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Chapter 8

Chapter 8. Empirical & Molecular Formulas. Types of Formulas. The molecular formula of a compound indicates the actual number of each atom present in a compound.

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Chapter 8

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  1. Chapter 8 Empirical & Molecular Formulas

  2. Types of Formulas • The molecular formula of a compound indicates the actual number of each atom present in a compound. • Anempirical formula of a compound indicates the smallest whole number ratio of the elements of each atom present in a compound.

  3. Empirical and Molecular Formulas

  4. Chemical Formulas of Compounds • Chemical formulas give the relative numbers of atoms or moles of each element in a compound which is always a whole number ratio. • Example CO2 2 atoms of O for every 1 atom of C or 2 moles of O for every 1 mole of C • If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.

  5. To Calculate an Empirical Formula 1. Determine the mass in grams of each element present, if necessary. 2. Calculate the number of moles of each element. 3. Divide each answer from step 2 by the smallest number of moles to obtain the simplest whole number ratio. • If whole numbers are not obtained* in step 3), multiply through by the smallest number that will give all whole numbers *Be careful! Do not round off numbers prematurely

  6. A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance.

  7. A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance. 2.34g N 5.34g O

  8. A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance. 2.34g N mol = 0.167 mol N 14.0g 5.34g O mol = 0.334 mol O 16.0g

  9. A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance. 2.34g N mol = 0.167 mol N 14.0g N0.167O0.334 5.34g O mol = 0.334 mol O 16.0g

  10. A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance. 2.34g N mol = 0.167 mol N 14.0g N0.167O0.334 5.34g O mol = 0.334 mol O 16.0g 0.167 mol N 0.334 mol O = 1 mol N = 2 mol O 0.167 0.167

  11. A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance. 2.34g N mol = 0.167 mol N 14.0g N0.167O0.334 5.34g O mol = 0.334 mol O 16.0g 0.167 mol N 0.334 mol O = 1 mol N = 2 mol O 0.167 0.167 NO2

  12. To Calculate a Molecular Formula 1. Determine the empirical formula and calculate its mass. 2. Divide the molecular mass by the mass of the empirical formula you determined in step 1. Your answer should be rounded to the nearest whole number. 3. Multiply the empirical formula by the whole number answer found in step 2.

  13. A colorless liquid, used in rocket engines has a molar mass of 92.0 g/mol. What is the molecular formulaof the compound if the empirical formula is NO2?

  14. A colorless liquid, used in rocket engines has a molar mass of 92.0 g/mol. What is the molecular formulaof the compound if the empirical formula is NO2? NO2 = 14.0 + 2(16.0) = 46.0 g/mol

  15. A colorless liquid, used in rocket engines has a molar mass of 92.0 g/mol. What is the molecular formulaof the compound if the empirical formula is NO2? NO2 = 14.0 + 2(16.0) = 46.0 g/mol 92.0 g/mol = 2 46.0 g/mol

  16. A colorless liquid, used in rocket engines has a molar mass of 92.0 g/mol. What is the molecular formulaof the compound if the empirical formula is NO2? NO2 = 14.0 + 2(16.0) = 46.0 g/mol 92.0 g/mol = 2 46.0 g/mol 2 x NO2 = N2O4

  17. What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S?

  18. What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S? 53.73g Fe 46.27g S

  19. What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S? 53.73g Fe mol 46.27g S mol = 0.963 mol Fe = 1.44 mol S 55.8g 32.1g

  20. What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S? 53.73g Fe mol 46.27g S mol = 0.963 mol Fe = 1.44 mol S 55.8g 32.1g 0.963 mol Fe 1.44 mol S = 1 Fe = 1.5 S 0.963 mol 0.963 mol

  21. What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S? 53.73g Fe mol 46.27g S mol = 0.963 mol Fe = 1.44 mol S 55.8g 32.1g 0.963 mol Fe 1.44 mol S = 1 Fe = 1.5 S 0.963 mol 0.963 mol FeS1.5 x 2 = Fe2S3

  22. What is the molecular formula of a substance with the empirical formula Fe2S3 and a gram formula mass of 624 g/mol?

  23. What is the molecular formula of a substance with the empirical formula Fe2S3 and a gram formula mass of 624 g/mol? Fe6S9

  24. What is the molecular formula of a substance with the empirical formula Fe2S3 and a gram formula molecular mass of 624 g/mol? 2Fe + 3S = 2(55.8) + 3(32.1) = 207.9 g/mol 624 g/mol 207.9 g/mol 3 x Fe2S3 = Fe6S9 ≈ 3

  25. What is the empirical formula for Ibuprofen if the percent composition is 75.7% C, 8.80% H, and 15.5% O?

  26. What is the empirical formula for Ibuprofen if the percent composition is 75.7% C, 8.80% H, and 15.5% O? 75.7g C 8.80g H 15.5g O

  27. What is the empirical formula for Ibuprofen if the percent composition is 75.7% C, 8.80% H, and 15.5% O? 75.7g C mol C = 6.31 mol C 12.0g C 8.80g H mol H = 8.80 mol H 1.0 g H 15.5g O mol O = 0.969 mol O 16.0g O

  28. What is the empirical formula for Ibuprofen if the percent composition is 75.7% C, 8.80% H, and 15.5% O? 75.7g C mol C 6.31 mol C = 6.31 mol C = 6.5 C 12.0g C 0.969 mol 8.80g H mol H 8.80 mol H = 9 H = 8.80 mol H 1.0 g H 0.969 mol 15.5g O mol O 0.969 mol H = 1 O = 0.969 mol O 16.0g O 0.969 mol

  29. What is the empirical formula for Ibuprofen if the percent composition is 75.7% C, 8.80% H, and 15.5% O? 75.7g C mol C 6.31 mol C = 6.31 mol C = 6.5 C 12.0g C 0.969 mol 8.80g H mol H 8.80 mol H = 9 H = 8.80 mol H 1.0 g H 0.969 mol 15.5g O mol O 0.969 mol H = 1 O = 0.969 mol O 16.0g O 0.969 mol C6.5H9O x 2 = C13H18O2

  30. What is the molecular formula for Ibuprofen if the molar mass is 206 g/mole? C13H18O2 13(12.0) + 18(1.0) + 2(16.0) = 206g/mol Therefore the molecular formula is also C13H18O2

  31. Homework • Percentage Composition Worksheet (due tomorrow). • Lab Summary: Determination of the Empirical Formula of a Compound (due tomorrow). • Empirical & Molecular Formulas Worksheet • Due in two days

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