Nomenclature i.e. Naming chemical compounds

Binary compounds Compounds consisting of a metal and a nonmetal Compounds consisting of two nonmetals

Type I Metals Group 1, 2, 3, & 4 –Group 1, 1 + oxidation number –Group 2, 2 + oxidation number –Group 3, 3+ oxidation number –Group 4, 4+ oxidation number Ge 4+,Al 3+, Ga 3+, In 3+,Zn 2+,Cd 2+, Ag 1+,Ta 5+,Tc 7+

Binary compounds Metal & Nonmetal Identify the cation –Type I metal Single oxidation number –Type II metal Various oxidation numbers Write the cation’s name first, unchanged Identify the anion –nonmetal The anion is named by keeping the root and adding the ending -ide

Naming Type I MgCl 2 Na 2 O Ag 2 O Al 2 O 3 ZnI 2 Ca 3 P 2 Magnesium chloride Sodium oxide Silver oxide Aluminum oxide Zinc iodide Calcium phosphide

Type II Metals Most transition and heavy metals Have multiple oxidation numbers Oxidation numbers found mathematically or given

Determining Type II Oxidation Numbers Anion oxidation Known Find Cation by working from Anion and subscripts

Example Fe 2 O 3 O 2- 3 oxygen 6 - total 2 Fe 6  2 = 3 so Fe 3+ oxidation state MnO 3 3 (2-) = 6- 6  1 = 6 + So Mn 6+ oxidation state

Naming Type II Identify the cation Identify the anion Determine the total charge of the anion Calculate the oxidation number of the cation Cation ( roman numeral) anion-ide –Roman numeral is the oxidation number of the cation

Naming Type II cont. Example Fe 2 O 3 –Fe = Iron –O = oxide –O 2-, 6 - total / 2 = Fe 3+ –Name - Iron (III) oxide

Type II Practice Cu 2 O NiN CoCl 2 Mn 2 S 5 CrP 2 Cr 2 Se 3 VI 5 Copper (I) oxide Nickel (III) nitride Cobalt (II) chloride Manganese (V) sulphide Chromium (VI) phosphide Chromium (III) selenide Vanadium (V) iodide

Rules for Type III Identify the first element Identify the second element Determine the prefix for the first and second element based on the number of each in the compound (DO NOT Use the prefix mono- for the first element)

Type III Prefixes Mono - 1 Di - 2 Tri - 3 Tetra - 4 Penta - 5 Hexa - 6 Hepta - 7 Octa - 8 Nona - 9 Deca - 10

example NO 3 Nitrogen - 1- mono- Oxygen - 3- tri- Do Not use Mono- Nitrogen trioxide N 2 O Nitrogen 2- di Oxygen 1- mono Dinitrogen monoxide

Type III Practice CO 2 CO SO 3 P 2 O 5 SF 2 SF 6 Carbon dioxide Carbon monoxide Sulfur trioxide Diphosphorus pentaoxide Sulfur difloride Sulfur hexafloride

Binary Compound Flow Chart Type of Cation? Type I Cation Anion-ide Type III Prefix cation Prefix anion-ide Type II Cation (roman numeral) anion - ide

Ternary Compound Naming Type of Cation ? Type I Cation polyatomic ion Type II Cation (Roman Numeral) polyatomic ion

example Na 2 (CO 3 ) Na 1+ - sodium (CO 3 ) 2- - carbonate Sodium carbonate Fe 2 (SO 4 ) 3 Fe 3+ = Iron (III) (SO 4 ) 2- = sulfate Iron (III) sulfate

PRACTICE CsClO 4 KMnO 4 Co(ClO 4 ) 2 Zn(ClO 3 ) 2 Cu(NO 3 ) 2 (NH 4 )(ClO 2 ) Cesium perchlorate Potassium permanganate Cobalt (II) perchlorate Zinc chlorate Copper (II) nitrate Ammonium chlorite

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Naming Acids Does the anion contain oxygen? No Hydro + root-ic Hydro(anionroot-ic) acid Yes Anion ends in -ate Replace with -ic Anion-ic acid Anion ends in -ite Replace with -ous Anion-ous acid

example HCl No oxygen Hydro- Chlorine - -chloric Hydrochloric acid H 2 SO 4 Contains oxygen SO 4 - sulfate -ic replaces -ate Sulfuric acid

Practice HBr HClO HNO 3 H 2 SO 3 HC 2 H 3 O 2 HF Hydrobromic acid Hypochlorous acid Nitric acid Sulfurous acid Acetic acid Hydrofloric acid

PERCENT COMPOSITION Calculate the Molar mass of the Compound Divide the total mass of each element of the compound by the Molar mass of the compound

Example of formula to percent BaCl 2 MW = 207 g/mol % Ba = 137/207 –66.2% % Cl = 70 /207 –33.8% Fe 3 (PO 4 ) 2 MW = 358 g/mol % Fe = 168/ % % P = 62/ % % O = 128/ %

Practice CaCl 2 FeO P 2 O 5 H 2 SO 4 NaHCO % Ca, 63.6% Cl 77.7%Fe, 22.2% O 43.7% P, 56.3% O 2.0%H, 32.7%S,65.3%O 27.3%Na,1.2%H, 14.3%C, 57.1%O

Percent to Formula Convert percents to grams Find Moles of Each Element Mole to Mole ratio Reduce to Smallest Whole Number ratio

Example 18.7% Zr, 28.6% Cl, 52.7% O 18.7% = 18.7 g 18.7/91 =0.205 mol Zr 28.6% = 28.6 g 28.6/35 = mol Cl 52.7% = 52.7g 52.7/16 = 3.29 mol O Ratio : : 3.29 Divide by Smallest value 1 : 4:16 Zr(ClO 4 ) 4