8 Ions When atoms lose or gain electrons, they become ions. – Cations are positive and are formed by elements on the left side of the periodic chart. – Anions are negative and are formed by elements on the right side of the periodic chart.
16 Would you expect the following to be ionic or molecular: N 2 O Na 2 O CaCl 2 SF 4 CBr 4 FeS P 4 O 6 PbF 2
17 Types of Formulas Empirical formulas give the lowest whole- number ratio of atoms of each element in a compound. Molecular formulas give the exact number of atoms of each element in a compound. Structural Formulas show the order in which atoms are bonded
18 Types of Formulas Structural formulas show the order in which atoms are bonded. Perspective drawings also show the three-dimensional array of atoms in a compound.
Types of Formulas Formula Unit: the smallest neutral collection of ions (most reduced, for ionic)
Formula Mass vs. Molar Mass vs. Molecular Mass Formula Mass: the mass of a formula unit (amu) Molecular Mass: the mass of a molecule (amu) Molar Mass: the mass of one mole of a compound (grams) H 2 O: Molecule Mass 18.0153 amu (1 molecule) Molar Mass 18.0153 grams (1 mole)
Molecular Formulas Diatomic molecules: H 2 O 2 N 2 F 2 Cl 2 Br 2 I 2 Molecules: P 4 (White Phosphorus) S 8 (Sulfur) Distinguish between molecule and atom!
Example 3-1A How many grams of MgCl 2 would you need to obtain 5.0 x 10 -23 Cl - ions?
Example 3-2A Gold has a density of 19.32 g/cm 3. A piece of gold foil is 2.50 cm on each side and 0.100mm thick. How many atoms of gold are in this piece of gold foil?
Example 3-3A Halothane: C 2 HBrClF 3 How many grams of Br are contained in 25.00 mL of halothane (d=1.871g/mL)
31 Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: % element = (number of atoms)(MM of element) (MM of the compound) x 100
32 Percent Composition So the percentage of carbon in ethane is… %C = (2)(12.0 g/mol) (30.0 g/mol) 24.0 g/mol 30.0 g/mol = x 100 = 80.0%
Example 3-4A What are the mass percent composition in C 10 H 16 N 5 P 3 O 13 ?
34 Calculating Empirical Formulas One can calculate the empirical formula from the percent composition
Establishing Formulas From % Comp Assume 100g, % grams Grams Moles Divide by smallest moles – Get whole numbers in most reduced form Ration= MM Molecular/ MM Empirical Multiply Ratio by Empirical to get Molecular Formula
Example 3-5A Sorbitol is a sweetener that has a molecular mass of 182 u and percent composition of 39.56% C 7.74% H 52.70 % O What are the empirical and molecular formulas?
37 Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.
38 Calculating Empirical Formulas Assuming 100.00 g of para-aminobenzoic acid, C:61.31 g x = 5.105 mol C H: 5.14 g x= 5.09 mol H N:10.21 g x= 0.7288 mol N O:23.33 g x = 1.456 mol O 1 mol 12.01 g 1 mol 14.01 g 1 mol 1.01 g 1 mol 16.00 g
39 Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C:= 7.005 7 H:= 6.984 7 N:= 1.000 O:= 2.001 2 5.105 mol 0.7288 mol 5.09 mol 0.7288 mol 1.458 mol 0.7288 mol
40 Calculating Empirical Formulas These are the subscripts for the empirical formula: C 7 H 7 NO 2
41 Combustion Analysis Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this – C is determined from the mass of CO 2 produced – H is determined from the mass of H 2 O produced – O is determined by difference after the C and H have been determined
42 Example The combustion of 5.00 grams of an alcohol produces 9.55 g of CO 2 and 5.87 g of H 2 O. Find the empirical formula. Solution: First you need to find the individual masses of the elements. The general equation would look like this: C x H y O z + O 2 CO 2 + H 2 O
43 The mass of the carbon in the CO 2 all came from the alcohol so… 9.55 gCO 2 X 12.01 g C 2.61 g C 44.01 gCO 2 The mass of the hydrogen in H 2 O all came from the alcohol so… 5.87 g H 2 O X 2.02 g H 0.658 g H 18.02 g H 2 O
44 Because the mass of the oxygen in both CO 2 and H 2 O is derived from both the alcohol and the oxygen from combustion you need to find the mass of oxygen only in the alcohol…. By subtracting the mass of the carbon and hydrogen from the total mass of the compound. 5.00 g – (2.61 g + 0.658 g) = 1.73 g O
45 Now you can determine the number of moles of each element in the formula….. Moles of C = 2.61 g X 1.00 mole C 0.217 mol C 12.01 g C Moles of H = 0.658 g X 1.00 mole H 0.651 mol H 1.01 g H Moles of O = 1.73 g O X 1.00 mole O 0.108 mol O 16.00 g O
46 From the number of moles we find the mole ratios,…. C = 0.217/0.108 = 2.01 H = 0.651/ 0.108 = 6.03 O = 0.108/ 0.108 = 1.00 Thus the formula will be C 2 H 6 O or C 2 H 5 OH
47 Practice Problem 1.540 g of an organic acid burns completely to produce 2.257 g CO 2 and 0.9241 g H 2 O. Find the empirical formula. If the molecular mass is 90.0 grams what is the molecular formula? C 2.257 gCO 2 X 12.01/44.01 = 0.6159g H 0.9241g H 2 O X 2.02/18.02 = 0.1036g O 1.540 –(0.6159 + 0.1036) = 0.8205g C 0.6159/12.01= 0.0513 H 0.1026/ 1.01 = 0.1026 O 0.8205/16.00= 0.05128 Empirical formula= CH 2 O Molecular formula = (CH 2 O)x X= molecular mass/empirical mass X= 90/30= 3 C 3 H 6 O 3
Example 3-6B Combustion of a 1.505 g sample of thiophene, a carbon-hydrogen-sulfur compound yields 3.149 g CO 2, 0.645 g H 2 O and 1.146 g SO 2 as the only combustion products. What is the empirical formula?
