A compound has the empirical formula CHCl. A 256 mL flask, at 373 K and 750 torr, contains 0.800 g of the compound, what is the molecular formula?

1 Answer
Nov 14, 2015

Answer:

Assuming ideality, the molecular formula is #C_2H_2Cl_2#

Explanation:

From the ideal gas law, #MM = (mRT)/(PV)#; where #MM# #=# molecular mass; #m# #=# mass; #P# #=# pressure in atmospheres; #V =# volume in litres; #R# #=# gas constant with appropriate units.

So, #(0.800*gxx0.0821*cancelL*cancel(atm)*cancelK^(-1)*mol^(-1)xx373*cancelK)/(0.256*cancelLxx0.987*cancel(atm))# #=# #97.0# #g*mol^(-1)#.

#nxx(12.01+1.01+2xx35.45)*g*mol^(-1)# #=# #97.0*g*mol^(-1)#.

Clearly, #n# #=# #1#. And molecular formula #=# #C_2H_2Cl_2#.

I seem to recall (but can't be bothered to look up) that vinylidene chloride, #H_2C=C(Cl)_2# is a low boiling point gas, whereas the 1,2 dichloro species is a volatile liquid. At any rate we have supplied the molecular formula as required.