1. Timberlake LecturePLUS1 Determining formulae The percentage composition of a compound leads directly to its empirical formula. Recall: An empirical formula for a compound is the formula of a substance written with the smallest integer subscripts. Eg. Consider hydrogen peroxide: Molecular formula = H 2 O 2 Empirical formula = HO

2. Timberlake LecturePLUS2 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. Timberlake LecturePLUS3 Compounds with different molecular formulae can have the same empirical formula, and such substances will have the same percentage composition. Eg. acetylene = C 2 H 2 benzene = C 6 H 6 both have the empirical formula = ?

4. Timberlake LecturePLUS4 Empirical Formulas Write your own one-sentence definition for each of the following: Empirical formula Molecular formula

5. Timberlake LecturePLUS5 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.

6. Timberlake LecturePLUS6 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

7. Timberlake LecturePLUS7 Solution EF-1 A. What is the empirical formula for C 4 H 8 ? 2) CH 2 B. What is the empirical formula for C 8 H 14 ? 1) C 4 H 7 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

8. Timberlake LecturePLUS8 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

9. Timberlake LecturePLUS9 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.

10. Timberlake LecturePLUS10 Empirical and Molecular Formulas molar mass = a whole number = n simplest mass n = 1 molar mass = empirical mass molecular formula = empirical formula n = 2 molar mass = 2 x empirical mass molecular formula = 2 x empirical formula molecular formula = or > empirical formula

11. Timberlake LecturePLUS11 Empirical Formula Empirical Mass Molecular Formula Molecular Mass

12. Timberlake LecturePLUS12 Empirical formula from Composition Consider the following flow-diagram: Percent composition Mass Composition Number of moles of each element Divide by smallest number of moles to find the molar ratios Multiply by appropriate number to get whole number subscripts

13. Timberlake LecturePLUS13 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

14. Timberlake LecturePLUS14 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/EF 176.0 g = 2.00 88.0

15. Timberlake LecturePLUS15 Learning Check EF-4 If there are 192.0 g of O in the molecular formula, what is the true formula if the EF is C 7 H 6 O 4 ? 1) C 7 H 6 O 4 2) C 14 H 12 O 8 3) C 21 H 18 O 12

16. Timberlake LecturePLUS16 Solution EF-4 If there are 192.0 g of O in the molecular formula, what is the true formula if the EF is C 7 H 6 O 4 ? 3) C 21 H 18 O 12 192 g O = 3 x O 4 or 3 x C 7 H 6 O 4 64.0 g O in EF

17. Timberlake LecturePLUS17 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 gC 24.27 g H 4.07 g

18. Timberlake LecturePLUS18 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

19. Timberlake LecturePLUS19 Why moles? Why do you need the number of moles of each element in the compound?

20. Timberlake LecturePLUS20 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 4. Write the simplest or empirical formula CH 2 Cl

21. Timberlake LecturePLUS21 5. EM (empirical mass) = 1(C) + 2(H) + 1(Cl) = 49.5 6. n = molar mass/empirical mass Molar mass = 99.0 g/mol = n = 2 E M 49.5 g/EM 7.Molecular formula (CH 2 Cl) 2 = C 2 H 4 Cl 2

22. Timberlake LecturePLUS22 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.

23. Timberlake LecturePLUS23 Solution EF-5 60.0 g C x ___________= ______ mol C 4.5 g H x ___________ = _______mol H 35.5 g O x ___________ = _______mol O

24. Timberlake LecturePLUS24 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

25. Timberlake LecturePLUS25 Divide by the smallest # of moles. 5.00 mol C = ________________ ______ mol O 4.5 mol H = ________________ ______ mol O 2.22 mol O = ________________ ______ mol O Are are the results whole numbers?_____

26. Timberlake LecturePLUS26 Divide by the smallest # of moles. 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?_____

27. Timberlake LecturePLUS27 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

28. Timberlake LecturePLUS28 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

29. Timberlake LecturePLUS29 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?

30. Timberlake LecturePLUS30 Solution EF 6 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/EF Mol. Mass/ Empirical mass 351/117.1 = 3 Molecular formula = S 3 N 3 Cl 6

31. Timberlake LecturePLUS31 STRUCTURAL FORMULA The atoms in a molecule are connected or chemically bonded in a precise way. A SF. Shows how the atoms in a molecule are arranged. For ex: H 2 O H-O-H C 2 H 6 CH 3 H-C- C- H H

32. Timberlake LecturePLUS32 Empirical formula The simplest whole number ratio of atoms of elements in a compound, described with the use of subscripts. Ionic compounds are always shown as empirical formulas. Molecular Formula The actual numbers of atoms in a molecule. Structural Formula Show the relative arrangements of atoms in a molecule

33. Timberlake LecturePLUS33

34. Timberlake LecturePLUS34 HYDRATES Solids which are found in combined form with water in definite proportion are called as HYDRATES. When hydrates are heated, H 2 O evaporates, and only solid is obtained in amorphous. (w/o a certain geometric structure, generally in powdered form. H 2 O molecules surround ionic substances with certain amounts.

35. Timberlake LecturePLUS35 WATER OF HYDRATION : Water molecules of a hydrate. Na 2 CO 3.10H 2 O Na 2 CO 3(s) + 10H 2 O (g) DEHYDRATION: Evaporation of water of hydration. Na 2 CO 3.10H 2 O, CaSO 4.2H 2 O, CuSO 4.5H 2 O

36. Timberlake LecturePLUS36