© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 2.1 Atomic Mass Units: Gram-to-Atom Conversions We know the sample mass in grams and the mass of one atom in atomic mass units. To find the number of atoms in the sample, two conversions are needed, the first between grams and atomic mass units and the second between atomic mass units and the number of atoms. The conversion factor between atomic mass units and grams is Analysis How many atoms are in a small piece of aluminum foil with a mass of g? The mass of an atom of aluminum is 27.0 amu. STEP 1: Identify known information. STEP 2: Identify unknown answer and units. STEP 3: Identify needed conversion factors. Knowing the mass of foil (in g) and the mass of individual atoms (in amu) we need to convert from atoms/amu to atoms/g. STEP 4: Solve. Set up an equation using known information and conversion factors so that unwanted units cancel. Solution Ballpark Estimate An atom of aluminum has a mass of 27.0 amu; since 1 amu ~10 24 g, the mass of a single aluminum atom is very small (≈ 10 –23 g). A very large number of atoms, therefore, (10 22 ?) is needed to obtain a mass of g. Ballpark Check Our estimate was 10 22, which is within a factor of 10.

© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 2.2 Atomic Structure: Protons, Neutrons, and Electrons The atomic number gives the number of protons, which is the same as the number of electrons, and the mass number gives the total number of protons plus neutrons. Analysis Phosphorus has the atomic number Z = 15. How many protons, electrons, and neutrons are there in phosphorus atoms, which have mass number A = 31? Solution Phosphorus atoms, with Z = 15, have 15 protons and 15 electrons. To find the number of neutrons, subtract the atomic number from the mass number:

© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 2.3 Atomic Structure: Atomic Number and Atomic Mass The number of protons and the number of electrons are the same and are equal to the atomic number Z, 28 in this case. Subtracting the number of protons (28) from the total number of protons plus neutrons (60) gives the number of neutrons. Analysis An atom contains 28 protons and has A = 60. Give the number of electrons and neutrons in the atom, and identify the element. Solution The atom has 28 electrons and = 32 neutrons. The list of elements inside the front cover shows that the element with atomic number 28 is nickel (Ni).

© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 2.4 Average Atomic Mass: Weighted-Average Calculation We can calculate the average atomic mass for the element by summing up the contributions from each of the naturally occurring isotopes. Analysis Gallium is a metal with a very low melting point—it will melt in the palm of your hand. It has two naturally occurring isotopes: 60.4% is Ga-69 (mass = amu), and 39.6% is Ga-71 (mass = amu). Calculate the atomic weight for gallium. STEP 1: Identify known information. STEP 2: Identify unknown answer and units. STEP 3: Identify needed conversion factors. This equation calculates the average atomic weight as a weighted average of all naturally occurring isotopes. STEP 4: Solve. Substitute known information and solve. Solution Ballpark Estimate The masses of the two naturally occurring isotopes of gallium differ by 2 amu (68.9 and 70.9 amu). Since slightly more than half of the Ga atoms are the lighter isotope (Ga-69), the average mass will be slightly less than halfway between the two isotopic masses; estimate = 69.8 amu. Ballpark Check Our estimate (69.8 amu) is close!

© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 2.5 Identifying Isotopes from Atomic Mass and Atomic Number The identity of the atom corresponds to the atomic number—78. Analysis Identify element X in the symbol, and give its atomic number, mass number, number of protons, number of electrons, and number of neutrons. Solution Element X has Z = 78, which shows that it is platinum. (Look inside the front cover for the list of elements.) The isotope has a mass number of 194, and we can subtract the atomic number from the mass number to get the number of neutrons. This platinum isotope therefore has 78 protons, 78 electrons, and 194 – 78 = 116 neutrons.

© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 2.6 Atomic Structure: Electron Shells The number of electrons in the atom is calculated by adding the total electrons in each shell. We can identify the element from the number of protons in the nucleus, which is equal to the number of electrons in the atom. Analysis How many electrons are present in an atom that has its first and second shells filled and has 4 electrons in its third shell? Name the element. Solution The first shell of an atom holds 2 electrons in its 1s orbital, and the second shell holds 8 electrons (2 in a 2s orbital and 6 in three 2p orbitals). Thus, the atom has a total of = 14 electrons and must be silicon (Si).

© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 2.7 Atomic Structure: Electron Configurations Magnesium, Z = 12, has 12 electrons to be placed in specific orbitals. Assignments are made by putting 2 electrons in each orbital, according to the order shown in Figure 2.6. Analysis Show how the electron configuration of magnesium can be assigned. Solution Magnesium has the configuration.

© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 2.8 Electron Configurations: Orbital-Filling Diagrams Phosphorus has 15 electrons, which occupy orbitals according to the order shown in Figure 2.6. Analysis Write the electron configuration of phosphorus, Z = 15, using up and down arrows to show how the electrons in each orbital are paired. Solution

© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 2.9 Electron Configurations: Valence Electrons Locate the row and the block in which each of the elements is found in Figure 2.7. The location can be used to determine the complete electron configuration and to identify the valence electrons. Analysis Write the electron configuration for the following elements, using both the complete and the shorthand notations. Indicate which electrons are the valence electrons. (a) Na (b) Cl (c) Zr Solution (a) Na (sodium) is located in the third row and in the first column of the s-block. Therefore, all orbitals up to the 3 s are completely filled, and there is one electron in the 3s orbital. or (valence electrons are underlined) (b) Cl (chlorine) is located in the third row and in the fifth column of the p-block. or (c) Zr (zirconium) is located in the fifth row and in the second column of the d-block. All orbitals up to the 4d are completely filled, and there are 2 electrons in the 4d orbitals. Note that the 4d orbitals are filled after the 5s orbitals in both Figures 2.6 and 2.7. or

© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 2.10 Electron Configurations: Valence-Shell Configurations The elements in group 6A have 6 valence electrons. In each element, the first two of these electrons are in the valence s subshell, giving, and the next four electrons are in the valence p subshell, giving. Analysis Using n to represent the number of the valence shell, write a general valence-shell configuration for the elements in group 6A. Solution For group 6A, the general valence-shell configuration is.

© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 2.11 Electron Configurations: Inner Shells versus Valence Shell The total number of electrons will be the same as the atomic number for tin (Z = 50). The number of valence electrons will equal the number of electrons in the valence shell. Analysis How many electrons are in a tin atom? Give the number of electrons in each shell. How many valence electrons are there in a tin atom? Write the valence-shell configuration for tin. Solution Checking the periodic table shows that tin has atomic number 50 and is in group 4A. The number of electrons in each shell is As expected from the group number, tin has 4 valence electrons. They are in the 5s and 5p subshells and have the configuration.

© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 2.12 Electron Configurations: Electron-Dot Symbols The group number, 5A, indicates 5 valence electrons. The first four are distributed singly around the four sides of the element symbol, and any additional are placed to form electron pairs. Analysis Write the electron-dot symbol for any element X in group 5A. Solution (5 electrons)