element are in the molecule.
For example, ($\ce{CH4}$) is a molecule formula of methane which means there is one carbon and four hydrogen atoms in a methane molecule. $\ce{C2H6}$ is a molecule formula of ethane, which means the ethane molecule has two carbon and six hydrogen atoms. Molecular formulas do not tell about the molecule's bonds and shapes.
Lewis formula
The Lewis structure or Lewis formula shows all the bonding electron pairs as lines (bonds) and lone pairs (non-bonding electron pairs) as pairs of dots around each atom in a molecule.
For example, Lewis formula of ethane ($\ce{C2H6}$) is: $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$ that shows each carbon is bonded with one $\ce{C}$ and three $\ce{H}'s$ by single bonds. Similarly, Lewis formula of formaldehyde ($\ce{CH2O}$) is: $\ce{\small{H}-{\underset{\underset{\Large{H}} |}{C}}=\overset{\Large{\cdot\cdot}}{O}\!:}$ that shows $\ce{C}$ is bonded with two $\ce{H}'s$ by single bonds and with one $\ce{O}$ by a double bond and $\ce{O}$ has two lone pairs on it. The lone pairs are usually omitted from Lewis structures except when needed to emphasize their presence. For example, Lewis formula for methanol ($\ce{CH4O}$) with lone pairs is: $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot\cdot}}{O}}\!-H}$, but it can also be shown as $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-{O}\!-H}$ where the lone pairs are not shown on $\ce{O}$ but it is understood that two lone pairs are there.
Condensed formula
The Lewis formulas become complicated and time-consuming for larger organic compounds.
Condensed formula simplifies the Lewis formula by writing each $\ce{C}$ followed by $\ce{H's}$ attached with it. Subscripts are used to show more than one $\ce{H's}$. If there is a heteroatom, i.e., any atom other than $\ce{C}$ or $\ce{H}$ in the chain, it is condensed like $\ce{C}$, except for halogens which are condensed like $\ce{H}$.
For example, the Lewis formula of ethane $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$ is condensed as $\ce{CH3CH3}$. Lewis formula for methanol $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-{O}\!-H}$ is condensed as $\ce{CH3OH}$. Lewis formula for ethanamine $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-{\underset{\underset{\Large{H}} |}{N}}\!-H}$ is condensed as $\ce{CH3CH2NH2}$. Lewis formula for 2-chloropropane $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{Cl}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$, is condensed as $\ce{CH3CHClCH3}$.
Skeletal formula
The Lewis and condensed formulas do not show the geometry of the organic compounds. Further, the Lewis and condensed formulas become complicated and time-consuming for large organic compounds.
Skeletal formulas or line-angle formulas overcome these drawbacks by simplifying the representation of organic compounds by omitting $\ce{C's}$ and $\ce{H's}$ in the formula and showing only the skeleton of the compound by $\ce{C}$-to-$\ce{C}$ bonds as lines in a geometry that is closer to the actual geometry. Any heteroatom and $\ce{H's}$ attached to the heteroatom are shown in the skeletal formula.
Some terms related to the primary classification of organic compounds
• Before learning skeletal formulas, it is essential to understand the following terms related to the primary type of organic compounds.
• Organic compounds containing only $\ce{C's}$ and $\ce{H's}$ are called hydrocarbons. For example, ethane ($\ce{CH3CH3}$), ethene ($\ce{CH2CH2}$), and ethyne ($\ce{CHCH}$) described in previous sections are hydrocarbons.
• The hydrocarbons contain only $\sigma$-bonds are called alkanes. For example, methane ($\ce{CH4}$) and ethane ($\ce{CH3CH3}$) are alkanes.
• Alkanes containing a chain of $\ce{C's}$ where $\ce{C's}$ are connected with either one or two $\ce{C's}$ are called straight chain alkanes or normal-chain alkanes (n-alkanes). For example, n-alkane having four carbons is n-butane (($\ce{CH3CH2CH2CH3}$), five carbons is n-pentane ($\ce{CH3CH2CH2CH2CH3}$), six carbons is n-hexane ($\ce{CH3CH2CH2CH2CH2CH3}$), and so on.
• Alkanes in which at least on $\ce{C}$ connected with three or four other $\ce{C's}$ are called branched-chain alkanes. For example, isopropane $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\!\!\!\overset{\overset{\Large{H-\overset{\overset{\Large{H}}|}{C}\!-H}}|}{\underset{\underset{\Large{H}} |}{C}}\!\!\!-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$ is a branched-chain alkane
• The hydrocarbons containing at least one double bond, i.e., a $\sigma$-bond and a $\pi$-bond together, are called alkenes. For example, ethene
$\ce{\small{H}-{\underset{\underset{\Large{H}} |}{C}}={\underset{\underset{\Large{H}} |}{C}}\!-H}$ is an alkene.
• The hydrocarbons containing at least one triple-bond, i.e., a $\sigma$-bond and two $\pi$-bonds together, are called alkynes. For example, ethyne $\ce{\small{H-C≡C-H}}$ is an alkyne.
• The hydrocarbons with a planer cyclic structure having alternating odd numbers of double bonds are a special class of hydrocarbons called aromatic hydrocarbons that will be described later.
• Organic compounds containing at least one heteroatom, i.e., $\ce{O}$, $\ce{N}$, $\ce{S}$, $\ce{P}$, etc. are not hydrocarbons. For example, methanol $\ce{CH3OH}$ is not a hydrocarbon. Several classes of organic compounds are not hydrocarbons, which will be described later.
Skeletal formulas of n-alkanes
Methane ($\ce{CH4}$) is the simplest alkane that has a tetrahedral geometry around its $\ce{C}$, as illustrated in Figure $1$a. Gray lines in Figure $1$a show the outline of the tetrahedron shape, and the black lines show $\ce{C-H}$ bonds. The plane defined by $\ce{C}$, $\ce{H}$ on the top, and $\ce{H}$ on the right is in the plane of the paper in this perspective drawing (Figure $1$a). The $\ce{H}$ on the hashed wedge is going below, and the $\ce{H}$ on the solid wedge is coming above the plane of the paper. The plane of the paper cuts through the middle of two $\ce{H's}$ on the left of the drawing. The point of view is slightly above $\ce{C}$ towards the top-right corner.
Figure $1$b shows the same structure without the tetrahedron layout drawing. All bonds are equal, and all bond angles are 109.5o. Figure $1$c shows the model of $\ce{CH4}$ molecule from approximately the same view. Note that the geometry of $\ce{CH4}$ molecule is two V's of 109.5o internal angle, placed perpendicular to each other, and joined at the vertex. Figure $1$d shows the same structure rotated such that the two $\ce{H's}$ in the plane of the paper are on a straight line at the bottom of the drawing. Figure $1$e shows the model rotated in the same orientation as the perspective drawing in Figure $1$d.
Replacing any $\ce{H}$ with another sp3 hybridized $\ce{C}$ results in ethane ($\ce{CH3CH3}$) as shown in row two of Table 1. A line drawn to represent $\ce{C-C}$ bond in ethane is the skeletal formula of ethane, as shown in row two of Table 1. When 2nd $\ce{H}$ of methane is also replaced with another sp3 hybridized $\ce{C}$, it results in propane ($\ce{CH3CH2CH3}$), as shown in row three of Table 1. Two $\ce{H's}$ of methane in the plane of the page have been replaced with $\ce{C's}$ in this case resulting in $\ce{C-C-C}$ bonds in an inverted V-shape in the plane of the page. The skeletal formula representing a propane molecule is two lines connected in an inverted V-shape, as shown in row three of Table 1.
Replacing any $\ce{H}$ of a terminal $\ce{C}$ of a propane with another sp3 hybridized $\ce{C}$ results in butane ($\ce{CH3CH2CH2CH3}$) as shown in 4th row of Table 1. Remember rotation around $\ce{C-C}$ single bond happens. Therefore, the 4th $\ce{C}$ of butane can be placed at any angle relative to the plane defined by the other three $\ce{C's}$. However, placing all four $\ce{C's}$ of butane in the same plane is the most stable arrangement because the terminal $\ce{C's}$, which are the bulkiest groups attached to the internal $\ce{C's}$, are farthest apart in this arrangement. Three lines connected by zigzag is the skeletal formula representing butane, as shown in row 4th of Table 1. The zigzag lines representing a chain of 4 $\ce{C's}$ can be extended to represent a chain of 5 $\ce{C's}$, by adding one line to represent a chain of 6 $\ce{C's}$ by adding two lines, and so on, as shown in Table 1 for the cases of n-alkanes having a chain of 2 to 12 $\ce{C's}$.
In summary, the skeletal formula of n-alkane is a line or lines connected zigzag representing $\ce{C-C}$ bonds. It is understood that:
• the terminals (end) and corners (bends) of the lines are $\ce{C's}$, and
• each $\ce{C}$ has four bonds, so the bonds which are not shown by the lines are the bonds to $\ce{H's}$.
With this knowledge, it is clear that a line or lines connected zigzag way are the structural formulas that represent structures of n-alkanes without showing $\ce{C's}$ and $\ce{H's}$ in the formula.
Table 1: Names, molecular formulas, condensed formulas, models, and skeletal formulas of n-alkanes containing 2 to 12 $\ce{C's}$ in a chain. (Note: Methane has no structural formula as it has no $\ce{C-C}$ bond. The $\ce{C's}$ are black and $\ce{H's}$ are white color in the models))
# of $\ce{C's}$ Name Molecular formula
Model,
Condensed formula, and
Structural formula
1 Methane $\ce{CH4}$
$\ce{CH4}$
2 Ethane $\ce{C2H6}$
$\ce{CH3CH3}$
3 Propane $\ce{C3H8}$
$\ce{CH3CH2CH3}$
4 Butane $\ce{C4H10}$
$\ce{CH3CH2CH2CH3}$
5 Pentane $\ce{C5H12}$
$\ce{CH3CH2CH2CH2CH3}$
6 Hexane $\ce{C6H14}$
$\ce{CH3CH2CH2CH2CH2CH3}$
7 Heptane $\ce{C7H16}$
$\ce{CH3CH2CH2CH2CH2CH2CH3}$
8 Octane $\ce{C8H18}$
$\ce{CH3CH2CH2CH2CH2CH2CH2CH3}$
9 Nonane $\ce{C9H20}$
$\ce{CH3CH2CH2CH2CH2CH2CH2CH2CH3}$
10 Decane $\ce{C10H22}$
$\ce{CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3}$
11 Undecane $\ce{C11H24}$
$\ce{CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3}$
12 Dodeane $\ce{C12H26}$
$\ce{CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3}$
Stem names of organic compounds
Names of n-alkanes without the last syllable, i.e., without -ane, are the stem names that represent the number of $\ce{C's}$ in the organic compound. For example, meth- from methane represents one $\ce{C}$, eth- from ethane represents two $\ce{C's}$, prop- from propane represents three $\ce{C's}$, and so on. These stem names will be described later in reference to name organic compounds.
Homologous series
It is evident from comparing the molecular formals of n-alkanes shown in Table 1 that n-alkanes differ from each other by a $\ce{CH2}$ or a multiple of $\ce{CH2}$ units. For example, add a $\ce{CH2}$ to methane ($\ce{CH4}$) to convert it to propane ($\ce{C2H6}$), and two $\ce{CH2}$ units to convert it to butane ($\ce{C4H10}$ ) and so on. Series of organic compounds that differ from each other by a $\ce{CH2}$ or multiple of $\ce{CH2}$'s are homologous series. All of the n-Alkanes shown in Table 1 are members of a homologous series.
General formula of n-alkanes
The general formula of n-alkanes is $\ce{C_{n}H_{2n + 2}}$ where n is a counting number, i.e., 1, 2, 3,... For example, when n = 1, it is methane $\ce{C_{1}H_{2\times{1} + 2}}$ = $\ce{CH4}$ (Recall: that when the subscript to the element symbol in the formal is 1, it is not written.). When n = 2, it is ethane $\ce{C_{2}H_{2\times{2} + 2}}$ = $\ce{C2H6}$, and so on
Variations in the structural formula to represent different configurations
The structural formulas in Table 1 represent the linear conformation of n-alkanes which is the most stable confirmation. Remember: rotation is possible around any single bond. For example, n-hexane shown in Figure $2$a is rotated around the middle $\ce{C-C}$ bond to acquire new confirmation shown in Figure $2$b and rotated further along the 2nd $\ce{C-C}$ from left to acquire another confirmation shown in Figure $2$c. All three structural formulas are shown in Figure $2$a, b, and c represent the same molecule. Remember: different shapes of the same molecule obtained by rotation around a single bond are different configurations of the same molecule.
Condensed and skeletal formulas of branched-alkanes
Replacing one or both $\ce{H's}$ of a non-terminal carbon of straight chain alkane (n-alkane) with a $\ce{C}$ or a $\ce{C}$ chain results in a branched-alkane. Replacing hydrogen on the middle carbon of propane ( $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$) with a $\ce{C}$ results in isopropane ( $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\!\!\!\overset{\overset{\Large{H-\overset{\overset{\Large{H}}|}{C}\!-H}}|}{\underset{\underset{\Large{H}} |}{C}}\!\!\!-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$) which is a branched chain alkane.
There are two ways to show the condensed formula of the branched alkane:
• show the condensed formula of the branch within small brackets next to the carbon it is attached to, e.g., $\ce{CH3CH(CH3)CH3}$ is the condensed formula of isopropane;
• show the condensed formula of the branch hanging above or below the carbon it is attached, e.g., $\ce{\small{CH3}-\overset{\overset{\LARGE{CH3\!\!\!\!\!\!\!}}|}{C}H-CH3}$ is the condensed formula of isopropane.
The skeletal formula of the branched alkane shows the skeletal formula of the branch with the terminal connected to the carbon in the main branch to which it is attached. For example, is the skeletal formula of isopropane. Figure $3$ shows another example of branched alkane and its condensed and skeletal formulas.
Skeletal formulas of alkenes
Hydrocarbons containing at least one double bond ($\ce{C=C}$ bond) are called alkene.
Since the two sp2 $\ce{C's}$ at the double bond have trigonal planer geometry with bond angles 120o, the double bond between the two sp2 $\ce{C's}$ fit in the zigzag skeletal structure, like alkanes, except that two lines are drawn where there is a double bond in the chain.
Table 2 shows the names, Lewis structures, models, condensed formulas, and skeletal formulas of two alkene examples. Their nomenclature will be explained later.
Table 2: Names, Lewis structures, models, condensed formulas, and skeletal formulas of some alkene examples. (Note: $\ce{C's}$ are black and $\ce{H's}$ are white color in the models)
Name Lewis structure
Model,
Condensed formula, and
Structural formula
Hept-2-ene $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-{\underset{\underset{\Large{H}} |}{C}}={\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$
$\ce{CH3CHCHCH2CH2CH2CH3}$
Propene $\ce{\small{H}-{\underset{\underset{\Large{H}} |}{C}}={\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$
$\ce{CH2CHCH3}$
Skeletal formulas of alkynes
Hydrocarbons containing at least one triple bond ($\ce{C≡C}$) bond are called alkyne.
Since the two sp $\ce{C's}$ at the triple-bond have linear geometry, three lines show the triple bond and the two single bonds to the other atoms attached to them are drawn in line with the triple bond. The zigzag skeletal structure shows the rest of the structure as usual.
Table 3 shows some alkyne examples' names, Lewis structures, models, condensed formulas, and skeletal formulas. Their nomenclature will be explained later.
