ATAR Notes: Forum

VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: danielf on March 02, 2008, 11:47:42 am

Title: Gravimetric Analysis
Post by: danielf on March 02, 2008, 11:47:42 am: New to the site and hopin someone could lend a hand for my first SAC. Part 1 is a gravimetric analysis, one of the questions is 'why does the crucible have to be cool when weighed?' I've been told it has something to to with water adsorption to the porcelain however some more definite explanations would be greatly appreciated.
Thanks heaps
Title: Re: Gravimetric Analysis
Post by: Four on March 02, 2008, 12:36:45 pm: My gut feeling here is that a hot crucible would damage the scales; electronic scales are very sensitive, and putting a hot crucible on it would be a no-no. The crucible is heated to remove any water content in the salt, sure, but keeping the water out would require a desiccator rather than a cool crucible.
Title: Re: Gravimetric Analysis
Post by: iamdan08 on March 02, 2008, 05:10:07 pm: The crucible has to be cooled to room temperature because if it was still hot, the higher temperature would cause air currents around the balance, giving inaccurate results.
Title: Re: Gravimetric Analysis
Post by: danielf on March 02, 2008, 06:10:01 pm: Good point, thanks
Title: Re: Gravimetric Analysis
Post by: purple_rose on March 08, 2008, 01:14:08 pm: Wot questions would u expect to be on a PRAC SAC involving, finding the sulfate content in fertiliser by the means of gravimetric analysis (I don't mean conclusion, obervation, aim...etc)?
Title: Re: Gravimetric Analysis
Post by: chid on March 08, 2008, 02:10:20 pm: Possibly questions about the procedure such as:

Why does the precipitate need to be washed?
Why was an excess of barium chloride solution added?
Why must the precipitate be heated to constant mass?
List two reasons why barium sulfate is a suitable product in this gravimetric anaysis?

Also, calculations based on results, and writing an ionic equation for the reaction.

Hope this is helpful. :)
Title: Re: Gravimetric Analysis
Post by: danielf on March 08, 2008, 09:06:37 pm: There will probably be a question on sources of uncertainty too. make sure you write about possible impurities in the lawn feed (may form additional precipitate) as well as the various forms of mechanical losses. Also, if the precipitate was formed rapidly, some of the solution (water) may have become trapped in the crystal grain structure of the precipitate (thus inreasing its weight).
Title: Re: Gravimetric Analysis
Post by: purple_rose on March 08, 2008, 10:41:06 pm: i noe the observation part of the prac is usually one of the most easily parts to write, but for this prac i just don't noe wot 2 write. i mean there wasn't that much to observe except for the colour change and weight change, which is already in my results section. Wot do u think i should note. my prac is on http://resources.mhs.vic.edu.au/chemistry/year12/3pracproc.pdf , its this first and thrid prac (we're doing a combined prac, how fun).
Title: Re: Gravimetric Analysis
Post by: lanvins on March 09, 2008, 04:15:03 pm: 1. Lawn fertiliser contains nitrogen. A 1.3g sample of lawn fertiliser was weighed and carefully transferred to a 250ml volumetric flask. The weighing bottle was washed with a little de-ionised water and added to the content, until it reached the calibration line. A 20ml aliquot of this solution was added to a flask containing 20ml of .1M sodium hydroxide solution. 50ml of de-ionised water was than added. The flask was heated until the reaction NH4+ (aq) +OH-(aq) = NH3 (aq) +H20) (l) was complete. The burette was than titrated with .1M hydrochloric acid, using methyl red as indicator. The end point was reached when 44.3ml had been added.

Calculated the amount of NH4+ ions in the 1.3g fertiliser sample and the percentage by mass of nitrogen in the fertiliser, assuming all the nitrogen is present as ammonium ions.

2. 1.4g of ground fertiliser was placed in a 100ml beaker. 50ml of de-ionised water was added and filtered into a 600ml beaker. 3ml of 2M hydrochloric acid and water was added, until the total volume was 200ml. The solution was than boiled and 15ml of .5M barium chloride solution was added drop by drop. The precipate was colected filtrated, dried and weighed. A mass of 1.56g was obtained. How do u find the proportion of sulfate in the fertiliser?

