1. Mass Relationships in Chemical Reactions Chapter 3 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 2 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

3. 3 The average atomic mass is the weighted average of all of the naturally occurring isotopes of the element.

4. 4 Naturally occurring 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 Average atomic mass of lithium:

5. Calcium has 6 isotopes. Their masses & percentage abundances are given in the table below. Calculate the average atomic mass or atomic weight of Ca. IsotopeIsotope mass (amu) Percentage Abundance Ca-4039.9625996.941 Ca-4241.958620.647 Ca-4342.958770.135 Ca-4443.955482.086 Ca-4645.953690.004 Ca-4847.952530.187 {(39.96259 x 96.941) + (41.95862 x 0.647) + (0.135 x 42.95877) + (2.086 x 43.95548) + (.004 x 45.95369) + (47.95253 x 0.187)}amu/100 =4007.8amu/100 = 40.078 amu

6. 6 Average atomic mass (6.941)

7. 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 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

8. Molar mass is the mass of 1 mole of in grams eggs shoes marbles atoms 1 12 C atom = 12.00 amu 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 mole 12 C atoms = 12.00 g 12 C Average mass of 1 Li atom = 6.941 amu Molar mass of Li = mass of 1 mole lithium atoms = 6.941 g For any element, numerical value of atomic mass (amu) = numerical value of molar mass (grams) 3.2

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

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

11. 1. 15.0 g Mercury has a.How many moles of mercury? 0.0748 mol b.How many atoms of Mercury? 4.50 x 10 22 atoms 2. 5.16 x 10 22 atoms of Fe is a.How many moles of Fe? 0.0857 mol Fe b.What is the corresponding mass? 4.79 g Fe 3. 15.0 g Zinc has a.How many moles of Zinc? b.How many atoms of Zinc? 4. 5.16 x 10 22 atoms of Cu is a.How many moles of Cu? b.What is the corresponding mass?

12. 12 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

13. 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 The numerical value of molecular mass in amu = the numerical value of molar mass in grams 1 molecule of SO 2 weighs 64.07 amu 1 mole of SO 2 weighs 64.07 g 1 mole of SO 2 =6.022 x 10 23 molecules of SO 2 3.3

14. You cannot go directly from grams of a compound to moles of A CONSTITUENT ATOM You have to first calculate the moles of the compound and then from that calculate the moles of a constituent atom: Grams of compound  Moles of Compound  Moles of a constituent atom The molecular formula not only tells us how many atoms of each type are present in one molecule of a compound but also how many MOLES of atoms of each type are present in 1MOLE OF THE COMPOUND. For example, the molecular formula of sugar, C 12 H 22 O 11, tells us that 1 mole of sugar has 12 moles of C atoms, 22 moles of H atoms & 11 moles of O atoms besides the fact that 12 atoms of C, 22 H atoms & 11 O atoms are present in 1 sugar molecule.

15. 1.Consider a sample of 15.0 g Borazine (B 3 N 3 H 6 ) a.How many moles of borazine are present? 0.186 mol b.How many molecules of borazine are present? 1.12 x 10 23 molecules c.How many moles of N atoms are present? 0.558 mol N d.How many N atoms are present? 3.36 x 10 23 N atoms 2. A sample of ethanol, C 2 H 5 OH has 6.00 x 10 23 H atoms. a.How many ethanol molecules are present? 1.00 x 10 23 molecules b.How many moles of ethanol are present?0.166 mol c.What is the mass of the sample? 7.65 g

16. Complete the following conversions; 1.Consider 0.00500 g of H 2 O Calculate the following: a.The moles of H 2 O: 2.77 × 10 -4 b.The molecules of H 2 O: 1.67 × 10 20 c.The moles of H atoms: 5.55 × 10 -4 d.The number of H atoms present: 3.34 × 10 20 2. Na atoms in 3.50 mole NaOH 3. grams mass of 2.50 mole O 2 4. carbon atoms in 10.5 g C 6 H 12 O 6 5. A sample of benzene, C 6 H 6 has 3.00 x 10 22 H atoms. a.How many benzene molecules are present? 5.00 × 10 21 b.How many moles of benzene are present? 8.30 × 10 -3 c.What is the mass of the sample? 0.649 g

17. Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C 3 H 8 O ? moles of C 3 H 8 O = 72.5 g / 60.095 g/mol = 1.21 mol 1 mol of H atoms is 6.022 x 10 23 H atoms 5.82 x 10 24 H atoms 3.3 1 mol C 3 H 8 O molecules contains 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 = Steps: 1. Convert grams of C 3 H 8 O to moles of C 3 H 8 O. 2. Convert moles of C 3 H 8 O to moles of H atoms. 3. Convert moles of H atoms to number of H atoms.

