1. Chapter 3 sample problems

2. Average atomic mass Calculate the average atomic mass of magnesium given the following isotopic mass and mass percent data. Calculate the average atomic mass of magnesium given the following isotopic mass and mass percent data. Mg-24: 23.98504 amu, 78.99% Mg-24: 23.98504 amu, 78.99% Mg-25: 24.98584 amu, 10.00% Mg-25: 24.98584 amu, 10.00% Mg-26: 25.98259 amu, 11.01% Mg-26: 25.98259 amu, 11.01%

3. Magnesium average atomic mass (23.98504 amu x 0.7899) + (24.98584 amu x 0.1000) + (25.98259 amu x 0.1101) = 24.31 amu 24.31 amu

4. Calculate the number of moles in 508 grams of AlCl 3. 508 g AlCl 3 x 1 mole AlCl 3 133.33 g AlCl 3 133.33 g AlCl 3 = 3.81 moles AlCl 3

5. Calculate the mass of 0.433 moles of Ca(NO 3 ) 2. 0.433 moles Ca(NO 3 ) 2 x 164.10 g Ca(NO 3 ) 2 1 mole Ca(NO 3 ) 2 1 mole Ca(NO 3 ) 2 = 71.1 g Ca(NO 3 ) 2 = 71.1 g Ca(NO 3 ) 2

6. Calculate the number of oxygen atoms in 24.76 grams of Na 2 CO 3. Roadmap: g  mol  molecules  atoms 24.76 grams Na 2 CO 3 x 1 mol Na 2 CO 3 x 6.022 E23 molecules Na 2 CO 3 x 3 oxygen atoms 105.99 g Na 2 CO 3 1 mol Na 2 CO 3 1 molecule Na 2 CO 3 = 4.220 E23 oxygen atoms (E = exponential)

7. Calculate the % by mass of each element in glucose, C 6 H 12 O 6. %C: 6 x 12.01 g/mol (carbon) x 100 180.16 g/mol (glucose) 180.16 g/mol (glucose) = 39.98 % carbon by mass = 39.98 % carbon by mass

8. Calculate the empirical formula of ascorbic acid, given the following % by mass composition data. C: 40.92 % H: 4.58 % O: 54.50 % Since the above numbers are mass percents, in 100 grams of compound we have: C: 40.92 g H: 4.58 g O: 54.5O g

9. Empirical formula of ascorbic acid Roadmap: g  mol  ratio C: 40.92 g C x 1 mol C = 3.407 mol C 12.01 g C 12.01 g C H: 4.58 g H x 1 mol H = 4.54 mol H 1.008 g H O: 54.50 g O x 1 mol O = 3.406 mol O 16.00 g O 16.00 g O 3.407/3.406 = 1 x 3 = 3 4.54/3.406 = 1.33 x 3 = 4 3.406/3.406 = 1 x 3 = 3 Empirical formula is C3H4O3

10. Calculate the empirical and molecular formulas of a compound given the following data. C: 38.7 % H: 9.7 % O: 51.6 % Molar mass: 62.1 g/mol

11. Empirical & molecular formulas In 100 grams: 38.7 g C, or 38.7 g x 1 mol/12.01 g = 3.22 mol C 9.7 g H, or 9.7 g x 1 mol/1.008 g = 9.6 mol H 51.6 g O, or 51.6 g x 1 mol/16.00 g = 3.23 mol O __________________________________________ C: 3.22/3.22 = 1 H: 9.6/3.22 = 3 So, empirical formula is CH 3 O. O: 3.23/3.22 = 1

12. Empirical and molecular formulas EF = CH 3 O MF? EW = MW = 62.1 EF = CH 3 O MF = C 2 H 6 O 2 EW = 31.03 MW = 62.1 EW = Empirical weight Divide MW by EW Divide MW by EW to get an integer (2). to get an integer (2). Multiply EF through Multiply EF through by the integer to by the integer to get MF. get MF. EF = CH 3 O MF = ? EW = 31.03 MW = 62.1