John A. SchreifelsChem 211*Chapter 3Calculations involving Chemical Formulae and Equations

John A. SchreifelsChem 211*ContentsMass and Moles of a SubstanceMolecular weightMolesDetermining Chemical FormulaeMass % from formulaElemental Analysis: gives % C, H, ODetermining formula from elemental analysisStoichiometryAmounts of substances in a reactionLimiting reagents

John A. SchreifelsChem 211*CHEMICAL EQUATIONSChemical reaction indicates the reactants and products. Reactants on left; Products on right 2H2(g) + O2(g) 2H2O(l); 4Fe(s) + 3O2(g) 2Fe2O3(s)NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq) Total number of each atom should be the same on each side of the reaction. Law of Conservation of mass should guide us.Steps to Balancing an Equation (simple):Write an unbalanced equation with all reactants and products:Balance by changing co-efficientsReduce coefficientsE.g. the combustion of methanol produces water and carbon dioxide:CH3OH(l) + O2(g) CO2(g) + H2O(l)E.g. Balance: P4 + N2O(g) P4O6(s) + N2(g) P2O5 + H2O H3PO4 Quiz

John A. SchreifelsChem 211*Atomic and Molecular WeightsMolecular Weight (Formula weight): Summed masses of all atoms in the formula unit of the compound.How?For the general formula AaBbCcDd... fm= formula mass = aMA + bMB + cMC + dMD + ...E.g. determine the formula mass (formula weight) of the following:NaOH, H2O, CaCO3, C2H6OPercentage composition from formula (mass % contribution of each element to the total mass of the compound). For AaBbCc

E.g. determine the mass % composition of each element in: NaOH, H2O, CaCO3, C2H6O

John A. SchreifelsChem 211*The MoleMole: Number of atoms (or molecules) as there are in 12.00 g Carbon-12; 1 mol = 6.02x1023 and is called Avogadro's number.Mole is used to indicate a certain number of particles (like a dozen or bushel might be used).1 mol of water occupies approximately 18 mL and has 6.02x1023 molecules. gold occupies approximately 10 mL and has 6.02x1023 atoms. Since individual atoms are very small, they weigh very little. It is not convenient deal with masses that small, since we can typically measure gram amounts in the lab.

John A. SchreifelsChem 211*Molar mass1 mol = FM = 6.02x1023 particlesThe definition of a mol allows us to perform a variety of conversions. Let n = # mol, n = # atoms (molecules), m = mass and FM = formula mass then we can convert to: m from n, FME.g. Determine the mass of NaCl needed to have 5.0 mol of it.n from m, FME.g. Determine the number of mol of NaCl in 15.00 g of it.n from nE.g. Determine the number of molecules in 3.222 mol of NaCln from m, FME.g. Determine the number of atoms in 4.32 g of NaClm from n, FME.g. Determine the mass of one formula unit of NaCl.

John A. SchreifelsChem 211*EMPIRICAL FORMULAEmpirical Formula: Simplest formula where all coefficients are integers.In earlier example Fe2O3, Fe4O6 Fe6O9 Fe8O12 possible formulas for iron oxide. Its empirical formula Fe2O3. Often obtained from % composition data.Assume 100 g of compound- Determine mol of each element.Divide by smallest # of moles.Multiply by smallest whole number to make all coefficients whole numbers.E.g. Determine the empirical formula of a compound with the following % compositions:Mass%O = 34.7%Mass%C = 52.1%Mass%H = 13.1%

John A. SchreifelsChem 211*Combustion Analysis: Equipment Organic compound placed in Oven Compound heated in O2 and burned converting it to H2O and CO2. H2O, CO2 collected separately.

John A. SchreifelsChem 211*COMBUSTION ANALYSIS: Data AnalysisExperimental mass % also determined. For organic compounds the sample is combusted. Analyzes the amount of C,H, and O in compounds.C converted to CO2 and mass measured.H converted to H2O and mass measuredE.g.: Combustion analysis of a 1.621 g sample of ethanol, which contains only C,H,O, was performed. Calculate the mass % composition of each element in ethanol.Results: Mass CO2 = 3.095 gMass of H2O = 1.902 g.

John A. SchreifelsChem 211*MOLECULAR FORMULAMeasure molecular mass Multiply empirical mass by integer to obtain molecular mass.Multiply all the co-efficients by this integer.E.g. Determine the molecular mass of a compound with the empirical formula NO2 and a molecular mass of 92.00 g/mol.

John A. SchreifelsChem 211*STOICHIOMETRYThe relative amounts of reactants and products in a reaction are given by the ratio of stoichiometric coefficients. (Conservation of mass).E.g. For the reaction : 2Na(s) + Cl2(g) 2NaCl(s)2 mol Na = 1 mol of Cl2 = 2 mol of NaCl.E.g. 2 Determine # mol of Cl2 needed to react with 4.2 mol of Na. How many mol of NaCl will form?Mol of Cl2

Mol NaCl:

Summary: For aA + bB cC

John A. SchreifelsChem 211*STOICHIOMETRY2

Mass of one reactant can be related to the mass of another reactant or product. (Law of definite proportions).E.g. Determine the mass of Na needed to react with 34.45 g of Cl2 and the maximum mass of NaCl that could be produced.Write reaction and express the moles of one compared to the moles of the other.

