Principles of Chemistry I
Principles of Chemistry I CHM 211
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Ionic CationAnion Covalent Nonmetals Only Compounds Containing Hydrogen Rule Name of cation name of anion word ion dropped Examples NaCl sodium chloride MgClz magnesium chloride Fe3N2 ironII nitride Na2C03 sodium carbonate CaHz calcium hydride CaN032 calcium nitrate Comment The name does not indicate the numbers of cations and anions because there is only one possibility for the ions to combine to form a compound Rule a Less electronegative l HNonmetal l element rst I 1 when one of the elements is hydrogen b Number of atoms of each element is speci ed by Greek pre xes c Pre x mono at the beginning is dropped Pre xes 1 mono 6 hexa 2 di 7 hepta 3 tri 8 octa 4tetra 9nona 5penta 10 deca Examples N204 dinitrogen tetroxide CO carbon monoxide C02 Carbon dioxide N20 dinitrogen monoxide Comment Tetraoxide becomes tetroxide monoxide becomes monoxide Rule 1 Without the presence of H20 Hydrogen iide Examples HCl hydrogen chloride HF hydrogen uoride HZS hydrogen sulfide Rule 2 When dissolved in HZO Hydroiic acid Examples HClaq hydrochloric acid HFaq hydro uoric acid HZSaq hydrosulfuric acid Comment a These Hcontaining compounds are named as if they were ionic compounds b The aq in the formula of the acids is often omitted when it is obvious from the context that they are acids etc H 39 do not follow a rule concerning the order in which the elements are written and should be memorized H20 NH3 etc Rule 1 Without HZO Cation anion Hydrogen hypoiite Hydrogen iite Hydrogen fate Hydrogen periate Rule 2 When dissolved in HZO Hypoious acid ious acid iic acid periic acid Exam ple s HClO hypochlorous ac1 HClOz chlorous HC103 chloric acid HClO4 perchloric acid HNOZ nitrous acid HNO3 nitric acid H2503 sulfurous ac1d H2504 sulfuric aci H3PO4 phosporic acid 115 Vapor Pressure The pressure of a gas over the same substance as a liquid is called its vapor pressure At constant temperature when a liquid and gas of the same substance are in a sealed container they will achieve equilibrium The vapor pressure remains constant because as many molecules gain enough energy through collisions to leave the liquid phase as stick to the liquid when gas molecules collide with the liquid surface Vapor pressures decrease as intermolecular forces become stronger Substances that evaporate easily are said to be m Volatile substances have high vapor pressures and evaporate readily Boiling and evaporation both result in the conversion of a liquid to a gas When the external pressure equals the vapor pressure of a substance the substance m When the external pressure is greater than the vapor pressure the substance evaporates The normal boiling point is the temperature at which a substance boils when the external pressure is 1 atmosphere 116 Phase Diagrams Phase diagrams are a plot of melting boiling and sublimation temperatures as a lnction of pressure A number of such points are measured and plotted on a graph like that below The points are then connected by a straight or curved line as appropriate allowing the melting point boiling point or sublimation point of a substance to be determined at a pressure that had not before been examined critical point DHsrJVJOHDU Temperature SO2 g H20 gt HZSO3 aq a major component of acid rain Metalloids have intermediate properties They usually look like metals but are brittle They conduct heat and electricity but poorly Their oxides are amphoteric can act both acidic and basic depending on the other reagent or nearly so 77 Group Trends for the Active Metals Group IA The Alkali Metals These are so metals with low melting points Their appearance and conductivity are typical of metals All have low densities and several would oat on water if they didn t react with it Like nearly all metals they cannot exist in nature as the ee element because of high reactivity Alkali metals readily react with both oxygen and water Each alkali metal has the lowest ionization energy and is the A Sheet Ofmetal atoms most active metal in its period Very high second ionization energies lead exclusively to 1 ions in all of their compounds Each alkali metal is obtained by a process called electrolysis which occurs when an electrical current is passed through a salt solution or molten salt This process forces an electron onto a metal ion and pulls one olT a halide ion to yield the respective elements For example 2 NaClw m 2 Na C12 9 Each reacts with nearly all nonmetals to form salts For example 2 M H2 gt 2 MHS H39 hydride 2 M X2 gt 2 MXS X halogen or with water to produce hydrogen which may ignite because the reaction is very exothermic So ll electron af nity and because there is no shielding it has an unusually high ionization energy This results in hydrogen being fairly nonreactive For example despite the stability of water a mixture of hydrogen and oxygen will not combine without a spark