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Applications of the Mole Concept Percent Composition Empirical Formula

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Applications of the Mole Concept Percent Composition Empirical Formula The makeup of a compound by mass How does it work? Determine the molar mass of the compound ... – PowerPoint PPT presentation

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Title: Applications of the Mole Concept Percent Composition Empirical Formula


1
Applications of the Mole Concept Percent
CompositionEmpirical Formula
  • The makeup of a compound by mass

2
How does it work?
  • Determine the molar mass of the compound
  • Divide each part by the molar mass
  • Multiply by 100

3
Examples
  • H2O
  • Molar mass 18.02g
  • H2 2.02g
  • O 16.00g
  • H2 2.02/18.02 x 100 11.11
  • O 16.00/18.02 x 100 88.89
  • Or as a shortcut, once you have all but one
    component remaining subtract what has already
    been calculated from 100, i.e, 100 11.11
    88.89

4
Example 2
  • (NH4)2CO3 MM 96.09
  • N 14.01 x 2 28.02/96.09 29.17
  • H 1.01 x 8 8.06/96.09 8.33
  • C 12.01 x 1 12.01/96.09 12.50
  • O 16.00 x 3 48.00/96.09 50.00

  • 100.00

5
Empirical Formula
  • Recall that empirical formula is the simplest
    whole-number ratio of elements in a compound.
  • If we know how much mass of each element is
    present, the empirical formula can be calculated.
  • If we know the percentage composition, we can
    calculate the empirical formula.

6
Mass of each component known
  • A 9.2g sample of a substance contains 2.8g
    Nitrogen and 6.4g oxygen. What is the empirical
    formula?
  • Divide each by its molar mass
  • 2.8g N / 14 g/mol N 0.20 mol N
  • 6.4g O / 16g/mol O 0.40 mol O
  • Divide each molar amount by the smallest to make
    whole-number ratios
  • 0.2/0.2 1 mol N
  • 0.4/0.2 2 mol O
  • Empirical formula NO2

7
Percentage composition known
  • What is the empirical formula if a compound is
    65.2 As and 34.8 O by mass?
  • Assume 100g of substance
  • Divide by molar mass
  • 65.2g As/ 74.9g/mol As 0.870mol As
  • 34.8g O / 16.0g/mol O 2.18 mol O
  • Divide each by .870
  • 0.870/0.870 1.00mol As 2.18/0.870 2.50mol O
  • Since fractions cannot be used just double these
    subscripts to obtain As2O5

8
Molecular Formula
  1. Determine empirical formula
  2. Determine empirical formula mass
  3. Compare to given mass
  4. If not the same, divide given molecular mass by
    empirical formula mass to determine factor to
    multiply subscripts

9
Molecular formula example
  • Benzene is a common component of gasoline and
    other fuels. Its molecular weight is 78.11. It
    contains 92.26 Carbon by weight, the rest is
    hydrogen. What are its empirical and molecular
    formulas?

10
Solution
  • Assume 100g
  • 92.26g/12.01g/mol C 7.68mol C
  • 7.74g/1.008g/mol H 7.68mol H
  • Divide by the smallest of moles
  • Empirical Formula CH
  • Mass of CH 13.02g
  • Divide benzene mass by this mass
  • 78.11/13.02 6 so molecular formula of benzene
    is C6H6

11
Assignment
  • Percent Composition/Empirical Formula Problem Set
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