Chapter 3 Molecules, Ions and Compounds Molecules A molecule is the smallest particle of an element that can have a stable independent existence. Usually have 2 or more atoms bonded together Examples of molecules H 2 O 2 S 8 H 2 O CH 4 C 2 H 5 OH Chemical Formulas Chemical formula shows the chemical composition of the substance. ratio of the elements present in the molecule or compound He, Au, Na monatomic elements O 2, H 2, Cl 2, F 2, I 2, N 2, Br 2 diatomic elements (molecules) O 3, S 4, P 8 - more complex molecules H 2 O, C 12 H 22 O 11 compounds Substance consists of two or more different elements 1
Chemical Formulas Compound 1 Molecule Contains HCl 1 H atom & 1 Cl atom H 2 O NH 3 C 3 H 8 2 H atoms & 1 O atom 1 N atom & 3 H atoms 3 C atoms & 8 H atoms Chemical bonds Attractive forces that hold atoms together in compounds are called chemical bonds. Chemical bonds are classified into two types: o Ionic bonding results from electrostatic attractions among ions, which are formed by the transfer of one or more electrons from one atom to another. o Covalent bonding results from sharing one or more electron pairs between two atoms. Ionic Bonding Formation of Ionic Compounds An ion is an atom or a group of atoms possessing a net electrical charge. 1. positive (+) ions or cations These atoms have lost 1 or more electrons. 2. negative (-) ions or anions These atoms have gained 1 or more electrons. 2
Ions and Ionic Compounds Ions are atoms or groups of atoms that possess an electric charge. Two basic types of ions: Positive ions or cations one or more electrons less than neutral Na +, Ca 2+, Al 3+ NH 4+ - polyatomic cation Negative ions or anions one or more electrons more than neutral F -, O 2-, N 3- SO 4 2-, PO 4 3-, HCO 3- - polyatomic anions You must know the names, formulas, and charges of the elements and common ions in the handout. Formation of Ionic Compounds Ionic bonds are formed by the attraction of cations for anions usually to form solids. Commonly, metals react with nonmetals to form ionic compounds. The formation of NaCl is one example of an ionic compound formation. Formation of Ionic Compounds Reaction of Group IA Metals with Group VIIA Nonmetals G -1metal 2 Li silver solid (s) + G -17 nometal F yellow gas 2(g) 2 LiF (s) white solid with an 842 o melting point C 3
Formation of Ionic Compounds The reaction of potassium with bromine is a second example of a group IA metal with a Group IIA non metal. IA metal VIIA nonmetal 2 K (s) + Br2( l ) 2 KBr (s) ionic solid Formation of Ionic Compounds Simple Binary Ionic Compounds Table Reacting Groups Compound General Formula Example IA + VIIA MX NaF IIA + VIIA MX 2 BaCl 2 IIIA + VIIA MX 3 AlF 3 IA + VIA M 2 X Na 2 O IIA + VIA MX BaO IIIA + VIA M 2 X 3 Al 2 S 3 4
Formation of Ionic Compounds Reacting Groups Compound General Formula Example IA + VA M 3 X Na 3 N IIA + VA M 3 X 2 Mg 3 P 2 IIIA + VA MX AlN H, a nonmetal, forms ionic compounds with IA and IIA metals for example, LiH, KH, CaH 2, and BaH 2. Other hydrogen compounds are covalent. Binary compounds are made of two elements. metal + nonmetal = ionic compound nonmetal + nonmetal = covalent compound Name the more metallic element first. Use the element s name. Name the less metallic element second. Add the suffix ide to the element s stem. Nonmetal Stems Element Stem Boron (B) bor Carbon (C) carb Silicon (Si) silic Nitrogen (N) nitr Fluorine(F) fluor Chlorine (Cl) chlor 5
