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EIT Review S2007 Dr. J.A. Mack Atom: The smallest divisible unit of an element Compound: A substance made of two or more atoms Ion: A charged atom or molecule Cation: Positive ion Anion: Negative ion www.csus.edu/indiv/m/mackj/ NOMENCLATURE Format for naming chemical compounds using prefixes, suffixes, and other modifications of the names of elements which constitute compounds. Part 1 1 2 Sometimes things get confusing… Elements: inert or “noble” gasses: H = hydrogen O = oxygen C = carbon Compounds: H2 = hydrogen O2 = oxygen H2O = water Metals form Cations 3 non-metals form anions Metalloids can do both 4 1 Ion Charges: Compounds fall into one of two classes: +1 +3 -3 -2 -1 Inorganic Salts Molecules metal cation non-metal + + non-metal or polyatomic anion non-metal (no cations or anions) +2 Variable The two use different formalisms for naming… 5 Binary Compounds: Metal & non-Metal Metals of variable charge (transition) with a non-metal Metal of fixed oxidation (charge) state combined with a non-metal. Examples: Cation non-metal takes on “ide” suffix Examples: Cation Anion Formula K+ Cl− KCl Potassium chloride Ca2+ O2- CaO Calcium Oxide Na+ S2- Na2S Sodium sulfide Al3+ S2- Al2S3 Aluminum sulfide 6 Pb2+ Name modify transition metal name with roman numeral Anion Formula Name Cl− PbCl2 lead (II) chloride pronounced: lead - two - chloride 7 Pb4+ Cl− PbCl4 lead (IV) chloride Fe3+ O2− Fe2O3 Iron (III) oxide 8 2 Some common polyatomic ions: Ternary Compounds: Those with three different elements Type A: metal of fixed charge with a complex ion NH4+ ammonium H3O+ hydronium CO32– carbonate HCO32– hydrogen or bicarbonate NO2– nitrite NO3– nitrate SO42– sulfate SO32– sulfite PO43– phosphate C2H3O2– acetate Cation Anion Formula Name K+ OH− KOH Potassium hydroxide Ca2+ OH− Ca(OH)2 Calcium hydroxide Na+ SO 24− Na2SO4 Sodium sulfate Al3+ SO 24− Al2(SO4)2 Aluminum sulfate 9 Metal of variable charge transition) with a complex ion Cation Anion Formula Fe3+ NO3− Fe(NO3)3 Iron (III) nitrate Fe2+ NO −2 Fe(NO2)2 Iron (II) nitrite 10 NonNon-metal with a nonnon-metal When non-metals combine, they form molecules. They may do so in multiple forms: Name CO CO2 Because of this we need to specify the number of each atom by way of a prefix. 1 = mono 5 = penta 11 2 = di 3 = tri 6 = hexa 4 = tetra 7 = hepta 12 3 Examples: Formula D) Writing formulas for acids and Bases Name: BCl3 boron trichloride SO3 sulfur trioxide NO nitrogen monoxide we don’t write: N2O4 •An acid is a substance that produces H+ when dissolved in water. •Certain gaseous molecules become acids when dissolved in water. nitrogen monooxide or mononitrogen monoxide •A base produces OH− when dissolved in water. •Bases often are Group I and Group II hydroxide salts. dinitrogen tetraoxide 13 14 H+ and S2- Type I Acids: Acids derived from –ide anions. it takes 2 H+ to cancel one S2- The names for these acids follows the formula: “hydro” + the root of the ide anion + ic “acid” Anion: Acid: chloride HCl hydrochloric acid fluoride HF hydrofluoric acid H2S Name: hydro sulfuric acid 15 16 4 Examples: Practice: Anion: (nitrate) NO3− (sulfate) 2− 4 (acetate) SO Acid: Name: HNO3 nitric acid C2 H 3O 2− H2SO4 N2SO4 sodium sulfate barium carbonate BaCO3 FeO Iron (II) oxide zinc phosphide Zn3P2 NiBr2 nickel (II) bromide sulfuric acid HC2H3O2 acetic acid vinegar 17 18 Common Names: CO2 carbon dioxide carbon monoxide CO P2O5 diphosphorous pentaoxide nitrogen trihydride NH3 (ammonia) 19 H2O water ammonia NH3 CH4 methane NO nitric oxide N2O nitrous oxide 20 5 Modern Atomic Theory: •Atoms are made of protons, neutrons and electrons. please go to my chem. 1A web site and download a nomenclature practice worksheet if you need more review. http://www.csus.edu/indiv/m/mackj/chem1a/chem1A_lab.html •The nucleus of the atom carries most of the mass. • It consists of the protons and neutrons surrounded by a cloud of electrons. The charge on the electron is –1 The charge on the proton is +1 The link to the worksheet is at the top of the page. There is no charge on the neutron The Atomic Number or number of protons in the nucleus defines an element. 