# Determine what the unknown element is

A piece of electronic equipment originally contained $\pu{1.250 g}$ of a valuable metal. Corrosion over time has given a compound of formula $\ce{M(OH)3}$, which is isolated but not weighed. This compound is treated with sulfuric acid to give a compound that has beautiful red crystals and is found to have a molecular formula of $\ce{M2(SO4)2}$ and a mass of $\pu{1.860 g}$. What is this valuable metal?

My attempt:
$$1.860 ~\mathrm{g} - 1.250~\mathrm{g}= 0.61~\mathrm{g}\, \ce{SO4}$$

I took the $\pu{0.61 g}$ and converted them to moles of $\ce{SO4}$, then I converted 2 moles $\ce{SO4}$ to 2 moles $\ce{M}$ and got $0.0064~\mathrm{M}$. Then I divided that by $\pu{1.250 g}$.

I got 195.31, so is platinum the correct answer?

• I got Bromine for my answer? I used conversion table from SO4 in grams to moles all the way to the metal in grams? – Mason Oct 23 '14 at 8:31
• Bromine is not a metal, so this is no possible answer. However, it would be very nice if you could edit your question to contain information in the scope of the aforementioned homework policy. Thank you. – Martin - マーチン Oct 23 '14 at 8:39
• Oops thats embarressing its not even a metal lol i edited with more of my thinking. How about Platinum? – Mason Oct 23 '14 at 8:52
• Now, if you could format everything so it would be easier on the eyes, I'd give you an upvote. Check out the comment by Philipp. – tschoppi Oct 23 '14 at 9:41

The mass of the sulfate is $m_{\ce{SO4}} = 0.610~\mathrm{g}$, the mass of the metal is $m_\text{M} = 1.250~\mathrm{g}$.

With the molar mass of the sulfate anion $M_{\ce{SO4}} = 96.061~\mathrm{g\, mol^{-1}}$ we can calculate the amount of sulfate present in the final compound: $$n_{\ce{SO4}} = \frac{m_{\ce{SO4}}}{M_{\ce{SO4}}} = 6.35~\mathrm{mmol}$$

Since the sulfate has the formula $\ce{M2(SO4)2}$ we know that we have the same amount of metal as sulfate: $n_\text{M} = n_{\ce{SO4}}$.

We can now calculate the molar mass of the metal using the information above: $$M_\text{M} = \frac{m_\text{M}}{n_\text{M}} = 196.85~\mathrm{g\, mol^{-1}}$$

This number resembles most closely gold with a molar mass of $M_\text{Au} = 196.97~\mathrm{g\, mol^{-1}}$.

Note: It is important that you don't round your numbers while doing the calculations, but round them at the end. So you should keep as much precision as possible in between to minimize rounding errors along the way.

If you divide $\mathrm{1.250\ g}$ by $\mathrm{6.35\ mmol}$ you will get $\mathrm{196.85\ g/mol}$ (gold).
If you divide $\mathrm{1.250\ g}$ by $\mathrm{6.4\ mmol}$ you will get $\mathrm{195.31 g/mol}$ (platinum).