1. Chapter 3: STOICHIOMETRY Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

2. Atomic Masses IUPAC define: The system of atomic masses is based on 12 C as the standard, 12 C has 12 atomic mass units (amu). (Mass 13 C)/Mass 12 C = 1.0836129 Mass 13 C = (12 amu )(1.0836129) = 13.003355 Elements occur in nature as mixtures of isotopes Carbon =98.89% 12 C0.9889 x 12amu 1.11% 13 C.0111 x 13.0034 amu <0.01% 14 C~ Carbon atomic mass = 12.01 amu Average atomic mass = Atomic mass

3. Figure 3.1: (left) A scientist injecting a sample into a mass spectrometer. (right) Schematic diagram of a mass spectrometer. zV = ( ½ )mv 2; v = (2zV/m) 0.5

4. Figure 3.2: (a) Neon gas glowing in a discharge tube. (b) "peaks" and (c) a bar graph. 20 Ne = 90.92%, 21 Ne = 0.257%, 22 Ne = 8.82%

5. Figure 3.3: Mass spectrum of natural copper.

6. The Mole The number equals 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 Avogadro ’ s number equals 6.022  10 23 units

7. Figure 3.4: Proceeding clockwise from the top samples containing one mole each of copper, aluminum, iron, sulfur, iodine, and (in the center) mercury.

8. The mass of 1mole of an element = atomic mass (g)

9. 12 C: (6.022 x 10 23 atoms) ( ) = 12 g 6.022 x 10 23 amu = 1 g

10. 12 C: Ex 3.2 : The mass in grams of a sample of Am containing 6 atoms. 6 atoms x 243 = 1.46 x 10 3 amu ∵ 6.022 x 10 23 amu = 1 g  1.46 x 10 3 amu = 2.42 x 10 -21 g

11. Pure aluminum.

12. Aluminum alloys are used for many high- quality bicycle components, such as this chain wheel.

13. Ex 3.3 Determining Moles of Atoms The number of moles of atoms and the number of atoms in a 10.0g Al sample. (10.0 g Al) x = 0.371 mol Al atoms 0.371 mol Al x = 2.33 x 10 23 atoms

14. Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CH 4 = 16.04 grams per mole

15. Ex 3.6 Calculating Moles Mass (a) Calculate the molar mass of C 10 H 6 O 3 of juglone (b) The number of moles of 1.56 x 10 -2 g juglone sample. 10 C:10 x 12.01 g = 120.1 g 6H : 6 x 1.008 g = 6.048 g 3O :3 x 16.000 g = 48.00 g Mass of 1 mole juglone = 174.1 g (b) 1.56 x 10 -2 g x = 8.96 x 10 -5 moles

16. Percent Composition Mass percent of an element: For iron in iron (III) oxide, (Fe 2 O 3 ) x 100% = 70.07 %

17. Figure 3.5: A schematic diagram of the combustion device used to analyze substances for carbon and hydrogen. Determining the Formula of a compound

18. A 0.1156 sample, containing C, H, and N only, reacts with excess O, 0.1638 g CO 2 and 0.1676 H 2 O are collected. Determining the formula of the compound. 0.1638 g CO 2 x = 0.04470 g C, »= 38.87 % C = 38.67 g C/100g 0.1676 g H 2 O x = 0.01875 g H, »= 16.22 % H = 16.22 g H, Then,= 45.11% N = 45.11 g N 38.67 g C x = 3.220 mol C, = 16.09 mol H, = 3.219 mol N

19. C:= 1.00 = 1 H: = 4.997 = 5 N: = 1.000 = 1 ⇒ CH 5 N (empirical formula) To specify the molecular formula, we must know the molecular mass.

20. Determining the Formula of a compound molecular formula = (empirical formula) n [n = integer] molecular formula = C 6 H 6 = (CH) 6 empirical formula = CH

21. Ex. 3.11 Determine the empirical and molecular formula for a compound that gives the following percentages upon analysis (in mass percents):71.65% Cl, 24.27% C, 4.07% H. The molar mass is known to be 98.96 g/mol 71. 65 g Cl x = 2.021 mol Cl, 24.27 g C x = 2.021 mol C 4.07 g H x = 4.04 mol H, ⇒ ClCH 2 (empirical formula) Empirical formula mass = 49.48 g/mol = =2 Molecular formula = (ClCH 2 ) 2 = Cl 2 C 2 H 4

22. Figure 3.7: The two forms of dichloroethane.

23. Figure 3.8: The structure of P 4 O 10.

24. Computer-generated molecule of caffeine.

25. Empirical Formula Determination 1.Base calculation on 100 grams of compound. 2.Determine moles of each element in 100 grams of compound. 3.Divide each value of moles by the smallest of the values. 4.Multiply each number by an integer to obtain all whole numbers.

26. Molecular Formula Determination Method I 1.Obtain the empirical formula. 2.Calculate the empirical formula mass. 3.Calculate the ratio (n) = 4. Molecular formula = (empirical formula) n Method II 1.Using the mass percentage and molar mass to determine the mass of each element per mole of compound. 2.Determine the # of moles of each element /mol compound. 3.The integers of # of moles of each element are the subscript in the molecular formula.

27. Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances.

28. Chemical Equation A representation of a chemical reaction: CH 4 + O 2  CO 2 + H 2 O reactants products

29. Molecular rep of a rxn

31. Hydrochloric acid reacts with solid sodium hydrogen carbonate to produce gaseous carbon dioxide.

33. Copyright © Houghton Mifflin Company. All rights reserved.3a–33 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

34. Decomposition of ammonium dichromate.

35. Decomposition of ammonium dichromate. (cont'd)

36. 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.

37. Figure 3.9: Three different stoichiometric mixtures of methane and water, which react one-to-one.

38. Figure 3.10: A mixture of CH 4 and H 2 O molecules.

39. Figure 3.11: Methane and water have reacted to form products according to the equation CH 4 + H 2 O 3H 2 + CO.

40. Figure 3.12: Hydrogen and nitrogen react to form ammonia according to the equation N 2 + 3H 2 2NH 3.

41. Limiting Reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed.

42. 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.

43. Solving stoichiometry problems flow chart.