1. LECTURE 7 CHM 151 ©slg 1. The Mole: Molecules and Compounds 2. % Composition from Formulas 3. Empirical Formula from % Topics:

2. Mercury is a liquid metal with a density of 13.534 g/cm 3. If you measured out 75.0 mL of Hg into a graduated cylinder, how many atoms of Hg would be in the sample? Question: 75.0 mL Hg = ? Atoms Hg Relationships: 1 mL or cm 3 Hg = 13.534 g Hg 200.59 g Hg = 1 mol Hg 1 mol Hg = 6.022 x 10 23 atoms Hg Setup and solve: mL---> g ---> mol --->atoms Warm-up:

4. Molecules, Compounds, and the Mole Let us now extend the use of molar mass, M, to include all particles chemists need to measure: not just atoms but also especially ions and molecules.... The basic principle is this: whenever you weigh out the“formula weight” of any substance or species in grams, you have A’s number of particles of that species, and the molar mass of that species...

5. Molar Mass of Molecules The formula of any molecule describes the number of atoms making up one unit of that molecule: Br 2 The diatomic bromine molecule, as bromine is found in nature: the formula tells us that 2 atoms of bromine are contained in every molecule. By extension, 2 moles of bromine atoms are contained in every 1 mole of bromine molecules. The calculation of the molar mass of molecular bromine then looks like this:

6. The atomic weight of Br, from the PT, is 79.904 amu’s. Therefore: 2 moles of Br = 2 X 79.904 g = 159.808 g And the molar mass, M, of Br 2 is 159.808 g/mol Now let’s try the molar mass of CH 3 CH 2 OH, ethyl alcohol:

7. Molar Mass, M, CH 3 CH 2 OH M, CH 3 CH 2 OH, 46.07 g/mol

8. Molar Mass of Ionic Compounds The formula of an ionic compound indicates the simplest ratio of ions present in any sample of the compound. It is this “formula unit” that we use for calculating the molar mass. Actually, we needn’t ask what kind of compound we are getting the M for; we simply calculate for all atoms found in the formula of any species! Cl 2 (2 Cl) Fe(CN) 2 (1Fe 2C 2N) (NH 4 ) 2 CO 3 (2N 8H 1C 3O)

9. M, CH 3 CH 2 OH, 46.07 g/mol, use in problems: Given a mass, or volume and density, solve for: a) moles of compound or individual atoms b) grams of individual atoms c) number of molecules or atoms

10. How many moles of ethyl alcohol are contained in a sample that weighs 33.95 g? (CH 3 CH 2 OH, 46.07 g/mol). Question: 33.95 g CH 3 CH 2 OH = ? mol CH 3 CH 2 OH Relationship: 46.07 g CH 3 CH 2 OH = 1 mol Setup and Solve: ( g ---> mol)

11. g ----------> mol

14. How many moles of hydrogen are contained in 33.95 g CH 3 CH 2 OH? Question: 33.95 g CH 3 CH 2 OH = ? mol H Relationship: 46.07 g CH 3 CH 2 OH = 1 mol CH 3 CH 2 OH 1 mol CH 3 CH 2 OH = 6 mol H Setup and Solve: ( g ---> mol CH 3 CH 2 OH ---> mol H)

16. How many grams of hydrogen are contained in 33.95 g CH 3 CH 2 OH? Question: 33.95 g CH 3 CH 2 OH = ? g H Relationship: 46.07 g CH 3 CH 2 OH = 1 mol CH 3 CH 2 OH 1 mol CH 3 CH 2 OH = 6 mol H 1 mol H = 1.008 g H Setup and Solve: ( g CH 3 CH 2 OH ---> mol CH 3 CH 2 OH ---> mol H -----> g H)

19. How many grams of carbon are contained in 33.95 g CH 3 CH 2 OH? Question: 33.95 g CH 3 CH 2 OH = ? g C Relationship: 46.07 g CH 3 CH 2 OH = 1 mol CH 3 CH 2 OH 1 mol CH 3 CH 2 OH = 2 mol C 1 mol C = 12.01 g C Setup and Solve: ( g CH 3 CH 2 OH ---> mol CH 3 CH 2 OH ---> mol C -----> g C) Group Work