or ame to initiate the reaction The explosive power of hydrogen results from the energy released when it forms strong bonds to other elements such as oxygen Hydrogen reacts with metals to from anionic hydrides and nonmetals to form molecular compounds 2 Na H2 a 2 NaHS H39 hydride H2 X2 a 2 HXg X halogen Group VIA The Chalco gens This group contains all three types of elements nonmetals O S Se a metalloid Te and a metal Po thus generalizations are dif cult All are relatively dense We will discuss only oxygen and sulfur Oxygen exists in two forms naturally dioxygen OZ and ozone H O3 These are called allotropes different forms of an element under Dioxygen the same cond1tlons Dloxygen 1s a colorless odorless react1ve gas Ozone is a colorless pungent very reactive gas Its extra reactivity originates in the energy required to make it from dioxygen Oxygen usually forms the 0239 ion in ionic compounds although two other Ozone oxidation states occasionally occur Reactions of oxygen and ozone with metals result in oxidation of the metal Interestingly we know today that ozone is damaging to lung tissue ie ozone alerts but less than a hundred years ago physicians thought that the tineg sensation in the lungs was bene cial and they advised patients to breathe ozone from special generators Sul lr has several naturally occurring allotropes of which 88 is the most stable It is a Sul lr yellow low melting solid In addition to the expected 2 ion sulfur also has positive oxidation states 2 4 and 6 available In ancient times it was called brimstone Group VIIA The Halo gens All halogens exist as colored diatomic molecules F2 is a yellow gas C12 is a yellowgreen gas Br2 is a deep redbrown liquid and 12 is a grayishblack solid All possess high electron A halogen molecule a 39mities which causes them to have high reactivities and to have a stable 1 oxidation state They react with metals to form ionic compounds and nonmetals to form molecular compounds Like the alkali metals and alkaline earth metals they are too reactive to exist in nature as free elements All except uorine have the 1 3 5 and 7 oxidation states available Group VIIIA The Noble Gases The noble gases have physical properties typical of nonmetals All are colorless monoatomic nonreactive gases with very low melting points Like hydrogen helium does not solidify at any temperature at atmospheric pressure He Ne and Ar form no compounds whatsoever All have high ionization energies and positive electron a 39mities Krypton does form a single compound with uorine KrFZ while xenon forms a number of compounds with uorine and oxygen Xenon does so because although it is a noble gas its rst ionization energy is not very large This is because its outer electrons are so far from the nucleus that the latter s pull on the former is diminished Helium holds the distinction of being the only element not discovered on Earth August 25 2008 AHm AHWiAHE AHMfAHBAH As you can see from Table 82 p 305 lattice energies are all very endothermic Why First lattice energy is really a measure of the attraction ions have for each other in the solid state The magnitude of the attraction arises from the construction of the crystal and the nature of ionic interactions The simplest crystal pattern to visualize is that of sodium chloride so we ll use it as an example but all other patterns work the same way Consider an isolated sodium ion All nearby chloride ions will attract it since they each have a negative charge Unlike in a molecule though ions do not bind to a speci c neighbor they are attracted to several neighbors in this case 6 X I X xrx Likewise each chloride ion is surrounded by and attracted to 6 sodium cations This explains why it is so dif cult to separate an ion from the lattice The lattice holds each ion in place by strong interactions at several sites The energy of attraction for any two ions is given by EkQ1Q2 d where Q1 Q2 are the charges on the ions and d is their distance of separation This equation is consistent with what we would expect Lattice energy increases in magnitude as the charges on the ions increase and increases as the ions get smaller and hence pack closer together Ionic compounds have the following general properties High melting points crystalline hard brittle All of these properties derive om the crystal lattice and the requirement of large amounts of energy to displace ions located in that lattice We ll discuss this thher in Chapter 11 9 Groups 5 amp 6 subtract 04 going down for Groups 3 amp 4 subtract 03 going down the group and for Groups 2 and l subtract 02 and 01 respectively While this gives values different from your book i electronegativity is a qualitative concept and ii it works fairly well in predicting relative bond polarities 85 Drawing Lewis Structures I ve drawn a few Lewis structures for you on the board but have not discussed a procedure for how to do so The following rules will allow you to draw them This