1. Atoms, Molecules, and Ions
Chapter Two
Atoms, Molecules, and Ions

2. Laws of Chemical Combination
Law of Conservation of Mass
The total mass remains constant during a chemical reaction.
Law of Definite Proportions
All samples of a compound have the same composition, or …
All samples have the same proportions, by mass, of the elements present.

3. Example 2.1 A Conceptual Example
Jan Baptista van Helmont (1579–1644) first measured the mass of a young willow tree and, separately, the mass of a bucket of soil and then planted the tree in the bucket. After five years, he found that the tree had gained 75 kg in mass even though the soil had lost only kg. He had added only water to the bucket, and so he concluded that all the mass gained by the tree had come from the water. Explain and criticize his conclusion.

4. Example 2.1 A Conceptual Example
Jan Baptista van Helmont (1579–1644) first measured the mass of a young willow tree and, separately, the mass of a bucket of soil and then planted the tree in the bucket. After five years, he found that the tree had gained 75 kg in mass even though the soil had lost only kg. He had added only water to the bucket, and so he concluded that all the mass gained by the tree had come from the water. Explain and criticize his conclusion.
Analysis and Conclusions
Van Helmont’s explanation anticipated the law of conservation of mass by dismissing the possibility that the increased mass of the tree could have been created from nothing. His main failure, however, was in not identifying all the substances involved. He did not know about the role of carbon dioxide gas in the growth of plants. In applying the law of conservation of mass, we must focus on all the substances involved in a chemical reaction.
A sealed photographic flashbulb containing magnesium and oxygen has a mass of
45.07 g. On firing, a brilliant flash of white light is emitted and a white powder is formed inside the bulb. What should the mass of the bulb be after firing? Explain.
Exercise 2.1A
Is the mass of the burned match pictured in the photograph the same as, less than, or more than the mass of the burning match? Explain your answer.
Exercise 2.1B

5. The Law of Definite Proportions
… have the same composition.
Three different sources of a compound …

6. Example 2.2
The mass ratio of oxygen to magnesium in the compound magnesium oxide is :1. What mass of magnesium oxide will form when g of magnesium is completely converted to magnesium oxide by burning in pure oxygen gas?

7. Example 2.2 Strategy Solution
The mass ratio of oxygen to magnesium in the compound magnesium oxide is :1. What mass of magnesium oxide will form when g of magnesium is completely converted to magnesium oxide by burning in pure oxygen gas?
Strategy
We usually write mass ratios in a form such as “the ratio of O to Mg is :1.” The first number represents the mass of the first element named—in this case, a mass of oxygen, say g oxygen—and the second number represents the mass of the second element named—here a mass of magnesium. Although written only as “1,” we assume that the second number is as precisely known as the first, that is, g magnesium. According to the law of constant composition, the mass of oxygen that combines with the g of magnesium must be just the right amount that makes the mass ratio of oxygen to magnesium in the product :1. According to the law of conservation of mass, the mass of magnesium oxide must equal the sum of the masses of the magnesium and oxygen that react.
Solution
We can state the mass ratio in the form of a conversion factor and then determine the required mass of oxygen.
The mass of the sole product, magnesium oxide, equals the total of the masses of the substances entering into the reaction.

8. Example 2.2 continued Assessment Exercise 2.2A Exercise 2.2B
As a simple check, compare the calculated mass with the starting mass of magnesium. From the law of conservation of mass, the product of the reaction
of magnesium with oxygen must have a mass greater than the starting mass of magnesium; and g magnesium oxide is indeed greater than g magnesium.
What mass of magnesium oxide is formed when g of oxygen combines with magnesium?
Exercise 2.2A
When a strip of magnesium metal was burned in pure oxygen gas, g of oxygen was consumed and the only product formed was magnesium oxide. What must have been the masses of magnesium metal burned and magnesium oxide formed?
Exercise 2.2B

9. Law of Multiple Proportions
When two or more different compounds of the same two elements are compared, the masses of one element that combine with a fixed mass of the second element are in the ratio of small whole numbers.