21 22 75 The isotope 34 Se is used medically for diagnosis of Isotopes, Atomic Numbers, and Mass Numbers pancreatic disorders. How many protons, neutrons, and electrons does an atom of •Atomic number (Z) = number of protons in the nucleus. 75 34 Se have? •Mass number (A) = total number of nucleons in the nucleus (i.e., protons and neutrons). 75 protons + neutrons •One nucleon has a mass of 1 amu (Atomic Atomic Mass Unit) Dalton” Unit a.k.a “Dalton 75 34 Se 34 protons •Isotopes have the same Z but different A. •The elements are arranged by Z on the periodic table. By convention, for element X, we write protons = 34 A ZX electrons = 34 neutrons = 75-34 = 41 23 24 6 Molar Masses (Molecular Weights) of Compounds: Avagadro’ Avagadro’s Number The molar mass of a molecular compound is the sum of the molar masses of its atoms. Since one mole of 12C has a mass of 12g (exactly), 12g of 12C contains 6.022142 x 1023 12C-atoms. Example: But carbon exists as 3 isotopes: C-12, C-13 &C-14 The molar mass of CO2 is: The average atomic mass of carbon is 12.011 u. From this we conclude that 12.011g of carbon contains 1 x (12.01 g/mol) 6.022142 x 1023 C-atoms + 2 x (16.00 g/mol) Yes, since NA is so large, the statistics hold. Is this a valid assumption? = 44.01 g/mol 25 How many oxygen atoms are there in 25.1g of chromium (III) acetate? Cr(C2H3O2)3 step 2: calculate the molar mass… Question: What is the weight % of each element in C2H6? 229.13 g/mol First determine the molar mass of C2H6: 1 mol of C 2 H 6 has a mass of 30.07 g step 3: use dimensional analysis to solve the problem… 25.1g Cr(C2 H 3O 2 )3 × 1mol Cr(C2 H 3O 2 )3 229.13g Cr(C2 H 3O 2 )3 × 6.022 × 1023 O − atoms = 1mol O 3.96 × 1023 × Percent Composition: The relative amounts of each atom in a molecule or compound can be represented fraction of the whole. step 1: write the correct chemical formula… Cr3+ & C2H3O2− 26 (2×12.01 + 6 ×1.008) g/mol Next determine the mass of hydrogen in 1 mol of the compound: 6mol O 1mol Cr(C2 H3 O 2 )3 1 mol C2 H 6 × O-atoms 27 6 mol H 1.0079 g H × = 6.047 g H 1 mol C2 H 6 1 mol H 28 7 Determining a Formula from Percent Composition: Now relate the mass of H in one mol of the compound to the molar mass of the compound Given the relative percentages of each element in a compound, 1 × 100 = 20.11% H 6.047 g H × 30.07g C2 H 6 10 % X, Since there is only C as the remaining element: % C = 100 − % H 20 % Y, 30 % Z etc… one can find the empirical formula of the compound. = 79.89 % C The empirical formula of a compound or molecule represents the simplest ratio of each element in 1 mol of the compound or molecule. The compound C2H6 is 20.11% H & 79.89% C 29 Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? Solution: Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? 