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AU / Chemistry / CHEM 1030 / Ionization energy refers to what?

Ionization energy refers to what?

Ionization energy refers to what?

Description

School: Auburn University
Department: Chemistry
Course: Fundamentals Chemistry I
Professor: John gorden
Term: Fall 2015
Tags: Chemistry
Cost: 50
Name: Chemistry Exam 2 Study Guide
Description: This is the revised and complete study guide for exam 2. It covers everything from Chapters 4, 5, and 6. It includes many diagrams and the worked examples from the book at the end of the document.
Uploaded: 03/07/2016
24 Pages 80 Views 8 Unlocks
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1


Ionization energy refers to what?



Chemistry Exam 2 Study Guide  

Ionization Energy- minimum energy required to remove an electron from an atom in  a gas phase, resulting in an ion (chemical species with net charge

POSITIVE charge means it is a CATION


What is the meaning of electron affinity?



A. Example: Na (g) —> Na+ (g) + e 

B. Therefore, sodium has an ionization energy of 495.8 kJ/mol (first ionization   energy of sodium).  

C. IE1 (Na) which corresponds to the removal of the most loosely held electrons  In general, as Zeff increases, ionization energy also increases Higher atomic radius = lower Zeff = lower ionization energy  Lower energy = closer to the radius = more stable


What is the characteristic of metal?



We also discuss several other topics like How is epithelial tissue classified?
If you want to learn more check out What is the definition of rapport building?

NOTE: removing a paired electron is easier because of the repulsion forces between  2 electrons If you want to learn more check out What are the four layers of the atmosphere?

NOTE: Removing electrons can  

lower energy  

IE1 values for main group  Don't forget about the age old question of What is the effect of marriage on wages for women?

elements (kJ/mol) are to the right  

It is possible to remove  Don't forget about the age old question of What is cellular respiration & fermentation?

additional electrons in  

subsequent ionizations, giving  

IE1, IE2, etc.

2

It takes more energy to remove the 2nd/3rd/4th electrons because it’s harder to  remove core electrons than valence electrons

IE1 (Mg) > IE1 (Na) because Mg is to the right so if it has a greater Zeff and more  difficult to remove (496 kJ/mol < 738 kJ/mol)

IE2 (Na) > IE2 (Mg) because the second ionization of Mg removes a valence electron  where the second ionization of Na removes a core electron If you want to learn more check out What is the steady-state distribution?

Electron affinity (EA) is the energy  

released when an atom in the gas  

phase accepts an electron

Cl (g) + e- —> Cl- (g)  

Like ionization energy, electron affinity  

increases from left to right across a  

period as Zeff increases

NOTE: It’s easier to add an electron to an empty orbital than to add an electron into  an full orbital

More than 1 electron may be added to an atom.

O (g) + e^- —> O^ - (g) (EA1 = -141 kJ/mol)  

O^ - + e^ - —> O^ 2- (g) (EA2 = -741 kJ/mol)  

While many 1st electron affinities are positive, subsequent EA’s are ALWAYS  NEGATIVE. Considerable forces must be used to overpower compulsion energy

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EA1 (Si) > EA1 (Al) because Si is more to the right therefore it has a greater Zeff

EA1 (Si) > EA1 (P) because even though P is more to the right it involves putting it in a  3p orbital. The energy it costs of PAIRING outweighs advantage of adding electrons to  atoms with easier Zeff

Metallic Character

A. Metals tend to be shiny, lustrous, malleable, ductile, and  

 good conductors of heat and electricity  

B. They have low IE (because they have CATIONS and not  

 ANIONS)

C. Many of the periodic trends of elements can be  

 explained using Coulombs Law which states that the  

 force (F) between two charged objects (Q1 and Q2) is  

 directly proportional to the product of the two charges  

 and inversely proportional to the distance (d) between  

 the objects squared  

Ions of main group element

A. Species with identical electron configurations to the noble gas to the right are   called isoelectronic

B. Common monatomic ions are arranged by their positions in the periodic table  C. Exceptions: Mercury is actually a polyatomic ion (Hg2^2+)

