Question

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

Answer 1

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`