Using Stoichiometry to Determine Chemical

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Using Stoichiometry to Determine Chemical Powered By Docstoc
					Q3U2 :Theoretical Yield, Mass Percents, and Empirical Formulas
 Theoretical Yield: The amount of product
  formed through a chemical reaction, as
  predicted by Stoichiometry
 Actual Yield: The amount of product formed
  when a reaction is performed in a lab, usually
  less than theoretical yield.
Why are they different?
 The collection techniques, apparatus used,
  time, and skill of the chemist can effect the
  actual yield.
 When actual and theoretical yields are
  different, you express the efficiency of the
  reaction using percent yield.
 Percent yield is the ratio of actual yield to
  theoretical yield, expressed as a percent.
Percent yield = (actual yield) X 100
               (theoretical yield)
   Chemical formulas tell you the elements
    involved in a reaction, the number of atoms
    of each and the number of moles of reactants
    and products
   What is the major element in a reaction, by
    mass?
   To answer this, determine the mass percents
    of each element
 Steps
1. Determine the molar mass of the element in
   the compound
2. Multiply the mass of 1 mole times the
   number of moles in the compound
3. Now use that mass to find the mass percent
   in the sample
In 154 grams of C10H18O (Geraniol), How many grams does
   the Carbon atom contribute?
 1 molecule of C10H18O contains 10 atoms of Carbon
 Therefore 1 mole of C10H18O contains 10 moles of carbon
   atoms
 Multiply the mass of 1 mole of carbon by 10 to get the
   mass of carbon in one mole C10H18O
    (10 mol) (12.0gC) = 120g C
           (1 mol)
   Use mass to determine mass percent
     mass % of C = ( 120 gC )      X 100 = 77.9%
                   (154 g C10H18O)
Given 154 g of C10H18O

 What is mass% of H?
18 mol H X 1.01 g/mole = 18 g H

Mass% H = mass H X 100
         mass C10H18O
          = 18.0 g H     X 100 = 11.7% H
           154 g C10H18O

   What is the Mass% of O?
What is the Mass% of O?
 (1 mol O) (16 g O) = 16 g O
            ( 1 mol)

  16 g O     X 100 = 10.4% O
154g C10H18O
   The previous steps will allow you to determine
    the composition of any compound, as long as
    you know the formula.
Try the Following:
1. Hydrogen fuels are rated with respect to their
   hydrogen content. Determine the percent
   hydrogen for the following fuels.
    a) Ethane, C2H6
    b) Methane, CH4
    c) Whale Oil, C32H64O2
a)      Ethane, C2H6 2 mol C, 6 mole H
                      X 12.0g X 1.0 g
                        24.0g    6.0 g   = 30.0g
     (6.0 g/30.0 g ) X 100 = 20%

b)      Methane, CH4 1 mol C , 4 mol H
                       X 12.0g X 1.0 g
                        12.0 g   4.0 g
     (4.0 g/16.0 g ) X 100 = 25%

c)     Whale Oil, C32H64O2 32 mol C, 64 mol H, 2 mol O
                           X 12.0g X 1.0 g X 16.0 g
                          384. 0g   64.0 g 32.0 g
 (64.0 g/480.0 g ) X 100 = 13%
  Calculate the Percent Composition of oxygen
   in the following compounds
a. SO3
b. CH3COOH
c. Ca(NO3)2
d. Ammonium Sulfate, (NH3) SO2
a.   60.00%
b.   53.29%
c.   58.50%
d.   48.43%
   Hydrogen peroxide (H2O2), what is the mass
    % of Oxygen?


   Sodium nitrate (NaNO3), what is the mass %
    of Oxygen?
     a) H2O2             2 mol O, 2 mole H
                        X 16.0g X 1.0 g
                          32.0g    2.0 g     = 34.0g
     (32.0 g/34.0 g ) X 100 = 94.1%

b)    NaNO3          1 mol Na, 1 mol N, 3 mole O
                    X 23.0g X 14.0 g X 16.0 g
                      23.0 g 14.0g       48.0 g
 (48.0 g/85.0 g ) X 100 = 56.4%
   Fish in some lakes have been found to contain a mercury
    compound, possibly a contaminant from the making of
    paper.
   Analysis of this compound gives the following mass
    percentages: Carbon, 5.57% ; hydrogen, 1.40% ; and
    mercury, 93.03%.
   Using this information, determine the Empirical
    Formula of the compound.
   Empirical Formula: the formula of a compound that
    has the smallest whole-number ratio of atoms in the
    compound
 Steps
1. Fin d the relative numbers of atoms in the
     formula unit of the compound
2.   Use the molar mass to find the number of
     moles of each element
3.   Divide the mole numbers by the smallest one,
     to find the whole number ratios