Electron Configuration of Ions

A. Write the original configuration

B. Add/remove to make appropriate number of electrons

C. Note: in CATIONS, electrons are removed from the occupied orbitals with the   highest value of n, but in ANIONS, electrons are added to the empty/partially   filled orbitals with the lowest value of n

4

NOTE:

A. Low electron affinity- hard to accept electrons

B. High electron affinity- easy to accept electrons  

C. Low ionization energy- easy to lose electrons

D. High ionization energy- hard to lose electrons

E. Noble gases have lower ionization energies  

Ions of d-block elements  

A. Ions of d block elements are formed by removing electrons first from the shell   with the highest value of n. (When you're removing electrons from the final   configuration, you’re making it more stable by lessening the repulsion forces.) B. Always go to fill s first, then p.  

C. Exception: when you're trying to fill the p blocks  

Worked example 4.8: write the electron configuration for the following ions of d block  elements  

A. Zn^2+ —> [Ar] 3d^10  

B. Mn ^2+ —> [Ar] 3d^5  

C. Cr ^3+ —> [Ar] 3d^5  

Ionic radius- the radius of a cation or an anion (radius- distance between valence  shell and nucleus)  

A. When an atom loses an electron to become a CATION, its radius decreases   due in part to a reduction on electron-electron repulsions in the valence shell  B. A significant decreases in radius occurs when ALL of an atoms valence   electrons are removed  

C. When an atom gains one or more electrons and becomes an ANION, its   radius increases due to increased electron-electron repulsions D. When referring to radius, a cation < neutral atom < anion

5

Isoelectronic Series  

A. An isoelectronic series is a series of two or more species that have identical   electron configurations but different nuclear charges  

B. Ionic radius and attraction force are difference  

C. Example: O^2- and F^- have the same electron configuration but   different attraction forces

NOTE: Stronger attraction forces make ionic radius shrink

Worked example 4.9- know how to rank them in order of increasing radius  

A compound is a substance composed of two or more elements combined in a  specific ratio and held together by chemical bonds, such as salt (NaCl) or water (H20)

Lewis dot symbols  

A. When atoms form compounds, their valence electrons actually interact B. A LDS consists of the elements symbol with dots (do not pair until needed)  C. Examples

1. Boron —> 1s2/2s2/2p1 = 3 valence electrons

2. Carbon—> 4 valence electrons  

3. Nitrogen—> 5 valence electrons  

D. The atoms combine in order to achieve a more stable electron configuration E. Maximum stability results when a chemical specifies is isoelectronic with a   noble gas

For main group metals such as Na the number of dots is the number of electrons  that are lost

For nonmetals in the second period the number of unpaired dots is the number of  bonds the atom can form

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Ions may also be represented by Lewis dot symbols (O^2-) which can be represented  by writing the element with correct number of electrons then brackets with the charge  on the outside  

Ionic Compound Bonding- refers to the electrostatic attraction that holds  oppositely charged ions together in an ionic compound

A. Metals- low ionization energies lose electrons very easily  

B. Nonmetals- high electron affinity accepts electrons very easily  

C. Ionic Compounds  

1. Lewis Dot Symbol  

2. Na (dot) —> Na+ + e 

Lattice Energy  

A. A 3-dimensional array of oppositely-charged ions is called a lattice  

B. Lattice energy is the amount of energy required to convert a mole of ionic   solid to its constituent ions in the gas phase

C. The magnitude of lattice energy is a measure of an ionic compounds stability D. Lattice energy depends on magnitudes of the charge and on the distance   between them  

