Student Name ______________________________________________________________
ADVANCED PLACEMENT CHEMISTRY
WORKBOOK AND NOTE SET
Student Name ______________________________________________________________
ADVANCED PLACEMENT CHEMISTRY
WORKBOOK AND NOTE SET
II
Fundamental Review 2
Atomic Character and Structure 2
Atomic Number 2
Mass Number and Isotopes 3
Ions and Electrical Charge 5
Molecules and Molecular Compounds; Ions and Ionic Compounds 6
Quantitative Aspects of Compounds 7
Percentage Composition 7
Empirical and Molecular Formulas 7
Hydrated Compounds and their Formulas 10
Mass Ratios 11
Stoichiometry Review 12
Limiting Reactants 12
Theoretical and Percent Yield 13
Percent Error 14
Review of the Periodic Table 15
Oxidation Numbers 16
Naming Compounds and Writing Formulas 19
Introduction to Descriptive Chemistry: Solutions and Introductory Solution Chemistry 22
The Solution Process for Ionic Compounds 22
The Solution Process for Molecular Compounds 23
Electrolytic Properties of Solutions 24
General Aqueous Solubility Guidelines 22
Solution Concentration 27
Dilutions 28
Reactions in Aqueous Solution 30
Writing Equations in Aqueous Solution 32
Using the Ideal Gas Equation and Gas Stoichiometry 34
Periodic Table of the Elements 35
Atomic Radius 36
Ionization Energy 37
III
Reduction Potentials 44
Oxidizing and Reducing Agents 47
Common Oxidizing and Reducing Agents 48
Balancing Redox Reactions 50
Electrochemistry 54
Voltaic Cells 54
Chemical Kinetics and Reaction Mechanisms 59
Collision Theory 60
Activation Energy 61
Collision Orientation 62
Chemical Kinetics and Rates of Reaction 63
Definition of Reaction Rate 63
Rates in Terms of Concentration 65
Stoichiometric Differences 66
Concentration Dependence of Rate 67
Reaction Order 69
Method of Initial Rates 70
Concentration Changes over Time 72
First-Order Reactions 72
First-Order Half-Life 73
Second-Order Reactions 74
Second-Order Half-Life 75
Rate and Temperature 75
The Arrhenius Equation 75
Reaction Mechanisms 77
IV
Quantitatively Expressing Equilibrium 88
The Equilibrium Constant 89
The Magnitude of the Equilibrium Constant 91
Manipulating Equations and Kc Values 92
Heterogeneous Equilibria 93
Calculating Equilibrium Constants 94
Reaction Quotients 98
Calculating Equilibrium Concentrations 100
Le Chatelier’s Principle 101
Chemical Equilibrium II: Acid-Base Equilibrium and Solubility Equilibrium 104
The Arrhenius Acid and the Arrhenius Base 105
The Brønsted-Lowry Acid and the Brønsted-Lowry Base 105
Conjugate Acid-Base Pairs 107
Relative Strengths of Acids and Bases 108
The Autoionization of Water 110
The pH Scale 112
Measuring pH 112
Strong Acids and Strong Bases 114
Equilibrium Expressions of Strong Acids and Strong Bases 115
Weak Acids and Weak Bases 116
Equilibrium Expressions of Weak Acids 116
Calculating Ka from pH 117
Determining the Validity of Assumptions and Percent Ionization 119
Calculating pH from Ka 120
Equilibrium Expressions of Weak Bases 121
Calculating [OH-] of a Weak Base Solution 121
The Equilibrium Concentrations of Polyprotic Acids 123
The Relationship Between Ka and Kb 124
Acid-Base Properties of Salt Solutions/Hydrolysis 125
Acid-Base Strength and Chemical Structure 127
V
Buffering Capacity 131
The pH of Buffered Solutions and the Henderson-Hasselbalch Equation 132 The pH of Buffered Solutions when Strong Acids or Strong Bases are Added 135
Acid-Base Titrations 136
Titration Curves 137
Strong Acid – Strong Base Titrations 137
Strong Acid-Weak Base or Strong Base-Weak Acid Titrations 140
Summary of Acid-Base Titration Characteristics 144
The Value of the Solubility-Product Constant, Ksp 147
Factors Affecting Solubility 150
Thermochemistry, Thermodynamics and Electrochemistry II 153
The Nature of Energy 154
Distinguishing a System from the Surroundings 155
The First Law of Thermodynamics 155
The Internal Energy 156
State Functions 159 Enthalpy 160 Reaction Enthalpies 161 Hess’s Law 162 Bond Enthalpies 164 Heats of Formation 167 Calorimetry 169 Spontaneous Processes 173
Entropy and the Second and Third Laws of Thermodynamics 173
Gibbs Free Energy 177
The Electrochemistry-Redox-Thermochemistry-Equilibrium Relationship 183
Concentration Cells 185
Electrolysis 186
Electroplating 187
Electrolysis and Corrosion 187
VI
Covalent Bonding 194
Review of Typical Geometry and Bond Number 195
Polar Bonds and Polar Molecules 196
Multiple Bonds 197
Lewis Structures 198
Drawing Lewis Structures 198
Deciding Between Several Lewis Structures: Formal Charge 202
Resonance Structures 203
Exceptions to the Octet Rule 205
VSEPR Theory 207
Electron Density Geometries 209
Molecular Geometries 210
Bond Angles 218
Hybridization Theory 218
The Behavior of Solutions, Liquids and Gases; Phase Change and Intermolecular Forces 225
Intramolecular Review: Ionic, Covalent and Metallic Bonding 226
Electronegativity and Polarity 227
Intermolecular Forces 228
Ion-Dipole Forces 229
Dipole-Dipole Forces 229
London Dispersion Forces 230
Hydrogen Bonding 232
Relative Strength of Intermolecular Forces 233
Vapor Pressure and Boiling Point 234
Properties and Intermolecular Forces 235
Change of Phase 238
Heating Curves 239
Critical Temperature and Critical Pressure 242
Phase Diagrams 243
Process of Solution 245
Degree of Saturation and Solubility 246
VII
Boiling Point Elevation 250
Freezing Point Depression 251
Osmotic Pressure 252
The Gas Laws and Gas Behavior 254
Boyle’s Law 254
Charles’s Law 255
Combined Gas Law 256
Avogadro’s Law 256
Gay-Lussac’s Law 257
Ideal Gas Equation 257
Variations of the Ideal Gas Equation 258
Dalton’s Law of Partial Pressures 259
Mole Fractions 261
Kinetic Molecular Theory of Gases 261
Deviations from Ideal Behavior 262
Van der Waals Equation of State 264
Diffusion and Effusion 264
Collecting Gas Over Water 265
Descriptive Chemistry 267
Introduction 268
Synthesis Reactions 269
Decomposition Reactions 270
Single-Replacement 271
Double-Replacement (Metathesis) Reactions 272
Hydrolysis Reactions 273
Outline of Aqueous Metathesis Reactions 273
Redox Reactions 274
Oxidizing and Reducing Agents 275
Electrolysis 276
Naming Common Complexes 276
Coordination Chemistry 277
Organic Reactions 278
Fundamental Review
Students will be able to:
identify the character, location and nature of the subatomic particles in the atom describe the relationship between the subatomic particles and atomic number, mass
number and ion charge
calculate average atomic mass from natural abundances
describe the formation of molecular compounds, ionic compounds and compare the properties of these classes of compounds
determine empirical and molecular formulas from percent composition data & reaction analysis data calculate percent composition
determine the formulas and describe the nature of hydrated compounds discuss the relationships given by chemical equati
use chemical equations to determine limiting reactants and product yields in chemical reactions name compounds using the stock system of naming
name acids, polyatomic ions and hydrated compounds predict valence electron structure from elements’
assign and predict oxidation states to atoms, ions and compounds describe the solution process for molecular and ionic compounds
describe and predict the electrolytic properties of solutions based upon the character of the s predict the solubility of ionic and molecular compounds, including acids and bases
write ionization/dissociation equations for appropriate compounds, and write full molecular, ionic and net ionic equations fo reactions in aqueous solution
calculate solution concentration, and determine volumes of solutions required based upon concentration data calculate required amounts for and perform dilutions
predict the occurrence of metathesis reactions and write their chemical equations using appropriate phase use the ideal gas equation and perform gas stoichiometry
Fundamental Review
identify the character, location and nature of the subatomic particles in the atom relationship between the subatomic particles and atomic number, mass
calculate average atomic mass from natural abundances
