1. Empirical and Molecular Formulas

2. CH 2 O CH 3 OOCH = C 2 H 4 O 2 CH 3 O Empirical Formula A formula that gives the simplest whole-number ratio of the atoms of each element in a compound. Molecular Formula Empirical Formula H2O2H2O2H2O2H2O2HO C 6 H 12 O 6 CH 2 O

3. Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca. Steps 1.Find mole amounts. 2.Divide each mole by the smallest mole.

4. 1.Find mole amounts. 2.128 g Cl x 1 mol Cl = 0.0600 mol Cl 35.45 g Cl 35.45 g Cl 1.203 g Ca x 1 mol Ca = 0.0300 mol Ca 40.08 g Ca

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

6. A compound weighing 298.12 g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Hint “ Percent to mass Mass to mole Divide by small Multiply ‘ til whole ”

7. A compound weighing 298.12 g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Percent to mass: Mg – (72.2%/100)*298.12 g = 215.24 g N – (27.8%/100)*298.12 g = 82.88 g Mass to mole: Mg – 215.24 g * ( 1 mole ) = 8.86 mole 24.3 g N – 82.88 g * ( 1 mole ) = 5.92 mole 14.01 g Divide by small: Mg - 8.86 mole/5.92 mole = 1.50 N - 5.92 mole/5.92 mole = 1.00 mole Multiply ‘ til whole: Mg – 1.50 x 2 = 3.00 N – 1.00 x 2 = 2.00 Mg 3 N 2

8. 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 Mass (EFM). 3.Divide the molar mass by the “ EFM ”. 4.Multiply empirical formula by factor. Find the molecular formula for a compound whose molar mass is ~124.06 and empirical formula is CH 2 O 3. 2. “ EFM ” = 62.03 g 3.124.06/62.03 = 2 4.2(CH 2 O 3 ) = C 2 H 4 O 6

9. Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol. Steps 1.Find the empirical formula. 2.Calculate the Empirical Formula Mass. 3.Divide the molar mass by the “ EFM ”. 4.Multiply empirical formula by factor.

10. Empirical formula. A.Find mole amounts. 4.90 g N x 1 mol N = 0.350 mol N 14.01 g N 11.2 g O x 1 mol O = 0.700 mol O 16.00 g O

11. B.Divide each mole by the smallest mole. B.Divide each mole by the smallest mole. N = 0.350 = 1.00 mol N 0.350 O = 0.700 = 2.00 mol O 0.350 Empirical Formula = NO 2 Empirical Formula Mass = 46.01 g/mol

12. Molecular formula Molar Mass = 92.0 g/mol = 2.00 Emp. Formula Mass46.01 g/mol Molecular Formula = 2 x Emp. Formula = N 2 O 4

13. A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula?

14. g C – (48.38/100)*528.39 g = 255.64 g g H – (8.12/100)*528.39 g = 42.91 g g O – (43.5/100)*528.39 g = 229.85 g mole C - 255.64 g * ( 1 mole ) = 21.29 mol 12.01 g mole H – 42.91 g * ( 1 mole ) = 42.49 mol 1.01 g mole O – 229.85 g * ( 1 mole ) = 14.37 mol 16.00 g

15. A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula? From last slide: 21.29 mol C, 42.49 mol H, 14.27 mol O C – 21.29/14.27 = 1.49 H – 42.49/14.27 = 2.98 (esentially 3) O – 14.27/14.27 = 1.00 C – 1.49 x 2 = 3 H – 3 x 2 = 6 O – 1 x 2 = 2 C3H6O2C3H6O2

16. A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula? From last slide: Empirical formula = C 3 H 6 O 2 “ EFM ” = 74.09 Molar mass = 222.24 = ~3 EFM 74.09 3(C 3 H 6 O 2 ) = C 9 H 18 O 6