Learning Chemical Formulas Simple Ionic Compounds Concept Ionic

Learning Chemical Formulas

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: Na 1+ Mg 2+ A 3+ P 4+

Simple Ionic Compounds z. Fact 2: Negative ions are formed by nonmetals that gained electrons. Examples: Cl 1 - O 2 N 3 - Si 4 -

Simple Ionic Compounds Fact 3: The overall charge of ionic compounds must equal zero. Examples : Na 1+ Cl 1 - Ca 2+Br 1 Al 3+O 2 -

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: (SO 4) = S +6 and O -8 = -2 (NO 3)= N +5 and O -6 = -1 (NH 4) = N -3 and H +4 = +1

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: Ca 2+(NO 3) 1 -2 Ga 3+ (SO 4) 2(NH 4) 1+2 O 2 -

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: (NO 3)1 - is nitrate (SO 4)2 - is sulfate (Cl. O 3) 1 - is chlorate

Polyatomic Ion Formulas Fact 7: Those polyatomic ions with one oxygen less than the "ate" ions are named "----ite" ions. Examples: (NO 2)1 - is nitrite (SO 3)2 - is sulfite (Cl. O 2)1 - is chlorite

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: (SO 4)2 -becomes H(SO 4)1(SO 3)2 - becomes H(SO 3)-1 (CO 3)2 - becomes H(CO 31(PO 4)3 - becomes H(PO 4)2 -

Polyatomic Ion Formulas Fact 9: These ion groups are named hydrogen ____ or bi ______. Examples: H(CO 3)1 - is called hydrogen carbonate or bicarbonate H(SO 4) 1 - is called hydrogen sulfate or bisulfate

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: Cu 1+is Copper I Pb 2+is Lead II Fe 3+is Iron III Sn 4+s Tin IV

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: Cu 1+ Cl 1 Pb 2+O 2 Fe 3+Br 1 -3 Sn 4+O 2 -

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 1+ Zn 2+ Cd 2+

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: Na. Cl has one atom of sodium and one atom of chlorine. H 2 SO 4 has two atoms of hydrogen, one atom of sulfur and four atoms of oxygen

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 = 23 g/mol and Cl = 35. 5 g/mol so Na. Cl has a molar mass of 58. 5 g/mol Nitric acid is HNO 3 so its molar mass is H=1 x 1 = 1 N = 14 x 1 = 14 O = 16 x 3 = 48 Total = 63 g/mol

Using Formulas in Problem Solving Fact 15: From the formula ion charges, also called oxidation numbers, can be determined using simple math processes. Example: Fe 2 O 3 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--2 O-23 It is clear that the missing charge is +3. Thus the complete name of this compound is iron III oxide.

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 Na. Cl Na = 23. 0 % Na = 23. 0 x 100 = 39. 3%Na Cl = 35. 5 58. 5 g/mol % Cl = 35. 5 x 100 = 60. 7% Cl 58. 5 Notice that the total of the percentages is always equal or very close to 100%.