Chapter 3 - Calculations with Chemical Formulas

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Chapter 3 - Calculations with Chemical Formulas
Eqns

• Online HW 3 due Friday, 4/27/2012
• Equation Practice HW Available until 4/30
• Online Stoichiometry Lab due Weds, 4/25
• - Exam 2 will cover chapters 3 4

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End-of-ChapterSuggested Problems, Chapter 3, pp
114-123

• 3, 5, 6, 7, 14, 15, 16, 25, 26, 27,
• 33, 35, 37, 43, 47, 51, 57d, 61,
• 65, 67, 71, 73, 75, 79, 81, 83, 89,
• 91, 95, 97

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I. Mole A. Introduction

• Chemists devised a new term called a mole or
  mol.
• A mole has two definitions
• Definition 1 6.02x1023 number of that object
• 1.00 mole 6.02x1023
• Definition 2 The formula weight (FW) in grams
• 1.00 mole FW in g (from periodic chart)
• The above two definitions work because 6.02x1023
  protons or neutrons weights 1.00 gram.
• Following is true for chemicals 6.02x1023
  FW in g

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I. Mole A. Introduction
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I. Mole A. Introduction

• Mole Conversion Factors to know
• A) 1.00 mole Formula Weight in Grams (from
  Periodic Chart)
• B) 1.00 mole 6.02x1023
• A few examples of conversion factors from H2O
• 1.00 mole H2O 18.0 g H2O 1.00 mole H2O
• 18.0 g H2O 1.00 mole H2O
  6.02x1023 molecules
• 1.00 mole H2O 1.00 mole H2O
• 12x1023 H atoms 18x1023 total atoms

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I. Mole B. Calculations

• 1. How many grams are in 1.0 mole of H2SO4.
• 2x1 32 4x16 98 g H2SO4
• 2. How many grams are in 1.00 mole of Ca(NO2)2.
  
• 40 2x14 4x16 132 g Ca(NO2)2
• 3. How many moles are in 0.36 g Be.
• 0.36g Be x 1.0 mole Be 0.040 mole Be
• 9.0g Be
• 4. How many moles are in 36 g of H2O.
• 36g H2O x 1.0 mole H2O 2.0 moles H2O
• 18 g H2O

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I. Mole B. Calculations continued

• 5. How many g present in 1.5x10-3 moles CaCO3?
• 1.5x10-3 moles CaCO3 x 100. g CaCO3
  0.15 g CaCO3
• 1.00 mole CaCO3
• 6. How many molecules are in 0.20 moles H20
• 0.20 mol H2O x 6x1023 molecule
  1.2x1023 molecules
• 1 mole H2O
• 7. How many atoms are in 0.20 moles H2O.
• 0.20 mol H2O x 6x1023 molecule x 3
  atoms 3.6x1023 atom
• 1 mole H2O 1 molecule H2O

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I. Mole B. Calculations continued

• 8. How many moles (m) of O are in 2.0 moles of
  H3PO4?
• 2.0 m H3PO4 x 4 m O 8.0 mole of O
• 1 m H3PO4
• 9. How many grams of O atoms are in 0.60 moles
  of H3PO4 ?
• 0.60 m H3PO4 x 4 m O x 16 g O
  38 g O
• 1 m H3PO4 1 m O

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I. Mole C. Mass Composition

• Note Part x 100. - Know
• Total
• composition g element x
  100. .
• total g compound
• Example 1 Calculate composition of C in C4H10
• - Pick mass of 1 mole of compound for the total.
  Why?
• - Total g (1 mole) of C4H10 (4 x 12 g C) (10
  x 1 g H) 58 g
• - g Carbon 4 x 12 g C 48 g C
• C 48 g C x 100 83 C
• 58 g C4H10

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I. Mole C. Mass Composition

• Example 2 A compound contains 53.3 S. How
  many g of S are in 71 g of the compound?
• (53.3 S 53.3 g S / 100. g compd
  conversion factor)
• 71 g compd x 53.3 g S
  38 g S
• 100. g compd
• Example 3 56 mg (0.056g) of a hydrocarbon gave
  0.082 g of CO2 upon burning. Calculate w/w C.
• CxHy O2 -------) CO2 H2O
  (unbalanced)
• g C / g compound x 100 C
• 0.082g CO2 x 1m CO2 x 1m C x
  12g C 0.022g C
• 44g CO2 1m CO2
  1 m C
• 0.022 g C / 0.056 g compd x 100
  39 C

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II. Empirical Formulas A. Definition

• - The empirical formula (EF) is the simplest
  whole number ratio of atoms (or mols) in the
  unit.
• Examples Unit EF
• H2O2 HO
• C6H6 CH
• N2O4 NO2
• P4S10 P2S5
• - Obtain the EF from experimental or gram (g)
  data.
• Procedure
• g mols
  simplest whole ratio

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II. Empirical Formulas B. Calculation

• Example Calculation from grams or from w/w
• A compound was found to be 50.0 S and 50.0
  O. Calculate EF. Formula SxOy Pick 100 g
  of compound, then get mols x and y.
• 50.0 g S x (1 mol S / 32 g S) 1.56 mol S
• 50.0 g O x (1 mol O / 16 g O) 3.12 mol O
• Convert to simplest whole ratio by dividing
  both by the smaller number of mols
• S1.56 O3.12 EF S1.56/1.56O3.12/1.56
  S1O2 or SO2
• - Note if get something like S1.00O1.33 then
  what do you do?

