2. EMPIRICAL FORMULA empirical formulaThe empirical formula represents the smallest ratio of atoms present in a compound. molecular formulaThe molecular formula gives the total number of atoms of each element present in one molecule of a compound. The empirical formula is the simplest formula and the molecular formula is the “true” formula. The empirical formula is the simplest formula and the molecular formula is the “true” formula.

3. Empirical Formula The empirical formula gives the ratio of the number of atoms of each element in a compound. Compound Molecular FormulaEmpirical Formula Hydrogen peroxideH 2 O 2 OH BenzeneC 6 H 6 CH EthyleneC 2 H 4 CH 2 PropaneC 3 H 8 C 3 H 8

4. Percentage Composition Glucose has the molecular formula C 6 H 12 O 6. What is its empirical formula, and what is the percentage composition of glucose? Empirical Formula = smallest whole number ratio CH 2 O

5. Percentage Composition CH 2 O Total mass = 12.01 + 2.02 + 16.00 = 30.03 %C = 12.01/30.03 x 100% = 39.99% %H = 2.02/30.03 x 100% = 6.73% %O = 16.00/30.03 x 100% = 53.28%

6. Empirical Formula A compound’s empirical formula can be determined from its percent composition. A compound’s molecular formula is determined from the molar mass and empirical formula.

7. Example: A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. 1- What is the empirical formula for the compound? 2- What is the molecular formula if the Molecular mass is 58.12 g/mole?

8. Answer: Step1:Assume 100 g of sample, then 82.67 g are C, and 17.33 g are H. Step 2:Convert masses to moles: 82.67 g C x mole/12.011 g = 6.88 moles C 17.33 g H x mole/1.008 g = 17.19 mole H Step 3:Find relative # of moles (divide by smallest number) i.e. Convert moles to ratios: 6.88/6.88 = 1 C 17.19/6.88 = 2.50 H Or 2 carbons for every 5 hydrogens Then, the empirical formula is: C 2 H 5

9. Step 4: Formula weight is: 29.06 g/mole If the molecular weight is known to be 58.12 g/mole Then the molecular formula must be: C 4 H 10

10. 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? (MW= 60) 1) CH 2 O2) C 2 H 4 O 2 3) C 3 H 6 O 3

11. 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

12. Empirical and Molecular Formulas molar mass = a whole number = n simplest mass When: n = 1 Then, molar mass = empirical mass molecular formula = empirical formula When: n = 2 molar mass = 2 x empirical mass molecular formula = 2 x empirical formula *Always … molecular formula=or> empirical formula

13. 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. 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. 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. 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. 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. 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. Why moles? Why do you need the number of moles of each element in the compound?

20. 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. 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. 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. 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. 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. 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 the results whole numbers?_____

26. 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 the results whole numbers?_____

27. 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. 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. 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. 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. Ascorbic acid (vitamin C) cures scurvy and may help prevent the common cold. It is composed of 40.92% C, 4.58% H, and 54.50% O. Determine the empirical formula of vitamin C.?

32. MOLARITY

33. Molarity (M): Is the number of moles of solute dissolved in one liter of solution. To make a 0.5-molar (0.5M) solution: a)Add 0.5 mol of solute to a 1-L volumetric flask half filled with distilled water. b)Swirl the flask carefully to dissolve the solute. c)Fill the flask with water exactly to the 1-L mark. 16.2

34. Units of Molarity 2.0 M HCl = 2.0 moles HCl 1 L HCl solution 6.0 M HCl= 6.0 moles HCl 1 L HCl solution

35. 2. Calculate the molarity of a solution prepared by mixing 23.0 g of NaCl in 500.0 mL of water. 1. Calculate the moles necessary to make 50.0 mL of a 3.0 M solution of hydrochloric acid. 3. How many grams of sodium sulfate will be required to make 150.0 mL of a 0.25 M solution? To calculate the molarity of a solution, divide the moles of solute by the volume of the solution.

36. 16.2

37. Molarity Calculation Q: If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution? Soln: 1) 4.0 g NaOH x 1 mole NaOH = 0.10 mole NaOH 40.0 g NaOH 2) 500. mL x 1 L _ = 0.500 L 1000 mL 3. 0.10 mole NaOH = 0.20 mole NaOH 0.500 L 1 L = 0.20 M NaOH

38. Learning Check M1 A KOH solution with a volume of 400 mL contains 2 mole KOH. What is the molarity of the solution? 1) 8 M 2) 5 M 3) 2 M

39. Solution M1 A KOH solution with a volume of 400 mL contains 2 moles of KOH. What is the molarity of the solution? 2) 5 M M = 2 mole KOH = 5 M 0.4 L

40. Learning Check M2 Stomach acid is a 0.10 M HCl solution. How many moles of HCl are in 1500 mL of stomach acid solution? 1) 15 moles HCl 2) 1.5 moles HCl 3) 0.15 moles HCl

41. Solution M2 3) 1500 mL x 1 L = 1.5 L 1000 mL 1.5 L x 0.10 mole HCl = 0.15 mole HCl 1 L (Molarity factor)

42. Learning Check M3 How many grams of KCl are present in 2.5 L of 0.50 M KCl? 1) 1.3 g 2) 5.0 g 3) 93 g

43. Solution M3 3) 2.5 L x 0.50 mole x 74.6 g KCl = 93 g KCl 1 L 1 mole KCl

44. –problem: –problem: the concentration of NaCl in blood serum is approximately 0.14 M. What volume of serum contains 2.0 g of NaCl? –first find the number of moles NaCl in 2.0 g NaCl –next find the volume in liters that contains this many moles of NaCl Learning Check M4

45. MOLARITY & DILUTION M 1 V 1 = M 2 V 2 The act of diluting a solution is to simply add more water (or solvent) thus leaving the amount of solute unchanged.

46. Making Dilutions The total number of moles of solute remains unchanged upon dilution, so you can write this equation. M 1 and V 1 are the molarity and volume of the initial solution, and M 2 and V 2 are the molarity and volume of the diluted solution.

47. –Q: How do you prepare 200 mL of 3.5 M aqueous solution of acetic acid if you have a bottle of 6.0 M acetic acid –Soln: First find the number of L of 6.0 M acetic acid needed –to prepare the desired solution, put 120 mL of 6.0 M acetic in a 200 mL volumetric flask and fill to the mark Learning Check dilutions

48. 1- How would you prepare a 2 L of 3M HCl from a 12 M solution? (0.5 L diluted to 2 L) 2- How would you prepare 300 mL of a 0.6 N solution of H 2 SO 4 from a 36 N solution? (5 mL diluted to 300 mL)

49. إن الله وملائكتــه يصلــون على النبي يــأيــها الذيـن آمنوا صلــوا عليه وسلمـــوا تسليمـــا