MgCl2 = magnesium chloride

Transcription

1 Chapters 8.1 plus 7.3 and : Nomenclature, Writing Formulas, and Percent Composition Nomenclature Names and formulas for ionic compounds The smallest unit of an ionic compound is the formula unit (expresses an empirical formula) Monatomic ions (either cations or anions) are single atoms with charges Examples: Na +, Mg 2+, Al 3+, Cl, O 2, N 3 Note that metals formed cations (lost e ) and nonmetals formed anions (gained e ) Note the relationship between group numbers and charges Na in Group IA lost 1e, Mg in Group IIA lost 2 e, and Al in Group IIIA lost 3e Cl in Group VIIA gained 1e, O in Group VIA gained 2 e, and N in Group VA gained 3e Polyatomic ions (either cations or anions) are groups of atoms bonded together and charged Examples: NH4 + and CO3 2 Note that metals formed cations (lost e ) and nonmetals formed anions (gained e ) Naming monatomic ions Cations: name the element H + = hydrogen ion, Na + = sodium ion Anions: change element ending to the suffix ide F = fluoride, O 2 = oxide Trivial system some cations may have two charges (e.g.: Fe 2+ and Fe 3+ ) Higher oxidation state change ending to ic : Fe 3+ = ferric ion Lower oxidation state change ending to ous : Fe 2+ = ferrous ion Note use of older name for Fe, ferrum IUPAC system newer method for cations having two charges: use roman numerals Higher oxidation state Fe 3+ = iron (III) Lower oxidation state Fe 2+ = iron (II) Note use of English names for metals (Fe = iron ) Note: cations with only one common oxidation state ignore trivial and IUPAC names Binary ionic compounds combine cation and anion names Al2O3 = aluminum oxide MgCl2 = magnesium chloride Fe2O3 = ferric oxide FeO = ferrous oxide Fe2O3 = iron (III) oxide FeO = iron (II) oxide Polyatomic ions Cations: most polyatomic cations have trivial names NH4 + = ammonium ion H3O + = hydronium ion Anions with trivial names: some polyatomic cations have trivial names and sound binary OH = hydroxide ion CN = cyanide ion O2 2 = peroxide ion SCN = thiocyanide ion Anions: most polyatomic anions are oxyanions which use prefixes and suffixes Chlorine Greatest number of oxygens ClO4 = perchlorate Greater number of oxygens ClO3 = chlorate Lesser number of oxygens ClO4 = chlorite Least number of oxygens ClO4 = hypochlorite

2 Phosphorus Trivial IUPAC Greater number of oxygens PO4 3 = phosphate phosphate Lesser number of oxygens HPO3 2 = phosphite phosphonate Least number of oxygens H2PO2 1 = hypophosphite phosphinate Binary molecular compounds use two systems, older prefix system and newer Stock system IUPAC system never use mono for the first element if singular, otherwise use prefixes (mono, di, tri, tetra, penta, hexa, hepta, octa, nona, and deca 1 thru 10, respectively) N2O = dinitrogen monoxide NO2 = nitrogen dioxide N2O3 = dinitrogen trioxide N2O4 = dinitrogen tetroxide N2O5 = dinitrogen pentoxide Stock system use roman numerals for the oxidation state of the first element N2O = nitrogen (I) oxide NO2 = nitrogen (IV) oxide N2O3 = nitrogen (III) oxide N2O4 = dinitrogen (II) oxide (-I) [or nitrogen peroxide] N2O5 = nitrogen (V) oxide (which should persuade anyone to use IUPAC nomenclature) Network compounds use the prefix system SiC = silicon carbide SiO2 = silicon dioxide Si3N4 = trisilicon tetranitride Oxyacids change the suffixes from polyatomic oxyanions ( ate to ic and ite to ous ) Chlorine Greatest number of oxygens HClO4 = perchloric acid Greater number of oxygens ClO3 = chloric acid Lesser number of oxygens ClO4 = chlorous acid Least number of oxygens ClO4 = hypochlorous acid Partial salts of multiprotic acids there are trivial and IUPAC names Trivial IUPAC NaHCO3 = sodium bicarbonate NaHCO3 = sodium hydrogen carbonate NaHSO3 = sodium bisulfite NaHSO3 = sodium hydrogen sulfite NaH2PO4 = monosodium phosphate NaH2PO4 = sodium dihydrogen phosphate Na2HPO4 = disodium phosphate Na2HPO4 = sodium hydrogen phosphate Naming hydrates Hydrates use the same IUPAC prefixes used for molecules (hemi, mono, di, tri, tetra, penta, hexa, hepta, octa, nona, and deca ½ thru 10, respectively) CaSO4 ½H2O = calcium sulfate hemihydrate (commonly: plaster of Paris) (NH4)2C2O2 H2O = ammonium oxalate monohydrate CaSO4 2H2O = calcium sulfate dihydrate (commonly: gypsum) NaC2H3O2 3H2O = sodium acetate trihydrate FePO4 4H2O = iron(iii) phosphate tetrahydrate CuSO4 5H2O = copper(ii) sulfate pentahydrate CoCl2 6H2O = cobalt(ii) chloride hexahydrate MgSO4 7H2O = magnesium sulfate heptahydrate (commonly: Epsom salt) Ba(OH)2 8H2O = barium hydroxide octahydrate Fe(NO3)3 9H2O = iron(iii) nitrate nonahydrate Na2CO3 10H2O = sodium carbonate decahydrate

