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Hints for Exam 3--covering Chapter 8 1. 10 points Avogadro’s number calculation

Hints for Exam 3--covering Chapter 8 1. 10 points Avogadro’s number calculation 2. 15 points Determine empirical formula 3. 35 points Limiting Reagent (ding ding ding) 4. 40 points Determine mass percent and empirical formula from combustion data.

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Hints for Exam 3--covering Chapter 8 1. 10 points Avogadro’s number calculation

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  1. Hints for Exam 3--covering Chapter 8 1. 10 points Avogadro’s number calculation 2. 15 points Determine empirical formula 3. 35 points Limiting Reagent (ding ding ding) 4. 40 points Determine mass percent and empirical formula from combustion data.

  2. Two Compounds with Molecular Formula C2H6O Property Ethanol Dimethyl Ether M (g/mol) 46.07 46.07 Color Colorless Colorless Melting point - 117oC - 138.5oC Boiling point 78.5oC - 25oC Density (at 20oC) 0.789 g/mL 0.00195 g/mL Use Intoxicant in In refrigeration alcoholic beverages H H H H H C C O H H C O C H H H H H

  3. Ascorbic Acid ( Vitamin C ) - I Contains C , H , and O • Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO2 and 2.64 mg H2O • Calculate it’s Empirical formula! • C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2) = 2.65 x 10-3 g C • H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O) = 2.92 x 10-4 g H • Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg = 3.54 mg O

  4. Vitamin C Combustion - II • C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) = = 2.21 x 10-4 mol C • H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) = = 2.92 x 10-4 mol H • O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) = = 2.21 x 10-4 mol O • Divide each by 2.21 x 10-4 • C = 1.00 Multiply each by 3 = 3.00 = 3.0 • H = 1.32 = 3.96 = 4.0 • O = 1.00= 3.00 = 3.0 C3H4O3

  5. When 66.6 g of O2 gas is mixed with 27.8 g of NH3 gas and 25.1 g of CH4 gas, 36.4 g of HCN gas is produced by the following reaction: 2CH4 + 2NH3 + 3O2 2HCN + 6H2O Given : 17.03 g NH3/mol, 32.00 g O2/mol, 27.03 g HCN/mol, 16.04 g CH4/mol What is the percent yield of HCN in this reaction?

  6. Theoretical Yield: Which Reactant is Limiting? 1) calculate moles (or mass) of product formed by complete reaction of each reactant. 2) the reactant that yields the least product is the limiting reactant. 3) the theoretical yield for a reaction is the maximum amount of product that could be generated by complete consumption of the limiting reactant.

  7. 2CH4 + 2NH3 + 3O2 2HCN + 6H2O 66.6 g of O2 = 2.08 mol O2 27.8 g of NH3 = 1.63 mol NH3 25.1 g of CH4 = 1.56 mol CH4 Which reactant is limiting? 2.08 mol O2 can yield 1.39 mol (or 37.5 g) HCN 1.63 mol NH3 can yield 1.63 mol (or 44.1 g) HCN 1.56 mol CH4 can yield 1.56 mol (or 42.2 g) HCN O2 is the limiting reactant.

  8. O2 is the limiting reagent; thus, the theoretical yield is based on 100% consumption of O2. 2.08 mol O2 can yield 1.39 mol (or 37.5 g) HCN % yield = actual yield * 100 theoretical yield % yield = 36.4 g HCN = 97.1% * 100 37.5 g HCN

  9. When 66.6 g of O2 gas is mixed with 27.8 g of NH3 gas and 25.1 g of CH4 gas, 36.4 g of HCN gas is produced by the following reaction: 2CH4 + 2NH3 + 3O2 2HCN + 6H2O Given : 17.03 g NH3/mol, 32.00 g O2/mol, 27.03 g HCN/mol, 16.04 g CH4/mol What is the percent yield of HCN in this reaction? How many grams of NH3 remain?

  10. Determining a Chemical Formula from Combustion Analysis - I Problem: Erthrose (M = 120 g/mol) is an important chemical compound as a starting material in chemical synthesis, and contains Carbon Hydrogen, and Oxygen. Combustion analysis of a 700.0 mg sample yielded 1.027 g CO2 and 0.4194 g H2O. Plan: We find the masses of Hydrogen and Carbon using the mass fractions of H in H2O, and C in CO2. The mass of Carbon and Hydrogen are subtracted from the sample mass to get the mass of Oxygen. We then calculate moles, and construct the empirical formula, and from the given molar mass we can calculate the molecular formula.

  11. Determining a Chemical Formula from Combustion Analysis - II Calculating the mass fractions of the elements: Mass fraction of C in CO2 = = = = 0.2729 g C / 1 g CO2 Mass fraction of H in H2O = = = = 0.1119 g H / 1 g H2O Calculating masses of C and H Mass of Element = mass of compound x mass fraction of element mol C xM of C mass of 1 mol CO2 1 mol C x 12.01 g C/ 1 mol C 44.01 g CO2 mol H xM of H mass of 1 mol H2O 2 mol H x 1.008 g H / 1 mol H 18.02 g H2O

  12. Determining a Chemical Formula from Combustion Analysis - III 0.2729 g C 1 g CO2 Mass (g) of C = 1.027 g CO2 x = 0.2803 g C Mass (g) of H = 0.4194 g H2O x = 0.04693 g H Calculating the mass of O: Mass (g) of O = Sample mass -( mass of C + mass of H ) = 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g O Calculating moles of each element: C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O C0.02334H0.04656O0.02330 = CH2O formula weight = 30 g / formula 120 g /mol / 30 g / formula = 4 formula units / cpd = C4H8O4 0.1119 g H 1 g H2O

  13. Some Compounds with Empirical Formula CH2O (Composition by Mass 40.0% C, 6.71% H, 53.3%O) Molecular M Formula (g/mol) Name Use or Function CH2O 30.03 Formaldehyde Disinfectant; Biological preservative C2H4O2 60.05 Acetic acid Acetate polymers; vinegar ( 5% solution) C3H6O3 90.08 Lactic acid Causes milk to sour; forms in muscle during exercise C4H8O4 120.10 Erythrose Forms during sugar metabolism C5H10O5 150.13 Ribose Component of many nucleic acids and vitamin B2 C6H12O6 180.16 Glucose Major nutrient for energy in cells

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