1. Dr. Laila Mohammed Al-Harbi Assistant professor Contact Info: Lalhrbi@kau.edu.sa Lalhrbi@kau.edu.sa Web Site: http://lalhrbi.kau.edu.sa

2.  3.1 atomic mass  3.2 Avogadro’s number and molar mass of an element  3.3 molecular mass  3.5 percent composition of compounds  3.6 experimental determination of empirical formula  3.9 limiting reagents  3.10 reaction yield  Homework p110: 3.5, 3.6, 3.7, 3.8p72: 2.26, 2.34, 2.36, 2.44, 2.46, 2.50 p110: 3.14, 3.16, 3.18, 3.20, 3.22p111: 3.24, 3.26, 3.28 p111:, 3.40, 3.42, 3.44, 3.46, 3.48, 3.50, 3.52p111: 3.44, 3.46, 3.48, 3.50, 3.52, 3.60 p114: 3.84, 3.86

3. By definition: 1 atom 12 C “weighs” 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00 amu  Atomic mass is the mass of an atom in atomic mass units (amu) Average atomic Mass: the average mass of all of the isotopes of an element, each one weighted by its proportionate abundance

4. Average atomic mass

5. Average atomic mass of Lithium Average atomic mass of carbon  Natural lithium is:  7.42% 6 Li (6.015 amu)  92.58% 7 Li (7.016 amu  Natural Carbon is:  1.1% 13 C (6.015 amu)  98.9% 12 C (7.016 amu (7.42% x 6.015) + (92.58% x 7.016) 100 = 6.941 amu (98.9 % x 12) + (1.18% x 13) 100 = 12.01 amu The average atomic mass is between the atomic masses of the isotopes And near the value of the highest abundance weighted average of the mixture of isotopes of carbon

6. EXAMPLE 3.1 PRACTIES EXERICISE 3.1  65 Cu (30.91percent) Atomic mass 64.9278  63 Cu (69.091percent) Atomic mass 62.93  10 B (19.78 percent) Atomic mass 10.0129  11 B (80.78percent) Atomic mass 11.0093 ( 30.91% x 64.9278) + 69.091% x 62.93) 100 = 63.55 amu ( 19.78 % x 10.0129) + 80.78% 11.0093) 100 =10.81amu

7.  The atomic masses of 6 Li and 7 Li are 6.0151amu and 7.0160 amu, respectively. Calculate the natural abundances of these two isotopes. The average atomic mass of Li is 6.941  The sum of natural abundances of isotopes should equal = 1  If the natural abundance of isotope 1 = x 1  then the natural abundance of isotope 2 = x 2  So x 1 + x 2 = 1 → x 2 = 1- x 1  Average atomic mass =(natural abundance × atomic mass) isotope 1 + (natural abundance × atomic mass) isotope 2  6.941 = 6.0151× x 1 + 7.0160 × (1- x 1 ) = 6.0151x 1 + 7.0160 - 7.0160 x 1  -7.0160 + 6.941 = (6.0151x 1 - 7.0160 x 1 ) = 0.1 = x  → x 1 = 7.5 5 → x 2 = 1- x 1  X 2 = 92.5

8.  Avogadro's Number Is the number of atoms in exactly 12 grams of carbon-12  NA = 6.022 x 10 23   The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C  One mole of a substance contains an Avogadro's Number of units

9. Molar mass is the mass of 1 mole of atoms and molecules in grams 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu 1 mole 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams) molar mass (grams) = 1 mole = = 6.022 x 10 23

10. 1 amu = 1.66 x 10 -24 g or 1 g = 6.022 x 10 23 amu 1 12 C atom 12.00 amu x 12.00 g 6.022 x 10 23 12 C atoms = 1.66 x 10 -24 g 1 amu

11. C S Cu Fe Hg C One mole of these substances contain = 6.022 x 10 23 atoms but is not equal because they have different molar masses

12. Atom № of molesAvogadro's Number Molar mass weight of one atom Carbon C 1 mole6.022 x 10 23 12.011.994 x 10 -23 g Sulfur S 1 mole6.022 x 10 23 32.075.325 x 10 -23 g Iron Fe 1 mole6.022 x 10 23 55.589.229 x 10 -23 g Copper Cu 1 mole6.022 x 10 23 63.551.055 x 10 -23 g Mercury Hg 1 mole6.022 x 10 23 200.63.331 x 10 -23 g

13. How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 10 23 atoms K 0.551 g K 1 mol K 39.10 g K x x 6.022 x 10 23 atoms K 1 mol K = = 8.49 x 10 21 atoms K

