1. LECTURE 6 CHM 151 ©slg TOPICS: 1. Ionic Nomenclature, Completion 2. Naming Binary Molecular Compounds 3. “The Mole”: Introduction

2. Cr 3+ CO 3 2- Ni 2+ CN - Zn 2+ NO 2 - Bi 3+ H 2 PO 4 - Pb 2+ N 3- Cr 2 (CO 3 ) 3 Chromium(III) carbonate Ni(CN) 2 Nickel(II) cyanide Zn(NO 2 ) 2 Zinc Nitrite Bi(H 2 PO 4 ) 3 Bismuth(III) dihydrogen phosphate Pb 3 N 2 Lead(II) Nitride Tues. Group Work:

3. Polyatomic Anions of O, S (6A) Cr (6B) Oxygen: OH - hydroxide Remember also: O 2- oxide Sulfur: SO 4 2- sulfate SO 3 2- sulfite HSO 4 - hydrogen sulfate HSO 3 - hydrogen sulfite Remember also: S 2-, sulfide Chromium: CrO 4 2- chromate

4. The “hydroxides” and “oxides” of the metallic elements are referred to as “bases”; all other ionic combinations are referred to as “salts” “BASES” and “SALTS” Bases: Mg(OH) 2 NaOH CaO Fe(OH) 3 Salts: MgCl 2 MgHSO 4 MgCO 3 Na 3 PO 4 NaNO 2 Na 2 SO 3 Ca(NO 3 ) 2 Ca 3 N 2 CaSO 4 Fe(CN) 2 Fe(CH 3 CO 2 ) 3 Fe(H 2 PO 4 ) 2

5. Polyatomic Anions of Cl, Br, I (7A) Mn (7B) Fluorine, F forms only the monatomic anion F - ; Bromine, Br and Iodine, I form the same ions as chlorine, Cl: Chlorine: ClO - hypochlorite ClO 2 - chlorite ClO 3 - chlorate ClO 4 - perchlorate Remember also: Cl -, Chloride Manganese: MnO 4 - permanganate

6. SUMMARY, NAMING IONIC SALTS AND BASES State name of the cation, then name of the anion. Cations with a variable charge are named by adding a Roman numeral Monoatomic anions are named by changing their elemental name to end in “ide” Polyatomic anions (memorized) end in “ite” or “ate”...

7. GROUP WORK FORMULA NH 4 ClO Cd(BrO 2 ) 2 Co(IO 3 ) 3 Ba(ClO 4 ) 2 KMnO 4 Ag 2 CrO 4 NAME

8. FORMULA NH 4 ClO Cd(BrO 2 ) 2 Co(IO 3 ) 3 Ba(ClO 4 ) 2 KMnO 4 Ag 2 CrO 4 NAME Ammonium hypochlorite Cadmium bromite Cobalt(III) Iodate Barium perchlorate Potassium permanganate Silver chromate

9. Naming Binary Molecular Compounds All compounds beginning with a metal or ammonium are named as “ ionic compounds.” Compounds containing only two elements (“binary”) in which both elements in the formula are a non-metal or metalloid are named in a different manner... The change in nomenclature reflects the fact that these compounds are “molecular” and not “ionic” in nature!

10. Name the first element in the formula Name the second element in the formula to end in “ide”: carbide, nitride, phosphide, oxide, sulfide, fluoride, bromide, chloride, iodide Add numerical prefixes to indicate more than one atom of the element in the formula: di (2), tri (3), tetra (4), penta (5), hexa (6), hepta (7), octa(8) Binary Molecular Nomenclature Method:

11. Typical Nomenclature NO 2 SF 6 ICl 5 N 2 O 5 CBr 4 SO 3 P 2 O 3 nitrogen dioxide sulfur hexafluoride iodine pentachloride dinitrogen pentoxide carbon tetrabromide sulfur trioxide diphosphorus trioxide

12. BH 3 CH 4 SiH 4 NH 3 PH 3 borane methane* silane ammonia* phosphine COMMON NAMES, BINARY MOLECULES ENDING IN H

13. From PARTICULATE (“too small to touch”) to MACROSCOPIC (amounts we can handle): THE MOLE Kotz & Treichel, Chapter 3, 3.6-3.8

14. Many different items we encounter in our daily lives come packaged in set amounts described by various “counting terms”. Let’s consider a few of them: Many different items we encounter in our daily lives come packaged in set amounts described by various “counting terms”. Let’s consider a few of them:

15. shoes and socks and earrings come in pairs (2), eggs come in dozens (12), pencils are wholesaled by the gross (144), donuts and sweet corn are often sold as “the baker’s dozen” or “the farmer’s dozen” (13), and diet pop and beer by the 6- pack or case (24).....