Table 3: Names, Lewis structures, models, condensed formulas, and skeletal formulas of some alkyne examples. (Note: $\ce{C's}$ are black and $\ce{H's}$ are white color in the models)
Name Lewis structure
Model,
Condensed formula, and
Structural formula
Propyne $\ce{\small{H}-{C}≡{C}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$
$\ce{CHCCH3}$
Pent-2-yne $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-{C}≡{C}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$
$\ce{CH3CCCH2CCH3}$
Non-3-yne $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-{C}≡{C}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$
$\ce{CH3CH2CCCH2CH2CH2CH2CH3}$
Skeletal formulas of organic compounds containing heteroatoms
If any heteroatom, i.e., an atom other than $\ce{C}$ or $\ce{H}$, is present in an organic compound:
• its symbol in the skeletal formula shows it and
• any $\ce{H}$ present on the heteroatom is also shown next to the heteroatom as in the condensed formulas.
The skeletal structure shows the rest of the structure as usual. Table 4 shows the names, Lewis structures, models, condensed formulas, and skeletal formulas of some example organic compounds containing heteroatoms. Their nomenclature will be explained later.
Table 4: Names, Lewis structures, models, condensed formulas, and skeletal formulas of some organic compounds containing heteroatoms. (Note: $\ce{C's}$ are black, $\ce{H's}$ are white, and heteroatom are in other colors in the models)
Name Lewis structure
Model,
Condensed formula, and
Structural formula
2-Chloropropane $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{Cl}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$
$\ce{CH3CHClCH3}$
Ethanamine $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-{\underset{\underset{\Large{H}} |}{N}}\!-H}$
$\ce{CH3CH2NH2}$
Diethyl ether $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-{O}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$
$\ce{CH3CH2OCH2CH3}$
Acetone $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{O\!}}||}{C}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$
$\ce{CH3COCH3}$ or $\ce{\small{CH3}\overset{\overset{\Large{O\!}}||}{C}{CH3}}$
Variations in the formulas representing organic compounds
The condensed and the skeletal formulas are usually presented in organic chemistry, as described in the previous sections. However, they may be varied slightly according to the need.
If needed, mixed Lewis, condensed, and skeletal structures emphasize a particular part of the molecule.
For example, the $\ce{H's}$ are usually written to the right of $\ce{C}$ or to the right of heteroatoms in the condensed formula, but sometimes this order may be reversed. For example, Lewis formula for methanol $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-{O}\!-H}$ is presented in condensed form as $\ce{CH3OH}$, but it may be condensed as $\ce{H3COH}$, or as $\ce{H3C-OH}$ to emphasize the $\ce{C-O}$-bond. Similarly, Lewis formula of acetone $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{O\!}}||}{C}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$ is changed to skeletal formula as , but it may also be shown as to emphasize the $\ce{C-H}$ bond.
Sometimes the formula is further condensed by placing the repeating units within a bracket and subscript outside the bract to show the number of repeat units. For example, the condensed formula of isopropane is $\ce{\small{CH3}-\overset{\overset{\LARGE{CH3\!\!\!\!\!\!\!}}|}{C}H-CH3}$ that has three $\ce{CH3}$ units attached to the central $\ce{C}$. It can be further condensed as $\ce{(CH3)3CH}$. Similarly, the condensed formula of decane $\ce{CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3}$ contains eight $\ce{CH2}$ units and can be condensed further as $\ce{CH3(CH2)8CH3}$.
Examples of drawing Lewis, condensed, and skeletal formulae
Example $1$
Draw molecular, Lewis, condensed, and skeletal formulas of butane.
Solution
But- in butane means four $\ce{C's}$ and -ane means it is an alkane. Substituting n = 4 in the general formula of alkanes ($\ce{C_{n}H_{n\times{2} + 2}}$) gives:
molecular formula = ($\ce{C_{4}H_{4\times{2} + 2} = C4H10}$.
For Lewis, condensed, and skeletal formulas follow the steps shown in the figure below.
Example $2$
Draw Lewis, condensed, and skeletal formulas for methylethylamine
Solution
Meth- means one $\ce{C}$, eth-means two $\ce{C's}$, and aminie indicates a $\ce{N}$ two which the carbon chains are attached. With this information, follow the steps shown in the figure below.
Example $3$
Draw Lewis, condensed, and skeletal formula of 1-chlorobutane
Solution
But- means four $\ce{C's}$ chain and 1-chloro- means there is chlorine at the terminal $\ce{C}$. The nomenclature will be explained later. With this information, follow the steps shown in the figure below. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/01%3A_Bonding_in_organic_compounds/1.04%3A_Representing_organic_compounds.txt |
Learning Objectives
• Calculate the formal charge on atoms in organic molecules.
• Understand the geometry of and the location of valence electrons in the hybrid orbitals of species with formal positive charge, formal negative charge, and organic radicals.
Prelude to a formal charge
Isolated atoms are neutral because they have equal protons and electrons. When an atom loses a a valence electron it becomes +ve species or cation, e.g., $\ce{Na}$ after loosing an electron becomes $\ce{Na^{+}}$, and $\ce{Ca}$ after loosing two electron becomes $\ce{Ca^{2+}}$. When an atom gains a valence electron, it becomes -ve species or anion, e.g., $\ce{Cl}$ after gaining an electron becomes $\ce{Cl^{-}}$, and $\ce{O}$ after gaining two electrons becomes $\ce{O^{2-}}$. Covalently bonded species can also be ions, e.g., $\ce{NH3^{+}}$, and $\ce{H3O^{+}}$ are cations, and $\ce{HO^{-}}$, and $\ce{CO2^{2-}}$ are anions. Organic compounds often become ions as intermediates in reactions.
What is the formal charge?
A formal charge is assigned to an atom in a molecule based on the assumption that bonding electrons are shared equally.
How to calculate the formal charge?
The formal charge is calculated by subtracting an atom's valence electrons in a molecule from the valence electron of the isolated atom. Valence electrons of an isolated atom are equal to the first digit of the group number in the periodic table. For example, $\ce{H}$ is in group 1 and has one valence electron shown as: $\ce{\overset{\Large{\cdot}}{H}}$ in Lewis symbol. $\ce{C}$ is in group 14 and has four valence electrons shown as: $\ce{.\overset{\Large{\cdot}}{\underset{\Large{\cdot}}{C}}.}$, i.e., four unpaired dots in its Lewis symbol. $\ce{N}$ is in group 15 and has five valence electrons shown as: $\ce{.\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot}}{N}}.}$, i.e., five dots (one paired and three unpaired) in its Lewis symbol. Halogens are in group 17 and have seven valence electrons, e.g. $\ce{.\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot\cdot}}{Cl}}\!:}$, i.e., shown as three paired and one unpaired dots. Valence electrons in a molecule are assigned to an atom, assuming that the bonding electrons are equally shared, i.e., one electron to the atom per one bond. All the nonbonding electrons are assigned to the atom they are on.
The formula for calculating formal charge is:
$Fc = Ve - (B + Nb)\nonumber$
, where $Fc$ is the formal charge, $Ve$ is the valence electrons in an isolated atom, $B$ is the number of bonds attached to the atom, and $Nb$ is nonbonding electrons on the atom in the molecule.
Example $1$
What is the formal charge on $\ce{C}$ in methane ($\ce{ \small {H}-\overset{\overset{\LARGE{H}}|}{\underset{\underset{\LARGE{H}} |}{C}}\!-H}$).
Solution
Isolated $\ce{C}$ has four valance electrons, there are four bonds and no nonbonding electron in the molecule, so: $Ve$ = 4, $B$ = 4, $Nb$ = 0
$Fc = Ve - (B + Nb)$ = 4 - (4 + 0) = 0
Answer: 0 formal charge, i.e., $\ce{C}$ in methane ($\ce{ \small {H}-\overset{\overset{\LARGE{H}}|}{\underset{\underset{\LARGE{H}} |}{C}}{^{0}}\!\!-H}$), where superscript over the atom ($\ce{C}$ in this case) show the formal charge
Reactive intermediates and their formal charges
If one of the $\ce{C-H}$ bonds in an organic compound breaks, three situations may arise, e.g., for $\ce{ \small {H}-\overset{\overset{\LARGE{H}}|}{\underset{\underset{\LARGE{H}} |}{C}}-H}$ the situations are:
1. one of the bonding electrons may remain on the $\ce{C}$, i.e., ($\ce{\small{H}-\overset{\LARGE{\cdot}}{\underset{\underset{\LARGE{H}} |}{C}}-H}$),
2. none of the bonding electrons may remain on the $\ce{C}$, i.e, ($\ce{\small{H}-{\underset{\underset{\LARGE{H}} |}{C}}-H}$), and
3. both the bonding electrons may remain on the $\ce{C}$, i.e., ($\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{C}}-H}$).
Each of these situations results in a $\ce{C}$ with three bonds which are reactive intermediates. They are short-lived and tend to react to re-establish regular four bonds around the $\ce{C}$. Figure $1$ shows the structure and hybridization of the reactive carbon species. A specie with a single unshared electron on an atom is a free radical that tends to make a covalent bond by sharing the electron. A $\ce{\small{H}-\overset{ \LARGE {\cdot}}{\underset{\underset{ \LARGE {H}} |}{C}}-H}$ is an example of a free. The ($\ce{\small{H}-\overset{ \LARGE {\cdot}}{\underset{\underset{ \LARGE {H}} |}{C}}-H}$), is sp2 hybridized with three bonds and an unshared electron in its p-orbital, as illustrated in Figure $1$a.
Example $2$
What is the formal charge on $\ce{C}$ in $\ce{\small{H}-\overset{\LARGE{\cdot}}{\underset{\underset{\LARGE{H}} |}{C}}\!-H}$.
Solution
Isolated $\ce{C}$ has four valance electrons, there are three bonds and one nonbonding electron on $\ce{C}$ in the molecule, so: $Ve$ = 4, $B$ = 3, $Nb$ = 1
$Fc = Ve - (B + Nb)$ = 4 - (3 + 1) = 0
Answer: 0 formal charge, i.e. $\ce{C}$ in $\ce{\small{H}-\overset{\LARGE{\cdot}}{\underset{\underset{\LARGE{H}} |}{C}}^{0}-H}$.
Example $3$
What is the formal charge on $\ce{C}$ in $\ce{\small{H}-{\underset{\underset{\LARGE{H}} |}{C}}\!-H}$.
Solution
Isolated $\ce{C}$ has four valance electrons, there are three bonds and no nonbonding electron on $\ce{C}$ in the molecule, so: $Ve$ = 4, $B$ = 3, $Nb$ = 0
$Fc = Ve - (B + Nb)$ = 4 - (3 + 1) = +1
Answer: +1 formal charge, i.e. ($\ce{\small{H}-\overset{+1}{\underset{\underset{\LARGE{H}} |}{C}}-H}$).
Example $4$
What is the formal charge on $\ce{C}$ in $\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{C}}\!-H}$.
Solution
Isolated $\ce{C}$ has four valance electrons, and there are three bonds and two nonbonding electrons on $\ce{C}$ in the molecule, so: Ve = 4, B = 3, Nb = 2
$Fc = Ve - (B + Nb)$ = 4 - (3 + 2) = -1
Answer: 0 formal charge, i.e. $\ce{\small{H}-\overset{ \LARGE {\cdot\cdot}}{\underset{\underset{ \LARGE {H}} |}{C}}^{-1}\!\!\!-H}$.
• A radical is a $\ce{C}$ with three bonds and one unshared electron. $\ce{C}$ in $\ce{\small{H}-\overset{\LARGE{\cdot}}{\underset{\underset{\LARGE{H}} |}{C}}\!-H}$ is an example of a radical that is sp2 hybridized with three bonds and a p-orbital is partially filled by a single electron, as illustrated in Figure $1$a. A radical tends to make a bond by accepting an electron in its partially filled p-orbital.
• A carbocation is a $\ce{C}$ with three bonds and no unshared electrons. A $\ce{\small{H}-\overset{+1}{\underset{\underset{\LARGE{H}} |}{C}}-H}$ is an example of carbonation that is sp2 hybridized with three bonds and a vacant p-orbital, as illustrated in Figure $1$b. A carbocation tends to make a bond by accepting a lone pair in its vacant p-orbital.
• A carbanion is a $\ce{C}$ with three bonds and two unshared electrons. $\ce{\small{H}-\overset{ \LARGE {\cdot\cdot}}{\underset{\underset{ \LARGE {H}} |}{C}}^{-1}\!\!\!-H}$ is an example of a carbanion that is sp3 hybridized with three bonds and the fourth sp3 orbital occupied by a lone pair, as illustrated in Figure $1$c. A carbanion tends to make a bond by donating the lone pair.
Species containing $\ce{N}$ and $\ce{O}$ with formal charges
$\ce{N}$ has three bonds and one lone pair, e.g., in ammonia $\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{N}}\!-H}$ and $\ce{O}$ has two bonds and two lone pairs, e.g., water $\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{ \LARGE {H}} |}{O}}\large{:}}$ molecule. One more bond results in a cation, and one fewer bond results in anion species, as shown by solving the formal charges in the following examples.
Example $5$
What is the formal charge on a) $\ce{N}$ in $\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{N}} -H }$ and b) on $\ce{O}$ in $\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{O}} \large{:} }$
Solution
a) Isolated $\ce{N}$ has five valance electrons, there are three bonds and two nonbonding electrons (a lone pair) in the molecule, so: $Ve$ = 5, $B$ = 3, $Nb$ = 2
$Fc = Ve - (B + Nb)$ = 5 - (3 + 2) = 0
Answer: 0 formal charge, i.e., $\ce{N}$ in $\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{N}}^{0}\!\!\!-H}$
a) Isolated $\ce{O}$ has six valance electrons, there are two bonds and four nonbonding electrons (two lone pairs) in the molecule, so: $Ve$ = 6, $B$ = 2, $Nb$ = 4
$Fc = Ve - (B + Nb)$ = 6 - (2 + 4) = 0
Answer: 0 formal charge, i.e., $\ce{N}$ in $\ce{O}$ in $\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{O}}^{0}\!\! \large{:} }$
Example $6$
What is the formal charge on a) $\ce{N}$ in ($\ce{ \small {H}-\overset{\overset{\LARGE{H}}|}{\underset{\underset{\LARGE{H}} |}{N}}\!-H}$) and b) on $\ce{O}$ in $\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{O}}-H}$
Solution
a) Isolated $\ce{N}$ has five valance electrons, there are four bonds and no nonbonding electrons in the molecule, so: $Ve$ = 5, $B$ = 4, $Nb$ = 0
$Fc = Ve - (B + Nb)$ = 5 - (4 + 0) = +1
Answer: +1 formal charge, i.e., $\ce{N}$ in $\ce{N}$ in ($\ce{ \small {H}-\overset{\overset{ \LARGE {H}}|}{\underset{\underset{\LARGE{H}} |}{N}}{^{+1}}\!\!\!\!-H}$)
a) Isolated $\ce{O}$ has six valance electrons, there are three bonds and two nonbonding electrons (one lone pair) in the molecule, so: $Ve$ = 6, $B$ = 3, $Nb$ = 2
$Fc = Ve - (B + Nb)$ = 6 - (3 + 2) = +1
Answer: +1 formal charge, i.e., $\ce{O}$ in $\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{O}}^{+1}\!\!\!\!-H}$
Example $7$
What is the formal charge on a) $\ce{N}$ in $\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{N}}\large{:}}$ and b) on $\ce{O}$ in $\ce{\small{H}-\overset{\LARGE{\cdot\cdot}} {\underset{\LARGE{\cdot\cdot}}{O}} \large{:} }$
Solution
a) Isolated $\ce{N}$ has five valance electrons, there are two bonds and four nonbonding electrons (two lone pairs) in the molecule, so: $Ve$ = 5, $B$ = 2, $Nb$ = 4
$Fc = Ve - (B + Nb)$ = 5 - (2 + 4) = -1
Answer: -1 formal charge, i.e. $\ce{N}$ in $\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{N}}^{-1} \!\!\!\!\!\large{:} }$
a) Isolated $\ce{O}$ has six valance electrons, there is one bond and six nonbonding electrons (three lone pair) in the molecule, so: $Ve$ = 6, $B$ = 1, $Nb$ = 6
$Fc = Ve - (B + Nb)$ = 6 - (1 + 6) = -1
Answer: +1 formal charge, i.e., $\ce{O}$ in $\ce{\small{H}-\overset{\LARGE{\cdot\cdot}} {\underset{\LARGE{\cdot\cdot}}{O}} \large{:} ^{-1}}$ | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/01%3A_Bonding_in_organic_compounds/1.05%3A_Formal_Charge.txt |
Learning Objectives
• Understand the concept of resonance and how it stabilizes molecules.
