Chemical Calculations

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Chapter 6

• Chemical Calculations

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Sec 6.1 Formula Masses

• Formula mass the sum total of the atomic
  weights for the atoms in the formula
• Examples C6H12O6, Mg(NO2)2
• Some rounding is acceptable
• Example KOH (39.1 15.99 1.01) 56.1
• KOH (39 16 1) 56

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Sec 6.2 The Mole

• The mole is a measuring concept used to handle
  the tiny size of an atom
• The weight of one atom is unable to be measured
  in a lab
• Mole is defined as the number of atoms in exactly
  12 grams of Carbon-12
• Carbon was chosen as the reference point

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Sec 6.2 The Mole

• Avogadros number is the name for the value of
  the number of objects in a mole
• Mole is always 6.02x1023 atoms or molecules
• This is a conversion you must know!

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Sec 6.3 The Mass of a Mole

• How much does a mole weigh? How much does a dozen
  weigh?
• These terms are counts of objects, it depends on
  the object in question
• Molar mass is the formula weight expressed in
  grams.
• Essentially the mass of one mole for every
  element is the atomic weight listed on the
  periodic table

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Sec 6.3 The Mass of a Mole
Figure 6.5 Page 129. 1 mole of various substances
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Sec 6.3 The Mass of a Mole

• Molar mass is the formula weight expressed in
  grams.
• Example NO2
• Molecular weight is 14 16 16 46 amu
• Molar mass 46 grams/mole
• Example
• 1 mole of Boron is 10.8 grams

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Sec 6.4 Chemical Formulas

• The chemical formula of a compound can be thought
  of as the ratios of moles
• For example H2SO4
• There are 2 moles of Hydrogen atoms in every 1
  mole of H2SO4
• (and similarly, 1 mole of S atoms and 4 moles of
  O atoms)

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Sec 6.5 The Mole and Calculations

• So, going from mass to moles or moles to mass, we
  always use molar mass from the periodic table

Figure 6.7 Page 132
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Sec 6.6 Balancing Chemical Reactions

• Law of Conservation of mass (mass cannot be
  created or destroyed)
• The matter on the products side must equal the
  matter on the reactants side
• Chemical Equations must be balanced
• We balance equations using coefficients in front
  of the formulas

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Sec 6.6 Balancing Chemical Reactions
Figure 6.8 Page 136 Showing the Law of
Conservation of Mass
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Sec 6.6 Balancing Chemical Reactions

• Steps to balancing
• 1. Count the number of atoms on both sides of the
  reaction for each element
• 2. Multiply the coefficient to equalize one
  element, remember that this coefficient applies
  to the entire formula
• 3. Continue going back and forth until all
  elements are equal
• 4. As a guideline, choose Oxygen and Hydrogen
  last

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Sec 6.6 Balancing Chemical Reactions

• Examples
• Ca(OH)2 HCl ? CaCl2 H2O
• C4H10 O2 ? CO2 H2O
• Na2SO3 H3PO4 ? H2SO3 Na3PO4
• Lead(II)Nitrate plus Potassium gives Potassium
  Nitrate and elemental lead

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Sec 6.7 Equations and the Mole

• The coefficients in a balanced chemical reaction
  give us molar ratios
• These ratios can be made for any substances
  within the balanced reaction
• Think of a reaction as similar to the recipe for
  baking

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Sec 6.7 Equations and the Mole

• The following chemical reaction
• N2 3H2 ? 2NH3
• Is really saying that for every 1 mole of N2, we
  expect to get 2 moles of NH3
• Or that to get 2 moles of NH3 it takes 3 moles of
  H2
• Balanced Chemical reactions give us a mole to
  mole ratio between the different substances

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Sec 6.7 Equations and the Mole

• Keeping the baking analogy, we can either
  increase or decrease the size of our recipe while
  maintaining the proper ratio
• For 4Fe 3O2 ? 2Fe2O3
• How much will we make with 2 moles Fe?
• How much O2 do we need in order to make 6 moles
  of Fe2O3
• How much Fe is needed to complete react with 0.75
  moles of O2

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Sec 6.8 Chemical Calculations

• Stoichiometry using mass relationships in
  chemical reactions
• Stoichiometry combines several concepts from our
  previous knowledge
• Molar mass
• Balancing equations
• Ratios/Conversion factors

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Sec 6.8 Chemical Calculations

• Four types of common problems
• 1. How much of a particular product are formed if
  we have a give amount of a reactant?
• 2. How much of one reactant are needed to
  completely react with a given amount of another
  reactant?
• 3. How much of one reactant is needed if we want
  to form a given amount of a specific product?
• 4. How much of a specific product is formed when
  a given amount of another product is produced?

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Sec 6.8 Chemical Calculations

• Common tactical approach

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Sec 6.8 Chemical Calculations

• Do lots of problems

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Sec 6.8 Chemical Calculations

• Percent Yield
• Often we dont get the full mass of product of
  what we expect, it is diminished
• The actual yield is the true experimental mass
• Theoretical yield is the mass predicted by
  stoichiometry

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Sec 6.8 Chemical Calculations

• Formula for percent yield

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Sec 6.8 Chemical Calculations

• Reactions can either give off or absorb heat as
  part of the reaction
• We merely write heat on the products or reactants
  side
• Reactants heat ? products
• Or Reactants ? products heat
• We express heat in units of kcal or kJoules

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Sec 6.8 Chemical Calculations

• Exothermic if the reaction gives off heat (heat
  is on the products side) it is exothermic
• Endothermic if the reaction absorbs heat (heat
  is on the reactants side)
• Note Treating heat as a reactant or product, we
  can do simple stoichiometric problems

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Sec 6.8 Chemical Calculations

• Example
• 2NH3 22.0 kcal ? N2 3H2
• Exothermic or endothermic?
• How much heat needed to make 9 moles of H2 ?

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Problems

• Assigned problems pages 144 - 147
• 6.1, 6.9, 6.11, 6.17, 6.21, 6.25
• 6.31 through 6.40
• 6.43, 6.45 through 6.52
• Practice Test Page 146