topic is extraordinarily important if you plan to take organic chemistry 1 Add up the total number of valence electrons on all atoms If you are working with a polyatomic ion eg acetate or ammonium ions add an electron for each negative charge or subtract an electron for each positive charge N Write the atomic symbols of the atoms in the correct spatial arrangement 3 Draw a line to connect each pair of bonded atoms Subtract 2 electrons from the total for each bond drawn 4 If the total from 3 equals that from 1 stop If not see if the octets can be completed by placing the remaining electrons on the atoms If so do it and stop 5 If not change one or more single bonds to multiple bonds and recalculate until all atoms except H have an octet It is important to remember that hydrogen always forms exactly one bond NH3 5 3l 8 electrons H N H I 8e 32 e 2e 1 H lTil H The nitrogen now has H an octet of electrons HCN l 4 5 10 electrons H C N lOe 22 e 6e 1 4e 2 e39 2e H CEN H CN 6e 2 e 4e The carbon and nitrogen now have an octet of electrons H CEN Formal Charge It is important to keep in mind that formal charges are not real charges They are a bookkeeping system for electrons When two atoms bind together covalently each atom usually contributes one electron to the bond However occasionally one of the atoms contributes both electrons to the bond These bonds are called coordinate covalent or dative bonds When this happens electron density shi s away from the atom which contributes both electrons towards the atom that contributes none Formal charge is calculated as follows i Calculate the total number of valence electrons on all atoms ii Subtract from that the number of bonds to the atom m the number of electrons not involved in bonding on the atom Formal charge total valence e39 7 bonds 7 e39 in lone pairs The sum of the formal charges on each atom in a molecule or ion will equal the overall charge on the atom or ion Let s look at the ammonia example from above In Section 87 we ll work 3 more interesting examples In general atoms with large negative formal charges are more electron rich than one might expect initially while those with large positive formal charges have less electron density than would initially be expected NH3 Each hydrogen has one valence electron forms one bond and has no unbound electrons FC l l 0 0 The nitrogen has 5 valence electrons 3 bonds and 2 unbound electrons FC5320 Now consider the compound H3NBF3 It is sometimes written H3N gtBF3 because the electrons for the bond between nitrogen and boron come only from nitrogen Each hydrogen has one valence electron forms one bond and has no unbound electrons FC l l 0 0 Each uorine has seven valence electrons forms one bond and has six unbound electrons FC 7 l 6 0 The boron has 3 valence electrons forms 4 bonds and has no unbound electrons FC 3 4 0 l The nitrogen has 5 valence electrons forms 4 bonds and has no unbound electrons FC 5 4 0 1 So what does this mean It certame does not mean that the boron atom is a lot like an anion and the nitrogen atom is a lot like a cation It doesn t necessarily even mean that the boron has a partial negative charge and the nitrogen a partial positive charge What it does mean is that the boron atom has more electron density on it than you might have at first guessed based only on their positions on the periodic table or electronegativities vide infra and the nitrogen atom less electron density than a first guess 86 Resonance Structures The examples you ve seen so far are typical but it turns out that there are some molecules for which it is possible to write 2 or more dilTerent Lewis structures An example is ozone o o Q Q 9 Q Your initial response is probably that these molecules are identical because ipping the first one over would yield the second But the double bond connects dilTerent atoms in the structures 12 and the difference is real imagine the oxygens on the ends as different isotopes In reality only one structure represents ozone and it is an average of the two you see here The double headed O arrow indicates there are 2 energetically reasonable structures not that the molecule switches back and forth between the structures It does not Evidence for this comes from bond length data If either of the structures was correct we would eXpect to see 2 bonds of different lengths In fact both distances are the same and they are intermediate between the length of a single and double bond Energetically similar structures with the same atom connectivity are called resonance structures The word resonance is an unfortunate one because it implies the molecules resonate between different forms Again they don t there is only one structure Resonance forms simply show the inadequacy of attempts to depict them In general resonance forms with smaller formal charges are favored over those with larger ones Likewise the closer formal charges are to each other on the molecule the better In