Answer to Question #186295 in Chemistry for Shawn Elbert F.Cabiles

Question

A sample containing NaBr and KBr only weighs 255 mg.  The sample was dissolved in water and treated with excess AgNO₃.  The precipitate formed was found to weigh 420 mg.  Calculate the % NaBr in the sample. 

Answer 1

The molar mass of NaBr is 102.894 g/mol.

The molar mass of KBr is 119.002 g/mol.

The molar mass of AgBr is 187.77 g/mol.

Solution:

If the solution of the mixture of NaBr and KBr is treated with AgNO₃, an AgBr precipitate will be formed.

The chemical reactions for the formation of AgBr are written as follows:

NaBr + AgNO₃ → AgBr + NaNO₃

KBr + AgNO₃ → AgBr + KNO₃

Let, mass of NaBr = a, mass of KBr = b.

Here, mass of NaBr + KBr = 255 mg

Thus, a + b = 255

The amount of AgBr formed from NaBr is:

Mass of AgBr from NaBr = Mass of NaBr × (Molar mass of AgBr / Molar mass of NaBr) × (1/1)

Mass of AgBr from NaBr = (a) × (187.77 g mol^-1 / 102.894 g mol^-1) × (1/1) = 1.8249a

Mass of AgBr from NaBr = 1.8249a

The amount of AgBr formed from KBr is:

Mass of AgBr from KBr = Mass of KBr × (Molar mass of AgBr / Molar mass of KBr) × (1/1)

Mass of AgBr from KBr = (b) × (187.77 g mol^-1 / 119.002 g mol^-1) × (1/1) = 1.5779b

Mass of AgBr from KBr = 1.5779b

Here, mass of AgBr = 420 mg

Thus, 1.8249a + 1.5779b = 420

Let's consider the two equations:

a + b = 255

1.8249a + 1.5779b = 420

Multiply the first equation by -1.5779 and then add it to the second equation, we have

-1.5779a - 1.5779b = -402.3645

1.8249a + 1.5779b = 420

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

0.247a = 17.6355

a = 71.3988 = 71.4

Thus, mass of NaBr = a = 71.4 mg

%NaBr = (Mass of NaBr / Mass of sample) × 100%

%NaBr = (71.4 mg / 255 mg) × 100% = 28.0%

%NaBr = 28.0%

Answer: %NaBr in the sample is 28.0%.