Question

Question #e2fd1

Answer 1

`C_7H_7NO`

Explanation:

As with all of these problems, we assume a `100*g` mass of unknown compound, and we interrogate its atomic composition...

`"Moles of carbon"=(69.40*g)/(12.011*g*mol^-1)=5.78*mol`.

`"Moles of hydrogen"=(5.825*g)/(1.00794*g*mol^-1)=5.78*mol`.

`"Moles of oxygen"=(13.21*g)/(15.999*g*mol^-1)=0.826*mol`.

`"Moles of nitrogen"=(11.57*g)/(14.01*g*mol^-1)=0.826*mol`.

And if we divide thru by the SMALLEST molar quantity, i.e. `0.826*mol` to get a trial empirical formula of......

`C_7H_7NO`............