Bromine (Br) Iodine (I) brom iod Oxygen (O) ox Sulfur (S) sulf Selenium (Se) selen Phosphorus (P) phosph Hydrogen (H) hydr Ionic Compounds are made of a metal cation and a nonmetal anion. Cation named first Anion named second Stem plus IDE ending LiBr lithium bromide MgCl 2 magnesium chloride Li 2 S lithium sulfide Al 2 O 3 You do it! LiBr MgCl 2 Li 2 S Al 2 O 3 Na 3 P lithium bromide magnesium chloride lithium sulfide aluminum oxide You do it! 6
LiBr MgCl 2 Li 2 S Al 2 O 3 Na 3 P Mg 3 N 2 lithium bromide magnesium chloride lithium sulfide aluminum oxide sodium phosphide You do it! LiBr lithium bromide MgCl 2 magnesium chloride Li 2 S lithium sulfide Al 2 O 3 aluminum oxide Na 3 P sodium phosphide Mg 3 N 2 magnesium nitride Notice that binary ionic compounds with metals having one oxidation state (representative metals) do not use prefixes or Roman numerals. ionic compounds containing metals that exhibit more than one oxidation state (charge) memorize them, on your handout Metals exhibiting multiple oxidation states are: 1. most of the transition metals 2. metals in groups IIIA (except Al), IVA, & VA 7
There are two methods to name these compounds. 1. Older method add suffix ic to element s Latin name for higher oxidation state add suffix ous to element s Latin name for lower oxidation state 2. Modern method use Roman numerals in parentheses to indicate metal s oxidation state Compound Old System Modern System FeBr 2 ferrous bromide iron(ii) bromide FeBr 3 ferric bromide iron(iii) bromide SnO stannous oxide tin(ii) oxide SnO 2 stannic oxide tin(iv) oxide CoCl 2 cobaltous chloride cobalt(ii) chloride CoCl 3 You do it! Compound Old System Modern System FeBr 2 ferrous bromide iron(ii) bromide FeBr 3 ferric bromide iron(iii) bromide SnO stannous oxide tin(ii) oxide SnO 2 stannic oxide tin(iv) oxide CoCl 2 cobaltous chloride cobalt(ii) chloride CoCl 3 cobaltic chloride cobalt(iii) chloride PbS You do it! PbS 2 You do it! 8
Compound Old System Modern System FeBr 2 ferrous bromide iron(ii) bromide FeBr 3 ferric bromide iron(iii) bromide SnO stannous oxide tin(ii) oxide SnO 2 stannic oxide tin(iv) oxide CoCl 2 cobaltous chloride cobalt(ii) chloride CoCl 3 cobaltic chloride cobalt(iii) chloride PbS plumbous sulfide lead(ii) sulfide PbS 2 plumbic sulfide lead(iv) sulfide There are polyatomic ions that commonly form binary ionic compounds. 1. OH - hydroxide 2. CN - cyanide 3. NH 4+ ammonium Use binary ionic compound naming system. KOH potassium hydroxide Ba(OH) 2 barium hydroxide Al(OH) 3 aluminum hydroxide Fe(OH) 2 You do it! KOH Ba(OH) 2 Al(OH) 3 Fe(OH) 2 Fe(OH) 3 potassium hydroxide barium hydroxide aluminum hydroxide iron (II) hydroxide You do it! Formula writing the the polyatomic ion stays together, if more than one, need parenthesis 9
KOH Ba(OH) 2 Al(OH) 3 Fe(OH) 2 Fe(OH) 3 Ba(CN) 2 potassium hydroxide barium hydroxide aluminum hydroxide iron (II) hydroxide iron (III) hydroxide You do it! KOH Ba(OH) 2 Al(OH) 3 Fe(OH) 2 Fe(OH) 3 Ba(CN) 2 (NH 4 ) 2 S potassium hydroxide barium hydroxide aluminum hydroxide iron (II) hydroxide iron (III) hydroxide barium cyanide You do it! KOH Ba(OH) 2 Al(OH) 3 Fe(OH) 2 Fe(OH) 3 Ba(CN) 2 (NH 4 ) 2 S NH 4 CN potassium hydroxide barium hydroxide aluminum hydroxide iron (II) hydroxide iron (III) hydroxide barium cyanide ammonium sulfide You do it! 10
