Slide 1

FOUNDATION CHEMISTRY ICHM092CHAPTER 1: STOICHIOMETRY

Subtopic1.0Molecular View of Elements and Compounds1.1Chemical Formulas1.2Naming of Compounds1.3Balanced Chemical Equations1.4Stoichiometry1.5Mole Concept1.6Relative Atomic Mass1.7Molar Mass and Avogadros Number

LESSON OUTCOMESAt the end of this lecture, students should be able toWrite and name chemical formulas of ionic and covalent compounds according to IUPAC Construct and balance chemical equationDefine and calculate mole and Avogadros numberDefine and calculate relative atomic mass and relative formula mass

1.0Molecular View of Elements and Compounds

4Classifying Elements & Compounds5Tro: Chemistry: A Molecular Approach, 2/eElements = the simplest type of substance with unique physical and chemical properties. An element consists of only one type of atom. It cannot be broken down into any simpler substances by physical or chemical means.

Atomic elements = elements whose particles are single atomsMolecular elements = elements whose particles are multi-atom molecules56

Atomic ElementsClassifying Elements & Compounds

7Tro: Chemistry: A Molecular Approach, 2/eClassifying Elements & Compounds

7Compound - a substance composed of two or more elements which are chemically combined.

Molecular compounds - compounds whose particles are molecules made of only nonmetals e.g. dry ice, quartz, Ionic compounds - compounds whose particles are cations (metals) and anions (nonmetals) e.g. lithium fluoride, calcium chloride

Mixture - a group of two or more elements and/o compounds that are physically intermingled.

Classifying Elements & Compounds

Molecular Compounds vs. Ionic ComoundsPropane contains individual C3H8 moleculesTable salt containsan array of Na+ ionsand Cl- ions9Tro: Chemistry: A Molecular Approach, 2/eClassifying Elements & Compounds9Classify Each of the Following as Either an Atomic Element, Molecular Element, Molecular Compound, or Ionic Compound Aluminum, AlAluminum chloride, AlCl3Chlorine, Cl2Acetone, C3H6OCarbon monoxide, COCobalt, Co

10atomic elementionic compoundmolecular elementmolecular compoundmolecular compoundatomic elementTro: Chemistry: A Molecular Approach, 2/eExample 1Chemical Formulas1.1

Types of Chemical FormulasAn empirical formula indicates the relative number of atoms of each element in the compound. It is the simplest type of formula.A molecular formula shows the actual number of atoms of each element in a molecule of the compound.A structural formula shows the number of atoms and the bonds between them, that is, the relative placement and connections of atoms in the molecule.A chemical formula is comprised of element symbols and numerical subscripts that show the type and number of each atom present in the smallest unit of the substance.The empirical formula for hydrogen peroxide is HO.The molecular formula for hydrogen peroxide is H2O2.The structural formula for hydrogen peroxide is H-O-O-H.1.1Chemical Formulas121.1Chemical Formulas

2009, Prentice-Hall, Inc.Writing Formulas for Ionic CompoundsBecause compounds are electrically neutral, one can determine the formula of a compound this way:The charge on the cation becomes the subscript on the anion.The charge on the anion becomes the subscript on the cation.If these subscripts are not in the lowest whole-number ratio, divide them by the greatest common factor. The formula is usually the same as the empirical formula Compound must have no total charge, therefore we must balance the numbers of cations and anions in a compound to get 0 charge

141.1Chemical FormulasAl2O32 x +3 = +63 x -2 = -6Al3+O2-CaBr21 x +2 = +22 x -1 = -2Ca2+Br-Na2CO32 x +1 = +21 x -2 = -2Na+CO32-Compounds that Contain Ions1.1Chemical FormulasWriting the formula for ionic compounds containing polyatomic ionIron(III) phosphate1.Write the symbol for the cation and its charge2.Write the symbol for the anion and its charge3.Charge (without sign) becomes subscript for other ion4.Reduce subscripts to smallest whole number ratio5.Check that the total charge of the cations cancels the total charge of the anions Fe3+ PO43Fe3+ PO43Fe3(PO4)3Fe = (1)(3+) = +3PO4 = (1)(3) = 3FePO4Writing the chemical formula for a binary ionic compound containing variable charge metalmanganese(IV) sulfide1.Write the symbol for the cation and its charge2.Write the symbol for the anion and its charge3.Charge (without sign) becomes subscript for other ion4.Reduce subscripts to smallest whole number ratio5.Check that the total charge of the cations cancels the total charge of the anions Mn4+ S2-Mn4+ S2Mn2S4Mn = (1)(4+) = +4S = (2)(2) = 4MnS2171.1Chemical FormulasExercise 2: Find the empirical formula for each of the following The ionic compound that has two aluminum ions for every three oxide ionsarabinose, C5H10O5pyrimidineethylene glycol

Al2O3CH2OC2H2NCH3OExercise 3 : Write the empirical formulas for the following molecules:

acetylene (C2H2), which is used in welding torchesglucose (C6H12O6), a substance known as blood sugarnitrous oxide (N2O), a gas that is used as an anesthetic gas (laughing gas) and as an aerosol propellant for whipped creams.1.1Chemical FormulasSolutionThere are two carbon atoms and two hydrogen atoms in acetylene. Dividing the subscripts by 2, we obtain the empirical formula CH.

In glucose there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. Dividing the subscripts by 6, we obtain the empirical formula CH2O.

Because the subscripts in N2O are already the smallest possible whole numbers, the empirical formula for nitrous oxide is the same as its molecular formula.

1.1Chemical Formulas1.1Chemical FormulasExercise 4: Write a formula for an ionic compound between calcium and oxygen

Exercise 5 : What are the formulas for compounds made from the following ions?Aluminum ion with a sulfate ion

Chromium(II) with hydrogen carbonate Cr2+ with HCO3 Cr(HCO3)2Al3+ with SO42Al2(SO4)3TopicNaming of Compounds

1.21.2Naming of CompoundsRules for naming Ionic CompoundsConsists of cation and anionSome have one or more nicknames that are only learned by experienceNaCl = table salt, NaHCO3 = baking sodaWrite systematic name by simply naming the ionsif cation is:metal with invariant charge = metal namemetal with variable charge = metal name(charge)polyatomic ion = name of polyatomic ionif anion is:nonmetal = stem of nonmetal name + idepolyatomic ion = name of polyatomic ion1.2Naming of Compounds

Naming Metal CationsMetals with invariant charge metals whose ions can only have one possible charge cation name = metal name1.2Naming of CompoundsNaming Metal Cations

Metals with variable Chargesmetals whose ions can have more than one possible chargedetermine charge by charge on anion and cation name = metal name with Roman numeral charge in parentheses

1.2Naming of CompoundsNaming Monatomic Nonmetal Anion

Determine the charge from position on the Periodic TableTo name anion, change ending on the element name to ide1.2Naming of CompoundsNaming Binary Ionic Compounds forMetals with Invariant Charge

1.name metal cation first, name nonmetal anion second2.cation name is the metal name3.nonmetal anion named by changing the ending on the nonmetal name to -ide1.2Naming of CompoundsExample: Naming Binary Ionic with Invariant Charge Metal, CsF1.Identify cation and anionCs = Cs+ because it is Group 1AF = F because it is Group 7A

2.Name the cationCs+ = cesium

3.Name the anionF = fluoride

4.Write the cation name first, then the anion namecesium fluoride Exercise 6 : Name the ionic compound formed from each of the following pairs of elements:(a) magnesium and nitrogen(b) iodine and barium(c) strontium and fluorine(d) sulfur and cesium1.2Naming of Compounds291.2Naming of CompoundsNaming Binary Ionic Compounds forMetals with Variable Charge

