KEY for Unit 1 Your Chemical Toolbox: Scientific Concepts, Fundamentals of Typical Calculations, the Atom and Much More The Modern Periodic Table The Periodic Law - when elements are arranged according to their atomic number, their chemical and physical properties show repeatable trends. The vertical columns are called groups and the horizontal rows are called periods. There are four categories of groups (although some people do not separate the Noble gases from the Main Group elements): 1. Representative (also Main Group Elements) 2. Transition elements 3. Inner transition (also Lanthanide and Actinide) 4. Noble gasses The letters A and B in the group distinguish families: A = representative and B = transition. One of the most useful features of the periodic table is that it summarizes properties in a succinct manner. Take a look at a periodic table and think about the following. Where are metals found? See periodic table below elements in various shades of blue. Where are nonmetals found? See periodic table below elements in yellow. Where are the metalloids found? See periodic table below elements in green. Are all main group elements metals, nonmetals, metalloids? All Group 1A (except H) and 2A elements are metals. In groups 3A 8A degree of metallic character increasing down the groups (with groups 7A and 8A all being listed as nonmetals). Metalloids are in the p-block of the periodic table running in a staircase pattern from boron (B) down and to the right.. Locate the following groups in the periodic table. Alkali metals (Group 1A or simply Group 1), alkaline earth metals (Group 2A or simply Group 2), noble gases [in the past called the inert gases] (Group 8A or simply Group 18), halogens (Group 7A or simply Group 17), chalcogens (Group 6A or simply Group 16), coinage metals (Group 1B or simply Group 11), and list the semimetals [the metalloids] boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), and tellurium (Te). Polonium (Po) is often considered a metalloid, too). 1
Check three different periodic tables and see if they all list the same metalloids. Think about why there might not be universal agreement about which elements to call metalloids (think about the definition of metalloids, or semi-metals as they are sometimes called). Metalloids are intermediate in behavior between metals and non-metals, and since metallic behavior varies across a period and up (or down) a group, we should not be surprised if there is not universal agreement on EXACTLY where the dividing line should be drawn. Some elements that are listed as metalloids look quite metallic in appearance (like antimony, Sb) whereas others display properties more similar to nonmetals (example the brittle nature of silicon, Si). Figure: a reasonably current modern periodic table new elements are added above atomic number 109, but that area of the periodic table has no naturally occurring elements. 2
Some Practice Problems with Atomic Numbers, etc. Neutral Atoms Isotope Mass # #n #p #e C-12 12 6 6 6 Full Isotopic Symbol Ions Mass # #n #p #e 55 2+ 26 Fe 55 29 26 24 C-13 13 7 6 6 12 2 6 C 12 6 6 8 Pb-206 206 124 82 82 16 2 8O 16 8 8 10 F-19 19 10 9 9 209 3+ 83 Bi 209 126 83 80 Fe-56 56 30 26 26 81 1 Br 81 46 36 Na-23 23 12 11 11 79 1 Br 79 44 36 Cu-61 61 32 29 29 2 6+ 92U 2 143 92 86 Cl- 18 17 17 108 1+ 47 Ag 108 61 47 46 Cl-37 37 19 17 17 14 3 7 N 14 7 7 10 O-18 18 10 8 8 32 2 16 S 32 16 16 18 3