each case the goal is to minimize the separation of charge We ll see an example at the end of Section 87 There are three requirements for resonance structures i the atom connectivity must be the same for each ii the bonding must be reasonable and iii they must be energetically similar The nitrate ion has three resonance forms 0 6 b III 39I N As you can see it appears that the double bond rotates around the ion Thus all three criteria are met The same atoms bind to each other 3 N O bonds no others Each atom has a complete l3 octet so the bonding is likely to be acceptable Since each structure possesses two NO single bonds and one NO double bond and the spatial arrangement of the atoms is identical in all three structures the energy of each structure must be the same as the others As we shall see shortly not all resonance structures will be so similar Read Resonance in Benzene on your own 87 Exceptions to the Octet Rule Odd Number of Electrons A relatively small number of compounds have an odd number of valence electrons These compounds are sometimes called free radicals or more simply radicals You may have heard of these compounds from advertisements for vitamins or skin care products In general odd electron compounds display very high reactivity and can do great damage in biological systems Normally when odd electron compounds form two of the molecules will combine and the odd electrons will pair in a bond eg Fluorine atoms have 7 electrons The unpaired electrons on two uorine atoms pair to form a single bond This kind of molecule is rarely observed in General Chemistry but will be a little more common in later years Nitric oxide is a fairly typical odd electron compound N N Which of these forms is favored and why Less Than an Octet This is very rare if one excludes molecules that also have an odd number of electrons It occurs when one of the atoms can t get 8 electrons in its valence shell r F 14 Like odd electron compounds those with less than an octet will usually react to raise their electron count For example aluminum chloride AlCl3 in the gas phase has the structure 1146 o Al Al 1 l You The extra bonds come from lone pairs on chloride ions providing the electrons for a new bond between the two AlCl3 units More Than an Octet This is the most common of the exceptions and is sometimes called an expanded octet In most cases a central atom possesses the extra electrons Sulfur has 12 valence electrons Expanded octets only occur for elements on the third period and lower Two factors account for this First completing an octet lls all low energy orbitals s amp p Beginning with the third row d orbitals are available for population Thus paired s and p electrons can be unpaired with one of the electrons om each orbital promoted to the d set The d orbitals are close enough in energy to the p orbitals for this to be possible T Tt T T T T T T T T 3s 3p 3d 3s 3 17 3d The other reason is that elements on the second row are too small to t more than four substituents around them Expanded octets occur most frequently when highly electronegative elements are bound to a central atom eg F 0 Cl etc Expanded octets occur because of the energy derived from the additional bonds more than compensates for the energy required to promote the electron to a higher energy orbital 6 2 O 2 O 2 Q Q l Q l Q l resonance form 4 resonance forms 6 resonance forms FCS6402 FCS6501 FCS6600 FCO6l6l FCO6l6l FCO6l6l singlebond FCO6240 FCO6240 doublebond This ion has a total of 11 resonance forms The first resonance structure has the advantage of all atoms possessing an octet of electrons but is raised in energy by all atoms possessing a formal charge one of which is large The final structures have the lowest formal charges but are raised in energy by having a sul Jr atom with 12 valence electrons The sulfate ion would have an intermediate structure with all bond lengths equal and somewhat shorter than an SO single bond 88 Strengths of Covalent Bonds The energy required to break a particular bond within a gas phase molecule is called its m mgy D It is sometimes called bond enthalpy One would eXpect that the energy required to break an XY bond would depend the atoms bound to X and Y While true to some extent the attached groups usually have only a small effect on bond energy Table 84 p 326 lists bond energies for numerous common bonds Bond energies tell us something about reactivity As bond energies increase stronger bonds reactivity decreases Reactions frequently happen at the weakest bond in a molecule Bond quot 39 39 and the F quot 39 39 of Reaction You learned in Chapter 5 that almost all reactions have a heat change associated with them and that those reactions almost always release heat exothermic Where does that energy come 16 from In all reactions existing bonds are broken and new bonds are formed Almost invariably the new bonds are stronger than the old ones and that energy dilTerence accounts