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

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

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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.