E. If they have the same charges, the only thing they have that differs one from   another is the distance between them  

F. *Know how to arrange ionic compounds in order of increasing/decreasing   lattice energy (opposite of atomic radius)*  

G. NOTE: big atomic radius means low lattice energy

H. The formation of ionic bonds RELEASES a large amount of energy

The resulting electrically  

neutral compound, sodium  

chloride, is represented with  

the chemical formula NaCl

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Naming Ions and Ionic Compounds  

A. A monoatomic cation is named by adding the word ION to the name of the   element  

B. A monoatomic anion is named by changing the ending of the elements name   to IDE, such as oxide, carbide, sulfide

C. Some metals can form cations of more than one possible charge (especially   true for elements of D-block)

1. Fe2+ : ferrous ion [Fe(II)]

2. Fe3+ : ferric ion [Fe(III)]

3. Mn2+ : manganese (II) ion

4. Mn3+ : manganese (III) ion

5. Mn4+ : manganese (IV) ion

Formulas of Ionic Compounds  

A. Ionic compounds are electronegatively neutral  

1. In order for ionic compounds to be electrically neutral,

 the sum of the charges of the cation and anion in each  

 formula must be zero

2. Example: aluminum oxide. 2(+3) + 3(-2) = 0

B. To name ions and ionic compounds:  

1. Name the cation by omitting the word ion and using a roman numeral if   the cation can have more than one charge  

2. Name the anion by adding the word IDE

3. Examples

a. NaBr —> Sodium Bromide

b. CaO —> Calcium Oxide

c. Mg3N2 —> Manganese Nitride  

d. Fe2S3 —> Iron(III) Sulfide  

C. NOTE: the subscript of the anion is the charge of the cation  

D. NOTE: the subscript of the cation is the charge of the anion

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Covalent Bonding and Molecules  

A. When compounds form between elements with similar properties with similar   properties, electrons are not transferred from one electron to another but   instead are shared in order to give each atoms a noble gas configuration  B. This approach is known as the Lewis Theory of Bonding, named for its  proponent, Gilbert Lewis  

C. Lewis Theory depicts bond formation in H2 (right)

D. A molecule may be an element or a compound  

E. Different samples of a given compound always contain the  

 same ratio, known as the law of definite proportions  

F. A molecule is a combination of at least two atoms in  

 specific arrangement held together by chemical forces  

 (chemical bonds)

G. A molecule may be an element or a compound

H. Different samples of a given compound always  

 contain the same ratio. This is known as the law  

 of definite proportions.  

I. If two elements can form two or more different  

 compounds, the law of multiple proportions tells  

 us that when the masses of two elements when  

 combined with each other to form more than one  

 compound are in a ratio of small whole numbers

J. The mass ratio of oxygen to carbon dioxide is 2.66:1, and the ratio of   oxygen to carbon in carbon monoxide is 1.33:1

K. The ratio of two such mass rations can be expressed as small whole numbers L. Diatomic molecules contain two atoms and may either be heteronuclear or  homonuclear

(a) would be homonuclear  

diatomic but it has more than  

two so it is not diatomic. Imagine  

the same colors but with only  

two atoms bonded.

(b) is polyatomic  

(c) is heteronuclear diatomic

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Formulas  

A. A chemical formula denotes the composition of the substance  B. A molecular formula shows the exact number of atoms in each element in a   molecule  

C. Some elements have two or more distant forms known as allotropes, such as   oxygen (O2) and ozone (O3) are allotropes of oxygen  

D. A structural formula shows not only the elemental composition but also the   general arrangements

E. An empirical formula uses whole-number ratios of elements to get the   smallest bit of information that we can get from observation. While the   molecular formulas tell us the actual number of atoms (the true formula), the   empirical formula gives the simplest formula. Sometimes the true formula IS   the empirical formula, like with H2O.  

