Formula for Calculating Percentages - PowerPoint by ueh10012


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     Percent Composition
oNew food labels are required to
 describe the ingredients using
 percents of the daily reccom-
 mended allowance
 • These numbers tell
   what part of the total #
   of calories can be ob-
   tained from a product
 • AKA percent composition
     Percent Composition
oTo get the information found on
 food labels the chemists had to
 know what fraction of the
 whole was each component
  • Component/total and then
    multiply by 100
  • There are a couple of
    procedures used to calculate
    percent compositions
 Calculating PC given formula
   What percentage of
  Hydrogen and Oxygen is
      in Water (H2O)?
Assume you have 1 mole of
water, and calculate its molar
(2•1.008g) + (1•15.994g) =
                     18.01g      4
 Calculating PC given formula
oThere are 2 mols of H atoms for
 every 1 mol of Water molecules
oHow much do 2 mols of H atoms
     H: (2•1.008g)= 2.016g H
oPercent of H in Water?
   2.016g H
                X 100%= 11.2%
  18.01 g H2O
 Calculating PC given formula
oThere is 1 mol of O atoms for
 every 1 mol of Water molecules
oHow much does 1 mol of O
 atoms weigh?
    O: (1•15.994g)= 15.994g O
oPercent of O in Water?
    15.994 O
                X 100%= 88.8%
  18.01 g H2O
      Percent Composition
o Another method of calculating
  the percent composition is by
  experimental analysis.
   • the overall mass of the sample
     is measured.
   • then the sample is decomp-
     osed or separated into its
     component elements
     Percent Composition
o The masses of the component
  elements are then determined
  and the percent composition is
  calculated as before
   • by dividing the mass of each
     element by the total mass of
     the sample
   • then multiplying by 100
Calculating PC given sample

Find the percent composition
  of a compnd that contains
  1.94g of carbon, 0.48g of
   Hydrogen, and 2.58g of
Sulfur in a 5.0g sample of the
 Calculating PC given sample
o Calculate the percents for
  each element much like you
  would calculate the percents
  for anything.
 C: 1.94g/5.0g X 100% = 38.8%
 H: 0.48g/5.0g X 100% = 9.6%
 S: 2.58g/5.0g X 100% = 51.6%
       Empirical Formulas
o Once the percent compositions
  are determined then they can
  be used to calculate a simple
  chem formula for the compnd
  • key is to convert the percents
    by mass into amounts in moles
  • Then, compare the moles using
    ratios to determine coefficients
Calculating Empirical Formulas

 What is the empirical formula
  of a compound that is 80%C
       and 20%H by mass
oSince we have been given per-
 cents rather than masses we
 need to make an assumption.
 • Let’s suppose we have a total
   sample that weighs 100 g.
Calculating Empirical Formulas
o This allows us to say that if we
  had a 100 grams of sample,
  • 80 g is Carbon
  • 20 g is Hydrogen
o Now that we have a set of
  masses we need to convert
  them to moles
   • Divide by the molar masses
     from the Periodic Table
Calculating Empirical Formulas
        1 mole C
80g C              = 6.7mol C
         12 g C
        1 mole H
20g H              = 20 mol H
• Now calculate the simplest ratio
  of each by dividing both values
  by the smallest value              14
Calculating Empirical Formulas
 Divide each mole value by the
   smaller of the two values:

 C: 6.7/6.7=1
 H: 20/6.7 = 2.98  3
Ratio is 1 C’s for every 3 H’s;
so the formula is = CH
Calculating Empirical Formulas

  Determine the empirical
  formula of a compound
 containing 25.9g of N and
        74.1g of O.
    Notice we have masses
    this time not percents,
    we can convert masses
       directly to moles
Calculating Empirical Formulas
          1 mol N
25.9g N             = 1.85 mol N
          14 g N
                       1.85 mol

          1 mol O
74.1g O           = 4.63 mol O
           16 g O
                     1.85 mol

Calculating Empirical Formulas
  Is the final answer N1O2.5?
         Of course not!
   We need a whole number
Each part of the ratio is multiplied
 by a number that converts the
   fraction to a whole number

      N2(1)O2(2.5)= N2O5
       Molecular Formulas
o The empirical formula indicates
  the simplest ratio of the atoms
  in the compnd
  •However, it does not tell you the
   actual numbers of atoms in
   each molecule of the compnd
  •For instance, glucose has the
   molecular formula of C6H12O6
  •Empirical form would be CH2O
      Molecular Formulas
o The empirical formula of CH2O,
  could be several compnds.
  •C2H4O2 or C3H6O3 or
o It’s more important to know the
  exact numbers of atoms
   The numbers of atoms define
    the properties of the compnd    20
       Molecular Formulas
o The molecular formula is
  always a whole-number multiple
  of the emp. formula
o In order to calculate the
  molecular formula you must
  have 2 pieces of information
 • Empirical formula
 • Molar mass of the unknown
   compound (must be given)
Calculating Molecular Formulas
 Find the molecular formula of a
     compound that contains
   56.36 g of O and 54.6 g of P.
     If the molar mass of the
    compound is 189.5 g/mol.
1) Find the Empirical Formula
2) Find the MM of the Emp. Form.
3) Find the ratio of the 2 molar
   masses (Mol MM/Emp MM)
 1)Find the Empirical Formula

56.36g O 1 mol O
                 = 3.5 mol O
          16 g O    1.8 mol
          1 mol P
54.6g P         = 1.8 mol P
          31g P
                   1.8 mol

    Empirical formula: P1O2     23
2) Find the MM of the Emp
MM of PO2: (1•31g P) + (2•16g O)
                       = 63g/mol
3)Find the ratio of the 2 molar
   masses (mol MM/emp MM)
     GIVEN   189.5 g/mol
                           = 3.00
  CALCULATED   63 g/mol

Calculating Molecular Formulas
o So the Molecular formula is
  3 times heavier than the
  Empirical formula
  • Therefore, the molecular
    formula has 3 times more
    atoms than the emp. formula

      P3(1)O2(3)= P3O6

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