20. Solution

21. How many atoms of carbon are contained in 33.95 g CH 3 CH 2 OH? Question: 33.95 g CH 3 CH 2 OH = ? atoms C Relationship: 46.07 g CH 3 CH 2 OH = 1 mol CH 3 CH 2 OH 1 mol CH 3 CH 2 OH = 2 mol C 1 mol C = 6.02 x 10 23 atoms C Setup and Solve: ( g CH 3 CH 2 OH ---> mol CH 3 CH 2 OH ---> mol C -----> atoms C)

24. What mass of ethyl alcohol, CH 3 CH 2 OH, would contain 2.06 X 10 24 atoms of carbon? Question: 2.06 X 10 24 atoms C = ? g CH 3 CH 2 OH Relationship: 6.02 x 10 23 atoms C = 1 mol C 2 mol C = 1 mol CH 3 CH 2 OH 1 mol CH 3 CH 2 OH = 46.07 g CH 3 CH 2 OH Setup and Solve: (atoms C-----> mol C ---> mol CH 3 CH 2 OH ---> g CH 3 CH 2 OH ) Group Work

25. SOLUTION

26. New TOPIC: % Composition from Formula of the Compound The formula of a compound can be used to determine the mass % of each element by using the molar relationships we have learned... Consider the following: What is the % by mass of Carbon, Hydrogen and Oxygen in a sample of ethyl alcohol, CH 3 CH 2 OH?

27. Mass % of Elements in a Compound This is done using the formula weight calculations and the approach: % by mass = total mass of one element (the part) X 100 % total mass of compound (the whole) based on one mole of the compound

28. % Composition of Ethyl Alcohol, CH 3 CH 2 OH Molar Mass Calculation: 2 C = 2 X 12.011 = 24.022 g C 6 H = 6 X 1.008 = 6.048 g H 1 O = 1 X 15.999 = 15.999 g O 46.069 g/mol CH 3 CH 2 OH %C = 24.022 g C X 100 = 52.144% C 46.069 g CH 3 CH 2 OH %H = 6.048 g H X 100 = 13.13% H 46.069 g CH 3 CH 2 OH %O = 15.999 g O X 100 = 34.728% O 46.069 g CH 3 CH 2 OH

29. CHECKING..... 52.144% C 13.13 % H 34.728% O 100.002 % CH 3 CH 2 OH = 100.00% (2 digits allowed after decimal)

30. Formula from % Composition If one can go from formula to % composition, one should be able to go from % composition to the formula of the compound. This is quite true (almost): We can take % composition back to the “empirical formula”, which describes the simplest mole ratio of atoms in the formula... That is not always the same thing...

31. Empirical Vs Molecular Formulas

32. For “organic” or “molecular carbon containing compounds”, there exist a list of compounds which share almost every conceivable empirical formula. To determine which compound one has from analytical data, one needs the molar mass as well, as we will see... For most ionic compounds and simple molecular compounds, the empirical and molecular formulas are identical.

33. Empirical Formula from % Composition: Let’s take our ethyl alcohol compound back to its formula from its percent composition.The trick is to use 100 grams of sample whenever one is calculating from mass %. Then we have: 52.144% C = 52.14 g carbon 13.13 % H = 13.13 g hydrogen 34.728% O = 34.73 g Oxygen Our procedure will be: % element ---> g element ---> mol element ---> simplest mole ratio

34. Method of choice: make a chart:

35. Chart out your information in this fashion to determine formula: Moles: 52.14 g C X 1 mol C = 4.341 mol C 12.01 g C

36. Your empirical formula is the simplest ratio of moles of the elements in the formula, for which you have determined that: For every 1.00 mole of O atoms, you have 6.002 mole of H atoms and 2.000 mole of C atoms Empirical formula: C 2 H 6 O

37. Empirical Formula using Experimental Data A new compound weighing 0.678 g, containing xenon and fluorine, was made from a mixture of the gases in sunlight. If the xenon showed a mass of 0.526 g, what is the empirical formula of the compound? Note: 0.678 g compound - 0.526 g Xe =.152 g F 2 gas Although the gas itself is diatomic, when combined into a compound it exists as individual atoms, and the atomic weight of F. 19.0 g/mol, is used for formula calculation, not 38.0 g/mol F 2.

