1. Mass Relationships in Chemical Reactions Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Chapter 3 (80-107) 3.1 Atomic mass. 3.2 Avogadro’s number and the molar mass of an element. 3.3 Molecular mass. 3.5 Percent composition of compounds. 3.6 Experimental determination of empirical and molecular formulas. 3.7 Chemical reactions. and chemical equations. 3.8 Amounts of reactants and products. 3.9 Limiting reagents. 3.10 Reaction yield.

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) One atomic mass unit is defined as a mass exactly equal to one – twelfth the mass of one carbon-12 atom. Micro World atoms & molecules Macro World grams 3.1

4. Natural lithium is: 7.42% 6 Li (6.015 amu) 92.58% 7 Li (7.016 amu) 7.42 x 6.015 + 92.58 x 7.016 100 = 6.941 amu 3.1 Average atomic mass of lithium:

7. 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 isotope 3.2 1 mol = N A = 6.0221367 x 10 23 Avogadro’s number (N A )

8. Molar mass M 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) 3.2

9. One Mole of: C S Cu Fe Hg 3.2

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

11. x 6.022 x 10 23 atoms K 1 mol K = Do You Understand Molar Mass? 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 3.2

14. Worked Example 3.4

15. 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 3.3

17. Worked Example 3.6

18. Worked Example 3.7

19. Do You Understand Molecular Mass? 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 C 3 H 8 O 1 mol H = 6.022 x 10 23 atoms H 5.82 x 10 24 atoms H 3.3 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 =

20. Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. 1Na22.99 amu 1Cl + 35.45 amu NaCl 58.44 amu For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl 3.3 NaCl

21. Do You Understand Formula Mass? What is the formula mass of Ca 3 (PO 4 ) 2 ? 3.3 1 formula unit of Ca 3 (PO 4 ) 2 3 Ca 3 x 40.08 2 P2 x 30.97 8 O + 8 x 16.00 310.18 amu

22. Percent composition of an element in a compound = is the percent by mass of each 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% 3.5

24. 2.What information would you need to calculate the average atomic mass of an element? A)The number of neutrons in the element. B)The atomic number of the element. C)The mass and abundance of each isotope of the element. D)The position in the periodic table of the element. 3.The atomic masses of Cl (75.53 percent) and Cl (24.47 percent) are 34.968 amu and 36.956 amu, respectively. Calculate the average atomic mass of chlorine. The percentages in parentheses denote the relative abundances. A)35.96 amu B)35.45 amu C)36.47 amu D)71.92 amu 6.How many amu are there in 8.4 g? A)8.4 x10 23 amu B)1.4 x 10 -23 amu C)8.4 amu D)5.1 x 10 24 amu 9.How many atoms are there in 5.10 moles of sulfur (S)? A)3.07 x 10 24 B)9.59 x 10 22 C)6.02 x 10 23 D)9.82 x 10 25

25. 14.What is the mass in grams of a single atom of As? A)1.244 x 10 -22 g B)2.217 x 10 -26 g C)8.039 x 10 21 g D)4.510 x10 25 g 16.How many atoms are present in 3.14 g of copper (Cu)? A)2.98 x 10 22 B)1.92 x 10 23 C)1.89 x 10 24 D)6.02 x 10 23 19.Calculate the molar mass of Li 2 CO 3. A)73.89 g B)66.95 g C)41.89 g D)96.02 g 22.How many molecules of ethane (C 2 H 6 ) are present in 0.334 g of C 2 H 6 ? A)2.01 x 10 23 B)6.69 x 10 21 C)4.96 x 10 22 D)8.89 x 10 20

26. 33.Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4 percent; H: 6.21 percent; S: 39.5 percent; O: 9.86 percent. What is its molecular formula given that its molar mass is about 162 g? A)C 12 H 20 S 4 O 2 B)C 7 H 14 SO C)C 6 H 10 S 2 O D)C 5 H 12 S 2 O 2 35.The formula for rust can be represented by Fe 2 O 3. How many moles of Fe are present in 24.6 g of the compound? A)2.13 mol B)0.456 mol C)0.154 mol D)0.308 mol

27. KE = 1/2 x m x v 2 v = (2 x KE/m) 1/2 F = q x v x B 3.4 Light Heavy

28. 3.5 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

29. 3.5 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

30. Worked Example 3.9

31. 3.6 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

32. Worked Example 3.11

33. 3.7 3 ways of representing the reaction of H 2 with O 2 to form H 2 O 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 reactantsproducts

34. How to “Read” Chemical Equations 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 IS NOT 2 grams Mg + 1 gram O 2 makes 2 g MgO 3.7

35. Balancing Chemical Equations 1.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 2.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. 3.7 2C 2 H 6 NOT C 4 H 12

36. Balancing Chemical Equations 3.Start by balancing those elements that appear in only one reactant and one product. C 2 H 6 + O 2 CO 2 + H 2 O 3.7 start with C or H but not O 2 carbon on left 1 carbon on right multiply CO 2 by 2 C 2 H 6 + O 2 2CO 2 + H 2 O 6 hydrogen on left 2 hydrogen on right multiply H 2 O by 3 C 2 H 6 + O 2 2CO 2 + 3H 2 O

