Calculating the number of water molecules in trans. metal complex

In summary, the conversation discusses a lab experiment involving the creation of VO(acac)2 from VOSO4⋅xH2O and the calculation of percentage yield. The student is unsure of how to determine the value of x, the number of water molecules coordinated with the vanadyl sulfate, in order to accurately calculate the yield. They provide detailed information on the reagents used and their attempts at solving the problem, but ultimately conclude that the low yield obtained may have affected their results. They seek advice on how to approach the problem and determine the correct value of x.
  • #1
Ryaners
50
2

Homework Statement


[/B]
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!]

Homework Equations



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

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? :redface: Can I extract anything meaningful from that result?

Thanks in advance for any help.
 
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  • #2
Hi theere, cute avatar :rolleyes:
  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 ?
Ryaners said:
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
 
  • #3
BvU said:
  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 ?

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
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
Sorry, nothing valuable to add.

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

Ryaners said:
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

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
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 :)
 

FAQ: Calculating the number of water molecules in trans. metal complex

How do you calculate the number of water molecules in a trans. metal complex?

The number of water molecules in a trans. metal complex can be calculated by using the formula [M(H2O)n]z, where M is the metal ion, n is the number of water molecules, and z is the charge of the complex.

What is the significance of calculating the number of water molecules in a trans. metal complex?

Calculating the number of water molecules in a trans. metal complex is important because it helps us understand the structure and properties of the complex. It also allows us to determine the coordination number of the metal ion and the stability of the complex.

Can the number of water molecules in a trans. metal complex vary?

Yes, the number of water molecules in a trans. metal complex can vary depending on factors such as the type of metal ion, the oxidation state of the metal, and the surrounding environment.

What techniques are used to determine the number of water molecules in a trans. metal complex?

Various techniques such as X-ray crystallography, NMR spectroscopy, and mass spectrometry can be used to determine the number of water molecules in a trans. metal complex. These techniques provide information about the structure and composition of the complex.

How does the number of water molecules affect the properties of a trans. metal complex?

The number of water molecules in a trans. metal complex can significantly impact its properties. It can affect the solubility, reactivity, and stability of the complex. In general, a higher number of water molecules can increase the stability of the complex, while a lower number can increase its reactivity.

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