Chapter 3 Chemical Compounds

3.1 Types of Chemical Compounds and Their Formulas

Molecular Compounds Molecular compounds are composed of molecules and contain only nonmetals. Electrons are shared AKA Covalent Compounds Represented by chemical formulas

The ratio of the masses of carbon and hydrogen, C:H in methane is 1. 4:1 2. 1:4 3. 3:1 4. 1:3 5. 1:1 Slide 4 of 49

The ratio of the masses of carbon and hydrogen, C:H in methane is 1. 4:1 2. 1:4 3. 3:1 4. 1:3 5. 1:1 Slide 5 of 49

The ratio of the masses of carbon and hydrogen, C:H in ethane is 1. 4:1 2. 1:4 3. 3:1 4. 1:3 5. 1:1 Slide 6 of 49

The ratio of the masses of carbon and hydrogen, C:H in ethane is 1. 4:1 2. 1:4 3. 3:1 4. 1:3 5. 1:1 Slide 7 of 49

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.

Figure: 02-21-02UN Title: Ionization of sodium. Caption: A sodium atom loses an electron to form a sodium ion.

Figure: 02-21-03UN Title: Formation of a chloride ion. Caption: A chlorine atom gains an electron to form a chloride ion.

Ionic Bonds Ionic compounds (such as NaCl) are generally formed between metals and nonmetals.

A mole of solid sodium chloride, salt, contains 22.99 g of sodium and 34.45 g of chlorine A B 2. 6.02x1023 g of sodium and 6.02x1023 g of chloride 3. 22.99 u of sodium and 34.45 u of chloride 4. 22.99 g of sodium and 34.45 g of chloride. Slide 12 of 49

A mole of solid sodium chloride, salt, contains 22.99 g of sodium and 34.45 g of chlorine A B 2. 6.02x1023 g of sodium and 6.02x1023 g of chloride 3. 22.99 u of sodium and 34.45 u of chloride 4. 22.99 g of sodium and 34.45 g of chloride. Slide 13 of 49

A molecule of solid sodium chloride, salt, contains 22.99 g of Na and 34.45 g of Cl A B 2. 22.99 u of Na+ and 34.45 u of Cl- 3. 22.99 g of Na and 34.45 g of Cl 4. None of the above Slide 14 of 49

A molecule of solid sodium chloride, salt, contains 22.99 g of sodium and 34.45 g of chlorine A B 2. 22.99 u of sodium and 34.45 u of chloride 3. 22.99 g of sodium and 34.45 g of chloride. 4. None of the above; sodium chloride is an ionic compound, as such, it does not exist in discrete molecular units as is characteristic of covalent compounds. Slide 15 of 49

Would you expect the following to be ionic or molecular: N2O Na2O CaCl2 SF4 CBr4 FeS P4O6 PbF2

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

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)

For the ball and stick model of naphthalene to the right, the empirical formula is 1. C10H8 2. C4H5 3. C5H4 4. CH Slide 20 of 49

For the ball and stick model of naphthalene to the right, the empirical formula is 1. C10H8 2. C4H5 3. C5H4 4. CH Slide 21 of 49

For the ball and stick model of pyrimidine to the right, the molecular formula is: 1. C4N2H4 2. C2N2H2 3. C2NH2 4. (CH)4N2 Slide 22 of 49

For the ball and stick model of pyrimidine to the right, the molecular formula is: 1. C4N2H4 2. C2N2H2 3. C2NH2 4. (CH)4N2 Slide 23 of 49

3.2 The Mole Concept and Chemical Compounds

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) H2O: Molecule Mass 18.0153 amu (1 molecule) Molar Mass 18.0153 grams (1 mole)

Molecular Formulas Diatomic molecules: H2 O2 N2 F2 Cl2 Br2 I2 Molecules: P4 (White Phosphorus) S8 (Sulfur) Distinguish between molecule and atom!

Example 3-1A How many grams of MgCl2 would you need to obtain 5.0 x 10-23 Cl- ions?

Example 3-2A Gold has a density of 19.32 g/cm3. 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: C2HBrClF3 How many grams of Br are contained in 25.00 mL of halothane (d=1.871g/mL)

3.3 Composition of Chemical Compounds

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

Percent Composition So the percentage of carbon in ethane is… (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 C10H16N5P3O13?

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?

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.

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 14.01 g 1.01 g 16.00 g

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 1.458 mol

Calculating Empirical Formulas These are the subscripts for the empirical formula: C7H7NO2

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 CO2 produced H is determined from the mass of H2O produced O is determined by difference after the C and H have been determined

Example The combustion of 5.00 grams of an alcohol produces 9.55 g of CO2 and 5.87 g of H2O. Find the empirical formula.   Solution: First you need to find the individual masses of the elements. The general equation would look like this: CxHyOz + O2  CO2 + H2O

The mass of the carbon in the CO2 all came from the alcohol so… 9.55 gCO2 X 12.01 g C 2.61 g C 44.01 gCO2 The mass of the hydrogen in H2O all came from the alcohol so… 5.87 g H2O X 2.02 g H 0.658 g H 18.02 g H2O

Because the mass of the oxygen in both CO2 and H2O 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

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

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 C2H6O or C2H5OH

Practice Problem 1.540 g of an organic acid burns completely to produce 2.257 g CO2 and 0.9241 g H2O. Find the empirical formula. If the molecular mass is 90.0 grams what is the molecular formula? C 2.257 gCO2 X 12.01/44.01 = 0.6159g H 0.9241g H2O X 2.02/18.02 = 0.1036g O 1.540 –(0.6159 + 0.1036) = 0.8205g Empirical formula= CH2O Molecular formula = (CH2O)x X= molecular mass/empirical mass X= 90/30= 3 C3H6O3 C 0.6159/12.01= 0.0513 H 0.1026/ 1.01 = 0.1026 O 0.8205/16.00= 0.05128

