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Stoichiometry

Stoichiometry. Chapter 3. Chemical Stoichiometry. Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions. Schematic diagram of a mass spectrometer. Average Atomic Mass. Elements occur in nature as mixtures of isotopes

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Stoichiometry

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  1. Stoichiometry • Chapter 3

  2. Chemical Stoichiometry • Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

  3. Schematic diagram of a mass spectrometer.

  4. Average Atomic Mass • Elements occur in nature as mixtures of isotopes • Carbon = 98.89% 12C • 1.11% 13C • <0.01% 14C • (0.9889)(12.0000 amu) = 11.87 amu • (0.0111)(13.0034 amu) = 0.144 amu • 12.01 amu

  5. The Mole • The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. • 1 mole of anything = 6.022  1023units of that thing Equal moles of substances have equal numbers of atoms, molecules, ions, formula units, etc.

  6. Counting Atoms • Atoms are too small to be seen or counted individually. • Atoms can only be counted by weighing them. • all jelly beans are not identical. • jelly beans have an average mass of 5 g. • How could 1000 jelly beans be counted?

  7. Jelly Beans & Mints • Mints have an average mass of 15 g. • How would you count out 1000 mints? • Why do 1000 mints have a mass greater than 1000 jelly beans?

  8. Atomic Mass Unit • Atoms are so tiny that the gram is much too large to be practical. • The mass of a single carbon atom is 1.99 • x 10-23 g. • The atomic mass unit (amu) is used for atoms and molecules.

  9. AMU’s and Grams • 1 amu = 1.661 x 10 -24 g • Conversion Factors • 1.661 x 10-24g/amu • 6.022 x 1023amu/g

  10. The Mole • One mole of rice grains is more than the number of grains of all rice grown since the beginning of time! • A mole of marshmallows would cover the U.S. to a depth of 600 miles! • A mole of hockey pucks would be equal in mass to the moon.

  11. The Mole • Substance Average Atomic Mass # Moles# Atoms • (g) • Na22.9916.022 x 1023 • Cu63.55 16.022 x 1023 • S32.061 6.022 x 1023 • Al26.98 16.022 x 1023

  12. Measurements • dozen = 12 • gross = 12 dozen = 144 • ream = 500 • mole = 6.022 x 1023

  13. Avogadro’s number equals6.022  1023 units

  14. Unit Cancellation • How many dozen eggs would 36 eggs be? • (36 eggs)(1 dozen eggs/12 eggs) = 3 dozen eggs • How many eggs in 5 dozen? • (5 dozen eggs)(12 eggs/1 dozen eggs) = 60 eggs

  15. Calculating Mass Using AMU’S • 1 N atom = 14.01 amu • (23 N atoms)(14.01 amu/1N atom) = 322.2 amu

  16. Calculating Number of Atoms from Mass • 1 O atom = 16.00 amu • (288 amu)(1 O atom/16.00 amu) = 18 atoms O

  17. AMU’s & Grams • 1 atom C = 12.011 amu = 1.99 x 10-23g • 1 mol C = 12.011 g • Use TI-83 or TI-83 Plus to store 6.022 x 1023 to A.

  18. Calculating Moles & Number of Atoms • 1 mol Co = 58.93 g • (5.00 x 1020 atoms Co)(1mol/6.022 x 1023 atoms) • = 8.30 x 10-4 mol Co • (8.30 x 10-4 mol)(58.93g/1 mol) = 0.0489 g Co • Moles are the doorway • grams <---> moles <---> atoms

  19. Molar Mass • A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. • CO2 = 44.01 grams per mole

  20. Calculating Mass from Moles • CaCO3 • 1 Ca = 1 (40.08 g) = 40.08 g • 1 C = 1 (12.01 g) = 12.01 g • 3 O = 3 (16.00 g) = 48.00 g • 100.09 g • (4.86 molCaCO3)(100.09 g/1 mol) = 486 g CaCO3

