Determine the simplest formula for gold chloride if a 0.303 g sample of gold chloride reacts with agno3 to produce gold nitrate and 0.430 g of agcl. (hint: given that 0.430 g of agcl are formed in the reaction one must assume all the chloride in 0.430 g of agcl comes from the gold chloride. )

Au Cl₃

Explanation:

1) Determine the mass of chloride (Cl) in 0.430 g of AgCl

a) atomic and molar masses (from a periodic table)

b) Use proportions

35.453 g Cl / 143.321 g AgCl = x / 0.430 g AgCl ⇒

x = 0.430 g Cl × 35.453 g Cl / 143.321 g AgCl = 0.106 g Cl

2) Determine the mass of Au

Mass balance:

mass of the samle =  mass of Au in the sample + mass of Cl in the sample ⇒

mass of Au = mass of sample - mass of Cl = 0.303 g - 0.106 g = 0.197 g

3) Convert the masses in grams of Au and Cl into number of moles

4) Determine the ratio of moles:

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