Example 3-6B Combustion of a 1.505 g sample of thiophene, a carbon-hydrogen-sulfur compound yields 3.149 g CO2, 0.645 g H2O and 1.146 g SO2 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

Common Oxidation States

Rules for Assigning Oxidation States (OS) The OS of a free element is 0 The total OS of all atoms in a compound is 0 The total OS of all atoms in a ion is equal to the charge of the ion Group 1 = +1, Group 2 = +2 Fluorine = -1 Hydrogen usually +1 Oxygen usually -2 In Binary Ionic : Group 17 = -1, Group 16 = -2, Group 15 = -3

Exceptions Rule 6: H bonded to metals; LiH, NaH, CaH2 Rule 7: O-F Bonds OF2, H2O2, and KO2

Example What are the oxidation states of: S8 Cr2O72- MgCl2 Al2O3 MnO4-

The oxidation number for N in nitric acid, HNO3, is 1. 1 2. 2 3. 3 4. 4 5. 5 Slide 54 of 49

The oxidation number for N in nitric acid, HNO3, is 1. 1 2. 2 3. 3 4. 4 5. 5 Slide 55 of 49

3.5 Naming Compounds: Organic and Inorganic Compounds

Cations 1. Cations from metal atoms have the same name as the metal Na+ sodium ion Ba2+ barium ion 2. Use Roman numerals to indicate the charge on a metal ion that can form more than one positive charge Fe2+ Iron (II) ion Fe3+ Iron(III) ion

Cations Latin names for some metals still are used and the ending –ous and –ic are used to indicate the charge Fe2+ ferrous ion Fe3+ ferric Sn2+ stannous ion Sn4+ stannic ion Other latin root names include: Plumbous Auric Cuprous

Cations Cations from nonmetal atoms end in-ium NH4+ ammonium ion H3O+ hydronium ion

Anions Monatomic anions are named are named by adding –ide at the end H- Hydride S2- sulfide P3- phosphide Some polyatomics have –ide endings OH- hydroxide CN- cyanide O22- peroxide

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. NO3- nitrate NO2- nitrite SO42- sulfate SO32- sulfite

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. ClO4- perchlorate ClO3- chlorate ClO2- chlorite ClO- hypochlorite

Anions Anions with H+ are named by adding the prefix hydrogen or dihydrogen HCO3- hydrogen carbonate (bicarbonate) HSO4- dihydrogen sulfate (bisulfate)

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.

Potassium dichromate is used in breathalyzers Potassium dichromate is used in breathalyzers. When it comes in contact with alcohol vapor it turns from orange to green. It is an ionic compound where the polyatomic anion has the formula Cr2O72-. What is the chemical formula for potassium dichromate? 1. KCr2O7 2. K(Cr2O7)2 3. K2Cr2O7 4. K2(Cr2O7)3 5. K3(Cr2O7)2 Slide 65 of 49

Potassium dichromate is used in breathalyzers Potassium dichromate is used in breathalyzers. When it comes in contact with alcohol vapor it turns from orange to green. It is an ionic compound where the polyatomic anion has the formula Cr2O72-. What is the chemical formula for potassium dichromate? 1. KCr2O7 2. K(Cr2O7)2 3. K2Cr2O7 4. K2(Cr2O7)3 5. K3(Cr2O7)2 Slide 66 of 49

Sodium phosphate is an active component in some constipation medicines and enemas. The chemical formula for sodium phosphate is 1. NaPO4 2. Na2PO3 3. Na2PO4 4. Na3PO4 5. Na3(PO4)2 Slide 67 of 49

Sodium phosphate is an active component in some constipation medicines and enemas. The chemical formula for sodium phosphate is 1. NaPO4 2. Na2PO3 3. Na2PO4 4. Na3PO4 5. Na3(PO4)2 Slide 68 of 49

Marble consists of primarily of calcium carbonate Marble consists of primarily of calcium carbonate. Acids react with and dissolve marble as evident by this marble statue eroded by acid rain. The chemical formula for calcium carbonate is 1. CaCO3 2. Ca2CO3 3. CaCO4 4. Ca3(CO3)2 5. Ca(CO3)2 Slide 69 of 49

Marble consists of primarily of calcium carbonate Marble consists of primarily of calcium carbonate. Acids react with and dissolve marble as evident by this marble statue eroded by acid rain. The chemical formula for calcium carbonate is 1. CaCO3 2. Ca2CO3 3. CaCO4 4. Ca3(CO3)2 5. Ca(CO3)2 Slide 70 of 49

Common Cations

Common Anions

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.

Name the following compounds: K2SO4 Ba(OH)2 FeCl3 Write the chemical formulas for the following compounds: potassium sulfide, calcium hydrogen carbonate, nickel(II) perchlorate.

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

Acid Nomenclature If the anion in the acid ends in -ite, change the ending to -ous acid: HClO: hypochlorous acid HClO2: chlorous acid

Acid Nomenclature If the anion in the acid ends in -ate, change the ending to -ic acid: HClO3: chloric acid HClO4: perchloric acid

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.)

Nomenclature of Binary Compounds The ending on the more electronegative element is changed to -ide. CO2: carbon dioxide CCl4: carbon tetrachloride

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: N2O5: dinitrogen pentoxide

3.6 Names and Formulas of Inorganic Compounds

Organic compounds Alkanes-Carbon atoms bonded to four other atoms Methane CH4 Ethane C2H6 Propane C3H8 Butane C4H10 Alkanes with 5 or more carbons use the following prefixes Penta Hexa Hepta Octa Nona Deca

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