Hello,
I need to calculate the yield % of [(Ph)₃PH]₂[CoCl₄]

The actual yield is 2.715g
My issue is with the theoretical yield, here is how I did it:

I'm following this reaction that was given in the explanations of my homework; (C₆H₅)₃P + HCl => (C₆H₅)₃PH+ + Cl-

Molar mass of (C₆H₅)₃P: 262.29 g/mol
Molar mass of (C₆H₅)₃PH⁺: 263.30 g/mol
Mass of Triphenylphosphine measured: 1.062 g

n(triphenylphosphine)=m(triphenylphosphine)/M(triphenylphosphine) =1.062g/(262.29 g/mol)=4.049×10^-3 mol

Molar mass of [(Ph)₃PH]₂[CoCl₄] : 727.33 g/mol
Since there are 2 molecules of (Ph)₃PH in the [(Ph)₃PH]₂[CoCl₄], the n(triphenylphosphine) needs to be divided by  2.

n([(Ph)₃PH]₂[CoCl₄])=n(triphenylphosphine)/2=2.025×10^-3 mol

m[(Ph)₃PH]₂[CoCl₄] =n[(Ph)₃PH]₂[CoCl₄] ×M[(Ph)₃PH]₂[CoCl₄]

m[(Ph)₃PH]₂[CoCl₄] =(2.025x10^-3 mol)×(727.33 g⁄mol)
m[(Ph)₃PH]₂[CoCl₄] =1.479 g = Theoretical yield

You see, my theoretical yield is smaller than my actual yield.
If someone sees my mistake, please let me know.

Thank you,
MV101

It would help the clarity of your question if you defined n, m, and M clearly and unambiguously.

n is the # of moles.

m is the mass of the measured samples

M is the Molar mass of the comound.

So far, I have not found an error.  I did see one thing that bothered me a bit, and it could be a clue.  It is difficult to see how the protonated form of a phosphine could be a ligand to the metal ion.  The unprotonated form would be fine.