Question

An organic compound containing 78.5% carbon, 13.1% nitrogen, 8.4% hydrogen. If the molar mass of the compound is 107 g/mol, what is the empirical and molecular formulas?

Answer 1

`"Empirical formula"` `=` `"Molecular formula"` `=` `C_7H_9N`.

Explanation:

With all these problems, we assume a mass of `100*g` and divide the individual elemental masses thru by the atomic mass, i.e.

`C:` `(78.5*g)/(12.011*g*mol^-1)=6.54*mol.`

`H:` `(8.4*g)/(1.00794*g*mol^-1)=8.33*mol.`

`N:` `(13.1*g)/(14.01*g*mol^-1)=0.935*mol.`

We divide thru by the smallest molar quantity, that of nitrogen, to get an empirical formula of `C_7H_9N`. The hydrogen ratio took a bit of rounding up.

Now the `"molecular formula"` is always a multiple of the `"empirical formula"`. So we solve for `n` in the equation `nxx(7xx12.011+9xx1.00794+1xx14.01)*g*mol^-1=107*g*mol^-1.`

Clearly, `n=1`, and the `"molecular formula"` `-=` `C_7H_9N`.