(Can u plz show working?) thanks so much
Title: Re: Gravimetric Analysis
Post by: Collin Li on March 09, 2008, 05:29:24 pm: Question 1

First, recognise that after the preparatory reaction (the given reaction), we are left with a base: NH3.

Strategy:
- Calculate the number of mole of hydrochloric acid consumed per aliquot.
- Realise there will be no excess $\mbox{NaOH}$ , hence ammonia will react with hydrochloric acid.
- Write an equation for the reaction between hydrochloric acid and ammonia (acid + base).
- Hence, find the number of mole of ammonia that is present per aliquot.
- Using this, calculate the number of mole of ammonium present (using the given reaction) per aliquot.
- Scale this number to the number of moles of ammonium present in the 1.3 gram sample.
- Hence find the mass of nitrogen (N only), and the percentage mass of nitrogen

Solution:
$\mbox{n(HCl)} = 0.0443\times 0.1 = 0.00443\mbox{ mol}$

There is no excess $\mbox{NaOH}$ , because it is all neutralised with the ammonium, so the reaction that takes place is:

Chemical equation: $\mbox{HCl + NH}_3 \rightarrow \mbox{NH4}^+\mbox{ + Cl}^-$

$\implies \mbox{n(NH4}^+\mbox{)}_{\mbox{aliquot}} = \mbox{n(NH3)} = \mbox{n(HCl)} = 0.00443\mbox{ mol}$

$\implies \mbox{n(NH4}^+\mbox{)}_{\mbox{sample}} = 0.00443 \times \frac{250}{20} = 0.0554\mbox{ mol}$

$\implies \mbox{m(N)} = \mbox{n(NH4}^+\mbox{)} \times \mbox{M(N)} = 0.0554 \times 14 = 0.775\mbox{ g}$

$\implies \mbox{\%N by mass} = \frac{0.775}{1.3}\times 100 = 59.6\mbox{\% (w/w)}$

(It should be 1 significant figure, but since that's ridiculous, I've given it to three)
Title: Re: Gravimetric Analysis
Post by: Collin Li on March 09, 2008, 05:46:52 pm: Question 2

To find the proportion of sulfate in the fertiliser, you recognise that $\mbox{BaSO}_4$ will precipitate upon addition of barium chloride. We should assume that this is the sole precipitate (as we do in any gravimetric analysis - an implicit assumption that most people overlook), and therefore that:

$\mbox{m(BaSO}_4\mbox{)} = 1.05\mbox{ g}$

No "un-dilutions" need to take place as was the case in question 1, because the entire sample was used (wasn't split up into aliquots). Therefore, we can go straight ahead and calculate how much sulfate there is:

$\mbox{m(SO}_4\,^{2-}\mbox{)} = 1.05 \times \frac{\mbox{MW(SO}_4\,^{2-}\mbox{)}}{\mbox{MW(BaSO}_4\mbox{)}} = 1.05 \times \frac{96.1}{233.4} = 0.432\mbox{ g}$

This gives us the proportion of the mass which is solely sulfate, and hence we can find the percentage mass of it:

$\implies \mbox{\%SO}_4\mbox{ by mass} = \frac{0.432}{1.0}\times 100 = 43.2\mbox{\% (w/w)}$

I hope I'm right, I've got a massive headache today :(
Title: Re: Gravimetric Analysis
Post by: lanvins on March 09, 2008, 06:52:45 pm: thanks coblin
Title: Re: Gravimetric Analysis
Post by: lanvins on March 10, 2008, 04:52:59 pm: looking back, i think u made a mistake. For the step where u had to - Scale this number to the number of moles of ammonium present in the 1.3 gram sample- isn't the molar mass of NH4+ 18?
Title: Re: Gravimetric Analysis
Post by: Collin Li on March 10, 2008, 07:45:41 pm: Nah, you're supposed to find the mass of nitrogen only. I sort of skipped the obvious step:

$\mbox{n(N)} = \mbox{n(NH4}^+\mbox{)} \implies \mbox{m(N)} = 14\times \mbox{n(NH4}^+\mbox{)}$