Chapter 3 sample problems

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: amu, 78.99% Mg-24: amu, 78.99% Mg-25: amu, 10.00% Mg-25: amu, 10.00% Mg-26: amu, 11.01% Mg-26: amu, 11.01%

Magnesium average atomic mass ( amu x ) + ( amu x ) + ( amu x ) = amu amu

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

Calculate the mass of moles of Ca(NO 3 ) moles Ca(NO 3 ) 2 x 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

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

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

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

Empirical formula of ascorbic acid Roadmap: g  mol  ratio C: g C x 1 mol C = mol C g C g C H: 4.58 g H x 1 mol H = 4.54 mol H g H O: g O x 1 mol O = mol O g O g O 3.407/3.406 = 1 x 3 = /3.406 = 1.33 x 3 = /3.406 = 1 x 3 = 3 Empirical formula is C3H4O3

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

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

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 = 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 = MW = 62.1