A Professor treats a water-soluble compound of gold and chlorine with silver nit

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.

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