14. EXAMPLE 3.2 How many moles of He atoms are in 6.46 g of He ? 1 mol He = 4.003 g K x mol He = 6.46 g He =6.461 g He 1 mol He 4.003 g He x moles of He = 1.161 mol He

15. Conversions with Molar Mass and Avogadro’s Number Example a) What is the atomic mass of Fe in amu? 55.85 amu b ) What is the molar mass of Fe? 55.85 g c) How many moles of Fe atoms are in 35.6 g? 0.6 mol d) How many atoms of Fe are in 35.6 g? 3.86 x 10 23 e)What is the mass of 1 atom of Fe in units of g? 9.229 x 10 -23 g

16. Since one mol equal 65.39 g so logically 0.356 mol weight less

17. What you think??

18. Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO 2 1S32.07 amu 2O+ 2 x 16.00 amu SO 2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2

19. For any element atomic mass (amu) = molar mass (grams) molar mass (grams) = 1 mole = = 6.022 x 10 23 Note 12amu of Carbon = 12 g = 1mol = 6.022 x 10 23 atoms of carbon But 18 amu of H 2 O = 18 g = 1mole = = 6.022 x 10 23 molecules of H 2 O 18 amu of O = 18 g = 1mole = = 6.022 x 10 23 atoms of O 18 amu of H = 18 g = 2 mole = = 2× 6.022 x 10 23 atoms of H

20.  Calculate the molecular masses ( in amu) of the following compounds ?  Sulfur dioxide SO 2 = 32.07+2 (16) = 64.07 amu  Caffeine C 8 H 10 N 4 O 2 = 8(12.01)+ 10 (1.008) + 4(14.01)+ 2(16) = 194.20amu Practice exercise3.5 Calculate the molecular masses of methanol?  methanol C H 4 O = 1(12.01)+ 4 (1.008) + 1(16) = 32.4 amu

23. Percent composition of an element in compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound C2H6OC2H6O %C = 2 x (12.01 g) 46.07 g x 100% = 52.14%H = 6 x (1.008 g) 46.07 g x 100% = 13.13%O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0% 3.5 Percent composition of compounds

24. Example 3.8PRACTIES EXERICISE 3.8  Calculate the percent composition by mass of H, P, and O in H 3 PO 4 acid ?  Molar mass of H 3 PO 4  = 3(1.008)+ 1 (30.97) + 4(16) = 97.99 amu  Calculate the percent composition by mass of H, P, and O in H 2 SO 4 acid ?  Molar mass of H 2 SO 4  = 2(1.008)+ 1 (32.7) + 4(16) = 98.72 amu %H = 3(1.008) 97.99 x 100% = 3.0864% %P = 1(30.97) 97.99 x 100% = 31.61 % %O= 4(16) 97.99 x 100% = 65.31% %H = 2(1.008) 98.72 x 100% = 2.026 % %S = 1(32.07) 98.72 x 100% = 32.486 % %O= 4(16) 98.72 x 100% = 64.83%

25. Percent Composition and Empirical Formulas Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent. n K = 24.75 g K x = 0.6330 mol K 1 mol K 39.10 g K n Mn = 34.77 g Mn x = 0.6329 mol Mn 1 mol Mn 54.94 g Mn n O = 40.51 g O x = 2.532 mol O 1 mol O 16.00 g O n X = % of element x 1 mol element Molar mass of element

26. Percent Composition and Empirical Formulas K : ~ ~ 1.0 0.6330 0.6329 Mn : 0.6329 = 1.0 O : ~ ~ 4.0 2.532 0.6329 n K = 0.6330, n Mn = 0.6329, n O = 2.532 KMnO 4

27.  Ascorbic acid (vitamin C) cures scurvy. It is composed of 40.92 percent carbon C 4.58 percent hydrogen (H) and 54.5 percent (O) mass determine its empirical formula n C = 40.92 g C x =3.407mol C 1 mol C 12.01 g C n H = 4.58g H x = 4.54 mol H 1 mol H 1.008 g H n O = 54.5 g O x = 3.406 mol O 1 mol O 16 g O C : ~ ~ 1.0 3.408 3.406 H : 4.54 0.6329 = 1.33 O : = 1.0 3.406 We need to convert 1.33 into integral no, this can be done by trail-and error procedure 1.33 × 1 = 1.33 1.33 × 2 = 2.66 1.33 × 3 = 3.99 ≈ 4 C3H4O3C3H4O3 C = 1 × 3= 1.33 H =1.33 × 3 = 2 3.99 ≈ 4 O = 1 × 3 = 3

