Notes

Outline

Chemical Formulas and Equations Stoichiometry and Calculations

Laws Of Chemistry Law of Mass Conservation Law of Constant Composition Law of Multiple Proportions

Law of Mass Conservation The total mass of the initial substances will equal the total mass of the end products Mass can neither be created or destroyed Proposed by Antoine Lavoisier Reacted Mercury with Oxygen to produce red powder Mercury II Oxide Decomposed the red powder and trapped the Oxygen gas showing that the total mass of Mercury and Oxygen combined equaled the mass of the Mercury II Oxide

Law of Constant Composition Proposed by Proust Also called the Law of Definite Proportions CuCO3 ---àCuO   +   CO2 The elements in a compound are combined in fixed numbers of atoms and in fixed  mass proportions(mass percent composition is fixed)

Law of Multiple Proportions Proposed by John Dalton When two elements form more than one compound, the masses of one element in these compounds combine with a fixed mass of the other element are in a ratio of simple whole numbers Examples—CO vs CO2 or PCl3 vs PCl5

Terminology of Chemical Equations Chemical Equation-A symbolic representation of a Physical or Chemical Process aA(l)   +   bB(s) ---à cC(g)    +  dD(aq) Reactants                  products Where  l = liquid; s = solid; g = gas;             aq = aqueous

Balancing Chemical Equations Methods of Balancing Balancing By Inspection Systematic Balancing

T-18 Balancing Chemical Equations

Tips on Balancing By Inspection Number of atoms of each element must be the same on both sides of equation Only coefficients may be adjusted;Never the subscripts To determine total atoms of an element in a formula multiply the coefficient by the subscript of the element in the formula If more than one source of an element exists on the same side of equation must consider the total number of the element on that side

Isotopes Isotopes-two or more forms of an element whose atoms have the same number of protons and electrons but differ in the number of neutrons Isotopic name is the name of the element with mass number suffixed to it Carbon-12, Carbon-13 Oxygen-15, Oxygen-16, Oxygen-18 Protium, Deuterium, Tritium

Distribution of Isotopes In An Element Natural Abundance-The proportion of an isotope in a randomly selected sample of an element Natural Abundances of each isotope remain fixed Expressed as a percentage or a fraction Number of atoms of that isotope / total number of atoms in sample Multiplied by 100 makes it Percent Natural Abundance

T-2.11 Determining Isotopic Masses and Natural Abundances

T-54 A Mass Spectrometer

T-55 Mass spectrum For Mercury

Atomic Mass Of An Element Not the same as mass number Atomic mass- weighted average of all isotopic masses based on Carbon –12 as a standard Weighting Factor is the fractional Natural Abundance Grade analogy

Determining Atomic Mass  of An Element Mass spectrometer provides isotopic masses and fractional abundances At. Mass =(natural abundance of isotope 1) (isotopic mass of isotope 1) + (natural abundance of isotope 2) (isotopic mass of isotope 2) + …. Example

Molecular Mass and Formula Mass Use molecular mass when referring to a molecular compound Use Formula Mass when referring to an ionic compound No difference in the way they are determined

Determining Molecular(Formula)Mass Requires the formula Molecular (formula) mass = subscript of element x atomic mass Example: HNO3  molecular mass = (subscript H x at mass H) + (subscript for N x at mass of N) + (subscript of O x at mass of O) Molecular mass = (1 x 1.0) + (1 x 14.0) + (3 x 16.0) = 1.0 + 14.0 + 48.0 = 63.0 amu Example

Mole Concept 1 mol = 6.023 X 10 23 = avagadro’s number = N 1 mol eggs = 6.023 X 10 23 eggs 1 mol of molecules = 6.023 X 10 23 molecules Required because atoms and molecules are too small to deal with hands on

Relationship Between Mol Unit and Atoms and Molecules Atomic Substances (most elements) 1 mol Ar = 6.023 X 10 23 atoms Ar Ionic Compounds 1 mol NaCl = 6.023 X 10 23 formula units of NaCl Molecular compound 1 mol H2O = 6.023 X 10 23 molecules H2O

Relationship Between Mols of a Compound and Mols of atoms or Ions Within The Compound 1 mol H2SO4 = 2 mols Hydrogen 1 mol H2SO4 = 1 mol Sulfur 1 mol H2SO4 = 4 mols Oxygen 1 mol CaCl2 = 1 mol Ca +2 ions 1 mol CaCl2 = 2 mols Cl –

Molar Mass The mass in grams of 1 mol of a compound Synonymous with gram molecular mass Units are grams / mol Molar mass = molecular mass in grams Establishes the relationship between grams of a compound and mols For CO2   molecular mass = 44.0 a.m.u                   molar mass = 44.0 grams / mol