2. Convert the grams of each element to moles. determine X, Y & Z in (CXHYOZ) (g element X → mole X etc…) 1. Since the percentages for each element sum to 100%, if one equates % to grams (g), the sum of the masses must be 100g. (i.e. one can assume 100g of the compound) 64.82 g C 21.59 g O 30 13.59 g H 31 64.82g C × 1 mol C 12.011 g C = 5.397 mol C 21.59g O × 1 mol O 16.00 g O = 1.349 mol O 13.59g H × 1 mol H = 13.48 mol H 1.0079 g H 32 8 3. Divide each of the individual moles by the smallest number of Rounding to the nearest whole numbers: moles to gain the molar ratios for each element in the compound. These are the formula subscripts. (X2Y3 etc…) Subscript for C Subscript for H X = 4.001 = 4 **If the ratios are fractional (0.5, 1.5 or 0.333) multiply each ratio by a whole number to get even number formula subscripts. 0.5 × 2 = 1 Z = 1.000 = 1 Subscript for O 13.48 1.349 5.397 = 9.992 Z = = 1.000 X= = 4.001 Y = 1.349 1.349 1.349 Examples: Y = 9.992 =10 0.25 × 4 = 1 C4H10O The empirical formula is: The results of this calculation tells us only about the empirical formula of the compound. To determine the molecular formula, we need more information. This will be shown in a later example. 0.333 × 3 = 1 33 34 Balancing Chemical Reactions: Example Chemical Equations: C2H6 + Mass is conserved in a chemical reaction. → O2 2 C’s & 6 H’s 2 O’s balance last CO2 + H2O 1 C & 2 O’s 2 H’s & 1 O balance H first Total mass of reactants = Total mass of products 2 2H6 ___C + → O2 CO2 3 H2O 6 ___ + balance C next Chemical equations must therefore be balance for mass 2C2H6 + balance O The number and type of atoms on either side of the equation must be equal! 2C2H6 + O2 7 O2 ____ 4 C’s 12 H’s 14 O’s 35 → 4 CO2 ___ → + 4CO2 + 6H2O 6H2O 4 C’s 12 H’s 14 O’s 36 9 Reduction and Oxidation Reactions: RedOx Oxidation Numbers +1 +2 +3 -3 -2 -1 Oxidation involves an atom or compound losing electrons Variable Reduction involves an atom or substance gaining electrons Niether process can occur alone… that is, there must be an exchange of electrons in the process. Metals take on their formal charge: Non-metals do the same The substance that is oxidized is the reducing agent The substance that is reduced is the oxidizing agent Chemists use oxidation numbers to account for the transfer of electrons in a RedOx reaction. 37 Balancing REDOX reactions: Electrochemistry: Oxidation numbers Fe + O2 → Fe2O3 In the compound potassium bromate (KBrO3), the oxidation number of bromine (Br) is? The compound is neutral so the sum of the oxidation numbers should be zero. +1 5 ?? 3×(-2) = -6 KBrO3 1+ ?? + (-6) = 0 38 oxidation states: 0 0 +3 -2 oxidation half reaction: {Fe → Fe+3 reduction half reaction: {O2 + 3e-} x 4 + 4e- → 2O2-} x 3 Balance electrons transferred then sum the half RXN’s: 4Fe + 3O2 + 12e- → 2Fe2O3 + 12e- ?? = 5 4Fe + 3O2 → 2Fe2O3 39 40 10 Quantitative calculations: Mass and moles 2HCl (aq) + Ba(OH)2 (aq) Æ 2H2O (l) + BaCl2(aq) Consider the following reaction: HCl (aq) + Ba(OH)2 (aq) Æ H2O (l) + BaCl2(aq) g BaCl2 Æ mol BaCl2 Balacing: Æ mol HCl 2 HCl (aq) + Ba(OH)2 (aq) Æ 2 H2O (l) + BaCl2(aq) How many moles of HCl are consumed if 1.50 g of BaCl2 are produced assuming that Ba(OH)2 is in excess? Solution: convert to g BaCl2 then convert Æ using the Molar mass of BaCl2 mol BaCl2 1.50 g BaCl 2 × 1 mol BaCl 2 2 mol HCl × = 0.0144 mol HCl 208.24 g BaCl 2 1 mol BaCl 2 to Æ Molar mass BaCl2 mol HCl Molar ratio for equation using the Molar ratio for equation 41 42 Stoichiometry: Conversions between masses & moles in chemical reactions. Limiting Reactant: ÖWhen one reactant is present in an amount such that it is completely consumed before all other reactants, we say that it limits the reaction. Ö The other reactants are said to be in excess. excess Ö The Theoretical Yield is determined by the stoichiometry of the limiting reactant. Ö The limiting reactant can only be determined through molar ratios. It cannot be identified by mass. 43 11