Yield % Problem

[Missvici101]

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

[BabcockHall]

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

[Missvici101]

n is the # of moles.

m is the mass of the measured samples

M is the Molar mass of the comound.

[BabcockHall]

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.