The mineral mangan neptunite has the empirical formula **K _{2}Na_{4}Li_{2}Mn_{3}Fe^{2+}Ti_{4}Si_{16}O_{48}**

It’s been found locally here in St.Hilaire and samples on ebay are not cheap. The compound fascinates me because it consists of seven oxides, all of which are binary compounds. Can you give formulas for each of the oxides? We expect you to get the answer from a basic knowledge of the periodic table and from a bit of logic, not from knowing geology.

Here’s the solution:

In the compound **K _{2}Na_{4}Li_{2}Mn_{3}Fe^{2+}Ti_{4}Si_{16}O_{48}**

we find:

K^{+}, O^{-2} = K_{2}O

Na^{+} , O^{-2} = Na_{2}O

Li^{+}, O^{-2} = Li_{2}O

Fe^{+2}(given as such) = FeO

SiO_{2} (same ratio as family member carbon)

To figure out Mn and Ti’s charges, we apply some logic:

So for two K and two Li we need a total of only **2** oxygens. Four sodiums require **2** oxygens. Sixteen Si’s(see formula) need **32** oxygens. One Fe^{+2} needs **1** oxygen. There are 48 oxygens in all in mangan’s formula. So far we have accounted for 37. The remaining 11 oxygens bond to 3Mn’s and 4 titaniums.

Let x = Mn’s charge

Let y = Ti’s charge

3x + 4y = 22

x = (22 – 4y)/3

From trial and error we see that the only whole numbers that satisfy the above equation are

y = 1 and x = 6.(but 6 is too high an oxidation # for Mn in a mineral)

y = 4 and x = 2.

So Mn^{+2} and O^{-2} = MnO.

Three Mn’s require 3 oxygens, which leaves (11-3 =) 8 for titanium, which is exactly what four Ti^{+4} need.

Ti ^{+4}, O^{-2} = TiO_{2.}