Chemistry 12th Edition Chang Solutions Manual

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CHAPTER 2

ATOMS, MOLECULES, AND IONS
Problem Categories
Biological: 2.79, 2.80.
Conceptual: 2.31, 2.32, 2.33, 2.34, 2.61, 2.65, 2.71, 2.72, 2.81, 2.82, 2.93, 2.103, 2.113.
Descriptive: 2.24, 2.25, 2.26, 2.49, 2.50, 2.66, 2.70, 2.76, 2.84, 2.85, 2.86, 2.87, 2.88, 2.90, 2.91, 2.92, 2.94, 2.98,
2.101, 2.114.
Environmental: 2.122.
Organic: 2.47, 2.48, 2.69, 2.105, 2.107, 2.108, 2.115.
Difficulty Level
Easy: 2.7, 2.8, 2.13, 2.14, 2.15, 2.16, 2.23, 2.31, 2.32, 2.33, 2.43, 2.44, 2.45, 2.46, 2.47, 2.48, 2.62, 2.91, 2.92, 2.99,
2.100.
Medium: 2.17, 2.18, 2.24. 2.26, 2.34, 2.35, 2.36, 2.49, 2.50, 2.57, 2.58, 2.59, 2.60, 2.61, 2.63, 2.64, 2.65, 2.66, 2.67,
2.68, 2.69, 2.70, 2.71, 2.72, 2.73, 2.74, 2.75, 2.76, 2.79, 2.80, 2.81, 2.82, 2.83, 2.84, 2.85, 2.86, 2.88, 2.89, 2.90, 2.93,
2.94, 2.95, 2.96, 2.98, 2.101, 2.102, 2.103, 2.110, 2.112, 2.114, 2.115.
Difficult: 2.25, 2.77, 2.78, 2.87, 2.97, 2.104, 2.105, 2.106, 2.107, 2.108, 2.109, 2.111, 2.113.
2.7 First, convert 1 cm to picometers.
10
12
0.01 m 1 pm
1 cm 1 10 pm
1 cm 1 10 m 
   

10

(1 10 pm)
  ? He atoms 2.8 Note that you are given information to set up the unit factor relating meters and miles.
4 4
nucleus
1 m 1 mi
10 10 2.0 cm
100 cm 1609 m
ratom       r 0.12 mi
2.13 For iron, the atomic number Z is 26. Therefore the mass number A is:
A  26  28  54
2.14 Strategy: The 239 in Pu-239 is the mass number. The mass number (A) is the total number of neutrons
and protons present in the nucleus of an atom of an element. You can look up the atomic number (number of
protons) on the periodic table.
Solution:
mass number  number of protons  number of neutrons
number of neutrons  mass number  number of protons  239  94  145
2.15 Isotope 23He 4 2He 24 12Mg 12 25Mg 48 22Ti 79 35Br 195 78Pt
No. Protons 2 2 12 12 22 35 78
No. Neutrons 1 2 12 13 26 44 117
CHAPTER 2: ATOMS, MOLECULES, AND IONS 29

AZ
X
where,
A  mass number
Z  atomic number
(a) 186 74W (b) 201 80 Hg

(b) Metallic character decreases from the left side of the table (where the metals are located) to the right
side of the table (where the nonmetals are located).

(b) This is a polyatomic molecule that is a compound.
(c) This is a diatomic molecule that is a compound.

(c) This is a polyatomic molecule that is the elemental form of the substance. It is not a compound.

(d) H2O and C12H22O11 (sucrose)
30 CHAPTER 2: ATOMS, MOLECULES, AND IONS

2.43 (a) Sodium ion has a 1 charge and oxide has a 2 charge. The correct formula is Na2O.
(b) The iron ion has a 2 charge and sulfide has a 2 charge. The correct formula is FeS.
(c) The correct formula is Co2(SO4)3
(d) Barium ion has a 2 charge and fluoride has a 1 charge. The correct formula is BaF2.
2.44 (a) The copper ion has a 1 charge and bromide has a 1 charge. The correct formula is CuBr.
(b) The manganese ion has a 3 charge and oxide has a 2 charge. The correct formula is Mn2O3.
(c) We have the Hg22 ion and iodide (I). The correct formula is Hg2I2.
(d) Magnesium ion has a 2 charge and phosphate has a 3 charge. The correct formula is Mg3(PO4)2.