KOH Ba(OH) 2 Al(OH) 3 Fe(OH) 2 Fe(OH) 3 Ba(CN) 2 (NH 4 ) 2 S NH 4 CN potassium hydroxide barium hydroxide aluminum hydroxide iron (II) hydroxide iron (III) hydroxide barium cyanide ammonium sulfide ammonium cyanide Names and Formulas of Some Ionic Compounds Formulas of ionic compounds are determined by the charges of the ions. Charge on the cations must equal the charge on the anions. The compound must be neutral. NaCl sodium chloride (Na 1+ & Cl 1- ) KOH potassium hydroxide(k 1+ & OH 1- ) CaSO 4 calcium sulfate (Ca 2+ & SO 4 2- ) Al(OH) 3 aluminum hydroxide (Al 3+ & 3 OH 1- ) Writing binary chemical formulas Charge becomes subscript. Then give the subscript as lowest common denominator Li Ca Al 1+ 2+ 3+ Mg Ba K 2+ 2+ + Cu Fe 2+ 3+ Cl F Br O N I O 2 3 Cl 2 LiCl CaF AlBr 3?????? 2 2 3 MgO BaN 2 2 MgO 11
Names and Formulas What is the formula of nitric acid? What is the formula of sulfur trioxide? Names and Formulas of Some Ionic Compounds What is the name of K 2 SO 3? What is charge on sulfite ion? What is the formula of ammonium sulfide? What is charge on ammonium ion? What is the formula of aluminum sulfate? What is charge on both ions? Name or Write the formula: Formula Name 1.Cu(OH) 2 2. CuOH 3.MgCl 2 4.Li 2 O 5.Zn 3 N 2 6. calcium bromide 7. sodium hydroxide 8. aluminum phosphide 9. barium iodide 10. magnesium cyanide 12
Name or Write the formula: Formula Name 1. iron(ii) bromide 2. iron(iii) hydroxide 3. copper(ii) oxide 4. lead(iv) cyanide Binary Acids are binary compounds consisting of hydrogen and a nonmetal. Compounds are usually gases at room temperature and pressure. Nomenclature for the gaseous compounds is hydrogen (stem)ide. Formula HF HCl HBr H 2 S Name hydrogen fluoride hydrogen chloride hydrogen bromide You do it! 13
Formula HF HCl HBr H 2 S Name hydrogen fluoride hydrogen chloride hydrogen bromide hydrogen sulfide Salts of poly atomic ions name metal then ion Salt NaNO 2 sodium nitrite NaNO 3 sodium nitrate Na 2 SO 3 sodium sulfite Na 2 SO 4 sodium sulfate Na 3 PO 4 sodium phosphate Name these salts MgSO 4 Ca(NO 3 ) 2 BaCO 3 K 3 PO 4 14
Acidic Salts are made from ternary acids that retain one or more of their acidic hydrogen atoms. Made from acid base reactions where there is an insufficient amount of base to react with all of the hydrogen atoms. Modern system uses prefixes and the word hydrogen. NaHCO 3 Old system Modern system KHSO 4 Modern system KH 2 PO 4 Modern system K 2 HPO 4 sodium bicarbonate sodium hydrogen carbonate potassium hydrogen sulfate potassium dihydrogen phosphate You do it! Covalent Bonding Covalent bonds are formed when two nonmetals combine The atoms share electrons. If the atoms share 2 electrons a single covalent bond is formed. If the atoms share 4 electrons a double covalent bond is formed. If the atoms share 6 electrons a triple covalent bond is formed. 15
Formation of Covalent Bonds Representation of the formation of an H 2 molecule from H atoms. Covalent molecular compounds composed of two nonmetals other than hydrogen Nomenclature must include prefixes that specify the number of atoms of each element in the compound. Use the minimum number of prefixes necessary to specify the compound. Frequently drop the prefix mono-. 16