describe the formation of molecular compounds, ionic compounds and compare the ompounds
determine empirical and molecular formulas from percent composition data & reaction analysis data
determine the formulas and describe the nature of hydrated compounds discuss the relationships given by chemical equations
use chemical equations to determine limiting reactants and product yields in chemical reactions name compounds using the stock system of naming
name acids, polyatomic ions and hydrated compounds
predict valence electron structure from elements’ locations on the periodic table assign and predict oxidation states to atoms, ions and compounds
describe the solution process for molecular and ionic compounds
describe and predict the electrolytic properties of solutions based upon the character of the solute predict the solubility of ionic and molecular compounds, including acids and bases
write ionization/dissociation equations for appropriate compounds, and write full molecular, ionic and net ionic equations fo
solution concentration, and determine volumes of solutions required based upon concentration data calculate required amounts for and perform dilutions
predict the occurrence of metathesis reactions and write their chemical equations using appropriate phase use the ideal gas equation and perform gas stoichiometry
determine empirical and molecular formulas from percent composition data & reaction analysis data
write ionization/dissociation equations for appropriate compounds, and write full molecular, ionic and net ionic equations for
solution concentration, and determine volumes of solutions required based upon concentration data
2 Figure 2. The atomic number of carbon is shown as 6. This tells you that an atom of
carbon has 6 protons in its nucleus. The atomic number of hydrogen is shown as 1, which tells you that an atom of hydrogen has 1 proton in its nucleus. The proton number of an atom cannot change or the identity of the atom has also changed. Atomic number is often represented with an italicized Z.
Part 1: ATOMIC CHARACTER AND STRUCTURE
By the time of Bohr's model of the atom, much was in fact known about the physical properties of atoms. The electron had been discovered, followed by protons and the nucleus, and finally, the neutron. This section discusses many of the properties of the atom fundamental to the study of chemistry.
Many of the points below are extended upon in the sections of the pages that follow. The figure on the right represents a "textbook" view of the atom. Realize that the atom is far more complicated than this figure, and we will add detail during our study this year.
The atom is made of protons, neutrons and electrons.
The protons and neutrons are located in a dense core, and all of the mass of the atom is located here. Electrons surround the nucleus in a region of space.
Protons have a 1+ electrical charge, electrons have a 1- electrical charge, and neutrons have no electrical charge. Because the protons are positively charged and the neutrons have no electrical charge, the overall charge on the nucleus is positive.
The atomic number of an atom is the number of protons in its nucleus. The atomic number of an element is unique.
The mass number of an atom is the sum of its protons and neutrons; recall, only protons and neutrons have mass. The mass number is central in determining the average atomic mass of an element, and atoms of an element that have different numbers of neutrons are called isotopes.
An atom with equal numbers of protons and electrons is electrically neutral; each 1+ proton "cancels" the charge on a 1- electron. When this is not the case
the atom is more specifically called an ion.
ATOMIC NUMBER
Each element has an atomic number, which is given on nearly every periodic table. The atomic number of an element is equal to the number of protons in the nucleus of an atom of the element. Atomic number defines an element. No two elements have the same atomic number, and to change the atomic number is to change the identity of an atom. See Figure 2.
Figure 1. A textbook model of an atom. The darker area surrounding the nucleus
represents the region occupied by electrons as they orbit.
3 The atomic number can provide many details when
attempting to characterize the physical properties of atoms:
You will read soon about atomic mass and mass number, both of which are dependent upon the number of protons in an atom; thus, they are related to atomic number.
Your determination of other characteristics will be made easier if you remember that atomic number does not change for an element. Electron number and neutron number may vary, but the number of protons is constant for an element.
MASS NUMBER & ISOTOPES
Protons and neutrons are the only subatomic particles that have mass, and these are located in the nucleus. Together, these particles contribute to the atomic mass of an atom. The mass number of an atom is the number of neutrons and protons in its nucleus. Figure 4
shows the atomic mass of three atoms.
Atoms of an element can have different numbers of neutrons in their nuclei. Isotopes of atoms contain different numbers of neutrons in their nuclei. Another common term used to describe a specific nucleus is nuclide. For example, the isotope carbon-12’s nucleus could be described as the carbon-12 nuclide.
The name of a particular isotope is the element name plus its mass number. The name of the isotope on the far left in Figure 4 is carbon-12, the middle isotope is carbon-13 and the isotope on the right is carbon-14. This system of naming isotopes helps distinguish between the varying isotopes of an element that are present in nature. A shorthand notation for naming isotopes is to indicate the mass number of the isotope to the upper-left of its symbol. This notation is shown in Figure 4. The number in the lower-left corner is the atomic number of the atom.
Figure 3. The three atoms above show atomic mass clearly - it is the number of protons plus the number of neutrons. You can see that the hydrogen atom has a mass of 1 (one p+ plus zero n0), the carbon atom has a
mass of 12 (six p+ plus six n0), and the uranium atom has a mass of 238
(ninety-two p+ plus one hundred forty-six n0). The electrons are shown, but
recall that they are massless and do not contribute to atomic mass.
Figure 4. You can see in the figure above that there are three different isotopes of carbon atoms present in the lump of coal. The atoms do not differ in atomic number (because atomic number defines the atoms as carbon atoms), but they do have different numbers of neutrons; these represent the isotopes of carbon. These isotopes have the same chemical properties, but they do have a significant difference in the physical property of mass. The atomic mass of the first isotope is 12, which is the sum of the 6 p+
and 6 n0 in its nucleus. The atomic mass of the second isotope shown is 13, the sum of 6
4 The average atomic mass of an element is the weighted-average mass of all of the naturally-occurring isotopes of an element. As an analogy, consider a class where your grade is made up of several categories. For example, perhaps tests are 70% of your grade, homework is 10% and laboratory work is 20%. You can see that your performance on tests will have a greater impact on your final grade than a few homework assignments will. In a similar manner, if there are three isotopes of an element in nature and one of those isotopes makes up 70% of all naturally-occurring isotopes, then we would expect that the average atomic mass of the element would be near the mass of this particular isotope (much in the same way that your final grade would be near your test average). The average atomic mass is often simply called the atomic mass.
Practice 1.1
From the table you can see that the mass of silicon-28 contributes more than the other two naturally-occurring isotopes to the average atomic mass of silicon, which is 28.09. We can calculate the average atomic mass by determining the weighted average of the values shown.