• Draw resonance contributors and predict equal, major, and minor resonance contributors.
• Predict resonance stabilization and factors that affect it.
• Predict the relative strength of a $\pi$-bond based on orientation and sizes of p-orbitals.
Prelude to resonance
$\ce{C}$ not only makes straight chains and branched chains but makes cyclic chains, as illustrated in Figure $1$. Benzene is a member of a class of hydrocarbons called aromatic hydrocarbons. A set of characteristics of aromatic compounds will be described later. Here only the nature of bonds in benzene is described to introduce the concept of resonance.
More than one correct Lewis structure can be drawn for some compounds or polyatomic ions.
If more than one Lewis structure can be drawn for a compound, these are called contributing structures or resonance contributors.
The contributing structures of a compound are separated from each other by double-headed arrows, as shown in Figure $2$, for the case of benzene molecule and carbonate ($\ce{CO3^{2-}}$) polyatomic anion.
Are the compounds or polyatomic ions with more than one contributing structure a mixture of these structures? The answer is no; none of the contributing structures exist. For example, both the contributing structures of benzene imply that three are three $\ce{C-C}$ single bonds and three $\ce{C=C}$ double bonds. Experimental results show that all six $\ce{C-C}$ bonds in benzene are equal. Similarly, all three contributing structures of $\ce{CO3^{2-}}$ imply one double bond and two single bonds. Experimental results show that all three $\ce{C-O}$ bonds in $\ce{CO3^{2-}}$ are equal. The concept of resonance was introduced to deduce the actual structure in the cases of compounds having two or more contributing structures.
What is resonance?
• Resonance is a way of describing bonding in molecules and polyatomic ions by a combination of contributing structures into the actual structure.
• The hybrid of the contributing structures -the actual structure, is called a resonance hybrid.
Consider the structure of a benzene molecule: every $\ce{C}$ in benzene is sp2 hybridized with three $\sigma$-bonds in a trigonal planar geometry and a p-orbital perpendicular to the plane, as illustrated in Figure $3$a. All $\ce{C's}$ and $\ce{H's}$ are in one plane, and all the p-orbitals are perpendicular to them and parallel to each other. Assuming p-orbitals in benzene simultaneously make a half-$\pi$-bond with two adjacent p-orbitals, it can be concluded that all of the $\ce{C}$-to-$\ce{C}$ bonds are equal, i.e., each one is composed of one $\sigma$ and a half $\pi$-bond, as illustrated in Figure $3$b. The same structure is assumed by a fifty%-fifty% hybrid of its two contributing structures. The bonding in the resonance hybrid of benzene agrees with the experimental results, i.e., all six $\ce{C-C}$ bonds in benzene are equal. Similarly, the resonance hybrid of $\ce{CO3^{2-}}$, arrived at by assuming it is a hybrid of its three equal contributing structures, is illustrated in Figure $3$c. The bonding in the resonance hybrid of $\ce{CO3^{2-}}$ agrees with the experimental results, i.e., all three $\ce{C-O}$ bonds in $\ce{CO3^{2-}}$ are equal.
Note that the $\pi$ bond is not localized between two adjacent atoms in resonance hybrids; it is spread over three or more adjacent atoms.
The $\pi$-bond spread over three or more adjacent atoms is called a delocalized $\pi$-bond and the elections in the delocalied $\pi$-bond are called delocalized electrons.
For example, the delocalized $\pi$-bonds are represented by a dashed-lined circle inside the benzene ring or by dashed lines in the case of $\ce{CO3^{2-}}$ resonance hybrids.
How is an sp3 atom bearing a lone pair involved in resonance?
An sp3-hybridized atom with lone pair changes its hybridization from sp3 to sp2 and places its lone pair in the p-orbital for resonance to happen when it is adjacent to an sp2- or sp-hybridized atom.
Delocalization of $\pi$- or nonbonding electrons results in more bonding, i.e., delocalized electrons are spread over more than two nuclei. It results in lowering the potential energy, called resonance stabilization. The resonance hybrid is always more stable than the predicted stability of any of its contributing structures.
Summary of facts about resonance
1. When more than one correct Lewis structure can be drawn for a compound, these are called resonance contributing structures.
2. The actual structure of the compound that is deduced by assuming it is a hybrid of the contributing structures is called the resonance hybrid.
3. The extra stability of the resonance hybrid due to the resonance is called resonance stabilization.
4. The contributing structures do not exist; they are just the correct Lewis structures that help deduce the resonance hybrid. Consequently, the double-headed arrow does not show a chemical reaction; it is used only to separate the contributing structures of a compound.
How to draw a resonance contributor?
If one Lewis structure is already drawn, its resonance contributor can be drawn by the following rules, as illustrated with examples:
• Move $\pi$-electrons or lone pair only (do not move $\sigma$-electrons) to create a new $\pi$-bond or lowne pair.
• Curved arrows show the movement of electrons, the arrow starts where the electrons are located, and the arrowhead points where they are going to create a new $\pi$ bond or a new nonbinding electron (do not make a new $\sigma$-bond).
• The electron movement often creates formal charges. Calculate the formal charge and show it if it is not zero. For example, $\pi$-electrons are moved to create a new lone pair and formal charges in the following example.
• For resonance to happen, there must be at least two adjacent sp2 or sp atoms. Resonance can not happen involving sp3 atom.
• An exception is an sp3 atom carrying a lone pair; it changes its hybridization to sp2 and places its lone pair in the p-orbital for resonance, as shown in the following example.
• Atoms do not move, i.e., a $\sigma$-bond is not broken or created during regular resonance. The first example below is not resonance because a $\ce{C-H}$ $\sigma$-bond from a $\ce{-CH2{-}}$,is broken and a $\ce{O-H}$ $\sigma$-bond is formed. It is a chemical reaction, not a resonance. The second example is resonance, as only a lone pair and a $\pi$-bond pair are moved.
• A $\ce{H}$ can not have more than two valence electrons (duet rule), and second-row elements, like $\ce{C}$, $\ce{N}$, $\ce{O}$, $\ce{F}$, etc. can not have more than eight valence electrons (octet rule) in any of the resonance contributors. (Third-row and higher-row elements are exceptions and may have up to twelve valence electrons in molecules). For example, the movement of a lone pair to create a $\pi$-bond in the following example is not allowed as it makes a $\ce{C}$ with ten valence electrons (five bonds). Simultaneous movement of the $\ce{C=O}$ $\pi$-bond avoids this problem, as in the previous example.
The relative importance of the contributing structures
• There is more resonance stabilization for equal resonance contributors than for unequal resonance contributors. The first example below has more resonance stabilization due to equal contributors than the second one with unequal contributors.
• More the number of equal contributors means more resonance stabilization. For example, resonance stabilization is more for $\ce{CO3^{2-}}$ with three equal contributors than for $\ce{CH3COO^{-}}$ with two equal contributors.
In the cases of unequal contributors, the following factors determine the major contributor:
• The negative charge on a more electronegative atom or positive charge on a less electronegative atom is more stable than otherwise. The contributor on the right in the following example with a negative charge on more electronegative $\ce{O}$ is a major contributor, and the other is minor.
• A contributor with an octet incomplete on an atom is significantly less stable than the contributor with all atoms octet complete.
• A contributor with more bonds is significantly more stable than one with fewer bonds. In the following example, the contributor on the left with octet incomplete on one $\ce{C}$ and fewer bonds is a negligible contributor relative to the other.
• A contributor with more formal charges is significantly less stable than a contributor without formal charges. The contributor on the left, with two formal charges and fewer bonds, is a negligible contributor compared to the other.
Strength of the $\pi$-bond and resonance in relation to size and orientation of the p-orbitals
Both the $\pi$-bond and, consequently, the resonance are strong when the two p-orbitals are of the same size and parallel. The strength of the $\pi$-bond and the resonances become weaker when
• the two p-orbitals are out of parallel orientation and become zero when the two p-orbitals are perpendicular to each other, and
• when the size of the two p-orbitals differ, the more significant the difference, the weaker the overlap.
These two effects are illustrated in the Figure below.
The two p-orbitals are parallel, resulting in the maximum $\pi$-bond strength and the maximum resonance
The two p-orbitals are perpendicular, with no $\pi$-bond and no resonance
The two p-orbitals are parallel but of different sizes resulting in a weaker $\pi$-bond and weaker resonance.
• In addition to the two effects described above, the third factor is electronegativity, i.e., the less electronegative the atom donating a lone pair of electrons for the resonance, the stronger the resonance.
The following examples support the three effects described above.
The $\ce{N}$ in pyridine can donate its lone pair of electrons to a proton, i.e., it is a base ($\ce{pKb}$ 8.8) because the lone pair is in an sp2-orbital which is perpendicular to the other p-orbitals and not occupied in resonance. The $\ce{N}$ in pyrrole is not able to donate its lone pair of electrons to a proton, i.e., it is a very weak base ($\ce{pKb}$ 17.8) because the lone pair is in a p-orbital parallel to other p-orbitals and occupied in resonance and not that much available to protons, as shown in Figure $4$.
Three examples of resonance are shown below, where the second contributor in each case is minor because there is charge separation, and +ve charge is placed on an electronegative atom. However, there are differences in these three cases, i.e., i) the second contributor is minor in the 1st example where 2p-orbital of $\ce{O}$ is in resonance with 2p-orbital of $\ce{C}$ which are similar in sizes, but negligible in the second example because 3p-orbital of $\ce{Cl}$ is in resonance with 2p-orbital of $\ce{C}$ which are different in sizes, ii) the second contributor is minor in the 1st example but significant in the 3rd example because the p-orbitals are similar in size but $\ce{N}$ is less electronegative and more willing to donate its lone pair than the $\ce{O}$ in the 1st example. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/01%3A_Bonding_in_organic_compounds/1.06%3A_Resonance.txt |
• 2.1: Hydrocarbons
Organic compounds containing only carbons and hydrogens, i.e., hydrocarbons, their subclasses, some physical properties, IUPAC nomenclature of hydrocarbons, and reading their electrostatic potential maps to identify reactive sites in them are also described.
• 2.2: Functional group
Two modes of bond breaking and making in organic reactions and how to show electron movement are described. A functional group is defined and major classes of functional groups are listed.
• 2.3: Functional groups containing sp3-hybridized heteroatom
Functional groups containing sp3-hybridized heteroatoms, including haloalkanes, alcohol, ethers, amines, etc., describe their electrostatic potential maps and physical properties. These functional groups' primary, secondary, tertiary, and quaternary designations are explained.
• 2.4: Functional groups containing sp2-hybridized heteroatom
Aldehydes, ketones, and imines are described. Their bond polarities and how polarities affect their reactivities and physical properties, like solubility in water, boiling point, etc., are described.
• 2.5: Functional groups containing mix of sp3- and sp2-, or sp-hybridized heteroatom
IUPACE nomenclature of and nature of bonding and its effect on the reactivity of carboxylic acids and their derivates, including acid halides, acid anhydrides, esters, amides, and nitriles, is described. Anhydrides and esters of phosphoric acid are introduced.
02: Nomenclature and physical properties of organic compounds
Learning Objectives
• Identify hydrocarbons and their subclasses: alkane, alkene, alkyne, and aromatic.
• Write IUPAC name for a skeletal structure and draw the skeletal structure from the IUPAC name for simple hydrocarbons.
• Understand some physical properties like melting points, boiling points, and solubilities of hydrocarbons.
• Predict the reactive sites, i.e., $\delta$- (nucleophilic) and $\delta$+ (electrophilic) regions in the structures of simple hydrocarbons.
• Identify primary, secondary, tertiary, or quaternary $\ce{C's}$ and $\ce{H's}$.
Classification of hydrocarbons
What are hydrocarbons
Organic compounds composed of $\ce{C's}$ and $\ce{H's}$ only are called hydrocarbons.
The hydrocarbons are divided into two major classes based on the absence or presence of an aromatic ring.
Aromatic ring
An aromatic ring is a cyclic chain of only sp2-hybridized atoms in planer geometry having $\pi$-electrons in the ring equal to 4n+2, where n is a positive integer, i.e., 0, 1, 2, 3, ...,. Benzene is an example of an aromatic ring that is a cycle of six sp2-hybridized $\ce{C's}$ having six $\pi$-electrons, which is a number equal to 4n+2 for n = 1.
Benzene
• Hydrocarbons with at least one aromatic ring in their structure are called aromatic hydrocarbons.
• Hydrocarbons that do not have an aromatic ring in their structure are called aliphatic hydrocarbons. The aliphatic hydrocarbons are subdivided into i) alkanes, ii) alkenes, and iii) alkynes. Figure $1$ illustrates the classification of hydrocarbons.
• Alkanes have all $\ce{C-C}$ single bonds, i.e., all the $\ce{C's}$ are sp3-hybridized, e.g., propane ( $\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$).
• Alkenes have at least one $\ce{C=C}$ double bond, e.g., propene ( $\ce{\small{H}-{\underset{\underset{\Large{H}} |}{C}}={\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$).
• Alkynes have at least one $\ce{C≡C}$ triple bond, e.g., propyne ( $\ce{\small{H}-{C}≡{C}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}$).
Alkanes
Nomenclature of alkanes
Organic compounds have trivial names as well as systematic names. Systematic names are based on the International Union of Pure and Applied Chemistry (IUPAC) recommendations, referred to as IUPAC nomenclature. Table 1 lists stem names that represent the number of $\ce{C's}$ in a continuous chain of $\ce{C's}$ (straight chain) or a cyclic chain part of an organic compound.
Table 1: Stem names representing from one to hundred carbons
# of $\ce{C’s}$ Stem name # of $\ce{C’s}$ Stem name # of $\ce{C’s}$ Stem name
1 Meth 11 Undec 30 Triacont
2 Eth 12 Dodec 40 Tetracont
3 Prop 13 Tridec 50 Pentacont
4 But 14 Tetradec 60 Hexacont
5 Pent 15 Pentadec 70 Heptacont
6 Hex 16 Hexadec 80 Octacont
7 Hept 17 Heptadec 90 Nonacont
8 Oct 18 Octadec 100 hect
9 Non 19 Nonadec
10 Dec 20 Icos
Straight chain alkanes
To name a straight chain alkane, the first syllable, i.e., alk- is replaced with the stem name that tells the number of $\ce{C's}$ in the chain, and the last syllable, i.e., -ane is retained as a suffix in which 'an' tells all bonds are single bonds, and 'e' tells it is a hydrocarbon, i.e., -ane means an alkane.