their atoms. Can you divide the subscripts in the formula by some factor to end up with smaller wholenumber subscripts?
Solution:
(a) Dividing both subscripts by 2, the simplest whole number ratio of the atoms in Al2Br6 is AlBr3.
(b) Dividing all subscripts by 2, the simplest whole number ratio of the atoms in Na2S2O4 is NaSO2.
(c) The molecular formula as written, N2O5, contains the simplest whole number ratio of the atoms present.
In this case, the molecular formula and the empirical formula are the same.
(d) The molecular formula as written, K2Cr2O7, contains the simplest whole number ratio of the atoms
present. In this case, the molecular formula and the empirical formula are the same.
2.47 The molecular formula of glycine is C2H5NO2.

CHAPTER 2: ATOMS, MOLECULES, AND IONS 31
2.57

2.58 Strategy: When naming ionic compounds, our reference for the names of cations and anions is Table 2.3 of
the text. Keep in mind that if a metal can form cations of different charges, we need to use the Stock system.
In the Stock system, Roman numerals are used to specify the charge of the cation. The metals that have only
one charge in ionic compounds are the alkali metals (1), the alkaline earth metals (2), Ag, Zn2, Cd2, and
Al3.
When naming acids, binary acids are named differently than oxoacids. For binary acids, the name is based on
the nonmetal. For oxoacids, the name is based on the polyatomic anion. For more detail, see Section 2.7 of
the text.
Solution:
(a) This is an ionic compound in which the metal cation (K) has only one charge. The correct name is
potassium hypochlorite. Hypochlorite is a polyatomic ion with one less O atom than the chlorite ion,
ClO2.
(b) silver carbonate
(c) This is an ionic compound in which the metal can form more than one cation. Use a Roman numeral to
specify the charge of the Fe ion. Since the chloride ion has a 1 charge, the Fe ion has a 2 charge.
The correct name is iron(II) chloride.
(d) potassium permanganate (e) cesium chlorate (f) hypoiodous acid
(g) This is an ionic compound in which the metal can form more than one cation. Use a Roman numeral to
specify the charge of the Fe ion. Since the oxide ion has a 2 charge, the Fe ion has a 2 charge. The
correct name is iron(II) oxide.
(h) iron(III) oxide

(m) This is an ionic compound in which the metal cation (Na) has only one charge. The O22 ion is called
the peroxide ion. Each oxygen has a 1 charge. You can determine that each oxygen only has a 1
charge, because each of the two Na ions has a 1 charge. Compare this to sodium oxide in part (l). The
correct name is sodium peroxide.
(n) iron(III) chloride hexahydrate
2.59

