CHM1045 Lecture 8 Outline of Last LectureI. Atomic Mass1. Example 1II. Avogrado’s Number and the Molar Mass1. Example 22. Example 33. Example 4III. Molecular Mass1. Example 52. Example 63. Example 7Outline of Current LectureI. Percent Composition of Compounds1. Example 12. Example 2II. Percent composition & Empirical Formulas1. Example 32. Example 43. Example 54. Example 6Current Lecture:Percent Composition of an element in a compound= n x molar mass of element Molar mass of compoundN= # of moles f the element in 1 mole of the compoundExample 1 C2H6 O % C = 2 x (12.01 g) X 100% = 52.14 % 46.07 g %H= 6 x (1.008g) X 100 % = 13.13 % 46.07% O= 1 x (16.0g) X 100 % = 34.73 % 46.07 52.14% + 13.13% + 34.73% = 100.00 %Example 2 Phosphoric acid ( H3PO4) is a colorless, syrupy liquid used in detergents, fertilizers, toothpastes, and in carbonated beverages for a “tangy” flavor. Calculate the percent composition be mass H,P, and O in this compound.H3PO4(3X1.008)+30.97+(4x16.00)= 97.99g/mol% H= (3 x 1.008) X 100% = 3.09%97.99% P= 30.97 X 100 %= 31.6 % 97.99%O= (4x16) X 100%= 65.3% 97.993.09%+31.6%+65.3%= 99.99%Percent composition and Empirical FormulasMass % (Convert to grams & ÷ by molar mass)Moles of each element (÷ by smallest # of moles)Mole ratios by elements (change to integer subscripts)Empirical Formula Example 3: Ascorbic acid (Vitamin C) cures scurvy. It is composed of 40.92 percent carbon (C). 4.58 percent hydrogen (H), and 54.50 percent oxygen (O) by mass. Determine its empirical formula. Assume 100g of ascorbic acid Step 1: C 40.92 g X 1 mol C = 3.41 Mol C12.01g CH 4.58 g X 1 mol H = 4.54 Mol H 1.008 g H O 54.54 g X 1mol O = 3.41 Mol O16.00 gOStep 2:C 3.41÷ 3.41 = 1H 4.54÷ 3.41 = 1.33 C1H1.33O1 O 3.41÷ 3.41= 1We need to convert 1.33, the subscript for H, into an integer. This can be done by trial and error.1.33 × 1 = 1.331.33 × 2 = 2.661.33 × 3= 3.99 ≈ 4Because 1.33 × 3 gives us an integer (4) we multiply all the subscripts by 3 and you’ll get C3H4O3Example 4: Chalcopyrite (CuFeS2) is a principal mineral of copper. Calculate the number of kilogramsof Cu in 3.71 × 103 kg of chalcopyrite.CuFeS2Cu 63.55 g/molFe 55.85 g/mol +S2 (2 × 32.07 g/mol) 183.54 g/mol %Cu = 63.55 X 100% = 34.62% Cu 183.54(3.71 × 103 kg) × ( 0.3462) = 1.28 × 103 kg CuExample 5: Empirical Formula of ethanol? Combust 11.5 g Ethanol. Collect 22g CO2 & 13.5g H2OStep 1:22.0 g CO2 × 1mol CO2 × 1 molC × 12.01gC = 6g C 44.01 g CO2 1 mol CO2 1 mole C13.5 g H2O × 1 mol H20 × 2 mol H × 1.008 H = 1.51g H 18.02g H20 1 mol H2O 1 mol HStep 2: 11.5 g Ethanol − 6.09 g C1.51g H3.99 OxygenStep 3: 6.00 g C × 1 mol = .5 mol C 12.01 g C1.51 g H × 1 mol H = 1.50 mol H 1.008 g H3.99 g O × 1 mol O = .25 Mol O 16.0 g OStep 4: Divide by the smallest numberC. .5 =2 H. 1.5 = 6 O. .25 =1 .25 .25 .25C2H6OExample 6:A sample of a compound contains 30.46 percent nitrogen and 69.54 percent oxygen by mass, as determined by a mass spectrometer. In a separate experiment, the molar mass of the compound is found to be between 90g and 95g. Determine the molecular formulaand the accurate molar mass of the compound.Assume 100g 30.46% = 30.46 g N69.54 % = 69.54 g O30.46 g N × 1 mol N = 2.18 Mol N 14.00 g N69.54 g O × 1 mol O = 4.35 Mol O 16.0 g ON2.18 O 4.35N: 2.18 = 1 O. 4.35 = 2 NO2 Empirical Formula 2.18 2.18 (14.00 + 2(16)) = 46.00 g/mol ( molar mass ) 90 ≈ 2 95 ≈ 2(empirical molar mass) 46 412(NO2) = N2O4 Molecular
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