10. Law of Multiple Proportions (cont’d)
Ratio of oxygen-to-carbon in CO2 is exactly twice the ratio in CO.

11. Law of Multiple Proportions (cont’d)
Four different oxides of nitrogen can be formed by combining 28 g of nitrogen with:
16 g oxygen, forming Compound I
48 g oxygen, forming Compound II
64 g oxygen, forming Compound III
80 g oxygen, forming Compound IV
What is the ratio 16:48:64:80 expressed as small whole numbers?
Compounds I–IV are N2O, N2O3, N2O4, N2O5

12. Dalton’s Atomic Theory
Proposed in 1803 to explain the law of conservation of mass, law of definite proportions, and law of multiple proportions.
Matter is composed of atoms: tiny, indivisible particles.
All atoms of a given element are the same.
Atoms of one element differ from atoms of other elements.
Compounds are formed when atoms of different elements unite in fixed proportions.
A chemical reaction involves rearrangement of atoms. No atoms are created, destroyed, or broken apart.

13. Dalton’s Atomic Theory: Conservation of Mass and Definite Proportions
… six fluorine atoms and four hydrogen atoms after reaction. Mass is conserved.
Six fluorine atoms and four hydrogen atoms before reaction …
HF always has one H atom and one F atom; always has the same proportions (1:19) by mass.

14. Subatomic Particles
Protons and neutrons are located at the center of an atom (at the nucleus).
Electrons are dispersed around the nucleus.

15. Isotopes
Atoms that have the same number of protons but different numbers of neutrons are called isotopes.
The atomic number (Z) is the number of protons in the nucleus of a given atom of a given element.
The mass number (A) is an integral number that is the sum of the numbers of protons and neutrons in an atom.
The number of neutrons = A – Z.

16. Isotopes (cont’d)
Atoms can be represented using the element’s symbol and the mass number (A) and atomic number (Z): 35 Cl 17 37 Cl 17 A E Z
How many protons are in chlorine-35?
How many protons are in chlorine-37?
How many neutrons are in chlorine-37?

17. Example 2.3
How many protons, neutrons, and electrons are present in a 81Br atom?

18. Example 2.3 Strategy Solution
How many protons, neutrons, and electrons are present in a 81Br atom?
Strategy
We can use the identity of the element and several simple relationships between the subatomic particles described above to determine the numbers of these particles.
Solution
The atomic number of bromine Z = 35 = number of protons
is not given here, but we can
obtain it from the list of elements
on the inside front cover.
In the bromine atom, the number Number of protons = number of electrons = 35
of positively charged protons
equals the number of negatively
charged electrons.
From mass number 81 and Number of neutrons = A – Z = 81 – 35 = 46
atomic number 35, we calculate
the number of neutrons, using
Equation (2.1).

19. Atomic Mass
Atoms are very tiny, so a tiny unit is needed to express the mass of an atom or molecule.
One atomic mass unit (u) = 1/12 the mass of a C-12 atom.
1 u = × 10–24 g
The mass of an atom is not exactly the sum of the masses of the protons + neutrons + electrons (we will see why in Chapter 19).

20. Atomic Mass (cont’d)
Question: do all isotopes of an element have the same mass? Why or why not?
The atomic mass given on the periodic table is the weighted average of the masses of the naturally occurring isotopes of that element.

21. Example 2.5 An Estimation Example
Use the data cited above to determine the weighted average atomic mass of carbon.
Example 2.5 An Estimation Example
Indium has two naturally occurring isotopes and a weighted average atomic mass of u. One of the isotopes has a mass of u. Which is likely to be the second isotope: 111In, 112In, 114In, or 115In?

22. Example 2.4 Strategy Solution Assessment Exercise 2.4A Exercise 2.4B
Use the data cited above to determine the weighted average atomic mass of carbon.
Strategy
The contribution each isotope makes to the weighted average atomic mass is given by Equation (2.2). The weighted average atomic mass is the sum of the two contributions.
Solution
The contributions are
Contribution of carbon-12 = x u = u
Contribution of carbon-13 = x u = u
The weighted average mass is
Atomic mass of carbon = u u = u
Assessment
This is the value listed in a table of atomic masses. As expected, the atomic mass of carbon is much closer to 12 u than to 13 u.
There are three naturally occurring isotopes of neon. Their percent abundances and atomic masses are neon-20, 90.51%, u; neon-21, 0.27%, u; neon-22, 9.22%, u. Calculate the weighted average atomic mass of neon.
Exercise 2.4A
The two naturally occurring isotopes of copper are copper-63, mass u, and copper-65, mass u. What must be the percent natural abundances of the two isotopes if the atomic mass of copper listed in a table of atomic masses is u?
Exercise 2.4B

