[fenor]

Hi,

I have taken 1.9812g of CuCl2 and have added 3 mol equivalent of DMSO (1.104g cm-3).
Now I have to find the theoretical yield. I have done the calculation for it but it does not seem right to me.

Steps:

CuCl2 + 2DMSO → CuCl2*2DMSO

1.9812gCuCl2 / 134.45g * 290.5g CuCl2·2DMSO( molecular weight) which comes to 4.280g

am I doing this right for the theoretical yield?
I have to get the percent yield.

thank you

[Borek]

Assuming you added excess DMSO (and CuCl₂ was a limiting reagent, your wording is not clear to me) that's OK.

[fenor]

I have to calculate the percent yield, the theoretical yield that i got does not seem right to me : 4.280g, since my actual yield is 4.787g. Is this more clear?

In the lab 1.9812 g of CUCl2 mixed with ethanol to be completly mixed then 3 mol equivalent of DMSO was added. The density of DMSO is (1.104g cm-3). Then precipitate was formed and then filtered. The final product that was obtained was 4.787g.