For example, $\ce{CH4}$ is called methane, $\ce{CH3-CH3}$ is called ethane, and $\ce{CH3-CH2-CH3}$ is called propane, where meth-, eth-, and prop- are the stem names representing one, two, and three $\ce{C's}$. Table 2 shows the formulas, IUPAC names of some straight-chain alkanes, and some of their physical properties. The physical properties will be discussed in a later section.
Table 2: Formulas, IUPAC names, melting points, boiling points, densities, and physical states of some straight-chain alkanes.
Formula IUPAC Name Melting point (oC) Boiling point (oC) Density (g/ml) Physical state at 20 oC
$\ce{CH4}$ Methane −182 −162 0.000656 Gas
$\ce{C2H6}$ Ethane −183 −89 0.00126 Gas
$\ce{C3H8}$ Propane −188 −42 0.00201 Gas
$\ce{C4H10}$ Butane −138 0 0.00248 Gas
$\ce{C5H12}$ Pentane −130 36 0.626 Liquid
$\ce{C6H14}$ Hexane −95 69 0.659 Liquid
$\ce{C7H16}$ Heptane −91 98 0.684 Liquid
$\ce{C8H18}$ Octane −57 126 0.703 Liquid
$\ce{C9H20}$ Nonane −54 151 0.718 Liquid
$\ce{C10H22}$ Decane −30 174 0.730 Liquid
$\ce{C11H24}$ Undecane 196 0.740 Liquid
$\ce{C12H26}$ Dodecane −10 216 0.749 Liquid
$\ce{C13H28}$ Tridecane −5.4 235 0.756 Liquid
$\ce{C14H30}$ Tetradecane 5.9 253 0.763 Liquid
$\ce{C15H32}$ Pentadecane 10 270 0.769 Liquid
$\ce{C16H34}$ Hexadecane 18 287 0.773 Liquid
$\ce{C17H36}$ Heptadecane 22 303 0.777 Liquid
$\ce{C18H38}$ Octadecane 28 317 0.781 Solid
$\ce{C19H40}$ Nonadecane 32 330 0.785 Solid
$\ce{C20H42}$ Icosane 37 343 0.789 Solid
Continuous $\ce{C}$ chain can be straight or wiggle waggle
A regular zigzag line, also called a straight chain, usually represents the hydrocarbons, but it may be a wiggle waggle because of rotation allowed around each $\ce{C-C}$ single bond. Figure $2$ shows the semi-condensed formulas of decane presented in five different configurations and the corresponding skeletal formulas. The terminal $\ce{C}$ in a chain is the carbon at the ends connected with only one other $\ce{C}$. The internal carbon in a chain is the $\ce{C}$ connected with at least two other $\ce{C's}$.
• A continuous $\ce{C}$ chain starts from a terminal carbon and reaches the other terminal $\ce{C}$ through the connected continuous $\ce{C-C}$ single bonds.
• A straight-chain hydrocarbon has all the internal $\ce{C's}$ connected with only two other $\ce{C's}$, no matter how much wiggle waggle the chain may be.
Branched alkanes
If at least one $\ce{C}$ in an alkane is connected with three or four other $\ce{C's}$, it is a branched alkane. The parent chain is the longest continues $\ce{C}$ chain. The smaller chains branching from the parent chain are called branches, substituents, or alkyl groups. R- represents a general alkyl group. Rules for naming the branched alkanes are the following.
• Find the most extended continuous $\ce{C}$ chain, called the parent chain. Name it by using the stem name that tells the number of $\ce{C's}$ in it with the suffix -ane that means it is an alkane.
Not
• Name the branch chain by using the stem name that tells the number of $\ce{C's}$ in the continues $\ce{C}$ chain of the branch with the suffix -yl that tells it is a branch. Use the name of the branch as a prefix to the name of the parent chain.
• If there are two or more longest chains of equal length, choose the parent chain with the highest number of branches on it.
not
• If there is more than one branch, arrange their names alphabetically as prefixes to the name of the parent chain.
• If there is more than one branch with the same name, list it once with the prefix di-, tri-, tetra-, etc., added to the branch name to represent two, three, four, etc. of the same branches. The alphabetization of the branch names is based on the branch name without the prefixes di-, tri-, tetra-, etc. For example, dimethyl, trimethyl, and tetramethyl are all alphabetized based on the 'm' of methyl, not based on the 'd' or 't' first alphabets of the prefixes.
not
• Number the $\ce{C's}$ of the parent chain starting from the end that gives the lowest number to whichever branch appears first on the parent chain. not
• If numbering from left or right has the same number for the first branch, start the numbering from the side that gives a lower number to the branch that comes alphabetically first.
not
• List the number to which the branch is attached to the parent chain before the branch name, separated by a hyphen. This is the IUPAC name of the alkane. ,, and
• If two or more branches have the same name, list their numbers in increasing order, separated by commas, and the final number separated by a hyphen from the branch name.
• If two branches are on the same $\ce{C}$, use the same number twice, i.e., once for each branch.
• If a branch is further branched, name it as if it is a parent chain following the above rules, except that i) the suffix of the brach is -yl instead of -ane, ii) insert a number from which the beach is attached to the parent chain before -yl, iii) Enclose the branch name in small brackets and place it as a prefix to the name of the parent chain, iv) insert a number before the bracket to which the beach is attached on the parent change, as shown in the following example.
Primary, secondary, and tertiary $\ce{C's}$ and $\ce{H's}$
• The terminal $\ce{C's}$, i.e., the $\ce{C's}$ with only one single bond with another $\ce{C}$ are primary carbon.
• The $\ce{H's}$ bonded to primary $\ce{C's}$ are primary hydrogen.
• The $\ce{C's}$ with two single bonds with other $\ce{C's}$ are secondary carbon.
• The $\ce{H's}$ bonded to secondary $\ce{C's}$ are secondary hydrogen.
• The $\ce{C's}$ with three single bonds with other $\ce{C's}$ are tertiary carbon.
• The $\ce{H's}$ bonded to tertiary $\ce{C's}$ are tertiary hydrogen.
• The $\ce{C's}$ with four single bonds with other $\ce{C's}$ are quaternary carbon.
The quaternary $\ce{C's}$ do not have any $\ce{H}$ bonded to them. The following figure labels the primary, secondary, and tertiary carbons and hydrogen with different colors.
Cycloalkanes
An alkane with one ring of $\ce{C's}$ in its structure is called cycloalkane. IUPAC rules for naming cycloalkanes are the following.
• If there is one ring of $\ce{C's}$ without any substituent on it, it is named by using the stem name that tells the number of $\ce{C's}$ in the ring with prefix cyclo- and suffix -ane, as shown in the following examples.
• If there is only one branch or a substituent on the ring of $\ce{C's}$, then the ring supplies the parent name, and the branch name is added as a prefix, as for alkanes but without any number preceding the branch name. Some examples of IUPAC names of branched cycloalkanes are shown below.
, , and
• If there are two substitutes, they are listed alphabetically, preceded by location number, as in the case of alkanes. The numbering of $\ce{C's}$ in the chain starts from the point of attachment of the substituent that comes alphabetically first. The numbering goes clockwise or counterclockwise, whichever gives the second substitute the lowest number, as shown in the following examples.
, , and
• If there are more than two substituents, they are listed as prefixes in alphabetic order as for alkanes. The ring $\ce{C's}$ attached to one of the substituents is assigned number 1, and the numbering is continued clockwise or counterclockwise so that the addition of numbers assigned to the substituents is the lowest possible. Some examples are shown below.
, , and
Sometimes the ring name is treated as a branch with the last syllable changed from -ane to -yl, e.g., cyclopropane becomes cyclopropyl as a branch. This change is applied particularly when the chain attached to the ring has more number of $\ce{C's}$ than the number of $\ce{C's}$ in the ring.
Cis, trans isomerism in cycloalkanes
Cycloalkanes with two alkyl groups on different carbons have two configurations: i) both alkyl groups pointing in the same direction are called 'cis' orientation, and ii) the two bulky groups pointing in the opposite direction are called 'trans' orientation, as shown below for the case of 1,2-dimethyl cyclopropane and 1-ethyl-3-methylcyclohexane examples.
The cis- and trans- configuration can not inter-convert by rotation around $\ce{C-C}$ bond as the rotation is restricted by the carbon bridge in the case of cycloalkanes. Therefore, the cis- and trans-configuration of cycloalkane are isomers of each other, i.e., the same formula but different compounds. For example, cis-1,2-dimethylcyclopropane has a boiling point of 37 C, and its isomer, i.e., trans-1,2-dimethylcylopropane has a boiling point of 28 oC. This phenomenon, i.e., cis-, trans- isomerism, is common in all cyclic compounds.
Bonding and physical properties of alkanes
Bonding
The general formula of straight chain and branched alkanes is $C_{n}H_{(2n+2)}$, and that of cycloalkanes having one ring is $C_{n}H_{(2n)}$ where n is a positive integer. All $\ce{C's}$ in alkanes are sp3 hybridized, i.e., tetrahedral geometry with all bond angles about 109.5o, as illustrated in Figure $3$ for the case of ethane molecule. The bond energy for $\ce{C-C}$ bond is ~3.8 eV and for $\ce{C-H}$ bond is ~4.4 eV.
Polarity
The electronegative difference for the $\ce{C-H}$ bond is 0.4, which falls in the nonpolar bond category. The electrostatic potential map of a molecule shows $\delta$+ region in blue, $\delta$- region in red, and neutral or nonpolar region in green color. Alkanes are nonpolar, as shown in green in the electrostatic map of ethane in Figure $3$. The strong $\ce{C-C}$ and $\ce{C-H}$ bonds make the alkanes stable and less reactive. The nonpolar nature of alkanes makes them more inert chemically.
Intermolecular forces and boiling points
Alkanes are nonpolar molecules where the only intermolecular interaction is London dispersion forces. Straight-chain alkanes are a homologous series.
As the molar mass, and, consequently, the surface area and the number of electrons in the electron cloud, gradually increase, the London dispersion forces increase, resulting in a gradual increase in melting and boiling points, as shown in Table 2.
The first four members for the alkane series, i.e., $\ce{CH4}$ to $\ce{C4H10}$ are gases, the next from $\ce{C5H12}$ to $\ce{C17H36}$ are liquid and the larger alkanes are wax or soft solids.
Natural gas is the primary source of methane and ethane. Petroleum, which in the crude oil form is a viscous liquid composed of thousands of organic compounds, primarily hydrocarbons, is the primary source of alkanes and also the main source of organic raw materials. The differences in the boiling points allow the segregation of crude oil into different fractions, as illustrated in Figure $4$.
Solubility
Alkanes are not soluble in water because alkanes are nonpolar, and water is a polar solvent. Alkanes are soluble in each other or other nonpolar solvents. Alkanes float on water because they are less dense than water. The density of liquid alkanes varies in the range of 0.6 g/ml to 0.8 g/ml, which is less than the density of water, which is ~1 g/ml at room temperature. Petroleum and crude oil are mainly composed of alkanes, which is why oil spills make a layer of oil float on the water's surface.
Changes in physical properties with branching
The physical properties of branched alkanes are almost the same as that of straight chain alkanes, except that:
• the polling point increases as the molar mass increases, and
• for the same molar mass, the boiling point decreases as the branching increases, as shown in Table 1.
Table 1: Examples of hydrocarbons with increasing molar mass and increasing branches and their corresponding boiling points
Number of $\ce{C's}$ Name Condensed formula Boiling point
5 Pentane $\ce{CH3CH2CH2CH2CH3}$ 36oC
6 Hexane $\ce{CH3CH2CH2CH2CH2CH3}$ 69oC
6 2-Methylpentane $\ce{(CH3)2CHCH2CH2CH3}$ 60oC
6 2,2-Dimethylbutane $\ce{(CH3)3CCH2CH3}$ 50oC
7 Heptane $\ce{CH3CH2CH2CH2CH2CH2CH3}$ 98oC
Straight-chain alkanes are long and cylindrical, as illustrated in Figure $5$. Branching makes them spherical, resulting in less contact area between molecules, fewer London dispersion forces, and lower boiling points. Note that the effect of molar mass is dominant over that of branching.
Properties of cycloalkanes
The physical properties of cycloalkanes are almost similar to alkanes, except that bonds have to bend to acquire ring-shape that causes angle strain. Further, the molecule can not acquire the most stable configuration due to no or limited rotation around $\ce{C-C}$ single bond. Therefore, the cycloalkanes, mainly the smaller three or four-member rings, are relatively unstable. Except for cyclopropane, there is a limited rotation possible around $\ce{C-C}$ bond that allows the molecule to acquire bent-shape, rather than a planer shape that releases some of the strains, as shown by models in Figure $6$. The strain decreases, and the stability increases in this order: cyclopropane > cyclobutane > cyclopentane > cyclohexane < cycloheptane > cyclooctane. Notably, the five and six-membered rings, e.g., cyclopentane and cyclohexane, have little or no strain, so they are the most common cyclic compounds found in nature.
Some uses of alkanes
The first four members, i.e., methane, ethane, propane, and butane, are gases used as fuels. The next four are components of gasoline, and higher than that are components of kerosene and diesel, used as fuels.
Alkanes and other hydrocarbons are part of almost all organic compounds but are not so common in pure form in living things. Some examples of uses in biological systems are the following. Long-alkanes having eighteen or more $\ce{C,s}$ are semisolids found in waxes as in vaseline used in ointments and cosmetics. The waxes are found on surfaces of leaves and fruits to protect against water loss, prevent the leaching of essential minerals by rain, and protect against bacteria, fungi, and harmful insects. Alkanes are also found in some pheromones, e.g., sand bees use tricosane ($\ce{C23H48}$), pentacosane ($\ce{C25H53}$), and heptacosane ($\ce{C27H56}$) to identify a mate.
Alkenes
Nomenclature of alkenes
The procedure for IUPAC naming alkenes is the same as alkanes, except for the following changes.
• The suffix to the stem name of the parent chain is -ene, where 'en' tells its an alkene, and 'e' means it is a hydrocarbon, e.g., $\ce{CH2=CH2}$ is ethene.
• A location number of the first $\ce{C}$ of $\ce{C=C}$ bond is inserted between the stem- name and -ene suffix, separated by hyphens, as in the following example, except for propene that does not need to show the location number.
3-ethylpent-2-ene
• The parent chain numbering starts from the end that gives the lowest number to the first $\ce{C}$ of $\ce{C=C}$ bond.
• If the first $\ce{C}$ of $\ce{C=C}$ bond receives the same number from either side, then the rule for alkane applies, i.e., start numbering from the side that gives the lower number to the branch that appears first, as shown in the following examples.
• If two or more double bonds exist, the suffix -ene is changed to -diene, triene-, etc. The rest of the nomenclature remains the same.
• In cycloalkenes, the $\ce{C's}$ of $\ce{C=C}$ are #1 and #2 and do not need to show the location number.
• If there is a branch on cycloalkene, the $\ce{C'}$ of $\ce{C=C}$ bond are #1 and #2, and the numbering is continued clockwise or counterclockwise to give the lowest number to the branch, as shown in the following examples.