Defining the Relationship Between Grams and Mols Using Formula 1 mol of compound = molecular mass in grams For P2O5   1 mol P2O5 = 142.0 grams P2O5 Examples

Relationship Between Mol Units and Basic Particles 1 mol N2O = 6.023 X 10 23 molecules N2O 1 mol N2O = 2 mols Nitrogen atoms = 2 x 6.023 X 10 23 atoms Nitrogen 1 mol N2O = 1 mol Oxygen = 1 x 6.023 X 10 23 atoms Oxygen Example

T-24 Procedure For Mass-Molecules Interconversions

Mass Percent Composition Mass Percent X element = (mass of X element in compound / mass of compound) x 100

Determining Mass percent Composition From A Formula Mass % element Y = (subscript of Y in formula x atomic mass of Y / Formula Mass) x 100 Formula Mass = Molecular Mass Example

Determining the Mass of an Element In A Sample From Compound sample and Mass Percent Given Mass of compound sample x mass percent of element / 100 = mass of element in sample Example

T-25 Procedure For Calculating Empirical Formulas

Determining Empirical Formula With Mass Percent Data Assume 100 grams sample Convert grams of each element to mols using atomic mass of elements Divide each mol figure by the smallest mol figure in step 2 to get simplest ratio If ratios are not whole numbers or nearly whole numbers (within a tenth of the next whole number) multiply each by simplest whole number factor that results in whole number mols for step 3 Resulting simplest whole number ratio become subscripts Example

Relationship Between Molecular Formula and Simple Formula Molecular Mass / Empirical Mass = n (AxBy)n = AnxBny = Molecular Formula

Determining Molecular Formula From Empirical Formula and Molecular Mass Determine the simple mass based on given simple formula Divide the given molecular mass by the simple mass from step 1 to get whole number factor Multiply each subscript in the given simple formula by the factor from step 2 to get subscripts for molecular formula Example

T-3.6 Combustion Method For Determining Mass Percent

Experimental Determination of Mass Percent From Sample Mass and Mass of Combustion Products Convert mass of CO2 combustion product to mass of Carbon Divide mass of carbon from step 1 by the given sample mass and then multiply by 100 to get mass % Carbon in compound Convert mass of H2O combustion product to mass of Hydrogen Divide the mass of Hydrogen from step 3 by the given sample mass and then multiply by 100 to get mass % Hydrogen in compound Example

Determination of Simple(Empirical) Formula From the Sample mass and Mass of Each Element In Sample (Binary Compounds) Divide the mass of each element given by the atomic mass of element to get mols each element Divide each mol figure by the smallest mol figure in step 1 to get simplest ratio If ratios are not whole numbers or nearly whole numbers (within a tenth of the next whole number) multiply each by simplest whole number factor that results in whole number mols for step 2 Resulting simplest whole number ratio become subscripts Example

Relationship Between Stoicheometric Coefficients And Mols For 2H2   +   O2  --à  2H2O 2 mols H2 = 1 mol O2 2 mols H2 = 2 mols H2O 1 mol O2 = 2 mols H2O

Determining Mols of One Component From Mols Of A Given Component In A Balanced Equation Mols of Given --à Mols of Requested Mols of Given   x  coeff of requested / coeff of given = mols requested Example

T-26 Procedure For Relating Quantities In Chemical products

Determining The Mass of One Component From The Mass Of Another Component                                             1                                      2 Grams of given--àmols of given-àmols of                   3 requested--àgrams of requested Grams of given  x 1 mol given / molecular mass in grams = mols of given Mols of given   x   coeff of requested / coeff of given = mols of requested Mols requested  x molecular mass of requested in grams / 1 mol requested = grams of requested Example

Limiting Reagent And Theoretical Yield Limiting Reagent- Reagent that is used up first thereby stopping the reaction and limiting the product formation Reagent that produces the least amount of product will be the limiting reagent

T-3.16 Limiting Reagent Analogy Using Sandwiches

Determining Limiting Reagent and Theoretical Yield Given the grams or mols of two reagents Convert mass of each reagent to mols(step 1 not necessary if given mols of reagent) Convert mols of reactant 1 to mols of product using coefficients Convert mols of reagent 2 to mols of product using coefficients If mols of product from reagent 1 < mols of product from reagent 2 Then reagent 1 is the limiting reagent and mols of product from reagent 1 is theoretical yield Example

Theoretical Yield vs Actual Yield Actual Yield-The amount of product that is actually isolated in the lab Theoretical Yield-The maximum amount of product that can be ideally produced under perfect conditions

Percent Yield % Yield = (Actual Yield / Theoretical Yield) x 100 Why not 100 % Yield? Side reactions taking place beyond the experimenter’s control that distracts from product formation Experimenter’s error in procedure Example