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Section 6.2 Page 268 Empirical and Molecular Formulas

Section 6.2 Page 268 Empirical and Molecular Formulas. Empirical vs. Molecular Formula. Empirical Formula (E f ) A formula that gives the simplest whole-number ratio of the atoms of each element in a compound. Empirical vs. Molecular Formula. Empirical Formula (E f )

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Section 6.2 Page 268 Empirical and Molecular Formulas

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  1. Section 6.2 Page 268Empirical and Molecular Formulas

  2. Empirical vs. Molecular Formula Empirical Formula (Ef) A formula that gives the simplest whole-number ratio of the atoms of each element in a compound.

  3. Empirical vs. Molecular Formula Empirical Formula (Ef) A formula that gives the simplest whole-number ratio of the atoms of each element in a compound. Molecular Formula (Mf) A formula that gives the actual number of atoms of each element in a compound.

  4. CH3O CH3O CH3OOCH = C2H4O2 CH2O

  5. Steps to determining the empirical formula. Step 1. Find mole amounts.

  6. Steps to determining the empirical formula. Step 1. Find mole amounts. Step 2. Divide each mole amount by the smallest mole amount.

  7. Example 1: Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca.

  8. 1. Find mole amounts.

  9. 1. Find mole amounts. nCl = 2.128 g = 0.0600 mol Cl 35.45 g/mol

  10. 1. Find mole amounts. nCl = 2.128 g = 0.0600 mol Cl 35.45 g/mol nCa = 1.203 g = 0.0300 mol Ca 40.08 g/mol

  11. 1. Find mole amounts. nCl = 2.128 g = 0.0600 mol Cl 35.45 g/mol nCa = 1.203 g = 0.0300 mol Ca 40.08 g/mol

  12. 2. Divide each mole by the smallest mole.

  13. 2. Divide each mole by the smallest mole. Cl = 0.0600 mol Cl = 2.00 mol Cl 0.0300 Ca = 0.0300 mol Ca = 1.00 mol Ca 0.0300

  14. 2. Divide each mole by the smallest mole. Cl = 0.0600 mol Cl = 2.00 mol Cl 0.0300 Ca = 0.0300 mol Ca = 1.00 mol Ca 0.0300 Ratio – 1 Ca : 2 Cl Empirical Formula = CaCl2

  15. Example 2:A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula?

  16. Example 2:A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? • Always assume that you have a 100g sample. • That way you can convert percent directly to grams... • In this case you would have 50.81g of Zn, 16.04g of P, and 33.15g of O.

  17. Rhyme “Percent to mass Mass to mole Divide by small Multiply ‘til whole”

  18. A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula?

  19. A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula?

  20. A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula?

  21. A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula?

  22. A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula?

  23. A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula?

  24. A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula?

  25. A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula?

  26. A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula?

  27. A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula?

  28. A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula?

  29. A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? Therefore empirical formula would be Zn3P2O8

  30. Example 3:The percentage composition of a compound is 48.63% carbon, 21.59% oxygen, 18.90% nitrogen, and the rest hydrogen. Find the empirical formula for the compound.

  31. Molecular Formula The molecular formula gives the actual number of atoms of each element in a molecular compound.

  32. Molecular Formula The molecular formula gives the actual number of atoms of each element in a molecular compound. Steps 1. Find the empirical formula. 2. Calculate the Empirical Formula Molar Mass. (MEf) 3. Divide the Molecular Formula Molar Mass (MMf) by the “MEf”. 4. Multiply empirical formula by factor.

  33. Molecular Formula The molecular formula gives the actual number of atoms of each element in a molecular compound. Steps 1. Find the empirical formula. 2. Calculate the Empirical Formula Molar Mass. (MEf) 3. Divide the Molecular Formula Molar Mass (MMf) by the “MEf”. 4. Multiply empirical formula by factor. Note: The molecular formula molar mass (MMf) is always given in the question.

  34. Example 1: Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH2O3.

  35. Example 1: Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH2O3. Step 2. “MEf” = 62.03 g

  36. Example 1: Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH2O3. Step 2. “MEf” = 62.03 g Step 3. MMf/MEf = 124.06/62.03 = 2

  37. Example 1: Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH2O3. Step 2. “MEf” = 62.03 g Step 3. MMf/MEf = 124.06/62.03 = 2 Step 4. 2(CH2O3) = C2H4O6

  38. Example 1: Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH2O3. Step 2. “MEf” = 62.03 g Step 3. MMf/MEf = 124.06/62.03 = 2 Step 4. 2(CH2O3) = C2H4O6  Therefore the Molecular formula (Mf) is C2H4O6

  39. Example 2:A compound containing carbon, hydrogen, iodine and oxygen is found to be 18.0% carbon, 2.5% hydrogen, 63.5% iodine, and 16.0% O. The molar mass of this compound is known to be 400 g/mol. What is its molecular formula? • Find the empirical formula. • Then you can find the molecular formula.

  40. Mass Spectrometry Demo!

  41. Poison Identification. Acetone: C3H6O Hydrogen cyanide: HCN Ethylene Glycol: C2H6O2 Isopropanol: C3H8O Methanol: CH4O Sodium hypochlorite: NaOCl Acetaminophen: C8H9NO2

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