1.name metal cation first, name nonmetal anion second2.metal cation name is the metal name followed by a Roman numeral in parentheses to indicate its chargedetermine charge from anion chargecommon ions Table 3.43.nonmetal anion named by changing the ending on the nonmetal name to -ide1.2Naming of CompoundsExample: Naming binary ionic with variable charge metal, CuF21.Identify the cation and anionF = F because it is Group 7Cu = Cu2+ to balance the two () charges from 2 F

2.Name the cationCu2+ = copper(II)

3.Name the anionF = fluoride

4.Write the cation name first, then the anion namecopper(II) fluoride Exercise 7 : Name the following compounds1.TiCl42.PbBr23.Fe2S3

titanium(IV) chloridelead(II) bromideiron(III) sulfide1.2Naming of Compounds321.2Naming of CompoundsExercise 8 : Give the systematic name for each formula or the formula for each name for the following compounds:tin(II) fluorideCrI3ferric oxideCoS

1.2Naming of Compounds(a) Tin(II) is Sn2+; fluoride is F-; so the formula is SnF2.(b) The anion I- is iodide; 3I- means that Cr (chromium) is +3. CrI3 is chromium(III) iodide.(c) Ferric is a common name for Fe3+; oxide is O2-; therefore the formula is Fe2O3.(d) Co is cobalt; the anion S2- is sulfide; the compound is cobalt(II) sulfide.Solution 1.2Naming of CompoundsCompounds Containing Polyatomic IonsPolyatomic ions are single ions that contain more than one atomOften identified by parentheses around ion in formulaName and charge of polyatomic ion do not changeName any ionic compound by naming cation first and then anion

Periodic Pattern of Polyatomic Ions Containing-ate groups

1.2Naming of Compounds36Patterns for Polyatomic Ions-ate ionchlorate = ClO3-ate ion + 1 O same charge, per- prefixperchlorate = ClO4-ate ion 1 O same charge, -ite suffixchlorite = ClO2-ate ion 2 O same charge, hypo- prefix, -ite suffixhypochlorite = ClO1.2Naming of Compounds37Patterns for Polyatomic Ions1.Elements in the same column form similar polyatomic ionssame number of Os and same charge ClO3 = chlorate \ BrO3 = bromate

2.If the polyatomic ion starts with H, add hydrogen- prefix before name and add 1 to the chargeCO32 = carbonate \ HCO3 = hydrogen carbonate

1.2Naming of Compounds381.2Naming of CompoundsFormulaNameFormulaNameCationsNH4+ammoniumH3O+hydroniumCommon AnionsCH3COO-CN-OH-ClO-ClO2-ClO3-NO2-NO3-MnO4-acetatecyanidehydroxidehypochloritechloritechloratenitritenitratepermanganateCO32-HCO3-CrO42-Cr2O72-O22-PO43-HPO42-SO32-SO42-carbonatebicarbonatechromatedichromateperoxidephosphatehydrogen phosphatesulfitesulfateNaming Polyatomic ion Example: Naming ionic compounds containing a polyatomic ion, Na2SO41.Identify the ionsNa = Na+ because in Group 1ASO4 = SO42 a polyatomic ion2.Name the cationNa+ = sodium, metal with invariant charge3.Name the anionSO42 = sulfate4.Write the name of the cation followed by the name of the anionsodium sulfate

1.2Naming of Compounds40Example: Naming ionic compounds containing a polyatomic ion, Fe(NO3)31.Identify the ionsNO3 = NO3 a polyatomic ionFe = Fe3+ to balance the charge of the 3 NO32.Name the cationFe3+ = iron(III), metal with variable charge3.Name the anionNO3 = nitrateWrite the name of the cation followed by the name of the anion iron(III) nitrate1.2Naming of Compounds41Exercise 9 : Name the Following Compounds1.NH4Cl2.Ca(C2H3O2)23.Cu(NO3)2

ammonium chloridecalcium acetatecopper(II) nitrate1.2Naming of Compounds42Hydrates

Hydrates are ionic compounds containing a specific number of waters for each formula unitWater of hydration often driven off by heatingIn formula, attached waters follow CoCl26H2OIn name attached waters indicated by prefix+hydrate after name of ionic compoundCoCl26H2O = cobalt(II) chloride hexahydrateCaSO4H2O = calcium sulfate hemihydrate1.2Naming of Compounds43Cobalt(II) chloride hexahydrate

1.2Naming of Compounds45BaCl22H2OLiClH2OMgSO47H2OSr(NO3)2 4H2Obarium chloride dihydratelithium chloride monohydratemagnesium sulfate heptahydratestrontium nitrate tetrahydrate

CuSO45H2OCuSO41.2Naming of CompoundsExercise 20 : What is the formula of magnesium sulfateheptahydrate?

Exercise 21 : What is the name of NiCl26H2O?Mg2+ + SO42MgSO4MgSO47H2OCl + Ni2+nickel(II) chloridenickel(II) chloride hexahydrate1.2Naming of Compounds46Writing Names of Binary Molecular Compounds of Two Nonmetals

1.Write name of first element in formulaa)element furthest left and down on the Periodic Tableuse the full name of the element

2.Writes name the second element in the formula with an -ide suffixas if it were an anion, however, remember these compounds do not contain ions!

3.Use a prefix in front of each name to indicate the number of atomsa)Never use the prefix mono- on the first element47Subscript Prefixes Drop last a if name begins with a vowel1 = mono- not used on first nonmetal2 = di-3 = tri-4 = tetra-5 = penta- 6 = hexa- 7 = hepta- 8 = octa- 9 = nona-10 = deca-1.2Naming of Compounds48Example: Naming a binary molecular compound, BF31.Name the first elementboron2.Name the second element with an idefluorine fluoride3.Add a prefix to each name to indicate the subscriptmonoboron, trifluoride4.Write the first element with prefix, then the second element with prefixdrop prefix mono from first elementboron trifluoride1.2Naming of Compounds49Exercise 10 : Name the FollowingNO2PCl5I2F7Nitrogen dioxidePhosphorus pentachlorideDiiodine heptafluoride1.2Naming of Compounds50Exercise 11 : Write Formulas for the Followingdinitrogen tetroxidesulfur hexafluoridediarsenic trisulfideN2O4SF6As2S31.2Naming of Compounds51

52Tro: Chemistry: A Molecular Approach, 2/e1.2Naming of Compounds Acids are molecular compounds that form H+ when dissolved in waterAcid Nomenclature52Binary Acid NomenclatureIf the anion in the acid ends in -ide, change the ending to -ic acid and add the prefix hydro- .HCl: hydrochloric acidHBr: hydrobromic acidHI: hydroiodic acid

1.2Naming of Compounds53If the anion in the acid ends in -ate, change the ending to -ic acid.HClO3: chloric acidHClO4: perchloric acid

Oxyacid Nomenclature1.2Naming of Compounds54If the anion in the acid ends in -ite, change the ending to -ous acid.HClO: hypochlorous acidHClO2: chlorous acid

Oxyacid Nomenclature1.2Naming of Compounds5556

1.2Naming of CompoundsExercise 12 : Name the Following anions and give the name and formula of the acid derived from itH2SO4HClO2HNO2Sulfuric acidchlorous acidnitrous acid1.2Naming of CompoundsSO42-

ClO2-

NO2-Sulfatechloritenitrite57Exercise 13 : Name the Following anions and give the name and formula of the acid derived from itHNO3H2CrO4H2Cr2O7Nitric acidChromic acidDichromic acid1.2Naming of CompoundsNO3-

