Atomic Structure; Isotopes

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Atomic Structure; Isotopes CHEMISTRY 103 – Help Sheet #2 Atoms, Molecules, and Ions Do the topics appropriate for your course Prepared by Dr. Tony Jacob https://clc.chem.wisc.edu (Resources page) Nuggets: Atomic structure; Isotopes; Counting neutrons, protons, electrons; Average Atomic Mass; Mass Spectroscopy (Part I); Ionic vs molecular compounds; Naming chemicals; Writing formulas from names; Polyatomic ions; Coulomb’s law ATOMIC STRUCTURE Rutherford's Au Foil Experiment: Mass of atom is concentrated in a highly dense nucleus that was positively charged; the rest of the atom has only a small amount of mass; alpha particles are positively charged so they are repelled by the nucleus comprised of protons (positively charged) and are small compared to the size of the atom allowing them to pass through the atom Nucleus: contains nearly all the atom’s mass; neutrons and protons located in the nucleus and the nucleus is a very small volume of the atom; extremely dense; atom is mostly empty space Elements: determined by the number of protons they contain Masses of all atoms are compared to 12C which is the standard reference; 12C is defined as 12.0000... atomic mass units (exact value) = 12.0000...amu; (amu = atomic mass units) 1 23 -24 1 amu = /6.022 x 10 = 1.6605 x 10 g Neutron (n): has a neutral charge; mass ≈ 1amu ≈ 1 x 10-24g (1.6749 x 10-24g) Proton (p+): has a charge of +1; mass ≈ 1amu ≈ 1 x 10-24g (1.6726 x 10-24g) Electron (e-): has a charge of -1; mass ≈ 0.0005amu ≈ 1 x 10-27g (9.1094 x 10-28g) Relative masses: n > p+ >> e-; mass of n ≈ p+ ≈ 1 with e- ≈ 1/2000 ISOTOPES: elements with the same #p+ (same element) but different #n (different mass) A + + Isotope Symbol: ZElement Symbol A = mass number = #p + #n; Z = atomic number = #p 12 13 14 + C isotopes: 6C , 6C , and 6C; the 3 isotopes of C each have Z = 6 (6p ) but different #n (6, 7, and 8) Example 1: What is mass of one atom of 16O-2 in kg? ⎛ −24 ⎞ -26 16 -2 16 -2 16 −2 1.6605 x 10 g ⎛ 1kg ⎞ −26 Answer 1:€ 2.7 x 10 kg { O = 16amu; mass of 1 atom O = 16u; O = 16amu⎜ ⎟ ⎜ ⎟ = 2.6568 x 10 kg } € € € ⎝⎜ 1amu ⎠⎟ ⎝ 1000g⎠ AVERAGE ATOMIC MASS (AAM): weighted average of all isotopes of one element taking into account the abundance of each isotope; units = amu; e.g., average atomic mass Br = 79.90amu Abundance: percent of one isotope as compared to all atoms of that given element AVERAGE ATOMIC MASS (AAM) = S(mass of isotope) x (fractional abundance) AAM = (massiso1)(fractional abundanceiso1) + (massiso2)(fractional abundanceiso2) + … = AAM = (M1)(FA1) + (M2)(FA2) + … Sum of fractional abundances = 1 (i.e., 100%); for elements with only 2 isotopes: FA1 + FA2 = 1 Example 2: Calculate the average atomic mass of an element given following data: isotope 1: 27.977amu, 92.21% abundance; isotope 2: 28.976amu, 4.70% abundance; and isotope 3: 29.974amu, 3.09% abundance. Answer 2: 28.09amu {AAM = (M1)(FA1) + (M2)(FA2) + (M3)(FA3) = (27.977)(0.9221) + (28.976)(0.0470) + (29.974)(0.0309) = 28.09amu} Example 3: What are the abundances of the 2 B isotopes (10B mass = 10.0129u, 11B mass = 11.0093u), if the average atomic mass of B is 10.811amu? Answer 3: 10B = 19.90% and 11B = 80.10% AAM = (M1)(FA1) + (M2)(FA2); 10.811 = 10.0129x + 11.0093y; since there are 2 unknowns 2 equations are needed; second equation: FA1 + FA2 = 1 (which is 100%); x + y = 1 ® y = 1 - x; substitute this into the first equation: 10.811 = 10.0129x + 11.0093(1-x); solve for x: 10.811 = 10.0129x +11.0093 – 11.0093x; 0.1983 = 0.9964x; x = 0.1990 = 19.90%; y = 1-x = 1-0.1990 = 0.8010 = 80.10%; since x was the abundance for 10B ® abundance of 10B = 19.90% and abundance of 11B = y = 80.10%} MASS SPECTROSCOPY: single atoms experimental technique that yields a plot called a mass spectrum of mass to charge ratio, (m/z), versus abundance (or intensity); see the class video M2Q3 for a wonderful description of a mass spectrometer Example 4: Mg has 3 isotopes: 24Mg (79% abundance), 25Mg (10% abundance), 26Mg (11% abundance). Draw the mass spectRum foR this element? 79% Mg 10% Abundance Intensity or 11% 10 20 30 40 24 25 26 Answer 4: m/z MASS SPECTROSCOPY: compounds (Not all instructors will be covering mass spectroscopy of compounds; instructors will be covering mass spectroscopy of single atom elements or molecules like Br2, Cl2, etc., but some sections may not cover compounds like C2H6O.) Hard Mass Spectroscopy: This technique breaks the molecule into individual atoms Soft Mass Spectroscopy: This technique ionizes the molecule and the one peak shown represents the molar mass of the molecule (molar mass is the sum of the average atomic mass of the individual elements within the molecule; e.g., for C3H6O1 the molar mass = 3(mass of C) + 6(mass of H) + 1(mass of O) = 3(12) + 