Chemistry Problem Solutions: Mass-Mole Conversions and Empirical Formulas, Exams of Chemistry

Download Chemistry Problem Solutions: Mass-Mole Conversions and Empirical Formulas and more Exams Chemistry in PDF only on Docsity! Solutions 1b (suggested problems before Exam #1) Chem151 [Kua] 2.34 Mass-mole-number conversions require the use of masses in grams, molar masses, and/or the Avogadro constant. To determine the number of moles, convert the mass into grams and divide by the molar mass. (a) n = MM m = !! " # $$ % & !! " # $$ % & g 47.88 mol 1 kg 1 g10 kg 67.3 3 = 76.6 mol; (b) n = MM m = !! " # $$ % & !! " # $$ % & ' g 40.08 mol 1 mg 1 g10 mg .97 3 = 2.0 x 10-4 mol; (c) n = MM m = !! " # $$ % & g 101.07 mol 1 g 1.56 = 1.54 x 10-2 mol; and (d) n = MM m = !! " # $$ % & !! " # $$ % & ' g 98 mol 1 pg 1 g10 pg 9.63 12 = 9.8 x 10-14 mol. 3.34 All parts of this question involve mass-mole-number conversions. Moles and mass in grams are related through the equation, n = MM m . Use the Avogadro constant to convert between number and moles. When masses are not given in grams, unit conversions must be made. The chemical formula states the number of atoms of each element per molecule of substance. MM (C20H24O2) = 20(12.01 g/mol) + 24(1.008 g/mol) + 2(16.00g/mol) = 296.4 g/mol; n = MM m = !! " # $$ % & !! " # $$ % & g 296.4 mol 1 mg 1 g10 mg 035.0 -3 = 1.18 x 10-7 mol; (note an additional sig. fig. is added here to avoid rounding errors later on) # = n NA = (1.2 x 10-7 mol)(6.022 x 1023 molecules/mol) = 7.2 x 1016 molecules; # atoms = (atoms/molecule)(# molecules) #C atoms = 7.2 x 1016molecules !! " # $$ % & 22420 OHC molecule 1 C atom 20 = 1.4 x 1018 atoms C; m = n MM = ! " # $ % & !! " # $$ % &' mol 1 g 12.01 OHC mole 1 C mol 20 mol10 x 18.1 22420 7 = 2.9 x 10-5 g. (remember there should only be two sig figs here because the number of moles is only known to 2 sig figs) 3.38 When working with mass percent compositions, it is convenient to consider 100 g of substance. Divide the mass of each element by its elemental molar mass to obtain relative amounts of each element in the substance. O: !! " # $$ % & !! " # $$ % & g 6.001 mol 1 g 100 g 41.41 = 2.588 mol O/100 g; P: !! " # $$ % & !! " # $$ % & g 0.973 mol 1 g 100 g 18.50 = 0.5974 mol P/100 g; H: !! " # $$ % & !! " # $$ % & g .0081 mol 1 g 100 g 0.20 = 0.20 mol H/100 g; Ca: !! " # $$ % & !! " # $$ % & g 0.084 mol 1 g 100 g 39.89 = 0.9953 mol Ca/100 g; To obtain the empirical formula divide each amount by the smallest among them (hydrogen with 0.20 mol/100 g:) and round to whole numbers. This tells us the number of each element relative to one nitrogen atom in the compound: O: ! " # $ % & !! " # $$ % & H mol .200 g 100 g 100 O mol 2.588 = 13 mol O/mol H, round to 13 O/H; P: ! " # $ % & !! " # $$ % & H mol .200 g 100 g 100 P mol 0.5974 = 3.0 mol P/mol H, round to 3 P/H; Ca: ! " # $ % & !! " # $$ % & H mol .200 g 100 g 100 Ca mol 0.9953 = 5.0 mol Ca/mol H, round to 5 Ca/H; Empirical formula: Ca5HP3O13, empirical MM = 502.3 g/mol; To convert empirical formula to molecular formula, multiply subscripts in the empirical formula by the ratio of the compound's molar mass to its empirical formula mass. MM/empirical MM = !! " # $$ % & g/mol 502.3 g/mol 1004 = 2; Molecular formula: Ca10H2P6O26. 