3Scientific MethodYou are given a computer and asked to make a graph. After booting the computer, opening excel and entering data, the screen goes blank.Oh Gees! Now What!
4Units of Measurement Expect you to know Units used in science pico to gigaAnd be able to convertUnits used in scienceKilograms, meters, seconds, kelvins, amps, moles
5Significant Figures There is more than one convention! AP Chemistry allows for some variationIf you are within one sig fig, it is OKWe will follow thisRules are on Pg 23 of your bookWe will use these for every calculationYou lose a point for incorrect sig figs on test
6Calculations Adding and subtraction Answer has the same number of decimal places as the least precise measurement.12.1118.01.01331.12331.1Multiplication and DivisionAnswer has the same number of significant figuresas the least precise measurement4.56 x = 6.38corrected6.4pHThe number to the left of the decimal is the exponentThe number to the right of the decimal contains thecorrect number of sig figs.pH = has 2 sig figs
7Dimensional Analysis Do I really have to? It’s way easier! No, but it will cost you extra work explaining yourselfUnits written out in Dim Analysis are self explanatoryIt’s way easier!Way, way easier!!Just Do it!
8Mercury poisoning is a debilitating disease that is often fatal Mercury poisoning is a debilitating disease that is often fatal. In the human body, mercury reacts with essential enzymes leading to irreversible inactivity of these enzymes. If the amount of mercury in a polluted lake is micrograms Hg per milliliter, what is the total mass in kilograms of mercury in the lake. The lake has a surface area of mi2 and an average depth of 20.1 ft. (5280. ft in a mile, 12 in in a foot, 2.54 cm in an inch, 106 micrograms in a gram)
9Classification of Matter What is a mixture?Name two typesHow can we separate hetero?Homo?If I say something is a pure substance, what does that mean?What is the difference between an element and a compound?What is an element made up of?
10It’s the Law Explain the following laws: Conservation of MassDefinite proportionMultiple proportionName four parts of Dalton’s Atomic TheoryAtomsAll atoms of same element are identicalSame compound always has same elements in same proportionsAtoms themselves do not change in chemical reactions
11Famous Atomic Experiments Describe the Experiment JJ Thompson and CRT’sUsed CRT to determine charge to mass ratioDiscovered electronRutherford’s Gold FoilUsed alpha particles and gold foilDiscovered a dense, positive nucleusMillakan’s Oil DropletDiscovered the charge of an electronCalculated the mass of an electron with JJ’s reults
12Modern Theory Subatomic particles are? Nucleus is composed of ? Electron, neutron and protonNucleus is composed of ?Neutron and protonElectrons are in “clouds”What does that mean?
13F Symbol -1 19 How many protons? 9 How many neutrons? How many electrons?
15Modern periodic table * * + + H He Li Be B C N O F Ne Na Mg Al Si P S I A II A III A IV A V A VI A VIIA 0HHe1234567LiBeBCNOFNeIII B IVB V B VIB VIIB VIII B IB IIBNaMgAlSiPSClArKCaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKrRbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe*CsBaLuHfTaWReOsIrPtAuHgTlPbBiPoAtRn+FrRaLr*GdCmTbBkSmPuEuAmNdUPmNpCeThPrPaYbNoLaAcErFmTmMdDyCfHoEs+
16Metals * * + + H B He Ne F O N C Li Be Na Mg Al As Si Kr Br Se Ar Cl S PGeKCaScTiVCrMnFeCoNiCuZnGaRbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe*CsBaLuHfTaWReOsIrPtAuHgTlPbBiPoAtRn+FrRaLr*LaCePrNdPmSmEuGdTbDyHoErTmYb+AcThPaUNpPuAmCmBkCfEsFmMdNo
17Nonmetals * * + + H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca TiVCrMnFeCoNiCuZnGaGeAsSeBrKrRbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe*CsBaLyHfTaWReOsIrPtAuHgTlPbBiPoAtRn+FrRaLr*LaCePrNdPmSmEuGdTbDyHoErTmYb+AcThPaUNpPuAmCmBkCfEsFmMdNo
18Semimetals or Metalloids HHeLiBeBCNOFNeNaMgAlSiPSClArKCaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKrRbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe*CsBaLuHfTaWReOsIrPtAuHgTlPbBiPoAtRn+FrRaLr*LaCePrNdPmSmEuGdTbDyHoErTmYb+AcThPaUNpPuAmCmBkCfEsFmMdNo
19Bonds and Stuff Explain the following Ion Cation and anion Ionic Bond Covalent BondMoleculeFormula UnitChemical FormulaStructural Formula
20Ions and Ionic Compounds Important: note that there are no easily identified NaCl molecules in the ionic lattice. Therefore, we cannot use molecular formulas to describe ionic substances.