EXAMPLE: Average Atomic Masses Atomic and Formula Masses Isotopes and Atomic Masses Cl- 75.76 % 34.969 amu Cl-37 24.24 % 36.966 amu Ans. Ave. Cl = (34.969)(0.7576)amu + (36.966)(24.24)amu =.45 3 amu Guided Discovery for Unit Five: Average Atomic Masses 1. A particular element is found to exist as two isotopes. (a) One has a mass that is 6.7405 times that of C-12, the other is 6.5787 times the mass of C-12. What are the masses of these isotopes in atomic mass units (amu)? ans.: (6.7405)(12 amu exactly) = 80.886 0 amu (6.5787)(12 amu exactly) = 78.944 4 amu (b) The isotope that has the lower mass has a relative abundance of 50.57 percent. What is the relative abundance of the isotope with the larger mass? ans.: 100 % 50.57 % = 49.43 % (c) What is the average atomic mass for this element? ans.: (0.5057)(78.944 4 )amu + (0.4943)(80.886 0 )amu = 79.90 4 amu (d) Write the full isotopic symbols for these isotopes (you know, the elemental symbol with the mass numbers AND atomic numbers). ans.: 79 Br 81 Br 2. The element discussed in question number one is a halogen. As a pure element, it exists as a diatomic molecule (at room temperature and 1 atm pressure it is most stable as a liquid). (a) What is the average mass of each of these diatomic molecules (in amu)? ans.: if two average atoms combine they will create an average molecule with an average mass = 2*79.90 4 amu = 79.90 4 amu + 79.90 4 amu = 159.80 8 amu (b) Do any of the individual diatomic molecules have this mass? NO!!! 4
(c) Calculate the three masses that these diatomic molecules can actually have based on the masses of the two isotopes used to form the molecules. ans.: lightest = 2 Br-79, (2)(78.944 4 )amu = 157.888 8 amu medium = Br-79+Br-81 or Br-81+Br-79, (78.944 4 )amu + (80.886 0 )amu = 159.830 4 amu heaviest = 2 Br-81, (2)(80.886 0 )amu = 161.772 0 amu (d) 39.95 grams of this element would contain how many atoms of this element? ans.: 23 6.022x10 Br atoms 1mole Br atoms x x39.95g Br atoms = 1mole Br atoms 79.90 g Br atoms 3.011x10 23 Br atoms 4 (e) 39.95 grams of this element would contain how many diatomic molecules of the element? Since each diatomic molecule is made up of two atoms, the number of molecules is half the number of atoms from which they can be built = 1.505x10 23 Br 2 molecules Mass Relations in Chemical Formulas You can try this one: find the mass percent of each of the elements in the molecule C 2 HBrClF 3 % C = 100*(2*12.011 amu C)/(197.39 amu cmpd.) = 12.17 % % F = 100*(3*19.00 amu F)/(197.39 amu cmpd.) = 28.88 % % Br = 100*(1*79.90 4 amu Br)/(197.39 amu cmpd.) = 40.48 % % Cl = 100*(1*.453 amu Cl)/(197.39 amu cmpd.) = 17.96 % % H = 100*(1*1.008 amu H)/(197.39 amu cmpd.) = 0.510% check: the sum of these mass % = 100 examples (a) 7.985 g of an ionic compound that is composed of sodium, chlorine, and oxygen is found to contain 1.725 g of sodium and 2.659 g of chlorine. What is the simplest (empirical formula)? strategy: (1) find the grams of oxygen present, (2) convert the mass of each element to moles, and (3) divide by the smallest number of moles to get a whole number ratio. (1) 7.985 g cmpd. 1.725 g Na 2.659 g Cl = 3.601 g O (2) mole Na = 1.725 g Na/23.00 gmole -1 = 0.07500 mole Na mole Cl = 2.659 g Cl/.453 gmole -1 = 0.07500 mole Cl 5