Substitute for mol in terms of mass and formula mass in each part.Let n = mol; FM = formula mass then

E.g. 1: Determine the mass of oxygen consumed when it reacts with 10 g CH3CHO.E.g. 2. Calculate the mass of oxygen needed to react with 100 g Al to for Al2O3. Quiz

John A. SchreifelsChem 211*LIMITING REAGENTS

Limiting reagent: the reactant which limits the maximum amount of product that can be produced. It will be completely consumed in the reaction before any other reactants. Limiting reagent must be known before theoretical yield can be determined. E.g. 1 Determine limiting reagent if 3.00 moles of Al react with 2.15 moles of O2 to form Al2O3.Strategy: Determine mol of Al2O3 that could be produced from AlDetermine mol of Al2O3 that could be produced from O2Reactant producing smallest amount of Al2O3 is limiting reagent. E.g.2 Calculate the theoretical yield when 20 g Al react with 25 g O2. Strategy: Determine expected yield from each reactant.Compare. If mass of Al2O3 from produced from Al is less than from O2, then Al is limiting. Otherwise O2 is limiting.

John A. SchreifelsChem 211*YIELDS IN CHEMICAL REACTIONSTheoretical yield: the maximum amount of product that can be obtained from given amounts of reactants.Actual Yield : the actual amount of product obtained from a reaction.Theoretical yield calculated from amount of reactants and stoichiometry Actual yield is the amount of product actually obtained.% yield is the yield as a percent of the theoretical yield:

E.g. The reaction for the synthesis of acetic acid from methanol is shown below. What is the % yield if 15.0 g of methanol were combined with a stoichiometric amount of carbon monoxide to form19.1 g of product?CH3OH(l) + CO(g) HC2H3O2(l)Quiz

John A. SchreifelsChemistry 211-notes*Solution Composition, MolarityMost reactions performed in solution (homogeneous mixture) since reactants mobile.Solute dissolved substance.Solvent substance in which solute is dissolved.Concentration: amount of solute dissolved in a given amount of solvent.Concentrated solution: large amount of solute in solvent.Dilute solution: very little solute in solvent. Often obtained by dilution.Molar concentration ( Molarity, M ): moles of solute dissolved in a liter of solution. E.g. An aqueous solution of 0.25 M NaCl can be prepared by dissolving:0.25 mol NaCl in a 1-Liter flask0.50 mol NaCl in a 2-Liter flask0.125 mol in 1/2 liter flask (500 mL). E.g. 1 Determine mass needed to prepare exactly 2 liters of 0.150M NaCl. E.g. 2 Determine the concentration when 12.5 g NaCl is dissolved and diluted to 500.0 mL.

John A. SchreifelsChemistry 211-notes*Ion Concentrations in SolutionsConcentrations of ions after dissolution depends on formulaDetermine concentration of each ion in the following solutions: 0.100 M CaSO4, 0.100 M Cu(NO3)2, 0.100 M FeCl3

John A. SchreifelsChemistry 211-notes*Mass % To MolarityOften manufacturers provide us with mass % of a compound in solution, but it is more convenient to use molarity.E.g. Determine the molarity of NH3(aq) if the mass% = 28.0 % NH3 and the density = 0.898 g/mL. Assume 100g of solution From mass of solute (28 g NH3) in 100 g determine .From the mass of solution (100 g) and the density, determine V = the volume of the solution.From above steps determine the molar concentration: [NH3] = n(NH3)/V = 1.65 mol NH3/ 0.111L = 14.96 M NH3.

John A. SchreifelsChemistry 211-notes*DILUTIONSDilute solutions are prepared from more concentrated ones by adding solvent to the concentrated one.The concentration of the dilute solution can be determined if we know:The volume of the concentrated solution, Vi.The concentration of the concentrated solution, Mi.The volume of the dilute solution, Vf.The relationship between the molarities and volumes is:

E.g. Determine the volume needed to prepare 500.0 mL of a 0.100 M HCl solution from a 12.40 M stock solution.Quiz

John A. SchreifelsChemistry 211-notes*REACTIONS IN SOLUTIONReactions usually carried out in solution. Amounts of reactants and products (m or n) must be determined from the volume and molarity of the solution. Start with the stoichiometric relationship for any reaction:aA + bB cC

Depending upon what is given in the problem substitute for mol A, B or C. E.g., if we are dealing with a solution we substitute MAVA for the mol A.E.g. Calculate the volume of 0.200 M KI required to react with 50.0 mL of 0.300 M Pb(NO3)2.Strategy:Balance reaction: Pb(NO3)2 + 2KI PbI2 + 2KNO3.From stoichiometry:

Substitute for mol: Solve

John A. SchreifelsChemistry 211-notes*TITRATIONSTitration: a procedure for determining the amount of one substance A by adding a carefully measured amount of a solution B until A is just consumed.Calculations are the same as in the last overhead.E.g. What is the molarity of HCl if 25.00 mL of it was titrated to the equivalence point with 33.33 mL of 0.1000 M Ba(OH)2?

The stoichiometric relationship is:Substitute the given quantities and solve for the [HCl].The ratio of stoichiometric coefficients tells how much of one compound will react if we know the amount of the other:aA + bB cC

Solutions: n = MV; solids: n = m/FM

John A. SchreifelsChem 211*SummaryMoles are used to describe measurable quantities of a substance (1 mol = 6.02x1023 particles).Reactions occur in well defined proportions and are represented by balanced chemical equations.The ratio of stoichiometric coefficients tells how much of one compound will react if we know the amount of the other: aA + bB cC Limiting reagent is one that limits the amount of product. Use it to determine the theoretical yield.Empirical formula: simplest formula; all integers.Molecular formula: actual formula of a compound.

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