CrO42-

Cr2O72-NitrateChromateDichromate58Exercise 14 What are the formulas for the following acids?H+ with ClO2 HClO2 H+ with PO43 H3PO4H+ with Br HBrchlorous acidphosphoric acidhydrobromic acid59Tro: Chemistry: A Molecular Approach, 2/e59Example: Binary Acidshydrosulfuric acid1.Write the symbol for the cation and its charge2.Write the symbol for the anion and its charge3.Charge (without sign) becomes subscript for other ion4.Add (aq) to indicate dissolved in water5.Check that the total charge of the cations cancels the total charge of the anions H+ S2H+ S2H2SH = (2)(1+) = +2S = (1)(2) = 2H2S(aq)in all acids the cation is H+hydro meansbinary1.2Naming of Compounds60Example: OxyacidsPerbromic acid1.Write the symbol for the cation and its charge2.Write the symbol for the anion and its charge3.Charge (without sign) becomes subscript for other ion4.Add (aq) to indicate dissolved in water5.Check that the total charge of the cations cancels the total charge of the anions H+ BrO4H+ BrO4HBrO4H = (1)(1+) = +1BrO4 = (1)(1) = 1HBrO4(aq)in all acids the cation is H+no hydro meanspolyatomic ion-ic means -ate ion1.2Naming of Compounds61Example: Oxyacidssulfurous acid1.Write the symbol for the cation and its charge2.Write the symbol for the anion and its charge3.Charge (without sign) becomes subscript for other ion4.Add (aq) to indicate dissolved in water5.Check that the total charge of the cations cancels the total charge of the anions H+ SO32H+ SO32H2SO3H = (2)(1+) = +2SO3 = (1)(2) = 2H2SO3(aq)in all acids the cation is H+no hydro meanspolyatomic ion-ous means -ite ion1.2Naming of Compounds6263Additional ProblemsExplain what is wrong with the name or formula at the end of each statement and correct it:

Ba(C2H3O2)2 barium diacetateAmmonium phosphate (NH3)4PO4Iron (II) sulfate Fe2(SO4)3Cr(NO3)3 - chromic (III) nitrideDichlorine heptaoxide Cl2O6BrCl3 Trichlorine bromide1.2Naming of CompoundsBalanced Chemical Equations

1.3Chemical EquationsShorthand way of describing a reactionProvides information about the reactionformulas of reactants and productsstates of reactants and productsrelative numbers of reactant and product molecules that are requiredcan be used to determine weights of reactants used and products that can be made1. 3 Balanced Chemical Equations65 2009, Prentice-Hall, Inc.Anatomy of a Chemical EquationCH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Chemical equations are concise representations of chemical reactions66 2009, Prentice-Hall, Inc.Anatomy of a Chemical EquationReactants appear on the left side of the equation.CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

67 2009, Prentice-Hall, Inc.Anatomy of a Chemical EquationProducts appear on the right side of the equation.CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

68 2009, Prentice-Hall, Inc.Anatomy of a Chemical EquationThe states of the reactants and products are written in parentheses to the right of each compound.CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

69 2009, Prentice-Hall, Inc.Anatomy of a Chemical EquationCoefficients are inserted to balance the equation.CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

70 2009, Prentice-Hall, Inc.Subscripts and Coefficients Give Different InformationSubscripts tell the number of atoms of each element in a molecule.

71 2009, Prentice-Hall, Inc.Subscripts and Coefficients Give Different InformationSubscripts tell the number of atoms of each element in a moleculeCoefficients tell the number of molecules.

72Balancing Chemical EquationsWrite the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and waterC2H6 + O2CO2 + H2OChange the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C2H6NOTC4H12Balancing Chemical EquationsStart by balancing those elements that appear in only one reactant and one product. C2H6 + O2CO2 + H2Ostart with C or H but not O2 carbonon left1 carbonon rightmultiply CO2 by 2C2H6 + O22CO2 + H2O6 hydrogenon left2 hydrogenon rightmultiply H2O by 3C2H6 + O22CO2 + 3H2OBalancing Chemical EquationsBalance those elements that appear in two or more reactants or products. 2 oxygenon left4 oxygen(2x2)C2H6 + O22CO2 + 3H2O+ 3 oxygen(3x1)multiply O2 by 72= 7 oxygenon rightC2H6 + O22CO2 + 3H2O72remove fractionmultiply both sides by 22C2H6 + 7O24CO2 + 6H2O76Balancing Chemical EquationsCheck to make sure that you have the same number of each type of atom on both sides of the equation. 2C2H6 + 7O24CO2 + 6H2OReactantsProducts4 C12 H14 O4 C12 H14 O4 C (2 x 2)4 C12 H (2 x 6)12 H (6 x 2)14 O (7 x 2)14 O (4 x 2 + 6)translate the statementbalance the atomsspecify states of matter (solid (s), liquid (l), gas (g), aquous (aq))adjust the coefficientscheck the atom balanceBalancing Chemical Equation77Example : Write a balanced equation for the combustion of butane, C4H108 C 8; 20 H 20; 26 O 26{C4H10(l) + 13/2 O2(g) 4 CO2(g) + 5 H2O(g)}x 22 C4H10(l) + 13 O2(g) 8 CO2(g) + 10 H2O(g)13/2 x 2 O 13C4H10(l) + 13/2 O2(g) 4 CO2(g) + 5 H2O(g)

4 C 1 x 4C4H10(l) + O2(g) 4 CO2(g) + H2O(g)10 H 2 x 5C4H10(l) + O2(g) 4 CO2(g) + 5 H2O(g)C4H10(l) + O2(g) CO2(g) + H2O(g)CheckIf fractional coefficients, multiply thru by denominatorBalance free elements by adjusting coefficient in front of free elementBalance atoms in complex substances firstWrite a skeletal equation78Exercise 15 : when aluminum metal reacts with air, it produces a white, powdery compound, aluminum oxidereacting with air means reacting with O2aluminum(s) + oxygen(g) aluminum oxide(s)Al(s) + O2(g) Al2O3(s)

4 Al(s) + 3 O2(g) 2 Al2O3(s)1. 3 Balanced Chemical EquationsExercise 16 : Acetic acid reacts with the metal aluminum to make aqueous aluminum acetate and gaseous hydrogenacids are always aqueousmetals are solid except for mercuryAl(s) + HC2H3O2(aq) Al(C2H3O2)3(aq) + H2(g) 2 Al(s) + 6 HC2H3O2(aq) 2 Al(C2H3O2)3(aq) + 3 H2(g)79Mole ConceptTopic:

1.481By definition: 1 atom 12C weighs 12 amuOn this scale1H = 1.008 amu 16O = 16.00 amuAtomic mass is the mass of an atom in atomic mass units (amu)Micro Worldatoms & moleculesMacro WorldgramsAtomic Mass

82The average atomic mass is the weightedaverage of all of the naturally occurringisotopes of the element.

Element having the same atomic number but difference atomic mass number83Naturally occurring lithium is:7.42% 6Li (6.015 amu)92.58% 7Li (7.016 amu)0.0742 x 6.015 + 0.9258 x 7.016= 6.941 amuAverage atomic mass of lithium:(fraction of isotope,n) x (mass of isotope, n)

84

Average atomic mass (6.941)What is the difference between Atomic Mass, Formula Mass and Molecular Mass ??

85 2009, Prentice-Hall, Inc.Formula Mass (FM)Formula mass are generally reported for ionic compounds.A formula mass is the sum of the atomic masses for the atoms in a chemical formula.So, the formula mass of calcium chloride, CaCl2, would be Ca: 1(40.1 amu) + Cl: 2(35.5 amu) 111.1 amu86 2009, Prentice-Hall, Inc.Molecular Mass (MM)A molecular mass is the sum of the atomic masses of the atoms in a molecule.For the molecule ethane, C2H6, the molecular mass would beC: 2(12.0 amu)30.0 amu+ H: 6(1.0 amu)8788The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C

1 mol = NA = 6.0221367 x 1023Avogadros number (NA)

Dozen = 12

Pair = 2The Mole (mol): A unit to count numbers of particlesThe Mole89Molar mass is the mass of 1 mole of in gramseggsshoesmarblesatoms1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g1 12C atom = 12.00 amu1 mole 12C atoms = 12.00 g 12C1 mole lithium atoms = 6.941 g of LiFor any element atomic mass (amu) = molar mass (grams)1.4 Mole ConceptMole and Mass Relationships

1 molesulfur32.06 g

1 molecarbon12.01 g

90901.4 Mole ConceptThe following relationships supply the conversion factors for the conversions among mass in grams, amount in moles, and number of elementary units, Mass (gram) of substance Mol substance = molar mass of substance (g mol-1) Number of elementary units = mol X Avogadros No (atoms / molecules / formula units)Examples : 1 mol of copper, Cu contains 6.022 x 1023 of Cu atoms. 1 mol of water , H2O contains 6.022 x 1023 of H2O molecules. 1 mol of MgCl2 crystal contains 6.022 x 1023 of MgCl2 formula units. 2009, Prentice-Hall, Inc.Avogadros Number6.02 x 10231 mole of 12C has a mass of 12 g.