28.  Example 3-11  A sample of compound contains 1.25 g of nitrogen(N) and 3.47g of oxygen (O). The molar mass of this compound is between 90g and 95 g. Determine the molecular formula and the accurate molar mass of the compound?  n N = 1.52 g N x =0.108 mol N 1 mol N 14.01 g N n O = 3.47 g O x =0.217 mol O 1 mol O 16.00 g O N : =1.0 0.108 O : 0.217 0.108 = 2 NO 2 Molar mass of empirical formula = 14.01 + 2 x 16 = 46.01 g The ratio between molar mass and the molar mass of empirical formula = molar mass / empirical formula = 90 g / 46.01 g ≈ 2 N2O4N2O4 Molar mass = 2 14.01 + 4 16 = 92.02 g

29.  A sample of a compound containing born (B) and hydrogen (H) contains 6.444g of B and 1.803 g of (H). The molar mass of the compound is about 30g. What is its molecular formula? n B = 6.444 g B x =0.5961 mol B 1 mol B 10.81 g B n H = 1.803g H x =1.7888 mol H 1 mol H 1.008 g H B : =1.0 0.5961 H : 1.7888 0.5961 = 3 BH 3 Molar mass of empirical formula = 10.81 + 3 x 1.008 = 13.834g The ratio between molar mass and the molar mass of empirical formula = molar mass / empirical formula = 30 g / 13.834 g ≈ 2 B2H6B2H6

30.  A process in which one or more substances is changed into one or more new substances is a chemical reaction  A chemical equation uses chemical symbols to show what happens during a chemical reaction 3 ways of representing the reaction of H 2 with O 2 to form H 2 O reactantsproducts

31. 2 Mg + O 2 2 MgO 2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O 2 makes 80.6 g MgO 2 grams Mg + 1 gram O 2 makes 2 g MgO IS NOT

32.  Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.  Ethane reacts with oxygen to form carbon dioxide and water C 2 H 6 + O 2 CO 2 + H 2 O Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts 2C 2 H 6 NOT C 4 H 12

33.  Start by balancing those elements that appear in only one reactant and one product.

34.  Balance those elements that appear in two or more reactants or products.

35.  Check to make sure that you have the same number of each type of atom on both sides of the equation.

36. Example 3.12 PRACTIES EXERICISE 3.12 Al + O 2 → Al 2 O 3 2Al + O 2 → Al 2 O 3 2Al + 3/2 O 2 → Al 2 O 3 2(2Al + 3/2O 2 → Al 2 O 3 ) 4Al + 3O 2 → 2Al 2 O 3  Fe 2 O 3 + CO → Fe +CO 2  Fe 2 O 3 + CO → 2Fe +CO 2  Fe 2 O 3 + 1/3 CO → 2Fe+1/3CO 2  3(Fe 2 O 3 + 1/3 CO → 2Fe+1/3CO 2 )  Fe 2 O 3 + 3CO → 2Fe +3CO 2

37. 1.Write balanced chemical equation 2.Convert quantities of known substances into moles 3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4.Convert moles of sought quantity into desired units Amounts of Reactants and Products

38. Mass to mass conversions:  STEP 1: Write the balanced chemical equation.  STEP 2: Choose molar masses and mole ratios to convert known information into needed information.  STEP 3: Set up the factor- label expression, and calculate the answer.  STEP 4: Estimate or check the answer using a ballpark solution.

39.  The food we eat is degraded, or broken down, in our bodies to provide energy for growth and function. A general over all equation for this very complex process represents the degradation of glucose (C 6 H 12 O 6 ) to CO 2 and water If 856 g of C 6 H 12 O 6 is consumed by person over a certain period, what is the mass of CO 2 produced?  From the equation mole of C 6 H 12 O 6 → produce 6CO 2  From the equation 180.2g C 6 H 12 O 6 → 6× 44.01g CO 2  From the equation 856g C 6 H 12 O 6 → x CO 2  the mass of CO 2 produced = 856 × 6× 44.01g / 180.2 = 1254.35 g C 6 H 12 O 6 + 6O 2 → 6 H 2 O +6CO 2

40. Methanol burns in air according to the equation If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OHmoles CH 3 OHmoles H 2 Ograms H 2 O molar mass CH 3 OH coefficients chemical equation molar mass H 2 O 209 g CH 3 OH 1 mol CH 3 OH 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH x 18.0 g H 2 O 1 mol H 2 O x = = 235 g H 2 O 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O PRACTIES EXERICISE 3.13