1. Molecular formula: N2H2

2. Empirical formula: NH2

3. Look at the worked example 5.6  

Naming Molecular Compounds  

A. Remember: binary molecular compounds are substances that consist of just   two different elements

B. Nomenclature  

1. Name the first element that appears in the formula  

2. Name the second element that appears in the formula, change ending   to IDE

C. Greek prefixes are used to denote the number of atoms of each element 1. mono- is usually omitted for the first element

2. For ease of pronunciation, we usually eliminate the last letter of a prefix   that ends in “o” or “a” when naming an oxide

3. Worked example  

a. NF3 ???? nitrogen trifluoride

b. N2O4 ???? dinitrogen tetroxide

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D. Compounds containing hydrogen

1. The names of molecular compounds containing hydrogen do not   usually conform to the systematic nomenclature guidelines

2. Many are called by the common, nonsystematic names or by names   that do not indicate explicitly the number of H atoms present

a. B2H6 ???? Diborane

b. SiH4 ???? Silane

c. NH3 ???? Ammonia

d. PH3 ???? Phosphine

e. H2O ???? water  

f. H2S ???? hydrogen sulfide  

E. One definition of an acid is a substance that produces hydrogen ions (H+)   when dissolved in water (HCl is an example of a binary compound that is an   acid when dissolved in water)  

1. To name these acids:  

a. Remove the –gen ending from hydrogen

b. Change the –ide ending to on the second element to –ic  

 (Hydrogen chloride ???? hydrochloric acid)  

2. A compound must contain at least one ionizable hydrogen atom to be   an acid upon dissolving

Organic Compounds  

A. Our nomenclature discussion so far as focused on inorganic compounds,   generally defined as those without carbon

B. Organic compounds contain carbon and hydrogen, sometimes in combination   with other atoms

C. Hydrocarbons contain only carbon and hydrogen  

D. The simplest hydrocarbons are called alkanes

 E. Many organic compounds contain groups of atoms known as functional   groups, which often determine a molecule’s reactivity

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Covalent Bonding in Ionic Species  

A. Polyatomic ions consist of a combination of two or more atoms B. Formulas are determined following the same rule as for ionic compounds   containing only monatomic ions: ions must combine in a ratio that gives a   neutral formula overall  

C. Oxoanions are polyatomic anions that contain one or more oxygen atoms   and one atom (the central atom) of another element  

D. Starting with the oxoanions that end in –ate, we can name these ions: 1. The ion with one MORE O atom atom than the –ate ion is called the   per…ate ion. (ie ClO3- = a chlorate ion, so CLO4- = a perchlorate ion)  2. The ion with one LESS O atom than the -ate ion is called the –ite ion.   (ie ClO2- is the chlorite ion)

3. The in with TWO FEWER O atom than the -ate ion is called the   hypo…ite ion. (ie ClO- is the hypochlorite ion)

E. Oxoanions  

1. percholate  ClO4

2. chlorate  ClO3-

3. chlorite ClO2-

4. hypochlorite  ClO

5. nitrate  NO3-

6. nitrite  NO2-

7. phosphate  PO43-

8. phosphite  PO33-

9. sulfate  SO42

10. sulfite  SO32-

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F. Oxoacids, when dissolved in water, produce hydrogen ions and the   corresponding oxoanions  

1. An–ate ion is called ……. ic acid (HClO3 ???? chloric acid)

2. An –ite ion is called …….. ous acid (HClO2 ???? chlorous acid)  3. Prefixes in oxoanions names are retained in naming oxoacids

4. Oxoacids, can be monoprotic (one ionizable hydrogen) or polyprotic  (more than one ionizable hydrogen)

Hydrates

A. A compound that has a specific number of water molecules within its solid   structure  

B. For example, in its normal state, copper(II) sulfate has five water molecules   associated with it

C. Copper(II) sulfate pentahydrate —> CuSO4 x 2H2O

D. When the water molecules are driven off by heating, the resulting compound,   Cu(SO)4 is sometimes called anhydrous copper(II) sulfate

E. Anhydrous means the compound no longer has water molecules associated   with it  

F. Cu(SO)4 is white but Cu(SO)4 x 5H2O is blue  

Molecular and Formula Mass

A. The molecular mass is the mass in atomic mass units (amu) of an individual   molecule

B. To calculate the molecular mass, multiply the atomic mass for each element in   a molecule by the number of atoms of that element and total the masses C. Molecular mass of H2O = 2(atomic mass of H) + atomic mass of 0 D. Although the ionic compound does not have a molecular mass, we can use   its empirical formula to calculate its formula mass  

E. Because the atomic masses on the periodic table are average atomic   masses, the result of such a determination is an average molecular mass,   sometimes referred to as the molecular weight.

