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0.87 g of Au +sulfur = 1.00 g of silver sulfide. Formula?
Submitted by Anonymous on 21 November 2005 - 1:00am.
In an experiment, 0.87 g of silver reacted completely with sulfure and formed 1.00 g of silver sulfide. Find the simplest formula of the product formed. If someone could tell me the answer and explain how to obtain that answer I would appreciate it. Thnx beforehand. |
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Ag2S
I did it in kind of a roundabout way, but the answer should stil be good.
I got Ag2[/sub]S
by first guessing that the charges would best cancel out if there are two Ag+[/sup]
and one S-
Since all .87g of the Ag was used, we can easily find out how many moles that is by using the molar mass.
If you use the (number of moles of Ag) times (the molar mass) times two, and then add (the number of moles of Ag) times the (molar mass of S), you get 1.00g!
If you do it a different way in class, do it that way because I know this was mostly good guess work.
If there's someone else out there who knows a better way, I'd like to know too, because I know I've seen this same kind of problem before and I've forgotten all about them. :roll: :lol:
Thnx for the answer + your 'round about' work.
I couldn't figure out how the mass worked in, so I could not do the problem lol.
Convert mass to molar mass (by using what's off the periodic table) for Ag. Subtract the amount of silver (0.87 g) from the total mass of the compound (1.00 g). This will give you the mass of the sulfur. Convert that to moles of sulfur and you should get the answer, which I believe would be Ag2S, at least according to the charges on the ions.
And if you're taking a multiple choice test (either the SAT 2's or AP), a quick way to do those problems (as long as the compound is ionic) is to use the charges to figure out the formula. This is a heck of a lot faster than doing the math- especially when you don't have a calculator! ;)
Oh, by the way, in the title, it says "Au" and not "Ag". You might want to fix that because there's quite a difference between the two!