2 Simple Ionic Compounds Concept:Ionic Formulas are formed from positive and negative ions.Fact 1:Positive ions are formed by metals that lost electrons.Examples: Na1+ Mg2+ A3+ P4+
3 Simple Ionic Compounds Fact 2: Negative ions are formed by nonmetals that gained electrons Examples: Cl O2- N3- Si4-
4 Simple Ionic Compounds Fact 3:The overall charge of ionic compounds must equal zero.Examples : Na1+ Cl1- Ca2+Br Al3+O2-
5 Lesson Two--Polyatomic Ion Formulas Concept:Polyatomic ions are groups of atoms that behave as one unit.Fact 4:Some ions have more that one atom but their overall charge can be determined.Examples: (SO4) = S +6 and O -8 = -2(NO3)= N +5 and O -6 = -1(NH4) = N -3 and H +4 = +1
6 Polyatomic Ion Formulas Fact 5: These group ions, polyatomic ions, are treated like single ions in formulas, but must have parentheses when more than one is used in a formula.Examples: Ca2+(NO3) Ga3+ (SO4) 2-(NH4) 1+2O2-
7 Polyatomic Ion Formulas Fact 6:The polyatomic ions are named based on the atoms that they contain. Those with oxygen and another nonmetal are often name "____ate" with the root of the other nonmetal in the blank.Examples: (NO3)1- is nitrate(SO4)2- is sulfate(ClO3) 1- is chlorate
8 Polyatomic Ion Formulas Fact 7:Those polyatomic ions with one oxygen less than the "ate" ions are named "----ite" ions.Examples: (NO2) is nitrite(SO3)2- is sulfite(ClO2)1- is chlorite
9 Polyatomic Ion Formulas Fact 8:Some polyatomic ions contain a regular polyatomic ion and one or more hydrogen ions. Each hydrogen that is present add one positive charge to the total charge.Examples: (SO4)2-becomes H(SO4)1-(SO3)2- becomes H(SO3)-1(CO3)2- becomes H(CO31-(PO4)3- becomes H(PO4)2-
10 Polyatomic Ion Formulas Fact 9:These ion groups are named hydrogen ____or bi ______.Examples: H(CO3)1- is called hydrogen carbonate or bicarbonateH(SO4) 1- is called hydrogen sulfate or bisulfate
11 Lesson Three--Transition Metal Compounds Concept:Transition metals have electrons in d orbitals and can donate different numbers of electrons, thus giving them several different positive charges.Fact 10:These can be determined from the Roman numeral which is written next to the metal's name.Example: Cu1+is Copper IPb2+is Lead IIFe3+is Iron IIISn4+s Tin IV
12 Transition Metal Compounds Fact 11:These transition metals are used in formulas just like other metals, once the charge is determined from the Roman numeral in the name.Example: Cu1+ Cl Pb2+O2-Fe3+Br Sn4+O2-
13 Transition Metal Compounds Fact 12:A few transition metal ions only have one charge and never change so they can be written without a Roman numeral in their formula name.Example: Ag Zn2+ Cd2+
14 Lesson Four--Using Formulas in Problem Solving Concept: Correctly written chemical formulas hold a large amount of information for the prepared student to find.Fact 13:The subscripts tell us the number of atoms of each kind that is present in the compound.Example: NaCl has one atom of sodium and one atom of chlorine.H2SO4 has two atoms of hydrogen, one atom of sulfur and four atoms of oxygen
15 Using Formulas in Problem Solving Fact 14:The sum of the atomic masses times the number of atoms of each kind of element is equal to the mass of one mole of the substance.Examples: Na = 23g/mol and Cl = 35.5g/mol so NaCl has a molar mass of 58.5g/molNitric acid is HNO3so its molar mass isH = 1 x 1 = 1N = 14 x 1 = 14O = 16 x 3 = 48Total = 63g/mol
16 Using Formulas in Problem Solving Fact 15:From the formula ion charges, also called oxidation numbers, can be determined using simple math processes.Example:Fe2O3 is iron oxide, but what is the charge of the iron in this formula? It is known that the oxygen's ionic charge is -2 so substitute in that charge and find the iron's ionic charge.Fe--2O It is clear that the missing charge is +3.Thus the complete name of this compound isiron III oxide.
17 Using Formulas in Problem Solving Fact 16: The percentage composition of each element in a compound can be determined using only the correct formula and the atomic masses.Example: Sodium chloride or NaClNa = % Na = x 100 = 39.3%NaCl =58.5g/mol% Cl = x 100 = 60.7% Cl58.5Notice that the total of the percentages is always equal or very close to 100%.