The weighted average – the average atomic mass – is calculated by summing the results of the following calculation for each isotope:
൬percent abundance100 ൰ x isotope mass = contribution of isotope to average atomic mass Determine the average atomic mass of the silicon atom based on the data above.
The average atomic mass of an element is 58.69, and two naturally-occurring isotopes of the element are known to exist. If the mass of the isotope with an abundance of 78.3% is found to be 58.4, what is the mass of the second isotope?
Isotope Atomic mass Percent abundance
silicon-28 28 92.23
silicon-29 29 4.67
5 IONS AND ELECTRICAL CHARGE
Atoms that contain equal numbers of electrons and protons are electrically neutral. However, atoms can easily lose or gain electrons and form ions, which are electrically-charged chemical species. The identity of the element does not change – only the electrical charge of the species changes. Recall from first-year chemistry that atoms lose or gain electrons based upon their metal or nonmetal character, and that the number of electrons gained or lost is related to valence structure.
Metals tend to lose electrons to form positive ions called cations. For example, an atom of sodium will readily lose its outermost electron and form the sodium ion, which has a charge of 1+. The charge is due to the difference between positive charges (protons) and negative charges (electrons): an atom of sodium contains 11 protons and 11 electrons; an ion of sodium contains 11 protons and 10 electrons – this gives a charge of (+11) + (–10) = +1. Metal ions are named just as their atoms are. We represent the ion by placing its charge to the upper-right of the symbol: sodium ion is represented as Na+.
Nonmetals tend to gain electrons to form negative ions called anions. For example, an atom of nitrogen will readily gain three electrons to fill its valence shell and form the nitrogen ion, which has a charge of 3-. The charge is due to the difference between positive charges (protons) and negative charges (electrons): an atom of nitrogen contains 7 protons and 7 electrons; an ion of nitrogen contains 7 protons and 10 electrons – this gives a charge of (+7) + (–10) = –3. Nonmetal ions are named by changing their name ending to “–ide.” The nitrogen ion is called “nitride ion.” We represent the ion by placing its charge to the upper-right of the symbol: nitrogen ion is represented as N3–.
Practice 1.2
Answer the following questions about atoms, isotopes and ions.
What is the number of protons, neutrons and electrons in a neutral atom of silicon-31?
How many protons and neutrons are in an isotope of lead-204?
82; n0 = 122
What is the mass number of an isotope of carbon that contains 8 neutrons and 6 electrons?
Identify the charge and total number of electrons in the most common ion of the following elements. Write the chemical symbols for the ions you select.
6 Part 2: MOLECULES AND MOLECULAR COMPOUNDS; IONS AND IONIC COMPOUNDS
Although we tend to use the term loosely, a molecule is a discrete collection of covalently-bonded atoms that form a single unit – thus, we should not use the term molecule to describe ionic compounds, metallic compounds or alloys. Molecules include water (H2O), hexane (C6H14) and ammonia (NH3). From
these examples we might conclude the general statement:
Molecules are units of matter composed of a definite number of (usually) nonmetallic atoms bonded to one another. Unlike ionic compounds – which can be broken along their ionic bonds and maintain their identity – the bonds of molecules are covalent bonds that cannot be broken unless a chemical change occurs.
Ionic compounds are made of ions that are attracted by electrostatic attraction due to their opposite charges – the ions do not share electrons. Thus, because the ions of ionic compounds are not sharing electrons, we do not refer to them as molecules; instead, they are referred to as formula units.
Notice in Figure 5 that breaking the bonds in the histidine molecule changes the chemical structure – it is no longer histidine. However, because the formula of NaCl is just that – NaCl – there is no change in chemical nature if a “layer” of Na–Cl ions is removed – the sample is simply smaller (Figure 6).
It is important to note that polyatomic ions are charge-carrying covalently-bonded ions. The bonds between the atoms in the polyatomic ion are covalent, while the attraction that holds the ion as a whole to another ion is an ionic bond. Compounds that form in this manner – all of which are ionic compounds – separate upon dissolving along the ionic bonds and never along their covalent bonds within the polyatomic ion. Again, to break a covalent bond is to form a new substance.
Figure 5. A molecule of histidine. The bonds in a molecule are covalent bonds.
Figure 6. A unit of NaCl. The crystal is a series of repeating formula units rather than a "molecule."
7 Part 3: QUANTITATIVE ASPECTS OF COMPOUNDS
PERCENTAGE COMPOSITION
Percentage composition indicates the percent composition by mass of the elements in a compound. It is easily determined when the formula of a compound is known: divide the mass of an element of interest by the molar mass of a compound.
Practice 1.3
What is the percent composition by mass sodium in sodium carbonate?
A 2.424 g sample of a hydrocarbon is burned in excess oxygen to produce 3.756 g water and 7.342 g carbon dioxide. Use this information to determine the percent composition carbon and hydrogen in the hydrocarbon.
EMPIRICAL AND MOLECULAR FORMULAS
Empirical formulas provide one with the relative numbers and identities of the atoms in a molecular compound. However, they do not necessarily represent the actual number of each atom – they only give a whole-number ratio of the atoms in the compound. A molecular formula, by comparison, gives the actual number and identities of the atoms in a compound – it is a multiple of the empirical formula. The empirical formula can be a molecular formula if the actual numbers of atoms in a molecule is represented by the empirical formula.
If we know the molecular formula for a compound, we can easily determine its empirical formula, but other methods are required to determine a molecular formula if we know the empirical formula.
8
Divide the percent composition or mass of each element by its
molar mass to determine mol
Divide the larger mol values by the smallest – multiply to eliminate decimal values if
necessary
Write the chemical formula using the mol ratios determined
in the previous step
Formulas are mol ratios of the atoms in the compound – thus, in order to determine the empirical formula, simply think about how to find the ratio of mol in the compound from the data provided.
You should know how to determine the empirical formula of a compound by the following methods: mass composition, percent composition and combustion analysis
Practice 1.4
A certain compound is found to be 71.40% by mass carbon, 9.59% by mass hydrogen and 19.02% by mass oxygen. Determine the empirical formula of the compound.
Convert all percents into moles by dividing by the atomic mass of the element; divide by smallest; change to whole numbers
52.14 / 12.011 © = 4.34 mol C 13.13 / 1.0079 = 13.02 mol H 34.73 / 15.9994 = 2.17 m4 / 2.17 = 2 mol C
Imagine a 14.50 g sample of a compound containing only nitrogen and oxygen was decomposed into its elements; a mass of 3.76 grams of nitrogen was collected. What is the empirical formula of the compound?
9 If we know the products of a reaction that a particular compound undergoes, then we can analyze the products to determine the composition of the original compound. This allows us to determine the original compound’s empirical formula. For example, imagine that an unknown compound containing only carbon, hydrogen and oxygen (e.g., a sugar) is burned. The products of this reaction are carbon dioxide and water. We will perform 4 steps to determine the empirical formula of the sugar:
• Determine the percent composition carbon in carbon dioxide and the percent composition hydrogen in water.
• Use the percent composition of carbon and hydrogen to determine the mass of carbon and hydrogen in the original compound. • Subtract the mass of carbon and the mass of hydrogen from the mass of the original compound to obtain the mass of oxygen. • Solve for empirical formula.