for most of the heat released Thus the dilTerence in bond energies for reactants and products gives a quick estimate of the heat of reaction Aern z 2Dbonds broken 2Dbonds formed Note the order It occurs this way because bond energies are positive H H H H cc H2 4 HiciciH H H H H AHIXH z DCC DHH DCC 2DCH z 1 mol6l4 kJmol 1 mol436 kJmol l mol348 kJmol 2 mol4l3 kJmol 2 124 kJ actual value is l37 1d ca 10 higher than the estimate For any two elements the bond length decreases as bond energy increases August 25 2008 Only One Ion Possible Monatomic Positive Ions Cations More Than One Ion Possible transition metals Rule Cations formed from nonmetal atoms have names that end in iium Rule I Name of element ion Rule Examples Positive charges NH4 ammonium ion Examples indicated by a Roman H3O hydronium i011 Na SOdium ion numeral Hg2 mercurYG i011 Mg magnes1um lon H hydrogen ion Examples Fe2 ironH ion Fe3 ironIII ion Cu copperI ion Cu2 copperII ion Negative Ions Anions Monatomic Oxyanions Oxyanions Others and Containing Oxygen Containing Hydrogen Exceptions Rule Rule Rule Rule Stem of the Least oxygen hypoiite ion H oxyanion hydrogen These items do not element name Less oxygen iite ion name of oxyanion or follow any rules ide More oxygen fate ion bi oxyanion they must be Most oxygen periate ion Hz oxyanion dihyrogen memorized Examples name of oxyanion H39 hydride ion Examples Examples F39 uoride ion ClO39 hypochlorite ion Examples OH39 hydroxide O392 oxide ion ClOz39 chlorite ion HCOg39 hydrogen carbonate ion N393 nitride ion ClO339 chlorate ion ion or bicarbonate ion CN39 cyanide ion ClO439 perchlorate ion HSO439 hydrogen sulfate ion SCN39 thiocyanate SOg392 sulfate ion or bisulfate ion ion SO4392 sul te ion HzPO439 dihydrogen Oz392 peroxide phosphate ion ion Comment Halogens except Comment CzHgOz39 acetate F form all four ions Most HzCO3 is not named ion others only form two ions according to this rule because the iite and the fate ions it is a compound not an ion Chanler 3 r Sloichiome rl Stoichiometgy refers to the quantity relationship between reacting chemical species For example when hydrogen and oxygen gases react to form water two hydrogen molecules react with one oxygen molecule to form two watermolecules 0 H H H H o o gt H H o H H 2HZ oZ 4 2HZO The 2 to 1 to 2 relationship is the stoichiometry ofthe reaction Stoichiometry also allows the number of atoms or molecules involved in a reaction to be related to the measured quantities mass and volume for gases or liquids ofthe substances Thus 1 gram ofhydrogen reacts with 8 grams ofoxygen to yield 9 grams ofvwater 31 chemical Equations By far and away the most convenient way of referring to chemical reactions is the chemical equation The generic form is given as aA bB 4 CC dD where a b c and d are whole number balancing coef cients 1 is not used A amp B are reactants while c amp D are products Sometimes the symbols A heat orhv light are placed above the arrow 39 439 T 394 39 39 39 Also sometimes the solvent is placed below the arrow Usually the physical state ofeach species will be indicated by placing one ofthe following symbols after each reactant and product g gas 2 liquid s solid aq aqueous dissolved in water A chemical equation does not tell you 1 Reaction rate how fast the reaction goes 2 Reaction pathway Most reactions go through several steps on the way to completion We ll discuss both ofthese topics in CHM 212 How does one balance a chemical equation determine the coef cients We begin by assuming you know the reactants and products in the reaction mixture The steps are 1 Write out all reactants and products on the correct side of the arrow 2 Count out how many atoms of each element or group appear on either side of the arrow 3 Choose the element or group that appears in the fewest number of places d has different numbers on either side of the arrow Balance using coef cients 4 Repeat with the remaining atoms groups 5 Make all coef cients whole numbers by multiplying through by the least common denominator of the existing coef cients Examples The blue atoms are being balanced iPfClz gtiPC13 iC7H14ioz gtiC02iH20 P 3012 gt 213013 C7H14 02 gt 7002 H20 2P3C12 gt2PC13 C7H1402 gt7C027H20 C7H14 212 02 H 7C02 7H20 2C7H14 2102 gt 14C02 14H20 There are a couple of points worth mentioning 1 Never change subscripts in formulas Doing so changes the identity of the substance For example H20 is water but H202 is hydrogen peroxide and you wouldn t want to mix these up if you re thirsty 2 Coef cients 3 should not have a common divisor That is reaction between hydrogen and oxygen to form water is not properly balanced as 4 H2 2 02 gt 4 H20 because each coefficient is an even number To balance it correctly you must divide by 2 point 5 above One common mistake when writing out consecutive reactions is to write the first reaction then add a reactant to the product draw an arrow then the nal reaction For example let s say that you