3 Assigning oxidation numbers 1. The oxidation number of any free (or uncombined) element is zero 2. A monatomic ion has an oxidation number equal to its charge 3. Fluorine has an oxidation number of -1 in all of its compounds 4. In all compounds, group 1 and group 2 metals have positive oxidation numbers equal to their group number. 5. Oxygen will have an oxidation number of -2 in almost all of its compounds. Exceptions: For each F atom oxygen combines with, add +1 to the O oxidation state. In peroxides, oxygen has an oxidation number of -1. In superoxides, oxygen has an oxidation number of -½. 6. Hydrogen has an oxidation number of +1 when combined with more electronegative elements and an oxidation number of -1 when combined with metals (hydrides) 7. For compounds of other elements, the most electronegative element is assigned an oxidation number as if it were an ion. 8. The algebraic sum of the oxidation numbers of all the atoms in a neutral compound or polyatomic ion is equal to the charge on the compound or ion. Examples: Ca loses 2e to form Ca 2+ with an oxidation number of +2 N gains 3e to form N 3 with an oxidation number of 3 Writing chemical formulas Chemical Formulas Molecular formulas show the total numbers of each element: H2O2 and C6H12O6 Empirical formulas show the simplest ratio of each atom: Molecular: N2O4 Empirical: NO2 Ionic compounds always show empirical formulas: Al2(SO4)3 Using oxidation numbers Example: write the correct formula for iron (III) oxide iron (III) = Fe +3 oxide = O 2 But formulas of compounds cannot be charged (although polyatomic ions will be charged) The least common multiple of 3 and 2 is 6, so two Fe +3 and three O 2 will be required to balance the charges: Fe +3 2O 2 3 or Fe2O3

4 Percent composition (Note: always by mass unless otherwise stated) mass of the element Percent by mass of an element = mass of the compound Example I: What is the mass percent of oxygen in sodium bicarbonate? Sodium bicarbonate formula is NaHCO3 μμ of NaHCO3 Na: 1( u) H: 1( u) C: 1( u) O: 3( u) = u (or g mol 1 ) Percent by mass of an element = = 3 ( u) u = % = % oxygen 100% mass of the element mass of the compound 100% Empirical formula from mass analysis data 1. If analysis is in percent, change units to grams 2. Convert mass to moles (since formulas always represent mole ratios) 3. Normalize molar ratios (divide all data by the smallest number) 4. Multiply to remove any fractional data 5. Write the formula Example: Alumina is known to contain only aluminum and oxygen. If the compound is found to be % oxygen by mass, what is the empirical formula of alumina? Since alumina contains only Al and O, if % is O the remainder ( %) is Al, so express units as mass instead of percents Al: % or gal O: % or go Convert mass to mole ratios: Al: gal 1 mol = molal g O: go 1 mol = molo g Normalize mole ratios: Al: molal = molal O: molo = molo Remove fractions (in this case multiply both values by 2 to remove the ½ mole oxygen: Al: molal 2 = molal O: molo 2 = molo Write the formula: Al2O3

5 Finding molecular formulas from empirical formulas 1. Find the μμ of the empirical formula 2. Compare the mass of the molecular formula to the empirical formula (division problem) 3. Distribute the resulting ratio to all elements in the empirical formula Example: Mass analysis of glucose gives the empirical formula CH2O. Subsequent analysis shows glucose to have a molar mass of g mol 1. Find the molecular formula of glucose. The molar mass of CH2O is g mol 1 Dividing the molar mass of the molecule by the molar mass of the empirical formula: g/mol g/mol Distribute: 6 CH2O = C6H12O6 = empirical formulas per molecule