37. Balancing Chemical Equations 4.Balance those elements that appear in two or more reactants or products. 3.7 2 oxygen on left 4 oxygen (2x2) C 2 H 6 + O 2 2CO 2 + 3H 2 O + 3 oxygen (3x1) multiply O 2 by 7 2 = 7 oxygen on right C 2 H 6 + O 2 2CO 2 + 3H 2 O 7 2 remove fraction multiply both sides by 2 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O

38. Balancing Chemical Equations 5.Check to make sure that you have the same number of each type of atom on both sides of the equation. 3.7 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O ReactantsProducts 4 C 12 H 14 O 4 C 12 H 14 O 4 C (2 x 2)4 C 12 H (2 x 6)12 H (6 x 2) 14 O (7 x 2)14 O (4 x 2 + 6)

39. Worked Example 3.12

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 3.8

41. 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 3.8

42. Worked Example 3.13a

43. Worked Example 3.13b

45. Limiting Reagents is the reactant used up first In a reaction Excess reagents are the Reactants present in quantities Greater than necessary to react With the present quantity of The limiting reagent. 3.9 2NO + 2O 2 2NO 2 NO is the limiting reagent O 2 is the excess reagent

46. Do You Understand Limiting Reagents? 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 3.9

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 3.9

48. Worked Example 3.15a

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 The percent Yield is the proportion of the actually yield to the theoretical yield which can be obtained from the following relation:

53. 36.How many grams of sulfur (S) are needed to react completely with 246 g of mercury (Hg) to form HgS? A)39.3 g B)24.6 g C)9.66  10 3 g D)201 g 38.Tin(II) fluoride (SnF 2 ) is often added to toothpaste as an ingredient to prevent tooth decay. What is the mass of F in grams in 24.6 g of the compound? A)18.6 g B)24.3 g C)5.97 g D)75.7 g 39.What is the empirical formula of the compound with the following composition? 2.1 percent H, 65.3 percent O, 32.6 percent S. A)H 2 SO 4 B)H 2 SO 3 C)H2S2O3H2S2O3 D)HSO 3 42.Which of the following equations is balanced? A)2C + O 2 CO B)2CO + O 2 2CO 2 C)H 2 + Br 2 HBr D)2K + H 2 O 2KOH + H 2

54. 48.Consider the combustion of carbon monoxide (CO) in oxygen gas: 2CO(g) + O 2 (g)  2CO 2 (g) Starting with 3.60 moles of CO, calculate the number of moles of CO 2 produced if there is enough oxygen gas to react with all of the CO. A)7.20 mol B)44.0 mol C)3.60 mol D)1.80 mol 57.Nitrous oxide (N 2 O) is also called “laughing gas.” It can be prepared by the thermal decomposition of ammonium nitrate (NH 4 NO 3 ). The other product is H 2 O. The balanced equation for this reaction is: NH 4 NO 3  N 2 O + 2H 2 O How many grams of N 2 O are formed if 0.46 mole of NH 4 NO 3 is used in the reaction? A)2.0 g B)3.7  10 1 g C)2.0  10 1 g D)4.6  10 -1 g 58.The fertilizer ammonium sulfate [(NH 4 ) 2 SO 4 ] is prepared by the reaction between ammonia (NH 3 ) and sulfuric acid: 2NH 3 (g) + H 2 SO 4 (aq)  (NH 4 ) 2 SO 4 (aq) How many kilograms of NH 3 are needed to produce 1.00  10 5 kg of (NH 4 ) 2 SO 4 ? A)1.70  10 4 kg B)3.22  10 3 kg C)2.58  10 4 kg D)7.42  10 4 kg

55. 60.Nitric oxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO 2 ), a dark-brown gas: 2NO(g) + O 2 (g)  2NO 2 (g) In one experiment 0.886 mole of NO is mixed with 0.503 mole of O 2. Calculate the number of moles of NO 2 produced (note: first determine which is the limiting reagent). A)0.886 mol B)0.503 mol C)1.01 mol D)1.77 mol 64.Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of aluminum metal. It is prepared by the reaction CaF 2 + H 2 SO 4  CaSO 4 + 2HF In one process 6.00 kg of CaF 2 are treated with an excess of H 2 SO 4 and yield 2.86 kg of HF. Calculate the percent yield of HF. A)93.0 % B)95.3 % C)47.6 % D)62.5 %

56. Chemistry In Action: Chemical Fertilizers Plants need: N, P, K, Ca, S, & Mg 3H 2 (g) + N 2 (g) 2NH 3 (g) NH 3 (aq) + HNO 3 (aq) NH 4 NO 3 (aq) 2Ca 5 (PO 4 ) 3 F (s) + 7H 2 SO 4 (aq) 3Ca(H 2 PO 4 ) 2 (aq) + 7CaSO 4 (aq) + 2HF (g) fluorapatite

57. Answer Key 1-C 2-B 3-D 4-A 5-A 6-A 7-A 8-B 9-C 10-D 11-A 12-C 13-A 14-B 15-C 16-C 17-C 18-A 19-A Problems 3.5 – 3.6 – 3.7 – 3.8 3.14 – 3.16 – 3.18 – 3.20 – 3.22 3.24 – 3.26 – 3.28 – 3.40 – 3.42 3.44 – 3.46 – 3.48 – 3.50 – 3.52 – 3.60 3.84 – 3.86