13

Percent Composition of Compounds  

A. A list of the percent by mass of each element in a compound is known as the   compounds percent composition by mass

B. Percent mass of an element = (n x atomic mass of element) / (molecular   or formula mass of compound) x 100%  

C. We could have also used the empirical formula of hydrogen peroxide (HO)   for the calculation  

D. In this case, we could have used the empirical formula mass to find the   mass in amu of one of the compounds

E. Worked Example 5.13  

Molar mass (M )

A. In a substance, the molar mass is the mass in grams of one mole of the   substance

B. The molar mass of an element is numerically equal to its atomic mass  1. 1 mol C = 12.01 g  

2. 1 C atom = 12.01 amu

C. The molar mass of a compound us the sum of the molar masses of the   elements it contains  

1. 1 mol H2O = 2 x 1.008 g + 16g = 18.02 g

2. 1 mol NaCl = 22.99 g + 35.45g = 58.44g  

The Octet Rule  

A. According to the octet rule, atoms will lose, gain, or share electrons in order to   achieve a noble gas electron configuration  

B. Only valence electrons contribute to bonding

C. Only two valence electrons participate in the formation of the F2 bond

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Lewis Structure

A. A lewis structure is a  

 representation of covalent  

 bonding

B. Shared electron pairs are  

 shown either as dashes or as pairs of dots  

C. Lone pairs are shown as pairs of dots on individual atoms

D. In a single bond, atoms are held together by one electron pair  E. In a double bond, atoms share two pairs of electrons  

F. In a triple bond, atoms are held together by three electron pairs  G. Bond length is defined as the distance between the nuclei of two covalently   bonded atoms  

H. Multiple bonds are shorter than single bonds

I. For a given pair of atoms, triple bonds are shorter than double bonds   which are shorter than single bonds

J. We quantify bond strength by measuring the quantity of energy required to   break it

Electronegativity and Polarity  

A. There are two extremes in the spectrum of bonding:  

1. Covalent bonds occur between atoms that SHARE electrons

2. Ionic bonds occur between a metal and nonmetal and involve ions B. Bonds that fall between these extremes are polar  

C. In polar covalent bonds, electrons are shared but not shared equally (the   delta is used to denote partial charges on the atoms)  

E. Pure covalent bonds- neutral atoms held together by equally shared   electrons  

G. Ionic bonds- oppositely charged ions held together by electrostatic attraction  H. Electron density maps show the distributions of charge

I. Typically, electrons spend a lot of time in red and very little time in blue

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Electronegativity  

A. Electronegativity is the ability of an atom in a compound to draw electrons to   itself  

B. There is no sharp distinction between non-polar covalent and polar covalent   or between polar covalent and ionic  

C. The following rules help distinguish among them:

1. A bond between atoms whose electronegativities differ by less than 0.5  is generally considered purely covalent or nonpolar

2. A bond between atoms who's electronegativities differ by the range of   0.5 to 2.0 is generally considered polar covalent  

3. A bond between atoms whose electronegativities differ by 2.0 or more  is generally considered ionic  

Dipole Movement, Partial Charges, and Percent Ionic Character  A. Regions where electrons spend little time is typically blue while regions where   electrons spend a lot of time is tropically red

B. An arrow is used to indicate the direction of electron shift

C. A quantitative measure of the polarity of a bond is its dipole movement (mu)  D. mu = Q x r (Q is the charge; r is the distance between the charges, and mu is   always positive and expressed in debye units denoted by the letter D)  E. 1 D = 3.336 × 10-30 coulomb meter.

