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AP CHEMISTRY CHAPTER 3 STOICHIOMETRY

AP CHEMISTRY CHAPTER 3 STOICHIOMETRY. Average atomic mass - weighted average based on isotopic composition To calculate : % Isotope A (mass of A) + % Isotope B (mass of B) + …= avg. atomic mass.

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AP CHEMISTRY CHAPTER 3 STOICHIOMETRY

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  1. AP CHEMISTRYCHAPTER 3STOICHIOMETRY

  2. Average atomic mass- weighted average based on isotopic compositionTo calculate : % Isotope A (mass of A) + % Isotope B (mass of B) + …= avg. atomic mass

  3. What is the average atomic mass in grams of lithium if 7.42% exists as 6Li (6.015 g/mol) and 92.58% exists as 7Li (7.016 g/mol)? (0.0742)(6.015) + (0.9258)(7.016) = ? 6.94 g/mol Don’t forget to change percents to decimals! Don’t divide by the number of different isotopes! Don’t round until the end. Your answer must be between the masses of the individual isotopes. Check!

  4. Mass Spectrometer A mass spectrometer is an instrument used to determine the relative masses of atoms by the deflection of their ions in a magnetic field. One use is to determine the isotopic abundance of a sample of an element. Samples are vaporized and ionized. The ions are then separated by mass and the data is graphed.

  5. Figure 3.2Neon Gas These show the Mass spec data for samples of neon. Using this data, what would the approximate average atomic mass of neon be?

  6. Figure 3.3MassSpectrum of Natural Copper

  7. What element results in this mass spec data?

  8. Mole -the # of C atoms in 12g of pure carbon-12Avogadro’s Number = 6.022 × 1023 The mass of one mole of an element is equal to its atomic mass in grams.

  9. Cody found a gold nugget that had a mass of 1.250 oz. How many moles was this? How many atoms? (1 lb = 16 oz, 453.59g = 1 lb) 1.250 oz Au 1 lb 453.59g 1 mol Au 16 oz 1 lb 196.97g Au = 0.1799 moles 0.1799 moles Au 6.022 × 1023 atoms 1 mol Au = 1.083 × 1023 atoms

  10. Molar Mass = mass in grams of one mole of a substance

  11. Calculate the molecular mass of cisplantin, Pt(NH3)2Cl2. Pt 1 × 195.08 = 195.08 N 2 × 14.01 = 28.02 H 6 × 1.01 = 6.06 Cl 2 × 35.45 = 70.90 300.06 g

  12. How many grams are 3.25 moles of cisplantin? 3.25 mol cisplantin 300.06g 1 mol cisplantin = 975g

  13. Percent Composition -“mass percent” or “percent by mass”Total mass of element× 100 = % comp of element Total mass of compound

  14. Find the percent composition of all elements in cisplantin, Pt(NH3)2Cl2. Pt 195.08 × 100 = 65.01% 300.06 N 28.02 × 100 = 9.34% 300.06 H 6.06 × 100 = 2.02% 300.06 Cl 70.90 × 100 = 23.63% 300.06

  15. Determining the Formula of a CompoundEmpirical formula- simplest whole number ratio of the various types of atoms in a compoundMolecular formula- the exact formula (empirical formula)x

  16. A sample of a compound contains 11.66g of iron and 5.01g of oxygen. What is the empirical formula of this compound? 11.66g Fe 1 mol Fe = 0.2088 mol Fe 55.85g Fe 5.01g O 1 mol O = 0.3131 mol O 16.00g O Fe:O 0.2088 : 0.3131 1:1.5 or 2:3 0.2088 0.2088 Fe2O3

  17. What is the empirical formula of hydrazine, which contains 87.5% N and 12.5%H? 87.5gN 1 mol N = 6.246 mol N 14.01g N 12.5g H 1 mol H = 12.38 mol H 1.01 g H N:H = 6.246 : 12.38 1 : 1.98 = 1 : 2 6.246 6.246 NH2

  18. Combustion Analysis Additional O2 is added to burn the sample of a compound containing C, H, and sometimes O. All of the H2O is absorbed in the first chamber and all of the CO2 produced is absorbed in the second chamber. Increases in the masses of the two chambers are used to determine the mass of CO2 and H2O produced.

