Chapter 9 Stoichiometry

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					 Chapter 9:
Stoichiometry
   Coach Kelsoe
    Chemistry
  Pages 298–325
        Section 9-1:
Introduction to Stoichiometry
          Coach Kelsoe
           Chemistry
         Pages 299–301
          Section 9-1 Objectives
n   Define stoichiometry
n   Describe the importance of the mole ratio in
    stoichiometric calculations.
n   Write a mole ratio relating two substances in
    a chemical equation
                 Stoichi-what?
n   Stoichiometry (pronounced “sto-key-ometry”)
    comes from two Greek words:
    n Stoicheion means “element”
    n Metron means “measure”
      Introduction to Stoichiometry
n   A lot of our knowledge of chemistry is based
    on the careful quantitative analysis of
    substances involved in chemical reactions.
n   Composition stoichiometry deals with the
    mass relationships of elements in compounds.
n   Reaction stoichiometry involves the mass
    relationships between reactants and products
    in a chemical reaction.
      Introduction to Stoichiometry
n   The good news: this is not entirely new
    material. We’ve already learned about
    composition stoichiometry in Section 3-2. We
    just never called it that!
n   Reaction stoichiometry is based on chemical
    equations and the law of conservation of
    matter.
n   ALL reaction stoichiometry calculations start
    with a balanced equation.
    Reaction-Stoichiometry Problems
n   The problems in this chapter can be classified
    according to the information given in the problem
    and the information you are expected to find, called
    the unknown.
n   The given and unknown may both be reactants,
    they may both be products, or one may be a
    reactant and the other a product.
n   Masses should be expressed in grams.
n   Stoichiometric problems are solved by using ratios
    from the balanced equation to convert the given
    quantity using the following methods.
Reaction-Stoichiometry Problems
n   Problem Type 1:
    Given and unknown quantities are in moles.
    n When you are given the amount of substance in
      moles and asked to calculate the amount in moles
      of another substance in the chemical reaction, the
      general plan is
    n Amount of given               Amount of unknown
      substance (in mol)      substance (in mol)
Reaction-Stoichiometry Problems
n   Problem Type 2:
    Given is an amount in moles and the
    unknown is a mass that is often expressed in
    grams.
    n   When you are given the amount in moles of one
        substance and asked to calculate the mass of
        another substance in the chemical reaction, the
        general plan is
    n   Amount of            Amount of        Mass of unknown
        given substance   unknown substance     substance
        (in mol)              (in mol)          (in grams)
Reaction-Stoichiometry Problems
n   Problem Type 3:
    Given is a mass in grams and the unknown is
    an amount in moles.
    n   When you are given the mass of one substance
        and asked to calculate the amount in moles of
        another substance in the chemical reaction, the
        general plan is
    n   Mass of              Amount of      Amount of unknown
        given substance   given substance       substance
        (in grams)            (in mol)           (in mol)
Reaction-Stoichiometry Problems
n   Problem Type 4:
    Given is a mass in grams and the unknown is
    a mass in grams.
    n   When you are given the mass of one substance
        and asked to calculate the mass of another
        substance in a chemical reaction, the general plan
        is
    n   Mass of      Amount of   Amount of    Mass of
        given          given     unknown     unknown
        substance    subtance    substance   substance
        (in grams)   (in mol)     (in mol)   (in grams)
                  Mole Ratio
n   Solving any reaction-stoichiometry problem
    requires the use of a mole ratio to convert
    from moles or grams of one substance in a
    reaction to moles or grams of another
    substance.
n   A mole ratio is a conversion factor that
    relates the amounts in moles of any two
    substances involved in a chemical reaction. It
    is obtained directly from the balanced
    chemical equation.
                  Mole Ratio
n   Consider the following equation:
            2Al2O3(l) à 4Al(s) + 3O2(g)
n   Remember that the coefficients in a chemical
    equations satisfy the law of conservation of
    matter and represent the relative amounts of
    moles of reactants and products.
n   Therefore, 2 mol of aluminum oxide
    decompose to produce 4 mol of aluminum
    and 3 mol of oxygen gas. See board for the
    six ratios.
                   Mole Ratio
n   For the decomposition of aluminum oxide, the
    appropriate mole ratio would be used as a
    conversion factor to convert a given amount in
    moles of one substance to the corresponding
    amount in moles of another substance.
n   For example, if given 13 mol of aluminum oxide
    and need to find out how much oxygen it
    decomposes into:

    13.0 mol Al2O3 x 3 mol O2 = 19.5 mol O2
                    2 mol Al2O3
                      Mole Ratio
n   Think about this in terms of making a PB&J
    sandwich:
    4 slices of bread + 1 cup PB + 1 cup J à 2 PB&J sandwiches
n   Let’s assume this equation is balanced.
n   If we have 16 slices of bread, how much PB
    will we need?
n   16 bread x 1 cup PB = 4 cups PB
                4 Bread
n   We can do this for any ingredient!
                   Mole Ratio
n   Mole ratios are exact, so they do not limit the
    number of significant figures in a calculation.
n   The number of significant figures in the
    answer is therefore determined only by the
    number of significant figures of any measured
    quantities in a particular problem.
n   In layman’s terms, you should have the same
    number of sig figs when you multiply by your
    mole ratio as you had in your given amount.
                 Molar Mass
n   Remember that the molar mass is the mass,
    in grams, of one mole of a substance.
n   The molar mass is the conversion factor that
    relates the mass of a substance to the
    amount in moles of that substance.
n   To solve reaction-stoichiometry problems, you
    will need to determine the molar masses
    using the periodic table.
                   Molar Mass
n   Returning to the aluminum oxide problem:
          2Al2O3(l) à 4Al(s) + 3O2(g)
n   The molar masses can be expressed as so:
        Aluminum      101.96 g          1 mol Al2O3
    n
                     1 mol Al2O3   OR    101.96 g
        oxide

    n   Aluminum      26.98 g      OR   1 mol Al
                      1 mol Al          26.98 g

                      32.00 g      OR   1 mol O2
    n   Oxygen
                      1 mol O2           32.00 g
                 Molar Mass
n   To find the number of grams of aluminum
    equivalent to 26.0 mol of aluminum, the
    calculation would be as follows:
n    26.0 mol Al x 26.98 g Al = 701 g Al
                    1 mol Al
                 Vocabulary
n   Composition stoichiometry
n   Mole ratio
n   Reaction stoichiometry

				
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posted:7/18/2013
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