23. Example 2.5 An Estimation Example
Indium has two naturally occurring isotopes and a weighted average atomic mass of u. One of the isotopes has a mass of u. Which is likely to be the second isotope: 111In, 112In, 114In, or 115In?
Analysis and Conclusions
The masses of isotopes differ only slightly from whole numbers, which tells us that the isotope with mass u is 113In. To account for the observed weighted average atomic mass of u, the second isotope must have a mass number greater than 114. It can be only 115In.
The masses of the three naturally occurring isotopes of magnesium are 24Mg,
u; 25Mg, u; 26Mg, u. Use the atomic mass given inside the front cover to determine which of the three is the most abundant. Can you determine which is the second most abundant? Explain.
Exercise 2.5A
For the three magnesium isotopes described in Exercise 2.5A, (a) could the percent natural abundance of 24Mg be 60.00%? (b) What is the smallest possible value for the percent natural abundance of 24Mg?
Exercise 2.5B

24. Mendeleev’s Periodic Table
Mendeleev arranged the known elements in order of increasing atomic weight from left to right and from top to bottom in groups.
Elements that closely resembled one another were arranged in the same vertical group.
Gaps were left where undiscovered elements should appear.
From the locations of the gaps, he was able to predict properties of some of the undiscovered elements.

25. Germanium: Prediction vs. Observation

26. The Modern Periodic Table
Except for H, elements left of the zigzag line are metals.
To the right of the line we find nonmetals, including the noble gases.
Some elements adjacent to the line are called metalloids.

27. Molecules and Formulas
A molecule is a group of two or more atoms held together by covalent bonds.
A molecular formula gives the number of each kind of atom in a molecule.
An empirical formula simply gives the (whole number) ratio of atoms of elements in a compound.
Compound
Molecular formula
Empirical formula
Hydrogen peroxide
H2O2 HO
Octane
C8H18
????

28. Structural Formulas and Models
Structural formulas and models show how atoms are attached to one another.
The condensed structural formula for acetic acid is CH3COOH.
C2H4O2: two C atoms, four H atoms, two O atoms.
CH3COOH shows how the atoms are arranged.
Ball-and-stick model Space-filling model

29. Nomenclature
… is the method for naming compounds and writing formulas for compounds.
We could have a specific name for each compound—but we would have to memorize each one!
Can you imagine having to memorize the names of half a million different inorganic compounds? Twenty million organic compounds??
Instead we have a systematic method— conventions and rules—for naming compounds and writing formulas.

30. Nomenclature of Binary Molecular Compounds
Binary compounds contain ___ elements.
Molecular compounds exist as ________.

31. Naming Binary Molecular Compounds
The name consists of two words.
First word: name of the element that appears first in the formula.
Second word: stem of the name of the second element, ending with -ide.
Names are further modified by adding prefixes to denote the numbers of atoms of each element in the molecule.

32. Which element is named first?
Begin with boron and follow the line to determine the order of naming.
Rule of thumb: the element that is farthest down and to the left on the periodic table is usually written first.
In a compound consisting of arsenic (As) and sulfur (S), which element is named first?

34. Example 2.6
Write the molecular formula and name of a compound for which each molecule contains six oxygen atoms and four phosphorus atoms.
Example 2.7
Write (a) the molecular formula of phosphorus pentachloride and (b) the name of S2F10.

35. Example 2.6 Strategy Solution Exercise 2.6A Exercise 2.6B
Write the molecular formula and name of a compound for which each molecule contains six oxygen atoms and four phosphorus atoms.
Strategy
We need to write the chemical symbols of the two elements and use the stated numbers of atoms as subscripts following the symbols. We then determine which element to place first in the molecular formula.
Solution
We represent the six atoms of oxygen as O6 and the four atoms of phosphorus as P4. According to the scheme in Figure 2.9, the element O is followed only by F. Phosphorus therefore comes first in the formula; we write P4O6.
The name of a compound with four (tetra-) P atoms and six (hexa-) oxygen atoms in its molecules is tetraphosphorus hexoxide.
Write the molecular formula and name of a compound for which each molecule contains four fluorine atoms and two nitrogen atoms.
Exercise 2.6A
Write the molecular formula and name of a compound the molecules of which each contain one oxygen atom and eight sulfur atoms.
Exercise 2.6B