41.  All alkali metals react with water to produce hydrogen gas and the corresponding alkali metal hydroxide. A typical reaction is that between lithium and water How many grams of Li are needed to produce 9.89g of H 2 ?  From the equation 2 mole of Li → produce mole of H 2  From the equation 2× 6.941 g Li → 2.016g H 2  From the equation x g Li → 9.89 g CO 2  the mass of CO 2 produced = 2× 6.941× 9.89 g / 2.016g = 68.1g Li Li (s) + 2 H 2 O (l) → 2 Li OH (aq) + H 2 (g)

42. 2NO (g) + O 2 (g) → 2 NO 2 (g) How many grams of O 2 are needed to produced 2.21g of NO 2 From the equation mole of O 2 → produce 2mole of NO 2 From the equation 32 g O 2 → 2 ×46.01 NO 2 From the equation x g O 2 → 2.21g NO 2 the mass of CO 2 produced = 32× 2.21g / 2 ×46.01 g = 0.769 g O 2

43. 3.9 Limiting Reagents

44. A molecular view of a Limiting reactant situation for the ammonia Synthesis. To make 4 molecules of NH 3 requires 2 molecules of N 2 and 6 molecules of H 2. If we start with 4 molecules of N 2 and 6 molecules of H 2, H 2 is the limiting reactant.

45. Limiting Reagent The number of bicycles that can be assembled is limited by whichever part runs out first. In the inventory shown in this figure, wheels are that part. The reactant that is completely consumed by the reaction

46.  10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl 2  2 AlCl 3  Start with Al:  Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl 3 133.5 g AlCl 3 27.0 g Al 2 mol Al 1 mol AlCl 3 = 49.4g AlCl 3 35.0g Cl 2 1 mol Cl 2 2 mol AlCl 3 133.5 g AlCl 3 71.0 g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9g AlCl 3 We found that chlorine is the limiting reactant, and 43.8 g of aluminum chloride are produced.

47. 10.0g of aluminum reacts with 35.0 grams of chlorine gas 2 Al + 3 Cl 2  2 AlCl 3  We found that chlorine is the limiting reactant, and 43.8 g of aluminum chloride are produced. 35.0 g Cl 2 1 mol Cl 2 2 mol Al 27.0 g Al 71 g Cl 2 3 mol Cl 2 1 mol Al = 8.8 g Al USED! 10.0 g Al – 8.8 g Al = 1.2 g Al EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

48.  Urea (NH 2 ) 2 CO is prepared by reacting ammonia with carbon dioxide 2NH 3 (g) + CO 2 (g) → (NH 2 ) 2 CO (aq) + H2O (ι)  In on process 637.2 g of NH 3 are treated with 1142 g of CO 2 a) which of the two limiting reagents? b) calculate the mass of (NH 2 ) 2 CO formed ? C) how much excess reagent ( in gram) is left at the end of the reaction  Start with NH 3 : 637.2 g NH 3 1 mol NH 3 mol (NH 2 ) 2 CO 60.06 g (NH 2 ) 2 CO 17.03 g NH 3 2 mol NH 3 1 mol (NH 2 ) 2 CO = 1124g (NH 2 ) 2 CO 1142 g CO 2 1 mol CO 2 1mol (NH 2 ) 2 CO 60.06 g g (NH 2 ) 2 CO 44.01.0 g CO 2 1mol CO 2 1 mol (NH 2 ) 2 CO = 1558.8g (NH 2 ) 2 CO We found that ammonia is the limiting reactant, because it produces a smaller amount of urea and 1124g of urea are produced. Now CO 2 : Example 3.15

49. PRACTIES EXERICISE 3.15 In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. Al is limiting reagent and 234g of Al 2 O 3 is produced Start with Al Start with Fe 2 O 3 601g Fe 2 O 3 1 mol Al mol Al 2 O 3 102 g Al 2 O 3 159.16 g Al mol Al 1 mol Al 2 O 3 124g Al 1 mol Al mol Al 2 O 3 102 g Al 2 O 3 27.0 g Al 2mol Al 1 mol Al 2 O 3 = 234g Al 2 O 3 = 358.5g Al 2 O 3

50. Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100 3.10 Reaction Yield Actual Yield is always lees.