Practice 1.5
Imagine a 0.5540 gram sample of an unknown sugar is burned. Sugars contain only carbon, hydrogen and oxygen. The combustion of the sugar produces 0.6793 g water and 1.383 g carbon dioxide. What is the empirical formula of the sugar?
Molecular formulas are whole number multiples of empirical formulas; that is, a molecular formula and empirical formula are related by the ratio of their masses – thus, in order to determine the molecular formula, simply divide the experimentally determined molar mass by the mass of the empirical formula, and multiply the empirical formula by the result.
Determine the mass of the empirical formula
Divide the mass of the molecular formula by the mass of the
empirical formula
Multiply the empirical formula by the value of the ratio obtained
10 Practice 1.6
The major ingredient in antifreeze is composed of 38.7 g carbon, 9.70 g hydrogen and 51.6 g oxygen per 100.0 grams of antifreeze. Its molar mass has been found to be 62.1 g/mol. What is the molecular formula of antifreeze?
HYDRATED COMPOUNDS AND THEIR FORMULAS
A hydrated compound is an ionic compound that contains water molecules in positions of the crystalline lattice not occupied by ions. For example, the compound CoCl2, a purple compound, is the dehydrated form of the compound CoCl2 • 6H2O, which contains six
water molecules in various lattice positions. The term for a compound that is the dehydrated form of a hydrated compound is anhydrous. In the examples provided, the purple cobalt(II) chloride is called anhydrous cobalt(II) chloride, while the hydrated form is called cobalt(II) chloride hexahydrate. The naming of a hydrate involves adding the prefixed form of hydrate after the name of the ionic compound. We can determine the number of water molecules that are contained in the hydrated form by removing the water through dehydration, which involves driving off the water by heating. You can often distinguish between the hydrated and anhydrous forms of most ionic compounds because they are different colors.
Practice 1.7
A 1.023 g sample of a hydrated copper(II) sulfate salt was dehydrated and found to have a mass of 0.6540 g. What is the percent composition water of the hydrate, and what is the number of molecules of water in the hydrate’s formula? Name the compound.
11 MASS RATIOS
Mass ratios indicate the ratio of the masses of the elements in a chemical compound. They can be easily determined by determining the mass of each element and establishing the ratios between the elements. Although it may seem elementary, it is worth noting that the mass ratio of “A to B” places A in the numerator and B in the denominator!
Practice 1.8
12 Part 4: STOICHIOMETRY REVIEW: LIMITING REACTANTS & THEORETICAL YIELDS
The following relationships can be made between the species in a stoichiometric set-up:
When considering stoichiometry questions, be sure to consider:
For what are you being asked to solve? Is the chemical equation balanced? Are there any prior problems that must be
solved (empirical or molecular formula, percentage composition or limiting reactant)?
Check the units in the set-up to ensure that all of the units except the one desired in the answer cancel.
Is there a problem that must be solved afterward (percent yield, percent error)? Are the conversions correctly set up – for
example, did you indicate “there are 6.022 x 1023 atoms in a mol” and not incorrectly
make the relationship “there are 6.022 x 1023
mol in an atom?”
If you are looking for atoms, did you correctly consider the number of atoms in the formula of interest?
Did you remember to correctly calculate the molar masses of diatomic molecules and polyatomic ions?
LIMITING REACTANTS
A limiting reactant is a reactant in a chemical equation that is not present in sufficient stoichiometric quantity to react with another chemical. You will know that a calculation requires a limiting reactant calculation prior to stoichiometric determinations because the question will provide the mass of more than one reactant – this clues you that you must determine which reactant will be used-up first. Alternatively, most questions that do not require a limiting reactant determination expressly state that one of the reactants is “in excess.” Determining the limiting reactant requires simply determining the stoichiometry between the quantities available, and then using the reactant of which there is limiting quantity in the problem of interest.
Ratio from chemical equation Grams of Substance A Mol of Substance A Use molar mass
of A
Mol of Substance B
Grams of Substance B
Use molar mass of B
Through the use of Avogadro’s number,
you can move to atoms or molecules
from mole values
[One mole of any particle is 6.022 x 1023 of the particles.
The particles could be atoms, molecules or ions.]
13 Practice 1.9
A 35.0 gram of nitrogen is placed in a sealed container with a 5.00 gram sample of hydrogen. Assuming 100% reaction, what mass of ammonia can be produced?
THEORETICAL YIELD AND PERCENT YIELD
A theoretical yield is the stoichiometric amount of product that one could calculate as expected from a chemical reaction that occurs to 100% completion. However, most reactions do not yield the theoretical yield of product for various reasons that we will discuss later. The amount of product that is actually collected during a chemical reaction is called the actual yield or laboratory yield. Like all percent calculations, the percent yield is an expression of the ratio between an actual quantity and a maximum quantity:
From laboratory → laboratory yield
x 100 = percent yield From calculation → theoretical yield
Practice 1.10
A student collected a mass of 0.2850 grams of lead(II) iodide when she reacted a 0.2500 gram sample of lead(II) nitrate with an excess sample of potassium iodide according to the double-displacement reaction shown here:
Pb(NO3)2 + 2 KI → PbI2 + 2 KNO3
14 A 200.0 g sample of CH4 was burned in 35.6% yield, and the H2O product collected. What mass of H2O was collected?
The reaction between methane and a halogen is called a substitution reaction – one in which a hydrogen on methane is replaced by a halogen atom.
Write the reaction between bromine and methane.
If the reaction occurs to about 29.5%, what mass of bromine is required to prepare 500.0 g of HBr?
PERCENT ERROR
A percent error provides a measure of the closeness of a laboratory value to the accepted value of a measurement – it is not a measure of the “correctness” of an answer!
lab value: recorded lab value |lab value – accepted value|
x 100 = percent error accepted value: value from reference work, table of data, etc. accepted value
Practice 1.11
15 The combustion of octane, C8H18, produces the typical products of hydrocarbon combustion.
Write the balanced chemical equation for the combustion of octane.
How many grams H2O are produced when 265.5 g C8H18 is burned in the presence of 685.0 g O2?
How many atoms of oxygen are used each time 1.00 mol octane burns?
How many molecules of carbon dioxide are produced when a 4.00 mol sample of octane burns completely?
Part 5: REVIEW OF THE PERIODIC TABLE
Electrons are responsible for the chemical behavior of atoms. And, we will see here, that it is only the electrons in the highest-occupied quantum level that are of real significance for us. Why should only a few electrons be responsible for behavior? There are several explanations that will make your understanding on this point more clear:
• In order to react with one another, atoms must lose, gain or share electrons. We should not be surprised, then, that the electrons farthest out on the atom are those that will most likely "tangle" up with the electron clouds of other atoms. Thus, these electrons are the electrons responsible for the observed chemical behavior of atoms.
• Full quantum levels are not very reactive. All the quantum levels before an atom's outermost quantum are full. Thus, only the partially-filled outer quantum is responsible for most of the chemical behavior of atoms. This is why noble gases (Group 18 atoms) are not very reactive; the Group 18 atoms have a full quantum level.
As an example, let’s review oxygen and sulfur. Oxygen has electrons in two quantum levels: 1 and 2. Sulfur has electrons in t quantum levels: 1, 2 & 3. Look at the electrons in the
quantum three for sulfur. Oxygen has 6 electrons in the highest quantum (2s
The electrons in the highest quantum for each element are an element's valence electrons, and you can s
column on the periodic table have the same number of valence electrons; the electrons are just in different quantum levels. O sulfur both have six valence electrons, and we say that their valence shells are "s
• Valence electrons are the electrons that are in an atom's highest quantum level. They are the
of highest energy. Because the transition metals are filling a previous quantum's energy level, they are not part of the vale structure. Instead, the valence shell of most transition metals is
electrons.