2.60 Strategy: When writing formulas of molecular compounds, the prefixes specify the number of each type of
atom in the compound.
When writing formulas of ionic compounds, the subscript of the cation is numerically equal to the charge of
the anion, and the subscript of the anion is numerically equal to the charge on the cation. If the charges of the
cation and anion are numerically equal, then no subscripts are necessary. Charges of common cations and
32 CHAPTER 2: ATOMS, MOLECULES, AND IONS
anions are listed in Table 2.3 of the text. Keep in mind that Roman numerals specify the charge of the cation,
not the number of metal atoms. Remember that a Roman numeral is not needed for some metal cations,
because the charge is known. These metals are the alkali metals (1), the alkaline earth metals (2), Ag,
Zn2, Cd2, and Al3.
When writing formulas of oxoacids, you must know the names and formulas of polyatomic anions (see Table
2.3 of the text).
Solution:
(a) The Roman numeral I tells you that the Cu cation has a 1 charge. Cyanide has a 1 charge. Since, the
charges are numerically equal, no subscripts are necessary in the formula. The correct formula is
CuCN.
(b) Strontium is an alkaline earth metal. It only forms a 2 cation. The polyatomic ion chlorite, ClO2, has
a 1 charge. Since the charges on the cation and anion are numerically different, the subscript of the
cation is numerically equal to the charge on the anion, and the subscript of the anion is numerically
equal to the charge on the cation. The correct formula is Sr(ClO2)2.
(c) Perbromic tells you that the anion of this oxoacid is perbromate, BrO4. The correct formula is
HBrO4(aq). Remember that (aq) means that the substance is dissolved in water.
(d) Hydroiodic tells you that the anion of this binary acid is iodide, I. The correct formula is HI(aq).
(e) Na is an alkali metal. It only forms a 1 cation. The polyatomic ion ammonium, NH4, has a 1 charge
and the polyatomic ion phosphate, PO43, has a 3 charge. To balance the charge, you need 2 Na
cations. The correct formula is Na2(NH4)PO4.
(f) The Roman numeral II tells you that the Pb cation has a 2 charge. The polyatomic ion carbonate,
CO32, has a 2 charge. Since, the charges are numerically equal, no subscripts are necessary in the
formula. The correct formula is PbCO3.
(g) The Roman numeral II tells you that the Sn cation has a 2 charge. Fluoride has a 1 charge. Since the
charges on the cation and anion are numerically different, the subscript of the cation is numerically
equal to the charge on the anion, and the subscript of the anion is numerically equal to the charge on the
cation. The correct formula is SnF2.
(h) This is a molecular compound. The Greek prefixes tell you the number of each type of atom in the
molecule. The correct formula is P4S10.
(i) The Roman numeral II tells you that the Hg cation has a 2 charge. Oxide has a 2 charge. Since, the
charges are numerically equal, no subscripts are necessary in the formula. The correct formula is HgO.
(j) The Roman numeral I tells you that the Hg cation has a 1 charge. However, this cation exists as
Hg22. Iodide has a 1 charge. You need two iodide ion to balance the 2 charge of Hg22. The
correct formula is Hg2I2.
(k) This is a molecular compound. The Greek prefixes tell you the number of each type of atom in the
molecule. The correct formula is SeF6.
2.61 Let’s compare the ratio of the fluorine masses in the two compounds.
3.55 g F
1.50
2.37 g F

This calculation indicates that there is 1.5 times more fluorine by mass in SF6 compared to the other
compound. The value of n is:
6
1.5
n   4
This is consistent with the Law of Multiple Proportions which states that if two elements combine to form
more than one compound, the masses of one element that combine with a fixed mass of the other element are
in ratios of small whole numbers. In this case, the ratio of the masses of fluorine in the two compounds is 6:4
or 3:2.
CHAPTER 2: ATOMS, MOLECULES, AND IONS 33

2.64 (a) 52 25Mn (b) 22 10 Ne (c) 107 47Ag (d) 127 53I (e) 239 94Pu
2.65 Uranium is radioactive. It loses mass because it constantly emits alpha () particles.

three, seven, and ten sulfur atoms, respectively. If the atom ratios are in small whole number ratios, then the
mass ratios must also be in small whole number ratios.

34 CHAPTER 2: ATOMS, MOLECULES, AND IONS
2.76 (a) Cu (b) P (c) Kr (d) Cs (e) Al (f) Sb (g) Cl (h) Sr
2.77 When an anion is formed from an atom, you have the same number of protons attracting more electrons. The
electrostatic attraction is weaker, which allows the electrons on average to move farther from the nucleus. An
anion is larger than the atom from which it is derived. When a cation is formed from an atom, you have the
same number of protons attracting fewer electrons. The electrostatic attraction is stronger, meaning that on
average, the electrons are pulled closer to the nucleus. A cation is smaller than the atom from which it is
derived.
2.78 (a) Rutherford’s experiment is described in detail in Section 2.2 of the text. From the average magnitude of
scattering, Rutherford estimated the number of protons (based on electrostatic interactions) in the
nucleus.
(b) Assuming that the nucleus is spherical, the volume of the nucleus is:
4 4 3 13 3 37 3
(3.04 10 cm) 1.177 10 cm

 
V r       



  

m V
To calculate the density of the space occupied by the electrons, we need both the mass of 11 electrons,
and the volume occupied by these electrons.
The mass of 11 electrons is:
28
9.1095 10 g 26
11 electrons 1.00205 10 g
1 electron

 
  
The volume occupied by the electrons will be the difference between the volume of the atom and the
volume of the nucleus. The volume of the nucleus was calculated above. The volume of the atom is
calculated as follows:
12
8
2
1 10 m 1 cm
186 pm 1.86 10 cm
1 pm 1 10 m




   

3 8 3 23 3
atom
4 4
(1.86 10 cm) 2.695 10 cm
3 3
 
V r       
Velectrons  Vatom  Vnucleus  (2.695  1023 cm3)  (1.177  1037 cm3)  2.695  1023 cm3
As you can see, the volume occupied by the nucleus is insignificant compared to the space occupied by
the electrons.
The density of the space occupied by the electrons can now be calculated.




