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

## How would you determine the molecular formula of $N P C {l}_{2} \left(347.64 \frac{g}{m o l}\right)$?

Nov 7, 2015

I presume you have quoted the empirical formula in $P N C {l}_{2}$. The molecular formula is ALWAYS a multiple of the empirical formula

#### Explanation:

Given that ${\left(E F\right)}_{n} = M F$, all we have to do is to use atomic masses and solve for $n$.

So $347.64 \cdot g \cdot m o {l}^{- 1} = n \times \left(30.9747 + 14.01 + 2 \times 35.45\right) \cdot g \cdot m o {l}^{-} 1$.

And $347.64 \cdot g \cdot m o {l}^{- 1} = n \left(115.88\right) \cdot g \cdot m o {l}^{- 1}$

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.