Solutions Manual
Fundamentals of Analytical Chemistry 9th Edition
By
Douglas A. Skoog,
Donald M. West,
F. James Holler,
Stanley R. Crouch
( All Chapters Included - 100% Verified Solutions )
1
,Fundamentals of Analytical Chemistry: 9th ed. Chapter 3
Chapter 3
3-1. (a) SQRT returns the square root of a number or result of a calculation.
(b) AVERAGE returns the arithmetic mean of a series of numbers.
(c) PI returns the value of pi accurate to 15 digits
(d) FACT returns the factorial of a number, equal to 1 × 2 × 3 × … × number.
(e) EXP returns e raised to the value of a given number.
(f) LOG returns the logarithm of a number to a base specified by the user.
1
2
Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
,Fundamentals of Analytical Chemistry: 9th ed. Chapter 4
Chapter 4
4-1. (a) The millimole is an amount of a chemical species, such as an atom, an ion, a
molecule or an electron. There are
particles mole particles
6.02 × 1023 × 10−3 = 6.02 × 10 20
mole millimole millimole
(c) The millimolar mass is the mass in grams of one millimole of a chemical species.
3
1000 mL 1 cm 3 ⎛ 1 m ⎞
4-3. The liter: 1 L = × ×⎜ ⎟ = 10−3 m 3
1L 1 mL ⎝ 100 cm ⎠
1 mol 1L 1 mol
Molar concentration: 1 M = × =
1L 10−3 m 3 10 −3 m 3
1 MHz
4-4. (a) 3.2 × 108 Hz × = 320 MHz
106 Hz
1 mol
(c) 8.43 ×107 μmol × = 84.3 mol
106 μmol
1 mm
(e) 8.96 × 106 nm × = 8.96 mm
106 nm
4-5. For oxygen, for example 15.999 u/atom = 15.999 g/6.022 ×1023 atoms = 15.999 g/mol.
So 1 u = 1 g/mol.
Thus, 1g = 1mol u.
1 mol Na 3 PO 4 3 mol Na + 6.022 × 1023 Na +
4-7. 2.92 g Na 3 PO 4 × × × = 3.22 × 1022 Na +
163.94 g 1 mol Na 3 PO 4 1 mol Na +
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3
Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
, Fundamentals of Analytical Chemistry: 9th ed. Chapter 4
2 mol B 1 mol B2O3
4-9. (a) 8.75 g B2O3 × × = 0.251 mol B
1 mol B2O3 69.62 g B2O3
1g 7 mol O
167.2 mg Na 2 B4O7 • 10H 2 O × ×
1000 mg 1 mol Na 2 B4 O7 • 10H 2 O
(b)
1 mol Na 2 B4O7 • 10H 2O
× = 3.07 × 10−3 mol O = 3.07 mmol
381.37 g
1 mol Mn 3O4 3 mol Mn
(c) 4.96 g Mn 3O4 × × = 6.50 × 10−2 mol Mn
228.81 g Mn 3O4 1 mol Mn 3O4
1g mol CaC2O 4 2 mol C
333 mg CaC2 O 4 × × × = 5.20 × 10−3 mol C
(d) 1000 mg 128.10 g CaC 2O 4 1 mol CaC2O 4
= 5.20 mmol
0.0555 mol KMnO4 1000 mmol
4-11. (a) × × 2.00 L = 111 mmol KMnO4
L 1 mol
3.25 × 10−3 M KSCN 1000 mmol L
× × ×750 mL
(b) L 1 mol 1000 mL
= 2.44 mmol KSCN
3.33 mg CuSO 4 1g 1 mol CuSO 4 1000 mmol
× × × × 3.50 L
(c) 1L 1000 mg 159.61 g CuSO 4 1 mol
= 7.30×10−2 mmol CuSO 4
0.414 mol KCl 1000 mmol 1L
(d) × × × 250 mL = 103.5 mmol KCl
1L 1 mol 1000 mL
63.01 g HNO3 1000 mg
4-13. (a) 0.367 mol HNO3× × = 2.31 × 104 mg HNO3
1 mol HNO3 1g
1 mol 40.30 g MgO 1000 mg
(b) 245 mmol MgO× × × = 9.87 × 103mg MgO
1000 mmol 1 mol MgO 1g
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4
Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
Fundamentals of Analytical Chemistry 9th Edition
By
Douglas A. Skoog,
Donald M. West,
F. James Holler,
Stanley R. Crouch
( All Chapters Included - 100% Verified Solutions )
1
,Fundamentals of Analytical Chemistry: 9th ed. Chapter 3
Chapter 3
3-1. (a) SQRT returns the square root of a number or result of a calculation.
