**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 V_{2}O_{3}

The same calculation type (using V = 2.573 g and O = 2.016 g) for the first oxide yields V_{2}O_{5} 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 V_{2}O_{3} 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*(SO_{4})_{2} · 12H_{2}O, 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 H_{2}O in hydrate. 20.000 g minus 11.123 g = 8.877 g of H_{2}O lost.

8.877 g ÷ 18.015 g/mol = 0.493 mol H_{2}O

(2) Use a molar proportion to determine moles of the anhydrous compound present. The molar ratio between MM*(SO_{4})_{2} and H_{2}O is 1/12. Set this equal to x / 0.493 and solve.

x = 0.0410 moles of MM*(SO_{4})_{2} present.

(3) Calculate the molecular weight of the anhydrous compound MM*(SO_{4})_{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*(SO_{4})_{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.