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Background terms Oxidation: loss of electron(s) from neutral atom Oxidation state (valence): "charge" on atom General rules for determination of oxidation states: Group A elements 1. positive oxidation state is equal to column number 2. negative oxidation state is equal to column number minus 8 (beginning with column IVA) 3. if an atom is in an even numbered column, it will have an even number oxidation state; similarly with odd numbered columns 4. maximum change is going to be "8" (magic number) Group B elements 1. one possible positive oxidation state is the column number except for column VIIIB 2. another possible positive oxidation state is "2+" except for IIIB which will only have "3+" More rules: always IA +1 IIA +2 Al, Sc +3 Zn, Cd +2 Metallic endings new system old system iron (III) Fe+3 ferric iron (II) Fe+2 ferrous -ic suffix usually corresponds to highest oxidation state element usually takes Example: Cr (group VIB) maximum + charge= group number 6+ 3+ chromic (usually exists this way) 2+ chromous Sn tin (IV) or stannic SnF4 stannic fluoride tin (II) or stannous SnF2 stannous fluoride BINARY COMPOUNDS are compounds made up of two elements metal + non-metal The metal always takes its given name; a common suffix for a metal is -ium non-metal suffixes are usually -on, -gen, and -ine The metal takes its given name, the non-metal suffix changes to -ide ex. NaCl KF AlN sodium chloride potassium fluoride aluminum nitride Cross-over process write the symbols and charges for the ions next to each other, always writing the cation first 2. cross over the charges by using the absolute value of each ion’s charge as the subscript for the other ion 3. check the subscripts and divide them by their largest common factor to give the smallest possible wholenumber ratio of ions. 1. The purpose of this process is to ensure that the compound is electrically neutral. Binary two non-metals Here we need the use of various prefixes to tell us how many of each kind of atom we have present mono- 1 hexa- 6 di- 2 hepta- 7 tri- 3 octa- 8 tetra- 4 nona- 9 penta- 5 deca- 10 When we name a binary compound composed of two non-metals we state how many atoms of each kind we have by using the above prefixes. If we only have one of the first element listed, we do not need to state that by using the prefix mono-. However, we do need to state any other quantity of the elements. Examples: ex. NO N2O NO2 N2O3 N2O4 mononitrogen monooxide dinitrogen monoxide nitrogen dioxide dinitrogen trioxide dinitrogen tetraoxide When we have similar vowels together, such as two O's or an A and an O we cancel out the first vowel to make for more sensible spelling and pronunciation. Ternary compounds Compounds made of 3 or more elements, one of which is usually oxygen. metal (takes its name) + radical (polyatomic ion) (2 or more elements joined together with a residual charge; when written as the second component of a compound it usually has a negative charge) The radical name is based on the amount of oxygen it contains and is named according to this definition even if it doesn't contain any oxygen. RADICAL NAMING CHART EXPLANATION EXAMPLE SUFFIX FORMAT NAME 1 more O ClO4than normal per-rootate perchlorate normal ClO3amount of O -ate chlorate 1 less O than ClO2normal -ite chlorite 2 less O than ClOnormal hyporoot-ite hypochlorite More “radical” rules In group VIIA: Cl, Br, and I follow the preceding rules; F does not In group VIA: S and those elements below it follow the above rules (4 oxygens is the "normal" amount) In group VA: P and As follow the above rules (4 oxygens is "normal“ for P and As) [3 oxygens is the “normal“ amount for N] An important point to remember is that the oxidation state of the radical does not change as the amount of oxygen changes. Examples that do not follow rules: CNcyanide CNOcyanate SCNthiocyanate (thio- tells us there is sulfur present) Examples using polyatomic ions: NaClO3 MgSO4 AlPO3 sodium chlorate magnesium sulfate aluminum phosphite ACID NAMING binary acids To recognize a binary acid, it must be made up of two elements (one of which is hydrogen) and must have the word aqueous, abbreviated, following the formula. The name must include the prefix hydro- and the word acid, i.e. Hydro -root-ic acid ex. HCl(aq) hydrochloric acid H2S(aq) hydrosulfuric acid Ternary acids ORIGINAL RADICAL SUFFIX per-rootate -ate -ite Hypo-rootite EXAMPLE HClO4(aq) HClO3(aq) HClO2(aq) HClO (aq) ACID SUFFIX NAME Per-root-ic Perchloric acid -ic Chloric acid -ous Chlorous acid Hypo-root- Hypochlorous ous acid Bases In general, the presence of the hydroxide ion and the term aqueous are sufficient to denote a base. ex. NaOH(aq) Bi(OH)3(aq) sodium hydroxide (present in lye, Drano, liquid plumber) bismuth hydroxide (present in pepto bismol) Mg(OH) 2(aq) + Al(OH) 3(aq) magnesium hydroxide and aluminum hydroxide (present in mylanta) Crossing over process Ex: magnesium phosphate We know that Mg has a +2 charge, and PO4 has a -3 charge. These two numbers do not add up to zero. Thus, we find a least common denominator and find out what we must multiply each number by to get this result. Out LCD is 6, thus we multiply +2 by 3 and -3 by 2. This results in +6 and -6 cancelling out to zero. Mg3(PO4) 2 lead (II) nitrite = Pb(NO2) 2 A mole represents 6.02 x 1023 particles of a substance Type of matter Type of particle Elements ionic compounds atoms formula units molecular compounds molecules Amadeo Avogadro di Quarengo (Italian 1776-1856) is credited with discovering this number; therefore it is called Avogadro's number Examples: 1 mole NaCl 1 mole N2O4 1 mole Fe 1 mole Ca2+ = 6.02 x 1023 formula units = 6.02 x 1023 molecules = 6.02 x 1023 atoms = 6.02 x 1023 ions (A mole of molecules contains more than a mole of atoms) Calculation example: H2O 1 mole molecules = 6.02 x 1023 molecules 1 molecule = 3 atoms 6.02 x 1023 molecules x 3 atoms = 1.80 x 1024 1 mol H2O molecule atoms/mole H2O The mass of a mole varies with the substance 1 mole C = 6.02 x 1023 atoms of C = 12 g C Formula mass: the sum of the average atomic masses of all atoms represented in its formula (in a.m.u.) CO2 (12.01 amu + 2(16.00 amu) = 44.01 amu Molar mass: the mass of 1 mole of any element or compound (in grams/mole) [numerically equal to formula mass] For an element the molar mass is the atomic mass found on the periodic table. H2O 2(1) + 16 = 18 g/mol CO2 12 + 2(16) = 44 g/mol Percent composition: % by mass of each element in a compound For H2O %H = 2 g x 100 = 11.11% 18 g %O = 16 g x 100 = 88.89% 18 g For CO2 %C = 12 g x 100 = 27.27% 44 g %O = 32 g x 100 = 73.73% 44 g Empirical formula/simplest formula gives the lowest whole-number ratio of the elements in a compound (or the lowest whole-number ratio of moles of atoms in a compound) ex. Empirical formulas CH CH2O Molecular formulas C2H2 acetylene C6H6 benzene CH2O formaldehyde C2H4O2 acetic acid C6H12O6 glucose Determining the empirical/simplest formula Using the % composition, determine the mass of each of the elements in 100 g of that compound Convert the masses in grams to moles Determine the simplest whole number mole ratio between elements and set that equal to the atoms ratio in the simplest formula Example: What is the empirical formula of a compound containing 25.9% N and 74.1 % O? 25.9 g N x 1 mole N = 1.85 mole N ÷ 1.85 = 1 x 2 = 2 14 g N 74.1 g O x 1 mole O = 4.63 mole O÷1.85 = 2.5 x 2 = 5 16 g O So, the simplest formula is N2O5 Alternate method If you are not provided with the percent composition of the elements in the compound, you may follow this method: 1. Divide the grams provided of each element by its molar mass (atomic mass) 2. Divide all mole values by the smallest value to obtain the smallest whole number ratio possible. 3. Set those values equal to the subscripts in the chemical formula. To calculate the molecular formula from simplest formula you need the molecular weight of the compound and the simplest formula. Ex. If the simplest formula for acetic acid is CH2O and the molecular mass is 60, what is the molecular formula? CH2O Simplest formula mass = 30 g/mole 30 X = 60 X=2 so multiply all subscripts in the empirical formula by X to get C2H4O2 A chemical equation allows us to describe in a concise manner, on paper, a chemical reaction that has taken place. It represents, with symbols and formulas, the identities and relative molecular or molar amounts of the reactants and products in a chemical reaction. Evidence for a chemical reaction: Energy release as heat or light. Color change Evolution of gas (bubbles and/or odor) Appearance of a solid (precipitate) Parts of a chemical equation: The items on the left of the arrow are called reactants; (arrow) means "yields; the right products States of matter are described: s (↓), l, g (↑), aq Other symbols used in chemical equations are on p. 266, Table 2 Coefficients represent relative numbers of particles that take part in the reaction #atoms are conserved mass is conserved Equations must be balanced; the number of atoms on both sides of the arrow must be equal. Basics for balancing: 1. Start from left to right. 