36. Example 2.7 Solution Exercise 2.7A Exercise 2.7B
Write (a) the molecular formula of phosphorus pentachloride and (b) the name of S2F10.
Solution
(a) Choosing which element symbol goes first. The order of elements shown in the molecular formula must be the same as the order in the name.
Writing subscripts. The lack of a prefix on phosphorus signifies one P atom per molecule. The prefix penta- indicates five chlorine atoms. The molecular formula is PCl5.
(b) The subscripts indicate two (di-) sulfur atoms and ten (deca-) fluorine atoms. The compound is disulfur decafluoride.
Write (a) the molecular formula of tetraphosphorus decoxide and (b) the name of S7O2.
Exercise 2.7A
Write a plausible molecular formula for a compound that has one sulfur atom, two oxygen atoms, and two fluorine atoms in each of its molecules. Comment on any ambiguity that exists in this case.
Exercise 2.7B

37. Ions and Ionic Compounds
An atom that either gains or loses electron(s) is an ion.
There is no change in the number of protons or neutrons in the nucleus of the atom.
Cation – has a positive charge from loss of electron(s).
Anion – has a negative charge from gain of electron(s).

38. Ions and Ionic Compounds (cont’d)
In an ionic compound, oppositely charged ions are attracted to each other such that the compound has no net charge.
There are no distinct molecules of sodium chloride. I
Sodium chloride simply consists of sodium ions and chloride ions, regularly arranged.

39. Example 2.8
Determine the formula for (a) calcium chloride and (b) magnesium oxide.
Example 2.9
What are the names of (a) MgS and (b) CrCl3?

40. Example 2.8 Solution Exercise 2.8A Exercise 2.8B
Determine the formula for (a) calcium chloride and (b) magnesium oxide.
Solution
(a) First we write the symbols for the ions, with the cation first: Ca2+ and Cl–. The simplest combination of these ions that gives an electrically neutral formula unit is one Ca2+ ion for every two Cl– ions. The formula is CaCl2 .
Ca Cl– = CaCl2
(b) Figure 2.10 tells us that the ions are Mg2+ and O2–. The simplest ratio for an electrically neutral formula unit is 1:1. The formula of this binary ionic compound is MgO.
Mg2+ + O2– = MgO
Give the formula for each of the following ionic compounds:
(a) potassium sulfide (b) lithium oxide (c) aluminum fluoride
Exercise 2.8A
Give the formula for each of the following ionic compounds:
(a) chromium(III) oxide (b) iron(II) sulfide (c) lithium nitride
Exercise 2.8B

41. Example 2.9 Solution Exercise 2.9A Exercise 2.9B
What are the names of (a) MgS and (b) CrCl3?
Solution
(a) MgS is made up of Mg2+ and S2– ions. Its name is magnesium sulfide.
(b) From Figure 2.10 we see that there are two simple ions of chromium, Cr3+ and Cr2+. Because there are three Cl– ions in the formula unit CrCl3, the cation must have a charge of 3+, that is, it must be Cr3+. Because there are two chromium cations, there are two chlorides, CrCl2 and CrCl3, and we must assign a different name to each. Therefore, the name of our compound cannot be simply chromium chloride; instead, to indicate the 3+ cation, we say the name is chromium(III) chloride.
Name the following compounds:
(a) CaBr (b) Li2S (c) FeBr (d) CuI
Exercise 2.9A
Write the name and formula for each of the following compounds:
(a) the sulfide of copper(I) (b) the oxide of cobalt(III) (c) the nitride of magnesium
Exercise 2.9B

42. Monatomic Ions Group IA metals form ions of 1+ charge.
Group IIA metals form ions of 2+ charge.
Aluminum, a group IIIA metal, forms ions with a 3+ charge.
Nonmetal ions of groups V, VI, and VII usually have charges of 3–, 2–, and 1–, respectively.
Group B metal ions (transition metal ions) often have more than one possible charge. A Roman numeral is used to indicate the actual charge.
A few transition elements have only one common ion (Ag, Zn, Cd), and a Roman numeral is not often used.

