1. General Chemistry 101 CHEM د. عبد العزيز بن علي الغامدي 1434-1435 هـ

2. References Chemistry the General Science 11E T. L. Brown, H. E. LeMay, B. E. Bursten and C. J. Murphy. الكيمياء العامة. د.أحمد العويس، د.سليمان الخويطر، د.عبدالعزيز الواصل، د.عبدالعزيز السحيباني.

3. الاختبار الفصلي الأول : لم يحدد بعد. الاختبار الفصلي الثاني : لم يحدد بعد. المكتب 2 أ 134 و سيكون قريبا 2 أ 146 الأيميل aalghamdia@ksu.edu.sa

4. MATTER We define matter as anything that has mass and takes up space.

5. Matter Atoms are the building blocks of matter. Each element is made of the same kind of atom. A compound is made of two or more different kinds of elements.

6. States of Matter

7. Units of Measurement SI Units Système International d’Unités A different base unit is used for each quantity.

8. Metric System Prefixes convert the base units into units that are appropriate for the item being measured.

9. Volume The most commonly used metric units for volume are the liter (L) and the milliliter (mL). – A liter is a cube 1 dm long on each side. – A milliliter is a cube 1 cm long on each side.

10. Density Density is a physical property of a substance. d = mVmV Density = mass Volume

11. Chemical Equations Chemical equations are concise representations of chemical reactions. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

12. Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) Reactants appear on the left side of the equation.

13. Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) Products appear on the right side of the equation.

14. Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) The states of the reactants and products are written in parentheses to the right of each compound.

15. Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) Coefficients are inserted to balance the equation.

16. Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule.

17. Subscripts and Coefficients Give Different Information Coefficients tell the number of molecules.

18. Examples of balancing chemical equation HCl + Zn ZnCl 2 + H 2 C 3 H 8 + O 2 CO 2 + H 2 O Zn + HNO 3 Zn(NO 3 ) 2 + H 2 2 5 3 4 2

19. Reaction Types Combination & decomposition reactions Combustion in Air

20. Combination Reactions In this type of reaction two or more substances react to form one product. Examples: – 2 Mg (s) + O 2 (g)  2 MgO (s) – N 2 (g) + 3 H 2 (g)  2 NH 3 (g) – C 3 H 6 (g) + Br 2 (l)  C 3 H 6 Br 2 (l)

21. Decomposition Reactions In a decomposition one substance breaks down into two or more substances. Examples: – CaCO 3 (s)  CaO (s) + CO 2 (g) – 2 KClO 3 (s)  2 KCl (s) + O 2 (g) – 2 NaN 3 (s)  2 Na (s) + 3 N 2 (g)

22. Combustion Reactions These are generally rapid reactions that produce a flame. Most often involve hydrocarbons reacting with oxygen in the air. Examples: – CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g) – C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g)

23. Formula Weights(FW) A formula weight is the sum of the atomic weights for the atoms in a chemical formula. So, the formula weight of calcium chloride, CaCl 2, would be Ca: 1(40.1 amu*) + Cl: 2(35.5 amu) 111.1 amu Formula weights are generally reported for ionic compounds. * atomic mass unit

24. Molecular Weight (MW) A molecular weight is the sum of the atomic weights of the atoms in a molecule. For the molecule ethane, C 2 H 6, the molecular weight would be C: 2(12.0 amu) + H: 6(1.0 amu) 30.0 amu

25. Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: % element = (number of atoms)(atomic weight) (FW of the compound) x 100

26. Percent Composition So the percentage of carbon in ethane is… %C = (2)(12.0 amu) (30.0 amu) 24.0 amu 30.0 amu = x 100 = 80.0%

27. Avogadro’s Number 6.02 x 10 23 1 mole of 12 C has a mass of 12 g. 1 mole of H 2 O has a mass of 18g.

28. Molar Mass By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol). – The molar mass of an element is the mass number for the element that we find on the periodic table. – The formula weight (in amu’s) will be the same number as the molar mass (in g/mol).

29. Using Moles Moles provide a bridge from the molecular scale to the real-world scale.

30. Mole Relationships One mole of atoms, ions, or molecules contains Avogadro’s number of those particles. One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.

