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when a 328 mg sample of gold and chlorine was treated with silver nitrate, Cl was converted to AgCl producing 464 mg of silver chloride. what is the empirical formula?

AuCl + Ag(NO3) -> AgCl + Au(NO3)

1 answer

First, you can NOT write an equation because you don't know the formula of gold chloride not the formula of gold nitrate..

The Cl in gold chloride was converted to AgCl. So how much Cl must have been there with the gold? That's 464 mg x (atomic mass Cl/molar mass AgCl) = approx 115 mg Cl.
That leaves 328-115 = approx 213 mg for Au.
mols Au = grams/atomic mass Au = ?
mols Cl = grams/atomic mass Cl = ?

Now find the ratio of Au to Cl with the lowest number being 1.00. The easy way to do that is to divide the smaller number by itself which makes sure that is 1.00. Then divide the other number by the same small number. Round to a whole number. That gives you the x and y to make the empirical formula AuxCly. I get AuCl3.