I'm a gold smith and I want to start experimenting with non-standard metal-alloys. Now high school is a long time ago for me so while I think I've got this right, I'm just not as sure as I should be when working with precious metals costing 40 dollar a gram. So could anyone confirm that what I'm doing is right:

I have two metals Metal X, with an atomic-weight of 20 Metal Y, with an atomic-weight of 3

If I want to create an intermetallic alloy of $\ce{XY2}$, where one atom of X is combined with two atoms of Y and I'd want 10 grams of that alloy. I would alloy (20*10)/26=7.692g of metal X and (6*10)/26=2.308g of metal Y, correct?

  • $\begingroup$ I think this is perfectly fine here, great question! $\endgroup$
    – jonsca
    Nov 27, 2012 at 21:09

1 Answer 1


Yes, you are correct.

7.692 g / (20 g/mol) = 0.3846 mol X (or 2.316 * 10^23 atoms X)

2.308 g / (3 g/mol) = 0.7693 mol Y (or 4.633 * 10^23 atoms Y)

Moles are just a count of atoms, so if you have two moles of Y per mole of X, you have two Y atoms per X atom.


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