  

4 3
m 3.72 10 g/cm
V
The above results do support Rutherford’s model. Comparing the space occupied by the electrons to the
volume of the nucleus, it is clear that most of the atom is empty space. Rutherford also proposed that
the nucleus was a dense central core with most of the mass of the atom concentrated in it. Comparing
CHAPTER 2: ATOMS, MOLECULES, AND IONS 35
the density of the nucleus with the density of the space occupied by the electrons also supports
Rutherford’s model.
2.79 The molecular formula of caffeine is C8H10N4O2. The empirical formula is C4H5N2O.

(b) The charge on the copper ion is +1. The correct name is copper(I) oxide.
(c) The charge on the cobalt cation needs to be specified. The correct name is cobalt(II) nitrate.
(d) Cr O 2 72 is the dichromate ion. The correct name is sodium dichromate.

(c) I, halogens have a 1 charge in ionic compounds
(d) N3
(e) Al3, aluminum always has a 3 charge in ionic compounds
(f) Cs, alkali metals always have a 1 charge in ionic compounds
(g) Mg2, alkaline earth metals always have a 2 charge in ionic compounds.

acid; HI, hydroiodic acid. Oxoacids containing Group 7A elements (using the specific examples for chlorine)
are: HClO4, perchloric acid; HClO3, chloric acid; HClO2, chlorous acid: HClO, hypochlorous acid.
Examples of oxoacids containing other Group A-block elements are: H3BO3, boric acid (Group 3A); H2CO3,
carbonic acid (Group 4A); HNO3, nitric acid and H3PO4, phosphoric acid (Group 5A); and H2SO4, sulfuric
acid (Group 6A). Hydrosulfuric acid, H2S, is an example of a binary Group 6A acid while HCN,
hydrocyanic acid, contains both a Group 4A and 5A element.
36 CHAPTER 2: ATOMS, MOLECULES, AND IONS
2.88 Mercury (Hg) and bromine (Br2)

The neutron/proton ratio increases with increasing atomic number.
2.90 H2, N2, O2, F2, Cl2, He, Ne, Ar, Kr, Xe, Rn

2.92 They do not have a strong tendency to form compounds. Helium, neon, and argon are chemically inert.

(b) Einsteinium (Albert Einstein); Fermium (Enrico Fermi); Curium (Marie and Pierre Curie);
Mendelevium (Dmitri Mendeleev); Lawrencium (Ernest Lawrence).
(c) Arsenic, Cesium, Chlorine, Chromium, Iodine.

hydrogen (or deuterium) is 1:1 in both compounds. This does not violate the law of definite proportions.
When the law of definite proportions was formulated, scientists did not know of the existence of isotopes.

2.100 All of these are molecular compounds. We use prefixes to express the number of each atom in the molecule.
The names are nitrogen trifluoride (NF3), phosphorus pentabromide (PBr5), and sulfur dichloride (SCl2).
CHAPTER 2: ATOMS, MOLECULES, AND IONS 37
2.101
The metalloids are shown in gray.
2.102

2.103 (a) CO2(s), solid carbon dioxide

(b) NaCl, sodium chloride 2.104 The change in energy is equal to the energy released. We call this E. Similarly, m is the change in mass.
Because
 2
E
m
c
, we have
3

1000 J
(1.715 10 kJ)
 

     