16. Chemists deal in atoms, molecules and ions, which need to be counted and measured as well. BUT: The mass of one atom of the 19 F isotope is 3.156X10 -23 g. The radius of a nucleus is about.001 pm and the radius of an atom about 100 pm. (10 12 picometers or 1,000,000,000,000 pm = 1 m) THEREFORE......

17. Chemists need their own unit for counting and weighing amounts of substances which come in particle size too tiny to be seen or weighed on any balance. For convenience in describing amounts of atoms, molecules, and ions, chemists have a unique unit of measure, THE MOLE

18. The MOLE The chemist’s counting number Comes from the Latin meaning “whole heap or pile of” SI base unit for measuring amount of substance Defined as the number of atoms in exactly 12 grams of 12 C, 6.022 X 10 23

19. 1 Mole always contains the same number of particles of whatever is being described, that is, Avogadro’s number of particles: 1 Mole = Avogadro’s number of particles = 6.022136736 X 10 23 particles = 602,213,673,600,000,000,000,000 particles If one used A’s number of particles to describe macroscopic objects, one would be overwhelmed: I mole of green peas would cover the entire United States to a depth of 3 miles!

20. How do we get “a mole” of a substance ??? One mole is defined as the number of particles in exactly 12 g of the 12 C isotope of carbon. Carbon was used as the standard for the amu scale, where the mass of one atom of 12 C was defined as 12 amu. The mole answers the question: “How many atoms would you have if you took the amu scale (which describes mass of one atom) and use it as grams instead? Mass, amu, = 1 atom Mass, g = ? atoms = 6.02 X 10 23 atoms (1 mole)

21. So, How to Get a Mole: We consult the periodic table, obtain the atomic mass of an element in amu’s, the relative mass of one atom..... We weigh this amount out in grams..... We now have one mole of atoms, A’s number, 6.02 X 10 23 atoms, a convenient “package of atoms”..... We have gone into the chemist’s counting system and can deliver not a dozen eggs but a mole of atoms....

22. This system works because of the relative nature of the atomic mass unit scale, in which all atoms were assigned a mass relative to 12 C, the mole standard: “One mole is the number of atoms in exactly 12 g of 12 C” The mole “pile or heap” of atoms for each element will weigh more or less than the mole “pile” for carbon, depending on whether the individual atoms weigh more or less than carbon.

23. If one mole of carbon atoms weighs 12.0 g, then one mole of oxygen atoms, which weighs 1.33 times more than carbon, would be: 1.33 X 12.0 g =16.0 g = 1 mol O Since H atom is 1/12 the mass of a carbon atom, the matching pile of hydrogen atoms would be 1/12 X 12.0 g = 1.00 g = 1 mol H

24. For any element, the molar mass, M, is the mass in grams of a mole of atoms, “#g/mole” M, molar mass, is NUMERICALLY equal to the mass of one atom in amu’s as given on your PT. If, however, one weighs out the molar mass, one has 6.022 x 10 23 atoms every time The “Molar Mass, M”

25. MOLES of ATOMS: the MOLAR MASS PT: amu’s / 1 atom M, g /mol, A’s # atoms Li, 6.941 amu/atomLi, 6.941 g/ mol Pb, 207.2 amu/atom Pb, 207.2 g/ mol Zn, 65.39 amu/atom Zn, 65.39 g/ mol

28. Using this knowledge, the chemist can interconvert grams, moles, and atoms of any element. The molar mass, “g/mol”, like density, “g/cm 3 ”, is a convenient conversion factor: For any element: 1mole = atomic weight, grams = 6.022 X 10 23 atoms

29. Suppose you weighed out 35.89 g of aluminum metal. How many moles and how many atoms of aluminum would be contained in this sample? Question: 35.89 g Al = ? mol Al = ? atoms Al Relationships: 1 mol Al = 26.98 g Al = 6.022 X 10 23 atoms Al Setup and Solve: g ---> mol

31. g --------> mol ---------> atoms

32. What would 9.00 x 10 24 atoms of mercury weigh in grams? Question: 9.00 x 10 24 atoms Hg = ? g Hg Relationships: 1 mol Hg = 200.59 g Hg = 6.022 X 10 23 atoms Hg Setup and Solve: atoms -----> mol -----> g

34. 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: 13.534 g Hg = 1cm 3 or mL 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

36. 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...

37. 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:

38. 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:

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

40. 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!

41. Molar Mass, M, NaCl M, NaCl = 58.44 g/mol

42. 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

43. 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)

44. g ----------> mol

45. How many moles of hydrogen atoms 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 1 mol CH 3 CH 2 OH = 6 mol H Setup and Solve: ( g ---> mol CH 3 CH 2 OH ---> mol H)

47. 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 1 mol CH 3 CH 2 OH = 6 mol H 1 mol H = 1.008 g Setup and Solve: ( g CH 3 CH 2 OH ---> mol CH 3 CH 2 OH ---> mol H -----> g H)

49. END, Lecture 6