Microanalytical data give #C, H, N: 65.73%; 15.06%; 19.21%#. What is the molecular formula if the #"vapour density=37"#?

1 Answer
Dec 21, 2016

Answer:

We assume that we have been given #C; H; N# #"microanalysis"# #=# #65.73%; 15.06%; 19.21%.#
We get (eventually) #"molecular formula"=C_4H_11N#

Explanation:

We find the empirical formula in the usual way; that is assume that there are #100*g# of unknown compound, and divide thru by the atomic masses of each constituent:

#C:# #(65.73*g)/(12.011*g*mol^-1)=5.47*mol#

#H:# #(15.06*g)/(1.00794*g*mol^-1)=14.94*mol#

#N:# #(19.21*g)/(14.01*g*mol^-1)=1.37*mol#

And now we divide thru by the smallest molar quantity, that of #N#, to give the empirical formula:

#C_4H_11N#

Now, vapour density is the density of a vapour in relation to that of dihydrogen.

And thus here, #"Molar mass of gas"/"Molar mass of dihydrogen"=37#.

And thus #"Molar mass of gas"=74*g*mol^-1#.

As is typical, the molecular formula is a whole number multiple of the empirical formula, and thus,

#74*g*mol^-1~=nxx(4xx12.011+11xx1.00794+14.01)*g*mol^-1#

The vapour density was a bit out, but in fact this persuades me that these data came from an actual experiment, and not a problem someone pulled out of their posterior. Typically, vapour density measurements are not perfect.

Clearly, #n=1#, and the #"empirical formula"# #=# #"molecular formula"# #=# #C_4H_11N#