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.)

Answer: 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)

Ag: 107.868 g/mol Cl: 35.453 g/mol AgCl: 107.868 g/mol + 35.453 g/mol = 143.321 g/mol

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

Formula: number of moles = mass in grams / atomic mass atomic mass of Au: 196.967 g/mol number of moles of Au = 0.1967 g / 196.967 g/mol = 0.000999 mol number of moles of Cl = 0.106 g / 36.453 g/mol = 0.0029 mol

4) Determine the ratio of moles:

divide by the least number of moles: 0.000999 Au: 0.000999 / 0.000999 = 1 Cl: 0.0029 / 0.000999 = 2.9 ≈ 3 Ratio: 1 mole Au : 3 mole Cl Simplest formula: Au Cl₃ ← answer