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Gravimetric analysis

 
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Apr3-09, 01:22 PM   #1
 

Gravimetric analysis


Im working on the below problem

An impure sample of table salt that weighed .8421 g, when dissoved in water and treated with exess AgNO3, formed 2.044g of AgCL. What is the percentage of NaCl in the impure sample..

Im trying to figure the approach to this problem....Any ideas?
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Apr3-09, 04:16 PM   #2
 
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Sodium Chloride is dissolved in water. What does that mean?
Silver Nitrate is added and insoluble Silver Chloride is isolated and weighed.

What is the reaction? You just work backward to find amount of sodium chloride in the original "impure" sample. What is the expected impurity? How do you know that some other chloride compound is not also present?
Apr3-09, 09:53 PM   #3
 
NaCl dissolved in water means that the salt will completely disolve into the water.....the impurities wouldnt dissolve......Silver Nitrate added to water forms a precipitate...hence the Silver Chloride....But what about the sodium?

Im just not sure how to start....
Apr3-09, 11:54 PM   #4
 
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Gravimetric analysis


Quoting from original poster Jason03:

.....the impurities wouldnt dissolve......
How do you know that they would or would not dissolve? What are the impurities? Do you need to assume anything about them?

hence the Silver Chloride
Where is the written reaction which forms Silver Nitrate? Write it and use it.

You also asked about Sodium. What form of it is present in the solution? What do you know about solubility of sodium compounds?
Apr4-09, 11:14 AM   #5
 
This would be the reaction


AgNO3 + NaCl ----------------> AgCl + NaNO3
Apr4-09, 07:27 PM   #6
 
Ok I think I got it.....

From the reaction we know that 1 mole of AgCl is is equal to 1 mole of NaCl

so .......2.044g of AgCl/(143.33g/mole) = 0.01426 moles of AgCl.......

than........(0.01426)(58.44g/mole) = .8334g NaCl

so...... .8334/.8421 = 98.79%
Apr4-09, 10:36 PM   #7
 
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Jason03,
Your work appears good. I did not check your numeric values, but your units are well handled. Your logical steps seem good.
Apr5-09, 07:24 AM   #8
 
Thanks for the Help!
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