92StoichiometryTopic:

1.5

1.5Stoichiometry

Quantities in Chemical ReactionsThe amount of every substance used and made in a chemical reaction is related to the amounts of all the other substances in the reactionLaw of Conservation of MassBalancing equations by balancing atomsThe study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry94

1.5Stoichiometry

UnitsAlways write every number with its associated unitAlways include units in your calculationsyou can do the same kind of operations on units as you can on numberscm cm = cm2cm + cm = cmcm cm = 1using units as a guide to problem solving is called dimensional analysis9595

1.5Stoichiometry

Dimensional Analysis: Converting One Unit to AnotherDimensional analysis method is emphasized, because

Provide a systematic, straightforward way to set up problems Gives a clear understanding of the principle involved Trains you to organize evaluate data Helps to identify errors, since unwanted units are not eliminated if the setup of the problem is incorrect

1.5Stoichiometry

Reaction StoichiometryThe coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)2 molecules of C8H18 react with 25 molecules of O2to form 16 molecules of CO2 and 18 molecules of H2O2 moles of C8H18 react with 25 moles of O2to form 16 moles of CO2 and 18 moles of H2O2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O9797

1.5Stoichiometry

Predicting Amounts from Stoichiometry The amounts of any other substance in a chemical reaction can be determined from the amount of just one substance. A chemically balanced equation in the calculationExample 2.6.1How many moles of CO2 can be produced in the combustion of 22.0 moles of C8H18 ?The equation of reaction is2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)The mole: mole relationship from the balanced equation2 moles C8H18 : 16 moles CO29898

Example 1.5.1: Calculate the number of atoms in 2.45 mol of copperbecause atoms are small, the large number of atoms makes sense1 mol = 6.022 x 1023 atoms2.45 mol Cuatoms CuCheck:Solution:Conceptual Plan:

Relationships:Given:Find:mol Cuatoms Cu

9999

Example 1.5.2: How many copper atoms are in a penny weighing 3.10 g?because the given amount is much less than 1 mol Cu, the number makes sense1 mol Cu = 63.55 g, 1 mol = 6.022 x 10233.10 g Cuatoms CuCheck:Solution:Conceptual Plan:

Relationships:Given:Find:g Cumol Cuatoms Cu

100100Example 1.5.3: A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?because the number of atoms given is less than Avogadros number, the answer makes sense1 mol = 6.022 x 1023 atoms1.1 x 1022 atoms Agmoles AgCheck:Solution:Conceptual Plan:

Relationships:Given:Find:atoms Agmol Ag

101101Practice How many moles are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00)because the given amount is less than 239.2 g, the moles being < 1 makes sense1 mol PbO2 = 239.2 g50.0 g mol PbO2moles PbO2Check:Solution:Conceptual Plan:

Relationships:Given:Find:g PbO2mol PbO2102

Tro: Chemistry: A Molecular Approach, 2/e102Example: Find the number of CO2 molecules in 10.8 g of dry icebecause the given amount is much less than 1 mol CO2, the number makes sense1 mol CO2 = 44.01 g, 1 mol = 6.022 x 102310.8 g CO2molecules CO2Check:Solution:Conceptual Plan:

Relationships:Given:Find:g CO2mol CO2molec CO2103

Tro: Chemistry: A Molecular Approach, 2/e103Practice How many formula units are in 50.0 g of PbO2? (PbO2 = 239.2)because the given amount is less than 1 mol PbO2, the number makes sense1 mol PbO2 = 239.2 g,1 mol = 6.022 x 102350.0 g PbO2formula units PbO2Check:Solution:Conceptual Plan:

Relationships:Given:Find:g PbO2mol PbO2units PbO2104

Tro: Chemistry: A Molecular Approach, 2/e104Practice What is the mass of 4.78 x 1024 NO2 molecules?because the given amount is more than Avogadros number, the mass > 46 g makes sense1 mol NO2 = 46.01 g, 1 mol = 6.022 x 10234.78 x 1024 NO2 moleculesg NO2Check:Solution:Conceptual Plan:

Relationships:Given:Find:moleculesmol NO2g NO2105

Tro: Chemistry: A Molecular Approach, 2/e105

1 mol NaCl = 58.44 g, 1 mol Na = 22.99 g, 1 mol Na : 1 mol NaClExercise 1.8.4 : Find the mass of sodium in 6.2 g of NaCl (Na = 22.99; Cl = 35.45)because the amount of Na is less than the amount of NaCl, the answer makes sense 6.2 g NaClg NaCheck:Solution:Conceptual Plan:

Relationships:Given:Find:g NaClmol NaClmol Nag Na

106106Example 1.5.5: Calculate the number of (i) H2O molecules (ii) H atoms and (iii) all the atoms in 0.25 mol of water.because the number of molecules is less than that of 1 mol, the answer makes sense1 mol = 6.022 x 1023 molecules0.25 mol H2O(i) H2O molecules Check:Solution:Conceptual Plan:

Relationships:Given:Find:Mol H2OMolecules of H2O6.022 x 1023 molecules 1 mol

107107Given 0.25 mol H2O, to find (ii) No. of H atoms (iii) total atoms Conceptual plan :Taking the answer from (i), we proceed to (ii) & (iii)(ii) No. H atoms = No. of H2O molecules x 2 H atoms 1 molecule = 3.01 x 1023 H atoms(iii) Total atoms = No. of H2O molecules x 3 atoms 1 molecule = 4.52 x 1023 atomsMol H2O Molecules H2OH atoms6.022 x 1023 molecules 1 mol2 H atoms1 moleculeTotal atoms 3 atoms 1 molecule

Example 1.4.2 : (b) Estimate the mass of CO2 produced by the combustion of 3.5 x 1015 g gasoline, C8H18.because 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18:16 mol CO23.4 x 1015 g C8H18g CO2Check:Solution:Conceptual Plan:

Relationships:Given:Find:g C8H18mol CO2g CO2mol C8H18

109109110x6.022 x 1023 atoms K1 mol K=How many atoms are in 0.551 g of potassium (K) ?1 mol K = 39.10 g K 1 mol K = 6.022 x 1023 atoms K0.551 g K1 mol K39.10 g Kx8.49 x 1021 atoms K111How many H atoms are in 72.5 g of C3H8O ?1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O1 mol H = 6.022 x 1023 atoms H5.82 x 1024 atoms H1 mol C3H8O molecules = 8 mol H atoms72.5 g C3H8O1 mol C3H8O60 g C3H8Ox8 mol H atoms1 mol C3H8Ox6.022 x 1023 H atoms1 mol H atomsx=Practice 2Calculate the mass, in grams, of each of the following : (a) 1.5 mol of C9H8O3. (b) 2.2 x 1020 of Ag atom(c) 3.0 x 1022 of HI molecules { Ans (a) 246 g (b) 0.0432 g (c) 6.28 g }Practice 3A sample of suphur dioxide gas, SO2 , contains 6.6 x 1022 of oxygen, O, atoms. Calculate (i) the number of S atoms (ii) the number of moles of SO2 (iii) the mass of SO2 in the above sample. { Ans 3.3 x 1022 S atom , (ii) 0.055 mol (iii) 3.52 g }Practice 4Calculate the number of (i) formula units of AlF3 , (ii) F- ions and (iii) Al3+ ions and (iv) all the ions in 0.02 mol of the compound AlF3. { Ans : (i) 1.204 x 1022 formula units , (ii) 3.612 x 1022 F ions (iii) 1.204 x 1022 Al3+ ions (iv) 4.82 x 1022 ions } Additional Problems1. The carat is the unit of mass used by jewelers. One carat is exactly 200mg. How many carbon atoms are present in a 24-carat diamond?

