2. Empirical Formula Is the formula with the smallest whole-number mole ratio of the elements in a compound.

3. Examples: 1) Glucose – Molecular formula C 6 H 12 O 6. Empirical formula

4. Examples: 1) Glucose – Molecular formula C 6 H 12 O 6. Empirical formula CH 2 O.

5. Examples: 1) Glucose – Molecular formula C 6 H 12 O 6. Empirical formula CH 2 O. 2)Succinic acid– Molecular formula C 4 H 6 O 4. Empirical formula

6. Examples: 1) Glucose – Molecular formula C 6 H 12 O 6. Empirical formula CH 2 O. 2)Succinic acid– Molecular formula C 4 H 6 O 4. Empirical formula C 2 H 3 O 2.

7. Examples: 1) Glucose – Molecular formula C 6 H 12 O 6. Empirical formula CH 2 O. 2)Succinic acid– Molecular formula C 4 H 6 O 4. Empirical formula C 2 H 3 O 2. 3) Sulfuric acid -- Molecular formula H 2 SO 4. Empirical formula

8. Examples: 1) Glucose – Molecular formula C 6 H 12 O 6. Empirical formula CH 2 O. 2)Succinic acid– Molecular formula C 4 H 6 O 4. Empirical formula C 2 H 3 O 2. 3) Sulfuric acid -- Molecular formula H 2 SO 4. Empirical formula H 2 SO 4.

9. 8 Practice: Determine the Empirical Formula of Benzopyrene, C 20 H 12. Empirical formula (C 5 H 3 ) 4 = Find the greatest common factor (GCF) of the subscripts: 20 factors = 12 factors = 201,2,4,5,10, 1,24,3,6,12 GCF =4 Divide each subscript by the GCF to find the empirical formula: C 20/4 H 12/4 = C 5 H 3 C 20 H 12

10. 1) What is the empirical formula for a compound which contains 0.0134 g of iron, 0.00769 g of sulfur and 0.0115 g of oxygen? Fe = 0.0134 g x S = O = 0.00769 g x 0.0115 g x = 0.00072 mol (O) = 0.00024 mol (Fe) = 0.00024 mol (S) 1 st – Determine the # of moles of each element:

11. FeSO 3 Fe = S = O = = 3 mol (O) = 1 mol (Fe) = 1 mol (S) Empirical Formula 2 nd – Calculate the simplest ratio (dividing each value by the smallest one):

12. 2) Determine the empirical formula for a compound that contains 10.89 % magnesium, 31.77 % chlorine, and 57,34 % oxygen?

13. 1 st – Determine the # of moles of each element: (Mg = 10.89 % ; Cl = 31.77 % & O = 57.34 %) Mg = 10.89 g x Cl = O = 31.77 g x 57.34 g x = 3.59 mol (O) = 0.448 mol (Mg) = 0.896 mol (Cl)

14. 2 nd – Calculate the simplest ratio (dividing each value by the smallest one): (Mg = 0.448 ; Cl = 0.896 & O = 3.59) = 8 mol (O) = 1.000 mol (Mg) = 2.000 mol (Cl) = 8.013 mol (O) = 1 mol (Mg) = 2 mol (Cl) Empirical Formula MgCl 2 O 8

15. 3) A laboratory analysis of aspirin determined the following mass percent composition carbon (60.00 %), hydrogen (4.48 %) and oxygen (35.53 %). Calculate its empirical formula. = 4.996 mol (C) = 2.221 mol (O) = 4.440 mol (H) O = 35.53 g x C = 60.00 g x H = 4.48 g x

16. 3) A laboratory analysis of aspirin determined the following mass percent composition carbon (60.00 %), hydrogen (4.48 %) and oxygen (35.53 %). Calculate its empirical formula. C = 4.996 mol H = 4.44 mol O = 2.221 mol Dividing by the smallest one:

17. 3) A laboratory analysis of aspirin determined the following mass percent composition carbon (60.00 %), hydrogen (4.48 %) and oxygen (35.53 %). Calculate its empirical formula. C = 4.996 mol H = 4.44 mol O = 2.221 mol C = 4.996 ÷ 2.221 = 2.249 H = 4.44 ÷ 2.221 = 2.000 O = 2.221 ÷ 2.221 = 1.000 Dividing by the smallest one:

18. 3) A laboratory analysis of aspirin determined the following mass percent composition carbon (60.00 %), hydrogen (4.48 %) and oxygen (35.53 %). Calculate its empirical formula. C = 4.996 mol H = 4.44 mol O = 2.221 mol C = 4.996 ÷ 2.221 = 2.249 H = 4.44 ÷ 2.221 = 2.00 multiply by 4 O = 2.221 ÷ 2.221 = 1.000 Dividing by the smallest one:

19. 3) A laboratory analysis of aspirin determined the following mass percent composition carbon (60.00 %), hydrogen (4.48 %) and oxygen (35.53 %). Calculate its empirical formula. C = 4.996 ÷ 2.221 = 2.249 x 4 = 9 H = 4.44 ÷ 2.221 = 2.00 x 4 = 8 O = 2.221 ÷ 2.221 = 1.000 x 4 = 4 C9H8O4C9H8O4 (Empirical Formula)