  19. Suppose you isolate an acid from clover leaves and know that it contains only the elements C, H, and O. Heating 0.513g of the acid in oxygen produces 0.501g of CO2 and 0.103g of H2O. What is the empirical formula of the acid? Given that another experiment has shown that the molar mass of the acid is 90.04g/mol, what is its molecular formula? C + O2  CO2 0.501g CO2 1 mol CO2 1 mol C 12.01gC 44.01g CO2 1 mol CO2 1 mol C = 0.1367 g C in compound

  20. 2H2 + O2 2H2O 0.103g H2O 1 mol H2O 2 mol H 1.01g H 18.02g H2O 1 mol H2O 1 mol H = 0.01155 g H 0.513g cmpd  (0.137 + 0.012)= 0.364g O 0.364g O 1 mol O = 0.02275 mol O 16.00g O 0.137g C 1 mol C = 0.01141 mol C 12.01g C 0.01155g H 1 mol H = 0.01144 mol H 1.01 g H

  21. C:H:O 0.01141: 0.01144 : 0.02275 0.01141: 0.01144 : 0.02275 0.01141 0.01141 0.01141 1: 1: 1.98 1:1:2 CHO2 efm = 12.01 + 1.01 + 32.00 = 45.02 90.04/45.02 = 2 C2H2O4

  22. Chemical EquationsReactants  Products Yields A balanced equation must have the same # of atoms of each element on each side.Symbols representing physical states: (s) (l) (g) (aq)

  23. Balancing Chemical EquationsWe balance equations by adding coefficients, never by changing formulas.Most equations can be balanced by inspection. Some redox reactions require a different method.

  24. Ex. Al(s) + Cl2 (g)  AlCl3(s) 2Al(s) + 3 Cl2 (g) 2AlCl3(s) N2O5(s) + H2O(l)  HNO3(l) N2O5(s) + H2O(l)  2 HNO3(l) Pb(NO3)2(aq) + NaCl(aq)  PbCl2(s) + NaNO3(aq) Pb(NO3)2(aq) + 2 NaCl(aq)  PbCl2(s) + 2 NaNO3(aq)

  25. Phosphine, PH3(g) is combusted in air to form gaseous water and solid diphosphorus pentoxide. 2PH3(g) + 4O2(g)  3H2O(g) + P2O5(s)

  26. When ammonia gas is passed over hot liquid sodium metal, hydrogen is released and sodium amide, NaNH2, is formed as a solid product. 2NH3(g) + 2Na(l)  H2(g) + 2NaNH2(s)

  27. Stoichiometric CalculationsMole ratio = moles required/moles given

  28. What mass of NH3 is formed when 5.38g of Li3N reacts with water according to the equation: Li3N(s) + 3H2O  3LiOH(s) + NH3(g)? 5.38g Li3N 1 mol Li3N 1 mol NH3 17.04g NH3 34.83g Li3N 1 mol Li3N 1 mol NH3 = 2.63g NH3

  29. Limiting reagent- reagent that limits or determines the amount of product that can be formedAnimationIf you are given amounts of two or more reactants in a stoichiometry problem and asked to determine how much product forms, the easiest thing to do is to work a problem with each reactant and take the smaller of the answers.

  30. You work in a sandwich shop. You MUST make sandwiches according to the following formula: You use two pieces of bread, 4 pieces of meat, 2 pieces of cheese, 3 pieces of bacon and 2 ounces of special sauce. If you have 20 pieces of bread, 32 pieces of meat, 20 pieces of cheese, 21 pieces of bacon and one quart of special sauce, how many sandwiches can you make?

  31. X + 2Y  XY2 X Y XY2 Before reaction After reaction Assume that the reaction goes to completion. Draw the resulting particles in the right-hand box. What is the limiting reagent?

  32. X + 2Y  XY2 X Y XY2 Before reaction After reaction Assume that the reaction goes to completion. Draw the resulting particles in the right-hand box. What is the limiting reagent?

  33. How many moles of Fe(OH)3(s) can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O and 3.0 mol O2 to react? 2Fe2S3(s) + 6H2O(l) + 3O2(g)  4Fe(OH)3(s) +6S(s) 1.0 mol 2.0 mol 3.0 mol limiting 2.0 mol H2O 4 mol Fe(OH)3 6 mol H2O = 1.3 mol Fe(OH)3

  34. If 17.0g of NH3(g) were reacted with 32.0g of oxygen in the following reaction, how many grams of NO(g) would be formed?4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l) 17.0g 32.0g Xg 17.0g NH3 1 mol NH3 4 mol NO 30.01 g NO = 29.9g NO 17.04g NH3 4 mol NH3 1 mol NO 32.0 g O2 1 mol O2 4 mol NO 30.01 g NO = 24.0g NO 32.00g O2 5 mol O2 1 mol NO 24.0g is the smaller yield, so O2 is limiting and 24.0 g is the correct answer.

  35. Theoretical yield- amount of product that should form according to stoichiometric calculationsActual yield- experimental yieldPercent yield = actual yield× 100 Theoretical yield

  36. In the reaction of 1.00 mol of CH4 with an excess of Cl2, 83.5g of CCl4 is obtained. What is the theoretical yield, actual yield and % yield? Actual yield = 83.5g CCl4 CH4 + 2Cl2 CCl4 + 2H2 1.00 mol CH4 1 mol CCl4 153.81g CCl4 1 mol CH4 1 mol CCl4 = 154g CCl4 (theoretical yield) 83.5× 100 = 54.2% yield 154

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