(b) AVERAGE returns the arithmetic mean of a series of numbers.
(c) PI returns the value of pi accurate to 15 digits
(d) FACT returns the factorial of a number, equal to 1 × 2 × 3 × … × number.
(e) EXP returns e raised to the value of a given number.
(f) LOG returns the logarithm of a number to a base specified by the user.
1
2
Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
,Fundamentals of Analytical Chemistry: 9th ed. Chapter 4
Chapter 4
4-1. (a) The millimole is an amount of a chemical species, such as an atom, an ion, a
molecule or an electron. There are
particles mole particles
6.02 × 1023 × 10−3 = 6.02 × 10 20
mole millimole millimole
(c) The millimolar mass is the mass in grams of one millimole of a chemical species.
3
1000 mL 1 cm 3 ⎛ 1 m ⎞
4-3. The liter: 1 L = × ×⎜ ⎟ = 10−3 m 3
1L 1 mL ⎝ 100 cm ⎠
1 mol 1L 1 mol
Molar concentration: 1 M = × =
1L 10−3 m 3 10 −3 m 3
1 MHz
4-4. (a) 3.2 × 108 Hz × = 320 MHz
106 Hz
1 mol
(c) 8.43 ×107 μmol × = 84.3 mol
106 μmol
1 mm
(e) 8.96 × 106 nm × = 8.96 mm
106 nm
4-5. For oxygen, for example 15.999 u/atom = 15.999 g/6.022 ×1023 atoms = 15.999 g/mol.
So 1 u = 1 g/mol.
Thus, 1g = 1mol u.
1 mol Na 3 PO 4 3 mol Na + 6.022 × 1023 Na +
4-7. 2.92 g Na 3 PO 4 × × × = 3.22 × 1022 Na +
163.94 g 1 mol Na 3 PO 4 1 mol Na +
2
3
Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
, Fundamentals of Analytical Chemistry: 9th ed. Chapter 4
2 mol B 1 mol B2O3
4-9. (a) 8.75 g B2O3 × × = 0.251 mol B
1 mol B2O3 69.62 g B2O3
1g 7 mol O
167.2 mg Na 2 B4O7 • 10H 2 O × ×
1000 mg 1 mol Na 2 B4 O7 • 10H 2 O
(b)
1 mol Na 2 B4O7 • 10H 2O
× = 3.07 × 10−3 mol O = 3.07 mmol
381.37 g
1 mol Mn 3O4 3 mol Mn
(c) 4.96 g Mn 3O4 × × = 6.50 × 10−2 mol Mn
228.81 g Mn 3O4 1 mol Mn 3O4
1g mol CaC2O 4 2 mol C
333 mg CaC2 O 4 × × × = 5.20 × 10−3 mol C
(d) 1000 mg 128.10 g CaC 2O 4 1 mol CaC2O 4
= 5.20 mmol
0.0555 mol KMnO4 1000 mmol
4-11. (a) × × 2.00 L = 111 mmol KMnO4
L 1 mol
3.25 × 10−3 M KSCN 1000 mmol L
× × ×750 mL
(b) L 1 mol 1000 mL
= 2.44 mmol KSCN
3.33 mg CuSO 4 1g 1 mol CuSO 4 1000 mmol
× × × × 3.50 L
(c) 1L 1000 mg 159.61 g CuSO 4 1 mol
= 7.30×10−2 mmol CuSO 4
0.414 mol KCl 1000 mmol 1L
(d) × × × 250 mL = 103.5 mmol KCl
1L 1 mol 1000 mL
63.01 g HNO3 1000 mg
4-13. (a) 0.367 mol HNO3× × = 2.31 × 104 mg HNO3
1 mol HNO3 1g
1 mol 40.30 g MgO 1000 mg
(b) 245 mmol MgO× × × = 9.87 × 103mg MgO
1000 mmol 1 mol MgO 1g
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4
Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