3.4 Oxidation States Tell the number of electrons gained or lost when forming compounds Handout
Rules for Assigning Oxidation States (OS) 1.The OS of a free element is 0 2.The total OS of all atoms in a compound is 0 3.The total OS of all atoms in a ion is equal to the charge of the ion 4.Group 1 = +1, Group 2 = +2 5.Fluorine = -1 6.Hydrogen usually +1 7.Oxygen usually -2 8.In Binary Ionic : Group 17 = -1, Group 16 = -2, Group 15 = -3
Exceptions Rule 6: H bonded to metals; LiH, NaH, CaH 2 Rule 7: O-F Bonds OF 2, H 2 O 2, and KO 2
Example What are the oxidation states of: S 8 Cr 2 O 7 2- MgCl 2 Al 2 O 3 MnO 4 -
3.5 Naming Compounds: Organic and Inorganic Compounds
57 Cations 1. Cations from metal atoms have the same name as the metal Na + sodium ion Ba 2+ barium ion 2. Use Roman numerals to indicate the charge on a metal ion that can form more than one positive charge Fe 2+ Iron (II) ion Fe 3+ Iron(III) ion
58 Cations Latin names for some metals still are used and the ending –ous and –ic are used to indicate the charge Fe 2+ ferrous ion Fe 3+ ferric Sn 2+ stannous ion Sn 4+ stannic ion Other latin root names include: Plumbous Auric Cuprous
59 Cations Cations from nonmetal atoms end in-ium NH 4 + ammonium ion H 3 O + hydronium ion
60 Anions Monatomic anions are named are named by adding –ide at the end H - Hydride S 2- sulfide P 3- phosphide Some polyatomics have –ide endings OH - hydroxide CN - cyanide O 2 2- peroxide
61 Anions Polyatomic anions containing oxygen end in –ite or -ate The –ate ending is used for the most common oxyanion of the element. The –ite ending is used for the oxyanion that has the same charge but one O atom less. NO 3 - nitrate NO 2 - nitrite SO 4 2- sulfate SO 3 2- sulfite
62 Anions Prefixes are used when there is a series of four oxyanions. (usually the halogens) Per- is used to indicate one more O than the –ate ending and hypo- is used for one less O than the –ite ending. ClO 4 - perchlorate ClO 3 - chlorate ClO 2 - chlorite ClO - hypochlorite
63 Anions Anions with H + are named by adding the prefix hydrogen or dihydrogen HCO 3 - hydrogen carbonate (bicarbonate) HSO 4 - dihydrogen sulfate (bisulfate)
64 Writing Formulas Because compounds are electrically neutral, one can determine the formula of a compound this way: – The charge on the cation becomes the subscript on the anion. – The charge on the anion becomes the subscript on the cation. – If these subscripts are not in the lowest whole-number ratio, divide them by the greatest common factor.
73 Inorganic Nomenclature Write the name of the cation. If the anion is an element, change its ending to -ide; if the anion is a polyatomic ion, simply write the name of the polyatomic ion. If the cation can have more than one possible charge, write the charge as a Roman numeral in parentheses.
74 Name the following compounds: (a)K 2 SO 4 (b) Ba(OH) 2 (c) FeCl 3 Write the chemical formulas for the following compounds: (a)potassium sulfide, (b) calcium hydrogen carbonate, (c) nickel(II) perchlorate.
75 Acid Nomenclature If the anion in the acid ends in -ide, change the ending to -ic acid and add the prefix hydro- : – HCl: hydrochloric acid – HBr: hydrobromic acid – HI: hydroiodic acid
76 Acid Nomenclature If the anion in the acid ends in -ite, change the ending to -ous acid: – HClO: hypochlorous acid – HClO 2 : chlorous acid
77 Acid Nomenclature If the anion in the acid ends in -ate, change the ending to -ic acid: – HClO 3 : chloric acid – HClO 4 : perchloric acid
78 Nomenclature of Binary Covalent Compounds The less electronegative atom is usually listed first. A prefix is used to denote the number of atoms of each element in the compound (mono- is not used on the first element listed, however.)
79 Nomenclature of Binary Compounds The ending on the more electronegative element is changed to -ide. – CO 2 : carbon dioxide – CCl 4 : carbon tetrachloride
80 Nomenclature of Binary Compounds If the prefix ends with a or o and the name of the element begins with a vowel, the two successive vowels are often elided into one: N 2 O 5 : dinitrogen pentoxide
Organic compounds Alkanes-Carbon atoms bonded to four other atoms Methane CH 4 Ethane C 2 H 6 Propane C 3 H 8 Butane C 4 H 10 Alkanes with 5 or more carbons use the following prefixes Penta Hexa Hepta Octa Nona Deca 82
Organic functional groups An alcohol is an alkane with an –OH group and is named by adding an –ol ending Methanol In naming the carbon with the functional group is identified by a number 2-Propanol 83