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Stoichiometry

Stoichiometry. Dr. Ron Rusay. Chemical Stoichiometry. Stoichiometry is the study of mass in chemical reactions. It deals with both reactants and products.

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Stoichiometry

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  1. Stoichiometry Dr. Ron Rusay

  2. Chemical Stoichiometry • Stoichiometry is the study of mass in chemical reactions. It deals with both reactants and products. • It quantitatively and empirically relates the behavior of atoms and molecules in a balanced chemical equation to observable chemical change and measurable mass effects. • It accounts for mass and the conservation of mass, just as the conservation of atoms in a balanced chemical equation.

  3. Chemical ReactionsAtoms, Mass & Balance: eg. Zn(s) + S(s)

  4. Stoichiometry _ C2H5OH + _ O2_ CO2 + _ H2O Reactants Products • C=2; H =5+1=6; O=2+1 C=1; H=2; O=2+1 • 1 C2H5OH + 3 O22 CO2 + 3 H2O Must begin with a correctly balanced equation:

  5. https://phet.colorado.edu/sims/html/balancing-chemical-equations/latest/balancing-chemical-equations_en.htmlhttps://phet.colorado.edu/sims/html/balancing-chemical-equations/latest/balancing-chemical-equations_en.html

  6. Chemical Equation The balanced equation can be completely stated as: • 1 mole of ethanolreacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water. C2H5OH +3 O22 CO2+3 H2O

  7. Chemical Equation All Balanced Equations relate on a molar mass basis. For example the combustion of octane: 2 C8H18(l)+ 25 O2(g)16 CO2(g) +18 H2O(l) • moles of octane react with 25 moles of oxygento produce 16 moles of carbon dioxide and 18 moles of water.

  8. The Chemical Equation: Mole & Masses C2H5OH +3 O22 CO2+3 H2O • 46g (1 mole) of ethanolreacts with 3 moles of oxygen (96g) to produce 2 moles of carbon dioxide and 3 moles of water. • How many grams of carbon dioxide and water are respectively produced from 46g (1 mole) of ethanol? 2 mol x 44 g/mol = 88g 3 mol x 18 g/mol = 54 g

  9. The Chemical Equation: Moles & Masses • C2H5OH +3 O22 CO2+3 H2O • How many grams of oxygen are needed to react with 15.3g of ethanol in a 12oz. beer? 15.3gethanol x molethanol /46.0gethanol = 0.333molethanol 0.333molethanol x 3moloxygen / 1molethanol = 0.999moloxygen 32.0goxygen /moloxygen x 0.999moloxygen = 32.0goxygen NOTE: It takes approximately 1 hour for the biologically equivalent amount of oxygen available from cytochrome p450 to consume the alcohol in a human in 1 beer to a level below the legal limit of 0.08%.

  10. Chemical Stoichiometry • Epsom salt (magnesium sulfate heptahydrate) is one of five possible hydrates: mono-, di-, tri-, hexa-, or hepta- hydrate. • How can stoichiometry be used to determine, which hydrate is present in a pure unknown sample, by heating the sample in a kitchen oven at 400 o C for 45 minutes? MgSO4. x H2O(s) MgSO4(s) + x H2O(g) Refer to % Hydrate Lab Experiment for answer.

  11. Mass Calculations All Balanced Equations relate on a molar and mass basis. For the combustion of octane: 2 C8H18(l)+ 25 O2(g)16 CO2(g) +18 H2O(l) 228 g of octane (2 moles)* will react with 800 g of oxygen (25 moles) to produce (16 moles) 704 g of carbon dioxide and (18 moles) 324 g of water. *(2 moles octane x 114 g/mol = 228 g )

  12. Mass Calculations: Reactants Products

  13. Mass Calculations: Reactants Products • 1. Balance the chemical equation. • 2. Convert mass of reactant or product • to moles. • 3. Identify mole ratios in balanced equation: • They serve as the “Gatekeeper”. • 4. Calculate moles of desired product or • reactant. • 5. Convert moles to grams.

  14. Mass Calculations: Reactants Products

  15. QUESTION • The fuel in small portable lighters is butane (C4H10). After using a lighter for a few minutes, 1.0 gram of fuel was used. How many moles of carbon dioxide would it produce? • 58 moles • 0.017 moles • 1.7  10–24 moles • 0.068 moles

  16. ANSWER • The fuel in small portable lighters is butane (C4H10). After using a lighter for a few minutes, 1.0 gram of fuel was used. How many moles of carbon dioxide would it produce? • 58 moles • 0.017 moles • 1.7  10–24 moles • 0.068 moles The molar mass of butane: 4  12g/mol = 48 for carbon + 10  1g/mol = 10.0 for hydrogen. Total = 48.0 + 10.0 = 58.0 g/mol. Next; 1.0 gram of butane  1 mol/58.0 g = 0.017 mol

  17. Mass Calculations: Reactants Products • How many grams of salicylic acid are theoretically needed to produce 1.80 kg of aspirin? • Balanced Equation: 1 1 1 1