18. 18 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 NaCl

19. Do You Understand Formula Mass? What is the formula mass of Ca 3 (PO 4 ) 2 ? 310.18 amu 3.3 1 formula unit of Ca 3 (PO 4 ) 2 3 Ca 3 x 40.08 g/mol 2 P2 x 30.97 g/mol 8 O + 8 x 16.00 g/mol 310.18 g/mol = molar mass Units of grams per mole are the most practical for chemical calculations!

20. 1.You have a 10.0 g of Ca 3 (PO 4 ) 2. (Molar mass = 310.18 g/mol) a.How many moles are present? 0.0322mol b.How many moles of Ca +2 ions are present? 0.0966 mol c. How many phosphate ions are present? =0.0644 mol = 3.88 x 10 22 ions. 2. You have a 10.0 g of (NH 4 ) 3 PO 4 (molar mass = 149.12g/mol) a.How many moles are present? 0.0671 mol b.How many moles of phosphate ions are present? 0.0671 mol c. How many ammonium ions are present? 1.21 × 10 23

21. 21 Heavy Light Mass Spectrometer Mass Spectrum of Ne

22. 22 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%

23. Calculate percentage composition of a.Borazine (B 3 N 3 H 6 ; molar mass = 80.52g/mol) 40.28%B, 52.20% N, 7.52% H b.Ca 3 (PO 4 ) 2 (Molar mass = 310.18 g/mol) 38.76 % Ca, 19.97 % P, 41.27 % O

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 To begin, assume for simplicity that you have 100 g of compound!

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. Determine the empirical formula of the compound whose percentage composition is 62.1% C, 5.21% H, 12.1% N, and 20.7% O Step 1: assume 100 g: 62.1 g C, 5.21 g H, 12.1 g N, 20.7 g O Step 2: (62.1/12.01)mol C = 5.17 mol C; (5.21/1.01) mol H = 5.16 mol H; (12.1/14.01)mol N = 0.864 mol N; (20.7/16.00) mol O = 1.29 mol O Step 3: C:H:N:O = 5.17 : 5.16 : 0.864 : 1.29 = (5.17/0.864) : (5.16/0.864) : 1 : (1.29/0.864) = = 5.98 : 5.97 : 1: 1.50 = 6 : 6 : 1 : 1.50 (DO NOT ROUND OFF NUMBERS WITH.5; MULTIPLY BY 2 BEFORE ROUNDING OFF) = 12 : 12 : 2 : 3; C 12 H 12 N 2 O 3

27. A sample of an organic compound was found to contain 9.01g C, 1.01g H, 12.0 g O. Calculate the empirical formula of this compound. 9.01g C = (9.01/12.01) mol C = 0.750 mol C 1.01 g H = (1.01/1.01)mol H = 1.00 mol H 12.0 g O = (12.0/16.00) mol O = 0.750 mol O C : H : O = 0.750 : 1.00 : 0.750 = (0.750/0.750) : (1.00/0.750) : (0.750/0.750) = 1.00 : 1.33 : 1.00 = 1 : 1.33 : 1 DO NOT ROUND OFF ANY NO. WITH 0.3 OR 0.33 OR 0.333 OR 0.5 OR ANY THAT CAN BE ROUNDED OFF TO 0.5. MULTIPLY THOSE WITH.3,.33 etc. WITH 3 (REMEMBER TO MULTIPLY ALL THE NUMBERS IN THE RATIO) MULTIPLY THOSE WITH.5 WITH 2 (REMEMBER TO MULTIPLY ALL THE NUMBERS IN THE RATIO) C : H : O = 1 : 1.33 : 1 = 3 x (1 : 1.33 : 1) = 3 : 3.99 : 3 = 3: 4 : 3 Empirical formula = C 3 H 4 O 3