mole O = 3.601 g O/16.00 gmole -1 = 0.2251 mole O (3) divide by moles of Na (or Cl in this case too) to find the simplest formula is NaClO 3 (b) A molecular compound is found to contain 40.0 % C, 6.73 % H, and 53.3 % O. What is the simplest formula of the compound AND if it is known to have a molar mass of 181 g/mole, what is the molecular formula. strategy: (1) assume 100 g sample and convert % to grams, (2) convert the mass of each element to moles, and (3) divide by the smallest number of moles to get a whole number ratio. To find the molecular formula, find the mass of the empirical formula and determine how many of the empirical units are needed to have a mass equal to the molecular mass. (1) 40.0 g C, 6.73 g H, and 53.3 g O (2) mole C = 40.0 g C/12.011 gmole -1 = 3.33 mole C mole H = 6.73 g H/1.008 gmole -1 = 6.66 mole H mole O = 53.3 g O/16.00 gmole -1 = 3.33 mole O (3) divide by the mole of C (or also O in this case) empirical formula = CH 2 O [which has an emp. mass of ~30.01 amu] since 181/30.01 = about 6, the molecule is made up of 6 empirical units giving us a molecular formula = 6(CH 2 O) = C 6 H 12 O 6 (c) Cadaverine is a molecular compound formed in decomposing fish. It has a molar mass of 102.18 and is composed solely of C, H, and N. When 0.060 g of cadaverine is combusted in an excess of oxygen, 0.07665 g of CO 2 and 0.04392 g of H 2 O are formed. What are the simplest formula and molecular formula of cadaverine. strategy: (1) find moles of carbon dioxide and water, (2) convert to moles of carbon and hydrogen, (3) find the mass of carbon and hydrogen, (4) find the mass of nitrogen by difference, (5) find moles of nitrogen, and (6) divide by the smallest number of moles (C, H, or N) to get a whole number ratio. To find the molecular formula, find the mass of the empirical formula and determine how many of the empirical units are needed to have a mass equal to the molecular mass. (1) mole of CO 2 = 0.07665 g CO 2 /44.01 gmole -1 = 0.001742 mole CO 2 mole of H 2 O = 0.04392 g H 2 O/18.02 gmole -1 = 0.002437 mole H 2 O (2) mole C = mole CO 2 = 0.001742 mole C mole H = 2*mole H 2 O = (2moleH/mole H 2 O)*0.002437 mole H 2 O = 0.004874 mole H (3) mass C = mole C*12.011g/mole = 0.02092 g C mass H = mole H*1.008 g/mole = 0.004913 g H mass N = mass cmpd mass C mass H = 0.060 g 0.02092 g C 0.004913 g H mass N = 0.00977 g N (4) mole N = 0.00977 g N/14.01 gmole -1 = 0.000697 mole N (5) divide by the mole of N to find empirical formula = C 2.5 H 7 N 1 which must be doubled to find C 5 H 14 N 2 the empirical mass is 102.19 amu thus the molecule consists of a single empirical unit giving molecular formula = C 5 H 14 N 2 6
Some Practice Problems Molecular Weight and Formula Weight (to find MW or FW add together atomic masses in formula) 1. NaCl has a formula weight = 58.44 amu and a molar mass = 58.44 g/mole 2. K2CO3 has a formula weight = 138.206 amu and a molar mass = 138.206 g/mole 3. CuSO4 5H2O has a formula weight = 249.69 amu and a molar mass = 249.69 g/mole 4. CH4 has a molecular weight = 16.04 amu and a molar mass = 16.04 g/mole 5. C6H13OH has a molecular weight = 102.18 amu and a molar mass = 102.18 g/mole 6. HCl has a molecular weight = 36.461 amu and a molar mass = 36.461 g/mole Mole and Molar Mass 1. 79.63 g NaCl = (1mole NaCl/58.44 g)*(79.63 g) = 1.363 mole NaCl 2. 2.694 mole CH4 = (16.04 g CH 4 /1 mole)*(2.694 mole) = 43.21 g CH4 3. 476.92 g CuSO4 5H2O = (476.92 g)/(249.69 gmole -1 ) = 1.910 mole CuSO4 5H2O 5 moles of water per mole of hydrate yields = 5*1.910 mole H2O = 9.550 mole H 2 O Mole and Avogadro s Number (info needed 1 mole = 6.022 x10 23 items) use molar mass for g to mole or mole to gram conversions use Avogadro s number for number of things to mole or mole to number of things conversions 1. 475.9 g Na (sodium) = 20.70 mole Na = 1.247x10 25 Na atoms 2. 2.74 mole Al (aluminum) = 73.93 g Al = 1.65x10 24 Al atoms 3. 7.46 x10 21 Bi (bismuth) atoms = 0.0124 mole Bi = 2.59 g Bi End coverage at page 53.... DO NOT FORGET know ALL chemical symbols and names through atomic number 30 [that s from #1 hydrogen (H) through #30 zinc (Zn)] 7