51.  Titanium is strong light corrosion-resistant metal that is used in rockets aircraft, In one process, 3.54×10 7 g of TiCl 4 are reacted with 1.13×10 7 g of Mg TiCl 4 (g) + 2Mg (ι)→ Ti (s)+ 2MgCl 2 (ι) a) Calculate the theoretical yield of the Ti? b) calculate the percent yield if 7.91×10 6 g of Ti are obtained ? Start with TiCl 4 3.54×10 7 g TiCl 4 1 mol TiCl 4 mol Ti 102 g Ti 47.88 g TiCl 4 mol TiCl 4 1 mol Ti = 8.93×10 6 g Ti 1.13×10 7 Mg 1 mol Mg mol Ti 102 g Ti 47.88 g Mg mol Mg 1 mol Ti = 1.11×10 7 g Ti Start with Mg % Yield = Actual Yield Theoretical Yield x 100 % Yield = 7.91×10 6 g 8.93×10 6 x 100 = 88.4% Theoretical Yield

52.  In one process, 1.54×10 3 g of V 2 O 5 are reacted with 1.96×10 3 g of Ca V 2 O 5 + 5Ca → CaO + 2V a) Calculate the theoretical yield of the V? b) calculate the percent yield if 803 g of V are obtained ? Start with V 2 O 5 1.54×10 3 g V 2 O 5 mol V 2 O 5 2mol V 50.9 g V 181.8 g V 2 O 5 mol V 2 O 5 1 mol V = 996.9g V Start with Ca 803g Ca mol Ca 2mol V 50.9 g V 40.03 g Ca 5mol Ca 1 mol V = 862.33g V % Yield = Actual Yield Theoretical Yield x 100 = 803 g 862.33 x 100 = 93.11% Ca is limiting reagent and 862.33g of Vis produced(theoretical yield)

53.  1. What is the mass, in grams, of one copper atom?  A.1.055  10 -22 g  B.63.55 g  C.1 amu  D.1.66  10 -24 g  E.9.476  10 21 g  2.Determine the number of moles of aluminum in 96.7 g of Al.  A. 0.279 mol  B.3.58 mol  C.7.43 mol  D.4.21 mol  E.6.02  10 23 mol

54.  3.Which of the following samples contains the greatest number of atoms?  A. 100 g of Pb  B.2.0 mole of Ar  C.0.1 mole of Fe  D.5 g of He  E.20 million O 2 molecules  A) 100× 6.022 x 10 23 / 207.2 =  B) 2 × 6.022 x 10 23 =  C) 0.1 × 6.022 x 10 23 =  D) 5× 6.022 x 10 23 / 4.003  =  E) 2 × 10 7 × 6.022 x 10 23

55.  4.Formaldehyde has the formula CH 2 O. How many molecules are there in 0.11 g of formaldehyde? A. 6.1  10 -27  B.3.7  10 -3  C.4  D.2.2  10 21  E.6.6  10 22  5.How many sulfur atoms are present in 25.6 g of Al 2 (S 2 O 3 ) 3 ?  A. 0.393  B.6  C.3.95  10 22  D.7.90  10 22  E.2.37  10 23 

56. How many H atoms are in 72.5 g of C 3 H 8 O ? 1 mol C 3 H 8 O = (3 x 12) + (8 x 1) + 16 = 60 g 1 mol H = 6.022 x 10 23 atoms H = 5.82 x 10 24 atoms H 1 mol C 3 H 8 O molecules = 8 mol H atoms 72.5 g C 3 H 8 O 1 mol C 3 H 8 O 60 g C 3 H 8 O x 8 mol H atoms 1 mol C 3 H 8 O x 6.022 x 10 23 H atoms 1 mol H atoms x =

57. x 6.022 x 10 23 atoms K 1 mol K = 0.551 g K 1 mol K 39.10 g K x = 8.49 x 10 21 atoms K  How many atoms are in 0.551 g of potassium (K) ?  What is the formula mass of Ca 3 (PO 4 ) 2 ? 1 formula unit of Ca 3 (PO 4 ) 2 1 mol K = 39.10 g K 1 mol K = 6.022 x 10 23 atoms K 3 Ca 3 x 40.08 2 P2 x 30.97 8 O + 8 x 16.00 310.18 amu

58. A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H What is the empirical formula of the substance? Consider a sample size of 100 grams This will contain: 60.80 grams of Na, 28.60 grams of B, and 10.60 grams H Determine the number of moles of each Determine the simplest whole number ratio

59. A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance. require mole ratios so convert grams to moles moles of N = 2.34g of N = 0.167 moles of N 14.01 g/mole 14.01 g/mole moles of O = 5.34 g = 0.334 moles of O 16.00 g/mole 16.00 g/mole Formula: Formula: