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 %