2. One molecule of the antibiotic known as penicillin G has a mass of 5.342 x 10-21g.What is the molar mass of penicillin G?Amount-mass-number relationships for compounds.

114114Interconverting Moles, Mass, and Number of Chemical EntitiesMass (g) = no. of moles x no. of grams1 molNo. of moles = mass (g) xno. of grams1 molNo. of entities = no. of moles x6.022x1023 entities1 molNo. of moles = no. of entities x 6.022x1023 entities1 molgM115Percent CompositionTopic

1.61.6 Percent CompositionPercent (%) means parts per 100 partsThe percent composition of a compound is the mass percent of each element in the compoundThe molar mass represents the total mass, or 100%, of the compoundAccording to the Law of Definite Composition, the percent composition must be the same no matter what size sample is takenE.g. the percent composition of water, H2O, is 11.19% and 88.79% O by massPercent composition of a compound can be determined:From knowing its formulaFrom experimental data1.6 Percent CompositionProblem-solving strategy1) Calculate the molar mass2) Divide the total mass of each element in the formula by the molar mass and multiply by 100Percent Composition from FormulaCompound: XaYb

Mass % of element X = (atomic mass of X) x (a) Molar mass of XaYb

Mass % of element Y = (atomic mass of Y) x (b) Molar mass of XaYb

x 100% x 100% Percent composition of an element in a compound =n x molar mass of elementmolar mass of compoundx 100%n is the number of moles of the element in 1 mole of the compound

C2H6O%C =2 x (12.01 g)46.07 gx 100% = 52.14%%H =6 x (1.008 g)46.07 gx 100% = 13.13%%O =1 x (16.00 g)46.07 gx 100% = 34.73%52.14% + 13.13% + 34.73% = 100.0%Percent CompositionExercise 1.8.1: Find the mass percent of Cl in C2Cl4F2because the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes senseC2Cl4F2% Cl by massCheck:Solution:Conceptual Plan:

Relationships:Given:Find:

121121 Example 2.10: Determine the percent composition of CaCl2 (Ca = 40.08, Cl = 35.45)

1.6 Percent Composition122122Percent Composition from Experimental Data Problem-solving strategy

1) Calculate the mass of the compound formed2) Divide the mass of each element by the total mass of the compound and multiply by 100Practice: Zinc oxide is a compound with many uses from preventing sunburn to a pigment in white paint. When heated in air, 1.63 g of zinc (Zn) combines with 0.40 g of oxygen (O2) to form zinc oxide. Calculate the percent composition of the compound formedGiven : 1.63 g Zn 0.40 g O2

Mass of the product formed (ZnO) = 163 g Zn + 0.40 g O2 = 2.03 g ZnO

Percent for each element

Zn:

O:1.63 g 2.03 g x100 = 80.3% Zn0.40 g2.03 g x100 = 20% O

Mass Percent as a Conversion FactorThe mass percent also tells you the mass of a constituent element in 100 g of the compoundthe fact that CCl2F2 is 58.64% Cl by mass means that 100 g of CCl2F2 contains 58.64 g ClThis can be used as a conversion factor100 g CCl2F2 : 58.64 g Cl

1.6 Percent Composition125125Exercise 1.8.2 : Table salt contains 39% of Na. Find the mass of table salt containing 2.4 g of Nabecause the mass of NaCl is more than 2x the mass of Na, the number makes sense100 g NaCl : 39 g Na2.4 g Na, 39% Nag NaClCheck:Solution:Conceptual Plan:

Relationships:Given:Find:g Nag NaCl

126126Exercise 1.8.3: Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C?because the mass of benzaldehyde is more than the mass of C, the number makes sense100 g benzaldehyde : 79.2 g C19.8 g C, 79.2% Cg benzaldehydeCheck:Solution:Conceptual Plan:

Relationships:Given:Find:g Cg benzaldehyde

127127Conversion Factors in Chemical FormulasChemical formulas have inherent in them relationships between numbers of atoms and moleculesor moles of atoms and moleculesThese relationships can be used to convert between amounts of constituent elements and moleculeslike percent composition1.6 Percent Composition128128Additional ProblemsMyoglobin stores oxygen for metabolic processes in muscle. Chemical analysis shows that it contains 0.34% Fe by mass. What is the molar mass of myoglobin? (There is one Fe atom per molecule).The aluminium sulfate hydrate [Al2(SO4)3.xH2O] contains 8.10 % Al by mass. Calculate x, that is the number of water molecules associated with each Al2(SO4)3 unit.An oxybromate compound KBrOx, where x is unknown, is analyzed and found to contain 52.92% Br. What is the value of x?

Additional ProblemsAn element X forms an iodide (XI3) and a chloride (XCl3). The iodide is quantitatively converted to the chloride when it is heated in a stream of chlorine: 2 XI3 + 3Cl2 2 XCl3 + 3I2 If 0.5000 g of XI3 is treated, 0.2360 g of XCl3 is obtained.Calculate the atomic weight of the element xIdentify the element X2. A mixture of NaBr and Na2SO4 contains 29.96 % Na by mass. Calculate the percent by mass of each compound in the mixture.Additional ProblemsA mixture of CUSO4.5H2O and MgSO4.7H2O is heated until all the water is lost. If 5.020g of mixture gives 2.988g of the anhydrous salts, what is the percent by mass of CuSO4.5H2O.

Empirical FormulaandMolecular FormulaTopic

1.7Empirical FormulaSimplest, whole-number ratio of the atoms of elements in a compoundCan be determined from elemental analysismasses of elements formed when a compound is decompose, or that react together to form a compoundcombustion analysispercent composition1.7 Empirical Formula andMolecular Formula133133Finding an Empirical Formula1.Convert the percentages to gramsa)assume you start with 100 g of the compoundb)skip if already in grams2.Convert grams to molesa)use molar mass of each element3.Divide all by smallest number of molesa)if result is within 0.1 of whole number, round to whole number4.Multiply all mole ratios by a number to make all whole numbersa)if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3; if ratio 0.25 or 0.75, multiply all by 4; etc. b)skip if already whole numbers1.7 Empirical Formula andMolecular Formula134134Empirical Formula and Molecular Formulag Cmol Cg Hmol Hempiricalformulamoleratiowholenumberratiog Omol OExample 2.5.1: Laboratory analysis of aspirin determined the following mass percent composition as follows: C = 60.00%, H = 4.48% and O = 35.53%. Calculate the empirical formula of aspirin. Solution: In 100 g of aspirin there are 60.00 g C, 4.48 g H, 35.53 g O

The conceptual plan is1351351.7 Empirical Formula andMolecular FormulaGiven the molar mass: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 gcalculate the moles of each element

The mole ratio of C : H : O = 4.996 : 4.44 : 2.220 Divide the mole ratio by the smallest number of mole in the above mole ratio to get the whole number mole ratio1361361.7 Empirical Formula andMolecular FormulaWhole number mole ratio:C : H : O = 4.996 : 4.44 : 2.220 = 2.25 : 2 : 1 2.220 2.220 2.220

Choose a small number to multiply the mole ratio in order to get the whole number mole ratioMultiply by 4

