A Professor treats a water-soluble compound of gold and chlorine with silver nit
ID: 882397 • Letter: A
Question
A Professor treats a water-soluble compound of gold and chlorine with silver nitrate to convert the chlorine completely to silver chloride, AgCl. In an experiment, 0.328 g of the compound gave 0.464 g of silver chloride. What is the empirical formula of the compound?
b) Calculate the molar concentration of chloride ions (Cl-) present in a solution formed by dissolving 6.35 g of the water-soluble compound from part a in sufficient water to form 750.0 ml of solution. (4 pts)
c) Name the water-soluble compound from part a. (1 pt)
Explanation / Answer
Solution :-
Lets first calculate the moles of the silver chloride
Moles = mass / molar mass
Moles of AgCl = 0.464 g / 143.32 g per mol = 0.003238 mol AgCl
Since mole ratio of the AgCl to Cl is 1 :1 therefore moles of Cl = 0.003238 moles
Now lets convert moles of Cl to its mass
Mass = moles * molar mass
Mass of Cl = 0.003238 mol * 35.453 g per mol = 0.1148 g
Now lets calculate the mass of the gold
Mass of gold = mass of sample – mass of Cl
= 0.328 g – 0.1148 g
= 0.2132 g
Now lets calculate moles of the gold
Moles of gold = 0.2132 g / 197 g peer mol = 0.001082 mol Au
Now lets find the mole ratio of the Au and Cl
Cl = 0.003238 / 0.001082 = 2.99 rounded to 3
Au = 0.001082 /0.001082 = 1
Therefore the empirical formula of the gold compound is AuCl3
b) Calculate the molar concentration of chloride ions (Cl-) present in a solution formed by dissolving 6.35 g of the water-soluble compound from part a in sufficient water to form 750.0 ml of solution. (4 pts)
Solution :-
6.35 g AuCl3 in 750.0 ml water
Concentration= ?
Moles of AuCl3 = 6.35 g / 303.325 g per mol = 0.021 mol
Now lets calculate the molarity
Molarity = moles / volume in liter
Lets put the values in the formula
Molartiy of AuCl3 = 0.021 mol / 0.750 L = 0.028 M
Therefore the molar concentration of the AuCl3 = 0.028 M
c) Name the water-soluble compound from part a.
Solution :-
As we determined the formula of the gold compound is AuCl3 threfore its name is written as
AuCl3 = Gold (lll) chloride
Because gold is the transition metal therefore we write its oxidation state in the roman numbers using the parentheses.