1. 1 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) Micro World atoms & molecules Macro World grams

2. 2 The average atomic mass is the weighted average of all of the naturally occurring isotopes of the element.

3. Copyright © Cengage Learning. All rights reserved. 3 | 3 Molecular Mass The sum of the atomic masses of all the atoms in a molecule of the substance. Formula Mass The sum of the atomic masses of all atoms in a formula unit of the compound, whether molecular or not.

4. Copyright © Cengage Learning. All rights reserved. 3 | 4 Calculate the formula weight of the following compounds from their formulas. Report your answers to three significant figures. calcium hydroxide, Ca(OH) 2 methylamine, CH 3 NH 2

5. 2 decimal places 74.10 amu Copyright © Cengage Learning. All rights reserved. 3 | 5 2 significant figures 31.06 amu Total74.096 2 O2(16.00) =32.00 amu Ca(OH) 2 1 Ca1(40.08) =40.08 amu 2 H2(1.008) =2.016 amu CH 3 NH 2 1 C1(12.01) =12.01 amu 5 H5(1.008) =5.040 amu 1 N1(14.01) =14.01 amu Total31.060

6. 6 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 1 mol = N A = 6.0221367 x 10 23 Avogadro’s number (N A ) Dozen = 12 Pair = 2 The Mole (mol): A unit to count numbers of particles

7. 7 One Mole of: C S Cu Fe Hg

8. 8 Molar mass is the mass of 1 mole of in grams eggs shoes marbles atoms 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)

9. M = molar mass in g/mol N A = Avogadro’s number

10. 10 x 6.022 x 10 23 atoms K 1 mol K = 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 8.49 x 10 21 atoms K

11. Copyright © Cengage Learning. All rights reserved. 3 | 11 A sample of nitric acid, HNO 3, contains 0.253 mol HNO 3. How many grams is this? First, find the molar mass of HNO 3 : 1 H1(1.008) =1.008 1 N1(14.01) =14.01 3 O3(16.00) =48.00 63.018 (2 decimal places) 63.02 g/mol Note: We need one digit more in the molar mass than in the measured quantity.

12. Copyright © Cengage Learning. All rights reserved. 3 | 12 Next, using the molar mass, find the mass of 0.253 mole: = 15.94406 g

13. Copyright © Cengage Learning. All rights reserved. 3 | 13 Calcite is a mineral composed of calcium carbonate, CaCO 3. A sample of calcite composed of pure calcium carbonate weighs 23.6 g. How many moles of calcium carbonate is this? First, find the molar mass of CaCO 3 : 1 Ca1(40.08) =40.08 1 C1(12.01) =12.01 3 O3(16.00) =48.00 100.09 2 decimal places 100.09 g/mol

14. Copyright © Cengage Learning. All rights reserved. 3 | 14 Next, find the number of moles in 23.6 g:

15. Copyright © Cengage Learning. All rights reserved. 3 | 15 The average daily requirement of the essential amino acid leucine, C 6 H 14 O 2 N, is 2.2 g for an adult. What is the average daily requirement of leucine in moles? First, find the molar mass of leucine: 6 C6(12.01) =72.06 2 O2(16.00) =32.00 1 N1(14.01) =14.01 14 H14(1.008) =14.112 132.182 2 decimal places 132.18 g/mol

16. Copyright © Cengage Learning. All rights reserved. 3 | 16 Next, find the number of moles in 2.2 g:

17. Copyright © Cengage Learning. All rights reserved. 3 | 17 The daily requirement of chromium in the human diet is 1.0 × 10 -6 g. How many atoms of chromium does this represent?

18. Copyright © Cengage Learning. All rights reserved. 3 | 18 First, find the molar mass of Cr: 1 Cr1(51.996) =51.996 Now, convert 1.0 x 10 -6 grams to moles: =1.158166  10 16 atoms 1.2  10 16 atoms (2 significant figures)

19. 19 Percent composition of an element in a 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%

20. Copyright © Cengage Learning. All rights reserved. 3 | 20 Lead(II) chromate, PbCrO 4, is used as a paint pigment (chrome yellow). What is the percentage composition of lead(II) chromate? First, find the molar mass of PbCrO 4 : 1 Pb1(207.2) =207.2 1 Cr1(51.996) =51.996 4 O4(16.00) =64.00 323.196 (1 decimal place) 323.2 g/mol

21. Copyright © Cengage Learning. All rights reserved. 3 | 21 Now, convert each to percent composition: Check: 64.11 + 16.09 + 19.80 = 100.00