Formula CO CO 2 SO 3 OF 2 P 4 O 6 Name carbon monoxide carbon dioxide sulfur trioxide oxygen difluoride You do it! 17
Inorganic Compounds Formula Name CO carbon monoxide CO 2 carbon dioxide SO 3 sulfur trioxide OF 2 oxygen difluoride P 4 O 6 tetraphosphorus hexoxide P 4 O 10 You do it! Formula CO?? SO 3?? P 4 O 6?? Name carbon monoxide carbon dioxide sulfur trioxide oxygen difluoride tetraphosphorus hexoxide tetraphosphorus decoxide Formula N 2 O NO N 2 O 3 NO 2???? Modern Name dinitrogen monoxide nitrogen monoxide dinitrogen trioxide nitrogen dioxide dinitrogen tetroxide dinitrogen pentoxide 18
Naming Review Formula Weights, Molecular Weights, and Moles of compounds Add atomic weights of each atom The molar mass of propane, C 3 H 8, is: 3 C= 3 12.01amu 8 H= 8 1.01amu Molar mass = 36.03amu = 8.08 amu = 44.11amu Formula Weights, Molecular Weights, and Moles The molar mass of calcium nitrate, Ca(NO 3 ) 2, is: 19
Formula Weights, Molecular Weights, and Moles 1 Ca = 1 40.08 amu 2 N = 2 14.01 amu 6 O = 6 16.00 amu Molar mass = 40.08 amu = 28.02 amu = 96.00 amu = 164.10 amu Formula Weights, Molecular Weights, and Moles One Mole of Contains Cl 2 or 70.90g 6.022 x 10 23 Cl 2 molecules C 3 H 8 2(6.022 x 10 23 ) Cl atoms Formula Weights, Molecular Weights, and Moles One Mole of Contains Cl 2 or 70.90g 6.022 x 10 23 Cl 2 molecules 2(6.022 x 10 23 ) Cl atoms C 3 H 8 or 44.11 g 6.022 x 10 23 C 3 H 8 molecules 3 (6.022 x 10 23 ) C atoms 8 (6.022 x 10 23 ) H atoms 20
Formula Weights, Molecular Weights, and Moles Calculate the number of C 3 H 8 molecules in 74.6 g of propane.? C 3H8 molecules= 74.6 g C3H8 Formula Weights, Molecular Weights, and Moles Calculate the number of C 3 H 8 molecules in 74.6 g of propane.? C H 3 8 molecules= 74.6 g C H 23 1mole C H 3 8 6.022 10 C H molecules 3 8 = 44.11 g C3H 8 44.11 g C3H 8 1.02 10 24 molecules 3 8 Formula Weights, Molecular Weights, and Moles Calculate the number of O atoms in 26.5 g of Li 2 CO 3. 21
Formula Weights, Molecular Weights, and Moles Calculate the number of O atoms in 26.5 g of Li 2 CO 3. 1 mol Li2CO3? O atoms= 26.5 g Li2CO3 73.8 g Li CO 23 6.022 10 form.units Li2CO 1 mol Li CO 2 3 6.49 10 23 3 3 O atoms 1formula unit Li CO O atoms 2 3 2 3 = Formula Weights, Molecular Weights, and Moles Occasionally, we will use millimoles. Symbol - mmol 1000 mmol = 1 mol For example: oxalic acid (COOH) 2 1 mol = 90.04 g 1 mmol = 0.09004 g or 90.04 mg Percent Composition and Formulas of Compounds % composition = mass of an individual element in a compound divided by the total mass of the compound x 100% Determine the percent composition of C in C 3 H 8. mass C % C= 100% mass C H 3 12.01 g = 100% 44.11 g = 81.68% 3 8 22
Percent Composition and Formulas of Compounds What is the percent composition of H in C 3 H 8? What is the percent composition of H in H 2 O? You do it! Percent Composition and Formulas of Compounds What is the percent composition of H in C 3 H 8? mass H % H = 100% mass C H 8 H = 100% C H 8 1.01 g = 100% = 18.32 % 44.11 g or 3 18.32% = 100% 81.68% 8 3 8 Percent Composition Example 2-10: Calculate the percent composition of Fe 2 (SO 4 ) 3 to 3 sig. fig. 2 Fe 2 55.8 g % Fe= 100% = 100% = 27.9% Fe Fe (SO ) 399.9 g 2 3 S 3 32.1g % S = 100% = 100% = 24.1% S Fe (SO ) 399.9 g 2 12 O 12 16.0 g % O = 100% = 100% = 48.0% O Fe (SO ) 399.9 g 2 4 3 4 3 4 3 Total = 100% 23