43. Symbols and Periodic Table Locations of Some Monatomic Ions
Copper forms either copper(I) or copper(II) ions.
Titanium forms both titanium(II) and titanium(IV) ions.
What is the charge on a zirconium(IV) ion?

44. Formulas and Names of Binary Ionic Compounds
Binary ionic compounds are made up of monatomic cations and anions.
These combinations must be electrically neutral.
The formula unit is the simplest collection of cations and anions that represents an electrically neutral unit.
Formula unit is to ion as ________ is to atom.
To write a formula, combine the proper number of each ion to form a neutral unit.
To name a binary ionic compound, name the cation, then the anion.
Monatomic anion names end in -ide.

45. Polyatomic Ions
A polyatomic ion is a charged group of covalently bonded atoms.
There are many more polyatomic anions than there are polyatomic cations.
You should (eventually!) commit to memory much of Table 2.4
hypo- and per- are sometimes seen as prefixes in oxygen-containing polyatomic ions (oxoanions).
-ite and -ate are commonly found as suffixes in oxygen-containing polyatomic ions.

47. Example 2.10
Write the formula for (a) sodium sulfite and (b) ammonium sulfate.
Example 2.11
What is the name of (a) NaCN and (b) Mg(ClO4)2?

48. Example 2.10 Solution Exercise 2.10A Exercise 2.10B
Write the formula for (a) sodium sulfite and (b) ammonium sulfate.
Solution
(a) It is possible to identify the sulfite ion without memorizing all the ions in Table 2.4. If you remember the name and formula of one of the sulfur–oxygen polyatomic anions, you should be able to deduce the names of others. Suppose you remember that sulfate is SO42–. The -ite anion has one fewer oxygen atom, 3 instead of 4, and so it is SO32–. The charges of the two ions in a formula unit must balance, which means the formula unit of sodium sulfite must have Na+ and SO32– in the ratio 2:1. The formula is therefore Na2SO3.
(b) The ammonium ion is NH4+, and the sulfate ion is SO42–. A formula unit of ammonium sulfate has two NH4+ ions and one SO42– ion. To represent the two NH4+ ions, we place parentheses around the NH4, followed by a subscript 2, (NH4)2, and thus arrive at the formula (NH4)2SO4.
What is the formula for (a) ammonium carbonate, (b) calcium hypochlorite, and
(c) chromium(III) sulfate?
Exercise 2.10A
Write a plausible formula for
(a) potassium aluminum sulfate (b) magnesium ammonium phosphate
Exercise 2.10B

49. Example 2.11 Solution Exercise 2.11A Exercise 2.11B
What is the name of (a) NaCN and (b) Mg(ClO4)2?
Solution
(a) The ions in this compound are Na+, sodium ion, and CN–, cyanide ion
(see Table 2.4). The name of the compound is sodium cyanide.
(b) The ions present are Mg2+, magnesium ion, and ClO4–, perchlorate ion. The name of the compound is magnesium perchlorate.
Name each of the following compounds:
(a) KHCO (b) FePO (c) Mg(H2PO4)2
Exercise 2.11A
Give a plausible name for the following:
(a) Na2SeO (b) FeAs (c) Na2HPO3
Exercise 2.11B

50. Hydrates
A hydrate is an ionic compound in which the formula unit includes a fixed number of water molecules associated with cations and anions.
To name a hydrate, the compound name is followed by “___hydrate” where the blank is a prefix to indicate the number of water molecules.
The number of water molecules associated with each formula unit is written as an appendage to the formula unit name separated by a dot.
Examples: BaCl2 . 2 H2O; CuSO4 . 5 H2O

51. Hydrates (cont’d)
How many atoms are in one formula unit of copper(II) sulfate pentahydrate?

52. Acids … Taste sour, if diluted with enough water to be tasted safely.
May produce a pricking or stinging sensation on the skin.
Turn the color of litmus or indicator paper from blue to red.
React with many metals to produce ionic compounds and hydrogen gas.
Also react with bases, thus losing their acidic properties.

53. Bases …
Taste bitter, if diluted with enough water to be tasted safely.
Feel slippery or soapy on the skin.
Turn the color of litmus or indicator paper from red to blue.
React with acids, thus losing their basic properties.