C : H : O = 2.25 x 4 : 2 x 4 : 1 x 4 = 9 : 8 : 4

The empirical formula = C9H8O4

Once youve learnt how to convert one unit to another according to the conceptual plan, you can evaluate the empirical formula in a table format.137137Element ABC% composition by mass (g)Relative atomic massNumber of molesMole ratiosSimplest ratio(whole number)Empirical and molecular formulaElement CHO% composition by mass (g)60.004.4835.53Relative atomic mass12.011.00816.00Number of moles4.9964.442.220Mole ratios4.9962.220= 2.254.442.220= 22.2202.220= 1Simplest ratio(x4)984Empirical formulaPractice 1: Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)1.7 Empirical Formula andMolecular Formula140140Molecular Formulas

The molecular formula is the true formula , representing the total number of atom of each element present in one molecule of a compound

1.7 Empirical Formula andMolecular Formula- The molecular formula is (AxBy)n 141141Example 1.9.2: Find the molecular formula of butanedionethe molar mass of the calculated formula is in agreement with the given molar massemp. form. = C2H3O; MM = 86.03 g/molmolecular formulaCheck:Solution:Conceptual Plan: and Relationships:Given:

Find:

142142Example 1.9.3: Benzopyrene has a molar mass of 252 g and an empirical formula of C5H3. What is its molecular formula? (C = 12.01, H=1.01)C5 =5(12.01 g) = 60.05 gH3 =3(1.01 g) = 3.03 gC5H3 = 63.08 gMolecular formula = {C5H3} x 4 = C20H12

143Practice 2Methyl salicylate has a mass percent composition of 63.2% C, 31.6% O and 5.26% H. The mass of one molecule of the compound is 2.53 x 10-22 g. Determine the empirical formula and the molecular formula of methyl salicylate. { Ans : C3H8O3 }Practice 3The mass percent composition of an organic acid is given as 41.4% C, 3.47% H, 55.1% O. 0.05 mol of this acid has a mass of 5.80g. Determine the empirical formula and the molecular formula of this acid. { Ans : CHO; C4H4O4 }Practice 4An 0.648 g sample of isobutene contains 0.556g of carbon and the rest is hydrogen. 0.5 mol of the isobutene has a mass of 28.5g. What is the empirical formula and molecular formula of isobutene. { Ans : CH2 ; MW = 57 g mol-1 ; C4H8 }Combustion AnalysisCompounds containing C, H and O are routinely analyzed through combustion in a chamber like this.C is determined from the mass of CO2 produced.H is determined from the mass of H2O produced.O is determined by difference after the C and H have been determined.Once the masses of all the constituent elements in the original compound have been determined, the empirical formula can be found

A common technique for analyzing compounds is to burn a known mass of compound and weigh the amounts of product made145gCO2, H2OmolratioempiricalformulamolCO2, H2OmolC, HgC, HgOmolOmolC, H, OExample 1.9.4Combustion of a 0.8233 g sample of a compound containing only carbon, hydrogen, and oxygen produced 2.445 g CO2 and 0.6003 g H2O. Determine the empirical formula of the compound

Solution:Let the empirical formula to be CxHyOzWrite a conceptual plan

146146

Convert : g CO2 mol CO2 mol C 9 H2O mol H2O 2 mol HUse the following relationship:Molar mass of CO2 = 12.01 + (16.00)2 = 44.01 g/molMolar mass of H2O = 2(1.008) + 16.00 = 18.02 g/mol 1mol CO2 = 1 mol C; 1mol H2O = 2 mol H147147

Calculate the grams of C and H using molar mass of C and HCalculate the grams and moles of OMass of O = Mass of compound (mass of C + mass of H) =

0.8233 g (0.6673 g + 0.06715 g) = 0.0889 g O

148148Mole ratio C : H : O = 0.05556 : 0.06662 : 0.00556Divide by the smallest mole C : H : O = 0.05556 : 0.06662 : 0.00556 0.00556 0.00556 0.00556 = 10 : 12 : 1 The empirical formula = C10H12O Practice 1The smell of dirty gym socks is caused by the compound caproic acid. Combustion of 0.844 g of caproic acid produced 0.784 g of H2O and 1.92 g of CO2. If the molar mass of caproic acid is 116.2 g/mol, what is the molecular formula of caproic acid? (Molar mass C = 12.01, H = 1.008, O = 16.00) {Ans : C3H6O ; C6H12O2 }149149Additional ExerciseAn organic compound was found to contain only C, H, and Cl. When a 1.50 g sample of compound was completely combusted in air 3.52g of CO2 was formed. In a separate experiment the chlorine in a 1.0 g sample of the compound was converted to 1.27 g of AgCl. Determine the empirical formula of the compound. Ferrocene, first synthesized in 1951, was the first organic iron compound with Fe-C bonds. An understanding of the structure of ferrocene gave rise to new ideas about chemical bonding and led to the preparation of many useful compounds, In combustion analysis of ferrocene, which contains only Fe, C and H, a 0.9437 g of sample produced 2.233 g of CO2 and 0.457 g of H2O. What is the empirical formula of ferrocene?150150Additional ExerciseLysine, an essential amino acid in the human body, contains C,H,O and N. In one experiment, the complete combustion of 2.175 g of lysine gave 3.94 g Co2 and 1.89 g H2O. In a separate experiment, 1.873 g of lysine gave 0.43g g of NH3.(a) Calculate the empirical formula of lysine.(b) The approximate molar mass of lysine is 150 g. What is the molecular formula of the compound?151151 2009, Prentice-Hall, Inc.Limiting Reactants1.8152 2009, Prentice-Hall, Inc.How Many Cookies Can I Make?You can make cookies until you run out of one of the ingredients.Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).

153 2009, Prentice-Hall, Inc.How Many Cookies Can I Make?In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.

154An ice cream sundae analogy for limiting reactant.

1.8 Limiting Reactants155Limiting ReactantsThe limiting reactant is the reactant present in the smallest stoichiometric amount.In other words, its the reactant that will be completely used up in the reaction thus, limit the amount of product formThe reactant that is not limiting is in excess some of this reactant will be left over.

156 2009, Prentice-Hall, Inc.Theoretical YieldThe theoretical yield is the maximum amount of product that can be made.In other words its the amount of product possible as calculated through the stoichiometry problem.This is different from the actual yield, which is the amount one actually produces and measures.157 2009, Prentice-Hall, Inc.Percent YieldOne finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield).Actual YieldTheoretical YieldPercent Yield = x 100158Problem-Solving Strategy for Limiting Reactant Problems

Calculate the amount of product (moles or grams as needed) formed from each reactantsDetermine which reactant is limiting (the reactant that gives the least amount of product is the limiting reactant; the other reactant is in excess. Once we know the limiting reactant, the amount of product formed can be determined. It is the amount determined by the limiting reactantIf we need to know how much the other reactant remains, we calculate the amount of the other reactant required to react with the limiting reactant , then subtract this amount from the starting quantity of the reactant. This gives the amount of that substance that remains unreacted

Limitingreactant

2 mol N2 : 1 Si3N4; 3 mol Si : 1 Si3N41.20 mol Si, 1.00 mol N2mol Si3N4Solution:Conceptual Plan:

Relationships:Given:Find:mol N2mol Si3N4mol Simol Si3N4Pick leastamountLimiting reactant andtheoreticalyieldTheoretical yield

Example 2.6.3How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 moles of N2 in the reaction 3 Si + 2 N2 Si3N4?

160160160Practice 2.6.3Consider the reaction between iron (III) oxide, Fe2O3 and carbon monoxide, CO. Fe2O3 + 3CO 2Fe + 3CO2In one process, 213 g of Fe2O3 are reacted with 140 g of CO. (a) calculate the mass (in grams) of Fe formed. (b) How many moles of the excess reagent is left at the end of the reaction? { Ans : (a) 2.66 mol = 149g (b) 1 mol }1.8 Limiting ReactantsPractice 2.6.4In one experiment x grams of CS2 is mixed with y grams of NaOH. The reaction is represented by the unbalanced equation below.

