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

Dec 21, 2016

We assume that we have been given C; H; N $\text{microanalysis}$ $=$ 65.73%; 15.06%; 19.21%.
We get (eventually) $\text{molecular formula} = {C}_{4} {H}_{11} N$

#### Explanation:

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

$C :$ $\frac{65.73 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 5.47 \cdot m o l$

$H :$ $\frac{15.06 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 14.94 \cdot m o l$

$N :$ $\frac{19.21 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1} = 1.37 \cdot m o l$

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

${C}_{4} {H}_{11} N$

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

And thus here, $\text{Molar mass of gas"/"Molar mass of dihydrogen} = 37$.

And thus $\text{Molar mass of gas} = 74 \cdot g \cdot m o {l}^{-} 1$.

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

$74 \cdot g \cdot m o {l}^{-} 1 \cong n \times \left(4 \times 12.011 + 11 \times 1.00794 + 14.01\right) \cdot g \cdot m o {l}^{-} 1$

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 $\text{empirical formula}$ $=$ $\text{molecular formula}$ $=$ ${C}_{4} {H}_{11} N$