Empirical and Molecular Formulas Empirical Formula - smallest whole-number ratio of atoms present in a compound Molecular Formula - actual numbers of atoms of each element present in a molecule of the compound We determine the empirical and molecular formulas of a compound from the percent composition of the compound. Empirical And Molecular Formulas Empirical Formulas Example 2-11: A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula? Make the simplifying assumption that we have 100.0 g of compound. In 100.0 g of compound there are: 24.74 g of K 34.76 g of Mn 40.50 g of O 24
Empirical Formulas 1 mol K? mol K= 24.74 g K = 0.6327 mol K 39.10 g K? mol K? mol Mn Empirical Formulas = 24.74 g K = 34.76 g Mn 1 mol K 39.10 g K 1 mol Mn 54.94 g Mn = 0.6327 mol K = 0.6327 mol Mn? mol K? mol Mn? mol O Empirical Formulas = 24.74 g K = 34.76 g Mn = 40.50 g O 1 mol K 39.10 g K 1 mol Mn 54.94 g Mn 1mol O 16.00 g O = 2.531mol O obtain smallest whole number ratio = 0.6327 mol K = 0.6327 mol Mn 25
? mol K? mol Mn? mol O Empirical Formulas 1 mol K = 24.74 g K = 0.6327 mol K 39.10 g K = 34.76 g Mn = 40.50 g O obtain smallest whole number ratio 0.6327 for K = 1 K 0.6327 1 mol Mn 54.94 g Mn 1mol O 16.00 g O = 0.6327 mol Mn = 2.531mol O 0.6327 for Mn = 1 Mn 0.6327 Empirical Formulas? mol? mol? mol K for K 1 mol Mn Mn = 34.76 g Mn = 0.6327 mol Mn 54.94 g Mn O 1 mol K = 24.74 g K = 0.6327 mol K 39.10 g K obtain 1mol O = 40.50 g O = 2.531 mol O 16.00 g O smallest whole number ratio 0.6327 0.6327 = 1 K for Mn 0.6327 0.6327 2.531 for O = 4 O 0.6327 thus the chemical formula is KMnO 4 = 1 Mn Empirical Formulas Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? You do it! 26
Empirical Formulas Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound?? mol Co? mol O = 6.541g Co = 2.368 g O 1 mol Co 58.93 gco 1mol O 16.00 g O = 0.1480 mol O find smallest whole number ratio = 0.1110 mol Co Empirical Formulas Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? 0.1110 for Co 0.1110 multipy both by = 1 Co 3 to turn 0.1480 for O = 1.333O 0.1110 fraction to whole number 1 Co 3= 3 Co 1.333 O 3= 4 O Thus the compound' s formula is : Co 3 O 4 Molecular Formulas Example 2-13: A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? short cut method 1 mol contains 56.1g 85.63% is C and 14.37% is H 56.1 g 0.8563= 48.0 g of C 56.1 g 0.1437= 8.10 g of H 27
Determination of Molecular Formulas convert masses to moles 1 mol C 48.0 g of C = 4 mol C 12.0 g C 1 mol H 8.10 g of H = 8 mol H 1.01 g H Thus the formula is : C 4 H 8 Some Other Interpretations of Chemical Formulas What mass of ammonium phosphate, (NH 4 ) 3 PO 4, would contain 15.0 g of N? Some Other Interpretations of Chemical Formulas What mass of ammonium phosphate, (NH 4 ) 3 PO 4, would contain 15.0 g of N? 1.07 mol N molar massof 4 3 1mol(NH4) 3PO 3 mol N (NH ) PO = 149.0g/mol 1mol N? mol N= 15.0g of N = 1.07 mol N 14.0g N 149.0g (NH4) 3PO 0.357mol(NH4) 3PO4 1mol(NH ) PO 4 4 = 0.357mol(NH ) PO 4 3 4 4 4 3 = 53.2g (NH ) PO 4 4 3 4 28