28. 28 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

29. Determination of Molecular Formulas from Empirical Formulas 1.Determine the ratio of the molecular weight to empirical formula weight (the molecular weight HAS to be provided): molecular weight/empirical formula weight 2.Multiply all subscripts in the empirical formula by the above ratio to obtain the molecular formula. 1.A hydrocarbon has 12.01g C, 3.03g H. Its molecular weight = 30.08 amu. Calculate the molecular formula of the compound. 12.01 g C, 3.03 g H = (12.01/12.01) mol C, (3.03/1.01) mol H = 1.000 mol C, 3.00 mol H = 1 mol C, 3 mol H Empirical formula =CH 3 Empirical formula weight = (12.01) +(3 x 1.01) amu = 15.04 amu Molecular weight/empirical formula weight = 30.08/15.04 = 2 Therefore molecular formula : C (1 x 2) H (3 x 2) = C 2 H 6

30. 30 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

31. Balance equations ONLY by putting numbers IN FRONT OF formulas. Such numbers are called COEFFICIENTS NEVER CHANGE SUBSCRIPTS – doing so will change the identity of reactants, products and your equation will no longer represent the reaction that you are considering

32. 32 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 NOT 2 grams Mg + 1 gram O 2 makes 2 g MgO

33. 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. 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. Start by balancing those elements that appear in only one reactant and one product. 4. Balance those elements that appear in two or more reactants or products. 5. Coefficients should be whole numbers 6. Check to make sure that you have the same number of each type of atom on both sides of the equation. Balancing Chemical Equations

34. 34 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. 2C 2 H 6 NOT C 4 H 12

35. 35 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 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

36. 36 Balancing Chemical Equations 4.Balance those elements that appear in two or more reactants or products. 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

37. 37 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. 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)

38. 38 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

39. 39 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

40. 40 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.

41. Limiting Reactant: Is that reactant that is first used up. When it is used up, the reaction stops because all reactants have to be present for the reaction to occur. Excess Reactant: A reactant that is remaining even after the reaction has stopped Theoretical Yield: The maximum mass of a product that is CALCULATED to form when all of the LIMITING REACTANT is consumed. (The limiting reactant predicts the smallest yield) Actual Yield: An EXPERIMENTAL value: the mass of a product ACTUALLY formed when the experiment is performed Actual yield ≤ Theoretical Yield Percentage Yield = (Actual/Theoretical) x 100; ≤ 100%

42. Do You Understand Limiting Reactants? In a reaction, 124 g of Al are reacted with 601 g of Fe 2 O 3. 2 Al + Fe 2 O 3 Al 2 O 3 + 2 Fe a.Calculate the thoeretical mass of Al 2 O 3 that can form in grams. b.What mass of excess reactant remains? c.If the actual yield of Al 2 O 3 is 220g, what is the percent yield 3.9 1.Balanced reaction: Done. 2.Moles of “given” reactants. Moles of Al = 124 g / 26.9815 g/mol = 4.60 mol Moles of Fe 2 O 3 = 601 g / 159.6882 g/mol = 3.76 mol Solution to a: Fabulous Four Steps

43. 3. Moles of “desired” product, Al 2 O 3. 3.9 Moles of Al 2 O 3 = 4.60 mol Al X 1 mol Al 2 O 3 = 2.30 mol Al 2 O 3 based on Al 1 2 mol Al 2 Al + Fe 2 O 3 Al 2 O 3 + 2 Fe Moles of Al 2 O 3 = 3.76 mol Fe 2 O 3 X 1 mol Al 2 O 3 = 3.76 mole Al 2 O 3 based on Fe 2 O 3 1 1 mol Fe 2 O 3 Keep the smaller answer! Al is the limiting reactant. 4. Grams of Al 2 O 3. Grams of Al 2 O 3 = 2.30 mol X 101.9612 g/mol = 235 g

44. In a certain experiment 2.50 g of NH 3 and 2.85 g of O 2 are taken. The following reaction occurs: 4NH 3 + 5 O 2  4NO + 6 H 2 O 1. Determine the theoretical yield of NO? 2.14 g not 4.40g Limiting reactant = O 2, excess reactant: NH 3 2. How much excess reactant remains after the liming reactant is completely consumed? 1.29 g 3. If the actual yield is 2.00g of NO, what is the percentage yield? 93.5% If only 1 reactant amount is given, it is implied that that reactant is the limiting reactant. There is no need for any further identification.

45. Consider the reaction between benzene, C6H6 and bromine, Br2, forming bromobenzene, C6H5Br: C6H6 + Br2 , C6H5Br + HBr 1.If 30.0g of C6H6 is used. What is theoretical yield of C6H5Br? 60.30g 2. If the actual yield of C6H5Br is 56.7 g, what is the % yield? 94.0 %

46. 46 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% Reaction Yield

47. 47 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