21Naming CompoundsMemorize all polyatomic ions pg 63 and how to determine the restMemorize names of elementss block, p block = allTransition metalsNeed to know common metalsCharges on Al+3, Zn+2, Ag+1, Cd+2
22Ions Cation - a positively charged ion. Mg Mg2+ + 2e- Ions are charged particles formed by the transfer of electrons between elements or combinations of elements.Cation - a positively charged ion.Mg Mg e-Anion - a negatively charged ion.F2 + 2e F-
23Writing Formulas Al3+ O2- 2 3 Al2O3 All compounds are electrically neutral. The sum of the positive and negative charges must add up to zero.Al3+O2-23Use subscripts to indicate how many of each ion is used.Al2O3
24Naming inorganic compounds When an element forms only one compound with a given anion.name the cationname the anion using the ending (-ide) for monatomic ionsNaCl sodium chlorideMgBr2 magnesium bromideAl2O3 aluminum oxideK3N potassium nitride
25Naming ionic compounds Many metals form more than one compound with some anions.For these, Roman numerals are used in the name to indicate the charge on the metal.Cu O2- = Cu2Ocopper(I) oxide copper(I) oxideCu O = CuOcopper(II) oxide copper(II) oxide
26Naming ionic compounds Since the charge of some metal ions can vary, look at everything else first.What ever is left is the charge on the metal!FeBr3The three bromides are each 1- so iron must be 3+ for the compound to have zero net charge.Iron (III) bromide
27Examples FeCl2 iron (II) chloride FeCl3 iron (III) chloride SnS AgClZnSiron (II) chlorideiron (III) chloridetin (II) sulfidetin (IV) sulfidesilver chloridezinc sulfideNote: Some transition metals have only one oxidation state, so Roman numbers are omitted.
28Metals with multiple charges Transition metals.Here it is easier to list some of the common elements that only have a single oxidation state.All Group 3B are 3+Zn and Cd are 2+Ag is 1+
29Oxidation numbers and the P.T. Some observed trends in compounds.Metals have positive oxidation numbers.Transition metals typically have more than one oxidation number.Nonmetals and semimetals have both positive and negative oxidation numbers.No element exists in a compound with an oxidation number greater than +8.The most negative oxidation numbers equals the group number - 8
31Polyatomic ions NH4+ ammonium NO3- nitrate SO42- sulfate OH- hydroxide A special class of ions where a group of atoms tend to stay together.NH4+ ammoniumNO3- nitrateSO42- sulfateOH- hydroxideO22- peroxideYour book contains a more complete list.
32Polyatomic ionsFor compounds that contain 1 or 2 polyatomic ions, base the formulas upon the given ion name(s).ammonium chloride NH4Clsodium hydroxide NaOHpotassium permanganate KMnO4ammonium sulfate (NH4)2SO4
33Naming Inorganic Compounds Names and Formulas of Ionic CompoundsPolyatomic anions containing oxygen with more than two members in the series are named as follows (in order of decreasing oxygen):per- …. -ate ClO41-…. -ate ClO31-…. -ite ClO21-hypo- …. -ite ClO1-
34Oxidation number and nomenclature Polyatomic anions containing oxygen rely on a modification of the name of the other element to indicate the oxidation number.Anionsper ________ate________ate________itehypo ________iteIncreased #oxygen andOxidation number
35Oxidation number and nomenclature ExamplesCl oxidationnumber Formula Name+7 NaClO4 sodium perchlorate+5 NaClO3 sodium chlorate+3 NaClO2 sodium chlorite+1 NaClO sodium hypochlorite-1 NaCl sodium chlorideUsually, the overall charges of all ions for a nonmetal are the same. Sometimes the -ates and -ites have a different charge than the -ide ions.