• there are two alkyl groups, one on each $\ce{C}$ of $\ce{C=C}$ bond (or third or fourth group different from the other group on the same $\ce{C}$) then a descriptor E indicating cis, or Z indicating the trans orientation of the bulkier groups, preceded by the location number of the first $\ce{C}$ of the $\ce{C=C}$ bond, is added as a prefix enclosed within brackets.
• if the double bond is in a branch, the -ene suffix is replaced with -enyl, where 'en' tells it is an alkene and 'yl' means it is a branch.
Caution
In the older system:
• The parent chain may not be the longest, but the parent chain is the longest chain counting the double bond, and
• The location number of the double bond is inserted before the stem name, e.g., but-2-ene ($\ce{CH3CH=CHCH3}$) is written as 2-butene in the old IUPAC nomenclature.
Some common names of alkenes
Some common names are still being used in the chemical literature and industries. For example, ethylene, propylene, isobutylene, and isoprene are often used in place of IUPAC names ethene, propene, 2-methylpropene, and 2-methylbuta-1,3-diene.
Structure and IUPAC name
ethene
prop-1-ene
2-methylprop-1-ene
2-methylbuta-1,3-diene
Common name ethylene propylene isobutylene isoprene
Bonding and physical properties of alkenes
Alkenes are nonpolar compounds with only London dispersion forces as intermolecular forces, like alkanes. Therefore, the physical properties of alkenes are similar to alkanes with the same $\ce{C}$ skeleton, i.e., they are insoluble in water, soluble in nonpolar solvents, and have densities lower than that of water. The significant difference between alkanes and alkenes is the $\ce{C=C}$ bond in alkenes described below.
bonding
The general formula of alkenes with one double bond is $C_{n}H_{(2n)}$, where n is a positive integer. The $\ce{C's}$ in $\ce{C=C}$ bond are sp2 hybridized with trigonal planer geometry around each of the sp2 $\ce{C's}$ with bond angles around 120o, as illustrated in Figure $7$. Except for $\ce{C=C}$, all other $\ce{C's}$ in alkenes are sp3 hybridized, i.e., tetrahedral geometry with all bond angles about 109.5o. Although alkenes are nonpolar overall, the Electrostatic potential map in Figure $7$ shows a red region where $\pi$-electrons are located above and below the $\sigma$ bond and slightly blush region where $\ce{H's}$ of sp2 $\ce{C's}$ are located. These $\delta {-}$ and $\delta {+}$ regions impart reactivity to alkenes that will be described in later sections.
Degree of unsaturation or index of hydrogen deficiency (IHD)
The general formula of an alkane is $C_{n}H_{(2n+2)}$, where n is a positive integer. One degree of unsaturation is two less $\ce{H's}$ than in an alkane of the same carbon structure. For example, one ring in cycloalkane or one double bond in alkenes reduces two $\ce{H's}$, one unsaturation degree. The degree of unsaturation is also known as the index of hydrogen deficiency (IHD). Remember, if a halogen with one bond is added to an alkane, it replaces one $\ce{H}$, an $\ce{O}$ with two bonds does not change the number of $\ce{H's}$, and a $\ce{N}$ with three bonds add one $\ce{H}$. Based on this information, the degree of unsaturation, i.e., number of double bonds + rings, can be calculated by the following formula.
$IHD = (C's + 1) - {\frac{H's-N's+X's}{2}}\nonumber$
, where $\ce{IHD}$ is an index of hydrogen deficiency, which is equivalent to the number of $\pi$ bonds + rings, $\ce{C's}$ is the number of $\ce{C}$ atoms, $\ce{H's}$ is the number of $\ce{H}$ atoms, $\ce{N's}$ is the number of $\ce{N}$ atoms, and $\ce{X's}$ is number of halogen atoms in the molecular formula.
If oxygen is in the molecule, it is not included in the IHD calculation.
Example $1$
What is the degree of saturation in cyclohexane ($\ce{C6H10}$?
Solution
$\ce{C's}$ = 6, $\ce{H's}$ = 10, $\ce{N's}$ = 0, $\ce{X's}$ = 0.
$IHD = (C's + 1) - {\frac{H's-N's+X's}{2}}$ = $(6 + 1) - {\frac{10-0+0}{2}}$ = 2, which is one $\pi$-bond and one ring as shown below.
Substituents attached to sp2 hybridized $\ce{C's}$ of alkenes are also called vinylic. The bond energy for $\ce{C=C}$ bond is ~7.4 eV which is less than two times of ~3.8 eV of $\ce{C-C}$ bond, and the vinylic $\ce{=C-H}$ bond is ~4.8 eV, which is stronger than alkane $\ce{-C-H}$~4.4 eV. Comparison ~7.4 eV for $\ce{C=C}$ bond vs 3.8 eV for $\ce{C-C}$ bond shows that, although a double bond (($\sigma$ + $\pi$) is stronger than a single bond, a $\pi$ bond is weaker than a $\sigma$ bond.
Cis, trans isomerism in alkenes
If there are two alkyl groups, one on each $\ce{C}$ of a double bond, they can be pointing in the same direction, called cis orientation, or in the opposite direction, called trans orientation, as shown in the Figure below for the case of but-2-ne. The cis and trans alkenes of the same molecular formula are different compounds related as isomers of each other. This is because three is no rotation around $\ce{C=C}$ bond without breaking and re-making the $\pi$ bond.
Cis trans isomerism plays a vital role in biological systems. For example, fatty acids are significant components of fats and vegetable oils. Fatty acid contains a long alkyl chain attached to a carboxylic acid group. The alkyl chain is usually alkane in the cases of animal fats or includes one or more double bonds, almost always in the cis configuration in the cases of vegetable oils and fish oil. For example, models of steric acid with no double bonds, oleic acid with one cis double bond, and linoleic acid with two cis double bonds are shown below.
The cis-double bonds change the long cylindrical shape of fatty acids to a bent or more spherical shape with less contact area with the neighboring molecules. It results in lower intermolecular forces and melting points than the same chain with a trans or no double bond. That is why vegetable oils with more cis-double bonds are liquids, while animal fats with little to no double bonds are solids at room temperature. Another example is vitamin A having five trans double bonds, as shown in the model drawing below. Vitamin A leads to retinol synthesis, which plays a crucial role in our vision based on switching one of its double bonds between cis and trans configurations.
Model of vitamin A molecule with five trans double bonds.
Polarity and chemical reactivity
Although $\ce{C=C}$ bond is nonpolar, the electrostatic potential map of ethene in Figure $7$ shows there is an electron-rich region (red color in the map) above and below the axis joining the nuclei. This is because the $\sigma$ bond is along the axis of nuclei, and the $\pi$ bond places electrons above and below the $\sigma$ bond. Molecules with negative charge or electron-rich ($\delta$-) regions, as in the case of $\ce{C=C}$ bond in alkene, are nucleophiles. Molecules with positive charge or electron-deficient ($\delta$+) region are electrophiles. Chemical reactions between nucleophiles and electrophiles are facilitated by the attraction between the opposite charges that allow them to come closer to each other for bond-making and bond-breaking to happen.
Electrophile/nucleophile or acid/base?
The term acid is used for electrophile when $\delta$+ part is a proton, and the base is used in the place of nucleophile when $\delta$- region is a lone pair of electrons interacting with a proton.
Alkenes uses and importance in biological systems
Alkenes are among the raw materials for the synthesis of several organic chemicals. Mainly, ethene and propene are the raw material for synthesizing commercially important polymers: polyethylene and polypropylene.
Ethene is a ripening agent for fruits. It allows fruit growers to pick fruits while they are green and less susceptible to bruising and then treat them with ethene gas for ripening when ready for sale. Lycopene, -a red pigment in tomatoes, and carotene, -a yellow pigment in carrots, are polyenes, shown below.
Terpenes -a class of natural products built from an alkene
Terpenes are a class of natural products having a general formula $\ce{(C_{5}H_{8})n}$, where n is a positive integer. Isoprene ($\ce{(C_{5}H_{8})}$, i.e., 2-methylbuta-1,3-diene is the building block of terpene.
isoprene, IUPAC name: 2-methylbuta-1,3-diene
Terpenoids are derivatives of terpenes in which heteroatoms/functional group is added. Natural rubber, essential oils, and steroids are examples of terpenes or terpenoids. They play important roles in living things, e.g., in defense against diseases. Some examples of terpenes and terpenoids are shown below, with the isoprene skeleton highlighted.
Alkynes
Nomenclature
The IUPAC nomenclature of alkynes followes the same rule as alkenes, except for the following changes.
• The suffix to the stem name is -yne where 'yn' tells it is an alkyne, and 'e' means it is a hydrocarbon, e.g., $\ce{CH≡CH}$ is ethyne.
• If a double and a triple bond are in the same molecule, they are treated equally. One exception is when the double receives the same number as the triple bond counted from the other end of the parent chain; preference is given to the double bond, as shown in the following examples.
Note: in the second example, the double bond received a lower number than the triple bond. The common name of ethyne is acetylene, which is also accepted as its IUPAC name.
Bonding and physical properties of alkynes
Alkynes are nonpolar compounds with only London dispersion forces as intermolecular forces, like alkenes and alkanes. Therefore, the physical properties of alkynes are similar to alkanes with the same $\ce{C}$ skeleton. They are insoluble in water, soluble in nonpolar solvents, and have lower densities than water's. The significant difference is the presence of a triple bond in alkynes described below.
Bonding
The general formula of alkyne with one triple bond is $C_{n}H_{(2n-2)}$, where n is a positive integer. The $\ce{C's}$ in $\ce{C≡C}$ bond are sp hybridized with a linear geometry around each of the sp $\ce{C's}$ with bond angles 180o, as illustrated in Figure $8$. Substituents attached to sp hybridized $\ce{C's}$ of alkynes are also called acetylenic. The bond energy for $\ce{C≡C}$ bond is ~10 eV which is about 2.5 times stronger than that of $\ce{C-C}$ bond. Due to the cylindrical shape of two $\pi$-bond clouds, there is free rotation possible around a $\ce{C≡C}$ bond, but due to the linear geometry, it means less.
polarity and chemical reactivity
Although $\ce{C≡C}$ bond is nonpolar, the electrostatic potential map of ethyne in Figure $8$ shows there is an electron-rich region (red color in the map) around the axis joining the nuclei. This is due to $\pi$-bonding electrons around the $\sigma$ bond. The $\pi$ bonds provide $\delta$- region in alkynes, making them nucleophiles, like alkenes. The $\ce{H's}$ are also blue in the electrostatic potential map of ethyne, showing that the $\ce{C-H}$ bond is polar with $\delta {+}$ charge on $\ce{H}$.
Dependence of electronegativity of $\ce{C}$ on hybridization
The $\ce{C-H}$ bond is nonpolar with an electronegativity difference of 0.4 between $\ce{C}$ and $\ce{H}$. The green color in the electrostatic potential map of ethane reflects it. However, the $\ce{C-H}$ bond becomes shorter, stronger, and polar with a change in the hybridization of the $\ce{C}$ from sp3 to sp2 and sp. It means the electronegativity of $\ce{C}$ increases with a change in the hybridization from sp3 to sp2 and sp. The reason is that the sp orbital is wider and shorter due to more s-orbital character (50% s-character) than sp2 orbital (33% s-character), which, in turn is wider and shorter the sp3 orbital (25% s-character), as illustrated in Fig. 7. The wide and shorter orbitals place the electron closer to the $\ce{C}$ nucleus for a stronger attraction to the nucleus, which means higher electronegativity.
Uses of alkynes
Alkynes are less common in biological systems or the chemical industry than alkenes and alkanes. The first member, i.e., acetylene, is used in oxyacetylene torches for cutting and welding metals, as illustrated in the figure on the right. Acetylene is also a starting material for several chemicals used in the chemical industry.
Aromatic hydrocarbons
Aromatic hydrocarbons contain a planer cycle or ring (not bent or puckered) in their structure with all atoms sp2-hybridized and $\pi\text{-electrons in the ring} = 4n +2$, where n is a positive integer. Benzene is an example of an aromatic ring that will be described here. Although aromatic hydrocarbons have systematic names like other hydrocarbons, some alternate and familiar names are so widely used that they are accepted as equivalent to IUPAC names.
Nomenclature of benzene derivatives
• The name of the six sp2 hybridized $\ce{C}$-chain is benzene. It is used as the parent name for the substituted benzene rings.
• If there is only one substituent, i.e., mono-substituted benzene, use the branch name as a prefix to the parent name 'benzene' without numbering.
• If the substituent is methyl ($\ce{-CH3}$), the parent name 'toluene' is often used for it, and if the substituent is ethenyl ($\ce{-CH=CH2}$), it is also called styrene. A few examples are given below.
• If there are two substituents (di-substituted benzene), start numbering the chain from the first substituent and go clockwise or counterclockwise so that the second substituent receives the lower number. List the substitutes alphabetically, preceded by their number, as prefixes to the parent name 'benzene.'
• If one of the two substitutes is methyl ($\ce{-CH3}$), start numbering the chain from the point of attachment of methyl and go clockwise or counterclockwise so that the second substitute receives the lower number. List the second substituent preceded by its number as a prefix to the parent name 'toluene.'
• If there are two methyls ($\ce{-CH3}$) substituents, the parent name 'xylene' is often used. Numbering starts from one of the methyls and goes clockwise or counterclockwise, so the second methyl receives the lower number. List the numbers as prefixes to the parent name 'xylene,' numbers separated by commas from each other and by a hyphen from the parent name, as usual.
• Alternate to the numbers, the word 'ortho' or 'o' for a 1,2 relationship, 'meta' or 'm' for a 1,3 relationship, and 'para' or 'p' for a 1,3 relationship of two substituents on a benzene ring are also accepted. A few examples are the following.
• If there are more than two substituents (poly-substituted benzene), start numbering the chain from the first substituent and go clockwise or counterclockwise so that the substituents receive an overall lower number. List the substitutes alphabetically, preceded by their number, as prefixes to the parent name 'benzene.'
• If one substituent is methyl ($\ce{-CH3}$), start numbering from the point of attachment of methyl ($\ce{-CH3}$) and go as usual to assign the lower overall number to the rest of the substituents, and use the parent name 'toluene.'
Bonding and physical properties of aromatic hydrocarbons
Aromatic hydrocarbons are nonpolar compounds with only London dispersion forces as intermolecular forces, like other hydrocarbons. Benzene has low solubility in water (~1.8 g/L at room temperature), melts at 5.53 oC, and boils at 80.1 oC, which are comparable to that of cyclohexane (immiscible in water, 6.5 oC, and boiling point is 80.7 oC). The primary difference is resonance resulting in significant stabilization of benzene compared to any other hydrocarbon, as described below.
Bonding
All $\ce{C's}$ in the benzene ring are sp2 hybridized, i.e., trigonal planer geometry with all bond angles about 120o, as illustrated in Figure $10$. All $\ce{C's}$ in benzene ring and $\ce{H's}$ or other substituents attached to them are in one plane. Substituents attached to the $\ce{C's}$ of the benzene ring are also called phenylic. The bond energy for $\ce{C\overset\cdots{-}C}$ bond in benzene ring is ~5.4 eV which is between a $\ce{C-C}$ and a $\ce{C=C}$ bond. the phenylic $\ce{C-H}$ bond is ~4.9 eV, almost the same as in alkenes. The $\ce{C\overset\cdots{-}C}$ bond in the benzene ring is (140 pm), i.e., about in the middle of a single and a double bond. This is explained by benzene having two equal resonance contributors in which a single and double bond alternates and the resonance hybrid has a 50% $\pi$-bond.
polarity and chemical reactivity
Although benzene is nonpolar, the electrostatic potential map of benzene in Figure $10$ shows there is a circle of an electron-rich region (red color in the map) above and below the ring. This is due to the delocalized $\pi$-bond extending over all $\ce{C's}$ of the benzene ring. The resonance in a cycle of aromatic rings imparts significantly more resonance stabilization, called aromatic stabilization, than in any other hydrocarbon.
