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 Determine the molar mass of the compound Divide each part by the molar mass Divide each part by the molar mass Multiply by 100 Multiply by 100

Examples H 2 O H 2 O Molar mass = 18.02g Molar mass = 18.02g H 2 = 2.02g H 2 = 2.02g O = 16.00g O = 16.00g %H 2 = 2.02/18.02 x 100 = 11.11% %H 2 = 2.02/18.02 x 100 = 11.11% %O = 16.00/18.02 x 100 = 88.89% %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% 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%

Example 2 (NH 4 ) 2 CO 3 MM = (NH 4 ) 2 CO 3 MM = %N = x 2 = 28.02/96.09 = 29.17% %N = x 2 = 28.02/96.09 = 29.17% %H = 1.01 x 8 = 8.06/96.09 = 8.33% %H = 1.01 x 8 = 8.06/96.09 = 8.33% %C = x 1 = 12.01/96.09 = 12.50% %C = x 1 = 12.01/96.09 = 12.50% %O = x 3 = 48.00/96.09 = 50.00% %O = x 3 = 48.00/96.09 = 50.00% % %

Empirical Formula Recall that empirical formula is the simplest whole-number ratio of elements in a compound. 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 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. If we know the percentage composition, we can calculate the empirical formula.

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? 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 Divide each by its molar mass 2.8g N / 14 g/mol N = 0.20 mol N 2.8g N / 14 g/mol N = 0.20 mol N 6.4g O / 16g/mol O = 0.40 mol O 6.4g O / 16g/mol O = 0.40 mol O Divide each molar amount by the smallest to make whole-number ratios Divide each molar amount by the smallest to make whole-number ratios 0.2/0.2 =1 mol N 0.2/0.2 =1 mol N 0.4/0.2 = 2 mol O 0.4/0.2 = 2 mol O Empirical formula = NO 2 Empirical formula = NO 2

Percentage composition known What is the empirical formula if a compound is 65.2% As and 34.8% O by mass? What is the empirical formula if a compound is 65.2% As and 34.8% O by mass? Assume 100g of substance Assume 100g of substance Divide by molar mass Divide by molar mass 65.2g As/ 74.9g/mol As = 0.870mol As 65.2g As/ 74.9g/mol As = 0.870mol As 34.8g O / 16.0g/mol O = 2.18 mol O 34.8g O / 16.0g/mol O = 2.18 mol O Divide each by.870 Divide each by /0.870 = 1.00mol As; 2.18/0.870 = 2.50mol O 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 As 2 O 5 Since fractions cannot be used just double these subscripts to obtain As 2 O 5

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

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

Solution 1) Assume 100g 2) 92.26g/12.01g/mol C = 7.68mol C 3) 7.74g/1.008g/mol H = 7.68mol H 4) Divide by the smallest # of moles 5) Empirical Formula = CH 6) Mass of CH = 13.02g 7) Divide benzene mass by this mass = 78.11/13.02 = 6 so molecular formula of benzene is C 6 H 6

Assignment Percent Composition/Empirical Formula Problem Set