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Chapter 3: Stoichiometry Notes

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  1. Atomic Masses
    Isotopes:
    The element indium exists naturally as two isotopes. 113In has a mass of 112.9043 amu, and 115In has a mass of 114.9041 amu. The average atomic mass of indium is 114.82 amu. Calculate the percent relative abundance of the two isotopes of indium.
    Solve by using two simultaneous equations:

    112.9043( x ) + 114.9041( y ) = 114.82
    x + y = 1
    Solve second equation in terms of y

    112.9043( x ) + 114.9041( y ) = 114.82
    y = 1 - x
    Divide first equation by 114.9041

    .9825960( x ) + y = 0.99927
    y = 1 - x
    Substitute second equation into first

    .9825960( x ) + 1 - x = 0.99927
    y = 1 - x
    Solve first equation in terms of x

    -.017404( x ) = -.00073
    x = 0.042
    y = 1 - x
    Replace x into second equation

    x = 0.042
    y = 1 - 0.042
    y = 0.958

    112.94.2%113In
    114.995.8%115In

  2. Mole
    Avogadro's number of anything: 6.022 x 1023
    1 amu is equal to 1.67 x 10-24 g
    12 amu = 12 g/mol

  3. Molar Mass = the sum of the masses of the individual atoms in the molecule.

    1. calculate the molar mass of potassium dichromate ( K2Cr2O7 ) ALIGN=RIGHT
      K239.10 g/mol78.2 g/mol
      Cr252.00 g/mol104.0 g/mol
      O716.00 g/mol112.0 g/mol
      Total:294.20 g/mol
    2. How many microg are there in 3.82x10-7 moles of sodium hydrogen sulfate (NaHSO4 120.06 g/mol)?

      45.9 micrograms NaHSO4

  4. Percent Composition of Compounds
    Calculate the mass percent of potassium in the compound potassium hexacyanoferrate III ( K3[Fe(CN)6] ).
    100(117.20/329.26)=35.6%

  5. Types of Formulas
    1. Empirical Formula: simplest formula of a compound, gives the mole ratio.
    2. Molecular Formula: gives the actual composition of a molecule. May be identical with the empirical formula or an integral multiple of it.
      NameMolecular FormulaEmpirical Formula
      glucoseC6H12O6CH2O
    3. Hydrate: solid substances that contain water molecules in their crystal lattice.
      BaCl2*2H2O
      CuSO4*5H2O

      -Empirical formula from percent composition.
      Find the mass of each element in the sample of compound, then find the number of moles of each element, and finally the mole ratio.

      -Molecular formula from simplest formula.
      You must know the molar mass.

      Empirical formula problems:

      1. 1. The analysis of a rocket fuel showed that it contained 87.4% nitrogen and 12.6% hydrogen by weight. Mass spectral analysis showed the fuel to have a molar mass of 32.05g. What are the empirical and molecular formulas of the fuel?

        Assume that 100 g of compound are present. The percentages can be used as gram masses.

        empirical formula: NH2
        molecular formula: N2H4
        (the compound is called hydrazine)

      2. The following reaction was performed:

        Fe2O3 (s) + 2X (s) --> 2Fe (s) + X2O3 (s)

        It was found that 79.847 g of Fe2O3 reacted with "X" to form 55.847 g of Fe and 50.982 g X2O3. Identify element X.

        X is Al

      3. A combustion device was used to determine the empirical formula of a compound containing only carbon, hydrogen, and oxygen. A .6349 g sample of the unknown produced 1.603 g of CO2 and .2810 g of H2O. Determine the empirical formula of the compound.

        Solve for grams
        Solve for O
        C7H6O2
        C7H6O2

  6. Chemical Equations
    1. Balancing: Must have same number of atoms of each type on both sides. Achieve this by adjusting coefficients in front of formulas.
    2. Mass/Mass problems
      Law of Conservation of Mass

      eg. How many grams of silver sulfide can be generated from the reaction of 3.94 g of silver nitrate with excess sodium sulfide?

      2AgNO3 + Na2S --> 2NaNO3 + Ag2S
      2.87 g Ag2S

  7. Yield of Product in Reaction
    1. Limiting reactant, theoretical yield:
      Reactants are not in exact ratios, one is in excess and one is limiting.
      The one in excess will have some reactant remaining after the reaction.
      The one that is limiting will be completely consumed in the reaction.
      Therefore, the limiting one is used to calculate the theoretical yield.

      To calculate the theoretical yield and identify the limiting reactant:

      1. Calculate he yield to be expected if the first reactant is limiting.
      2. Repeat this calculation for the second reactant.
      3. Theoretical yield is the smaller of these two quantities. The reactant that gives the smaller calculated yield is the limiting reactant.
      4. Actual yield, percent yield.
        The actual yield is the amount formed when the experiment is actually carried out. The percent yield is found by taking the actual yield and dividing it by the theoretical yield.

      eg. sodium hydroxide reacts with phosphoric acid in a neutralization reaction. If 17.82 g of sodium hydroxide are mixed with 15.40 g of phosphoric acid,

      1. How many grams of sodium phosphate can be formed?
      2. How many grams of the excess reactant remains unreacted?
      3. If the actual yield of sodium phosphate is 15.00 g, what is the percent yield of sodium phosphate?

      1. 3NaOH + H3PO4 --> Na3PO4 + 3H2O

        24.35 g Na3PO4
        NaOH is limiting.
        24.35 g Na3PO4

      2. .0086 mol Na3PO4
        .84 g H3PO4

      3. 61.60%


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