Although the $\delta^-$ region of delocalized elections makes benzene a nucleophile, its nucleophilic character is weak because donating these electrons causes a loss of aromatic stabilization. Therefore benzene needs stronger electrophiles to react with than an alkene and follows a different reaction route than an alkene does.
Cancer risk from polycyclic aromatic hydrocarbons (PAHs)
Benzene increases the risk of leukemia and other blood disorders. Polycyclic aromatic hydrocarbons (PAHs) contain two or more benzene rings fused edge to edge. Naphthalene has two benzene rings fused. Naphthalene is used in mothballs. Anthracene has there benzene rings linked together. Anthracene is used in the manufacture of dyes. Phenanthrene is also three benzene rings fused. Phenanthrene is known to cause cancer, i.e., it is a carcinogen. Benzo[a]pyrene has five benzene rings connected. Benzo[a]pyrene is a potent carcinogen found in tobacco smoke, barbecued meat, and automobile exhaust.
Uses of aromatic hydrocarbons
Benzene is used as a raw material for synthesizing several chemicals. Some familiar aromatic ring-containing compounds include explosives, e.g., trinitrotoluene (TNT), common medicines, e.g., aspirin, acetaminophen, ibuprofen, sulfanilamide, and flavoring agents, e.g., vanillin, shown below. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/02%3A_Nomenclature_and_physical_properties_of_organic_compounds/2.01%3A_Hydrocarbons.txt |
Learning Objectives
• Recognize the two basic mechanisms of bond breaking and making and draw curly arrows to show the electron movement in these steps.
• Define organic functional group and recognize it in an organic compound.
What is a chemical reaction?
A chemical reaction involves making and breaking chemical bonds. Often the elemental composition of the compound changes during a chemical reaction. Sometimes the composition remains the same; atoms rearrange, resulting in a new combination that is an isomer of the initial compound. Often making and breaking of a covalent bond is involved in organic reactions. Recall that a covalent bond is a shared pair of electrons. The bond breaking or making can happen in one of two modes: homolytic or heterolytic, as described below.
Homolytic breaking and bond-making
In homolytic bond breaking, each bonded atom receives one of the two electrons in a covalent bond, e.g.:
, where half-headed arrows show the movement of a single electron. The above reaction produces two atoms with one unpaired electron, i.e., free radicals. The reverse of it is hemolytic bond making, e.g.:
Heterolytic bond-making and breaking
In heterolytic bond breaking, the bonding electrons move to one of the bonded atoms, usually to a more electronegative atom, making a lone pair, e.g.:
, where the regular double-headed arrows show the movement of an electron pair. The formal charge changes in heterolytic bond-making and breaking. The bonded atom that receives the electrons usually becomes an anion, and the other becomes a cation in the above reaction. Revers of it is heterolytic bond making, e.g.:
What is a functional group?
A functional group is an atom or a group of atoms that imparts a characteristic set of physical and chemical properties to the compound.
For example, alkanes with no functional group are the least reactive classes of organic compounds. Alkanes have only $\ce{C-C}$ and $\ce{C-H}$ single bonds, which are nonpolar and strong bonds. Alkenes, alkynes, and aromatic compounds are relatively reactive classes of organic compounds because they have a $\ce{C=C}$ bond, $\ce{C≡C}$ bond, or a benzene ring as functional groups. Although alkenes, alkynes, and aromatic hydrocarbons are nonpolar, there is a partial negative ($\delta^{-}$) character in the $\pi$-bond region that attracts electrophilic reagents having $\delta^{+}$ regions. Further, the $\pi$-bond is weaker than a $\sigma$-bond making it easier to break during chemical reactions.
Classification of organic compounds based on the functional groups
One primary class of organic compounds is hydrocarbons that contain only $\ce{C's}$ and $\ce{H's}$, listed in Table 1, and the others are organic compounds with heteroatom/s, in their functional group/s. Hydrocarbons have been introduced in the previous section. Other classes of organic compounds containing heteratom\s in their functional groups are described next.
The heteroatom/s in the functional group can be a single bonded, i.e., sp3-hybridized halogen, $\ce{O}$, $\ce{N}$, $\ce{S}$, $\ce{P}$, etc., listed in Table 2; a double bonded, i.e., sp2-hybridized $\ce{O}$, $\ce{N}$, $\ce{S}$, $\ce{P}$, etc., listed in Table 3; or a combination of these, listed in Table 4.
Table 1: Classes of hydrocarbons based on their functional groups. (R-, R1-, R2, R3, and R4- represents a general alkyl group or hydrogen)
Class name Group name General structural formula General condensed formula Prefix Suffix Example
Alkane Alkyl R R alkyl- -ane
Alkene Alkenyl $\ce{R2C=CR2}$ alkenyl- -ene
Alkyne Alkynyl $\ce{RC≡CR2}$ alkynyl- -yne
Aromatic (benzene derivatives) Phenyl $\ce{R-C6H5}$ phenyl- -benzene
Table 2: Classes of organic compounds containing an sp3-hybridized heteroatom in functional groups. (R-, R1-, R2, R3, and R4- represent a general alkyl group or hydrogen.)
Class name Group name General structural formula General condensed formula Prefix Suffix Example
Haloalkane Halo $\ce{R-X}$, i.e., $\ce{R-F}$, $\ce{R-Cl}$, $\ce{R-Br}$, or $\ce{R-I}$ $\ce{R-X}$ i.e., $\ce{R-F}$, $\ce{R-Cl}$, $\ce{R-Br}$, or $\ce{R-I}$ halo- i.e., fluoro-, chloro-, bromo-, or iodo-) - $\ce{CH3-CH2-Cl}$ (chloroethane)
Alcohol Alcohol $\ce{R-OH}$ $\ce{ROH}$ hydroxy- -ol $\ce{CH3-CH2-OH}$ (ethanol)
Phenol Phenol $\ce{C6H5OH}$ - -phenol
Ether Ether $\ce{R-O-R'}$ $\ce{ROR'}$ Alkoxy - $\ce{CH3-CH2-O-CH2-CH3}$ (ethoxyethane)
Epoxide Epoxide epoxy- -oxirane
Peroxide Peroxy $\ce{R-O-O-R'}$ $\ce{ROOR'}$ peroxy- alkyl peroxide
Thiol Sulfhydryl $\ce{R-S-H}$ $\ce{RSH}$ sulfanyl- -thiol
Sulfide (Thioether) Sulfide $\ce{R-S-R'}$ $\ce{RSR'}$ sulfanyl- -sulfide
Disulfide Disulfide $\ce{R-S-S-R'}$ $\ce{RSSR'}$ disulfanyl- -disulfide
Amine Amine $\ce{R1R2R3N}$ amino- -amine
Aniline Aniline $\ce{C6H5NH2}$ - -aniline
Table 3: Classes of organic compounds containing an sp2-hybridized heteroatom in functional groups. (R-, R1-, R2, R3, and R4- represents a general alkyl group or hydrogen)
Class name Group name General structural formula General condensed formula Prefix Suffix Example
Aldehyde Aldehyde $\ce{RCHO}$ oxo- -al
Benzaldehyde Benzaldehyde $\ce{C6H5COH}$ - -benzaldehyde
Ketone Ketone $\ce{RCOR'}$ oxo- -one
Imine Imine $\ce{R1C(=NR3)R2}$ imino- -imine
Table 4: Classes of organic compounds containing a mix of sp3-, sp2-, and sp-hybridized heteroatoms in functional groups. (R-, R1-, R2, R3, and R4- represents a general alkyl group or hydrogen)
Class name Group name General structural formula General condensed formula Prefix Suffix Example
Carboxylic acid Carboxyl $\ce{R-COOH}$ carboxy- oic acid
Acyl halide Acyl halide $\ce{R-COX}$ - oil halide
Acid anhydride Acid anhydride $\ce{R1-(CO)O(CO)R2}$ - R1-oil R2-oat, or R1-oic R2-oic anhydride
Easter Easter $\ce{R1COORR'}$ - alkyl -alkanoate
Amide Amide $\ce{RCONR'R"}$ alkyl carbamoyl -amide
Carboxylate Carboxylate $\ce{RCOO^{-}}$ -carboxy -oate
Nitrile Nitrile $\ce{R-C≡N}$ $\ce{RCN}$ cyano- -nitrile
Phosphate Phosphate $\ce{ROP(=O)(OH)2}$ - phosphate
Nitro Nitro $\ce{RNO2}$ nitro- - | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/02%3A_Nomenclature_and_physical_properties_of_organic_compounds/2.02%3A_Functional_group.txt |
Learning Objectives
• Recognize and assign IUPAC name to functional groups containing sp3-hybridized heteroatoms, including haloalkanes, alcohols, ethers, and amines.
• Predict the polarity of the bonds in the functional groups of haloalkanes, alcohols, ethers, and amines and recognize the creative sites based on the partial charges in the functional group regions.
• Identify primary, secondary, tertiary, and quaternary designation of haloalkanes, alcohols, ethers, and amines.
Halogenated hydrocarbons
When a $\ce{H}$ in a hydrocarbon is replaced an $\ce{X}$, where $\ce{X }$ is a halogen, i.e., fluorine ($\ce{F}$), chlorine ($\ce{Cl}$), bromine ($\ce{Br}$), or iodine ($\ce{I}$), it is a halogenated hydrocarbon. A halogenated alkane is called haloalkane.
Nomenclature of halogenated hydrocarbons
Halogen is treated as a branch of a hydrocarbon. The parent chain is numbers following the same rules as for branched hydrocarbons, and the halogens are listed in alphabetic order, preceded by location number, like other alkyl branches.
Example $1$
What is the IUPAC name of the following compound?
Solution
• The longest chain is the parent name: the longest chain is 6-$\ce{C}$ long, so the parent name is hexane.
• List the branches before the parent name in alphabetic order: the branches are chloro- and methyl-, so: chloromethylhexane
• Number the parent chain from the end that gives the lowest number to the first branch: $\ce{Cl}$ is at #2 from the right, and $\ce{CH3}$ is at #3 from the left, so number the parent chain from the right side that gives #2 to $\ce{Cl}$ and #4 to methyl.
Answer: 2-chloro-4-methyl hexane
Example $2$
Wright the skeletal structure of 5-chlorohex-2-yne?
Solution
Find the parent name: hex-2-yne. Hex tells six $\ce{C}$ chain and 2-yne tells there is a triple bond at $\ce{C}$#2, i.e. .
Attach the branches to the parent chain according to their location numbers: the only branch in this chase is chloro- at $\ce{C}$#5.
Answer:
Example $3$
What is the IUPAC name of the following compound?
Solution
• The longest chain is the parent name: the longest chain is 6-$\ce{C}$ cycle with alternate double bonds, i.e., all the $\ce{C's}$ are sp2-hybridized, i.e., an aromatic hydrocarbon with parent name benzene.
• List the branches before the parent name in alphabetic order: the branches are chloro- and bromo-, so: bromochlorobenzene
• Number the benzene ring starting from the 1st branch and going in the direction that gives the lowest number to the 2nd branch: $\ce{Br}$ is at #1, and $\ce{Cl}$ is at #3.
Answer: 1-bromo-3-chlorobenzene
Physical properties of halogenated hydrocarbons
The halogen atom is sp3-hybridized with three sp3 orbitals occupied by lone pairs, and one makes bonds with the $\ce{C}$, i.e., $\ce{R-\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot\cdot}}{X}}\normalsize{:}}$, where $\ce{X}$ is a halogen: $\ce{F}$, $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$. The lone pairs are usually not shown except when needed. The $\ce{C-X}$ bond is polar, i.e., $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{X}}$, because halogens are more electronegative than $\ce{C}$. It makes the $\ce{C}$ a $\delta^{+}$, i.e., an electrophile in reactivity. The lone pairs also add to the polarity because these electrons are localized on one side of the nucleus in sp3 orbitals. The bond polarity can be observed from electrostatic potential maps, shown in Table 5. Recall that green is neutral, red is $\delta^{-}$ and blue is $\delta^{+}$ in electrostatic potential map. Note that the listed electronegativity values of $\ce{C}$ and $\ce{I}$ are the same, but the bond is still polar. This is because the listed values are slightly different from the actual values. Further, the polarizability increases in this order: $\ce{F}$<$\ce{Cl}$<$\ce{Br}$<$\ce{I}$, i.e., $\ce{I}$ is the most polarizable.
What is polarizability
Polarizability is the tendency of a matter to acquire dipole-moment when subjected to an electric field. Larger atoms have more loosely held outermost electrons, which makes them more polarizable. Therefore, the polarizability increases in going from top to bottom in a group, e.g., the polarizability order of halogen is: $\ce{F}$<$\ce{Cl}$<$\ce{Br}$<$\ce{I}$.
Bond length increases in this order: $\ce{F}$<$\ce{Cl}$<$\ce{Br}$<$\ce{I}$ and electronegativity has the opposite trend, i.e., decreases in this order: $\ce{F}$>$\ce{Cl}$>$\ce{Br}$>$\ce{I}$. The dipole moment depends on both the electronegativity difference of the bonded atoms and the bond length; the two effects contradict each other. As a result, the dipole moment does not change much by changing the halogen. The bond dissociation energy decrease in this order: $\ce{F}$>$\ce{Cl}$>$\ce{Br}$>$\ce{I}$, which makes $\ce{C-F}$ the strongest and the least reactive and $\ce{C-I}$ the weakest and the most reactive group.
Table 5: Physical characteristics of $\ce{C-X}$ bond in comparison with $\ce{C-H}$ bond.
Halogen $\ce{H}$ for reference $\ce{F}$ $\ce{Cl}$ $\ce{Br}$ $\ce{I}$
Electrostatic potential map
$\ce{C-X}$ bond length (pm) 109 139 178 193 214
$\ce{C-X}$ bond dissociation energy (kJ/mol) 440 464 355 309 228
Electronegativity of $\ce{X}$ or $\ce{H}$ 2.2 4.0 3.0 2.8 2.5
Dipole moment (D) 0.33 1.85 1.87 1.81 1.62
Solubility in water and boiling points of fluoro- and chloro-alkanes are comparable to alkanes of the same molar mass. Although the $\ce{C-F}$ and $\ce{C-Cl}$ are polar that tend to increase intermolecular forces due to dipole-dipole interaction, the volume of the molecule is small compared to the alkane of similar molar mass due to $\ce{F}$ or $\ce{Cl}$ being heavier atoms. Smaller volume means less contact area and fewer intermolecular forces. The opposing factors cancel each other.
Haloalkanes are denser than alkanes. Mono-fluoroalkanes and mono-chloroalkanes are less dense than water, but bromo- and iodo-alkanes are denser than water. For example, $\ce{CH3CH2Cl}$ and $\ce{CH3CH2Br}$ are liquids having densities of 0.891 g/mL and 1.354 g/mL at 25 oC. Although mono-chloroalkanes are less dense than water, di-, tri-, and tetra-chloroalkanes are denser than water. For example, $\ce{CH3Cl}$ is a gas at room temperature with a density of 1.003 g/mL at its boiling point -23.8 oC, while $\ce{CH2Cl2}$, $\ce{CHCl3}$, and $\ce{CCl4}$ are liquids with density higher than water, i.e., 1.327 g/mL, 1.483 g/mL, and 1.594 g/mL, respectively. So, hydrocarbons float on water while the denser haloalkanes sink in water.