F. Although the designations “covalent,” “polar covalent,” and “ionic” can be   useful, sometimes chemists wish to describe and compare chemical bonds   with more precision.

G. Comparing the calculated dipole moment with the measured values gives us a   quantitative way to describe the nature of a bond using the term percent   ionic character.

16

Lewis Structures and Formal Charge  

A. Formal charge can be used to determine the most plausible Lewis Structure   when more than one possibility exists for a compound.

B. To determine associated electrons:  

1. All the atom’s nonbonding electrons are associated with the atom  2. Half the atom’s bonding electrons are associated with the atom  C. When there is more than one possible structure, the best arrangement is   determined by the following guidelines:

1. A Lewis structure in which all formal charges are zero is preferred  2. Small formal charges are preferred to large formal charges  

3. Formal charges are associated with electronegativities.

Resonance  

A. A resonance structure is one of two or more Lewis structures for a single   molecule that cannot be represented accurately by only one Lewis structure B. Resonance structures are a human invention  

C. Resonance structures differ only in the positions of their electrons  

Exceptions to the Octet Rule: Incomplete Octets  

A. The central atom has fewer than eight electrons due to a shortage of electrons  1. Elements in group 3A also tend to form compounds surrounded by   fewer than eight electrons  

2. Boron, for example, reacts with halogens to form compounds of the   general formula BX3 having six electrons around the boron atom  B. The central atom has fewer than eight electrons due to an odd number of   electrons. (Molecules with an odd number of electrons are sometimes referred   to as free radicals)  

C. The central atom has more than eight electrons  

1. Atoms in and beyond the third period can have more than 8 valence e’s 2. In addition to the 3s and 3p orbitals, elements in the third period also   have 3d orbitals than can be used in bonding

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D. The bond between B-N has both the electrons contributed by the N atom  1. This type of bond is a coordinate covalent or dative bond

2. This type of bond formation is an example of a Lewis acid-base process BF3 is a Lewis acid: it can accept a pair of electrons

NH3 is a Lewis Base: it donates a pair of electrons

18

WORKED EXAMPLES  

Worked Example 5.2

Arrange MgO, CaO, and SrO in order of increasing lattice energy.  

Consider the charges on the ions and the distances between them. Apply Coulomb’s  law to determine the relative lattice energies. All three compounds contain O2- and all  three cations are +2. Recalling that lattice energy increases as the distance between  ions decreases, we need only consider the radii of the cations as all three contain the  same anion. From Figure 4.13, the ionic radii are 0.72 Å (Mg2+), 1.00 Å (Ca2+), and  1.18 Å (Sr2+).

Worked example 5.3

Name the following ionic compounds: (a) CaO, (b) Mg3N2, and (c) Fe2S3.  

Begin by identifying the cation and anion in each compound, and then combine the  names for each, eliminating the word ion.  

Solution  

(a) CaO is  

(b) Mg3N2 is  

(c) Fe2S3 is  

Think About It Be careful not to confuse the subscript in the formula with the charge in  the metal ion. In part (c), for example, the subscript on Fe is 2, but this is an iron(III)  compound.  

Worked Example 5.6

Write the empirical formulas for the following molecules: (a) glucose (C6H12O6), a  substance known as blood sugar; (b) adenine (C5H5N5), also known as vitamin B4;  and (c) nitrous oxide (N2O), a gas that is used as an anesthetic (“laughing gas”) and as  an aerosol propellant for whipped cream.

19

Worked Example 5.7  

Name the following binary molecular compounds: (a) NF3 and (b) N2O4.  

Strategy Each compound will be named using the systematic nomenclature including,  where necessary, appropriate Greek prefixes.  

Think About It Make sure that the prefixes match the subscripts in the molecular  formulas and that the word oxide is not preceded immediately by an “a” or an “o”.  