36Slivka’s Square -ate has 4 Oxygens -ate has 3 Oxygens Polyatomic Ions I A II A III A IV A VA VI A VIIA 0HHe-ate & -ite charges usually = -ide charge1234567LiBeBCNOFNeIII B IVB V B VIB VIIB VIII B IB IIBNaMgSlivka’s SquareAlSiPSClArKCaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKrRbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe*CsBaLuHfTaWReOsIrPtAuHgTlPbBiPoAtRn+FrRaLr*GdCmTbBkSmPuEuAmNdUPmNpCeThPrPaYbNoLaAcErFmTmMdDyCfHoEs+
37Polyatomic Ions“ate”4 Oxygens .. Inside Slivka’s Squareex: SO42- sulfate3 Oxygens .. Borders the outside of the squareex: NO31- nitrate
381 less Oxygen compared to the -ate ex: ClO21- chlorite SO32- sulfite Polyatomic Ions“ite”1 less Oxygen compared to the -ateex: ClO21- chloriteSO32- sulfite
39“per” root name “ate” has 1 more O than the “ate” ex: IO41- periodate Polyatomic Ions“per” root name “ate” has 1 more O than the “ate” ex: IO41- periodate“hypo” root name “ite” has 2 less O than the “ate”ex: ClO1- hypochlorite
40“Per”-“ate” 1 more O - “ate” - “ite” 1 less O “hypo”-“ite” 2 less O Polyatomic Ions“Per”-“ate” 1 more O- “ate”- “ite” 1 less O“hypo”-“ite” 2 less O(also notice oxidation # of nonmetal changes)
41follow Group A patterns Polyatomic IonsGroup B Elementsfollow Group A patternsI A II A VIIIACrO chromateMnO41- permanganateHHeIII A IV A VA VI A VIIA1234567LiBeBCNOFNeIII B IVB V B VIB VIIB VIII B IB IIBNaMgAlSiPSClArKCaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKrRbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe*CsBaLuHfTaWReOsIrPtAuHgTlPbBiPoAtRn+FrRaLr*GdCmTbBkSmPuEuAmNdUPmNpCeThPrPaYbNoLaAcErFmTmMdDyCfHoEs+
42Other Methods of Naming - Latin For ionic compounds containing a metal and a nonmetal, the Latin root word for the metal is sometimes used with an -ous or -ic suffix.The -ous suffix indicates a lower oxidation state, -ic a higher one.Ex. ferrous = Fe2+ferric = Fe3+
43Latin Root Words plumbous = Pb2+ cuprous = Cu1+ plumbic = Pb4+ cupric = Cu2+stannous = Sn2+stannic = Sn4+plumbous = Pb2+plumbic = Pb4+aurous = Au1+auric = Au3+Note that there is no pattern between the -ous and -ic suffixes and the actual charge of the ions.
44Naming Covalent Compounds What is the difference betweenSO32- and SO3polyatomic ion vs. neutral compoundsulfite ion vs. sulfur (VI) oxide
45Naming Covalent Compounds Some nonmetals can have more than one positive oxidation state when they share electrons to form molecules called covalent compounds. Therefore Roman numbers must be used.SCl4 sulfur (IV) chlorideSCl sulfur (VI) chlorideCO carbon (II) oxideCO2 carbon (IV) oxide
46Naming Inorganic Compounds Names and Formulas of Binary Molecular CompoundsBinary molecular compounds are composed of two nonmetallic elements.The element with the positive oxidation number (the one closest to the lower left corner on the periodic table) is usually written first Exception: NH3.Greek prefixes are used to indicate the number of atoms in the molecule(subscripts).PCl5 is phosphorus pentachloride
47Naming Inorganic Compounds Names and Formulas of Binary Molecular CompoundsRoman numerals can be used to indicate the positive oxidation number, but sometimes prefixes more accurately describe the actual composition of the molecule.Example: sulfur (V) fluoride exists as disulfur decafluoride molecules.FFFFSSFFFFFF
48Other Methods of Naming molecules For binary molecular compounds composed of two nonmetals, prefixes are sometimes used to indicate the number of atoms of each element present.Common prefixesmono = di = tri = 3tetra = penta = hexa = 6hepta = octa = deca = 10
49Other Methods of Naming molecules name elements in the formula.use prefixes to indicate how many atoms there are of each type.N2O5CO2COCCl4dinitrogen pentoxidecarbon dioxidecarbon monoxidecarbon tetrachlorideThe rule may be modified to improve how a name sounds.Example - use monoxide not monooxide.
50-ide becomes hydro-….-ic acid -ate becomes -ic acid Other Naming -AcidsAcids are substances that produce H+ ions in water solutions (aqueous).The names of acids are related to the names of the anions to which H+ is bonded:-ide becomes hydro-….-ic acidH2S is hydrosulfuric acid-ate becomes -ic acidH3PO4 is phosphoric acid-ite becomes -ous acidHNO2 is nitrous acid
52Naming Inorganic Acids Salt Name FormulaAcid Name FormulaSodium acetate NaC2H3O2Acetic acid HC2H3O2Sodium chloride NaClHydrochloric acid HClSodium hyponitrite NaNOHyponitrous acid HNOSodium phosphite Na3PO3Phosphorous acid H3PO3Sodium sulfate Na2SO4Sulfuric acid H2SO4
53Naming Inorganic Compounds Names and Formulas of AcidsAcids contain hydrogen as the only cation. The names of acids are related to the names of the anions:-ide becomes hydro-….-ic acid;H2S is hydrosulfuric acid-ate becomes -ic acid;H3PO4 is phosphoric acid-ite becomes -ous acid.HNO2 is nitrous acid
54Other Naming -Double & Triple Salts Polyatomic anions containing oxygen with additional hydrogens are named by adding hydrogen (or bi-) for one extra H, dihydrogen (for two extra H), etc., to the name as follows:CO32- is named carbonate, but HCO3- is hydrogen carbonate (or bicarbonate)H2PO4- dihydrogen phosphate anion.Note that these are not named as acids, since another cation is still needed to balance the charge.