• Valence electrons are responsible for chemical behavior, and that is why the elements in the same column on the periodic tabl have similar chemical behavior - they have the same valence structure, which is shown in the periodic table below.
• An important aspect of valence electron configuration
chemical behavior and physical properties of elements. Data suggests that each orbital in a sublevel (e.g., the three orbital sublevel) possesses one electron before any of the orbitals possess two. This leads to a regular pattern of
orbitals for the elements. For example, the nitrogen atom’s three ‘p’ electrons are not paired up in the first ‘p’ orbital; r of the three orbitals possesses one electron, which gives nitrogen three half
units of study.
OXIDATION NUMBERS
The valence structure of elements allows us to predict their will take, or the hypothetical charge that covalently
oxidation numbers is fundamental to understanding how compounds bond example, in the compound NaCl, the ionic charge on sodium ion is 1 has lost an electron and chlorine has gained an electron when ions
involve the production of ions because the atoms share electrons. How do we assign the “charge” on each species if there are present? This is where oxidation states (oxidation numbers) come in useful. They allow us to keep track of
covalent bond just as easily as we can keep track of transferred electrons in an ionic bond.
As an example, let’s review oxygen and sulfur. Oxygen has electrons in two quantum levels: 1 and 2. Sulfur has electrons in t quantum levels: 1, 2 & 3. Look at the electrons in the highest quantum level for each element, which is quantum level quantum three for sulfur. Oxygen has 6 electrons in the highest quantum (2s2 and 2p4), and sulfur also has 6 electrons (3s
The electrons in the highest quantum for each element are an element's valence electrons, and you can see that elements in the same column on the periodic table have the same number of valence electrons; the electrons are just in different quantum levels. O sulfur both have six valence electrons, and we say that their valence shells are "s2p4."
are the electrons that are in an atom's highest quantum level. They are the s or p electrons of the quantum level of highest energy. Because the transition metals are filling a previous quantum's energy level, they are not part of the vale
structure. Instead, the valence shell of most transition metals is s2. Electrons that are not valence electrons are called
Valence electrons are responsible for chemical behavior, and that is why the elements in the same column on the periodic tabl have the same valence structure, which is shown in the periodic table below.
configuration to consider is called Hund’s Rule, which can be useful in explaining the chemical behavior and physical properties of elements. Data suggests that each orbital in a sublevel (e.g., the three orbital
vel) possesses one electron before any of the orbitals possess two. This leads to a regular pattern of
orbitals for the elements. For example, the nitrogen atom’s three ‘p’ electrons are not paired up in the first ‘p’ orbital; r
of the three orbitals possesses one electron, which gives nitrogen three half-filled orbitals. We shall see the effect of this in later
The valence structure of elements allows us to predict their oxidation states in compounds, which are the actual charges that ions will take, or the hypothetical charge that covalently-bonded atoms would take if their bonds were completely ionic
oxidation numbers is fundamental to understanding how compounds bond and predicting the formulas for chemical compounds. For example, in the compound NaCl, the ionic charge on sodium ion is 1+, while the ionic charge on chlorine is 1
has lost an electron and chlorine has gained an electron when ions formed. However, think about the case of NH
involve the production of ions because the atoms share electrons. How do we assign the “charge” on each species if there are present? This is where oxidation states (oxidation numbers) come in useful. They allow us to keep track of
covalent bond just as easily as we can keep track of transferred electrons in an ionic bond.
16 As an example, let’s review oxygen and sulfur. Oxygen has electrons in two quantum levels: 1 and 2. Sulfur has electrons in three
which is quantum level two for oxygen and ), and sulfur also has 6 electrons (3s2 and 3p4).
ee that elements in the same column on the periodic table have the same number of valence electrons; the electrons are just in different quantum levels. Oxygen and
electrons of the quantum level of highest energy. Because the transition metals are filling a previous quantum's energy level, they are not part of the valence
. Electrons that are not valence electrons are called core
Valence electrons are responsible for chemical behavior, and that is why the elements in the same column on the periodic table have the same valence structure, which is shown in the periodic table below.
, which can be useful in explaining the chemical behavior and physical properties of elements. Data suggests that each orbital in a sublevel (e.g., the three orbitals of the ‘p’
vel) possesses one electron before any of the orbitals possess two. This leads to a regular pattern of half-filled and filled orbitals for the elements. For example, the nitrogen atom’s three ‘p’ electrons are not paired up in the first ‘p’ orbital; rather, each
We shall see the effect of this in later
the actual charges that ions were completely ionic. The use of and predicting the formulas for chemical compounds. For , while the ionic charge on chlorine is 1–. This is because sodium formed. However, think about the case of NH3, which does not
involve the production of ions because the atoms share electrons. How do we assign the “charge” on each species if there are no ions present? This is where oxidation states (oxidation numbers) come in useful. They allow us to keep track of the electrons shared in a
17 We assign oxidation numbers to the atoms or ions in a bond according to a set of rules, which are based upon experimental evidence. Moreover, they are assigned in the order shown here:
The oxidation number of a species in elemental form is zero The oxidation number of a monatomic ion is the charge on the ion
The oxidation number of oxygen is usually 2–; of hydrogen, usually 1+; and of the halogens, usually 1– The sum of the oxidation numbers of a charged species is the charge on the species
The sum of the oxidation numbers of a neutral compound is zero; and negative oxidation states are assigned to more electronegative elements
You should keep in mind that oxidation numbers are a man-made concept, and their use does not mean that actual charges are on the species being discussed, although this might be the case for ions. We shall use the concept of formal charge later to better describe the location of electrons, but even that concept ignores some information about how electrons are arranged – only sophisticated quantum mechanical calculations can provide definitive information about how electrons are arranged in bonds. However, we shall use both oxidation numbers and formal charge, as these are of great value in predicting and explaining chemical reactions.
Experimental evidence suggests the following oxidation numbers are common for atoms:
Metals may take on positive oxidation states equal to their valence electron count, their ‘s’ electron count, their ‘d’ electron count or the sum of their ‘s’ and ‘d’ electron count. They may also take on states related to any ‘p’ valence electron they possess. Typically, oxidation numbers we discuss are on the order of 1+ to 4+, with a few middle transition metals taking on larger states.
Nonmetals may exhibit positive oxidation numbers equal to their valence electron count
Nonmetals may exhibit positive oxidation numbers equal to the number of ‘s’ or ‘p’ electrons they possess
Nonmetals may exhibit negative oxidation numbers equal to the number of valence electrons required to fill their valence shells (which are the typical numbers seen when the nonmetal form ions)
At any rate, keep in mind the rules on assigning oxidation numbers before randomly-assigning oxidation sates! And, let the periodic table be your most valuable tool!
Practice 1.12
For each species below, determine the oxidation number on all elements:
1) CaBr2 2) H2O 3) NaNO3
18
7) Cr2O72- 8) NaNO2 9) SnBr4
10) P2O5 11) NCl3 12) PO43-
For the reaction below, identify the oxidation number for each element.
