Question #e2fd1

1 Answer
Jun 19, 2017

Answer:

#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#............