Some uses of halogenated hydrocarbons
$\ce{C-I}$, $\ce{C-Br}$, and $\ce{C-Cl}$ bond are polar and weaker than $\ce{C-H}$ that makes them reactive functional groups. The raw organic material, i.e., alkanes, is usually first converted to haloalkanes intermediates to synthesize other organic compounds. The $\ce{C-F}$ bond is stable, which is why Teflon, composed of $\ce{-(CF2)_{n}{-}}$ chains, is one of the most inert polymers. Haloalkenes with all of the $\ce{C-H}$ bonds replaced with $\ce{C-F}$ or $\ce{C-Cl}$, called chlorofluorocarbons (CFCs) are usually nontoxic, nonflammable, odorless, and noncorrosive liquids that make them ideal as solvents, degreasing agents, and heat transfer fluids in air condition and refrigerator systems. For example, $\ce{CCl2F2}$ (Freon-12) and $\ce{CCl3F}$ (Freon-11) were used as refrigerants, and $\ce{CCl4}$ common name carbon tetrachloride was used as a solvent, cleaning and degreasing agent. They are being phased out because Freons destroy the stratosphere's ozone layer, and $\ce{CCl4}$ is toxic and carcinogenic.
Role of CFCs in the ozone hole
The CFCs like $\ce{CCl2F2}$ and $\ce{CCl3F}$, being stable molecules, survive when released in the atmosphere, reach the stratosphere, and decompose when UV light shines on them in the stratosphere. Their by-products, particularly $\ce{Cl2}$, catalyze the destruction of the ozone layer in the stratosphere that filters out UV light. It resulted in an ozone-depleted region over the southern hemisphere called the ozone hole. Without the ozone layer, UV light reaches the earth's surface and causes damage to living things. Therefore, the CFCs are being replaced with haloalkanes with some $\ce{C-H}$ bonds left to make them less stable, so they may decompose before reaching the stratosphere. As a result of this worldwide effort, the ozone hole is recovering over time, as shown in Figure $1$.
The most promising replacements include hydrofluorocarbons (HFCs) and hydrochlorofluorocarbons (HCFCs), like $\ce{CF3-CH2F}$ called HFC-134a and $\ce{CH3-CCl2F}$ called HCFC-141b. Carbon tetrachloride $\ce{CCl4}$, which is toxic and carcinogenic, is also being replaced with dichloromethane $\ce{CH2Cl2}$ solvent.
Alcohols
An alkyl group attached with an alcohol ($\ce{-OH}$) group, i.e., $\ce{R-OH}$, is an alcohol. A benzene ring attached with an alcohol group is a phenol.
Nomenclature of alcohols and phenols
The nomenclature of hydrocarbons is followed for IUPAC names of alcohols and phenols with the following changes.
• The longest chain containing the $\ce{-OH}$ group is chosen for the parent name, and the 'e' in the suffix -ane, -ene, or -yne is replaced with -ol. For example, $\ce{CH3-OH}$ is methanol, and $\ce{CH3CH2-OH}$ is ethanol.
• Numbering starts from the end, giving the $\ce{-OH}$ group the lowest number in preference over hydrocarbon or halogen branches. This is demonstrated in the following example, where the parent chain is not heptane; it is hexane that contains $\ce{-OH}$ group, and the number starts from the end that gives $\ce{-OH}$ lowest number 2 not from the other end that gives $\ce{-Cl}$ the lowest number 1.
• In a cyclic compound containing only $\ce{-OH}$ group, numbering is not needed, e.g, number is not used in naming cyclohexanone:
• In a cyclic compound containing $\ce{-OH}$ group and a hydrocarbon or halogen group, numbering begins from $\ce{-OH}$ group, e.g.:
• A benzene ring containing an $\ce{-OH}$ group is given the parent name phenol. If there is another group present, the numbering starts from the $\ce{-OH}$ group, and phenol is used as the parent name, e.g.:
Example $1$
What is the IUPAC name of this compound?
Solution
• Find the parent name: four $\ce{C}$ chain is but, and double bond makes it an alkene, i.e., butene.
• Change the 'e' of the suffix with -ol that tells it is an alcohol, i.e., butenol.
• Start number from the end that gives $\ce{-OH}$ group the lowest number: $\ce{-OH}$ receives #1 and $\ce{C=C}$ receives #3.
Answer: but-3-en-1-ol
Example $2$
What is the IUPAC name of the following alcohol?
Solution
• Find the parent name: two $\ce{C}$ chain is ethane.
• Replace the last 'e' of the suffix with -ol, which tells it is an alcohol; in this case, there are two $\ce{-OH}$ groups, so add di, i.e., ethandiol
• The number is the same from either end, i.e., #1 and #2 for the two $\ce{-OH}$ groups.
Answer: ethane-1,2-diol
Ethane-1,2-diol () is known by its common name ethylene glycol. Similarly, propane-1,2,3-triol () is known by its common name, glycerol or glycerine.
Example $3$
What is the IUPAC name of this compound?
Solution
• Find the parent name: six $\ce{C}$ cycle is cyclohexane.
• Replace the last 'e' of the suffix with -ol which means it is an alcohol, i.e., cyclohexanol.
• A cycle with only $\ce{-OH}$ group do not need numbering.
Answer: cyclohexanol
Example $4$
What is the IUPAC name of this compound?
Solution
• Find the parent name: six $\ce{C}$ chain with all sp2-hybridized $\ce{C's}$ is benzene, and benzene with an $\ce{-OH}$ group is phenol.
• List the substituent other than $\ce{OH}$ as prefix: ethylphenol.
• Start numbering form the $\ce{-OH}$ group and continue in the direction that gives the lowest overall number to the other substituents: ethyl receives #3.
Answer: 3-ethylphenol
Physical properties of alcohol and phenols
Alcohols contain a sp3-hybridized $\ce{O}$ bonded to a $\ce{C}$ and a $\ce{H}$ and the remaining two sp3-orbitals occupied by lone pairs, i.e.,: $\ce{R-\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot\cdot}}{O}}-H}$. The $\ce{O}$ atom is more electronegative than $\ce{C}$ and $\ce{H}$, which means both the bonds are polar, i.e., $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{O}{-}\overset{\delta{+}}{H}}$, as observed in the electrostatic maps of some alcohols in Figure $2$.
An alcohol group has three reactive points: an electrophilic $\ce{\overset{\delta{+}}{C}}$, an acidic $\ce{\overset{\delta{+}}{H}}$, and nucleophylic/basic $\ce{\overset{\delta{-}}{O}}$. The acidity of an alcoholic $\ce{H}$ is obvious from the fact that it has a pKa ~16 that is about the same acid strength as $\ce{H2O}$.
Nucleophile/base and electrophile/acid?
Recall that $\ce{\overset{\delta{+}}{C}}$ is called an electrophile. At the same time, proton exchange reactions are acid/base reactions, i.e., $\ce{\overset{\delta{+}}{H}}$ is an acidic proton, and a $\delta{-}$ site when reacts with a $\ce{C}$ is called nucleophile, and when it donates its lone pair to a $\ce{H}$, it is called a base.
Alcohols have significantly higher boiling points compared to alkanes of comparable molar mass.
For example, the boiling point of methanol (molar mass 32 g/mol) is 65 oC which is significantly higher than the -89 oC boiling point of ethane of comparable molar mass (30 g/mol). It is explained by the fact that, in addition to London dispersion forces, alcohols have hydrogen bonding between $\ce{\overset{\delta{+}}{H}}$ of one molecule with $\ce{\overset{\delta{-}}{O}}$ of the neighboring molecule as illustrated in Figure $3$. It takes more thermal energy to break London dispersion forces + hydrogen bonding in alcohols than London dispersion forces alone in alkanes. The boiling points of alcohols increase as their molar mass increases, as shown in Table 6.
Smaller alcohols having up to three $\ce{C'}$ are entirely soluble in water, i.e., miscible, one with four $\ce{C'}$ is partially soluble, and those with longer than four $\ce{C'}$ are almost insoluble as listed in Table 6. This trend of water solubility change is explained by a balance tween hydrophilic and hydrophobic components in an alcohol molecule, explained later.
Table 6: Molecular weight, boiling points, and solubility in water of some alcohols
IUPAC name Molecular formula Molecular weight (g/mol) Boiling point (oC) Solubility in water
Methanol $\ce{CH3OH}$ 32 65 Miscible
Ethanol $\ce{CH3CH2OH}$ 46 78 Miscible
Propan-1-ol $\ce{CH3CH2CH2OH}$ 60 97 Miscible
Butan-1-ol $\ce{CH3CH2CH2CH2OH}$ 74 117 Slightly soluble
Pentan-1-ol $\ce{CH3CH2CH2CH2CH2OH}$ 88 138 Insoluble
Hexan-1-ol $\ce{CH3CH2CH2CH2CH2CH2OH}$ 102 157 Insoluble
Phenol has an $\ce{-OH}$ attached to an sp2-hybridized $\ce{C}$ of a benzene ring. Phenol is a volatile white crystalline solid that melts at 41 oC, and boils at 182 oC. Phenol is partially soluble in water.
$\ce{-OH}$ -An activating group on a benzene ring
When an atom with a lone pair of electrons on it is bonded to an sp2-hybridized $\ce{C}$, it changes its hybridization from sp3 to sp2 for resonance to happen. The driving force is the resonance stabilization effect. The same happens to $\ce{O}$ in phenol resulting in resonance that sends electrons from $\ce{O}$ into the benzene ring as illustrated below. Consequently, the benzene ring of phenol is more electron-rich, i.e., more nucleophilic than benzene without such a group attached. The groups like $\ce{-OH}$ that enhance the reactivity of the benzene ring by donating electrons to the ring are called activating groups.
Another consequence of this electron donation by $\ce{O}$ is that the $\ce{O}$ is more able to hold on to the electrons of $\ce{O-H}$ bond when the proton leaves acting as an acid, i.e., phenolic $\ce{O-H}$ is a stronger acid (pKa ~10) compared to an alcoholic $\ce{O-H}$ (pKa ~16).
The balance between the hydrophilic and hydrophobic character of organic compounds containing a polar group
Polar functional groups like $\ce{-OH}$ are attracted to water, i.e., water-loving or hydrophilic. This is based on the general principle "like dissolves like." On the other hand, Alkyl groups are nonpolar and repel water molecules, i.e., hydrophobic. Alkyl groups are more soluble in lipids and fats, which are also called lipophilic or nonpolar. Alcohols have both groups in them, and their properties depend on which group dominates. The hydrophilic character dominates when the alkyl chain is small, three ($\ce{C's}$ or small, and the hydrophobic character dominates when the alkyl chain is long, i.e., more than four ($\ce{C's}$ long making the compound slightly soluble or insoluble. In summary,
• increasing the alkyl-group increases hydrophobic character, e.g., propan-1-ol ($\ce{CH3CH2CH2OH}$) is water soluble, but butan-1-ol ($\ce{CH3CH2CH2CH2OH}$) is partially soluble, and
• increasing the polar group increases the hydrophilic character, e.g., butan-1-ol ($\ce{CH3CH2CH2CH2OH}$) with one ($\ce{-OH}$) group is partially soluble in water, but butane-1,2-diol ($\ce{CH3CH2CH(OH)CH2OH}$) with two ($\ce{-OH}$) groups is soluble.
This trend is generally applicable to all polar groups containing organic compounds.
Some examples of important alcohols and phenols
Methanol ($\ce{CH3-OH}$) is used as a solvent, paint remover, and fuel and converted to formaldehyde which is then used to make plastics. If methanol is ingested, it converts to formaldehyde, which causes headache and blindness and may cause death. That is why methanol is mixed with ethanol not meant for drinking.
Ethanol ($\ce{CH3-OH}$) is the active ingredient in alcoholic beverages. It is also used as a solvent for perfumes, varnishes, and medicines, e.g., in tincture iodine. The fermentation process obtains ethanol, but these days, it is mainly derived from ethene.
Isopropanol. 2-methylpropanol, commonly known as isopropanol, kills bacteria and viruses. Ethanol also has the same property. Ethanol or isopropanol are 60% to 80% fraction of hand sanitizers; the remaining is usually ethylene or glycerin to keep the skin soft. If ethanol is the active ingredient in the hand sanitizer, care is needed as ethanol is volatile and flammable.
Ethylene glycol, i.e., ethane-1,2-diol ($\ce{HO-CH2CH2-OH}$) is used as antifreeze in cooling systems of vehicles and as a solvent for paints, inks, and plastics. Ethylene glycol is poisonous because, if ingested, it converts to oxalic acid, which causes kidney stones.
Glycerin or glycerol, i.e., propane-1,2,3-triol ($\ce{HO-CH2CH(OH)CH2-OH}$) is obtained as a byproduct during the soap making from fats and oils. It is a viscous liquid used in skin lotions, cosmetics, shaving creams, and liquid soaps. Nitroglycerin, the active component of explosive dynamite, is derived from glycerol.
Bisphenol a (BPA) is used used in making polycarbonate that is used to manufacture beverage bottles. There is a concern that leaching out BPA from beverage bottles may cause harmful health effects.
Phenol is primarily used for the production of precursors of plastics. Phenol is used as an antiseptic and disinfectant. Phenol is also a part of the structure of essential oils responsible for plants' odor and flavor, e.g., isoeugenol from nutmeg, eugenol from clover, thymol from thyme, and vanillin from vanilla, shown in Table 1.
Table 3: Examples of phenol present in essential oils
Isoeugenol from nutmeg Eugenol from clover Thymol from thyme Vanillin from vanilla
Isoeugenol Eugenol Thymol Vanillin
Thiol
Thiols are sulfur analogs of alcohols, i.e., they contain the thiol ($\ce{-SH}$) group. Like alcohols, thiols contain an electrophilic $\ce{C}$ and acidic $\ce{H}$ attached to $\ce{S}$ atom as illustrated by the electrostatic potential map of methanethiol $\ce{CH3-SH}$ in the figure on the right, where $\ce{S}$ is shown as yellow, $\ce{H}$ as white, and $\ce{C}$ as gray sphere in the model.
The thiols are named following the rules of alcohols, except that suffix -thiol is added to the name of alkane, e.g., $\ce{CH3-SH}$ is methanethiol and (\ce{CH3CH2-SH}\)) is ethanethiol.
One characteristic of thiols is that they have a strong odor, e.g., methanthiol ($\ce{CH3-SH}$) is an odor-causing compound present in oysters, cheddar cheese, onions, and garlic. The garlic odor also contains prop-2-ene-1-thiol (\ce{CH2=CH-CH2-SH}\). Onion odor is also due to propane-1-thiol ((\ce{CH3CH2CH2-SH}\), which is a lachrymator, i.e., a substance that makes eyes tear.
Ethers
Ethers have $\ce{O}$ with two single bonds with two alkyl groups, i.e., $\ce{R-O-R'}$ group.
Nomenclature of ethers
IUPAC naming of ether follows the rules of branched alkanes. The large alkyl group is named as parent chain, and the smaller alkyl groups with oxygen are named as a branch, i.e., an alkoxy ($\ce{R-O{-}}$) group. For example, $\ce{CH3CH2-O-CH2CH3}$ is ethoxyethane and $\ce{CH3CH2CH2-O-CH3}$ is 1-methoxypropane . However, trivial names for ethers are often used. A common name is formed by listing the two alkyl groups in alphabetic order, followed by the word ether. For example, common name of $\ce{CH3CH2-O-CH2CH3}$ is diethylether and of $\ce{CH3CH2CH2-O-CH3}$ is propyl methyl ether.