We can add together the masses of hydrogen and oxygen to calculate the mass of water that should be
formed.
12.096 g  96.000  108.096 g
The predicted change (loss) in mass is only 1.91 × 108 g which is too small a quantity to measure. Therefore,
for all practical purposes, the law of conservation of mass is assumed to hold for ordinary chemical processes.
alkali
metals
alkaline
earth metals
2A 3A 4A 5A 6A
1A
7A
8A
halogens noble gases
38 CHAPTER 2: ATOMS, MOLECULES, AND IONS
2.105 CH4, C2H6, and C3H8 each only have one structural formula.
H C
H
H
H C
H
H
H
H C H
H C
H
H
H
H C H
H C
H
H
C4H10 has two structural formulas.
H C
H
H
H C H
H C
C
H
H
H H
C
H
H
H
C H
C
H
H
H
H C
H
H
C5H12 has three structural formulas.
H C
H
H
H C H
H C
C
H
C
H
H H
H
H
C
H
H C
H
C
H
C
H C
H
H
H
H
H H H
C
H
HH
C C
C
H
H
H
H C
H
H
H
H
H
2.106 (a) The volume of a sphere is
4 3
3
V r  
Volume is proportional to the number of nucleons. Therefore,
V  A (mass number)
r
3
 A
r  A1/3
(b) Using the equation given in the problem, we can first solve for the radius of the lithium nucleus and
then solve for its volume.
r  r0A1/3
r  (1.2 × 1015 m)(7)1/3
r  2.3 × 1015 m
4 3
3
V r  
4 15 3
(2.3 10 m)
3

Vnucleus     5.1 10 m  44 3
CHAPTER 2: ATOMS, MOLECULES, AND IONS 39
(c) In part (b), the volume of the nucleus was calculated. Using the radius of a Li atom, the volume of a Li
atom can be calculated.
3 12 3 29 3
atom
4 4
(152 10 m) 1.47 10 m
3 3
 
V r       
The fraction of the atom’s volume occupied by the nucleus is:
44 3
nucleus
29 3
atom
5.1 10 m
1.47 10 m
 

 

15
V 3.5 10
V


Yes, this calculation shows that the volume of the nucleus is much, much smaller than the volume of the
atom, which supports Rutherford’s model of an atom.
2.107 Two different structural formulas for the molecular formula C2H6O are:
H C
H
H
C O H
H H
H C
H
H
O C
H H
H
In the second hypothesis of Dalton’s Atomic Theory, he states that in any compound, the ratio of the number
of atoms of any two of the elements present is either an integer or simple fraction. In the above two
compounds, the ratio of atoms is the same. This does not necessarily contradict Dalton’s hypothesis, but
Dalton was not aware of chemical bond formation and structural formulas.
2.108

 
We can calculate the ratio of H in the two compounds.
1.14 g
3
0.383 g

This is consistent with the Law of Multiple Proportions which states that if two elements combine to
form more than one compound, the masses of one element that combine with a fixed mass of the other
element are in ratios of small whole numbers. In this case, the ratio of the masses of hydrogen in the
two compounds is 3:1.
(b) For a given amount of carbon, there is 3 times the amount of hydrogen in ethane compared to acetylene.
Reasonable formulas would be:

40 CHAPTER 2: ATOMS, MOLECULES, AND IONS

cm
3
 grams  atoms

22
6.6 10 Pt atoms 
(b) Since 74 percent of the available space is taken up by Pt atoms, 6.6  1022 atoms occupy the following
volume:
0.74  1.0 cm3  0.74 cm3
We are trying to calculate the radius of a single Pt atom, so we need the volume occupied by a single Pt
atom.
3

The volume of a sphere is 4 3
3
r . Solving for the radius:
23 3 3 4
1.12 10 cm
3

V r    
r
3
 2.67  1024 cm3
r  1.4  108 cm
Converting to picometers:
8
12
0.01 m 1 pm
(1.4 10 cm)
1 cm 1 10 m


    

