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Percent Composition Empirical Formulas and Molecular Formulas

Percent Composition Empirical Formulas and Molecular Formulas. Quantification in Chemistry. Homework: Chapter 11 pg 333 problems 46-50 pg 335 problems 51-53 pg 337 problems 54-57. Warm Up-10/17/13. 1. Determine the percent composition of phosphoric acid H 3 PO 4

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Percent Composition Empirical Formulas and Molecular Formulas

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  1. Percent CompositionEmpirical Formulas and Molecular Formulas Quantification in Chemistry

  2. Homework: Chapter 11 • pg 333 problems 46-50 • pg 335 problems 51-53 • pg 337 problems 54-57

  3. Warm Up-10/17/13 1. Determine the percent composition of phosphoric acid H3PO4 2. A compound contains 36.84% nitrogen and 63.16% oxygen. What is the empirical formula for this compound? 3. A liquid composed of 46.68% nitrogen and 53.32% oxygen has a molar mass of 60.01 g/mol. What is the molecular formula?

  4. Percent Composition Analytical chemist work to identify elements in compounds and to determine their percent by mass. To determine the percentage of an element in a compound you do the following: mass of element X 100 mass of compound

  5. For Example: A 100-gram sample of a compound contains 55 grams of element X and 45 grams of element Y. What are the percentages of mass for each element? X: 55g element XX 100 = .55 X 100 = 55% 100 g of compound and, Y: 45g element Y X 100 = .45 X 100 = 45% 100 g of compound

  6. Percent Composition: Example 2 Determine the Percent Composition in the Compound Sodium Hydrogen Carbonate (NaHCO3). 1 mol Na = 22.99 g Na 1 mol H = 1.008 g H 1 mol C = 12.01 g C 1 mol O = 15.99 x 3 g O (48g) 84.01 g/mol NaHCO3

  7. Percent Composition: Example 2 (continued) Percent Na = 22.99 g Na= .2737 X 100= 27.37% Na 84.01 g/mol NaHCO3 Percent H = 1.008 g H= .012 X 100 = 1.2% H 84.01 g/mol NaHCO3 Percent C = 12.01 g C = .143 X 100 = 14.3% C 84.01 g/mol NaHCO3 Percent O = 15.99 x 3 g O= .5714 X 100 = 57.1% O 84.01 g/mol NaHCO3

  8. Empirical Formulas An empirical formula is the formula for a compound with the smallest whole number ratio of its elements. • Example 1: The percent composition for sulfur oxide is 40.05 % sulfur and 59.95 % oxygen. What is the empirical formula for sulfur oxide? 40.05 g S X 1 mol = 1.29 mol S and 32.07g/mol S 59.95 g O X 1 mol = 3.747 mol O and 15.99g/mol O

  9. Empirical FormulasExample 1 (continued) 1.29 mol S / 1.29 = 1 mol S 3.747 mol O / 1.29 = 3 mol O The empirical formula for sulfur oxide is: SO3

  10. Empirical Formulas Example 2-Determine the empirical formula for methyl acetate (CHO) if the compound has the following percent composition: 48.64% C, 8.16% H and 43.20% O. 48.64g C x 1 mol C = 4.050 mol C 12.01g/mol C 8.16g H x 1 mol H = 8.10 mol H 1.008g/mol H 43.20g O x 1 mol O = 2.70 mol O 15.99g/mol O

  11. Empirical Formulas Example 2 (continued) 4.050 mol C / 2.70 = 1.5 mol C 8.10 mol H / 2.70 = 3 mol H and 2.70 mol O / 2.70 = 1 mol O Multiply the numbers of moles to turn them into whole number ratios 4.050 mol C / 2.70 = 1.5 mol C x 2 = 3 mol 8.10 mol H / 2.70 = 3 mol H x 2 = 6 mol 2.70 mol O / 2.70 = 1 mol O x 2 = 2 mol • The formula for methyl acetate is: C3H6O2

  12. Warm Up-10/15/13 • Rubbing alcohol was found to contain 60.0 % carbon, 13.4 % hydrogen, and the remaining mass was due to oxygen. What is the empirical formula of rubbing alcohol? 2. A sample of indium chloride weighing 0.5000 g is found to contain 0.2404 g of chlorine. What is the empirical formula of the indium compound?

  13. Molecular Formulas • Molecular Formulas specify the actual number of atoms that are in a given compound. • Example 1: What is the molecular formula for methyl acetate given that its experimental mass was determined to be 144 g/mol? mass of empirical formula : C3H6O2 C = 12.01 x 3 = 36.03 144 g/mol)/(72 g/mol) = 2 (n) H = 1.008 x 6 = 6.0482 x C3H6O2 = C6H12O4 O = 15.99 x 2 = 31.98 72 g/mol

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