31. Examples 1- Calculate how many atoms there are in 0.200 moles of copper. The number of atoms in one mole of Cu is equal to the Avogadro number = 6.02 x 10 23. Number of atoms in 0.200 moles of Cu = (0.200 mol) x (6.02x10 23 mol -1 ) = 1.20 x 10 23. 2- Calculate how molecules of H 2 O there are in 12.10 moles of water. Number of water molecules = (12.10 mol)x(6.02 x 10 23 ) = 7.287 x 10 24

32. 3- Calculate the number of moles of glucose (C 6 H 12 O 6 ) in 5.380 g of C 6 H 12 O 6. Moles of C 6 H 12 O 6 = = 0.02989 mol. 4- Calculate the mass, in grams, of 0.433 mol of Ca(NO 3 ) 2. Mass = o.433 mol x 164.1 g/mol = 71.1 g. 5.380 g 180.0 gmol -1

33. 5- How many glucose molecules are in 5.23 g of C 6 H 12 O 6 ? How many oxygen atoms are in this sample? Molecules of C 6 H 12 O 6 = x (6.02x10 23 ) = 1.75 x 10 22 molecules Atoms O = 1.75 x 10 22 x 6 = 1.05 x 10 23 5.23 g 180.0 gmol -1

34. Finding Empirical Formulas

35. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition.

36. Calculating Empirical Formulas The compound para-aminobenzoic acid is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

37. Calculating Empirical Formulas Assuming 100.00 g of para-aminobenzoic acid, C:61.31 g x = 5.105 mol C H: 5.14 g x= 5.09 mol H N:10.21 g x= 0.7288 mol N O:23.33 g x = 1.456 mol O 1 mol 12.01 g 1 mol 14.01 g 1 mol 1.01 g 1 mol 16.00 g

38. Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C: = 7.005  7 H: = 6.984  7 N:= 1.000 O:= 2.001  2 5.105 mol 0.7288 mol 5.09 mol 0.7288 mol 1.458 mol 0.7288 mol

39. Calculating Empirical Formulas These are the subscripts for the empirical formula: C 7 H 7 NO 2

40. Combustion Analysis Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this. – C is determined from the mass of CO 2 produced. – H is determined from the mass of H 2 O produced. – O is determined by difference after the C and H have been determined.

41. Determining Empirical Formula by Combustion Analysis *Combustion of 0.255 g of isopropyl alcohol produces 0.561 g of CO 2 and 0.306 g of H 2 O. Determine the empirical formula of isopropyl alcohol. Grams of C = x 1 x 12 gmol-1 = 0.153 g 0.561 g CO 2 44 gmol -1 Grams of H = x 2 x 1 gmol-1 = 0.0343 g 0.306 g H 2 O 18 gmol -1 Grams of O = mass of sample – (mass of C + mass of H) = 0.255 g – ( 0.153 g + 0.0343 g) = 0.068 g

42. Moles of C = = 0.0128 mol 0.153 g C 12 gmol -1 Moles of H = = 0.0343 mol Moles of O = = 0.0043 mol 0.0343 g H 1 gmol -1 0.068 g O 16 gmol -1 Calculate the mole ratio by dividing by the smallest number of moles: C:H:O 2.98:7.91:1.00 C3H8OC3H8O

43. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products.

44. Stoichiometric Calculations Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).

45. Stoichiometric Calculations C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O Starting with 1.00 g of C 6 H 12 O 6 … we calculate the moles of C 6 H 12 O 6 … use the coefficients to find the moles of H 2 O… and then turn the moles of water to grams.

46. Limiting Reactants The reactant that is completely consumed in a reaction is called the limiting reactant (limiting reagent). – In other words, it’s the reactant that run out first (in this case, the H 2 ).

47. In example below, O 2 would be the excess reactant (excess reagent). Limiting Reactants

48. Theoretical Yield The theoretical yield is the maximum amount of product that can be made. The amount of product actually obtained in a reaction is called the actual yield.

49. Theoretical Yield The actual yield is almost always less than the theoretical yield. Why? Part of the reactants may not react. Side reaction. Difficult recovery.