55Other Naming -Double & Triple Salts Two or three different positive ions can be attracted to the same negative ion to form a single compound. These are called double or triple salts. Each ion is named as it appears.NaHCO3 sodium hydrogen carbonate (or sodium bicarbonate)AlK(SO4)2 aluminum potassium sulfate
56Other Naming - Hydrates pentahydrate5H2OOther Naming - HydratesHydrated compounds physically trap water molecules as part of their structure. A prefix is used to indicate the relative number of water molecules present with the word hydrate added after the compound’s name.copper (II) sulfate pentahydrate CuSO45H2O
57Other Naming - Historical Names Sometimes the names of compounds are based upon their historical significance or derivation. There are no patterns or rules for determining these names, so they would have to be memorized.For example, H2O is called water, not dihydrogen monoxideCheck out
58A quick review of nomenclature Is a metal presentas the first element?NoNoIs a nonmetal thefirst element?Is hydrogenfirst element?YesYesYesCan the metal havemore than oneoxidation state?Use Roman numeralsor may use prefixes(mono, di, tri ...)Name asan acidNo-ides becomehydro- -ic acids-ates become-ic acids-ites become-ous acidsYesRoman numeralsare not needed.Use Roman numeralsor may use Latin namewith -ous/-ic suffixes
59A quick review of nomenclature Look up thename or formulaIs it a binarycompound?NoYesNoDoes it containone of the 8common ions?Does it have moreor less O atoms thanone of the -ate ions?NoUse the -idesuffix for thenegative ionYesYesName the commonpolyatomic ionUse per- -ate 1 more O-ite 1 less Ohypo- -ite 2 less O
60Naming Compounds Summary Simple rules that will keep you out of trouble most of the time.Groups IA, 2A, 3A (except Tl) only have a single oxidation state that is the same as the group number - don’t use numbers.Most other metals and semimetals have multiple oxidation states - use numbers.If you are sure that a transition group element only has a single state, don’t use a number.
61Average Atomic Mass What is the average atomic mass of carbon-12? Why is this a bad question?If I traveled to alpha centauri, would the average atomic mass of chlorine be 35.45?Can you calculate the AAM of the following:1H = 99%2H = 1%
62Moles and Moles How many atoms in a mole? What does the “mole” do? How do you calculate molar mass?What is an empirical formula?What is a molecular formula?
63Atomic masses Atoms are composed of protons, neutrons and electrons. Almost all of the mass of an atom comes from the protons and neutrons.All atoms of the same element will have the same number of protons. The number of neutrons may vary - isotopes.Most elements exist as a mixture of isotopes.
64IsotopesIsotopes Atoms of the same element but having different masses.Each isotope has a different number of neutrons.Isotopes of hydrogen H H HIsotopes of carbon C C C12131126136146
65Isotopes Most elements occur in nature as a mixture of isotopes. Element Number of stable isotopesHCOFeSnThis is one reason why atomic masses(weights) are not whole numbers. They are based on averages.
66Atomic massesAs a reference point, we use the atomic mass unit (u), which is equal to 1/12th of the mass of a 12C atom.(One atomic mass unit (u) = 1.66 x gram)Using this relative system, the mass of all other atoms can be assigned.Examples 7Li = u14N = u29Si = u
67Average atomic massesOne can calculate the average atomic weight of an element if the abundance of each isotope for that element is known.Silicon exists as a mixture of three isotopes. Determine it’s average atomic mass based on the following data.Isotope Mass (u) Abundance28Si %29Si %30Si %
68Average atomic masses 92.23 (27.9769265 u) = 25.80 u 28Si 100 4.67 3.10( u) = u28Si29Si30SiAverage atomic mass for silicon = u
69The mole The number of atoms in 12.000 grams of 12C can be calculated. One atom 12C = u = 12 x (1.661 x g)= x g / atom# atoms = g (1 atom / x g)= x 1023 atomsThe number of atoms of any element needed to equal its atomic mass in grams will always be 6.02 x 1023 atoms - the mole.
70Moles and masses Atoms come in different sizes and masses. A mole of atoms of one type would have a different mass than a mole of atoms of another type.H grams / molO grams / molMo grams / molPb grams / molWe rely on a straight forward system to relate mass and moles.
711 mole of any element = 6.02 x 1023 atoms The mole1 mole of any element = 6.02 x 1023 atoms= gram atomic massAtoms, ions and molecules are too small to directly measure in u.Using moles gives us a practical unit.We can then relate atoms, ions and molecules, using an easy to measure unit - the gram.
