A item of electronic equipment originally included $\pu1.250 g$ that a valuable metal. Corrosion with time has given a compound of formula $\ceM(OH)3$, i beg your pardon is isolated however not weighed. This compound is treated v sulfuric acid to offer a compound that has beautiful red crystals and also is uncovered to have a molecule formula of $\ceM2(SO4)2$ and a massive of $\pu1.860 g$. What is this valuable metal?

My attempt:$$ 1.860 ~\mathrmg - 1.250~\mathrmg= 0.61~\mathrmg\, \ceSO4 $$

I took the $\pu0.61 g$ and also converted lock to moles of $\ceSO4$, then ns converted 2 moles $\ceSO4$ come 2 moles $\ceM$ and also got $0.0064~\mathrmM$. Climate I divided that through $\pu1.250 g$.

I gained 195.31, so is platinum the correct answer?


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Gaurang Tandon
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The mass of the sulfate is $m_\ceSO4 = 0.610~\mathrmg$, the fixed of the metal is $m_\textM = 1.250~\mathrmg$.

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With the molar massive of the sulfate anion $M_\ceSO4 = 96.061~\mathrmg\, mol^-1$ we can calculate the quantity of sulfate current in the final compound:$$ n_\ceSO4 = \fracm_\ceSO4M_\ceSO4 = 6.35~\mathrmmmol $$

Since the sulfate has actually the formula $\ceM2(SO4)2$ we understand that we have actually the same amount of metal as sulfate: $n_\textM = n_\ceSO4$.

We have the right to now calculation the molar mass of the steel using the info above:$$ M_\textM = \fracm_\textMn_\textM = 196.85~\mathrmg\, mol^-1 $$

This number resembles most very closely gold through a molar massive of $M_\textAu = 196.97~\mathrmg\, mol^-1$.

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Note: the is important that friend don"t round your numbers when doing the calculations, but round them at the end. So you must keep as lot precision as possible in in between to minimization rounding errors follow me the way.

If you divide $\mathrm1.250\ g$ by $\mathrm6.35\ mmol$ girlfriend will obtain $\mathrm196.85\ g/mol$ (gold).If you divide $\mathrm1.250\ g$ through $\mathrm6.4\ mmol$ friend will acquire $\mathrm195.31 g/mol$ (platinum).