Question

How would you determine the molecular formula of `NPCl_2` (347.64 g/mol)?

How would you determine the molecular formula of `NPCl_2 (347.64 g/(mol))`?

Answer 1

I presume you have quoted the empirical formula in `PNCl_2`. The molecular formula is ALWAYS a multiple of the empirical formula

Explanation:

Given that `(EF)_n = MF`, all we have to do is to use atomic masses and solve for `n`.

So `347.64*g*mol^(-1) = nxx(30.9747+14.01+2xx35.45)*g*mol^-1`.

And `347.64*g*mol^(-1) = n(115.88)*g*mol^(-1)`

`n = ??` And molecular formula `= (PNCl_2)xxn = P_31N_5Cl_102?`

Note that sometimes (but not here). the empirical formula is the same as the molecular formula.