# Question #e2fd1

##### 1 Answer
Jun 19, 2017

${C}_{7} {H}_{7} N O$

#### Explanation:

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

$\text{Moles of carbon} = \frac{69.40 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 5.78 \cdot m o l$.

$\text{Moles of hydrogen} = \frac{5.825 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 5.78 \cdot m o l$.

$\text{Moles of oxygen} = \frac{13.21 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 0.826 \cdot m o l$.

$\text{Moles of nitrogen} = \frac{11.57 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1} = 0.826 \cdot m o l$.

And if we divide thru by the SMALLEST molar quantity, i.e. $0.826 \cdot m o l$ to get a trial empirical formula of......

${C}_{7} {H}_{7} N O$............