72Masses of atoms and molecules Atomic massThe average, relative mass of an atom in an element. Can be expressed in relative amtomic mass units (u) or grams / mole.Molecular or formula massThe total mass for all atoms in a compound.
73Masses of atoms and molecules H2O - water2 hydrogen 2 x u1 oxygen 1 x umass of molecule u18.02 g / molRounded off basedon significant figures
74The moleIf we had one mole of water and one mole of hydrogen, we would have the same number of molecules of each.1 mol H2O = x 1023 molecules1 mol H2 = x 1023 moleculesWe can’t weigh out moles-we use grams.We would need to weigh out a different number of grams to have the same number of molecules
75Converting units Factor label method Regardless of conversion, keeping track of units makes thing come out rightMust use conversion factors- The relationship between two unitsCanceling out units is a way of checking that your calculation is set up right!
76Molecular mass vs. formula mass Formula mass - Add the masses of all the atoms in formula; for molecular and ionic compounds.Molecular mass - Calculated the same as formula mass; only valid for molecules.Both have units of either u or grams / mole.Molar mass is the generic term for the mass of one mole of anything.
77Formula mass, FMThe sum of the atomic masses of all elements in a compound based on the chemical formula.You must use the atomic masses of the elements listed in the periodic table.CO atom of C and 2 atoms of O1 atom C x u = u2 atoms O x u = uFormula mass = uor g / mol
78Another example CH3CH2OH - ethyl alcohol 6 hydrogen 6 x 1.008 u 2 carbon 2 x u6 hydrogen 6 x u1 oxygen 1 x umass of molecule u46.07 g /mol
79Molar massesOnce you know the mass of an atom, ion, or molecule, just remember:Mass of one unit - use amuMass of one mole of units - use grams / moleThe numbers DON’T change -- just the units.
80Example - (NH4)2SO4How many atoms are in 20.0 grams of ammonium sulfate?Formula weight = grams/ 1 molAtoms in formula = 15 atoms / 1 formula unitX moles = 20.0 g x = mol1 molgatoms = mol x x6.02 x1023 units1 mol15 atoms1 unitatoms = 1.36 x1024
81Formula weight = 132.14 g / 1 mol (NH4)2SO4 Example - (NH4)2SO4Other information can be derived from the chemical formula of a compound.How many moles of ammonium ions are in 20.0 grams of ammonium sulfate?Formula weight = g / 1 mol (NH4)2SO42 moles NH4 / 1 mol (NH4)2SO4x moles = 20.0 g x = mol (NH4)2SO41 molgx moles NH4 = mol (NH4)2SO4 x2 moles NH41 mol (NH4)2SO4moles NH4 = 0.302
82Formula weight = 132.14 g / 1 mol (NH4)2SO4 Example - (NH4)2SO4How many grams of sulfate ions are in 20.0 grams of ammonium sulfate?Formula weight = g / 1 mol (NH4)2SO496.06 grams SO4 / 1 mol (NH4)2SO4x moles = 20.0 g x = mol (NH4)2SO41 molgx grams SO4 = mol (NH4)2SO4 x96.06 g SO41 mol (NH4)2SO4grams SO4 = 14.5
83Masses of atoms and molecules Law of Definite Composition - compounds always have a definite proportion of the elements that make it up.These proportions can be expressed as ratios of atoms, equivalent mass values, percentage by mass or volumes of gaseous elements.Ex. Water always contains 2 H atoms for every O atom, which is 2 g H for every 16 g O or 11.1% H and 88.9% O by mass.
84Percent Composition by Mass Percent composition can also be determined from experimental data.Example: When 2.47 g KClO3 is heated strongly, 0.96 g of O2 gas is driven off. What is the % by mass of oxygen in KClO3?% 0 = g O x 100 = % O2.47 g KClO3Based upon the formula mass:% O = u O x 100 = % Ou KClO3
85Gay-Lussac’s Law Law of of Combining Volumes. At constant temperature and pressure, the volumes of gases involved in a chemical reaction are in the ratios of small whole numbers.Studies by Joseph Gay-Lussac led to a better understanding of molecules and their reactions.
86Gay-Lussac’s Law Example. Reaction of hydrogen and oxygen gases. Two ‘volumes’ of hydrogen will combine with one ‘volume’ of oxygen to produce two volumes of water.We now know that the equation is:2 H2 (g) + O2 (g) H2O (g)+H2O2H2O
87Contain same number of moles of molecules Avogadro’s lawEqual volumes of gas at the same temperature and pressure contain equal numbers of molecules (or moles of molecules).Contain same number of moles of molecules
88Standard conditions (STP) Remember the following standard conditions.Standard temperature = Kelvin(the normal freezing point of water, 0ºC)Standard pressure = 1 atmosphere(the normal air pressure at sea level, 14.7 psi)At these conditions:One mole of any gas has a volume of 22.4 liters at STP.
