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Empirical and Molecular Formulas: Calculations and Examples

Learn how to calculate empirical and molecular formulas using mass ratios and molar masses. Practice problems included.

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Empirical and Molecular Formulas: Calculations and Examples

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  1. BELLWORK 2/04/19 • What is the mass of 2 moles of MgCl2?

  2. ANSWER • Find the formula mass of the ionic compound first. • Mg = 24.3amu • 2 Cl = 35.5 x 2 =71.0 amu • Formula Mass = Molar Mass • Molar Mass = 71.0 + 24.3 = 95.3 g/mol • Multiply Molar Mass x 2 = 190.6g

  3. Types of Formulas • The molecular formula is the true or actual number of the atoms in a molecule. • The empirical formula is the simplest whole number ratio of the atoms. • The empirical formula is calculated by dividing the subscripts in the molecular formula by a whole number to give the lowest ratio.C5H10O5  5 = C1H2O1 = CH2Omolecular empirical formulaformula

  4. Some Molecular and Empirical Formulas

  5. 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. Which is a possible molecular formula for CH2O? 1) C4H4O4 2) C2H4O2 3) C3H6O3

  6. Solution A. What is the empirical formula for C4H8? 2) CH2C4H8  4 B. What is the empirical formula for C8H14? 1) C4H7C8H14 2 C. Which is a possible molecular formula for CH2O? 2) C2H4O2 3) C3H6O3

  7. Learning Check If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 1) SN 2) SN4 3) S4N4

  8. Solution If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 3) S4N4 If the molecular formula has 4 atoms of N, and S and N are related 1:1, then there must also be 4 atoms of S.

  9. Relating Empirical and Molecular Formulas • A molecular formula is equal to or a multiple of the empirical formula. • Thus, the molar mass is equal to or a multiple of the empirical mass. molar mass = a whole number empirical mass • Multiply the empirical formula by the whole numberto determine the molecular formula.

  10. Finding the Molecular Formula Determine the molecular formula of a compound that has a molar mass of 78.0g/mol and an empirical formula of CH. 1. Empirical mass of CH = 13.0 g/mol 2. Divide the molar mass by the empirical mass. 3. 78.0 g/mol = 6.00 13.0 g/mol 4. Multiply the subscripts in CH by 6. 5. Molecular formula = (CH)6 = C6H6

  11. Learning Check A compound has a formula mass of 176.0 g/mol and an empirical formula of C3H4O3. What is the molecular formula?

  12. Solution A compound has a formula mass of 176.0g/mol and an empirical formula of C3H4O3. What is the molecular formula? C3H4O3 = 3(12.0) + 4(1.0) + 3(16.0) = 88.0 g/mol 176.0 g/mol (molar mass) = 2.00 88.0 g/mol (empirical mass) Molecular formula = 2 (empirical formula) (C3H4O3 )2 = C6H8O6

  13. Finding the Molecular Formula A compound is C 24.27%, H 4.07%, and Cl 71.65%. The molar mass is known to be 99.0 g. What are the empirical and molecular formulas? 1. Write the mass percents as the grams in a 100.00-g sample of the compound. C 24.27 g H 4.07 g Cl 71.65 g

  14. Finding the Molecular FormulaContinued 2. Calculate the number of moles of each element. 24.27 g C x 1 mole C = 2.02 moles C 12.0 g C 4.07 g H x 1 mole H = 4.03 moles H 1.01 g H 71.65 g Cl x 1 mole Cl = 2.02 moles Cl 35.5 g Cl

  15. Finding the Molecular Formula(continued) 3. Divide each by the smallest 2.02 moles C = 1 mole C 2.024.03 moles H = 2 moles H 2.022.02 moles Cl = 1 mole Cl 2.02 Empirical formula = C1H2Cl1 = CH2Cl

  16. Finding the Molecular Formula(continued) 4. Calculate empirical mass (EM)empirical mass CH2Cl = 49.5 g/mol 5. Divide molar mass by empirical mass Molar mass = 99.0 g/mol = 2Empirical mass 49.5 g/mol 6. Determine Molecular formula(CH2Cl)2 = C2H4Cl2

  17. Learning Check Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula.

  18. Solution (continued) In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O. 60.0 g C x 1 mole C = 5.00 moles C 12.0 g C 4.5 g H x 1 mole H= 4.5 moles H 1.01 g H 35.5 g O x 1mole O= 2.22 moles O 16.0 g O

  19. Solution (continued) Divide by the smallest number of moles. 5.00 moles C = 2.25 moles C 2.22 4.5 moles H = 2.0 moles H 2.22 2.22 moles O = 1.00 mole O 2.22 Note that the results are not all whole numbers. To obtain whole numbers, multiply by a factor to give whole numbers.

  20. Solution (continued) Multiply each number of moles by 4 C: 2.25 moles C x 4 = 9 moles C H: 2.0 moles H x 4 = 8 moles H O: 1.00 mole O x 4 = 4 moles O Use the whole numbers as subscripts to obtain the simplest formula C9H8O4

  21. 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, what is the molecular formula?

  22. Solution In 100.0 g, there are 27.4 g S, 12.0 g N and 60.6 g Cl. 27.4 g S x 1 mole S =0.854 mole S 32.1 g S 12.0 g N x 1 mole N= 0.857moles N 14.0 g N 60.6 g Cl x 1mole Cl = 1.71 moles Cl 35.5 g Cl

  23. Solution (continued) Dividing by the smallest number of moles 0.854 mole S /0.854 = 1.00 mole S 0.857 mole N/0.854 = 1.00 mole N 1.71 moles Cl/0.854 = 2.00 moles Cl Empirical formula = SNCl2 = 117.1 g/mol Molar Mass/ Empirical mass 351 = 3117.1 Molecular formula = (SNCl2)3 = S3N3Cl6

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