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KEY for Unit 5 - Chemical Concepts Formulas of Ionic Compounds Ex. Determine the formula of an ionic compound made up of the following pairs of ions: (a) Al3+ with O2-, forms Al2O3 (b) Mg2+ with N3-, forms Mg3N2 (c) Na+ with O22-, forms Na2O2 (d) Fe3+ with P3-, forms FeP and (e) Ba2+ with Br-, forms BaBr2 Atomic and Formula Masses Isotopes and Atomic Masses Cl-35 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 = 35.453 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) = 80.8860 amu (6.5787)(12 amu exactly) = 78.9444 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)(78.9444)amu + (0.4943)(80.8860)amu = 79.904 amu 87 (d) Write the full isotopic symbols for these isotopes (you know, the elemental symbol with the mass numbers AND atomic numbers). ans.: 2. 79 35 Br 81 35 Br 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*79.904 amu = 79.904 amu + 79.904 amu = 159.808 amu (b) Do any of the individual diatomic molecules have this mass? Ans.: NO !!! (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)(78.9444)amu = 157.8888 amu medium = Br-79+Br-81 or Br-81+Br-79, (78.9444)amu + (80.8860)amu = 159.8304 amu heaviest = 2 Br-81, (2)(80.8860)amu = 161.7720 amu (d) 39.95 grams of this element would contain how many atoms of this element? ans.: 6.022 x10 23 Br atoms 1 mole Br atoms x x39.95g Br atoms = 1 mole Br atoms 79.90 4 g Br atoms 3.011x10 23 Br atoms (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 23 the number of atoms from which they can be built = 1.505x10 Br2 molecules Mass Relations in Chemical Formulas You can try this one: find the mass percent of each of the elements in the molecule C2HBrClF3 Ans.: % 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 % 88 % Br = 100*(1*79.904 amu Br)/(197.39 amu cmpd.) = 40.48 % % Cl = 100*(1*35.453 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. Ans.: (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/35.453 gmole-1 = 0.07500 mole Cl 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 NaClO3 (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. Ans.: (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 = CH2O [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(CH2O) = C6H12O6 (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 0.03560 g of cadaverine is combusted in an excess of oxygen, 0.07665 g of CO2 and 0.04392 g of H2O 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 89 determine how many of the empirical units are needed to have a mass equal to the molecular mass. Ans.: (1) mole of CO2 = 0.07665 g CO2/44.01 gmole-1 = 0.001742 mole CO2 mole of H2O = 0.04392 g H2O/18.02 gmole-1 = 0.002437 mole H2O (2) mole C = mole CO2 = 0.001742 mole C mole H = 2*mole H2O = (2moleH/mole H2O)*0.002437 mole H2O = 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 = 0.03560 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 = C2.5H7N1 which must be doubled to find C5H14N2 the empirical mass is 102.19 amu thus the molecule consists of a single empirical unit giving molecular formula = C5H14N2 Some Practice Problems Molecular Weight and Formula Weight (to find MW or FW add together atomic masses in formula) 1. 2. 3. 4. 5. NaCl has a formula weight K2CO3 has a formula weight CuSO4·5H2O has a formula weight CH4 has a molecular weight C6H13OH has a molecular weight 6. HCl has a molecular weight = = = = = 58.44 amu 138.206 amu 249.69 amu 16.04 amu 102.18 amu = 36.461 amu and a molar mass and a molar mass and a molar mass and a molar mass and a molar mass = = = = = 58.44 g/mole 138.206 g/mole 249.69 g/mole 16.04 g/mole 102.18 g/mole 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 CH4/1 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 H2O Mole and Avogadro’s Number (info needed 1 mole = 6.022 x1023 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 2. 2.74 mole Al (aluminum) = 73.93 g Al 21 3. 7.46 x10 Bi (bismuth) atoms = 0.0124 mole Bi 90 = 1.247x1025 Na atoms = 1.65x1024 Al atoms = 2.59 g Bi