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Topic: Calculating mass % from precipitation extraction  (Read 9263 times)

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Calculating mass % from precipitation extraction
« on: June 05, 2009, 02:04:50 AM »
1) You have a sample of rat poison whose active ingredient is thallium (I) sulphate. You analyse this sample for the mass percentage of active ingredient by adding potassium iodide to precipitate yellow thallium (I) iodide. If the same of rat poison weighed 759.0mg and you obtained 212.2mg of the dry precipitate, what is the mass percentage of the thallium (I) sulphate in the rat poison?

I worked out the mass percentage of Tl to be 0.617 or 130.92mg Tl
Used conversion factors to work out the mass of equivalent sulphate for 130.92mg Tl to be 20.5mg sulphate.
Mass of Tl and sulphate is 151.42mg and percentage is 20% but the answer is 12.68%

I don't know the steps to get that answer.

2) A sample of copper sulphate pentahydrate was heated to 110 deg C, where it lost water and gave another hydrate of copper (II) ion that contains 32.50% Cu. A 98.77mg sample of this new hydrate gave 116.66mg of barium sulphate preciptate when treated with a barium nitrate solution. What is the formula of the new hydrate?

Worked out the difference in Cu % and proportioned new molecular weight to be 195.4amu which is three hydrate molecules difference with the old molecular weight so I thought the answer is copper (II) sulphate trihydrate but its dihydrate.

Then tried working out the sulphate percentate of barium sulphate to find mass of sulphate in 116.66mg barium sulphate to be 48mg sulphate.
Used conversion factors for 48mg sulphate to get 31.5mg Cu
Added 48mg sulphate and 31.5mg sulphate to get 79.5mg Cu (II) sulphate and remaining 19.3mg of hydrate worked out to be Cu (II) sulphate monohydrate.
Where did I go wrong?

Borek

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Re: Calculating mass % from precipitation extraction
« Reply #1 on: June 05, 2009, 03:33:24 AM »
Used conversion factors to work out the mass of equivalent sulphate for 130.92mg Tl to be 20.5mg sulphate.

You did something wrong here. Mass of Tl is correct, mass of sulfate is not.

Quote
Worked out the difference in Cu % and proportioned new molecular weight to be 195.4amu which is three hydrate molecules difference with the old molecular weight so I thought the answer is copper (II) sulphate trihydrate but its dihydrate.

195.4 is OK, show how you got trihydrate from here.

Quote
Then tried working out the sulphate percentate of barium sulphate to find mass of sulphate in 116.66mg barium sulphate to be 48mg sulphate.

No, it is not 48 mg.

Note: these calculations are much easier to do in moles. Calculate number of moles of BaSO4 - that gives you immediately number of moles of CuSO4, water excluded. Calculate mass of CuSO4 - and you have mass of water. Convert it to moles and you have what you have been looking for - molar ratio water/sulfate.
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Re: Calculating mass % from precipitation extraction
« Reply #2 on: June 13, 2009, 02:00:28 AM »
Thanks, I got the Cu sulphate dihyrate, was just short of one step.

The Tl2SO4 I worked out 212.2mg TlI = 0.64mmol TlI
0.64mmol TlI by conversion factors = 0.32mmol Tl2SO4
0.32mmol Tl2SO4 = 161.66mg Tl2SO4
161.66mg Tl2SO4 is 21.3% of 759mg rat poison but the answer is 12.68%

Borek

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Re: Calculating mass % from precipitation extraction
« Reply #3 on: June 13, 2009, 05:46:33 AM »
21.3% seems correct to me.
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Re: Calculating mass % from precipitation extraction
« Reply #4 on: June 14, 2009, 10:05:20 PM »
I thought so, the textbook answer isn't always right if there's formatting and human error in the editing, right?

I don't know if this question could be another one. The active ingredients of an antacid tablet contained only magnesium hydroxide and aluminium hydroxide. Complete neutralisation of a sample of the active ingredients require 48.5mL of 0.187M hydrochloric acid. The chloride salts from this neutralisation were obtained by evaporation of the filtrate from the titration; they weighed 0.4200g. What was the percentage by mass of magnesium hydroxide in the active ingredients of the anatacid tablet?

Mg(OH)2 + Al(OH)3 + HCl = MgCl2 + AlCl3 + H2O
mol HCl = 0.0090695mol HCl
g Mg(OH)2 = 0.1058g Mg(OH)2
g Al(OH)3 = 0.1415g Al(OH)3
total g = 0.2473g
g % Mg(OH)2 = 42.8%

But answer is 61.7%. Have I used the right qualities to calculate the right mass percentage of magnesium hydroxide in the combined mass of the active salt ingredients?

Borek

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Re: Calculating mass % from precipitation extraction
« Reply #5 on: June 15, 2009, 02:56:34 AM »
Mg(OH)2 + Al(OH)3 + HCl = MgCl2 + AlCl3 + H2O

This is not a correct reaction equation. You have two separate reactions taking place. You can't assume 1:1 ratio of Mg/Al.

61.7% is a correct result.
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Re: Calculating mass % from precipitation extraction
« Reply #6 on: June 18, 2009, 04:19:18 AM »
Sorry I forgot to write Mg(OH)2 + Al(OH)3 + 5HCl = MgCl2 + AlCl3 + 5H2O

I thought I could summarise the two separate reactions as one. Even now I haven't done it correctly, have I?

Other than the molar ratioes, were the steps, correct?

Borek

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Re: Calculating mass % from precipitation extraction
« Reply #7 on: June 18, 2009, 04:48:10 AM »
Adding 5 doesn't make this reaction correct. You can't combine reactions, they are separate processes, each going according to its own reaction equation.

You have two unknowns - number of moels of Mg and number of moles of Al.

Hint: if there is nMg of Mg and nAl of Al, what is mass of chlorides? Number of moles of HCl that have to react with both hydroxides?
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