Empirical Formula: the formula of a compound
  that has the smallest whole-number ratio of
  atoms in the compound
   Analysis of this compound gives the following mass
    percentages: Carbon, 5.57% ; hydrogen, 1.40% ; and mercury,
    93.03%.
     Using this information, determine the Empirical Formula of
      the compound.
Carbon: 5.57g
       12.01 g/mol = 0.464 mol           ratio = 1
Hydrogen: 1.40 g
         1.008 g/mol = 1.39 mol          ratio: 1.39/ 0.464 = 3
Mercury: 93.03g
        200.59 g/mol = 0.464 mol         ratio = 1
       Empirical formula is CH3Mg, methyl mercury
   An unknown compound is 18.8% Na, 29.0% Cl
    and 52.2% O, what is the empirical formula of
    the unknown?
   Use molar mass to find the number of moles of each element.
     (18.8 g Na) (1mol Na) = 0.817 mol Na
                  (23.0 g Na)
     (29.0 g Cl) (1mol Cl) = 0.817 mol Cl
                 (35.5 g Cl)
     (52.2 g O) (1mol O) = 3.26 mol O
                 (16.0 g O)
   Divide the mole numbers by the smallest one.
     0.817/ 0.817 = 1 mol Na ratio = 1
     0.817/ 0.817 = 1 mol Cl  ratio = 1
     3.26/ 0.817 = 3.99 mol O ratio = 4

                         The Empirical formula is NaClO4
   Hydrogen = 5.17%, nitrogen is 35.9% and
    sodium is 58.9%, what is the empirical
    formula?
   5.17 g H X 1 mol H/ 1.01 g H = 5.12 mol H
   35.9 g N X 1 mol N/ 14.0 g N = 2.56 mol N
   58.9 g Na X 1 mol Na/ 23.0 g Na = 2.56 mol Na

Ratio of H:N:Na is 2:1:1

Empirical formula is NaNH2
a.   15.8% carbon and 84.2% sulfur

b.   43.6% phosphorus and 56.4% oxygen

c.   28.7% K, 1.5% H, 22.8% P and 47.0% O
a.   CS2
b.   P2O5
c.   KH2PO4
Calculate the empirical formula for the following
 compounds from percent composition

a. 0.0130 mol C, 0.0390 mol H, 0.0065 mol O

b. 11.66 g iron, 5.01 g oxygen

c. 40.0 percent C, 6.7 percent H, and 53.3
  percent O by mass
a.   C2H6O
b.   Fe2O3
c.   CH2O
   For most ionic compounds the two are the same
   Covalent compounds can be a very different
    case.
     The ratio between moles is the same, but the atoms
     can share electrons differently, so the formula may be
     different.
      ▪ Ex. C6H12O6 and C3H6O3 and C2H4O2
            glucose       lactic acid  acetic acid
      Different compounds with different properties, same molar
        ratios, different chemical formulas!
   You need the molar mass of the compound,
    change to molecular mass (amu)
   Divide the molecular mass of the compound
    by the molecular mass of the empirical
    formula unit
   This will tell you how many of the empirical
    formula units are in the chemical formula
 40.0% C, 6.7% H, 53.3% O
Find the empirical formula


Is this the same as the chemical formula, if the
   molar mass is 90.0 g/mol
3.33 mol C = 1 mol C
6.7o mol H = 2.01 mol H
3.33 mol O = 1.o mol O
Empirical formula is CH2O

90.0 g/mol = 90.0 amu molecular mass
CH2O molecular mass= 12+2 + 16= 30 amu

90.0/30.0 =3 3 empirical formula units
  C3H6O3, lactic acid
a. empirical formula CH , molar mass = 78 g/mol


b. empirical formula NO2 , molar mass = 92.02 g/mol


c    caffeine, 49.5% C , 5.15% H , 28.9% N , 16.5% O
    by mass, molar mass = 195 g.
   a. C6H6

   b. N2O4


    c. Use mlar mass to find ratios, and empirical
    formula first, divide molecular and empirical
    formula masses to determine number of
    empirical formula units in sample grams
    C8H10N4O2

				
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posted:11/23/2012
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