Errata to accompany
Chemistry: The Molecular Nature of Matter and Change, 6e
Martin Silberberg
Page 49: Table 2.2 should read “Properties of the Three Key Subatomic Particles” and
the absolute mass of the electron should be 9.10939x10-28
Prob. 2.50: Should read “… 63Cu (isotopic mass = 62.9296 amu)” instead of 62.6396
Prob. 7.34: Should read “… (c) Rank the absorptions in terms of increasing wavelength
of light absorbed.”
Page 839: Follow-up problem 20.9 should read … “(a) If K = 20 …”
Appendix E
Problem 6.60a should read:
C2H4(g) + 3O2(g) → 2CO2(g) 1 2H2O(g); ∆Hrxn = −1411 kJ
Problem 18.134a should read:
Ni(H2O)62+(aq) + 6NH3(aq)
Ni(NH3)62+ + 6H2O(l)
Problems 19.50 and 19.52
The answer should read “bromothymol”
Student Solutions Manual
15.116 Plan: Convert each mass to moles, and divide the moles by the smallest number to determine
molar ratio, and thus relative numbers of the amino acids. To find the minimum molar mass, add
the products of the moles of each amino acid and its molar mass to find the total mass of the amino
acids and subtract the total mass of water that is eliminated during the formation of the peptide
bonds.
Solution:
a) The hydrolysis process requires the addition of water to break the peptide bonds.
b)
(3.00 g gly)/(75.07 g/mol) = 0.0399627 mol glycine
(0.90 g ala)/(89.10 g/mol) = 0.010101010 mol alanine
(3.70 g val)/(117.15 g/mol) = 0.0315834 mol valine
(6.90 g pro)/(115.13 g/mol) = 0.0599323 mol proline
(7.30 g ser)/(105.10 g/mol) = 0.0694577 mol serine
(86.00 g arg)/(174.21 g/mol) = 0.493657 mol arginine
Divide by the smallest value (0.010101010 mol alanine), and round to a whole number.
(0.0399627 mol glycine)/(0.010101010 mol) = 4
(0.010101010 mol alanine)/(0.010101010 mol) = 1
(0.0315834 mol valine)/(0.010101010 mol) = 3
(0.0599323 mol proline)/(0.010101010 mol) = 6
(0.0694577 mol serine)/(0.010101010 mol) = 7
(0.493657 mol arginine)/(0.010101010 mol) = 49
Total moles of amino acids = 4 + 1 + 3 + 6 + 7 + 49 = 70 moles
c) Mass of 70 moles of amino acids = (4 x 75.07 g/mol) + (1 x 89.09 g/mol) + (3 x 117.15 g/mol)
+ (6 x 115.13 g/mol) + (7 x 105.09 g/mol) + (49 x 174.20 g/mol) = 10,703.59 g
To link 70 moles of amino acids, 69 peptide bonds are formed by the elimination of 69 moles of H 2O.
 18.02 g H 2 O 
Mass of H2O eliminated =  69 mol H 2 O  
 = 1243.38 g
 1 mol H 2 O 
Minimum M of peptide = 10,703.59 g – 1243.38 g = 9,460 g/mol