22. Copyright © Cengage Learning. All rights reserved. 3 | 22 The chemical name of table sugar is sucrose, C 12 H 22 O 11. How many grams of carbon are in 68.1 g of sucrose? First, find the molar mass of C 12 H 22 O 11 : 12 C12(12.01) =144.12 11 O11(16.00) =176.00 22 H22(1.008) =22.176 342.296 (2 decimal places) 342.30 g/mol

23. Copyright © Cengage Learning. All rights reserved. 3 | 23 Now, find the mass of carbon in 68.1 g sucrose:

24. 24 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

25. 25 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

26. Copyright © Cengage Learning. All rights reserved. 3 | 26 Percentage Composition The composition is determined by experiment, often by combustion. When a compound is burned, its component elements form oxides—for example, CO 2 and H 2 O. The CO 2 and H 2 O are captured and weighed to determine the amount of C and H in the original compound.

27. 27 g CO 2 mol CO 2 mol Cg C g H 2 O mol H 2 Omol Hg H g of O = g of sample – (g of C + g of H) Combust 11.5 g ethanol Collect 22.0 g CO 2 and 13.5 g H 2 O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H 4.0 g O = 0.25 mol O Empirical formula C 0.5 H 1.5 O 0.25 Divide by smallest subscript (0.25) Empirical formula C 2 H 6 O

28. Copyright © Cengage Learning. All rights reserved. 3 | 28 Empirical Formula (Simplest Formula) The formula of a substance written with the smallest integer subscripts. For example: The empirical formula for N 2 O 4 is NO 2. The empirical formula for H 2 O 2 is HO.

29. Copyright © Cengage Learning. All rights reserved. 3 | 29 Determining the Empirical Formula Beginning with percent composition: Assume exactly 100 g so percentages convert directly to grams. Convert grams to moles for each element. Manipulate the resulting mole ratios to obtain whole numbers.

30. Copyright © Cengage Learning. All rights reserved. 3 | 30 Manipulating the ratios: Divide each mole amount by the smallest mole amount. If the result is not a whole number: Multiply each mole amount by a factor to make whole numbers. For example: If the decimal portion is 0.5, multiply by 2. If the decimal portion is 0.33 or 0.67, multiply by 3. If the decimal portion is 0.25 or 0.75, multiply by 4.

31. Copyright © Cengage Learning. All rights reserved. 3 | 31 Benzene is composed of 92.3% carbon and 7.7% hydrogen. What is the empirical formula of benzene? Empirical formula: CH

32. Copyright © Cengage Learning. All rights reserved. 3 | 32 Molecular Formula A formula for a molecule in which the subscripts are whole-number multiples of the subscripts in the empirical formula.

33. Copyright © Cengage Learning. All rights reserved. 3 | 33 To determine the molecular formula: Compute the empirical formula weight. Find the ratio of the molecular weight to the empirical formula weight. Multiply each subscript of the empirical formula by n.

34. Copyright © Cengage Learning. All rights reserved. 3 | 34 Benzene has the empirical formula CH. Its molecular weight is 78.1 amu. What is its molecular formula? Molecular formula C 6 H 6

35. Copyright © Cengage Learning. All rights reserved. 3 | 35 Sodium pyrophosphate is used in detergent preparations. It is composed of 34.5% Na, 23.3% P, and 42.1% O. What is its empirical formula?

36. Copyright © Cengage Learning. All rights reserved. 3 | 36 Empirical formula Na 4 P 2 O 7  2 = 4  2 = 2  2 = 7

37. Copyright © Cengage Learning. All rights reserved. 3 | 37 Hexamethylene is one of the materials used to produce a type of nylon. It is composed of 62.1% C, 13.8% H, and 24.1% N. Its molecular weight is 116 amu. What is its molecular formula? Empirical formula C 3 H 8 N

38. Copyright © Cengage Learning. All rights reserved. 3 | 38 Stoichiometry The calculation of the quantities of reactants and products involved in a chemical reaction. Interpreting a Chemical Equation The coefficients of the balanced chemical equation may be interpreted in terms of either (1) numbers of molecules (or ions or formula units) or (2) numbers of moles, depending on your needs.

39. Copyright © Cengage Learning. All rights reserved. 3 | 39 1.Convert grams of A to moles of A  Using the molar mass of A 2.Convert moles of A to moles of B  Using the coefficients of the balanced chemical equation 3.Convert moles of B to grams of B  Using the molar mass of B To find the amount of B (one reactant or product) given the amount of A (another reactant or product):

40. 40 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

41. Copyright © Cengage Learning. All rights reserved. 3 | 41 Propane, C 3 H 8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation: C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) How many grams of CO 2 are produced when 20.0 g of propane is burned?