Cd(s) + NiO2(s) + H2O(l) → Cd(OH)2(s) + Ni(OH)2(s)
Write the full electron configuration for each of the elements below, and then use this information to predict oxidation states that should be observed.
S
Na
Cr
Si
This is the reaction that occurs in nickel-cadmium – “nicad” – batteries
19 Part 6: NAMING COMPOUNDS USING THE STOCK SYSTEM AND WRITING CHEMICAL FORMULAS
You likely have learned that molecular compounds are named by using prefixes, that ionic compounds might be named using Roman numerals to indicate charge and that cations forming only a single charge ion cannot be named with Roman numerals.
You must believe that simple binary compounds (those containing only two elements or two ions) can ALL be named using the Roman numeral system, and it will be beneficial to you to simply do this! Very few elements only take on a single oxidation state; it is easier to always indicate the state than to consider which elements do or do not exhibit variable oxidation states.
• Naming a binary compound that does not contain a polyatomic ion.
1. Identify the two elements in the compound.
2. Identify the oxidation state of the first element – this is the Roman numeral to be enclosed in parentheses and used after the element name. (No space goes between the element and the Roman numeral.) It would be unusual, however, to indicate the oxidation state on hydrogen.
3. Change the name of the second element with the “-ide” ending, and put the two names together.
• Naming a binary compound that contains a polyatomic anion.
1. Identify the two components in the compound.
2. Identify the oxidation state of the first element – this is the Roman numeral to be enclosed in parentheses and used after the element name. (No space goes between the element and the Roman numeral.)
3. Do not change the name of the polyatomic anion and include it as the second part of the name of the compound.
• Naming a binary compound that contains a polyatomic cation.
1. Identify the two components in the compound.
2. Name the cation directly.
3. Change the name of the second element with the “-ide” ending, and put the two names together.
• Naming a binary compound that contains two polyatomic ions.
1. Identify the two components in the compound.
2. Name the cation and anion directly, and put the names together.
• Writing the formula of an ionic compound.
1. Identify the ions in the compound and their charges – the Roman numeral of a monatomic ion is its charge.
2. Select the lowest whole-number ratio of ions that provides for a neutral compound.
• Writing the formula of a molecular compound.
1. If the compound is named with the prefix system, simply use the indicated number of each atom. (Except H2O is always
“water” and NH3 is always “ammonia.”)
2. If the compound is named with the stock system, determine the symbol and oxidation state of each element. Select the ratio of atoms that provides for a neutral molecule.
20
• Naming Acids
Acids are named according to the anion from which an acid is derived.
Practice 1.13
Name the acids below.
HClO
HClO2
HClO3
HClO4
Write the formulas for the following acids.
phosphoric acid
phosphorus acid
Select two sulfur polyatomic ions and name their acids. Anion ends in
“-ide”
Corresponding acid has the prefix “hydro-” and the suffix “-ic”
Anion ends in “-ate”
Corresponding acid has the suffix “-ic” (Retain any prefixes on the anion)
Anion ends in “-ite”
21 Practice 1.14
Name the following compounds, or write their formulas, as appropriate. Include the stock and prefix names of molecular compounds. iron(III) nitrate carbon(IV) oxide nitrogen(V) oxide sodium chloride PF3 ammonium carbonate CuSO4 Cu2SO4 CoCl6 • 5H2O H2O2 sodium acetate HCl xenon(VI) chloride
22 Part 7: INTRODUCTION TO DESCRIPTIVE CHEMISTRY: SOLUTIONS AND INTRODUCTORY SOLUTION CHEMISTRY
Many of the reactions that we will discuss – and most of those that we will perform in the lab – will be done in aqueous solution, which is a solution made when a soluble compound is dissolved in water. Although we will not discuss solutions in detail until later in the year, there are several points that we should make now to help us understand the nature of aqueous chemistry:
• A solution is a homogeneous mixture made up of two or more substances that do not chemically combine. Instead, the substances mix uniformly in the solution.
• The solvent in a solution is the substance present in largest quantity by volume. Usually, the solvent of interest is water.
• The solute is the substance that is dissolved in the solvent.
In Figure 7 above, you can see that the ionic compound (represented by the +/– spheres) is dissolving in water. The positive ion is attracted to the negative end of the polar water molecule, while the negative ion is attracted to the positive end of the water molecule. Solutions generally form when the substances of which the solutions are made are of similar nature in terms of polarity: that is, polar substances dissolve polar substances (or ionic substances), while nonpolar substances dissolve nonpolar substances.
THE SOLUTION PROCESS FOR IONIC COMPOUNDS
To illustrate the process of solution, we will look at the solution of NaCl in water, which is represented in the figure above.
Recall that the water molecule is polar, with a concentration of negative charge on the more electronegative oxygen atom. This leaves the hydrogen ‘ends’ with partial positive charge. The negative ion in a soluble ionic compound is strongly attracted to the positive regions of water molecules, while the positive ion is strongly attracted to the negative end of the water molecules. When the conditions are right, this attraction exceeds any attraction between the ions themselves and causes dissociation. For an ionic compound like NaCl, the process of solution occurs because the solvent-solute attractions between water and salt are greater than the lattice energy of NaCl.
An ionic compound that dissociates considerably into its component ions in solution is termed soluble. For example, the compound NaCl separates into sodium ions and chloride ions in solution – the process of solution for ionic compounds involves the breaking of chemical bonds.
23 Practice 1.15
Into what ions do the following compounds dissociate?
NH4OH
NaCH3COO
H2SO4
THE SOLUTION PROCESS FOR MOLECULAR COMPOUNDS
If a molecular compound is water-soluble, then it will dissolve in water. However, unlike ionic compounds, the individual atoms of most molecular compounds do not separate in solution. That is, the molecules of most molecular compounds stay together – they are bonded covalently and would require a much stronger “pull” from water molecules to separate into ions. A solution of methanol, CH3OH, for
example, does not separate into OH- and CH3+ or any other combination in solution. A solution of methanol is molecules of water and
molecules of methanol.
For example, look at the figure below, which shows a model of an alcohol (1-propanol). When this substance dissolves in water, the individual atoms that make up the molecules do not separate as they do when ionic compounds dissolve.
Some common exceptions to this general statement about molecular compounds’ behavior in water include hydrochloric acid, which completely ionizes into hydrogen ions and chloride ions, and several chemicals that react with water to form ions, including ammonia. (Notice that the term “dissociate” is used for ionic compounds’ separation, while the term “ionize” is used to describe molecular compounds’ separation.)
Figure 8. The characterization of the solution process for a molecular compound. In a molecular compound
like this alcohol, the hydrogen bonding that occurs between the molecules hold the liquid together. In the presence of water, the molecules hydrogen bond to the water as this hydrogen bond is disrupted.
The hydrogen bond between the alcohol molecules is being disrupted by the water molecule’s attraction – a new hydrogen bond is forming between water and the alcohol.
24 ELECTROLYTIC PROPERTIES OF SOLUTIONS
Pure water cannot conduct electricity, but the solutions formed when compounds dissociate or ionize in solution do conduct electricity. Electrical conductivity of solutions is directly proportional to the concentration of ions in solution. Compounds that ionize or dissociate in solution – thus, causing the solution to conduct electrical current – are called electrolytes.
Strong electrolytes are ionic compounds that dissociate (or molecular compounds that ionize) nearly 100% in solution. Strong bases, strong acids and soluble salts are strong electrolytes.
