1. Empirical and Molecular Formulas

2. Formaldehyde CH 2 O Acetic acid C 2 H 4 O 2 Gylceradehyde C 3 H 6 O 3 40% C; 6.7% H; 53.3% O

3. Empirical Formula Analysis of chemical compounds gives the % composition of each element. From this we can determine the Empirical Formula. The empirical formula shows the smallest whole number mole ratio of the atoms in a compound.

4. CH 2 O CH 3 OOCH = C 2 H 4 O 2 CH 3 O Empirical Formula. For ionic compounds, the formula unit is usually the compound’s empirical formula. For molecular compounds, the molecular formula and the empirical formula can be different. Molecular Formula Empirical Formula H2O2H2O2H2O2H2O2HO C 6 H 12 O 6 CH 2 O

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

6. Example:

7. Example: 2.Divide each mole by the smallest mole amount. 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 is 1 Ca: 2 Cl Empirical Formula = CaCl 2

8. Example 2: A compound is analyzed and found to contain 36.70% potassium, 33.27% chlorine, and 30.03% oxygen. What is the empirical formula of the compound? KClO 2

9. 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

10. 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

11. 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 ” to get the “factor”. 4.Multiply empirical formula by factor to get molecular formula.

12. Molecular Formula 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 Factor Empirical Formula Molecular Formula

13. 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.

14. Example: 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

15. Example 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

16. Example: 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

17. 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?

18. 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

19. 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

20. 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