3.66 The chemical formula of a substance provides all the information needed to compute its molar characteristics. (a) MM = 58.69 + 32.07 + 4(16.00) + 6[(2)(1.008) + 16.00] = 262.9 g/mol; (b) n = MM m = !! " # $$ % & g 262.9 mol 1 g .025 = 9.51 x 10-2 mol; (c) To determine the percent composition, work with one mole of substance. Take the ratio of the mass of each element that one mole contains to the mass of one mole (molar mass). % Ni = 100% g/mol 262.9 g/mol 58.69 ! = 22.33 %; 3.50 This is a dilution type problem. Begin by deriving the expression to solve for the volume: MiVi = MfVf, or Vi = i ff M VM Vi = ( )( ) M 14.8 L 1.25M 0.500 = 0.0422 L or 42.2 mL 3.52 In each of the following, remember that concentration is M = V n . Begin each problem by identifying the major ionic species present. (b) Major ionic species: Fe3+, Cl- Determine the number of moles in 1.44 mg of the salt and use stoichiometric ratios to determine the number of moles of each ion: MM(FeCl3) = 55.85 g/mol + 3(35.45 g/mol) = 162.2 g/mol n(FeCl3) = !! " # $$ % & !! " # $$ % & g 162.2 mol 1 mg 1 g10 mg 1.44 -3 = 8.88 x 10-6 mol n(Fe3+) = !! " # $$ % & + 3 3 3 6- FeCl mol 1 Fe mol 1 FeCl mol 10 x 8.88 = 8.88 x 10-6 mol n(Cl-) = !! " # $$ % & ' 3 3 6- FeCl mol 1 Cl mol 3 FeCl mol 10 x 8.88 = 2.66 x 10-5 mol Obtain the molarities by dividing the number of moles of each ion by the volume: V = 2.75 mL !! " # $$ % & mL 1 L10 -3 = 2.75 x 10-3 L; [Fe3+] = !! " # $$ % & L 00275.0 mol10 x 88.8 -6 = 3.23 x 10-3 M [Cl-] = !! " # $$ % & L 00275.0 mol10 x .662 -5 = 9.69 x 10-3 M (c) KNO3 contains both a Group 1 metal ion and a polyatomic ion. The major ionic species are: K+, NO3 - Since all stoichiometric ratios are 1 for this compound, the moles of each ion equal the moles of the compound: MM = 39.10 g/mol + 14.01 g/mol + 3(16.00 g/mol) = 101.1 g/mol n(K+) = n(NO3-) = n(KNO3) = !! " # $$ % & !! " # $$ % & g 101.1 mol 1 kg 1 g10 kg 8.75 3 = 86.5 mol; Determine the molarities by dividing the number of moles by the volume: [K+] = [NO3 -] = L 235 mol 86.5 =0.368 M 3.84 (a) MM = 3(22.99 g/mol Na) + 30.97 g/mol P + 4(16.00 g/mol O) = 163.94 g/mol n = mol 0.549 g 163.94 mol 1 g 0.09 =!! " # $$ % & M = ! " # $ % & L 5.1 mol 549.0 = 0.37 M Na3PO4 (use 0.366 M in b and c to avoid round off error) (b) mol 10 x 1.52 g 163.94 mol 1 g 2.50 2!="" # $ %% & ' volume = M n = ! " # $ % & olm .3660 L 1 mol 10 x .521 2- = 4.2 x 10-2 L or 42 mL (c) 2(0.705 M) ! " # $ % & Lm 125 mL 0.05 +3(0.366 M) ! " # $ % & Lm 125 mL 5.07 = 1.22 M