  21. Calculating Moles from Mass • Juglone • 10 C = 10(12.01g) = 120.1 g • 6 H = 6(1.008 g) = 6.048 g • 3 O = 3(16.00 g) = 48.00 g • 174.1 g • (1.56 g juglone)(1 mol/174.1 g) = 0.00896mol juglone

  22. Percent Composition • Mass percent of an element: • For iron in iron (III) oxide, (Fe2O3)

  23. % Composition • CuSO4. 5 H2O • 1 Cu = 1 (63.55 g) = 63.55 g • 1 S = 1 (32.06 g) = 32.06 g • 4 O = 4 (16.00 g) = 64.00 g • 5 H2O = 5 (18.02 g) = 90.10 g • 249.71 g

  24. % Composition (Continued) • % Cu = 63.55 g/249.71g (100 %) = 25.45 % Cu • % S = 32.06 g/249.71 g (100 %) = 12.84 % S • % O = 64.00 g/249.71 g (100 %) = 25.63 % O • % H2O = 90.10 g/249.71 g (100 %) = 36.08 % H2O Check: Total percentages. Should be equal to 100 % plus or minus 0.01 %.

  25. Formulas • Ionic compounds -- empirical formula • NaCl • CaCl2 • Covalent compounds -- molecular formula - • C6H12O6 • C2H6

  26. Formula of a Compound Calculations • Example for Problems 75-78 on page 125. • A compound contains only carbon, hydrogen, and nitrogen. When 0.1156 g of it is reacted with oxygen, 0.1638 g of CO2 and 0.1676 g of water are collected. • Molar mass of CO2 is 44.01 g • Molar mass of HOH is 18.02 g

  27. Formula of a Compound Calculations • (0.1638g CO2)(12.01g C/44.01g CO2) = 0.04470g C • %C = (0.04470g/0.1156g)(100%) = 38.67% C • (0.1676 g HOH)(2.016 g H/18.02g HOH) = 0.01875g H • %H = (0.01875g/0.1156g)(100%) = 16.22% H • %H + %C + %N = 100% • %N = 100% -38.67% -16.22% = 45.11% N

  28. Formula of a Compound Calculations • (38.67g C)(1 mol/12.01g) = 3.220 mol C = 1 C • 3.219 mol • (16.22g H)(1 mol/1.008g) = 16.09 mol H = 5 H • 3.219 mol • (45.11g N)(1 mol/14.01g) = 3.219 mol N = 1 N • 3.219 mol • Empirical Formula is CH5N.

  29. Formulas • molecular formula = (empirical formula)n [n = integer] • molecular formula = C6H6 = (CH)6 • empirical formula = CH

  30. Empirical Formula Determination • 1. Base calculation on 100 grams of compound. • 2. Determine moles of each element in 100 grams of compound. • 3. Divide each value of moles by the smallest of the values. • 4. Multiply each number by an integer to obtain all whole numbers.

  31. Calculating Empirical Formulas • 4.151 g Al & 3.692 g O • (4.151 g Al)(1 mol/26.98 g) = 0.1539 mol Al/0.1539 mol = 1.000 • (3.692 g O)(1 mol/16.00 g) = 0.2308 mol O/0.1539 mol = 1.500 • 1 Al (2) = 2 Al • 1.5 O (2) = 3 O • Al2O3

  32. Molecular Formulas • 71.65 % Cl, 24.27 % C, & 4.07 % H • (71.65g Cl)(1 mol/35.45g) = 2.021 mol/2.021 mol = 1 • (24.27 g C)(1 mol/12.01g) = 2.021 mol/2.021 mol = 1 • (4.07 g H )(1 mol/1.01g) = 4.03 mol/2.021 mol = 2 • (EM)x = (MM) • (49.46)x = (98.96) • x = 2 (EF)x = (MF) (ClCH2)2 = Cl2C2H4

  33. Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances.

