CHEM 105 1nd Edition Lecture 5Outline of Last Lecture • Limiting Reactants and Percent yield • Calculations Involving Limiting Reactants • Actural Yields versus Yields Outline of Current Lecture • Theoretical Yield• Empirical Formula• Formula unit• Molecular FormulaCurrent Lecture• Stoichiometric calculations tell us the maximum amount of product that can be made from the available reactants. This is called the theoretical yield.Q: 15.00g of octane and 60.00g of oxygen react to form 34.875g of carbon dioxide. Whatis the % yield of this reaction?A: 2C8H18 (L) +25O2 (g) -----> 16CO2 (g)+ 18H2O (g)C8H18 : 8x12.01 g/mol +18x1.008 g/mol= 114.224 g/molO2 : 2x16.00 g/mol= 32.00 g/molCO2: 12.01 g/mol +16.00x2= 44.01 g/mol15.00 g C8H18 x1mol C8H18/114.224g x16mol CO2/2 mol CO2 x44.01g/ 1 mol CO2 = 46.2355g CO260.00g O2 x 1mol O2/32.00g x16mol CO2/25mol O2 x44.01g/1mol Co2= 52.812g CO2Octane is the limiting reactant. The theoretical Yield is 46.2355g 34.875g/46.2355g x100.00% =75.43% • On Page 56 you will look into the following: Empirical Formula: shows the smallest whole number ratios of the elements in a compound.Formula unit: The formula for ionic compounds are usually presented as empirical formula.The Molecular Formula: will be a whole number multiple of the empirical formula.• (page 100)Percent Composition: A self-explanatory concept.Q: What is the percent composition of chalcopyrite (CuFeS2), a cp[[er ora? A: 1 mole CuFeS2 has a mass of 183.54g % Cu : 63.55g/183.54g x100.00% = 34.62%% Fe: 55.85g/183.54g x100.00% =30.43%% S2 : 2x32.07g/183.54 x100.00% =34.95% • Empirical formula from % mass data:Q: A compound is 50.0% carbon, 5.6% H, and 44.4% oxygen by mass. What is the empirical formula of this compound?A: 100.00g of this compound contain 50.0%g of C, 5.6%g of H, and 44.4%g of O50.0g of C x 1mol/12.01g = 4.163197 mol of C5.6g H x 1mol/1.008g = 5.5656 mol of H44.4g O x 1mol/16.00g = 2.775 mol of O4.163197 mol of C/2.775 mol of O= 1.50625 C/O5.55656 mol of H/ 2.775 mol of O= 2.00200 H/O2.775 mole of O/2.2775 mol of O= 1Q: An compound is a colorless gas at room temperature. It is 69.55% oxygen by mass and 30.45% nitrogen by mass. Its molecular mass has been determined to be 92.02g/mol. Calculate the empirical and molecular formulae of this compound.A: 69.55g of O x1mol/ 16.00g =4.346875 mol O30.45g of N x1mol/14.01g =2.1734475 mol N4.346875 mol O/2.1734475 mol N =1.99999076 O/NEmpirical formula: NO2Empirical molsr mass: 14.01g/mol+ 2x16.00g/mol92.02g/mol /46.01g/mol =2 empirical formula in the molecular formulaMolecular Formul: N2O4• Combustion analysis:Q: 0.956g of a compound containing only C, H, and O burns completely in oxygen producing 2.692g of carbon dioxide and 1.102g water.What is the empirical formula of this compound?A: CxHyOz + nO2 -----> xCO2 + y/2 H2O2.69g CO2 x 1mol CO2/44.01g x 1mol C/1mol CO2 =0.06116792 mol C1.102g H2O x1mol/18.016g x2mol H/1mol H2O =0.1223357 mol H0.06116792 mol x 12.01g/1mol +0.1223357 mol x1.008g/1mol =0.858g C+H0.956g - 0.858g =0.098g
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