burn phosphorus in oxygen to yield diphosphorus pentaoxide then react it with water to produce phosphoric acid It is incorrect to write 4 PS 5 02 g gt 2 P205 5 6 H20 2 gt 4 H3PO4 aq The reason for this is that it implies that the first reaction produces the water which then further reacts with the P205 The correct way to write this is 4 PS 5 02 g H 2 P205 5 P205 5 3 H20 2 gt 2 H3PO4 aq 32 Some Simple Patterns of Chemical Reactivity In Chapter 2 you learned about the derivation of the names of lithium sodium and potassium p 7 of notes Throughout the periodic table one nds patterns of similar reactivity down a group Examples include 2 MS 2 H20 2 MOHGMD H2 g M all alkali metals CS 02 g gt C02 g amp Sis 02 g gt Si02 s The last two reactions illustrate an important point While great similarities in reactivity frequently exist the products may have radically different properties Common Types of Reactions Combination reactions occur when 2 or more substances react to form one The reactions that occurred in an oldfashioned ash bulb were combination reactions 3MgN2 gtMg3N2 amp 2Mg02 gt2Mg0 Decomposition reactions occur when one substance breaks up into more than one substance They are the reverse of combination reactions The following reaction shows how limestone decomposes to lime CaCO3 Ca0 C02 with heating Combustion reactions are rapid reactions that produce a ame Oxygen is usually a reactant although sometimes another substance such as nitrogen in the magnesium example above can also participate CH4 2 02 a C02 2 H20 burning of natural gas 33 Formula Weights Your book draws a correct distinction between molecular and formula weights In real life though the latter term is rarely used and the former does double duty Both can be thought of as the sum of the masses of the atoms that compose a compound Thus for glucose C6H1206 molecular weight 6120ll amu 12l0079 amu 6159994 amu 180 157 amu Note that I don t round off to the tenths place like the book does You shouldn t either The reason is twofold With calculators it isn t worth the trouble and in some cases you may make a significant figure error Percent Composition This is a common and very useful calculation in chemistry When a new compound is prepared its elemental composition is usually determined The following calculation provides the expected composition based on molecular formula For glucose 612011amu C x 100 4000 180157 amu H W x 100 671 180157 amu 0 W x100 5328 180157 amu 9999 There are three points worth noting here 1 Substituting grams for amu gives exactly the same answers see the next section for why this is so Use whichever unit the problem requires 2 You should add the percentages together and make sure they total 100 i 002 There will frequently be a very small deviation from 100 because of rounding but it should never be much larger than a couple hundredths of a percent 3 By convention percent composition is almost always reported to a maximum accuracy of hundredths of a percent even though the calculation may be far more accurate This is because percent composition calculations are usually used in the determination of the identity of unknown materials It is quite difficult to obtain materials more than 999 pure and combined with experimental error in making the measurements hundredths of a percent does quite well for almost all purposes 34 Avogadro s Number and the Mole As you ve already surmised on the macroscopic scale the amu is as inconvenient as the gram is at the atomic scale so we use the gram for normal scale situations However it would be nice to be able to move easily between these scales The conversion factor 6022 x 1023 allows for such easy movement As your book points out even relatively small amounts of material contain unimaginably large numbers of atoms For example one teaspoon of water holds 1023 molecules For that 6 reason a collective term was developed The mole contains 6022 x 1023 items Avogadro s number Memorize this value The mole is similar to the dozen 12 items and gross 144 items in that it is a counting term Technically a mole contains the number of atoms in exactly 12 g of 12C Exactly means there is no error in this number Using the gamu conversion factor means if one carbon12 atom weighs exactly 12 amu then one mole of 12C weighs exactly 12 g Therefore one chooses the mass units based on the scale one works on This is why the periodic table does not have units next to the masses Example How many moles of glucose C6H1206 are in 100 g How many molecules How many carbon atoms MWC6H1206 180157 amumolecule AMWc HuO 180157 gmole mac mo6 100 g L01 555 x10393 mol 180157 g 23 moleculescmno 100 g 1mol 16022 x10 molecules 180157 g mol 334 x 1021 molecules 23 atomsc 100 g 1mol 16022 x10 molecules 6 atomsC 180157 g mol molecule 200 x 1022 atomsC On p 92 your book makes a very valuable observation That is when discussing materials be careful to think about what is really being asked for in