6(1) +1(16) = 58g/mol; this will be revisited in HelpSheet #4!) Empirical formula: the simplest formula that shows the ratio between atoms within the compound (this will be revisited in HelpSheet #4!) Molecular formula: the actual formula of the compound; (e.g., molecular formula = N2O4 while the empirical formula = NO2; i.e., the simplest ratio between N and O atoms in N2O4) Example 5: A chemical compound was analyzed by mass spectroscopy and both “hard” and “soft” mass spectra were obtained. a. What is the empirical formula for the unknown compound? b. What is the molecular formula for the unknown compound? Hard MS Soft MS 100% 60.0% 30.0% Intensity or Abundance Intensity or 10.0% Abundance Intensity or 1 12 16 116 m/z m/z Answer 5: a. empirical formula = C3H6O1 {from the hard mass spectrum (Hard MS) it can be determined that the compound contains H (mass = 1), C (mass = 12), and O (mass = 16). From the percent intensity or abundance the number of each element can be determined by dividing each element by the smallest percent; 10/0% in this example. 60% 30% 10% H = = 6 ; C = = 3 ; O = = 1 ; this yields an empirical formula = C H O } 10% 10% 10% 3 6 1 b. molecular formula = C6H12O2 {from the soft mass spectrum (Soft MS) the peak represents the compound’s molar mass; molar mass = 116g/mol; the empirical formula (C3H6O1) molar mass = 58g/mol [3(mass of C) + 6(mass of H) + 1(mass of O) = 3(12) + 6(1) +1(16) = 58g/mol] molar massmolecular formula 116g / mol to determine the molecular formula, determine ratio = ; ratio = = 2 ; molar massempiricalformula 58g / mol use the ratio just determined and multiply the empirical formula by 2: C(3 x 2)H(6 x 2)O(1 x 2) = C6H12O2} IONIC COMPOUNDS: contain metal + nonmetal (can substitute a polyatomic ion for eitheR or both) Properties: extended solids; not discrete molecules; high melting/boiling points; conducts electricity in molten state; Examples: NaCl(s), Ca(NO3)2(s), NH4NO3(s), AgCl(s) Name Formula - Ammonium + Ions: atoms or molecules that have lost/gained e (charges are not changed NH4 Hydronium + by gaining/losing protons!) H3O Cations: positively charged atoms (electrons have been lost) Hydroxide OH- Acetate - Anions: negatively charged atoms (electrons have been gained) CH3COO Polyatomic Ions: “many-atom ions”; these compounds are not broken up Nitrate - NO3 and are treated as a group: Permanganate - MnO4 Hydrogen Carbonate - HCO3 Charges on elements in Ionic Compounds based on the column of Periodic Carbonate -2 Table the element resides in: CO3 Hydrogen Sulfate - + + +2 +2 HSO4 Group IA: +1 (H , Li , ...) Group IIA: +2 (Be , Mg , ...) Sulfate -2 +3 +3 SO4 Group IIIA: +3 (B , Al , ...) Phosphate -3 -3 -3 -2 -2 PO4 Group VA: -3 (N , P , ...) Group VIA: -2 (O , S , ...) Group VIIA: -1 (F-, Cl-, ...) Naming • Metal + Nonmetal ® ionic compound; metal name followed by nonmetal “root”+"ide" Example 6: What is the name for CaCl2? ® Answer 6: calcium chloride (note: not dichloride!) • Contains one oR two Polyatomic Ions ® ionic compound; use polyatomic ion name Example 7: What is the name for Ca3(PO4)2? ® Answer 7: calcium phosphate (note: not tricalcium nor diphosphate!) Example 8: What is the name for (NH4)2SO4? ® Answer 8: ammonium sulfate (note: not diammonium!) • Contains Transition Metal ® ionic compound; metal name (charge in Roman numerals) Example 9: What is the name for Fe2(SO4)3? ® Answer 9: iron(III) sulfate (see below for process to follow) 1. Transition metals have charges that vary so the metal charge needs to be determined and specified in the name; the charge is written in Roman numerals in parentheses after the metal name -2 2. Find charge on Fe. First, SO4 charge is -2 (memorized); total negative charge: 3(-2) = -6 from the 3 SO4 ; since the compound is neutral the 2 Fe must total +6 ® therefore, each Fe = +3 = Fe+3; the Roman numeral for 3 = III IONIC COMPOUNDS (continued): Writing Formulas 1. Take metal name or polyatomic ion and write formula including charge • If metal has a Roman numeral, then charge is determined from that numeral (e.g., copper(II) = Cu+2) + • If it does not have a Roman numeral, it is a memorized polyatomic ion (e.g., ammonium = NH4 ) or determined from the column in the Periodic Table it resides in (e.g., magnesium = Mg+2 since it is in column IIA) 2. Take nonmetal name or polyatomic ion and write formula including charge • If nonmetal name ends with “ide” it is a single atom ion with a negative charge with the charge determined from column in Periodic Table it resides in (e.g., chloride = Cl- since column VIIA has a -1 charge); exceptions: hydroxide (OH-) and cyanide (CN-) are polyatomic ions with a name that ends in “ide” • If name ends with “ite” or “ate”, or starts with “hypo” or “per” it is a memorized polyatomic ion 3.
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