89Applying Law of Definite Composition In an expermiment, 10.0 grams of water is decomposed by electrolysis.Problem: How many liters of O2 gas will be formed at STP? How many grams of H2 gas will be formed?X L O2 =10.0g H2O x x x1 mol H2O18.0 g H2O0.5 mol O21 mol H2O22.4 L O21 mol O2liters O2 = Note that 12.4 L H2 will also be formed.X g H2 =10.0g H2O x2.02 g H218.02 g H2O= g H2
90Empirical formulaThis type of formula shows the ratios of the number of atoms of each kind in a compound.For organic compounds, the empirical formula can be determined by combustion analysis.Elemental analyzerAn instrument in which an organic compound is quantitatively converted to carbon dioxide and water -- both of which are then measured.
91Elemental analyzerfurnaceCO2trapH2OO2sampleA sample is ‘burned,’ completely converting it to CO2 and H2O. Each is collected and measured as a weight gain. By adding other traps elements like oxygen, nitrogen, sulfur and halogens can also be determined.
92Elemental analysisExample: A compound known to contain only carbon, hydrogen and nitrogen is examined by elemental analysis. The following information is obtained.Original sample mass = gMass of CO2 collected= gMass of H2O collected= gDetermine the % of each element in the compound.
93Elemental analysis Mass of carbon 12.01 g C = 0.04470 g C 44.01 g CO2 Mass of hydrogenMass of nitrogeng CO212.01 g C44.01 g CO2= g Cg H2O2.016 g H18.01 g H2O= g Hg sample g C g H= g N
94Elemental analysisSince we know the total mass of the original sample, we can calculate the % of each element.% C = x 100% = %% H = x 100% = %% N = x 100% = %gggg
95Empirical formula Empirical formula The simplest formula that shows the ratios of the number of atoms of each element in a compound.Example - the empirical formula for hydrogen peroxide (H2O2) is HO.We can use either our mass or our percent composition information from the earlier example to determine an empirical formula.
96Empirical formula 1 mol C = 0.003722 mol C 12.01 g C 1 mol H g H1 mol H1.008 g H= mol Hg C1 mol C12.01 g C= mol Cg N1 mol N14.01 g N= mol N
97Empirical formulaThe empirical formula is then found by looking for the smallest whole number mole ratio.C / = 1.000H / = 4.998N / = 1.000The empirical formula is CH5N
98Empirical formula % C = 38.67 % % H = 16.22 % % N = 45.11 % From experimental analysis, we found that a compound had a composition of:If we assume that we have a gram sample, then we can divide each percentage by the elements atomic mass and determine the relative number of moles of each.% C = %% H = %% N = %
99Empirical formula 16.22 g H 1 mol H 1.008 g H = 16.09 mol H 38.67 g C 1 mol C12.01 g C= mol C45.11 g N1 mol N14.01 g N= mol N
100Empirical formulaThe empirical formula is found by looking for the smallest whole number ratio.C / = 1.000H / = 4.997N / = 1.000The empirical formula is determined to be the same, CH5N, whether using actual masses of the elements present in the sample or by using their % composition by mass.
101Molecular formulaMolecular formula - shows the actual number of each type of atom in a molecule.They are multiples of the empirical formula.If you know the molecular mass, then the molecular formula can be found.For our earlier example, what would be the molecular formula if you knew that the molecular mass was 62.12?
102Molecular formula Empirical formula CH5N Empirical formula mass uMolecular mass 62.12Ratio: / = 2The molecular formula is C2H10N2Note: This does not tell you how the atoms are arranged in the compound!
103Hydrated CompoundsThe formula for hydrated compounds are solved in a similar fashion as empirical formulas.Example: When a g sample of hydrated barium chloride is heated todryness, g H2O is lost.5.000g hydrate g H2O= 4.262g BaCl2
104Hydrated Compounds BaCl2 0.0205 / 0.0205 = 1.00 4.262g BaCl2 (1 mol BaCl2) = mol BaCl2(208.2 g BaCl2)0.738 g H2O ( 1 mol H2O ) = mol H2O(18.0 g H2O )BaCl / = 1.00H2O / = 2.00The compound’s formula is BaCl22H2O
105Molarity M = moles solute mol liters of solution L = Molarity Recognizes that compounds have differentformula weights.A 1 M solution of sucrose contains thesame number of molecules as 1 M ethanol.[ ] - special symbol which means molar( mol/L )
106Molarity M NaOH = 10 mol NaOH / 2.0 L = 5.0 M Calculate the molarity of a 2.0 L solution that contains 10 moles of NaOH.M NaOH = 10 mol NaOH / 2.0 L = 5.0 M
107Solution preparation Solutions are typically prepared by: Dissolving the proper amount of solute and diluting to volume.Dilution of a concentrated solution.Lets look at an example of the calculations required to prepare known molar solutions using both approaches.