the text The book notes that if nitrogen is listed as a product of reaction what is typically meant is molecular nitrogen or N2 not atomic nitrogen N There is similar ambiguity in other materials e g oxygen and chlorine and you should learn to anticipate what is expected in these situations 35 Empirical Formulas from Analyses A short time ago you learned how to calculate the percent composition of a material At that time you learned this was a very use ll calculation for identifying a substance In principle the experiment is simple For example we can burn a sample containing C amp H and collect and weigh the C02 amp H20 produced From the weights the amount of C and H in the original sample can be determined Similar experiments are performed for other elements The calculation follows the following steps 1 Confirm the percentages total to 100 2 Assume there are 100 g of compound present Then the amount of each element present in grams will equal the percentage of that element present L V Calculate the number of moles present of each element Make sure to use the element in elemental form eg oxygen as 159994 gmol not 319998 gmol A C Calculate the molar ratios diViding by the element with the smallest number of moles present Retain units throughout the process Example An unknown compound is found to be composed of 400 carbon 67 hydrogen and 533 oxygen Its molecular weight was measured as 180 gmol What are its empirical and molecular formulas 1mm mol 400 79 333 mol c gc12011gC C 1mm mol 67 H 665 mol H gH10079 gH H lmd mol 533 0 333 mol 0 g0159994g0 0 molar ratios Note C 333molc 100 333molc 665 molH 333 molC H 200 molH per molC O M 100 molO per molC 333 molc The empirical formula CHZO that with one exception all of the numbers in this example have units Be careful in rounding the molar ratios You should round numbers very close to integers e g 097 rounds to l but numbers close to 025 033 and 05 should be treated differently In these cases all ratios are multiplied by a factor to yield whole number values ratios Fe Eg a compound has the following 100 O 133 oxygen atoms per iron atom Multiply through by 3 to get Fe3O4 If you also have the molecular weight of the material you can determine the molecular formula When going from empirical formula to molecular formula all of the subscripts are multiplied by a whole number Example In the preVious problem you were told the molecular weight was about 180 gmol you calculate the molecular formula as follows The empirical weight of CHZO is 30 gmol MW 180 gmol 60 EW 30 gmol Thus there are 6 empirical units in each molecule 3 C6H1206 Read Combustion Analysis on your own 36 Information from Balanced Pmm nnc In the real world when we want to know how much of a substance we have we usually 9 measure its mass or volume We never count the number of molecules present On the other hand when we discuss a chemical reaction we discuss the reaction of individual particles How do we reconcile these views In a balanced equation the coef cients tell us the relative number of atoms or molecules reacting Because the mole is a number term the coef cients also tell us the relative number of moles of atomsmolecules involved in a reaction Finally molecular weights allow us to relate numbers of species with their masses This allows the amount of all substances produced or consumed in a reaction to be calculated from only one reaction species Two frequently asked questions in a chemistry lab are 1 if I have x grams of reactant A what mass of reactant B will I need to react with it and 2 if I have x grams of reactant A what mass of product C will be generated in the reaction The general solution to these problems is 1 Write out the balanced equation 2 Convert all quantities to moles 3 Using the balanced equation determine the molar ratio of desired to known species 4 Convert the molar amount of desired species into the appropriate units 5 Check for reasonability This is important Example Consider the reaction C82 3 02 gt C02 2 802 What mass of 02 is required to react completely with 934 g carbon disulfrde lmolcSZ 3 m01023200 goZ 118 gm mass 934 02 gCSZ 7613gcs2 lmol lmolOZ csz In a similar manner the masses of C02 and 02 produced can be calculated 540 gcoz and 157 gsoz 37 Limiting Reactants In practice reactions are rarely conducted with exact stoichiometric amounts of each reagent for three reasons First if one of the reactants is a gas it may be dif cult to measure out that reactant Second one reactant is nearly always more expensive than the others In this case one wants to use all of that reagent Adding small excesses of the others does this Finally some reactions can proceed through multiple paths and using excess amounts of some reagents can maximize the yield of a desired product When some reagents are used in excess the completely consumed reagent is called the limiting reagent or less commonly limiting reactant When doing a limiting reagent