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Chapter Three: Stoichiometry: Calculations with Chemical Formulas and Equations

Chapter Three: Stoichiometry: Calculations with Chemical Formulas and Equations. Overview. Chemical Equations Patterns/Reactions Atomic/Molecular Weights Moles/Molar Mass Empirical/Molecular Formulas Quantitative Relationships Limiting Reactants/Theoretical Yields. Chemical Equations.

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Chapter Three: Stoichiometry: Calculations with Chemical Formulas and Equations

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  1. Chapter Three: Stoichiometry: Calculations with Chemical Formulas and Equations

  2. Overview • Chemical Equations • Patterns/Reactions • Atomic/Molecular Weights • Moles/Molar Mass • Empirical/Molecular Formulas • Quantitative Relationships • Limiting Reactants/Theoretical Yields

  3. Chemical Equations • chemical ‘sentences’ • reactants and products described by formulas or symbols combined with “punctuation” reactant formulas product formula ‘react to form’ 2 H2(g) + O2(g) 2 H2O(l) coefficients physical state

  4. “atoms can be neither created nor destroyed” • all equations must be ‘balanced’ with the same number of atoms on both sides of the reaction arrow unbalanced H2O + O2Ù H2O2 2 H & 3 O ¹ 2 H & 2 O balanced 2H2O + O2Ù 2H2O2 4 H & 4 O = 4 H & 4 O

  5. H2O O2 H2O2 unbalanced balanced one formula unit two formula units

  6. Examples 2 3 2 4 • CH3OH(l) + O2(g)Ù CO2(g) + H2O(l) • Na(s) + H2O(l) Ù NaOH(aq) + H2(g) • HBr(aq)+ Ba(OH)2(aq)Ù H2O(l) + BaBr2(aq) 2 2 2 2 2

  7. specific 2K(s) + 2H2O(l)Ù 2KOH(aq) + H2(g) Patterns of Chemical Reactivity • Because elements are grouped by chemical properties, their reactions can also be grouped: • alkali metals and water general 2M(s) + 2H2O(l)Ù 2MOH(aq) + H2(g)

  8. specific C3H8(g) + O2(g)Ù CO2(g) + H2O(l) 5 3 4 • Combustion in air general CxHy + O2(g)Ù CO2(g) + H2O(l) hydrocarbon

  9. specific 2Mg(s) + O2(g) Ù 2MgO(s) • Combination Reactions general X + Y Ù XY

  10. specific CaCO3(s)Ù CaO(s) + CO2(g) • Decomposition Reactions general XY Ù X + Y

  11. Name the Reaction decomposition • PbCO3(s)Ù PbO(s) + CO2(g) • C(s) + O2(g)Ù CO2(g) combination • 2NaN3(s)Ù 2Na(s) + 3N2(g) decomposition • 2C2H6(g) + 7O2(g)Ù 4CO2(g) + 6H2O(l) combustion

  12. Atomic and Molecular Masses • Amu scale • defined by assigning the mass of 12C as 12 amu exactly • 1 amu = 1.66054 x 10-24 g • 1 g = 6.02214 x 1023 amu • Average Atomic Masses • 12C 98.892% abundant 13C 1.1108% abundant (0.98892)(12 amu) + (0.01108)(13.00335 amu) = 12.011 amu atomic mass

  13. vitamin C C6H8O6 Formula and Molecular Masses • sum of all atomic masses in the formula of an ionic or molecular compound 6 x 12.0 = 72.0 amu 8 x 1.0 = 8.0 amu 6 x 16.0 = 96.0 amu 176.0 amu formula mass of vitamin C (often called molecular mass)

  14. Percentage Composition • Calculate the percent mass that each type of atom contributes to a molecule • % X = (no. X atoms)(X amu) x 100 formula mass cmpd • C6H8O6 % C = (6)(12.01amu) x 100 = 40.94% C 176.0 amu % H = (8)(1.01amu) x 100 = 4.59% H 176.0 amu % O = (6)(16.00 amu) x 100 = 54.55% O 176.0 amu