Worked Example 5.9  

Name the following ionic compounds: (a) Fe2(SO4)3, (b) Al(OH)3, and (c) Hg2O.  Strategy Begin by identifying the cation and anion in each compound, and then  

combine the names for each, eliminating the word ion.  

Think About It Be careful not to confuse the subscript in the formula with the charge in  the metal ion. In part (a), for example, the subscript on Fe is 2, but this is an iron(III)  compound.  

Worked Example 5.10

Name the following species: (a) BrO4-, (b) HCO3-, and (c) H2CO3.  

Strategy Each species is either an oxoanion or an oxoacid. Identify the “reference  oxidation” (the one with the –ate ending) for each, and apply the rules to determine  appropriate names.  

Think About It Make sure that the charges sum to zero in each compound formula. In  part (a), for example, Hg2+ + 2Cl- = (+2) + 2(-1) = 0; in part (b), (+2) + 2(-1) = 0; and in  part (c), 3(+1) + (-3) = 0.  

Worked Example 5.11

Determine the formula of sulfurous acid.  

Strategy The –ous ending in the name of an acid indicates that the acid is derived from  an oxoanion ending in –ite. The oxoanion must be sulfite, SO32-, so add enough  hydrogen ions to make a neutral formula.

20

Worked Example 5.12

Calculate the molecular mass or the formula mass, as appropriate, for each of the  following corresponds: (a) propane, C3H8, (b) lithium hydroxide, LiOH, and (c) barium  acetate, Ba(C2H3O2)2.  

Strategy Determine the molecular mass (for each molecular compound) or formula  mass (for each ionic compound) by summing all the atomic masses.  

Worked Example 5.13

Lithium carbonate, Li2CO3, was the first “mood-stabilizing” drug approved by the FDA  for the treatment of mania and manic-depressive illness, also known as bipolar disorder.  Calculate the percent composition by mass of lithium carbonate.  

Strategy Use Equation 5.1 to determine the percent by mass contributed by each  element in the compound.  

Worked Example 5.14

Determine (a) the number of moles of CO2 in 10.00 g of carbon dioxide and (b) the  mass of 0.905 mole of sodium chloride.  

Strategy Use molar mass to convert from mass to moles and to convert from moles to  mass. The molar mass of carbon dioxide (CO2) is 44.01 g/mol and the molar mass of  sodium chloride (NaCl) is 58.44 g/mol.  

Think About It Always double-check unit cancellations in problems such as these–errors  are common when molar mass is used as a conversion factor. Also make sure that the  results make sense. In both cases, a mass smaller than the molar mass corresponds to  less than a mole of substance.

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Worked Example 5.15

(a) Determine the number of water molecules and the numbers of H and O atoms in  3.26 g of water.  

(b) Determine the mass of 7.92×1019 carbon dioxide molecules.  

Strategy Use molar mass and Avogadro’s number to convert from mass to molecules,  and vice versa. Use the molecular formula of water to determine the numbers of H and  O atoms. 2 23

(b) 7.92×10 CO molecules × 6.022×10 CO molecules × 1 mol CO  

= 5.79×10-3 g CO2  

 

Think About It Again, check the cancellation of units carefully and make sure that the  magnitudes of your results are reasonable.  

Worked Example 6.1

Classify the following bonds as nonpolar, polar, or ionic: (a) the bond in ClF, (b) the bond  in CsBr, and (c) the carbon-carbon double bond in C2H4.

Strategy Electronegativity values are: Cl (3.0), F (4.0), Cs (0.7), Br (2.8), C (2.5).  Use this information to determine which bonds have identical, similar, and widely  different electronegativities.

Solution  

(a) The difference between the electronegativies of F and Cl is 4.0 – 3.0 = 1.0,  making the bond in ClF polar.