Weak electrolytes are ionic compounds that dissociate (or molecular compounds that ionize) very little in solution. Weak soluble bases, weak acids and sparingly soluble salts are weak electrolytes.
The strong acids, some common weak acids and the strong bases are provided below, and a list of the guidelines for solubility of ionic compounds is provided on Page 25. You should memorize these lists, and assume that all other acids and bases are weak. Also, assume that any compound not listed as soluble is insoluble.
Common strong acids – The common strong acids are soluble and 100% dissociated or ionized in solution; they are strong electrolytes
HCl hydrochloric acid
HBr hydrobromic acid
HI hydroiodic acid
HNO3 nitric acid
HClO4 perchloric acid
HClO3 chloric acid
H2SO4 sulfuric acid
Some common weak acids – The common weak acids are soluble but are not significantly dissociated or ionized in solution; they are weak electrolytes
HF hydrofluoric acid
CH3COOH acetic acid
HCN hydrocyanic acid
HNO2 nitrous acid
H2CO3 carbonic acid
H2SO3 sulfurous acid
H3PO4 phosphoric acid
Common strong bases – The common strong bases are 100% dissociated in solution; they are strong electrolytes NaOH sodium hydroxide
LiOH lithium hydroxide KOH potassium hydroxide Ca(OH)2 calcium hydroxide
Sr(OH)2 strontium hydroxide
25 GENERAL AQUEOUS SOLUBILITY GUIDELINES
1. The common inorganic acids are soluble. Low-molecular-weight organic acids, e.g., acetic acid, are soluble. 2. All common compounds of the Group 1 metals are soluble.
3. All common ammonium ion compounds are soluble.
4. All common nitrates, acetates, chlorates and perchlorates are soluble.
5. The common halide compounds are soluble except as noted: these fluoride compounds are insoluble: MgF2, CaF2, SrF2, BaF2 &
PbF2; these halide compounds are insoluble: AgX, Hg2X2, PbX2.
6. The common sulfates are soluble except barium sulfate, lead sulfates and mercury sulfates
7. The common metal hydroxides are insoluble except for those of the Group 1 metals and Ca, Sr and Ba from Group 2.
8. The common carbonates, phosphates and arsenates are insoluble except for those of Group 1 metals and those of ammonium. 9. The common sulfides are insoluble except those of Group 1 metals, Group 2 metals and those of ammonium.
10. Polar compounds are generally soluble.
11. Nonpolar compounds are generally insoluble, or are soluble to very low concentrations. Summary of solubility and solution character –
Soluble ionic compound or strong base? Is the compound an acid? Is it a strong acid? Is the compound ammonia or molecular base? Strong Electrolyte YES YES NO NO NO Weak electrolyte Nonelectrolyte YES NO Strong electrolyte YES
26 Practice 1.16
For the following compounds:
- identify the compound as ionic or molecular - identify any as acids, bases or salts
- identify any acids or bases as strong or weak - identify salts as soluble or insoluble
- label all acids, bases and salts as weak or strong electrolytes
- for all compounds that dissociate or ionize, write the dissociation or ionization equation
CaCl2 HNO3 C2H5OH LiOH HCl C6H12O6 CH3COOH HBr HF NaF PbF2 Ca(OH)2 Mg(OH)2
27 SOLUTION CONCENTRATION
The concentration of a solution is measured by the amount of solute per given quantity of solution. We generally use the molarity of a solution to discuss concentration. Molarity is a measure of mol solute per liter of solution. The unit of molarity is shown with an italicized uppercase em, M. Spoken, molarity can be described as “mol per liter.”
molarity = amount of solute (mole) volume of solution (liter)=
ࡸ
You can rearrange the molarity equation to solve for mol of solute or for liters of total solution. You must remember that the concentration of a solution expressed in molarity is based on the unit liter – this means that any milliliter measurements must be converted in order to calculate correct values. Additionally, the solute amount is in mol – not mass. Thus, any mass amounts must be converted to mol in order to use the concentration unit of molarity.
Practice 1.17
Answer the following questions.
How many mol of NaCl are in 750. mL of a solution that is described as 2.50 M?
What is the molarity of 2.25 L of solution in which 0.0550 mol silver nitrate is dissolved?
What mass of CaBr2 is required to make 0.650 L of 0.500 M solution?
How many mol of NaCl are in 35.0 mL of a solution that is described as 2.35 M?
28 It is important to note that compounds that ionize or dissociate in solution will contain a greater molarity of ions than is expressed by simply indicating the molarity of the solution. For example, consider a solution that is 1 M NaCl. This measurement tells us that there is 1 mol of sodium chloride ion each liter of solution. However, sodium chloride is an ionic compound that dissociates in solution into sodium ions and chloride ions. Thus, the total mol ions in the solution is 2 mol : 1 mol of sodium ions and 1 mol chloride ions – the solution is, therefore, 2 M in ions. This would not be observed for nonionizing molecular compounds.
Practice 1.18
How many mol species (ions or molecules) are present in 0.750 L of the following solutions? Where appropriate, identify the ions. Give the molarity of ions for the solutions that dissociate.
0.50 M NaCl 2.3 M glucose 1.2 M sulfuric acid 0.750 M (NH4)3PO4
DILUTIONS
Common laboratory dilutions are performed with stock solution, which is concentrated solution. To determine the final molarity of a dilute solution, the following equation can be used:
M1V1 = M2V2, where
M1 and V1 are the molarity and volume of the stock solution
M2 and V2 are the molarity and volume of the desired solution
The equation will provide the amount of stock solution you need to dilute to achieve a total solution volume, V2, of the molarity desired,
M2.
Practice 1.19
29 It is often that you will want to dilute a prepared volume of dilute solution to a lower molarity. In this case, do not forget to account for the volume of solution already present – remember, the dilution equation gives the total final volume of the dilute solution.
Practice 1.20
How much water should be added to 110.0 mL of 0.5000 M HCl to make a solution of 0.0850 M?
Imagine that you have found 45.0 mL 2.00 M calcium acetate solution.
What volume of water should be added to make 0.350 M solution?
How many total mol of ions are present in the new solution?
Part 8: REACTIONS IN AQUEOUS SOLUTIONS
Imagine the following scenario for the reaction on the right. Before this picture was taken, two solutions were prepared: 0.70 M Pb(NO3)2 and 1.4 M KI.
In the sample of the first solution, Pb(NO3)2,
there are two ion species present: Pb2+
and NO3–. In the sample of the second solution there are also two ion species present: K
mixture initially contains all four species. Immediately, however, the Pb
an insoluble compound. The formation of an insoluble compound is called precipitation, and the insoluble compound is called a precipitate. The remaining two ions – K+ and NO
anion "partners;" i.e., lead ion is now associated with the iodide ion and the potassium ion is now associated with the nitra
Why did this mixture of solutions form lead to a chemical reaction
reaction – a reaction in which cations and anions exchange bonding partners ions and iodide ions were removed: the compound is more
more stable in solution than the compound KNO evaporated away, then the two ions will form a compo
Gas-forming Reactions
When one of the products of a double-displacement reaction is a gas, then the reaction will occur. Most commonly, the gases that form are NH3 (when “NH4OH” decomposes into NH
decomposes into SO2 and HOH) or H2S (directly from the combination of hydrogen ion and sulfide ion). Gas
commonly those that involve a solid ionic compound and an acid.
