[Tarondar]

A particular deposit of ore contains 32.6% Gold(I) Oxide (Let's ignore the patent instability of such a molecule).  How many Kilos of Ore would have to be mined in order to collect 120 grams of gold?

I have established the formula as Au2O.  I know there are going to be 2 moles of Gold per mole of Au2O.  I know the molar mass of Au2O is 409.9326 g/mole, and the mass of the Gold within is 393.9331 g/mole.  Finally, I know that 120 grams divided by 393.9331 g/mol means that it's .3046 moles. 

I have ground to a halt here.  I can't make the mental leap needed to finish the problem, if I knew what to find out next, I might be able to solve this, but right now I can't seem to do this myself.

[sjb]

A particular deposit of ore contains 32.6% Gold(I) Oxide (Let's ignore the patent instability of such a molecule).  How many Kilos of Ore would have to be mined in order to collect 120 grams of gold?

I have established the formula as Au2O.  I know there are going to be 2 moles of Gold per mole of Au2O.  I know the molar mass of Au2O is 409.9326 g/mole, and the mass of the Gold within is 393.9331 g/mole.  Finally, I know that 120 grams divided by 393.9331 g/mol means that it's .3046 moles. 

I have ground to a halt here.  I can't make the mental leap needed to finish the problem, if I knew what to find out next, I might be able to solve this, but right now I can't seem to do this myself.

So what would be the mass of 0.3046 moles of gold(I) oxide? This assumes that your ore is pure oxide, which it isn't, so how do you correct for that?

[Tarondar]

Thanks!  I ended up getting 41 Kilos of ore after rounding for significant figures (it actually wanted to know about 1.2*10^2 grams of Au)
Enjoy the mole snack!

[sjb]

Thanks!  I ended up getting 41 Kilos of ore after rounding for significant figures (it actually wanted to know about 1.2*10^2 grams of Au)
Enjoy the mole snack!

Thanks for the snack, but are you sure of your answer? Just under a third of your ore is Au₂O, which is nothing like 120g as a proportion of 41 kg.

[Tarondar]

Aw...

Dammit...

 

Thanks for the help anyway, it's too late to take that answer back (It was for an assignment)

I think I see my problem, I multiplied 120 by the percentage, in retrospect, I think 120 + (120(100- x percent)) would have been the correct approach, limiting the answer to... one or two kilos?

[sjb]

Not even that, simply that you have 124.886 g of Au₂O, which divided by your purity of 32.6% gives ... ?

[Tarondar]

It's too late to take back my dreadful conclusion on my paper, but I truly appreciate the *delete me*
Thanks for the help and enjoy that second snack!

By the way that's about .38 Kilograms...