What is the molecular formula of a compound with the empirical formula #CH# and a molar mass of 39.01 grams per mole?

1 Answer
May 26, 2016

Answer:

#"Molecular formula"# #=# #("Empirical formula")xxn#

Explanation:

And thus the molecular mass must be a whole number multiple of the empirical formula mass.

Here, clearly, #39.01*g*mol^-1# #=# #3xx(12.011+1.008)*g*mol^-1#

Thus the molecular formula #=# #C_3H_3#.

As far as I know, this result make no sense at all, inasmuch as there is no molecule with a formula of #C_3H_3#. (If I am not seeing something bleeding obvious, then someone will tell me!)

There is a cyclopropenyl cation, #C_3H_3^+#, but this would have a different molecular mass; why? Allene, #C_3H_4# does not fit this description.