(b) In CsBr, the difference is 2.8 – 0.7 = 2.1, making the bond ionic. (c) In C2H4, the two atoms are identical. (Not only are they the same element, but  each C atom is bonded to two H atoms.) The carbon-carbon double bond is C2H4 is nonpolar.

22

Worked Example 6.2  

Burns caused by hydrofluoric acid [HF(aq)] are unlike any other acid burns and present  unique medical complications. HF solutions typically penetrate the skin and damage  internal tissues, including bone, often with minimal surface damage. Less concentrated  solutions actually can cause greater injury than more concentrated ones by penetrating  more deeply before causing injury, thus delaying the onset of symptoms and preventing  timely treatment. Determine the magnitude of the partial positive and partial negative  charges in the HF molecule.

Strategy Solve for Q. Convert the resulting charge in coulombs to units of electronic  charge. According to Table 6.2, μ = 1.82 D and r = 0.92 Å for HF. The dipole moment  must be converted from debye to C·m and the distance between ions must be  converted to meters.

Worked Example 6.4  

Draw the Lewis structure for carbon disulfide (CS2).

Setup  

Step 1: C and S have identical electronegativities. We will draw the skeletal structure  with the unique atom, C, at the center.

Step 2: The total number of valence electrons is 16: 6 from each S atom and 4 from the  C atom [2(6) + 4 = 16].

Step 3: Subtract 4 electrons to account for the bonds in the skeletal structure, leaving us  12 electrons to distribute.

Step 4: Distribute the 12 remaining electrons as 3 lone pairs on each S atom. Step 5: There are no electrons remaining after step 4, so step 5 does not apply.

Step 6: To complete carbon’s octet, use one lone pair from each S atom to make a  double bond to the C atom.

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Worked Example 6.5  

The widespread use of fertilizers has resulted in the contamination of some  groundwater with nitrates, which are potentially harmful. Nitrate toxicity is due primarily  to its conversion in the body to nitrite (NO2-), which interferes with the ability of  hemoglobin to transport oxygen. Determine the formal charges on each atom in the  nitrate ion (NO3-).

Strategy Follow the six steps to draw the Lewis structure of (NO3-). For each atom,  subtract the associated electrons from the valence electrons.

Setup The N atom has five valence electrons and four associated electrons (one from  each single bond and two from the double bond). Each singly bonded O atom has six  valence electrons and seven associated electrons (six in three lone pairs and one from  the single bond). The doubly bonded O atom has six valence electrons and six  associated electrons (four in two lone pairs and two from the double bond.)

Solution  

The formal charges are as follows: +1 (N atom), -1 (singly bonded O atoms), and 0  (doubly bonded O atom).

Worked Example 6.6

Formaldehyde (CH2O), which can be used to preserve biological specimens, is  commonly sold as a 37 percent aqueous solution. Use formal charges to determine  which skeletal arrangements of atoms shown here is the best choice for the Lewis  structure of CH2O.

Solution Of the two possible arrangements, the structure on the left has an O atom  with a positive formal charge, which is inconsistent with oxygen’s high electronegativity.  Therefore, the structure on the right, in which both H atoms are attached directly to the  C atoms and all atoms have a formal charge of zero, is the better choice for the Lewis  structure of CH2O.

O

H C O H

H C H

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Worked Example 6.7  

High oil and gasoline prices have renewed interest in alternative methods of producing  energy, including the “clean” burning of coal. Part of what makes “dirty” coal dirty is its  high sulfur content. Burning dirty coal produces sulfur dioxide (SO2), among other  pollutants. Sulfur dioxide is oxidized in the atmosphere to form sulfur trioxide (SO3),  which subsequently combines with water to produce sulfuric acid – a major component  of acid rain. Draw all possible resonance structures of sulfur trioxide.

Strategy Following the steps for drawing Lewis structures, we determine that a correct  Lewis structure for SO3 contains two sulfur-oxygen single bonds and one sulfur-oxygen  double bond.

O

O S O

O

O S O

O

O S O

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