108Making a solutionYou are assigned the task of preparing mL of a 1.00 M solution of sodium hydroxide.What do you do?First, you need to know how many moles of NaOH are in mL of a 1.00 M solution.mol = M x V (in liters)= 1.00 M x liters= moles NaOH
109Making a solutionNext, we need to know how many grams of NaOH to weigh out.g NaOH = mol x molar massNaOH= mol x 40.0 g/mol= 10.0 grams NaOH
110Making a solution Finally, you’re ready to make the solution. Weigh out exactly 10.0 grams of dry, pure NaOH and transfer it to a volumetric flask, (or some other containing where the exact volume can be accurately measured.)Fill the flask about 1/2 of the way with distilled water and gently swirl until the solid dissolves.Now, dilute exactly to the mark, cap and mix.
111DilutionOnce you have a solution, it can be diluted by adding more solvent. This is also important for materials only available as solutionsM1V1 = M2V21 = initial = finalAny volume or concentration unit can be used as long as you use the same units on both sides of the equation.
112DilutionHow many ml of concentrated 12.0 M HCl must be diluted to produce mL of 1.00 M HCl?M1V1 = M2V2M1 = 12.0 M M2 = 1.00V1 = ??? ml V2 = mLV1 = M2V2 / M1M2 = (1.00 M) (250.0 mL) = 20.8 mL(12.0 M)
113Diluting an AcidWhen diluting concentrated acids, ALWAYS add the acid to water to help dissipate the heat released.Fill the volumetric flask (or some other containing where the exact volume can be accurately measured.) about 1/2 of the way with distilled water.Measure out exactly 20.8 mL of 12.0 M HCl and transfer it to a volumetric flask. Gently swirl to mix.Now, dilute exactly to the mark, cap and mix.
114Other Methods of Expressing Concentration When making different solutions with a specific molarity, the number of milliliters of solvent needed to prepare 1 liter of solution will vary.Sometimes it is necessary to know the exact proportions of solute to solvent that are in a particular solution.Various methods have been devised to express these proportions.
115Molality moles solute mol Molality (m) = = kilograms of solvent kg Recognizes that the ratio between moles of solute and kg of solvent can vary.A 1 m solution of sucrose contains the same number of molecules as 1 m ethanol.The freezing point of water is lowered by 1.86ºC/ m and the boiling point is raised by 0.51ºC/ m.
116Density grams of solution g Density (D) = = milliliters of solution mL Focuses on the total solution and does not emphasize either the solute or solvent.g solution = g solute + g solventUnits may be expressed as other mass per volume ratios.
117Percent Composition value of the part Percent Composition = x 100 Value of the wholePercent Composition = x 100% by Mass = g solute / g solution x 100% by Volume = mL solute / mL solution x100% by Mass per Volume =g solute/mL solution x 100Must specify which type of % composition.
118Mole Fraction moles of solute or solvent Mole fraction = total moles of solute & solventMole fraction =Often used to compare ratio between moles of gases in a mixture.The mole ratio of gases in a mixture is equal to their pressure ratio and their volume ratio.
119Parts per Million or Billion # grams of solute1,000,000 g solutionParts per million (ppm) =Parts per Million or BillionUsed to express concentrations for verydilute solutions.For aqueous solutions, the mixture ismostly water. Therefore, the density ofthe solution = 1 g/mL, and1 ppm = 1 g/1000 L.
120Stoich Baby Given the following equation ___N2 + ___H2 ___NH3 Given one mole of nitrogen gas, how many moles of ammonia would form?Assuming gases at STP, given one mole of nitrogen, how many liters of ammonia would form?Given 2 liters of nitrogen and 5 liters of hydrogen, how may liters of ammonia are formed? What is left over?
121Stoichiometry Stoichiometry The study of quantitative relationships between substances undergoing chemical changes.Law of Conservation of MatterIn chemical reactions, the quantity of matter does not change.The total mass of the products must equal that of the reactants.
122Chemical equations Chemist’s shorthand to describe a reaction. It shows:All reactants and productsThe state of all substancesAny conditions used in the reactionCaCO3 (s) CaO (s) CO2 (g)Reactant ProductsA balanced equation shows the relationshipbetween the quantities of all reactants and products.
