CHEM 101 1st Edition Lecture 11 Outline of Last Lecture I. StoichiometryII. Limiting ReactantsOutline of Current Lecture I. Quantitative Chemical AnalysisII. CombustionIII. Analysis for Hydrocarbons by CombustionIV. Volumetric Chemical AnalysisCurrent Lecture- Quantitative Chemical Analysiso A material of unknown composition can be converted to one (or more) substance(s) of known compositiono The amounts of this/these substances(s) can then be determined and related to the amount of the original, unknown substanceo You have 2 reactants which unknown masses. They produce a product with a precipitate. You can filter the solution from the precipitate and weigh the precipitate to determine the mass of the precipitate and help determine the original masses as well.o Ex: A 0.123 g sample of the mineral thenardite contains sodium sulfate along with several inert compounds. The sodium sulfate in the sample is converted to insoluble barium sulfate by adding aqueous barium chloride in excess to the dissolved thenardite sample. The mass of recovered barium sulfate is 0.177g. What is the mass percent of sodium sulfate in the mineral?Na2SO4 (Thenardite)(aq) + BaCl2(aq) BaSO4(aq) + 2NaCl(aq) l-----------Stoichiometric Ration 1:1-----------lStep 1: mass(g)BaSO4 mols BaSO40.177(g) BaSO4 x 1 mol BaSO 4 = .000758 233.4G BaSo2 Step 2: mols BaSo4 mols Na2So4.000758 x 1 mol Na 2 SO 4 = .000758 These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.1 mol BaSo4Step 3: mols Na2SO4 mass(g) Na2SO4.000758 x 142.0g Na2So4 = 0.108g Na2So4 1 mol Na2So4Step 4: mass(g) Na2SO4 mass percent Na2So4 mass Na 2 So 4 x 100 = 0.108 g x 100 = 87.8% Na2So4 (by mass)mass Sample Thenardite 0.123 g- Combustion involves the addition oxygen to another elemento When hydrocarbon molecules burn completely, the products are carbon dioxide gas and watero The heat from the furnace converts the carbon in the compound to CO2 and the hydrogen into H2O. The CO2 and H2O are collected in pre-weighed traps.o …o Ex: 1.516 g of a compound containing carbon, hydrogen and oxygen (CxHyOz) is subjected to combustion analysis. The results show that 2.082g of CO2 and 1.705g of H2O were produced. What is the empirical formula for the compound?Data information: mass of products Step 1: Calculate moles of CO2 and H2OAmounts of Co2 and H2O in molesStep 2: Convert moles of CO2 and H2O into moles of C and H of the sampleAmount of C and H in the sample in molesStep 3: Convert moles of C and H to mass in gramsMass of C and H in the sample in gramsStep 4: Subtract the mass of C and H from the sample massMass of O in the sampleStep 5: Convert mass of O to moles of OAmounts of C, H and O in moles = empirical formulaStep 1: 2.082g CO2 x 1 mol CO 2 = 0.04731 mol CO2 44.01g CO2 1Step 2: 0.04731 mol CO2 x 1 mol C = 0.04731 mol C 1 mol CO2 0.09462 mol H2O x 2 mol H = 0.1892 mol H 1 mol H2OStep 3: 0.04731 mol C x 12.01 g C = 0.5682 g C 1 mol C.1892 mol H x 1.0079 g H = 0.1907 g H 1 mol HStep 4: 1.516 g – 0.5682 g – 0.1907 g = 0.7571g (which is the mass of O)Step 5: 0.7571g O x 1 mol O = 0.04732 mol O 16.00 g OEmpirical formula: 0.04731 mol C x= 1.000 0.1892 mol H = CXHYOZ y= 3.999 = CH4O 0.04732 mol O z= 1.000- Analysis for Hydrocarbons by Combustiono Why can’t we determine the oxygen from CO2 and H2O directly?CXHYOZ + O2(g) COZ (g) + H2O(l)o The oxygen in the CO2 and H2O originates from both the sample and the O2 in the furnace!o Empirical Formula Molecular formulan x CXHYOZ = CnXHnYOnZ (n= 1,2,3,4…)o Since the molecular formula is a multiple of the empirical formula, scaled by a factor “n”, the molecular and empirical molar masses must also scale by the same ratio:molar formula mass (g/mol) = nempirical formula mass (g/mol)o Ex: The empirical formula of the compound is CH4O. The molecular mass of this compound, 160.2g/mol, was independently determined. What is the molecular formula for the compound?[12.01 + 4 x (1.160.2 g/mol = 5.00032.04 g/molThe molecular formula is 5 x CH4O so the answer is C5H20O5- Volumetric Chemical Analysiso Most chemical studies require quantitative measurementso Experiments involving aqueous solutions are measured in volumes of solutions rather than masses of solids, liquids, or gases.o Solution concentration, expressed as molarity relates the volume of solution in liters to the amount of substance in moleso Solution = solute + solvent Solute- component which is dissolved (lesser amount) Solvent- component which dissolves (greater amount)o Volumetric flask is better than a graduated cylinder fro volumetric chemical analysiso Molarity = moles of solute per liter of solutionMolarity of X (cX) = moles of solute (units: mol/L) L of solutionCNaCL = (NaCl) = 1.00 1 mol NaCl = 1.00 M 1L MolarityMolesLiterso If molarity and volume are known, moles can be determined:molarity x volume = molesmols x L = moles moles x 1 = volume mol x L = L L M
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