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II. Empirical Formulas C. Molecular Weight

• - If one has the empirical formula and the
  molecular weight (MW), then one can calculate the
  molecular formula (MF).
• MW / EW Then multiply EF by to get the
  MF.
• 1) The MW of a compound is 284 g and the EF is
  P2O5 (142 g). Calculate the MF.
• 284g / 142 g 2.00 MF 2 x P2O5
  P4O10
• 2) The MW 78 g and the EF C1H1. Obtain the
  MF.
• 78 g / 13 g 6.0 MF 6 x CH C6H6

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III. Stoichiometry A. Introduction

• Mole Conversion Factors from Equations
• - The balancing coefficients in a chemical
  equation can be assigned units of moles used as
  conversion factors.
• - Example 2 H2 1 O2 ---) 2 H2O
• (2 mol of H2 will react with 1 mol of O2 to
  give 2 mol of H2O)
• a) How many moles of H2 will react with 5.5 mol
  of O2?
• 5.5 m O2 x 2 m H2 11 moles of H2
• 1 m O2
• b) How many moles of H2 are needed to produce 13
  mol H2O?
• 13 mol H2O x 2 m H2 13 mol H2
• 2 m H2O

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III. Stoichiometry B. Methodology

• - The coefficients in a balanced chemical
  reaction can be used as conversion factors in
  calculating the amount of product produced or the
  amount of reactant used.
• - The problem solving scheme is
• a b
  c
• g known -----) moles known -----)
  moles unknown -----) g unknown
• - Conversion factors for a c are from Periodic
  Chart (1 mole FW in grams) conversion factor
  for step b is from the balanced chemical equation.

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III. Stoichiometry B. Regular Problems

• - In regular problems, one is given the amount
  of ONE reagent assumes there is enough of the
  other reagents for complete reaction.
• Examples Given 2 Al(OH)3 3 H2SO4
  -----) 6 H2O 1 Al2(SO4)3
• 1. How many moles of H2SO4 are needed to produce
  8.0 moles of Al2(SO4)3
• 8.0 moles Al2(SO4)3 x 3.0 moles H2SO4
  24 moles H2SO4
• 1.0 mole Al2(SO4)3
• 2. How many moles of H2O will be produced from
  156 g of Al(OH)3 ?
• 156 g Al(OH)3 1 mole Al(OH)3 6 mole
  H2O 6.00 mole H2O
• 78.0 g Al(OH)3 2 mole
  Al(OH)3

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III. Stoichiometry B. Regular Problems
Continued

• Given 2 Al(OH)3 3 H2SO4 -----) 6
  H2O 1 Al2(SO4)3
• Example
• 3. How many g of Al(OH)3 will react with 59 g of
  H2SO4 ?
• 59 g H2SO4 1 mol H2SO4 2 mol Al(OH)3 78 g
  Al(OH)3 31 g Al(OH)3
• 98 g H2SO4 3 mol H2SO4
  1 mol Al(OH)3

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III. Stoichiometry C. Limiting Reagent
Problems

• You are given TWO starting materials and one will
  limit the amount of product. Either a) determine
  the limiting reagent ahead of time or b)
  calculate the result for each reagent use the
  one which gives the smallest mols or g of
  product.
• 1. How many g of HgS will be produced from 20. g
  of Hg and 64 g of S ?
• 1 Hg 1 S -----) 1 HgS
• 64 g S 1 mol S 1 mol HgS
  232 g HgS 464 g HgS
• 32 g S 1 mole S
  1 mol HgS
• 20. g Hg 1 mol Hg 1 mol HgS 232 g
  HgS 23 g HgS
• 200. g Hg 1 mole Hg
  1 mole HgS

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III. Stoichiometry C. Limiting Reagent
Problems

• 2. How many g of CS2 can be made from 20. g C and
  30. g SO2?
• 5 C 2 SO2 -----) 1 CS2 4 CO
• 20.g C 1 mol C 1 mol CS2 76 g CS2
  25 g CS2
• 12 g C 5 mol C 1 mol CS2
• 30.g SO2 1mol SO2 1 mol CS2 76g CS2
  18 g CS2 (Correct)
• 64g SO2 2 mol SO2 1 mol CS2
• 3. 11 moles of Hg are mixed with 9 moles of S.
  How many moles of excess reagent are left
  over? 1 Hg 1 S -----) 1 HgS
• 11 m Hg gives 11 m HgS 9 m S gives 9 m
  HgS S is LR
• 11 m Hg m Hg that reacts m Hg left over
• 9 mol S x 1 mol Hg 9 mols of Hg
  react.
• 1 mol S
• 11 mols Hg - 9 mols Hg 2 mol Hg Left Over

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III. Stoichiometry C. Yield

• Example 48 mol of O2 reacts with excess Mg 39
  moles of MgO are
• obtained in the lab. Calculate the Yield.
  2 Mg 1O2 -----) 2 MgO
• - Yield Actual Yield x 100
  39 moles MgO x 100
• Theoretical Yield
  Theoretical Yield
• - Are given the actual yield, 39 moles MgO
  calculate the theoretical yield.
• 48. mol O2 x 2 mol MgO 96 mol MgO
  (Theoretical Yield in moles)
• 1 mol O2
• Yield 39. mol MgO x 100
  41
• 96 mol MgO
• - Note Can use either g or moles in problems
  get same answer.