Physical properties of ethers
Ethers contain a sp3-hybridized $\ce{O}$ bonded to two $\ce{C's}$ and the remaining two sp3-orbitals occupied by lone pairs, i.e.,: $\ce{R-\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot\cdot}}{O}}-R}$. The $\ce{O}$ atom is more electronegative than $\ce{C}$, which means both the bonds are polar, i.e., $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{O}{-}\overset{\delta{+}}{C}}$, as observed in the electrostatic maps of diethylether shown on in figure on the right.
Since there is no $\ce{O-H}$ bond, pure ethers do not have hydrogen bonding; they have a weak dipole-dipole interaction and London dispersion forces. Therefore, the boiling points of ether are comparable to alkanes of the same molecular mass, e.g., diethyl ether (molar mass 74 g/mol) has a boiling point of 35 oC, which is almost the same as the boiling point of pentane (molar mass 72 g/mol, boiling point 36 oC). However, ethers do have $\ce{\overset{\delta{-}}{O}}$ and that can establish hydrogen bonding with $\ce{\overset{\delta{+}}{H}}$ of water molecule. So, ethers have solubility in water comparable to alcohols of the same molar mass, i.e., ethers with three or less $\ce{C's}$ are miscible in water, and those with four or more $\ce{C's}$ are slightly soluble or insoluble in water.
Ethers are less reactive than alcohols because alcohols have an acidic $\ce{\overset{\delta{+}}{H}}$, which is missing in ethers. Because of relatively low chemical reactivity, ethers are used as solvents to conduct other compounds' reactions.
Epoxides -reactive ethers and sterilizing agents
Epoxides are cyclic ethers in which one atom in a three-membered ring is $\ce{O}$. Epoxides take the parent name oxirane, and numbering, if needed, starts from the $\ce{O}$ atom. Three examples are shown below.
Epoxides are unstable and highly reactive due to angle strain as the bonds bent from a regular angle of 109.5o to 60o. They are used as intermediates in organic synthesis. Oxirane is a room-temperature gas that reacts fast with the compounds in microorganisms, causing their death. So, it is used as a fumigant in foodstuffs and textiles and as a hospital sterilizer for surgical instruments.
Ethers as anesthetics
Diethyl ether has been used as an anesthetic agent for a long time. Its use in anesthesia has been discontinued because it is a volatile and flammable gas posing a fire or explosion risk in the surgery room. Further, it has an irritating effect on the respiratory passages and causes nausea. Fluorinated or chlorofluorinated ethers, such as desflurane, sevoflurane, and isoflurane listed below, are used as anesthetic agents because they are less volatile, less flammable, and have fewer other undesirable side effects.
Peroxides, sulfides, and disulfides
Peroxides have $\ce{O-O}$ bond, which is one of the weak bonds with bond dissociation energy ~200 kJ/mol, about half of the strength of a $\ce{C-C}$ or a $\ce{C-H}$ bond. This bond is found in hydroperoxy group ($\ce{R-O-O-H}$) and peroxy group ($\ce{R-O-O-R'}$. The peroxides are less common but more reactive due to weak $\ce{O-O}$ bond.
Sulfides are $\ce{S}$ analogs of ethers, i.e, they have sulfide ($\ce{R-\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot\cdot}}{S}}-R}$) group. The names of sulfides are similar to those of ethers, i.e., list the two alkyl groups alphabetically, followed by the word sulfide. For example, $\ce{CH3CH2-S-CH2CH3}$ is diethyl sulfide and $\ce{CH3CH2CH2-S-CH3}$ is propyl methyl sulfide. Sulfides are less common in nature. Disulfide $\ce{R-S-S-R}$ group, i.e., a sulfur analog of peroxides found in proteins. The $\ce{S-S}$ is a relatively weak bond with a bond dissociation energy of ~250 kJ/mol.
Primary (p), secondary (s), tertiary (t), and quaternary classification
The p, s, t- classification of $\ce{C's}$ and $\ce{H's}$
In an organic compound:
• if a $\ce{C}$ is not bonded with any other $\ce{C}$ or bonded with only one other $\ce{C}$, it is a primary $\ce{C}$;
• if bonded with two other $\ce{C's}$, it is a secondary $\ce{C}$;
• if bonded with three other $\ce{C's}\}, it is a tertiary \(\ce{C}$; and
• if bonded with four other $\ce{C's}$, it is a quaternary $\ce{C}$.
The $\ce{H's}$ on a primary $\ce{C's}$ are primary $\ce{H's}$, those on a secondary $\ce{C's}$ are secondary $\ce{H's}$, and those on tertiary $\ce{C's}$ are tertiary $\ce{H's}$. Quaternary $\ce{C's}$ do not have any $\ce{H}$ on them. The figure on the right illustrates the primary, secondary, tertiary, and quaternary $\ce{C's}$ and $\ce{H's}$ marked in different colors and pointed by arrows. Table 1 illustrates the primary, secondary, and tertiary haloalkanes, alcohols, alkyl groups in ethers, and amine.
Table 1: Primary, secondary, and tertiary classifications of haloalkanes, alcohols, alkyl groups, and amines.
(Common names using secondary (s) and tertiary (t) designation or iso (for the case of secondary propyl) alkyl groups are used.)
Classification Primary Secondary Tertiary
Haloalkanes
ethyl chloride
isopropyl chloride
t-butyl chloride
Alcohols
ethanol
sec-butanol
t-butanol
Alkyl groups in ethers
diethyl ether
diisopropyl ether
ditertiary butyl ether
Amines
t-butyl amine
ethyl(methyl)amine
ethyl(methyl)propylamine
The p, s, t- classification of haloalkanes and alcohols
A halogen or an $\ce{-OH}$ group:
• if bonded with a primary $\ce{C}$, it is a primary haloalkane, or primary alcohol;
• if bonded with a secondary $\ce{C}$, it is a secondary haloalkane or secondary alcohol;
• if bonded with a tertiary $\ce{C}\}, it is a tertiary haloalkane or tertiary alcohol. Quaternary \(\ce{C's}$ do not have any halogen or $\ce{-OH}$ group on them.
The p, s, t- classification of alkyl groups
An alkyl group connected through:
• a primary $\ce{C}$ to a parent chain or a functional group is a primary alkyl group (p-alkyl);
• a secondary $\ce{C}$ is a secondary alky group (s-alkyl); and
• a tertiary $\ce{C}$ is a tertiary alkyl group (t-alkyl).
The primary (p), secondary (sec, or s), and tertiary (t) designation of alkyl groups (except for isopropyl used in the place of sec-propyl) are also accepted in IUPAC nomenclature, and these designations are often used in common names of organic compounds, as shown by the commons in Table 1.
The p, s, t- classification of amines
Amines are organic compounds that have one or more $\ce{H's}$ of ammonia ($\ce{NH3}$) replaced with an aliphatic hydrocarbon group.
• If there is only one alkyl group bonded with the $\ce{N}$, it is a primary amine group, i.e., $\ce{RNH2}$;
• if two alkyl groups are bonded with the $\ce{N}$, it is a secondary amine group, i.e., $\ce{RR'NH}$;
• if three alkyl groups are bonded with the $\ce{N}$, it is a tertiary amine group, i.e., $\ce{RR'R"N}$, and
• if four alkyl groups are bonded with the $\ce{N}$, it is not an amine but an ammonium ion, i.e., $\ce{RR'R"R'''N^{+}}$, e.g., tetramethylammonium: .
Caution-p, s, or t designation of amines is based on the $\ce{N}$ not on the $\ce{C}$
Unlike haloalkenes and alcohols, the primary, secondary, and tertiary classification of amines is based on whether there are one, two, or three alkyl groups bonded with the $\ce{N}$, irrespective of the $\ce{C}$ bonded with the $\ce{N}$ is primary, secondary, or tertiary. For example, as shown below, t-butylamine is a primary, and trimethylamine is a tertiary amine.
Amines
Ammonia has an sp3-hybridized $\ce{N}$ having three single bonds with nitrogen and one lone pair, i.e., $\ce{:\!NH3}$. If one or more $\ce{H's}$ of ammonia are replaced with an aliphatic hydrocarbon, it is an amine group, and the compound is an amine, i.e., $\ce{RNH2}$, $\ce{RR'NH}$, or $\ce{RR'R''N}$ are amine groups. If one or more $\ce{H's}$ of ammonia are replaced with a benzene ring, it is an aromatic amine.
Nomenclature of amines
IUPAC naming of amines follows the rules for naming alcohols with the following changes:
• The suffix's last letter 'e' is replaced with amine, e.g., $\ce{CH3NH2}$ is methanamine. A few examples are the following.
• If more than one hydrocarbon group is attached with the $\ce{N}$, the longest one is chosen as the parent name, and the smaller ones are listed alphabetically as substituents preceded by N-, where N tells the branch is attached to $\ce{N}$. For example, $\ce{CH3CH2NHCH3}$ is N-methylethanamine.
• An alternate and easier way is to list the alkyl groups as substituents in alphabetic order (the middle one is enclosed in small brackets for easy reading), followed by the word amine. For example, $\ce{CH3CH2NHCH3}$ can be named ethyl(methyl)amine. A few examples are listed below.
• A benzene ring containing an $\ce{−NH2}$ group is given the parent name aniline. If there is another group present, the numbering starts from the $\ce{−NH2}$ group, and aniline is used as the parent name, e.g.:
Example $1$
What is the IUPAC name of the compound shown in the figure on the right?
Solution
1. List the alkyl groups in alphabetic order (place the middle name in brackets) and end with the word amine: ethyl(methyl)amine, or
2. Choose the longest substituted as the parent name and the other as substitute preceded by N- to indicate the substituent is on $\ce{N}$, i.e.,: N-methylethanamine.
Example $2$
What is the IUPAC name of the compound shown in the figure on the right?
Solution
Ethane is the only alkyl group, so the stem name is ethane. It has two amine groups, so use di-preceded by number to tell the location of the amine groups: ethan-1,2-diamine.
Example $3$
What is the IUPAC name of the compound shown in the figure on the right?
Solution
A benzene ring attached to an amine group has the parent name aniline. Methyl substituent is on $\ce{N}$, so list it as a prefix branch name preceded by N-, i.e., N-methylaniline.
Example $4$
What is the IUPAC name of the compound shown in the figure on the right?
Solution
The alkyl group is cyclohexane; replace the last 'e' with amine: cyclohexanamine.
Physical properties of amines
Amines contain a sp3-hybridized $\ce{N}$ bonded to $\ce{C's}$ and $\ce{H's}$ and one sp3-orbitals occupied by lone pairs, i.e.,: $\ce{RR'R'N\!:}$. The $\ce{N}$ atom is more electronegative than $\ce{C}$, which means the $\ce{C-N}$ and $\ce{N-H}$ bonds are polar, i.e., $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{N}{-}\overset{\delta{+}}{H}}$, as observed in the electrostatic maps of methanamine shown in the figure on the right, where $\ce{N}$ is blue, $\ce{C}$ is gray, and $\ce{H}$ is a white sphere.
The $\ce{N}$ atom is less electronegative than $\ce{O}$, so the $\ce{C-N}$ bond (3.0-2.1 = 0.9) is less polar and less reactive than $\ce{C-O}$ (3.5-2.1 = 1.4). Due to the less electronegativity of $\ce{N}$, it is more willing to donate its lone pair to any proton around, i.e., amiens are basic compounds. Amines are also classified as organic bases.
Primary and secondary amines have hydrogen bonding, but the $\ce{N-H}$ bond is less polar than the $\ce{O-H}$ bond. Therefore, the boiling points of primary and secondary amines are higher than alkanes but lower than alcohol of comparable molar mass, as shown in Table 8.
Table 8: Molar masses and boiling points of some amines
Name Condensed Molar mass Boiling point
Ethane $\ce{CH3CH3}$ 30.1 g/mol -88.6 oC
Methylamine $\ce{CH3NH2}$ 31.1 g/mo -6.3 oC
Methanol $\ce{CH3OH}$ 32.0 g/mol 65.0 oC
Butane $\ce{CH3CH2CH2CH3}$ 58 g/mole 0 oC
Ttrimetylamine $\ce{(CH3)3N}$ 59 g/mole 3 oC
Propylamine $\ce{CH3CH2CH2NH2}$ 59 g/mol 48.0 oC
Tertiary amines have boiling points comparable with alkanes as they do not have $\ce{N-H}$ bond and do not make hydrogen bonding with each other, as shown in Table 8.
The primary, secondary, and tertiary amines make a hydrogen bonds with water. Therefore, amines containing up to five $\ce{C's}$ are soluble in water, and those with more than five $\ce{C's}$ are slightly soluble or insoluble.
Importance of amines in health and medicine
Amines are present in amino acids, the building blocks of proteins. Amines are present in several physiologically active compounds. The body releases the histamine in response to injury or allergic reactions. Histamine causes blood vessels to dilate, and redness and swelling occur in the area. Antihistamine administered to block the effects of histamine is another amine, diphenhydramine.
Epinephrine (adrenaline) and norepinephrine (noradrenaline) are released in a "fight-or-flight" situation. They increase blood glucose and move the blood to the muscles. Norepinephrine is a remedy for colds, hay fever, and asthma. It contracts the capillaries in the mucous membranes of the respiratory passage.
Tranquilizers are drugs that relieve the symptoms of anxiety and tension. These include diazepam (earlier name valium), Chlordiazepoxide (trade name Librium), and other benzodiazepines, which are sedative and hypnotic medicines that cause calming effects and drowsiness.
Alkaloids
Amines can donate a lone pair to a proton relatively easily because $\ce{N}$ is less electronegative than $\ce{O}$, i.e., amines are basic compounds with pKa value ~10. Alkaloids are nitrogen-containing basic compounds extracted from plants. Most alkaloids are physiologically active and used in anesthetics, antidepressants, and stimulants; many are habit-forming. For example, coniine extracted from "poison hemlock" can cause weakness, fast respiration, paralysis, and death. Nicotine found in tobacco is addictive, but in large doses, it causes depression, nausea, vomiting, or death. The solution of nicotine in water is used as an insecticide. Cocaine extracted from the coca plant is a stimulant for the central nervous system. Piperidine is responsible for the pungent smell and taste of black pepper. The Chemical structures of these alkaloids are shown below.
(S)-Conine
Nicotine
Cocaine
Piperidine
Caffeine is found in coffee and tea. Caffeine increases alertness and is used in some pain relievers to counter the drowsiness caused by antihistamines. Quinine from the bark of the cinchona tree is used to treat malaria. Atropine from belladonna accelerates slow heart rates and is anesthesia for eye exams. Their structures are shown below.
Caffeine
Quinine
Atropine
Morphine and codeine are obtained from the opium poppy plant and used as painkillers and cough syrup. Heroin, which is strongly addictive, is received by the chemical modification of morphine. OxyContin, which is a prescription drug for the relief of fever pain, is structurally similar to heroin and also has similar physiological effects. The structures of these alkaloids are shown below.
Seedhead of Opium Poppy
Morphine
Codeine
Heroin
OxyContin | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/02%3A_Nomenclature_and_physical_properties_of_organic_compounds/2.03%3A_Functional_groups_containing_sp3-hybridized_heteroatom.txt |
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