  15. The Mole • We can measure masses in amu but how do we relate that to mass in grams? We define a quantity of atoms – a mole – which has the same mass in grams as the mass of the element in amu. • So how many atoms does it take to make, say, 1.00 g of H? 1.0 g H x 1 atom H@6.0 x 1023 atoms of H 1.7 x 10-24 g H a mole 12.0 g C x 1 atom C@6.0 x 1023 atoms of C 2.0 x 10-23 g C

  16. Avogadro’s Number • 6.02214 x 1023 units/mole • No. of atoms per mole of an element • No. of molecules per mole of molecular cmpd. • No. of formula units per mole of ionic cmpd. • No. of cows per mole of cows Memorize this number & what it means!

  17. 1 C atom = 12 amu 1 mole C atoms = 12 g • 1 Mg atom = 24 amu 1 mole Mg atoms = 24g • 1 CO molecule = 28 amu 1 mole CO molecules = 28 g • 1 NaCl fm. unit = 58 amu 1 mole NaCl fm.units = 58g

  18. Molar Mass • From this information we can define something called the molar mass (MM) of an atom (or molecule or formula): from the equality: 1 mole C = 12.0 g C we define the molar mass of a substance12.0 g C = MM or Molar Mass1 mole C (Atomic Mass) (Molecular Mass) (Formula Mass) conversion factor

  19. Problems • Practice Ex. 3.9: • How many mole in 508 g of NaHCO3? Given: MM = 84.02 g/mol NaHCO3 508 g NaHCO3 508 g NaHCO3 x 1 mole = 6.05 mole NaHCO3 84.02 g NaHCO3 • How many formula units of NaHCO3? Given: 6.02 x 1023 form. units/mole NaHCO3 6.05 mole NaHCO3 x 6.02 x 1023 fm. units = 3.6 x 1024 fm. 1 mole units NaHCO3

  20. Molar Mass converts between moles and grams of a substance • Avogadro’s number converts between moles of a substance and atoms (or molecules or formula units) of that substance These are very important conversion factors, know & understand them!

  21. Problems • How many moles of vitamin C are contained in 5.00 g of vitamin C? C6H8O6 176.0 g/mol • 17.5 mg of cocaine (C17H21NO4) per kg of body weight is a lethal dose. How many moles is that? How many molecules? • In 25 g of C12H30O2 THC (tetrahydrocannibinol) how many moles are there? How many molecules are there? How many C atoms are there?

  22. How many moles of O are contained in 1.50 moles of C6H5NO3? • How many grams of nitrogen are contained in 70.0 g of C6H5NO3? How many atoms?

  23. Calculate the number of H atoms in 50.0 mg of acetominophen, C8H9O2N.

  24. Determination Empirical Formulas • simplest ratio of atoms • change g of each element to moles or • assume 100 grams of substance & change the % of each element to moles • change the mole ratio of atoms to the simplest ratio by dividing by the smallest number of moles

  25. Practice Ex. 3.12: • 5.325 g methyl benzoate contains 3.758 g C, 0.316 g H, 1.251 g O. Determine empirical formula. 3.758 g C x 1 mole = 0.313 mol C 12.01 g C0.313H0.313O0.0782 0.316 g H x 1 mole = 0.313 mol H 1.01 g C4H4O 1.251 g O x 1 mole = 0.0782 mol O 16.00 g

  26. Determination of Molecular Formulas • actual ratio of atoms • determine the empirical formula • divide the actual molar mass by the empirical formula mass to get ‘n’ • multiply the mole ratio in the empirical formula by ‘n’

