1. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chapter 3 Stoichiometric Atomic Masses, Mole concept, and Molar Mass (Average atomic mass). Number of atoms per amount of element. Percent composition and Empirical formula of molecules. Chemical equations, Balancing equations, and Stoichiometric calculations including limiting reagents.

2. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 2 Chemical Stoichiometry Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

3. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 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) Micro World atoms & molecules Macro World grams

4. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 4 Atomic Masses Elements occur in nature as mixtures of isotopes Carbon =98.89% 12 C 1.11% 13 C <0.01% 14 C Carbon atomic mass = 12.01 amu

5. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 5 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 Average atomic mass of lithium:

6. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 6 The Mole The number equal to the number of carbon atoms in exactly 12 grams of pure 12 C. 1 mole of anything = 6.022  10 23 units of that thing

7. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7 Avogadro’s number equals 6.022  10 23 units

8. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8 Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. C=12 O=16 CO 2 = 44.01 grams per mole

9. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 9 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)

10. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 10 1 amu = 1.66 x 10 -24 g or 1 g = 6.022 x 10 23 amu

11. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 11 Measuring Atomic Mass Figure 3.1: (left) A scientist injecting a sample into a mass spectrometer. (right) Schematic diagram of a mass spectrometer.

12. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 12 Spectrum Most Abundant Isotope

13. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 13 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 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 =

14. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 14 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%

15. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 15 Figure 3.5: A schematic diagram of the combustion device used to analyze substances for carbon and hydrogen. Determining Elemental Composition (Formula)

16. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 16 The masses obtained (mostly CO 2 and H 2 O and sometimes N 2 )) will be used to determine: 1.% composition in compound 2.Empirical formula 3.Chemical or molecular formula if the Molar mass of the compound is known or given.

17. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 17 Example of Combustion Combust 11.5 g ethanol Collect 22.0 g CO 2 and 13.5 g H 2 O

18. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 18 1mol CO 2 44gCO 2 1mol C12g C 22gCO 2 mcmc m c = Collect 22.0 g CO 2 and 13.5 g H 2 O 12 16 x 2 44gCO 2 22gCO 2 x 12gC % C in CO 2 collected = 6g C

19. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 19 1mol C12g C n mol C6g 6g x 1molC n c = 12 gC = 0.5 mol Convert g to mole: Repeat the same for H from H 2 O 1mol H 2 O18g H 2 O 2 mol H2g H 13.5g H 2 On mol H 2x13.5 n mol H = = 1.5 mol H 18 mol H Faster H but still need O mH

20. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 20 Empirical formula C 0.5 H 1.5 O 0.25 Divide by smallest subscript (0.25) Empirical formula C 2 H 6 O 2x13.5 mH = = 1.5 g H 18 m O = 11.5g – m C – m H = 11.5 – 6 – 1.5 = 4g m4 n O == = 0.25 mol O MM 16

21. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 21 OR for C m C 12.01 1. Fraction of C in CO 2 = == MM CO2 44.01 2. Mass of C in compound = mass of CO 2 x Fraction of C m C 3. % C in compound = x 100 m sample 4. If N then % N = 100 - % C - % H

22. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 22 Then Empirical Formula Using the previously calculated % in compound: % in gram a. Number of mole of C = Atomic mass of C % in gram b. Number of mole of H = Atomic mass of H abc Then divide by the smallest number: smallestsmallest smallest ::

23. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 23 Note If results are :0.99 : 2.01 : 1.00 Then you have to convert to whole numbers: 1: 2: 1 CH 2 N If results are :1.49: 3.01 : 0.99 Then you have to multiply by 2: 3: 6: 2 C 3 H 6 N 2 Hence, empirical formula is the simplest formula of a compound

24. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 24 Formulas molecular formula = (empirical formula) n [n = integer] molecular formula = C 6 H 6 = (CH) 6 empirical formula = CH Then Molecular Mass =n=n Empirical Mass

25. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 25 Figure 3.6: Examples of substances whose empirical and molecular formulas differ. Notice that molecular formula = (empirical formula)n, where n is a integer.

26. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 26 Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances.

27. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 27 Chemical Equation A representation of a chemical reaction: C 2 H 5 OH + O 2  CO 2 + H 2 O reactants products Unbalanced !

28. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 28 Chemical Equation C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O The equation is balanced. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

29. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 29

30. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 30 Calculating Masses of Reactants and Products 1.Balance the equation. 2.Convert mass to moles. 3.Set up mole ratios. 4.Use mole ratios to calculate moles of desired substituent. 5.Convert moles to grams, if necessary.

31. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 31 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

32. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 32 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O 2 mol 4 mol 2x(12+4+16) g4x(2+16) g 209 gm 209 x 4(2+16) m = 2(12+4+16) OR

33. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 33 Limiting Reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed.

34. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 34 Solving a Stoichiometry Problem 1.Balance the equation. 2.Convert masses to moles. 3.Determine which reactant is limiting. 4.Use moles of limiting reactant and mole ratios to find moles of desired product. 5.Convert from moles to grams.

35. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 35 6 green used up 6 red left over Limiting Reagents

36. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 36 Limiting Reactant Calculations What weight of molten iron is produced by 1 kg each of the reactants? Fe 2 O 3 (s) + 2 Al(s)  Al 2 O 3 (s) + 2 Fe( l ) 1 mol2 mol 6.26mol18.52 Ratio: 0.160 > 0.108 LimitingExcess The 6.26 mol Fe 2 O 3 will Disappear first

37. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 37 Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Its amount is Calculated using the balanced equation. Actual Yield is the amount of product actually obtained from a reaction. It is usually given.

38. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 38 Percent Yield Actual yield = quantity of product actually obtained Theoretical yield = quantity of product predicted by stoichiometry

39. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 39 Percent Yield Example 14.4 gexcess Actual yield = 6.26 g

40. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 40 Sample Exercise Titanium tetrachloride, TiCl 4, can be made by combining titanium-containing ore (which is often impure TiO 2 ) with carbon and chlorine - TiO 2(s) + 2 Cl 2(g) + C (s) TiCl 4(l) + CO 2(g) If one begins with 125 g each of Cl 2 and C, but plenty of titanium-containing ore, which is the limiting reagent in the reaction? What quantity of TiCl 4 can be produced?

41. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 41 Virtual Laboratory Project

42. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 42 Practice Example 1 A compound contains C, H, N. Combustion of 35.0mg of the compound produces 33.5mg CO 2 and 41.1mg H 2 O. What is the empirical formula of the compound? Solution: 1. Determine C and H, the rest from 33.5mg is N. 2. Determine moles from masses. 3. Divide by smallest number of moles.

43. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 43 Practice Example 2 Caffeine contains 49.48% C, 5.15% H, 28.87% N and 16.49% O by mass and has a molar mass of 194.2 g/mol. Determine the molecular formula. Solution: 1.Convert mass to moles. 2.Determine empirical formula. 3.Determine actual formula. C 8 H 10 N 4 O 2

44. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 44 Practice Example 3 Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1g of NH 3 is reacted with 90.4g of CuO, which is the limiting reactant? How many grams of N 2 will be formed.

45. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 45 Practice Example 4 Methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen. Suppose 68.5Kg CO(g) is reacted with 8.60Kg H 2(g). Calculate the theoretical yield of methanol. If 3.57x104g CH3OH is actually produced, what is the percent yield of methanol?

46. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 46 Practice Example 5 SnO 2 (s) + 2 H 2 (g)  Sn(s) + 2 H 2 O(l) a) the mass of tin produced from 0.211 moles of hydrogen gas. b) the number of moles of H 2 O produced from 339 grams of SnO 2. c) the mass of SnO 2 required to produce 39.4 grams of tin. d) the number of atoms of tin produced in the reaction of 3.00 grams of H 2. e) the mass of SnO 2 required to produce 1.20 x 10 21 molecules of water.