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Applications of the Mole Concept Percent Composition Empirical Formula

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 Divide each part by the molar mass Multiply by 100. Examples. H 2 O Molar mass = 18.02g H 2 = 2.02g O = 16.00g

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Applications of the Mole Concept Percent Composition Empirical Formula

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  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 • Determine empirical formula • Determine empirical formula mass • Compare to given mass • 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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