  27. Practice Ex. 3.13: • Ethylene glycol is composed of 38.7% C, 9.7% H & 51.6% O by mass. Its true molar mass is 62.1 g/mol. What are the empirical and molecular formulas? 38.7 g C x 1 mole = 3.23 mole C 12.0 g C3.23H9.60O3.22 9.7 g H x 1 mole = 9.60 mole H 1.01 g empirical formula CH3O n = 2 51.6 g O x 1 mole = 3.22 mole O 16.0 g molecular formulaC2H6O2

  28. Formulas from Combustion Data • Formulas determined from products of combustion products • Menthol is composed of C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g H2O. What is the empirical formula? CxHyOz + O2ÙCO2 + H2O0.1005 g 0.2829 g 0.1159 g • Calculate moles CO2 & C; moles H2O & H

  29. 0.2829 g CO2x 1 mol x 1 mol C = 0.00643 mol C 44.0 g 1 mol CO2 0.1159 g H2Ox 1 mol x 2 mol H = 0.0129 mol H 18.0 g 1 mol H2O total mass of C + H = 0.0902 g mass of O = 0.1005 g - 0.0902 g = 0.0103 g O x 1 mol = 16.0 g total mass of all C, H & O 6.44 x 10-4 mol O

  30. 0.00643 mol C 0.0129 mol H 6.44 x 10-4 mol O C0.00643H0.0129O0.000644 C10H20O (empirical formula mass 156 g/mol) If the MM is 156 g/mol, what is the molecular formula? n=1 therefore molecular formula is C10H20O

  31. Quantitative Stoichiometry • Determination of quantities from balanced chemical reaction equations • mole ratios from balanced chemical equation convert between species • if quantities are given for more than one reactant, the limiting reactant must be determined

  32. Given the following balanced equation:1Mg(OH)2 + 2HCl Ù1MgCl2 + 2H2O • Calculate the number of moles of HCl required to react completely with 0.42 mol of Mg(OH)2 0.42 mol Mg(OH)2 x 2 mol HCl = 0.84 mol HCl 1 mol Mg(OH)2 The mole ratio comes from the balanced chemical equation

  33. How many grams of MgCl2 can be produced?0.42 mol Mg(OH)2 x 1 mol MgCl2 x 95.3 g MgCl2 1 mol Mg(OH)2 1 mol Theoretical Yield -- maximum amount that can be produced = 40.0 g MgCl2

  34. moles of reactant grams of reactant grams of product moles of product General Sequence of Conversion: MM reactant mole ratio MM product

  35. Practice Ex. 3.14: • How many grams of O2 can be prepared from 4.50 g of KClO3?2KClO3Ù 2KCl + 3O2 4.50 g KClO3 x 1 mol x 3 mol O2 x 32.0 g O2 = 1.76 g O2 122.6 g 2 mol KClO3 1 mol

  36. Limiting Reactant • given a non-stoichiometric amount of both reactants, you will have to determine which is the limiting reagentor reactant • example:you have 10 bicycle frames and 16 bicycle wheels and you need to put them together to produce as many bicycles as possible, how many bicycles can be produced, what is the limiting “reagent”, and how much excess “reagent” do you have left over?

  37. Balanced ‘Equation’1 (mole) frame + 2 (moles) wheelsÙ1 (mole) bicycles[10 (moles) frames][16 (moles) wheels] [8(moles) bicycles] • Limiting Reactant -- will produce the least amount of product 10 mol frames x 1 mol bicycles = 10 bicycles 1 mol frames limiting reactant 16 mol wheels x 1 mol bicycles = 8 bicycles 2 mol wheels

  38. Practice Ex. 3.16: • A mixture of 1.5 mol of Al and 3.0 mol of Cl2 react. What is limiting & how many moles of AlCl3 are formed? 2Al(s) + 3Cl2(g)Ù2AlCl3(s)1.5 mol 3.0 mol 1.5 mol 1.5 mol Al x 2 mol AlCl3 = 1.5 mol AlCl3 2 mol Al 3.0 mol Cl2 x 2 mol AlCl3 = 2.0 mol AlCl3 3 mol Cl2

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