1. Empirical Formulas

2. Types of Formulas The formulas for compounds can be expressed as an empirical formula and as a molecular(true) formula. Empirical Molecular (true)Name CHC 2 H 2 acetylene CHC 6 H 6 benzene CO 2 CO 2 carbon dioxide CH 2 OC 5 H 10 O 5 ribose

3. Empirical Formulas Write your own one-sentence definition for each of the following: Empirical formula Molecular formula

4. An empirical formula represents the simplest whole number ratio of the atoms in a compound. The molecular formula is the true or actual ratio of the atoms in a compound.

5. Learning Check EF-1 A. What is the empirical formula for C 4 H 8 ? 1) C 2 H 4 2) CH 2 3) CH B. What is the empirical formula for C 8 H 14 ? 1) C 4 H 7 2) C 6 H 12 3) C 8 H 14 C. What is a molecular formula for CH 2 O? 1) CH 2 O2) C 2 H 4 O 2 3) C 3 H 6 O 3

6. Solution EF-1 A. What is the empirical formula of C 4 H 8 ? 2) CH 2 B. What is the empirical formula of C 8 H 14 ? 1) C 4 H 7 C. What is a molecular formula of CH 2 O? 1) CH 2 O 2) C 2 H 4 O 2 3) C 3 H 6 O 3

7. Learning Check EF-2 If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 1) SN 2) SN 4 3) S 4 N 4

8. Solution EF-2 If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 3) S 4 N 4 If the actual formula has 4 atoms of N, and S is related 1:1, then there must also be 4 atoms of S.

9. Empirical and Molecular Formulas mm molecular formula = Multipling = n mm empirical formula empirical formula x multipling factor = molecular formula

10. Learning Check EF-3 A compound has a formula mass of 176.0 and an empirical formula of C 3 H 4 O 3. What is the molecular formula? 1) C 3 H 4 O 3 2) C 6 H 8 O 6 3) C 9 H 12 O 9

11. Solution EF-3 A compound has a formula mass of 176.0 and an empirical formula of C 3 H 4 O 3. What is the molecular formula? 2)C 6 H 8 O 6 C 3 H 4 O 3 = 88.0 g/mol 176.0 g = 2.00 88.0

12. Finding the Molecular Formula A compound is Cl 71.65%, C 24.27%, and H 4.07%. What are the empirical and molecular formulas? The molar mass is known to be 99.0 g/mol. 1. State mass percents as grams in a 100.00-g sample of the compound. Cl 71.65 g C 24.27 g H 4.07 g

13. 2. Calculate the number of moles of each element. 71.65 g Cl x 1 mol Cl = 2.02 mol Cl 35.5 g Cl 24.27 g C x 1 mol C = 2.02 mol C 12.0 g C 4.07 g H x 1 mol H = 4.04 mol H 1.01 g H

14. Why moles? Why do you need the number of moles of each element in the compound? The subscripts in the chemical formula will equal the mole to mole ratio of the elements present.

15. 3. Find the smallest whole number ratio by dividing each mole value by the smallest mole values: Cl: 2.02 = 1 Cl 2.02 C: 2.02 = 1 C 2.02 H: 4.04 = 2 H 2.02

16. 4. Write the simplest or empirical formula = CH 2 Cl 5. Determine molecular formula a. Find formula mass of emp form. 1(C) + 2(H) + 1(Cl) = 1(12.0) + 2(1.0) + 1(35.5) = 49.5g/mol b. Determine mult factor 99.0 g/mol 49.5 g/mol CH 2 Cl x 2 = C 2 H 4 Cl 2 = Mult factor = 2

17. Learning Check EF-5 Aspirin is 60.0% C, 4.5 % H and 35.5 O. Calculate its simplest formula. In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O.

18. Solution EF-5 60.0 g C x ______ = ______ mol C 4.5 g H x _______ = ______ mol H 35.5 g O x ______ = ______ mol O

19. Solution EF-5 60.0 g C x 1 mol C = 5.00 mol C 12.0 g C 4.5 g H x 1 mol H = 4.5 mol H 1.01 g H 35.5 g O x 1mol O= 2.22 mol O 16.0 g O

20. 5.00 mol C= 2.25 2.22 mol O 4.5 mol H = 2.00 2.22 mol O 2.22 mol O = 1.00 2.22 mol O Are are the results whole numbers?_____ Divide by the smallest # of moles.

21. Finding Subscripts A fraction between 0.1 and 0.9 must not be rounded. Multiply all results by an integer to give whole numbers for subscripts. (1/2) 0.5 x 2 = 1 (1/3)0.333 x 3 = 1 (1/4)0.25 x 4 = 1 (3/4)0.75 x 4 = 3

22. Multiply everything x 4 C: 2.25 mol C x 4 = 9 mol C H: 2.0 mol Hx 4 = 8 mol H O: 1.00 mol O x 4 = 4 mol O Use the whole numbers of mols as the subscripts in the simplest formula C 9 H 8 O 4

23. Learning Check EF-6 A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g/mol, what is the molecular formula?

24. Change % to g and determine moles of each element 27.4 g S x 1 mol C = 5.00 mol C 32.1 g C 12.0 g N x 1 mol H = 4.5 mol H 14.0 g H 60.6 g Cl x 1mol O= 2.22 mol O 35.5 g O

25. Divide by smaller # of moles 0.853 mol S /0.853 = 1 S 0.857 mol N /0.853 = 1 N 1.71 mol Cl /0.853 = 2 Cl Empirical formula = SNCl 2 =117.1 g/mol

26. Determine Molecular Formula Mol. Mass 351 = 3 Empirical mass 117.1 3 x SNCl 2 = Molecular formula Molecular formula = S 3 N 3 Cl 6