2. Balance polyatomic ions as a single entity whenever possible. 3. Balance the H's and O's last, and 4. NEVER manipulate subscripts in a formula. 5. Try and keep the coefficients to the smallest whole numbers possible (fractional coefficients are acceptable) Combination/synthesis reactions: 2 or more substances react to form a single substance. Reactants are often elements and/or simple compounds, often H2O Products are compounds Difficult to guess product of nonmetals (must be told) Often liberates energy General format: A + X AX Examples: 8Ca (s) + S8 (s) 8CaS (s) 2Mg (s) + O2 (g) 2MgO (s) 2Fe (s) + O2 (g) 2FeO (s) Na2O (s) + H2O (l) 2NaOH (aq) Decomposition/analysis reactions: one compound broken down into two or more simpler products Products are often elements and/or compounds in any combination; difficult to predict Binary compounds break down into their elements Energy is required for the reaction to take place General format: AX A + X Decomposition/analysis reactions Examples: a. metallic carbonates metallic oxides + CO2 (g) CaCO3 (s) CaO (s) + CO2 (g) (NH4) 2CO3(s)2NH3(g)+H2O(g) +CO2(g) b. metallic hydroxides metallic oxides + H2O Ca (OH) 2 (s) CaO (s) + H2O (g) NaOH and KOH are exceptions c. metallic chlorates metallic chlorides + O2 2KClO3 (s) 2KCl (s) + 3O2 (g) d. some acids nonmetallic oxides + H2O H2CO3 (aq) H2O(l) + CO2 (g) H2SO3 (aq) H2O(l) + SO2 (g) e. some oxides decompose upon heating 2HgO (s) 2Hg (l) + O2 (g) 2PbO2 (s) 2PbO (s) + O2 (g) f. electric current - electrolysis 2H2O(l) 2H2 (g) + O2 (g) 2NaCl (s) 2Na (s) + Cl2 (g) 2HI (g) H2 (g) + I2 (g) Single replacement reactions: atoms of one element replace atoms of a 2nd similar element in a compound determined by relative reactivities of the 2 metals (Activity Series is a list of elements arranged according to the ease with which the elements undergo certain chemical reactions.) Halogens are nonmetals that are replaced (activity decreases going down Group VIIA) General format: A + BX AX + B Y + BX BY + X Displacement of Hydrogen in water or acid by a metal Double replacement reactions: exchange of positive ions between 2 compounds Generally are reactions between ionic compounds in aqueous solutions. General format: AX + BY AY + BX At least one statement below is usually true of one of the products Only slightly soluble and ppt. from solution is a gas and bubbles out of solution is a molecular compound such as H2O Examples: 2NaOH (aq) + H2SO4 (aq) Na2SO4 (aq) + 2H2O (l) (neutralization rxn) AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) (precipitation rxn) ionic reactions: Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl-(aq) AgCl(s) + Na+ (aq) + NO3-(aq) net ionic: Ag+ (aq) + Cl-(aq) AgCl(s) Net ionic reactions DO NOT include spectator ions! Combustion reactions: oxygen reacts with another substance often producing energy in the form of heat and/or light Commonly involve hydrocarbons Complete combustion produces water and carbon dioxide Incomplete combustion produces CO and C in addition to CO2 and H2O due to decreased amounts of O2 Examples: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O(g) Sub-category – Redox reactions Oxidation-reduction reactions, typically called “redox rxns” involve two simultaneous processes Oxidation – loss of electrons Reduction – gain of electrons Mnemonic device to remember this: OILRIG – oxidation is loss, reduction is gain Identifying Redox rxns 0 +2 -2 0 +1 -2 H2 (g) + CuO (s) Cu (s) + H2O (l) Hydrogen is oxidized; copper is reduced Disproportionation This occurs when one substance is both oxidized and reduced during the same chemical reaction. +1 -1 +1 -2 0 2H2O2 (aq) 2H2O (l) + O2 (g) Oxygen is both oxidized and reduced in this reaction.