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Stoichiometry

Stoichiometry. Mass Relationships in Chemical Processes. Stoichiometry. “ Stoi ” = mass, heaviness, solidity ( stoic ) “ Chio ”= chemistry, chemical, component “ Metry ” = measure. Hence, Stoichiometry = the measurement of chemical masses. Stoichiometry.

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Stoichiometry

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  1. Stoichiometry Mass Relationships in Chemical Processes

  2. Stoichiometry • “Stoi” = mass, heaviness, solidity (stoic) • “Chio”= chemistry, chemical, component • “Metry” = measure Hence, Stoichiometry = the measurement of chemical masses

  3. Stoichiometry • “Stoicheîon” = element, component • “Metry” = measure Hence, Stoichiometry = the measurement of elements or components Oh, well...

  4. Stoichiometry • Equations provide the necessary relationships • Calculations require assumption of certain chemical process(es), identified by the balanced equation • Fundamental quantity of chemical substance in the SI system is the mole.

  5. Mole (chemical context) • Not the blind, long-nosed animal that burrows into the ground, leaving a pile (or mole-hill) of dirt at its entrance • Not an infiltrator into espionage organization • Not a skin blemish • Not the spicy, chocolaty-peanut sauce so good on chicken and enchiladas

  6. Mole • A certain number of “things” • Like “dozen,” except lots more than 12 things • 602 sextillion (approximately) things, or 6.02 x 1023 things • So many things that only atomic- or molecular-sized things exist in mole quantities (a mole of paper clips, for example, has the earth’s total mass)

  7. Mole: Official Definition • The precise number of C-12 atoms that have total mass exactly 12 grams • Must be determined experimentally • Known to several significant figures, but not exactly

  8. Mole: Working Definition • Quantity of substance equal to the substance’s formula weight (molar mass) in grams. • Equivalent to a mass amount • Conversion factors of form: one mole X = certain grams X • one mole Cl = 35.45 g Cl • one mole H2O = 18.0 g H2O • one mole NaCl = 58.5 g NaCl

  9. Mole: Formula Weight • FW = the sumof the atomic weights of all atoms in the formula (as often as they appear) • FW H2O = 18.0 • 2 x 1.0 (H’s) + 16.0 (oxygen) • one mole H2O = 18.0 g H2O • FW NaCl = 58.5 • 23.0 (Na) + 35.5 (Cl) • one mole NaCl = 58.5 g NaCl

  10. Mole: You Do One... Formula weight of CaCl2: • At wt Ca = 40.0 • At wt Cl = 35.5 • FW CaCl2 = 40.0 + (2 x 35.5) = 111.0 • One mole CaCl2 = 111.0 g One mole CaCl2 = 111.0 grams

  11. Mole: You Do Another One... One mole (NH4)2SO4 = 132.1 grams Formula weight of (NH4)2SO4: • At wt N = 14.0 • At wt H = 1.0 • At wt S = 32.1 • At wt O = 16.0 • FW (NH4)2SO4 = (2 x 14.0) + (8 x 1.0) + 32.1 + (4 x 16.0) = 132.1 • One mole (NH4)2SO4 = 132.1 g

  12. 1 mole Mg 24.305 g Mg Moles: Counting Atoms and Molecules 0.514 g Mg= 0.0211 mole(s) 0.514 g Mg x = 0.0211 moles

  13. Moles: Counting Atoms and Molecules Formula weight of H2O: • At wt H = 1.0 • At wt O = 16.0 • FW H2O = (2 x 1.0) + 16.0 = 18.0 • One mole H2O = 18.0 g 75.0 g H2O= mole(s)

  14. 75.0 g H2O x = Moles: Counting Atoms and Molecules 75.0 g H2O= 4.17 mole(s) 4.17 moles