42. Copyright © Cengage Learning. All rights reserved. 3 | 42 Molar masses C 3 H 8 : 3(12.01) + 8(1.008) = 44.094 g CO 2 : 1(12.01) + 2(16.00) = 44.01 g 59.9 g CO 2 (3 significant figures)

43. Copyright © Cengage Learning. All rights reserved. 3 | 43 Molar masses: O 2 2(16.00) = 32.00 g C 3 H 8 3(12.01) + 8(1.008) = 44.094 g 72.6 g O 2 (3 significant figures) How many grams of O 2 are required to burn 20.0 g of propane?

44. 44 Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O 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

45. 45 Limiting Reagent: 2NO + O 2 2NO 2 NO is the limiting reagent O 2 is the excess reagent Reactant used up first in the reaction.

46. 46 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. g Almol Almol Fe 2 O 3 neededg Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al neededg Al needed 124 g Al 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al x 160. g Fe 2 O 3 1 mol Fe 2 O 3 x = 367 g Fe 2 O 3 Start with 124 g Alneed 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent

47. 47 Use limiting reagent (Al) to calculate amount of product that can be formed. g Almol Almol Al 2 O 3 g Al 2 O 3 124 g Al 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al x 102. g Al 2 O 3 1 mol Al 2 O 3 x = 234 g Al 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe At this point, all the Al is consumed and Fe 2 O 3 remains in excess.

48. Copyright © Cengage Learning. All rights reserved. 3 | 48 All amounts produced and reacted are determined by the limiting reactant. How can we determine the limiting reactant? 1.Use each given amount to calculate the amount of product produced. 2.The limiting reactant will produce the lesser or least amount of product.

49. Copyright © Cengage Learning. All rights reserved. 3 | 49 Magnesium metal is used to prepare zirconium metal, which is used to make the container for nuclear fuel (the nuclear fuel rods): ZrCl 4 (g) + 2Mg(s)  2MgCl 2 (s) + Zr(s) How many moles of zirconium metal can be produced from a reaction mixture containing 0.20 mol ZrCl 4 and 0.50 mol Mg?

50. Copyright © Cengage Learning. All rights reserved. 3 | 50 Since ZrCl 4 gives the lesser amount of Zr, ZrCl 4 is the limiting reactant. 0.20 mol Zr will be produced.

51. Copyright © Cengage Learning. All rights reserved. 3 | 51 Urea, CH 4 N 2 O, is used as a nitrogen fertilizer. It is manufactured from ammonia and carbon dioxide at high pressure and high temperature: 2NH 3 + CO 2 (g)  CH 4 N 2 O + H 2 O In a laboratory experiment, 10.0 g NH 3 and 10.0 g CO 2 were added to a reaction vessel. What is the maximum quantity (in grams) of urea that can be obtained? How many grams of the excess reactant are left at the end of the reactions?

52. Copyright © Cengage Learning. All rights reserved. 3 | 52 Molar masses NH 3 1(14.01) + 3(1.008) = 17.02 g CO 2 1(12.01) + 2(16.00) = 44.01 g CH 4 N 2 O1(12.01) + 4(1.008) + 2(14.01) + 1(16.00) = 60.06 g CO 2 is the limiting reactant since it gives the lesser amount of CH 4 N 2 O. 13.6 g CH 4 N 2 O will be produced.

53. Copyright © Cengage Learning. All rights reserved. 3 | 53 10.0at start −7.73 reacted 2.27 g remains 2.3 g NH 3 is left unreacted. (1 decimal place) To find the excess NH 3, we need to find how much NH 3 reacted. We use the limiting reactant as our starting point. Now subtract the amount reacted from the starting amount:

54. Copyright © Cengage Learning. All rights reserved. 3 | 54 Theoretical Yield Theorerical yield is the maximum amount of product that can be obtained by a reaction from given amounts of reactants. This is a calculated amount.

55. Copyright © Cengage Learning. All rights reserved. 3 | 55 Actual Yield The amount of product that is actually obtained. This is a measured amount. Percentage Yield

56. Copyright © Cengage Learning. All rights reserved. 3 | 56 2NH 3 + CO 2 (g)  CH 4 N 2 O + H 2 O When 10.0 g NH 3 and 10.0 g CO 2 are added to a reaction vessel, the limiting reactant is CO 2. The theoretical yield is 13.6 of urea. When this reaction was carried out, 9.3 g of urea was obtained. What is the percent yield?

57. Copyright © Cengage Learning. All rights reserved. 3 | 57 Theoretical yield = 13.6 g Actual yield = 9.3 g Percent yield = = 68% yield (2 significant figures)