Answer to Question #180705 in Chemistry for Kyle

Question

A mixture of mercurous chloride (MW 472.09) and mercurous bromide (MW 560.99) weighs 1.50 g. the mixture is quantitatively reduced to mercury metal (MW 200.59) which weighs 1.50 g. Calculate the % mercurous chloride and mercurous bromide in the original mixture. Show your solution.

Answer 1

Solution:

mercurous chloride - Hg₂Cl₂

The molar mass of Hg₂Cl₂ is 472.09 g mol^-1.

mercurous bromide - Hg₂Br₂

The molar mass of Hg₂Br₂ is 560.99 g mol^-1.

mercury metal - Hg

The molar mass of Hg is 200.59 g mol^-1.

Hg₂Cl₂ → 2Hg

GF = (200.59 g Hg mol^-1) / 472.09 g Hg₂Cl₂ mol^-1) × (2/1) = 0.8498 g Hg / g Hg₂Cl₂ 

Gravimetric factor (GF) = 0.8498

Hg₂Br₂ → 2Hg

GF = (200.59 g Hg mol^-1) / 560.99 g Hg₂Br₂ mol^-1) × (2/1) = 0.71513 g Hg / g Hg₂Br₂ 

Gravimetric factor (GF) = 0.71513

Mass of mixture = Mass of Hg₂Cl₂ + Mass of Hg₂Br₂ = 1.50 g

Let's assume that:

Mass of Hg₂Cl₂ = y

Mass of Hg₂Br₂ = 1.50 - y

Hence,

1.50 = (0.8498 × y) + (0.71513 × (1.50-y))

1.50 = 0.8498y + 1.072695 - 0.71513y

0.427305 = 0.13467y

y = 3.173 - This value has no mathematical meaning, since 'y' can not be more than 1.5.

Therefore, a mistake was made in the problem statement.

However, if a mixture of mercury chloride Hg₂Cl₂ and mercury bromide Hg₂Br₂ weighs 2.00 g.

Mass of mixture = Mass of Hg₂Cl₂ + Mass of Hg₂Br₂ = 2.00 g

Let's assume that:

Mass of Hg₂Cl₂ = y

Mass of Hg₂Br₂ = 2.00 - y

Hence,

1.50 = (0.8498 × y) + (0.71513 × (2.00-y))

1.50 = 0.8498y + 1.43026 - 0.71513y

0.06974 = 0.13467y

y = 0.5179 - Mass of Hg₂Cl₂

2.00 - y = 2.00 - 0.5179 = 1.4821 - Mass of Hg₂Br₂

%Hg₂Cl₂ = (Mass of Hg₂Cl₂ / Mass of mixture) × 100%

%Hg₂Cl₂ = (0.5179 g / 2.00 g) × 100% = 25.895% = 25.90%

%Hg₂Cl₂ = 25.90%

%Hg₂Br₂ = 100% - %Hg₂Cl₂ = 100% - 25.90% = 74.10%

%Hg₂Br₂ = 74.10%

Answer:

%Hg₂Cl₂ = 25.90%

%Hg₂Br₂ = 74.10%