50. Percent Yield The percent yield of a reaction relates to the actual yield to the theoretical (calculated) yield. Actual Yield Theoretical Yield Percent Yield =x 100

51. Examples Fe 2 O 3 (s) + 3 CO(g) 2 Fe(s) + 3 CO 2 (g) If we start with 150 g of Fe 2 O 3 as the limiting reactant, and found actual yield of Fe was 87.9 g, what is the percent yield? The percent yield = x 100 Actual Yield Theoretical Yield

52. Theoretical yield = 105 g. The percent yield = x 100 = 83.7 % Fe 2 O 3 (s) + 3 CO(g) 2 Fe(s) + 3 CO 2 (g) 150 g Fe 2 O 3 150 g 159 g mol- 1 0.943 mol Fe 2 O 3 2 mol Fe 1 mol Fe 2 O 3 X 1.887 mol Fe 1.887 mol x 55.85 g mol -1 105 g Fe 87.9 g 105 g

53. Solutions Solutions are defined as homogeneous mixtures of two or more pure substances. The solvent is present in greatest abundance. All other substances are solutes.

54. Molarity Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different. Molarity is one way to measure the concentration of a solution. moles of solute volume of solution in liters Molarity (M) =

55. Mixing a Solution To create a solution of a known molarity, one weighs out a known mass (and, therefore, number of moles) of the solute. The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask.

56. Examples Calculate the molarity of a solution made by dissolving 0.750 g of sodium sulfate (Na 2 SO 4 ) in enough water to form 850 mL of solution. Moles of Na 2 SO 4 = = 0.0053 mol Molarity = = 0.0062 M 0.750 g 142 g mol -1 0.0053 mol 0.850 L

57. How many moles of KMnO 4 are present in 250 mL of a 0.0475 M solution? M = Moles of KMnO 4 = 0.0475 x 0.25 L = 0.012 mol mol L mol L

58. How many milliliters of 11.6 M HCl solution are needed to obtain 0.250 mol of HCl? moles of solute volume of solution in liters Molarity (M) = volume in liters = 0.022 L = 22 mL. 11.6 M = 0.250 mol volume in liters

59. What are the molar concentrations of each of the ions present in a 0.025 M aqueous solution of Ca(NO 3 ) 2 ? Ca 2+ = 0.025 M NO 3 - = 0.025 x 2 = 0.05 M

60. Dilution One can also dilute a more concentrated solution by – Using a pipet to deliver a volume of the solution to a new volumetric flask, and – Adding solvent to the line on the neck of the new flask.

61. Dilution The molarity of the new solution can be determined from the equation M c  V c = M d  V d where M c and M d are the molarity of the concentrated and dilute solutions, respectively, and V c and V d are the volumes of the two solutions.

62. Dilution M c  V c = M d  V d Moles solute before dilution = moles solute after dilution

63. Examples How many milliliters of 3.0 M H 2 SO 4 are needed to make 450 mL of 0.10 M H 2 SO 4 ? M c  V c = M d  VV 3.0 M x V c = 0.10 M x 450 mL V c = 15 mL

64. Ways of Expressing Concentrations of Solutions mass of A in solution total mass of solution  100 Mass % of A = Mass Percentage, ppm, and ppb  Mass Percentage Example: 36% HCl by mass contains 36 g of HCl for each 100 g of solution (64 g H 2 O)

65.  Parts per Million (ppm) ppm =  Parts per Billion (ppb) mass of A in solution total mass of solution  10 6 ppb = mass of A in solution total mass of solution  10 9

66. Examples Calculate the mass percentage of Na 2 SO 4 in a solution containing 10.6 g Na 2 SO 4 in 483 g water. = 2.15 % Mass % of Na 2 SO 4 = 10.6 g (483 + 10.6) g  100

67. An ore contains 2.86 g of silver per ton of ore. What is the concentration of silver in ppm? = 2.86 ppm ppm = 2.86 g 10 6 g  10 6

68. Mole Fraction, Molarity, and Molality  Mole Fraction (X)  Molarity (M) moles of A total moles in solution X A = moles of solute volume of solution in liters Molarity (M) =

69.  Molality (m) moles of solute Kilograms of solvent m = Since both moles and mass do not change with temperature, molality (unlike molarity) is not temperature-dependent. Since volume is temperature-dependent, molarity can change with temperature.

70. Examples An aqueous solution of hydrochloric acid contains 36 % HCl by mass. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution. Moles HCl = = 0.99 mol Moles H 2 O = = 3.6 mol 36 g 36.5 g mol -1 64 g 18 g mol -1

71. X HCl = = moles HCl moles H 2 O + moles HCl 0.99 mol HCl 0.064 kg H 2 O 0.99 3.6 + 0.99 = 0.22 Molality of HCl = = 15.5 m