ZnS(s) +
Na2CO3(aq) +
REACTIONS IN AQUEOUS SOLUTIONS
. In the sample of the second solution there are also two ion species present: K+ and I-. When the two solutions are mixed, the
mixture initially contains all four species. Immediately, however, the Pb2+ ions and I- ions react with one another to bond and form PbI
an insoluble compound. The formation of an insoluble compound is called precipitation, and the insoluble compound is called a and NO3– – do not react in solution. In this reaction the cations of each compound switched
anion "partners;" i.e., lead ion is now associated with the iodide ion and the potassium ion is now associated with the nitra
Why did this mixture of solutions form lead to a chemical reaction while other combinations do not? The driving force for a metathesis a reaction in which cations and anions exchange bonding partners – is the removal of ions from solution
ions and iodide ions were removed: the compound is more stable than the free ions in solution. Because the other ions, K more stable in solution than the compound KNO3, these two ions do not form a compound in solution. However, if the water is
evaporated away, then the two ions will form a compound because the ions are less stable out of solution than is the compound KNO
displacement reaction is a gas, then the reaction will occur. Most commonly, the gases that form OH” decomposes into NH3 and HOH), CO2 (when “H2CO3” decomposes into CO2 and HOH), SO
S (directly from the combination of hydrogen ion and sulfide ion). Gas solid ionic compound and an acid.
HCl(aq) →→ →→ H2S(g) + HCl(aq) →→→→ H2CO3(aq) + ↑ ↑↑ ↑ H2CO3(H2O + CO2) H2SO3 (H2O + SO2) NH4OH (H2O + NH3) 30 . When the two solutions are mixed, the ions react with one another to bond and form PbI2,
an insoluble compound. The formation of an insoluble compound is called precipitation, and the insoluble compound is called a In this reaction the cations of each compound switched anion "partners;" i.e., lead ion is now associated with the iodide ion and the potassium ion is now associated with the nitrate ion.
The driving force for a metathesis is the removal of ions from solution. Here, the lead stable than the free ions in solution. Because the other ions, K+ and NO3–
, are , these two ions do not form a compound in solution. However, if the water is
und because the ions are less stable out of solution than is the compound KNO3.
displacement reaction is a gas, then the reaction will occur. Most commonly, the gases that form and HOH), SO2 (when “H2SO3”
S (directly from the combination of hydrogen ion and sulfide ion). Gas-forming reactions are
ZnCl2(aq)
31 Molecular Compound-formation reactions
A double-replacement will occur when a non-ionizing molecular compound is formed. The molecular compound is usually water but can be others, including weak acids. Common examples of this reaction type are acid-base reactions, which are also called acid-base neutralization reactions. In these reactions an acid and a base react to form a salt and the non-ionizing molecule H2O. The general form
of these double-replacement reactions is shown here:
CH3COOH(aq) + NaOH(aq) → →→→ NaCH3COO(aq) + HOH(l)
↑ ↑↑ ↑ water
Precipitation Reactions
A double-replacement precipitation reaction is one in which two soluble, dissociated compounds are mixed and one - or both - of the exchange species is an insoluble salt. You need to know what compounds are soluble and what compounds are insoluble in order to predict the results of any particular reaction attempt - this information is contained in the solubility rules. The general form of double-replacement precipitation reactions is shown here:
AgNO3(aq) + KCl(aq) →→→→ AgCl(s) + KNO3(aq)
↑ ↑ ↑ ↑ precipitate
32 WRITING EQUATIONS IN AQUEOUS SOLUTION
Practice 1.21
For the following proposed reactions write molecular and net ionic equations. Give a reason if no reaction should occur.
barium chloride solution is added to sodium sulfate solution
potassium chloride solution is mixed with sodium sulfate solution YES
YES YES
NO Are both reactants soluble
ionic compounds?
Is one of the reactants a non-oxidizing acid and the
other a solid carbonate, hydroxide, sulfite or
sulfide?
NO
STOP.
No chemical reaction will occur.
Step 1:
Check the reactants
Is one of the potential products: a molecular compound (e.g., H2O,
NH3, or weak acid)
a decomposing compound an insoluble gas
an insoluble ionic compound?
NO
STOP.
No chemical reaction will occur.
Step 2:
Check the products
1. Identify the decomposition products of NH4OH, H2CO3 or
H2SO3, if present
2. Write the equation in molecular form
3. Write all soluble compounds, strong acids and strong bases in dissociated form with appropriate electrical charges
4. Cancel all species that remain unchanged from reactant-side to product-reactant-side
Step 3:
Write the net ionic equation
Stop at 2 for the full molecular
equation
Stop at 3 for the full ionic equation
33 sulfuric acid is dropped onto a solid sample of calcium carbonate
a solution of HCl is added to a sample of aqueous sodium hydroxide
potassium hydroxide and cobalt(III) nitrate are mixed together
aqueous preparations of calcium chloride and sodium phosphate are mixed
pieces of sodium acetate are dropped into hydrochloric acid
small chunks of lithium metal are added to cold water
34 Part 9: USING THE IDEAL GAS EQUATION AND GAS STOICHIOMETRY
While we will explore the use of the ideal gas equation in more detail later in a formal unit of study, we need to use it in its fundamental form throughout the year. Thus, we will review it here and leave additional details to be added later. According the kinetic molecular theory of gases, an ideal gas behaves according to the ideal gas law:
ࡼࢂ = ࡾࢀ
P = the pressure of the gas (in atmospheres, kilopascals, torr or mm Hg); V = the volume of the gas (in liters when used with any value of R); T = the temperature of the gas (in kelvin when used with any value of R: °C + 273 = K); R = 0.0821 when pressure is in atm, 8.314 when pressure is in kPa Use the relationships 760 torr = 760 mm Hg = 1 atm = 101.325 kPa to convert pressure.
Using the above equation, we can easily show that 1 mol of any gas occupies 22.414 L at standard temperature and pressure, STP, which is 273 K (0°C) and 1 atm (101.325 kPa).
Moreover, the equation and the relationship above allow us to convert between mol of gas and volume of gas in order to perform stoichiometry using gases. This is essential to the course.
Practice 1.22
Convert the following pressure values to the unit requested using dimensional analysis.
1.25 atm to torr 85.0 kPa to atm
The reaction between hydrogen gas and oxygen gas produces liquid water.
Write the balanced equation for the reaction
What mass of water can be produced when 2.5 L of hydrogen is burned in excess oxygen at STP?
Use the ideal gas equation to find the pressure of the water gas above when it is transferred to a container with a volume of 0.500 L at a temperature of 200°C.
ADVANCED PLACEMENT CHEMISTRY
Periodic Table of the Elements
Students will be able to:
compare the relative sizes of the atoms of elements based upon their location on the periodic table and explain why & how the radii of atoms changes across a period and down a group
identify the property of ionization energy and discuss and periodic trend of ionization energy
identify the property of electron affinity and discuss and explain the periodic trend of electron affinity
describe the effects of atomic radius, ionization energy and electron affinity on the observed properties of atoms; e.g., formation of ions and reactivity
Periodic Table of the Elements
compare the relative sizes of the atoms of elements based upon their location on the periodic table and explain why & how the radii of atoms changes across a period and down a group
identify the property of ionization energy and discuss and explain the
identify the property of electron affinity and discuss and explain the
describe the effects of atomic radius, ionization energy and electron affinity ties of atoms; e.g., formation of ions and reactivity
