Nomenclature Naming Compounds. Binary Compounds Compounds with only two elements in any ratio.

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Presentation transcript:

Nomenclature Naming Compounds

Binary Compounds

Compounds with only two elements in any ratio

KClNO 2 CaCl 2 P 2 O 5 Al 2 O 3 H 2 S Na 2 Oetc

Binary compounds can be either ionic or covalent

Naming Binary Compounds

Naming Ionic Compounds

Name positive element first with its normal name Name negative element last & change its ending to -ide

KCl CaCl 2 Al 2 O 3 Na 2 O

Name Each: CaCl 2 CaS K 2 OLiF

Name Each: FeCl 2 FeCl 3

If the Positive element is not from columns I or II its ox # must be determined and written in roman numerals

Determining the Charge 1)Add up the oxidation numbers of all the negative elements 2)The positive portion must balance out the negative portion 3)Divide the positive portion by the metal subscript

CuClFe 2 O 3 MnO 2 CrO 3 Name Each:

Molecule

A covalent compound that can exist as a separate unit Non-metals bond to form molecules

Naming Molecules or Covalent Compounds

Same rules as ionic compounds except: use geometric prefixes to determine the # of each atom

Geometric Prefixes 1-mono5-penta 2-di6-hexa 3-tri7-hepta 4-tetraetc

COS 2 O 3 N 2 H 4 SO 3 Name Each:

CaOP 2 O 5 Al 2 S 3 SO 2 Drill: Name Each:

Deriving Formulas 1)Write the symbol for each element 2)Determine ox #s for each 3)Determine lowest common multiple to balance the charge 4)Apply subscripts

Write formulas for: Sodium sulfide Lead (II) iodide Diphosphorus pentoxide

Write formulas for: Chromium(III) oxide Aluminum carbide

Polyatomic Ion

A group of atoms chemically combined that together have a charge

Most are oxoanions PO 4 -3 SO 4 -2 A root element bound to oxygen

Drill: Name each CuCl 2 KBr PCl 5 MgO Mn 2 O 7 S 2 O 3

Naming Polyatomic Ions

Name the root element Change the ending to -ate PO 4 -3 = phosphate Some are unusual

CN -1 OH -1 C 2 H 3 O 2 -1 C 2 O 4 -2

H 2 O H 3 O +1 NH 3 NH 4 +1

Polyatomic Ion Endings Maximum O = -ate 1 less than max O = -ite SO 4 -2 = sulfate SO 3 -2 = sulfite `

Naming Ternary Compounds

Ternary Compounds Compounds containing more than two different elements Most contain polyatomic ions

Follow ionic rules for naming the compound Name the polyatomic ion as the positive or negative portion

CaCO 3 K 2 SO 4 Name Each:

Pb(NO 3 ) 2 MgSO 3 Drill: Name:

Write Formulas For: Lead (II) nitrate Aluminum sulfate Potassium chlorate Ammonium phosphite

Name the Following: BaSO 4 CuNO 3 SO 2 (NH 4 ) 3 PO 4

Naming Acids

Binary acids become: Hydro _____ ic acids HCl - Hydrochloric acid

Ternary acids become: _____ ic acids or _____ ous acids H 2 SO 4 - Sulfuric acid H 2 SO 3 - Sulfurous acid

____ ic acids form from polyatomic ions ending with ___ ate ____ ous acids form from polyatomic ions ending with ___ ite

___ ide ions become: hydro ___ ic acids ___ ate ions become: ___ ic acids ___ ite ions become: ___ ous acids

Name or Give Formulas For: HBr (aq) H 2 SO 4(aq) NaVO 3 (NH 4 ) 3 PO 3 Phosphoric acid Nitric acid Chloric acid

Drill: Name each: KBrMgS BaF 2 K 3 P K 2 OLiH Al 2 O 3 H 2 S

Derive formulas for each: Cesium oxide Barium chloride Calcium phosphide Aluminum sulfide

Name each of the following: SeOCS 2 NO 2 Cl 2 O N 2 O 4 PCl 3

Derive formulas for each: Silicon dioxide phosphorus trichloride Sulfur hexafluoride Iodine trifluoride

Name each of the following: SO 4 -2 SO 3 -2 PO 4 -3 NO 3 -1 ClO 4 -1 ClO 3 -1 ClO 2 -1 ClO -1

Derive formulas for each: ChromateSulfate ArsenateNitrate ArseniteCyanide BromiteAcetate

Name each of the following: HClH 2 SO 4 H 2 S H 3 PO 3 HNO 3 HBrO H 2 CO 3 HBrO 4

Derive formulas for each: Chromic acid Hydroiodic acid Sulfurous acid Bromic acid

Name each of the following: CuOMnS PbO 2 Cu 2 O CrCl 2 MnF 2 CrCl 3 SnCl 4

Percent Composition by Mass

Determine the atomic mass of each element in the compound Determine the molecular mass of the compound by adding Divide each elemental mass by molecular mass Multiply by 100 %

MgCl 2 Mg = 24.3 g/mole 2 Cl = 2 x 35.5 = 71.0 g/mole MgCl 2 = total = 95.3 g/mole % Mg =24.3/95.3 x 100% % Cl = 71.0/95.3 x 100 %

Determine % Comp for Each: Fe 2 O 3 C 3 H 6 O 3 CuSO 4 *5H 2 O

Drill: Determine % Comp for Each: C 3 H 6 O 3 CuSO 4 *5H 2 O

Name each of the following: FeO Fe 2 O 3

Derive formulas for each: Lead(IV)oxide Copper(II)sulfide Manganese(VII)oxide Nickel(II)fluoride

Name each of the following: BaCO 3 KNO 2 CuClO 3 Al 2 (SO 4 ) 3

Derive formulas for each: Potassium sulfate Lead(II)chromate Aluminum hydroxide Ammonium cyanide

Name each of the following: Cl 2 OSO 2 N 2 O 3 P 2 O 5 CO 2 CO SO 3 N 2 H 4

Name each of the following: NH 4 ClBaSO 4 KC 2 H 3 O 2 K 2 HPO 3 KNO 3 CuBrO Li 2 CO 3 MgC 2 O 4

Name each of the following: KClMnSO 4 SO 2 HI (aq) NaNO 3 HClO H 2 CO 3(aq) NH 4 BrO 4

Name each of the following: NaCl MnSO 4 S 2 O 3 HBr (aq) Na 2 CO 3 HClO HNO 3(aq) NH 4 IO 4