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(4.6/4.7) Empirical and Molecular Formulas SCH 3U

(4.6/4.7) Empirical and Molecular Formulas SCH 3U. An empirical formula represents the simplest whole number ratio of the atoms in a compound. The molecular formula is the true or actual ratio of the atoms in a compound. Types of Formulas.

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(4.6/4.7) Empirical and Molecular Formulas SCH 3U

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  1. (4.6/4.7) Empirical and Molecular Formulas SCH 3U

  2. An empirical formula represents the simplest whole number ratio of the atoms in a compound. • The molecular formula is the true or actual ratio of the atoms in a compound.

  3. Types of Formulas The formulas for compounds can be expressed as an empirical formula and as a molecular (true) formula. eg. empirical formula = CH2O

  4. Learning Check A. What is the empirical formula for C4H8? 1) C2H4 2) CH2 3) CH B. What is the empirical formula for C8H14? 1) C4H7 2) C6H12 3) C8H14 C. What is a molecular formula for CH2O? 1) CH2O 2) C2H4O2 3) C3H6O3

  5. Finding the Empirical Formula a) A compound is 71.65% Cl, 24.27% C, and 4.07% H. What is the empirical formula? 1. Assume a 100 g sample. Determine the mass, in grams, of each element present. Cl 71.65 g C 24.27 g H 4.07 g

  6. 2. Calculate the number of moles of each element. nCl = 71.65 g = 2.021 mol 35.45 g/mol nC = 24.27 g = 2.021 mol 12.01 g/mol nH = 4.07 g = 4.03 mol 1.01 g/mol

  7. Divide each by the smallest number of moles to obtain the simplest whole number ratio. **If whole numbers are not obtained, multiply subscripts by the smallest number that will give whole numbers** Cl: 2.021 mol = 1.000 Cl (1) 2.021 mol C: 2.021 mol = 1.000 C (1) 2.021 mol H: 4.04 mol = 2.00 H (2) 2.021 mol 4. Write the simplest or empirical formula. CH2Cl

  8. Learning Check Aspirin is 60.0% C, 4.5 % H and 35.5% O. Calculate the empirical formula.

  9. nC = 60.0 g = 5.00 mol 12.01 g/mol nH = 4.5 g = 4.5 mol 1.01 g/mol nO = 35.5 g = 2.22 mol 16.00 g/mol

  10. X 4 = 9 mol C 5.00 mol C = 2.25 mol 2.22 4.5 mol H = 2.0 mol 2.22 2.22 mol O = 1.00 mol 2.22 X 4 = 8 mol H X 4 = 4 mol O Therefore, the Empirical Formula (EF) = C9H8O4

  11. Finding the molecular formula a) A compound is Cl 71.65%, C 24.27%, and H 4.07%. What is the empirical formula? (from yesterday, CH2Cl) b) The molar mass is known to be 99.0 g/mol. What is the molecular formula? 1. Calculate EFM (empirical formula mass) 1(12.01g/mol) + 2(1.01 g/mol) + 1(35.45g/mol) = 49.48 g/mol 2. Calculate Multiplier: Molar mass (M) = 99.0 g/mol = 2.00 EFM 49.48 g/mol 3. Multiply the empirical formula subscripts by the multiplier (CH2Cl)x 2 = C2H4Cl2

  12. Learning Check A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g/mol, what is the molecular formula?

  13. Solution nS = 27.4 g = 0.855 mol 32.06 g/mol nN = 12.0 g = 0.857 mol 14.01 g/mol nCl = 60.6 g = 1.71 mol 35.45 g/mol

  14. Solution 0.855 mol S = 1.00 mol 0.855 0.857 mol N = 1.00 mol 0.855 1.71 mol Cl = 2.00 mol 0.855 Therefore, the Empirical Formula (EF) = SNCl2

  15. Solution empirical formula mass (EFM) = 32.06 g/mol + 14.01 g/mol + (2)(35.45 g/mol) = 116.97 g/mol Molar Mass = 351 g/mol Multiplier = 351 g/mol = 3.00 116.97 g/mol  so MF is S3N3Cl6

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