CS2 + NaOH Na2CS3 + Na2CO3 + H2O

At the end of the reaction, 7.5 g of Na2CO3 is formed while 3.3 g of CS2 is left unreacted in the reaction vessel. All the NaOH in the vessel is completely consumed. a)Balance the above equationb)Calculate the amount of CS2 (x grams) and NaOH (y grams) added to the reaction vessel at the beginning of the reaction. { Ans : x = 19.49 g ; y =17.0 g } 1.8 Limiting Reactants

Example 2.6.4:When 28.6 kg of C are allowed to react with 88.2 kg of TiO2 in the reaction below, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield.

kgTiO2kgC}smallestamount isfrom limitingreactantsmallestmol Ti

163163Relationship required : 1000 g = 1 kg Molar Mass TiO2 = 79.87 g/mol Molar Mass Ti = 47.87 g/mol Molar Mass C = 12.01 g/mol 1 mole TiO2 : 1 mol Ti (from the chem. equation) 2 mole C : 1 mol Ti (from the chem. equation)

smallest moles of Tilimiting reactant164164

theoretical yield

limiting reactant = TiO2, theoretical yield = 52.9 kgpercent yield = 80.9% Because Ti has lower molar mass than TiO2, the T.Y. makes sense and the percent yield makes sense as it is less than 100%165165Practice 2.6.5How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield?

Solutions 1.9 2009, Prentice-Hall, Inc.1.9 SolutionsSolutions are defined as homogeneous mixtures of two or more pure substances.The solvent is present in greatest abundance.All other substances are solutes (present in the smallest amount).

168169

Solutions 2009, Prentice-Hall, Inc.Solution ChemistryIt is helpful to pay attention to exactly what species are present in a reaction mixture (i.e., solid, liquid, gas, aqueous solution).If we are to understand reactivity, we must be aware of just what is changing during the course of a reaction.170 2009, Prentice-Hall, Inc.How Does a Solution Form?As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.

171 2009, Prentice-Hall, Inc.SolutionsThe intermolecular forces between solute and solvent particles must be strong enough to compete with those between solute particles and those between solvent particles.

172 2009, Prentice-Hall, Inc.DissociationWhen an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them.This process is called dissociation.

173DissociationAn electrolyte is a substances that dissociates into ions when dissolved in water.

A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.174

Concentration of Solution

ConcentrationTo describe a solution, you need to describe the components and their relative amountsConcentration = amount of solute in a given amount of solution175

1.9 Concentration of Solutions

Solution ConcentrationQualitatively, solutions are often described as dilute or concentratedDilute solutions have a small amount of solute compared to solventConcentrated solutions have a large amount of solute compared to solvent176176 2009, Prentice-Hall, Inc.MolarityMolarity is one way to measure the concentration of a solution.Since volume is temperature-dependent, molarity can change with temperaturemoles of solutevolume of solution in litersMolarity (M) =177

1.9 Concentration of Solutions

Molarity and DissociationThe molarity of the ionic compound allows you to determine the molarity of the dissolved ionsCaCl2(aq) = Ca2+(aq) + 2 Cl(aq)A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 in each liter of solution1 L = 1.0 moles CaCl2, 2 L = 2.0 moles CaCl2Because each CaCl2 dissociates to give one Ca2+, a 1.0 M CaCl2 solution is 1.0 M Ca2+1 L = 1.0 moles Ca2+, 2 L = 2.0 moles Ca2+Because each CaCl2 dissociates to give 2 Cl, a 1.0 M CaCl2 solution is 2.0 M Cl1 L = 2.0 moles Cl, 2 L = 4.0 moles Cl178178 2009, Prentice-Hall, Inc.Mixing a SolutionTo create a solution of a known molarity, one weighs out a known mass (and, therefore, number of moles) of the solute.The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask.

179

1.9 Concentration of Solutions

Preparing 1 L of a 1.00 M NaCl Solution180180Example 2.6.5: Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solutionbecause most solutions are between 0 and 18 M, the answer makes sense1 mol KBr = 119.00 g, M = moles/L25.5 g KBr, 1.75 L solutionmolarity, MCheck:Solution:Conceptual Plan:

Relationships:Given:Find:g KBrmol KBrL solnM

181181Example 2.6.6: How many liters of 0.125 M NaOH contain 0.255 mol NaOH?because each L has only 0.125 mol NaOH, it makes sense that 0.255 mol should require a little more than 2 L0.125 mol NaOH = 1 L solution0.125 M NaOH, 0.255 mol NaOHliters, LCheck:Solution:Conceptual Plan:

Relationships:Given:Find:mol NaOHL soln

182182Example 2.6.7: Determine the mass of CaCl2 (MM = 110.98) in 1.75 L of 1.50 M solutionbecause each L has 1.50 mol CaCl2, it makes sense that 1.75 L should have almost 3 moles1.50 mol CaCl2 = 1 L solution; 110.98 g CaCl2 = 1 mol1.50 M CaCl2, 1.75 Lmass CaCl2, gCheck:Solution:Conceptual Plan:

Relationships:Given:Find:mol CaCl2L solng CaCl2

183183DilutionDilution is the procedure for preparing a less concentrated solution from a more concentrated solution.One can also dilute a more concentrated solution byUsing a pipet to deliver a volume of the solution to a new volumetric flask, andAdding solvent to the line on the neck of the new flask.

184DilutionAdd SolventMoles of solutebefore dilution (i)Moles of soluteafter dilution (f)=MiViMfVf=

The molarity of the new solution can be determined from the equationDilutionExample 2.6.9: To what volume should you dilute 0.200 L of 15.0 M NaOH to make 3.00 M NaOH?because the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it doesM1V1 = M2V2V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 MV2, LCheck:Solution:Conceptual Plan:

Relationships:Given:Find:V1, M1, M2V2

186186Example 2.23: What is the concentration of a solution prepared by diluting 45.0 mL of 8.25 M HNO3 to 135.0 mL?because the solution is diluted by a factor of 3, the molarity should decrease by a factor of 3, and it doesM1V1 = M2V2V1 = 45.0 mL, M1 = 8.25 M, V2 = 135.0 mLM2, LCheck:Solution:Conceptual Plan:

Relationships:Given:Find:V1, M1, V2M2

187187

1.9 Concentration of Solutions

Practice 2.6.6How would you prepare 200.0 mL of 0.25 M NaCl solution from a 2.0 M solution? { Ans: 25 mL}Practice 2.6.7A 200.0 cm3 sample of oxalic acid solution contains 8.584 g of H2C2O4 . 25.0 cm3 of this acid solution is withdrawn and diluted to 500.0 cm3 by adding water in a 500ml volumetric flask. What is the molarity of the final solution? { Ans : 0.48 M; 2.383 x 10-2 M } 2009, Prentice-Hall, Inc.Using Molarities inStoichiometric Calculations

1891.9 Solution StoichiometryExample 2.24 3 Cu + 8 HNO3 3 Cu(NO3)2 + 2 NO + 4 H2OIn an experiment x grams of Cu reacted completely with 40 cm3 of 0.5 M HNO3 solution. a) Calculate the value of x. b) How many moles of NO will be formed in the above reaction ? 1901901.9 Solution StoichiometrySolution: Conceptual Plan Vol, Molarity HNO3 Mol HNO3 Mol Cu g CuMol = M x V g Cu = mol Cu x MM CuMol Cu = 3Mol HNO3 8MM Cu = 63.55 g/mol(a) g Cu = (0.5 mol/L x 0.04 L)HNO3 x 3 mol Cu x 63.55 g 8 mol HNO3 1 mol = 0.48 g `(b) Mol HNO3 Mol NOMol NO = 2 = 1Mol HNO3 8 4Mol NO formed = (0.5 x 0.04)mol HNO3 x 1 mol NO 4 mol HNO3 = 0.005 mol3 Cu + 8 HNO3 3 Cu(NO3)2 + 2 NO + 4 H2O;1.9 Solution StoichiometryPractice 2.8111.9 g of NH3 is produced when x grams of (NH4)2SO4 reacted completely in v cm3 of 2.5 M NaOH according to the equation below : (NH4)2SO4 + 2NaOH Na2SO4 + 2H2O + 2NH3Calculate the values of x and v. { Ans : x = 46.2 g ; v = 280 cm3 } 2009, Prentice-Hall, Inc.TitrationTitration is an analytical technique in finding the conc. of a solute in a solution