calculation rst convert the masses of all reactants to moles which then must be converted to a common scale using stoichiometric coef cients Example Reaction of tungsten with chlorine gas yields tungstenVI chloride Find the mass of unreacted starting material when 126 g of tungsten is treated with 136 g of chlorine gas How much tungstenVI chloride is formed We begin by generating a balanced chemical equation Ws 1 3 C12 g gt S 1 mol a mol 126 00685 mol w gw 1839 gw w 1 mol mol 136 0192 mol C12 gc12 7090 gm C12 Now calculate how much chlorine is needed to completely react with tungsten 3 mol 00685 molw 1 1C 0206 molcu but you only have 0192 mol of chlorine so mo w it is the limiting reagent Calculate the mass of tungsten consumed massw136 gc12amp 1m 1w KM 118 gw 7090 gch 3 molch molW massWexcess 126 g 118 g 08 gw IImOI wCl6 254 gwc16 b mass 136 g WC16 C12 7090 gC12 3 mol mol C12 WC16 Theoretical Yields In real life reactions rarely generate all of the product you d expect from the reaction stoichiometry because there are frequently side reactions that use up the starting materials to make other products Equilibria can also occur Also sometimes the products or reactants aren t extremely stable under the reaction conditions The theoretical yield for a reaction is the amount of product expected based on stoichiometry Part b of the preVious calculation was a theoretical yield calculation The amount of product obtained is called the actual yield The ratio of actual to theoretical yields is called the percent yield actual yield percent yield x 100 theoretica 1 yield Using the preVious example what would the percent yield be if 176 g of tungstenVI chloride actually formed 17 percent yield g 100 693 254 g In this case the side reactions might be production of WC12 or WCl4 resulting from incomplete reaction of the starting materials Septermber 5 2008 Chapter 6 Electronic Structure of Atoms The last two chapters discussed types of reactions and the energy changes associated with them We now begin a look at atoms and molecules at a fundamental level Earlier we saw that energy is consumed or released when a chemical reaction occurs Why Because during reactions some bonds brake and others form and all will have diiferent strengths For example typically stronger bonds replace weak bonds and energy is released Chemical bonds involve the interaction of electrons between atoms For this reason we must understand the interaction of electrons with nuclei and each other That is what this chapter is all about 61 The Wave Nature of Light When a fire burns you can feel the heat from the ames without touching them The reaction between wood and oxygen releases this energy Electromagnetic radiation carries energy through space and is sometimes called radiant energy There are a number of diiferent kinds of such energy including visible light Xrays and radiowaves as well as the infrared radiation coming from the re They all share the property of traveling in waves at 300 x 108 ms in a vacuum you should remember this value wavelength 7 m A amphtude V V Wavelength X is the distance between successive peaks Amplitude is the distance from the centerpoint to the top of the wavecrest The distance covered by one wavelength is sometimes called a wavecycle or more simply a M The wavecycle is frequently used as a counting term For example 5 wavecycles passed this point in one second It is important to remember that when waves move through a medium they don t carry the Be or ls2 2s2 or He 2s2 ls 2s B g T or ls2 2s2 2P1 or He 2s2 2p1 ls 2s 2p The next electron going into the atom will go into a 2p orbital the question is Does it go into a vacant orbital or spin pair with the other electron Two electrons in the same orbital spend much of the time relatively close to one another Since electrons repel one another their energy increases Electrons placed in other orbitals have no such energy increase so electrons pair only after each orbital in the subshell contains one electron A second question is will the spins align or oppose Remember that a spinning charge has an associated magnetic eld The magnetic elds of electrons spinning in the same direction reinforce one another while those spinning in opposite directions offset one another Thus unpaired electrons will align their spins to maximize the resulting magnetic eld Hund s Rule C E E or ls2 2s2 2p2 or He 2s2 2p2 ls 2s 2p N E E or ls2 2s2 2p3 or He 2s2 2p3 ls 2s 2p 0 E E or ls2 2s2 2p4 or He 2s2 2p4 ls 2s 2p F E E or ls2 2s2 2p5 or He 2s2 2P5 ls 2s 2p Electrons in the outer shell are referred to as valence electrons Once a shell is completed the electrons are referred to as core electrons except the outermost electrons in the noble gases and anions which are still valence electrons Valence electrons are the chemically active electrons while core electrons are chemically inert Ne 3s1 core 639 J Lvalence e39
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