Bonus Empirical Formula Problem Answers 5 & 6

Problem #5: A sample of an oxide of vanadium weighing 4.589 g was heated with hydrogen gas to form water and another oxide of vanadium weighing 3.782 g. The second oxide was treated further with hydrogen until only 2.573 g of vanadium metal remained.

(a) What are the simplest formulas of the two oxides?
(b) What is the total mass of water formed in the successive reactions?

Solution

(1) oxygen in second oxide is 3.782 g minus 2.573 g - 1.209 g

(2) calculate moles of each element present: V is 2.573 g ÷ 50.94 g/mol = 0.0505104 mol and O is 1.209 ÷ 16.00 g/mol = 0.0755625 mol

(3) dividing by the smaller gives V = 1.00 and O = 1.50. Therefore, the empirical formula for the second oxide is V2O3

The same calculation type (using V = 2.573 g and O = 2.016 g) for the first oxide yields V2O5 as the answer.

A more traditional solution style would get the percent composition after step one above. V is 68.03% (2.573/3.872) and O is 31.97%. From there, the molar amounts (step two above) are calculated with V = 1.33 (68.03/50.94) and O = 2 (31.97/16). When you divide through by 1.33, you get V = 1 and O = 1.5, whence the formula V2O3 immediately follows.

Sometimes it is helpful to remember that 1.33 is 4/3, although it is not needed in this case. Notice the reasoning: V = 4/3 and O = 2. Multiply by 3 to get V = 4 and O = 6. Reducing gives V = 2 and O = 3. Yes, it is a bit redundant in context, but it is another valid way to think about the solution.


Problem #6: The term "alum" refers to a class of compounds of general formula MM*(SO4)2 · 12H2O, where M and M* are different metals. A 20.000 g sample of a certain alum is heated to drive off the water; the anhydrous residue weighs 11.123 g. Treatment of the residue with excess NaOH precipitates all the M* as M*(OH)3, which weighs 4.388 g. Calculate the molar mass of the alum and identity the two metals, M and M*.

Solution

(1) Calculate moles of H2O in hydrate. 20.000 g minus 11.123 g = 8.877 g of H2O lost.

8.877 g ÷ 18.015 g/mol = 0.493 mol H2O

(2) Use a molar proportion to determine moles of the anhydrous compound present. The molar ratio between MM*(SO4)2 and H2O is 1/12. Set this equal to x / 0.493 and solve.

x = 0.0410 moles of MM*(SO4)2 present.

(3) Calculate the molecular weight of the anhydrous compound MM*(SO4)2. 11.123 g ÷ 0.0410 mol = 270.9 g/mol. Since the two sulfates contribute 192, the combined weight of M and M* is 78.9.

(4) Identify M*. 4.388 g ÷ 0.0410 mol = 107, the molecular weight of M*(OH)3. 0.0410 was used because of the one-to-one ratio between M* and MM*(SO4)2. 107 minus (17 x 3) = 56, the atomic weight of Fe. 17 is the molecular weight of the hydroxide.

(5) Identify M. 78.9 - 56 = 22.9, the atomic weight of sodium.