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Topic: Yield question  (Read 2594 times)

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Offline earthnation112

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Yield question
« on: November 04, 2016, 09:16:23 PM »
Hi, just did an experiment titled "Preparation of oxobis(pentane 2,4 dionato)vanadium (IV)".

The method of the experiment was:

In a fume cupboard, carefully add 4 cm3 of concentrated sulphuric acid to a mixture of 12 cm3 ethanol and 6 cm3 water in a small conical flask or beaker. 

Add 2.5 g of vanadium pentoxide and heat the mixture gently on hot plate until the reduction to V(IV) has been observed and a blue-green colour is seen. This should take about 30-60 minutes but take care not to boil the solution dry!  (If the solution is boiled to dryness, seek assistance from a demonstrator.)

When this is complete, filter off the unreacted vanadium pentoxide and collect the filtrate in a flask. In a fume cupboard, add 6 cm3 of cold acetylacetone to the filtrate. Leave the mixture to stand for a few minutes, with a rubber bung loosely in the top of the flask, and then neutralise as instructed below.

Make up a solution of 10 g of sodium carbonate in 60 cm3 of water.  Add portions of this sodium carbonate solution slowly to your filtrate, stirring until there is no more effervescence.  At this point, check that your solution is neutral to litmus paper.  DO NOT ADD TOO MUCH SODIUM CARBONATE SOLUTION!

Filter off the precipitate using a Buchner funnel and dry by drawing air through the product. The turquoise precipitate is oxobis(pentane 2,4 dionato)vanadium (IV), otherwise known as “Vanadyl acac”, or VO(acac)2. Weigh your product and record the mass gained for use in percentage yield calculations.


One of the questions was, "Calculate the percentage yield based on V2O5 taken"

I have had an attempt below, have I made any mistakes?

Weight of V2O5 used: 2.5 g
Weight of hydrated Na2CO3: 10 g
Weight of dried VO (acac)2: 5.7387 g
V2O5  +  4H+   ---------> 2(VO)2+  +  2H2O  +  1/2O2    (1)
(VO)2+ + 2 acacH ------->2H+  +  VO (acac)2          (2)
rmm of V2O5 = (50.94 x 2) + (5 x 16.00) = 181.88
No. of moles of V2O5 used = 2.5 / 181.88 = 0.0137 mol
Using equation 1 and 2, each mole of V2O5 yields 2 moles of (VO)2+ and hence 2 moles of VO (acac)2.
No. of mole of VO (acac)2 = 0.0137 mol × 2 = 0.0274 mol
rmm of VO(acac)2 = (50.94 + 16.00) + 2× (5 × 12.01 + 7 × 1.008 + 2 × 16.00) = 265.15 g mol
Theoretical yield of VO(acac)2 = 265.15 g/mol × 0.0274 mol = 7.2651
Percentage yield of VO (acac)2 = 5.7387 / 7.2651 x 100 ≈ 78.99 %

Offline AWK

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Re: Yield question
« Reply #1 on: November 05, 2016, 03:14:17 AM »
Calculate mass of V2O5 needed for reaction using stoichiometry scheme"
1/2V2O5 => VO(acac)2
equivalent to
V2O5 => 2VO(acac)2
The calculation of yield is a simple division.
AWK

Offline Borek

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Re: Yield question
« Reply #2 on: November 05, 2016, 03:55:30 AM »
What you did looks OK to me.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline AWK

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Re: Yield question
« Reply #3 on: November 05, 2016, 05:19:43 AM »
I did not write that solution is wrong. But without balanced reaction or at least schematic reaction with balanced atoms in consideration calculations are less worthy.
Such a stoichiometry calculations without reaction were performed before 1792 year (Richter).
AWK

Offline Borek

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Re: Yield question
« Reply #4 on: November 05, 2016, 08:33:22 AM »
No idea what you mean. OP posted reactions and a correct stoichiometry scheme and used them to calculate the final answer. I don't see what makes the OP solution less worthy than possible other ones.

There is a slight problem with formatting ((VO)2+ instead of a VO2+) but it is an obvious typo.
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Offline earthnation112

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Re: Yield question
« Reply #5 on: November 05, 2016, 10:29:59 AM »
Thanks for the input Borek and AWK, cheers :)

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