54. Acids and Bases: The Arrhenius Concept
There are several definitions which may be used to describe acids and bases.
An Arrhenius acid is a compound that ionizes in water to form a solution of H+ ions and anions.
An Arrhenius base is a compound that ionizes in water to form solutions of OH– and cations.
Neutralization is the process of an acid reacting with a base to form water and a salt.
A salt is the combination of the cation from a base and the anion from an acid.

56. Acid Nomenclature
Notice that the acid name is related to the anion name.
Hydrochloric acid, chloride ion
Hydrosulfuric acid, sulfide ion
Phosphoric acid, phosphate ion
Nitric acid, nitrate ion
Nitrous acid, nitrite ion

57. Organic Compounds
Organic chemistry is the study of carbon and its compounds.
Carbon compounds can have an almost unlimited diversity, because carbon atoms can bond to one another, and to other atoms, to form chains and rings.
Carbon compounds containing one or more of the elements H, O, N, or S are especially common.
Many organic compounds have common names as well as systematic names.

58. Alkanes
Hydrocarbons are molecules that contain only hydrogen and carbon atoms.
Alkanes are saturated (have the maximum number of hydrogen atoms possible for the number of carbon atoms).

59. Alkanes
Isomers are compounds with the same molecular formula but different structural formulas.
Alkane molecules with ring structures are named with the prefix cyclo- and are called cycloalkanes.

60. Propane, used in gas grills, is an alkane with three carbon atoms
Butyric acid, which gives rancid butter its “fragrance,” contains four carbon atoms.
Octane, a component of gasoline, is a(n) ______ which contains _____ carbon atoms.

61. Types of Organic Compounds
Many organic compounds contain a functional group.
A functional group is an atom or group of atoms attached to the hydrocarbon chain, which confers particular physical and/or chemical properties upon the compound.
Compounds with the same functional group often undergo similar reactions.
A list of common functional groups is found in Table D.1.

62. Types of Organic Compounds (cont’d)
For alcohols, the functional group is a hydroxyl group attached to the carbon chain.
Carboxylic acids have a carboxyl group (–COOH) attached to the carbon chain; they are acidic (of course! Why else would they be called carboxylic acids??).

63. Cumulative Example
Show that the following experiment is consistent with the law of conservation of mass (within the limits of experimental error): A g sample of calcium carbonate was dissolved in mL of hydrochloric acid solution (d = g/mL). The products were g of solution (a mixture of hydrochloric acid and calcium chloride) and 2.22 L of carbon dioxide gas (d = g/mL).

64. Cumulative Example Strategy Solution
Show that the following experiment is consistent with the law of conservation of mass (within the limits of experimental error): A g sample of calcium carbonate was dissolved in mL of hydrochloric acid solution (d = g/mL). The products were g of solution (a mixture of hydrochloric acid and calcium chloride) and 2.22 L of carbon dioxide gas (d = g/mL).
Strategy
This problem may give the initial impression of being quite formidable because several data are given. Upon examination, it is much less challenging than you might think. We must show that mass is conserved in the experiment. That means that we must compare the mass of the starting materials to the mass of the end products of the chemical change. If mass is conserved, the two masses should be identical, within the limits of experimental error. Our job then is to find the masses of the starting materials and of the end products.
Starting mass: The mass of calcium carbonate is given. We can use the density and the volume of the HCl solution to find its mass.
Solution
Let’s begin by identifying the starting materials and end products. The context of the problem makes it clear that calcium carbonate reacts with a hydrochloric acid solution, and so calcium carbonate and the HCl solution are the starting materials. The end products are another solution and carbon dioxide gas.

65. Cumulative Example continued
Solution continued
Then we can add the masses of the two starting materials.
Mass of products: This time, the mass of the solution is given. We must use volume and density of carbon dioxide gas to find its mass. However, we must first convert the volume in liters to milliliters because the density is given in grams per milliliter. Then we can add the masses of the two products.
Assessment
We note that the masses of reactants and products are not exactly the same. However, the g of reactants is reported to four significant figures, and so it is precise only to 0.1 g. The difference between the masses of the starting materials and end products is less than 0.1 g. Clearly, the difference in masses is smaller than the uncertainty in the mass of the starting materials. We can therefore conclude that this experiment is consistent with the law of conservation of mass.