  15. Moles: Counting Atoms and Molecules Formula weight of C4H10: • At wt C = 12.0 • At wt H = 1.0 • FW C4H10 = (4 x 12.0) + (10 x 1.0) = 58.0 • One mole C4H10 = 58.0 g 1.97 g C4H10 = mole(s)

  16. 1.97 g C4H10 x = Moles: Counting Atoms and Molecules 1.97 g C4H10 = 0.0340 mole(s) 0.0340 moles

  17. Moles and Stoichiometry • Chemical equations specify quantities as well as substances • Quantities are indicated in mole amounts in balanced equations

  18. Moles and Stoichiometry Consider the reaction of methane with oxygen which we discussed previously: CH4 + 2 O2 CO2 + 2 H2O

  19. Two molecules of water One molecule of methane One molecule of carbon dioxide Two molecules of oxygen Moles and Stoichiometry CH4 + 2 O2 CO2 + 2 H2O Quantities are specified in terms of (a) individualmolecules or (b) moles of molecules

  20. Two moles of water One mole of methane One mole of carbon dioxide Two moles of oxygen Moles and Stoichiometry CH4 + 2 O2 CO2 + 2 H2O Quantities are specified in terms of (a) individual molecules or (b) moles of molecules

  21. Two moles of water One mole of methane One mole of carbon dioxide Two moles of oxygen Moles and Stoichiometry CH4 + 2 O2 CO2 + 2 H2O The mole scale is 602 sextillion times as big as the molecule scale!

  22. Quantities can be at various scalings: CH4 + 2 O2 CO2 + 2 H2O • 1 molecule + 2 molecules  1 molecule + 2 molecules • 1 mole + 2 moles  1 mole + 2 moles • 5 moles + 10 moles  5 moles + 10 moles • 0.2 mole + 0.4 mole  0.2 mole + 0.4 mole • 4 grams + 16 grams  11 grams + 9 grams (0.25 mole) (0.50 mole) (0.25 mole) (0.50 moles) • Last example represents usual information about reaction quantities.

  23. ? grams ? moles ? moles Practice Example Based on the equation below, how many grams of water should be formed by reaction of 21.0 grams of methane, CH4? CH4 + 2 O2 CO2 + 2 H2O 21.0 grams Use FW of CH4 for conversion factor Use FW of H2O for conversion factor Use equation coefficients for conversion factor Three-step conversion-factor sequence applies in most instances.

  24. ? grams ? moles ? moles 18.0g H2O 2 mole H2O 1 mole CH4 x x x 16.0 g CH4 1 mole H2O 1 mole CH4 Practice Example Based on the equation below, how many grams of water should be formed by reaction of 21.0 grams of methane, CH4? CH4 + 2 O2 CO2 + 2 H2O 21.0 grams Use FW of CH4 for conversion factor Use FW of H2O for conversion factor Use equation coefficients for conversion factor 21.0 g CH4 = 47.3 g H2O

  25. ? grams ? moles CH4 ? moles O2 16.0 g CH4 1 mole CH4 1 mole O2 x x x 32.0 g O2 1 mole CH4 2 mole O2 Practice Example Based on the equation below, how many grams of CH4 should be required to react with 125 grams of oxygen, O2? CH4 + 2 O2 CO2 + 2 H2O 125 grams Use FW of O2 for conversion factor Use FW of CH4 for conversion factor Use equation coefficients for conversion factor 125 gO2 = 31.3 g CH4

  26. ? grams ? moles ? moles 159.7 g Fe2O3 1 mole Fe2O3 1 mole Fe x x x 55.85 g Fe 1 mole Fe2O3 2 mole Fe Practice Example Based on the equation below, how many grams of Fe2O3 would be required to form 5.00 grams of iron, Fe? Fe2O3 + 3 CO  2 Fe + 3 CO2 5.00 grams Use FW of Fe for conversion factor Use FW of Fe2O3 for conversion factor Use equation coefficients for conversion factor = 7.15 g Fe2O3 5.00 g Fe

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