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

Question

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

5075 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-07T09:40:44

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

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.