2
radius Pt atom 1.4 10  pm
2.110 The mass number is the sum of the number of protons and neutrons in the nucleus.
Mass number  number of protons  number of neutrons
Let the atomic number (number of protons) equal A. The number of neutrons will be 1.2A. Plug into the
above equation and solve for A.
55  A  1.2A
A  25
The element with atomic number 25 is manganese, Mn.
2.111
2.112 The acids, from left to right, are chloric acid, nitrous acid, hydrocyanic acid, and sulfuric acid.
S N
B I
CHAPTER 2: ATOMS, MOLECULES, AND IONS 41
2.113 From the equation derived in Problem 2.106, we can first solve for the radius of the iron nucleus and then
solve for its volume.
r  r0A1/3
r = (1.2 × 1015 m)(56)1/3
r = 4.6 × 1015 m
4 3
3
V r  
15 3 43 3 37 3
nucleus
4
(4.6 10 m) 4.1 10 m 4.1 10 cm
3
V          
The density of an iron-56 nucleus is:
23
37 3
mass 9.229 10 g
volume 4.1 10 cm



  

14
density 2.3 10  g/cm3
The density of an iron nucleus is 230 trillion grams per 1 cm3. Because of this very high density, neutrons
are needed to stabilize the nucleus, keeping the protons from repelling each other. It has been postulated that
the exterior of the neutron is negatively charged and the interior is positively charged. The negative exteriors
of the neutrons attract the protons holding the particles together in this extremely dense arrangement.
2.114 The formula of the ionic compound is XY2. Element X is most likely in Group 4B and element Y is most
likely in Group 6A. A possible compound is TiO2, titanium(IV) oxide. Other choices are elements in
Group 4A: SnO2 [tin(IV) oxide] and PbO2 [lead(IV) oxide].
2.115 Let’s compare the ratio of the hydrogen masses in the three compounds. To do this, we need to start with the
same mass of carbon. If we were to start with 0.749 g of C in ethane, how much hydrogen would combine
with 0.749 g of carbon? If we were to start with 0.749 g of C in propane, how much hydrogen would
combine with 0.749 g of carbon?
0.749 g C
ethane : 0.201 g H 0.188 g H
0.799 g C
0.749 g C
propane : 0.183 g H 0.168 g H
0.817 g C
 
 
We can calculate the ratio of H between methane and propane, and ethane and propane.



This is consistent with the Law of Multiple Proportions which states that if two elements combine to
form more than one compound, the masses of one element (in this case hydrogen) that combine with a
fixed mass of the other element (in this case carbon) are in ratios of small whole numbers. The ratio of
the masses of hydrogen in the three compounds, propane, ethane, and methane is 1:1.12:1.49 or 8:9:12.
42 CHAPTER 2: ATOMS, MOLECULES, AND IONS
2.116 Basic approach:
 Determine the charge of an alpha particle (helium nucleus) and gold nucleus in coulombs.
 Use Coulomb’s law to solve for the distance at which the electrical potential energy is equal to the initial
kinetic energy.
When the α particle comes to a stop, its kinetic energy is converted to potential energy. The charge of a proton is
+1.6022 × 10-19 C. An α particle which contains 2 protons and 2 neutrons has a charge, Q1, of 3.2044 × 10-19 C.
A gold nucleus which contains 79 protons has a charge, Q2, of 1.2657 × 10-17 C. We solve for r, the distance of
separation in meters between the alpha particle and the gold nucleus.
3
9 19 17

kg m
9.0 10 (3.2044 10 C)(1.2657 10 C)
s
 
  
 
2.117 Basic approach:
 Consider that almost all of the volume in an atom or monatomic ion is occupied by the electrons.
 Determine the number of electrons in each species, and rank their size according the number of electrons.
The volume of each species depends on the outer boundary of the electrons. The number of protons remains
the same in all three cases. Li– has the most electrons and therefore the greatest repulsion among them,
leading to a larger outer boundary compared to the Li atom, which has one fewer electron. Li+ has the fewest
number of electrons and the smallest outer boundary. Thus, the size decreases as follows:
Li– > Li > Li+
In general for a given element, the anion is the largest, the atom is second, and the cation is the smallest. In
Chapter 8 we will consider a more sophisticated way of predicting the relative sizes of atoms and monatomic
ions based on electron configurations and nuclear charge.
2.118 Isotopes of the same element have the same number of electrons, hence following the reasoning in Problem
2.117, they would have the same size. The number of neutrons affects the size of the nucleus, but not the size
of the atom.

