Chapter 3
(a) 12.01 amu; (b) 12.01 amu; (c) 144.12 amu; (d) 60.05 amu
(a) 123.896 amu; (b) 18.015 amu; (c) 164.086 amu; (d) 60.052 amu; (e) 342.297 amu
(a) 56.107 amu; (b) 54.091 amu; (c) 199.9976 amu; (d) 97.9950 amu
Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.
Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.
The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of 6.022 10 23 molecules.
(a) 256.48 g/mol; (b) 72.150 g mol −1 ; (c) 378.103 g mol −1 ; (d) 58.080 g mol −1 ; (e) 180.158 g mol −1
(a) 197.382 g mol −1 ; (b) 257.163 g mol −1 ; (c) 194.193 g mol −1 ; (d) 60.056 g mol −1 ; (e) 306.464 g mol −1
(a) 0.819 g; (b) 307 g; (c) 0.23 g; (d) 1.235 10 6 g (1235 kg); (e) 765 g
(a) 99.41 g; (b) 2.27 g; (c) 3.5 g; (d) 222 kg; (e) 160.1 g
(a) 9.60 g; (b) 19.2 g; (c) 28.8 g
zirconium: 2.038 10 23 atoms; 30.87 g; silicon: 2.038 10 23 atoms; 9.504 g; oxygen: 8.151 10 23 atoms; 21.66 g
AlPO 4 : 1.000 mol, or 26.98 g Al; Al 2 Cl 6 : 1.994 mol, or 53.74 g Al; Al 2 S 3 : 3.00 mol, or 80.94 g Al; The Al 2 S 3 sample thus contains the greatest mass of Al.
3.113 10 25 C atoms
0.865 servings, or about 1 serving.
20.0 g H 2 O represents the least number of molecules since it has the least number of moles.
(a) % N = 82.24%, % H = 17.76%; (b) % Na = 29.08%, % S = 40.56%, % O = 30.36%; (c) % Ca 2 + = 38.76%
% NH 3 = 38.2%
(a) CS 2 ; (b) CH 2 O
C 6 H 6
Mg 3 Si 2 H 3 O 8 (empirical formula), Mg 6 Si 4 H 6 O 16 (molecular formula)
C 15 H 15 N 3
We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.
(a) 0.679 M ; (b) 1.00 M ; (c) 0.06998 M ; (d) 1.75 M ; (e) 0.070 M ; (f) 6.6 M
(a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g
(a) 37.0 mol H 2 SO 4 , 3.63 10 3 g H 2 SO 4 ; (b) 3.8 10 −7 mol NaCN, 1.9 10 −5 g NaCN; (c) 73.2 mol H 2 CO, 2.20 kg H 2 CO; (d) 5.9 10 −7 mol FeSO 4 , 8.9 10 −5 g FeSO 4
(a) Determine the molar mass of KMnO 4 ; determine the number of moles of KMnO 4 in the solution; from the number of moles and the volume of solution, determine the molarity; (b) 1.15 10 −3 M
(a) 5.04 10 −3 M ; (b) 0.499 M ; (c) 9.92 M ; (d) 1.1 10 −3 M
0.025 M
0.5000 L
1.9 mL
(a) 0.125 M ; (b) 0.04888 M ; (c) 0.206 M ; (d) 0.0056 M
11.9 M
1.6 L
(a) The dilution equation can be used, appropriately modified to accommodate mass-based concentration units: . This equation can be rearranged to isolate mass 1 and the given quantities substituted into this equation. (b) 58.8 g
114 g
1.75 10 −3 M
95 mg/dL
2.38 10 −4 mol
0.29 mol