38. Grams M Moles: #g/ M Simplest mole ratio Xe.526 131.29.00401.00401/.00401 = 1 F.152 19.00.00800.00800/.00401 = 2 EMPIRICAL FORMULA: XeF 2

39. Molecular and Empirical Formula Nicotine, a poisonous compound found in tobacco leaves, is 74.0% C, 8.65% H, and 17.35% N. Its molar mass is 162 g/mol. What are the empirical and molecular formulas of this compound? Note: When the molar mass is included in the problem, the exact molecular formula can be determined...

40. grams Molar mass Moles Simplest mole ratio C 74.0 g12.01 6.16 6.16/1.238 = 4.98 H 8.65 g1.008 8.58 8.58/1.238 = 6.93 N 17.35 g14.01 1.238 1.238/1.238= 1.000 Empirical Formula: C 5 H 7 N In order to obtain the molecular formula, you must divide the molar mass (mass of molecule) by the empirical formula mass (the mass of the simplest ratio of atoms...)

41. Empirical Formula Mass: 5C = 5 X 12.01 = 60.05 7H = 7 X 1.008 = 7.056 1N = 1 X 14.01 = 14.01 81.12 g/mol Molar mass = 162 = 2.00 = “n” Emp. Form. mass 81.12 (C 5 H 7 N) n = (C 5 H 7 N) 2 : C 10 H 14 N 2

42. Hydrated Compounds Crystalline ionic solids (salts!) are frequently found in nature or are produced from aqueous solutions with a specific number of water molecules associated with each set of formula ions: CuSO 4. 5H 2 O NiCl 2.6H 2 O CaSO 4.2H 2 O Frequently the color of the salt depends on the presence of these “waters of hydration.”

43. The number of water molecules associated with a particular salt is characteristic but not easy to predict: therefore the value is determined experimentally: The hydrated salt is weighed, heated carefully to drive off the water, and reweighed. The mass of the water driven off is calculated, converted to moles and compared to moles of the parent, anhydrous salt to determine the formula of the hydrate...

44. Ex 3.15: Naturally occurring hydrated copper(II) chloride is called eriochalcite. If 0.235 g of CuCl 2.xH 2 O is heated to drive off the water, 0.185 g residue remains. What is the value of x? 0.235 g hydrate - 0.185 g parent salt = 0.050 g H 2 O 1Cu= 1 X 63.55 = 63.55 2Cl = 2 X 35.45 =70.90 134.45 g/mol CuCl 2 2H= 2 x 1.008= 2.016 1O= 1 x 16.00= 16.00 18.02 g/mol H 2 O

45. grams Molar mass Moles Simplest mole ratio CuCl 2 0.185 g 134.45g/mol.00138.00138/.00138= 1 H 2 O 0.050 g 18.02 g/mol.00278.00278/.00138= 2 Formula of eriochalcite: CuCl 2. 2H 2 O

46. GROUP WORK: Do both problems, be sure all names are on sheet, hand in as you leave! 1. If 7.572 g Cu (63.546 g/mol) reacts with 1.910 g S (32.066 g/mol) to form a binary compound Cu x S y, what is the empirical formula of the compound formed? 2. If 1.023 g of a hydrated compound, CuSO 4. xH 2 O shows a mass of 0.654 g when dehydrated, what is the formula of the compound? (CuSO 4, 159.6 g/mol; H 2 O, 18.02 g/mol). The answers are Cu 2 S, CuSO 4. 5 H 2 O; show work to prove... End, Unit One material