 Calculate the number of silver atoms required to reach the minimum length visible to the human eye, and
then calculate the number of silver atoms required to give a surface area that would be visible.
A Web search shows that the atomic radius of an Ag atom is 144 pm or 1.44  10–8 cm. The number of Ag
atoms that must be lined up along one dimension is given by the minimum length humans can see divided by
the diameter of a silver atom
–5


But to actually be seen by the human eye, a grouping of silver atoms would need to have both length and
width. Assuming a square array of 700 Ag atoms by 700 Ag atoms
CHAPTER 2: ATOMS, MOLECULES, AND IONS 43
the number of silver atoms required to be visible would be 700  700 ≈ 500,000 Ag atoms.
2.120 Basic approach:
 Estimate the radius of a pea.
 Look up the radius of a typical atom and its nucleus. The radius of the atom provides a range of possible
distances of the electron from the nucleus.
 Use the ratio of the radii for the pea and the atom to calculate the distance of the electron from the nucleus
if the nucleus were the size of a pea.
The diameter of a pea is about 0.5 cm. Radius of nucleus is about 5  10–13 cm. Therefore, the expansion
factor for the radius of the nucleus is
11
–13
0.5 cm / 2
5 10 times
5 10 cm
 

A typical atom has a radius of about 100 pm (10–8 cm). Using that value as the distance of the electron from
the nucleus in an atom, then if the atom was scaled to the size of a pea, the electron would be (1 10–8 cm) 
(5  1011) = 5  103 cm or 50 m (roughly half a football field) from the nucleus!
2.121 Basic approach:
 Consider the relative composition of sodium and potassium ions in biological systems (for starters see the
Chemistry in Action on p. 49, and then look up the relative compositions in other biological systems).
Over long periods of time (on a geological scale), minerals containing potassium and sodium are slowly
decomposed by wind and rain, and their K+ and Na+ ions are converted to more soluble compounds.
Eventually, rain leaches these compounds out of the soil and carries them to the sea. Plants take up many of
the K+ ions along the way, while the Na+ ions are free to move on to the sea because they are not needed for
biological functions by the plants.
2.122 Basic approach:
 Estimate the diameter of typical carbon microsphere and calculate the volume.
 Estimate for the density of carbon in the microsphere.
 Look up the most common isotope of carbon and calculate the mass of that isotope.
 Calculate the number of atoms in the microsphere.
700 Ag atoms
700 Ag atoms
44 CHAPTER 2: ATOMS, MOLECULES, AND IONS
A typical microsphere appears to be about 3 m in diameter, giving a volume of
3 3 2 3
11 3
6
4 4 3 m 10 cm
1 10 cm
3 2 3 2 10 m
d 
 


     
         
     
It is reasonable to expect that the density of carbon in the microspheres would be somewhere between that of
graphite (2.2 g/cm3) and diamond (3.5 g/cm3), so we assume a density of 3 g/cm3 for the carbon in the
microsphere. The mass of typical carbon microsphere would be
11 3 11
3
3 g C
1 10 cm 3 10 g C
cm
 
   
While this may seem like an exceedingly small mass, it still represents a very large number of carbon atoms
which are much smaller and far less massive than the microspheres. In order to determine the number of
atoms in a typical microsphere, we need to approximate the mass of an individual carbon atom. An online
search tells us that the most common isotope of carbon is 12 6C . Taking this isotope as a typical carbon atom
and using the data in Table 2.1, we can calculate the mass of one atom by summing the masses of six
electrons, six protons, and six neutrons.
28 24
24 23
6 electron(9.10938 10 g/electron) 6 proton(1.67262 10 g/proton)
6 neutron(1.67493 10 g/neutron) 2 10 g
 
 
  
   
Therefore, the number of carbon atoms in a typical microsphere would be
11 12
23
1 C atom
3 10 g C 2 10 C atoms
2 10 g C


   

This calculation gives us a feel for the incredibly small size of an atom. It should be noted that the mass of the
carbon atom used in the calculation is only an approximation. Besides the assumption that all carbon atoms
are carbon-12 (only about 99% of carbon atoms are carbon-12), we will see in Chapter 19 that the mass of an
atom is actually somewhat less than the sum of the masses of the individual particles. In Chapter 3 we will
introduce “atomic mass” which is the average mass of the naturally occurring isotopes of a particular
element.
ANSWERS TO REVIEW OF CONCEPTS