# Calculating the number of water molecules in trans. metal complex

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1. Jan 20, 2017

### Ryaners

1. The problem statement, all variables and given/known data

I had an inorganic lab this week which involved making VO(acac)2 from VOSO4⋅xH2O. In order to calculate the percentage yield, I need to work out x, that is, the number of water molecules coordinated with the vanadyl sulfate n-hydrate before the reaction. I'm stuck, because I know we didn't get a very high yield as we only had a short time to crystallise, and the yield we did measure wasn't very accurate as we had little time to dry the product before weighing (from mixing reagents to weighing product: less than an hour). As far as I know, without knowing how many water molecules there were in the vanadyl reagent, I can't know which was the limiting reagent & therefore what the percentage yield is, but if the mass of product we got is not close to what it should be stoichiometrically, I can't calculate x.

Am I correct? Is there anything I can do? Detailed info below.

[By the by: while I do need to work this out for a lab report, I'm posting here because I'd really appreciate some insight into the problem - I know the demonstrator bent the rules & told some classmates what x is & I can just ask them, but I'd like to see if there's a way to work it out. Everyone else has been carrying out this experiment in the same timeframe which implies that there might be a way to reason it out, even with imperfect data. Thanks in advance!]

2. Relevant equations

Chemical equation:
VOSO4⋅xH2O + Na2CO3 + 2C5H8O2 → VO(C5H7O2)2 + Na2SO4 + (x+1)H2O + CO2

3. The attempt at a solution

Amounts of reagents used:
Molar mass of Na2CO3 = 105.99 g mol-1
2.5 g used = (2.5 / 105.99) mol = 0.0236 mol used

Molar mass C5H8O2 = 100.12 g mol-1
2.5 cm3 used
Density of liquid C5H8O2 = (970 g / 1000 cm3) = 0.970 g cm-3 ⇒ 2.5 cm3 = (2.5 x 0.97) g = 2.425 g
2.425 g used = (2.425 / 100.12) mol = 0.0242 mol used

Molar mass VOSO4⋅xH2O = [163.00 + 18.015x] g mol-1
2.5 g used
If x = 1: 2.5 g = (2.5 / 181.015) mol = 0.138 mol theoretical yield
If x = 2: 2.5 g = (2.5 / 199.03) mol = 0.126 mol theoretical yield
If x = 3: 2.5g = (2.5 / 217.045) mol = 0.0115 mol theoretical yield

(I asked the demonstrator at the start of the class if x was 4 (from my pre-lab reading, I thought that was most likely) & he said no, it was fewer than that, so I've just worked out x = 1,2,3 for the product.)

Product:
Molar mass VO(C5H7O2)2 = 265.157 g mol-1
1.66 g yielded = (1.66 / 265.157) mol = 0.00626 mol yielded

This is far less than any of the possibilities given above. Am I screwed? Can I extract anything meaningful from that result?

Thanks in advance for any help.

2. Jan 20, 2017

### BvU

Hi theere, cute avatar
1. How many mole of C5H8O2 per mole of VOSO4.xH2O ?
2. Don't you become even a little bit suspicious if x = 1,2,3 gives such vastly differing numbers of moles yield ?

3. Jan 21, 2017

### Ryaners

Thanks for the response! So there are 2 moles C5H8O2 per mole of VOSO4.xH2O, which tells me that
1. The Na2CO3 is in excess & won't affect yield
2. If the stoichiometric amount of VOSO4.xH2O was used, that would be (0.5 x 0.0242 mol) = 0.0121 mol of VOSO4.xH2O.
This is closest to x=2 in the formula, which I worked out as corresponding to 0.0126 mol of VOSO4.2H2O in the 2.5g we used. (Looks like I left out a decimal place in the x=... calculations - I'll fix that in the original post now.)

So... is that how I should go about this problem? It makes sense if the C5H8O2 was the limiting reagent - but what if the VOSO4.xH2O was? I think that's the part I'm not quite grasping.

4. Jan 21, 2017

### BvU

This is awkward -- it seems I read your original post only halfway (up to what I quoted) and missed the minimal yield you obtained.
I can't fathom why you only get about half of the expected yield, unless insufficient time for crystallisation and drying etc indeed messed things up. @Borek any ideas ?

5. Jan 22, 2017

### Staff: Mentor

The only thing that caught my attention is that of these three:

only the last one is correct (hint: limiting reagent is different for first two values of x). But it doesn't make any difference when it comes to calculating the x, which - if the numbers are treated blindly as they are - comes out as close to 13.

6. Jan 23, 2017

### Ryaners

Ok, so basically I'm scuppered by the low yield - that's what I suspected. Thanks for your input @Borek & @BvU , was good to think through the problem regardless :)