193

A solution of accurately known concentration is gradually added to another solution of unknown conc. until the chemical reaction between the two solutions is complete.Titration1.12 Solution StoichiometryExample 2.7.1: What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO3)2 in the reaction 2 KCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2 KNO3(aq)? 2 KCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2 KNO3(aq)? Let Molarity of KCl = Ma, Molarity of Pb(NO3)2 = Mb Volume of KCl = Va, Volume of Pb(NO3)2 = Vb Stoichiometry coeffient a, b = 2,1 for KCl,Pb(NO3)2, resp. Use the relatioship MaVa = a MbVb b Ma x 0.150 = 2 0.175 x 0.150 1 Ma = 2 x 0.175 = 0.350 M1951.9 Solution StoichiometryPractice 2.7.143.8 mL of 0.107 M HCl is to neutralize 37.6 mL of Ba(OH)2 solution. What is the molarity of the base? 2 HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2 H2O(aq) {Ans: 0.0623 M}

Practice 2.7.2Calculate the molarity of phosphoric acid, H3PO4 , if 40.0 cm3 of it requires 25.0 cm3 of 0.545 M potassium hydroxide, KOH for its neutralization. Balance the equation below for the neutralization reaction described above.

H3PO4 + KOH K3PO4 + H2O { Ans : 0.11 M }1961961.9 Solution StoichiometryPractice 2.7.3Aluminium reacts with sulphuric acid to produce hydrogen gas and aluminium sulphate.

2Al + 3H2SO4 Al2(SO4)3 + 3H2If a reaction vessel contains 2.4 g of Al and 200 cm3 of 0.5M H2SO4 , a)which compound is the limiting reactant? How many moles of H2 can be produced?How much of the excess reagent is left at the end of the reaction? { Ans : (a) H2SO4 (b) 0.1 mol (c) 0.022 mol = 0.594 g}1.9 Solution StoichiometryPractice 2.7.4In one experiment x grams of CS2 is mixed with y grams of NaOH. The reaction is represented by the unbalanced equation below. CS2 + NaOH Na2CS3 + Na2CO3 + H2O

At the end of the reaction, 7.5 g of Na2CO3 is formed while 3.3 g of CS2 is left unreacted in the reaction vessel. All the NaOH in the vessel is completely consumed. a) Balance the above equationb) Calculate the amount of CS2 (x grams) and NaOH (y grams) added to the reaction vessel at the beginning of the reaction. { Ans : x = 19.49 g ; y =17.0 g } 2009, Prentice-Hall, Inc.Other Ways of Expressing Concentrations of Solutions1991.9 Solution StoichiometryConcentration can be expressed in terms of:

1. Molality (mol/kg) 2. Mole Fraction (Xa) 3. Percent by Mass (w/w %)4. Percent by Volume (v/v %)5. Percent mass per volume (w/v %)

200Molality (m or mol/kg solvent)Moles of solute per 1 kilogram of solventdefined in terms of amount of solvent, not solution like the othersDoes not vary with temperaturebecause based on masses, not volumes

201201Mole Fraction, XAThe mole fraction is the fraction of the moles of one component in the total moles of all the components of the solutionTotal of all the mole fractions in a solution = 1Unitless The mole percentage is the percentage of the moles of one component in the total moles of all the components of the solution; (mole fraction x 100%)

,nA,ntotal202202Example 2.7Calculate the molarity and molality of a solution prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of H2O to make 515 mL of solution? What is the mol fraction of C2H6O2 in the solution?M = mol/L, 1 mol C2H6O2 = 62.07 g, 1 mL = 0.001 Lg C2H6O2mol C2H6O2mL solnL solnM

Solution: Conceptual Plan:203203

Total mol = 0.2771 + 27.753 mol = 28.03 mol

204205What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL?m =moles of solutemass of solvent (kg)M =moles of soluteliters of solutionAssume 1 L of solution:5.86 moles ethanol = 270 g ethanol927 g of solution (1000 mL x 0.927 g/mL)mass of solvent = mass of solution mass of solute= 927 g 270 g = 657 g = 0.657 kgm =moles of solutemass of solvent (kg)= 5.86 moles C2H5OH0.657 kg solvent= 8.92 mPractice 2.7.5A solution is made by dissolving 34.0 g of NH3 in 2.00 x 103 mL of water. Calculate thea) molality of the solutionb) the mole fraction of NH3 in the solution(MMNH3 = 17.04 g/mol, dH2O = 1.00 g/mL) { Ans : 1.02 m, 0.0180}Practice 2.7.6Calculate the Molarity and molality of solution (i) Molality of solution (ii) (i) An aqueous solution consists of 6.55% by mass of glucose (C6H12O6). (ii) 6.2 M H2SO4(aq) c) What is the mole fraction of the solute in both the solutions. (MMH2SO4 = 98.08 g/mol, dsoln = 1.80 g/mL)

206 2009, Prentice-Hall, Inc.Percent by Mass (%w/w)Mass % of A = (%w/w) mass of A in solutiontotal mass of solution 100Concentrations show the relationship between the amount of solute and the amount of solvent12% (w/w) sugar solution: 12 g sugar per 100 g solution or 12 kg sugar 1n 100 kg solution207 2009, Prentice-Hall, Inc.Percent by Volume (% v/v) Volume % of A =(%v/v) Volume of A in solutiontotal volume of solution 100Concentrations show the relationship between the volume of solute and the volume of solution 22%(v/v) ethanol solution: 22 mL EtOH per 100 mL solution2081.9 Solution StoichiometryPercent Concentration

209209Example 2.7.2: What volume of 10.5% by mass soda contains 78.5 g of sugar? Density of the solution 1.04 g/mLthe unit is correct, the magnitude seems reasonable as the mass of sugar 10% the volume of solutionCheck:Solve:100 g soln = 10.5 g sugar, 1 mL soln = 1.04 gConceptual Plan:Relationships:78.5 g sugarvolume, mLGiven:Find:g soluteg solnmL soln

210Parts Solute in Parts Solution Parts per million = parts of solute in every 1 million parts solutionif a solution is 36 ppm by volume, then there are 36 mL of solute in 1 million mL of solution1mg/L (1 mg of arsenic in 1L in drinking water or 1mg/Kg (1 mg of Ca in 1 Kg of soil)211211 2009, Prentice-Hall, Inc.Parts per Million andParts per Billionppm =mass of A in solutiontotal mass of solution 106Parts per Million (ppm)Parts per Billion (ppb)ppb =mass of A in solutiontotal mass of solution 109212Additional Problems

1. Calculate the mass of solute and mass of solvent (water) from each prepared solutiona). 125 g of 1.0% of NaNO3b). 300 g of 0.115m of C2H6O2 c). 125 mL of 0.1M of NaNO32. Calculate the amount of water (in grams) that must be added to 5.00g urea (NH2)2CO in preparation of a 16.2% by mass solution 3. The density of a 2.45 M aqueous solution of methanol (CH3OH) is 0.976 g/mL. What is the molality of the solution? (MW= 32.04g)213Thank You.

214I purposely reversed the order of reactants in the equation compared to the description to make the students ask me if the order they write the chemicals matters