  34. Chemical Equation • A representation of a chemical reaction: • C2H5OH + 3O22CO2 + 3H2O • reactants products

  35. Physical States • solid (s) • liquid (l) • gas (g) • aqueous (aq)

  36. Important Equation Symbols •  • heat -------> • light • light --------> • elect. • electricity ------> • yields ------> • cat. • catalyst -------> • H2SO4 • catalyst ------>

  37. Chemical EquationsQuantitative Significance • 4 Al(s) + 3 O2(g) ---> 2 Al2O3(s) • This equation means • 4 Al atoms + 3 O2 molecules ---give---> • 2 molecules of Al2O3 • 4 moles of Al + 3 moles of O2 ---give---> • 2 moles of Al2O3

  38. Balancing Equation Prerequisites • Student must have memorized: • 44 chemical symbols • Table 2.3 on page 61 in text • Table 2.4 on page 62 • Table 2.5 on page 66 • Table 2.6 on page 67 • Type I, II, III, and acid nomenclature • Count HOFBrINCl

  39. Four Steps in Balancing Equations • 1. Get the facts down. • 2. Check for diatomic molecules (subscripts). • 3. Balance charges on compounds containing a metal, ammonium compounds, and acids (subscripts). • 4. Balance the number of atoms (coefficients). • a. Balance most complicated molecule first. • b. Balance other elements. • c. Balance hydrogen next to last. • d. Balance oxygen last.

  40. Balancing Equations Caution The identities (formulas) of the compounds must never be changed in balancing a chemical equation! Only coefficients can be used to balance the equation-subscripts will not change!

  41. Chemical Equation • C2H5OH +3O22CO2+3H2O • The equation is balanced. • 1 mole of ethanol reacts with 3 moles of oxygen • to produce • 2 moles of carbon dioxide and 3 moles of water

  42. Balancing Equations • When solid ammonium dichromate decomposes, it produce solid chromium(III) oxide, nitrogen gas, and water vapor. •  • (NH4)2Cr2O7(s) ----> Cr2O3(s) + N2(g) + HOH(g) •  • (NH4)2Cr2O7(s) ----> Cr2O3(s) + N2(g) + 4HOH(g)

  43. Calculating Masses of Reactants and Products • 1. Balance the equation. • 2. Convert mass to moles. • 3. Set up mole ratios. • 4. Use mole ratios to calculate moles of desired substituent. • 5. Convert moles to grams, if necessary.

  44. Gram to Mole & Gram to Gram • __Al(s) + __I2(s) ---> __AlI3(s) • 2Al(s) + 3I2(s) ---> 2AlI3(s) • How many moles and how many grams of aluminum iodide can be produce from 35.0 g of aluminum?

  45. Gram to Mole & Gram to Gram • 2Al(s) + 3I2(s) ---> 2AlI3(s) • (35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol Al) = 1.30 mol AlI3 • (35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol Al)(407.68 g/1 mol) = 529 g AlI3

  46. Limiting Reactant The limiting reactant is the reactant that is consumed first,limiting the amounts of products formed.

  47. Solving a Stoichiometry Problem • 1. Balance the equation. • 2. Convert masses to moles. • 3. Determine which reactant is limiting. • 4. Use moles of limiting reactant and mole ratios to find moles of desired product. • 5. Convert from moles to grams.

  48. Limiting Reactant Problem • If 56.0 g of Li reacts with 56.0 g of N2, how many grams of Li3N can be produced? • __Li(s) + __N2(g) ---> __Li3N(s) • 6 Li(s) + N2(g) ---> 2 Li3N(s) • (56.0 g Li) (1 mol/6.94g)(1 mol N2/6 mol Li) (28.0 g/1 mol) = 37.7 g N2 • Since there were 56.0 g of N2 and only 37.7 g used, N2 is the excess and Li is the Limiting Reactant.

  49. Limiting Reactant Problem • 6 Li(s) + N2(g) ---> 2 Li3N(s) • (56.0 g Li)(1 mol/6.94g)(2 mol LiN3/6 mol Li) (34.8 g/1 mol) = 93.6 g Li3N • How many grams of nitrogen are left? • 56.0g N2 given - 37.7 g used = 18.3 g excessN2

  50. % Yield • Values calculated using stoichiometry are always theoretical yields! • Values determined experimentally in the laboratory are actual yields!

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