123Balancing chemical equations Each side of a chemical equation must have the same number of each type of atom.CaCO3 (s) CaO (s) + CO2 (g)Reactants Products1 Ca Ca1 C C3 O O
124Balancing chemical equations Step 1 Count the number of atoms of each element on each side of the equation.Step 2 Determine which atom numbers are not balanced.Step 3 Balance one atom at a time by using coefficients in front of one or more substances.Step 4 Repeat steps 1-3 until everything is balanced.
125Example. Decomposition of urea (NH2)2CO + H2O ______> NH3 + CO22 N 1 N < not balanced6 H 3 H < not balanced1 C 1 C2 O 2 OWe need to double NH3 on the right.(NH2)2CO + H2O ______> 2NH3 + CO2
126Mass relationships in chemical reactions Stoichiometry - The calculation of quantities of reactants and products in a chemical reaction.2 H2 + O > 2 H2OYou need a balancedequation and you WILLwork with moles.
127Stoichiometry, General steps. 1Balance the chemical equation.2Calculate formula masses.3Convert masses to moles.4Use chemical equation toget the needed answer.Convert back to mass if needed.5
128Mole calculationsThe balanced equation shows the reacting ratio between reactants and products.2C2H6 + 7O CO2 + 6H2OFor each chemical, you can determine themoles of each reactant consumedmoles of each product madeIf you know the formula mass,mass quantities can be used.
129Mole-gram conversion How many moles are in 14 grams of N2 ? Formula mass= 2 N x g/mol= g /molmoles N2= 14 g x 1 mol /28.02 g= moles
130Mass calculations We don’t directly weigh out molar quantities. We can use measured masses like kilograms, grams or milligrams.The formula masses and the chemical equations allow us to use either mass or molar quantities.
131Mass calculationsHow many grams of hydrogen will be produced if 10.0 grams of calcium is added to an excess of hydrochloric acid?2HCl + Ca ______> CaCl2 + H2Note:We produce one H2 for each calcium.There is an excess of HCl so we have all we need.
132Mass calculations = 10.0 g x = 0.25 mol Ca 2HCl + Ca ____> CaCl2 + H2First - Determine the number of moles of calcium available for the reaction.Moles Ca = grams Ca / FM Ca= g x= mol Ca1 mol40.08 g
133Mass calculations2HCl + Ca _____> CaCl2 + H210 g Ca = 0.25 mol CaAccording to the chemical equation, we get one mole of H2 for each mole of Ca.So we will make 0.25 moles of H2.grams H2 produced = moles x FW H2= mol x g/mol= grams
134Mass calculations OK, so how many grams of CaCl2 were made? 2HCl + Ca _____> CaCl2 + H210 g Ca = 0.25 mol CaWe would also make 0.25 moles of CaCl2.g CaCl2 = 0.25 mol x FM CaCl2= 0.25 mol x g / mol CaCl2= g CaCl2
135Limiting reactant In the last example, we had HCl in excess. Reaction stopped when we ran out of Ca.Ca is considered the limiting reactant.Limiting reactant - the material that is in the shortest supply based on a balanced chemical equation.
137ExampleFor the following reaction, which is limiting if you have 5.0 g of hydrogen and 10 g oxygen?Balanced Chemical Reaction2H2 + O2 ________> 2H2OYou need 2 moles of H2 for each mole of O2.Moles of H2 5 g = 2.5 molMoles of O g = 0.31 mol1 mol2.0 g1 mol32.0 g
138Example Balanced Chemical Reaction 2H2 + O2 2H2O You need 2 moles of H2 for each mol of O2You have 2.5 moles of H2 and 0.31 mol of O2Need a ratio of 2:1but we have a ratio of 2.5 : or 8.3 : 1.Hydrogen is in excess and oxygen is thelimiting reactant.
139Theoretical, actual, and percent yields Theoretical yieldThe amount of product that should be formed according to the chemical reaction and stoichiometry.Actual yieldThe amount of product actually formed.Percent yieldRatio of actual to theoretical yield, as a %.Quantitative reactionWhen the percent yield equals 100%.
140Yield Less product is often produced than expected. Possible reasons A reactant may be impure.Some product is lost mechanically since the product must be handled to be measured.The reactants may undergo unexpected reactions - side reactions.No reaction truly has a 100% yield due to the limitations of equilibrium.
141Percent yieldThe amount of product actually formed divided by the amount of product calculated to be formed, times 100.% yield = x 100In order to determine % yield, you must be able to recover and measure all of the product in a pure form.Actual yieldTheoretical yield
142Stoichiometry Step 1 Step 2 Step 3 Step 4 Step 5 Identify species present in solution and determine the reaction that occursStep 2Write the balanced net ionic equationStep 3Calculate the moles of reactionsolution = molarity x volumeheterogeneous = grams molar massStep 4Consider the limiting reactantStep 5Answer the question using stoich!