  18. Mass Calculations: Product Reactant • How many grams of salicylic acid are theoretically needed to produce 1.80 kg of aspirin? • Balanced Equation: 1 1 1

  19. Mass Calculations: Product Reactant grams (Aspirin) grams (Salicylic Acid) Avogadro's Number Atoms C7H6O3MW = 138.12 Molecules Stoichiometry grams (SA) (Molecular 1 mol (A) 1800 grams (A) 1 mol SA Weight SA) ? (SA) 1 mol A grams (A) 1 mol (SA) (Molecular Weight A) "Gatekeeper" C9H8O4 MW = 180.15

  20. Mass Calculations: How many grams of salicylic acid are needed to produce 1.80 kg of aspirin? • Balanced Equation: 1 1 1 1 ?g C7 = 1.80 x 103 gC9x [molC9 /180.15gC9] x [1 molC7/ 1 molC9] x138.12 gC7/molC7 ?g C7 = 1380 grams

  21. Mass Calculations: Product Reactant grams (Aspirin) grams (Salicylic Acid) ?g C7 = 1.80 x 103 gC9x [molC9 /180.15gC9] x [1 molC7/ 1 molC9] x138.12 gC7/molC7 138.12grams (SA) (Molecular 1 mol (A) 1800 grams (A) 1 mol SA Weight SA) ? (SA) 1 mol A 180.15grams (A) 1380 grams 1 mol (SA) (Molecular "Gatekeeper" Weight A)

  22. Mass Calculations: Reactants Products

  23. QUESTION The fuel in small portable lighters is butane (C4H10). After using a lighter for a few minutes, 1.0 gram (0.017 moles)of fuel was used. How many grams of carbon dioxide would it produce? How many grams of carbon dioxide would this produce? A.) 750 mg B.) 6.0 g C) 1.5 g D.) 3.0 g

  24. ANSWER D.) 3.0 g 1.0 gC4H10 ?gCO2 mol C4H10 44 gCO2 8 mol CO2 C4H10 3.0 g CO2 ?g CO2 1.0 g 2 mol C4H10 molCO2 58 gC4H10

  25. Percent Yield • In synthesis as in any experimentation, it is very difficult and at most times impossible to be perfect. Therefore the actual yield (g) is measured and compared to the theoretical calculated yield (g). This is the percent yield: • % Yield = actual (g) / theoretical (g) x 100 Some DVC students may report percent yields greater than 100% in their first synthesis experiment. Hmmm?....... Why is this not possible? How would Walter White do?

  26. Theoretical (Yield) Mass Calculations Reactant Product grams (Reactant) grams (Product) Moles Molar Mass grams (P) 1 mol (R) grams (R) ? mol (P) ? grams (P) ? mol (R) grams (R) 1 mol (P) (Divide) Mass (R) (Multiply) Mass (P) by Molar "Gatekeepers” from Balanced reaction by Molar

  27. QUESTION • A synthetic reaction produced 2.45g of Ibogaine, C20H26N2O, a natural product with strong promise in treating heroin addiction (at least in Europe), the calculated theoretical yield was 3.05g, what is the % yield? A) 19.7% B) 39.4% C) 80.3% D) 160.6%

  28. ANSWER • If a reaction produced 2.45g of Ibogaine, C20H26N2O, a natural product with strong promise in treating heroin addiction, and the theoretical yield was 3.05g, what is the % yield? • 19.7% B) 39.4% C) 80.3% D) 160.6% % yield = 2.45g / 3.05g x 100 = 80.3%

  29. Combustion Analysis (Chem 120 Prep) • CxHy+(x+y/4)O2x CO2+y/2 H2O • Molecules with oxygen in their formula are more difficult to solve for Oz knowing only the respective masses of CxHyOz sample, CO2 and H2O. • CxHyOz+(x + y/4 - z) O2xCO2+y/2 H2O SEE: COMPARISON to wt % CALCULATIONS

  30. Combustion Analysis CalculationAscorbic Acid ( Vitamin C ) • Combustion of a 6.49 mg sample in excess oxygen, 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

  31. Vitamin C: Calculation(continued) • 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

  32. QUESTION Erythrose contains carbon, hydrogen and oxygen (MM = 120.0 g/mol). It is an important sugar that is used in many chemical syntheses. Combustion analysis of a 700.0 mg sample yielded 1.027 g CO2 and 0.4194 g H2O. Mass Spectrometry produced a molecular ion @ 120 mass units (m/z). What is the molecular formula of erythrose? • CH2O • C6H12O6 • C3H6O3 • C4H8O4

  33. ANSWER D) C4H8O4 Alternative Solution Method Mass ratio of C in CO2 = = = = 0.2729 g C / 1 g CO2 Mass ratio 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 ratio 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

  34. Alternative Method 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 / cmpd = C4H8O4 0.1119 g H 1 g H2O

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